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Page 1: hmt chap 1

HEAT AND MASS TRANSFER

SYLLABUS

It is branch of science which deals with the study of heat transfer rate and the mechanism of heat transfer

Thermodynamics Vs Heat transfer

Thermodynamics tells usbull How much heat is transferred (δQ)bull How much work is done (δW)bull Final state of the systemHeat transfer tells usbull How (with what modes) δQ is transferredbull At what rate δQ is transferredbull Temperature distribution inside the body

MODE OF HEAT TRANSFER

1048633ConductionConduction is the flow of heat in a substance due to exchange of energy between molecules having more energy and molecules having less energy

[ solids Lattice vibrations (ii) motion of free electrons][ fluids conduction is due to collision between the molecules caused by the random motion]

bull -needs matter (solid liquid gas)bull -molecular phenomenon (diffusion process)bull -without bulk motion of matter

1048633ConvectionThe transfer of energy from one region to another region due to macroscopic motion in fluid added on to the energy transfer by conduction is called heat transfer by conduction

bull -heat carried away by bulk motion of fluidbull -needs fluid matter

Forced Convection fluid motion is caused by an external agencyNatural Convection fluid motion occurs due to density variations caused by temperature differences

1048633RadiationAll the physical matter emits thermal radiation in the form of electromagnetic waves because of vibrational and rotational movements of the molecules and atoms which make up the matter

bull -does not needs matterbull -transmission of energy by electromagnetic waves bull Radiation increases with the temperature level

MODE OF HEAT TRANSFER

CONDUCTION- MECHANISM

bull CONDUCTION Conduction is the transfer of energy from more energetic

particles of a substance to the adjacent less energetic one as a result of interactions between the particles

In gases and liquids conduction is due to the collisions and diffusion of the molecules during their random motion

In solids it is due to the combination of vibrations of the molecules in a lattice and the energy transport by free electrons

Micro form of heat transferFOURIERrsquoS LAW

q =-KA( dT dX)

CONCECTION-MECHANISMbull Convection Heat transfer between a solid surface and an adjacent

gas or liquid It is the combination of conduction and flow motion Heat transferred from a solid surface to a liquid adjacent is conduction And then heat is brought away by the flow motion

bull Newtons law of cooling

where h = Convection heat transfer coefficient Ts = Temperature of the solid surface Tf = Temperature of the fluid

bull Macroform of heat transfer CONVECTION NEWTONrsquoS LAW OF COOLING

q = hA (Ts-Tf)

RADIATION-MECHANISM

bull Radiation The energy emitted by matter in the form of electromagnetic waves as a result of the changes in the electronic configurations of the atoms or molecules

bull Stefan - Boltzmann law

where σ = Stefan - Boltzmann constant ε = emissivity Ts = Surface temperature of the object

q =σ ε AT4

Conduction- Convection- Radiation

bull The three modes of heat transfer always exist simultaneously For example the heat transfer associated with double pane windows are

bull Conduction Hotter (cooler) air outside each pane causes conduction through solid glass

bull Convection Air between the panes carries heat from hotter pane to cooler pane

bull Radiation Sunlight radiation passes through glass to be absorbed on other side

Examples for different mode of heat transfer

bull Conduction

Heat loss through thermal insulation on steam pipe

Convection

Heat transfer to water flowing through a pipe in condenser

Radiation

Heat transfer in an electric furnace

Conduction Convection and Radiation

Solar energy used water heater

Fourierrsquos Law of Heat Conduction

Newtonrsquos Law of Cooling

Laws of Thermal Radiation

LAWS FOR HEAT TRANSFERCONDUCTION

4Tq

CONVECTION

NEWTONrsquoS LAW OF COOLING

RADIATION

STEFAN BOLTZMAN LAW

q=-KA( dT dX)

Assumptionsbull Steady statebull Unidirectional heat flowbull Const temperature gradientbull Linear profilebull No internal heat generationbull Isothermal bounding surfacebull Homogeneousbull Isotropic bull Applicable to solid liquid gas

q=σAT4q= hA (Tw-Tα)

Thq

XT

Kq

FOURIERrsquoS LAW

q =σ ε AT4

ELECTRICAL ampTHERMAL ANALOGY

THERMAL RESISTANCE(ELECTRICAL ANALOGY)

bull OHMrsquos LAW Flow of Electricity

V=I R elect

Voltage Drop = Current flowtimesResistance Temp Drop=Heat FlowtimesResistance

ΔT= q Rtherm

Thermal Analogy To OHMrsquoS Law

PLANE WALL-UNIDIRECTIONAL

PLANE WALL -SERIES

PLANE WALL IN SERIES

PLANE WALL -PARALLEL

PLANE WALL-PARALLEL

PLANE WALL-COMBINATION

PLANE WALL-FILM RESISTANCE

PLANE WALL- THERMAL CONTACT RESISTANCE

PLANE WALL- THERMAL CONTACT RESISTANCE

At the joining surface of the two slabs air is trapped in voids due to surface irregularities

CYLINDER-SOLID

HEAT TRANSFER IN SOLID CYLINDER

HEAT TRANSFER RATE IN HOLLOW CYLINDER

HEAT TRANSFER RATE IN HOLLOW CYLINDER-COMPOSITE

HEAT TRANSFER RATE IN HOLLOW CYLINDER-MULTY LAYER

HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER

HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER

HEAT TRANSFER IN SPHERE

HEAT TRANSFER IN SPHERE -MULTY LAYER

HEAT TRANSFER IN SPHERE -MULTY LAYER

bull Problem 1 The wall in a furnace consists of 125 mm thick refractory bricks and

125 mm thick insulating fire bricks separated by an air gap A 12 mm thick plaster covers the outer wall The inner surface of wall is at 1100oC and the ambient temperature 25oC the heat transfer coefficient on the outer wall to the air is 17 Wm2K and the resistance to heat flow of the air gap is 016 KW the thermal conductivities of refractory brick insulating firebrick and plaster are 16 03 and 014 WmK respectively

Calculatebull (a) the rate of heat loss per unit area of the wall surfacebull (b) the interface temperatures throughout the wallbull (c) the temperature at the outside surface of the wall

bull Problem 2Two large aluminum plates (k=240 WmK) each 2 cm thick with 10 mm surface roughness are placed in contact under 105 Nm2 pressure in air The temperature at the outside surfaces are 390 oC and 406 oCCalculate

A) the heat flux B) the temperature drop due to the contact resistance C) the contact temperatures Thermal contact resistance with air as the interfacial fluid for

10 mm roughness is 27510-4 m2KW

bull Problem 3bull Steam at 350 oC flowing in a pipe (k=80 WmK) of 5 cm

inner diameter and 56 cm outer diameter is covered with 3 cm thick insulation of k=005 WmK Heat is lost to surroundings at 5 oC by natural convection and radiation the combined h being 20 Wm2K Taking the heat transfer coefficient inside the pipe as 60 Wm2K

bull Determinebull A) the rate of heat loss from the steam per unit length of

the pipebull B) the temperature drop across the pipe and the insulation

bull PROBLEM4bull A nuclear reactor has a sperical pressure vessel 750 mm inside

diameter 80 mm wall thickness The temperature at the inner surface is 500 oC If the tempertarue at the outer surface is 495 oC

bull Calculatea) The rate of heat loss through the metal wallb) If the rate of heat loss is to limited to 305Whow much inulation

thickness needs to be applied Thermal conductivity of steel wall=46 WmoC Thermal conductivity of insulation=004 WmoC Film coefficient of heat tranfer at inner and outer surface are 200

Wm2 oC and 15 Wm2 oC Ambient air temperature=28 oC

CRITICAL RADIUS

bull The radius upto which heat flow increases and after decreases At critical radius q becomes qmax

bull The addition of insulation always increases (Rth)condbull Rtotal= (Rth)cond + (Rth)conv

bull The addition of insulation may reduce the (Rth)conv due to increase in surface area

bull r1ltrc -heat transfer increases by adding insulation till r1=rc If

insulation thickness is further increased the rate of heat loss will decrease

r1 gt rc

- heat transfer decreases by adding insulation above the rc

Plane wall- Insulation

thickness -study

Cylinder amp Sphere- Insulation thickness -

study

Problem5bull A 3mm diameter and 5 mm long electric wire is tightly

wrapped with a 2 mm thick plastic cover whose K=015 wmoC Electrical measurements indicate that a current of 10 A passes through the wire and there is a voltage drop of 8 V along the wire If the insulated wire is exposed to a medium at Ta=30oC with a heat transfer coefficient of h=12 wm2oC Determine the temperature at the interface of the wire and the plastic cover in steady operation Also determine whether doubling the thickness of the plastic cover will increase or decrease this interface temperature

dx

TdzdyKQ xx )(

dxQx

QQ xxdxx ][

dxdx

TdzdyK

xQQdQ xdxxxx ])([)(

)1()]([)(

ddzdydx

x

TK

xdQ xx

GENERAL HEAT CONDUCTION EQUATION IN CARTESIAN COORDINATES

Consider an unsteady (transient) three dimensional heat conduction with internal heat generation in heterogeneous amp anisotropic PLANE WALL

Cartesian co-ordinates x y z

Elemental length dxdydz

Consider an elemental volume ABCDEFGH = (dxdydz)

Let Temperature distribution is a function of T=T(xyz)

Let

=internal heat generation per unit volume per unit time (W m3)q= rate of heat flow (W)Q= qdt=total heat flow (J)r=mass density of the material element(kgm3)Cp=specific heat of the material element (Jkg K)

gq

Plane (y - z) lsquoxrsquo direction

Quantity of heat leaving from the face EFGH in lsquoXrsquo direction

Quantity of heat flowing into the face ABCD in lsquoXrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoXrsquo direction

dy

TdzdxKQ yy )(

dyQy

QQ yydyy ][

dydy

TdzdxK

yQQdQ ydyyyy ])([)(

)2()]([)(

ddzdydx

y

TK

ydQ yy

Plane (x - z) lsquoyrsquo direction

Quantity of heat flowing into the face ABFE in lsquoYrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoYrsquo direction

dz

TdydxKQ zz )(

dzQz

QQ zzdzz ][

dzdz

TdydxK

zQQdQ zdzzzz ])([)(

)3()]([)(

ddzdydx

z

TK

zdQ zz

Plane (x - y) lsquozrsquo direction

Quantity of heat flowing into the face ADHE in lsquoZrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoZrsquo direction

Quantity of heat leaving from the face DCGH in lsquoYrsquo direction

Quantity of heat leaving from the face BCGF in lsquoZrsquo direction

)4()()(

ddzdydxqQgg

Total amount of internal heat generation in the element

Ein + Egen = Eout+Est (Ein- Eout)+ Egen=Est

T

Cddzdydxddzdydxqddzdydxz

TK

zy

TK

yx

TK

x pgzyx )()())](()()([

T

pC

gq

z

TzK

zy

Ty

Kyx

TxK

x][][][

stgzyx QQdQdQdQ

)(

)5()()(

dT

CdzdydxQ pst

Total heat energy stored in the element

(ie thermal energy of the element can be increased because of heat accumulation in the element due to heat flow amp internal heat generation)

)( volumem d

TmCQ pst )(

T

CqTK pg)(

KKKyKx z If material is homogeneous amp isotropic

T

pC

gq

z

T

zy

T

yx

T

xK ][][][

T

Kp

C

Kgq

z

T

y

T

x

T2

2

2

2

2

2

T

Kgq

z

T

y

T

x

T 12

2

2

2

2

2

Thermal diffusivity pC

K

T

K

qT g

12

according to thermodynamics first law

)( zr )( dzrddr

)( dzrddr

Cylindrical polar co-ordinates

Elemental length

Consider an elemental volume ABCDEFGH =

GENERAL HEAT CONDUCTION EQUATION IN CYLINDRICAL COORDINATES

gq

gq

Consider an unsteady (transient) three dimensional heat conduction with internal heat generation in heterogeneous amp anisotropic cylindrical geometries (eg rods and pipes)

)( zrTLetTemperature distribution is a function of T=

Let

=internal heat generation per unit volume per unit time (W m3)q= rate of heat flow (W)Q= qdt=total heat flow (J)r=mass density of the material element(kgm3)Cp=specific heat of the material element (Jkg K)

planerdirectionz

planezrdirection

planezdirectionr

)(

)(

)(

LetHeat flow in

drdr

TdzrdK

rdQ

QQdQ

rr

drrrr

])([

)1(][

r

Tr

rddrdzdKdQ rr

Heat accumulation in the element due to heat flow by conduction in lsquorrsquo direction

rddr

TdzdrK

rdQ

QQdQ d

])([

][

2

T

r

ddrdzrdKdQ

)2(][

2

2

2

T

r

ddrdzrdKdQ

Heat accumulation in the element due to heat flow by conduction in lsquoqrsquo direction

Plane (z- q ) lsquorrsquo direction

dr

TdzrdKQ rr

)(

drQr

QQ rrdrr ][

Quantity of heat leaving from the face EFGH in lsquorrsquo direction

Quantity of heat flowing into the face ABCD in lsquorrsquo direction

Plane (r-z) lsquoq lsquodirection

Quantity of heat leaving from the face AEGC in lsquoqrsquo direction

d

r

TdzdrKQ )(

rdQ

rQQ d ][

Quantity of heat flowing into the face BFHD in lsquoqrsquo direction

)3())((2

2

z

TdzdrrdKdQ zz

)5()()(

dT

CdzdrrdQ pst

)4()()(

ddzdrrdqQgg

Total amount of internal heat generation in the element

Total heat energy stored in the element

(ie thermal energy of the element can be increased because of heat accumulation in the element due to heat flow amp internal heat generation)

according to thermodynamics first law

Ein + Egen = Eout + Est (Ein ndash Eout) + Egen = Est

)( volumem

)()(z

T

zdzdrrdKdQ zz

Heat accumulation in the element due to heat flow by conduction in lsquozrsquo direction

dzdz

TdrrdK

zdQ

QQdQ

zz

dzzzz

])([

d

TmCQ pst )(

Plane (r-q) lsquozrsquo direction

dz

TdrrdKQ zz )(

dzQz

QQ zzdzz ][ Quantity of heat leaving from the face CDHG in lsquozrsquo direction

Quantity of heat flowing into the face ABFE in lsquozrsquo direction

stgzr QQdQdQdQ

)(

r

r ][)(rT

rr

ddrdzdKr

][)(

2

2

2

T

r

ddrdzrdK ))((2

2

z

TddzdrrdK z

ddzdrrdq g )(

dT

Cdzdrrd p )(

+ + +

=

T

CqTK pg)(KKKK zr

[gq

)(1

r

Tr

rrK r

)(1

2

2

2 T

rK )(

2

2

z

TK z

)( ddzdrrd )(

ddzdrrd

TC p

+ + + ] =

If material is homogeneous amp isotropic

)(1

r

Tr

rr

)(1

2

2

2 T

r)(

2

2

z

T

K

q g

T

K

C p

)(1

r

Tr

rr

)(1

2

2

2 T

r )(2

2

z

T

K

q g

T

1

+ + + =

)(1

r

Tr

rrK r

)(1

2

2

2 T

rK )(

2

2

z

TK z

gq

T

C p + + + =

+ + + = Thermal diffusivity

T

K

qT g

12

pC

K

(Ein ndash Eout) + Egen = Est

)( rT

)( r

)sin( drrddr)sin( drrddr

planerdirection

planerdirection

planedirectionr

)(

)(

)(

Spherical polar co-ordinates

Elemental length

Consider an elemental volume ABCDEFGH =

GENERAL HEAT CONDUCTION EQUATION IN SPHERICAL COORDINATES

Let

=internal heat generation per unit volume per unit time (W m3)q= rate of heat flow (W)Q= qdt=total heat flow (J)r=mass density of the material element(kgm3)Cp=specific heat of the material element (Jkg K)

gq

Consider an unsteady (transient) three dimensional heat conduction with internal heat generation in heterogeneous amp anisotropic spherical geometries

LetTemperature distribution is a function of T=

X

Y

Z

drT

drrdKQ rr )sin(

drQr

QQ rrdrr ][

drdr

TdrrdK

rdQ

QQdQ

rr

drrrr

])sin([

)1(]([sin 2

rT

rr

ddrddKdQ rr

Plane ( -q f) lsquorrsquo directionQuantity of heat flowing into the face ABCD in lsquorrsquo direction

Quantity of heat leaving from the face EFGH in lsquorrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquorrsquo direction

drT

drdrKQ )sin(

rdQ

rQQ d ][

rddr

TdrdrK

rdQ

QQdQ d

])sin([

])[(sin

2

T

r

drdrdrKdQ

)2(])(sin[

2

T

r

drdrddrKdQ

Quantity of heat leaving from the face BFHD in lsquoqrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoqrsquo direction

X

Y

Z

Plane (r- f ) lsquoq lsquodirection

Quantity of heat flowing into the face AEGC in lsquoqrsquo direction

dr

TdrrdKQ

sin)(

drQ

rQQ d sin][

sin

drdr

TdrrdK

rdQ

QQdQ d

sin]sin

)([sin

][sinsin

)sin(

T

rr

ddrdrrdKdQ

)3()(sin

)sin(2

2

22

2

T

r

ddddrrKdQ

Plane (r- ) q lsquofrsquo directionQuantity of heat flowing into the face ABFE in lsquofrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquozrsquo direction

)4()sin()(

ddrrddrqQgg

)5()sin()(

dT

CdrrddrQ pst

Total amount of internal heat generation in the element

Total heat energy stored in the element

(Ein ndash Eout) + Egen = Est

d

TmCQ pst )(

)( volumem

Quantity of heat leaving from the face CDHG in lsquofrsquo direction

according to thermodynamics first law

Ein + Egen = Eout + Est

X

Y

Z

stgr QQdQdQdQ

)(

]([sin 2

r

Tr

rddrddKr

])(sin[

2

T

rdrdrddrK

)(sin

)sin(2

2

22

2

T

r

ddddrrK

ddrrddrq g )sin(

dT

Cdrrddr p )sin(

2

2

r

r

sin

sin

]([ 22 r

Tr

rr

Kr

])(sin[sin2

TrK

)(sin 2

2

22

T

r

Kgq

T

Cp

]([1

[ 22 r

Tr

rrK

])(sin[sin1

2

T

r)](

sin

12

2

22 T

r gq

T

Cp

]([1

[ 22 r

Tr

rr

])(sin[sin1

2

T

r)](

sin

12

2

22 T

rK

q g

T

K

Cp

]([1

[ 22 r

Tr

rr

])(sin[sin1

2

T

r)](

sin

12

2

22 T

rK

q g

T1

KKKKr If material is homogeneous amp isotropic

TCqTK pg)(

T

K

qT g

12 pC

K

Thermal diffusivity

(Ein ndash Eout) + Egen = Est

General Heat Conduction Equation

PLANE WALL CYLINDER SPHERE

Temperature profile

Plane wall SphereCylinder

bull A copper conductor (k = 380 WmoC resistivity ρ = 2x10-8 Ω m) having inner and outer radii 10 cm and 225 cm respectively is carrying a current density of 4800 amperes cm2 the conductor is internally cooled and a constant temperature of 65oC is maintained at the inner surface and there is no heat transfer through the insulation surrounding the conductor

Determinebull (i) The maximum temperature of the conductor and the radius at

which it occursbull (ii) The internal heat transfer rate

bull A chemical reaction takes place in a packed bed (k=06 WmoC) between two coaxial cylinders with radii 15 mm and 45 mm the inner surface is at 580oC and it is insulated Assuming the reaction rate of 055 MWm3 in the reactor volume find the temperature at the outer surface of the reactor

bull A current of 300 amperes passes through a stainless steel wire of 25 mm diameter and k=20 WmoC The resistivity of the wire is 70x10-8 ohm-m and length of the wire is 2 m if the wire is submerged in a fluid maintained at 50oC and h at the wire surface is 4000 Wm2oC calculate the steady state temperature at the centre and at the surface of the wire

bull An approximately spherical shaped orange ( k=023 WmoC) 90 mm diameter undergoes ripping process and generates 5100 Wm3 of energy If external surface the orange is at 8oC determine

bull (i) temperature at the centre of the orange and

bull (ii) heat flow from the outer surface of the orange

bull FINS External surfaces to used to cooling area and hence convection

heat transfer can be improvedbull APPLICATIONS IC engine cylinder ( air cooling) Heat exchangers automobile

radiators computer chip refrigerators electric motors generators and transformers etc

bull ASSUMPTIONS MADE IN ANALYSIS OF HEAT TRANFER THROUGH FIN

1 Fin material is homogenous and isotropic constthermal conductivity

2 Temperature at any cs of the fin is uniform3 lsquohrsquo is uniform over entire surface4 No internal heat generation5 Thermal contact resistance is negligible6 Radiation heat transfer is negligible7 Heat conduction is steady state and one dimensional

bull Problem 1

bull A stainless steel fin (k=20 WmK) having a diameter of 20 mm and a length of 01 m is attached to a wall at 300oC The ambient temperature is 50oC and the heat transfer coefficient is 10 Wm2K The fin tip is insulated Determine

bull (A) The rate of heat dissipation from the finbull (B) The temperature at the fin tipbull (C) The rate of heat transfer from the wall area covered by the fin if

the fin was not usedbull (D) The heat transfer rate from the same fin geometry if the

stainless steel fin is replaced by a fictitious fin with infinite thermal conductivity

bull Problem2bull One end of a long rod is inserted into a furnace while the other

projects into ambient temperature Under steady state the temperature of the rod is measured at two points 75mm apart and found to be 125oC and 885oC respectively while the ambient temperature is 20oC If the rod is 25mm in diameter and h is 2336 Wm2K

bull Find the thermal conductivity of the rod material

  • HEAT AND MASS TRANSFER
  • Thermodynamics Vs Heat transfer
  • MODE OF HEAT TRANSFER
  • MODE OF HEAT TRANSFER (2)
  • CONDUCTION- MECHANISM
  • CONCECTION-MECHANISM
  • RADIATION-MECHANISM
  • Conduction- Convection- Radiation
  • Slide 9
  • Slide 10
  • Fourierrsquos Law of Heat Conduction
  • Slide 12
  • Newtonrsquos Law of Cooling
  • Laws of Thermal Radiation
  • LAWS FOR HEAT TRANSFER
  • ELECTRICAL ampTHERMAL ANALOGY
  • Slide 17
  • Slide 18
  • PLANE WALL-UNIDIRECTIONAL
  • PLANE WALL -SERIES
  • PLANE WALL IN SERIES
  • PLANE WALL -PARALLEL
  • PLANE WALL-PARALLEL
  • PLANE WALL-COMBINATION
  • PLANE WALL-FILM RESISTANCE
  • PLANE WALL- THERMAL CONTACT RESISTANCE
  • PLANE WALL- THERMAL CONTACT RESISTANCE (2)
  • CYLINDER-SOLID
  • HEAT TRANSFER IN SOLID CYLINDER
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER-COMPOSITE
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER-MULTY LAYER
  • HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER
  • HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER (2)
  • HEAT TRANSFER IN SPHERE
  • HEAT TRANSFER IN SPHERE -MULTY LAYER
  • HEAT TRANSFER IN SPHERE -MULTY LAYER (2)
  • Slide 38
  • Slide 39
  • Slide 40
  • Slide 41
  • CRITICAL RADIUS
  • Plane wall- Insulation thickness -study
  • Problem5
  • Slide 45
  • Slide 46
  • Slide 47
  • GENERAL HEAT CONDUCTION EQUATION IN CYLINDRICAL COORDINATES
  • Slide 49
  • Slide 50
  • Slide 51
  • GENERAL HEAT CONDUCTION EQUATION IN SPHERICAL COORDINATES
  • Slide 53
  • Slide 54
  • Slide 55
  • General Heat Conduction Equation
  • Temperature profile
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Slide 67
Page 2: hmt chap 1

Thermodynamics Vs Heat transfer

Thermodynamics tells usbull How much heat is transferred (δQ)bull How much work is done (δW)bull Final state of the systemHeat transfer tells usbull How (with what modes) δQ is transferredbull At what rate δQ is transferredbull Temperature distribution inside the body

MODE OF HEAT TRANSFER

1048633ConductionConduction is the flow of heat in a substance due to exchange of energy between molecules having more energy and molecules having less energy

[ solids Lattice vibrations (ii) motion of free electrons][ fluids conduction is due to collision between the molecules caused by the random motion]

bull -needs matter (solid liquid gas)bull -molecular phenomenon (diffusion process)bull -without bulk motion of matter

1048633ConvectionThe transfer of energy from one region to another region due to macroscopic motion in fluid added on to the energy transfer by conduction is called heat transfer by conduction

bull -heat carried away by bulk motion of fluidbull -needs fluid matter

Forced Convection fluid motion is caused by an external agencyNatural Convection fluid motion occurs due to density variations caused by temperature differences

1048633RadiationAll the physical matter emits thermal radiation in the form of electromagnetic waves because of vibrational and rotational movements of the molecules and atoms which make up the matter

bull -does not needs matterbull -transmission of energy by electromagnetic waves bull Radiation increases with the temperature level

MODE OF HEAT TRANSFER

CONDUCTION- MECHANISM

bull CONDUCTION Conduction is the transfer of energy from more energetic

particles of a substance to the adjacent less energetic one as a result of interactions between the particles

In gases and liquids conduction is due to the collisions and diffusion of the molecules during their random motion

In solids it is due to the combination of vibrations of the molecules in a lattice and the energy transport by free electrons

Micro form of heat transferFOURIERrsquoS LAW

q =-KA( dT dX)

CONCECTION-MECHANISMbull Convection Heat transfer between a solid surface and an adjacent

gas or liquid It is the combination of conduction and flow motion Heat transferred from a solid surface to a liquid adjacent is conduction And then heat is brought away by the flow motion

bull Newtons law of cooling

where h = Convection heat transfer coefficient Ts = Temperature of the solid surface Tf = Temperature of the fluid

bull Macroform of heat transfer CONVECTION NEWTONrsquoS LAW OF COOLING

q = hA (Ts-Tf)

RADIATION-MECHANISM

bull Radiation The energy emitted by matter in the form of electromagnetic waves as a result of the changes in the electronic configurations of the atoms or molecules

bull Stefan - Boltzmann law

where σ = Stefan - Boltzmann constant ε = emissivity Ts = Surface temperature of the object

q =σ ε AT4

Conduction- Convection- Radiation

bull The three modes of heat transfer always exist simultaneously For example the heat transfer associated with double pane windows are

bull Conduction Hotter (cooler) air outside each pane causes conduction through solid glass

bull Convection Air between the panes carries heat from hotter pane to cooler pane

bull Radiation Sunlight radiation passes through glass to be absorbed on other side

Examples for different mode of heat transfer

bull Conduction

Heat loss through thermal insulation on steam pipe

Convection

Heat transfer to water flowing through a pipe in condenser

Radiation

Heat transfer in an electric furnace

Conduction Convection and Radiation

Solar energy used water heater

Fourierrsquos Law of Heat Conduction

Newtonrsquos Law of Cooling

Laws of Thermal Radiation

LAWS FOR HEAT TRANSFERCONDUCTION

4Tq

CONVECTION

NEWTONrsquoS LAW OF COOLING

RADIATION

STEFAN BOLTZMAN LAW

q=-KA( dT dX)

Assumptionsbull Steady statebull Unidirectional heat flowbull Const temperature gradientbull Linear profilebull No internal heat generationbull Isothermal bounding surfacebull Homogeneousbull Isotropic bull Applicable to solid liquid gas

q=σAT4q= hA (Tw-Tα)

Thq

XT

Kq

FOURIERrsquoS LAW

q =σ ε AT4

ELECTRICAL ampTHERMAL ANALOGY

THERMAL RESISTANCE(ELECTRICAL ANALOGY)

bull OHMrsquos LAW Flow of Electricity

V=I R elect

Voltage Drop = Current flowtimesResistance Temp Drop=Heat FlowtimesResistance

ΔT= q Rtherm

Thermal Analogy To OHMrsquoS Law

PLANE WALL-UNIDIRECTIONAL

PLANE WALL -SERIES

PLANE WALL IN SERIES

PLANE WALL -PARALLEL

PLANE WALL-PARALLEL

PLANE WALL-COMBINATION

PLANE WALL-FILM RESISTANCE

PLANE WALL- THERMAL CONTACT RESISTANCE

PLANE WALL- THERMAL CONTACT RESISTANCE

At the joining surface of the two slabs air is trapped in voids due to surface irregularities

CYLINDER-SOLID

HEAT TRANSFER IN SOLID CYLINDER

HEAT TRANSFER RATE IN HOLLOW CYLINDER

HEAT TRANSFER RATE IN HOLLOW CYLINDER-COMPOSITE

HEAT TRANSFER RATE IN HOLLOW CYLINDER-MULTY LAYER

HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER

HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER

HEAT TRANSFER IN SPHERE

HEAT TRANSFER IN SPHERE -MULTY LAYER

HEAT TRANSFER IN SPHERE -MULTY LAYER

bull Problem 1 The wall in a furnace consists of 125 mm thick refractory bricks and

125 mm thick insulating fire bricks separated by an air gap A 12 mm thick plaster covers the outer wall The inner surface of wall is at 1100oC and the ambient temperature 25oC the heat transfer coefficient on the outer wall to the air is 17 Wm2K and the resistance to heat flow of the air gap is 016 KW the thermal conductivities of refractory brick insulating firebrick and plaster are 16 03 and 014 WmK respectively

Calculatebull (a) the rate of heat loss per unit area of the wall surfacebull (b) the interface temperatures throughout the wallbull (c) the temperature at the outside surface of the wall

bull Problem 2Two large aluminum plates (k=240 WmK) each 2 cm thick with 10 mm surface roughness are placed in contact under 105 Nm2 pressure in air The temperature at the outside surfaces are 390 oC and 406 oCCalculate

A) the heat flux B) the temperature drop due to the contact resistance C) the contact temperatures Thermal contact resistance with air as the interfacial fluid for

10 mm roughness is 27510-4 m2KW

bull Problem 3bull Steam at 350 oC flowing in a pipe (k=80 WmK) of 5 cm

inner diameter and 56 cm outer diameter is covered with 3 cm thick insulation of k=005 WmK Heat is lost to surroundings at 5 oC by natural convection and radiation the combined h being 20 Wm2K Taking the heat transfer coefficient inside the pipe as 60 Wm2K

bull Determinebull A) the rate of heat loss from the steam per unit length of

the pipebull B) the temperature drop across the pipe and the insulation

bull PROBLEM4bull A nuclear reactor has a sperical pressure vessel 750 mm inside

diameter 80 mm wall thickness The temperature at the inner surface is 500 oC If the tempertarue at the outer surface is 495 oC

bull Calculatea) The rate of heat loss through the metal wallb) If the rate of heat loss is to limited to 305Whow much inulation

thickness needs to be applied Thermal conductivity of steel wall=46 WmoC Thermal conductivity of insulation=004 WmoC Film coefficient of heat tranfer at inner and outer surface are 200

Wm2 oC and 15 Wm2 oC Ambient air temperature=28 oC

CRITICAL RADIUS

bull The radius upto which heat flow increases and after decreases At critical radius q becomes qmax

bull The addition of insulation always increases (Rth)condbull Rtotal= (Rth)cond + (Rth)conv

bull The addition of insulation may reduce the (Rth)conv due to increase in surface area

bull r1ltrc -heat transfer increases by adding insulation till r1=rc If

insulation thickness is further increased the rate of heat loss will decrease

r1 gt rc

- heat transfer decreases by adding insulation above the rc

Plane wall- Insulation

thickness -study

Cylinder amp Sphere- Insulation thickness -

study

Problem5bull A 3mm diameter and 5 mm long electric wire is tightly

wrapped with a 2 mm thick plastic cover whose K=015 wmoC Electrical measurements indicate that a current of 10 A passes through the wire and there is a voltage drop of 8 V along the wire If the insulated wire is exposed to a medium at Ta=30oC with a heat transfer coefficient of h=12 wm2oC Determine the temperature at the interface of the wire and the plastic cover in steady operation Also determine whether doubling the thickness of the plastic cover will increase or decrease this interface temperature

dx

TdzdyKQ xx )(

dxQx

QQ xxdxx ][

dxdx

TdzdyK

xQQdQ xdxxxx ])([)(

)1()]([)(

ddzdydx

x

TK

xdQ xx

GENERAL HEAT CONDUCTION EQUATION IN CARTESIAN COORDINATES

Consider an unsteady (transient) three dimensional heat conduction with internal heat generation in heterogeneous amp anisotropic PLANE WALL

Cartesian co-ordinates x y z

Elemental length dxdydz

Consider an elemental volume ABCDEFGH = (dxdydz)

Let Temperature distribution is a function of T=T(xyz)

Let

=internal heat generation per unit volume per unit time (W m3)q= rate of heat flow (W)Q= qdt=total heat flow (J)r=mass density of the material element(kgm3)Cp=specific heat of the material element (Jkg K)

gq

Plane (y - z) lsquoxrsquo direction

Quantity of heat leaving from the face EFGH in lsquoXrsquo direction

Quantity of heat flowing into the face ABCD in lsquoXrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoXrsquo direction

dy

TdzdxKQ yy )(

dyQy

QQ yydyy ][

dydy

TdzdxK

yQQdQ ydyyyy ])([)(

)2()]([)(

ddzdydx

y

TK

ydQ yy

Plane (x - z) lsquoyrsquo direction

Quantity of heat flowing into the face ABFE in lsquoYrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoYrsquo direction

dz

TdydxKQ zz )(

dzQz

QQ zzdzz ][

dzdz

TdydxK

zQQdQ zdzzzz ])([)(

)3()]([)(

ddzdydx

z

TK

zdQ zz

Plane (x - y) lsquozrsquo direction

Quantity of heat flowing into the face ADHE in lsquoZrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoZrsquo direction

Quantity of heat leaving from the face DCGH in lsquoYrsquo direction

Quantity of heat leaving from the face BCGF in lsquoZrsquo direction

)4()()(

ddzdydxqQgg

Total amount of internal heat generation in the element

Ein + Egen = Eout+Est (Ein- Eout)+ Egen=Est

T

Cddzdydxddzdydxqddzdydxz

TK

zy

TK

yx

TK

x pgzyx )()())](()()([

T

pC

gq

z

TzK

zy

Ty

Kyx

TxK

x][][][

stgzyx QQdQdQdQ

)(

)5()()(

dT

CdzdydxQ pst

Total heat energy stored in the element

(ie thermal energy of the element can be increased because of heat accumulation in the element due to heat flow amp internal heat generation)

)( volumem d

TmCQ pst )(

T

CqTK pg)(

KKKyKx z If material is homogeneous amp isotropic

T

pC

gq

z

T

zy

T

yx

T

xK ][][][

T

Kp

C

Kgq

z

T

y

T

x

T2

2

2

2

2

2

T

Kgq

z

T

y

T

x

T 12

2

2

2

2

2

Thermal diffusivity pC

K

T

K

qT g

12

according to thermodynamics first law

)( zr )( dzrddr

)( dzrddr

Cylindrical polar co-ordinates

Elemental length

Consider an elemental volume ABCDEFGH =

GENERAL HEAT CONDUCTION EQUATION IN CYLINDRICAL COORDINATES

gq

gq

Consider an unsteady (transient) three dimensional heat conduction with internal heat generation in heterogeneous amp anisotropic cylindrical geometries (eg rods and pipes)

)( zrTLetTemperature distribution is a function of T=

Let

=internal heat generation per unit volume per unit time (W m3)q= rate of heat flow (W)Q= qdt=total heat flow (J)r=mass density of the material element(kgm3)Cp=specific heat of the material element (Jkg K)

planerdirectionz

planezrdirection

planezdirectionr

)(

)(

)(

LetHeat flow in

drdr

TdzrdK

rdQ

QQdQ

rr

drrrr

])([

)1(][

r

Tr

rddrdzdKdQ rr

Heat accumulation in the element due to heat flow by conduction in lsquorrsquo direction

rddr

TdzdrK

rdQ

QQdQ d

])([

][

2

T

r

ddrdzrdKdQ

)2(][

2

2

2

T

r

ddrdzrdKdQ

Heat accumulation in the element due to heat flow by conduction in lsquoqrsquo direction

Plane (z- q ) lsquorrsquo direction

dr

TdzrdKQ rr

)(

drQr

QQ rrdrr ][

Quantity of heat leaving from the face EFGH in lsquorrsquo direction

Quantity of heat flowing into the face ABCD in lsquorrsquo direction

Plane (r-z) lsquoq lsquodirection

Quantity of heat leaving from the face AEGC in lsquoqrsquo direction

d

r

TdzdrKQ )(

rdQ

rQQ d ][

Quantity of heat flowing into the face BFHD in lsquoqrsquo direction

)3())((2

2

z

TdzdrrdKdQ zz

)5()()(

dT

CdzdrrdQ pst

)4()()(

ddzdrrdqQgg

Total amount of internal heat generation in the element

Total heat energy stored in the element

(ie thermal energy of the element can be increased because of heat accumulation in the element due to heat flow amp internal heat generation)

according to thermodynamics first law

Ein + Egen = Eout + Est (Ein ndash Eout) + Egen = Est

)( volumem

)()(z

T

zdzdrrdKdQ zz

Heat accumulation in the element due to heat flow by conduction in lsquozrsquo direction

dzdz

TdrrdK

zdQ

QQdQ

zz

dzzzz

])([

d

TmCQ pst )(

Plane (r-q) lsquozrsquo direction

dz

TdrrdKQ zz )(

dzQz

QQ zzdzz ][ Quantity of heat leaving from the face CDHG in lsquozrsquo direction

Quantity of heat flowing into the face ABFE in lsquozrsquo direction

stgzr QQdQdQdQ

)(

r

r ][)(rT

rr

ddrdzdKr

][)(

2

2

2

T

r

ddrdzrdK ))((2

2

z

TddzdrrdK z

ddzdrrdq g )(

dT

Cdzdrrd p )(

+ + +

=

T

CqTK pg)(KKKK zr

[gq

)(1

r

Tr

rrK r

)(1

2

2

2 T

rK )(

2

2

z

TK z

)( ddzdrrd )(

ddzdrrd

TC p

+ + + ] =

If material is homogeneous amp isotropic

)(1

r

Tr

rr

)(1

2

2

2 T

r)(

2

2

z

T

K

q g

T

K

C p

)(1

r

Tr

rr

)(1

2

2

2 T

r )(2

2

z

T

K

q g

T

1

+ + + =

)(1

r

Tr

rrK r

)(1

2

2

2 T

rK )(

2

2

z

TK z

gq

T

C p + + + =

+ + + = Thermal diffusivity

T

K

qT g

12

pC

K

(Ein ndash Eout) + Egen = Est

)( rT

)( r

)sin( drrddr)sin( drrddr

planerdirection

planerdirection

planedirectionr

)(

)(

)(

Spherical polar co-ordinates

Elemental length

Consider an elemental volume ABCDEFGH =

GENERAL HEAT CONDUCTION EQUATION IN SPHERICAL COORDINATES

Let

=internal heat generation per unit volume per unit time (W m3)q= rate of heat flow (W)Q= qdt=total heat flow (J)r=mass density of the material element(kgm3)Cp=specific heat of the material element (Jkg K)

gq

Consider an unsteady (transient) three dimensional heat conduction with internal heat generation in heterogeneous amp anisotropic spherical geometries

LetTemperature distribution is a function of T=

X

Y

Z

drT

drrdKQ rr )sin(

drQr

QQ rrdrr ][

drdr

TdrrdK

rdQ

QQdQ

rr

drrrr

])sin([

)1(]([sin 2

rT

rr

ddrddKdQ rr

Plane ( -q f) lsquorrsquo directionQuantity of heat flowing into the face ABCD in lsquorrsquo direction

Quantity of heat leaving from the face EFGH in lsquorrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquorrsquo direction

drT

drdrKQ )sin(

rdQ

rQQ d ][

rddr

TdrdrK

rdQ

QQdQ d

])sin([

])[(sin

2

T

r

drdrdrKdQ

)2(])(sin[

2

T

r

drdrddrKdQ

Quantity of heat leaving from the face BFHD in lsquoqrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoqrsquo direction

X

Y

Z

Plane (r- f ) lsquoq lsquodirection

Quantity of heat flowing into the face AEGC in lsquoqrsquo direction

dr

TdrrdKQ

sin)(

drQ

rQQ d sin][

sin

drdr

TdrrdK

rdQ

QQdQ d

sin]sin

)([sin

][sinsin

)sin(

T

rr

ddrdrrdKdQ

)3()(sin

)sin(2

2

22

2

T

r

ddddrrKdQ

Plane (r- ) q lsquofrsquo directionQuantity of heat flowing into the face ABFE in lsquofrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquozrsquo direction

)4()sin()(

ddrrddrqQgg

)5()sin()(

dT

CdrrddrQ pst

Total amount of internal heat generation in the element

Total heat energy stored in the element

(Ein ndash Eout) + Egen = Est

d

TmCQ pst )(

)( volumem

Quantity of heat leaving from the face CDHG in lsquofrsquo direction

according to thermodynamics first law

Ein + Egen = Eout + Est

X

Y

Z

stgr QQdQdQdQ

)(

]([sin 2

r

Tr

rddrddKr

])(sin[

2

T

rdrdrddrK

)(sin

)sin(2

2

22

2

T

r

ddddrrK

ddrrddrq g )sin(

dT

Cdrrddr p )sin(

2

2

r

r

sin

sin

]([ 22 r

Tr

rr

Kr

])(sin[sin2

TrK

)(sin 2

2

22

T

r

Kgq

T

Cp

]([1

[ 22 r

Tr

rrK

])(sin[sin1

2

T

r)](

sin

12

2

22 T

r gq

T

Cp

]([1

[ 22 r

Tr

rr

])(sin[sin1

2

T

r)](

sin

12

2

22 T

rK

q g

T

K

Cp

]([1

[ 22 r

Tr

rr

])(sin[sin1

2

T

r)](

sin

12

2

22 T

rK

q g

T1

KKKKr If material is homogeneous amp isotropic

TCqTK pg)(

T

K

qT g

12 pC

K

Thermal diffusivity

(Ein ndash Eout) + Egen = Est

General Heat Conduction Equation

PLANE WALL CYLINDER SPHERE

Temperature profile

Plane wall SphereCylinder

bull A copper conductor (k = 380 WmoC resistivity ρ = 2x10-8 Ω m) having inner and outer radii 10 cm and 225 cm respectively is carrying a current density of 4800 amperes cm2 the conductor is internally cooled and a constant temperature of 65oC is maintained at the inner surface and there is no heat transfer through the insulation surrounding the conductor

Determinebull (i) The maximum temperature of the conductor and the radius at

which it occursbull (ii) The internal heat transfer rate

bull A chemical reaction takes place in a packed bed (k=06 WmoC) between two coaxial cylinders with radii 15 mm and 45 mm the inner surface is at 580oC and it is insulated Assuming the reaction rate of 055 MWm3 in the reactor volume find the temperature at the outer surface of the reactor

bull A current of 300 amperes passes through a stainless steel wire of 25 mm diameter and k=20 WmoC The resistivity of the wire is 70x10-8 ohm-m and length of the wire is 2 m if the wire is submerged in a fluid maintained at 50oC and h at the wire surface is 4000 Wm2oC calculate the steady state temperature at the centre and at the surface of the wire

bull An approximately spherical shaped orange ( k=023 WmoC) 90 mm diameter undergoes ripping process and generates 5100 Wm3 of energy If external surface the orange is at 8oC determine

bull (i) temperature at the centre of the orange and

bull (ii) heat flow from the outer surface of the orange

bull FINS External surfaces to used to cooling area and hence convection

heat transfer can be improvedbull APPLICATIONS IC engine cylinder ( air cooling) Heat exchangers automobile

radiators computer chip refrigerators electric motors generators and transformers etc

bull ASSUMPTIONS MADE IN ANALYSIS OF HEAT TRANFER THROUGH FIN

1 Fin material is homogenous and isotropic constthermal conductivity

2 Temperature at any cs of the fin is uniform3 lsquohrsquo is uniform over entire surface4 No internal heat generation5 Thermal contact resistance is negligible6 Radiation heat transfer is negligible7 Heat conduction is steady state and one dimensional

bull Problem 1

bull A stainless steel fin (k=20 WmK) having a diameter of 20 mm and a length of 01 m is attached to a wall at 300oC The ambient temperature is 50oC and the heat transfer coefficient is 10 Wm2K The fin tip is insulated Determine

bull (A) The rate of heat dissipation from the finbull (B) The temperature at the fin tipbull (C) The rate of heat transfer from the wall area covered by the fin if

the fin was not usedbull (D) The heat transfer rate from the same fin geometry if the

stainless steel fin is replaced by a fictitious fin with infinite thermal conductivity

bull Problem2bull One end of a long rod is inserted into a furnace while the other

projects into ambient temperature Under steady state the temperature of the rod is measured at two points 75mm apart and found to be 125oC and 885oC respectively while the ambient temperature is 20oC If the rod is 25mm in diameter and h is 2336 Wm2K

bull Find the thermal conductivity of the rod material

  • HEAT AND MASS TRANSFER
  • Thermodynamics Vs Heat transfer
  • MODE OF HEAT TRANSFER
  • MODE OF HEAT TRANSFER (2)
  • CONDUCTION- MECHANISM
  • CONCECTION-MECHANISM
  • RADIATION-MECHANISM
  • Conduction- Convection- Radiation
  • Slide 9
  • Slide 10
  • Fourierrsquos Law of Heat Conduction
  • Slide 12
  • Newtonrsquos Law of Cooling
  • Laws of Thermal Radiation
  • LAWS FOR HEAT TRANSFER
  • ELECTRICAL ampTHERMAL ANALOGY
  • Slide 17
  • Slide 18
  • PLANE WALL-UNIDIRECTIONAL
  • PLANE WALL -SERIES
  • PLANE WALL IN SERIES
  • PLANE WALL -PARALLEL
  • PLANE WALL-PARALLEL
  • PLANE WALL-COMBINATION
  • PLANE WALL-FILM RESISTANCE
  • PLANE WALL- THERMAL CONTACT RESISTANCE
  • PLANE WALL- THERMAL CONTACT RESISTANCE (2)
  • CYLINDER-SOLID
  • HEAT TRANSFER IN SOLID CYLINDER
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER-COMPOSITE
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER-MULTY LAYER
  • HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER
  • HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER (2)
  • HEAT TRANSFER IN SPHERE
  • HEAT TRANSFER IN SPHERE -MULTY LAYER
  • HEAT TRANSFER IN SPHERE -MULTY LAYER (2)
  • Slide 38
  • Slide 39
  • Slide 40
  • Slide 41
  • CRITICAL RADIUS
  • Plane wall- Insulation thickness -study
  • Problem5
  • Slide 45
  • Slide 46
  • Slide 47
  • GENERAL HEAT CONDUCTION EQUATION IN CYLINDRICAL COORDINATES
  • Slide 49
  • Slide 50
  • Slide 51
  • GENERAL HEAT CONDUCTION EQUATION IN SPHERICAL COORDINATES
  • Slide 53
  • Slide 54
  • Slide 55
  • General Heat Conduction Equation
  • Temperature profile
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Slide 67
Page 3: hmt chap 1

MODE OF HEAT TRANSFER

1048633ConductionConduction is the flow of heat in a substance due to exchange of energy between molecules having more energy and molecules having less energy

[ solids Lattice vibrations (ii) motion of free electrons][ fluids conduction is due to collision between the molecules caused by the random motion]

bull -needs matter (solid liquid gas)bull -molecular phenomenon (diffusion process)bull -without bulk motion of matter

1048633ConvectionThe transfer of energy from one region to another region due to macroscopic motion in fluid added on to the energy transfer by conduction is called heat transfer by conduction

bull -heat carried away by bulk motion of fluidbull -needs fluid matter

Forced Convection fluid motion is caused by an external agencyNatural Convection fluid motion occurs due to density variations caused by temperature differences

1048633RadiationAll the physical matter emits thermal radiation in the form of electromagnetic waves because of vibrational and rotational movements of the molecules and atoms which make up the matter

bull -does not needs matterbull -transmission of energy by electromagnetic waves bull Radiation increases with the temperature level

MODE OF HEAT TRANSFER

CONDUCTION- MECHANISM

bull CONDUCTION Conduction is the transfer of energy from more energetic

particles of a substance to the adjacent less energetic one as a result of interactions between the particles

In gases and liquids conduction is due to the collisions and diffusion of the molecules during their random motion

In solids it is due to the combination of vibrations of the molecules in a lattice and the energy transport by free electrons

Micro form of heat transferFOURIERrsquoS LAW

q =-KA( dT dX)

CONCECTION-MECHANISMbull Convection Heat transfer between a solid surface and an adjacent

gas or liquid It is the combination of conduction and flow motion Heat transferred from a solid surface to a liquid adjacent is conduction And then heat is brought away by the flow motion

bull Newtons law of cooling

where h = Convection heat transfer coefficient Ts = Temperature of the solid surface Tf = Temperature of the fluid

bull Macroform of heat transfer CONVECTION NEWTONrsquoS LAW OF COOLING

q = hA (Ts-Tf)

RADIATION-MECHANISM

bull Radiation The energy emitted by matter in the form of electromagnetic waves as a result of the changes in the electronic configurations of the atoms or molecules

bull Stefan - Boltzmann law

where σ = Stefan - Boltzmann constant ε = emissivity Ts = Surface temperature of the object

q =σ ε AT4

Conduction- Convection- Radiation

bull The three modes of heat transfer always exist simultaneously For example the heat transfer associated with double pane windows are

bull Conduction Hotter (cooler) air outside each pane causes conduction through solid glass

bull Convection Air between the panes carries heat from hotter pane to cooler pane

bull Radiation Sunlight radiation passes through glass to be absorbed on other side

Examples for different mode of heat transfer

bull Conduction

Heat loss through thermal insulation on steam pipe

Convection

Heat transfer to water flowing through a pipe in condenser

Radiation

Heat transfer in an electric furnace

Conduction Convection and Radiation

Solar energy used water heater

Fourierrsquos Law of Heat Conduction

Newtonrsquos Law of Cooling

Laws of Thermal Radiation

LAWS FOR HEAT TRANSFERCONDUCTION

4Tq

CONVECTION

NEWTONrsquoS LAW OF COOLING

RADIATION

STEFAN BOLTZMAN LAW

q=-KA( dT dX)

Assumptionsbull Steady statebull Unidirectional heat flowbull Const temperature gradientbull Linear profilebull No internal heat generationbull Isothermal bounding surfacebull Homogeneousbull Isotropic bull Applicable to solid liquid gas

q=σAT4q= hA (Tw-Tα)

Thq

XT

Kq

FOURIERrsquoS LAW

q =σ ε AT4

ELECTRICAL ampTHERMAL ANALOGY

THERMAL RESISTANCE(ELECTRICAL ANALOGY)

bull OHMrsquos LAW Flow of Electricity

V=I R elect

Voltage Drop = Current flowtimesResistance Temp Drop=Heat FlowtimesResistance

ΔT= q Rtherm

Thermal Analogy To OHMrsquoS Law

PLANE WALL-UNIDIRECTIONAL

PLANE WALL -SERIES

PLANE WALL IN SERIES

PLANE WALL -PARALLEL

PLANE WALL-PARALLEL

PLANE WALL-COMBINATION

PLANE WALL-FILM RESISTANCE

PLANE WALL- THERMAL CONTACT RESISTANCE

PLANE WALL- THERMAL CONTACT RESISTANCE

At the joining surface of the two slabs air is trapped in voids due to surface irregularities

CYLINDER-SOLID

HEAT TRANSFER IN SOLID CYLINDER

HEAT TRANSFER RATE IN HOLLOW CYLINDER

HEAT TRANSFER RATE IN HOLLOW CYLINDER-COMPOSITE

HEAT TRANSFER RATE IN HOLLOW CYLINDER-MULTY LAYER

HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER

HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER

HEAT TRANSFER IN SPHERE

HEAT TRANSFER IN SPHERE -MULTY LAYER

HEAT TRANSFER IN SPHERE -MULTY LAYER

bull Problem 1 The wall in a furnace consists of 125 mm thick refractory bricks and

125 mm thick insulating fire bricks separated by an air gap A 12 mm thick plaster covers the outer wall The inner surface of wall is at 1100oC and the ambient temperature 25oC the heat transfer coefficient on the outer wall to the air is 17 Wm2K and the resistance to heat flow of the air gap is 016 KW the thermal conductivities of refractory brick insulating firebrick and plaster are 16 03 and 014 WmK respectively

Calculatebull (a) the rate of heat loss per unit area of the wall surfacebull (b) the interface temperatures throughout the wallbull (c) the temperature at the outside surface of the wall

bull Problem 2Two large aluminum plates (k=240 WmK) each 2 cm thick with 10 mm surface roughness are placed in contact under 105 Nm2 pressure in air The temperature at the outside surfaces are 390 oC and 406 oCCalculate

A) the heat flux B) the temperature drop due to the contact resistance C) the contact temperatures Thermal contact resistance with air as the interfacial fluid for

10 mm roughness is 27510-4 m2KW

bull Problem 3bull Steam at 350 oC flowing in a pipe (k=80 WmK) of 5 cm

inner diameter and 56 cm outer diameter is covered with 3 cm thick insulation of k=005 WmK Heat is lost to surroundings at 5 oC by natural convection and radiation the combined h being 20 Wm2K Taking the heat transfer coefficient inside the pipe as 60 Wm2K

bull Determinebull A) the rate of heat loss from the steam per unit length of

the pipebull B) the temperature drop across the pipe and the insulation

bull PROBLEM4bull A nuclear reactor has a sperical pressure vessel 750 mm inside

diameter 80 mm wall thickness The temperature at the inner surface is 500 oC If the tempertarue at the outer surface is 495 oC

bull Calculatea) The rate of heat loss through the metal wallb) If the rate of heat loss is to limited to 305Whow much inulation

thickness needs to be applied Thermal conductivity of steel wall=46 WmoC Thermal conductivity of insulation=004 WmoC Film coefficient of heat tranfer at inner and outer surface are 200

Wm2 oC and 15 Wm2 oC Ambient air temperature=28 oC

CRITICAL RADIUS

bull The radius upto which heat flow increases and after decreases At critical radius q becomes qmax

bull The addition of insulation always increases (Rth)condbull Rtotal= (Rth)cond + (Rth)conv

bull The addition of insulation may reduce the (Rth)conv due to increase in surface area

bull r1ltrc -heat transfer increases by adding insulation till r1=rc If

insulation thickness is further increased the rate of heat loss will decrease

r1 gt rc

- heat transfer decreases by adding insulation above the rc

Plane wall- Insulation

thickness -study

Cylinder amp Sphere- Insulation thickness -

study

Problem5bull A 3mm diameter and 5 mm long electric wire is tightly

wrapped with a 2 mm thick plastic cover whose K=015 wmoC Electrical measurements indicate that a current of 10 A passes through the wire and there is a voltage drop of 8 V along the wire If the insulated wire is exposed to a medium at Ta=30oC with a heat transfer coefficient of h=12 wm2oC Determine the temperature at the interface of the wire and the plastic cover in steady operation Also determine whether doubling the thickness of the plastic cover will increase or decrease this interface temperature

dx

TdzdyKQ xx )(

dxQx

QQ xxdxx ][

dxdx

TdzdyK

xQQdQ xdxxxx ])([)(

)1()]([)(

ddzdydx

x

TK

xdQ xx

GENERAL HEAT CONDUCTION EQUATION IN CARTESIAN COORDINATES

Consider an unsteady (transient) three dimensional heat conduction with internal heat generation in heterogeneous amp anisotropic PLANE WALL

Cartesian co-ordinates x y z

Elemental length dxdydz

Consider an elemental volume ABCDEFGH = (dxdydz)

Let Temperature distribution is a function of T=T(xyz)

Let

=internal heat generation per unit volume per unit time (W m3)q= rate of heat flow (W)Q= qdt=total heat flow (J)r=mass density of the material element(kgm3)Cp=specific heat of the material element (Jkg K)

gq

Plane (y - z) lsquoxrsquo direction

Quantity of heat leaving from the face EFGH in lsquoXrsquo direction

Quantity of heat flowing into the face ABCD in lsquoXrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoXrsquo direction

dy

TdzdxKQ yy )(

dyQy

QQ yydyy ][

dydy

TdzdxK

yQQdQ ydyyyy ])([)(

)2()]([)(

ddzdydx

y

TK

ydQ yy

Plane (x - z) lsquoyrsquo direction

Quantity of heat flowing into the face ABFE in lsquoYrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoYrsquo direction

dz

TdydxKQ zz )(

dzQz

QQ zzdzz ][

dzdz

TdydxK

zQQdQ zdzzzz ])([)(

)3()]([)(

ddzdydx

z

TK

zdQ zz

Plane (x - y) lsquozrsquo direction

Quantity of heat flowing into the face ADHE in lsquoZrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoZrsquo direction

Quantity of heat leaving from the face DCGH in lsquoYrsquo direction

Quantity of heat leaving from the face BCGF in lsquoZrsquo direction

)4()()(

ddzdydxqQgg

Total amount of internal heat generation in the element

Ein + Egen = Eout+Est (Ein- Eout)+ Egen=Est

T

Cddzdydxddzdydxqddzdydxz

TK

zy

TK

yx

TK

x pgzyx )()())](()()([

T

pC

gq

z

TzK

zy

Ty

Kyx

TxK

x][][][

stgzyx QQdQdQdQ

)(

)5()()(

dT

CdzdydxQ pst

Total heat energy stored in the element

(ie thermal energy of the element can be increased because of heat accumulation in the element due to heat flow amp internal heat generation)

)( volumem d

TmCQ pst )(

T

CqTK pg)(

KKKyKx z If material is homogeneous amp isotropic

T

pC

gq

z

T

zy

T

yx

T

xK ][][][

T

Kp

C

Kgq

z

T

y

T

x

T2

2

2

2

2

2

T

Kgq

z

T

y

T

x

T 12

2

2

2

2

2

Thermal diffusivity pC

K

T

K

qT g

12

according to thermodynamics first law

)( zr )( dzrddr

)( dzrddr

Cylindrical polar co-ordinates

Elemental length

Consider an elemental volume ABCDEFGH =

GENERAL HEAT CONDUCTION EQUATION IN CYLINDRICAL COORDINATES

gq

gq

Consider an unsteady (transient) three dimensional heat conduction with internal heat generation in heterogeneous amp anisotropic cylindrical geometries (eg rods and pipes)

)( zrTLetTemperature distribution is a function of T=

Let

=internal heat generation per unit volume per unit time (W m3)q= rate of heat flow (W)Q= qdt=total heat flow (J)r=mass density of the material element(kgm3)Cp=specific heat of the material element (Jkg K)

planerdirectionz

planezrdirection

planezdirectionr

)(

)(

)(

LetHeat flow in

drdr

TdzrdK

rdQ

QQdQ

rr

drrrr

])([

)1(][

r

Tr

rddrdzdKdQ rr

Heat accumulation in the element due to heat flow by conduction in lsquorrsquo direction

rddr

TdzdrK

rdQ

QQdQ d

])([

][

2

T

r

ddrdzrdKdQ

)2(][

2

2

2

T

r

ddrdzrdKdQ

Heat accumulation in the element due to heat flow by conduction in lsquoqrsquo direction

Plane (z- q ) lsquorrsquo direction

dr

TdzrdKQ rr

)(

drQr

QQ rrdrr ][

Quantity of heat leaving from the face EFGH in lsquorrsquo direction

Quantity of heat flowing into the face ABCD in lsquorrsquo direction

Plane (r-z) lsquoq lsquodirection

Quantity of heat leaving from the face AEGC in lsquoqrsquo direction

d

r

TdzdrKQ )(

rdQ

rQQ d ][

Quantity of heat flowing into the face BFHD in lsquoqrsquo direction

)3())((2

2

z

TdzdrrdKdQ zz

)5()()(

dT

CdzdrrdQ pst

)4()()(

ddzdrrdqQgg

Total amount of internal heat generation in the element

Total heat energy stored in the element

(ie thermal energy of the element can be increased because of heat accumulation in the element due to heat flow amp internal heat generation)

according to thermodynamics first law

Ein + Egen = Eout + Est (Ein ndash Eout) + Egen = Est

)( volumem

)()(z

T

zdzdrrdKdQ zz

Heat accumulation in the element due to heat flow by conduction in lsquozrsquo direction

dzdz

TdrrdK

zdQ

QQdQ

zz

dzzzz

])([

d

TmCQ pst )(

Plane (r-q) lsquozrsquo direction

dz

TdrrdKQ zz )(

dzQz

QQ zzdzz ][ Quantity of heat leaving from the face CDHG in lsquozrsquo direction

Quantity of heat flowing into the face ABFE in lsquozrsquo direction

stgzr QQdQdQdQ

)(

r

r ][)(rT

rr

ddrdzdKr

][)(

2

2

2

T

r

ddrdzrdK ))((2

2

z

TddzdrrdK z

ddzdrrdq g )(

dT

Cdzdrrd p )(

+ + +

=

T

CqTK pg)(KKKK zr

[gq

)(1

r

Tr

rrK r

)(1

2

2

2 T

rK )(

2

2

z

TK z

)( ddzdrrd )(

ddzdrrd

TC p

+ + + ] =

If material is homogeneous amp isotropic

)(1

r

Tr

rr

)(1

2

2

2 T

r)(

2

2

z

T

K

q g

T

K

C p

)(1

r

Tr

rr

)(1

2

2

2 T

r )(2

2

z

T

K

q g

T

1

+ + + =

)(1

r

Tr

rrK r

)(1

2

2

2 T

rK )(

2

2

z

TK z

gq

T

C p + + + =

+ + + = Thermal diffusivity

T

K

qT g

12

pC

K

(Ein ndash Eout) + Egen = Est

)( rT

)( r

)sin( drrddr)sin( drrddr

planerdirection

planerdirection

planedirectionr

)(

)(

)(

Spherical polar co-ordinates

Elemental length

Consider an elemental volume ABCDEFGH =

GENERAL HEAT CONDUCTION EQUATION IN SPHERICAL COORDINATES

Let

=internal heat generation per unit volume per unit time (W m3)q= rate of heat flow (W)Q= qdt=total heat flow (J)r=mass density of the material element(kgm3)Cp=specific heat of the material element (Jkg K)

gq

Consider an unsteady (transient) three dimensional heat conduction with internal heat generation in heterogeneous amp anisotropic spherical geometries

LetTemperature distribution is a function of T=

X

Y

Z

drT

drrdKQ rr )sin(

drQr

QQ rrdrr ][

drdr

TdrrdK

rdQ

QQdQ

rr

drrrr

])sin([

)1(]([sin 2

rT

rr

ddrddKdQ rr

Plane ( -q f) lsquorrsquo directionQuantity of heat flowing into the face ABCD in lsquorrsquo direction

Quantity of heat leaving from the face EFGH in lsquorrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquorrsquo direction

drT

drdrKQ )sin(

rdQ

rQQ d ][

rddr

TdrdrK

rdQ

QQdQ d

])sin([

])[(sin

2

T

r

drdrdrKdQ

)2(])(sin[

2

T

r

drdrddrKdQ

Quantity of heat leaving from the face BFHD in lsquoqrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoqrsquo direction

X

Y

Z

Plane (r- f ) lsquoq lsquodirection

Quantity of heat flowing into the face AEGC in lsquoqrsquo direction

dr

TdrrdKQ

sin)(

drQ

rQQ d sin][

sin

drdr

TdrrdK

rdQ

QQdQ d

sin]sin

)([sin

][sinsin

)sin(

T

rr

ddrdrrdKdQ

)3()(sin

)sin(2

2

22

2

T

r

ddddrrKdQ

Plane (r- ) q lsquofrsquo directionQuantity of heat flowing into the face ABFE in lsquofrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquozrsquo direction

)4()sin()(

ddrrddrqQgg

)5()sin()(

dT

CdrrddrQ pst

Total amount of internal heat generation in the element

Total heat energy stored in the element

(Ein ndash Eout) + Egen = Est

d

TmCQ pst )(

)( volumem

Quantity of heat leaving from the face CDHG in lsquofrsquo direction

according to thermodynamics first law

Ein + Egen = Eout + Est

X

Y

Z

stgr QQdQdQdQ

)(

]([sin 2

r

Tr

rddrddKr

])(sin[

2

T

rdrdrddrK

)(sin

)sin(2

2

22

2

T

r

ddddrrK

ddrrddrq g )sin(

dT

Cdrrddr p )sin(

2

2

r

r

sin

sin

]([ 22 r

Tr

rr

Kr

])(sin[sin2

TrK

)(sin 2

2

22

T

r

Kgq

T

Cp

]([1

[ 22 r

Tr

rrK

])(sin[sin1

2

T

r)](

sin

12

2

22 T

r gq

T

Cp

]([1

[ 22 r

Tr

rr

])(sin[sin1

2

T

r)](

sin

12

2

22 T

rK

q g

T

K

Cp

]([1

[ 22 r

Tr

rr

])(sin[sin1

2

T

r)](

sin

12

2

22 T

rK

q g

T1

KKKKr If material is homogeneous amp isotropic

TCqTK pg)(

T

K

qT g

12 pC

K

Thermal diffusivity

(Ein ndash Eout) + Egen = Est

General Heat Conduction Equation

PLANE WALL CYLINDER SPHERE

Temperature profile

Plane wall SphereCylinder

bull A copper conductor (k = 380 WmoC resistivity ρ = 2x10-8 Ω m) having inner and outer radii 10 cm and 225 cm respectively is carrying a current density of 4800 amperes cm2 the conductor is internally cooled and a constant temperature of 65oC is maintained at the inner surface and there is no heat transfer through the insulation surrounding the conductor

Determinebull (i) The maximum temperature of the conductor and the radius at

which it occursbull (ii) The internal heat transfer rate

bull A chemical reaction takes place in a packed bed (k=06 WmoC) between two coaxial cylinders with radii 15 mm and 45 mm the inner surface is at 580oC and it is insulated Assuming the reaction rate of 055 MWm3 in the reactor volume find the temperature at the outer surface of the reactor

bull A current of 300 amperes passes through a stainless steel wire of 25 mm diameter and k=20 WmoC The resistivity of the wire is 70x10-8 ohm-m and length of the wire is 2 m if the wire is submerged in a fluid maintained at 50oC and h at the wire surface is 4000 Wm2oC calculate the steady state temperature at the centre and at the surface of the wire

bull An approximately spherical shaped orange ( k=023 WmoC) 90 mm diameter undergoes ripping process and generates 5100 Wm3 of energy If external surface the orange is at 8oC determine

bull (i) temperature at the centre of the orange and

bull (ii) heat flow from the outer surface of the orange

bull FINS External surfaces to used to cooling area and hence convection

heat transfer can be improvedbull APPLICATIONS IC engine cylinder ( air cooling) Heat exchangers automobile

radiators computer chip refrigerators electric motors generators and transformers etc

bull ASSUMPTIONS MADE IN ANALYSIS OF HEAT TRANFER THROUGH FIN

1 Fin material is homogenous and isotropic constthermal conductivity

2 Temperature at any cs of the fin is uniform3 lsquohrsquo is uniform over entire surface4 No internal heat generation5 Thermal contact resistance is negligible6 Radiation heat transfer is negligible7 Heat conduction is steady state and one dimensional

bull Problem 1

bull A stainless steel fin (k=20 WmK) having a diameter of 20 mm and a length of 01 m is attached to a wall at 300oC The ambient temperature is 50oC and the heat transfer coefficient is 10 Wm2K The fin tip is insulated Determine

bull (A) The rate of heat dissipation from the finbull (B) The temperature at the fin tipbull (C) The rate of heat transfer from the wall area covered by the fin if

the fin was not usedbull (D) The heat transfer rate from the same fin geometry if the

stainless steel fin is replaced by a fictitious fin with infinite thermal conductivity

bull Problem2bull One end of a long rod is inserted into a furnace while the other

projects into ambient temperature Under steady state the temperature of the rod is measured at two points 75mm apart and found to be 125oC and 885oC respectively while the ambient temperature is 20oC If the rod is 25mm in diameter and h is 2336 Wm2K

bull Find the thermal conductivity of the rod material

  • HEAT AND MASS TRANSFER
  • Thermodynamics Vs Heat transfer
  • MODE OF HEAT TRANSFER
  • MODE OF HEAT TRANSFER (2)
  • CONDUCTION- MECHANISM
  • CONCECTION-MECHANISM
  • RADIATION-MECHANISM
  • Conduction- Convection- Radiation
  • Slide 9
  • Slide 10
  • Fourierrsquos Law of Heat Conduction
  • Slide 12
  • Newtonrsquos Law of Cooling
  • Laws of Thermal Radiation
  • LAWS FOR HEAT TRANSFER
  • ELECTRICAL ampTHERMAL ANALOGY
  • Slide 17
  • Slide 18
  • PLANE WALL-UNIDIRECTIONAL
  • PLANE WALL -SERIES
  • PLANE WALL IN SERIES
  • PLANE WALL -PARALLEL
  • PLANE WALL-PARALLEL
  • PLANE WALL-COMBINATION
  • PLANE WALL-FILM RESISTANCE
  • PLANE WALL- THERMAL CONTACT RESISTANCE
  • PLANE WALL- THERMAL CONTACT RESISTANCE (2)
  • CYLINDER-SOLID
  • HEAT TRANSFER IN SOLID CYLINDER
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER-COMPOSITE
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER-MULTY LAYER
  • HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER
  • HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER (2)
  • HEAT TRANSFER IN SPHERE
  • HEAT TRANSFER IN SPHERE -MULTY LAYER
  • HEAT TRANSFER IN SPHERE -MULTY LAYER (2)
  • Slide 38
  • Slide 39
  • Slide 40
  • Slide 41
  • CRITICAL RADIUS
  • Plane wall- Insulation thickness -study
  • Problem5
  • Slide 45
  • Slide 46
  • Slide 47
  • GENERAL HEAT CONDUCTION EQUATION IN CYLINDRICAL COORDINATES
  • Slide 49
  • Slide 50
  • Slide 51
  • GENERAL HEAT CONDUCTION EQUATION IN SPHERICAL COORDINATES
  • Slide 53
  • Slide 54
  • Slide 55
  • General Heat Conduction Equation
  • Temperature profile
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Slide 67
Page 4: hmt chap 1

MODE OF HEAT TRANSFER

CONDUCTION- MECHANISM

bull CONDUCTION Conduction is the transfer of energy from more energetic

particles of a substance to the adjacent less energetic one as a result of interactions between the particles

In gases and liquids conduction is due to the collisions and diffusion of the molecules during their random motion

In solids it is due to the combination of vibrations of the molecules in a lattice and the energy transport by free electrons

Micro form of heat transferFOURIERrsquoS LAW

q =-KA( dT dX)

CONCECTION-MECHANISMbull Convection Heat transfer between a solid surface and an adjacent

gas or liquid It is the combination of conduction and flow motion Heat transferred from a solid surface to a liquid adjacent is conduction And then heat is brought away by the flow motion

bull Newtons law of cooling

where h = Convection heat transfer coefficient Ts = Temperature of the solid surface Tf = Temperature of the fluid

bull Macroform of heat transfer CONVECTION NEWTONrsquoS LAW OF COOLING

q = hA (Ts-Tf)

RADIATION-MECHANISM

bull Radiation The energy emitted by matter in the form of electromagnetic waves as a result of the changes in the electronic configurations of the atoms or molecules

bull Stefan - Boltzmann law

where σ = Stefan - Boltzmann constant ε = emissivity Ts = Surface temperature of the object

q =σ ε AT4

Conduction- Convection- Radiation

bull The three modes of heat transfer always exist simultaneously For example the heat transfer associated with double pane windows are

bull Conduction Hotter (cooler) air outside each pane causes conduction through solid glass

bull Convection Air between the panes carries heat from hotter pane to cooler pane

bull Radiation Sunlight radiation passes through glass to be absorbed on other side

Examples for different mode of heat transfer

bull Conduction

Heat loss through thermal insulation on steam pipe

Convection

Heat transfer to water flowing through a pipe in condenser

Radiation

Heat transfer in an electric furnace

Conduction Convection and Radiation

Solar energy used water heater

Fourierrsquos Law of Heat Conduction

Newtonrsquos Law of Cooling

Laws of Thermal Radiation

LAWS FOR HEAT TRANSFERCONDUCTION

4Tq

CONVECTION

NEWTONrsquoS LAW OF COOLING

RADIATION

STEFAN BOLTZMAN LAW

q=-KA( dT dX)

Assumptionsbull Steady statebull Unidirectional heat flowbull Const temperature gradientbull Linear profilebull No internal heat generationbull Isothermal bounding surfacebull Homogeneousbull Isotropic bull Applicable to solid liquid gas

q=σAT4q= hA (Tw-Tα)

Thq

XT

Kq

FOURIERrsquoS LAW

q =σ ε AT4

ELECTRICAL ampTHERMAL ANALOGY

THERMAL RESISTANCE(ELECTRICAL ANALOGY)

bull OHMrsquos LAW Flow of Electricity

V=I R elect

Voltage Drop = Current flowtimesResistance Temp Drop=Heat FlowtimesResistance

ΔT= q Rtherm

Thermal Analogy To OHMrsquoS Law

PLANE WALL-UNIDIRECTIONAL

PLANE WALL -SERIES

PLANE WALL IN SERIES

PLANE WALL -PARALLEL

PLANE WALL-PARALLEL

PLANE WALL-COMBINATION

PLANE WALL-FILM RESISTANCE

PLANE WALL- THERMAL CONTACT RESISTANCE

PLANE WALL- THERMAL CONTACT RESISTANCE

At the joining surface of the two slabs air is trapped in voids due to surface irregularities

CYLINDER-SOLID

HEAT TRANSFER IN SOLID CYLINDER

HEAT TRANSFER RATE IN HOLLOW CYLINDER

HEAT TRANSFER RATE IN HOLLOW CYLINDER-COMPOSITE

HEAT TRANSFER RATE IN HOLLOW CYLINDER-MULTY LAYER

HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER

HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER

HEAT TRANSFER IN SPHERE

HEAT TRANSFER IN SPHERE -MULTY LAYER

HEAT TRANSFER IN SPHERE -MULTY LAYER

bull Problem 1 The wall in a furnace consists of 125 mm thick refractory bricks and

125 mm thick insulating fire bricks separated by an air gap A 12 mm thick plaster covers the outer wall The inner surface of wall is at 1100oC and the ambient temperature 25oC the heat transfer coefficient on the outer wall to the air is 17 Wm2K and the resistance to heat flow of the air gap is 016 KW the thermal conductivities of refractory brick insulating firebrick and plaster are 16 03 and 014 WmK respectively

Calculatebull (a) the rate of heat loss per unit area of the wall surfacebull (b) the interface temperatures throughout the wallbull (c) the temperature at the outside surface of the wall

bull Problem 2Two large aluminum plates (k=240 WmK) each 2 cm thick with 10 mm surface roughness are placed in contact under 105 Nm2 pressure in air The temperature at the outside surfaces are 390 oC and 406 oCCalculate

A) the heat flux B) the temperature drop due to the contact resistance C) the contact temperatures Thermal contact resistance with air as the interfacial fluid for

10 mm roughness is 27510-4 m2KW

bull Problem 3bull Steam at 350 oC flowing in a pipe (k=80 WmK) of 5 cm

inner diameter and 56 cm outer diameter is covered with 3 cm thick insulation of k=005 WmK Heat is lost to surroundings at 5 oC by natural convection and radiation the combined h being 20 Wm2K Taking the heat transfer coefficient inside the pipe as 60 Wm2K

bull Determinebull A) the rate of heat loss from the steam per unit length of

the pipebull B) the temperature drop across the pipe and the insulation

bull PROBLEM4bull A nuclear reactor has a sperical pressure vessel 750 mm inside

diameter 80 mm wall thickness The temperature at the inner surface is 500 oC If the tempertarue at the outer surface is 495 oC

bull Calculatea) The rate of heat loss through the metal wallb) If the rate of heat loss is to limited to 305Whow much inulation

thickness needs to be applied Thermal conductivity of steel wall=46 WmoC Thermal conductivity of insulation=004 WmoC Film coefficient of heat tranfer at inner and outer surface are 200

Wm2 oC and 15 Wm2 oC Ambient air temperature=28 oC

CRITICAL RADIUS

bull The radius upto which heat flow increases and after decreases At critical radius q becomes qmax

bull The addition of insulation always increases (Rth)condbull Rtotal= (Rth)cond + (Rth)conv

bull The addition of insulation may reduce the (Rth)conv due to increase in surface area

bull r1ltrc -heat transfer increases by adding insulation till r1=rc If

insulation thickness is further increased the rate of heat loss will decrease

r1 gt rc

- heat transfer decreases by adding insulation above the rc

Plane wall- Insulation

thickness -study

Cylinder amp Sphere- Insulation thickness -

study

Problem5bull A 3mm diameter and 5 mm long electric wire is tightly

wrapped with a 2 mm thick plastic cover whose K=015 wmoC Electrical measurements indicate that a current of 10 A passes through the wire and there is a voltage drop of 8 V along the wire If the insulated wire is exposed to a medium at Ta=30oC with a heat transfer coefficient of h=12 wm2oC Determine the temperature at the interface of the wire and the plastic cover in steady operation Also determine whether doubling the thickness of the plastic cover will increase or decrease this interface temperature

dx

TdzdyKQ xx )(

dxQx

QQ xxdxx ][

dxdx

TdzdyK

xQQdQ xdxxxx ])([)(

)1()]([)(

ddzdydx

x

TK

xdQ xx

GENERAL HEAT CONDUCTION EQUATION IN CARTESIAN COORDINATES

Consider an unsteady (transient) three dimensional heat conduction with internal heat generation in heterogeneous amp anisotropic PLANE WALL

Cartesian co-ordinates x y z

Elemental length dxdydz

Consider an elemental volume ABCDEFGH = (dxdydz)

Let Temperature distribution is a function of T=T(xyz)

Let

=internal heat generation per unit volume per unit time (W m3)q= rate of heat flow (W)Q= qdt=total heat flow (J)r=mass density of the material element(kgm3)Cp=specific heat of the material element (Jkg K)

gq

Plane (y - z) lsquoxrsquo direction

Quantity of heat leaving from the face EFGH in lsquoXrsquo direction

Quantity of heat flowing into the face ABCD in lsquoXrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoXrsquo direction

dy

TdzdxKQ yy )(

dyQy

QQ yydyy ][

dydy

TdzdxK

yQQdQ ydyyyy ])([)(

)2()]([)(

ddzdydx

y

TK

ydQ yy

Plane (x - z) lsquoyrsquo direction

Quantity of heat flowing into the face ABFE in lsquoYrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoYrsquo direction

dz

TdydxKQ zz )(

dzQz

QQ zzdzz ][

dzdz

TdydxK

zQQdQ zdzzzz ])([)(

)3()]([)(

ddzdydx

z

TK

zdQ zz

Plane (x - y) lsquozrsquo direction

Quantity of heat flowing into the face ADHE in lsquoZrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoZrsquo direction

Quantity of heat leaving from the face DCGH in lsquoYrsquo direction

Quantity of heat leaving from the face BCGF in lsquoZrsquo direction

)4()()(

ddzdydxqQgg

Total amount of internal heat generation in the element

Ein + Egen = Eout+Est (Ein- Eout)+ Egen=Est

T

Cddzdydxddzdydxqddzdydxz

TK

zy

TK

yx

TK

x pgzyx )()())](()()([

T

pC

gq

z

TzK

zy

Ty

Kyx

TxK

x][][][

stgzyx QQdQdQdQ

)(

)5()()(

dT

CdzdydxQ pst

Total heat energy stored in the element

(ie thermal energy of the element can be increased because of heat accumulation in the element due to heat flow amp internal heat generation)

)( volumem d

TmCQ pst )(

T

CqTK pg)(

KKKyKx z If material is homogeneous amp isotropic

T

pC

gq

z

T

zy

T

yx

T

xK ][][][

T

Kp

C

Kgq

z

T

y

T

x

T2

2

2

2

2

2

T

Kgq

z

T

y

T

x

T 12

2

2

2

2

2

Thermal diffusivity pC

K

T

K

qT g

12

according to thermodynamics first law

)( zr )( dzrddr

)( dzrddr

Cylindrical polar co-ordinates

Elemental length

Consider an elemental volume ABCDEFGH =

GENERAL HEAT CONDUCTION EQUATION IN CYLINDRICAL COORDINATES

gq

gq

Consider an unsteady (transient) three dimensional heat conduction with internal heat generation in heterogeneous amp anisotropic cylindrical geometries (eg rods and pipes)

)( zrTLetTemperature distribution is a function of T=

Let

=internal heat generation per unit volume per unit time (W m3)q= rate of heat flow (W)Q= qdt=total heat flow (J)r=mass density of the material element(kgm3)Cp=specific heat of the material element (Jkg K)

planerdirectionz

planezrdirection

planezdirectionr

)(

)(

)(

LetHeat flow in

drdr

TdzrdK

rdQ

QQdQ

rr

drrrr

])([

)1(][

r

Tr

rddrdzdKdQ rr

Heat accumulation in the element due to heat flow by conduction in lsquorrsquo direction

rddr

TdzdrK

rdQ

QQdQ d

])([

][

2

T

r

ddrdzrdKdQ

)2(][

2

2

2

T

r

ddrdzrdKdQ

Heat accumulation in the element due to heat flow by conduction in lsquoqrsquo direction

Plane (z- q ) lsquorrsquo direction

dr

TdzrdKQ rr

)(

drQr

QQ rrdrr ][

Quantity of heat leaving from the face EFGH in lsquorrsquo direction

Quantity of heat flowing into the face ABCD in lsquorrsquo direction

Plane (r-z) lsquoq lsquodirection

Quantity of heat leaving from the face AEGC in lsquoqrsquo direction

d

r

TdzdrKQ )(

rdQ

rQQ d ][

Quantity of heat flowing into the face BFHD in lsquoqrsquo direction

)3())((2

2

z

TdzdrrdKdQ zz

)5()()(

dT

CdzdrrdQ pst

)4()()(

ddzdrrdqQgg

Total amount of internal heat generation in the element

Total heat energy stored in the element

(ie thermal energy of the element can be increased because of heat accumulation in the element due to heat flow amp internal heat generation)

according to thermodynamics first law

Ein + Egen = Eout + Est (Ein ndash Eout) + Egen = Est

)( volumem

)()(z

T

zdzdrrdKdQ zz

Heat accumulation in the element due to heat flow by conduction in lsquozrsquo direction

dzdz

TdrrdK

zdQ

QQdQ

zz

dzzzz

])([

d

TmCQ pst )(

Plane (r-q) lsquozrsquo direction

dz

TdrrdKQ zz )(

dzQz

QQ zzdzz ][ Quantity of heat leaving from the face CDHG in lsquozrsquo direction

Quantity of heat flowing into the face ABFE in lsquozrsquo direction

stgzr QQdQdQdQ

)(

r

r ][)(rT

rr

ddrdzdKr

][)(

2

2

2

T

r

ddrdzrdK ))((2

2

z

TddzdrrdK z

ddzdrrdq g )(

dT

Cdzdrrd p )(

+ + +

=

T

CqTK pg)(KKKK zr

[gq

)(1

r

Tr

rrK r

)(1

2

2

2 T

rK )(

2

2

z

TK z

)( ddzdrrd )(

ddzdrrd

TC p

+ + + ] =

If material is homogeneous amp isotropic

)(1

r

Tr

rr

)(1

2

2

2 T

r)(

2

2

z

T

K

q g

T

K

C p

)(1

r

Tr

rr

)(1

2

2

2 T

r )(2

2

z

T

K

q g

T

1

+ + + =

)(1

r

Tr

rrK r

)(1

2

2

2 T

rK )(

2

2

z

TK z

gq

T

C p + + + =

+ + + = Thermal diffusivity

T

K

qT g

12

pC

K

(Ein ndash Eout) + Egen = Est

)( rT

)( r

)sin( drrddr)sin( drrddr

planerdirection

planerdirection

planedirectionr

)(

)(

)(

Spherical polar co-ordinates

Elemental length

Consider an elemental volume ABCDEFGH =

GENERAL HEAT CONDUCTION EQUATION IN SPHERICAL COORDINATES

Let

=internal heat generation per unit volume per unit time (W m3)q= rate of heat flow (W)Q= qdt=total heat flow (J)r=mass density of the material element(kgm3)Cp=specific heat of the material element (Jkg K)

gq

Consider an unsteady (transient) three dimensional heat conduction with internal heat generation in heterogeneous amp anisotropic spherical geometries

LetTemperature distribution is a function of T=

X

Y

Z

drT

drrdKQ rr )sin(

drQr

QQ rrdrr ][

drdr

TdrrdK

rdQ

QQdQ

rr

drrrr

])sin([

)1(]([sin 2

rT

rr

ddrddKdQ rr

Plane ( -q f) lsquorrsquo directionQuantity of heat flowing into the face ABCD in lsquorrsquo direction

Quantity of heat leaving from the face EFGH in lsquorrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquorrsquo direction

drT

drdrKQ )sin(

rdQ

rQQ d ][

rddr

TdrdrK

rdQ

QQdQ d

])sin([

])[(sin

2

T

r

drdrdrKdQ

)2(])(sin[

2

T

r

drdrddrKdQ

Quantity of heat leaving from the face BFHD in lsquoqrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoqrsquo direction

X

Y

Z

Plane (r- f ) lsquoq lsquodirection

Quantity of heat flowing into the face AEGC in lsquoqrsquo direction

dr

TdrrdKQ

sin)(

drQ

rQQ d sin][

sin

drdr

TdrrdK

rdQ

QQdQ d

sin]sin

)([sin

][sinsin

)sin(

T

rr

ddrdrrdKdQ

)3()(sin

)sin(2

2

22

2

T

r

ddddrrKdQ

Plane (r- ) q lsquofrsquo directionQuantity of heat flowing into the face ABFE in lsquofrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquozrsquo direction

)4()sin()(

ddrrddrqQgg

)5()sin()(

dT

CdrrddrQ pst

Total amount of internal heat generation in the element

Total heat energy stored in the element

(Ein ndash Eout) + Egen = Est

d

TmCQ pst )(

)( volumem

Quantity of heat leaving from the face CDHG in lsquofrsquo direction

according to thermodynamics first law

Ein + Egen = Eout + Est

X

Y

Z

stgr QQdQdQdQ

)(

]([sin 2

r

Tr

rddrddKr

])(sin[

2

T

rdrdrddrK

)(sin

)sin(2

2

22

2

T

r

ddddrrK

ddrrddrq g )sin(

dT

Cdrrddr p )sin(

2

2

r

r

sin

sin

]([ 22 r

Tr

rr

Kr

])(sin[sin2

TrK

)(sin 2

2

22

T

r

Kgq

T

Cp

]([1

[ 22 r

Tr

rrK

])(sin[sin1

2

T

r)](

sin

12

2

22 T

r gq

T

Cp

]([1

[ 22 r

Tr

rr

])(sin[sin1

2

T

r)](

sin

12

2

22 T

rK

q g

T

K

Cp

]([1

[ 22 r

Tr

rr

])(sin[sin1

2

T

r)](

sin

12

2

22 T

rK

q g

T1

KKKKr If material is homogeneous amp isotropic

TCqTK pg)(

T

K

qT g

12 pC

K

Thermal diffusivity

(Ein ndash Eout) + Egen = Est

General Heat Conduction Equation

PLANE WALL CYLINDER SPHERE

Temperature profile

Plane wall SphereCylinder

bull A copper conductor (k = 380 WmoC resistivity ρ = 2x10-8 Ω m) having inner and outer radii 10 cm and 225 cm respectively is carrying a current density of 4800 amperes cm2 the conductor is internally cooled and a constant temperature of 65oC is maintained at the inner surface and there is no heat transfer through the insulation surrounding the conductor

Determinebull (i) The maximum temperature of the conductor and the radius at

which it occursbull (ii) The internal heat transfer rate

bull A chemical reaction takes place in a packed bed (k=06 WmoC) between two coaxial cylinders with radii 15 mm and 45 mm the inner surface is at 580oC and it is insulated Assuming the reaction rate of 055 MWm3 in the reactor volume find the temperature at the outer surface of the reactor

bull A current of 300 amperes passes through a stainless steel wire of 25 mm diameter and k=20 WmoC The resistivity of the wire is 70x10-8 ohm-m and length of the wire is 2 m if the wire is submerged in a fluid maintained at 50oC and h at the wire surface is 4000 Wm2oC calculate the steady state temperature at the centre and at the surface of the wire

bull An approximately spherical shaped orange ( k=023 WmoC) 90 mm diameter undergoes ripping process and generates 5100 Wm3 of energy If external surface the orange is at 8oC determine

bull (i) temperature at the centre of the orange and

bull (ii) heat flow from the outer surface of the orange

bull FINS External surfaces to used to cooling area and hence convection

heat transfer can be improvedbull APPLICATIONS IC engine cylinder ( air cooling) Heat exchangers automobile

radiators computer chip refrigerators electric motors generators and transformers etc

bull ASSUMPTIONS MADE IN ANALYSIS OF HEAT TRANFER THROUGH FIN

1 Fin material is homogenous and isotropic constthermal conductivity

2 Temperature at any cs of the fin is uniform3 lsquohrsquo is uniform over entire surface4 No internal heat generation5 Thermal contact resistance is negligible6 Radiation heat transfer is negligible7 Heat conduction is steady state and one dimensional

bull Problem 1

bull A stainless steel fin (k=20 WmK) having a diameter of 20 mm and a length of 01 m is attached to a wall at 300oC The ambient temperature is 50oC and the heat transfer coefficient is 10 Wm2K The fin tip is insulated Determine

bull (A) The rate of heat dissipation from the finbull (B) The temperature at the fin tipbull (C) The rate of heat transfer from the wall area covered by the fin if

the fin was not usedbull (D) The heat transfer rate from the same fin geometry if the

stainless steel fin is replaced by a fictitious fin with infinite thermal conductivity

bull Problem2bull One end of a long rod is inserted into a furnace while the other

projects into ambient temperature Under steady state the temperature of the rod is measured at two points 75mm apart and found to be 125oC and 885oC respectively while the ambient temperature is 20oC If the rod is 25mm in diameter and h is 2336 Wm2K

bull Find the thermal conductivity of the rod material

  • HEAT AND MASS TRANSFER
  • Thermodynamics Vs Heat transfer
  • MODE OF HEAT TRANSFER
  • MODE OF HEAT TRANSFER (2)
  • CONDUCTION- MECHANISM
  • CONCECTION-MECHANISM
  • RADIATION-MECHANISM
  • Conduction- Convection- Radiation
  • Slide 9
  • Slide 10
  • Fourierrsquos Law of Heat Conduction
  • Slide 12
  • Newtonrsquos Law of Cooling
  • Laws of Thermal Radiation
  • LAWS FOR HEAT TRANSFER
  • ELECTRICAL ampTHERMAL ANALOGY
  • Slide 17
  • Slide 18
  • PLANE WALL-UNIDIRECTIONAL
  • PLANE WALL -SERIES
  • PLANE WALL IN SERIES
  • PLANE WALL -PARALLEL
  • PLANE WALL-PARALLEL
  • PLANE WALL-COMBINATION
  • PLANE WALL-FILM RESISTANCE
  • PLANE WALL- THERMAL CONTACT RESISTANCE
  • PLANE WALL- THERMAL CONTACT RESISTANCE (2)
  • CYLINDER-SOLID
  • HEAT TRANSFER IN SOLID CYLINDER
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER-COMPOSITE
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER-MULTY LAYER
  • HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER
  • HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER (2)
  • HEAT TRANSFER IN SPHERE
  • HEAT TRANSFER IN SPHERE -MULTY LAYER
  • HEAT TRANSFER IN SPHERE -MULTY LAYER (2)
  • Slide 38
  • Slide 39
  • Slide 40
  • Slide 41
  • CRITICAL RADIUS
  • Plane wall- Insulation thickness -study
  • Problem5
  • Slide 45
  • Slide 46
  • Slide 47
  • GENERAL HEAT CONDUCTION EQUATION IN CYLINDRICAL COORDINATES
  • Slide 49
  • Slide 50
  • Slide 51
  • GENERAL HEAT CONDUCTION EQUATION IN SPHERICAL COORDINATES
  • Slide 53
  • Slide 54
  • Slide 55
  • General Heat Conduction Equation
  • Temperature profile
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Slide 67
Page 5: hmt chap 1

CONDUCTION- MECHANISM

bull CONDUCTION Conduction is the transfer of energy from more energetic

particles of a substance to the adjacent less energetic one as a result of interactions between the particles

In gases and liquids conduction is due to the collisions and diffusion of the molecules during their random motion

In solids it is due to the combination of vibrations of the molecules in a lattice and the energy transport by free electrons

Micro form of heat transferFOURIERrsquoS LAW

q =-KA( dT dX)

CONCECTION-MECHANISMbull Convection Heat transfer between a solid surface and an adjacent

gas or liquid It is the combination of conduction and flow motion Heat transferred from a solid surface to a liquid adjacent is conduction And then heat is brought away by the flow motion

bull Newtons law of cooling

where h = Convection heat transfer coefficient Ts = Temperature of the solid surface Tf = Temperature of the fluid

bull Macroform of heat transfer CONVECTION NEWTONrsquoS LAW OF COOLING

q = hA (Ts-Tf)

RADIATION-MECHANISM

bull Radiation The energy emitted by matter in the form of electromagnetic waves as a result of the changes in the electronic configurations of the atoms or molecules

bull Stefan - Boltzmann law

where σ = Stefan - Boltzmann constant ε = emissivity Ts = Surface temperature of the object

q =σ ε AT4

Conduction- Convection- Radiation

bull The three modes of heat transfer always exist simultaneously For example the heat transfer associated with double pane windows are

bull Conduction Hotter (cooler) air outside each pane causes conduction through solid glass

bull Convection Air between the panes carries heat from hotter pane to cooler pane

bull Radiation Sunlight radiation passes through glass to be absorbed on other side

Examples for different mode of heat transfer

bull Conduction

Heat loss through thermal insulation on steam pipe

Convection

Heat transfer to water flowing through a pipe in condenser

Radiation

Heat transfer in an electric furnace

Conduction Convection and Radiation

Solar energy used water heater

Fourierrsquos Law of Heat Conduction

Newtonrsquos Law of Cooling

Laws of Thermal Radiation

LAWS FOR HEAT TRANSFERCONDUCTION

4Tq

CONVECTION

NEWTONrsquoS LAW OF COOLING

RADIATION

STEFAN BOLTZMAN LAW

q=-KA( dT dX)

Assumptionsbull Steady statebull Unidirectional heat flowbull Const temperature gradientbull Linear profilebull No internal heat generationbull Isothermal bounding surfacebull Homogeneousbull Isotropic bull Applicable to solid liquid gas

q=σAT4q= hA (Tw-Tα)

Thq

XT

Kq

FOURIERrsquoS LAW

q =σ ε AT4

ELECTRICAL ampTHERMAL ANALOGY

THERMAL RESISTANCE(ELECTRICAL ANALOGY)

bull OHMrsquos LAW Flow of Electricity

V=I R elect

Voltage Drop = Current flowtimesResistance Temp Drop=Heat FlowtimesResistance

ΔT= q Rtherm

Thermal Analogy To OHMrsquoS Law

PLANE WALL-UNIDIRECTIONAL

PLANE WALL -SERIES

PLANE WALL IN SERIES

PLANE WALL -PARALLEL

PLANE WALL-PARALLEL

PLANE WALL-COMBINATION

PLANE WALL-FILM RESISTANCE

PLANE WALL- THERMAL CONTACT RESISTANCE

PLANE WALL- THERMAL CONTACT RESISTANCE

At the joining surface of the two slabs air is trapped in voids due to surface irregularities

CYLINDER-SOLID

HEAT TRANSFER IN SOLID CYLINDER

HEAT TRANSFER RATE IN HOLLOW CYLINDER

HEAT TRANSFER RATE IN HOLLOW CYLINDER-COMPOSITE

HEAT TRANSFER RATE IN HOLLOW CYLINDER-MULTY LAYER

HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER

HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER

HEAT TRANSFER IN SPHERE

HEAT TRANSFER IN SPHERE -MULTY LAYER

HEAT TRANSFER IN SPHERE -MULTY LAYER

bull Problem 1 The wall in a furnace consists of 125 mm thick refractory bricks and

125 mm thick insulating fire bricks separated by an air gap A 12 mm thick plaster covers the outer wall The inner surface of wall is at 1100oC and the ambient temperature 25oC the heat transfer coefficient on the outer wall to the air is 17 Wm2K and the resistance to heat flow of the air gap is 016 KW the thermal conductivities of refractory brick insulating firebrick and plaster are 16 03 and 014 WmK respectively

Calculatebull (a) the rate of heat loss per unit area of the wall surfacebull (b) the interface temperatures throughout the wallbull (c) the temperature at the outside surface of the wall

bull Problem 2Two large aluminum plates (k=240 WmK) each 2 cm thick with 10 mm surface roughness are placed in contact under 105 Nm2 pressure in air The temperature at the outside surfaces are 390 oC and 406 oCCalculate

A) the heat flux B) the temperature drop due to the contact resistance C) the contact temperatures Thermal contact resistance with air as the interfacial fluid for

10 mm roughness is 27510-4 m2KW

bull Problem 3bull Steam at 350 oC flowing in a pipe (k=80 WmK) of 5 cm

inner diameter and 56 cm outer diameter is covered with 3 cm thick insulation of k=005 WmK Heat is lost to surroundings at 5 oC by natural convection and radiation the combined h being 20 Wm2K Taking the heat transfer coefficient inside the pipe as 60 Wm2K

bull Determinebull A) the rate of heat loss from the steam per unit length of

the pipebull B) the temperature drop across the pipe and the insulation

bull PROBLEM4bull A nuclear reactor has a sperical pressure vessel 750 mm inside

diameter 80 mm wall thickness The temperature at the inner surface is 500 oC If the tempertarue at the outer surface is 495 oC

bull Calculatea) The rate of heat loss through the metal wallb) If the rate of heat loss is to limited to 305Whow much inulation

thickness needs to be applied Thermal conductivity of steel wall=46 WmoC Thermal conductivity of insulation=004 WmoC Film coefficient of heat tranfer at inner and outer surface are 200

Wm2 oC and 15 Wm2 oC Ambient air temperature=28 oC

CRITICAL RADIUS

bull The radius upto which heat flow increases and after decreases At critical radius q becomes qmax

bull The addition of insulation always increases (Rth)condbull Rtotal= (Rth)cond + (Rth)conv

bull The addition of insulation may reduce the (Rth)conv due to increase in surface area

bull r1ltrc -heat transfer increases by adding insulation till r1=rc If

insulation thickness is further increased the rate of heat loss will decrease

r1 gt rc

- heat transfer decreases by adding insulation above the rc

Plane wall- Insulation

thickness -study

Cylinder amp Sphere- Insulation thickness -

study

Problem5bull A 3mm diameter and 5 mm long electric wire is tightly

wrapped with a 2 mm thick plastic cover whose K=015 wmoC Electrical measurements indicate that a current of 10 A passes through the wire and there is a voltage drop of 8 V along the wire If the insulated wire is exposed to a medium at Ta=30oC with a heat transfer coefficient of h=12 wm2oC Determine the temperature at the interface of the wire and the plastic cover in steady operation Also determine whether doubling the thickness of the plastic cover will increase or decrease this interface temperature

dx

TdzdyKQ xx )(

dxQx

QQ xxdxx ][

dxdx

TdzdyK

xQQdQ xdxxxx ])([)(

)1()]([)(

ddzdydx

x

TK

xdQ xx

GENERAL HEAT CONDUCTION EQUATION IN CARTESIAN COORDINATES

Consider an unsteady (transient) three dimensional heat conduction with internal heat generation in heterogeneous amp anisotropic PLANE WALL

Cartesian co-ordinates x y z

Elemental length dxdydz

Consider an elemental volume ABCDEFGH = (dxdydz)

Let Temperature distribution is a function of T=T(xyz)

Let

=internal heat generation per unit volume per unit time (W m3)q= rate of heat flow (W)Q= qdt=total heat flow (J)r=mass density of the material element(kgm3)Cp=specific heat of the material element (Jkg K)

gq

Plane (y - z) lsquoxrsquo direction

Quantity of heat leaving from the face EFGH in lsquoXrsquo direction

Quantity of heat flowing into the face ABCD in lsquoXrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoXrsquo direction

dy

TdzdxKQ yy )(

dyQy

QQ yydyy ][

dydy

TdzdxK

yQQdQ ydyyyy ])([)(

)2()]([)(

ddzdydx

y

TK

ydQ yy

Plane (x - z) lsquoyrsquo direction

Quantity of heat flowing into the face ABFE in lsquoYrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoYrsquo direction

dz

TdydxKQ zz )(

dzQz

QQ zzdzz ][

dzdz

TdydxK

zQQdQ zdzzzz ])([)(

)3()]([)(

ddzdydx

z

TK

zdQ zz

Plane (x - y) lsquozrsquo direction

Quantity of heat flowing into the face ADHE in lsquoZrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoZrsquo direction

Quantity of heat leaving from the face DCGH in lsquoYrsquo direction

Quantity of heat leaving from the face BCGF in lsquoZrsquo direction

)4()()(

ddzdydxqQgg

Total amount of internal heat generation in the element

Ein + Egen = Eout+Est (Ein- Eout)+ Egen=Est

T

Cddzdydxddzdydxqddzdydxz

TK

zy

TK

yx

TK

x pgzyx )()())](()()([

T

pC

gq

z

TzK

zy

Ty

Kyx

TxK

x][][][

stgzyx QQdQdQdQ

)(

)5()()(

dT

CdzdydxQ pst

Total heat energy stored in the element

(ie thermal energy of the element can be increased because of heat accumulation in the element due to heat flow amp internal heat generation)

)( volumem d

TmCQ pst )(

T

CqTK pg)(

KKKyKx z If material is homogeneous amp isotropic

T

pC

gq

z

T

zy

T

yx

T

xK ][][][

T

Kp

C

Kgq

z

T

y

T

x

T2

2

2

2

2

2

T

Kgq

z

T

y

T

x

T 12

2

2

2

2

2

Thermal diffusivity pC

K

T

K

qT g

12

according to thermodynamics first law

)( zr )( dzrddr

)( dzrddr

Cylindrical polar co-ordinates

Elemental length

Consider an elemental volume ABCDEFGH =

GENERAL HEAT CONDUCTION EQUATION IN CYLINDRICAL COORDINATES

gq

gq

Consider an unsteady (transient) three dimensional heat conduction with internal heat generation in heterogeneous amp anisotropic cylindrical geometries (eg rods and pipes)

)( zrTLetTemperature distribution is a function of T=

Let

=internal heat generation per unit volume per unit time (W m3)q= rate of heat flow (W)Q= qdt=total heat flow (J)r=mass density of the material element(kgm3)Cp=specific heat of the material element (Jkg K)

planerdirectionz

planezrdirection

planezdirectionr

)(

)(

)(

LetHeat flow in

drdr

TdzrdK

rdQ

QQdQ

rr

drrrr

])([

)1(][

r

Tr

rddrdzdKdQ rr

Heat accumulation in the element due to heat flow by conduction in lsquorrsquo direction

rddr

TdzdrK

rdQ

QQdQ d

])([

][

2

T

r

ddrdzrdKdQ

)2(][

2

2

2

T

r

ddrdzrdKdQ

Heat accumulation in the element due to heat flow by conduction in lsquoqrsquo direction

Plane (z- q ) lsquorrsquo direction

dr

TdzrdKQ rr

)(

drQr

QQ rrdrr ][

Quantity of heat leaving from the face EFGH in lsquorrsquo direction

Quantity of heat flowing into the face ABCD in lsquorrsquo direction

Plane (r-z) lsquoq lsquodirection

Quantity of heat leaving from the face AEGC in lsquoqrsquo direction

d

r

TdzdrKQ )(

rdQ

rQQ d ][

Quantity of heat flowing into the face BFHD in lsquoqrsquo direction

)3())((2

2

z

TdzdrrdKdQ zz

)5()()(

dT

CdzdrrdQ pst

)4()()(

ddzdrrdqQgg

Total amount of internal heat generation in the element

Total heat energy stored in the element

(ie thermal energy of the element can be increased because of heat accumulation in the element due to heat flow amp internal heat generation)

according to thermodynamics first law

Ein + Egen = Eout + Est (Ein ndash Eout) + Egen = Est

)( volumem

)()(z

T

zdzdrrdKdQ zz

Heat accumulation in the element due to heat flow by conduction in lsquozrsquo direction

dzdz

TdrrdK

zdQ

QQdQ

zz

dzzzz

])([

d

TmCQ pst )(

Plane (r-q) lsquozrsquo direction

dz

TdrrdKQ zz )(

dzQz

QQ zzdzz ][ Quantity of heat leaving from the face CDHG in lsquozrsquo direction

Quantity of heat flowing into the face ABFE in lsquozrsquo direction

stgzr QQdQdQdQ

)(

r

r ][)(rT

rr

ddrdzdKr

][)(

2

2

2

T

r

ddrdzrdK ))((2

2

z

TddzdrrdK z

ddzdrrdq g )(

dT

Cdzdrrd p )(

+ + +

=

T

CqTK pg)(KKKK zr

[gq

)(1

r

Tr

rrK r

)(1

2

2

2 T

rK )(

2

2

z

TK z

)( ddzdrrd )(

ddzdrrd

TC p

+ + + ] =

If material is homogeneous amp isotropic

)(1

r

Tr

rr

)(1

2

2

2 T

r)(

2

2

z

T

K

q g

T

K

C p

)(1

r

Tr

rr

)(1

2

2

2 T

r )(2

2

z

T

K

q g

T

1

+ + + =

)(1

r

Tr

rrK r

)(1

2

2

2 T

rK )(

2

2

z

TK z

gq

T

C p + + + =

+ + + = Thermal diffusivity

T

K

qT g

12

pC

K

(Ein ndash Eout) + Egen = Est

)( rT

)( r

)sin( drrddr)sin( drrddr

planerdirection

planerdirection

planedirectionr

)(

)(

)(

Spherical polar co-ordinates

Elemental length

Consider an elemental volume ABCDEFGH =

GENERAL HEAT CONDUCTION EQUATION IN SPHERICAL COORDINATES

Let

=internal heat generation per unit volume per unit time (W m3)q= rate of heat flow (W)Q= qdt=total heat flow (J)r=mass density of the material element(kgm3)Cp=specific heat of the material element (Jkg K)

gq

Consider an unsteady (transient) three dimensional heat conduction with internal heat generation in heterogeneous amp anisotropic spherical geometries

LetTemperature distribution is a function of T=

X

Y

Z

drT

drrdKQ rr )sin(

drQr

QQ rrdrr ][

drdr

TdrrdK

rdQ

QQdQ

rr

drrrr

])sin([

)1(]([sin 2

rT

rr

ddrddKdQ rr

Plane ( -q f) lsquorrsquo directionQuantity of heat flowing into the face ABCD in lsquorrsquo direction

Quantity of heat leaving from the face EFGH in lsquorrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquorrsquo direction

drT

drdrKQ )sin(

rdQ

rQQ d ][

rddr

TdrdrK

rdQ

QQdQ d

])sin([

])[(sin

2

T

r

drdrdrKdQ

)2(])(sin[

2

T

r

drdrddrKdQ

Quantity of heat leaving from the face BFHD in lsquoqrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoqrsquo direction

X

Y

Z

Plane (r- f ) lsquoq lsquodirection

Quantity of heat flowing into the face AEGC in lsquoqrsquo direction

dr

TdrrdKQ

sin)(

drQ

rQQ d sin][

sin

drdr

TdrrdK

rdQ

QQdQ d

sin]sin

)([sin

][sinsin

)sin(

T

rr

ddrdrrdKdQ

)3()(sin

)sin(2

2

22

2

T

r

ddddrrKdQ

Plane (r- ) q lsquofrsquo directionQuantity of heat flowing into the face ABFE in lsquofrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquozrsquo direction

)4()sin()(

ddrrddrqQgg

)5()sin()(

dT

CdrrddrQ pst

Total amount of internal heat generation in the element

Total heat energy stored in the element

(Ein ndash Eout) + Egen = Est

d

TmCQ pst )(

)( volumem

Quantity of heat leaving from the face CDHG in lsquofrsquo direction

according to thermodynamics first law

Ein + Egen = Eout + Est

X

Y

Z

stgr QQdQdQdQ

)(

]([sin 2

r

Tr

rddrddKr

])(sin[

2

T

rdrdrddrK

)(sin

)sin(2

2

22

2

T

r

ddddrrK

ddrrddrq g )sin(

dT

Cdrrddr p )sin(

2

2

r

r

sin

sin

]([ 22 r

Tr

rr

Kr

])(sin[sin2

TrK

)(sin 2

2

22

T

r

Kgq

T

Cp

]([1

[ 22 r

Tr

rrK

])(sin[sin1

2

T

r)](

sin

12

2

22 T

r gq

T

Cp

]([1

[ 22 r

Tr

rr

])(sin[sin1

2

T

r)](

sin

12

2

22 T

rK

q g

T

K

Cp

]([1

[ 22 r

Tr

rr

])(sin[sin1

2

T

r)](

sin

12

2

22 T

rK

q g

T1

KKKKr If material is homogeneous amp isotropic

TCqTK pg)(

T

K

qT g

12 pC

K

Thermal diffusivity

(Ein ndash Eout) + Egen = Est

General Heat Conduction Equation

PLANE WALL CYLINDER SPHERE

Temperature profile

Plane wall SphereCylinder

bull A copper conductor (k = 380 WmoC resistivity ρ = 2x10-8 Ω m) having inner and outer radii 10 cm and 225 cm respectively is carrying a current density of 4800 amperes cm2 the conductor is internally cooled and a constant temperature of 65oC is maintained at the inner surface and there is no heat transfer through the insulation surrounding the conductor

Determinebull (i) The maximum temperature of the conductor and the radius at

which it occursbull (ii) The internal heat transfer rate

bull A chemical reaction takes place in a packed bed (k=06 WmoC) between two coaxial cylinders with radii 15 mm and 45 mm the inner surface is at 580oC and it is insulated Assuming the reaction rate of 055 MWm3 in the reactor volume find the temperature at the outer surface of the reactor

bull A current of 300 amperes passes through a stainless steel wire of 25 mm diameter and k=20 WmoC The resistivity of the wire is 70x10-8 ohm-m and length of the wire is 2 m if the wire is submerged in a fluid maintained at 50oC and h at the wire surface is 4000 Wm2oC calculate the steady state temperature at the centre and at the surface of the wire

bull An approximately spherical shaped orange ( k=023 WmoC) 90 mm diameter undergoes ripping process and generates 5100 Wm3 of energy If external surface the orange is at 8oC determine

bull (i) temperature at the centre of the orange and

bull (ii) heat flow from the outer surface of the orange

bull FINS External surfaces to used to cooling area and hence convection

heat transfer can be improvedbull APPLICATIONS IC engine cylinder ( air cooling) Heat exchangers automobile

radiators computer chip refrigerators electric motors generators and transformers etc

bull ASSUMPTIONS MADE IN ANALYSIS OF HEAT TRANFER THROUGH FIN

1 Fin material is homogenous and isotropic constthermal conductivity

2 Temperature at any cs of the fin is uniform3 lsquohrsquo is uniform over entire surface4 No internal heat generation5 Thermal contact resistance is negligible6 Radiation heat transfer is negligible7 Heat conduction is steady state and one dimensional

bull Problem 1

bull A stainless steel fin (k=20 WmK) having a diameter of 20 mm and a length of 01 m is attached to a wall at 300oC The ambient temperature is 50oC and the heat transfer coefficient is 10 Wm2K The fin tip is insulated Determine

bull (A) The rate of heat dissipation from the finbull (B) The temperature at the fin tipbull (C) The rate of heat transfer from the wall area covered by the fin if

the fin was not usedbull (D) The heat transfer rate from the same fin geometry if the

stainless steel fin is replaced by a fictitious fin with infinite thermal conductivity

bull Problem2bull One end of a long rod is inserted into a furnace while the other

projects into ambient temperature Under steady state the temperature of the rod is measured at two points 75mm apart and found to be 125oC and 885oC respectively while the ambient temperature is 20oC If the rod is 25mm in diameter and h is 2336 Wm2K

bull Find the thermal conductivity of the rod material

  • HEAT AND MASS TRANSFER
  • Thermodynamics Vs Heat transfer
  • MODE OF HEAT TRANSFER
  • MODE OF HEAT TRANSFER (2)
  • CONDUCTION- MECHANISM
  • CONCECTION-MECHANISM
  • RADIATION-MECHANISM
  • Conduction- Convection- Radiation
  • Slide 9
  • Slide 10
  • Fourierrsquos Law of Heat Conduction
  • Slide 12
  • Newtonrsquos Law of Cooling
  • Laws of Thermal Radiation
  • LAWS FOR HEAT TRANSFER
  • ELECTRICAL ampTHERMAL ANALOGY
  • Slide 17
  • Slide 18
  • PLANE WALL-UNIDIRECTIONAL
  • PLANE WALL -SERIES
  • PLANE WALL IN SERIES
  • PLANE WALL -PARALLEL
  • PLANE WALL-PARALLEL
  • PLANE WALL-COMBINATION
  • PLANE WALL-FILM RESISTANCE
  • PLANE WALL- THERMAL CONTACT RESISTANCE
  • PLANE WALL- THERMAL CONTACT RESISTANCE (2)
  • CYLINDER-SOLID
  • HEAT TRANSFER IN SOLID CYLINDER
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER-COMPOSITE
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER-MULTY LAYER
  • HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER
  • HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER (2)
  • HEAT TRANSFER IN SPHERE
  • HEAT TRANSFER IN SPHERE -MULTY LAYER
  • HEAT TRANSFER IN SPHERE -MULTY LAYER (2)
  • Slide 38
  • Slide 39
  • Slide 40
  • Slide 41
  • CRITICAL RADIUS
  • Plane wall- Insulation thickness -study
  • Problem5
  • Slide 45
  • Slide 46
  • Slide 47
  • GENERAL HEAT CONDUCTION EQUATION IN CYLINDRICAL COORDINATES
  • Slide 49
  • Slide 50
  • Slide 51
  • GENERAL HEAT CONDUCTION EQUATION IN SPHERICAL COORDINATES
  • Slide 53
  • Slide 54
  • Slide 55
  • General Heat Conduction Equation
  • Temperature profile
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Slide 67
Page 6: hmt chap 1

CONCECTION-MECHANISMbull Convection Heat transfer between a solid surface and an adjacent

gas or liquid It is the combination of conduction and flow motion Heat transferred from a solid surface to a liquid adjacent is conduction And then heat is brought away by the flow motion

bull Newtons law of cooling

where h = Convection heat transfer coefficient Ts = Temperature of the solid surface Tf = Temperature of the fluid

bull Macroform of heat transfer CONVECTION NEWTONrsquoS LAW OF COOLING

q = hA (Ts-Tf)

RADIATION-MECHANISM

bull Radiation The energy emitted by matter in the form of electromagnetic waves as a result of the changes in the electronic configurations of the atoms or molecules

bull Stefan - Boltzmann law

where σ = Stefan - Boltzmann constant ε = emissivity Ts = Surface temperature of the object

q =σ ε AT4

Conduction- Convection- Radiation

bull The three modes of heat transfer always exist simultaneously For example the heat transfer associated with double pane windows are

bull Conduction Hotter (cooler) air outside each pane causes conduction through solid glass

bull Convection Air between the panes carries heat from hotter pane to cooler pane

bull Radiation Sunlight radiation passes through glass to be absorbed on other side

Examples for different mode of heat transfer

bull Conduction

Heat loss through thermal insulation on steam pipe

Convection

Heat transfer to water flowing through a pipe in condenser

Radiation

Heat transfer in an electric furnace

Conduction Convection and Radiation

Solar energy used water heater

Fourierrsquos Law of Heat Conduction

Newtonrsquos Law of Cooling

Laws of Thermal Radiation

LAWS FOR HEAT TRANSFERCONDUCTION

4Tq

CONVECTION

NEWTONrsquoS LAW OF COOLING

RADIATION

STEFAN BOLTZMAN LAW

q=-KA( dT dX)

Assumptionsbull Steady statebull Unidirectional heat flowbull Const temperature gradientbull Linear profilebull No internal heat generationbull Isothermal bounding surfacebull Homogeneousbull Isotropic bull Applicable to solid liquid gas

q=σAT4q= hA (Tw-Tα)

Thq

XT

Kq

FOURIERrsquoS LAW

q =σ ε AT4

ELECTRICAL ampTHERMAL ANALOGY

THERMAL RESISTANCE(ELECTRICAL ANALOGY)

bull OHMrsquos LAW Flow of Electricity

V=I R elect

Voltage Drop = Current flowtimesResistance Temp Drop=Heat FlowtimesResistance

ΔT= q Rtherm

Thermal Analogy To OHMrsquoS Law

PLANE WALL-UNIDIRECTIONAL

PLANE WALL -SERIES

PLANE WALL IN SERIES

PLANE WALL -PARALLEL

PLANE WALL-PARALLEL

PLANE WALL-COMBINATION

PLANE WALL-FILM RESISTANCE

PLANE WALL- THERMAL CONTACT RESISTANCE

PLANE WALL- THERMAL CONTACT RESISTANCE

At the joining surface of the two slabs air is trapped in voids due to surface irregularities

CYLINDER-SOLID

HEAT TRANSFER IN SOLID CYLINDER

HEAT TRANSFER RATE IN HOLLOW CYLINDER

HEAT TRANSFER RATE IN HOLLOW CYLINDER-COMPOSITE

HEAT TRANSFER RATE IN HOLLOW CYLINDER-MULTY LAYER

HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER

HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER

HEAT TRANSFER IN SPHERE

HEAT TRANSFER IN SPHERE -MULTY LAYER

HEAT TRANSFER IN SPHERE -MULTY LAYER

bull Problem 1 The wall in a furnace consists of 125 mm thick refractory bricks and

125 mm thick insulating fire bricks separated by an air gap A 12 mm thick plaster covers the outer wall The inner surface of wall is at 1100oC and the ambient temperature 25oC the heat transfer coefficient on the outer wall to the air is 17 Wm2K and the resistance to heat flow of the air gap is 016 KW the thermal conductivities of refractory brick insulating firebrick and plaster are 16 03 and 014 WmK respectively

Calculatebull (a) the rate of heat loss per unit area of the wall surfacebull (b) the interface temperatures throughout the wallbull (c) the temperature at the outside surface of the wall

bull Problem 2Two large aluminum plates (k=240 WmK) each 2 cm thick with 10 mm surface roughness are placed in contact under 105 Nm2 pressure in air The temperature at the outside surfaces are 390 oC and 406 oCCalculate

A) the heat flux B) the temperature drop due to the contact resistance C) the contact temperatures Thermal contact resistance with air as the interfacial fluid for

10 mm roughness is 27510-4 m2KW

bull Problem 3bull Steam at 350 oC flowing in a pipe (k=80 WmK) of 5 cm

inner diameter and 56 cm outer diameter is covered with 3 cm thick insulation of k=005 WmK Heat is lost to surroundings at 5 oC by natural convection and radiation the combined h being 20 Wm2K Taking the heat transfer coefficient inside the pipe as 60 Wm2K

bull Determinebull A) the rate of heat loss from the steam per unit length of

the pipebull B) the temperature drop across the pipe and the insulation

bull PROBLEM4bull A nuclear reactor has a sperical pressure vessel 750 mm inside

diameter 80 mm wall thickness The temperature at the inner surface is 500 oC If the tempertarue at the outer surface is 495 oC

bull Calculatea) The rate of heat loss through the metal wallb) If the rate of heat loss is to limited to 305Whow much inulation

thickness needs to be applied Thermal conductivity of steel wall=46 WmoC Thermal conductivity of insulation=004 WmoC Film coefficient of heat tranfer at inner and outer surface are 200

Wm2 oC and 15 Wm2 oC Ambient air temperature=28 oC

CRITICAL RADIUS

bull The radius upto which heat flow increases and after decreases At critical radius q becomes qmax

bull The addition of insulation always increases (Rth)condbull Rtotal= (Rth)cond + (Rth)conv

bull The addition of insulation may reduce the (Rth)conv due to increase in surface area

bull r1ltrc -heat transfer increases by adding insulation till r1=rc If

insulation thickness is further increased the rate of heat loss will decrease

r1 gt rc

- heat transfer decreases by adding insulation above the rc

Plane wall- Insulation

thickness -study

Cylinder amp Sphere- Insulation thickness -

study

Problem5bull A 3mm diameter and 5 mm long electric wire is tightly

wrapped with a 2 mm thick plastic cover whose K=015 wmoC Electrical measurements indicate that a current of 10 A passes through the wire and there is a voltage drop of 8 V along the wire If the insulated wire is exposed to a medium at Ta=30oC with a heat transfer coefficient of h=12 wm2oC Determine the temperature at the interface of the wire and the plastic cover in steady operation Also determine whether doubling the thickness of the plastic cover will increase or decrease this interface temperature

dx

TdzdyKQ xx )(

dxQx

QQ xxdxx ][

dxdx

TdzdyK

xQQdQ xdxxxx ])([)(

)1()]([)(

ddzdydx

x

TK

xdQ xx

GENERAL HEAT CONDUCTION EQUATION IN CARTESIAN COORDINATES

Consider an unsteady (transient) three dimensional heat conduction with internal heat generation in heterogeneous amp anisotropic PLANE WALL

Cartesian co-ordinates x y z

Elemental length dxdydz

Consider an elemental volume ABCDEFGH = (dxdydz)

Let Temperature distribution is a function of T=T(xyz)

Let

=internal heat generation per unit volume per unit time (W m3)q= rate of heat flow (W)Q= qdt=total heat flow (J)r=mass density of the material element(kgm3)Cp=specific heat of the material element (Jkg K)

gq

Plane (y - z) lsquoxrsquo direction

Quantity of heat leaving from the face EFGH in lsquoXrsquo direction

Quantity of heat flowing into the face ABCD in lsquoXrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoXrsquo direction

dy

TdzdxKQ yy )(

dyQy

QQ yydyy ][

dydy

TdzdxK

yQQdQ ydyyyy ])([)(

)2()]([)(

ddzdydx

y

TK

ydQ yy

Plane (x - z) lsquoyrsquo direction

Quantity of heat flowing into the face ABFE in lsquoYrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoYrsquo direction

dz

TdydxKQ zz )(

dzQz

QQ zzdzz ][

dzdz

TdydxK

zQQdQ zdzzzz ])([)(

)3()]([)(

ddzdydx

z

TK

zdQ zz

Plane (x - y) lsquozrsquo direction

Quantity of heat flowing into the face ADHE in lsquoZrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoZrsquo direction

Quantity of heat leaving from the face DCGH in lsquoYrsquo direction

Quantity of heat leaving from the face BCGF in lsquoZrsquo direction

)4()()(

ddzdydxqQgg

Total amount of internal heat generation in the element

Ein + Egen = Eout+Est (Ein- Eout)+ Egen=Est

T

Cddzdydxddzdydxqddzdydxz

TK

zy

TK

yx

TK

x pgzyx )()())](()()([

T

pC

gq

z

TzK

zy

Ty

Kyx

TxK

x][][][

stgzyx QQdQdQdQ

)(

)5()()(

dT

CdzdydxQ pst

Total heat energy stored in the element

(ie thermal energy of the element can be increased because of heat accumulation in the element due to heat flow amp internal heat generation)

)( volumem d

TmCQ pst )(

T

CqTK pg)(

KKKyKx z If material is homogeneous amp isotropic

T

pC

gq

z

T

zy

T

yx

T

xK ][][][

T

Kp

C

Kgq

z

T

y

T

x

T2

2

2

2

2

2

T

Kgq

z

T

y

T

x

T 12

2

2

2

2

2

Thermal diffusivity pC

K

T

K

qT g

12

according to thermodynamics first law

)( zr )( dzrddr

)( dzrddr

Cylindrical polar co-ordinates

Elemental length

Consider an elemental volume ABCDEFGH =

GENERAL HEAT CONDUCTION EQUATION IN CYLINDRICAL COORDINATES

gq

gq

Consider an unsteady (transient) three dimensional heat conduction with internal heat generation in heterogeneous amp anisotropic cylindrical geometries (eg rods and pipes)

)( zrTLetTemperature distribution is a function of T=

Let

=internal heat generation per unit volume per unit time (W m3)q= rate of heat flow (W)Q= qdt=total heat flow (J)r=mass density of the material element(kgm3)Cp=specific heat of the material element (Jkg K)

planerdirectionz

planezrdirection

planezdirectionr

)(

)(

)(

LetHeat flow in

drdr

TdzrdK

rdQ

QQdQ

rr

drrrr

])([

)1(][

r

Tr

rddrdzdKdQ rr

Heat accumulation in the element due to heat flow by conduction in lsquorrsquo direction

rddr

TdzdrK

rdQ

QQdQ d

])([

][

2

T

r

ddrdzrdKdQ

)2(][

2

2

2

T

r

ddrdzrdKdQ

Heat accumulation in the element due to heat flow by conduction in lsquoqrsquo direction

Plane (z- q ) lsquorrsquo direction

dr

TdzrdKQ rr

)(

drQr

QQ rrdrr ][

Quantity of heat leaving from the face EFGH in lsquorrsquo direction

Quantity of heat flowing into the face ABCD in lsquorrsquo direction

Plane (r-z) lsquoq lsquodirection

Quantity of heat leaving from the face AEGC in lsquoqrsquo direction

d

r

TdzdrKQ )(

rdQ

rQQ d ][

Quantity of heat flowing into the face BFHD in lsquoqrsquo direction

)3())((2

2

z

TdzdrrdKdQ zz

)5()()(

dT

CdzdrrdQ pst

)4()()(

ddzdrrdqQgg

Total amount of internal heat generation in the element

Total heat energy stored in the element

(ie thermal energy of the element can be increased because of heat accumulation in the element due to heat flow amp internal heat generation)

according to thermodynamics first law

Ein + Egen = Eout + Est (Ein ndash Eout) + Egen = Est

)( volumem

)()(z

T

zdzdrrdKdQ zz

Heat accumulation in the element due to heat flow by conduction in lsquozrsquo direction

dzdz

TdrrdK

zdQ

QQdQ

zz

dzzzz

])([

d

TmCQ pst )(

Plane (r-q) lsquozrsquo direction

dz

TdrrdKQ zz )(

dzQz

QQ zzdzz ][ Quantity of heat leaving from the face CDHG in lsquozrsquo direction

Quantity of heat flowing into the face ABFE in lsquozrsquo direction

stgzr QQdQdQdQ

)(

r

r ][)(rT

rr

ddrdzdKr

][)(

2

2

2

T

r

ddrdzrdK ))((2

2

z

TddzdrrdK z

ddzdrrdq g )(

dT

Cdzdrrd p )(

+ + +

=

T

CqTK pg)(KKKK zr

[gq

)(1

r

Tr

rrK r

)(1

2

2

2 T

rK )(

2

2

z

TK z

)( ddzdrrd )(

ddzdrrd

TC p

+ + + ] =

If material is homogeneous amp isotropic

)(1

r

Tr

rr

)(1

2

2

2 T

r)(

2

2

z

T

K

q g

T

K

C p

)(1

r

Tr

rr

)(1

2

2

2 T

r )(2

2

z

T

K

q g

T

1

+ + + =

)(1

r

Tr

rrK r

)(1

2

2

2 T

rK )(

2

2

z

TK z

gq

T

C p + + + =

+ + + = Thermal diffusivity

T

K

qT g

12

pC

K

(Ein ndash Eout) + Egen = Est

)( rT

)( r

)sin( drrddr)sin( drrddr

planerdirection

planerdirection

planedirectionr

)(

)(

)(

Spherical polar co-ordinates

Elemental length

Consider an elemental volume ABCDEFGH =

GENERAL HEAT CONDUCTION EQUATION IN SPHERICAL COORDINATES

Let

=internal heat generation per unit volume per unit time (W m3)q= rate of heat flow (W)Q= qdt=total heat flow (J)r=mass density of the material element(kgm3)Cp=specific heat of the material element (Jkg K)

gq

Consider an unsteady (transient) three dimensional heat conduction with internal heat generation in heterogeneous amp anisotropic spherical geometries

LetTemperature distribution is a function of T=

X

Y

Z

drT

drrdKQ rr )sin(

drQr

QQ rrdrr ][

drdr

TdrrdK

rdQ

QQdQ

rr

drrrr

])sin([

)1(]([sin 2

rT

rr

ddrddKdQ rr

Plane ( -q f) lsquorrsquo directionQuantity of heat flowing into the face ABCD in lsquorrsquo direction

Quantity of heat leaving from the face EFGH in lsquorrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquorrsquo direction

drT

drdrKQ )sin(

rdQ

rQQ d ][

rddr

TdrdrK

rdQ

QQdQ d

])sin([

])[(sin

2

T

r

drdrdrKdQ

)2(])(sin[

2

T

r

drdrddrKdQ

Quantity of heat leaving from the face BFHD in lsquoqrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoqrsquo direction

X

Y

Z

Plane (r- f ) lsquoq lsquodirection

Quantity of heat flowing into the face AEGC in lsquoqrsquo direction

dr

TdrrdKQ

sin)(

drQ

rQQ d sin][

sin

drdr

TdrrdK

rdQ

QQdQ d

sin]sin

)([sin

][sinsin

)sin(

T

rr

ddrdrrdKdQ

)3()(sin

)sin(2

2

22

2

T

r

ddddrrKdQ

Plane (r- ) q lsquofrsquo directionQuantity of heat flowing into the face ABFE in lsquofrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquozrsquo direction

)4()sin()(

ddrrddrqQgg

)5()sin()(

dT

CdrrddrQ pst

Total amount of internal heat generation in the element

Total heat energy stored in the element

(Ein ndash Eout) + Egen = Est

d

TmCQ pst )(

)( volumem

Quantity of heat leaving from the face CDHG in lsquofrsquo direction

according to thermodynamics first law

Ein + Egen = Eout + Est

X

Y

Z

stgr QQdQdQdQ

)(

]([sin 2

r

Tr

rddrddKr

])(sin[

2

T

rdrdrddrK

)(sin

)sin(2

2

22

2

T

r

ddddrrK

ddrrddrq g )sin(

dT

Cdrrddr p )sin(

2

2

r

r

sin

sin

]([ 22 r

Tr

rr

Kr

])(sin[sin2

TrK

)(sin 2

2

22

T

r

Kgq

T

Cp

]([1

[ 22 r

Tr

rrK

])(sin[sin1

2

T

r)](

sin

12

2

22 T

r gq

T

Cp

]([1

[ 22 r

Tr

rr

])(sin[sin1

2

T

r)](

sin

12

2

22 T

rK

q g

T

K

Cp

]([1

[ 22 r

Tr

rr

])(sin[sin1

2

T

r)](

sin

12

2

22 T

rK

q g

T1

KKKKr If material is homogeneous amp isotropic

TCqTK pg)(

T

K

qT g

12 pC

K

Thermal diffusivity

(Ein ndash Eout) + Egen = Est

General Heat Conduction Equation

PLANE WALL CYLINDER SPHERE

Temperature profile

Plane wall SphereCylinder

bull A copper conductor (k = 380 WmoC resistivity ρ = 2x10-8 Ω m) having inner and outer radii 10 cm and 225 cm respectively is carrying a current density of 4800 amperes cm2 the conductor is internally cooled and a constant temperature of 65oC is maintained at the inner surface and there is no heat transfer through the insulation surrounding the conductor

Determinebull (i) The maximum temperature of the conductor and the radius at

which it occursbull (ii) The internal heat transfer rate

bull A chemical reaction takes place in a packed bed (k=06 WmoC) between two coaxial cylinders with radii 15 mm and 45 mm the inner surface is at 580oC and it is insulated Assuming the reaction rate of 055 MWm3 in the reactor volume find the temperature at the outer surface of the reactor

bull A current of 300 amperes passes through a stainless steel wire of 25 mm diameter and k=20 WmoC The resistivity of the wire is 70x10-8 ohm-m and length of the wire is 2 m if the wire is submerged in a fluid maintained at 50oC and h at the wire surface is 4000 Wm2oC calculate the steady state temperature at the centre and at the surface of the wire

bull An approximately spherical shaped orange ( k=023 WmoC) 90 mm diameter undergoes ripping process and generates 5100 Wm3 of energy If external surface the orange is at 8oC determine

bull (i) temperature at the centre of the orange and

bull (ii) heat flow from the outer surface of the orange

bull FINS External surfaces to used to cooling area and hence convection

heat transfer can be improvedbull APPLICATIONS IC engine cylinder ( air cooling) Heat exchangers automobile

radiators computer chip refrigerators electric motors generators and transformers etc

bull ASSUMPTIONS MADE IN ANALYSIS OF HEAT TRANFER THROUGH FIN

1 Fin material is homogenous and isotropic constthermal conductivity

2 Temperature at any cs of the fin is uniform3 lsquohrsquo is uniform over entire surface4 No internal heat generation5 Thermal contact resistance is negligible6 Radiation heat transfer is negligible7 Heat conduction is steady state and one dimensional

bull Problem 1

bull A stainless steel fin (k=20 WmK) having a diameter of 20 mm and a length of 01 m is attached to a wall at 300oC The ambient temperature is 50oC and the heat transfer coefficient is 10 Wm2K The fin tip is insulated Determine

bull (A) The rate of heat dissipation from the finbull (B) The temperature at the fin tipbull (C) The rate of heat transfer from the wall area covered by the fin if

the fin was not usedbull (D) The heat transfer rate from the same fin geometry if the

stainless steel fin is replaced by a fictitious fin with infinite thermal conductivity

bull Problem2bull One end of a long rod is inserted into a furnace while the other

projects into ambient temperature Under steady state the temperature of the rod is measured at two points 75mm apart and found to be 125oC and 885oC respectively while the ambient temperature is 20oC If the rod is 25mm in diameter and h is 2336 Wm2K

bull Find the thermal conductivity of the rod material

  • HEAT AND MASS TRANSFER
  • Thermodynamics Vs Heat transfer
  • MODE OF HEAT TRANSFER
  • MODE OF HEAT TRANSFER (2)
  • CONDUCTION- MECHANISM
  • CONCECTION-MECHANISM
  • RADIATION-MECHANISM
  • Conduction- Convection- Radiation
  • Slide 9
  • Slide 10
  • Fourierrsquos Law of Heat Conduction
  • Slide 12
  • Newtonrsquos Law of Cooling
  • Laws of Thermal Radiation
  • LAWS FOR HEAT TRANSFER
  • ELECTRICAL ampTHERMAL ANALOGY
  • Slide 17
  • Slide 18
  • PLANE WALL-UNIDIRECTIONAL
  • PLANE WALL -SERIES
  • PLANE WALL IN SERIES
  • PLANE WALL -PARALLEL
  • PLANE WALL-PARALLEL
  • PLANE WALL-COMBINATION
  • PLANE WALL-FILM RESISTANCE
  • PLANE WALL- THERMAL CONTACT RESISTANCE
  • PLANE WALL- THERMAL CONTACT RESISTANCE (2)
  • CYLINDER-SOLID
  • HEAT TRANSFER IN SOLID CYLINDER
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER-COMPOSITE
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER-MULTY LAYER
  • HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER
  • HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER (2)
  • HEAT TRANSFER IN SPHERE
  • HEAT TRANSFER IN SPHERE -MULTY LAYER
  • HEAT TRANSFER IN SPHERE -MULTY LAYER (2)
  • Slide 38
  • Slide 39
  • Slide 40
  • Slide 41
  • CRITICAL RADIUS
  • Plane wall- Insulation thickness -study
  • Problem5
  • Slide 45
  • Slide 46
  • Slide 47
  • GENERAL HEAT CONDUCTION EQUATION IN CYLINDRICAL COORDINATES
  • Slide 49
  • Slide 50
  • Slide 51
  • GENERAL HEAT CONDUCTION EQUATION IN SPHERICAL COORDINATES
  • Slide 53
  • Slide 54
  • Slide 55
  • General Heat Conduction Equation
  • Temperature profile
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Slide 67
Page 7: hmt chap 1

RADIATION-MECHANISM

bull Radiation The energy emitted by matter in the form of electromagnetic waves as a result of the changes in the electronic configurations of the atoms or molecules

bull Stefan - Boltzmann law

where σ = Stefan - Boltzmann constant ε = emissivity Ts = Surface temperature of the object

q =σ ε AT4

Conduction- Convection- Radiation

bull The three modes of heat transfer always exist simultaneously For example the heat transfer associated with double pane windows are

bull Conduction Hotter (cooler) air outside each pane causes conduction through solid glass

bull Convection Air between the panes carries heat from hotter pane to cooler pane

bull Radiation Sunlight radiation passes through glass to be absorbed on other side

Examples for different mode of heat transfer

bull Conduction

Heat loss through thermal insulation on steam pipe

Convection

Heat transfer to water flowing through a pipe in condenser

Radiation

Heat transfer in an electric furnace

Conduction Convection and Radiation

Solar energy used water heater

Fourierrsquos Law of Heat Conduction

Newtonrsquos Law of Cooling

Laws of Thermal Radiation

LAWS FOR HEAT TRANSFERCONDUCTION

4Tq

CONVECTION

NEWTONrsquoS LAW OF COOLING

RADIATION

STEFAN BOLTZMAN LAW

q=-KA( dT dX)

Assumptionsbull Steady statebull Unidirectional heat flowbull Const temperature gradientbull Linear profilebull No internal heat generationbull Isothermal bounding surfacebull Homogeneousbull Isotropic bull Applicable to solid liquid gas

q=σAT4q= hA (Tw-Tα)

Thq

XT

Kq

FOURIERrsquoS LAW

q =σ ε AT4

ELECTRICAL ampTHERMAL ANALOGY

THERMAL RESISTANCE(ELECTRICAL ANALOGY)

bull OHMrsquos LAW Flow of Electricity

V=I R elect

Voltage Drop = Current flowtimesResistance Temp Drop=Heat FlowtimesResistance

ΔT= q Rtherm

Thermal Analogy To OHMrsquoS Law

PLANE WALL-UNIDIRECTIONAL

PLANE WALL -SERIES

PLANE WALL IN SERIES

PLANE WALL -PARALLEL

PLANE WALL-PARALLEL

PLANE WALL-COMBINATION

PLANE WALL-FILM RESISTANCE

PLANE WALL- THERMAL CONTACT RESISTANCE

PLANE WALL- THERMAL CONTACT RESISTANCE

At the joining surface of the two slabs air is trapped in voids due to surface irregularities

CYLINDER-SOLID

HEAT TRANSFER IN SOLID CYLINDER

HEAT TRANSFER RATE IN HOLLOW CYLINDER

HEAT TRANSFER RATE IN HOLLOW CYLINDER-COMPOSITE

HEAT TRANSFER RATE IN HOLLOW CYLINDER-MULTY LAYER

HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER

HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER

HEAT TRANSFER IN SPHERE

HEAT TRANSFER IN SPHERE -MULTY LAYER

HEAT TRANSFER IN SPHERE -MULTY LAYER

bull Problem 1 The wall in a furnace consists of 125 mm thick refractory bricks and

125 mm thick insulating fire bricks separated by an air gap A 12 mm thick plaster covers the outer wall The inner surface of wall is at 1100oC and the ambient temperature 25oC the heat transfer coefficient on the outer wall to the air is 17 Wm2K and the resistance to heat flow of the air gap is 016 KW the thermal conductivities of refractory brick insulating firebrick and plaster are 16 03 and 014 WmK respectively

Calculatebull (a) the rate of heat loss per unit area of the wall surfacebull (b) the interface temperatures throughout the wallbull (c) the temperature at the outside surface of the wall

bull Problem 2Two large aluminum plates (k=240 WmK) each 2 cm thick with 10 mm surface roughness are placed in contact under 105 Nm2 pressure in air The temperature at the outside surfaces are 390 oC and 406 oCCalculate

A) the heat flux B) the temperature drop due to the contact resistance C) the contact temperatures Thermal contact resistance with air as the interfacial fluid for

10 mm roughness is 27510-4 m2KW

bull Problem 3bull Steam at 350 oC flowing in a pipe (k=80 WmK) of 5 cm

inner diameter and 56 cm outer diameter is covered with 3 cm thick insulation of k=005 WmK Heat is lost to surroundings at 5 oC by natural convection and radiation the combined h being 20 Wm2K Taking the heat transfer coefficient inside the pipe as 60 Wm2K

bull Determinebull A) the rate of heat loss from the steam per unit length of

the pipebull B) the temperature drop across the pipe and the insulation

bull PROBLEM4bull A nuclear reactor has a sperical pressure vessel 750 mm inside

diameter 80 mm wall thickness The temperature at the inner surface is 500 oC If the tempertarue at the outer surface is 495 oC

bull Calculatea) The rate of heat loss through the metal wallb) If the rate of heat loss is to limited to 305Whow much inulation

thickness needs to be applied Thermal conductivity of steel wall=46 WmoC Thermal conductivity of insulation=004 WmoC Film coefficient of heat tranfer at inner and outer surface are 200

Wm2 oC and 15 Wm2 oC Ambient air temperature=28 oC

CRITICAL RADIUS

bull The radius upto which heat flow increases and after decreases At critical radius q becomes qmax

bull The addition of insulation always increases (Rth)condbull Rtotal= (Rth)cond + (Rth)conv

bull The addition of insulation may reduce the (Rth)conv due to increase in surface area

bull r1ltrc -heat transfer increases by adding insulation till r1=rc If

insulation thickness is further increased the rate of heat loss will decrease

r1 gt rc

- heat transfer decreases by adding insulation above the rc

Plane wall- Insulation

thickness -study

Cylinder amp Sphere- Insulation thickness -

study

Problem5bull A 3mm diameter and 5 mm long electric wire is tightly

wrapped with a 2 mm thick plastic cover whose K=015 wmoC Electrical measurements indicate that a current of 10 A passes through the wire and there is a voltage drop of 8 V along the wire If the insulated wire is exposed to a medium at Ta=30oC with a heat transfer coefficient of h=12 wm2oC Determine the temperature at the interface of the wire and the plastic cover in steady operation Also determine whether doubling the thickness of the plastic cover will increase or decrease this interface temperature

dx

TdzdyKQ xx )(

dxQx

QQ xxdxx ][

dxdx

TdzdyK

xQQdQ xdxxxx ])([)(

)1()]([)(

ddzdydx

x

TK

xdQ xx

GENERAL HEAT CONDUCTION EQUATION IN CARTESIAN COORDINATES

Consider an unsteady (transient) three dimensional heat conduction with internal heat generation in heterogeneous amp anisotropic PLANE WALL

Cartesian co-ordinates x y z

Elemental length dxdydz

Consider an elemental volume ABCDEFGH = (dxdydz)

Let Temperature distribution is a function of T=T(xyz)

Let

=internal heat generation per unit volume per unit time (W m3)q= rate of heat flow (W)Q= qdt=total heat flow (J)r=mass density of the material element(kgm3)Cp=specific heat of the material element (Jkg K)

gq

Plane (y - z) lsquoxrsquo direction

Quantity of heat leaving from the face EFGH in lsquoXrsquo direction

Quantity of heat flowing into the face ABCD in lsquoXrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoXrsquo direction

dy

TdzdxKQ yy )(

dyQy

QQ yydyy ][

dydy

TdzdxK

yQQdQ ydyyyy ])([)(

)2()]([)(

ddzdydx

y

TK

ydQ yy

Plane (x - z) lsquoyrsquo direction

Quantity of heat flowing into the face ABFE in lsquoYrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoYrsquo direction

dz

TdydxKQ zz )(

dzQz

QQ zzdzz ][

dzdz

TdydxK

zQQdQ zdzzzz ])([)(

)3()]([)(

ddzdydx

z

TK

zdQ zz

Plane (x - y) lsquozrsquo direction

Quantity of heat flowing into the face ADHE in lsquoZrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoZrsquo direction

Quantity of heat leaving from the face DCGH in lsquoYrsquo direction

Quantity of heat leaving from the face BCGF in lsquoZrsquo direction

)4()()(

ddzdydxqQgg

Total amount of internal heat generation in the element

Ein + Egen = Eout+Est (Ein- Eout)+ Egen=Est

T

Cddzdydxddzdydxqddzdydxz

TK

zy

TK

yx

TK

x pgzyx )()())](()()([

T

pC

gq

z

TzK

zy

Ty

Kyx

TxK

x][][][

stgzyx QQdQdQdQ

)(

)5()()(

dT

CdzdydxQ pst

Total heat energy stored in the element

(ie thermal energy of the element can be increased because of heat accumulation in the element due to heat flow amp internal heat generation)

)( volumem d

TmCQ pst )(

T

CqTK pg)(

KKKyKx z If material is homogeneous amp isotropic

T

pC

gq

z

T

zy

T

yx

T

xK ][][][

T

Kp

C

Kgq

z

T

y

T

x

T2

2

2

2

2

2

T

Kgq

z

T

y

T

x

T 12

2

2

2

2

2

Thermal diffusivity pC

K

T

K

qT g

12

according to thermodynamics first law

)( zr )( dzrddr

)( dzrddr

Cylindrical polar co-ordinates

Elemental length

Consider an elemental volume ABCDEFGH =

GENERAL HEAT CONDUCTION EQUATION IN CYLINDRICAL COORDINATES

gq

gq

Consider an unsteady (transient) three dimensional heat conduction with internal heat generation in heterogeneous amp anisotropic cylindrical geometries (eg rods and pipes)

)( zrTLetTemperature distribution is a function of T=

Let

=internal heat generation per unit volume per unit time (W m3)q= rate of heat flow (W)Q= qdt=total heat flow (J)r=mass density of the material element(kgm3)Cp=specific heat of the material element (Jkg K)

planerdirectionz

planezrdirection

planezdirectionr

)(

)(

)(

LetHeat flow in

drdr

TdzrdK

rdQ

QQdQ

rr

drrrr

])([

)1(][

r

Tr

rddrdzdKdQ rr

Heat accumulation in the element due to heat flow by conduction in lsquorrsquo direction

rddr

TdzdrK

rdQ

QQdQ d

])([

][

2

T

r

ddrdzrdKdQ

)2(][

2

2

2

T

r

ddrdzrdKdQ

Heat accumulation in the element due to heat flow by conduction in lsquoqrsquo direction

Plane (z- q ) lsquorrsquo direction

dr

TdzrdKQ rr

)(

drQr

QQ rrdrr ][

Quantity of heat leaving from the face EFGH in lsquorrsquo direction

Quantity of heat flowing into the face ABCD in lsquorrsquo direction

Plane (r-z) lsquoq lsquodirection

Quantity of heat leaving from the face AEGC in lsquoqrsquo direction

d

r

TdzdrKQ )(

rdQ

rQQ d ][

Quantity of heat flowing into the face BFHD in lsquoqrsquo direction

)3())((2

2

z

TdzdrrdKdQ zz

)5()()(

dT

CdzdrrdQ pst

)4()()(

ddzdrrdqQgg

Total amount of internal heat generation in the element

Total heat energy stored in the element

(ie thermal energy of the element can be increased because of heat accumulation in the element due to heat flow amp internal heat generation)

according to thermodynamics first law

Ein + Egen = Eout + Est (Ein ndash Eout) + Egen = Est

)( volumem

)()(z

T

zdzdrrdKdQ zz

Heat accumulation in the element due to heat flow by conduction in lsquozrsquo direction

dzdz

TdrrdK

zdQ

QQdQ

zz

dzzzz

])([

d

TmCQ pst )(

Plane (r-q) lsquozrsquo direction

dz

TdrrdKQ zz )(

dzQz

QQ zzdzz ][ Quantity of heat leaving from the face CDHG in lsquozrsquo direction

Quantity of heat flowing into the face ABFE in lsquozrsquo direction

stgzr QQdQdQdQ

)(

r

r ][)(rT

rr

ddrdzdKr

][)(

2

2

2

T

r

ddrdzrdK ))((2

2

z

TddzdrrdK z

ddzdrrdq g )(

dT

Cdzdrrd p )(

+ + +

=

T

CqTK pg)(KKKK zr

[gq

)(1

r

Tr

rrK r

)(1

2

2

2 T

rK )(

2

2

z

TK z

)( ddzdrrd )(

ddzdrrd

TC p

+ + + ] =

If material is homogeneous amp isotropic

)(1

r

Tr

rr

)(1

2

2

2 T

r)(

2

2

z

T

K

q g

T

K

C p

)(1

r

Tr

rr

)(1

2

2

2 T

r )(2

2

z

T

K

q g

T

1

+ + + =

)(1

r

Tr

rrK r

)(1

2

2

2 T

rK )(

2

2

z

TK z

gq

T

C p + + + =

+ + + = Thermal diffusivity

T

K

qT g

12

pC

K

(Ein ndash Eout) + Egen = Est

)( rT

)( r

)sin( drrddr)sin( drrddr

planerdirection

planerdirection

planedirectionr

)(

)(

)(

Spherical polar co-ordinates

Elemental length

Consider an elemental volume ABCDEFGH =

GENERAL HEAT CONDUCTION EQUATION IN SPHERICAL COORDINATES

Let

=internal heat generation per unit volume per unit time (W m3)q= rate of heat flow (W)Q= qdt=total heat flow (J)r=mass density of the material element(kgm3)Cp=specific heat of the material element (Jkg K)

gq

Consider an unsteady (transient) three dimensional heat conduction with internal heat generation in heterogeneous amp anisotropic spherical geometries

LetTemperature distribution is a function of T=

X

Y

Z

drT

drrdKQ rr )sin(

drQr

QQ rrdrr ][

drdr

TdrrdK

rdQ

QQdQ

rr

drrrr

])sin([

)1(]([sin 2

rT

rr

ddrddKdQ rr

Plane ( -q f) lsquorrsquo directionQuantity of heat flowing into the face ABCD in lsquorrsquo direction

Quantity of heat leaving from the face EFGH in lsquorrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquorrsquo direction

drT

drdrKQ )sin(

rdQ

rQQ d ][

rddr

TdrdrK

rdQ

QQdQ d

])sin([

])[(sin

2

T

r

drdrdrKdQ

)2(])(sin[

2

T

r

drdrddrKdQ

Quantity of heat leaving from the face BFHD in lsquoqrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoqrsquo direction

X

Y

Z

Plane (r- f ) lsquoq lsquodirection

Quantity of heat flowing into the face AEGC in lsquoqrsquo direction

dr

TdrrdKQ

sin)(

drQ

rQQ d sin][

sin

drdr

TdrrdK

rdQ

QQdQ d

sin]sin

)([sin

][sinsin

)sin(

T

rr

ddrdrrdKdQ

)3()(sin

)sin(2

2

22

2

T

r

ddddrrKdQ

Plane (r- ) q lsquofrsquo directionQuantity of heat flowing into the face ABFE in lsquofrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquozrsquo direction

)4()sin()(

ddrrddrqQgg

)5()sin()(

dT

CdrrddrQ pst

Total amount of internal heat generation in the element

Total heat energy stored in the element

(Ein ndash Eout) + Egen = Est

d

TmCQ pst )(

)( volumem

Quantity of heat leaving from the face CDHG in lsquofrsquo direction

according to thermodynamics first law

Ein + Egen = Eout + Est

X

Y

Z

stgr QQdQdQdQ

)(

]([sin 2

r

Tr

rddrddKr

])(sin[

2

T

rdrdrddrK

)(sin

)sin(2

2

22

2

T

r

ddddrrK

ddrrddrq g )sin(

dT

Cdrrddr p )sin(

2

2

r

r

sin

sin

]([ 22 r

Tr

rr

Kr

])(sin[sin2

TrK

)(sin 2

2

22

T

r

Kgq

T

Cp

]([1

[ 22 r

Tr

rrK

])(sin[sin1

2

T

r)](

sin

12

2

22 T

r gq

T

Cp

]([1

[ 22 r

Tr

rr

])(sin[sin1

2

T

r)](

sin

12

2

22 T

rK

q g

T

K

Cp

]([1

[ 22 r

Tr

rr

])(sin[sin1

2

T

r)](

sin

12

2

22 T

rK

q g

T1

KKKKr If material is homogeneous amp isotropic

TCqTK pg)(

T

K

qT g

12 pC

K

Thermal diffusivity

(Ein ndash Eout) + Egen = Est

General Heat Conduction Equation

PLANE WALL CYLINDER SPHERE

Temperature profile

Plane wall SphereCylinder

bull A copper conductor (k = 380 WmoC resistivity ρ = 2x10-8 Ω m) having inner and outer radii 10 cm and 225 cm respectively is carrying a current density of 4800 amperes cm2 the conductor is internally cooled and a constant temperature of 65oC is maintained at the inner surface and there is no heat transfer through the insulation surrounding the conductor

Determinebull (i) The maximum temperature of the conductor and the radius at

which it occursbull (ii) The internal heat transfer rate

bull A chemical reaction takes place in a packed bed (k=06 WmoC) between two coaxial cylinders with radii 15 mm and 45 mm the inner surface is at 580oC and it is insulated Assuming the reaction rate of 055 MWm3 in the reactor volume find the temperature at the outer surface of the reactor

bull A current of 300 amperes passes through a stainless steel wire of 25 mm diameter and k=20 WmoC The resistivity of the wire is 70x10-8 ohm-m and length of the wire is 2 m if the wire is submerged in a fluid maintained at 50oC and h at the wire surface is 4000 Wm2oC calculate the steady state temperature at the centre and at the surface of the wire

bull An approximately spherical shaped orange ( k=023 WmoC) 90 mm diameter undergoes ripping process and generates 5100 Wm3 of energy If external surface the orange is at 8oC determine

bull (i) temperature at the centre of the orange and

bull (ii) heat flow from the outer surface of the orange

bull FINS External surfaces to used to cooling area and hence convection

heat transfer can be improvedbull APPLICATIONS IC engine cylinder ( air cooling) Heat exchangers automobile

radiators computer chip refrigerators electric motors generators and transformers etc

bull ASSUMPTIONS MADE IN ANALYSIS OF HEAT TRANFER THROUGH FIN

1 Fin material is homogenous and isotropic constthermal conductivity

2 Temperature at any cs of the fin is uniform3 lsquohrsquo is uniform over entire surface4 No internal heat generation5 Thermal contact resistance is negligible6 Radiation heat transfer is negligible7 Heat conduction is steady state and one dimensional

bull Problem 1

bull A stainless steel fin (k=20 WmK) having a diameter of 20 mm and a length of 01 m is attached to a wall at 300oC The ambient temperature is 50oC and the heat transfer coefficient is 10 Wm2K The fin tip is insulated Determine

bull (A) The rate of heat dissipation from the finbull (B) The temperature at the fin tipbull (C) The rate of heat transfer from the wall area covered by the fin if

the fin was not usedbull (D) The heat transfer rate from the same fin geometry if the

stainless steel fin is replaced by a fictitious fin with infinite thermal conductivity

bull Problem2bull One end of a long rod is inserted into a furnace while the other

projects into ambient temperature Under steady state the temperature of the rod is measured at two points 75mm apart and found to be 125oC and 885oC respectively while the ambient temperature is 20oC If the rod is 25mm in diameter and h is 2336 Wm2K

bull Find the thermal conductivity of the rod material

  • HEAT AND MASS TRANSFER
  • Thermodynamics Vs Heat transfer
  • MODE OF HEAT TRANSFER
  • MODE OF HEAT TRANSFER (2)
  • CONDUCTION- MECHANISM
  • CONCECTION-MECHANISM
  • RADIATION-MECHANISM
  • Conduction- Convection- Radiation
  • Slide 9
  • Slide 10
  • Fourierrsquos Law of Heat Conduction
  • Slide 12
  • Newtonrsquos Law of Cooling
  • Laws of Thermal Radiation
  • LAWS FOR HEAT TRANSFER
  • ELECTRICAL ampTHERMAL ANALOGY
  • Slide 17
  • Slide 18
  • PLANE WALL-UNIDIRECTIONAL
  • PLANE WALL -SERIES
  • PLANE WALL IN SERIES
  • PLANE WALL -PARALLEL
  • PLANE WALL-PARALLEL
  • PLANE WALL-COMBINATION
  • PLANE WALL-FILM RESISTANCE
  • PLANE WALL- THERMAL CONTACT RESISTANCE
  • PLANE WALL- THERMAL CONTACT RESISTANCE (2)
  • CYLINDER-SOLID
  • HEAT TRANSFER IN SOLID CYLINDER
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER-COMPOSITE
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER-MULTY LAYER
  • HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER
  • HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER (2)
  • HEAT TRANSFER IN SPHERE
  • HEAT TRANSFER IN SPHERE -MULTY LAYER
  • HEAT TRANSFER IN SPHERE -MULTY LAYER (2)
  • Slide 38
  • Slide 39
  • Slide 40
  • Slide 41
  • CRITICAL RADIUS
  • Plane wall- Insulation thickness -study
  • Problem5
  • Slide 45
  • Slide 46
  • Slide 47
  • GENERAL HEAT CONDUCTION EQUATION IN CYLINDRICAL COORDINATES
  • Slide 49
  • Slide 50
  • Slide 51
  • GENERAL HEAT CONDUCTION EQUATION IN SPHERICAL COORDINATES
  • Slide 53
  • Slide 54
  • Slide 55
  • General Heat Conduction Equation
  • Temperature profile
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Slide 67
Page 8: hmt chap 1

Conduction- Convection- Radiation

bull The three modes of heat transfer always exist simultaneously For example the heat transfer associated with double pane windows are

bull Conduction Hotter (cooler) air outside each pane causes conduction through solid glass

bull Convection Air between the panes carries heat from hotter pane to cooler pane

bull Radiation Sunlight radiation passes through glass to be absorbed on other side

Examples for different mode of heat transfer

bull Conduction

Heat loss through thermal insulation on steam pipe

Convection

Heat transfer to water flowing through a pipe in condenser

Radiation

Heat transfer in an electric furnace

Conduction Convection and Radiation

Solar energy used water heater

Fourierrsquos Law of Heat Conduction

Newtonrsquos Law of Cooling

Laws of Thermal Radiation

LAWS FOR HEAT TRANSFERCONDUCTION

4Tq

CONVECTION

NEWTONrsquoS LAW OF COOLING

RADIATION

STEFAN BOLTZMAN LAW

q=-KA( dT dX)

Assumptionsbull Steady statebull Unidirectional heat flowbull Const temperature gradientbull Linear profilebull No internal heat generationbull Isothermal bounding surfacebull Homogeneousbull Isotropic bull Applicable to solid liquid gas

q=σAT4q= hA (Tw-Tα)

Thq

XT

Kq

FOURIERrsquoS LAW

q =σ ε AT4

ELECTRICAL ampTHERMAL ANALOGY

THERMAL RESISTANCE(ELECTRICAL ANALOGY)

bull OHMrsquos LAW Flow of Electricity

V=I R elect

Voltage Drop = Current flowtimesResistance Temp Drop=Heat FlowtimesResistance

ΔT= q Rtherm

Thermal Analogy To OHMrsquoS Law

PLANE WALL-UNIDIRECTIONAL

PLANE WALL -SERIES

PLANE WALL IN SERIES

PLANE WALL -PARALLEL

PLANE WALL-PARALLEL

PLANE WALL-COMBINATION

PLANE WALL-FILM RESISTANCE

PLANE WALL- THERMAL CONTACT RESISTANCE

PLANE WALL- THERMAL CONTACT RESISTANCE

At the joining surface of the two slabs air is trapped in voids due to surface irregularities

CYLINDER-SOLID

HEAT TRANSFER IN SOLID CYLINDER

HEAT TRANSFER RATE IN HOLLOW CYLINDER

HEAT TRANSFER RATE IN HOLLOW CYLINDER-COMPOSITE

HEAT TRANSFER RATE IN HOLLOW CYLINDER-MULTY LAYER

HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER

HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER

HEAT TRANSFER IN SPHERE

HEAT TRANSFER IN SPHERE -MULTY LAYER

HEAT TRANSFER IN SPHERE -MULTY LAYER

bull Problem 1 The wall in a furnace consists of 125 mm thick refractory bricks and

125 mm thick insulating fire bricks separated by an air gap A 12 mm thick plaster covers the outer wall The inner surface of wall is at 1100oC and the ambient temperature 25oC the heat transfer coefficient on the outer wall to the air is 17 Wm2K and the resistance to heat flow of the air gap is 016 KW the thermal conductivities of refractory brick insulating firebrick and plaster are 16 03 and 014 WmK respectively

Calculatebull (a) the rate of heat loss per unit area of the wall surfacebull (b) the interface temperatures throughout the wallbull (c) the temperature at the outside surface of the wall

bull Problem 2Two large aluminum plates (k=240 WmK) each 2 cm thick with 10 mm surface roughness are placed in contact under 105 Nm2 pressure in air The temperature at the outside surfaces are 390 oC and 406 oCCalculate

A) the heat flux B) the temperature drop due to the contact resistance C) the contact temperatures Thermal contact resistance with air as the interfacial fluid for

10 mm roughness is 27510-4 m2KW

bull Problem 3bull Steam at 350 oC flowing in a pipe (k=80 WmK) of 5 cm

inner diameter and 56 cm outer diameter is covered with 3 cm thick insulation of k=005 WmK Heat is lost to surroundings at 5 oC by natural convection and radiation the combined h being 20 Wm2K Taking the heat transfer coefficient inside the pipe as 60 Wm2K

bull Determinebull A) the rate of heat loss from the steam per unit length of

the pipebull B) the temperature drop across the pipe and the insulation

bull PROBLEM4bull A nuclear reactor has a sperical pressure vessel 750 mm inside

diameter 80 mm wall thickness The temperature at the inner surface is 500 oC If the tempertarue at the outer surface is 495 oC

bull Calculatea) The rate of heat loss through the metal wallb) If the rate of heat loss is to limited to 305Whow much inulation

thickness needs to be applied Thermal conductivity of steel wall=46 WmoC Thermal conductivity of insulation=004 WmoC Film coefficient of heat tranfer at inner and outer surface are 200

Wm2 oC and 15 Wm2 oC Ambient air temperature=28 oC

CRITICAL RADIUS

bull The radius upto which heat flow increases and after decreases At critical radius q becomes qmax

bull The addition of insulation always increases (Rth)condbull Rtotal= (Rth)cond + (Rth)conv

bull The addition of insulation may reduce the (Rth)conv due to increase in surface area

bull r1ltrc -heat transfer increases by adding insulation till r1=rc If

insulation thickness is further increased the rate of heat loss will decrease

r1 gt rc

- heat transfer decreases by adding insulation above the rc

Plane wall- Insulation

thickness -study

Cylinder amp Sphere- Insulation thickness -

study

Problem5bull A 3mm diameter and 5 mm long electric wire is tightly

wrapped with a 2 mm thick plastic cover whose K=015 wmoC Electrical measurements indicate that a current of 10 A passes through the wire and there is a voltage drop of 8 V along the wire If the insulated wire is exposed to a medium at Ta=30oC with a heat transfer coefficient of h=12 wm2oC Determine the temperature at the interface of the wire and the plastic cover in steady operation Also determine whether doubling the thickness of the plastic cover will increase or decrease this interface temperature

dx

TdzdyKQ xx )(

dxQx

QQ xxdxx ][

dxdx

TdzdyK

xQQdQ xdxxxx ])([)(

)1()]([)(

ddzdydx

x

TK

xdQ xx

GENERAL HEAT CONDUCTION EQUATION IN CARTESIAN COORDINATES

Consider an unsteady (transient) three dimensional heat conduction with internal heat generation in heterogeneous amp anisotropic PLANE WALL

Cartesian co-ordinates x y z

Elemental length dxdydz

Consider an elemental volume ABCDEFGH = (dxdydz)

Let Temperature distribution is a function of T=T(xyz)

Let

=internal heat generation per unit volume per unit time (W m3)q= rate of heat flow (W)Q= qdt=total heat flow (J)r=mass density of the material element(kgm3)Cp=specific heat of the material element (Jkg K)

gq

Plane (y - z) lsquoxrsquo direction

Quantity of heat leaving from the face EFGH in lsquoXrsquo direction

Quantity of heat flowing into the face ABCD in lsquoXrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoXrsquo direction

dy

TdzdxKQ yy )(

dyQy

QQ yydyy ][

dydy

TdzdxK

yQQdQ ydyyyy ])([)(

)2()]([)(

ddzdydx

y

TK

ydQ yy

Plane (x - z) lsquoyrsquo direction

Quantity of heat flowing into the face ABFE in lsquoYrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoYrsquo direction

dz

TdydxKQ zz )(

dzQz

QQ zzdzz ][

dzdz

TdydxK

zQQdQ zdzzzz ])([)(

)3()]([)(

ddzdydx

z

TK

zdQ zz

Plane (x - y) lsquozrsquo direction

Quantity of heat flowing into the face ADHE in lsquoZrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoZrsquo direction

Quantity of heat leaving from the face DCGH in lsquoYrsquo direction

Quantity of heat leaving from the face BCGF in lsquoZrsquo direction

)4()()(

ddzdydxqQgg

Total amount of internal heat generation in the element

Ein + Egen = Eout+Est (Ein- Eout)+ Egen=Est

T

Cddzdydxddzdydxqddzdydxz

TK

zy

TK

yx

TK

x pgzyx )()())](()()([

T

pC

gq

z

TzK

zy

Ty

Kyx

TxK

x][][][

stgzyx QQdQdQdQ

)(

)5()()(

dT

CdzdydxQ pst

Total heat energy stored in the element

(ie thermal energy of the element can be increased because of heat accumulation in the element due to heat flow amp internal heat generation)

)( volumem d

TmCQ pst )(

T

CqTK pg)(

KKKyKx z If material is homogeneous amp isotropic

T

pC

gq

z

T

zy

T

yx

T

xK ][][][

T

Kp

C

Kgq

z

T

y

T

x

T2

2

2

2

2

2

T

Kgq

z

T

y

T

x

T 12

2

2

2

2

2

Thermal diffusivity pC

K

T

K

qT g

12

according to thermodynamics first law

)( zr )( dzrddr

)( dzrddr

Cylindrical polar co-ordinates

Elemental length

Consider an elemental volume ABCDEFGH =

GENERAL HEAT CONDUCTION EQUATION IN CYLINDRICAL COORDINATES

gq

gq

Consider an unsteady (transient) three dimensional heat conduction with internal heat generation in heterogeneous amp anisotropic cylindrical geometries (eg rods and pipes)

)( zrTLetTemperature distribution is a function of T=

Let

=internal heat generation per unit volume per unit time (W m3)q= rate of heat flow (W)Q= qdt=total heat flow (J)r=mass density of the material element(kgm3)Cp=specific heat of the material element (Jkg K)

planerdirectionz

planezrdirection

planezdirectionr

)(

)(

)(

LetHeat flow in

drdr

TdzrdK

rdQ

QQdQ

rr

drrrr

])([

)1(][

r

Tr

rddrdzdKdQ rr

Heat accumulation in the element due to heat flow by conduction in lsquorrsquo direction

rddr

TdzdrK

rdQ

QQdQ d

])([

][

2

T

r

ddrdzrdKdQ

)2(][

2

2

2

T

r

ddrdzrdKdQ

Heat accumulation in the element due to heat flow by conduction in lsquoqrsquo direction

Plane (z- q ) lsquorrsquo direction

dr

TdzrdKQ rr

)(

drQr

QQ rrdrr ][

Quantity of heat leaving from the face EFGH in lsquorrsquo direction

Quantity of heat flowing into the face ABCD in lsquorrsquo direction

Plane (r-z) lsquoq lsquodirection

Quantity of heat leaving from the face AEGC in lsquoqrsquo direction

d

r

TdzdrKQ )(

rdQ

rQQ d ][

Quantity of heat flowing into the face BFHD in lsquoqrsquo direction

)3())((2

2

z

TdzdrrdKdQ zz

)5()()(

dT

CdzdrrdQ pst

)4()()(

ddzdrrdqQgg

Total amount of internal heat generation in the element

Total heat energy stored in the element

(ie thermal energy of the element can be increased because of heat accumulation in the element due to heat flow amp internal heat generation)

according to thermodynamics first law

Ein + Egen = Eout + Est (Ein ndash Eout) + Egen = Est

)( volumem

)()(z

T

zdzdrrdKdQ zz

Heat accumulation in the element due to heat flow by conduction in lsquozrsquo direction

dzdz

TdrrdK

zdQ

QQdQ

zz

dzzzz

])([

d

TmCQ pst )(

Plane (r-q) lsquozrsquo direction

dz

TdrrdKQ zz )(

dzQz

QQ zzdzz ][ Quantity of heat leaving from the face CDHG in lsquozrsquo direction

Quantity of heat flowing into the face ABFE in lsquozrsquo direction

stgzr QQdQdQdQ

)(

r

r ][)(rT

rr

ddrdzdKr

][)(

2

2

2

T

r

ddrdzrdK ))((2

2

z

TddzdrrdK z

ddzdrrdq g )(

dT

Cdzdrrd p )(

+ + +

=

T

CqTK pg)(KKKK zr

[gq

)(1

r

Tr

rrK r

)(1

2

2

2 T

rK )(

2

2

z

TK z

)( ddzdrrd )(

ddzdrrd

TC p

+ + + ] =

If material is homogeneous amp isotropic

)(1

r

Tr

rr

)(1

2

2

2 T

r)(

2

2

z

T

K

q g

T

K

C p

)(1

r

Tr

rr

)(1

2

2

2 T

r )(2

2

z

T

K

q g

T

1

+ + + =

)(1

r

Tr

rrK r

)(1

2

2

2 T

rK )(

2

2

z

TK z

gq

T

C p + + + =

+ + + = Thermal diffusivity

T

K

qT g

12

pC

K

(Ein ndash Eout) + Egen = Est

)( rT

)( r

)sin( drrddr)sin( drrddr

planerdirection

planerdirection

planedirectionr

)(

)(

)(

Spherical polar co-ordinates

Elemental length

Consider an elemental volume ABCDEFGH =

GENERAL HEAT CONDUCTION EQUATION IN SPHERICAL COORDINATES

Let

=internal heat generation per unit volume per unit time (W m3)q= rate of heat flow (W)Q= qdt=total heat flow (J)r=mass density of the material element(kgm3)Cp=specific heat of the material element (Jkg K)

gq

Consider an unsteady (transient) three dimensional heat conduction with internal heat generation in heterogeneous amp anisotropic spherical geometries

LetTemperature distribution is a function of T=

X

Y

Z

drT

drrdKQ rr )sin(

drQr

QQ rrdrr ][

drdr

TdrrdK

rdQ

QQdQ

rr

drrrr

])sin([

)1(]([sin 2

rT

rr

ddrddKdQ rr

Plane ( -q f) lsquorrsquo directionQuantity of heat flowing into the face ABCD in lsquorrsquo direction

Quantity of heat leaving from the face EFGH in lsquorrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquorrsquo direction

drT

drdrKQ )sin(

rdQ

rQQ d ][

rddr

TdrdrK

rdQ

QQdQ d

])sin([

])[(sin

2

T

r

drdrdrKdQ

)2(])(sin[

2

T

r

drdrddrKdQ

Quantity of heat leaving from the face BFHD in lsquoqrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoqrsquo direction

X

Y

Z

Plane (r- f ) lsquoq lsquodirection

Quantity of heat flowing into the face AEGC in lsquoqrsquo direction

dr

TdrrdKQ

sin)(

drQ

rQQ d sin][

sin

drdr

TdrrdK

rdQ

QQdQ d

sin]sin

)([sin

][sinsin

)sin(

T

rr

ddrdrrdKdQ

)3()(sin

)sin(2

2

22

2

T

r

ddddrrKdQ

Plane (r- ) q lsquofrsquo directionQuantity of heat flowing into the face ABFE in lsquofrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquozrsquo direction

)4()sin()(

ddrrddrqQgg

)5()sin()(

dT

CdrrddrQ pst

Total amount of internal heat generation in the element

Total heat energy stored in the element

(Ein ndash Eout) + Egen = Est

d

TmCQ pst )(

)( volumem

Quantity of heat leaving from the face CDHG in lsquofrsquo direction

according to thermodynamics first law

Ein + Egen = Eout + Est

X

Y

Z

stgr QQdQdQdQ

)(

]([sin 2

r

Tr

rddrddKr

])(sin[

2

T

rdrdrddrK

)(sin

)sin(2

2

22

2

T

r

ddddrrK

ddrrddrq g )sin(

dT

Cdrrddr p )sin(

2

2

r

r

sin

sin

]([ 22 r

Tr

rr

Kr

])(sin[sin2

TrK

)(sin 2

2

22

T

r

Kgq

T

Cp

]([1

[ 22 r

Tr

rrK

])(sin[sin1

2

T

r)](

sin

12

2

22 T

r gq

T

Cp

]([1

[ 22 r

Tr

rr

])(sin[sin1

2

T

r)](

sin

12

2

22 T

rK

q g

T

K

Cp

]([1

[ 22 r

Tr

rr

])(sin[sin1

2

T

r)](

sin

12

2

22 T

rK

q g

T1

KKKKr If material is homogeneous amp isotropic

TCqTK pg)(

T

K

qT g

12 pC

K

Thermal diffusivity

(Ein ndash Eout) + Egen = Est

General Heat Conduction Equation

PLANE WALL CYLINDER SPHERE

Temperature profile

Plane wall SphereCylinder

bull A copper conductor (k = 380 WmoC resistivity ρ = 2x10-8 Ω m) having inner and outer radii 10 cm and 225 cm respectively is carrying a current density of 4800 amperes cm2 the conductor is internally cooled and a constant temperature of 65oC is maintained at the inner surface and there is no heat transfer through the insulation surrounding the conductor

Determinebull (i) The maximum temperature of the conductor and the radius at

which it occursbull (ii) The internal heat transfer rate

bull A chemical reaction takes place in a packed bed (k=06 WmoC) between two coaxial cylinders with radii 15 mm and 45 mm the inner surface is at 580oC and it is insulated Assuming the reaction rate of 055 MWm3 in the reactor volume find the temperature at the outer surface of the reactor

bull A current of 300 amperes passes through a stainless steel wire of 25 mm diameter and k=20 WmoC The resistivity of the wire is 70x10-8 ohm-m and length of the wire is 2 m if the wire is submerged in a fluid maintained at 50oC and h at the wire surface is 4000 Wm2oC calculate the steady state temperature at the centre and at the surface of the wire

bull An approximately spherical shaped orange ( k=023 WmoC) 90 mm diameter undergoes ripping process and generates 5100 Wm3 of energy If external surface the orange is at 8oC determine

bull (i) temperature at the centre of the orange and

bull (ii) heat flow from the outer surface of the orange

bull FINS External surfaces to used to cooling area and hence convection

heat transfer can be improvedbull APPLICATIONS IC engine cylinder ( air cooling) Heat exchangers automobile

radiators computer chip refrigerators electric motors generators and transformers etc

bull ASSUMPTIONS MADE IN ANALYSIS OF HEAT TRANFER THROUGH FIN

1 Fin material is homogenous and isotropic constthermal conductivity

2 Temperature at any cs of the fin is uniform3 lsquohrsquo is uniform over entire surface4 No internal heat generation5 Thermal contact resistance is negligible6 Radiation heat transfer is negligible7 Heat conduction is steady state and one dimensional

bull Problem 1

bull A stainless steel fin (k=20 WmK) having a diameter of 20 mm and a length of 01 m is attached to a wall at 300oC The ambient temperature is 50oC and the heat transfer coefficient is 10 Wm2K The fin tip is insulated Determine

bull (A) The rate of heat dissipation from the finbull (B) The temperature at the fin tipbull (C) The rate of heat transfer from the wall area covered by the fin if

the fin was not usedbull (D) The heat transfer rate from the same fin geometry if the

stainless steel fin is replaced by a fictitious fin with infinite thermal conductivity

bull Problem2bull One end of a long rod is inserted into a furnace while the other

projects into ambient temperature Under steady state the temperature of the rod is measured at two points 75mm apart and found to be 125oC and 885oC respectively while the ambient temperature is 20oC If the rod is 25mm in diameter and h is 2336 Wm2K

bull Find the thermal conductivity of the rod material

  • HEAT AND MASS TRANSFER
  • Thermodynamics Vs Heat transfer
  • MODE OF HEAT TRANSFER
  • MODE OF HEAT TRANSFER (2)
  • CONDUCTION- MECHANISM
  • CONCECTION-MECHANISM
  • RADIATION-MECHANISM
  • Conduction- Convection- Radiation
  • Slide 9
  • Slide 10
  • Fourierrsquos Law of Heat Conduction
  • Slide 12
  • Newtonrsquos Law of Cooling
  • Laws of Thermal Radiation
  • LAWS FOR HEAT TRANSFER
  • ELECTRICAL ampTHERMAL ANALOGY
  • Slide 17
  • Slide 18
  • PLANE WALL-UNIDIRECTIONAL
  • PLANE WALL -SERIES
  • PLANE WALL IN SERIES
  • PLANE WALL -PARALLEL
  • PLANE WALL-PARALLEL
  • PLANE WALL-COMBINATION
  • PLANE WALL-FILM RESISTANCE
  • PLANE WALL- THERMAL CONTACT RESISTANCE
  • PLANE WALL- THERMAL CONTACT RESISTANCE (2)
  • CYLINDER-SOLID
  • HEAT TRANSFER IN SOLID CYLINDER
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER-COMPOSITE
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER-MULTY LAYER
  • HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER
  • HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER (2)
  • HEAT TRANSFER IN SPHERE
  • HEAT TRANSFER IN SPHERE -MULTY LAYER
  • HEAT TRANSFER IN SPHERE -MULTY LAYER (2)
  • Slide 38
  • Slide 39
  • Slide 40
  • Slide 41
  • CRITICAL RADIUS
  • Plane wall- Insulation thickness -study
  • Problem5
  • Slide 45
  • Slide 46
  • Slide 47
  • GENERAL HEAT CONDUCTION EQUATION IN CYLINDRICAL COORDINATES
  • Slide 49
  • Slide 50
  • Slide 51
  • GENERAL HEAT CONDUCTION EQUATION IN SPHERICAL COORDINATES
  • Slide 53
  • Slide 54
  • Slide 55
  • General Heat Conduction Equation
  • Temperature profile
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Slide 67
Page 9: hmt chap 1

Examples for different mode of heat transfer

bull Conduction

Heat loss through thermal insulation on steam pipe

Convection

Heat transfer to water flowing through a pipe in condenser

Radiation

Heat transfer in an electric furnace

Conduction Convection and Radiation

Solar energy used water heater

Fourierrsquos Law of Heat Conduction

Newtonrsquos Law of Cooling

Laws of Thermal Radiation

LAWS FOR HEAT TRANSFERCONDUCTION

4Tq

CONVECTION

NEWTONrsquoS LAW OF COOLING

RADIATION

STEFAN BOLTZMAN LAW

q=-KA( dT dX)

Assumptionsbull Steady statebull Unidirectional heat flowbull Const temperature gradientbull Linear profilebull No internal heat generationbull Isothermal bounding surfacebull Homogeneousbull Isotropic bull Applicable to solid liquid gas

q=σAT4q= hA (Tw-Tα)

Thq

XT

Kq

FOURIERrsquoS LAW

q =σ ε AT4

ELECTRICAL ampTHERMAL ANALOGY

THERMAL RESISTANCE(ELECTRICAL ANALOGY)

bull OHMrsquos LAW Flow of Electricity

V=I R elect

Voltage Drop = Current flowtimesResistance Temp Drop=Heat FlowtimesResistance

ΔT= q Rtherm

Thermal Analogy To OHMrsquoS Law

PLANE WALL-UNIDIRECTIONAL

PLANE WALL -SERIES

PLANE WALL IN SERIES

PLANE WALL -PARALLEL

PLANE WALL-PARALLEL

PLANE WALL-COMBINATION

PLANE WALL-FILM RESISTANCE

PLANE WALL- THERMAL CONTACT RESISTANCE

PLANE WALL- THERMAL CONTACT RESISTANCE

At the joining surface of the two slabs air is trapped in voids due to surface irregularities

CYLINDER-SOLID

HEAT TRANSFER IN SOLID CYLINDER

HEAT TRANSFER RATE IN HOLLOW CYLINDER

HEAT TRANSFER RATE IN HOLLOW CYLINDER-COMPOSITE

HEAT TRANSFER RATE IN HOLLOW CYLINDER-MULTY LAYER

HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER

HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER

HEAT TRANSFER IN SPHERE

HEAT TRANSFER IN SPHERE -MULTY LAYER

HEAT TRANSFER IN SPHERE -MULTY LAYER

bull Problem 1 The wall in a furnace consists of 125 mm thick refractory bricks and

125 mm thick insulating fire bricks separated by an air gap A 12 mm thick plaster covers the outer wall The inner surface of wall is at 1100oC and the ambient temperature 25oC the heat transfer coefficient on the outer wall to the air is 17 Wm2K and the resistance to heat flow of the air gap is 016 KW the thermal conductivities of refractory brick insulating firebrick and plaster are 16 03 and 014 WmK respectively

Calculatebull (a) the rate of heat loss per unit area of the wall surfacebull (b) the interface temperatures throughout the wallbull (c) the temperature at the outside surface of the wall

bull Problem 2Two large aluminum plates (k=240 WmK) each 2 cm thick with 10 mm surface roughness are placed in contact under 105 Nm2 pressure in air The temperature at the outside surfaces are 390 oC and 406 oCCalculate

A) the heat flux B) the temperature drop due to the contact resistance C) the contact temperatures Thermal contact resistance with air as the interfacial fluid for

10 mm roughness is 27510-4 m2KW

bull Problem 3bull Steam at 350 oC flowing in a pipe (k=80 WmK) of 5 cm

inner diameter and 56 cm outer diameter is covered with 3 cm thick insulation of k=005 WmK Heat is lost to surroundings at 5 oC by natural convection and radiation the combined h being 20 Wm2K Taking the heat transfer coefficient inside the pipe as 60 Wm2K

bull Determinebull A) the rate of heat loss from the steam per unit length of

the pipebull B) the temperature drop across the pipe and the insulation

bull PROBLEM4bull A nuclear reactor has a sperical pressure vessel 750 mm inside

diameter 80 mm wall thickness The temperature at the inner surface is 500 oC If the tempertarue at the outer surface is 495 oC

bull Calculatea) The rate of heat loss through the metal wallb) If the rate of heat loss is to limited to 305Whow much inulation

thickness needs to be applied Thermal conductivity of steel wall=46 WmoC Thermal conductivity of insulation=004 WmoC Film coefficient of heat tranfer at inner and outer surface are 200

Wm2 oC and 15 Wm2 oC Ambient air temperature=28 oC

CRITICAL RADIUS

bull The radius upto which heat flow increases and after decreases At critical radius q becomes qmax

bull The addition of insulation always increases (Rth)condbull Rtotal= (Rth)cond + (Rth)conv

bull The addition of insulation may reduce the (Rth)conv due to increase in surface area

bull r1ltrc -heat transfer increases by adding insulation till r1=rc If

insulation thickness is further increased the rate of heat loss will decrease

r1 gt rc

- heat transfer decreases by adding insulation above the rc

Plane wall- Insulation

thickness -study

Cylinder amp Sphere- Insulation thickness -

study

Problem5bull A 3mm diameter and 5 mm long electric wire is tightly

wrapped with a 2 mm thick plastic cover whose K=015 wmoC Electrical measurements indicate that a current of 10 A passes through the wire and there is a voltage drop of 8 V along the wire If the insulated wire is exposed to a medium at Ta=30oC with a heat transfer coefficient of h=12 wm2oC Determine the temperature at the interface of the wire and the plastic cover in steady operation Also determine whether doubling the thickness of the plastic cover will increase or decrease this interface temperature

dx

TdzdyKQ xx )(

dxQx

QQ xxdxx ][

dxdx

TdzdyK

xQQdQ xdxxxx ])([)(

)1()]([)(

ddzdydx

x

TK

xdQ xx

GENERAL HEAT CONDUCTION EQUATION IN CARTESIAN COORDINATES

Consider an unsteady (transient) three dimensional heat conduction with internal heat generation in heterogeneous amp anisotropic PLANE WALL

Cartesian co-ordinates x y z

Elemental length dxdydz

Consider an elemental volume ABCDEFGH = (dxdydz)

Let Temperature distribution is a function of T=T(xyz)

Let

=internal heat generation per unit volume per unit time (W m3)q= rate of heat flow (W)Q= qdt=total heat flow (J)r=mass density of the material element(kgm3)Cp=specific heat of the material element (Jkg K)

gq

Plane (y - z) lsquoxrsquo direction

Quantity of heat leaving from the face EFGH in lsquoXrsquo direction

Quantity of heat flowing into the face ABCD in lsquoXrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoXrsquo direction

dy

TdzdxKQ yy )(

dyQy

QQ yydyy ][

dydy

TdzdxK

yQQdQ ydyyyy ])([)(

)2()]([)(

ddzdydx

y

TK

ydQ yy

Plane (x - z) lsquoyrsquo direction

Quantity of heat flowing into the face ABFE in lsquoYrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoYrsquo direction

dz

TdydxKQ zz )(

dzQz

QQ zzdzz ][

dzdz

TdydxK

zQQdQ zdzzzz ])([)(

)3()]([)(

ddzdydx

z

TK

zdQ zz

Plane (x - y) lsquozrsquo direction

Quantity of heat flowing into the face ADHE in lsquoZrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoZrsquo direction

Quantity of heat leaving from the face DCGH in lsquoYrsquo direction

Quantity of heat leaving from the face BCGF in lsquoZrsquo direction

)4()()(

ddzdydxqQgg

Total amount of internal heat generation in the element

Ein + Egen = Eout+Est (Ein- Eout)+ Egen=Est

T

Cddzdydxddzdydxqddzdydxz

TK

zy

TK

yx

TK

x pgzyx )()())](()()([

T

pC

gq

z

TzK

zy

Ty

Kyx

TxK

x][][][

stgzyx QQdQdQdQ

)(

)5()()(

dT

CdzdydxQ pst

Total heat energy stored in the element

(ie thermal energy of the element can be increased because of heat accumulation in the element due to heat flow amp internal heat generation)

)( volumem d

TmCQ pst )(

T

CqTK pg)(

KKKyKx z If material is homogeneous amp isotropic

T

pC

gq

z

T

zy

T

yx

T

xK ][][][

T

Kp

C

Kgq

z

T

y

T

x

T2

2

2

2

2

2

T

Kgq

z

T

y

T

x

T 12

2

2

2

2

2

Thermal diffusivity pC

K

T

K

qT g

12

according to thermodynamics first law

)( zr )( dzrddr

)( dzrddr

Cylindrical polar co-ordinates

Elemental length

Consider an elemental volume ABCDEFGH =

GENERAL HEAT CONDUCTION EQUATION IN CYLINDRICAL COORDINATES

gq

gq

Consider an unsteady (transient) three dimensional heat conduction with internal heat generation in heterogeneous amp anisotropic cylindrical geometries (eg rods and pipes)

)( zrTLetTemperature distribution is a function of T=

Let

=internal heat generation per unit volume per unit time (W m3)q= rate of heat flow (W)Q= qdt=total heat flow (J)r=mass density of the material element(kgm3)Cp=specific heat of the material element (Jkg K)

planerdirectionz

planezrdirection

planezdirectionr

)(

)(

)(

LetHeat flow in

drdr

TdzrdK

rdQ

QQdQ

rr

drrrr

])([

)1(][

r

Tr

rddrdzdKdQ rr

Heat accumulation in the element due to heat flow by conduction in lsquorrsquo direction

rddr

TdzdrK

rdQ

QQdQ d

])([

][

2

T

r

ddrdzrdKdQ

)2(][

2

2

2

T

r

ddrdzrdKdQ

Heat accumulation in the element due to heat flow by conduction in lsquoqrsquo direction

Plane (z- q ) lsquorrsquo direction

dr

TdzrdKQ rr

)(

drQr

QQ rrdrr ][

Quantity of heat leaving from the face EFGH in lsquorrsquo direction

Quantity of heat flowing into the face ABCD in lsquorrsquo direction

Plane (r-z) lsquoq lsquodirection

Quantity of heat leaving from the face AEGC in lsquoqrsquo direction

d

r

TdzdrKQ )(

rdQ

rQQ d ][

Quantity of heat flowing into the face BFHD in lsquoqrsquo direction

)3())((2

2

z

TdzdrrdKdQ zz

)5()()(

dT

CdzdrrdQ pst

)4()()(

ddzdrrdqQgg

Total amount of internal heat generation in the element

Total heat energy stored in the element

(ie thermal energy of the element can be increased because of heat accumulation in the element due to heat flow amp internal heat generation)

according to thermodynamics first law

Ein + Egen = Eout + Est (Ein ndash Eout) + Egen = Est

)( volumem

)()(z

T

zdzdrrdKdQ zz

Heat accumulation in the element due to heat flow by conduction in lsquozrsquo direction

dzdz

TdrrdK

zdQ

QQdQ

zz

dzzzz

])([

d

TmCQ pst )(

Plane (r-q) lsquozrsquo direction

dz

TdrrdKQ zz )(

dzQz

QQ zzdzz ][ Quantity of heat leaving from the face CDHG in lsquozrsquo direction

Quantity of heat flowing into the face ABFE in lsquozrsquo direction

stgzr QQdQdQdQ

)(

r

r ][)(rT

rr

ddrdzdKr

][)(

2

2

2

T

r

ddrdzrdK ))((2

2

z

TddzdrrdK z

ddzdrrdq g )(

dT

Cdzdrrd p )(

+ + +

=

T

CqTK pg)(KKKK zr

[gq

)(1

r

Tr

rrK r

)(1

2

2

2 T

rK )(

2

2

z

TK z

)( ddzdrrd )(

ddzdrrd

TC p

+ + + ] =

If material is homogeneous amp isotropic

)(1

r

Tr

rr

)(1

2

2

2 T

r)(

2

2

z

T

K

q g

T

K

C p

)(1

r

Tr

rr

)(1

2

2

2 T

r )(2

2

z

T

K

q g

T

1

+ + + =

)(1

r

Tr

rrK r

)(1

2

2

2 T

rK )(

2

2

z

TK z

gq

T

C p + + + =

+ + + = Thermal diffusivity

T

K

qT g

12

pC

K

(Ein ndash Eout) + Egen = Est

)( rT

)( r

)sin( drrddr)sin( drrddr

planerdirection

planerdirection

planedirectionr

)(

)(

)(

Spherical polar co-ordinates

Elemental length

Consider an elemental volume ABCDEFGH =

GENERAL HEAT CONDUCTION EQUATION IN SPHERICAL COORDINATES

Let

=internal heat generation per unit volume per unit time (W m3)q= rate of heat flow (W)Q= qdt=total heat flow (J)r=mass density of the material element(kgm3)Cp=specific heat of the material element (Jkg K)

gq

Consider an unsteady (transient) three dimensional heat conduction with internal heat generation in heterogeneous amp anisotropic spherical geometries

LetTemperature distribution is a function of T=

X

Y

Z

drT

drrdKQ rr )sin(

drQr

QQ rrdrr ][

drdr

TdrrdK

rdQ

QQdQ

rr

drrrr

])sin([

)1(]([sin 2

rT

rr

ddrddKdQ rr

Plane ( -q f) lsquorrsquo directionQuantity of heat flowing into the face ABCD in lsquorrsquo direction

Quantity of heat leaving from the face EFGH in lsquorrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquorrsquo direction

drT

drdrKQ )sin(

rdQ

rQQ d ][

rddr

TdrdrK

rdQ

QQdQ d

])sin([

])[(sin

2

T

r

drdrdrKdQ

)2(])(sin[

2

T

r

drdrddrKdQ

Quantity of heat leaving from the face BFHD in lsquoqrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoqrsquo direction

X

Y

Z

Plane (r- f ) lsquoq lsquodirection

Quantity of heat flowing into the face AEGC in lsquoqrsquo direction

dr

TdrrdKQ

sin)(

drQ

rQQ d sin][

sin

drdr

TdrrdK

rdQ

QQdQ d

sin]sin

)([sin

][sinsin

)sin(

T

rr

ddrdrrdKdQ

)3()(sin

)sin(2

2

22

2

T

r

ddddrrKdQ

Plane (r- ) q lsquofrsquo directionQuantity of heat flowing into the face ABFE in lsquofrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquozrsquo direction

)4()sin()(

ddrrddrqQgg

)5()sin()(

dT

CdrrddrQ pst

Total amount of internal heat generation in the element

Total heat energy stored in the element

(Ein ndash Eout) + Egen = Est

d

TmCQ pst )(

)( volumem

Quantity of heat leaving from the face CDHG in lsquofrsquo direction

according to thermodynamics first law

Ein + Egen = Eout + Est

X

Y

Z

stgr QQdQdQdQ

)(

]([sin 2

r

Tr

rddrddKr

])(sin[

2

T

rdrdrddrK

)(sin

)sin(2

2

22

2

T

r

ddddrrK

ddrrddrq g )sin(

dT

Cdrrddr p )sin(

2

2

r

r

sin

sin

]([ 22 r

Tr

rr

Kr

])(sin[sin2

TrK

)(sin 2

2

22

T

r

Kgq

T

Cp

]([1

[ 22 r

Tr

rrK

])(sin[sin1

2

T

r)](

sin

12

2

22 T

r gq

T

Cp

]([1

[ 22 r

Tr

rr

])(sin[sin1

2

T

r)](

sin

12

2

22 T

rK

q g

T

K

Cp

]([1

[ 22 r

Tr

rr

])(sin[sin1

2

T

r)](

sin

12

2

22 T

rK

q g

T1

KKKKr If material is homogeneous amp isotropic

TCqTK pg)(

T

K

qT g

12 pC

K

Thermal diffusivity

(Ein ndash Eout) + Egen = Est

General Heat Conduction Equation

PLANE WALL CYLINDER SPHERE

Temperature profile

Plane wall SphereCylinder

bull A copper conductor (k = 380 WmoC resistivity ρ = 2x10-8 Ω m) having inner and outer radii 10 cm and 225 cm respectively is carrying a current density of 4800 amperes cm2 the conductor is internally cooled and a constant temperature of 65oC is maintained at the inner surface and there is no heat transfer through the insulation surrounding the conductor

Determinebull (i) The maximum temperature of the conductor and the radius at

which it occursbull (ii) The internal heat transfer rate

bull A chemical reaction takes place in a packed bed (k=06 WmoC) between two coaxial cylinders with radii 15 mm and 45 mm the inner surface is at 580oC and it is insulated Assuming the reaction rate of 055 MWm3 in the reactor volume find the temperature at the outer surface of the reactor

bull A current of 300 amperes passes through a stainless steel wire of 25 mm diameter and k=20 WmoC The resistivity of the wire is 70x10-8 ohm-m and length of the wire is 2 m if the wire is submerged in a fluid maintained at 50oC and h at the wire surface is 4000 Wm2oC calculate the steady state temperature at the centre and at the surface of the wire

bull An approximately spherical shaped orange ( k=023 WmoC) 90 mm diameter undergoes ripping process and generates 5100 Wm3 of energy If external surface the orange is at 8oC determine

bull (i) temperature at the centre of the orange and

bull (ii) heat flow from the outer surface of the orange

bull FINS External surfaces to used to cooling area and hence convection

heat transfer can be improvedbull APPLICATIONS IC engine cylinder ( air cooling) Heat exchangers automobile

radiators computer chip refrigerators electric motors generators and transformers etc

bull ASSUMPTIONS MADE IN ANALYSIS OF HEAT TRANFER THROUGH FIN

1 Fin material is homogenous and isotropic constthermal conductivity

2 Temperature at any cs of the fin is uniform3 lsquohrsquo is uniform over entire surface4 No internal heat generation5 Thermal contact resistance is negligible6 Radiation heat transfer is negligible7 Heat conduction is steady state and one dimensional

bull Problem 1

bull A stainless steel fin (k=20 WmK) having a diameter of 20 mm and a length of 01 m is attached to a wall at 300oC The ambient temperature is 50oC and the heat transfer coefficient is 10 Wm2K The fin tip is insulated Determine

bull (A) The rate of heat dissipation from the finbull (B) The temperature at the fin tipbull (C) The rate of heat transfer from the wall area covered by the fin if

the fin was not usedbull (D) The heat transfer rate from the same fin geometry if the

stainless steel fin is replaced by a fictitious fin with infinite thermal conductivity

bull Problem2bull One end of a long rod is inserted into a furnace while the other

projects into ambient temperature Under steady state the temperature of the rod is measured at two points 75mm apart and found to be 125oC and 885oC respectively while the ambient temperature is 20oC If the rod is 25mm in diameter and h is 2336 Wm2K

bull Find the thermal conductivity of the rod material

  • HEAT AND MASS TRANSFER
  • Thermodynamics Vs Heat transfer
  • MODE OF HEAT TRANSFER
  • MODE OF HEAT TRANSFER (2)
  • CONDUCTION- MECHANISM
  • CONCECTION-MECHANISM
  • RADIATION-MECHANISM
  • Conduction- Convection- Radiation
  • Slide 9
  • Slide 10
  • Fourierrsquos Law of Heat Conduction
  • Slide 12
  • Newtonrsquos Law of Cooling
  • Laws of Thermal Radiation
  • LAWS FOR HEAT TRANSFER
  • ELECTRICAL ampTHERMAL ANALOGY
  • Slide 17
  • Slide 18
  • PLANE WALL-UNIDIRECTIONAL
  • PLANE WALL -SERIES
  • PLANE WALL IN SERIES
  • PLANE WALL -PARALLEL
  • PLANE WALL-PARALLEL
  • PLANE WALL-COMBINATION
  • PLANE WALL-FILM RESISTANCE
  • PLANE WALL- THERMAL CONTACT RESISTANCE
  • PLANE WALL- THERMAL CONTACT RESISTANCE (2)
  • CYLINDER-SOLID
  • HEAT TRANSFER IN SOLID CYLINDER
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER-COMPOSITE
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER-MULTY LAYER
  • HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER
  • HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER (2)
  • HEAT TRANSFER IN SPHERE
  • HEAT TRANSFER IN SPHERE -MULTY LAYER
  • HEAT TRANSFER IN SPHERE -MULTY LAYER (2)
  • Slide 38
  • Slide 39
  • Slide 40
  • Slide 41
  • CRITICAL RADIUS
  • Plane wall- Insulation thickness -study
  • Problem5
  • Slide 45
  • Slide 46
  • Slide 47
  • GENERAL HEAT CONDUCTION EQUATION IN CYLINDRICAL COORDINATES
  • Slide 49
  • Slide 50
  • Slide 51
  • GENERAL HEAT CONDUCTION EQUATION IN SPHERICAL COORDINATES
  • Slide 53
  • Slide 54
  • Slide 55
  • General Heat Conduction Equation
  • Temperature profile
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Slide 67
Page 10: hmt chap 1

Fourierrsquos Law of Heat Conduction

Newtonrsquos Law of Cooling

Laws of Thermal Radiation

LAWS FOR HEAT TRANSFERCONDUCTION

4Tq

CONVECTION

NEWTONrsquoS LAW OF COOLING

RADIATION

STEFAN BOLTZMAN LAW

q=-KA( dT dX)

Assumptionsbull Steady statebull Unidirectional heat flowbull Const temperature gradientbull Linear profilebull No internal heat generationbull Isothermal bounding surfacebull Homogeneousbull Isotropic bull Applicable to solid liquid gas

q=σAT4q= hA (Tw-Tα)

Thq

XT

Kq

FOURIERrsquoS LAW

q =σ ε AT4

ELECTRICAL ampTHERMAL ANALOGY

THERMAL RESISTANCE(ELECTRICAL ANALOGY)

bull OHMrsquos LAW Flow of Electricity

V=I R elect

Voltage Drop = Current flowtimesResistance Temp Drop=Heat FlowtimesResistance

ΔT= q Rtherm

Thermal Analogy To OHMrsquoS Law

PLANE WALL-UNIDIRECTIONAL

PLANE WALL -SERIES

PLANE WALL IN SERIES

PLANE WALL -PARALLEL

PLANE WALL-PARALLEL

PLANE WALL-COMBINATION

PLANE WALL-FILM RESISTANCE

PLANE WALL- THERMAL CONTACT RESISTANCE

PLANE WALL- THERMAL CONTACT RESISTANCE

At the joining surface of the two slabs air is trapped in voids due to surface irregularities

CYLINDER-SOLID

HEAT TRANSFER IN SOLID CYLINDER

HEAT TRANSFER RATE IN HOLLOW CYLINDER

HEAT TRANSFER RATE IN HOLLOW CYLINDER-COMPOSITE

HEAT TRANSFER RATE IN HOLLOW CYLINDER-MULTY LAYER

HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER

HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER

HEAT TRANSFER IN SPHERE

HEAT TRANSFER IN SPHERE -MULTY LAYER

HEAT TRANSFER IN SPHERE -MULTY LAYER

bull Problem 1 The wall in a furnace consists of 125 mm thick refractory bricks and

125 mm thick insulating fire bricks separated by an air gap A 12 mm thick plaster covers the outer wall The inner surface of wall is at 1100oC and the ambient temperature 25oC the heat transfer coefficient on the outer wall to the air is 17 Wm2K and the resistance to heat flow of the air gap is 016 KW the thermal conductivities of refractory brick insulating firebrick and plaster are 16 03 and 014 WmK respectively

Calculatebull (a) the rate of heat loss per unit area of the wall surfacebull (b) the interface temperatures throughout the wallbull (c) the temperature at the outside surface of the wall

bull Problem 2Two large aluminum plates (k=240 WmK) each 2 cm thick with 10 mm surface roughness are placed in contact under 105 Nm2 pressure in air The temperature at the outside surfaces are 390 oC and 406 oCCalculate

A) the heat flux B) the temperature drop due to the contact resistance C) the contact temperatures Thermal contact resistance with air as the interfacial fluid for

10 mm roughness is 27510-4 m2KW

bull Problem 3bull Steam at 350 oC flowing in a pipe (k=80 WmK) of 5 cm

inner diameter and 56 cm outer diameter is covered with 3 cm thick insulation of k=005 WmK Heat is lost to surroundings at 5 oC by natural convection and radiation the combined h being 20 Wm2K Taking the heat transfer coefficient inside the pipe as 60 Wm2K

bull Determinebull A) the rate of heat loss from the steam per unit length of

the pipebull B) the temperature drop across the pipe and the insulation

bull PROBLEM4bull A nuclear reactor has a sperical pressure vessel 750 mm inside

diameter 80 mm wall thickness The temperature at the inner surface is 500 oC If the tempertarue at the outer surface is 495 oC

bull Calculatea) The rate of heat loss through the metal wallb) If the rate of heat loss is to limited to 305Whow much inulation

thickness needs to be applied Thermal conductivity of steel wall=46 WmoC Thermal conductivity of insulation=004 WmoC Film coefficient of heat tranfer at inner and outer surface are 200

Wm2 oC and 15 Wm2 oC Ambient air temperature=28 oC

CRITICAL RADIUS

bull The radius upto which heat flow increases and after decreases At critical radius q becomes qmax

bull The addition of insulation always increases (Rth)condbull Rtotal= (Rth)cond + (Rth)conv

bull The addition of insulation may reduce the (Rth)conv due to increase in surface area

bull r1ltrc -heat transfer increases by adding insulation till r1=rc If

insulation thickness is further increased the rate of heat loss will decrease

r1 gt rc

- heat transfer decreases by adding insulation above the rc

Plane wall- Insulation

thickness -study

Cylinder amp Sphere- Insulation thickness -

study

Problem5bull A 3mm diameter and 5 mm long electric wire is tightly

wrapped with a 2 mm thick plastic cover whose K=015 wmoC Electrical measurements indicate that a current of 10 A passes through the wire and there is a voltage drop of 8 V along the wire If the insulated wire is exposed to a medium at Ta=30oC with a heat transfer coefficient of h=12 wm2oC Determine the temperature at the interface of the wire and the plastic cover in steady operation Also determine whether doubling the thickness of the plastic cover will increase or decrease this interface temperature

dx

TdzdyKQ xx )(

dxQx

QQ xxdxx ][

dxdx

TdzdyK

xQQdQ xdxxxx ])([)(

)1()]([)(

ddzdydx

x

TK

xdQ xx

GENERAL HEAT CONDUCTION EQUATION IN CARTESIAN COORDINATES

Consider an unsteady (transient) three dimensional heat conduction with internal heat generation in heterogeneous amp anisotropic PLANE WALL

Cartesian co-ordinates x y z

Elemental length dxdydz

Consider an elemental volume ABCDEFGH = (dxdydz)

Let Temperature distribution is a function of T=T(xyz)

Let

=internal heat generation per unit volume per unit time (W m3)q= rate of heat flow (W)Q= qdt=total heat flow (J)r=mass density of the material element(kgm3)Cp=specific heat of the material element (Jkg K)

gq

Plane (y - z) lsquoxrsquo direction

Quantity of heat leaving from the face EFGH in lsquoXrsquo direction

Quantity of heat flowing into the face ABCD in lsquoXrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoXrsquo direction

dy

TdzdxKQ yy )(

dyQy

QQ yydyy ][

dydy

TdzdxK

yQQdQ ydyyyy ])([)(

)2()]([)(

ddzdydx

y

TK

ydQ yy

Plane (x - z) lsquoyrsquo direction

Quantity of heat flowing into the face ABFE in lsquoYrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoYrsquo direction

dz

TdydxKQ zz )(

dzQz

QQ zzdzz ][

dzdz

TdydxK

zQQdQ zdzzzz ])([)(

)3()]([)(

ddzdydx

z

TK

zdQ zz

Plane (x - y) lsquozrsquo direction

Quantity of heat flowing into the face ADHE in lsquoZrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoZrsquo direction

Quantity of heat leaving from the face DCGH in lsquoYrsquo direction

Quantity of heat leaving from the face BCGF in lsquoZrsquo direction

)4()()(

ddzdydxqQgg

Total amount of internal heat generation in the element

Ein + Egen = Eout+Est (Ein- Eout)+ Egen=Est

T

Cddzdydxddzdydxqddzdydxz

TK

zy

TK

yx

TK

x pgzyx )()())](()()([

T

pC

gq

z

TzK

zy

Ty

Kyx

TxK

x][][][

stgzyx QQdQdQdQ

)(

)5()()(

dT

CdzdydxQ pst

Total heat energy stored in the element

(ie thermal energy of the element can be increased because of heat accumulation in the element due to heat flow amp internal heat generation)

)( volumem d

TmCQ pst )(

T

CqTK pg)(

KKKyKx z If material is homogeneous amp isotropic

T

pC

gq

z

T

zy

T

yx

T

xK ][][][

T

Kp

C

Kgq

z

T

y

T

x

T2

2

2

2

2

2

T

Kgq

z

T

y

T

x

T 12

2

2

2

2

2

Thermal diffusivity pC

K

T

K

qT g

12

according to thermodynamics first law

)( zr )( dzrddr

)( dzrddr

Cylindrical polar co-ordinates

Elemental length

Consider an elemental volume ABCDEFGH =

GENERAL HEAT CONDUCTION EQUATION IN CYLINDRICAL COORDINATES

gq

gq

Consider an unsteady (transient) three dimensional heat conduction with internal heat generation in heterogeneous amp anisotropic cylindrical geometries (eg rods and pipes)

)( zrTLetTemperature distribution is a function of T=

Let

=internal heat generation per unit volume per unit time (W m3)q= rate of heat flow (W)Q= qdt=total heat flow (J)r=mass density of the material element(kgm3)Cp=specific heat of the material element (Jkg K)

planerdirectionz

planezrdirection

planezdirectionr

)(

)(

)(

LetHeat flow in

drdr

TdzrdK

rdQ

QQdQ

rr

drrrr

])([

)1(][

r

Tr

rddrdzdKdQ rr

Heat accumulation in the element due to heat flow by conduction in lsquorrsquo direction

rddr

TdzdrK

rdQ

QQdQ d

])([

][

2

T

r

ddrdzrdKdQ

)2(][

2

2

2

T

r

ddrdzrdKdQ

Heat accumulation in the element due to heat flow by conduction in lsquoqrsquo direction

Plane (z- q ) lsquorrsquo direction

dr

TdzrdKQ rr

)(

drQr

QQ rrdrr ][

Quantity of heat leaving from the face EFGH in lsquorrsquo direction

Quantity of heat flowing into the face ABCD in lsquorrsquo direction

Plane (r-z) lsquoq lsquodirection

Quantity of heat leaving from the face AEGC in lsquoqrsquo direction

d

r

TdzdrKQ )(

rdQ

rQQ d ][

Quantity of heat flowing into the face BFHD in lsquoqrsquo direction

)3())((2

2

z

TdzdrrdKdQ zz

)5()()(

dT

CdzdrrdQ pst

)4()()(

ddzdrrdqQgg

Total amount of internal heat generation in the element

Total heat energy stored in the element

(ie thermal energy of the element can be increased because of heat accumulation in the element due to heat flow amp internal heat generation)

according to thermodynamics first law

Ein + Egen = Eout + Est (Ein ndash Eout) + Egen = Est

)( volumem

)()(z

T

zdzdrrdKdQ zz

Heat accumulation in the element due to heat flow by conduction in lsquozrsquo direction

dzdz

TdrrdK

zdQ

QQdQ

zz

dzzzz

])([

d

TmCQ pst )(

Plane (r-q) lsquozrsquo direction

dz

TdrrdKQ zz )(

dzQz

QQ zzdzz ][ Quantity of heat leaving from the face CDHG in lsquozrsquo direction

Quantity of heat flowing into the face ABFE in lsquozrsquo direction

stgzr QQdQdQdQ

)(

r

r ][)(rT

rr

ddrdzdKr

][)(

2

2

2

T

r

ddrdzrdK ))((2

2

z

TddzdrrdK z

ddzdrrdq g )(

dT

Cdzdrrd p )(

+ + +

=

T

CqTK pg)(KKKK zr

[gq

)(1

r

Tr

rrK r

)(1

2

2

2 T

rK )(

2

2

z

TK z

)( ddzdrrd )(

ddzdrrd

TC p

+ + + ] =

If material is homogeneous amp isotropic

)(1

r

Tr

rr

)(1

2

2

2 T

r)(

2

2

z

T

K

q g

T

K

C p

)(1

r

Tr

rr

)(1

2

2

2 T

r )(2

2

z

T

K

q g

T

1

+ + + =

)(1

r

Tr

rrK r

)(1

2

2

2 T

rK )(

2

2

z

TK z

gq

T

C p + + + =

+ + + = Thermal diffusivity

T

K

qT g

12

pC

K

(Ein ndash Eout) + Egen = Est

)( rT

)( r

)sin( drrddr)sin( drrddr

planerdirection

planerdirection

planedirectionr

)(

)(

)(

Spherical polar co-ordinates

Elemental length

Consider an elemental volume ABCDEFGH =

GENERAL HEAT CONDUCTION EQUATION IN SPHERICAL COORDINATES

Let

=internal heat generation per unit volume per unit time (W m3)q= rate of heat flow (W)Q= qdt=total heat flow (J)r=mass density of the material element(kgm3)Cp=specific heat of the material element (Jkg K)

gq

Consider an unsteady (transient) three dimensional heat conduction with internal heat generation in heterogeneous amp anisotropic spherical geometries

LetTemperature distribution is a function of T=

X

Y

Z

drT

drrdKQ rr )sin(

drQr

QQ rrdrr ][

drdr

TdrrdK

rdQ

QQdQ

rr

drrrr

])sin([

)1(]([sin 2

rT

rr

ddrddKdQ rr

Plane ( -q f) lsquorrsquo directionQuantity of heat flowing into the face ABCD in lsquorrsquo direction

Quantity of heat leaving from the face EFGH in lsquorrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquorrsquo direction

drT

drdrKQ )sin(

rdQ

rQQ d ][

rddr

TdrdrK

rdQ

QQdQ d

])sin([

])[(sin

2

T

r

drdrdrKdQ

)2(])(sin[

2

T

r

drdrddrKdQ

Quantity of heat leaving from the face BFHD in lsquoqrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoqrsquo direction

X

Y

Z

Plane (r- f ) lsquoq lsquodirection

Quantity of heat flowing into the face AEGC in lsquoqrsquo direction

dr

TdrrdKQ

sin)(

drQ

rQQ d sin][

sin

drdr

TdrrdK

rdQ

QQdQ d

sin]sin

)([sin

][sinsin

)sin(

T

rr

ddrdrrdKdQ

)3()(sin

)sin(2

2

22

2

T

r

ddddrrKdQ

Plane (r- ) q lsquofrsquo directionQuantity of heat flowing into the face ABFE in lsquofrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquozrsquo direction

)4()sin()(

ddrrddrqQgg

)5()sin()(

dT

CdrrddrQ pst

Total amount of internal heat generation in the element

Total heat energy stored in the element

(Ein ndash Eout) + Egen = Est

d

TmCQ pst )(

)( volumem

Quantity of heat leaving from the face CDHG in lsquofrsquo direction

according to thermodynamics first law

Ein + Egen = Eout + Est

X

Y

Z

stgr QQdQdQdQ

)(

]([sin 2

r

Tr

rddrddKr

])(sin[

2

T

rdrdrddrK

)(sin

)sin(2

2

22

2

T

r

ddddrrK

ddrrddrq g )sin(

dT

Cdrrddr p )sin(

2

2

r

r

sin

sin

]([ 22 r

Tr

rr

Kr

])(sin[sin2

TrK

)(sin 2

2

22

T

r

Kgq

T

Cp

]([1

[ 22 r

Tr

rrK

])(sin[sin1

2

T

r)](

sin

12

2

22 T

r gq

T

Cp

]([1

[ 22 r

Tr

rr

])(sin[sin1

2

T

r)](

sin

12

2

22 T

rK

q g

T

K

Cp

]([1

[ 22 r

Tr

rr

])(sin[sin1

2

T

r)](

sin

12

2

22 T

rK

q g

T1

KKKKr If material is homogeneous amp isotropic

TCqTK pg)(

T

K

qT g

12 pC

K

Thermal diffusivity

(Ein ndash Eout) + Egen = Est

General Heat Conduction Equation

PLANE WALL CYLINDER SPHERE

Temperature profile

Plane wall SphereCylinder

bull A copper conductor (k = 380 WmoC resistivity ρ = 2x10-8 Ω m) having inner and outer radii 10 cm and 225 cm respectively is carrying a current density of 4800 amperes cm2 the conductor is internally cooled and a constant temperature of 65oC is maintained at the inner surface and there is no heat transfer through the insulation surrounding the conductor

Determinebull (i) The maximum temperature of the conductor and the radius at

which it occursbull (ii) The internal heat transfer rate

bull A chemical reaction takes place in a packed bed (k=06 WmoC) between two coaxial cylinders with radii 15 mm and 45 mm the inner surface is at 580oC and it is insulated Assuming the reaction rate of 055 MWm3 in the reactor volume find the temperature at the outer surface of the reactor

bull A current of 300 amperes passes through a stainless steel wire of 25 mm diameter and k=20 WmoC The resistivity of the wire is 70x10-8 ohm-m and length of the wire is 2 m if the wire is submerged in a fluid maintained at 50oC and h at the wire surface is 4000 Wm2oC calculate the steady state temperature at the centre and at the surface of the wire

bull An approximately spherical shaped orange ( k=023 WmoC) 90 mm diameter undergoes ripping process and generates 5100 Wm3 of energy If external surface the orange is at 8oC determine

bull (i) temperature at the centre of the orange and

bull (ii) heat flow from the outer surface of the orange

bull FINS External surfaces to used to cooling area and hence convection

heat transfer can be improvedbull APPLICATIONS IC engine cylinder ( air cooling) Heat exchangers automobile

radiators computer chip refrigerators electric motors generators and transformers etc

bull ASSUMPTIONS MADE IN ANALYSIS OF HEAT TRANFER THROUGH FIN

1 Fin material is homogenous and isotropic constthermal conductivity

2 Temperature at any cs of the fin is uniform3 lsquohrsquo is uniform over entire surface4 No internal heat generation5 Thermal contact resistance is negligible6 Radiation heat transfer is negligible7 Heat conduction is steady state and one dimensional

bull Problem 1

bull A stainless steel fin (k=20 WmK) having a diameter of 20 mm and a length of 01 m is attached to a wall at 300oC The ambient temperature is 50oC and the heat transfer coefficient is 10 Wm2K The fin tip is insulated Determine

bull (A) The rate of heat dissipation from the finbull (B) The temperature at the fin tipbull (C) The rate of heat transfer from the wall area covered by the fin if

the fin was not usedbull (D) The heat transfer rate from the same fin geometry if the

stainless steel fin is replaced by a fictitious fin with infinite thermal conductivity

bull Problem2bull One end of a long rod is inserted into a furnace while the other

projects into ambient temperature Under steady state the temperature of the rod is measured at two points 75mm apart and found to be 125oC and 885oC respectively while the ambient temperature is 20oC If the rod is 25mm in diameter and h is 2336 Wm2K

bull Find the thermal conductivity of the rod material

  • HEAT AND MASS TRANSFER
  • Thermodynamics Vs Heat transfer
  • MODE OF HEAT TRANSFER
  • MODE OF HEAT TRANSFER (2)
  • CONDUCTION- MECHANISM
  • CONCECTION-MECHANISM
  • RADIATION-MECHANISM
  • Conduction- Convection- Radiation
  • Slide 9
  • Slide 10
  • Fourierrsquos Law of Heat Conduction
  • Slide 12
  • Newtonrsquos Law of Cooling
  • Laws of Thermal Radiation
  • LAWS FOR HEAT TRANSFER
  • ELECTRICAL ampTHERMAL ANALOGY
  • Slide 17
  • Slide 18
  • PLANE WALL-UNIDIRECTIONAL
  • PLANE WALL -SERIES
  • PLANE WALL IN SERIES
  • PLANE WALL -PARALLEL
  • PLANE WALL-PARALLEL
  • PLANE WALL-COMBINATION
  • PLANE WALL-FILM RESISTANCE
  • PLANE WALL- THERMAL CONTACT RESISTANCE
  • PLANE WALL- THERMAL CONTACT RESISTANCE (2)
  • CYLINDER-SOLID
  • HEAT TRANSFER IN SOLID CYLINDER
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER-COMPOSITE
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER-MULTY LAYER
  • HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER
  • HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER (2)
  • HEAT TRANSFER IN SPHERE
  • HEAT TRANSFER IN SPHERE -MULTY LAYER
  • HEAT TRANSFER IN SPHERE -MULTY LAYER (2)
  • Slide 38
  • Slide 39
  • Slide 40
  • Slide 41
  • CRITICAL RADIUS
  • Plane wall- Insulation thickness -study
  • Problem5
  • Slide 45
  • Slide 46
  • Slide 47
  • GENERAL HEAT CONDUCTION EQUATION IN CYLINDRICAL COORDINATES
  • Slide 49
  • Slide 50
  • Slide 51
  • GENERAL HEAT CONDUCTION EQUATION IN SPHERICAL COORDINATES
  • Slide 53
  • Slide 54
  • Slide 55
  • General Heat Conduction Equation
  • Temperature profile
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Slide 67
Page 11: hmt chap 1

Newtonrsquos Law of Cooling

Laws of Thermal Radiation

LAWS FOR HEAT TRANSFERCONDUCTION

4Tq

CONVECTION

NEWTONrsquoS LAW OF COOLING

RADIATION

STEFAN BOLTZMAN LAW

q=-KA( dT dX)

Assumptionsbull Steady statebull Unidirectional heat flowbull Const temperature gradientbull Linear profilebull No internal heat generationbull Isothermal bounding surfacebull Homogeneousbull Isotropic bull Applicable to solid liquid gas

q=σAT4q= hA (Tw-Tα)

Thq

XT

Kq

FOURIERrsquoS LAW

q =σ ε AT4

ELECTRICAL ampTHERMAL ANALOGY

THERMAL RESISTANCE(ELECTRICAL ANALOGY)

bull OHMrsquos LAW Flow of Electricity

V=I R elect

Voltage Drop = Current flowtimesResistance Temp Drop=Heat FlowtimesResistance

ΔT= q Rtherm

Thermal Analogy To OHMrsquoS Law

PLANE WALL-UNIDIRECTIONAL

PLANE WALL -SERIES

PLANE WALL IN SERIES

PLANE WALL -PARALLEL

PLANE WALL-PARALLEL

PLANE WALL-COMBINATION

PLANE WALL-FILM RESISTANCE

PLANE WALL- THERMAL CONTACT RESISTANCE

PLANE WALL- THERMAL CONTACT RESISTANCE

At the joining surface of the two slabs air is trapped in voids due to surface irregularities

CYLINDER-SOLID

HEAT TRANSFER IN SOLID CYLINDER

HEAT TRANSFER RATE IN HOLLOW CYLINDER

HEAT TRANSFER RATE IN HOLLOW CYLINDER-COMPOSITE

HEAT TRANSFER RATE IN HOLLOW CYLINDER-MULTY LAYER

HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER

HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER

HEAT TRANSFER IN SPHERE

HEAT TRANSFER IN SPHERE -MULTY LAYER

HEAT TRANSFER IN SPHERE -MULTY LAYER

bull Problem 1 The wall in a furnace consists of 125 mm thick refractory bricks and

125 mm thick insulating fire bricks separated by an air gap A 12 mm thick plaster covers the outer wall The inner surface of wall is at 1100oC and the ambient temperature 25oC the heat transfer coefficient on the outer wall to the air is 17 Wm2K and the resistance to heat flow of the air gap is 016 KW the thermal conductivities of refractory brick insulating firebrick and plaster are 16 03 and 014 WmK respectively

Calculatebull (a) the rate of heat loss per unit area of the wall surfacebull (b) the interface temperatures throughout the wallbull (c) the temperature at the outside surface of the wall

bull Problem 2Two large aluminum plates (k=240 WmK) each 2 cm thick with 10 mm surface roughness are placed in contact under 105 Nm2 pressure in air The temperature at the outside surfaces are 390 oC and 406 oCCalculate

A) the heat flux B) the temperature drop due to the contact resistance C) the contact temperatures Thermal contact resistance with air as the interfacial fluid for

10 mm roughness is 27510-4 m2KW

bull Problem 3bull Steam at 350 oC flowing in a pipe (k=80 WmK) of 5 cm

inner diameter and 56 cm outer diameter is covered with 3 cm thick insulation of k=005 WmK Heat is lost to surroundings at 5 oC by natural convection and radiation the combined h being 20 Wm2K Taking the heat transfer coefficient inside the pipe as 60 Wm2K

bull Determinebull A) the rate of heat loss from the steam per unit length of

the pipebull B) the temperature drop across the pipe and the insulation

bull PROBLEM4bull A nuclear reactor has a sperical pressure vessel 750 mm inside

diameter 80 mm wall thickness The temperature at the inner surface is 500 oC If the tempertarue at the outer surface is 495 oC

bull Calculatea) The rate of heat loss through the metal wallb) If the rate of heat loss is to limited to 305Whow much inulation

thickness needs to be applied Thermal conductivity of steel wall=46 WmoC Thermal conductivity of insulation=004 WmoC Film coefficient of heat tranfer at inner and outer surface are 200

Wm2 oC and 15 Wm2 oC Ambient air temperature=28 oC

CRITICAL RADIUS

bull The radius upto which heat flow increases and after decreases At critical radius q becomes qmax

bull The addition of insulation always increases (Rth)condbull Rtotal= (Rth)cond + (Rth)conv

bull The addition of insulation may reduce the (Rth)conv due to increase in surface area

bull r1ltrc -heat transfer increases by adding insulation till r1=rc If

insulation thickness is further increased the rate of heat loss will decrease

r1 gt rc

- heat transfer decreases by adding insulation above the rc

Plane wall- Insulation

thickness -study

Cylinder amp Sphere- Insulation thickness -

study

Problem5bull A 3mm diameter and 5 mm long electric wire is tightly

wrapped with a 2 mm thick plastic cover whose K=015 wmoC Electrical measurements indicate that a current of 10 A passes through the wire and there is a voltage drop of 8 V along the wire If the insulated wire is exposed to a medium at Ta=30oC with a heat transfer coefficient of h=12 wm2oC Determine the temperature at the interface of the wire and the plastic cover in steady operation Also determine whether doubling the thickness of the plastic cover will increase or decrease this interface temperature

dx

TdzdyKQ xx )(

dxQx

QQ xxdxx ][

dxdx

TdzdyK

xQQdQ xdxxxx ])([)(

)1()]([)(

ddzdydx

x

TK

xdQ xx

GENERAL HEAT CONDUCTION EQUATION IN CARTESIAN COORDINATES

Consider an unsteady (transient) three dimensional heat conduction with internal heat generation in heterogeneous amp anisotropic PLANE WALL

Cartesian co-ordinates x y z

Elemental length dxdydz

Consider an elemental volume ABCDEFGH = (dxdydz)

Let Temperature distribution is a function of T=T(xyz)

Let

=internal heat generation per unit volume per unit time (W m3)q= rate of heat flow (W)Q= qdt=total heat flow (J)r=mass density of the material element(kgm3)Cp=specific heat of the material element (Jkg K)

gq

Plane (y - z) lsquoxrsquo direction

Quantity of heat leaving from the face EFGH in lsquoXrsquo direction

Quantity of heat flowing into the face ABCD in lsquoXrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoXrsquo direction

dy

TdzdxKQ yy )(

dyQy

QQ yydyy ][

dydy

TdzdxK

yQQdQ ydyyyy ])([)(

)2()]([)(

ddzdydx

y

TK

ydQ yy

Plane (x - z) lsquoyrsquo direction

Quantity of heat flowing into the face ABFE in lsquoYrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoYrsquo direction

dz

TdydxKQ zz )(

dzQz

QQ zzdzz ][

dzdz

TdydxK

zQQdQ zdzzzz ])([)(

)3()]([)(

ddzdydx

z

TK

zdQ zz

Plane (x - y) lsquozrsquo direction

Quantity of heat flowing into the face ADHE in lsquoZrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoZrsquo direction

Quantity of heat leaving from the face DCGH in lsquoYrsquo direction

Quantity of heat leaving from the face BCGF in lsquoZrsquo direction

)4()()(

ddzdydxqQgg

Total amount of internal heat generation in the element

Ein + Egen = Eout+Est (Ein- Eout)+ Egen=Est

T

Cddzdydxddzdydxqddzdydxz

TK

zy

TK

yx

TK

x pgzyx )()())](()()([

T

pC

gq

z

TzK

zy

Ty

Kyx

TxK

x][][][

stgzyx QQdQdQdQ

)(

)5()()(

dT

CdzdydxQ pst

Total heat energy stored in the element

(ie thermal energy of the element can be increased because of heat accumulation in the element due to heat flow amp internal heat generation)

)( volumem d

TmCQ pst )(

T

CqTK pg)(

KKKyKx z If material is homogeneous amp isotropic

T

pC

gq

z

T

zy

T

yx

T

xK ][][][

T

Kp

C

Kgq

z

T

y

T

x

T2

2

2

2

2

2

T

Kgq

z

T

y

T

x

T 12

2

2

2

2

2

Thermal diffusivity pC

K

T

K

qT g

12

according to thermodynamics first law

)( zr )( dzrddr

)( dzrddr

Cylindrical polar co-ordinates

Elemental length

Consider an elemental volume ABCDEFGH =

GENERAL HEAT CONDUCTION EQUATION IN CYLINDRICAL COORDINATES

gq

gq

Consider an unsteady (transient) three dimensional heat conduction with internal heat generation in heterogeneous amp anisotropic cylindrical geometries (eg rods and pipes)

)( zrTLetTemperature distribution is a function of T=

Let

=internal heat generation per unit volume per unit time (W m3)q= rate of heat flow (W)Q= qdt=total heat flow (J)r=mass density of the material element(kgm3)Cp=specific heat of the material element (Jkg K)

planerdirectionz

planezrdirection

planezdirectionr

)(

)(

)(

LetHeat flow in

drdr

TdzrdK

rdQ

QQdQ

rr

drrrr

])([

)1(][

r

Tr

rddrdzdKdQ rr

Heat accumulation in the element due to heat flow by conduction in lsquorrsquo direction

rddr

TdzdrK

rdQ

QQdQ d

])([

][

2

T

r

ddrdzrdKdQ

)2(][

2

2

2

T

r

ddrdzrdKdQ

Heat accumulation in the element due to heat flow by conduction in lsquoqrsquo direction

Plane (z- q ) lsquorrsquo direction

dr

TdzrdKQ rr

)(

drQr

QQ rrdrr ][

Quantity of heat leaving from the face EFGH in lsquorrsquo direction

Quantity of heat flowing into the face ABCD in lsquorrsquo direction

Plane (r-z) lsquoq lsquodirection

Quantity of heat leaving from the face AEGC in lsquoqrsquo direction

d

r

TdzdrKQ )(

rdQ

rQQ d ][

Quantity of heat flowing into the face BFHD in lsquoqrsquo direction

)3())((2

2

z

TdzdrrdKdQ zz

)5()()(

dT

CdzdrrdQ pst

)4()()(

ddzdrrdqQgg

Total amount of internal heat generation in the element

Total heat energy stored in the element

(ie thermal energy of the element can be increased because of heat accumulation in the element due to heat flow amp internal heat generation)

according to thermodynamics first law

Ein + Egen = Eout + Est (Ein ndash Eout) + Egen = Est

)( volumem

)()(z

T

zdzdrrdKdQ zz

Heat accumulation in the element due to heat flow by conduction in lsquozrsquo direction

dzdz

TdrrdK

zdQ

QQdQ

zz

dzzzz

])([

d

TmCQ pst )(

Plane (r-q) lsquozrsquo direction

dz

TdrrdKQ zz )(

dzQz

QQ zzdzz ][ Quantity of heat leaving from the face CDHG in lsquozrsquo direction

Quantity of heat flowing into the face ABFE in lsquozrsquo direction

stgzr QQdQdQdQ

)(

r

r ][)(rT

rr

ddrdzdKr

][)(

2

2

2

T

r

ddrdzrdK ))((2

2

z

TddzdrrdK z

ddzdrrdq g )(

dT

Cdzdrrd p )(

+ + +

=

T

CqTK pg)(KKKK zr

[gq

)(1

r

Tr

rrK r

)(1

2

2

2 T

rK )(

2

2

z

TK z

)( ddzdrrd )(

ddzdrrd

TC p

+ + + ] =

If material is homogeneous amp isotropic

)(1

r

Tr

rr

)(1

2

2

2 T

r)(

2

2

z

T

K

q g

T

K

C p

)(1

r

Tr

rr

)(1

2

2

2 T

r )(2

2

z

T

K

q g

T

1

+ + + =

)(1

r

Tr

rrK r

)(1

2

2

2 T

rK )(

2

2

z

TK z

gq

T

C p + + + =

+ + + = Thermal diffusivity

T

K

qT g

12

pC

K

(Ein ndash Eout) + Egen = Est

)( rT

)( r

)sin( drrddr)sin( drrddr

planerdirection

planerdirection

planedirectionr

)(

)(

)(

Spherical polar co-ordinates

Elemental length

Consider an elemental volume ABCDEFGH =

GENERAL HEAT CONDUCTION EQUATION IN SPHERICAL COORDINATES

Let

=internal heat generation per unit volume per unit time (W m3)q= rate of heat flow (W)Q= qdt=total heat flow (J)r=mass density of the material element(kgm3)Cp=specific heat of the material element (Jkg K)

gq

Consider an unsteady (transient) three dimensional heat conduction with internal heat generation in heterogeneous amp anisotropic spherical geometries

LetTemperature distribution is a function of T=

X

Y

Z

drT

drrdKQ rr )sin(

drQr

QQ rrdrr ][

drdr

TdrrdK

rdQ

QQdQ

rr

drrrr

])sin([

)1(]([sin 2

rT

rr

ddrddKdQ rr

Plane ( -q f) lsquorrsquo directionQuantity of heat flowing into the face ABCD in lsquorrsquo direction

Quantity of heat leaving from the face EFGH in lsquorrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquorrsquo direction

drT

drdrKQ )sin(

rdQ

rQQ d ][

rddr

TdrdrK

rdQ

QQdQ d

])sin([

])[(sin

2

T

r

drdrdrKdQ

)2(])(sin[

2

T

r

drdrddrKdQ

Quantity of heat leaving from the face BFHD in lsquoqrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoqrsquo direction

X

Y

Z

Plane (r- f ) lsquoq lsquodirection

Quantity of heat flowing into the face AEGC in lsquoqrsquo direction

dr

TdrrdKQ

sin)(

drQ

rQQ d sin][

sin

drdr

TdrrdK

rdQ

QQdQ d

sin]sin

)([sin

][sinsin

)sin(

T

rr

ddrdrrdKdQ

)3()(sin

)sin(2

2

22

2

T

r

ddddrrKdQ

Plane (r- ) q lsquofrsquo directionQuantity of heat flowing into the face ABFE in lsquofrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquozrsquo direction

)4()sin()(

ddrrddrqQgg

)5()sin()(

dT

CdrrddrQ pst

Total amount of internal heat generation in the element

Total heat energy stored in the element

(Ein ndash Eout) + Egen = Est

d

TmCQ pst )(

)( volumem

Quantity of heat leaving from the face CDHG in lsquofrsquo direction

according to thermodynamics first law

Ein + Egen = Eout + Est

X

Y

Z

stgr QQdQdQdQ

)(

]([sin 2

r

Tr

rddrddKr

])(sin[

2

T

rdrdrddrK

)(sin

)sin(2

2

22

2

T

r

ddddrrK

ddrrddrq g )sin(

dT

Cdrrddr p )sin(

2

2

r

r

sin

sin

]([ 22 r

Tr

rr

Kr

])(sin[sin2

TrK

)(sin 2

2

22

T

r

Kgq

T

Cp

]([1

[ 22 r

Tr

rrK

])(sin[sin1

2

T

r)](

sin

12

2

22 T

r gq

T

Cp

]([1

[ 22 r

Tr

rr

])(sin[sin1

2

T

r)](

sin

12

2

22 T

rK

q g

T

K

Cp

]([1

[ 22 r

Tr

rr

])(sin[sin1

2

T

r)](

sin

12

2

22 T

rK

q g

T1

KKKKr If material is homogeneous amp isotropic

TCqTK pg)(

T

K

qT g

12 pC

K

Thermal diffusivity

(Ein ndash Eout) + Egen = Est

General Heat Conduction Equation

PLANE WALL CYLINDER SPHERE

Temperature profile

Plane wall SphereCylinder

bull A copper conductor (k = 380 WmoC resistivity ρ = 2x10-8 Ω m) having inner and outer radii 10 cm and 225 cm respectively is carrying a current density of 4800 amperes cm2 the conductor is internally cooled and a constant temperature of 65oC is maintained at the inner surface and there is no heat transfer through the insulation surrounding the conductor

Determinebull (i) The maximum temperature of the conductor and the radius at

which it occursbull (ii) The internal heat transfer rate

bull A chemical reaction takes place in a packed bed (k=06 WmoC) between two coaxial cylinders with radii 15 mm and 45 mm the inner surface is at 580oC and it is insulated Assuming the reaction rate of 055 MWm3 in the reactor volume find the temperature at the outer surface of the reactor

bull A current of 300 amperes passes through a stainless steel wire of 25 mm diameter and k=20 WmoC The resistivity of the wire is 70x10-8 ohm-m and length of the wire is 2 m if the wire is submerged in a fluid maintained at 50oC and h at the wire surface is 4000 Wm2oC calculate the steady state temperature at the centre and at the surface of the wire

bull An approximately spherical shaped orange ( k=023 WmoC) 90 mm diameter undergoes ripping process and generates 5100 Wm3 of energy If external surface the orange is at 8oC determine

bull (i) temperature at the centre of the orange and

bull (ii) heat flow from the outer surface of the orange

bull FINS External surfaces to used to cooling area and hence convection

heat transfer can be improvedbull APPLICATIONS IC engine cylinder ( air cooling) Heat exchangers automobile

radiators computer chip refrigerators electric motors generators and transformers etc

bull ASSUMPTIONS MADE IN ANALYSIS OF HEAT TRANFER THROUGH FIN

1 Fin material is homogenous and isotropic constthermal conductivity

2 Temperature at any cs of the fin is uniform3 lsquohrsquo is uniform over entire surface4 No internal heat generation5 Thermal contact resistance is negligible6 Radiation heat transfer is negligible7 Heat conduction is steady state and one dimensional

bull Problem 1

bull A stainless steel fin (k=20 WmK) having a diameter of 20 mm and a length of 01 m is attached to a wall at 300oC The ambient temperature is 50oC and the heat transfer coefficient is 10 Wm2K The fin tip is insulated Determine

bull (A) The rate of heat dissipation from the finbull (B) The temperature at the fin tipbull (C) The rate of heat transfer from the wall area covered by the fin if

the fin was not usedbull (D) The heat transfer rate from the same fin geometry if the

stainless steel fin is replaced by a fictitious fin with infinite thermal conductivity

bull Problem2bull One end of a long rod is inserted into a furnace while the other

projects into ambient temperature Under steady state the temperature of the rod is measured at two points 75mm apart and found to be 125oC and 885oC respectively while the ambient temperature is 20oC If the rod is 25mm in diameter and h is 2336 Wm2K

bull Find the thermal conductivity of the rod material

  • HEAT AND MASS TRANSFER
  • Thermodynamics Vs Heat transfer
  • MODE OF HEAT TRANSFER
  • MODE OF HEAT TRANSFER (2)
  • CONDUCTION- MECHANISM
  • CONCECTION-MECHANISM
  • RADIATION-MECHANISM
  • Conduction- Convection- Radiation
  • Slide 9
  • Slide 10
  • Fourierrsquos Law of Heat Conduction
  • Slide 12
  • Newtonrsquos Law of Cooling
  • Laws of Thermal Radiation
  • LAWS FOR HEAT TRANSFER
  • ELECTRICAL ampTHERMAL ANALOGY
  • Slide 17
  • Slide 18
  • PLANE WALL-UNIDIRECTIONAL
  • PLANE WALL -SERIES
  • PLANE WALL IN SERIES
  • PLANE WALL -PARALLEL
  • PLANE WALL-PARALLEL
  • PLANE WALL-COMBINATION
  • PLANE WALL-FILM RESISTANCE
  • PLANE WALL- THERMAL CONTACT RESISTANCE
  • PLANE WALL- THERMAL CONTACT RESISTANCE (2)
  • CYLINDER-SOLID
  • HEAT TRANSFER IN SOLID CYLINDER
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER-COMPOSITE
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER-MULTY LAYER
  • HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER
  • HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER (2)
  • HEAT TRANSFER IN SPHERE
  • HEAT TRANSFER IN SPHERE -MULTY LAYER
  • HEAT TRANSFER IN SPHERE -MULTY LAYER (2)
  • Slide 38
  • Slide 39
  • Slide 40
  • Slide 41
  • CRITICAL RADIUS
  • Plane wall- Insulation thickness -study
  • Problem5
  • Slide 45
  • Slide 46
  • Slide 47
  • GENERAL HEAT CONDUCTION EQUATION IN CYLINDRICAL COORDINATES
  • Slide 49
  • Slide 50
  • Slide 51
  • GENERAL HEAT CONDUCTION EQUATION IN SPHERICAL COORDINATES
  • Slide 53
  • Slide 54
  • Slide 55
  • General Heat Conduction Equation
  • Temperature profile
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Slide 67
Page 12: hmt chap 1

Laws of Thermal Radiation

LAWS FOR HEAT TRANSFERCONDUCTION

4Tq

CONVECTION

NEWTONrsquoS LAW OF COOLING

RADIATION

STEFAN BOLTZMAN LAW

q=-KA( dT dX)

Assumptionsbull Steady statebull Unidirectional heat flowbull Const temperature gradientbull Linear profilebull No internal heat generationbull Isothermal bounding surfacebull Homogeneousbull Isotropic bull Applicable to solid liquid gas

q=σAT4q= hA (Tw-Tα)

Thq

XT

Kq

FOURIERrsquoS LAW

q =σ ε AT4

ELECTRICAL ampTHERMAL ANALOGY

THERMAL RESISTANCE(ELECTRICAL ANALOGY)

bull OHMrsquos LAW Flow of Electricity

V=I R elect

Voltage Drop = Current flowtimesResistance Temp Drop=Heat FlowtimesResistance

ΔT= q Rtherm

Thermal Analogy To OHMrsquoS Law

PLANE WALL-UNIDIRECTIONAL

PLANE WALL -SERIES

PLANE WALL IN SERIES

PLANE WALL -PARALLEL

PLANE WALL-PARALLEL

PLANE WALL-COMBINATION

PLANE WALL-FILM RESISTANCE

PLANE WALL- THERMAL CONTACT RESISTANCE

PLANE WALL- THERMAL CONTACT RESISTANCE

At the joining surface of the two slabs air is trapped in voids due to surface irregularities

CYLINDER-SOLID

HEAT TRANSFER IN SOLID CYLINDER

HEAT TRANSFER RATE IN HOLLOW CYLINDER

HEAT TRANSFER RATE IN HOLLOW CYLINDER-COMPOSITE

HEAT TRANSFER RATE IN HOLLOW CYLINDER-MULTY LAYER

HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER

HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER

HEAT TRANSFER IN SPHERE

HEAT TRANSFER IN SPHERE -MULTY LAYER

HEAT TRANSFER IN SPHERE -MULTY LAYER

bull Problem 1 The wall in a furnace consists of 125 mm thick refractory bricks and

125 mm thick insulating fire bricks separated by an air gap A 12 mm thick plaster covers the outer wall The inner surface of wall is at 1100oC and the ambient temperature 25oC the heat transfer coefficient on the outer wall to the air is 17 Wm2K and the resistance to heat flow of the air gap is 016 KW the thermal conductivities of refractory brick insulating firebrick and plaster are 16 03 and 014 WmK respectively

Calculatebull (a) the rate of heat loss per unit area of the wall surfacebull (b) the interface temperatures throughout the wallbull (c) the temperature at the outside surface of the wall

bull Problem 2Two large aluminum plates (k=240 WmK) each 2 cm thick with 10 mm surface roughness are placed in contact under 105 Nm2 pressure in air The temperature at the outside surfaces are 390 oC and 406 oCCalculate

A) the heat flux B) the temperature drop due to the contact resistance C) the contact temperatures Thermal contact resistance with air as the interfacial fluid for

10 mm roughness is 27510-4 m2KW

bull Problem 3bull Steam at 350 oC flowing in a pipe (k=80 WmK) of 5 cm

inner diameter and 56 cm outer diameter is covered with 3 cm thick insulation of k=005 WmK Heat is lost to surroundings at 5 oC by natural convection and radiation the combined h being 20 Wm2K Taking the heat transfer coefficient inside the pipe as 60 Wm2K

bull Determinebull A) the rate of heat loss from the steam per unit length of

the pipebull B) the temperature drop across the pipe and the insulation

bull PROBLEM4bull A nuclear reactor has a sperical pressure vessel 750 mm inside

diameter 80 mm wall thickness The temperature at the inner surface is 500 oC If the tempertarue at the outer surface is 495 oC

bull Calculatea) The rate of heat loss through the metal wallb) If the rate of heat loss is to limited to 305Whow much inulation

thickness needs to be applied Thermal conductivity of steel wall=46 WmoC Thermal conductivity of insulation=004 WmoC Film coefficient of heat tranfer at inner and outer surface are 200

Wm2 oC and 15 Wm2 oC Ambient air temperature=28 oC

CRITICAL RADIUS

bull The radius upto which heat flow increases and after decreases At critical radius q becomes qmax

bull The addition of insulation always increases (Rth)condbull Rtotal= (Rth)cond + (Rth)conv

bull The addition of insulation may reduce the (Rth)conv due to increase in surface area

bull r1ltrc -heat transfer increases by adding insulation till r1=rc If

insulation thickness is further increased the rate of heat loss will decrease

r1 gt rc

- heat transfer decreases by adding insulation above the rc

Plane wall- Insulation

thickness -study

Cylinder amp Sphere- Insulation thickness -

study

Problem5bull A 3mm diameter and 5 mm long electric wire is tightly

wrapped with a 2 mm thick plastic cover whose K=015 wmoC Electrical measurements indicate that a current of 10 A passes through the wire and there is a voltage drop of 8 V along the wire If the insulated wire is exposed to a medium at Ta=30oC with a heat transfer coefficient of h=12 wm2oC Determine the temperature at the interface of the wire and the plastic cover in steady operation Also determine whether doubling the thickness of the plastic cover will increase or decrease this interface temperature

dx

TdzdyKQ xx )(

dxQx

QQ xxdxx ][

dxdx

TdzdyK

xQQdQ xdxxxx ])([)(

)1()]([)(

ddzdydx

x

TK

xdQ xx

GENERAL HEAT CONDUCTION EQUATION IN CARTESIAN COORDINATES

Consider an unsteady (transient) three dimensional heat conduction with internal heat generation in heterogeneous amp anisotropic PLANE WALL

Cartesian co-ordinates x y z

Elemental length dxdydz

Consider an elemental volume ABCDEFGH = (dxdydz)

Let Temperature distribution is a function of T=T(xyz)

Let

=internal heat generation per unit volume per unit time (W m3)q= rate of heat flow (W)Q= qdt=total heat flow (J)r=mass density of the material element(kgm3)Cp=specific heat of the material element (Jkg K)

gq

Plane (y - z) lsquoxrsquo direction

Quantity of heat leaving from the face EFGH in lsquoXrsquo direction

Quantity of heat flowing into the face ABCD in lsquoXrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoXrsquo direction

dy

TdzdxKQ yy )(

dyQy

QQ yydyy ][

dydy

TdzdxK

yQQdQ ydyyyy ])([)(

)2()]([)(

ddzdydx

y

TK

ydQ yy

Plane (x - z) lsquoyrsquo direction

Quantity of heat flowing into the face ABFE in lsquoYrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoYrsquo direction

dz

TdydxKQ zz )(

dzQz

QQ zzdzz ][

dzdz

TdydxK

zQQdQ zdzzzz ])([)(

)3()]([)(

ddzdydx

z

TK

zdQ zz

Plane (x - y) lsquozrsquo direction

Quantity of heat flowing into the face ADHE in lsquoZrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoZrsquo direction

Quantity of heat leaving from the face DCGH in lsquoYrsquo direction

Quantity of heat leaving from the face BCGF in lsquoZrsquo direction

)4()()(

ddzdydxqQgg

Total amount of internal heat generation in the element

Ein + Egen = Eout+Est (Ein- Eout)+ Egen=Est

T

Cddzdydxddzdydxqddzdydxz

TK

zy

TK

yx

TK

x pgzyx )()())](()()([

T

pC

gq

z

TzK

zy

Ty

Kyx

TxK

x][][][

stgzyx QQdQdQdQ

)(

)5()()(

dT

CdzdydxQ pst

Total heat energy stored in the element

(ie thermal energy of the element can be increased because of heat accumulation in the element due to heat flow amp internal heat generation)

)( volumem d

TmCQ pst )(

T

CqTK pg)(

KKKyKx z If material is homogeneous amp isotropic

T

pC

gq

z

T

zy

T

yx

T

xK ][][][

T

Kp

C

Kgq

z

T

y

T

x

T2

2

2

2

2

2

T

Kgq

z

T

y

T

x

T 12

2

2

2

2

2

Thermal diffusivity pC

K

T

K

qT g

12

according to thermodynamics first law

)( zr )( dzrddr

)( dzrddr

Cylindrical polar co-ordinates

Elemental length

Consider an elemental volume ABCDEFGH =

GENERAL HEAT CONDUCTION EQUATION IN CYLINDRICAL COORDINATES

gq

gq

Consider an unsteady (transient) three dimensional heat conduction with internal heat generation in heterogeneous amp anisotropic cylindrical geometries (eg rods and pipes)

)( zrTLetTemperature distribution is a function of T=

Let

=internal heat generation per unit volume per unit time (W m3)q= rate of heat flow (W)Q= qdt=total heat flow (J)r=mass density of the material element(kgm3)Cp=specific heat of the material element (Jkg K)

planerdirectionz

planezrdirection

planezdirectionr

)(

)(

)(

LetHeat flow in

drdr

TdzrdK

rdQ

QQdQ

rr

drrrr

])([

)1(][

r

Tr

rddrdzdKdQ rr

Heat accumulation in the element due to heat flow by conduction in lsquorrsquo direction

rddr

TdzdrK

rdQ

QQdQ d

])([

][

2

T

r

ddrdzrdKdQ

)2(][

2

2

2

T

r

ddrdzrdKdQ

Heat accumulation in the element due to heat flow by conduction in lsquoqrsquo direction

Plane (z- q ) lsquorrsquo direction

dr

TdzrdKQ rr

)(

drQr

QQ rrdrr ][

Quantity of heat leaving from the face EFGH in lsquorrsquo direction

Quantity of heat flowing into the face ABCD in lsquorrsquo direction

Plane (r-z) lsquoq lsquodirection

Quantity of heat leaving from the face AEGC in lsquoqrsquo direction

d

r

TdzdrKQ )(

rdQ

rQQ d ][

Quantity of heat flowing into the face BFHD in lsquoqrsquo direction

)3())((2

2

z

TdzdrrdKdQ zz

)5()()(

dT

CdzdrrdQ pst

)4()()(

ddzdrrdqQgg

Total amount of internal heat generation in the element

Total heat energy stored in the element

(ie thermal energy of the element can be increased because of heat accumulation in the element due to heat flow amp internal heat generation)

according to thermodynamics first law

Ein + Egen = Eout + Est (Ein ndash Eout) + Egen = Est

)( volumem

)()(z

T

zdzdrrdKdQ zz

Heat accumulation in the element due to heat flow by conduction in lsquozrsquo direction

dzdz

TdrrdK

zdQ

QQdQ

zz

dzzzz

])([

d

TmCQ pst )(

Plane (r-q) lsquozrsquo direction

dz

TdrrdKQ zz )(

dzQz

QQ zzdzz ][ Quantity of heat leaving from the face CDHG in lsquozrsquo direction

Quantity of heat flowing into the face ABFE in lsquozrsquo direction

stgzr QQdQdQdQ

)(

r

r ][)(rT

rr

ddrdzdKr

][)(

2

2

2

T

r

ddrdzrdK ))((2

2

z

TddzdrrdK z

ddzdrrdq g )(

dT

Cdzdrrd p )(

+ + +

=

T

CqTK pg)(KKKK zr

[gq

)(1

r

Tr

rrK r

)(1

2

2

2 T

rK )(

2

2

z

TK z

)( ddzdrrd )(

ddzdrrd

TC p

+ + + ] =

If material is homogeneous amp isotropic

)(1

r

Tr

rr

)(1

2

2

2 T

r)(

2

2

z

T

K

q g

T

K

C p

)(1

r

Tr

rr

)(1

2

2

2 T

r )(2

2

z

T

K

q g

T

1

+ + + =

)(1

r

Tr

rrK r

)(1

2

2

2 T

rK )(

2

2

z

TK z

gq

T

C p + + + =

+ + + = Thermal diffusivity

T

K

qT g

12

pC

K

(Ein ndash Eout) + Egen = Est

)( rT

)( r

)sin( drrddr)sin( drrddr

planerdirection

planerdirection

planedirectionr

)(

)(

)(

Spherical polar co-ordinates

Elemental length

Consider an elemental volume ABCDEFGH =

GENERAL HEAT CONDUCTION EQUATION IN SPHERICAL COORDINATES

Let

=internal heat generation per unit volume per unit time (W m3)q= rate of heat flow (W)Q= qdt=total heat flow (J)r=mass density of the material element(kgm3)Cp=specific heat of the material element (Jkg K)

gq

Consider an unsteady (transient) three dimensional heat conduction with internal heat generation in heterogeneous amp anisotropic spherical geometries

LetTemperature distribution is a function of T=

X

Y

Z

drT

drrdKQ rr )sin(

drQr

QQ rrdrr ][

drdr

TdrrdK

rdQ

QQdQ

rr

drrrr

])sin([

)1(]([sin 2

rT

rr

ddrddKdQ rr

Plane ( -q f) lsquorrsquo directionQuantity of heat flowing into the face ABCD in lsquorrsquo direction

Quantity of heat leaving from the face EFGH in lsquorrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquorrsquo direction

drT

drdrKQ )sin(

rdQ

rQQ d ][

rddr

TdrdrK

rdQ

QQdQ d

])sin([

])[(sin

2

T

r

drdrdrKdQ

)2(])(sin[

2

T

r

drdrddrKdQ

Quantity of heat leaving from the face BFHD in lsquoqrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoqrsquo direction

X

Y

Z

Plane (r- f ) lsquoq lsquodirection

Quantity of heat flowing into the face AEGC in lsquoqrsquo direction

dr

TdrrdKQ

sin)(

drQ

rQQ d sin][

sin

drdr

TdrrdK

rdQ

QQdQ d

sin]sin

)([sin

][sinsin

)sin(

T

rr

ddrdrrdKdQ

)3()(sin

)sin(2

2

22

2

T

r

ddddrrKdQ

Plane (r- ) q lsquofrsquo directionQuantity of heat flowing into the face ABFE in lsquofrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquozrsquo direction

)4()sin()(

ddrrddrqQgg

)5()sin()(

dT

CdrrddrQ pst

Total amount of internal heat generation in the element

Total heat energy stored in the element

(Ein ndash Eout) + Egen = Est

d

TmCQ pst )(

)( volumem

Quantity of heat leaving from the face CDHG in lsquofrsquo direction

according to thermodynamics first law

Ein + Egen = Eout + Est

X

Y

Z

stgr QQdQdQdQ

)(

]([sin 2

r

Tr

rddrddKr

])(sin[

2

T

rdrdrddrK

)(sin

)sin(2

2

22

2

T

r

ddddrrK

ddrrddrq g )sin(

dT

Cdrrddr p )sin(

2

2

r

r

sin

sin

]([ 22 r

Tr

rr

Kr

])(sin[sin2

TrK

)(sin 2

2

22

T

r

Kgq

T

Cp

]([1

[ 22 r

Tr

rrK

])(sin[sin1

2

T

r)](

sin

12

2

22 T

r gq

T

Cp

]([1

[ 22 r

Tr

rr

])(sin[sin1

2

T

r)](

sin

12

2

22 T

rK

q g

T

K

Cp

]([1

[ 22 r

Tr

rr

])(sin[sin1

2

T

r)](

sin

12

2

22 T

rK

q g

T1

KKKKr If material is homogeneous amp isotropic

TCqTK pg)(

T

K

qT g

12 pC

K

Thermal diffusivity

(Ein ndash Eout) + Egen = Est

General Heat Conduction Equation

PLANE WALL CYLINDER SPHERE

Temperature profile

Plane wall SphereCylinder

bull A copper conductor (k = 380 WmoC resistivity ρ = 2x10-8 Ω m) having inner and outer radii 10 cm and 225 cm respectively is carrying a current density of 4800 amperes cm2 the conductor is internally cooled and a constant temperature of 65oC is maintained at the inner surface and there is no heat transfer through the insulation surrounding the conductor

Determinebull (i) The maximum temperature of the conductor and the radius at

which it occursbull (ii) The internal heat transfer rate

bull A chemical reaction takes place in a packed bed (k=06 WmoC) between two coaxial cylinders with radii 15 mm and 45 mm the inner surface is at 580oC and it is insulated Assuming the reaction rate of 055 MWm3 in the reactor volume find the temperature at the outer surface of the reactor

bull A current of 300 amperes passes through a stainless steel wire of 25 mm diameter and k=20 WmoC The resistivity of the wire is 70x10-8 ohm-m and length of the wire is 2 m if the wire is submerged in a fluid maintained at 50oC and h at the wire surface is 4000 Wm2oC calculate the steady state temperature at the centre and at the surface of the wire

bull An approximately spherical shaped orange ( k=023 WmoC) 90 mm diameter undergoes ripping process and generates 5100 Wm3 of energy If external surface the orange is at 8oC determine

bull (i) temperature at the centre of the orange and

bull (ii) heat flow from the outer surface of the orange

bull FINS External surfaces to used to cooling area and hence convection

heat transfer can be improvedbull APPLICATIONS IC engine cylinder ( air cooling) Heat exchangers automobile

radiators computer chip refrigerators electric motors generators and transformers etc

bull ASSUMPTIONS MADE IN ANALYSIS OF HEAT TRANFER THROUGH FIN

1 Fin material is homogenous and isotropic constthermal conductivity

2 Temperature at any cs of the fin is uniform3 lsquohrsquo is uniform over entire surface4 No internal heat generation5 Thermal contact resistance is negligible6 Radiation heat transfer is negligible7 Heat conduction is steady state and one dimensional

bull Problem 1

bull A stainless steel fin (k=20 WmK) having a diameter of 20 mm and a length of 01 m is attached to a wall at 300oC The ambient temperature is 50oC and the heat transfer coefficient is 10 Wm2K The fin tip is insulated Determine

bull (A) The rate of heat dissipation from the finbull (B) The temperature at the fin tipbull (C) The rate of heat transfer from the wall area covered by the fin if

the fin was not usedbull (D) The heat transfer rate from the same fin geometry if the

stainless steel fin is replaced by a fictitious fin with infinite thermal conductivity

bull Problem2bull One end of a long rod is inserted into a furnace while the other

projects into ambient temperature Under steady state the temperature of the rod is measured at two points 75mm apart and found to be 125oC and 885oC respectively while the ambient temperature is 20oC If the rod is 25mm in diameter and h is 2336 Wm2K

bull Find the thermal conductivity of the rod material

  • HEAT AND MASS TRANSFER
  • Thermodynamics Vs Heat transfer
  • MODE OF HEAT TRANSFER
  • MODE OF HEAT TRANSFER (2)
  • CONDUCTION- MECHANISM
  • CONCECTION-MECHANISM
  • RADIATION-MECHANISM
  • Conduction- Convection- Radiation
  • Slide 9
  • Slide 10
  • Fourierrsquos Law of Heat Conduction
  • Slide 12
  • Newtonrsquos Law of Cooling
  • Laws of Thermal Radiation
  • LAWS FOR HEAT TRANSFER
  • ELECTRICAL ampTHERMAL ANALOGY
  • Slide 17
  • Slide 18
  • PLANE WALL-UNIDIRECTIONAL
  • PLANE WALL -SERIES
  • PLANE WALL IN SERIES
  • PLANE WALL -PARALLEL
  • PLANE WALL-PARALLEL
  • PLANE WALL-COMBINATION
  • PLANE WALL-FILM RESISTANCE
  • PLANE WALL- THERMAL CONTACT RESISTANCE
  • PLANE WALL- THERMAL CONTACT RESISTANCE (2)
  • CYLINDER-SOLID
  • HEAT TRANSFER IN SOLID CYLINDER
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER-COMPOSITE
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER-MULTY LAYER
  • HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER
  • HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER (2)
  • HEAT TRANSFER IN SPHERE
  • HEAT TRANSFER IN SPHERE -MULTY LAYER
  • HEAT TRANSFER IN SPHERE -MULTY LAYER (2)
  • Slide 38
  • Slide 39
  • Slide 40
  • Slide 41
  • CRITICAL RADIUS
  • Plane wall- Insulation thickness -study
  • Problem5
  • Slide 45
  • Slide 46
  • Slide 47
  • GENERAL HEAT CONDUCTION EQUATION IN CYLINDRICAL COORDINATES
  • Slide 49
  • Slide 50
  • Slide 51
  • GENERAL HEAT CONDUCTION EQUATION IN SPHERICAL COORDINATES
  • Slide 53
  • Slide 54
  • Slide 55
  • General Heat Conduction Equation
  • Temperature profile
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Slide 67
Page 13: hmt chap 1

LAWS FOR HEAT TRANSFERCONDUCTION

4Tq

CONVECTION

NEWTONrsquoS LAW OF COOLING

RADIATION

STEFAN BOLTZMAN LAW

q=-KA( dT dX)

Assumptionsbull Steady statebull Unidirectional heat flowbull Const temperature gradientbull Linear profilebull No internal heat generationbull Isothermal bounding surfacebull Homogeneousbull Isotropic bull Applicable to solid liquid gas

q=σAT4q= hA (Tw-Tα)

Thq

XT

Kq

FOURIERrsquoS LAW

q =σ ε AT4

ELECTRICAL ampTHERMAL ANALOGY

THERMAL RESISTANCE(ELECTRICAL ANALOGY)

bull OHMrsquos LAW Flow of Electricity

V=I R elect

Voltage Drop = Current flowtimesResistance Temp Drop=Heat FlowtimesResistance

ΔT= q Rtherm

Thermal Analogy To OHMrsquoS Law

PLANE WALL-UNIDIRECTIONAL

PLANE WALL -SERIES

PLANE WALL IN SERIES

PLANE WALL -PARALLEL

PLANE WALL-PARALLEL

PLANE WALL-COMBINATION

PLANE WALL-FILM RESISTANCE

PLANE WALL- THERMAL CONTACT RESISTANCE

PLANE WALL- THERMAL CONTACT RESISTANCE

At the joining surface of the two slabs air is trapped in voids due to surface irregularities

CYLINDER-SOLID

HEAT TRANSFER IN SOLID CYLINDER

HEAT TRANSFER RATE IN HOLLOW CYLINDER

HEAT TRANSFER RATE IN HOLLOW CYLINDER-COMPOSITE

HEAT TRANSFER RATE IN HOLLOW CYLINDER-MULTY LAYER

HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER

HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER

HEAT TRANSFER IN SPHERE

HEAT TRANSFER IN SPHERE -MULTY LAYER

HEAT TRANSFER IN SPHERE -MULTY LAYER

bull Problem 1 The wall in a furnace consists of 125 mm thick refractory bricks and

125 mm thick insulating fire bricks separated by an air gap A 12 mm thick plaster covers the outer wall The inner surface of wall is at 1100oC and the ambient temperature 25oC the heat transfer coefficient on the outer wall to the air is 17 Wm2K and the resistance to heat flow of the air gap is 016 KW the thermal conductivities of refractory brick insulating firebrick and plaster are 16 03 and 014 WmK respectively

Calculatebull (a) the rate of heat loss per unit area of the wall surfacebull (b) the interface temperatures throughout the wallbull (c) the temperature at the outside surface of the wall

bull Problem 2Two large aluminum plates (k=240 WmK) each 2 cm thick with 10 mm surface roughness are placed in contact under 105 Nm2 pressure in air The temperature at the outside surfaces are 390 oC and 406 oCCalculate

A) the heat flux B) the temperature drop due to the contact resistance C) the contact temperatures Thermal contact resistance with air as the interfacial fluid for

10 mm roughness is 27510-4 m2KW

bull Problem 3bull Steam at 350 oC flowing in a pipe (k=80 WmK) of 5 cm

inner diameter and 56 cm outer diameter is covered with 3 cm thick insulation of k=005 WmK Heat is lost to surroundings at 5 oC by natural convection and radiation the combined h being 20 Wm2K Taking the heat transfer coefficient inside the pipe as 60 Wm2K

bull Determinebull A) the rate of heat loss from the steam per unit length of

the pipebull B) the temperature drop across the pipe and the insulation

bull PROBLEM4bull A nuclear reactor has a sperical pressure vessel 750 mm inside

diameter 80 mm wall thickness The temperature at the inner surface is 500 oC If the tempertarue at the outer surface is 495 oC

bull Calculatea) The rate of heat loss through the metal wallb) If the rate of heat loss is to limited to 305Whow much inulation

thickness needs to be applied Thermal conductivity of steel wall=46 WmoC Thermal conductivity of insulation=004 WmoC Film coefficient of heat tranfer at inner and outer surface are 200

Wm2 oC and 15 Wm2 oC Ambient air temperature=28 oC

CRITICAL RADIUS

bull The radius upto which heat flow increases and after decreases At critical radius q becomes qmax

bull The addition of insulation always increases (Rth)condbull Rtotal= (Rth)cond + (Rth)conv

bull The addition of insulation may reduce the (Rth)conv due to increase in surface area

bull r1ltrc -heat transfer increases by adding insulation till r1=rc If

insulation thickness is further increased the rate of heat loss will decrease

r1 gt rc

- heat transfer decreases by adding insulation above the rc

Plane wall- Insulation

thickness -study

Cylinder amp Sphere- Insulation thickness -

study

Problem5bull A 3mm diameter and 5 mm long electric wire is tightly

wrapped with a 2 mm thick plastic cover whose K=015 wmoC Electrical measurements indicate that a current of 10 A passes through the wire and there is a voltage drop of 8 V along the wire If the insulated wire is exposed to a medium at Ta=30oC with a heat transfer coefficient of h=12 wm2oC Determine the temperature at the interface of the wire and the plastic cover in steady operation Also determine whether doubling the thickness of the plastic cover will increase or decrease this interface temperature

dx

TdzdyKQ xx )(

dxQx

QQ xxdxx ][

dxdx

TdzdyK

xQQdQ xdxxxx ])([)(

)1()]([)(

ddzdydx

x

TK

xdQ xx

GENERAL HEAT CONDUCTION EQUATION IN CARTESIAN COORDINATES

Consider an unsteady (transient) three dimensional heat conduction with internal heat generation in heterogeneous amp anisotropic PLANE WALL

Cartesian co-ordinates x y z

Elemental length dxdydz

Consider an elemental volume ABCDEFGH = (dxdydz)

Let Temperature distribution is a function of T=T(xyz)

Let

=internal heat generation per unit volume per unit time (W m3)q= rate of heat flow (W)Q= qdt=total heat flow (J)r=mass density of the material element(kgm3)Cp=specific heat of the material element (Jkg K)

gq

Plane (y - z) lsquoxrsquo direction

Quantity of heat leaving from the face EFGH in lsquoXrsquo direction

Quantity of heat flowing into the face ABCD in lsquoXrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoXrsquo direction

dy

TdzdxKQ yy )(

dyQy

QQ yydyy ][

dydy

TdzdxK

yQQdQ ydyyyy ])([)(

)2()]([)(

ddzdydx

y

TK

ydQ yy

Plane (x - z) lsquoyrsquo direction

Quantity of heat flowing into the face ABFE in lsquoYrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoYrsquo direction

dz

TdydxKQ zz )(

dzQz

QQ zzdzz ][

dzdz

TdydxK

zQQdQ zdzzzz ])([)(

)3()]([)(

ddzdydx

z

TK

zdQ zz

Plane (x - y) lsquozrsquo direction

Quantity of heat flowing into the face ADHE in lsquoZrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoZrsquo direction

Quantity of heat leaving from the face DCGH in lsquoYrsquo direction

Quantity of heat leaving from the face BCGF in lsquoZrsquo direction

)4()()(

ddzdydxqQgg

Total amount of internal heat generation in the element

Ein + Egen = Eout+Est (Ein- Eout)+ Egen=Est

T

Cddzdydxddzdydxqddzdydxz

TK

zy

TK

yx

TK

x pgzyx )()())](()()([

T

pC

gq

z

TzK

zy

Ty

Kyx

TxK

x][][][

stgzyx QQdQdQdQ

)(

)5()()(

dT

CdzdydxQ pst

Total heat energy stored in the element

(ie thermal energy of the element can be increased because of heat accumulation in the element due to heat flow amp internal heat generation)

)( volumem d

TmCQ pst )(

T

CqTK pg)(

KKKyKx z If material is homogeneous amp isotropic

T

pC

gq

z

T

zy

T

yx

T

xK ][][][

T

Kp

C

Kgq

z

T

y

T

x

T2

2

2

2

2

2

T

Kgq

z

T

y

T

x

T 12

2

2

2

2

2

Thermal diffusivity pC

K

T

K

qT g

12

according to thermodynamics first law

)( zr )( dzrddr

)( dzrddr

Cylindrical polar co-ordinates

Elemental length

Consider an elemental volume ABCDEFGH =

GENERAL HEAT CONDUCTION EQUATION IN CYLINDRICAL COORDINATES

gq

gq

Consider an unsteady (transient) three dimensional heat conduction with internal heat generation in heterogeneous amp anisotropic cylindrical geometries (eg rods and pipes)

)( zrTLetTemperature distribution is a function of T=

Let

=internal heat generation per unit volume per unit time (W m3)q= rate of heat flow (W)Q= qdt=total heat flow (J)r=mass density of the material element(kgm3)Cp=specific heat of the material element (Jkg K)

planerdirectionz

planezrdirection

planezdirectionr

)(

)(

)(

LetHeat flow in

drdr

TdzrdK

rdQ

QQdQ

rr

drrrr

])([

)1(][

r

Tr

rddrdzdKdQ rr

Heat accumulation in the element due to heat flow by conduction in lsquorrsquo direction

rddr

TdzdrK

rdQ

QQdQ d

])([

][

2

T

r

ddrdzrdKdQ

)2(][

2

2

2

T

r

ddrdzrdKdQ

Heat accumulation in the element due to heat flow by conduction in lsquoqrsquo direction

Plane (z- q ) lsquorrsquo direction

dr

TdzrdKQ rr

)(

drQr

QQ rrdrr ][

Quantity of heat leaving from the face EFGH in lsquorrsquo direction

Quantity of heat flowing into the face ABCD in lsquorrsquo direction

Plane (r-z) lsquoq lsquodirection

Quantity of heat leaving from the face AEGC in lsquoqrsquo direction

d

r

TdzdrKQ )(

rdQ

rQQ d ][

Quantity of heat flowing into the face BFHD in lsquoqrsquo direction

)3())((2

2

z

TdzdrrdKdQ zz

)5()()(

dT

CdzdrrdQ pst

)4()()(

ddzdrrdqQgg

Total amount of internal heat generation in the element

Total heat energy stored in the element

(ie thermal energy of the element can be increased because of heat accumulation in the element due to heat flow amp internal heat generation)

according to thermodynamics first law

Ein + Egen = Eout + Est (Ein ndash Eout) + Egen = Est

)( volumem

)()(z

T

zdzdrrdKdQ zz

Heat accumulation in the element due to heat flow by conduction in lsquozrsquo direction

dzdz

TdrrdK

zdQ

QQdQ

zz

dzzzz

])([

d

TmCQ pst )(

Plane (r-q) lsquozrsquo direction

dz

TdrrdKQ zz )(

dzQz

QQ zzdzz ][ Quantity of heat leaving from the face CDHG in lsquozrsquo direction

Quantity of heat flowing into the face ABFE in lsquozrsquo direction

stgzr QQdQdQdQ

)(

r

r ][)(rT

rr

ddrdzdKr

][)(

2

2

2

T

r

ddrdzrdK ))((2

2

z

TddzdrrdK z

ddzdrrdq g )(

dT

Cdzdrrd p )(

+ + +

=

T

CqTK pg)(KKKK zr

[gq

)(1

r

Tr

rrK r

)(1

2

2

2 T

rK )(

2

2

z

TK z

)( ddzdrrd )(

ddzdrrd

TC p

+ + + ] =

If material is homogeneous amp isotropic

)(1

r

Tr

rr

)(1

2

2

2 T

r)(

2

2

z

T

K

q g

T

K

C p

)(1

r

Tr

rr

)(1

2

2

2 T

r )(2

2

z

T

K

q g

T

1

+ + + =

)(1

r

Tr

rrK r

)(1

2

2

2 T

rK )(

2

2

z

TK z

gq

T

C p + + + =

+ + + = Thermal diffusivity

T

K

qT g

12

pC

K

(Ein ndash Eout) + Egen = Est

)( rT

)( r

)sin( drrddr)sin( drrddr

planerdirection

planerdirection

planedirectionr

)(

)(

)(

Spherical polar co-ordinates

Elemental length

Consider an elemental volume ABCDEFGH =

GENERAL HEAT CONDUCTION EQUATION IN SPHERICAL COORDINATES

Let

=internal heat generation per unit volume per unit time (W m3)q= rate of heat flow (W)Q= qdt=total heat flow (J)r=mass density of the material element(kgm3)Cp=specific heat of the material element (Jkg K)

gq

Consider an unsteady (transient) three dimensional heat conduction with internal heat generation in heterogeneous amp anisotropic spherical geometries

LetTemperature distribution is a function of T=

X

Y

Z

drT

drrdKQ rr )sin(

drQr

QQ rrdrr ][

drdr

TdrrdK

rdQ

QQdQ

rr

drrrr

])sin([

)1(]([sin 2

rT

rr

ddrddKdQ rr

Plane ( -q f) lsquorrsquo directionQuantity of heat flowing into the face ABCD in lsquorrsquo direction

Quantity of heat leaving from the face EFGH in lsquorrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquorrsquo direction

drT

drdrKQ )sin(

rdQ

rQQ d ][

rddr

TdrdrK

rdQ

QQdQ d

])sin([

])[(sin

2

T

r

drdrdrKdQ

)2(])(sin[

2

T

r

drdrddrKdQ

Quantity of heat leaving from the face BFHD in lsquoqrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoqrsquo direction

X

Y

Z

Plane (r- f ) lsquoq lsquodirection

Quantity of heat flowing into the face AEGC in lsquoqrsquo direction

dr

TdrrdKQ

sin)(

drQ

rQQ d sin][

sin

drdr

TdrrdK

rdQ

QQdQ d

sin]sin

)([sin

][sinsin

)sin(

T

rr

ddrdrrdKdQ

)3()(sin

)sin(2

2

22

2

T

r

ddddrrKdQ

Plane (r- ) q lsquofrsquo directionQuantity of heat flowing into the face ABFE in lsquofrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquozrsquo direction

)4()sin()(

ddrrddrqQgg

)5()sin()(

dT

CdrrddrQ pst

Total amount of internal heat generation in the element

Total heat energy stored in the element

(Ein ndash Eout) + Egen = Est

d

TmCQ pst )(

)( volumem

Quantity of heat leaving from the face CDHG in lsquofrsquo direction

according to thermodynamics first law

Ein + Egen = Eout + Est

X

Y

Z

stgr QQdQdQdQ

)(

]([sin 2

r

Tr

rddrddKr

])(sin[

2

T

rdrdrddrK

)(sin

)sin(2

2

22

2

T

r

ddddrrK

ddrrddrq g )sin(

dT

Cdrrddr p )sin(

2

2

r

r

sin

sin

]([ 22 r

Tr

rr

Kr

])(sin[sin2

TrK

)(sin 2

2

22

T

r

Kgq

T

Cp

]([1

[ 22 r

Tr

rrK

])(sin[sin1

2

T

r)](

sin

12

2

22 T

r gq

T

Cp

]([1

[ 22 r

Tr

rr

])(sin[sin1

2

T

r)](

sin

12

2

22 T

rK

q g

T

K

Cp

]([1

[ 22 r

Tr

rr

])(sin[sin1

2

T

r)](

sin

12

2

22 T

rK

q g

T1

KKKKr If material is homogeneous amp isotropic

TCqTK pg)(

T

K

qT g

12 pC

K

Thermal diffusivity

(Ein ndash Eout) + Egen = Est

General Heat Conduction Equation

PLANE WALL CYLINDER SPHERE

Temperature profile

Plane wall SphereCylinder

bull A copper conductor (k = 380 WmoC resistivity ρ = 2x10-8 Ω m) having inner and outer radii 10 cm and 225 cm respectively is carrying a current density of 4800 amperes cm2 the conductor is internally cooled and a constant temperature of 65oC is maintained at the inner surface and there is no heat transfer through the insulation surrounding the conductor

Determinebull (i) The maximum temperature of the conductor and the radius at

which it occursbull (ii) The internal heat transfer rate

bull A chemical reaction takes place in a packed bed (k=06 WmoC) between two coaxial cylinders with radii 15 mm and 45 mm the inner surface is at 580oC and it is insulated Assuming the reaction rate of 055 MWm3 in the reactor volume find the temperature at the outer surface of the reactor

bull A current of 300 amperes passes through a stainless steel wire of 25 mm diameter and k=20 WmoC The resistivity of the wire is 70x10-8 ohm-m and length of the wire is 2 m if the wire is submerged in a fluid maintained at 50oC and h at the wire surface is 4000 Wm2oC calculate the steady state temperature at the centre and at the surface of the wire

bull An approximately spherical shaped orange ( k=023 WmoC) 90 mm diameter undergoes ripping process and generates 5100 Wm3 of energy If external surface the orange is at 8oC determine

bull (i) temperature at the centre of the orange and

bull (ii) heat flow from the outer surface of the orange

bull FINS External surfaces to used to cooling area and hence convection

heat transfer can be improvedbull APPLICATIONS IC engine cylinder ( air cooling) Heat exchangers automobile

radiators computer chip refrigerators electric motors generators and transformers etc

bull ASSUMPTIONS MADE IN ANALYSIS OF HEAT TRANFER THROUGH FIN

1 Fin material is homogenous and isotropic constthermal conductivity

2 Temperature at any cs of the fin is uniform3 lsquohrsquo is uniform over entire surface4 No internal heat generation5 Thermal contact resistance is negligible6 Radiation heat transfer is negligible7 Heat conduction is steady state and one dimensional

bull Problem 1

bull A stainless steel fin (k=20 WmK) having a diameter of 20 mm and a length of 01 m is attached to a wall at 300oC The ambient temperature is 50oC and the heat transfer coefficient is 10 Wm2K The fin tip is insulated Determine

bull (A) The rate of heat dissipation from the finbull (B) The temperature at the fin tipbull (C) The rate of heat transfer from the wall area covered by the fin if

the fin was not usedbull (D) The heat transfer rate from the same fin geometry if the

stainless steel fin is replaced by a fictitious fin with infinite thermal conductivity

bull Problem2bull One end of a long rod is inserted into a furnace while the other

projects into ambient temperature Under steady state the temperature of the rod is measured at two points 75mm apart and found to be 125oC and 885oC respectively while the ambient temperature is 20oC If the rod is 25mm in diameter and h is 2336 Wm2K

bull Find the thermal conductivity of the rod material

  • HEAT AND MASS TRANSFER
  • Thermodynamics Vs Heat transfer
  • MODE OF HEAT TRANSFER
  • MODE OF HEAT TRANSFER (2)
  • CONDUCTION- MECHANISM
  • CONCECTION-MECHANISM
  • RADIATION-MECHANISM
  • Conduction- Convection- Radiation
  • Slide 9
  • Slide 10
  • Fourierrsquos Law of Heat Conduction
  • Slide 12
  • Newtonrsquos Law of Cooling
  • Laws of Thermal Radiation
  • LAWS FOR HEAT TRANSFER
  • ELECTRICAL ampTHERMAL ANALOGY
  • Slide 17
  • Slide 18
  • PLANE WALL-UNIDIRECTIONAL
  • PLANE WALL -SERIES
  • PLANE WALL IN SERIES
  • PLANE WALL -PARALLEL
  • PLANE WALL-PARALLEL
  • PLANE WALL-COMBINATION
  • PLANE WALL-FILM RESISTANCE
  • PLANE WALL- THERMAL CONTACT RESISTANCE
  • PLANE WALL- THERMAL CONTACT RESISTANCE (2)
  • CYLINDER-SOLID
  • HEAT TRANSFER IN SOLID CYLINDER
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER-COMPOSITE
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER-MULTY LAYER
  • HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER
  • HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER (2)
  • HEAT TRANSFER IN SPHERE
  • HEAT TRANSFER IN SPHERE -MULTY LAYER
  • HEAT TRANSFER IN SPHERE -MULTY LAYER (2)
  • Slide 38
  • Slide 39
  • Slide 40
  • Slide 41
  • CRITICAL RADIUS
  • Plane wall- Insulation thickness -study
  • Problem5
  • Slide 45
  • Slide 46
  • Slide 47
  • GENERAL HEAT CONDUCTION EQUATION IN CYLINDRICAL COORDINATES
  • Slide 49
  • Slide 50
  • Slide 51
  • GENERAL HEAT CONDUCTION EQUATION IN SPHERICAL COORDINATES
  • Slide 53
  • Slide 54
  • Slide 55
  • General Heat Conduction Equation
  • Temperature profile
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Slide 67
Page 14: hmt chap 1

ELECTRICAL ampTHERMAL ANALOGY

THERMAL RESISTANCE(ELECTRICAL ANALOGY)

bull OHMrsquos LAW Flow of Electricity

V=I R elect

Voltage Drop = Current flowtimesResistance Temp Drop=Heat FlowtimesResistance

ΔT= q Rtherm

Thermal Analogy To OHMrsquoS Law

PLANE WALL-UNIDIRECTIONAL

PLANE WALL -SERIES

PLANE WALL IN SERIES

PLANE WALL -PARALLEL

PLANE WALL-PARALLEL

PLANE WALL-COMBINATION

PLANE WALL-FILM RESISTANCE

PLANE WALL- THERMAL CONTACT RESISTANCE

PLANE WALL- THERMAL CONTACT RESISTANCE

At the joining surface of the two slabs air is trapped in voids due to surface irregularities

CYLINDER-SOLID

HEAT TRANSFER IN SOLID CYLINDER

HEAT TRANSFER RATE IN HOLLOW CYLINDER

HEAT TRANSFER RATE IN HOLLOW CYLINDER-COMPOSITE

HEAT TRANSFER RATE IN HOLLOW CYLINDER-MULTY LAYER

HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER

HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER

HEAT TRANSFER IN SPHERE

HEAT TRANSFER IN SPHERE -MULTY LAYER

HEAT TRANSFER IN SPHERE -MULTY LAYER

bull Problem 1 The wall in a furnace consists of 125 mm thick refractory bricks and

125 mm thick insulating fire bricks separated by an air gap A 12 mm thick plaster covers the outer wall The inner surface of wall is at 1100oC and the ambient temperature 25oC the heat transfer coefficient on the outer wall to the air is 17 Wm2K and the resistance to heat flow of the air gap is 016 KW the thermal conductivities of refractory brick insulating firebrick and plaster are 16 03 and 014 WmK respectively

Calculatebull (a) the rate of heat loss per unit area of the wall surfacebull (b) the interface temperatures throughout the wallbull (c) the temperature at the outside surface of the wall

bull Problem 2Two large aluminum plates (k=240 WmK) each 2 cm thick with 10 mm surface roughness are placed in contact under 105 Nm2 pressure in air The temperature at the outside surfaces are 390 oC and 406 oCCalculate

A) the heat flux B) the temperature drop due to the contact resistance C) the contact temperatures Thermal contact resistance with air as the interfacial fluid for

10 mm roughness is 27510-4 m2KW

bull Problem 3bull Steam at 350 oC flowing in a pipe (k=80 WmK) of 5 cm

inner diameter and 56 cm outer diameter is covered with 3 cm thick insulation of k=005 WmK Heat is lost to surroundings at 5 oC by natural convection and radiation the combined h being 20 Wm2K Taking the heat transfer coefficient inside the pipe as 60 Wm2K

bull Determinebull A) the rate of heat loss from the steam per unit length of

the pipebull B) the temperature drop across the pipe and the insulation

bull PROBLEM4bull A nuclear reactor has a sperical pressure vessel 750 mm inside

diameter 80 mm wall thickness The temperature at the inner surface is 500 oC If the tempertarue at the outer surface is 495 oC

bull Calculatea) The rate of heat loss through the metal wallb) If the rate of heat loss is to limited to 305Whow much inulation

thickness needs to be applied Thermal conductivity of steel wall=46 WmoC Thermal conductivity of insulation=004 WmoC Film coefficient of heat tranfer at inner and outer surface are 200

Wm2 oC and 15 Wm2 oC Ambient air temperature=28 oC

CRITICAL RADIUS

bull The radius upto which heat flow increases and after decreases At critical radius q becomes qmax

bull The addition of insulation always increases (Rth)condbull Rtotal= (Rth)cond + (Rth)conv

bull The addition of insulation may reduce the (Rth)conv due to increase in surface area

bull r1ltrc -heat transfer increases by adding insulation till r1=rc If

insulation thickness is further increased the rate of heat loss will decrease

r1 gt rc

- heat transfer decreases by adding insulation above the rc

Plane wall- Insulation

thickness -study

Cylinder amp Sphere- Insulation thickness -

study

Problem5bull A 3mm diameter and 5 mm long electric wire is tightly

wrapped with a 2 mm thick plastic cover whose K=015 wmoC Electrical measurements indicate that a current of 10 A passes through the wire and there is a voltage drop of 8 V along the wire If the insulated wire is exposed to a medium at Ta=30oC with a heat transfer coefficient of h=12 wm2oC Determine the temperature at the interface of the wire and the plastic cover in steady operation Also determine whether doubling the thickness of the plastic cover will increase or decrease this interface temperature

dx

TdzdyKQ xx )(

dxQx

QQ xxdxx ][

dxdx

TdzdyK

xQQdQ xdxxxx ])([)(

)1()]([)(

ddzdydx

x

TK

xdQ xx

GENERAL HEAT CONDUCTION EQUATION IN CARTESIAN COORDINATES

Consider an unsteady (transient) three dimensional heat conduction with internal heat generation in heterogeneous amp anisotropic PLANE WALL

Cartesian co-ordinates x y z

Elemental length dxdydz

Consider an elemental volume ABCDEFGH = (dxdydz)

Let Temperature distribution is a function of T=T(xyz)

Let

=internal heat generation per unit volume per unit time (W m3)q= rate of heat flow (W)Q= qdt=total heat flow (J)r=mass density of the material element(kgm3)Cp=specific heat of the material element (Jkg K)

gq

Plane (y - z) lsquoxrsquo direction

Quantity of heat leaving from the face EFGH in lsquoXrsquo direction

Quantity of heat flowing into the face ABCD in lsquoXrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoXrsquo direction

dy

TdzdxKQ yy )(

dyQy

QQ yydyy ][

dydy

TdzdxK

yQQdQ ydyyyy ])([)(

)2()]([)(

ddzdydx

y

TK

ydQ yy

Plane (x - z) lsquoyrsquo direction

Quantity of heat flowing into the face ABFE in lsquoYrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoYrsquo direction

dz

TdydxKQ zz )(

dzQz

QQ zzdzz ][

dzdz

TdydxK

zQQdQ zdzzzz ])([)(

)3()]([)(

ddzdydx

z

TK

zdQ zz

Plane (x - y) lsquozrsquo direction

Quantity of heat flowing into the face ADHE in lsquoZrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoZrsquo direction

Quantity of heat leaving from the face DCGH in lsquoYrsquo direction

Quantity of heat leaving from the face BCGF in lsquoZrsquo direction

)4()()(

ddzdydxqQgg

Total amount of internal heat generation in the element

Ein + Egen = Eout+Est (Ein- Eout)+ Egen=Est

T

Cddzdydxddzdydxqddzdydxz

TK

zy

TK

yx

TK

x pgzyx )()())](()()([

T

pC

gq

z

TzK

zy

Ty

Kyx

TxK

x][][][

stgzyx QQdQdQdQ

)(

)5()()(

dT

CdzdydxQ pst

Total heat energy stored in the element

(ie thermal energy of the element can be increased because of heat accumulation in the element due to heat flow amp internal heat generation)

)( volumem d

TmCQ pst )(

T

CqTK pg)(

KKKyKx z If material is homogeneous amp isotropic

T

pC

gq

z

T

zy

T

yx

T

xK ][][][

T

Kp

C

Kgq

z

T

y

T

x

T2

2

2

2

2

2

T

Kgq

z

T

y

T

x

T 12

2

2

2

2

2

Thermal diffusivity pC

K

T

K

qT g

12

according to thermodynamics first law

)( zr )( dzrddr

)( dzrddr

Cylindrical polar co-ordinates

Elemental length

Consider an elemental volume ABCDEFGH =

GENERAL HEAT CONDUCTION EQUATION IN CYLINDRICAL COORDINATES

gq

gq

Consider an unsteady (transient) three dimensional heat conduction with internal heat generation in heterogeneous amp anisotropic cylindrical geometries (eg rods and pipes)

)( zrTLetTemperature distribution is a function of T=

Let

=internal heat generation per unit volume per unit time (W m3)q= rate of heat flow (W)Q= qdt=total heat flow (J)r=mass density of the material element(kgm3)Cp=specific heat of the material element (Jkg K)

planerdirectionz

planezrdirection

planezdirectionr

)(

)(

)(

LetHeat flow in

drdr

TdzrdK

rdQ

QQdQ

rr

drrrr

])([

)1(][

r

Tr

rddrdzdKdQ rr

Heat accumulation in the element due to heat flow by conduction in lsquorrsquo direction

rddr

TdzdrK

rdQ

QQdQ d

])([

][

2

T

r

ddrdzrdKdQ

)2(][

2

2

2

T

r

ddrdzrdKdQ

Heat accumulation in the element due to heat flow by conduction in lsquoqrsquo direction

Plane (z- q ) lsquorrsquo direction

dr

TdzrdKQ rr

)(

drQr

QQ rrdrr ][

Quantity of heat leaving from the face EFGH in lsquorrsquo direction

Quantity of heat flowing into the face ABCD in lsquorrsquo direction

Plane (r-z) lsquoq lsquodirection

Quantity of heat leaving from the face AEGC in lsquoqrsquo direction

d

r

TdzdrKQ )(

rdQ

rQQ d ][

Quantity of heat flowing into the face BFHD in lsquoqrsquo direction

)3())((2

2

z

TdzdrrdKdQ zz

)5()()(

dT

CdzdrrdQ pst

)4()()(

ddzdrrdqQgg

Total amount of internal heat generation in the element

Total heat energy stored in the element

(ie thermal energy of the element can be increased because of heat accumulation in the element due to heat flow amp internal heat generation)

according to thermodynamics first law

Ein + Egen = Eout + Est (Ein ndash Eout) + Egen = Est

)( volumem

)()(z

T

zdzdrrdKdQ zz

Heat accumulation in the element due to heat flow by conduction in lsquozrsquo direction

dzdz

TdrrdK

zdQ

QQdQ

zz

dzzzz

])([

d

TmCQ pst )(

Plane (r-q) lsquozrsquo direction

dz

TdrrdKQ zz )(

dzQz

QQ zzdzz ][ Quantity of heat leaving from the face CDHG in lsquozrsquo direction

Quantity of heat flowing into the face ABFE in lsquozrsquo direction

stgzr QQdQdQdQ

)(

r

r ][)(rT

rr

ddrdzdKr

][)(

2

2

2

T

r

ddrdzrdK ))((2

2

z

TddzdrrdK z

ddzdrrdq g )(

dT

Cdzdrrd p )(

+ + +

=

T

CqTK pg)(KKKK zr

[gq

)(1

r

Tr

rrK r

)(1

2

2

2 T

rK )(

2

2

z

TK z

)( ddzdrrd )(

ddzdrrd

TC p

+ + + ] =

If material is homogeneous amp isotropic

)(1

r

Tr

rr

)(1

2

2

2 T

r)(

2

2

z

T

K

q g

T

K

C p

)(1

r

Tr

rr

)(1

2

2

2 T

r )(2

2

z

T

K

q g

T

1

+ + + =

)(1

r

Tr

rrK r

)(1

2

2

2 T

rK )(

2

2

z

TK z

gq

T

C p + + + =

+ + + = Thermal diffusivity

T

K

qT g

12

pC

K

(Ein ndash Eout) + Egen = Est

)( rT

)( r

)sin( drrddr)sin( drrddr

planerdirection

planerdirection

planedirectionr

)(

)(

)(

Spherical polar co-ordinates

Elemental length

Consider an elemental volume ABCDEFGH =

GENERAL HEAT CONDUCTION EQUATION IN SPHERICAL COORDINATES

Let

=internal heat generation per unit volume per unit time (W m3)q= rate of heat flow (W)Q= qdt=total heat flow (J)r=mass density of the material element(kgm3)Cp=specific heat of the material element (Jkg K)

gq

Consider an unsteady (transient) three dimensional heat conduction with internal heat generation in heterogeneous amp anisotropic spherical geometries

LetTemperature distribution is a function of T=

X

Y

Z

drT

drrdKQ rr )sin(

drQr

QQ rrdrr ][

drdr

TdrrdK

rdQ

QQdQ

rr

drrrr

])sin([

)1(]([sin 2

rT

rr

ddrddKdQ rr

Plane ( -q f) lsquorrsquo directionQuantity of heat flowing into the face ABCD in lsquorrsquo direction

Quantity of heat leaving from the face EFGH in lsquorrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquorrsquo direction

drT

drdrKQ )sin(

rdQ

rQQ d ][

rddr

TdrdrK

rdQ

QQdQ d

])sin([

])[(sin

2

T

r

drdrdrKdQ

)2(])(sin[

2

T

r

drdrddrKdQ

Quantity of heat leaving from the face BFHD in lsquoqrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoqrsquo direction

X

Y

Z

Plane (r- f ) lsquoq lsquodirection

Quantity of heat flowing into the face AEGC in lsquoqrsquo direction

dr

TdrrdKQ

sin)(

drQ

rQQ d sin][

sin

drdr

TdrrdK

rdQ

QQdQ d

sin]sin

)([sin

][sinsin

)sin(

T

rr

ddrdrrdKdQ

)3()(sin

)sin(2

2

22

2

T

r

ddddrrKdQ

Plane (r- ) q lsquofrsquo directionQuantity of heat flowing into the face ABFE in lsquofrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquozrsquo direction

)4()sin()(

ddrrddrqQgg

)5()sin()(

dT

CdrrddrQ pst

Total amount of internal heat generation in the element

Total heat energy stored in the element

(Ein ndash Eout) + Egen = Est

d

TmCQ pst )(

)( volumem

Quantity of heat leaving from the face CDHG in lsquofrsquo direction

according to thermodynamics first law

Ein + Egen = Eout + Est

X

Y

Z

stgr QQdQdQdQ

)(

]([sin 2

r

Tr

rddrddKr

])(sin[

2

T

rdrdrddrK

)(sin

)sin(2

2

22

2

T

r

ddddrrK

ddrrddrq g )sin(

dT

Cdrrddr p )sin(

2

2

r

r

sin

sin

]([ 22 r

Tr

rr

Kr

])(sin[sin2

TrK

)(sin 2

2

22

T

r

Kgq

T

Cp

]([1

[ 22 r

Tr

rrK

])(sin[sin1

2

T

r)](

sin

12

2

22 T

r gq

T

Cp

]([1

[ 22 r

Tr

rr

])(sin[sin1

2

T

r)](

sin

12

2

22 T

rK

q g

T

K

Cp

]([1

[ 22 r

Tr

rr

])(sin[sin1

2

T

r)](

sin

12

2

22 T

rK

q g

T1

KKKKr If material is homogeneous amp isotropic

TCqTK pg)(

T

K

qT g

12 pC

K

Thermal diffusivity

(Ein ndash Eout) + Egen = Est

General Heat Conduction Equation

PLANE WALL CYLINDER SPHERE

Temperature profile

Plane wall SphereCylinder

bull A copper conductor (k = 380 WmoC resistivity ρ = 2x10-8 Ω m) having inner and outer radii 10 cm and 225 cm respectively is carrying a current density of 4800 amperes cm2 the conductor is internally cooled and a constant temperature of 65oC is maintained at the inner surface and there is no heat transfer through the insulation surrounding the conductor

Determinebull (i) The maximum temperature of the conductor and the radius at

which it occursbull (ii) The internal heat transfer rate

bull A chemical reaction takes place in a packed bed (k=06 WmoC) between two coaxial cylinders with radii 15 mm and 45 mm the inner surface is at 580oC and it is insulated Assuming the reaction rate of 055 MWm3 in the reactor volume find the temperature at the outer surface of the reactor

bull A current of 300 amperes passes through a stainless steel wire of 25 mm diameter and k=20 WmoC The resistivity of the wire is 70x10-8 ohm-m and length of the wire is 2 m if the wire is submerged in a fluid maintained at 50oC and h at the wire surface is 4000 Wm2oC calculate the steady state temperature at the centre and at the surface of the wire

bull An approximately spherical shaped orange ( k=023 WmoC) 90 mm diameter undergoes ripping process and generates 5100 Wm3 of energy If external surface the orange is at 8oC determine

bull (i) temperature at the centre of the orange and

bull (ii) heat flow from the outer surface of the orange

bull FINS External surfaces to used to cooling area and hence convection

heat transfer can be improvedbull APPLICATIONS IC engine cylinder ( air cooling) Heat exchangers automobile

radiators computer chip refrigerators electric motors generators and transformers etc

bull ASSUMPTIONS MADE IN ANALYSIS OF HEAT TRANFER THROUGH FIN

1 Fin material is homogenous and isotropic constthermal conductivity

2 Temperature at any cs of the fin is uniform3 lsquohrsquo is uniform over entire surface4 No internal heat generation5 Thermal contact resistance is negligible6 Radiation heat transfer is negligible7 Heat conduction is steady state and one dimensional

bull Problem 1

bull A stainless steel fin (k=20 WmK) having a diameter of 20 mm and a length of 01 m is attached to a wall at 300oC The ambient temperature is 50oC and the heat transfer coefficient is 10 Wm2K The fin tip is insulated Determine

bull (A) The rate of heat dissipation from the finbull (B) The temperature at the fin tipbull (C) The rate of heat transfer from the wall area covered by the fin if

the fin was not usedbull (D) The heat transfer rate from the same fin geometry if the

stainless steel fin is replaced by a fictitious fin with infinite thermal conductivity

bull Problem2bull One end of a long rod is inserted into a furnace while the other

projects into ambient temperature Under steady state the temperature of the rod is measured at two points 75mm apart and found to be 125oC and 885oC respectively while the ambient temperature is 20oC If the rod is 25mm in diameter and h is 2336 Wm2K

bull Find the thermal conductivity of the rod material

  • HEAT AND MASS TRANSFER
  • Thermodynamics Vs Heat transfer
  • MODE OF HEAT TRANSFER
  • MODE OF HEAT TRANSFER (2)
  • CONDUCTION- MECHANISM
  • CONCECTION-MECHANISM
  • RADIATION-MECHANISM
  • Conduction- Convection- Radiation
  • Slide 9
  • Slide 10
  • Fourierrsquos Law of Heat Conduction
  • Slide 12
  • Newtonrsquos Law of Cooling
  • Laws of Thermal Radiation
  • LAWS FOR HEAT TRANSFER
  • ELECTRICAL ampTHERMAL ANALOGY
  • Slide 17
  • Slide 18
  • PLANE WALL-UNIDIRECTIONAL
  • PLANE WALL -SERIES
  • PLANE WALL IN SERIES
  • PLANE WALL -PARALLEL
  • PLANE WALL-PARALLEL
  • PLANE WALL-COMBINATION
  • PLANE WALL-FILM RESISTANCE
  • PLANE WALL- THERMAL CONTACT RESISTANCE
  • PLANE WALL- THERMAL CONTACT RESISTANCE (2)
  • CYLINDER-SOLID
  • HEAT TRANSFER IN SOLID CYLINDER
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER-COMPOSITE
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER-MULTY LAYER
  • HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER
  • HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER (2)
  • HEAT TRANSFER IN SPHERE
  • HEAT TRANSFER IN SPHERE -MULTY LAYER
  • HEAT TRANSFER IN SPHERE -MULTY LAYER (2)
  • Slide 38
  • Slide 39
  • Slide 40
  • Slide 41
  • CRITICAL RADIUS
  • Plane wall- Insulation thickness -study
  • Problem5
  • Slide 45
  • Slide 46
  • Slide 47
  • GENERAL HEAT CONDUCTION EQUATION IN CYLINDRICAL COORDINATES
  • Slide 49
  • Slide 50
  • Slide 51
  • GENERAL HEAT CONDUCTION EQUATION IN SPHERICAL COORDINATES
  • Slide 53
  • Slide 54
  • Slide 55
  • General Heat Conduction Equation
  • Temperature profile
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Slide 67
Page 15: hmt chap 1

PLANE WALL-UNIDIRECTIONAL

PLANE WALL -SERIES

PLANE WALL IN SERIES

PLANE WALL -PARALLEL

PLANE WALL-PARALLEL

PLANE WALL-COMBINATION

PLANE WALL-FILM RESISTANCE

PLANE WALL- THERMAL CONTACT RESISTANCE

PLANE WALL- THERMAL CONTACT RESISTANCE

At the joining surface of the two slabs air is trapped in voids due to surface irregularities

CYLINDER-SOLID

HEAT TRANSFER IN SOLID CYLINDER

HEAT TRANSFER RATE IN HOLLOW CYLINDER

HEAT TRANSFER RATE IN HOLLOW CYLINDER-COMPOSITE

HEAT TRANSFER RATE IN HOLLOW CYLINDER-MULTY LAYER

HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER

HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER

HEAT TRANSFER IN SPHERE

HEAT TRANSFER IN SPHERE -MULTY LAYER

HEAT TRANSFER IN SPHERE -MULTY LAYER

bull Problem 1 The wall in a furnace consists of 125 mm thick refractory bricks and

125 mm thick insulating fire bricks separated by an air gap A 12 mm thick plaster covers the outer wall The inner surface of wall is at 1100oC and the ambient temperature 25oC the heat transfer coefficient on the outer wall to the air is 17 Wm2K and the resistance to heat flow of the air gap is 016 KW the thermal conductivities of refractory brick insulating firebrick and plaster are 16 03 and 014 WmK respectively

Calculatebull (a) the rate of heat loss per unit area of the wall surfacebull (b) the interface temperatures throughout the wallbull (c) the temperature at the outside surface of the wall

bull Problem 2Two large aluminum plates (k=240 WmK) each 2 cm thick with 10 mm surface roughness are placed in contact under 105 Nm2 pressure in air The temperature at the outside surfaces are 390 oC and 406 oCCalculate

A) the heat flux B) the temperature drop due to the contact resistance C) the contact temperatures Thermal contact resistance with air as the interfacial fluid for

10 mm roughness is 27510-4 m2KW

bull Problem 3bull Steam at 350 oC flowing in a pipe (k=80 WmK) of 5 cm

inner diameter and 56 cm outer diameter is covered with 3 cm thick insulation of k=005 WmK Heat is lost to surroundings at 5 oC by natural convection and radiation the combined h being 20 Wm2K Taking the heat transfer coefficient inside the pipe as 60 Wm2K

bull Determinebull A) the rate of heat loss from the steam per unit length of

the pipebull B) the temperature drop across the pipe and the insulation

bull PROBLEM4bull A nuclear reactor has a sperical pressure vessel 750 mm inside

diameter 80 mm wall thickness The temperature at the inner surface is 500 oC If the tempertarue at the outer surface is 495 oC

bull Calculatea) The rate of heat loss through the metal wallb) If the rate of heat loss is to limited to 305Whow much inulation

thickness needs to be applied Thermal conductivity of steel wall=46 WmoC Thermal conductivity of insulation=004 WmoC Film coefficient of heat tranfer at inner and outer surface are 200

Wm2 oC and 15 Wm2 oC Ambient air temperature=28 oC

CRITICAL RADIUS

bull The radius upto which heat flow increases and after decreases At critical radius q becomes qmax

bull The addition of insulation always increases (Rth)condbull Rtotal= (Rth)cond + (Rth)conv

bull The addition of insulation may reduce the (Rth)conv due to increase in surface area

bull r1ltrc -heat transfer increases by adding insulation till r1=rc If

insulation thickness is further increased the rate of heat loss will decrease

r1 gt rc

- heat transfer decreases by adding insulation above the rc

Plane wall- Insulation

thickness -study

Cylinder amp Sphere- Insulation thickness -

study

Problem5bull A 3mm diameter and 5 mm long electric wire is tightly

wrapped with a 2 mm thick plastic cover whose K=015 wmoC Electrical measurements indicate that a current of 10 A passes through the wire and there is a voltage drop of 8 V along the wire If the insulated wire is exposed to a medium at Ta=30oC with a heat transfer coefficient of h=12 wm2oC Determine the temperature at the interface of the wire and the plastic cover in steady operation Also determine whether doubling the thickness of the plastic cover will increase or decrease this interface temperature

dx

TdzdyKQ xx )(

dxQx

QQ xxdxx ][

dxdx

TdzdyK

xQQdQ xdxxxx ])([)(

)1()]([)(

ddzdydx

x

TK

xdQ xx

GENERAL HEAT CONDUCTION EQUATION IN CARTESIAN COORDINATES

Consider an unsteady (transient) three dimensional heat conduction with internal heat generation in heterogeneous amp anisotropic PLANE WALL

Cartesian co-ordinates x y z

Elemental length dxdydz

Consider an elemental volume ABCDEFGH = (dxdydz)

Let Temperature distribution is a function of T=T(xyz)

Let

=internal heat generation per unit volume per unit time (W m3)q= rate of heat flow (W)Q= qdt=total heat flow (J)r=mass density of the material element(kgm3)Cp=specific heat of the material element (Jkg K)

gq

Plane (y - z) lsquoxrsquo direction

Quantity of heat leaving from the face EFGH in lsquoXrsquo direction

Quantity of heat flowing into the face ABCD in lsquoXrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoXrsquo direction

dy

TdzdxKQ yy )(

dyQy

QQ yydyy ][

dydy

TdzdxK

yQQdQ ydyyyy ])([)(

)2()]([)(

ddzdydx

y

TK

ydQ yy

Plane (x - z) lsquoyrsquo direction

Quantity of heat flowing into the face ABFE in lsquoYrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoYrsquo direction

dz

TdydxKQ zz )(

dzQz

QQ zzdzz ][

dzdz

TdydxK

zQQdQ zdzzzz ])([)(

)3()]([)(

ddzdydx

z

TK

zdQ zz

Plane (x - y) lsquozrsquo direction

Quantity of heat flowing into the face ADHE in lsquoZrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoZrsquo direction

Quantity of heat leaving from the face DCGH in lsquoYrsquo direction

Quantity of heat leaving from the face BCGF in lsquoZrsquo direction

)4()()(

ddzdydxqQgg

Total amount of internal heat generation in the element

Ein + Egen = Eout+Est (Ein- Eout)+ Egen=Est

T

Cddzdydxddzdydxqddzdydxz

TK

zy

TK

yx

TK

x pgzyx )()())](()()([

T

pC

gq

z

TzK

zy

Ty

Kyx

TxK

x][][][

stgzyx QQdQdQdQ

)(

)5()()(

dT

CdzdydxQ pst

Total heat energy stored in the element

(ie thermal energy of the element can be increased because of heat accumulation in the element due to heat flow amp internal heat generation)

)( volumem d

TmCQ pst )(

T

CqTK pg)(

KKKyKx z If material is homogeneous amp isotropic

T

pC

gq

z

T

zy

T

yx

T

xK ][][][

T

Kp

C

Kgq

z

T

y

T

x

T2

2

2

2

2

2

T

Kgq

z

T

y

T

x

T 12

2

2

2

2

2

Thermal diffusivity pC

K

T

K

qT g

12

according to thermodynamics first law

)( zr )( dzrddr

)( dzrddr

Cylindrical polar co-ordinates

Elemental length

Consider an elemental volume ABCDEFGH =

GENERAL HEAT CONDUCTION EQUATION IN CYLINDRICAL COORDINATES

gq

gq

Consider an unsteady (transient) three dimensional heat conduction with internal heat generation in heterogeneous amp anisotropic cylindrical geometries (eg rods and pipes)

)( zrTLetTemperature distribution is a function of T=

Let

=internal heat generation per unit volume per unit time (W m3)q= rate of heat flow (W)Q= qdt=total heat flow (J)r=mass density of the material element(kgm3)Cp=specific heat of the material element (Jkg K)

planerdirectionz

planezrdirection

planezdirectionr

)(

)(

)(

LetHeat flow in

drdr

TdzrdK

rdQ

QQdQ

rr

drrrr

])([

)1(][

r

Tr

rddrdzdKdQ rr

Heat accumulation in the element due to heat flow by conduction in lsquorrsquo direction

rddr

TdzdrK

rdQ

QQdQ d

])([

][

2

T

r

ddrdzrdKdQ

)2(][

2

2

2

T

r

ddrdzrdKdQ

Heat accumulation in the element due to heat flow by conduction in lsquoqrsquo direction

Plane (z- q ) lsquorrsquo direction

dr

TdzrdKQ rr

)(

drQr

QQ rrdrr ][

Quantity of heat leaving from the face EFGH in lsquorrsquo direction

Quantity of heat flowing into the face ABCD in lsquorrsquo direction

Plane (r-z) lsquoq lsquodirection

Quantity of heat leaving from the face AEGC in lsquoqrsquo direction

d

r

TdzdrKQ )(

rdQ

rQQ d ][

Quantity of heat flowing into the face BFHD in lsquoqrsquo direction

)3())((2

2

z

TdzdrrdKdQ zz

)5()()(

dT

CdzdrrdQ pst

)4()()(

ddzdrrdqQgg

Total amount of internal heat generation in the element

Total heat energy stored in the element

(ie thermal energy of the element can be increased because of heat accumulation in the element due to heat flow amp internal heat generation)

according to thermodynamics first law

Ein + Egen = Eout + Est (Ein ndash Eout) + Egen = Est

)( volumem

)()(z

T

zdzdrrdKdQ zz

Heat accumulation in the element due to heat flow by conduction in lsquozrsquo direction

dzdz

TdrrdK

zdQ

QQdQ

zz

dzzzz

])([

d

TmCQ pst )(

Plane (r-q) lsquozrsquo direction

dz

TdrrdKQ zz )(

dzQz

QQ zzdzz ][ Quantity of heat leaving from the face CDHG in lsquozrsquo direction

Quantity of heat flowing into the face ABFE in lsquozrsquo direction

stgzr QQdQdQdQ

)(

r

r ][)(rT

rr

ddrdzdKr

][)(

2

2

2

T

r

ddrdzrdK ))((2

2

z

TddzdrrdK z

ddzdrrdq g )(

dT

Cdzdrrd p )(

+ + +

=

T

CqTK pg)(KKKK zr

[gq

)(1

r

Tr

rrK r

)(1

2

2

2 T

rK )(

2

2

z

TK z

)( ddzdrrd )(

ddzdrrd

TC p

+ + + ] =

If material is homogeneous amp isotropic

)(1

r

Tr

rr

)(1

2

2

2 T

r)(

2

2

z

T

K

q g

T

K

C p

)(1

r

Tr

rr

)(1

2

2

2 T

r )(2

2

z

T

K

q g

T

1

+ + + =

)(1

r

Tr

rrK r

)(1

2

2

2 T

rK )(

2

2

z

TK z

gq

T

C p + + + =

+ + + = Thermal diffusivity

T

K

qT g

12

pC

K

(Ein ndash Eout) + Egen = Est

)( rT

)( r

)sin( drrddr)sin( drrddr

planerdirection

planerdirection

planedirectionr

)(

)(

)(

Spherical polar co-ordinates

Elemental length

Consider an elemental volume ABCDEFGH =

GENERAL HEAT CONDUCTION EQUATION IN SPHERICAL COORDINATES

Let

=internal heat generation per unit volume per unit time (W m3)q= rate of heat flow (W)Q= qdt=total heat flow (J)r=mass density of the material element(kgm3)Cp=specific heat of the material element (Jkg K)

gq

Consider an unsteady (transient) three dimensional heat conduction with internal heat generation in heterogeneous amp anisotropic spherical geometries

LetTemperature distribution is a function of T=

X

Y

Z

drT

drrdKQ rr )sin(

drQr

QQ rrdrr ][

drdr

TdrrdK

rdQ

QQdQ

rr

drrrr

])sin([

)1(]([sin 2

rT

rr

ddrddKdQ rr

Plane ( -q f) lsquorrsquo directionQuantity of heat flowing into the face ABCD in lsquorrsquo direction

Quantity of heat leaving from the face EFGH in lsquorrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquorrsquo direction

drT

drdrKQ )sin(

rdQ

rQQ d ][

rddr

TdrdrK

rdQ

QQdQ d

])sin([

])[(sin

2

T

r

drdrdrKdQ

)2(])(sin[

2

T

r

drdrddrKdQ

Quantity of heat leaving from the face BFHD in lsquoqrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoqrsquo direction

X

Y

Z

Plane (r- f ) lsquoq lsquodirection

Quantity of heat flowing into the face AEGC in lsquoqrsquo direction

dr

TdrrdKQ

sin)(

drQ

rQQ d sin][

sin

drdr

TdrrdK

rdQ

QQdQ d

sin]sin

)([sin

][sinsin

)sin(

T

rr

ddrdrrdKdQ

)3()(sin

)sin(2

2

22

2

T

r

ddddrrKdQ

Plane (r- ) q lsquofrsquo directionQuantity of heat flowing into the face ABFE in lsquofrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquozrsquo direction

)4()sin()(

ddrrddrqQgg

)5()sin()(

dT

CdrrddrQ pst

Total amount of internal heat generation in the element

Total heat energy stored in the element

(Ein ndash Eout) + Egen = Est

d

TmCQ pst )(

)( volumem

Quantity of heat leaving from the face CDHG in lsquofrsquo direction

according to thermodynamics first law

Ein + Egen = Eout + Est

X

Y

Z

stgr QQdQdQdQ

)(

]([sin 2

r

Tr

rddrddKr

])(sin[

2

T

rdrdrddrK

)(sin

)sin(2

2

22

2

T

r

ddddrrK

ddrrddrq g )sin(

dT

Cdrrddr p )sin(

2

2

r

r

sin

sin

]([ 22 r

Tr

rr

Kr

])(sin[sin2

TrK

)(sin 2

2

22

T

r

Kgq

T

Cp

]([1

[ 22 r

Tr

rrK

])(sin[sin1

2

T

r)](

sin

12

2

22 T

r gq

T

Cp

]([1

[ 22 r

Tr

rr

])(sin[sin1

2

T

r)](

sin

12

2

22 T

rK

q g

T

K

Cp

]([1

[ 22 r

Tr

rr

])(sin[sin1

2

T

r)](

sin

12

2

22 T

rK

q g

T1

KKKKr If material is homogeneous amp isotropic

TCqTK pg)(

T

K

qT g

12 pC

K

Thermal diffusivity

(Ein ndash Eout) + Egen = Est

General Heat Conduction Equation

PLANE WALL CYLINDER SPHERE

Temperature profile

Plane wall SphereCylinder

bull A copper conductor (k = 380 WmoC resistivity ρ = 2x10-8 Ω m) having inner and outer radii 10 cm and 225 cm respectively is carrying a current density of 4800 amperes cm2 the conductor is internally cooled and a constant temperature of 65oC is maintained at the inner surface and there is no heat transfer through the insulation surrounding the conductor

Determinebull (i) The maximum temperature of the conductor and the radius at

which it occursbull (ii) The internal heat transfer rate

bull A chemical reaction takes place in a packed bed (k=06 WmoC) between two coaxial cylinders with radii 15 mm and 45 mm the inner surface is at 580oC and it is insulated Assuming the reaction rate of 055 MWm3 in the reactor volume find the temperature at the outer surface of the reactor

bull A current of 300 amperes passes through a stainless steel wire of 25 mm diameter and k=20 WmoC The resistivity of the wire is 70x10-8 ohm-m and length of the wire is 2 m if the wire is submerged in a fluid maintained at 50oC and h at the wire surface is 4000 Wm2oC calculate the steady state temperature at the centre and at the surface of the wire

bull An approximately spherical shaped orange ( k=023 WmoC) 90 mm diameter undergoes ripping process and generates 5100 Wm3 of energy If external surface the orange is at 8oC determine

bull (i) temperature at the centre of the orange and

bull (ii) heat flow from the outer surface of the orange

bull FINS External surfaces to used to cooling area and hence convection

heat transfer can be improvedbull APPLICATIONS IC engine cylinder ( air cooling) Heat exchangers automobile

radiators computer chip refrigerators electric motors generators and transformers etc

bull ASSUMPTIONS MADE IN ANALYSIS OF HEAT TRANFER THROUGH FIN

1 Fin material is homogenous and isotropic constthermal conductivity

2 Temperature at any cs of the fin is uniform3 lsquohrsquo is uniform over entire surface4 No internal heat generation5 Thermal contact resistance is negligible6 Radiation heat transfer is negligible7 Heat conduction is steady state and one dimensional

bull Problem 1

bull A stainless steel fin (k=20 WmK) having a diameter of 20 mm and a length of 01 m is attached to a wall at 300oC The ambient temperature is 50oC and the heat transfer coefficient is 10 Wm2K The fin tip is insulated Determine

bull (A) The rate of heat dissipation from the finbull (B) The temperature at the fin tipbull (C) The rate of heat transfer from the wall area covered by the fin if

the fin was not usedbull (D) The heat transfer rate from the same fin geometry if the

stainless steel fin is replaced by a fictitious fin with infinite thermal conductivity

bull Problem2bull One end of a long rod is inserted into a furnace while the other

projects into ambient temperature Under steady state the temperature of the rod is measured at two points 75mm apart and found to be 125oC and 885oC respectively while the ambient temperature is 20oC If the rod is 25mm in diameter and h is 2336 Wm2K

bull Find the thermal conductivity of the rod material

  • HEAT AND MASS TRANSFER
  • Thermodynamics Vs Heat transfer
  • MODE OF HEAT TRANSFER
  • MODE OF HEAT TRANSFER (2)
  • CONDUCTION- MECHANISM
  • CONCECTION-MECHANISM
  • RADIATION-MECHANISM
  • Conduction- Convection- Radiation
  • Slide 9
  • Slide 10
  • Fourierrsquos Law of Heat Conduction
  • Slide 12
  • Newtonrsquos Law of Cooling
  • Laws of Thermal Radiation
  • LAWS FOR HEAT TRANSFER
  • ELECTRICAL ampTHERMAL ANALOGY
  • Slide 17
  • Slide 18
  • PLANE WALL-UNIDIRECTIONAL
  • PLANE WALL -SERIES
  • PLANE WALL IN SERIES
  • PLANE WALL -PARALLEL
  • PLANE WALL-PARALLEL
  • PLANE WALL-COMBINATION
  • PLANE WALL-FILM RESISTANCE
  • PLANE WALL- THERMAL CONTACT RESISTANCE
  • PLANE WALL- THERMAL CONTACT RESISTANCE (2)
  • CYLINDER-SOLID
  • HEAT TRANSFER IN SOLID CYLINDER
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER-COMPOSITE
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER-MULTY LAYER
  • HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER
  • HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER (2)
  • HEAT TRANSFER IN SPHERE
  • HEAT TRANSFER IN SPHERE -MULTY LAYER
  • HEAT TRANSFER IN SPHERE -MULTY LAYER (2)
  • Slide 38
  • Slide 39
  • Slide 40
  • Slide 41
  • CRITICAL RADIUS
  • Plane wall- Insulation thickness -study
  • Problem5
  • Slide 45
  • Slide 46
  • Slide 47
  • GENERAL HEAT CONDUCTION EQUATION IN CYLINDRICAL COORDINATES
  • Slide 49
  • Slide 50
  • Slide 51
  • GENERAL HEAT CONDUCTION EQUATION IN SPHERICAL COORDINATES
  • Slide 53
  • Slide 54
  • Slide 55
  • General Heat Conduction Equation
  • Temperature profile
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Slide 67
Page 16: hmt chap 1

PLANE WALL -SERIES

PLANE WALL IN SERIES

PLANE WALL -PARALLEL

PLANE WALL-PARALLEL

PLANE WALL-COMBINATION

PLANE WALL-FILM RESISTANCE

PLANE WALL- THERMAL CONTACT RESISTANCE

PLANE WALL- THERMAL CONTACT RESISTANCE

At the joining surface of the two slabs air is trapped in voids due to surface irregularities

CYLINDER-SOLID

HEAT TRANSFER IN SOLID CYLINDER

HEAT TRANSFER RATE IN HOLLOW CYLINDER

HEAT TRANSFER RATE IN HOLLOW CYLINDER-COMPOSITE

HEAT TRANSFER RATE IN HOLLOW CYLINDER-MULTY LAYER

HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER

HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER

HEAT TRANSFER IN SPHERE

HEAT TRANSFER IN SPHERE -MULTY LAYER

HEAT TRANSFER IN SPHERE -MULTY LAYER

bull Problem 1 The wall in a furnace consists of 125 mm thick refractory bricks and

125 mm thick insulating fire bricks separated by an air gap A 12 mm thick plaster covers the outer wall The inner surface of wall is at 1100oC and the ambient temperature 25oC the heat transfer coefficient on the outer wall to the air is 17 Wm2K and the resistance to heat flow of the air gap is 016 KW the thermal conductivities of refractory brick insulating firebrick and plaster are 16 03 and 014 WmK respectively

Calculatebull (a) the rate of heat loss per unit area of the wall surfacebull (b) the interface temperatures throughout the wallbull (c) the temperature at the outside surface of the wall

bull Problem 2Two large aluminum plates (k=240 WmK) each 2 cm thick with 10 mm surface roughness are placed in contact under 105 Nm2 pressure in air The temperature at the outside surfaces are 390 oC and 406 oCCalculate

A) the heat flux B) the temperature drop due to the contact resistance C) the contact temperatures Thermal contact resistance with air as the interfacial fluid for

10 mm roughness is 27510-4 m2KW

bull Problem 3bull Steam at 350 oC flowing in a pipe (k=80 WmK) of 5 cm

inner diameter and 56 cm outer diameter is covered with 3 cm thick insulation of k=005 WmK Heat is lost to surroundings at 5 oC by natural convection and radiation the combined h being 20 Wm2K Taking the heat transfer coefficient inside the pipe as 60 Wm2K

bull Determinebull A) the rate of heat loss from the steam per unit length of

the pipebull B) the temperature drop across the pipe and the insulation

bull PROBLEM4bull A nuclear reactor has a sperical pressure vessel 750 mm inside

diameter 80 mm wall thickness The temperature at the inner surface is 500 oC If the tempertarue at the outer surface is 495 oC

bull Calculatea) The rate of heat loss through the metal wallb) If the rate of heat loss is to limited to 305Whow much inulation

thickness needs to be applied Thermal conductivity of steel wall=46 WmoC Thermal conductivity of insulation=004 WmoC Film coefficient of heat tranfer at inner and outer surface are 200

Wm2 oC and 15 Wm2 oC Ambient air temperature=28 oC

CRITICAL RADIUS

bull The radius upto which heat flow increases and after decreases At critical radius q becomes qmax

bull The addition of insulation always increases (Rth)condbull Rtotal= (Rth)cond + (Rth)conv

bull The addition of insulation may reduce the (Rth)conv due to increase in surface area

bull r1ltrc -heat transfer increases by adding insulation till r1=rc If

insulation thickness is further increased the rate of heat loss will decrease

r1 gt rc

- heat transfer decreases by adding insulation above the rc

Plane wall- Insulation

thickness -study

Cylinder amp Sphere- Insulation thickness -

study

Problem5bull A 3mm diameter and 5 mm long electric wire is tightly

wrapped with a 2 mm thick plastic cover whose K=015 wmoC Electrical measurements indicate that a current of 10 A passes through the wire and there is a voltage drop of 8 V along the wire If the insulated wire is exposed to a medium at Ta=30oC with a heat transfer coefficient of h=12 wm2oC Determine the temperature at the interface of the wire and the plastic cover in steady operation Also determine whether doubling the thickness of the plastic cover will increase or decrease this interface temperature

dx

TdzdyKQ xx )(

dxQx

QQ xxdxx ][

dxdx

TdzdyK

xQQdQ xdxxxx ])([)(

)1()]([)(

ddzdydx

x

TK

xdQ xx

GENERAL HEAT CONDUCTION EQUATION IN CARTESIAN COORDINATES

Consider an unsteady (transient) three dimensional heat conduction with internal heat generation in heterogeneous amp anisotropic PLANE WALL

Cartesian co-ordinates x y z

Elemental length dxdydz

Consider an elemental volume ABCDEFGH = (dxdydz)

Let Temperature distribution is a function of T=T(xyz)

Let

=internal heat generation per unit volume per unit time (W m3)q= rate of heat flow (W)Q= qdt=total heat flow (J)r=mass density of the material element(kgm3)Cp=specific heat of the material element (Jkg K)

gq

Plane (y - z) lsquoxrsquo direction

Quantity of heat leaving from the face EFGH in lsquoXrsquo direction

Quantity of heat flowing into the face ABCD in lsquoXrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoXrsquo direction

dy

TdzdxKQ yy )(

dyQy

QQ yydyy ][

dydy

TdzdxK

yQQdQ ydyyyy ])([)(

)2()]([)(

ddzdydx

y

TK

ydQ yy

Plane (x - z) lsquoyrsquo direction

Quantity of heat flowing into the face ABFE in lsquoYrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoYrsquo direction

dz

TdydxKQ zz )(

dzQz

QQ zzdzz ][

dzdz

TdydxK

zQQdQ zdzzzz ])([)(

)3()]([)(

ddzdydx

z

TK

zdQ zz

Plane (x - y) lsquozrsquo direction

Quantity of heat flowing into the face ADHE in lsquoZrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoZrsquo direction

Quantity of heat leaving from the face DCGH in lsquoYrsquo direction

Quantity of heat leaving from the face BCGF in lsquoZrsquo direction

)4()()(

ddzdydxqQgg

Total amount of internal heat generation in the element

Ein + Egen = Eout+Est (Ein- Eout)+ Egen=Est

T

Cddzdydxddzdydxqddzdydxz

TK

zy

TK

yx

TK

x pgzyx )()())](()()([

T

pC

gq

z

TzK

zy

Ty

Kyx

TxK

x][][][

stgzyx QQdQdQdQ

)(

)5()()(

dT

CdzdydxQ pst

Total heat energy stored in the element

(ie thermal energy of the element can be increased because of heat accumulation in the element due to heat flow amp internal heat generation)

)( volumem d

TmCQ pst )(

T

CqTK pg)(

KKKyKx z If material is homogeneous amp isotropic

T

pC

gq

z

T

zy

T

yx

T

xK ][][][

T

Kp

C

Kgq

z

T

y

T

x

T2

2

2

2

2

2

T

Kgq

z

T

y

T

x

T 12

2

2

2

2

2

Thermal diffusivity pC

K

T

K

qT g

12

according to thermodynamics first law

)( zr )( dzrddr

)( dzrddr

Cylindrical polar co-ordinates

Elemental length

Consider an elemental volume ABCDEFGH =

GENERAL HEAT CONDUCTION EQUATION IN CYLINDRICAL COORDINATES

gq

gq

Consider an unsteady (transient) three dimensional heat conduction with internal heat generation in heterogeneous amp anisotropic cylindrical geometries (eg rods and pipes)

)( zrTLetTemperature distribution is a function of T=

Let

=internal heat generation per unit volume per unit time (W m3)q= rate of heat flow (W)Q= qdt=total heat flow (J)r=mass density of the material element(kgm3)Cp=specific heat of the material element (Jkg K)

planerdirectionz

planezrdirection

planezdirectionr

)(

)(

)(

LetHeat flow in

drdr

TdzrdK

rdQ

QQdQ

rr

drrrr

])([

)1(][

r

Tr

rddrdzdKdQ rr

Heat accumulation in the element due to heat flow by conduction in lsquorrsquo direction

rddr

TdzdrK

rdQ

QQdQ d

])([

][

2

T

r

ddrdzrdKdQ

)2(][

2

2

2

T

r

ddrdzrdKdQ

Heat accumulation in the element due to heat flow by conduction in lsquoqrsquo direction

Plane (z- q ) lsquorrsquo direction

dr

TdzrdKQ rr

)(

drQr

QQ rrdrr ][

Quantity of heat leaving from the face EFGH in lsquorrsquo direction

Quantity of heat flowing into the face ABCD in lsquorrsquo direction

Plane (r-z) lsquoq lsquodirection

Quantity of heat leaving from the face AEGC in lsquoqrsquo direction

d

r

TdzdrKQ )(

rdQ

rQQ d ][

Quantity of heat flowing into the face BFHD in lsquoqrsquo direction

)3())((2

2

z

TdzdrrdKdQ zz

)5()()(

dT

CdzdrrdQ pst

)4()()(

ddzdrrdqQgg

Total amount of internal heat generation in the element

Total heat energy stored in the element

(ie thermal energy of the element can be increased because of heat accumulation in the element due to heat flow amp internal heat generation)

according to thermodynamics first law

Ein + Egen = Eout + Est (Ein ndash Eout) + Egen = Est

)( volumem

)()(z

T

zdzdrrdKdQ zz

Heat accumulation in the element due to heat flow by conduction in lsquozrsquo direction

dzdz

TdrrdK

zdQ

QQdQ

zz

dzzzz

])([

d

TmCQ pst )(

Plane (r-q) lsquozrsquo direction

dz

TdrrdKQ zz )(

dzQz

QQ zzdzz ][ Quantity of heat leaving from the face CDHG in lsquozrsquo direction

Quantity of heat flowing into the face ABFE in lsquozrsquo direction

stgzr QQdQdQdQ

)(

r

r ][)(rT

rr

ddrdzdKr

][)(

2

2

2

T

r

ddrdzrdK ))((2

2

z

TddzdrrdK z

ddzdrrdq g )(

dT

Cdzdrrd p )(

+ + +

=

T

CqTK pg)(KKKK zr

[gq

)(1

r

Tr

rrK r

)(1

2

2

2 T

rK )(

2

2

z

TK z

)( ddzdrrd )(

ddzdrrd

TC p

+ + + ] =

If material is homogeneous amp isotropic

)(1

r

Tr

rr

)(1

2

2

2 T

r)(

2

2

z

T

K

q g

T

K

C p

)(1

r

Tr

rr

)(1

2

2

2 T

r )(2

2

z

T

K

q g

T

1

+ + + =

)(1

r

Tr

rrK r

)(1

2

2

2 T

rK )(

2

2

z

TK z

gq

T

C p + + + =

+ + + = Thermal diffusivity

T

K

qT g

12

pC

K

(Ein ndash Eout) + Egen = Est

)( rT

)( r

)sin( drrddr)sin( drrddr

planerdirection

planerdirection

planedirectionr

)(

)(

)(

Spherical polar co-ordinates

Elemental length

Consider an elemental volume ABCDEFGH =

GENERAL HEAT CONDUCTION EQUATION IN SPHERICAL COORDINATES

Let

=internal heat generation per unit volume per unit time (W m3)q= rate of heat flow (W)Q= qdt=total heat flow (J)r=mass density of the material element(kgm3)Cp=specific heat of the material element (Jkg K)

gq

Consider an unsteady (transient) three dimensional heat conduction with internal heat generation in heterogeneous amp anisotropic spherical geometries

LetTemperature distribution is a function of T=

X

Y

Z

drT

drrdKQ rr )sin(

drQr

QQ rrdrr ][

drdr

TdrrdK

rdQ

QQdQ

rr

drrrr

])sin([

)1(]([sin 2

rT

rr

ddrddKdQ rr

Plane ( -q f) lsquorrsquo directionQuantity of heat flowing into the face ABCD in lsquorrsquo direction

Quantity of heat leaving from the face EFGH in lsquorrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquorrsquo direction

drT

drdrKQ )sin(

rdQ

rQQ d ][

rddr

TdrdrK

rdQ

QQdQ d

])sin([

])[(sin

2

T

r

drdrdrKdQ

)2(])(sin[

2

T

r

drdrddrKdQ

Quantity of heat leaving from the face BFHD in lsquoqrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoqrsquo direction

X

Y

Z

Plane (r- f ) lsquoq lsquodirection

Quantity of heat flowing into the face AEGC in lsquoqrsquo direction

dr

TdrrdKQ

sin)(

drQ

rQQ d sin][

sin

drdr

TdrrdK

rdQ

QQdQ d

sin]sin

)([sin

][sinsin

)sin(

T

rr

ddrdrrdKdQ

)3()(sin

)sin(2

2

22

2

T

r

ddddrrKdQ

Plane (r- ) q lsquofrsquo directionQuantity of heat flowing into the face ABFE in lsquofrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquozrsquo direction

)4()sin()(

ddrrddrqQgg

)5()sin()(

dT

CdrrddrQ pst

Total amount of internal heat generation in the element

Total heat energy stored in the element

(Ein ndash Eout) + Egen = Est

d

TmCQ pst )(

)( volumem

Quantity of heat leaving from the face CDHG in lsquofrsquo direction

according to thermodynamics first law

Ein + Egen = Eout + Est

X

Y

Z

stgr QQdQdQdQ

)(

]([sin 2

r

Tr

rddrddKr

])(sin[

2

T

rdrdrddrK

)(sin

)sin(2

2

22

2

T

r

ddddrrK

ddrrddrq g )sin(

dT

Cdrrddr p )sin(

2

2

r

r

sin

sin

]([ 22 r

Tr

rr

Kr

])(sin[sin2

TrK

)(sin 2

2

22

T

r

Kgq

T

Cp

]([1

[ 22 r

Tr

rrK

])(sin[sin1

2

T

r)](

sin

12

2

22 T

r gq

T

Cp

]([1

[ 22 r

Tr

rr

])(sin[sin1

2

T

r)](

sin

12

2

22 T

rK

q g

T

K

Cp

]([1

[ 22 r

Tr

rr

])(sin[sin1

2

T

r)](

sin

12

2

22 T

rK

q g

T1

KKKKr If material is homogeneous amp isotropic

TCqTK pg)(

T

K

qT g

12 pC

K

Thermal diffusivity

(Ein ndash Eout) + Egen = Est

General Heat Conduction Equation

PLANE WALL CYLINDER SPHERE

Temperature profile

Plane wall SphereCylinder

bull A copper conductor (k = 380 WmoC resistivity ρ = 2x10-8 Ω m) having inner and outer radii 10 cm and 225 cm respectively is carrying a current density of 4800 amperes cm2 the conductor is internally cooled and a constant temperature of 65oC is maintained at the inner surface and there is no heat transfer through the insulation surrounding the conductor

Determinebull (i) The maximum temperature of the conductor and the radius at

which it occursbull (ii) The internal heat transfer rate

bull A chemical reaction takes place in a packed bed (k=06 WmoC) between two coaxial cylinders with radii 15 mm and 45 mm the inner surface is at 580oC and it is insulated Assuming the reaction rate of 055 MWm3 in the reactor volume find the temperature at the outer surface of the reactor

bull A current of 300 amperes passes through a stainless steel wire of 25 mm diameter and k=20 WmoC The resistivity of the wire is 70x10-8 ohm-m and length of the wire is 2 m if the wire is submerged in a fluid maintained at 50oC and h at the wire surface is 4000 Wm2oC calculate the steady state temperature at the centre and at the surface of the wire

bull An approximately spherical shaped orange ( k=023 WmoC) 90 mm diameter undergoes ripping process and generates 5100 Wm3 of energy If external surface the orange is at 8oC determine

bull (i) temperature at the centre of the orange and

bull (ii) heat flow from the outer surface of the orange

bull FINS External surfaces to used to cooling area and hence convection

heat transfer can be improvedbull APPLICATIONS IC engine cylinder ( air cooling) Heat exchangers automobile

radiators computer chip refrigerators electric motors generators and transformers etc

bull ASSUMPTIONS MADE IN ANALYSIS OF HEAT TRANFER THROUGH FIN

1 Fin material is homogenous and isotropic constthermal conductivity

2 Temperature at any cs of the fin is uniform3 lsquohrsquo is uniform over entire surface4 No internal heat generation5 Thermal contact resistance is negligible6 Radiation heat transfer is negligible7 Heat conduction is steady state and one dimensional

bull Problem 1

bull A stainless steel fin (k=20 WmK) having a diameter of 20 mm and a length of 01 m is attached to a wall at 300oC The ambient temperature is 50oC and the heat transfer coefficient is 10 Wm2K The fin tip is insulated Determine

bull (A) The rate of heat dissipation from the finbull (B) The temperature at the fin tipbull (C) The rate of heat transfer from the wall area covered by the fin if

the fin was not usedbull (D) The heat transfer rate from the same fin geometry if the

stainless steel fin is replaced by a fictitious fin with infinite thermal conductivity

bull Problem2bull One end of a long rod is inserted into a furnace while the other

projects into ambient temperature Under steady state the temperature of the rod is measured at two points 75mm apart and found to be 125oC and 885oC respectively while the ambient temperature is 20oC If the rod is 25mm in diameter and h is 2336 Wm2K

bull Find the thermal conductivity of the rod material

  • HEAT AND MASS TRANSFER
  • Thermodynamics Vs Heat transfer
  • MODE OF HEAT TRANSFER
  • MODE OF HEAT TRANSFER (2)
  • CONDUCTION- MECHANISM
  • CONCECTION-MECHANISM
  • RADIATION-MECHANISM
  • Conduction- Convection- Radiation
  • Slide 9
  • Slide 10
  • Fourierrsquos Law of Heat Conduction
  • Slide 12
  • Newtonrsquos Law of Cooling
  • Laws of Thermal Radiation
  • LAWS FOR HEAT TRANSFER
  • ELECTRICAL ampTHERMAL ANALOGY
  • Slide 17
  • Slide 18
  • PLANE WALL-UNIDIRECTIONAL
  • PLANE WALL -SERIES
  • PLANE WALL IN SERIES
  • PLANE WALL -PARALLEL
  • PLANE WALL-PARALLEL
  • PLANE WALL-COMBINATION
  • PLANE WALL-FILM RESISTANCE
  • PLANE WALL- THERMAL CONTACT RESISTANCE
  • PLANE WALL- THERMAL CONTACT RESISTANCE (2)
  • CYLINDER-SOLID
  • HEAT TRANSFER IN SOLID CYLINDER
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER-COMPOSITE
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER-MULTY LAYER
  • HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER
  • HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER (2)
  • HEAT TRANSFER IN SPHERE
  • HEAT TRANSFER IN SPHERE -MULTY LAYER
  • HEAT TRANSFER IN SPHERE -MULTY LAYER (2)
  • Slide 38
  • Slide 39
  • Slide 40
  • Slide 41
  • CRITICAL RADIUS
  • Plane wall- Insulation thickness -study
  • Problem5
  • Slide 45
  • Slide 46
  • Slide 47
  • GENERAL HEAT CONDUCTION EQUATION IN CYLINDRICAL COORDINATES
  • Slide 49
  • Slide 50
  • Slide 51
  • GENERAL HEAT CONDUCTION EQUATION IN SPHERICAL COORDINATES
  • Slide 53
  • Slide 54
  • Slide 55
  • General Heat Conduction Equation
  • Temperature profile
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Slide 67
Page 17: hmt chap 1

PLANE WALL IN SERIES

PLANE WALL -PARALLEL

PLANE WALL-PARALLEL

PLANE WALL-COMBINATION

PLANE WALL-FILM RESISTANCE

PLANE WALL- THERMAL CONTACT RESISTANCE

PLANE WALL- THERMAL CONTACT RESISTANCE

At the joining surface of the two slabs air is trapped in voids due to surface irregularities

CYLINDER-SOLID

HEAT TRANSFER IN SOLID CYLINDER

HEAT TRANSFER RATE IN HOLLOW CYLINDER

HEAT TRANSFER RATE IN HOLLOW CYLINDER-COMPOSITE

HEAT TRANSFER RATE IN HOLLOW CYLINDER-MULTY LAYER

HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER

HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER

HEAT TRANSFER IN SPHERE

HEAT TRANSFER IN SPHERE -MULTY LAYER

HEAT TRANSFER IN SPHERE -MULTY LAYER

bull Problem 1 The wall in a furnace consists of 125 mm thick refractory bricks and

125 mm thick insulating fire bricks separated by an air gap A 12 mm thick plaster covers the outer wall The inner surface of wall is at 1100oC and the ambient temperature 25oC the heat transfer coefficient on the outer wall to the air is 17 Wm2K and the resistance to heat flow of the air gap is 016 KW the thermal conductivities of refractory brick insulating firebrick and plaster are 16 03 and 014 WmK respectively

Calculatebull (a) the rate of heat loss per unit area of the wall surfacebull (b) the interface temperatures throughout the wallbull (c) the temperature at the outside surface of the wall

bull Problem 2Two large aluminum plates (k=240 WmK) each 2 cm thick with 10 mm surface roughness are placed in contact under 105 Nm2 pressure in air The temperature at the outside surfaces are 390 oC and 406 oCCalculate

A) the heat flux B) the temperature drop due to the contact resistance C) the contact temperatures Thermal contact resistance with air as the interfacial fluid for

10 mm roughness is 27510-4 m2KW

bull Problem 3bull Steam at 350 oC flowing in a pipe (k=80 WmK) of 5 cm

inner diameter and 56 cm outer diameter is covered with 3 cm thick insulation of k=005 WmK Heat is lost to surroundings at 5 oC by natural convection and radiation the combined h being 20 Wm2K Taking the heat transfer coefficient inside the pipe as 60 Wm2K

bull Determinebull A) the rate of heat loss from the steam per unit length of

the pipebull B) the temperature drop across the pipe and the insulation

bull PROBLEM4bull A nuclear reactor has a sperical pressure vessel 750 mm inside

diameter 80 mm wall thickness The temperature at the inner surface is 500 oC If the tempertarue at the outer surface is 495 oC

bull Calculatea) The rate of heat loss through the metal wallb) If the rate of heat loss is to limited to 305Whow much inulation

thickness needs to be applied Thermal conductivity of steel wall=46 WmoC Thermal conductivity of insulation=004 WmoC Film coefficient of heat tranfer at inner and outer surface are 200

Wm2 oC and 15 Wm2 oC Ambient air temperature=28 oC

CRITICAL RADIUS

bull The radius upto which heat flow increases and after decreases At critical radius q becomes qmax

bull The addition of insulation always increases (Rth)condbull Rtotal= (Rth)cond + (Rth)conv

bull The addition of insulation may reduce the (Rth)conv due to increase in surface area

bull r1ltrc -heat transfer increases by adding insulation till r1=rc If

insulation thickness is further increased the rate of heat loss will decrease

r1 gt rc

- heat transfer decreases by adding insulation above the rc

Plane wall- Insulation

thickness -study

Cylinder amp Sphere- Insulation thickness -

study

Problem5bull A 3mm diameter and 5 mm long electric wire is tightly

wrapped with a 2 mm thick plastic cover whose K=015 wmoC Electrical measurements indicate that a current of 10 A passes through the wire and there is a voltage drop of 8 V along the wire If the insulated wire is exposed to a medium at Ta=30oC with a heat transfer coefficient of h=12 wm2oC Determine the temperature at the interface of the wire and the plastic cover in steady operation Also determine whether doubling the thickness of the plastic cover will increase or decrease this interface temperature

dx

TdzdyKQ xx )(

dxQx

QQ xxdxx ][

dxdx

TdzdyK

xQQdQ xdxxxx ])([)(

)1()]([)(

ddzdydx

x

TK

xdQ xx

GENERAL HEAT CONDUCTION EQUATION IN CARTESIAN COORDINATES

Consider an unsteady (transient) three dimensional heat conduction with internal heat generation in heterogeneous amp anisotropic PLANE WALL

Cartesian co-ordinates x y z

Elemental length dxdydz

Consider an elemental volume ABCDEFGH = (dxdydz)

Let Temperature distribution is a function of T=T(xyz)

Let

=internal heat generation per unit volume per unit time (W m3)q= rate of heat flow (W)Q= qdt=total heat flow (J)r=mass density of the material element(kgm3)Cp=specific heat of the material element (Jkg K)

gq

Plane (y - z) lsquoxrsquo direction

Quantity of heat leaving from the face EFGH in lsquoXrsquo direction

Quantity of heat flowing into the face ABCD in lsquoXrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoXrsquo direction

dy

TdzdxKQ yy )(

dyQy

QQ yydyy ][

dydy

TdzdxK

yQQdQ ydyyyy ])([)(

)2()]([)(

ddzdydx

y

TK

ydQ yy

Plane (x - z) lsquoyrsquo direction

Quantity of heat flowing into the face ABFE in lsquoYrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoYrsquo direction

dz

TdydxKQ zz )(

dzQz

QQ zzdzz ][

dzdz

TdydxK

zQQdQ zdzzzz ])([)(

)3()]([)(

ddzdydx

z

TK

zdQ zz

Plane (x - y) lsquozrsquo direction

Quantity of heat flowing into the face ADHE in lsquoZrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoZrsquo direction

Quantity of heat leaving from the face DCGH in lsquoYrsquo direction

Quantity of heat leaving from the face BCGF in lsquoZrsquo direction

)4()()(

ddzdydxqQgg

Total amount of internal heat generation in the element

Ein + Egen = Eout+Est (Ein- Eout)+ Egen=Est

T

Cddzdydxddzdydxqddzdydxz

TK

zy

TK

yx

TK

x pgzyx )()())](()()([

T

pC

gq

z

TzK

zy

Ty

Kyx

TxK

x][][][

stgzyx QQdQdQdQ

)(

)5()()(

dT

CdzdydxQ pst

Total heat energy stored in the element

(ie thermal energy of the element can be increased because of heat accumulation in the element due to heat flow amp internal heat generation)

)( volumem d

TmCQ pst )(

T

CqTK pg)(

KKKyKx z If material is homogeneous amp isotropic

T

pC

gq

z

T

zy

T

yx

T

xK ][][][

T

Kp

C

Kgq

z

T

y

T

x

T2

2

2

2

2

2

T

Kgq

z

T

y

T

x

T 12

2

2

2

2

2

Thermal diffusivity pC

K

T

K

qT g

12

according to thermodynamics first law

)( zr )( dzrddr

)( dzrddr

Cylindrical polar co-ordinates

Elemental length

Consider an elemental volume ABCDEFGH =

GENERAL HEAT CONDUCTION EQUATION IN CYLINDRICAL COORDINATES

gq

gq

Consider an unsteady (transient) three dimensional heat conduction with internal heat generation in heterogeneous amp anisotropic cylindrical geometries (eg rods and pipes)

)( zrTLetTemperature distribution is a function of T=

Let

=internal heat generation per unit volume per unit time (W m3)q= rate of heat flow (W)Q= qdt=total heat flow (J)r=mass density of the material element(kgm3)Cp=specific heat of the material element (Jkg K)

planerdirectionz

planezrdirection

planezdirectionr

)(

)(

)(

LetHeat flow in

drdr

TdzrdK

rdQ

QQdQ

rr

drrrr

])([

)1(][

r

Tr

rddrdzdKdQ rr

Heat accumulation in the element due to heat flow by conduction in lsquorrsquo direction

rddr

TdzdrK

rdQ

QQdQ d

])([

][

2

T

r

ddrdzrdKdQ

)2(][

2

2

2

T

r

ddrdzrdKdQ

Heat accumulation in the element due to heat flow by conduction in lsquoqrsquo direction

Plane (z- q ) lsquorrsquo direction

dr

TdzrdKQ rr

)(

drQr

QQ rrdrr ][

Quantity of heat leaving from the face EFGH in lsquorrsquo direction

Quantity of heat flowing into the face ABCD in lsquorrsquo direction

Plane (r-z) lsquoq lsquodirection

Quantity of heat leaving from the face AEGC in lsquoqrsquo direction

d

r

TdzdrKQ )(

rdQ

rQQ d ][

Quantity of heat flowing into the face BFHD in lsquoqrsquo direction

)3())((2

2

z

TdzdrrdKdQ zz

)5()()(

dT

CdzdrrdQ pst

)4()()(

ddzdrrdqQgg

Total amount of internal heat generation in the element

Total heat energy stored in the element

(ie thermal energy of the element can be increased because of heat accumulation in the element due to heat flow amp internal heat generation)

according to thermodynamics first law

Ein + Egen = Eout + Est (Ein ndash Eout) + Egen = Est

)( volumem

)()(z

T

zdzdrrdKdQ zz

Heat accumulation in the element due to heat flow by conduction in lsquozrsquo direction

dzdz

TdrrdK

zdQ

QQdQ

zz

dzzzz

])([

d

TmCQ pst )(

Plane (r-q) lsquozrsquo direction

dz

TdrrdKQ zz )(

dzQz

QQ zzdzz ][ Quantity of heat leaving from the face CDHG in lsquozrsquo direction

Quantity of heat flowing into the face ABFE in lsquozrsquo direction

stgzr QQdQdQdQ

)(

r

r ][)(rT

rr

ddrdzdKr

][)(

2

2

2

T

r

ddrdzrdK ))((2

2

z

TddzdrrdK z

ddzdrrdq g )(

dT

Cdzdrrd p )(

+ + +

=

T

CqTK pg)(KKKK zr

[gq

)(1

r

Tr

rrK r

)(1

2

2

2 T

rK )(

2

2

z

TK z

)( ddzdrrd )(

ddzdrrd

TC p

+ + + ] =

If material is homogeneous amp isotropic

)(1

r

Tr

rr

)(1

2

2

2 T

r)(

2

2

z

T

K

q g

T

K

C p

)(1

r

Tr

rr

)(1

2

2

2 T

r )(2

2

z

T

K

q g

T

1

+ + + =

)(1

r

Tr

rrK r

)(1

2

2

2 T

rK )(

2

2

z

TK z

gq

T

C p + + + =

+ + + = Thermal diffusivity

T

K

qT g

12

pC

K

(Ein ndash Eout) + Egen = Est

)( rT

)( r

)sin( drrddr)sin( drrddr

planerdirection

planerdirection

planedirectionr

)(

)(

)(

Spherical polar co-ordinates

Elemental length

Consider an elemental volume ABCDEFGH =

GENERAL HEAT CONDUCTION EQUATION IN SPHERICAL COORDINATES

Let

=internal heat generation per unit volume per unit time (W m3)q= rate of heat flow (W)Q= qdt=total heat flow (J)r=mass density of the material element(kgm3)Cp=specific heat of the material element (Jkg K)

gq

Consider an unsteady (transient) three dimensional heat conduction with internal heat generation in heterogeneous amp anisotropic spherical geometries

LetTemperature distribution is a function of T=

X

Y

Z

drT

drrdKQ rr )sin(

drQr

QQ rrdrr ][

drdr

TdrrdK

rdQ

QQdQ

rr

drrrr

])sin([

)1(]([sin 2

rT

rr

ddrddKdQ rr

Plane ( -q f) lsquorrsquo directionQuantity of heat flowing into the face ABCD in lsquorrsquo direction

Quantity of heat leaving from the face EFGH in lsquorrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquorrsquo direction

drT

drdrKQ )sin(

rdQ

rQQ d ][

rddr

TdrdrK

rdQ

QQdQ d

])sin([

])[(sin

2

T

r

drdrdrKdQ

)2(])(sin[

2

T

r

drdrddrKdQ

Quantity of heat leaving from the face BFHD in lsquoqrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoqrsquo direction

X

Y

Z

Plane (r- f ) lsquoq lsquodirection

Quantity of heat flowing into the face AEGC in lsquoqrsquo direction

dr

TdrrdKQ

sin)(

drQ

rQQ d sin][

sin

drdr

TdrrdK

rdQ

QQdQ d

sin]sin

)([sin

][sinsin

)sin(

T

rr

ddrdrrdKdQ

)3()(sin

)sin(2

2

22

2

T

r

ddddrrKdQ

Plane (r- ) q lsquofrsquo directionQuantity of heat flowing into the face ABFE in lsquofrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquozrsquo direction

)4()sin()(

ddrrddrqQgg

)5()sin()(

dT

CdrrddrQ pst

Total amount of internal heat generation in the element

Total heat energy stored in the element

(Ein ndash Eout) + Egen = Est

d

TmCQ pst )(

)( volumem

Quantity of heat leaving from the face CDHG in lsquofrsquo direction

according to thermodynamics first law

Ein + Egen = Eout + Est

X

Y

Z

stgr QQdQdQdQ

)(

]([sin 2

r

Tr

rddrddKr

])(sin[

2

T

rdrdrddrK

)(sin

)sin(2

2

22

2

T

r

ddddrrK

ddrrddrq g )sin(

dT

Cdrrddr p )sin(

2

2

r

r

sin

sin

]([ 22 r

Tr

rr

Kr

])(sin[sin2

TrK

)(sin 2

2

22

T

r

Kgq

T

Cp

]([1

[ 22 r

Tr

rrK

])(sin[sin1

2

T

r)](

sin

12

2

22 T

r gq

T

Cp

]([1

[ 22 r

Tr

rr

])(sin[sin1

2

T

r)](

sin

12

2

22 T

rK

q g

T

K

Cp

]([1

[ 22 r

Tr

rr

])(sin[sin1

2

T

r)](

sin

12

2

22 T

rK

q g

T1

KKKKr If material is homogeneous amp isotropic

TCqTK pg)(

T

K

qT g

12 pC

K

Thermal diffusivity

(Ein ndash Eout) + Egen = Est

General Heat Conduction Equation

PLANE WALL CYLINDER SPHERE

Temperature profile

Plane wall SphereCylinder

bull A copper conductor (k = 380 WmoC resistivity ρ = 2x10-8 Ω m) having inner and outer radii 10 cm and 225 cm respectively is carrying a current density of 4800 amperes cm2 the conductor is internally cooled and a constant temperature of 65oC is maintained at the inner surface and there is no heat transfer through the insulation surrounding the conductor

Determinebull (i) The maximum temperature of the conductor and the radius at

which it occursbull (ii) The internal heat transfer rate

bull A chemical reaction takes place in a packed bed (k=06 WmoC) between two coaxial cylinders with radii 15 mm and 45 mm the inner surface is at 580oC and it is insulated Assuming the reaction rate of 055 MWm3 in the reactor volume find the temperature at the outer surface of the reactor

bull A current of 300 amperes passes through a stainless steel wire of 25 mm diameter and k=20 WmoC The resistivity of the wire is 70x10-8 ohm-m and length of the wire is 2 m if the wire is submerged in a fluid maintained at 50oC and h at the wire surface is 4000 Wm2oC calculate the steady state temperature at the centre and at the surface of the wire

bull An approximately spherical shaped orange ( k=023 WmoC) 90 mm diameter undergoes ripping process and generates 5100 Wm3 of energy If external surface the orange is at 8oC determine

bull (i) temperature at the centre of the orange and

bull (ii) heat flow from the outer surface of the orange

bull FINS External surfaces to used to cooling area and hence convection

heat transfer can be improvedbull APPLICATIONS IC engine cylinder ( air cooling) Heat exchangers automobile

radiators computer chip refrigerators electric motors generators and transformers etc

bull ASSUMPTIONS MADE IN ANALYSIS OF HEAT TRANFER THROUGH FIN

1 Fin material is homogenous and isotropic constthermal conductivity

2 Temperature at any cs of the fin is uniform3 lsquohrsquo is uniform over entire surface4 No internal heat generation5 Thermal contact resistance is negligible6 Radiation heat transfer is negligible7 Heat conduction is steady state and one dimensional

bull Problem 1

bull A stainless steel fin (k=20 WmK) having a diameter of 20 mm and a length of 01 m is attached to a wall at 300oC The ambient temperature is 50oC and the heat transfer coefficient is 10 Wm2K The fin tip is insulated Determine

bull (A) The rate of heat dissipation from the finbull (B) The temperature at the fin tipbull (C) The rate of heat transfer from the wall area covered by the fin if

the fin was not usedbull (D) The heat transfer rate from the same fin geometry if the

stainless steel fin is replaced by a fictitious fin with infinite thermal conductivity

bull Problem2bull One end of a long rod is inserted into a furnace while the other

projects into ambient temperature Under steady state the temperature of the rod is measured at two points 75mm apart and found to be 125oC and 885oC respectively while the ambient temperature is 20oC If the rod is 25mm in diameter and h is 2336 Wm2K

bull Find the thermal conductivity of the rod material

  • HEAT AND MASS TRANSFER
  • Thermodynamics Vs Heat transfer
  • MODE OF HEAT TRANSFER
  • MODE OF HEAT TRANSFER (2)
  • CONDUCTION- MECHANISM
  • CONCECTION-MECHANISM
  • RADIATION-MECHANISM
  • Conduction- Convection- Radiation
  • Slide 9
  • Slide 10
  • Fourierrsquos Law of Heat Conduction
  • Slide 12
  • Newtonrsquos Law of Cooling
  • Laws of Thermal Radiation
  • LAWS FOR HEAT TRANSFER
  • ELECTRICAL ampTHERMAL ANALOGY
  • Slide 17
  • Slide 18
  • PLANE WALL-UNIDIRECTIONAL
  • PLANE WALL -SERIES
  • PLANE WALL IN SERIES
  • PLANE WALL -PARALLEL
  • PLANE WALL-PARALLEL
  • PLANE WALL-COMBINATION
  • PLANE WALL-FILM RESISTANCE
  • PLANE WALL- THERMAL CONTACT RESISTANCE
  • PLANE WALL- THERMAL CONTACT RESISTANCE (2)
  • CYLINDER-SOLID
  • HEAT TRANSFER IN SOLID CYLINDER
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER-COMPOSITE
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER-MULTY LAYER
  • HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER
  • HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER (2)
  • HEAT TRANSFER IN SPHERE
  • HEAT TRANSFER IN SPHERE -MULTY LAYER
  • HEAT TRANSFER IN SPHERE -MULTY LAYER (2)
  • Slide 38
  • Slide 39
  • Slide 40
  • Slide 41
  • CRITICAL RADIUS
  • Plane wall- Insulation thickness -study
  • Problem5
  • Slide 45
  • Slide 46
  • Slide 47
  • GENERAL HEAT CONDUCTION EQUATION IN CYLINDRICAL COORDINATES
  • Slide 49
  • Slide 50
  • Slide 51
  • GENERAL HEAT CONDUCTION EQUATION IN SPHERICAL COORDINATES
  • Slide 53
  • Slide 54
  • Slide 55
  • General Heat Conduction Equation
  • Temperature profile
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Slide 67
Page 18: hmt chap 1

PLANE WALL -PARALLEL

PLANE WALL-PARALLEL

PLANE WALL-COMBINATION

PLANE WALL-FILM RESISTANCE

PLANE WALL- THERMAL CONTACT RESISTANCE

PLANE WALL- THERMAL CONTACT RESISTANCE

At the joining surface of the two slabs air is trapped in voids due to surface irregularities

CYLINDER-SOLID

HEAT TRANSFER IN SOLID CYLINDER

HEAT TRANSFER RATE IN HOLLOW CYLINDER

HEAT TRANSFER RATE IN HOLLOW CYLINDER-COMPOSITE

HEAT TRANSFER RATE IN HOLLOW CYLINDER-MULTY LAYER

HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER

HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER

HEAT TRANSFER IN SPHERE

HEAT TRANSFER IN SPHERE -MULTY LAYER

HEAT TRANSFER IN SPHERE -MULTY LAYER

bull Problem 1 The wall in a furnace consists of 125 mm thick refractory bricks and

125 mm thick insulating fire bricks separated by an air gap A 12 mm thick plaster covers the outer wall The inner surface of wall is at 1100oC and the ambient temperature 25oC the heat transfer coefficient on the outer wall to the air is 17 Wm2K and the resistance to heat flow of the air gap is 016 KW the thermal conductivities of refractory brick insulating firebrick and plaster are 16 03 and 014 WmK respectively

Calculatebull (a) the rate of heat loss per unit area of the wall surfacebull (b) the interface temperatures throughout the wallbull (c) the temperature at the outside surface of the wall

bull Problem 2Two large aluminum plates (k=240 WmK) each 2 cm thick with 10 mm surface roughness are placed in contact under 105 Nm2 pressure in air The temperature at the outside surfaces are 390 oC and 406 oCCalculate

A) the heat flux B) the temperature drop due to the contact resistance C) the contact temperatures Thermal contact resistance with air as the interfacial fluid for

10 mm roughness is 27510-4 m2KW

bull Problem 3bull Steam at 350 oC flowing in a pipe (k=80 WmK) of 5 cm

inner diameter and 56 cm outer diameter is covered with 3 cm thick insulation of k=005 WmK Heat is lost to surroundings at 5 oC by natural convection and radiation the combined h being 20 Wm2K Taking the heat transfer coefficient inside the pipe as 60 Wm2K

bull Determinebull A) the rate of heat loss from the steam per unit length of

the pipebull B) the temperature drop across the pipe and the insulation

bull PROBLEM4bull A nuclear reactor has a sperical pressure vessel 750 mm inside

diameter 80 mm wall thickness The temperature at the inner surface is 500 oC If the tempertarue at the outer surface is 495 oC

bull Calculatea) The rate of heat loss through the metal wallb) If the rate of heat loss is to limited to 305Whow much inulation

thickness needs to be applied Thermal conductivity of steel wall=46 WmoC Thermal conductivity of insulation=004 WmoC Film coefficient of heat tranfer at inner and outer surface are 200

Wm2 oC and 15 Wm2 oC Ambient air temperature=28 oC

CRITICAL RADIUS

bull The radius upto which heat flow increases and after decreases At critical radius q becomes qmax

bull The addition of insulation always increases (Rth)condbull Rtotal= (Rth)cond + (Rth)conv

bull The addition of insulation may reduce the (Rth)conv due to increase in surface area

bull r1ltrc -heat transfer increases by adding insulation till r1=rc If

insulation thickness is further increased the rate of heat loss will decrease

r1 gt rc

- heat transfer decreases by adding insulation above the rc

Plane wall- Insulation

thickness -study

Cylinder amp Sphere- Insulation thickness -

study

Problem5bull A 3mm diameter and 5 mm long electric wire is tightly

wrapped with a 2 mm thick plastic cover whose K=015 wmoC Electrical measurements indicate that a current of 10 A passes through the wire and there is a voltage drop of 8 V along the wire If the insulated wire is exposed to a medium at Ta=30oC with a heat transfer coefficient of h=12 wm2oC Determine the temperature at the interface of the wire and the plastic cover in steady operation Also determine whether doubling the thickness of the plastic cover will increase or decrease this interface temperature

dx

TdzdyKQ xx )(

dxQx

QQ xxdxx ][

dxdx

TdzdyK

xQQdQ xdxxxx ])([)(

)1()]([)(

ddzdydx

x

TK

xdQ xx

GENERAL HEAT CONDUCTION EQUATION IN CARTESIAN COORDINATES

Consider an unsteady (transient) three dimensional heat conduction with internal heat generation in heterogeneous amp anisotropic PLANE WALL

Cartesian co-ordinates x y z

Elemental length dxdydz

Consider an elemental volume ABCDEFGH = (dxdydz)

Let Temperature distribution is a function of T=T(xyz)

Let

=internal heat generation per unit volume per unit time (W m3)q= rate of heat flow (W)Q= qdt=total heat flow (J)r=mass density of the material element(kgm3)Cp=specific heat of the material element (Jkg K)

gq

Plane (y - z) lsquoxrsquo direction

Quantity of heat leaving from the face EFGH in lsquoXrsquo direction

Quantity of heat flowing into the face ABCD in lsquoXrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoXrsquo direction

dy

TdzdxKQ yy )(

dyQy

QQ yydyy ][

dydy

TdzdxK

yQQdQ ydyyyy ])([)(

)2()]([)(

ddzdydx

y

TK

ydQ yy

Plane (x - z) lsquoyrsquo direction

Quantity of heat flowing into the face ABFE in lsquoYrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoYrsquo direction

dz

TdydxKQ zz )(

dzQz

QQ zzdzz ][

dzdz

TdydxK

zQQdQ zdzzzz ])([)(

)3()]([)(

ddzdydx

z

TK

zdQ zz

Plane (x - y) lsquozrsquo direction

Quantity of heat flowing into the face ADHE in lsquoZrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoZrsquo direction

Quantity of heat leaving from the face DCGH in lsquoYrsquo direction

Quantity of heat leaving from the face BCGF in lsquoZrsquo direction

)4()()(

ddzdydxqQgg

Total amount of internal heat generation in the element

Ein + Egen = Eout+Est (Ein- Eout)+ Egen=Est

T

Cddzdydxddzdydxqddzdydxz

TK

zy

TK

yx

TK

x pgzyx )()())](()()([

T

pC

gq

z

TzK

zy

Ty

Kyx

TxK

x][][][

stgzyx QQdQdQdQ

)(

)5()()(

dT

CdzdydxQ pst

Total heat energy stored in the element

(ie thermal energy of the element can be increased because of heat accumulation in the element due to heat flow amp internal heat generation)

)( volumem d

TmCQ pst )(

T

CqTK pg)(

KKKyKx z If material is homogeneous amp isotropic

T

pC

gq

z

T

zy

T

yx

T

xK ][][][

T

Kp

C

Kgq

z

T

y

T

x

T2

2

2

2

2

2

T

Kgq

z

T

y

T

x

T 12

2

2

2

2

2

Thermal diffusivity pC

K

T

K

qT g

12

according to thermodynamics first law

)( zr )( dzrddr

)( dzrddr

Cylindrical polar co-ordinates

Elemental length

Consider an elemental volume ABCDEFGH =

GENERAL HEAT CONDUCTION EQUATION IN CYLINDRICAL COORDINATES

gq

gq

Consider an unsteady (transient) three dimensional heat conduction with internal heat generation in heterogeneous amp anisotropic cylindrical geometries (eg rods and pipes)

)( zrTLetTemperature distribution is a function of T=

Let

=internal heat generation per unit volume per unit time (W m3)q= rate of heat flow (W)Q= qdt=total heat flow (J)r=mass density of the material element(kgm3)Cp=specific heat of the material element (Jkg K)

planerdirectionz

planezrdirection

planezdirectionr

)(

)(

)(

LetHeat flow in

drdr

TdzrdK

rdQ

QQdQ

rr

drrrr

])([

)1(][

r

Tr

rddrdzdKdQ rr

Heat accumulation in the element due to heat flow by conduction in lsquorrsquo direction

rddr

TdzdrK

rdQ

QQdQ d

])([

][

2

T

r

ddrdzrdKdQ

)2(][

2

2

2

T

r

ddrdzrdKdQ

Heat accumulation in the element due to heat flow by conduction in lsquoqrsquo direction

Plane (z- q ) lsquorrsquo direction

dr

TdzrdKQ rr

)(

drQr

QQ rrdrr ][

Quantity of heat leaving from the face EFGH in lsquorrsquo direction

Quantity of heat flowing into the face ABCD in lsquorrsquo direction

Plane (r-z) lsquoq lsquodirection

Quantity of heat leaving from the face AEGC in lsquoqrsquo direction

d

r

TdzdrKQ )(

rdQ

rQQ d ][

Quantity of heat flowing into the face BFHD in lsquoqrsquo direction

)3())((2

2

z

TdzdrrdKdQ zz

)5()()(

dT

CdzdrrdQ pst

)4()()(

ddzdrrdqQgg

Total amount of internal heat generation in the element

Total heat energy stored in the element

(ie thermal energy of the element can be increased because of heat accumulation in the element due to heat flow amp internal heat generation)

according to thermodynamics first law

Ein + Egen = Eout + Est (Ein ndash Eout) + Egen = Est

)( volumem

)()(z

T

zdzdrrdKdQ zz

Heat accumulation in the element due to heat flow by conduction in lsquozrsquo direction

dzdz

TdrrdK

zdQ

QQdQ

zz

dzzzz

])([

d

TmCQ pst )(

Plane (r-q) lsquozrsquo direction

dz

TdrrdKQ zz )(

dzQz

QQ zzdzz ][ Quantity of heat leaving from the face CDHG in lsquozrsquo direction

Quantity of heat flowing into the face ABFE in lsquozrsquo direction

stgzr QQdQdQdQ

)(

r

r ][)(rT

rr

ddrdzdKr

][)(

2

2

2

T

r

ddrdzrdK ))((2

2

z

TddzdrrdK z

ddzdrrdq g )(

dT

Cdzdrrd p )(

+ + +

=

T

CqTK pg)(KKKK zr

[gq

)(1

r

Tr

rrK r

)(1

2

2

2 T

rK )(

2

2

z

TK z

)( ddzdrrd )(

ddzdrrd

TC p

+ + + ] =

If material is homogeneous amp isotropic

)(1

r

Tr

rr

)(1

2

2

2 T

r)(

2

2

z

T

K

q g

T

K

C p

)(1

r

Tr

rr

)(1

2

2

2 T

r )(2

2

z

T

K

q g

T

1

+ + + =

)(1

r

Tr

rrK r

)(1

2

2

2 T

rK )(

2

2

z

TK z

gq

T

C p + + + =

+ + + = Thermal diffusivity

T

K

qT g

12

pC

K

(Ein ndash Eout) + Egen = Est

)( rT

)( r

)sin( drrddr)sin( drrddr

planerdirection

planerdirection

planedirectionr

)(

)(

)(

Spherical polar co-ordinates

Elemental length

Consider an elemental volume ABCDEFGH =

GENERAL HEAT CONDUCTION EQUATION IN SPHERICAL COORDINATES

Let

=internal heat generation per unit volume per unit time (W m3)q= rate of heat flow (W)Q= qdt=total heat flow (J)r=mass density of the material element(kgm3)Cp=specific heat of the material element (Jkg K)

gq

Consider an unsteady (transient) three dimensional heat conduction with internal heat generation in heterogeneous amp anisotropic spherical geometries

LetTemperature distribution is a function of T=

X

Y

Z

drT

drrdKQ rr )sin(

drQr

QQ rrdrr ][

drdr

TdrrdK

rdQ

QQdQ

rr

drrrr

])sin([

)1(]([sin 2

rT

rr

ddrddKdQ rr

Plane ( -q f) lsquorrsquo directionQuantity of heat flowing into the face ABCD in lsquorrsquo direction

Quantity of heat leaving from the face EFGH in lsquorrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquorrsquo direction

drT

drdrKQ )sin(

rdQ

rQQ d ][

rddr

TdrdrK

rdQ

QQdQ d

])sin([

])[(sin

2

T

r

drdrdrKdQ

)2(])(sin[

2

T

r

drdrddrKdQ

Quantity of heat leaving from the face BFHD in lsquoqrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoqrsquo direction

X

Y

Z

Plane (r- f ) lsquoq lsquodirection

Quantity of heat flowing into the face AEGC in lsquoqrsquo direction

dr

TdrrdKQ

sin)(

drQ

rQQ d sin][

sin

drdr

TdrrdK

rdQ

QQdQ d

sin]sin

)([sin

][sinsin

)sin(

T

rr

ddrdrrdKdQ

)3()(sin

)sin(2

2

22

2

T

r

ddddrrKdQ

Plane (r- ) q lsquofrsquo directionQuantity of heat flowing into the face ABFE in lsquofrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquozrsquo direction

)4()sin()(

ddrrddrqQgg

)5()sin()(

dT

CdrrddrQ pst

Total amount of internal heat generation in the element

Total heat energy stored in the element

(Ein ndash Eout) + Egen = Est

d

TmCQ pst )(

)( volumem

Quantity of heat leaving from the face CDHG in lsquofrsquo direction

according to thermodynamics first law

Ein + Egen = Eout + Est

X

Y

Z

stgr QQdQdQdQ

)(

]([sin 2

r

Tr

rddrddKr

])(sin[

2

T

rdrdrddrK

)(sin

)sin(2

2

22

2

T

r

ddddrrK

ddrrddrq g )sin(

dT

Cdrrddr p )sin(

2

2

r

r

sin

sin

]([ 22 r

Tr

rr

Kr

])(sin[sin2

TrK

)(sin 2

2

22

T

r

Kgq

T

Cp

]([1

[ 22 r

Tr

rrK

])(sin[sin1

2

T

r)](

sin

12

2

22 T

r gq

T

Cp

]([1

[ 22 r

Tr

rr

])(sin[sin1

2

T

r)](

sin

12

2

22 T

rK

q g

T

K

Cp

]([1

[ 22 r

Tr

rr

])(sin[sin1

2

T

r)](

sin

12

2

22 T

rK

q g

T1

KKKKr If material is homogeneous amp isotropic

TCqTK pg)(

T

K

qT g

12 pC

K

Thermal diffusivity

(Ein ndash Eout) + Egen = Est

General Heat Conduction Equation

PLANE WALL CYLINDER SPHERE

Temperature profile

Plane wall SphereCylinder

bull A copper conductor (k = 380 WmoC resistivity ρ = 2x10-8 Ω m) having inner and outer radii 10 cm and 225 cm respectively is carrying a current density of 4800 amperes cm2 the conductor is internally cooled and a constant temperature of 65oC is maintained at the inner surface and there is no heat transfer through the insulation surrounding the conductor

Determinebull (i) The maximum temperature of the conductor and the radius at

which it occursbull (ii) The internal heat transfer rate

bull A chemical reaction takes place in a packed bed (k=06 WmoC) between two coaxial cylinders with radii 15 mm and 45 mm the inner surface is at 580oC and it is insulated Assuming the reaction rate of 055 MWm3 in the reactor volume find the temperature at the outer surface of the reactor

bull A current of 300 amperes passes through a stainless steel wire of 25 mm diameter and k=20 WmoC The resistivity of the wire is 70x10-8 ohm-m and length of the wire is 2 m if the wire is submerged in a fluid maintained at 50oC and h at the wire surface is 4000 Wm2oC calculate the steady state temperature at the centre and at the surface of the wire

bull An approximately spherical shaped orange ( k=023 WmoC) 90 mm diameter undergoes ripping process and generates 5100 Wm3 of energy If external surface the orange is at 8oC determine

bull (i) temperature at the centre of the orange and

bull (ii) heat flow from the outer surface of the orange

bull FINS External surfaces to used to cooling area and hence convection

heat transfer can be improvedbull APPLICATIONS IC engine cylinder ( air cooling) Heat exchangers automobile

radiators computer chip refrigerators electric motors generators and transformers etc

bull ASSUMPTIONS MADE IN ANALYSIS OF HEAT TRANFER THROUGH FIN

1 Fin material is homogenous and isotropic constthermal conductivity

2 Temperature at any cs of the fin is uniform3 lsquohrsquo is uniform over entire surface4 No internal heat generation5 Thermal contact resistance is negligible6 Radiation heat transfer is negligible7 Heat conduction is steady state and one dimensional

bull Problem 1

bull A stainless steel fin (k=20 WmK) having a diameter of 20 mm and a length of 01 m is attached to a wall at 300oC The ambient temperature is 50oC and the heat transfer coefficient is 10 Wm2K The fin tip is insulated Determine

bull (A) The rate of heat dissipation from the finbull (B) The temperature at the fin tipbull (C) The rate of heat transfer from the wall area covered by the fin if

the fin was not usedbull (D) The heat transfer rate from the same fin geometry if the

stainless steel fin is replaced by a fictitious fin with infinite thermal conductivity

bull Problem2bull One end of a long rod is inserted into a furnace while the other

projects into ambient temperature Under steady state the temperature of the rod is measured at two points 75mm apart and found to be 125oC and 885oC respectively while the ambient temperature is 20oC If the rod is 25mm in diameter and h is 2336 Wm2K

bull Find the thermal conductivity of the rod material

  • HEAT AND MASS TRANSFER
  • Thermodynamics Vs Heat transfer
  • MODE OF HEAT TRANSFER
  • MODE OF HEAT TRANSFER (2)
  • CONDUCTION- MECHANISM
  • CONCECTION-MECHANISM
  • RADIATION-MECHANISM
  • Conduction- Convection- Radiation
  • Slide 9
  • Slide 10
  • Fourierrsquos Law of Heat Conduction
  • Slide 12
  • Newtonrsquos Law of Cooling
  • Laws of Thermal Radiation
  • LAWS FOR HEAT TRANSFER
  • ELECTRICAL ampTHERMAL ANALOGY
  • Slide 17
  • Slide 18
  • PLANE WALL-UNIDIRECTIONAL
  • PLANE WALL -SERIES
  • PLANE WALL IN SERIES
  • PLANE WALL -PARALLEL
  • PLANE WALL-PARALLEL
  • PLANE WALL-COMBINATION
  • PLANE WALL-FILM RESISTANCE
  • PLANE WALL- THERMAL CONTACT RESISTANCE
  • PLANE WALL- THERMAL CONTACT RESISTANCE (2)
  • CYLINDER-SOLID
  • HEAT TRANSFER IN SOLID CYLINDER
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER-COMPOSITE
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER-MULTY LAYER
  • HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER
  • HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER (2)
  • HEAT TRANSFER IN SPHERE
  • HEAT TRANSFER IN SPHERE -MULTY LAYER
  • HEAT TRANSFER IN SPHERE -MULTY LAYER (2)
  • Slide 38
  • Slide 39
  • Slide 40
  • Slide 41
  • CRITICAL RADIUS
  • Plane wall- Insulation thickness -study
  • Problem5
  • Slide 45
  • Slide 46
  • Slide 47
  • GENERAL HEAT CONDUCTION EQUATION IN CYLINDRICAL COORDINATES
  • Slide 49
  • Slide 50
  • Slide 51
  • GENERAL HEAT CONDUCTION EQUATION IN SPHERICAL COORDINATES
  • Slide 53
  • Slide 54
  • Slide 55
  • General Heat Conduction Equation
  • Temperature profile
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Slide 67
Page 19: hmt chap 1

PLANE WALL-PARALLEL

PLANE WALL-COMBINATION

PLANE WALL-FILM RESISTANCE

PLANE WALL- THERMAL CONTACT RESISTANCE

PLANE WALL- THERMAL CONTACT RESISTANCE

At the joining surface of the two slabs air is trapped in voids due to surface irregularities

CYLINDER-SOLID

HEAT TRANSFER IN SOLID CYLINDER

HEAT TRANSFER RATE IN HOLLOW CYLINDER

HEAT TRANSFER RATE IN HOLLOW CYLINDER-COMPOSITE

HEAT TRANSFER RATE IN HOLLOW CYLINDER-MULTY LAYER

HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER

HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER

HEAT TRANSFER IN SPHERE

HEAT TRANSFER IN SPHERE -MULTY LAYER

HEAT TRANSFER IN SPHERE -MULTY LAYER

bull Problem 1 The wall in a furnace consists of 125 mm thick refractory bricks and

125 mm thick insulating fire bricks separated by an air gap A 12 mm thick plaster covers the outer wall The inner surface of wall is at 1100oC and the ambient temperature 25oC the heat transfer coefficient on the outer wall to the air is 17 Wm2K and the resistance to heat flow of the air gap is 016 KW the thermal conductivities of refractory brick insulating firebrick and plaster are 16 03 and 014 WmK respectively

Calculatebull (a) the rate of heat loss per unit area of the wall surfacebull (b) the interface temperatures throughout the wallbull (c) the temperature at the outside surface of the wall

bull Problem 2Two large aluminum plates (k=240 WmK) each 2 cm thick with 10 mm surface roughness are placed in contact under 105 Nm2 pressure in air The temperature at the outside surfaces are 390 oC and 406 oCCalculate

A) the heat flux B) the temperature drop due to the contact resistance C) the contact temperatures Thermal contact resistance with air as the interfacial fluid for

10 mm roughness is 27510-4 m2KW

bull Problem 3bull Steam at 350 oC flowing in a pipe (k=80 WmK) of 5 cm

inner diameter and 56 cm outer diameter is covered with 3 cm thick insulation of k=005 WmK Heat is lost to surroundings at 5 oC by natural convection and radiation the combined h being 20 Wm2K Taking the heat transfer coefficient inside the pipe as 60 Wm2K

bull Determinebull A) the rate of heat loss from the steam per unit length of

the pipebull B) the temperature drop across the pipe and the insulation

bull PROBLEM4bull A nuclear reactor has a sperical pressure vessel 750 mm inside

diameter 80 mm wall thickness The temperature at the inner surface is 500 oC If the tempertarue at the outer surface is 495 oC

bull Calculatea) The rate of heat loss through the metal wallb) If the rate of heat loss is to limited to 305Whow much inulation

thickness needs to be applied Thermal conductivity of steel wall=46 WmoC Thermal conductivity of insulation=004 WmoC Film coefficient of heat tranfer at inner and outer surface are 200

Wm2 oC and 15 Wm2 oC Ambient air temperature=28 oC

CRITICAL RADIUS

bull The radius upto which heat flow increases and after decreases At critical radius q becomes qmax

bull The addition of insulation always increases (Rth)condbull Rtotal= (Rth)cond + (Rth)conv

bull The addition of insulation may reduce the (Rth)conv due to increase in surface area

bull r1ltrc -heat transfer increases by adding insulation till r1=rc If

insulation thickness is further increased the rate of heat loss will decrease

r1 gt rc

- heat transfer decreases by adding insulation above the rc

Plane wall- Insulation

thickness -study

Cylinder amp Sphere- Insulation thickness -

study

Problem5bull A 3mm diameter and 5 mm long electric wire is tightly

wrapped with a 2 mm thick plastic cover whose K=015 wmoC Electrical measurements indicate that a current of 10 A passes through the wire and there is a voltage drop of 8 V along the wire If the insulated wire is exposed to a medium at Ta=30oC with a heat transfer coefficient of h=12 wm2oC Determine the temperature at the interface of the wire and the plastic cover in steady operation Also determine whether doubling the thickness of the plastic cover will increase or decrease this interface temperature

dx

TdzdyKQ xx )(

dxQx

QQ xxdxx ][

dxdx

TdzdyK

xQQdQ xdxxxx ])([)(

)1()]([)(

ddzdydx

x

TK

xdQ xx

GENERAL HEAT CONDUCTION EQUATION IN CARTESIAN COORDINATES

Consider an unsteady (transient) three dimensional heat conduction with internal heat generation in heterogeneous amp anisotropic PLANE WALL

Cartesian co-ordinates x y z

Elemental length dxdydz

Consider an elemental volume ABCDEFGH = (dxdydz)

Let Temperature distribution is a function of T=T(xyz)

Let

=internal heat generation per unit volume per unit time (W m3)q= rate of heat flow (W)Q= qdt=total heat flow (J)r=mass density of the material element(kgm3)Cp=specific heat of the material element (Jkg K)

gq

Plane (y - z) lsquoxrsquo direction

Quantity of heat leaving from the face EFGH in lsquoXrsquo direction

Quantity of heat flowing into the face ABCD in lsquoXrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoXrsquo direction

dy

TdzdxKQ yy )(

dyQy

QQ yydyy ][

dydy

TdzdxK

yQQdQ ydyyyy ])([)(

)2()]([)(

ddzdydx

y

TK

ydQ yy

Plane (x - z) lsquoyrsquo direction

Quantity of heat flowing into the face ABFE in lsquoYrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoYrsquo direction

dz

TdydxKQ zz )(

dzQz

QQ zzdzz ][

dzdz

TdydxK

zQQdQ zdzzzz ])([)(

)3()]([)(

ddzdydx

z

TK

zdQ zz

Plane (x - y) lsquozrsquo direction

Quantity of heat flowing into the face ADHE in lsquoZrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoZrsquo direction

Quantity of heat leaving from the face DCGH in lsquoYrsquo direction

Quantity of heat leaving from the face BCGF in lsquoZrsquo direction

)4()()(

ddzdydxqQgg

Total amount of internal heat generation in the element

Ein + Egen = Eout+Est (Ein- Eout)+ Egen=Est

T

Cddzdydxddzdydxqddzdydxz

TK

zy

TK

yx

TK

x pgzyx )()())](()()([

T

pC

gq

z

TzK

zy

Ty

Kyx

TxK

x][][][

stgzyx QQdQdQdQ

)(

)5()()(

dT

CdzdydxQ pst

Total heat energy stored in the element

(ie thermal energy of the element can be increased because of heat accumulation in the element due to heat flow amp internal heat generation)

)( volumem d

TmCQ pst )(

T

CqTK pg)(

KKKyKx z If material is homogeneous amp isotropic

T

pC

gq

z

T

zy

T

yx

T

xK ][][][

T

Kp

C

Kgq

z

T

y

T

x

T2

2

2

2

2

2

T

Kgq

z

T

y

T

x

T 12

2

2

2

2

2

Thermal diffusivity pC

K

T

K

qT g

12

according to thermodynamics first law

)( zr )( dzrddr

)( dzrddr

Cylindrical polar co-ordinates

Elemental length

Consider an elemental volume ABCDEFGH =

GENERAL HEAT CONDUCTION EQUATION IN CYLINDRICAL COORDINATES

gq

gq

Consider an unsteady (transient) three dimensional heat conduction with internal heat generation in heterogeneous amp anisotropic cylindrical geometries (eg rods and pipes)

)( zrTLetTemperature distribution is a function of T=

Let

=internal heat generation per unit volume per unit time (W m3)q= rate of heat flow (W)Q= qdt=total heat flow (J)r=mass density of the material element(kgm3)Cp=specific heat of the material element (Jkg K)

planerdirectionz

planezrdirection

planezdirectionr

)(

)(

)(

LetHeat flow in

drdr

TdzrdK

rdQ

QQdQ

rr

drrrr

])([

)1(][

r

Tr

rddrdzdKdQ rr

Heat accumulation in the element due to heat flow by conduction in lsquorrsquo direction

rddr

TdzdrK

rdQ

QQdQ d

])([

][

2

T

r

ddrdzrdKdQ

)2(][

2

2

2

T

r

ddrdzrdKdQ

Heat accumulation in the element due to heat flow by conduction in lsquoqrsquo direction

Plane (z- q ) lsquorrsquo direction

dr

TdzrdKQ rr

)(

drQr

QQ rrdrr ][

Quantity of heat leaving from the face EFGH in lsquorrsquo direction

Quantity of heat flowing into the face ABCD in lsquorrsquo direction

Plane (r-z) lsquoq lsquodirection

Quantity of heat leaving from the face AEGC in lsquoqrsquo direction

d

r

TdzdrKQ )(

rdQ

rQQ d ][

Quantity of heat flowing into the face BFHD in lsquoqrsquo direction

)3())((2

2

z

TdzdrrdKdQ zz

)5()()(

dT

CdzdrrdQ pst

)4()()(

ddzdrrdqQgg

Total amount of internal heat generation in the element

Total heat energy stored in the element

(ie thermal energy of the element can be increased because of heat accumulation in the element due to heat flow amp internal heat generation)

according to thermodynamics first law

Ein + Egen = Eout + Est (Ein ndash Eout) + Egen = Est

)( volumem

)()(z

T

zdzdrrdKdQ zz

Heat accumulation in the element due to heat flow by conduction in lsquozrsquo direction

dzdz

TdrrdK

zdQ

QQdQ

zz

dzzzz

])([

d

TmCQ pst )(

Plane (r-q) lsquozrsquo direction

dz

TdrrdKQ zz )(

dzQz

QQ zzdzz ][ Quantity of heat leaving from the face CDHG in lsquozrsquo direction

Quantity of heat flowing into the face ABFE in lsquozrsquo direction

stgzr QQdQdQdQ

)(

r

r ][)(rT

rr

ddrdzdKr

][)(

2

2

2

T

r

ddrdzrdK ))((2

2

z

TddzdrrdK z

ddzdrrdq g )(

dT

Cdzdrrd p )(

+ + +

=

T

CqTK pg)(KKKK zr

[gq

)(1

r

Tr

rrK r

)(1

2

2

2 T

rK )(

2

2

z

TK z

)( ddzdrrd )(

ddzdrrd

TC p

+ + + ] =

If material is homogeneous amp isotropic

)(1

r

Tr

rr

)(1

2

2

2 T

r)(

2

2

z

T

K

q g

T

K

C p

)(1

r

Tr

rr

)(1

2

2

2 T

r )(2

2

z

T

K

q g

T

1

+ + + =

)(1

r

Tr

rrK r

)(1

2

2

2 T

rK )(

2

2

z

TK z

gq

T

C p + + + =

+ + + = Thermal diffusivity

T

K

qT g

12

pC

K

(Ein ndash Eout) + Egen = Est

)( rT

)( r

)sin( drrddr)sin( drrddr

planerdirection

planerdirection

planedirectionr

)(

)(

)(

Spherical polar co-ordinates

Elemental length

Consider an elemental volume ABCDEFGH =

GENERAL HEAT CONDUCTION EQUATION IN SPHERICAL COORDINATES

Let

=internal heat generation per unit volume per unit time (W m3)q= rate of heat flow (W)Q= qdt=total heat flow (J)r=mass density of the material element(kgm3)Cp=specific heat of the material element (Jkg K)

gq

Consider an unsteady (transient) three dimensional heat conduction with internal heat generation in heterogeneous amp anisotropic spherical geometries

LetTemperature distribution is a function of T=

X

Y

Z

drT

drrdKQ rr )sin(

drQr

QQ rrdrr ][

drdr

TdrrdK

rdQ

QQdQ

rr

drrrr

])sin([

)1(]([sin 2

rT

rr

ddrddKdQ rr

Plane ( -q f) lsquorrsquo directionQuantity of heat flowing into the face ABCD in lsquorrsquo direction

Quantity of heat leaving from the face EFGH in lsquorrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquorrsquo direction

drT

drdrKQ )sin(

rdQ

rQQ d ][

rddr

TdrdrK

rdQ

QQdQ d

])sin([

])[(sin

2

T

r

drdrdrKdQ

)2(])(sin[

2

T

r

drdrddrKdQ

Quantity of heat leaving from the face BFHD in lsquoqrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoqrsquo direction

X

Y

Z

Plane (r- f ) lsquoq lsquodirection

Quantity of heat flowing into the face AEGC in lsquoqrsquo direction

dr

TdrrdKQ

sin)(

drQ

rQQ d sin][

sin

drdr

TdrrdK

rdQ

QQdQ d

sin]sin

)([sin

][sinsin

)sin(

T

rr

ddrdrrdKdQ

)3()(sin

)sin(2

2

22

2

T

r

ddddrrKdQ

Plane (r- ) q lsquofrsquo directionQuantity of heat flowing into the face ABFE in lsquofrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquozrsquo direction

)4()sin()(

ddrrddrqQgg

)5()sin()(

dT

CdrrddrQ pst

Total amount of internal heat generation in the element

Total heat energy stored in the element

(Ein ndash Eout) + Egen = Est

d

TmCQ pst )(

)( volumem

Quantity of heat leaving from the face CDHG in lsquofrsquo direction

according to thermodynamics first law

Ein + Egen = Eout + Est

X

Y

Z

stgr QQdQdQdQ

)(

]([sin 2

r

Tr

rddrddKr

])(sin[

2

T

rdrdrddrK

)(sin

)sin(2

2

22

2

T

r

ddddrrK

ddrrddrq g )sin(

dT

Cdrrddr p )sin(

2

2

r

r

sin

sin

]([ 22 r

Tr

rr

Kr

])(sin[sin2

TrK

)(sin 2

2

22

T

r

Kgq

T

Cp

]([1

[ 22 r

Tr

rrK

])(sin[sin1

2

T

r)](

sin

12

2

22 T

r gq

T

Cp

]([1

[ 22 r

Tr

rr

])(sin[sin1

2

T

r)](

sin

12

2

22 T

rK

q g

T

K

Cp

]([1

[ 22 r

Tr

rr

])(sin[sin1

2

T

r)](

sin

12

2

22 T

rK

q g

T1

KKKKr If material is homogeneous amp isotropic

TCqTK pg)(

T

K

qT g

12 pC

K

Thermal diffusivity

(Ein ndash Eout) + Egen = Est

General Heat Conduction Equation

PLANE WALL CYLINDER SPHERE

Temperature profile

Plane wall SphereCylinder

bull A copper conductor (k = 380 WmoC resistivity ρ = 2x10-8 Ω m) having inner and outer radii 10 cm and 225 cm respectively is carrying a current density of 4800 amperes cm2 the conductor is internally cooled and a constant temperature of 65oC is maintained at the inner surface and there is no heat transfer through the insulation surrounding the conductor

Determinebull (i) The maximum temperature of the conductor and the radius at

which it occursbull (ii) The internal heat transfer rate

bull A chemical reaction takes place in a packed bed (k=06 WmoC) between two coaxial cylinders with radii 15 mm and 45 mm the inner surface is at 580oC and it is insulated Assuming the reaction rate of 055 MWm3 in the reactor volume find the temperature at the outer surface of the reactor

bull A current of 300 amperes passes through a stainless steel wire of 25 mm diameter and k=20 WmoC The resistivity of the wire is 70x10-8 ohm-m and length of the wire is 2 m if the wire is submerged in a fluid maintained at 50oC and h at the wire surface is 4000 Wm2oC calculate the steady state temperature at the centre and at the surface of the wire

bull An approximately spherical shaped orange ( k=023 WmoC) 90 mm diameter undergoes ripping process and generates 5100 Wm3 of energy If external surface the orange is at 8oC determine

bull (i) temperature at the centre of the orange and

bull (ii) heat flow from the outer surface of the orange

bull FINS External surfaces to used to cooling area and hence convection

heat transfer can be improvedbull APPLICATIONS IC engine cylinder ( air cooling) Heat exchangers automobile

radiators computer chip refrigerators electric motors generators and transformers etc

bull ASSUMPTIONS MADE IN ANALYSIS OF HEAT TRANFER THROUGH FIN

1 Fin material is homogenous and isotropic constthermal conductivity

2 Temperature at any cs of the fin is uniform3 lsquohrsquo is uniform over entire surface4 No internal heat generation5 Thermal contact resistance is negligible6 Radiation heat transfer is negligible7 Heat conduction is steady state and one dimensional

bull Problem 1

bull A stainless steel fin (k=20 WmK) having a diameter of 20 mm and a length of 01 m is attached to a wall at 300oC The ambient temperature is 50oC and the heat transfer coefficient is 10 Wm2K The fin tip is insulated Determine

bull (A) The rate of heat dissipation from the finbull (B) The temperature at the fin tipbull (C) The rate of heat transfer from the wall area covered by the fin if

the fin was not usedbull (D) The heat transfer rate from the same fin geometry if the

stainless steel fin is replaced by a fictitious fin with infinite thermal conductivity

bull Problem2bull One end of a long rod is inserted into a furnace while the other

projects into ambient temperature Under steady state the temperature of the rod is measured at two points 75mm apart and found to be 125oC and 885oC respectively while the ambient temperature is 20oC If the rod is 25mm in diameter and h is 2336 Wm2K

bull Find the thermal conductivity of the rod material

  • HEAT AND MASS TRANSFER
  • Thermodynamics Vs Heat transfer
  • MODE OF HEAT TRANSFER
  • MODE OF HEAT TRANSFER (2)
  • CONDUCTION- MECHANISM
  • CONCECTION-MECHANISM
  • RADIATION-MECHANISM
  • Conduction- Convection- Radiation
  • Slide 9
  • Slide 10
  • Fourierrsquos Law of Heat Conduction
  • Slide 12
  • Newtonrsquos Law of Cooling
  • Laws of Thermal Radiation
  • LAWS FOR HEAT TRANSFER
  • ELECTRICAL ampTHERMAL ANALOGY
  • Slide 17
  • Slide 18
  • PLANE WALL-UNIDIRECTIONAL
  • PLANE WALL -SERIES
  • PLANE WALL IN SERIES
  • PLANE WALL -PARALLEL
  • PLANE WALL-PARALLEL
  • PLANE WALL-COMBINATION
  • PLANE WALL-FILM RESISTANCE
  • PLANE WALL- THERMAL CONTACT RESISTANCE
  • PLANE WALL- THERMAL CONTACT RESISTANCE (2)
  • CYLINDER-SOLID
  • HEAT TRANSFER IN SOLID CYLINDER
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER-COMPOSITE
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER-MULTY LAYER
  • HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER
  • HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER (2)
  • HEAT TRANSFER IN SPHERE
  • HEAT TRANSFER IN SPHERE -MULTY LAYER
  • HEAT TRANSFER IN SPHERE -MULTY LAYER (2)
  • Slide 38
  • Slide 39
  • Slide 40
  • Slide 41
  • CRITICAL RADIUS
  • Plane wall- Insulation thickness -study
  • Problem5
  • Slide 45
  • Slide 46
  • Slide 47
  • GENERAL HEAT CONDUCTION EQUATION IN CYLINDRICAL COORDINATES
  • Slide 49
  • Slide 50
  • Slide 51
  • GENERAL HEAT CONDUCTION EQUATION IN SPHERICAL COORDINATES
  • Slide 53
  • Slide 54
  • Slide 55
  • General Heat Conduction Equation
  • Temperature profile
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Slide 67
Page 20: hmt chap 1

PLANE WALL-COMBINATION

PLANE WALL-FILM RESISTANCE

PLANE WALL- THERMAL CONTACT RESISTANCE

PLANE WALL- THERMAL CONTACT RESISTANCE

At the joining surface of the two slabs air is trapped in voids due to surface irregularities

CYLINDER-SOLID

HEAT TRANSFER IN SOLID CYLINDER

HEAT TRANSFER RATE IN HOLLOW CYLINDER

HEAT TRANSFER RATE IN HOLLOW CYLINDER-COMPOSITE

HEAT TRANSFER RATE IN HOLLOW CYLINDER-MULTY LAYER

HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER

HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER

HEAT TRANSFER IN SPHERE

HEAT TRANSFER IN SPHERE -MULTY LAYER

HEAT TRANSFER IN SPHERE -MULTY LAYER

bull Problem 1 The wall in a furnace consists of 125 mm thick refractory bricks and

125 mm thick insulating fire bricks separated by an air gap A 12 mm thick plaster covers the outer wall The inner surface of wall is at 1100oC and the ambient temperature 25oC the heat transfer coefficient on the outer wall to the air is 17 Wm2K and the resistance to heat flow of the air gap is 016 KW the thermal conductivities of refractory brick insulating firebrick and plaster are 16 03 and 014 WmK respectively

Calculatebull (a) the rate of heat loss per unit area of the wall surfacebull (b) the interface temperatures throughout the wallbull (c) the temperature at the outside surface of the wall

bull Problem 2Two large aluminum plates (k=240 WmK) each 2 cm thick with 10 mm surface roughness are placed in contact under 105 Nm2 pressure in air The temperature at the outside surfaces are 390 oC and 406 oCCalculate

A) the heat flux B) the temperature drop due to the contact resistance C) the contact temperatures Thermal contact resistance with air as the interfacial fluid for

10 mm roughness is 27510-4 m2KW

bull Problem 3bull Steam at 350 oC flowing in a pipe (k=80 WmK) of 5 cm

inner diameter and 56 cm outer diameter is covered with 3 cm thick insulation of k=005 WmK Heat is lost to surroundings at 5 oC by natural convection and radiation the combined h being 20 Wm2K Taking the heat transfer coefficient inside the pipe as 60 Wm2K

bull Determinebull A) the rate of heat loss from the steam per unit length of

the pipebull B) the temperature drop across the pipe and the insulation

bull PROBLEM4bull A nuclear reactor has a sperical pressure vessel 750 mm inside

diameter 80 mm wall thickness The temperature at the inner surface is 500 oC If the tempertarue at the outer surface is 495 oC

bull Calculatea) The rate of heat loss through the metal wallb) If the rate of heat loss is to limited to 305Whow much inulation

thickness needs to be applied Thermal conductivity of steel wall=46 WmoC Thermal conductivity of insulation=004 WmoC Film coefficient of heat tranfer at inner and outer surface are 200

Wm2 oC and 15 Wm2 oC Ambient air temperature=28 oC

CRITICAL RADIUS

bull The radius upto which heat flow increases and after decreases At critical radius q becomes qmax

bull The addition of insulation always increases (Rth)condbull Rtotal= (Rth)cond + (Rth)conv

bull The addition of insulation may reduce the (Rth)conv due to increase in surface area

bull r1ltrc -heat transfer increases by adding insulation till r1=rc If

insulation thickness is further increased the rate of heat loss will decrease

r1 gt rc

- heat transfer decreases by adding insulation above the rc

Plane wall- Insulation

thickness -study

Cylinder amp Sphere- Insulation thickness -

study

Problem5bull A 3mm diameter and 5 mm long electric wire is tightly

wrapped with a 2 mm thick plastic cover whose K=015 wmoC Electrical measurements indicate that a current of 10 A passes through the wire and there is a voltage drop of 8 V along the wire If the insulated wire is exposed to a medium at Ta=30oC with a heat transfer coefficient of h=12 wm2oC Determine the temperature at the interface of the wire and the plastic cover in steady operation Also determine whether doubling the thickness of the plastic cover will increase or decrease this interface temperature

dx

TdzdyKQ xx )(

dxQx

QQ xxdxx ][

dxdx

TdzdyK

xQQdQ xdxxxx ])([)(

)1()]([)(

ddzdydx

x

TK

xdQ xx

GENERAL HEAT CONDUCTION EQUATION IN CARTESIAN COORDINATES

Consider an unsteady (transient) three dimensional heat conduction with internal heat generation in heterogeneous amp anisotropic PLANE WALL

Cartesian co-ordinates x y z

Elemental length dxdydz

Consider an elemental volume ABCDEFGH = (dxdydz)

Let Temperature distribution is a function of T=T(xyz)

Let

=internal heat generation per unit volume per unit time (W m3)q= rate of heat flow (W)Q= qdt=total heat flow (J)r=mass density of the material element(kgm3)Cp=specific heat of the material element (Jkg K)

gq

Plane (y - z) lsquoxrsquo direction

Quantity of heat leaving from the face EFGH in lsquoXrsquo direction

Quantity of heat flowing into the face ABCD in lsquoXrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoXrsquo direction

dy

TdzdxKQ yy )(

dyQy

QQ yydyy ][

dydy

TdzdxK

yQQdQ ydyyyy ])([)(

)2()]([)(

ddzdydx

y

TK

ydQ yy

Plane (x - z) lsquoyrsquo direction

Quantity of heat flowing into the face ABFE in lsquoYrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoYrsquo direction

dz

TdydxKQ zz )(

dzQz

QQ zzdzz ][

dzdz

TdydxK

zQQdQ zdzzzz ])([)(

)3()]([)(

ddzdydx

z

TK

zdQ zz

Plane (x - y) lsquozrsquo direction

Quantity of heat flowing into the face ADHE in lsquoZrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoZrsquo direction

Quantity of heat leaving from the face DCGH in lsquoYrsquo direction

Quantity of heat leaving from the face BCGF in lsquoZrsquo direction

)4()()(

ddzdydxqQgg

Total amount of internal heat generation in the element

Ein + Egen = Eout+Est (Ein- Eout)+ Egen=Est

T

Cddzdydxddzdydxqddzdydxz

TK

zy

TK

yx

TK

x pgzyx )()())](()()([

T

pC

gq

z

TzK

zy

Ty

Kyx

TxK

x][][][

stgzyx QQdQdQdQ

)(

)5()()(

dT

CdzdydxQ pst

Total heat energy stored in the element

(ie thermal energy of the element can be increased because of heat accumulation in the element due to heat flow amp internal heat generation)

)( volumem d

TmCQ pst )(

T

CqTK pg)(

KKKyKx z If material is homogeneous amp isotropic

T

pC

gq

z

T

zy

T

yx

T

xK ][][][

T

Kp

C

Kgq

z

T

y

T

x

T2

2

2

2

2

2

T

Kgq

z

T

y

T

x

T 12

2

2

2

2

2

Thermal diffusivity pC

K

T

K

qT g

12

according to thermodynamics first law

)( zr )( dzrddr

)( dzrddr

Cylindrical polar co-ordinates

Elemental length

Consider an elemental volume ABCDEFGH =

GENERAL HEAT CONDUCTION EQUATION IN CYLINDRICAL COORDINATES

gq

gq

Consider an unsteady (transient) three dimensional heat conduction with internal heat generation in heterogeneous amp anisotropic cylindrical geometries (eg rods and pipes)

)( zrTLetTemperature distribution is a function of T=

Let

=internal heat generation per unit volume per unit time (W m3)q= rate of heat flow (W)Q= qdt=total heat flow (J)r=mass density of the material element(kgm3)Cp=specific heat of the material element (Jkg K)

planerdirectionz

planezrdirection

planezdirectionr

)(

)(

)(

LetHeat flow in

drdr

TdzrdK

rdQ

QQdQ

rr

drrrr

])([

)1(][

r

Tr

rddrdzdKdQ rr

Heat accumulation in the element due to heat flow by conduction in lsquorrsquo direction

rddr

TdzdrK

rdQ

QQdQ d

])([

][

2

T

r

ddrdzrdKdQ

)2(][

2

2

2

T

r

ddrdzrdKdQ

Heat accumulation in the element due to heat flow by conduction in lsquoqrsquo direction

Plane (z- q ) lsquorrsquo direction

dr

TdzrdKQ rr

)(

drQr

QQ rrdrr ][

Quantity of heat leaving from the face EFGH in lsquorrsquo direction

Quantity of heat flowing into the face ABCD in lsquorrsquo direction

Plane (r-z) lsquoq lsquodirection

Quantity of heat leaving from the face AEGC in lsquoqrsquo direction

d

r

TdzdrKQ )(

rdQ

rQQ d ][

Quantity of heat flowing into the face BFHD in lsquoqrsquo direction

)3())((2

2

z

TdzdrrdKdQ zz

)5()()(

dT

CdzdrrdQ pst

)4()()(

ddzdrrdqQgg

Total amount of internal heat generation in the element

Total heat energy stored in the element

(ie thermal energy of the element can be increased because of heat accumulation in the element due to heat flow amp internal heat generation)

according to thermodynamics first law

Ein + Egen = Eout + Est (Ein ndash Eout) + Egen = Est

)( volumem

)()(z

T

zdzdrrdKdQ zz

Heat accumulation in the element due to heat flow by conduction in lsquozrsquo direction

dzdz

TdrrdK

zdQ

QQdQ

zz

dzzzz

])([

d

TmCQ pst )(

Plane (r-q) lsquozrsquo direction

dz

TdrrdKQ zz )(

dzQz

QQ zzdzz ][ Quantity of heat leaving from the face CDHG in lsquozrsquo direction

Quantity of heat flowing into the face ABFE in lsquozrsquo direction

stgzr QQdQdQdQ

)(

r

r ][)(rT

rr

ddrdzdKr

][)(

2

2

2

T

r

ddrdzrdK ))((2

2

z

TddzdrrdK z

ddzdrrdq g )(

dT

Cdzdrrd p )(

+ + +

=

T

CqTK pg)(KKKK zr

[gq

)(1

r

Tr

rrK r

)(1

2

2

2 T

rK )(

2

2

z

TK z

)( ddzdrrd )(

ddzdrrd

TC p

+ + + ] =

If material is homogeneous amp isotropic

)(1

r

Tr

rr

)(1

2

2

2 T

r)(

2

2

z

T

K

q g

T

K

C p

)(1

r

Tr

rr

)(1

2

2

2 T

r )(2

2

z

T

K

q g

T

1

+ + + =

)(1

r

Tr

rrK r

)(1

2

2

2 T

rK )(

2

2

z

TK z

gq

T

C p + + + =

+ + + = Thermal diffusivity

T

K

qT g

12

pC

K

(Ein ndash Eout) + Egen = Est

)( rT

)( r

)sin( drrddr)sin( drrddr

planerdirection

planerdirection

planedirectionr

)(

)(

)(

Spherical polar co-ordinates

Elemental length

Consider an elemental volume ABCDEFGH =

GENERAL HEAT CONDUCTION EQUATION IN SPHERICAL COORDINATES

Let

=internal heat generation per unit volume per unit time (W m3)q= rate of heat flow (W)Q= qdt=total heat flow (J)r=mass density of the material element(kgm3)Cp=specific heat of the material element (Jkg K)

gq

Consider an unsteady (transient) three dimensional heat conduction with internal heat generation in heterogeneous amp anisotropic spherical geometries

LetTemperature distribution is a function of T=

X

Y

Z

drT

drrdKQ rr )sin(

drQr

QQ rrdrr ][

drdr

TdrrdK

rdQ

QQdQ

rr

drrrr

])sin([

)1(]([sin 2

rT

rr

ddrddKdQ rr

Plane ( -q f) lsquorrsquo directionQuantity of heat flowing into the face ABCD in lsquorrsquo direction

Quantity of heat leaving from the face EFGH in lsquorrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquorrsquo direction

drT

drdrKQ )sin(

rdQ

rQQ d ][

rddr

TdrdrK

rdQ

QQdQ d

])sin([

])[(sin

2

T

r

drdrdrKdQ

)2(])(sin[

2

T

r

drdrddrKdQ

Quantity of heat leaving from the face BFHD in lsquoqrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoqrsquo direction

X

Y

Z

Plane (r- f ) lsquoq lsquodirection

Quantity of heat flowing into the face AEGC in lsquoqrsquo direction

dr

TdrrdKQ

sin)(

drQ

rQQ d sin][

sin

drdr

TdrrdK

rdQ

QQdQ d

sin]sin

)([sin

][sinsin

)sin(

T

rr

ddrdrrdKdQ

)3()(sin

)sin(2

2

22

2

T

r

ddddrrKdQ

Plane (r- ) q lsquofrsquo directionQuantity of heat flowing into the face ABFE in lsquofrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquozrsquo direction

)4()sin()(

ddrrddrqQgg

)5()sin()(

dT

CdrrddrQ pst

Total amount of internal heat generation in the element

Total heat energy stored in the element

(Ein ndash Eout) + Egen = Est

d

TmCQ pst )(

)( volumem

Quantity of heat leaving from the face CDHG in lsquofrsquo direction

according to thermodynamics first law

Ein + Egen = Eout + Est

X

Y

Z

stgr QQdQdQdQ

)(

]([sin 2

r

Tr

rddrddKr

])(sin[

2

T

rdrdrddrK

)(sin

)sin(2

2

22

2

T

r

ddddrrK

ddrrddrq g )sin(

dT

Cdrrddr p )sin(

2

2

r

r

sin

sin

]([ 22 r

Tr

rr

Kr

])(sin[sin2

TrK

)(sin 2

2

22

T

r

Kgq

T

Cp

]([1

[ 22 r

Tr

rrK

])(sin[sin1

2

T

r)](

sin

12

2

22 T

r gq

T

Cp

]([1

[ 22 r

Tr

rr

])(sin[sin1

2

T

r)](

sin

12

2

22 T

rK

q g

T

K

Cp

]([1

[ 22 r

Tr

rr

])(sin[sin1

2

T

r)](

sin

12

2

22 T

rK

q g

T1

KKKKr If material is homogeneous amp isotropic

TCqTK pg)(

T

K

qT g

12 pC

K

Thermal diffusivity

(Ein ndash Eout) + Egen = Est

General Heat Conduction Equation

PLANE WALL CYLINDER SPHERE

Temperature profile

Plane wall SphereCylinder

bull A copper conductor (k = 380 WmoC resistivity ρ = 2x10-8 Ω m) having inner and outer radii 10 cm and 225 cm respectively is carrying a current density of 4800 amperes cm2 the conductor is internally cooled and a constant temperature of 65oC is maintained at the inner surface and there is no heat transfer through the insulation surrounding the conductor

Determinebull (i) The maximum temperature of the conductor and the radius at

which it occursbull (ii) The internal heat transfer rate

bull A chemical reaction takes place in a packed bed (k=06 WmoC) between two coaxial cylinders with radii 15 mm and 45 mm the inner surface is at 580oC and it is insulated Assuming the reaction rate of 055 MWm3 in the reactor volume find the temperature at the outer surface of the reactor

bull A current of 300 amperes passes through a stainless steel wire of 25 mm diameter and k=20 WmoC The resistivity of the wire is 70x10-8 ohm-m and length of the wire is 2 m if the wire is submerged in a fluid maintained at 50oC and h at the wire surface is 4000 Wm2oC calculate the steady state temperature at the centre and at the surface of the wire

bull An approximately spherical shaped orange ( k=023 WmoC) 90 mm diameter undergoes ripping process and generates 5100 Wm3 of energy If external surface the orange is at 8oC determine

bull (i) temperature at the centre of the orange and

bull (ii) heat flow from the outer surface of the orange

bull FINS External surfaces to used to cooling area and hence convection

heat transfer can be improvedbull APPLICATIONS IC engine cylinder ( air cooling) Heat exchangers automobile

radiators computer chip refrigerators electric motors generators and transformers etc

bull ASSUMPTIONS MADE IN ANALYSIS OF HEAT TRANFER THROUGH FIN

1 Fin material is homogenous and isotropic constthermal conductivity

2 Temperature at any cs of the fin is uniform3 lsquohrsquo is uniform over entire surface4 No internal heat generation5 Thermal contact resistance is negligible6 Radiation heat transfer is negligible7 Heat conduction is steady state and one dimensional

bull Problem 1

bull A stainless steel fin (k=20 WmK) having a diameter of 20 mm and a length of 01 m is attached to a wall at 300oC The ambient temperature is 50oC and the heat transfer coefficient is 10 Wm2K The fin tip is insulated Determine

bull (A) The rate of heat dissipation from the finbull (B) The temperature at the fin tipbull (C) The rate of heat transfer from the wall area covered by the fin if

the fin was not usedbull (D) The heat transfer rate from the same fin geometry if the

stainless steel fin is replaced by a fictitious fin with infinite thermal conductivity

bull Problem2bull One end of a long rod is inserted into a furnace while the other

projects into ambient temperature Under steady state the temperature of the rod is measured at two points 75mm apart and found to be 125oC and 885oC respectively while the ambient temperature is 20oC If the rod is 25mm in diameter and h is 2336 Wm2K

bull Find the thermal conductivity of the rod material

  • HEAT AND MASS TRANSFER
  • Thermodynamics Vs Heat transfer
  • MODE OF HEAT TRANSFER
  • MODE OF HEAT TRANSFER (2)
  • CONDUCTION- MECHANISM
  • CONCECTION-MECHANISM
  • RADIATION-MECHANISM
  • Conduction- Convection- Radiation
  • Slide 9
  • Slide 10
  • Fourierrsquos Law of Heat Conduction
  • Slide 12
  • Newtonrsquos Law of Cooling
  • Laws of Thermal Radiation
  • LAWS FOR HEAT TRANSFER
  • ELECTRICAL ampTHERMAL ANALOGY
  • Slide 17
  • Slide 18
  • PLANE WALL-UNIDIRECTIONAL
  • PLANE WALL -SERIES
  • PLANE WALL IN SERIES
  • PLANE WALL -PARALLEL
  • PLANE WALL-PARALLEL
  • PLANE WALL-COMBINATION
  • PLANE WALL-FILM RESISTANCE
  • PLANE WALL- THERMAL CONTACT RESISTANCE
  • PLANE WALL- THERMAL CONTACT RESISTANCE (2)
  • CYLINDER-SOLID
  • HEAT TRANSFER IN SOLID CYLINDER
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER-COMPOSITE
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER-MULTY LAYER
  • HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER
  • HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER (2)
  • HEAT TRANSFER IN SPHERE
  • HEAT TRANSFER IN SPHERE -MULTY LAYER
  • HEAT TRANSFER IN SPHERE -MULTY LAYER (2)
  • Slide 38
  • Slide 39
  • Slide 40
  • Slide 41
  • CRITICAL RADIUS
  • Plane wall- Insulation thickness -study
  • Problem5
  • Slide 45
  • Slide 46
  • Slide 47
  • GENERAL HEAT CONDUCTION EQUATION IN CYLINDRICAL COORDINATES
  • Slide 49
  • Slide 50
  • Slide 51
  • GENERAL HEAT CONDUCTION EQUATION IN SPHERICAL COORDINATES
  • Slide 53
  • Slide 54
  • Slide 55
  • General Heat Conduction Equation
  • Temperature profile
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Slide 67
Page 21: hmt chap 1

PLANE WALL-FILM RESISTANCE

PLANE WALL- THERMAL CONTACT RESISTANCE

PLANE WALL- THERMAL CONTACT RESISTANCE

At the joining surface of the two slabs air is trapped in voids due to surface irregularities

CYLINDER-SOLID

HEAT TRANSFER IN SOLID CYLINDER

HEAT TRANSFER RATE IN HOLLOW CYLINDER

HEAT TRANSFER RATE IN HOLLOW CYLINDER-COMPOSITE

HEAT TRANSFER RATE IN HOLLOW CYLINDER-MULTY LAYER

HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER

HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER

HEAT TRANSFER IN SPHERE

HEAT TRANSFER IN SPHERE -MULTY LAYER

HEAT TRANSFER IN SPHERE -MULTY LAYER

bull Problem 1 The wall in a furnace consists of 125 mm thick refractory bricks and

125 mm thick insulating fire bricks separated by an air gap A 12 mm thick plaster covers the outer wall The inner surface of wall is at 1100oC and the ambient temperature 25oC the heat transfer coefficient on the outer wall to the air is 17 Wm2K and the resistance to heat flow of the air gap is 016 KW the thermal conductivities of refractory brick insulating firebrick and plaster are 16 03 and 014 WmK respectively

Calculatebull (a) the rate of heat loss per unit area of the wall surfacebull (b) the interface temperatures throughout the wallbull (c) the temperature at the outside surface of the wall

bull Problem 2Two large aluminum plates (k=240 WmK) each 2 cm thick with 10 mm surface roughness are placed in contact under 105 Nm2 pressure in air The temperature at the outside surfaces are 390 oC and 406 oCCalculate

A) the heat flux B) the temperature drop due to the contact resistance C) the contact temperatures Thermal contact resistance with air as the interfacial fluid for

10 mm roughness is 27510-4 m2KW

bull Problem 3bull Steam at 350 oC flowing in a pipe (k=80 WmK) of 5 cm

inner diameter and 56 cm outer diameter is covered with 3 cm thick insulation of k=005 WmK Heat is lost to surroundings at 5 oC by natural convection and radiation the combined h being 20 Wm2K Taking the heat transfer coefficient inside the pipe as 60 Wm2K

bull Determinebull A) the rate of heat loss from the steam per unit length of

the pipebull B) the temperature drop across the pipe and the insulation

bull PROBLEM4bull A nuclear reactor has a sperical pressure vessel 750 mm inside

diameter 80 mm wall thickness The temperature at the inner surface is 500 oC If the tempertarue at the outer surface is 495 oC

bull Calculatea) The rate of heat loss through the metal wallb) If the rate of heat loss is to limited to 305Whow much inulation

thickness needs to be applied Thermal conductivity of steel wall=46 WmoC Thermal conductivity of insulation=004 WmoC Film coefficient of heat tranfer at inner and outer surface are 200

Wm2 oC and 15 Wm2 oC Ambient air temperature=28 oC

CRITICAL RADIUS

bull The radius upto which heat flow increases and after decreases At critical radius q becomes qmax

bull The addition of insulation always increases (Rth)condbull Rtotal= (Rth)cond + (Rth)conv

bull The addition of insulation may reduce the (Rth)conv due to increase in surface area

bull r1ltrc -heat transfer increases by adding insulation till r1=rc If

insulation thickness is further increased the rate of heat loss will decrease

r1 gt rc

- heat transfer decreases by adding insulation above the rc

Plane wall- Insulation

thickness -study

Cylinder amp Sphere- Insulation thickness -

study

Problem5bull A 3mm diameter and 5 mm long electric wire is tightly

wrapped with a 2 mm thick plastic cover whose K=015 wmoC Electrical measurements indicate that a current of 10 A passes through the wire and there is a voltage drop of 8 V along the wire If the insulated wire is exposed to a medium at Ta=30oC with a heat transfer coefficient of h=12 wm2oC Determine the temperature at the interface of the wire and the plastic cover in steady operation Also determine whether doubling the thickness of the plastic cover will increase or decrease this interface temperature

dx

TdzdyKQ xx )(

dxQx

QQ xxdxx ][

dxdx

TdzdyK

xQQdQ xdxxxx ])([)(

)1()]([)(

ddzdydx

x

TK

xdQ xx

GENERAL HEAT CONDUCTION EQUATION IN CARTESIAN COORDINATES

Consider an unsteady (transient) three dimensional heat conduction with internal heat generation in heterogeneous amp anisotropic PLANE WALL

Cartesian co-ordinates x y z

Elemental length dxdydz

Consider an elemental volume ABCDEFGH = (dxdydz)

Let Temperature distribution is a function of T=T(xyz)

Let

=internal heat generation per unit volume per unit time (W m3)q= rate of heat flow (W)Q= qdt=total heat flow (J)r=mass density of the material element(kgm3)Cp=specific heat of the material element (Jkg K)

gq

Plane (y - z) lsquoxrsquo direction

Quantity of heat leaving from the face EFGH in lsquoXrsquo direction

Quantity of heat flowing into the face ABCD in lsquoXrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoXrsquo direction

dy

TdzdxKQ yy )(

dyQy

QQ yydyy ][

dydy

TdzdxK

yQQdQ ydyyyy ])([)(

)2()]([)(

ddzdydx

y

TK

ydQ yy

Plane (x - z) lsquoyrsquo direction

Quantity of heat flowing into the face ABFE in lsquoYrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoYrsquo direction

dz

TdydxKQ zz )(

dzQz

QQ zzdzz ][

dzdz

TdydxK

zQQdQ zdzzzz ])([)(

)3()]([)(

ddzdydx

z

TK

zdQ zz

Plane (x - y) lsquozrsquo direction

Quantity of heat flowing into the face ADHE in lsquoZrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoZrsquo direction

Quantity of heat leaving from the face DCGH in lsquoYrsquo direction

Quantity of heat leaving from the face BCGF in lsquoZrsquo direction

)4()()(

ddzdydxqQgg

Total amount of internal heat generation in the element

Ein + Egen = Eout+Est (Ein- Eout)+ Egen=Est

T

Cddzdydxddzdydxqddzdydxz

TK

zy

TK

yx

TK

x pgzyx )()())](()()([

T

pC

gq

z

TzK

zy

Ty

Kyx

TxK

x][][][

stgzyx QQdQdQdQ

)(

)5()()(

dT

CdzdydxQ pst

Total heat energy stored in the element

(ie thermal energy of the element can be increased because of heat accumulation in the element due to heat flow amp internal heat generation)

)( volumem d

TmCQ pst )(

T

CqTK pg)(

KKKyKx z If material is homogeneous amp isotropic

T

pC

gq

z

T

zy

T

yx

T

xK ][][][

T

Kp

C

Kgq

z

T

y

T

x

T2

2

2

2

2

2

T

Kgq

z

T

y

T

x

T 12

2

2

2

2

2

Thermal diffusivity pC

K

T

K

qT g

12

according to thermodynamics first law

)( zr )( dzrddr

)( dzrddr

Cylindrical polar co-ordinates

Elemental length

Consider an elemental volume ABCDEFGH =

GENERAL HEAT CONDUCTION EQUATION IN CYLINDRICAL COORDINATES

gq

gq

Consider an unsteady (transient) three dimensional heat conduction with internal heat generation in heterogeneous amp anisotropic cylindrical geometries (eg rods and pipes)

)( zrTLetTemperature distribution is a function of T=

Let

=internal heat generation per unit volume per unit time (W m3)q= rate of heat flow (W)Q= qdt=total heat flow (J)r=mass density of the material element(kgm3)Cp=specific heat of the material element (Jkg K)

planerdirectionz

planezrdirection

planezdirectionr

)(

)(

)(

LetHeat flow in

drdr

TdzrdK

rdQ

QQdQ

rr

drrrr

])([

)1(][

r

Tr

rddrdzdKdQ rr

Heat accumulation in the element due to heat flow by conduction in lsquorrsquo direction

rddr

TdzdrK

rdQ

QQdQ d

])([

][

2

T

r

ddrdzrdKdQ

)2(][

2

2

2

T

r

ddrdzrdKdQ

Heat accumulation in the element due to heat flow by conduction in lsquoqrsquo direction

Plane (z- q ) lsquorrsquo direction

dr

TdzrdKQ rr

)(

drQr

QQ rrdrr ][

Quantity of heat leaving from the face EFGH in lsquorrsquo direction

Quantity of heat flowing into the face ABCD in lsquorrsquo direction

Plane (r-z) lsquoq lsquodirection

Quantity of heat leaving from the face AEGC in lsquoqrsquo direction

d

r

TdzdrKQ )(

rdQ

rQQ d ][

Quantity of heat flowing into the face BFHD in lsquoqrsquo direction

)3())((2

2

z

TdzdrrdKdQ zz

)5()()(

dT

CdzdrrdQ pst

)4()()(

ddzdrrdqQgg

Total amount of internal heat generation in the element

Total heat energy stored in the element

(ie thermal energy of the element can be increased because of heat accumulation in the element due to heat flow amp internal heat generation)

according to thermodynamics first law

Ein + Egen = Eout + Est (Ein ndash Eout) + Egen = Est

)( volumem

)()(z

T

zdzdrrdKdQ zz

Heat accumulation in the element due to heat flow by conduction in lsquozrsquo direction

dzdz

TdrrdK

zdQ

QQdQ

zz

dzzzz

])([

d

TmCQ pst )(

Plane (r-q) lsquozrsquo direction

dz

TdrrdKQ zz )(

dzQz

QQ zzdzz ][ Quantity of heat leaving from the face CDHG in lsquozrsquo direction

Quantity of heat flowing into the face ABFE in lsquozrsquo direction

stgzr QQdQdQdQ

)(

r

r ][)(rT

rr

ddrdzdKr

][)(

2

2

2

T

r

ddrdzrdK ))((2

2

z

TddzdrrdK z

ddzdrrdq g )(

dT

Cdzdrrd p )(

+ + +

=

T

CqTK pg)(KKKK zr

[gq

)(1

r

Tr

rrK r

)(1

2

2

2 T

rK )(

2

2

z

TK z

)( ddzdrrd )(

ddzdrrd

TC p

+ + + ] =

If material is homogeneous amp isotropic

)(1

r

Tr

rr

)(1

2

2

2 T

r)(

2

2

z

T

K

q g

T

K

C p

)(1

r

Tr

rr

)(1

2

2

2 T

r )(2

2

z

T

K

q g

T

1

+ + + =

)(1

r

Tr

rrK r

)(1

2

2

2 T

rK )(

2

2

z

TK z

gq

T

C p + + + =

+ + + = Thermal diffusivity

T

K

qT g

12

pC

K

(Ein ndash Eout) + Egen = Est

)( rT

)( r

)sin( drrddr)sin( drrddr

planerdirection

planerdirection

planedirectionr

)(

)(

)(

Spherical polar co-ordinates

Elemental length

Consider an elemental volume ABCDEFGH =

GENERAL HEAT CONDUCTION EQUATION IN SPHERICAL COORDINATES

Let

=internal heat generation per unit volume per unit time (W m3)q= rate of heat flow (W)Q= qdt=total heat flow (J)r=mass density of the material element(kgm3)Cp=specific heat of the material element (Jkg K)

gq

Consider an unsteady (transient) three dimensional heat conduction with internal heat generation in heterogeneous amp anisotropic spherical geometries

LetTemperature distribution is a function of T=

X

Y

Z

drT

drrdKQ rr )sin(

drQr

QQ rrdrr ][

drdr

TdrrdK

rdQ

QQdQ

rr

drrrr

])sin([

)1(]([sin 2

rT

rr

ddrddKdQ rr

Plane ( -q f) lsquorrsquo directionQuantity of heat flowing into the face ABCD in lsquorrsquo direction

Quantity of heat leaving from the face EFGH in lsquorrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquorrsquo direction

drT

drdrKQ )sin(

rdQ

rQQ d ][

rddr

TdrdrK

rdQ

QQdQ d

])sin([

])[(sin

2

T

r

drdrdrKdQ

)2(])(sin[

2

T

r

drdrddrKdQ

Quantity of heat leaving from the face BFHD in lsquoqrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoqrsquo direction

X

Y

Z

Plane (r- f ) lsquoq lsquodirection

Quantity of heat flowing into the face AEGC in lsquoqrsquo direction

dr

TdrrdKQ

sin)(

drQ

rQQ d sin][

sin

drdr

TdrrdK

rdQ

QQdQ d

sin]sin

)([sin

][sinsin

)sin(

T

rr

ddrdrrdKdQ

)3()(sin

)sin(2

2

22

2

T

r

ddddrrKdQ

Plane (r- ) q lsquofrsquo directionQuantity of heat flowing into the face ABFE in lsquofrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquozrsquo direction

)4()sin()(

ddrrddrqQgg

)5()sin()(

dT

CdrrddrQ pst

Total amount of internal heat generation in the element

Total heat energy stored in the element

(Ein ndash Eout) + Egen = Est

d

TmCQ pst )(

)( volumem

Quantity of heat leaving from the face CDHG in lsquofrsquo direction

according to thermodynamics first law

Ein + Egen = Eout + Est

X

Y

Z

stgr QQdQdQdQ

)(

]([sin 2

r

Tr

rddrddKr

])(sin[

2

T

rdrdrddrK

)(sin

)sin(2

2

22

2

T

r

ddddrrK

ddrrddrq g )sin(

dT

Cdrrddr p )sin(

2

2

r

r

sin

sin

]([ 22 r

Tr

rr

Kr

])(sin[sin2

TrK

)(sin 2

2

22

T

r

Kgq

T

Cp

]([1

[ 22 r

Tr

rrK

])(sin[sin1

2

T

r)](

sin

12

2

22 T

r gq

T

Cp

]([1

[ 22 r

Tr

rr

])(sin[sin1

2

T

r)](

sin

12

2

22 T

rK

q g

T

K

Cp

]([1

[ 22 r

Tr

rr

])(sin[sin1

2

T

r)](

sin

12

2

22 T

rK

q g

T1

KKKKr If material is homogeneous amp isotropic

TCqTK pg)(

T

K

qT g

12 pC

K

Thermal diffusivity

(Ein ndash Eout) + Egen = Est

General Heat Conduction Equation

PLANE WALL CYLINDER SPHERE

Temperature profile

Plane wall SphereCylinder

bull A copper conductor (k = 380 WmoC resistivity ρ = 2x10-8 Ω m) having inner and outer radii 10 cm and 225 cm respectively is carrying a current density of 4800 amperes cm2 the conductor is internally cooled and a constant temperature of 65oC is maintained at the inner surface and there is no heat transfer through the insulation surrounding the conductor

Determinebull (i) The maximum temperature of the conductor and the radius at

which it occursbull (ii) The internal heat transfer rate

bull A chemical reaction takes place in a packed bed (k=06 WmoC) between two coaxial cylinders with radii 15 mm and 45 mm the inner surface is at 580oC and it is insulated Assuming the reaction rate of 055 MWm3 in the reactor volume find the temperature at the outer surface of the reactor

bull A current of 300 amperes passes through a stainless steel wire of 25 mm diameter and k=20 WmoC The resistivity of the wire is 70x10-8 ohm-m and length of the wire is 2 m if the wire is submerged in a fluid maintained at 50oC and h at the wire surface is 4000 Wm2oC calculate the steady state temperature at the centre and at the surface of the wire

bull An approximately spherical shaped orange ( k=023 WmoC) 90 mm diameter undergoes ripping process and generates 5100 Wm3 of energy If external surface the orange is at 8oC determine

bull (i) temperature at the centre of the orange and

bull (ii) heat flow from the outer surface of the orange

bull FINS External surfaces to used to cooling area and hence convection

heat transfer can be improvedbull APPLICATIONS IC engine cylinder ( air cooling) Heat exchangers automobile

radiators computer chip refrigerators electric motors generators and transformers etc

bull ASSUMPTIONS MADE IN ANALYSIS OF HEAT TRANFER THROUGH FIN

1 Fin material is homogenous and isotropic constthermal conductivity

2 Temperature at any cs of the fin is uniform3 lsquohrsquo is uniform over entire surface4 No internal heat generation5 Thermal contact resistance is negligible6 Radiation heat transfer is negligible7 Heat conduction is steady state and one dimensional

bull Problem 1

bull A stainless steel fin (k=20 WmK) having a diameter of 20 mm and a length of 01 m is attached to a wall at 300oC The ambient temperature is 50oC and the heat transfer coefficient is 10 Wm2K The fin tip is insulated Determine

bull (A) The rate of heat dissipation from the finbull (B) The temperature at the fin tipbull (C) The rate of heat transfer from the wall area covered by the fin if

the fin was not usedbull (D) The heat transfer rate from the same fin geometry if the

stainless steel fin is replaced by a fictitious fin with infinite thermal conductivity

bull Problem2bull One end of a long rod is inserted into a furnace while the other

projects into ambient temperature Under steady state the temperature of the rod is measured at two points 75mm apart and found to be 125oC and 885oC respectively while the ambient temperature is 20oC If the rod is 25mm in diameter and h is 2336 Wm2K

bull Find the thermal conductivity of the rod material

  • HEAT AND MASS TRANSFER
  • Thermodynamics Vs Heat transfer
  • MODE OF HEAT TRANSFER
  • MODE OF HEAT TRANSFER (2)
  • CONDUCTION- MECHANISM
  • CONCECTION-MECHANISM
  • RADIATION-MECHANISM
  • Conduction- Convection- Radiation
  • Slide 9
  • Slide 10
  • Fourierrsquos Law of Heat Conduction
  • Slide 12
  • Newtonrsquos Law of Cooling
  • Laws of Thermal Radiation
  • LAWS FOR HEAT TRANSFER
  • ELECTRICAL ampTHERMAL ANALOGY
  • Slide 17
  • Slide 18
  • PLANE WALL-UNIDIRECTIONAL
  • PLANE WALL -SERIES
  • PLANE WALL IN SERIES
  • PLANE WALL -PARALLEL
  • PLANE WALL-PARALLEL
  • PLANE WALL-COMBINATION
  • PLANE WALL-FILM RESISTANCE
  • PLANE WALL- THERMAL CONTACT RESISTANCE
  • PLANE WALL- THERMAL CONTACT RESISTANCE (2)
  • CYLINDER-SOLID
  • HEAT TRANSFER IN SOLID CYLINDER
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER-COMPOSITE
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER-MULTY LAYER
  • HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER
  • HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER (2)
  • HEAT TRANSFER IN SPHERE
  • HEAT TRANSFER IN SPHERE -MULTY LAYER
  • HEAT TRANSFER IN SPHERE -MULTY LAYER (2)
  • Slide 38
  • Slide 39
  • Slide 40
  • Slide 41
  • CRITICAL RADIUS
  • Plane wall- Insulation thickness -study
  • Problem5
  • Slide 45
  • Slide 46
  • Slide 47
  • GENERAL HEAT CONDUCTION EQUATION IN CYLINDRICAL COORDINATES
  • Slide 49
  • Slide 50
  • Slide 51
  • GENERAL HEAT CONDUCTION EQUATION IN SPHERICAL COORDINATES
  • Slide 53
  • Slide 54
  • Slide 55
  • General Heat Conduction Equation
  • Temperature profile
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Slide 67
Page 22: hmt chap 1

PLANE WALL- THERMAL CONTACT RESISTANCE

PLANE WALL- THERMAL CONTACT RESISTANCE

At the joining surface of the two slabs air is trapped in voids due to surface irregularities

CYLINDER-SOLID

HEAT TRANSFER IN SOLID CYLINDER

HEAT TRANSFER RATE IN HOLLOW CYLINDER

HEAT TRANSFER RATE IN HOLLOW CYLINDER-COMPOSITE

HEAT TRANSFER RATE IN HOLLOW CYLINDER-MULTY LAYER

HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER

HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER

HEAT TRANSFER IN SPHERE

HEAT TRANSFER IN SPHERE -MULTY LAYER

HEAT TRANSFER IN SPHERE -MULTY LAYER

bull Problem 1 The wall in a furnace consists of 125 mm thick refractory bricks and

125 mm thick insulating fire bricks separated by an air gap A 12 mm thick plaster covers the outer wall The inner surface of wall is at 1100oC and the ambient temperature 25oC the heat transfer coefficient on the outer wall to the air is 17 Wm2K and the resistance to heat flow of the air gap is 016 KW the thermal conductivities of refractory brick insulating firebrick and plaster are 16 03 and 014 WmK respectively

Calculatebull (a) the rate of heat loss per unit area of the wall surfacebull (b) the interface temperatures throughout the wallbull (c) the temperature at the outside surface of the wall

bull Problem 2Two large aluminum plates (k=240 WmK) each 2 cm thick with 10 mm surface roughness are placed in contact under 105 Nm2 pressure in air The temperature at the outside surfaces are 390 oC and 406 oCCalculate

A) the heat flux B) the temperature drop due to the contact resistance C) the contact temperatures Thermal contact resistance with air as the interfacial fluid for

10 mm roughness is 27510-4 m2KW

bull Problem 3bull Steam at 350 oC flowing in a pipe (k=80 WmK) of 5 cm

inner diameter and 56 cm outer diameter is covered with 3 cm thick insulation of k=005 WmK Heat is lost to surroundings at 5 oC by natural convection and radiation the combined h being 20 Wm2K Taking the heat transfer coefficient inside the pipe as 60 Wm2K

bull Determinebull A) the rate of heat loss from the steam per unit length of

the pipebull B) the temperature drop across the pipe and the insulation

bull PROBLEM4bull A nuclear reactor has a sperical pressure vessel 750 mm inside

diameter 80 mm wall thickness The temperature at the inner surface is 500 oC If the tempertarue at the outer surface is 495 oC

bull Calculatea) The rate of heat loss through the metal wallb) If the rate of heat loss is to limited to 305Whow much inulation

thickness needs to be applied Thermal conductivity of steel wall=46 WmoC Thermal conductivity of insulation=004 WmoC Film coefficient of heat tranfer at inner and outer surface are 200

Wm2 oC and 15 Wm2 oC Ambient air temperature=28 oC

CRITICAL RADIUS

bull The radius upto which heat flow increases and after decreases At critical radius q becomes qmax

bull The addition of insulation always increases (Rth)condbull Rtotal= (Rth)cond + (Rth)conv

bull The addition of insulation may reduce the (Rth)conv due to increase in surface area

bull r1ltrc -heat transfer increases by adding insulation till r1=rc If

insulation thickness is further increased the rate of heat loss will decrease

r1 gt rc

- heat transfer decreases by adding insulation above the rc

Plane wall- Insulation

thickness -study

Cylinder amp Sphere- Insulation thickness -

study

Problem5bull A 3mm diameter and 5 mm long electric wire is tightly

wrapped with a 2 mm thick plastic cover whose K=015 wmoC Electrical measurements indicate that a current of 10 A passes through the wire and there is a voltage drop of 8 V along the wire If the insulated wire is exposed to a medium at Ta=30oC with a heat transfer coefficient of h=12 wm2oC Determine the temperature at the interface of the wire and the plastic cover in steady operation Also determine whether doubling the thickness of the plastic cover will increase or decrease this interface temperature

dx

TdzdyKQ xx )(

dxQx

QQ xxdxx ][

dxdx

TdzdyK

xQQdQ xdxxxx ])([)(

)1()]([)(

ddzdydx

x

TK

xdQ xx

GENERAL HEAT CONDUCTION EQUATION IN CARTESIAN COORDINATES

Consider an unsteady (transient) three dimensional heat conduction with internal heat generation in heterogeneous amp anisotropic PLANE WALL

Cartesian co-ordinates x y z

Elemental length dxdydz

Consider an elemental volume ABCDEFGH = (dxdydz)

Let Temperature distribution is a function of T=T(xyz)

Let

=internal heat generation per unit volume per unit time (W m3)q= rate of heat flow (W)Q= qdt=total heat flow (J)r=mass density of the material element(kgm3)Cp=specific heat of the material element (Jkg K)

gq

Plane (y - z) lsquoxrsquo direction

Quantity of heat leaving from the face EFGH in lsquoXrsquo direction

Quantity of heat flowing into the face ABCD in lsquoXrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoXrsquo direction

dy

TdzdxKQ yy )(

dyQy

QQ yydyy ][

dydy

TdzdxK

yQQdQ ydyyyy ])([)(

)2()]([)(

ddzdydx

y

TK

ydQ yy

Plane (x - z) lsquoyrsquo direction

Quantity of heat flowing into the face ABFE in lsquoYrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoYrsquo direction

dz

TdydxKQ zz )(

dzQz

QQ zzdzz ][

dzdz

TdydxK

zQQdQ zdzzzz ])([)(

)3()]([)(

ddzdydx

z

TK

zdQ zz

Plane (x - y) lsquozrsquo direction

Quantity of heat flowing into the face ADHE in lsquoZrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoZrsquo direction

Quantity of heat leaving from the face DCGH in lsquoYrsquo direction

Quantity of heat leaving from the face BCGF in lsquoZrsquo direction

)4()()(

ddzdydxqQgg

Total amount of internal heat generation in the element

Ein + Egen = Eout+Est (Ein- Eout)+ Egen=Est

T

Cddzdydxddzdydxqddzdydxz

TK

zy

TK

yx

TK

x pgzyx )()())](()()([

T

pC

gq

z

TzK

zy

Ty

Kyx

TxK

x][][][

stgzyx QQdQdQdQ

)(

)5()()(

dT

CdzdydxQ pst

Total heat energy stored in the element

(ie thermal energy of the element can be increased because of heat accumulation in the element due to heat flow amp internal heat generation)

)( volumem d

TmCQ pst )(

T

CqTK pg)(

KKKyKx z If material is homogeneous amp isotropic

T

pC

gq

z

T

zy

T

yx

T

xK ][][][

T

Kp

C

Kgq

z

T

y

T

x

T2

2

2

2

2

2

T

Kgq

z

T

y

T

x

T 12

2

2

2

2

2

Thermal diffusivity pC

K

T

K

qT g

12

according to thermodynamics first law

)( zr )( dzrddr

)( dzrddr

Cylindrical polar co-ordinates

Elemental length

Consider an elemental volume ABCDEFGH =

GENERAL HEAT CONDUCTION EQUATION IN CYLINDRICAL COORDINATES

gq

gq

Consider an unsteady (transient) three dimensional heat conduction with internal heat generation in heterogeneous amp anisotropic cylindrical geometries (eg rods and pipes)

)( zrTLetTemperature distribution is a function of T=

Let

=internal heat generation per unit volume per unit time (W m3)q= rate of heat flow (W)Q= qdt=total heat flow (J)r=mass density of the material element(kgm3)Cp=specific heat of the material element (Jkg K)

planerdirectionz

planezrdirection

planezdirectionr

)(

)(

)(

LetHeat flow in

drdr

TdzrdK

rdQ

QQdQ

rr

drrrr

])([

)1(][

r

Tr

rddrdzdKdQ rr

Heat accumulation in the element due to heat flow by conduction in lsquorrsquo direction

rddr

TdzdrK

rdQ

QQdQ d

])([

][

2

T

r

ddrdzrdKdQ

)2(][

2

2

2

T

r

ddrdzrdKdQ

Heat accumulation in the element due to heat flow by conduction in lsquoqrsquo direction

Plane (z- q ) lsquorrsquo direction

dr

TdzrdKQ rr

)(

drQr

QQ rrdrr ][

Quantity of heat leaving from the face EFGH in lsquorrsquo direction

Quantity of heat flowing into the face ABCD in lsquorrsquo direction

Plane (r-z) lsquoq lsquodirection

Quantity of heat leaving from the face AEGC in lsquoqrsquo direction

d

r

TdzdrKQ )(

rdQ

rQQ d ][

Quantity of heat flowing into the face BFHD in lsquoqrsquo direction

)3())((2

2

z

TdzdrrdKdQ zz

)5()()(

dT

CdzdrrdQ pst

)4()()(

ddzdrrdqQgg

Total amount of internal heat generation in the element

Total heat energy stored in the element

(ie thermal energy of the element can be increased because of heat accumulation in the element due to heat flow amp internal heat generation)

according to thermodynamics first law

Ein + Egen = Eout + Est (Ein ndash Eout) + Egen = Est

)( volumem

)()(z

T

zdzdrrdKdQ zz

Heat accumulation in the element due to heat flow by conduction in lsquozrsquo direction

dzdz

TdrrdK

zdQ

QQdQ

zz

dzzzz

])([

d

TmCQ pst )(

Plane (r-q) lsquozrsquo direction

dz

TdrrdKQ zz )(

dzQz

QQ zzdzz ][ Quantity of heat leaving from the face CDHG in lsquozrsquo direction

Quantity of heat flowing into the face ABFE in lsquozrsquo direction

stgzr QQdQdQdQ

)(

r

r ][)(rT

rr

ddrdzdKr

][)(

2

2

2

T

r

ddrdzrdK ))((2

2

z

TddzdrrdK z

ddzdrrdq g )(

dT

Cdzdrrd p )(

+ + +

=

T

CqTK pg)(KKKK zr

[gq

)(1

r

Tr

rrK r

)(1

2

2

2 T

rK )(

2

2

z

TK z

)( ddzdrrd )(

ddzdrrd

TC p

+ + + ] =

If material is homogeneous amp isotropic

)(1

r

Tr

rr

)(1

2

2

2 T

r)(

2

2

z

T

K

q g

T

K

C p

)(1

r

Tr

rr

)(1

2

2

2 T

r )(2

2

z

T

K

q g

T

1

+ + + =

)(1

r

Tr

rrK r

)(1

2

2

2 T

rK )(

2

2

z

TK z

gq

T

C p + + + =

+ + + = Thermal diffusivity

T

K

qT g

12

pC

K

(Ein ndash Eout) + Egen = Est

)( rT

)( r

)sin( drrddr)sin( drrddr

planerdirection

planerdirection

planedirectionr

)(

)(

)(

Spherical polar co-ordinates

Elemental length

Consider an elemental volume ABCDEFGH =

GENERAL HEAT CONDUCTION EQUATION IN SPHERICAL COORDINATES

Let

=internal heat generation per unit volume per unit time (W m3)q= rate of heat flow (W)Q= qdt=total heat flow (J)r=mass density of the material element(kgm3)Cp=specific heat of the material element (Jkg K)

gq

Consider an unsteady (transient) three dimensional heat conduction with internal heat generation in heterogeneous amp anisotropic spherical geometries

LetTemperature distribution is a function of T=

X

Y

Z

drT

drrdKQ rr )sin(

drQr

QQ rrdrr ][

drdr

TdrrdK

rdQ

QQdQ

rr

drrrr

])sin([

)1(]([sin 2

rT

rr

ddrddKdQ rr

Plane ( -q f) lsquorrsquo directionQuantity of heat flowing into the face ABCD in lsquorrsquo direction

Quantity of heat leaving from the face EFGH in lsquorrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquorrsquo direction

drT

drdrKQ )sin(

rdQ

rQQ d ][

rddr

TdrdrK

rdQ

QQdQ d

])sin([

])[(sin

2

T

r

drdrdrKdQ

)2(])(sin[

2

T

r

drdrddrKdQ

Quantity of heat leaving from the face BFHD in lsquoqrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoqrsquo direction

X

Y

Z

Plane (r- f ) lsquoq lsquodirection

Quantity of heat flowing into the face AEGC in lsquoqrsquo direction

dr

TdrrdKQ

sin)(

drQ

rQQ d sin][

sin

drdr

TdrrdK

rdQ

QQdQ d

sin]sin

)([sin

][sinsin

)sin(

T

rr

ddrdrrdKdQ

)3()(sin

)sin(2

2

22

2

T

r

ddddrrKdQ

Plane (r- ) q lsquofrsquo directionQuantity of heat flowing into the face ABFE in lsquofrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquozrsquo direction

)4()sin()(

ddrrddrqQgg

)5()sin()(

dT

CdrrddrQ pst

Total amount of internal heat generation in the element

Total heat energy stored in the element

(Ein ndash Eout) + Egen = Est

d

TmCQ pst )(

)( volumem

Quantity of heat leaving from the face CDHG in lsquofrsquo direction

according to thermodynamics first law

Ein + Egen = Eout + Est

X

Y

Z

stgr QQdQdQdQ

)(

]([sin 2

r

Tr

rddrddKr

])(sin[

2

T

rdrdrddrK

)(sin

)sin(2

2

22

2

T

r

ddddrrK

ddrrddrq g )sin(

dT

Cdrrddr p )sin(

2

2

r

r

sin

sin

]([ 22 r

Tr

rr

Kr

])(sin[sin2

TrK

)(sin 2

2

22

T

r

Kgq

T

Cp

]([1

[ 22 r

Tr

rrK

])(sin[sin1

2

T

r)](

sin

12

2

22 T

r gq

T

Cp

]([1

[ 22 r

Tr

rr

])(sin[sin1

2

T

r)](

sin

12

2

22 T

rK

q g

T

K

Cp

]([1

[ 22 r

Tr

rr

])(sin[sin1

2

T

r)](

sin

12

2

22 T

rK

q g

T1

KKKKr If material is homogeneous amp isotropic

TCqTK pg)(

T

K

qT g

12 pC

K

Thermal diffusivity

(Ein ndash Eout) + Egen = Est

General Heat Conduction Equation

PLANE WALL CYLINDER SPHERE

Temperature profile

Plane wall SphereCylinder

bull A copper conductor (k = 380 WmoC resistivity ρ = 2x10-8 Ω m) having inner and outer radii 10 cm and 225 cm respectively is carrying a current density of 4800 amperes cm2 the conductor is internally cooled and a constant temperature of 65oC is maintained at the inner surface and there is no heat transfer through the insulation surrounding the conductor

Determinebull (i) The maximum temperature of the conductor and the radius at

which it occursbull (ii) The internal heat transfer rate

bull A chemical reaction takes place in a packed bed (k=06 WmoC) between two coaxial cylinders with radii 15 mm and 45 mm the inner surface is at 580oC and it is insulated Assuming the reaction rate of 055 MWm3 in the reactor volume find the temperature at the outer surface of the reactor

bull A current of 300 amperes passes through a stainless steel wire of 25 mm diameter and k=20 WmoC The resistivity of the wire is 70x10-8 ohm-m and length of the wire is 2 m if the wire is submerged in a fluid maintained at 50oC and h at the wire surface is 4000 Wm2oC calculate the steady state temperature at the centre and at the surface of the wire

bull An approximately spherical shaped orange ( k=023 WmoC) 90 mm diameter undergoes ripping process and generates 5100 Wm3 of energy If external surface the orange is at 8oC determine

bull (i) temperature at the centre of the orange and

bull (ii) heat flow from the outer surface of the orange

bull FINS External surfaces to used to cooling area and hence convection

heat transfer can be improvedbull APPLICATIONS IC engine cylinder ( air cooling) Heat exchangers automobile

radiators computer chip refrigerators electric motors generators and transformers etc

bull ASSUMPTIONS MADE IN ANALYSIS OF HEAT TRANFER THROUGH FIN

1 Fin material is homogenous and isotropic constthermal conductivity

2 Temperature at any cs of the fin is uniform3 lsquohrsquo is uniform over entire surface4 No internal heat generation5 Thermal contact resistance is negligible6 Radiation heat transfer is negligible7 Heat conduction is steady state and one dimensional

bull Problem 1

bull A stainless steel fin (k=20 WmK) having a diameter of 20 mm and a length of 01 m is attached to a wall at 300oC The ambient temperature is 50oC and the heat transfer coefficient is 10 Wm2K The fin tip is insulated Determine

bull (A) The rate of heat dissipation from the finbull (B) The temperature at the fin tipbull (C) The rate of heat transfer from the wall area covered by the fin if

the fin was not usedbull (D) The heat transfer rate from the same fin geometry if the

stainless steel fin is replaced by a fictitious fin with infinite thermal conductivity

bull Problem2bull One end of a long rod is inserted into a furnace while the other

projects into ambient temperature Under steady state the temperature of the rod is measured at two points 75mm apart and found to be 125oC and 885oC respectively while the ambient temperature is 20oC If the rod is 25mm in diameter and h is 2336 Wm2K

bull Find the thermal conductivity of the rod material

  • HEAT AND MASS TRANSFER
  • Thermodynamics Vs Heat transfer
  • MODE OF HEAT TRANSFER
  • MODE OF HEAT TRANSFER (2)
  • CONDUCTION- MECHANISM
  • CONCECTION-MECHANISM
  • RADIATION-MECHANISM
  • Conduction- Convection- Radiation
  • Slide 9
  • Slide 10
  • Fourierrsquos Law of Heat Conduction
  • Slide 12
  • Newtonrsquos Law of Cooling
  • Laws of Thermal Radiation
  • LAWS FOR HEAT TRANSFER
  • ELECTRICAL ampTHERMAL ANALOGY
  • Slide 17
  • Slide 18
  • PLANE WALL-UNIDIRECTIONAL
  • PLANE WALL -SERIES
  • PLANE WALL IN SERIES
  • PLANE WALL -PARALLEL
  • PLANE WALL-PARALLEL
  • PLANE WALL-COMBINATION
  • PLANE WALL-FILM RESISTANCE
  • PLANE WALL- THERMAL CONTACT RESISTANCE
  • PLANE WALL- THERMAL CONTACT RESISTANCE (2)
  • CYLINDER-SOLID
  • HEAT TRANSFER IN SOLID CYLINDER
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER-COMPOSITE
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER-MULTY LAYER
  • HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER
  • HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER (2)
  • HEAT TRANSFER IN SPHERE
  • HEAT TRANSFER IN SPHERE -MULTY LAYER
  • HEAT TRANSFER IN SPHERE -MULTY LAYER (2)
  • Slide 38
  • Slide 39
  • Slide 40
  • Slide 41
  • CRITICAL RADIUS
  • Plane wall- Insulation thickness -study
  • Problem5
  • Slide 45
  • Slide 46
  • Slide 47
  • GENERAL HEAT CONDUCTION EQUATION IN CYLINDRICAL COORDINATES
  • Slide 49
  • Slide 50
  • Slide 51
  • GENERAL HEAT CONDUCTION EQUATION IN SPHERICAL COORDINATES
  • Slide 53
  • Slide 54
  • Slide 55
  • General Heat Conduction Equation
  • Temperature profile
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Slide 67
Page 23: hmt chap 1

PLANE WALL- THERMAL CONTACT RESISTANCE

At the joining surface of the two slabs air is trapped in voids due to surface irregularities

CYLINDER-SOLID

HEAT TRANSFER IN SOLID CYLINDER

HEAT TRANSFER RATE IN HOLLOW CYLINDER

HEAT TRANSFER RATE IN HOLLOW CYLINDER-COMPOSITE

HEAT TRANSFER RATE IN HOLLOW CYLINDER-MULTY LAYER

HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER

HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER

HEAT TRANSFER IN SPHERE

HEAT TRANSFER IN SPHERE -MULTY LAYER

HEAT TRANSFER IN SPHERE -MULTY LAYER

bull Problem 1 The wall in a furnace consists of 125 mm thick refractory bricks and

125 mm thick insulating fire bricks separated by an air gap A 12 mm thick plaster covers the outer wall The inner surface of wall is at 1100oC and the ambient temperature 25oC the heat transfer coefficient on the outer wall to the air is 17 Wm2K and the resistance to heat flow of the air gap is 016 KW the thermal conductivities of refractory brick insulating firebrick and plaster are 16 03 and 014 WmK respectively

Calculatebull (a) the rate of heat loss per unit area of the wall surfacebull (b) the interface temperatures throughout the wallbull (c) the temperature at the outside surface of the wall

bull Problem 2Two large aluminum plates (k=240 WmK) each 2 cm thick with 10 mm surface roughness are placed in contact under 105 Nm2 pressure in air The temperature at the outside surfaces are 390 oC and 406 oCCalculate

A) the heat flux B) the temperature drop due to the contact resistance C) the contact temperatures Thermal contact resistance with air as the interfacial fluid for

10 mm roughness is 27510-4 m2KW

bull Problem 3bull Steam at 350 oC flowing in a pipe (k=80 WmK) of 5 cm

inner diameter and 56 cm outer diameter is covered with 3 cm thick insulation of k=005 WmK Heat is lost to surroundings at 5 oC by natural convection and radiation the combined h being 20 Wm2K Taking the heat transfer coefficient inside the pipe as 60 Wm2K

bull Determinebull A) the rate of heat loss from the steam per unit length of

the pipebull B) the temperature drop across the pipe and the insulation

bull PROBLEM4bull A nuclear reactor has a sperical pressure vessel 750 mm inside

diameter 80 mm wall thickness The temperature at the inner surface is 500 oC If the tempertarue at the outer surface is 495 oC

bull Calculatea) The rate of heat loss through the metal wallb) If the rate of heat loss is to limited to 305Whow much inulation

thickness needs to be applied Thermal conductivity of steel wall=46 WmoC Thermal conductivity of insulation=004 WmoC Film coefficient of heat tranfer at inner and outer surface are 200

Wm2 oC and 15 Wm2 oC Ambient air temperature=28 oC

CRITICAL RADIUS

bull The radius upto which heat flow increases and after decreases At critical radius q becomes qmax

bull The addition of insulation always increases (Rth)condbull Rtotal= (Rth)cond + (Rth)conv

bull The addition of insulation may reduce the (Rth)conv due to increase in surface area

bull r1ltrc -heat transfer increases by adding insulation till r1=rc If

insulation thickness is further increased the rate of heat loss will decrease

r1 gt rc

- heat transfer decreases by adding insulation above the rc

Plane wall- Insulation

thickness -study

Cylinder amp Sphere- Insulation thickness -

study

Problem5bull A 3mm diameter and 5 mm long electric wire is tightly

wrapped with a 2 mm thick plastic cover whose K=015 wmoC Electrical measurements indicate that a current of 10 A passes through the wire and there is a voltage drop of 8 V along the wire If the insulated wire is exposed to a medium at Ta=30oC with a heat transfer coefficient of h=12 wm2oC Determine the temperature at the interface of the wire and the plastic cover in steady operation Also determine whether doubling the thickness of the plastic cover will increase or decrease this interface temperature

dx

TdzdyKQ xx )(

dxQx

QQ xxdxx ][

dxdx

TdzdyK

xQQdQ xdxxxx ])([)(

)1()]([)(

ddzdydx

x

TK

xdQ xx

GENERAL HEAT CONDUCTION EQUATION IN CARTESIAN COORDINATES

Consider an unsteady (transient) three dimensional heat conduction with internal heat generation in heterogeneous amp anisotropic PLANE WALL

Cartesian co-ordinates x y z

Elemental length dxdydz

Consider an elemental volume ABCDEFGH = (dxdydz)

Let Temperature distribution is a function of T=T(xyz)

Let

=internal heat generation per unit volume per unit time (W m3)q= rate of heat flow (W)Q= qdt=total heat flow (J)r=mass density of the material element(kgm3)Cp=specific heat of the material element (Jkg K)

gq

Plane (y - z) lsquoxrsquo direction

Quantity of heat leaving from the face EFGH in lsquoXrsquo direction

Quantity of heat flowing into the face ABCD in lsquoXrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoXrsquo direction

dy

TdzdxKQ yy )(

dyQy

QQ yydyy ][

dydy

TdzdxK

yQQdQ ydyyyy ])([)(

)2()]([)(

ddzdydx

y

TK

ydQ yy

Plane (x - z) lsquoyrsquo direction

Quantity of heat flowing into the face ABFE in lsquoYrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoYrsquo direction

dz

TdydxKQ zz )(

dzQz

QQ zzdzz ][

dzdz

TdydxK

zQQdQ zdzzzz ])([)(

)3()]([)(

ddzdydx

z

TK

zdQ zz

Plane (x - y) lsquozrsquo direction

Quantity of heat flowing into the face ADHE in lsquoZrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoZrsquo direction

Quantity of heat leaving from the face DCGH in lsquoYrsquo direction

Quantity of heat leaving from the face BCGF in lsquoZrsquo direction

)4()()(

ddzdydxqQgg

Total amount of internal heat generation in the element

Ein + Egen = Eout+Est (Ein- Eout)+ Egen=Est

T

Cddzdydxddzdydxqddzdydxz

TK

zy

TK

yx

TK

x pgzyx )()())](()()([

T

pC

gq

z

TzK

zy

Ty

Kyx

TxK

x][][][

stgzyx QQdQdQdQ

)(

)5()()(

dT

CdzdydxQ pst

Total heat energy stored in the element

(ie thermal energy of the element can be increased because of heat accumulation in the element due to heat flow amp internal heat generation)

)( volumem d

TmCQ pst )(

T

CqTK pg)(

KKKyKx z If material is homogeneous amp isotropic

T

pC

gq

z

T

zy

T

yx

T

xK ][][][

T

Kp

C

Kgq

z

T

y

T

x

T2

2

2

2

2

2

T

Kgq

z

T

y

T

x

T 12

2

2

2

2

2

Thermal diffusivity pC

K

T

K

qT g

12

according to thermodynamics first law

)( zr )( dzrddr

)( dzrddr

Cylindrical polar co-ordinates

Elemental length

Consider an elemental volume ABCDEFGH =

GENERAL HEAT CONDUCTION EQUATION IN CYLINDRICAL COORDINATES

gq

gq

Consider an unsteady (transient) three dimensional heat conduction with internal heat generation in heterogeneous amp anisotropic cylindrical geometries (eg rods and pipes)

)( zrTLetTemperature distribution is a function of T=

Let

=internal heat generation per unit volume per unit time (W m3)q= rate of heat flow (W)Q= qdt=total heat flow (J)r=mass density of the material element(kgm3)Cp=specific heat of the material element (Jkg K)

planerdirectionz

planezrdirection

planezdirectionr

)(

)(

)(

LetHeat flow in

drdr

TdzrdK

rdQ

QQdQ

rr

drrrr

])([

)1(][

r

Tr

rddrdzdKdQ rr

Heat accumulation in the element due to heat flow by conduction in lsquorrsquo direction

rddr

TdzdrK

rdQ

QQdQ d

])([

][

2

T

r

ddrdzrdKdQ

)2(][

2

2

2

T

r

ddrdzrdKdQ

Heat accumulation in the element due to heat flow by conduction in lsquoqrsquo direction

Plane (z- q ) lsquorrsquo direction

dr

TdzrdKQ rr

)(

drQr

QQ rrdrr ][

Quantity of heat leaving from the face EFGH in lsquorrsquo direction

Quantity of heat flowing into the face ABCD in lsquorrsquo direction

Plane (r-z) lsquoq lsquodirection

Quantity of heat leaving from the face AEGC in lsquoqrsquo direction

d

r

TdzdrKQ )(

rdQ

rQQ d ][

Quantity of heat flowing into the face BFHD in lsquoqrsquo direction

)3())((2

2

z

TdzdrrdKdQ zz

)5()()(

dT

CdzdrrdQ pst

)4()()(

ddzdrrdqQgg

Total amount of internal heat generation in the element

Total heat energy stored in the element

(ie thermal energy of the element can be increased because of heat accumulation in the element due to heat flow amp internal heat generation)

according to thermodynamics first law

Ein + Egen = Eout + Est (Ein ndash Eout) + Egen = Est

)( volumem

)()(z

T

zdzdrrdKdQ zz

Heat accumulation in the element due to heat flow by conduction in lsquozrsquo direction

dzdz

TdrrdK

zdQ

QQdQ

zz

dzzzz

])([

d

TmCQ pst )(

Plane (r-q) lsquozrsquo direction

dz

TdrrdKQ zz )(

dzQz

QQ zzdzz ][ Quantity of heat leaving from the face CDHG in lsquozrsquo direction

Quantity of heat flowing into the face ABFE in lsquozrsquo direction

stgzr QQdQdQdQ

)(

r

r ][)(rT

rr

ddrdzdKr

][)(

2

2

2

T

r

ddrdzrdK ))((2

2

z

TddzdrrdK z

ddzdrrdq g )(

dT

Cdzdrrd p )(

+ + +

=

T

CqTK pg)(KKKK zr

[gq

)(1

r

Tr

rrK r

)(1

2

2

2 T

rK )(

2

2

z

TK z

)( ddzdrrd )(

ddzdrrd

TC p

+ + + ] =

If material is homogeneous amp isotropic

)(1

r

Tr

rr

)(1

2

2

2 T

r)(

2

2

z

T

K

q g

T

K

C p

)(1

r

Tr

rr

)(1

2

2

2 T

r )(2

2

z

T

K

q g

T

1

+ + + =

)(1

r

Tr

rrK r

)(1

2

2

2 T

rK )(

2

2

z

TK z

gq

T

C p + + + =

+ + + = Thermal diffusivity

T

K

qT g

12

pC

K

(Ein ndash Eout) + Egen = Est

)( rT

)( r

)sin( drrddr)sin( drrddr

planerdirection

planerdirection

planedirectionr

)(

)(

)(

Spherical polar co-ordinates

Elemental length

Consider an elemental volume ABCDEFGH =

GENERAL HEAT CONDUCTION EQUATION IN SPHERICAL COORDINATES

Let

=internal heat generation per unit volume per unit time (W m3)q= rate of heat flow (W)Q= qdt=total heat flow (J)r=mass density of the material element(kgm3)Cp=specific heat of the material element (Jkg K)

gq

Consider an unsteady (transient) three dimensional heat conduction with internal heat generation in heterogeneous amp anisotropic spherical geometries

LetTemperature distribution is a function of T=

X

Y

Z

drT

drrdKQ rr )sin(

drQr

QQ rrdrr ][

drdr

TdrrdK

rdQ

QQdQ

rr

drrrr

])sin([

)1(]([sin 2

rT

rr

ddrddKdQ rr

Plane ( -q f) lsquorrsquo directionQuantity of heat flowing into the face ABCD in lsquorrsquo direction

Quantity of heat leaving from the face EFGH in lsquorrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquorrsquo direction

drT

drdrKQ )sin(

rdQ

rQQ d ][

rddr

TdrdrK

rdQ

QQdQ d

])sin([

])[(sin

2

T

r

drdrdrKdQ

)2(])(sin[

2

T

r

drdrddrKdQ

Quantity of heat leaving from the face BFHD in lsquoqrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoqrsquo direction

X

Y

Z

Plane (r- f ) lsquoq lsquodirection

Quantity of heat flowing into the face AEGC in lsquoqrsquo direction

dr

TdrrdKQ

sin)(

drQ

rQQ d sin][

sin

drdr

TdrrdK

rdQ

QQdQ d

sin]sin

)([sin

][sinsin

)sin(

T

rr

ddrdrrdKdQ

)3()(sin

)sin(2

2

22

2

T

r

ddddrrKdQ

Plane (r- ) q lsquofrsquo directionQuantity of heat flowing into the face ABFE in lsquofrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquozrsquo direction

)4()sin()(

ddrrddrqQgg

)5()sin()(

dT

CdrrddrQ pst

Total amount of internal heat generation in the element

Total heat energy stored in the element

(Ein ndash Eout) + Egen = Est

d

TmCQ pst )(

)( volumem

Quantity of heat leaving from the face CDHG in lsquofrsquo direction

according to thermodynamics first law

Ein + Egen = Eout + Est

X

Y

Z

stgr QQdQdQdQ

)(

]([sin 2

r

Tr

rddrddKr

])(sin[

2

T

rdrdrddrK

)(sin

)sin(2

2

22

2

T

r

ddddrrK

ddrrddrq g )sin(

dT

Cdrrddr p )sin(

2

2

r

r

sin

sin

]([ 22 r

Tr

rr

Kr

])(sin[sin2

TrK

)(sin 2

2

22

T

r

Kgq

T

Cp

]([1

[ 22 r

Tr

rrK

])(sin[sin1

2

T

r)](

sin

12

2

22 T

r gq

T

Cp

]([1

[ 22 r

Tr

rr

])(sin[sin1

2

T

r)](

sin

12

2

22 T

rK

q g

T

K

Cp

]([1

[ 22 r

Tr

rr

])(sin[sin1

2

T

r)](

sin

12

2

22 T

rK

q g

T1

KKKKr If material is homogeneous amp isotropic

TCqTK pg)(

T

K

qT g

12 pC

K

Thermal diffusivity

(Ein ndash Eout) + Egen = Est

General Heat Conduction Equation

PLANE WALL CYLINDER SPHERE

Temperature profile

Plane wall SphereCylinder

bull A copper conductor (k = 380 WmoC resistivity ρ = 2x10-8 Ω m) having inner and outer radii 10 cm and 225 cm respectively is carrying a current density of 4800 amperes cm2 the conductor is internally cooled and a constant temperature of 65oC is maintained at the inner surface and there is no heat transfer through the insulation surrounding the conductor

Determinebull (i) The maximum temperature of the conductor and the radius at

which it occursbull (ii) The internal heat transfer rate

bull A chemical reaction takes place in a packed bed (k=06 WmoC) between two coaxial cylinders with radii 15 mm and 45 mm the inner surface is at 580oC and it is insulated Assuming the reaction rate of 055 MWm3 in the reactor volume find the temperature at the outer surface of the reactor

bull A current of 300 amperes passes through a stainless steel wire of 25 mm diameter and k=20 WmoC The resistivity of the wire is 70x10-8 ohm-m and length of the wire is 2 m if the wire is submerged in a fluid maintained at 50oC and h at the wire surface is 4000 Wm2oC calculate the steady state temperature at the centre and at the surface of the wire

bull An approximately spherical shaped orange ( k=023 WmoC) 90 mm diameter undergoes ripping process and generates 5100 Wm3 of energy If external surface the orange is at 8oC determine

bull (i) temperature at the centre of the orange and

bull (ii) heat flow from the outer surface of the orange

bull FINS External surfaces to used to cooling area and hence convection

heat transfer can be improvedbull APPLICATIONS IC engine cylinder ( air cooling) Heat exchangers automobile

radiators computer chip refrigerators electric motors generators and transformers etc

bull ASSUMPTIONS MADE IN ANALYSIS OF HEAT TRANFER THROUGH FIN

1 Fin material is homogenous and isotropic constthermal conductivity

2 Temperature at any cs of the fin is uniform3 lsquohrsquo is uniform over entire surface4 No internal heat generation5 Thermal contact resistance is negligible6 Radiation heat transfer is negligible7 Heat conduction is steady state and one dimensional

bull Problem 1

bull A stainless steel fin (k=20 WmK) having a diameter of 20 mm and a length of 01 m is attached to a wall at 300oC The ambient temperature is 50oC and the heat transfer coefficient is 10 Wm2K The fin tip is insulated Determine

bull (A) The rate of heat dissipation from the finbull (B) The temperature at the fin tipbull (C) The rate of heat transfer from the wall area covered by the fin if

the fin was not usedbull (D) The heat transfer rate from the same fin geometry if the

stainless steel fin is replaced by a fictitious fin with infinite thermal conductivity

bull Problem2bull One end of a long rod is inserted into a furnace while the other

projects into ambient temperature Under steady state the temperature of the rod is measured at two points 75mm apart and found to be 125oC and 885oC respectively while the ambient temperature is 20oC If the rod is 25mm in diameter and h is 2336 Wm2K

bull Find the thermal conductivity of the rod material

  • HEAT AND MASS TRANSFER
  • Thermodynamics Vs Heat transfer
  • MODE OF HEAT TRANSFER
  • MODE OF HEAT TRANSFER (2)
  • CONDUCTION- MECHANISM
  • CONCECTION-MECHANISM
  • RADIATION-MECHANISM
  • Conduction- Convection- Radiation
  • Slide 9
  • Slide 10
  • Fourierrsquos Law of Heat Conduction
  • Slide 12
  • Newtonrsquos Law of Cooling
  • Laws of Thermal Radiation
  • LAWS FOR HEAT TRANSFER
  • ELECTRICAL ampTHERMAL ANALOGY
  • Slide 17
  • Slide 18
  • PLANE WALL-UNIDIRECTIONAL
  • PLANE WALL -SERIES
  • PLANE WALL IN SERIES
  • PLANE WALL -PARALLEL
  • PLANE WALL-PARALLEL
  • PLANE WALL-COMBINATION
  • PLANE WALL-FILM RESISTANCE
  • PLANE WALL- THERMAL CONTACT RESISTANCE
  • PLANE WALL- THERMAL CONTACT RESISTANCE (2)
  • CYLINDER-SOLID
  • HEAT TRANSFER IN SOLID CYLINDER
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER-COMPOSITE
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER-MULTY LAYER
  • HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER
  • HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER (2)
  • HEAT TRANSFER IN SPHERE
  • HEAT TRANSFER IN SPHERE -MULTY LAYER
  • HEAT TRANSFER IN SPHERE -MULTY LAYER (2)
  • Slide 38
  • Slide 39
  • Slide 40
  • Slide 41
  • CRITICAL RADIUS
  • Plane wall- Insulation thickness -study
  • Problem5
  • Slide 45
  • Slide 46
  • Slide 47
  • GENERAL HEAT CONDUCTION EQUATION IN CYLINDRICAL COORDINATES
  • Slide 49
  • Slide 50
  • Slide 51
  • GENERAL HEAT CONDUCTION EQUATION IN SPHERICAL COORDINATES
  • Slide 53
  • Slide 54
  • Slide 55
  • General Heat Conduction Equation
  • Temperature profile
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Slide 67
Page 24: hmt chap 1

CYLINDER-SOLID

HEAT TRANSFER IN SOLID CYLINDER

HEAT TRANSFER RATE IN HOLLOW CYLINDER

HEAT TRANSFER RATE IN HOLLOW CYLINDER-COMPOSITE

HEAT TRANSFER RATE IN HOLLOW CYLINDER-MULTY LAYER

HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER

HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER

HEAT TRANSFER IN SPHERE

HEAT TRANSFER IN SPHERE -MULTY LAYER

HEAT TRANSFER IN SPHERE -MULTY LAYER

bull Problem 1 The wall in a furnace consists of 125 mm thick refractory bricks and

125 mm thick insulating fire bricks separated by an air gap A 12 mm thick plaster covers the outer wall The inner surface of wall is at 1100oC and the ambient temperature 25oC the heat transfer coefficient on the outer wall to the air is 17 Wm2K and the resistance to heat flow of the air gap is 016 KW the thermal conductivities of refractory brick insulating firebrick and plaster are 16 03 and 014 WmK respectively

Calculatebull (a) the rate of heat loss per unit area of the wall surfacebull (b) the interface temperatures throughout the wallbull (c) the temperature at the outside surface of the wall

bull Problem 2Two large aluminum plates (k=240 WmK) each 2 cm thick with 10 mm surface roughness are placed in contact under 105 Nm2 pressure in air The temperature at the outside surfaces are 390 oC and 406 oCCalculate

A) the heat flux B) the temperature drop due to the contact resistance C) the contact temperatures Thermal contact resistance with air as the interfacial fluid for

10 mm roughness is 27510-4 m2KW

bull Problem 3bull Steam at 350 oC flowing in a pipe (k=80 WmK) of 5 cm

inner diameter and 56 cm outer diameter is covered with 3 cm thick insulation of k=005 WmK Heat is lost to surroundings at 5 oC by natural convection and radiation the combined h being 20 Wm2K Taking the heat transfer coefficient inside the pipe as 60 Wm2K

bull Determinebull A) the rate of heat loss from the steam per unit length of

the pipebull B) the temperature drop across the pipe and the insulation

bull PROBLEM4bull A nuclear reactor has a sperical pressure vessel 750 mm inside

diameter 80 mm wall thickness The temperature at the inner surface is 500 oC If the tempertarue at the outer surface is 495 oC

bull Calculatea) The rate of heat loss through the metal wallb) If the rate of heat loss is to limited to 305Whow much inulation

thickness needs to be applied Thermal conductivity of steel wall=46 WmoC Thermal conductivity of insulation=004 WmoC Film coefficient of heat tranfer at inner and outer surface are 200

Wm2 oC and 15 Wm2 oC Ambient air temperature=28 oC

CRITICAL RADIUS

bull The radius upto which heat flow increases and after decreases At critical radius q becomes qmax

bull The addition of insulation always increases (Rth)condbull Rtotal= (Rth)cond + (Rth)conv

bull The addition of insulation may reduce the (Rth)conv due to increase in surface area

bull r1ltrc -heat transfer increases by adding insulation till r1=rc If

insulation thickness is further increased the rate of heat loss will decrease

r1 gt rc

- heat transfer decreases by adding insulation above the rc

Plane wall- Insulation

thickness -study

Cylinder amp Sphere- Insulation thickness -

study

Problem5bull A 3mm diameter and 5 mm long electric wire is tightly

wrapped with a 2 mm thick plastic cover whose K=015 wmoC Electrical measurements indicate that a current of 10 A passes through the wire and there is a voltage drop of 8 V along the wire If the insulated wire is exposed to a medium at Ta=30oC with a heat transfer coefficient of h=12 wm2oC Determine the temperature at the interface of the wire and the plastic cover in steady operation Also determine whether doubling the thickness of the plastic cover will increase or decrease this interface temperature

dx

TdzdyKQ xx )(

dxQx

QQ xxdxx ][

dxdx

TdzdyK

xQQdQ xdxxxx ])([)(

)1()]([)(

ddzdydx

x

TK

xdQ xx

GENERAL HEAT CONDUCTION EQUATION IN CARTESIAN COORDINATES

Consider an unsteady (transient) three dimensional heat conduction with internal heat generation in heterogeneous amp anisotropic PLANE WALL

Cartesian co-ordinates x y z

Elemental length dxdydz

Consider an elemental volume ABCDEFGH = (dxdydz)

Let Temperature distribution is a function of T=T(xyz)

Let

=internal heat generation per unit volume per unit time (W m3)q= rate of heat flow (W)Q= qdt=total heat flow (J)r=mass density of the material element(kgm3)Cp=specific heat of the material element (Jkg K)

gq

Plane (y - z) lsquoxrsquo direction

Quantity of heat leaving from the face EFGH in lsquoXrsquo direction

Quantity of heat flowing into the face ABCD in lsquoXrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoXrsquo direction

dy

TdzdxKQ yy )(

dyQy

QQ yydyy ][

dydy

TdzdxK

yQQdQ ydyyyy ])([)(

)2()]([)(

ddzdydx

y

TK

ydQ yy

Plane (x - z) lsquoyrsquo direction

Quantity of heat flowing into the face ABFE in lsquoYrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoYrsquo direction

dz

TdydxKQ zz )(

dzQz

QQ zzdzz ][

dzdz

TdydxK

zQQdQ zdzzzz ])([)(

)3()]([)(

ddzdydx

z

TK

zdQ zz

Plane (x - y) lsquozrsquo direction

Quantity of heat flowing into the face ADHE in lsquoZrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoZrsquo direction

Quantity of heat leaving from the face DCGH in lsquoYrsquo direction

Quantity of heat leaving from the face BCGF in lsquoZrsquo direction

)4()()(

ddzdydxqQgg

Total amount of internal heat generation in the element

Ein + Egen = Eout+Est (Ein- Eout)+ Egen=Est

T

Cddzdydxddzdydxqddzdydxz

TK

zy

TK

yx

TK

x pgzyx )()())](()()([

T

pC

gq

z

TzK

zy

Ty

Kyx

TxK

x][][][

stgzyx QQdQdQdQ

)(

)5()()(

dT

CdzdydxQ pst

Total heat energy stored in the element

(ie thermal energy of the element can be increased because of heat accumulation in the element due to heat flow amp internal heat generation)

)( volumem d

TmCQ pst )(

T

CqTK pg)(

KKKyKx z If material is homogeneous amp isotropic

T

pC

gq

z

T

zy

T

yx

T

xK ][][][

T

Kp

C

Kgq

z

T

y

T

x

T2

2

2

2

2

2

T

Kgq

z

T

y

T

x

T 12

2

2

2

2

2

Thermal diffusivity pC

K

T

K

qT g

12

according to thermodynamics first law

)( zr )( dzrddr

)( dzrddr

Cylindrical polar co-ordinates

Elemental length

Consider an elemental volume ABCDEFGH =

GENERAL HEAT CONDUCTION EQUATION IN CYLINDRICAL COORDINATES

gq

gq

Consider an unsteady (transient) three dimensional heat conduction with internal heat generation in heterogeneous amp anisotropic cylindrical geometries (eg rods and pipes)

)( zrTLetTemperature distribution is a function of T=

Let

=internal heat generation per unit volume per unit time (W m3)q= rate of heat flow (W)Q= qdt=total heat flow (J)r=mass density of the material element(kgm3)Cp=specific heat of the material element (Jkg K)

planerdirectionz

planezrdirection

planezdirectionr

)(

)(

)(

LetHeat flow in

drdr

TdzrdK

rdQ

QQdQ

rr

drrrr

])([

)1(][

r

Tr

rddrdzdKdQ rr

Heat accumulation in the element due to heat flow by conduction in lsquorrsquo direction

rddr

TdzdrK

rdQ

QQdQ d

])([

][

2

T

r

ddrdzrdKdQ

)2(][

2

2

2

T

r

ddrdzrdKdQ

Heat accumulation in the element due to heat flow by conduction in lsquoqrsquo direction

Plane (z- q ) lsquorrsquo direction

dr

TdzrdKQ rr

)(

drQr

QQ rrdrr ][

Quantity of heat leaving from the face EFGH in lsquorrsquo direction

Quantity of heat flowing into the face ABCD in lsquorrsquo direction

Plane (r-z) lsquoq lsquodirection

Quantity of heat leaving from the face AEGC in lsquoqrsquo direction

d

r

TdzdrKQ )(

rdQ

rQQ d ][

Quantity of heat flowing into the face BFHD in lsquoqrsquo direction

)3())((2

2

z

TdzdrrdKdQ zz

)5()()(

dT

CdzdrrdQ pst

)4()()(

ddzdrrdqQgg

Total amount of internal heat generation in the element

Total heat energy stored in the element

(ie thermal energy of the element can be increased because of heat accumulation in the element due to heat flow amp internal heat generation)

according to thermodynamics first law

Ein + Egen = Eout + Est (Ein ndash Eout) + Egen = Est

)( volumem

)()(z

T

zdzdrrdKdQ zz

Heat accumulation in the element due to heat flow by conduction in lsquozrsquo direction

dzdz

TdrrdK

zdQ

QQdQ

zz

dzzzz

])([

d

TmCQ pst )(

Plane (r-q) lsquozrsquo direction

dz

TdrrdKQ zz )(

dzQz

QQ zzdzz ][ Quantity of heat leaving from the face CDHG in lsquozrsquo direction

Quantity of heat flowing into the face ABFE in lsquozrsquo direction

stgzr QQdQdQdQ

)(

r

r ][)(rT

rr

ddrdzdKr

][)(

2

2

2

T

r

ddrdzrdK ))((2

2

z

TddzdrrdK z

ddzdrrdq g )(

dT

Cdzdrrd p )(

+ + +

=

T

CqTK pg)(KKKK zr

[gq

)(1

r

Tr

rrK r

)(1

2

2

2 T

rK )(

2

2

z

TK z

)( ddzdrrd )(

ddzdrrd

TC p

+ + + ] =

If material is homogeneous amp isotropic

)(1

r

Tr

rr

)(1

2

2

2 T

r)(

2

2

z

T

K

q g

T

K

C p

)(1

r

Tr

rr

)(1

2

2

2 T

r )(2

2

z

T

K

q g

T

1

+ + + =

)(1

r

Tr

rrK r

)(1

2

2

2 T

rK )(

2

2

z

TK z

gq

T

C p + + + =

+ + + = Thermal diffusivity

T

K

qT g

12

pC

K

(Ein ndash Eout) + Egen = Est

)( rT

)( r

)sin( drrddr)sin( drrddr

planerdirection

planerdirection

planedirectionr

)(

)(

)(

Spherical polar co-ordinates

Elemental length

Consider an elemental volume ABCDEFGH =

GENERAL HEAT CONDUCTION EQUATION IN SPHERICAL COORDINATES

Let

=internal heat generation per unit volume per unit time (W m3)q= rate of heat flow (W)Q= qdt=total heat flow (J)r=mass density of the material element(kgm3)Cp=specific heat of the material element (Jkg K)

gq

Consider an unsteady (transient) three dimensional heat conduction with internal heat generation in heterogeneous amp anisotropic spherical geometries

LetTemperature distribution is a function of T=

X

Y

Z

drT

drrdKQ rr )sin(

drQr

QQ rrdrr ][

drdr

TdrrdK

rdQ

QQdQ

rr

drrrr

])sin([

)1(]([sin 2

rT

rr

ddrddKdQ rr

Plane ( -q f) lsquorrsquo directionQuantity of heat flowing into the face ABCD in lsquorrsquo direction

Quantity of heat leaving from the face EFGH in lsquorrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquorrsquo direction

drT

drdrKQ )sin(

rdQ

rQQ d ][

rddr

TdrdrK

rdQ

QQdQ d

])sin([

])[(sin

2

T

r

drdrdrKdQ

)2(])(sin[

2

T

r

drdrddrKdQ

Quantity of heat leaving from the face BFHD in lsquoqrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoqrsquo direction

X

Y

Z

Plane (r- f ) lsquoq lsquodirection

Quantity of heat flowing into the face AEGC in lsquoqrsquo direction

dr

TdrrdKQ

sin)(

drQ

rQQ d sin][

sin

drdr

TdrrdK

rdQ

QQdQ d

sin]sin

)([sin

][sinsin

)sin(

T

rr

ddrdrrdKdQ

)3()(sin

)sin(2

2

22

2

T

r

ddddrrKdQ

Plane (r- ) q lsquofrsquo directionQuantity of heat flowing into the face ABFE in lsquofrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquozrsquo direction

)4()sin()(

ddrrddrqQgg

)5()sin()(

dT

CdrrddrQ pst

Total amount of internal heat generation in the element

Total heat energy stored in the element

(Ein ndash Eout) + Egen = Est

d

TmCQ pst )(

)( volumem

Quantity of heat leaving from the face CDHG in lsquofrsquo direction

according to thermodynamics first law

Ein + Egen = Eout + Est

X

Y

Z

stgr QQdQdQdQ

)(

]([sin 2

r

Tr

rddrddKr

])(sin[

2

T

rdrdrddrK

)(sin

)sin(2

2

22

2

T

r

ddddrrK

ddrrddrq g )sin(

dT

Cdrrddr p )sin(

2

2

r

r

sin

sin

]([ 22 r

Tr

rr

Kr

])(sin[sin2

TrK

)(sin 2

2

22

T

r

Kgq

T

Cp

]([1

[ 22 r

Tr

rrK

])(sin[sin1

2

T

r)](

sin

12

2

22 T

r gq

T

Cp

]([1

[ 22 r

Tr

rr

])(sin[sin1

2

T

r)](

sin

12

2

22 T

rK

q g

T

K

Cp

]([1

[ 22 r

Tr

rr

])(sin[sin1

2

T

r)](

sin

12

2

22 T

rK

q g

T1

KKKKr If material is homogeneous amp isotropic

TCqTK pg)(

T

K

qT g

12 pC

K

Thermal diffusivity

(Ein ndash Eout) + Egen = Est

General Heat Conduction Equation

PLANE WALL CYLINDER SPHERE

Temperature profile

Plane wall SphereCylinder

bull A copper conductor (k = 380 WmoC resistivity ρ = 2x10-8 Ω m) having inner and outer radii 10 cm and 225 cm respectively is carrying a current density of 4800 amperes cm2 the conductor is internally cooled and a constant temperature of 65oC is maintained at the inner surface and there is no heat transfer through the insulation surrounding the conductor

Determinebull (i) The maximum temperature of the conductor and the radius at

which it occursbull (ii) The internal heat transfer rate

bull A chemical reaction takes place in a packed bed (k=06 WmoC) between two coaxial cylinders with radii 15 mm and 45 mm the inner surface is at 580oC and it is insulated Assuming the reaction rate of 055 MWm3 in the reactor volume find the temperature at the outer surface of the reactor

bull A current of 300 amperes passes through a stainless steel wire of 25 mm diameter and k=20 WmoC The resistivity of the wire is 70x10-8 ohm-m and length of the wire is 2 m if the wire is submerged in a fluid maintained at 50oC and h at the wire surface is 4000 Wm2oC calculate the steady state temperature at the centre and at the surface of the wire

bull An approximately spherical shaped orange ( k=023 WmoC) 90 mm diameter undergoes ripping process and generates 5100 Wm3 of energy If external surface the orange is at 8oC determine

bull (i) temperature at the centre of the orange and

bull (ii) heat flow from the outer surface of the orange

bull FINS External surfaces to used to cooling area and hence convection

heat transfer can be improvedbull APPLICATIONS IC engine cylinder ( air cooling) Heat exchangers automobile

radiators computer chip refrigerators electric motors generators and transformers etc

bull ASSUMPTIONS MADE IN ANALYSIS OF HEAT TRANFER THROUGH FIN

1 Fin material is homogenous and isotropic constthermal conductivity

2 Temperature at any cs of the fin is uniform3 lsquohrsquo is uniform over entire surface4 No internal heat generation5 Thermal contact resistance is negligible6 Radiation heat transfer is negligible7 Heat conduction is steady state and one dimensional

bull Problem 1

bull A stainless steel fin (k=20 WmK) having a diameter of 20 mm and a length of 01 m is attached to a wall at 300oC The ambient temperature is 50oC and the heat transfer coefficient is 10 Wm2K The fin tip is insulated Determine

bull (A) The rate of heat dissipation from the finbull (B) The temperature at the fin tipbull (C) The rate of heat transfer from the wall area covered by the fin if

the fin was not usedbull (D) The heat transfer rate from the same fin geometry if the

stainless steel fin is replaced by a fictitious fin with infinite thermal conductivity

bull Problem2bull One end of a long rod is inserted into a furnace while the other

projects into ambient temperature Under steady state the temperature of the rod is measured at two points 75mm apart and found to be 125oC and 885oC respectively while the ambient temperature is 20oC If the rod is 25mm in diameter and h is 2336 Wm2K

bull Find the thermal conductivity of the rod material

  • HEAT AND MASS TRANSFER
  • Thermodynamics Vs Heat transfer
  • MODE OF HEAT TRANSFER
  • MODE OF HEAT TRANSFER (2)
  • CONDUCTION- MECHANISM
  • CONCECTION-MECHANISM
  • RADIATION-MECHANISM
  • Conduction- Convection- Radiation
  • Slide 9
  • Slide 10
  • Fourierrsquos Law of Heat Conduction
  • Slide 12
  • Newtonrsquos Law of Cooling
  • Laws of Thermal Radiation
  • LAWS FOR HEAT TRANSFER
  • ELECTRICAL ampTHERMAL ANALOGY
  • Slide 17
  • Slide 18
  • PLANE WALL-UNIDIRECTIONAL
  • PLANE WALL -SERIES
  • PLANE WALL IN SERIES
  • PLANE WALL -PARALLEL
  • PLANE WALL-PARALLEL
  • PLANE WALL-COMBINATION
  • PLANE WALL-FILM RESISTANCE
  • PLANE WALL- THERMAL CONTACT RESISTANCE
  • PLANE WALL- THERMAL CONTACT RESISTANCE (2)
  • CYLINDER-SOLID
  • HEAT TRANSFER IN SOLID CYLINDER
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER-COMPOSITE
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER-MULTY LAYER
  • HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER
  • HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER (2)
  • HEAT TRANSFER IN SPHERE
  • HEAT TRANSFER IN SPHERE -MULTY LAYER
  • HEAT TRANSFER IN SPHERE -MULTY LAYER (2)
  • Slide 38
  • Slide 39
  • Slide 40
  • Slide 41
  • CRITICAL RADIUS
  • Plane wall- Insulation thickness -study
  • Problem5
  • Slide 45
  • Slide 46
  • Slide 47
  • GENERAL HEAT CONDUCTION EQUATION IN CYLINDRICAL COORDINATES
  • Slide 49
  • Slide 50
  • Slide 51
  • GENERAL HEAT CONDUCTION EQUATION IN SPHERICAL COORDINATES
  • Slide 53
  • Slide 54
  • Slide 55
  • General Heat Conduction Equation
  • Temperature profile
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Slide 67
Page 25: hmt chap 1

HEAT TRANSFER IN SOLID CYLINDER

HEAT TRANSFER RATE IN HOLLOW CYLINDER

HEAT TRANSFER RATE IN HOLLOW CYLINDER-COMPOSITE

HEAT TRANSFER RATE IN HOLLOW CYLINDER-MULTY LAYER

HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER

HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER

HEAT TRANSFER IN SPHERE

HEAT TRANSFER IN SPHERE -MULTY LAYER

HEAT TRANSFER IN SPHERE -MULTY LAYER

bull Problem 1 The wall in a furnace consists of 125 mm thick refractory bricks and

125 mm thick insulating fire bricks separated by an air gap A 12 mm thick plaster covers the outer wall The inner surface of wall is at 1100oC and the ambient temperature 25oC the heat transfer coefficient on the outer wall to the air is 17 Wm2K and the resistance to heat flow of the air gap is 016 KW the thermal conductivities of refractory brick insulating firebrick and plaster are 16 03 and 014 WmK respectively

Calculatebull (a) the rate of heat loss per unit area of the wall surfacebull (b) the interface temperatures throughout the wallbull (c) the temperature at the outside surface of the wall

bull Problem 2Two large aluminum plates (k=240 WmK) each 2 cm thick with 10 mm surface roughness are placed in contact under 105 Nm2 pressure in air The temperature at the outside surfaces are 390 oC and 406 oCCalculate

A) the heat flux B) the temperature drop due to the contact resistance C) the contact temperatures Thermal contact resistance with air as the interfacial fluid for

10 mm roughness is 27510-4 m2KW

bull Problem 3bull Steam at 350 oC flowing in a pipe (k=80 WmK) of 5 cm

inner diameter and 56 cm outer diameter is covered with 3 cm thick insulation of k=005 WmK Heat is lost to surroundings at 5 oC by natural convection and radiation the combined h being 20 Wm2K Taking the heat transfer coefficient inside the pipe as 60 Wm2K

bull Determinebull A) the rate of heat loss from the steam per unit length of

the pipebull B) the temperature drop across the pipe and the insulation

bull PROBLEM4bull A nuclear reactor has a sperical pressure vessel 750 mm inside

diameter 80 mm wall thickness The temperature at the inner surface is 500 oC If the tempertarue at the outer surface is 495 oC

bull Calculatea) The rate of heat loss through the metal wallb) If the rate of heat loss is to limited to 305Whow much inulation

thickness needs to be applied Thermal conductivity of steel wall=46 WmoC Thermal conductivity of insulation=004 WmoC Film coefficient of heat tranfer at inner and outer surface are 200

Wm2 oC and 15 Wm2 oC Ambient air temperature=28 oC

CRITICAL RADIUS

bull The radius upto which heat flow increases and after decreases At critical radius q becomes qmax

bull The addition of insulation always increases (Rth)condbull Rtotal= (Rth)cond + (Rth)conv

bull The addition of insulation may reduce the (Rth)conv due to increase in surface area

bull r1ltrc -heat transfer increases by adding insulation till r1=rc If

insulation thickness is further increased the rate of heat loss will decrease

r1 gt rc

- heat transfer decreases by adding insulation above the rc

Plane wall- Insulation

thickness -study

Cylinder amp Sphere- Insulation thickness -

study

Problem5bull A 3mm diameter and 5 mm long electric wire is tightly

wrapped with a 2 mm thick plastic cover whose K=015 wmoC Electrical measurements indicate that a current of 10 A passes through the wire and there is a voltage drop of 8 V along the wire If the insulated wire is exposed to a medium at Ta=30oC with a heat transfer coefficient of h=12 wm2oC Determine the temperature at the interface of the wire and the plastic cover in steady operation Also determine whether doubling the thickness of the plastic cover will increase or decrease this interface temperature

dx

TdzdyKQ xx )(

dxQx

QQ xxdxx ][

dxdx

TdzdyK

xQQdQ xdxxxx ])([)(

)1()]([)(

ddzdydx

x

TK

xdQ xx

GENERAL HEAT CONDUCTION EQUATION IN CARTESIAN COORDINATES

Consider an unsteady (transient) three dimensional heat conduction with internal heat generation in heterogeneous amp anisotropic PLANE WALL

Cartesian co-ordinates x y z

Elemental length dxdydz

Consider an elemental volume ABCDEFGH = (dxdydz)

Let Temperature distribution is a function of T=T(xyz)

Let

=internal heat generation per unit volume per unit time (W m3)q= rate of heat flow (W)Q= qdt=total heat flow (J)r=mass density of the material element(kgm3)Cp=specific heat of the material element (Jkg K)

gq

Plane (y - z) lsquoxrsquo direction

Quantity of heat leaving from the face EFGH in lsquoXrsquo direction

Quantity of heat flowing into the face ABCD in lsquoXrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoXrsquo direction

dy

TdzdxKQ yy )(

dyQy

QQ yydyy ][

dydy

TdzdxK

yQQdQ ydyyyy ])([)(

)2()]([)(

ddzdydx

y

TK

ydQ yy

Plane (x - z) lsquoyrsquo direction

Quantity of heat flowing into the face ABFE in lsquoYrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoYrsquo direction

dz

TdydxKQ zz )(

dzQz

QQ zzdzz ][

dzdz

TdydxK

zQQdQ zdzzzz ])([)(

)3()]([)(

ddzdydx

z

TK

zdQ zz

Plane (x - y) lsquozrsquo direction

Quantity of heat flowing into the face ADHE in lsquoZrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoZrsquo direction

Quantity of heat leaving from the face DCGH in lsquoYrsquo direction

Quantity of heat leaving from the face BCGF in lsquoZrsquo direction

)4()()(

ddzdydxqQgg

Total amount of internal heat generation in the element

Ein + Egen = Eout+Est (Ein- Eout)+ Egen=Est

T

Cddzdydxddzdydxqddzdydxz

TK

zy

TK

yx

TK

x pgzyx )()())](()()([

T

pC

gq

z

TzK

zy

Ty

Kyx

TxK

x][][][

stgzyx QQdQdQdQ

)(

)5()()(

dT

CdzdydxQ pst

Total heat energy stored in the element

(ie thermal energy of the element can be increased because of heat accumulation in the element due to heat flow amp internal heat generation)

)( volumem d

TmCQ pst )(

T

CqTK pg)(

KKKyKx z If material is homogeneous amp isotropic

T

pC

gq

z

T

zy

T

yx

T

xK ][][][

T

Kp

C

Kgq

z

T

y

T

x

T2

2

2

2

2

2

T

Kgq

z

T

y

T

x

T 12

2

2

2

2

2

Thermal diffusivity pC

K

T

K

qT g

12

according to thermodynamics first law

)( zr )( dzrddr

)( dzrddr

Cylindrical polar co-ordinates

Elemental length

Consider an elemental volume ABCDEFGH =

GENERAL HEAT CONDUCTION EQUATION IN CYLINDRICAL COORDINATES

gq

gq

Consider an unsteady (transient) three dimensional heat conduction with internal heat generation in heterogeneous amp anisotropic cylindrical geometries (eg rods and pipes)

)( zrTLetTemperature distribution is a function of T=

Let

=internal heat generation per unit volume per unit time (W m3)q= rate of heat flow (W)Q= qdt=total heat flow (J)r=mass density of the material element(kgm3)Cp=specific heat of the material element (Jkg K)

planerdirectionz

planezrdirection

planezdirectionr

)(

)(

)(

LetHeat flow in

drdr

TdzrdK

rdQ

QQdQ

rr

drrrr

])([

)1(][

r

Tr

rddrdzdKdQ rr

Heat accumulation in the element due to heat flow by conduction in lsquorrsquo direction

rddr

TdzdrK

rdQ

QQdQ d

])([

][

2

T

r

ddrdzrdKdQ

)2(][

2

2

2

T

r

ddrdzrdKdQ

Heat accumulation in the element due to heat flow by conduction in lsquoqrsquo direction

Plane (z- q ) lsquorrsquo direction

dr

TdzrdKQ rr

)(

drQr

QQ rrdrr ][

Quantity of heat leaving from the face EFGH in lsquorrsquo direction

Quantity of heat flowing into the face ABCD in lsquorrsquo direction

Plane (r-z) lsquoq lsquodirection

Quantity of heat leaving from the face AEGC in lsquoqrsquo direction

d

r

TdzdrKQ )(

rdQ

rQQ d ][

Quantity of heat flowing into the face BFHD in lsquoqrsquo direction

)3())((2

2

z

TdzdrrdKdQ zz

)5()()(

dT

CdzdrrdQ pst

)4()()(

ddzdrrdqQgg

Total amount of internal heat generation in the element

Total heat energy stored in the element

(ie thermal energy of the element can be increased because of heat accumulation in the element due to heat flow amp internal heat generation)

according to thermodynamics first law

Ein + Egen = Eout + Est (Ein ndash Eout) + Egen = Est

)( volumem

)()(z

T

zdzdrrdKdQ zz

Heat accumulation in the element due to heat flow by conduction in lsquozrsquo direction

dzdz

TdrrdK

zdQ

QQdQ

zz

dzzzz

])([

d

TmCQ pst )(

Plane (r-q) lsquozrsquo direction

dz

TdrrdKQ zz )(

dzQz

QQ zzdzz ][ Quantity of heat leaving from the face CDHG in lsquozrsquo direction

Quantity of heat flowing into the face ABFE in lsquozrsquo direction

stgzr QQdQdQdQ

)(

r

r ][)(rT

rr

ddrdzdKr

][)(

2

2

2

T

r

ddrdzrdK ))((2

2

z

TddzdrrdK z

ddzdrrdq g )(

dT

Cdzdrrd p )(

+ + +

=

T

CqTK pg)(KKKK zr

[gq

)(1

r

Tr

rrK r

)(1

2

2

2 T

rK )(

2

2

z

TK z

)( ddzdrrd )(

ddzdrrd

TC p

+ + + ] =

If material is homogeneous amp isotropic

)(1

r

Tr

rr

)(1

2

2

2 T

r)(

2

2

z

T

K

q g

T

K

C p

)(1

r

Tr

rr

)(1

2

2

2 T

r )(2

2

z

T

K

q g

T

1

+ + + =

)(1

r

Tr

rrK r

)(1

2

2

2 T

rK )(

2

2

z

TK z

gq

T

C p + + + =

+ + + = Thermal diffusivity

T

K

qT g

12

pC

K

(Ein ndash Eout) + Egen = Est

)( rT

)( r

)sin( drrddr)sin( drrddr

planerdirection

planerdirection

planedirectionr

)(

)(

)(

Spherical polar co-ordinates

Elemental length

Consider an elemental volume ABCDEFGH =

GENERAL HEAT CONDUCTION EQUATION IN SPHERICAL COORDINATES

Let

=internal heat generation per unit volume per unit time (W m3)q= rate of heat flow (W)Q= qdt=total heat flow (J)r=mass density of the material element(kgm3)Cp=specific heat of the material element (Jkg K)

gq

Consider an unsteady (transient) three dimensional heat conduction with internal heat generation in heterogeneous amp anisotropic spherical geometries

LetTemperature distribution is a function of T=

X

Y

Z

drT

drrdKQ rr )sin(

drQr

QQ rrdrr ][

drdr

TdrrdK

rdQ

QQdQ

rr

drrrr

])sin([

)1(]([sin 2

rT

rr

ddrddKdQ rr

Plane ( -q f) lsquorrsquo directionQuantity of heat flowing into the face ABCD in lsquorrsquo direction

Quantity of heat leaving from the face EFGH in lsquorrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquorrsquo direction

drT

drdrKQ )sin(

rdQ

rQQ d ][

rddr

TdrdrK

rdQ

QQdQ d

])sin([

])[(sin

2

T

r

drdrdrKdQ

)2(])(sin[

2

T

r

drdrddrKdQ

Quantity of heat leaving from the face BFHD in lsquoqrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoqrsquo direction

X

Y

Z

Plane (r- f ) lsquoq lsquodirection

Quantity of heat flowing into the face AEGC in lsquoqrsquo direction

dr

TdrrdKQ

sin)(

drQ

rQQ d sin][

sin

drdr

TdrrdK

rdQ

QQdQ d

sin]sin

)([sin

][sinsin

)sin(

T

rr

ddrdrrdKdQ

)3()(sin

)sin(2

2

22

2

T

r

ddddrrKdQ

Plane (r- ) q lsquofrsquo directionQuantity of heat flowing into the face ABFE in lsquofrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquozrsquo direction

)4()sin()(

ddrrddrqQgg

)5()sin()(

dT

CdrrddrQ pst

Total amount of internal heat generation in the element

Total heat energy stored in the element

(Ein ndash Eout) + Egen = Est

d

TmCQ pst )(

)( volumem

Quantity of heat leaving from the face CDHG in lsquofrsquo direction

according to thermodynamics first law

Ein + Egen = Eout + Est

X

Y

Z

stgr QQdQdQdQ

)(

]([sin 2

r

Tr

rddrddKr

])(sin[

2

T

rdrdrddrK

)(sin

)sin(2

2

22

2

T

r

ddddrrK

ddrrddrq g )sin(

dT

Cdrrddr p )sin(

2

2

r

r

sin

sin

]([ 22 r

Tr

rr

Kr

])(sin[sin2

TrK

)(sin 2

2

22

T

r

Kgq

T

Cp

]([1

[ 22 r

Tr

rrK

])(sin[sin1

2

T

r)](

sin

12

2

22 T

r gq

T

Cp

]([1

[ 22 r

Tr

rr

])(sin[sin1

2

T

r)](

sin

12

2

22 T

rK

q g

T

K

Cp

]([1

[ 22 r

Tr

rr

])(sin[sin1

2

T

r)](

sin

12

2

22 T

rK

q g

T1

KKKKr If material is homogeneous amp isotropic

TCqTK pg)(

T

K

qT g

12 pC

K

Thermal diffusivity

(Ein ndash Eout) + Egen = Est

General Heat Conduction Equation

PLANE WALL CYLINDER SPHERE

Temperature profile

Plane wall SphereCylinder

bull A copper conductor (k = 380 WmoC resistivity ρ = 2x10-8 Ω m) having inner and outer radii 10 cm and 225 cm respectively is carrying a current density of 4800 amperes cm2 the conductor is internally cooled and a constant temperature of 65oC is maintained at the inner surface and there is no heat transfer through the insulation surrounding the conductor

Determinebull (i) The maximum temperature of the conductor and the radius at

which it occursbull (ii) The internal heat transfer rate

bull A chemical reaction takes place in a packed bed (k=06 WmoC) between two coaxial cylinders with radii 15 mm and 45 mm the inner surface is at 580oC and it is insulated Assuming the reaction rate of 055 MWm3 in the reactor volume find the temperature at the outer surface of the reactor

bull A current of 300 amperes passes through a stainless steel wire of 25 mm diameter and k=20 WmoC The resistivity of the wire is 70x10-8 ohm-m and length of the wire is 2 m if the wire is submerged in a fluid maintained at 50oC and h at the wire surface is 4000 Wm2oC calculate the steady state temperature at the centre and at the surface of the wire

bull An approximately spherical shaped orange ( k=023 WmoC) 90 mm diameter undergoes ripping process and generates 5100 Wm3 of energy If external surface the orange is at 8oC determine

bull (i) temperature at the centre of the orange and

bull (ii) heat flow from the outer surface of the orange

bull FINS External surfaces to used to cooling area and hence convection

heat transfer can be improvedbull APPLICATIONS IC engine cylinder ( air cooling) Heat exchangers automobile

radiators computer chip refrigerators electric motors generators and transformers etc

bull ASSUMPTIONS MADE IN ANALYSIS OF HEAT TRANFER THROUGH FIN

1 Fin material is homogenous and isotropic constthermal conductivity

2 Temperature at any cs of the fin is uniform3 lsquohrsquo is uniform over entire surface4 No internal heat generation5 Thermal contact resistance is negligible6 Radiation heat transfer is negligible7 Heat conduction is steady state and one dimensional

bull Problem 1

bull A stainless steel fin (k=20 WmK) having a diameter of 20 mm and a length of 01 m is attached to a wall at 300oC The ambient temperature is 50oC and the heat transfer coefficient is 10 Wm2K The fin tip is insulated Determine

bull (A) The rate of heat dissipation from the finbull (B) The temperature at the fin tipbull (C) The rate of heat transfer from the wall area covered by the fin if

the fin was not usedbull (D) The heat transfer rate from the same fin geometry if the

stainless steel fin is replaced by a fictitious fin with infinite thermal conductivity

bull Problem2bull One end of a long rod is inserted into a furnace while the other

projects into ambient temperature Under steady state the temperature of the rod is measured at two points 75mm apart and found to be 125oC and 885oC respectively while the ambient temperature is 20oC If the rod is 25mm in diameter and h is 2336 Wm2K

bull Find the thermal conductivity of the rod material

  • HEAT AND MASS TRANSFER
  • Thermodynamics Vs Heat transfer
  • MODE OF HEAT TRANSFER
  • MODE OF HEAT TRANSFER (2)
  • CONDUCTION- MECHANISM
  • CONCECTION-MECHANISM
  • RADIATION-MECHANISM
  • Conduction- Convection- Radiation
  • Slide 9
  • Slide 10
  • Fourierrsquos Law of Heat Conduction
  • Slide 12
  • Newtonrsquos Law of Cooling
  • Laws of Thermal Radiation
  • LAWS FOR HEAT TRANSFER
  • ELECTRICAL ampTHERMAL ANALOGY
  • Slide 17
  • Slide 18
  • PLANE WALL-UNIDIRECTIONAL
  • PLANE WALL -SERIES
  • PLANE WALL IN SERIES
  • PLANE WALL -PARALLEL
  • PLANE WALL-PARALLEL
  • PLANE WALL-COMBINATION
  • PLANE WALL-FILM RESISTANCE
  • PLANE WALL- THERMAL CONTACT RESISTANCE
  • PLANE WALL- THERMAL CONTACT RESISTANCE (2)
  • CYLINDER-SOLID
  • HEAT TRANSFER IN SOLID CYLINDER
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER-COMPOSITE
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER-MULTY LAYER
  • HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER
  • HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER (2)
  • HEAT TRANSFER IN SPHERE
  • HEAT TRANSFER IN SPHERE -MULTY LAYER
  • HEAT TRANSFER IN SPHERE -MULTY LAYER (2)
  • Slide 38
  • Slide 39
  • Slide 40
  • Slide 41
  • CRITICAL RADIUS
  • Plane wall- Insulation thickness -study
  • Problem5
  • Slide 45
  • Slide 46
  • Slide 47
  • GENERAL HEAT CONDUCTION EQUATION IN CYLINDRICAL COORDINATES
  • Slide 49
  • Slide 50
  • Slide 51
  • GENERAL HEAT CONDUCTION EQUATION IN SPHERICAL COORDINATES
  • Slide 53
  • Slide 54
  • Slide 55
  • General Heat Conduction Equation
  • Temperature profile
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Slide 67
Page 26: hmt chap 1

HEAT TRANSFER RATE IN HOLLOW CYLINDER

HEAT TRANSFER RATE IN HOLLOW CYLINDER-COMPOSITE

HEAT TRANSFER RATE IN HOLLOW CYLINDER-MULTY LAYER

HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER

HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER

HEAT TRANSFER IN SPHERE

HEAT TRANSFER IN SPHERE -MULTY LAYER

HEAT TRANSFER IN SPHERE -MULTY LAYER

bull Problem 1 The wall in a furnace consists of 125 mm thick refractory bricks and

125 mm thick insulating fire bricks separated by an air gap A 12 mm thick plaster covers the outer wall The inner surface of wall is at 1100oC and the ambient temperature 25oC the heat transfer coefficient on the outer wall to the air is 17 Wm2K and the resistance to heat flow of the air gap is 016 KW the thermal conductivities of refractory brick insulating firebrick and plaster are 16 03 and 014 WmK respectively

Calculatebull (a) the rate of heat loss per unit area of the wall surfacebull (b) the interface temperatures throughout the wallbull (c) the temperature at the outside surface of the wall

bull Problem 2Two large aluminum plates (k=240 WmK) each 2 cm thick with 10 mm surface roughness are placed in contact under 105 Nm2 pressure in air The temperature at the outside surfaces are 390 oC and 406 oCCalculate

A) the heat flux B) the temperature drop due to the contact resistance C) the contact temperatures Thermal contact resistance with air as the interfacial fluid for

10 mm roughness is 27510-4 m2KW

bull Problem 3bull Steam at 350 oC flowing in a pipe (k=80 WmK) of 5 cm

inner diameter and 56 cm outer diameter is covered with 3 cm thick insulation of k=005 WmK Heat is lost to surroundings at 5 oC by natural convection and radiation the combined h being 20 Wm2K Taking the heat transfer coefficient inside the pipe as 60 Wm2K

bull Determinebull A) the rate of heat loss from the steam per unit length of

the pipebull B) the temperature drop across the pipe and the insulation

bull PROBLEM4bull A nuclear reactor has a sperical pressure vessel 750 mm inside

diameter 80 mm wall thickness The temperature at the inner surface is 500 oC If the tempertarue at the outer surface is 495 oC

bull Calculatea) The rate of heat loss through the metal wallb) If the rate of heat loss is to limited to 305Whow much inulation

thickness needs to be applied Thermal conductivity of steel wall=46 WmoC Thermal conductivity of insulation=004 WmoC Film coefficient of heat tranfer at inner and outer surface are 200

Wm2 oC and 15 Wm2 oC Ambient air temperature=28 oC

CRITICAL RADIUS

bull The radius upto which heat flow increases and after decreases At critical radius q becomes qmax

bull The addition of insulation always increases (Rth)condbull Rtotal= (Rth)cond + (Rth)conv

bull The addition of insulation may reduce the (Rth)conv due to increase in surface area

bull r1ltrc -heat transfer increases by adding insulation till r1=rc If

insulation thickness is further increased the rate of heat loss will decrease

r1 gt rc

- heat transfer decreases by adding insulation above the rc

Plane wall- Insulation

thickness -study

Cylinder amp Sphere- Insulation thickness -

study

Problem5bull A 3mm diameter and 5 mm long electric wire is tightly

wrapped with a 2 mm thick plastic cover whose K=015 wmoC Electrical measurements indicate that a current of 10 A passes through the wire and there is a voltage drop of 8 V along the wire If the insulated wire is exposed to a medium at Ta=30oC with a heat transfer coefficient of h=12 wm2oC Determine the temperature at the interface of the wire and the plastic cover in steady operation Also determine whether doubling the thickness of the plastic cover will increase or decrease this interface temperature

dx

TdzdyKQ xx )(

dxQx

QQ xxdxx ][

dxdx

TdzdyK

xQQdQ xdxxxx ])([)(

)1()]([)(

ddzdydx

x

TK

xdQ xx

GENERAL HEAT CONDUCTION EQUATION IN CARTESIAN COORDINATES

Consider an unsteady (transient) three dimensional heat conduction with internal heat generation in heterogeneous amp anisotropic PLANE WALL

Cartesian co-ordinates x y z

Elemental length dxdydz

Consider an elemental volume ABCDEFGH = (dxdydz)

Let Temperature distribution is a function of T=T(xyz)

Let

=internal heat generation per unit volume per unit time (W m3)q= rate of heat flow (W)Q= qdt=total heat flow (J)r=mass density of the material element(kgm3)Cp=specific heat of the material element (Jkg K)

gq

Plane (y - z) lsquoxrsquo direction

Quantity of heat leaving from the face EFGH in lsquoXrsquo direction

Quantity of heat flowing into the face ABCD in lsquoXrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoXrsquo direction

dy

TdzdxKQ yy )(

dyQy

QQ yydyy ][

dydy

TdzdxK

yQQdQ ydyyyy ])([)(

)2()]([)(

ddzdydx

y

TK

ydQ yy

Plane (x - z) lsquoyrsquo direction

Quantity of heat flowing into the face ABFE in lsquoYrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoYrsquo direction

dz

TdydxKQ zz )(

dzQz

QQ zzdzz ][

dzdz

TdydxK

zQQdQ zdzzzz ])([)(

)3()]([)(

ddzdydx

z

TK

zdQ zz

Plane (x - y) lsquozrsquo direction

Quantity of heat flowing into the face ADHE in lsquoZrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoZrsquo direction

Quantity of heat leaving from the face DCGH in lsquoYrsquo direction

Quantity of heat leaving from the face BCGF in lsquoZrsquo direction

)4()()(

ddzdydxqQgg

Total amount of internal heat generation in the element

Ein + Egen = Eout+Est (Ein- Eout)+ Egen=Est

T

Cddzdydxddzdydxqddzdydxz

TK

zy

TK

yx

TK

x pgzyx )()())](()()([

T

pC

gq

z

TzK

zy

Ty

Kyx

TxK

x][][][

stgzyx QQdQdQdQ

)(

)5()()(

dT

CdzdydxQ pst

Total heat energy stored in the element

(ie thermal energy of the element can be increased because of heat accumulation in the element due to heat flow amp internal heat generation)

)( volumem d

TmCQ pst )(

T

CqTK pg)(

KKKyKx z If material is homogeneous amp isotropic

T

pC

gq

z

T

zy

T

yx

T

xK ][][][

T

Kp

C

Kgq

z

T

y

T

x

T2

2

2

2

2

2

T

Kgq

z

T

y

T

x

T 12

2

2

2

2

2

Thermal diffusivity pC

K

T

K

qT g

12

according to thermodynamics first law

)( zr )( dzrddr

)( dzrddr

Cylindrical polar co-ordinates

Elemental length

Consider an elemental volume ABCDEFGH =

GENERAL HEAT CONDUCTION EQUATION IN CYLINDRICAL COORDINATES

gq

gq

Consider an unsteady (transient) three dimensional heat conduction with internal heat generation in heterogeneous amp anisotropic cylindrical geometries (eg rods and pipes)

)( zrTLetTemperature distribution is a function of T=

Let

=internal heat generation per unit volume per unit time (W m3)q= rate of heat flow (W)Q= qdt=total heat flow (J)r=mass density of the material element(kgm3)Cp=specific heat of the material element (Jkg K)

planerdirectionz

planezrdirection

planezdirectionr

)(

)(

)(

LetHeat flow in

drdr

TdzrdK

rdQ

QQdQ

rr

drrrr

])([

)1(][

r

Tr

rddrdzdKdQ rr

Heat accumulation in the element due to heat flow by conduction in lsquorrsquo direction

rddr

TdzdrK

rdQ

QQdQ d

])([

][

2

T

r

ddrdzrdKdQ

)2(][

2

2

2

T

r

ddrdzrdKdQ

Heat accumulation in the element due to heat flow by conduction in lsquoqrsquo direction

Plane (z- q ) lsquorrsquo direction

dr

TdzrdKQ rr

)(

drQr

QQ rrdrr ][

Quantity of heat leaving from the face EFGH in lsquorrsquo direction

Quantity of heat flowing into the face ABCD in lsquorrsquo direction

Plane (r-z) lsquoq lsquodirection

Quantity of heat leaving from the face AEGC in lsquoqrsquo direction

d

r

TdzdrKQ )(

rdQ

rQQ d ][

Quantity of heat flowing into the face BFHD in lsquoqrsquo direction

)3())((2

2

z

TdzdrrdKdQ zz

)5()()(

dT

CdzdrrdQ pst

)4()()(

ddzdrrdqQgg

Total amount of internal heat generation in the element

Total heat energy stored in the element

(ie thermal energy of the element can be increased because of heat accumulation in the element due to heat flow amp internal heat generation)

according to thermodynamics first law

Ein + Egen = Eout + Est (Ein ndash Eout) + Egen = Est

)( volumem

)()(z

T

zdzdrrdKdQ zz

Heat accumulation in the element due to heat flow by conduction in lsquozrsquo direction

dzdz

TdrrdK

zdQ

QQdQ

zz

dzzzz

])([

d

TmCQ pst )(

Plane (r-q) lsquozrsquo direction

dz

TdrrdKQ zz )(

dzQz

QQ zzdzz ][ Quantity of heat leaving from the face CDHG in lsquozrsquo direction

Quantity of heat flowing into the face ABFE in lsquozrsquo direction

stgzr QQdQdQdQ

)(

r

r ][)(rT

rr

ddrdzdKr

][)(

2

2

2

T

r

ddrdzrdK ))((2

2

z

TddzdrrdK z

ddzdrrdq g )(

dT

Cdzdrrd p )(

+ + +

=

T

CqTK pg)(KKKK zr

[gq

)(1

r

Tr

rrK r

)(1

2

2

2 T

rK )(

2

2

z

TK z

)( ddzdrrd )(

ddzdrrd

TC p

+ + + ] =

If material is homogeneous amp isotropic

)(1

r

Tr

rr

)(1

2

2

2 T

r)(

2

2

z

T

K

q g

T

K

C p

)(1

r

Tr

rr

)(1

2

2

2 T

r )(2

2

z

T

K

q g

T

1

+ + + =

)(1

r

Tr

rrK r

)(1

2

2

2 T

rK )(

2

2

z

TK z

gq

T

C p + + + =

+ + + = Thermal diffusivity

T

K

qT g

12

pC

K

(Ein ndash Eout) + Egen = Est

)( rT

)( r

)sin( drrddr)sin( drrddr

planerdirection

planerdirection

planedirectionr

)(

)(

)(

Spherical polar co-ordinates

Elemental length

Consider an elemental volume ABCDEFGH =

GENERAL HEAT CONDUCTION EQUATION IN SPHERICAL COORDINATES

Let

=internal heat generation per unit volume per unit time (W m3)q= rate of heat flow (W)Q= qdt=total heat flow (J)r=mass density of the material element(kgm3)Cp=specific heat of the material element (Jkg K)

gq

Consider an unsteady (transient) three dimensional heat conduction with internal heat generation in heterogeneous amp anisotropic spherical geometries

LetTemperature distribution is a function of T=

X

Y

Z

drT

drrdKQ rr )sin(

drQr

QQ rrdrr ][

drdr

TdrrdK

rdQ

QQdQ

rr

drrrr

])sin([

)1(]([sin 2

rT

rr

ddrddKdQ rr

Plane ( -q f) lsquorrsquo directionQuantity of heat flowing into the face ABCD in lsquorrsquo direction

Quantity of heat leaving from the face EFGH in lsquorrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquorrsquo direction

drT

drdrKQ )sin(

rdQ

rQQ d ][

rddr

TdrdrK

rdQ

QQdQ d

])sin([

])[(sin

2

T

r

drdrdrKdQ

)2(])(sin[

2

T

r

drdrddrKdQ

Quantity of heat leaving from the face BFHD in lsquoqrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoqrsquo direction

X

Y

Z

Plane (r- f ) lsquoq lsquodirection

Quantity of heat flowing into the face AEGC in lsquoqrsquo direction

dr

TdrrdKQ

sin)(

drQ

rQQ d sin][

sin

drdr

TdrrdK

rdQ

QQdQ d

sin]sin

)([sin

][sinsin

)sin(

T

rr

ddrdrrdKdQ

)3()(sin

)sin(2

2

22

2

T

r

ddddrrKdQ

Plane (r- ) q lsquofrsquo directionQuantity of heat flowing into the face ABFE in lsquofrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquozrsquo direction

)4()sin()(

ddrrddrqQgg

)5()sin()(

dT

CdrrddrQ pst

Total amount of internal heat generation in the element

Total heat energy stored in the element

(Ein ndash Eout) + Egen = Est

d

TmCQ pst )(

)( volumem

Quantity of heat leaving from the face CDHG in lsquofrsquo direction

according to thermodynamics first law

Ein + Egen = Eout + Est

X

Y

Z

stgr QQdQdQdQ

)(

]([sin 2

r

Tr

rddrddKr

])(sin[

2

T

rdrdrddrK

)(sin

)sin(2

2

22

2

T

r

ddddrrK

ddrrddrq g )sin(

dT

Cdrrddr p )sin(

2

2

r

r

sin

sin

]([ 22 r

Tr

rr

Kr

])(sin[sin2

TrK

)(sin 2

2

22

T

r

Kgq

T

Cp

]([1

[ 22 r

Tr

rrK

])(sin[sin1

2

T

r)](

sin

12

2

22 T

r gq

T

Cp

]([1

[ 22 r

Tr

rr

])(sin[sin1

2

T

r)](

sin

12

2

22 T

rK

q g

T

K

Cp

]([1

[ 22 r

Tr

rr

])(sin[sin1

2

T

r)](

sin

12

2

22 T

rK

q g

T1

KKKKr If material is homogeneous amp isotropic

TCqTK pg)(

T

K

qT g

12 pC

K

Thermal diffusivity

(Ein ndash Eout) + Egen = Est

General Heat Conduction Equation

PLANE WALL CYLINDER SPHERE

Temperature profile

Plane wall SphereCylinder

bull A copper conductor (k = 380 WmoC resistivity ρ = 2x10-8 Ω m) having inner and outer radii 10 cm and 225 cm respectively is carrying a current density of 4800 amperes cm2 the conductor is internally cooled and a constant temperature of 65oC is maintained at the inner surface and there is no heat transfer through the insulation surrounding the conductor

Determinebull (i) The maximum temperature of the conductor and the radius at

which it occursbull (ii) The internal heat transfer rate

bull A chemical reaction takes place in a packed bed (k=06 WmoC) between two coaxial cylinders with radii 15 mm and 45 mm the inner surface is at 580oC and it is insulated Assuming the reaction rate of 055 MWm3 in the reactor volume find the temperature at the outer surface of the reactor

bull A current of 300 amperes passes through a stainless steel wire of 25 mm diameter and k=20 WmoC The resistivity of the wire is 70x10-8 ohm-m and length of the wire is 2 m if the wire is submerged in a fluid maintained at 50oC and h at the wire surface is 4000 Wm2oC calculate the steady state temperature at the centre and at the surface of the wire

bull An approximately spherical shaped orange ( k=023 WmoC) 90 mm diameter undergoes ripping process and generates 5100 Wm3 of energy If external surface the orange is at 8oC determine

bull (i) temperature at the centre of the orange and

bull (ii) heat flow from the outer surface of the orange

bull FINS External surfaces to used to cooling area and hence convection

heat transfer can be improvedbull APPLICATIONS IC engine cylinder ( air cooling) Heat exchangers automobile

radiators computer chip refrigerators electric motors generators and transformers etc

bull ASSUMPTIONS MADE IN ANALYSIS OF HEAT TRANFER THROUGH FIN

1 Fin material is homogenous and isotropic constthermal conductivity

2 Temperature at any cs of the fin is uniform3 lsquohrsquo is uniform over entire surface4 No internal heat generation5 Thermal contact resistance is negligible6 Radiation heat transfer is negligible7 Heat conduction is steady state and one dimensional

bull Problem 1

bull A stainless steel fin (k=20 WmK) having a diameter of 20 mm and a length of 01 m is attached to a wall at 300oC The ambient temperature is 50oC and the heat transfer coefficient is 10 Wm2K The fin tip is insulated Determine

bull (A) The rate of heat dissipation from the finbull (B) The temperature at the fin tipbull (C) The rate of heat transfer from the wall area covered by the fin if

the fin was not usedbull (D) The heat transfer rate from the same fin geometry if the

stainless steel fin is replaced by a fictitious fin with infinite thermal conductivity

bull Problem2bull One end of a long rod is inserted into a furnace while the other

projects into ambient temperature Under steady state the temperature of the rod is measured at two points 75mm apart and found to be 125oC and 885oC respectively while the ambient temperature is 20oC If the rod is 25mm in diameter and h is 2336 Wm2K

bull Find the thermal conductivity of the rod material

  • HEAT AND MASS TRANSFER
  • Thermodynamics Vs Heat transfer
  • MODE OF HEAT TRANSFER
  • MODE OF HEAT TRANSFER (2)
  • CONDUCTION- MECHANISM
  • CONCECTION-MECHANISM
  • RADIATION-MECHANISM
  • Conduction- Convection- Radiation
  • Slide 9
  • Slide 10
  • Fourierrsquos Law of Heat Conduction
  • Slide 12
  • Newtonrsquos Law of Cooling
  • Laws of Thermal Radiation
  • LAWS FOR HEAT TRANSFER
  • ELECTRICAL ampTHERMAL ANALOGY
  • Slide 17
  • Slide 18
  • PLANE WALL-UNIDIRECTIONAL
  • PLANE WALL -SERIES
  • PLANE WALL IN SERIES
  • PLANE WALL -PARALLEL
  • PLANE WALL-PARALLEL
  • PLANE WALL-COMBINATION
  • PLANE WALL-FILM RESISTANCE
  • PLANE WALL- THERMAL CONTACT RESISTANCE
  • PLANE WALL- THERMAL CONTACT RESISTANCE (2)
  • CYLINDER-SOLID
  • HEAT TRANSFER IN SOLID CYLINDER
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER-COMPOSITE
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER-MULTY LAYER
  • HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER
  • HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER (2)
  • HEAT TRANSFER IN SPHERE
  • HEAT TRANSFER IN SPHERE -MULTY LAYER
  • HEAT TRANSFER IN SPHERE -MULTY LAYER (2)
  • Slide 38
  • Slide 39
  • Slide 40
  • Slide 41
  • CRITICAL RADIUS
  • Plane wall- Insulation thickness -study
  • Problem5
  • Slide 45
  • Slide 46
  • Slide 47
  • GENERAL HEAT CONDUCTION EQUATION IN CYLINDRICAL COORDINATES
  • Slide 49
  • Slide 50
  • Slide 51
  • GENERAL HEAT CONDUCTION EQUATION IN SPHERICAL COORDINATES
  • Slide 53
  • Slide 54
  • Slide 55
  • General Heat Conduction Equation
  • Temperature profile
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Slide 67
Page 27: hmt chap 1

HEAT TRANSFER RATE IN HOLLOW CYLINDER-COMPOSITE

HEAT TRANSFER RATE IN HOLLOW CYLINDER-MULTY LAYER

HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER

HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER

HEAT TRANSFER IN SPHERE

HEAT TRANSFER IN SPHERE -MULTY LAYER

HEAT TRANSFER IN SPHERE -MULTY LAYER

bull Problem 1 The wall in a furnace consists of 125 mm thick refractory bricks and

125 mm thick insulating fire bricks separated by an air gap A 12 mm thick plaster covers the outer wall The inner surface of wall is at 1100oC and the ambient temperature 25oC the heat transfer coefficient on the outer wall to the air is 17 Wm2K and the resistance to heat flow of the air gap is 016 KW the thermal conductivities of refractory brick insulating firebrick and plaster are 16 03 and 014 WmK respectively

Calculatebull (a) the rate of heat loss per unit area of the wall surfacebull (b) the interface temperatures throughout the wallbull (c) the temperature at the outside surface of the wall

bull Problem 2Two large aluminum plates (k=240 WmK) each 2 cm thick with 10 mm surface roughness are placed in contact under 105 Nm2 pressure in air The temperature at the outside surfaces are 390 oC and 406 oCCalculate

A) the heat flux B) the temperature drop due to the contact resistance C) the contact temperatures Thermal contact resistance with air as the interfacial fluid for

10 mm roughness is 27510-4 m2KW

bull Problem 3bull Steam at 350 oC flowing in a pipe (k=80 WmK) of 5 cm

inner diameter and 56 cm outer diameter is covered with 3 cm thick insulation of k=005 WmK Heat is lost to surroundings at 5 oC by natural convection and radiation the combined h being 20 Wm2K Taking the heat transfer coefficient inside the pipe as 60 Wm2K

bull Determinebull A) the rate of heat loss from the steam per unit length of

the pipebull B) the temperature drop across the pipe and the insulation

bull PROBLEM4bull A nuclear reactor has a sperical pressure vessel 750 mm inside

diameter 80 mm wall thickness The temperature at the inner surface is 500 oC If the tempertarue at the outer surface is 495 oC

bull Calculatea) The rate of heat loss through the metal wallb) If the rate of heat loss is to limited to 305Whow much inulation

thickness needs to be applied Thermal conductivity of steel wall=46 WmoC Thermal conductivity of insulation=004 WmoC Film coefficient of heat tranfer at inner and outer surface are 200

Wm2 oC and 15 Wm2 oC Ambient air temperature=28 oC

CRITICAL RADIUS

bull The radius upto which heat flow increases and after decreases At critical radius q becomes qmax

bull The addition of insulation always increases (Rth)condbull Rtotal= (Rth)cond + (Rth)conv

bull The addition of insulation may reduce the (Rth)conv due to increase in surface area

bull r1ltrc -heat transfer increases by adding insulation till r1=rc If

insulation thickness is further increased the rate of heat loss will decrease

r1 gt rc

- heat transfer decreases by adding insulation above the rc

Plane wall- Insulation

thickness -study

Cylinder amp Sphere- Insulation thickness -

study

Problem5bull A 3mm diameter and 5 mm long electric wire is tightly

wrapped with a 2 mm thick plastic cover whose K=015 wmoC Electrical measurements indicate that a current of 10 A passes through the wire and there is a voltage drop of 8 V along the wire If the insulated wire is exposed to a medium at Ta=30oC with a heat transfer coefficient of h=12 wm2oC Determine the temperature at the interface of the wire and the plastic cover in steady operation Also determine whether doubling the thickness of the plastic cover will increase or decrease this interface temperature

dx

TdzdyKQ xx )(

dxQx

QQ xxdxx ][

dxdx

TdzdyK

xQQdQ xdxxxx ])([)(

)1()]([)(

ddzdydx

x

TK

xdQ xx

GENERAL HEAT CONDUCTION EQUATION IN CARTESIAN COORDINATES

Consider an unsteady (transient) three dimensional heat conduction with internal heat generation in heterogeneous amp anisotropic PLANE WALL

Cartesian co-ordinates x y z

Elemental length dxdydz

Consider an elemental volume ABCDEFGH = (dxdydz)

Let Temperature distribution is a function of T=T(xyz)

Let

=internal heat generation per unit volume per unit time (W m3)q= rate of heat flow (W)Q= qdt=total heat flow (J)r=mass density of the material element(kgm3)Cp=specific heat of the material element (Jkg K)

gq

Plane (y - z) lsquoxrsquo direction

Quantity of heat leaving from the face EFGH in lsquoXrsquo direction

Quantity of heat flowing into the face ABCD in lsquoXrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoXrsquo direction

dy

TdzdxKQ yy )(

dyQy

QQ yydyy ][

dydy

TdzdxK

yQQdQ ydyyyy ])([)(

)2()]([)(

ddzdydx

y

TK

ydQ yy

Plane (x - z) lsquoyrsquo direction

Quantity of heat flowing into the face ABFE in lsquoYrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoYrsquo direction

dz

TdydxKQ zz )(

dzQz

QQ zzdzz ][

dzdz

TdydxK

zQQdQ zdzzzz ])([)(

)3()]([)(

ddzdydx

z

TK

zdQ zz

Plane (x - y) lsquozrsquo direction

Quantity of heat flowing into the face ADHE in lsquoZrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoZrsquo direction

Quantity of heat leaving from the face DCGH in lsquoYrsquo direction

Quantity of heat leaving from the face BCGF in lsquoZrsquo direction

)4()()(

ddzdydxqQgg

Total amount of internal heat generation in the element

Ein + Egen = Eout+Est (Ein- Eout)+ Egen=Est

T

Cddzdydxddzdydxqddzdydxz

TK

zy

TK

yx

TK

x pgzyx )()())](()()([

T

pC

gq

z

TzK

zy

Ty

Kyx

TxK

x][][][

stgzyx QQdQdQdQ

)(

)5()()(

dT

CdzdydxQ pst

Total heat energy stored in the element

(ie thermal energy of the element can be increased because of heat accumulation in the element due to heat flow amp internal heat generation)

)( volumem d

TmCQ pst )(

T

CqTK pg)(

KKKyKx z If material is homogeneous amp isotropic

T

pC

gq

z

T

zy

T

yx

T

xK ][][][

T

Kp

C

Kgq

z

T

y

T

x

T2

2

2

2

2

2

T

Kgq

z

T

y

T

x

T 12

2

2

2

2

2

Thermal diffusivity pC

K

T

K

qT g

12

according to thermodynamics first law

)( zr )( dzrddr

)( dzrddr

Cylindrical polar co-ordinates

Elemental length

Consider an elemental volume ABCDEFGH =

GENERAL HEAT CONDUCTION EQUATION IN CYLINDRICAL COORDINATES

gq

gq

Consider an unsteady (transient) three dimensional heat conduction with internal heat generation in heterogeneous amp anisotropic cylindrical geometries (eg rods and pipes)

)( zrTLetTemperature distribution is a function of T=

Let

=internal heat generation per unit volume per unit time (W m3)q= rate of heat flow (W)Q= qdt=total heat flow (J)r=mass density of the material element(kgm3)Cp=specific heat of the material element (Jkg K)

planerdirectionz

planezrdirection

planezdirectionr

)(

)(

)(

LetHeat flow in

drdr

TdzrdK

rdQ

QQdQ

rr

drrrr

])([

)1(][

r

Tr

rddrdzdKdQ rr

Heat accumulation in the element due to heat flow by conduction in lsquorrsquo direction

rddr

TdzdrK

rdQ

QQdQ d

])([

][

2

T

r

ddrdzrdKdQ

)2(][

2

2

2

T

r

ddrdzrdKdQ

Heat accumulation in the element due to heat flow by conduction in lsquoqrsquo direction

Plane (z- q ) lsquorrsquo direction

dr

TdzrdKQ rr

)(

drQr

QQ rrdrr ][

Quantity of heat leaving from the face EFGH in lsquorrsquo direction

Quantity of heat flowing into the face ABCD in lsquorrsquo direction

Plane (r-z) lsquoq lsquodirection

Quantity of heat leaving from the face AEGC in lsquoqrsquo direction

d

r

TdzdrKQ )(

rdQ

rQQ d ][

Quantity of heat flowing into the face BFHD in lsquoqrsquo direction

)3())((2

2

z

TdzdrrdKdQ zz

)5()()(

dT

CdzdrrdQ pst

)4()()(

ddzdrrdqQgg

Total amount of internal heat generation in the element

Total heat energy stored in the element

(ie thermal energy of the element can be increased because of heat accumulation in the element due to heat flow amp internal heat generation)

according to thermodynamics first law

Ein + Egen = Eout + Est (Ein ndash Eout) + Egen = Est

)( volumem

)()(z

T

zdzdrrdKdQ zz

Heat accumulation in the element due to heat flow by conduction in lsquozrsquo direction

dzdz

TdrrdK

zdQ

QQdQ

zz

dzzzz

])([

d

TmCQ pst )(

Plane (r-q) lsquozrsquo direction

dz

TdrrdKQ zz )(

dzQz

QQ zzdzz ][ Quantity of heat leaving from the face CDHG in lsquozrsquo direction

Quantity of heat flowing into the face ABFE in lsquozrsquo direction

stgzr QQdQdQdQ

)(

r

r ][)(rT

rr

ddrdzdKr

][)(

2

2

2

T

r

ddrdzrdK ))((2

2

z

TddzdrrdK z

ddzdrrdq g )(

dT

Cdzdrrd p )(

+ + +

=

T

CqTK pg)(KKKK zr

[gq

)(1

r

Tr

rrK r

)(1

2

2

2 T

rK )(

2

2

z

TK z

)( ddzdrrd )(

ddzdrrd

TC p

+ + + ] =

If material is homogeneous amp isotropic

)(1

r

Tr

rr

)(1

2

2

2 T

r)(

2

2

z

T

K

q g

T

K

C p

)(1

r

Tr

rr

)(1

2

2

2 T

r )(2

2

z

T

K

q g

T

1

+ + + =

)(1

r

Tr

rrK r

)(1

2

2

2 T

rK )(

2

2

z

TK z

gq

T

C p + + + =

+ + + = Thermal diffusivity

T

K

qT g

12

pC

K

(Ein ndash Eout) + Egen = Est

)( rT

)( r

)sin( drrddr)sin( drrddr

planerdirection

planerdirection

planedirectionr

)(

)(

)(

Spherical polar co-ordinates

Elemental length

Consider an elemental volume ABCDEFGH =

GENERAL HEAT CONDUCTION EQUATION IN SPHERICAL COORDINATES

Let

=internal heat generation per unit volume per unit time (W m3)q= rate of heat flow (W)Q= qdt=total heat flow (J)r=mass density of the material element(kgm3)Cp=specific heat of the material element (Jkg K)

gq

Consider an unsteady (transient) three dimensional heat conduction with internal heat generation in heterogeneous amp anisotropic spherical geometries

LetTemperature distribution is a function of T=

X

Y

Z

drT

drrdKQ rr )sin(

drQr

QQ rrdrr ][

drdr

TdrrdK

rdQ

QQdQ

rr

drrrr

])sin([

)1(]([sin 2

rT

rr

ddrddKdQ rr

Plane ( -q f) lsquorrsquo directionQuantity of heat flowing into the face ABCD in lsquorrsquo direction

Quantity of heat leaving from the face EFGH in lsquorrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquorrsquo direction

drT

drdrKQ )sin(

rdQ

rQQ d ][

rddr

TdrdrK

rdQ

QQdQ d

])sin([

])[(sin

2

T

r

drdrdrKdQ

)2(])(sin[

2

T

r

drdrddrKdQ

Quantity of heat leaving from the face BFHD in lsquoqrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoqrsquo direction

X

Y

Z

Plane (r- f ) lsquoq lsquodirection

Quantity of heat flowing into the face AEGC in lsquoqrsquo direction

dr

TdrrdKQ

sin)(

drQ

rQQ d sin][

sin

drdr

TdrrdK

rdQ

QQdQ d

sin]sin

)([sin

][sinsin

)sin(

T

rr

ddrdrrdKdQ

)3()(sin

)sin(2

2

22

2

T

r

ddddrrKdQ

Plane (r- ) q lsquofrsquo directionQuantity of heat flowing into the face ABFE in lsquofrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquozrsquo direction

)4()sin()(

ddrrddrqQgg

)5()sin()(

dT

CdrrddrQ pst

Total amount of internal heat generation in the element

Total heat energy stored in the element

(Ein ndash Eout) + Egen = Est

d

TmCQ pst )(

)( volumem

Quantity of heat leaving from the face CDHG in lsquofrsquo direction

according to thermodynamics first law

Ein + Egen = Eout + Est

X

Y

Z

stgr QQdQdQdQ

)(

]([sin 2

r

Tr

rddrddKr

])(sin[

2

T

rdrdrddrK

)(sin

)sin(2

2

22

2

T

r

ddddrrK

ddrrddrq g )sin(

dT

Cdrrddr p )sin(

2

2

r

r

sin

sin

]([ 22 r

Tr

rr

Kr

])(sin[sin2

TrK

)(sin 2

2

22

T

r

Kgq

T

Cp

]([1

[ 22 r

Tr

rrK

])(sin[sin1

2

T

r)](

sin

12

2

22 T

r gq

T

Cp

]([1

[ 22 r

Tr

rr

])(sin[sin1

2

T

r)](

sin

12

2

22 T

rK

q g

T

K

Cp

]([1

[ 22 r

Tr

rr

])(sin[sin1

2

T

r)](

sin

12

2

22 T

rK

q g

T1

KKKKr If material is homogeneous amp isotropic

TCqTK pg)(

T

K

qT g

12 pC

K

Thermal diffusivity

(Ein ndash Eout) + Egen = Est

General Heat Conduction Equation

PLANE WALL CYLINDER SPHERE

Temperature profile

Plane wall SphereCylinder

bull A copper conductor (k = 380 WmoC resistivity ρ = 2x10-8 Ω m) having inner and outer radii 10 cm and 225 cm respectively is carrying a current density of 4800 amperes cm2 the conductor is internally cooled and a constant temperature of 65oC is maintained at the inner surface and there is no heat transfer through the insulation surrounding the conductor

Determinebull (i) The maximum temperature of the conductor and the radius at

which it occursbull (ii) The internal heat transfer rate

bull A chemical reaction takes place in a packed bed (k=06 WmoC) between two coaxial cylinders with radii 15 mm and 45 mm the inner surface is at 580oC and it is insulated Assuming the reaction rate of 055 MWm3 in the reactor volume find the temperature at the outer surface of the reactor

bull A current of 300 amperes passes through a stainless steel wire of 25 mm diameter and k=20 WmoC The resistivity of the wire is 70x10-8 ohm-m and length of the wire is 2 m if the wire is submerged in a fluid maintained at 50oC and h at the wire surface is 4000 Wm2oC calculate the steady state temperature at the centre and at the surface of the wire

bull An approximately spherical shaped orange ( k=023 WmoC) 90 mm diameter undergoes ripping process and generates 5100 Wm3 of energy If external surface the orange is at 8oC determine

bull (i) temperature at the centre of the orange and

bull (ii) heat flow from the outer surface of the orange

bull FINS External surfaces to used to cooling area and hence convection

heat transfer can be improvedbull APPLICATIONS IC engine cylinder ( air cooling) Heat exchangers automobile

radiators computer chip refrigerators electric motors generators and transformers etc

bull ASSUMPTIONS MADE IN ANALYSIS OF HEAT TRANFER THROUGH FIN

1 Fin material is homogenous and isotropic constthermal conductivity

2 Temperature at any cs of the fin is uniform3 lsquohrsquo is uniform over entire surface4 No internal heat generation5 Thermal contact resistance is negligible6 Radiation heat transfer is negligible7 Heat conduction is steady state and one dimensional

bull Problem 1

bull A stainless steel fin (k=20 WmK) having a diameter of 20 mm and a length of 01 m is attached to a wall at 300oC The ambient temperature is 50oC and the heat transfer coefficient is 10 Wm2K The fin tip is insulated Determine

bull (A) The rate of heat dissipation from the finbull (B) The temperature at the fin tipbull (C) The rate of heat transfer from the wall area covered by the fin if

the fin was not usedbull (D) The heat transfer rate from the same fin geometry if the

stainless steel fin is replaced by a fictitious fin with infinite thermal conductivity

bull Problem2bull One end of a long rod is inserted into a furnace while the other

projects into ambient temperature Under steady state the temperature of the rod is measured at two points 75mm apart and found to be 125oC and 885oC respectively while the ambient temperature is 20oC If the rod is 25mm in diameter and h is 2336 Wm2K

bull Find the thermal conductivity of the rod material

  • HEAT AND MASS TRANSFER
  • Thermodynamics Vs Heat transfer
  • MODE OF HEAT TRANSFER
  • MODE OF HEAT TRANSFER (2)
  • CONDUCTION- MECHANISM
  • CONCECTION-MECHANISM
  • RADIATION-MECHANISM
  • Conduction- Convection- Radiation
  • Slide 9
  • Slide 10
  • Fourierrsquos Law of Heat Conduction
  • Slide 12
  • Newtonrsquos Law of Cooling
  • Laws of Thermal Radiation
  • LAWS FOR HEAT TRANSFER
  • ELECTRICAL ampTHERMAL ANALOGY
  • Slide 17
  • Slide 18
  • PLANE WALL-UNIDIRECTIONAL
  • PLANE WALL -SERIES
  • PLANE WALL IN SERIES
  • PLANE WALL -PARALLEL
  • PLANE WALL-PARALLEL
  • PLANE WALL-COMBINATION
  • PLANE WALL-FILM RESISTANCE
  • PLANE WALL- THERMAL CONTACT RESISTANCE
  • PLANE WALL- THERMAL CONTACT RESISTANCE (2)
  • CYLINDER-SOLID
  • HEAT TRANSFER IN SOLID CYLINDER
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER-COMPOSITE
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER-MULTY LAYER
  • HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER
  • HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER (2)
  • HEAT TRANSFER IN SPHERE
  • HEAT TRANSFER IN SPHERE -MULTY LAYER
  • HEAT TRANSFER IN SPHERE -MULTY LAYER (2)
  • Slide 38
  • Slide 39
  • Slide 40
  • Slide 41
  • CRITICAL RADIUS
  • Plane wall- Insulation thickness -study
  • Problem5
  • Slide 45
  • Slide 46
  • Slide 47
  • GENERAL HEAT CONDUCTION EQUATION IN CYLINDRICAL COORDINATES
  • Slide 49
  • Slide 50
  • Slide 51
  • GENERAL HEAT CONDUCTION EQUATION IN SPHERICAL COORDINATES
  • Slide 53
  • Slide 54
  • Slide 55
  • General Heat Conduction Equation
  • Temperature profile
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Slide 67
Page 28: hmt chap 1

HEAT TRANSFER RATE IN HOLLOW CYLINDER-MULTY LAYER

HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER

HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER

HEAT TRANSFER IN SPHERE

HEAT TRANSFER IN SPHERE -MULTY LAYER

HEAT TRANSFER IN SPHERE -MULTY LAYER

bull Problem 1 The wall in a furnace consists of 125 mm thick refractory bricks and

125 mm thick insulating fire bricks separated by an air gap A 12 mm thick plaster covers the outer wall The inner surface of wall is at 1100oC and the ambient temperature 25oC the heat transfer coefficient on the outer wall to the air is 17 Wm2K and the resistance to heat flow of the air gap is 016 KW the thermal conductivities of refractory brick insulating firebrick and plaster are 16 03 and 014 WmK respectively

Calculatebull (a) the rate of heat loss per unit area of the wall surfacebull (b) the interface temperatures throughout the wallbull (c) the temperature at the outside surface of the wall

bull Problem 2Two large aluminum plates (k=240 WmK) each 2 cm thick with 10 mm surface roughness are placed in contact under 105 Nm2 pressure in air The temperature at the outside surfaces are 390 oC and 406 oCCalculate

A) the heat flux B) the temperature drop due to the contact resistance C) the contact temperatures Thermal contact resistance with air as the interfacial fluid for

10 mm roughness is 27510-4 m2KW

bull Problem 3bull Steam at 350 oC flowing in a pipe (k=80 WmK) of 5 cm

inner diameter and 56 cm outer diameter is covered with 3 cm thick insulation of k=005 WmK Heat is lost to surroundings at 5 oC by natural convection and radiation the combined h being 20 Wm2K Taking the heat transfer coefficient inside the pipe as 60 Wm2K

bull Determinebull A) the rate of heat loss from the steam per unit length of

the pipebull B) the temperature drop across the pipe and the insulation

bull PROBLEM4bull A nuclear reactor has a sperical pressure vessel 750 mm inside

diameter 80 mm wall thickness The temperature at the inner surface is 500 oC If the tempertarue at the outer surface is 495 oC

bull Calculatea) The rate of heat loss through the metal wallb) If the rate of heat loss is to limited to 305Whow much inulation

thickness needs to be applied Thermal conductivity of steel wall=46 WmoC Thermal conductivity of insulation=004 WmoC Film coefficient of heat tranfer at inner and outer surface are 200

Wm2 oC and 15 Wm2 oC Ambient air temperature=28 oC

CRITICAL RADIUS

bull The radius upto which heat flow increases and after decreases At critical radius q becomes qmax

bull The addition of insulation always increases (Rth)condbull Rtotal= (Rth)cond + (Rth)conv

bull The addition of insulation may reduce the (Rth)conv due to increase in surface area

bull r1ltrc -heat transfer increases by adding insulation till r1=rc If

insulation thickness is further increased the rate of heat loss will decrease

r1 gt rc

- heat transfer decreases by adding insulation above the rc

Plane wall- Insulation

thickness -study

Cylinder amp Sphere- Insulation thickness -

study

Problem5bull A 3mm diameter and 5 mm long electric wire is tightly

wrapped with a 2 mm thick plastic cover whose K=015 wmoC Electrical measurements indicate that a current of 10 A passes through the wire and there is a voltage drop of 8 V along the wire If the insulated wire is exposed to a medium at Ta=30oC with a heat transfer coefficient of h=12 wm2oC Determine the temperature at the interface of the wire and the plastic cover in steady operation Also determine whether doubling the thickness of the plastic cover will increase or decrease this interface temperature

dx

TdzdyKQ xx )(

dxQx

QQ xxdxx ][

dxdx

TdzdyK

xQQdQ xdxxxx ])([)(

)1()]([)(

ddzdydx

x

TK

xdQ xx

GENERAL HEAT CONDUCTION EQUATION IN CARTESIAN COORDINATES

Consider an unsteady (transient) three dimensional heat conduction with internal heat generation in heterogeneous amp anisotropic PLANE WALL

Cartesian co-ordinates x y z

Elemental length dxdydz

Consider an elemental volume ABCDEFGH = (dxdydz)

Let Temperature distribution is a function of T=T(xyz)

Let

=internal heat generation per unit volume per unit time (W m3)q= rate of heat flow (W)Q= qdt=total heat flow (J)r=mass density of the material element(kgm3)Cp=specific heat of the material element (Jkg K)

gq

Plane (y - z) lsquoxrsquo direction

Quantity of heat leaving from the face EFGH in lsquoXrsquo direction

Quantity of heat flowing into the face ABCD in lsquoXrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoXrsquo direction

dy

TdzdxKQ yy )(

dyQy

QQ yydyy ][

dydy

TdzdxK

yQQdQ ydyyyy ])([)(

)2()]([)(

ddzdydx

y

TK

ydQ yy

Plane (x - z) lsquoyrsquo direction

Quantity of heat flowing into the face ABFE in lsquoYrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoYrsquo direction

dz

TdydxKQ zz )(

dzQz

QQ zzdzz ][

dzdz

TdydxK

zQQdQ zdzzzz ])([)(

)3()]([)(

ddzdydx

z

TK

zdQ zz

Plane (x - y) lsquozrsquo direction

Quantity of heat flowing into the face ADHE in lsquoZrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoZrsquo direction

Quantity of heat leaving from the face DCGH in lsquoYrsquo direction

Quantity of heat leaving from the face BCGF in lsquoZrsquo direction

)4()()(

ddzdydxqQgg

Total amount of internal heat generation in the element

Ein + Egen = Eout+Est (Ein- Eout)+ Egen=Est

T

Cddzdydxddzdydxqddzdydxz

TK

zy

TK

yx

TK

x pgzyx )()())](()()([

T

pC

gq

z

TzK

zy

Ty

Kyx

TxK

x][][][

stgzyx QQdQdQdQ

)(

)5()()(

dT

CdzdydxQ pst

Total heat energy stored in the element

(ie thermal energy of the element can be increased because of heat accumulation in the element due to heat flow amp internal heat generation)

)( volumem d

TmCQ pst )(

T

CqTK pg)(

KKKyKx z If material is homogeneous amp isotropic

T

pC

gq

z

T

zy

T

yx

T

xK ][][][

T

Kp

C

Kgq

z

T

y

T

x

T2

2

2

2

2

2

T

Kgq

z

T

y

T

x

T 12

2

2

2

2

2

Thermal diffusivity pC

K

T

K

qT g

12

according to thermodynamics first law

)( zr )( dzrddr

)( dzrddr

Cylindrical polar co-ordinates

Elemental length

Consider an elemental volume ABCDEFGH =

GENERAL HEAT CONDUCTION EQUATION IN CYLINDRICAL COORDINATES

gq

gq

Consider an unsteady (transient) three dimensional heat conduction with internal heat generation in heterogeneous amp anisotropic cylindrical geometries (eg rods and pipes)

)( zrTLetTemperature distribution is a function of T=

Let

=internal heat generation per unit volume per unit time (W m3)q= rate of heat flow (W)Q= qdt=total heat flow (J)r=mass density of the material element(kgm3)Cp=specific heat of the material element (Jkg K)

planerdirectionz

planezrdirection

planezdirectionr

)(

)(

)(

LetHeat flow in

drdr

TdzrdK

rdQ

QQdQ

rr

drrrr

])([

)1(][

r

Tr

rddrdzdKdQ rr

Heat accumulation in the element due to heat flow by conduction in lsquorrsquo direction

rddr

TdzdrK

rdQ

QQdQ d

])([

][

2

T

r

ddrdzrdKdQ

)2(][

2

2

2

T

r

ddrdzrdKdQ

Heat accumulation in the element due to heat flow by conduction in lsquoqrsquo direction

Plane (z- q ) lsquorrsquo direction

dr

TdzrdKQ rr

)(

drQr

QQ rrdrr ][

Quantity of heat leaving from the face EFGH in lsquorrsquo direction

Quantity of heat flowing into the face ABCD in lsquorrsquo direction

Plane (r-z) lsquoq lsquodirection

Quantity of heat leaving from the face AEGC in lsquoqrsquo direction

d

r

TdzdrKQ )(

rdQ

rQQ d ][

Quantity of heat flowing into the face BFHD in lsquoqrsquo direction

)3())((2

2

z

TdzdrrdKdQ zz

)5()()(

dT

CdzdrrdQ pst

)4()()(

ddzdrrdqQgg

Total amount of internal heat generation in the element

Total heat energy stored in the element

(ie thermal energy of the element can be increased because of heat accumulation in the element due to heat flow amp internal heat generation)

according to thermodynamics first law

Ein + Egen = Eout + Est (Ein ndash Eout) + Egen = Est

)( volumem

)()(z

T

zdzdrrdKdQ zz

Heat accumulation in the element due to heat flow by conduction in lsquozrsquo direction

dzdz

TdrrdK

zdQ

QQdQ

zz

dzzzz

])([

d

TmCQ pst )(

Plane (r-q) lsquozrsquo direction

dz

TdrrdKQ zz )(

dzQz

QQ zzdzz ][ Quantity of heat leaving from the face CDHG in lsquozrsquo direction

Quantity of heat flowing into the face ABFE in lsquozrsquo direction

stgzr QQdQdQdQ

)(

r

r ][)(rT

rr

ddrdzdKr

][)(

2

2

2

T

r

ddrdzrdK ))((2

2

z

TddzdrrdK z

ddzdrrdq g )(

dT

Cdzdrrd p )(

+ + +

=

T

CqTK pg)(KKKK zr

[gq

)(1

r

Tr

rrK r

)(1

2

2

2 T

rK )(

2

2

z

TK z

)( ddzdrrd )(

ddzdrrd

TC p

+ + + ] =

If material is homogeneous amp isotropic

)(1

r

Tr

rr

)(1

2

2

2 T

r)(

2

2

z

T

K

q g

T

K

C p

)(1

r

Tr

rr

)(1

2

2

2 T

r )(2

2

z

T

K

q g

T

1

+ + + =

)(1

r

Tr

rrK r

)(1

2

2

2 T

rK )(

2

2

z

TK z

gq

T

C p + + + =

+ + + = Thermal diffusivity

T

K

qT g

12

pC

K

(Ein ndash Eout) + Egen = Est

)( rT

)( r

)sin( drrddr)sin( drrddr

planerdirection

planerdirection

planedirectionr

)(

)(

)(

Spherical polar co-ordinates

Elemental length

Consider an elemental volume ABCDEFGH =

GENERAL HEAT CONDUCTION EQUATION IN SPHERICAL COORDINATES

Let

=internal heat generation per unit volume per unit time (W m3)q= rate of heat flow (W)Q= qdt=total heat flow (J)r=mass density of the material element(kgm3)Cp=specific heat of the material element (Jkg K)

gq

Consider an unsteady (transient) three dimensional heat conduction with internal heat generation in heterogeneous amp anisotropic spherical geometries

LetTemperature distribution is a function of T=

X

Y

Z

drT

drrdKQ rr )sin(

drQr

QQ rrdrr ][

drdr

TdrrdK

rdQ

QQdQ

rr

drrrr

])sin([

)1(]([sin 2

rT

rr

ddrddKdQ rr

Plane ( -q f) lsquorrsquo directionQuantity of heat flowing into the face ABCD in lsquorrsquo direction

Quantity of heat leaving from the face EFGH in lsquorrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquorrsquo direction

drT

drdrKQ )sin(

rdQ

rQQ d ][

rddr

TdrdrK

rdQ

QQdQ d

])sin([

])[(sin

2

T

r

drdrdrKdQ

)2(])(sin[

2

T

r

drdrddrKdQ

Quantity of heat leaving from the face BFHD in lsquoqrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoqrsquo direction

X

Y

Z

Plane (r- f ) lsquoq lsquodirection

Quantity of heat flowing into the face AEGC in lsquoqrsquo direction

dr

TdrrdKQ

sin)(

drQ

rQQ d sin][

sin

drdr

TdrrdK

rdQ

QQdQ d

sin]sin

)([sin

][sinsin

)sin(

T

rr

ddrdrrdKdQ

)3()(sin

)sin(2

2

22

2

T

r

ddddrrKdQ

Plane (r- ) q lsquofrsquo directionQuantity of heat flowing into the face ABFE in lsquofrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquozrsquo direction

)4()sin()(

ddrrddrqQgg

)5()sin()(

dT

CdrrddrQ pst

Total amount of internal heat generation in the element

Total heat energy stored in the element

(Ein ndash Eout) + Egen = Est

d

TmCQ pst )(

)( volumem

Quantity of heat leaving from the face CDHG in lsquofrsquo direction

according to thermodynamics first law

Ein + Egen = Eout + Est

X

Y

Z

stgr QQdQdQdQ

)(

]([sin 2

r

Tr

rddrddKr

])(sin[

2

T

rdrdrddrK

)(sin

)sin(2

2

22

2

T

r

ddddrrK

ddrrddrq g )sin(

dT

Cdrrddr p )sin(

2

2

r

r

sin

sin

]([ 22 r

Tr

rr

Kr

])(sin[sin2

TrK

)(sin 2

2

22

T

r

Kgq

T

Cp

]([1

[ 22 r

Tr

rrK

])(sin[sin1

2

T

r)](

sin

12

2

22 T

r gq

T

Cp

]([1

[ 22 r

Tr

rr

])(sin[sin1

2

T

r)](

sin

12

2

22 T

rK

q g

T

K

Cp

]([1

[ 22 r

Tr

rr

])(sin[sin1

2

T

r)](

sin

12

2

22 T

rK

q g

T1

KKKKr If material is homogeneous amp isotropic

TCqTK pg)(

T

K

qT g

12 pC

K

Thermal diffusivity

(Ein ndash Eout) + Egen = Est

General Heat Conduction Equation

PLANE WALL CYLINDER SPHERE

Temperature profile

Plane wall SphereCylinder

bull A copper conductor (k = 380 WmoC resistivity ρ = 2x10-8 Ω m) having inner and outer radii 10 cm and 225 cm respectively is carrying a current density of 4800 amperes cm2 the conductor is internally cooled and a constant temperature of 65oC is maintained at the inner surface and there is no heat transfer through the insulation surrounding the conductor

Determinebull (i) The maximum temperature of the conductor and the radius at

which it occursbull (ii) The internal heat transfer rate

bull A chemical reaction takes place in a packed bed (k=06 WmoC) between two coaxial cylinders with radii 15 mm and 45 mm the inner surface is at 580oC and it is insulated Assuming the reaction rate of 055 MWm3 in the reactor volume find the temperature at the outer surface of the reactor

bull A current of 300 amperes passes through a stainless steel wire of 25 mm diameter and k=20 WmoC The resistivity of the wire is 70x10-8 ohm-m and length of the wire is 2 m if the wire is submerged in a fluid maintained at 50oC and h at the wire surface is 4000 Wm2oC calculate the steady state temperature at the centre and at the surface of the wire

bull An approximately spherical shaped orange ( k=023 WmoC) 90 mm diameter undergoes ripping process and generates 5100 Wm3 of energy If external surface the orange is at 8oC determine

bull (i) temperature at the centre of the orange and

bull (ii) heat flow from the outer surface of the orange

bull FINS External surfaces to used to cooling area and hence convection

heat transfer can be improvedbull APPLICATIONS IC engine cylinder ( air cooling) Heat exchangers automobile

radiators computer chip refrigerators electric motors generators and transformers etc

bull ASSUMPTIONS MADE IN ANALYSIS OF HEAT TRANFER THROUGH FIN

1 Fin material is homogenous and isotropic constthermal conductivity

2 Temperature at any cs of the fin is uniform3 lsquohrsquo is uniform over entire surface4 No internal heat generation5 Thermal contact resistance is negligible6 Radiation heat transfer is negligible7 Heat conduction is steady state and one dimensional

bull Problem 1

bull A stainless steel fin (k=20 WmK) having a diameter of 20 mm and a length of 01 m is attached to a wall at 300oC The ambient temperature is 50oC and the heat transfer coefficient is 10 Wm2K The fin tip is insulated Determine

bull (A) The rate of heat dissipation from the finbull (B) The temperature at the fin tipbull (C) The rate of heat transfer from the wall area covered by the fin if

the fin was not usedbull (D) The heat transfer rate from the same fin geometry if the

stainless steel fin is replaced by a fictitious fin with infinite thermal conductivity

bull Problem2bull One end of a long rod is inserted into a furnace while the other

projects into ambient temperature Under steady state the temperature of the rod is measured at two points 75mm apart and found to be 125oC and 885oC respectively while the ambient temperature is 20oC If the rod is 25mm in diameter and h is 2336 Wm2K

bull Find the thermal conductivity of the rod material

  • HEAT AND MASS TRANSFER
  • Thermodynamics Vs Heat transfer
  • MODE OF HEAT TRANSFER
  • MODE OF HEAT TRANSFER (2)
  • CONDUCTION- MECHANISM
  • CONCECTION-MECHANISM
  • RADIATION-MECHANISM
  • Conduction- Convection- Radiation
  • Slide 9
  • Slide 10
  • Fourierrsquos Law of Heat Conduction
  • Slide 12
  • Newtonrsquos Law of Cooling
  • Laws of Thermal Radiation
  • LAWS FOR HEAT TRANSFER
  • ELECTRICAL ampTHERMAL ANALOGY
  • Slide 17
  • Slide 18
  • PLANE WALL-UNIDIRECTIONAL
  • PLANE WALL -SERIES
  • PLANE WALL IN SERIES
  • PLANE WALL -PARALLEL
  • PLANE WALL-PARALLEL
  • PLANE WALL-COMBINATION
  • PLANE WALL-FILM RESISTANCE
  • PLANE WALL- THERMAL CONTACT RESISTANCE
  • PLANE WALL- THERMAL CONTACT RESISTANCE (2)
  • CYLINDER-SOLID
  • HEAT TRANSFER IN SOLID CYLINDER
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER-COMPOSITE
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER-MULTY LAYER
  • HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER
  • HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER (2)
  • HEAT TRANSFER IN SPHERE
  • HEAT TRANSFER IN SPHERE -MULTY LAYER
  • HEAT TRANSFER IN SPHERE -MULTY LAYER (2)
  • Slide 38
  • Slide 39
  • Slide 40
  • Slide 41
  • CRITICAL RADIUS
  • Plane wall- Insulation thickness -study
  • Problem5
  • Slide 45
  • Slide 46
  • Slide 47
  • GENERAL HEAT CONDUCTION EQUATION IN CYLINDRICAL COORDINATES
  • Slide 49
  • Slide 50
  • Slide 51
  • GENERAL HEAT CONDUCTION EQUATION IN SPHERICAL COORDINATES
  • Slide 53
  • Slide 54
  • Slide 55
  • General Heat Conduction Equation
  • Temperature profile
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Slide 67
Page 29: hmt chap 1

HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER

HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER

HEAT TRANSFER IN SPHERE

HEAT TRANSFER IN SPHERE -MULTY LAYER

HEAT TRANSFER IN SPHERE -MULTY LAYER

bull Problem 1 The wall in a furnace consists of 125 mm thick refractory bricks and

125 mm thick insulating fire bricks separated by an air gap A 12 mm thick plaster covers the outer wall The inner surface of wall is at 1100oC and the ambient temperature 25oC the heat transfer coefficient on the outer wall to the air is 17 Wm2K and the resistance to heat flow of the air gap is 016 KW the thermal conductivities of refractory brick insulating firebrick and plaster are 16 03 and 014 WmK respectively

Calculatebull (a) the rate of heat loss per unit area of the wall surfacebull (b) the interface temperatures throughout the wallbull (c) the temperature at the outside surface of the wall

bull Problem 2Two large aluminum plates (k=240 WmK) each 2 cm thick with 10 mm surface roughness are placed in contact under 105 Nm2 pressure in air The temperature at the outside surfaces are 390 oC and 406 oCCalculate

A) the heat flux B) the temperature drop due to the contact resistance C) the contact temperatures Thermal contact resistance with air as the interfacial fluid for

10 mm roughness is 27510-4 m2KW

bull Problem 3bull Steam at 350 oC flowing in a pipe (k=80 WmK) of 5 cm

inner diameter and 56 cm outer diameter is covered with 3 cm thick insulation of k=005 WmK Heat is lost to surroundings at 5 oC by natural convection and radiation the combined h being 20 Wm2K Taking the heat transfer coefficient inside the pipe as 60 Wm2K

bull Determinebull A) the rate of heat loss from the steam per unit length of

the pipebull B) the temperature drop across the pipe and the insulation

bull PROBLEM4bull A nuclear reactor has a sperical pressure vessel 750 mm inside

diameter 80 mm wall thickness The temperature at the inner surface is 500 oC If the tempertarue at the outer surface is 495 oC

bull Calculatea) The rate of heat loss through the metal wallb) If the rate of heat loss is to limited to 305Whow much inulation

thickness needs to be applied Thermal conductivity of steel wall=46 WmoC Thermal conductivity of insulation=004 WmoC Film coefficient of heat tranfer at inner and outer surface are 200

Wm2 oC and 15 Wm2 oC Ambient air temperature=28 oC

CRITICAL RADIUS

bull The radius upto which heat flow increases and after decreases At critical radius q becomes qmax

bull The addition of insulation always increases (Rth)condbull Rtotal= (Rth)cond + (Rth)conv

bull The addition of insulation may reduce the (Rth)conv due to increase in surface area

bull r1ltrc -heat transfer increases by adding insulation till r1=rc If

insulation thickness is further increased the rate of heat loss will decrease

r1 gt rc

- heat transfer decreases by adding insulation above the rc

Plane wall- Insulation

thickness -study

Cylinder amp Sphere- Insulation thickness -

study

Problem5bull A 3mm diameter and 5 mm long electric wire is tightly

wrapped with a 2 mm thick plastic cover whose K=015 wmoC Electrical measurements indicate that a current of 10 A passes through the wire and there is a voltage drop of 8 V along the wire If the insulated wire is exposed to a medium at Ta=30oC with a heat transfer coefficient of h=12 wm2oC Determine the temperature at the interface of the wire and the plastic cover in steady operation Also determine whether doubling the thickness of the plastic cover will increase or decrease this interface temperature

dx

TdzdyKQ xx )(

dxQx

QQ xxdxx ][

dxdx

TdzdyK

xQQdQ xdxxxx ])([)(

)1()]([)(

ddzdydx

x

TK

xdQ xx

GENERAL HEAT CONDUCTION EQUATION IN CARTESIAN COORDINATES

Consider an unsteady (transient) three dimensional heat conduction with internal heat generation in heterogeneous amp anisotropic PLANE WALL

Cartesian co-ordinates x y z

Elemental length dxdydz

Consider an elemental volume ABCDEFGH = (dxdydz)

Let Temperature distribution is a function of T=T(xyz)

Let

=internal heat generation per unit volume per unit time (W m3)q= rate of heat flow (W)Q= qdt=total heat flow (J)r=mass density of the material element(kgm3)Cp=specific heat of the material element (Jkg K)

gq

Plane (y - z) lsquoxrsquo direction

Quantity of heat leaving from the face EFGH in lsquoXrsquo direction

Quantity of heat flowing into the face ABCD in lsquoXrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoXrsquo direction

dy

TdzdxKQ yy )(

dyQy

QQ yydyy ][

dydy

TdzdxK

yQQdQ ydyyyy ])([)(

)2()]([)(

ddzdydx

y

TK

ydQ yy

Plane (x - z) lsquoyrsquo direction

Quantity of heat flowing into the face ABFE in lsquoYrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoYrsquo direction

dz

TdydxKQ zz )(

dzQz

QQ zzdzz ][

dzdz

TdydxK

zQQdQ zdzzzz ])([)(

)3()]([)(

ddzdydx

z

TK

zdQ zz

Plane (x - y) lsquozrsquo direction

Quantity of heat flowing into the face ADHE in lsquoZrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoZrsquo direction

Quantity of heat leaving from the face DCGH in lsquoYrsquo direction

Quantity of heat leaving from the face BCGF in lsquoZrsquo direction

)4()()(

ddzdydxqQgg

Total amount of internal heat generation in the element

Ein + Egen = Eout+Est (Ein- Eout)+ Egen=Est

T

Cddzdydxddzdydxqddzdydxz

TK

zy

TK

yx

TK

x pgzyx )()())](()()([

T

pC

gq

z

TzK

zy

Ty

Kyx

TxK

x][][][

stgzyx QQdQdQdQ

)(

)5()()(

dT

CdzdydxQ pst

Total heat energy stored in the element

(ie thermal energy of the element can be increased because of heat accumulation in the element due to heat flow amp internal heat generation)

)( volumem d

TmCQ pst )(

T

CqTK pg)(

KKKyKx z If material is homogeneous amp isotropic

T

pC

gq

z

T

zy

T

yx

T

xK ][][][

T

Kp

C

Kgq

z

T

y

T

x

T2

2

2

2

2

2

T

Kgq

z

T

y

T

x

T 12

2

2

2

2

2

Thermal diffusivity pC

K

T

K

qT g

12

according to thermodynamics first law

)( zr )( dzrddr

)( dzrddr

Cylindrical polar co-ordinates

Elemental length

Consider an elemental volume ABCDEFGH =

GENERAL HEAT CONDUCTION EQUATION IN CYLINDRICAL COORDINATES

gq

gq

Consider an unsteady (transient) three dimensional heat conduction with internal heat generation in heterogeneous amp anisotropic cylindrical geometries (eg rods and pipes)

)( zrTLetTemperature distribution is a function of T=

Let

=internal heat generation per unit volume per unit time (W m3)q= rate of heat flow (W)Q= qdt=total heat flow (J)r=mass density of the material element(kgm3)Cp=specific heat of the material element (Jkg K)

planerdirectionz

planezrdirection

planezdirectionr

)(

)(

)(

LetHeat flow in

drdr

TdzrdK

rdQ

QQdQ

rr

drrrr

])([

)1(][

r

Tr

rddrdzdKdQ rr

Heat accumulation in the element due to heat flow by conduction in lsquorrsquo direction

rddr

TdzdrK

rdQ

QQdQ d

])([

][

2

T

r

ddrdzrdKdQ

)2(][

2

2

2

T

r

ddrdzrdKdQ

Heat accumulation in the element due to heat flow by conduction in lsquoqrsquo direction

Plane (z- q ) lsquorrsquo direction

dr

TdzrdKQ rr

)(

drQr

QQ rrdrr ][

Quantity of heat leaving from the face EFGH in lsquorrsquo direction

Quantity of heat flowing into the face ABCD in lsquorrsquo direction

Plane (r-z) lsquoq lsquodirection

Quantity of heat leaving from the face AEGC in lsquoqrsquo direction

d

r

TdzdrKQ )(

rdQ

rQQ d ][

Quantity of heat flowing into the face BFHD in lsquoqrsquo direction

)3())((2

2

z

TdzdrrdKdQ zz

)5()()(

dT

CdzdrrdQ pst

)4()()(

ddzdrrdqQgg

Total amount of internal heat generation in the element

Total heat energy stored in the element

(ie thermal energy of the element can be increased because of heat accumulation in the element due to heat flow amp internal heat generation)

according to thermodynamics first law

Ein + Egen = Eout + Est (Ein ndash Eout) + Egen = Est

)( volumem

)()(z

T

zdzdrrdKdQ zz

Heat accumulation in the element due to heat flow by conduction in lsquozrsquo direction

dzdz

TdrrdK

zdQ

QQdQ

zz

dzzzz

])([

d

TmCQ pst )(

Plane (r-q) lsquozrsquo direction

dz

TdrrdKQ zz )(

dzQz

QQ zzdzz ][ Quantity of heat leaving from the face CDHG in lsquozrsquo direction

Quantity of heat flowing into the face ABFE in lsquozrsquo direction

stgzr QQdQdQdQ

)(

r

r ][)(rT

rr

ddrdzdKr

][)(

2

2

2

T

r

ddrdzrdK ))((2

2

z

TddzdrrdK z

ddzdrrdq g )(

dT

Cdzdrrd p )(

+ + +

=

T

CqTK pg)(KKKK zr

[gq

)(1

r

Tr

rrK r

)(1

2

2

2 T

rK )(

2

2

z

TK z

)( ddzdrrd )(

ddzdrrd

TC p

+ + + ] =

If material is homogeneous amp isotropic

)(1

r

Tr

rr

)(1

2

2

2 T

r)(

2

2

z

T

K

q g

T

K

C p

)(1

r

Tr

rr

)(1

2

2

2 T

r )(2

2

z

T

K

q g

T

1

+ + + =

)(1

r

Tr

rrK r

)(1

2

2

2 T

rK )(

2

2

z

TK z

gq

T

C p + + + =

+ + + = Thermal diffusivity

T

K

qT g

12

pC

K

(Ein ndash Eout) + Egen = Est

)( rT

)( r

)sin( drrddr)sin( drrddr

planerdirection

planerdirection

planedirectionr

)(

)(

)(

Spherical polar co-ordinates

Elemental length

Consider an elemental volume ABCDEFGH =

GENERAL HEAT CONDUCTION EQUATION IN SPHERICAL COORDINATES

Let

=internal heat generation per unit volume per unit time (W m3)q= rate of heat flow (W)Q= qdt=total heat flow (J)r=mass density of the material element(kgm3)Cp=specific heat of the material element (Jkg K)

gq

Consider an unsteady (transient) three dimensional heat conduction with internal heat generation in heterogeneous amp anisotropic spherical geometries

LetTemperature distribution is a function of T=

X

Y

Z

drT

drrdKQ rr )sin(

drQr

QQ rrdrr ][

drdr

TdrrdK

rdQ

QQdQ

rr

drrrr

])sin([

)1(]([sin 2

rT

rr

ddrddKdQ rr

Plane ( -q f) lsquorrsquo directionQuantity of heat flowing into the face ABCD in lsquorrsquo direction

Quantity of heat leaving from the face EFGH in lsquorrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquorrsquo direction

drT

drdrKQ )sin(

rdQ

rQQ d ][

rddr

TdrdrK

rdQ

QQdQ d

])sin([

])[(sin

2

T

r

drdrdrKdQ

)2(])(sin[

2

T

r

drdrddrKdQ

Quantity of heat leaving from the face BFHD in lsquoqrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoqrsquo direction

X

Y

Z

Plane (r- f ) lsquoq lsquodirection

Quantity of heat flowing into the face AEGC in lsquoqrsquo direction

dr

TdrrdKQ

sin)(

drQ

rQQ d sin][

sin

drdr

TdrrdK

rdQ

QQdQ d

sin]sin

)([sin

][sinsin

)sin(

T

rr

ddrdrrdKdQ

)3()(sin

)sin(2

2

22

2

T

r

ddddrrKdQ

Plane (r- ) q lsquofrsquo directionQuantity of heat flowing into the face ABFE in lsquofrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquozrsquo direction

)4()sin()(

ddrrddrqQgg

)5()sin()(

dT

CdrrddrQ pst

Total amount of internal heat generation in the element

Total heat energy stored in the element

(Ein ndash Eout) + Egen = Est

d

TmCQ pst )(

)( volumem

Quantity of heat leaving from the face CDHG in lsquofrsquo direction

according to thermodynamics first law

Ein + Egen = Eout + Est

X

Y

Z

stgr QQdQdQdQ

)(

]([sin 2

r

Tr

rddrddKr

])(sin[

2

T

rdrdrddrK

)(sin

)sin(2

2

22

2

T

r

ddddrrK

ddrrddrq g )sin(

dT

Cdrrddr p )sin(

2

2

r

r

sin

sin

]([ 22 r

Tr

rr

Kr

])(sin[sin2

TrK

)(sin 2

2

22

T

r

Kgq

T

Cp

]([1

[ 22 r

Tr

rrK

])(sin[sin1

2

T

r)](

sin

12

2

22 T

r gq

T

Cp

]([1

[ 22 r

Tr

rr

])(sin[sin1

2

T

r)](

sin

12

2

22 T

rK

q g

T

K

Cp

]([1

[ 22 r

Tr

rr

])(sin[sin1

2

T

r)](

sin

12

2

22 T

rK

q g

T1

KKKKr If material is homogeneous amp isotropic

TCqTK pg)(

T

K

qT g

12 pC

K

Thermal diffusivity

(Ein ndash Eout) + Egen = Est

General Heat Conduction Equation

PLANE WALL CYLINDER SPHERE

Temperature profile

Plane wall SphereCylinder

bull A copper conductor (k = 380 WmoC resistivity ρ = 2x10-8 Ω m) having inner and outer radii 10 cm and 225 cm respectively is carrying a current density of 4800 amperes cm2 the conductor is internally cooled and a constant temperature of 65oC is maintained at the inner surface and there is no heat transfer through the insulation surrounding the conductor

Determinebull (i) The maximum temperature of the conductor and the radius at

which it occursbull (ii) The internal heat transfer rate

bull A chemical reaction takes place in a packed bed (k=06 WmoC) between two coaxial cylinders with radii 15 mm and 45 mm the inner surface is at 580oC and it is insulated Assuming the reaction rate of 055 MWm3 in the reactor volume find the temperature at the outer surface of the reactor

bull A current of 300 amperes passes through a stainless steel wire of 25 mm diameter and k=20 WmoC The resistivity of the wire is 70x10-8 ohm-m and length of the wire is 2 m if the wire is submerged in a fluid maintained at 50oC and h at the wire surface is 4000 Wm2oC calculate the steady state temperature at the centre and at the surface of the wire

bull An approximately spherical shaped orange ( k=023 WmoC) 90 mm diameter undergoes ripping process and generates 5100 Wm3 of energy If external surface the orange is at 8oC determine

bull (i) temperature at the centre of the orange and

bull (ii) heat flow from the outer surface of the orange

bull FINS External surfaces to used to cooling area and hence convection

heat transfer can be improvedbull APPLICATIONS IC engine cylinder ( air cooling) Heat exchangers automobile

radiators computer chip refrigerators electric motors generators and transformers etc

bull ASSUMPTIONS MADE IN ANALYSIS OF HEAT TRANFER THROUGH FIN

1 Fin material is homogenous and isotropic constthermal conductivity

2 Temperature at any cs of the fin is uniform3 lsquohrsquo is uniform over entire surface4 No internal heat generation5 Thermal contact resistance is negligible6 Radiation heat transfer is negligible7 Heat conduction is steady state and one dimensional

bull Problem 1

bull A stainless steel fin (k=20 WmK) having a diameter of 20 mm and a length of 01 m is attached to a wall at 300oC The ambient temperature is 50oC and the heat transfer coefficient is 10 Wm2K The fin tip is insulated Determine

bull (A) The rate of heat dissipation from the finbull (B) The temperature at the fin tipbull (C) The rate of heat transfer from the wall area covered by the fin if

the fin was not usedbull (D) The heat transfer rate from the same fin geometry if the

stainless steel fin is replaced by a fictitious fin with infinite thermal conductivity

bull Problem2bull One end of a long rod is inserted into a furnace while the other

projects into ambient temperature Under steady state the temperature of the rod is measured at two points 75mm apart and found to be 125oC and 885oC respectively while the ambient temperature is 20oC If the rod is 25mm in diameter and h is 2336 Wm2K

bull Find the thermal conductivity of the rod material

  • HEAT AND MASS TRANSFER
  • Thermodynamics Vs Heat transfer
  • MODE OF HEAT TRANSFER
  • MODE OF HEAT TRANSFER (2)
  • CONDUCTION- MECHANISM
  • CONCECTION-MECHANISM
  • RADIATION-MECHANISM
  • Conduction- Convection- Radiation
  • Slide 9
  • Slide 10
  • Fourierrsquos Law of Heat Conduction
  • Slide 12
  • Newtonrsquos Law of Cooling
  • Laws of Thermal Radiation
  • LAWS FOR HEAT TRANSFER
  • ELECTRICAL ampTHERMAL ANALOGY
  • Slide 17
  • Slide 18
  • PLANE WALL-UNIDIRECTIONAL
  • PLANE WALL -SERIES
  • PLANE WALL IN SERIES
  • PLANE WALL -PARALLEL
  • PLANE WALL-PARALLEL
  • PLANE WALL-COMBINATION
  • PLANE WALL-FILM RESISTANCE
  • PLANE WALL- THERMAL CONTACT RESISTANCE
  • PLANE WALL- THERMAL CONTACT RESISTANCE (2)
  • CYLINDER-SOLID
  • HEAT TRANSFER IN SOLID CYLINDER
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER-COMPOSITE
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER-MULTY LAYER
  • HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER
  • HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER (2)
  • HEAT TRANSFER IN SPHERE
  • HEAT TRANSFER IN SPHERE -MULTY LAYER
  • HEAT TRANSFER IN SPHERE -MULTY LAYER (2)
  • Slide 38
  • Slide 39
  • Slide 40
  • Slide 41
  • CRITICAL RADIUS
  • Plane wall- Insulation thickness -study
  • Problem5
  • Slide 45
  • Slide 46
  • Slide 47
  • GENERAL HEAT CONDUCTION EQUATION IN CYLINDRICAL COORDINATES
  • Slide 49
  • Slide 50
  • Slide 51
  • GENERAL HEAT CONDUCTION EQUATION IN SPHERICAL COORDINATES
  • Slide 53
  • Slide 54
  • Slide 55
  • General Heat Conduction Equation
  • Temperature profile
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Slide 67
Page 30: hmt chap 1

HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER

HEAT TRANSFER IN SPHERE

HEAT TRANSFER IN SPHERE -MULTY LAYER

HEAT TRANSFER IN SPHERE -MULTY LAYER

bull Problem 1 The wall in a furnace consists of 125 mm thick refractory bricks and

125 mm thick insulating fire bricks separated by an air gap A 12 mm thick plaster covers the outer wall The inner surface of wall is at 1100oC and the ambient temperature 25oC the heat transfer coefficient on the outer wall to the air is 17 Wm2K and the resistance to heat flow of the air gap is 016 KW the thermal conductivities of refractory brick insulating firebrick and plaster are 16 03 and 014 WmK respectively

Calculatebull (a) the rate of heat loss per unit area of the wall surfacebull (b) the interface temperatures throughout the wallbull (c) the temperature at the outside surface of the wall

bull Problem 2Two large aluminum plates (k=240 WmK) each 2 cm thick with 10 mm surface roughness are placed in contact under 105 Nm2 pressure in air The temperature at the outside surfaces are 390 oC and 406 oCCalculate

A) the heat flux B) the temperature drop due to the contact resistance C) the contact temperatures Thermal contact resistance with air as the interfacial fluid for

10 mm roughness is 27510-4 m2KW

bull Problem 3bull Steam at 350 oC flowing in a pipe (k=80 WmK) of 5 cm

inner diameter and 56 cm outer diameter is covered with 3 cm thick insulation of k=005 WmK Heat is lost to surroundings at 5 oC by natural convection and radiation the combined h being 20 Wm2K Taking the heat transfer coefficient inside the pipe as 60 Wm2K

bull Determinebull A) the rate of heat loss from the steam per unit length of

the pipebull B) the temperature drop across the pipe and the insulation

bull PROBLEM4bull A nuclear reactor has a sperical pressure vessel 750 mm inside

diameter 80 mm wall thickness The temperature at the inner surface is 500 oC If the tempertarue at the outer surface is 495 oC

bull Calculatea) The rate of heat loss through the metal wallb) If the rate of heat loss is to limited to 305Whow much inulation

thickness needs to be applied Thermal conductivity of steel wall=46 WmoC Thermal conductivity of insulation=004 WmoC Film coefficient of heat tranfer at inner and outer surface are 200

Wm2 oC and 15 Wm2 oC Ambient air temperature=28 oC

CRITICAL RADIUS

bull The radius upto which heat flow increases and after decreases At critical radius q becomes qmax

bull The addition of insulation always increases (Rth)condbull Rtotal= (Rth)cond + (Rth)conv

bull The addition of insulation may reduce the (Rth)conv due to increase in surface area

bull r1ltrc -heat transfer increases by adding insulation till r1=rc If

insulation thickness is further increased the rate of heat loss will decrease

r1 gt rc

- heat transfer decreases by adding insulation above the rc

Plane wall- Insulation

thickness -study

Cylinder amp Sphere- Insulation thickness -

study

Problem5bull A 3mm diameter and 5 mm long electric wire is tightly

wrapped with a 2 mm thick plastic cover whose K=015 wmoC Electrical measurements indicate that a current of 10 A passes through the wire and there is a voltage drop of 8 V along the wire If the insulated wire is exposed to a medium at Ta=30oC with a heat transfer coefficient of h=12 wm2oC Determine the temperature at the interface of the wire and the plastic cover in steady operation Also determine whether doubling the thickness of the plastic cover will increase or decrease this interface temperature

dx

TdzdyKQ xx )(

dxQx

QQ xxdxx ][

dxdx

TdzdyK

xQQdQ xdxxxx ])([)(

)1()]([)(

ddzdydx

x

TK

xdQ xx

GENERAL HEAT CONDUCTION EQUATION IN CARTESIAN COORDINATES

Consider an unsteady (transient) three dimensional heat conduction with internal heat generation in heterogeneous amp anisotropic PLANE WALL

Cartesian co-ordinates x y z

Elemental length dxdydz

Consider an elemental volume ABCDEFGH = (dxdydz)

Let Temperature distribution is a function of T=T(xyz)

Let

=internal heat generation per unit volume per unit time (W m3)q= rate of heat flow (W)Q= qdt=total heat flow (J)r=mass density of the material element(kgm3)Cp=specific heat of the material element (Jkg K)

gq

Plane (y - z) lsquoxrsquo direction

Quantity of heat leaving from the face EFGH in lsquoXrsquo direction

Quantity of heat flowing into the face ABCD in lsquoXrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoXrsquo direction

dy

TdzdxKQ yy )(

dyQy

QQ yydyy ][

dydy

TdzdxK

yQQdQ ydyyyy ])([)(

)2()]([)(

ddzdydx

y

TK

ydQ yy

Plane (x - z) lsquoyrsquo direction

Quantity of heat flowing into the face ABFE in lsquoYrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoYrsquo direction

dz

TdydxKQ zz )(

dzQz

QQ zzdzz ][

dzdz

TdydxK

zQQdQ zdzzzz ])([)(

)3()]([)(

ddzdydx

z

TK

zdQ zz

Plane (x - y) lsquozrsquo direction

Quantity of heat flowing into the face ADHE in lsquoZrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoZrsquo direction

Quantity of heat leaving from the face DCGH in lsquoYrsquo direction

Quantity of heat leaving from the face BCGF in lsquoZrsquo direction

)4()()(

ddzdydxqQgg

Total amount of internal heat generation in the element

Ein + Egen = Eout+Est (Ein- Eout)+ Egen=Est

T

Cddzdydxddzdydxqddzdydxz

TK

zy

TK

yx

TK

x pgzyx )()())](()()([

T

pC

gq

z

TzK

zy

Ty

Kyx

TxK

x][][][

stgzyx QQdQdQdQ

)(

)5()()(

dT

CdzdydxQ pst

Total heat energy stored in the element

(ie thermal energy of the element can be increased because of heat accumulation in the element due to heat flow amp internal heat generation)

)( volumem d

TmCQ pst )(

T

CqTK pg)(

KKKyKx z If material is homogeneous amp isotropic

T

pC

gq

z

T

zy

T

yx

T

xK ][][][

T

Kp

C

Kgq

z

T

y

T

x

T2

2

2

2

2

2

T

Kgq

z

T

y

T

x

T 12

2

2

2

2

2

Thermal diffusivity pC

K

T

K

qT g

12

according to thermodynamics first law

)( zr )( dzrddr

)( dzrddr

Cylindrical polar co-ordinates

Elemental length

Consider an elemental volume ABCDEFGH =

GENERAL HEAT CONDUCTION EQUATION IN CYLINDRICAL COORDINATES

gq

gq

Consider an unsteady (transient) three dimensional heat conduction with internal heat generation in heterogeneous amp anisotropic cylindrical geometries (eg rods and pipes)

)( zrTLetTemperature distribution is a function of T=

Let

=internal heat generation per unit volume per unit time (W m3)q= rate of heat flow (W)Q= qdt=total heat flow (J)r=mass density of the material element(kgm3)Cp=specific heat of the material element (Jkg K)

planerdirectionz

planezrdirection

planezdirectionr

)(

)(

)(

LetHeat flow in

drdr

TdzrdK

rdQ

QQdQ

rr

drrrr

])([

)1(][

r

Tr

rddrdzdKdQ rr

Heat accumulation in the element due to heat flow by conduction in lsquorrsquo direction

rddr

TdzdrK

rdQ

QQdQ d

])([

][

2

T

r

ddrdzrdKdQ

)2(][

2

2

2

T

r

ddrdzrdKdQ

Heat accumulation in the element due to heat flow by conduction in lsquoqrsquo direction

Plane (z- q ) lsquorrsquo direction

dr

TdzrdKQ rr

)(

drQr

QQ rrdrr ][

Quantity of heat leaving from the face EFGH in lsquorrsquo direction

Quantity of heat flowing into the face ABCD in lsquorrsquo direction

Plane (r-z) lsquoq lsquodirection

Quantity of heat leaving from the face AEGC in lsquoqrsquo direction

d

r

TdzdrKQ )(

rdQ

rQQ d ][

Quantity of heat flowing into the face BFHD in lsquoqrsquo direction

)3())((2

2

z

TdzdrrdKdQ zz

)5()()(

dT

CdzdrrdQ pst

)4()()(

ddzdrrdqQgg

Total amount of internal heat generation in the element

Total heat energy stored in the element

(ie thermal energy of the element can be increased because of heat accumulation in the element due to heat flow amp internal heat generation)

according to thermodynamics first law

Ein + Egen = Eout + Est (Ein ndash Eout) + Egen = Est

)( volumem

)()(z

T

zdzdrrdKdQ zz

Heat accumulation in the element due to heat flow by conduction in lsquozrsquo direction

dzdz

TdrrdK

zdQ

QQdQ

zz

dzzzz

])([

d

TmCQ pst )(

Plane (r-q) lsquozrsquo direction

dz

TdrrdKQ zz )(

dzQz

QQ zzdzz ][ Quantity of heat leaving from the face CDHG in lsquozrsquo direction

Quantity of heat flowing into the face ABFE in lsquozrsquo direction

stgzr QQdQdQdQ

)(

r

r ][)(rT

rr

ddrdzdKr

][)(

2

2

2

T

r

ddrdzrdK ))((2

2

z

TddzdrrdK z

ddzdrrdq g )(

dT

Cdzdrrd p )(

+ + +

=

T

CqTK pg)(KKKK zr

[gq

)(1

r

Tr

rrK r

)(1

2

2

2 T

rK )(

2

2

z

TK z

)( ddzdrrd )(

ddzdrrd

TC p

+ + + ] =

If material is homogeneous amp isotropic

)(1

r

Tr

rr

)(1

2

2

2 T

r)(

2

2

z

T

K

q g

T

K

C p

)(1

r

Tr

rr

)(1

2

2

2 T

r )(2

2

z

T

K

q g

T

1

+ + + =

)(1

r

Tr

rrK r

)(1

2

2

2 T

rK )(

2

2

z

TK z

gq

T

C p + + + =

+ + + = Thermal diffusivity

T

K

qT g

12

pC

K

(Ein ndash Eout) + Egen = Est

)( rT

)( r

)sin( drrddr)sin( drrddr

planerdirection

planerdirection

planedirectionr

)(

)(

)(

Spherical polar co-ordinates

Elemental length

Consider an elemental volume ABCDEFGH =

GENERAL HEAT CONDUCTION EQUATION IN SPHERICAL COORDINATES

Let

=internal heat generation per unit volume per unit time (W m3)q= rate of heat flow (W)Q= qdt=total heat flow (J)r=mass density of the material element(kgm3)Cp=specific heat of the material element (Jkg K)

gq

Consider an unsteady (transient) three dimensional heat conduction with internal heat generation in heterogeneous amp anisotropic spherical geometries

LetTemperature distribution is a function of T=

X

Y

Z

drT

drrdKQ rr )sin(

drQr

QQ rrdrr ][

drdr

TdrrdK

rdQ

QQdQ

rr

drrrr

])sin([

)1(]([sin 2

rT

rr

ddrddKdQ rr

Plane ( -q f) lsquorrsquo directionQuantity of heat flowing into the face ABCD in lsquorrsquo direction

Quantity of heat leaving from the face EFGH in lsquorrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquorrsquo direction

drT

drdrKQ )sin(

rdQ

rQQ d ][

rddr

TdrdrK

rdQ

QQdQ d

])sin([

])[(sin

2

T

r

drdrdrKdQ

)2(])(sin[

2

T

r

drdrddrKdQ

Quantity of heat leaving from the face BFHD in lsquoqrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoqrsquo direction

X

Y

Z

Plane (r- f ) lsquoq lsquodirection

Quantity of heat flowing into the face AEGC in lsquoqrsquo direction

dr

TdrrdKQ

sin)(

drQ

rQQ d sin][

sin

drdr

TdrrdK

rdQ

QQdQ d

sin]sin

)([sin

][sinsin

)sin(

T

rr

ddrdrrdKdQ

)3()(sin

)sin(2

2

22

2

T

r

ddddrrKdQ

Plane (r- ) q lsquofrsquo directionQuantity of heat flowing into the face ABFE in lsquofrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquozrsquo direction

)4()sin()(

ddrrddrqQgg

)5()sin()(

dT

CdrrddrQ pst

Total amount of internal heat generation in the element

Total heat energy stored in the element

(Ein ndash Eout) + Egen = Est

d

TmCQ pst )(

)( volumem

Quantity of heat leaving from the face CDHG in lsquofrsquo direction

according to thermodynamics first law

Ein + Egen = Eout + Est

X

Y

Z

stgr QQdQdQdQ

)(

]([sin 2

r

Tr

rddrddKr

])(sin[

2

T

rdrdrddrK

)(sin

)sin(2

2

22

2

T

r

ddddrrK

ddrrddrq g )sin(

dT

Cdrrddr p )sin(

2

2

r

r

sin

sin

]([ 22 r

Tr

rr

Kr

])(sin[sin2

TrK

)(sin 2

2

22

T

r

Kgq

T

Cp

]([1

[ 22 r

Tr

rrK

])(sin[sin1

2

T

r)](

sin

12

2

22 T

r gq

T

Cp

]([1

[ 22 r

Tr

rr

])(sin[sin1

2

T

r)](

sin

12

2

22 T

rK

q g

T

K

Cp

]([1

[ 22 r

Tr

rr

])(sin[sin1

2

T

r)](

sin

12

2

22 T

rK

q g

T1

KKKKr If material is homogeneous amp isotropic

TCqTK pg)(

T

K

qT g

12 pC

K

Thermal diffusivity

(Ein ndash Eout) + Egen = Est

General Heat Conduction Equation

PLANE WALL CYLINDER SPHERE

Temperature profile

Plane wall SphereCylinder

bull A copper conductor (k = 380 WmoC resistivity ρ = 2x10-8 Ω m) having inner and outer radii 10 cm and 225 cm respectively is carrying a current density of 4800 amperes cm2 the conductor is internally cooled and a constant temperature of 65oC is maintained at the inner surface and there is no heat transfer through the insulation surrounding the conductor

Determinebull (i) The maximum temperature of the conductor and the radius at

which it occursbull (ii) The internal heat transfer rate

bull A chemical reaction takes place in a packed bed (k=06 WmoC) between two coaxial cylinders with radii 15 mm and 45 mm the inner surface is at 580oC and it is insulated Assuming the reaction rate of 055 MWm3 in the reactor volume find the temperature at the outer surface of the reactor

bull A current of 300 amperes passes through a stainless steel wire of 25 mm diameter and k=20 WmoC The resistivity of the wire is 70x10-8 ohm-m and length of the wire is 2 m if the wire is submerged in a fluid maintained at 50oC and h at the wire surface is 4000 Wm2oC calculate the steady state temperature at the centre and at the surface of the wire

bull An approximately spherical shaped orange ( k=023 WmoC) 90 mm diameter undergoes ripping process and generates 5100 Wm3 of energy If external surface the orange is at 8oC determine

bull (i) temperature at the centre of the orange and

bull (ii) heat flow from the outer surface of the orange

bull FINS External surfaces to used to cooling area and hence convection

heat transfer can be improvedbull APPLICATIONS IC engine cylinder ( air cooling) Heat exchangers automobile

radiators computer chip refrigerators electric motors generators and transformers etc

bull ASSUMPTIONS MADE IN ANALYSIS OF HEAT TRANFER THROUGH FIN

1 Fin material is homogenous and isotropic constthermal conductivity

2 Temperature at any cs of the fin is uniform3 lsquohrsquo is uniform over entire surface4 No internal heat generation5 Thermal contact resistance is negligible6 Radiation heat transfer is negligible7 Heat conduction is steady state and one dimensional

bull Problem 1

bull A stainless steel fin (k=20 WmK) having a diameter of 20 mm and a length of 01 m is attached to a wall at 300oC The ambient temperature is 50oC and the heat transfer coefficient is 10 Wm2K The fin tip is insulated Determine

bull (A) The rate of heat dissipation from the finbull (B) The temperature at the fin tipbull (C) The rate of heat transfer from the wall area covered by the fin if

the fin was not usedbull (D) The heat transfer rate from the same fin geometry if the

stainless steel fin is replaced by a fictitious fin with infinite thermal conductivity

bull Problem2bull One end of a long rod is inserted into a furnace while the other

projects into ambient temperature Under steady state the temperature of the rod is measured at two points 75mm apart and found to be 125oC and 885oC respectively while the ambient temperature is 20oC If the rod is 25mm in diameter and h is 2336 Wm2K

bull Find the thermal conductivity of the rod material

  • HEAT AND MASS TRANSFER
  • Thermodynamics Vs Heat transfer
  • MODE OF HEAT TRANSFER
  • MODE OF HEAT TRANSFER (2)
  • CONDUCTION- MECHANISM
  • CONCECTION-MECHANISM
  • RADIATION-MECHANISM
  • Conduction- Convection- Radiation
  • Slide 9
  • Slide 10
  • Fourierrsquos Law of Heat Conduction
  • Slide 12
  • Newtonrsquos Law of Cooling
  • Laws of Thermal Radiation
  • LAWS FOR HEAT TRANSFER
  • ELECTRICAL ampTHERMAL ANALOGY
  • Slide 17
  • Slide 18
  • PLANE WALL-UNIDIRECTIONAL
  • PLANE WALL -SERIES
  • PLANE WALL IN SERIES
  • PLANE WALL -PARALLEL
  • PLANE WALL-PARALLEL
  • PLANE WALL-COMBINATION
  • PLANE WALL-FILM RESISTANCE
  • PLANE WALL- THERMAL CONTACT RESISTANCE
  • PLANE WALL- THERMAL CONTACT RESISTANCE (2)
  • CYLINDER-SOLID
  • HEAT TRANSFER IN SOLID CYLINDER
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER-COMPOSITE
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER-MULTY LAYER
  • HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER
  • HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER (2)
  • HEAT TRANSFER IN SPHERE
  • HEAT TRANSFER IN SPHERE -MULTY LAYER
  • HEAT TRANSFER IN SPHERE -MULTY LAYER (2)
  • Slide 38
  • Slide 39
  • Slide 40
  • Slide 41
  • CRITICAL RADIUS
  • Plane wall- Insulation thickness -study
  • Problem5
  • Slide 45
  • Slide 46
  • Slide 47
  • GENERAL HEAT CONDUCTION EQUATION IN CYLINDRICAL COORDINATES
  • Slide 49
  • Slide 50
  • Slide 51
  • GENERAL HEAT CONDUCTION EQUATION IN SPHERICAL COORDINATES
  • Slide 53
  • Slide 54
  • Slide 55
  • General Heat Conduction Equation
  • Temperature profile
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Slide 67
Page 31: hmt chap 1

HEAT TRANSFER IN SPHERE

HEAT TRANSFER IN SPHERE -MULTY LAYER

HEAT TRANSFER IN SPHERE -MULTY LAYER

bull Problem 1 The wall in a furnace consists of 125 mm thick refractory bricks and

125 mm thick insulating fire bricks separated by an air gap A 12 mm thick plaster covers the outer wall The inner surface of wall is at 1100oC and the ambient temperature 25oC the heat transfer coefficient on the outer wall to the air is 17 Wm2K and the resistance to heat flow of the air gap is 016 KW the thermal conductivities of refractory brick insulating firebrick and plaster are 16 03 and 014 WmK respectively

Calculatebull (a) the rate of heat loss per unit area of the wall surfacebull (b) the interface temperatures throughout the wallbull (c) the temperature at the outside surface of the wall

bull Problem 2Two large aluminum plates (k=240 WmK) each 2 cm thick with 10 mm surface roughness are placed in contact under 105 Nm2 pressure in air The temperature at the outside surfaces are 390 oC and 406 oCCalculate

A) the heat flux B) the temperature drop due to the contact resistance C) the contact temperatures Thermal contact resistance with air as the interfacial fluid for

10 mm roughness is 27510-4 m2KW

bull Problem 3bull Steam at 350 oC flowing in a pipe (k=80 WmK) of 5 cm

inner diameter and 56 cm outer diameter is covered with 3 cm thick insulation of k=005 WmK Heat is lost to surroundings at 5 oC by natural convection and radiation the combined h being 20 Wm2K Taking the heat transfer coefficient inside the pipe as 60 Wm2K

bull Determinebull A) the rate of heat loss from the steam per unit length of

the pipebull B) the temperature drop across the pipe and the insulation

bull PROBLEM4bull A nuclear reactor has a sperical pressure vessel 750 mm inside

diameter 80 mm wall thickness The temperature at the inner surface is 500 oC If the tempertarue at the outer surface is 495 oC

bull Calculatea) The rate of heat loss through the metal wallb) If the rate of heat loss is to limited to 305Whow much inulation

thickness needs to be applied Thermal conductivity of steel wall=46 WmoC Thermal conductivity of insulation=004 WmoC Film coefficient of heat tranfer at inner and outer surface are 200

Wm2 oC and 15 Wm2 oC Ambient air temperature=28 oC

CRITICAL RADIUS

bull The radius upto which heat flow increases and after decreases At critical radius q becomes qmax

bull The addition of insulation always increases (Rth)condbull Rtotal= (Rth)cond + (Rth)conv

bull The addition of insulation may reduce the (Rth)conv due to increase in surface area

bull r1ltrc -heat transfer increases by adding insulation till r1=rc If

insulation thickness is further increased the rate of heat loss will decrease

r1 gt rc

- heat transfer decreases by adding insulation above the rc

Plane wall- Insulation

thickness -study

Cylinder amp Sphere- Insulation thickness -

study

Problem5bull A 3mm diameter and 5 mm long electric wire is tightly

wrapped with a 2 mm thick plastic cover whose K=015 wmoC Electrical measurements indicate that a current of 10 A passes through the wire and there is a voltage drop of 8 V along the wire If the insulated wire is exposed to a medium at Ta=30oC with a heat transfer coefficient of h=12 wm2oC Determine the temperature at the interface of the wire and the plastic cover in steady operation Also determine whether doubling the thickness of the plastic cover will increase or decrease this interface temperature

dx

TdzdyKQ xx )(

dxQx

QQ xxdxx ][

dxdx

TdzdyK

xQQdQ xdxxxx ])([)(

)1()]([)(

ddzdydx

x

TK

xdQ xx

GENERAL HEAT CONDUCTION EQUATION IN CARTESIAN COORDINATES

Consider an unsteady (transient) three dimensional heat conduction with internal heat generation in heterogeneous amp anisotropic PLANE WALL

Cartesian co-ordinates x y z

Elemental length dxdydz

Consider an elemental volume ABCDEFGH = (dxdydz)

Let Temperature distribution is a function of T=T(xyz)

Let

=internal heat generation per unit volume per unit time (W m3)q= rate of heat flow (W)Q= qdt=total heat flow (J)r=mass density of the material element(kgm3)Cp=specific heat of the material element (Jkg K)

gq

Plane (y - z) lsquoxrsquo direction

Quantity of heat leaving from the face EFGH in lsquoXrsquo direction

Quantity of heat flowing into the face ABCD in lsquoXrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoXrsquo direction

dy

TdzdxKQ yy )(

dyQy

QQ yydyy ][

dydy

TdzdxK

yQQdQ ydyyyy ])([)(

)2()]([)(

ddzdydx

y

TK

ydQ yy

Plane (x - z) lsquoyrsquo direction

Quantity of heat flowing into the face ABFE in lsquoYrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoYrsquo direction

dz

TdydxKQ zz )(

dzQz

QQ zzdzz ][

dzdz

TdydxK

zQQdQ zdzzzz ])([)(

)3()]([)(

ddzdydx

z

TK

zdQ zz

Plane (x - y) lsquozrsquo direction

Quantity of heat flowing into the face ADHE in lsquoZrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoZrsquo direction

Quantity of heat leaving from the face DCGH in lsquoYrsquo direction

Quantity of heat leaving from the face BCGF in lsquoZrsquo direction

)4()()(

ddzdydxqQgg

Total amount of internal heat generation in the element

Ein + Egen = Eout+Est (Ein- Eout)+ Egen=Est

T

Cddzdydxddzdydxqddzdydxz

TK

zy

TK

yx

TK

x pgzyx )()())](()()([

T

pC

gq

z

TzK

zy

Ty

Kyx

TxK

x][][][

stgzyx QQdQdQdQ

)(

)5()()(

dT

CdzdydxQ pst

Total heat energy stored in the element

(ie thermal energy of the element can be increased because of heat accumulation in the element due to heat flow amp internal heat generation)

)( volumem d

TmCQ pst )(

T

CqTK pg)(

KKKyKx z If material is homogeneous amp isotropic

T

pC

gq

z

T

zy

T

yx

T

xK ][][][

T

Kp

C

Kgq

z

T

y

T

x

T2

2

2

2

2

2

T

Kgq

z

T

y

T

x

T 12

2

2

2

2

2

Thermal diffusivity pC

K

T

K

qT g

12

according to thermodynamics first law

)( zr )( dzrddr

)( dzrddr

Cylindrical polar co-ordinates

Elemental length

Consider an elemental volume ABCDEFGH =

GENERAL HEAT CONDUCTION EQUATION IN CYLINDRICAL COORDINATES

gq

gq

Consider an unsteady (transient) three dimensional heat conduction with internal heat generation in heterogeneous amp anisotropic cylindrical geometries (eg rods and pipes)

)( zrTLetTemperature distribution is a function of T=

Let

=internal heat generation per unit volume per unit time (W m3)q= rate of heat flow (W)Q= qdt=total heat flow (J)r=mass density of the material element(kgm3)Cp=specific heat of the material element (Jkg K)

planerdirectionz

planezrdirection

planezdirectionr

)(

)(

)(

LetHeat flow in

drdr

TdzrdK

rdQ

QQdQ

rr

drrrr

])([

)1(][

r

Tr

rddrdzdKdQ rr

Heat accumulation in the element due to heat flow by conduction in lsquorrsquo direction

rddr

TdzdrK

rdQ

QQdQ d

])([

][

2

T

r

ddrdzrdKdQ

)2(][

2

2

2

T

r

ddrdzrdKdQ

Heat accumulation in the element due to heat flow by conduction in lsquoqrsquo direction

Plane (z- q ) lsquorrsquo direction

dr

TdzrdKQ rr

)(

drQr

QQ rrdrr ][

Quantity of heat leaving from the face EFGH in lsquorrsquo direction

Quantity of heat flowing into the face ABCD in lsquorrsquo direction

Plane (r-z) lsquoq lsquodirection

Quantity of heat leaving from the face AEGC in lsquoqrsquo direction

d

r

TdzdrKQ )(

rdQ

rQQ d ][

Quantity of heat flowing into the face BFHD in lsquoqrsquo direction

)3())((2

2

z

TdzdrrdKdQ zz

)5()()(

dT

CdzdrrdQ pst

)4()()(

ddzdrrdqQgg

Total amount of internal heat generation in the element

Total heat energy stored in the element

(ie thermal energy of the element can be increased because of heat accumulation in the element due to heat flow amp internal heat generation)

according to thermodynamics first law

Ein + Egen = Eout + Est (Ein ndash Eout) + Egen = Est

)( volumem

)()(z

T

zdzdrrdKdQ zz

Heat accumulation in the element due to heat flow by conduction in lsquozrsquo direction

dzdz

TdrrdK

zdQ

QQdQ

zz

dzzzz

])([

d

TmCQ pst )(

Plane (r-q) lsquozrsquo direction

dz

TdrrdKQ zz )(

dzQz

QQ zzdzz ][ Quantity of heat leaving from the face CDHG in lsquozrsquo direction

Quantity of heat flowing into the face ABFE in lsquozrsquo direction

stgzr QQdQdQdQ

)(

r

r ][)(rT

rr

ddrdzdKr

][)(

2

2

2

T

r

ddrdzrdK ))((2

2

z

TddzdrrdK z

ddzdrrdq g )(

dT

Cdzdrrd p )(

+ + +

=

T

CqTK pg)(KKKK zr

[gq

)(1

r

Tr

rrK r

)(1

2

2

2 T

rK )(

2

2

z

TK z

)( ddzdrrd )(

ddzdrrd

TC p

+ + + ] =

If material is homogeneous amp isotropic

)(1

r

Tr

rr

)(1

2

2

2 T

r)(

2

2

z

T

K

q g

T

K

C p

)(1

r

Tr

rr

)(1

2

2

2 T

r )(2

2

z

T

K

q g

T

1

+ + + =

)(1

r

Tr

rrK r

)(1

2

2

2 T

rK )(

2

2

z

TK z

gq

T

C p + + + =

+ + + = Thermal diffusivity

T

K

qT g

12

pC

K

(Ein ndash Eout) + Egen = Est

)( rT

)( r

)sin( drrddr)sin( drrddr

planerdirection

planerdirection

planedirectionr

)(

)(

)(

Spherical polar co-ordinates

Elemental length

Consider an elemental volume ABCDEFGH =

GENERAL HEAT CONDUCTION EQUATION IN SPHERICAL COORDINATES

Let

=internal heat generation per unit volume per unit time (W m3)q= rate of heat flow (W)Q= qdt=total heat flow (J)r=mass density of the material element(kgm3)Cp=specific heat of the material element (Jkg K)

gq

Consider an unsteady (transient) three dimensional heat conduction with internal heat generation in heterogeneous amp anisotropic spherical geometries

LetTemperature distribution is a function of T=

X

Y

Z

drT

drrdKQ rr )sin(

drQr

QQ rrdrr ][

drdr

TdrrdK

rdQ

QQdQ

rr

drrrr

])sin([

)1(]([sin 2

rT

rr

ddrddKdQ rr

Plane ( -q f) lsquorrsquo directionQuantity of heat flowing into the face ABCD in lsquorrsquo direction

Quantity of heat leaving from the face EFGH in lsquorrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquorrsquo direction

drT

drdrKQ )sin(

rdQ

rQQ d ][

rddr

TdrdrK

rdQ

QQdQ d

])sin([

])[(sin

2

T

r

drdrdrKdQ

)2(])(sin[

2

T

r

drdrddrKdQ

Quantity of heat leaving from the face BFHD in lsquoqrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoqrsquo direction

X

Y

Z

Plane (r- f ) lsquoq lsquodirection

Quantity of heat flowing into the face AEGC in lsquoqrsquo direction

dr

TdrrdKQ

sin)(

drQ

rQQ d sin][

sin

drdr

TdrrdK

rdQ

QQdQ d

sin]sin

)([sin

][sinsin

)sin(

T

rr

ddrdrrdKdQ

)3()(sin

)sin(2

2

22

2

T

r

ddddrrKdQ

Plane (r- ) q lsquofrsquo directionQuantity of heat flowing into the face ABFE in lsquofrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquozrsquo direction

)4()sin()(

ddrrddrqQgg

)5()sin()(

dT

CdrrddrQ pst

Total amount of internal heat generation in the element

Total heat energy stored in the element

(Ein ndash Eout) + Egen = Est

d

TmCQ pst )(

)( volumem

Quantity of heat leaving from the face CDHG in lsquofrsquo direction

according to thermodynamics first law

Ein + Egen = Eout + Est

X

Y

Z

stgr QQdQdQdQ

)(

]([sin 2

r

Tr

rddrddKr

])(sin[

2

T

rdrdrddrK

)(sin

)sin(2

2

22

2

T

r

ddddrrK

ddrrddrq g )sin(

dT

Cdrrddr p )sin(

2

2

r

r

sin

sin

]([ 22 r

Tr

rr

Kr

])(sin[sin2

TrK

)(sin 2

2

22

T

r

Kgq

T

Cp

]([1

[ 22 r

Tr

rrK

])(sin[sin1

2

T

r)](

sin

12

2

22 T

r gq

T

Cp

]([1

[ 22 r

Tr

rr

])(sin[sin1

2

T

r)](

sin

12

2

22 T

rK

q g

T

K

Cp

]([1

[ 22 r

Tr

rr

])(sin[sin1

2

T

r)](

sin

12

2

22 T

rK

q g

T1

KKKKr If material is homogeneous amp isotropic

TCqTK pg)(

T

K

qT g

12 pC

K

Thermal diffusivity

(Ein ndash Eout) + Egen = Est

General Heat Conduction Equation

PLANE WALL CYLINDER SPHERE

Temperature profile

Plane wall SphereCylinder

bull A copper conductor (k = 380 WmoC resistivity ρ = 2x10-8 Ω m) having inner and outer radii 10 cm and 225 cm respectively is carrying a current density of 4800 amperes cm2 the conductor is internally cooled and a constant temperature of 65oC is maintained at the inner surface and there is no heat transfer through the insulation surrounding the conductor

Determinebull (i) The maximum temperature of the conductor and the radius at

which it occursbull (ii) The internal heat transfer rate

bull A chemical reaction takes place in a packed bed (k=06 WmoC) between two coaxial cylinders with radii 15 mm and 45 mm the inner surface is at 580oC and it is insulated Assuming the reaction rate of 055 MWm3 in the reactor volume find the temperature at the outer surface of the reactor

bull A current of 300 amperes passes through a stainless steel wire of 25 mm diameter and k=20 WmoC The resistivity of the wire is 70x10-8 ohm-m and length of the wire is 2 m if the wire is submerged in a fluid maintained at 50oC and h at the wire surface is 4000 Wm2oC calculate the steady state temperature at the centre and at the surface of the wire

bull An approximately spherical shaped orange ( k=023 WmoC) 90 mm diameter undergoes ripping process and generates 5100 Wm3 of energy If external surface the orange is at 8oC determine

bull (i) temperature at the centre of the orange and

bull (ii) heat flow from the outer surface of the orange

bull FINS External surfaces to used to cooling area and hence convection

heat transfer can be improvedbull APPLICATIONS IC engine cylinder ( air cooling) Heat exchangers automobile

radiators computer chip refrigerators electric motors generators and transformers etc

bull ASSUMPTIONS MADE IN ANALYSIS OF HEAT TRANFER THROUGH FIN

1 Fin material is homogenous and isotropic constthermal conductivity

2 Temperature at any cs of the fin is uniform3 lsquohrsquo is uniform over entire surface4 No internal heat generation5 Thermal contact resistance is negligible6 Radiation heat transfer is negligible7 Heat conduction is steady state and one dimensional

bull Problem 1

bull A stainless steel fin (k=20 WmK) having a diameter of 20 mm and a length of 01 m is attached to a wall at 300oC The ambient temperature is 50oC and the heat transfer coefficient is 10 Wm2K The fin tip is insulated Determine

bull (A) The rate of heat dissipation from the finbull (B) The temperature at the fin tipbull (C) The rate of heat transfer from the wall area covered by the fin if

the fin was not usedbull (D) The heat transfer rate from the same fin geometry if the

stainless steel fin is replaced by a fictitious fin with infinite thermal conductivity

bull Problem2bull One end of a long rod is inserted into a furnace while the other

projects into ambient temperature Under steady state the temperature of the rod is measured at two points 75mm apart and found to be 125oC and 885oC respectively while the ambient temperature is 20oC If the rod is 25mm in diameter and h is 2336 Wm2K

bull Find the thermal conductivity of the rod material

  • HEAT AND MASS TRANSFER
  • Thermodynamics Vs Heat transfer
  • MODE OF HEAT TRANSFER
  • MODE OF HEAT TRANSFER (2)
  • CONDUCTION- MECHANISM
  • CONCECTION-MECHANISM
  • RADIATION-MECHANISM
  • Conduction- Convection- Radiation
  • Slide 9
  • Slide 10
  • Fourierrsquos Law of Heat Conduction
  • Slide 12
  • Newtonrsquos Law of Cooling
  • Laws of Thermal Radiation
  • LAWS FOR HEAT TRANSFER
  • ELECTRICAL ampTHERMAL ANALOGY
  • Slide 17
  • Slide 18
  • PLANE WALL-UNIDIRECTIONAL
  • PLANE WALL -SERIES
  • PLANE WALL IN SERIES
  • PLANE WALL -PARALLEL
  • PLANE WALL-PARALLEL
  • PLANE WALL-COMBINATION
  • PLANE WALL-FILM RESISTANCE
  • PLANE WALL- THERMAL CONTACT RESISTANCE
  • PLANE WALL- THERMAL CONTACT RESISTANCE (2)
  • CYLINDER-SOLID
  • HEAT TRANSFER IN SOLID CYLINDER
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER-COMPOSITE
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER-MULTY LAYER
  • HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER
  • HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER (2)
  • HEAT TRANSFER IN SPHERE
  • HEAT TRANSFER IN SPHERE -MULTY LAYER
  • HEAT TRANSFER IN SPHERE -MULTY LAYER (2)
  • Slide 38
  • Slide 39
  • Slide 40
  • Slide 41
  • CRITICAL RADIUS
  • Plane wall- Insulation thickness -study
  • Problem5
  • Slide 45
  • Slide 46
  • Slide 47
  • GENERAL HEAT CONDUCTION EQUATION IN CYLINDRICAL COORDINATES
  • Slide 49
  • Slide 50
  • Slide 51
  • GENERAL HEAT CONDUCTION EQUATION IN SPHERICAL COORDINATES
  • Slide 53
  • Slide 54
  • Slide 55
  • General Heat Conduction Equation
  • Temperature profile
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Slide 67
Page 32: hmt chap 1

HEAT TRANSFER IN SPHERE -MULTY LAYER

HEAT TRANSFER IN SPHERE -MULTY LAYER

bull Problem 1 The wall in a furnace consists of 125 mm thick refractory bricks and

125 mm thick insulating fire bricks separated by an air gap A 12 mm thick plaster covers the outer wall The inner surface of wall is at 1100oC and the ambient temperature 25oC the heat transfer coefficient on the outer wall to the air is 17 Wm2K and the resistance to heat flow of the air gap is 016 KW the thermal conductivities of refractory brick insulating firebrick and plaster are 16 03 and 014 WmK respectively

Calculatebull (a) the rate of heat loss per unit area of the wall surfacebull (b) the interface temperatures throughout the wallbull (c) the temperature at the outside surface of the wall

bull Problem 2Two large aluminum plates (k=240 WmK) each 2 cm thick with 10 mm surface roughness are placed in contact under 105 Nm2 pressure in air The temperature at the outside surfaces are 390 oC and 406 oCCalculate

A) the heat flux B) the temperature drop due to the contact resistance C) the contact temperatures Thermal contact resistance with air as the interfacial fluid for

10 mm roughness is 27510-4 m2KW

bull Problem 3bull Steam at 350 oC flowing in a pipe (k=80 WmK) of 5 cm

inner diameter and 56 cm outer diameter is covered with 3 cm thick insulation of k=005 WmK Heat is lost to surroundings at 5 oC by natural convection and radiation the combined h being 20 Wm2K Taking the heat transfer coefficient inside the pipe as 60 Wm2K

bull Determinebull A) the rate of heat loss from the steam per unit length of

the pipebull B) the temperature drop across the pipe and the insulation

bull PROBLEM4bull A nuclear reactor has a sperical pressure vessel 750 mm inside

diameter 80 mm wall thickness The temperature at the inner surface is 500 oC If the tempertarue at the outer surface is 495 oC

bull Calculatea) The rate of heat loss through the metal wallb) If the rate of heat loss is to limited to 305Whow much inulation

thickness needs to be applied Thermal conductivity of steel wall=46 WmoC Thermal conductivity of insulation=004 WmoC Film coefficient of heat tranfer at inner and outer surface are 200

Wm2 oC and 15 Wm2 oC Ambient air temperature=28 oC

CRITICAL RADIUS

bull The radius upto which heat flow increases and after decreases At critical radius q becomes qmax

bull The addition of insulation always increases (Rth)condbull Rtotal= (Rth)cond + (Rth)conv

bull The addition of insulation may reduce the (Rth)conv due to increase in surface area

bull r1ltrc -heat transfer increases by adding insulation till r1=rc If

insulation thickness is further increased the rate of heat loss will decrease

r1 gt rc

- heat transfer decreases by adding insulation above the rc

Plane wall- Insulation

thickness -study

Cylinder amp Sphere- Insulation thickness -

study

Problem5bull A 3mm diameter and 5 mm long electric wire is tightly

wrapped with a 2 mm thick plastic cover whose K=015 wmoC Electrical measurements indicate that a current of 10 A passes through the wire and there is a voltage drop of 8 V along the wire If the insulated wire is exposed to a medium at Ta=30oC with a heat transfer coefficient of h=12 wm2oC Determine the temperature at the interface of the wire and the plastic cover in steady operation Also determine whether doubling the thickness of the plastic cover will increase or decrease this interface temperature

dx

TdzdyKQ xx )(

dxQx

QQ xxdxx ][

dxdx

TdzdyK

xQQdQ xdxxxx ])([)(

)1()]([)(

ddzdydx

x

TK

xdQ xx

GENERAL HEAT CONDUCTION EQUATION IN CARTESIAN COORDINATES

Consider an unsteady (transient) three dimensional heat conduction with internal heat generation in heterogeneous amp anisotropic PLANE WALL

Cartesian co-ordinates x y z

Elemental length dxdydz

Consider an elemental volume ABCDEFGH = (dxdydz)

Let Temperature distribution is a function of T=T(xyz)

Let

=internal heat generation per unit volume per unit time (W m3)q= rate of heat flow (W)Q= qdt=total heat flow (J)r=mass density of the material element(kgm3)Cp=specific heat of the material element (Jkg K)

gq

Plane (y - z) lsquoxrsquo direction

Quantity of heat leaving from the face EFGH in lsquoXrsquo direction

Quantity of heat flowing into the face ABCD in lsquoXrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoXrsquo direction

dy

TdzdxKQ yy )(

dyQy

QQ yydyy ][

dydy

TdzdxK

yQQdQ ydyyyy ])([)(

)2()]([)(

ddzdydx

y

TK

ydQ yy

Plane (x - z) lsquoyrsquo direction

Quantity of heat flowing into the face ABFE in lsquoYrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoYrsquo direction

dz

TdydxKQ zz )(

dzQz

QQ zzdzz ][

dzdz

TdydxK

zQQdQ zdzzzz ])([)(

)3()]([)(

ddzdydx

z

TK

zdQ zz

Plane (x - y) lsquozrsquo direction

Quantity of heat flowing into the face ADHE in lsquoZrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoZrsquo direction

Quantity of heat leaving from the face DCGH in lsquoYrsquo direction

Quantity of heat leaving from the face BCGF in lsquoZrsquo direction

)4()()(

ddzdydxqQgg

Total amount of internal heat generation in the element

Ein + Egen = Eout+Est (Ein- Eout)+ Egen=Est

T

Cddzdydxddzdydxqddzdydxz

TK

zy

TK

yx

TK

x pgzyx )()())](()()([

T

pC

gq

z

TzK

zy

Ty

Kyx

TxK

x][][][

stgzyx QQdQdQdQ

)(

)5()()(

dT

CdzdydxQ pst

Total heat energy stored in the element

(ie thermal energy of the element can be increased because of heat accumulation in the element due to heat flow amp internal heat generation)

)( volumem d

TmCQ pst )(

T

CqTK pg)(

KKKyKx z If material is homogeneous amp isotropic

T

pC

gq

z

T

zy

T

yx

T

xK ][][][

T

Kp

C

Kgq

z

T

y

T

x

T2

2

2

2

2

2

T

Kgq

z

T

y

T

x

T 12

2

2

2

2

2

Thermal diffusivity pC

K

T

K

qT g

12

according to thermodynamics first law

)( zr )( dzrddr

)( dzrddr

Cylindrical polar co-ordinates

Elemental length

Consider an elemental volume ABCDEFGH =

GENERAL HEAT CONDUCTION EQUATION IN CYLINDRICAL COORDINATES

gq

gq

Consider an unsteady (transient) three dimensional heat conduction with internal heat generation in heterogeneous amp anisotropic cylindrical geometries (eg rods and pipes)

)( zrTLetTemperature distribution is a function of T=

Let

=internal heat generation per unit volume per unit time (W m3)q= rate of heat flow (W)Q= qdt=total heat flow (J)r=mass density of the material element(kgm3)Cp=specific heat of the material element (Jkg K)

planerdirectionz

planezrdirection

planezdirectionr

)(

)(

)(

LetHeat flow in

drdr

TdzrdK

rdQ

QQdQ

rr

drrrr

])([

)1(][

r

Tr

rddrdzdKdQ rr

Heat accumulation in the element due to heat flow by conduction in lsquorrsquo direction

rddr

TdzdrK

rdQ

QQdQ d

])([

][

2

T

r

ddrdzrdKdQ

)2(][

2

2

2

T

r

ddrdzrdKdQ

Heat accumulation in the element due to heat flow by conduction in lsquoqrsquo direction

Plane (z- q ) lsquorrsquo direction

dr

TdzrdKQ rr

)(

drQr

QQ rrdrr ][

Quantity of heat leaving from the face EFGH in lsquorrsquo direction

Quantity of heat flowing into the face ABCD in lsquorrsquo direction

Plane (r-z) lsquoq lsquodirection

Quantity of heat leaving from the face AEGC in lsquoqrsquo direction

d

r

TdzdrKQ )(

rdQ

rQQ d ][

Quantity of heat flowing into the face BFHD in lsquoqrsquo direction

)3())((2

2

z

TdzdrrdKdQ zz

)5()()(

dT

CdzdrrdQ pst

)4()()(

ddzdrrdqQgg

Total amount of internal heat generation in the element

Total heat energy stored in the element

(ie thermal energy of the element can be increased because of heat accumulation in the element due to heat flow amp internal heat generation)

according to thermodynamics first law

Ein + Egen = Eout + Est (Ein ndash Eout) + Egen = Est

)( volumem

)()(z

T

zdzdrrdKdQ zz

Heat accumulation in the element due to heat flow by conduction in lsquozrsquo direction

dzdz

TdrrdK

zdQ

QQdQ

zz

dzzzz

])([

d

TmCQ pst )(

Plane (r-q) lsquozrsquo direction

dz

TdrrdKQ zz )(

dzQz

QQ zzdzz ][ Quantity of heat leaving from the face CDHG in lsquozrsquo direction

Quantity of heat flowing into the face ABFE in lsquozrsquo direction

stgzr QQdQdQdQ

)(

r

r ][)(rT

rr

ddrdzdKr

][)(

2

2

2

T

r

ddrdzrdK ))((2

2

z

TddzdrrdK z

ddzdrrdq g )(

dT

Cdzdrrd p )(

+ + +

=

T

CqTK pg)(KKKK zr

[gq

)(1

r

Tr

rrK r

)(1

2

2

2 T

rK )(

2

2

z

TK z

)( ddzdrrd )(

ddzdrrd

TC p

+ + + ] =

If material is homogeneous amp isotropic

)(1

r

Tr

rr

)(1

2

2

2 T

r)(

2

2

z

T

K

q g

T

K

C p

)(1

r

Tr

rr

)(1

2

2

2 T

r )(2

2

z

T

K

q g

T

1

+ + + =

)(1

r

Tr

rrK r

)(1

2

2

2 T

rK )(

2

2

z

TK z

gq

T

C p + + + =

+ + + = Thermal diffusivity

T

K

qT g

12

pC

K

(Ein ndash Eout) + Egen = Est

)( rT

)( r

)sin( drrddr)sin( drrddr

planerdirection

planerdirection

planedirectionr

)(

)(

)(

Spherical polar co-ordinates

Elemental length

Consider an elemental volume ABCDEFGH =

GENERAL HEAT CONDUCTION EQUATION IN SPHERICAL COORDINATES

Let

=internal heat generation per unit volume per unit time (W m3)q= rate of heat flow (W)Q= qdt=total heat flow (J)r=mass density of the material element(kgm3)Cp=specific heat of the material element (Jkg K)

gq

Consider an unsteady (transient) three dimensional heat conduction with internal heat generation in heterogeneous amp anisotropic spherical geometries

LetTemperature distribution is a function of T=

X

Y

Z

drT

drrdKQ rr )sin(

drQr

QQ rrdrr ][

drdr

TdrrdK

rdQ

QQdQ

rr

drrrr

])sin([

)1(]([sin 2

rT

rr

ddrddKdQ rr

Plane ( -q f) lsquorrsquo directionQuantity of heat flowing into the face ABCD in lsquorrsquo direction

Quantity of heat leaving from the face EFGH in lsquorrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquorrsquo direction

drT

drdrKQ )sin(

rdQ

rQQ d ][

rddr

TdrdrK

rdQ

QQdQ d

])sin([

])[(sin

2

T

r

drdrdrKdQ

)2(])(sin[

2

T

r

drdrddrKdQ

Quantity of heat leaving from the face BFHD in lsquoqrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoqrsquo direction

X

Y

Z

Plane (r- f ) lsquoq lsquodirection

Quantity of heat flowing into the face AEGC in lsquoqrsquo direction

dr

TdrrdKQ

sin)(

drQ

rQQ d sin][

sin

drdr

TdrrdK

rdQ

QQdQ d

sin]sin

)([sin

][sinsin

)sin(

T

rr

ddrdrrdKdQ

)3()(sin

)sin(2

2

22

2

T

r

ddddrrKdQ

Plane (r- ) q lsquofrsquo directionQuantity of heat flowing into the face ABFE in lsquofrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquozrsquo direction

)4()sin()(

ddrrddrqQgg

)5()sin()(

dT

CdrrddrQ pst

Total amount of internal heat generation in the element

Total heat energy stored in the element

(Ein ndash Eout) + Egen = Est

d

TmCQ pst )(

)( volumem

Quantity of heat leaving from the face CDHG in lsquofrsquo direction

according to thermodynamics first law

Ein + Egen = Eout + Est

X

Y

Z

stgr QQdQdQdQ

)(

]([sin 2

r

Tr

rddrddKr

])(sin[

2

T

rdrdrddrK

)(sin

)sin(2

2

22

2

T

r

ddddrrK

ddrrddrq g )sin(

dT

Cdrrddr p )sin(

2

2

r

r

sin

sin

]([ 22 r

Tr

rr

Kr

])(sin[sin2

TrK

)(sin 2

2

22

T

r

Kgq

T

Cp

]([1

[ 22 r

Tr

rrK

])(sin[sin1

2

T

r)](

sin

12

2

22 T

r gq

T

Cp

]([1

[ 22 r

Tr

rr

])(sin[sin1

2

T

r)](

sin

12

2

22 T

rK

q g

T

K

Cp

]([1

[ 22 r

Tr

rr

])(sin[sin1

2

T

r)](

sin

12

2

22 T

rK

q g

T1

KKKKr If material is homogeneous amp isotropic

TCqTK pg)(

T

K

qT g

12 pC

K

Thermal diffusivity

(Ein ndash Eout) + Egen = Est

General Heat Conduction Equation

PLANE WALL CYLINDER SPHERE

Temperature profile

Plane wall SphereCylinder

bull A copper conductor (k = 380 WmoC resistivity ρ = 2x10-8 Ω m) having inner and outer radii 10 cm and 225 cm respectively is carrying a current density of 4800 amperes cm2 the conductor is internally cooled and a constant temperature of 65oC is maintained at the inner surface and there is no heat transfer through the insulation surrounding the conductor

Determinebull (i) The maximum temperature of the conductor and the radius at

which it occursbull (ii) The internal heat transfer rate

bull A chemical reaction takes place in a packed bed (k=06 WmoC) between two coaxial cylinders with radii 15 mm and 45 mm the inner surface is at 580oC and it is insulated Assuming the reaction rate of 055 MWm3 in the reactor volume find the temperature at the outer surface of the reactor

bull A current of 300 amperes passes through a stainless steel wire of 25 mm diameter and k=20 WmoC The resistivity of the wire is 70x10-8 ohm-m and length of the wire is 2 m if the wire is submerged in a fluid maintained at 50oC and h at the wire surface is 4000 Wm2oC calculate the steady state temperature at the centre and at the surface of the wire

bull An approximately spherical shaped orange ( k=023 WmoC) 90 mm diameter undergoes ripping process and generates 5100 Wm3 of energy If external surface the orange is at 8oC determine

bull (i) temperature at the centre of the orange and

bull (ii) heat flow from the outer surface of the orange

bull FINS External surfaces to used to cooling area and hence convection

heat transfer can be improvedbull APPLICATIONS IC engine cylinder ( air cooling) Heat exchangers automobile

radiators computer chip refrigerators electric motors generators and transformers etc

bull ASSUMPTIONS MADE IN ANALYSIS OF HEAT TRANFER THROUGH FIN

1 Fin material is homogenous and isotropic constthermal conductivity

2 Temperature at any cs of the fin is uniform3 lsquohrsquo is uniform over entire surface4 No internal heat generation5 Thermal contact resistance is negligible6 Radiation heat transfer is negligible7 Heat conduction is steady state and one dimensional

bull Problem 1

bull A stainless steel fin (k=20 WmK) having a diameter of 20 mm and a length of 01 m is attached to a wall at 300oC The ambient temperature is 50oC and the heat transfer coefficient is 10 Wm2K The fin tip is insulated Determine

bull (A) The rate of heat dissipation from the finbull (B) The temperature at the fin tipbull (C) The rate of heat transfer from the wall area covered by the fin if

the fin was not usedbull (D) The heat transfer rate from the same fin geometry if the

stainless steel fin is replaced by a fictitious fin with infinite thermal conductivity

bull Problem2bull One end of a long rod is inserted into a furnace while the other

projects into ambient temperature Under steady state the temperature of the rod is measured at two points 75mm apart and found to be 125oC and 885oC respectively while the ambient temperature is 20oC If the rod is 25mm in diameter and h is 2336 Wm2K

bull Find the thermal conductivity of the rod material

  • HEAT AND MASS TRANSFER
  • Thermodynamics Vs Heat transfer
  • MODE OF HEAT TRANSFER
  • MODE OF HEAT TRANSFER (2)
  • CONDUCTION- MECHANISM
  • CONCECTION-MECHANISM
  • RADIATION-MECHANISM
  • Conduction- Convection- Radiation
  • Slide 9
  • Slide 10
  • Fourierrsquos Law of Heat Conduction
  • Slide 12
  • Newtonrsquos Law of Cooling
  • Laws of Thermal Radiation
  • LAWS FOR HEAT TRANSFER
  • ELECTRICAL ampTHERMAL ANALOGY
  • Slide 17
  • Slide 18
  • PLANE WALL-UNIDIRECTIONAL
  • PLANE WALL -SERIES
  • PLANE WALL IN SERIES
  • PLANE WALL -PARALLEL
  • PLANE WALL-PARALLEL
  • PLANE WALL-COMBINATION
  • PLANE WALL-FILM RESISTANCE
  • PLANE WALL- THERMAL CONTACT RESISTANCE
  • PLANE WALL- THERMAL CONTACT RESISTANCE (2)
  • CYLINDER-SOLID
  • HEAT TRANSFER IN SOLID CYLINDER
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER-COMPOSITE
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER-MULTY LAYER
  • HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER
  • HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER (2)
  • HEAT TRANSFER IN SPHERE
  • HEAT TRANSFER IN SPHERE -MULTY LAYER
  • HEAT TRANSFER IN SPHERE -MULTY LAYER (2)
  • Slide 38
  • Slide 39
  • Slide 40
  • Slide 41
  • CRITICAL RADIUS
  • Plane wall- Insulation thickness -study
  • Problem5
  • Slide 45
  • Slide 46
  • Slide 47
  • GENERAL HEAT CONDUCTION EQUATION IN CYLINDRICAL COORDINATES
  • Slide 49
  • Slide 50
  • Slide 51
  • GENERAL HEAT CONDUCTION EQUATION IN SPHERICAL COORDINATES
  • Slide 53
  • Slide 54
  • Slide 55
  • General Heat Conduction Equation
  • Temperature profile
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Slide 67
Page 33: hmt chap 1

HEAT TRANSFER IN SPHERE -MULTY LAYER

bull Problem 1 The wall in a furnace consists of 125 mm thick refractory bricks and

125 mm thick insulating fire bricks separated by an air gap A 12 mm thick plaster covers the outer wall The inner surface of wall is at 1100oC and the ambient temperature 25oC the heat transfer coefficient on the outer wall to the air is 17 Wm2K and the resistance to heat flow of the air gap is 016 KW the thermal conductivities of refractory brick insulating firebrick and plaster are 16 03 and 014 WmK respectively

Calculatebull (a) the rate of heat loss per unit area of the wall surfacebull (b) the interface temperatures throughout the wallbull (c) the temperature at the outside surface of the wall

bull Problem 2Two large aluminum plates (k=240 WmK) each 2 cm thick with 10 mm surface roughness are placed in contact under 105 Nm2 pressure in air The temperature at the outside surfaces are 390 oC and 406 oCCalculate

A) the heat flux B) the temperature drop due to the contact resistance C) the contact temperatures Thermal contact resistance with air as the interfacial fluid for

10 mm roughness is 27510-4 m2KW

bull Problem 3bull Steam at 350 oC flowing in a pipe (k=80 WmK) of 5 cm

inner diameter and 56 cm outer diameter is covered with 3 cm thick insulation of k=005 WmK Heat is lost to surroundings at 5 oC by natural convection and radiation the combined h being 20 Wm2K Taking the heat transfer coefficient inside the pipe as 60 Wm2K

bull Determinebull A) the rate of heat loss from the steam per unit length of

the pipebull B) the temperature drop across the pipe and the insulation

bull PROBLEM4bull A nuclear reactor has a sperical pressure vessel 750 mm inside

diameter 80 mm wall thickness The temperature at the inner surface is 500 oC If the tempertarue at the outer surface is 495 oC

bull Calculatea) The rate of heat loss through the metal wallb) If the rate of heat loss is to limited to 305Whow much inulation

thickness needs to be applied Thermal conductivity of steel wall=46 WmoC Thermal conductivity of insulation=004 WmoC Film coefficient of heat tranfer at inner and outer surface are 200

Wm2 oC and 15 Wm2 oC Ambient air temperature=28 oC

CRITICAL RADIUS

bull The radius upto which heat flow increases and after decreases At critical radius q becomes qmax

bull The addition of insulation always increases (Rth)condbull Rtotal= (Rth)cond + (Rth)conv

bull The addition of insulation may reduce the (Rth)conv due to increase in surface area

bull r1ltrc -heat transfer increases by adding insulation till r1=rc If

insulation thickness is further increased the rate of heat loss will decrease

r1 gt rc

- heat transfer decreases by adding insulation above the rc

Plane wall- Insulation

thickness -study

Cylinder amp Sphere- Insulation thickness -

study

Problem5bull A 3mm diameter and 5 mm long electric wire is tightly

wrapped with a 2 mm thick plastic cover whose K=015 wmoC Electrical measurements indicate that a current of 10 A passes through the wire and there is a voltage drop of 8 V along the wire If the insulated wire is exposed to a medium at Ta=30oC with a heat transfer coefficient of h=12 wm2oC Determine the temperature at the interface of the wire and the plastic cover in steady operation Also determine whether doubling the thickness of the plastic cover will increase or decrease this interface temperature

dx

TdzdyKQ xx )(

dxQx

QQ xxdxx ][

dxdx

TdzdyK

xQQdQ xdxxxx ])([)(

)1()]([)(

ddzdydx

x

TK

xdQ xx

GENERAL HEAT CONDUCTION EQUATION IN CARTESIAN COORDINATES

Consider an unsteady (transient) three dimensional heat conduction with internal heat generation in heterogeneous amp anisotropic PLANE WALL

Cartesian co-ordinates x y z

Elemental length dxdydz

Consider an elemental volume ABCDEFGH = (dxdydz)

Let Temperature distribution is a function of T=T(xyz)

Let

=internal heat generation per unit volume per unit time (W m3)q= rate of heat flow (W)Q= qdt=total heat flow (J)r=mass density of the material element(kgm3)Cp=specific heat of the material element (Jkg K)

gq

Plane (y - z) lsquoxrsquo direction

Quantity of heat leaving from the face EFGH in lsquoXrsquo direction

Quantity of heat flowing into the face ABCD in lsquoXrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoXrsquo direction

dy

TdzdxKQ yy )(

dyQy

QQ yydyy ][

dydy

TdzdxK

yQQdQ ydyyyy ])([)(

)2()]([)(

ddzdydx

y

TK

ydQ yy

Plane (x - z) lsquoyrsquo direction

Quantity of heat flowing into the face ABFE in lsquoYrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoYrsquo direction

dz

TdydxKQ zz )(

dzQz

QQ zzdzz ][

dzdz

TdydxK

zQQdQ zdzzzz ])([)(

)3()]([)(

ddzdydx

z

TK

zdQ zz

Plane (x - y) lsquozrsquo direction

Quantity of heat flowing into the face ADHE in lsquoZrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoZrsquo direction

Quantity of heat leaving from the face DCGH in lsquoYrsquo direction

Quantity of heat leaving from the face BCGF in lsquoZrsquo direction

)4()()(

ddzdydxqQgg

Total amount of internal heat generation in the element

Ein + Egen = Eout+Est (Ein- Eout)+ Egen=Est

T

Cddzdydxddzdydxqddzdydxz

TK

zy

TK

yx

TK

x pgzyx )()())](()()([

T

pC

gq

z

TzK

zy

Ty

Kyx

TxK

x][][][

stgzyx QQdQdQdQ

)(

)5()()(

dT

CdzdydxQ pst

Total heat energy stored in the element

(ie thermal energy of the element can be increased because of heat accumulation in the element due to heat flow amp internal heat generation)

)( volumem d

TmCQ pst )(

T

CqTK pg)(

KKKyKx z If material is homogeneous amp isotropic

T

pC

gq

z

T

zy

T

yx

T

xK ][][][

T

Kp

C

Kgq

z

T

y

T

x

T2

2

2

2

2

2

T

Kgq

z

T

y

T

x

T 12

2

2

2

2

2

Thermal diffusivity pC

K

T

K

qT g

12

according to thermodynamics first law

)( zr )( dzrddr

)( dzrddr

Cylindrical polar co-ordinates

Elemental length

Consider an elemental volume ABCDEFGH =

GENERAL HEAT CONDUCTION EQUATION IN CYLINDRICAL COORDINATES

gq

gq

Consider an unsteady (transient) three dimensional heat conduction with internal heat generation in heterogeneous amp anisotropic cylindrical geometries (eg rods and pipes)

)( zrTLetTemperature distribution is a function of T=

Let

=internal heat generation per unit volume per unit time (W m3)q= rate of heat flow (W)Q= qdt=total heat flow (J)r=mass density of the material element(kgm3)Cp=specific heat of the material element (Jkg K)

planerdirectionz

planezrdirection

planezdirectionr

)(

)(

)(

LetHeat flow in

drdr

TdzrdK

rdQ

QQdQ

rr

drrrr

])([

)1(][

r

Tr

rddrdzdKdQ rr

Heat accumulation in the element due to heat flow by conduction in lsquorrsquo direction

rddr

TdzdrK

rdQ

QQdQ d

])([

][

2

T

r

ddrdzrdKdQ

)2(][

2

2

2

T

r

ddrdzrdKdQ

Heat accumulation in the element due to heat flow by conduction in lsquoqrsquo direction

Plane (z- q ) lsquorrsquo direction

dr

TdzrdKQ rr

)(

drQr

QQ rrdrr ][

Quantity of heat leaving from the face EFGH in lsquorrsquo direction

Quantity of heat flowing into the face ABCD in lsquorrsquo direction

Plane (r-z) lsquoq lsquodirection

Quantity of heat leaving from the face AEGC in lsquoqrsquo direction

d

r

TdzdrKQ )(

rdQ

rQQ d ][

Quantity of heat flowing into the face BFHD in lsquoqrsquo direction

)3())((2

2

z

TdzdrrdKdQ zz

)5()()(

dT

CdzdrrdQ pst

)4()()(

ddzdrrdqQgg

Total amount of internal heat generation in the element

Total heat energy stored in the element

(ie thermal energy of the element can be increased because of heat accumulation in the element due to heat flow amp internal heat generation)

according to thermodynamics first law

Ein + Egen = Eout + Est (Ein ndash Eout) + Egen = Est

)( volumem

)()(z

T

zdzdrrdKdQ zz

Heat accumulation in the element due to heat flow by conduction in lsquozrsquo direction

dzdz

TdrrdK

zdQ

QQdQ

zz

dzzzz

])([

d

TmCQ pst )(

Plane (r-q) lsquozrsquo direction

dz

TdrrdKQ zz )(

dzQz

QQ zzdzz ][ Quantity of heat leaving from the face CDHG in lsquozrsquo direction

Quantity of heat flowing into the face ABFE in lsquozrsquo direction

stgzr QQdQdQdQ

)(

r

r ][)(rT

rr

ddrdzdKr

][)(

2

2

2

T

r

ddrdzrdK ))((2

2

z

TddzdrrdK z

ddzdrrdq g )(

dT

Cdzdrrd p )(

+ + +

=

T

CqTK pg)(KKKK zr

[gq

)(1

r

Tr

rrK r

)(1

2

2

2 T

rK )(

2

2

z

TK z

)( ddzdrrd )(

ddzdrrd

TC p

+ + + ] =

If material is homogeneous amp isotropic

)(1

r

Tr

rr

)(1

2

2

2 T

r)(

2

2

z

T

K

q g

T

K

C p

)(1

r

Tr

rr

)(1

2

2

2 T

r )(2

2

z

T

K

q g

T

1

+ + + =

)(1

r

Tr

rrK r

)(1

2

2

2 T

rK )(

2

2

z

TK z

gq

T

C p + + + =

+ + + = Thermal diffusivity

T

K

qT g

12

pC

K

(Ein ndash Eout) + Egen = Est

)( rT

)( r

)sin( drrddr)sin( drrddr

planerdirection

planerdirection

planedirectionr

)(

)(

)(

Spherical polar co-ordinates

Elemental length

Consider an elemental volume ABCDEFGH =

GENERAL HEAT CONDUCTION EQUATION IN SPHERICAL COORDINATES

Let

=internal heat generation per unit volume per unit time (W m3)q= rate of heat flow (W)Q= qdt=total heat flow (J)r=mass density of the material element(kgm3)Cp=specific heat of the material element (Jkg K)

gq

Consider an unsteady (transient) three dimensional heat conduction with internal heat generation in heterogeneous amp anisotropic spherical geometries

LetTemperature distribution is a function of T=

X

Y

Z

drT

drrdKQ rr )sin(

drQr

QQ rrdrr ][

drdr

TdrrdK

rdQ

QQdQ

rr

drrrr

])sin([

)1(]([sin 2

rT

rr

ddrddKdQ rr

Plane ( -q f) lsquorrsquo directionQuantity of heat flowing into the face ABCD in lsquorrsquo direction

Quantity of heat leaving from the face EFGH in lsquorrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquorrsquo direction

drT

drdrKQ )sin(

rdQ

rQQ d ][

rddr

TdrdrK

rdQ

QQdQ d

])sin([

])[(sin

2

T

r

drdrdrKdQ

)2(])(sin[

2

T

r

drdrddrKdQ

Quantity of heat leaving from the face BFHD in lsquoqrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoqrsquo direction

X

Y

Z

Plane (r- f ) lsquoq lsquodirection

Quantity of heat flowing into the face AEGC in lsquoqrsquo direction

dr

TdrrdKQ

sin)(

drQ

rQQ d sin][

sin

drdr

TdrrdK

rdQ

QQdQ d

sin]sin

)([sin

][sinsin

)sin(

T

rr

ddrdrrdKdQ

)3()(sin

)sin(2

2

22

2

T

r

ddddrrKdQ

Plane (r- ) q lsquofrsquo directionQuantity of heat flowing into the face ABFE in lsquofrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquozrsquo direction

)4()sin()(

ddrrddrqQgg

)5()sin()(

dT

CdrrddrQ pst

Total amount of internal heat generation in the element

Total heat energy stored in the element

(Ein ndash Eout) + Egen = Est

d

TmCQ pst )(

)( volumem

Quantity of heat leaving from the face CDHG in lsquofrsquo direction

according to thermodynamics first law

Ein + Egen = Eout + Est

X

Y

Z

stgr QQdQdQdQ

)(

]([sin 2

r

Tr

rddrddKr

])(sin[

2

T

rdrdrddrK

)(sin

)sin(2

2

22

2

T

r

ddddrrK

ddrrddrq g )sin(

dT

Cdrrddr p )sin(

2

2

r

r

sin

sin

]([ 22 r

Tr

rr

Kr

])(sin[sin2

TrK

)(sin 2

2

22

T

r

Kgq

T

Cp

]([1

[ 22 r

Tr

rrK

])(sin[sin1

2

T

r)](

sin

12

2

22 T

r gq

T

Cp

]([1

[ 22 r

Tr

rr

])(sin[sin1

2

T

r)](

sin

12

2

22 T

rK

q g

T

K

Cp

]([1

[ 22 r

Tr

rr

])(sin[sin1

2

T

r)](

sin

12

2

22 T

rK

q g

T1

KKKKr If material is homogeneous amp isotropic

TCqTK pg)(

T

K

qT g

12 pC

K

Thermal diffusivity

(Ein ndash Eout) + Egen = Est

General Heat Conduction Equation

PLANE WALL CYLINDER SPHERE

Temperature profile

Plane wall SphereCylinder

bull A copper conductor (k = 380 WmoC resistivity ρ = 2x10-8 Ω m) having inner and outer radii 10 cm and 225 cm respectively is carrying a current density of 4800 amperes cm2 the conductor is internally cooled and a constant temperature of 65oC is maintained at the inner surface and there is no heat transfer through the insulation surrounding the conductor

Determinebull (i) The maximum temperature of the conductor and the radius at

which it occursbull (ii) The internal heat transfer rate

bull A chemical reaction takes place in a packed bed (k=06 WmoC) between two coaxial cylinders with radii 15 mm and 45 mm the inner surface is at 580oC and it is insulated Assuming the reaction rate of 055 MWm3 in the reactor volume find the temperature at the outer surface of the reactor

bull A current of 300 amperes passes through a stainless steel wire of 25 mm diameter and k=20 WmoC The resistivity of the wire is 70x10-8 ohm-m and length of the wire is 2 m if the wire is submerged in a fluid maintained at 50oC and h at the wire surface is 4000 Wm2oC calculate the steady state temperature at the centre and at the surface of the wire

bull An approximately spherical shaped orange ( k=023 WmoC) 90 mm diameter undergoes ripping process and generates 5100 Wm3 of energy If external surface the orange is at 8oC determine

bull (i) temperature at the centre of the orange and

bull (ii) heat flow from the outer surface of the orange

bull FINS External surfaces to used to cooling area and hence convection

heat transfer can be improvedbull APPLICATIONS IC engine cylinder ( air cooling) Heat exchangers automobile

radiators computer chip refrigerators electric motors generators and transformers etc

bull ASSUMPTIONS MADE IN ANALYSIS OF HEAT TRANFER THROUGH FIN

1 Fin material is homogenous and isotropic constthermal conductivity

2 Temperature at any cs of the fin is uniform3 lsquohrsquo is uniform over entire surface4 No internal heat generation5 Thermal contact resistance is negligible6 Radiation heat transfer is negligible7 Heat conduction is steady state and one dimensional

bull Problem 1

bull A stainless steel fin (k=20 WmK) having a diameter of 20 mm and a length of 01 m is attached to a wall at 300oC The ambient temperature is 50oC and the heat transfer coefficient is 10 Wm2K The fin tip is insulated Determine

bull (A) The rate of heat dissipation from the finbull (B) The temperature at the fin tipbull (C) The rate of heat transfer from the wall area covered by the fin if

the fin was not usedbull (D) The heat transfer rate from the same fin geometry if the

stainless steel fin is replaced by a fictitious fin with infinite thermal conductivity

bull Problem2bull One end of a long rod is inserted into a furnace while the other

projects into ambient temperature Under steady state the temperature of the rod is measured at two points 75mm apart and found to be 125oC and 885oC respectively while the ambient temperature is 20oC If the rod is 25mm in diameter and h is 2336 Wm2K

bull Find the thermal conductivity of the rod material

  • HEAT AND MASS TRANSFER
  • Thermodynamics Vs Heat transfer
  • MODE OF HEAT TRANSFER
  • MODE OF HEAT TRANSFER (2)
  • CONDUCTION- MECHANISM
  • CONCECTION-MECHANISM
  • RADIATION-MECHANISM
  • Conduction- Convection- Radiation
  • Slide 9
  • Slide 10
  • Fourierrsquos Law of Heat Conduction
  • Slide 12
  • Newtonrsquos Law of Cooling
  • Laws of Thermal Radiation
  • LAWS FOR HEAT TRANSFER
  • ELECTRICAL ampTHERMAL ANALOGY
  • Slide 17
  • Slide 18
  • PLANE WALL-UNIDIRECTIONAL
  • PLANE WALL -SERIES
  • PLANE WALL IN SERIES
  • PLANE WALL -PARALLEL
  • PLANE WALL-PARALLEL
  • PLANE WALL-COMBINATION
  • PLANE WALL-FILM RESISTANCE
  • PLANE WALL- THERMAL CONTACT RESISTANCE
  • PLANE WALL- THERMAL CONTACT RESISTANCE (2)
  • CYLINDER-SOLID
  • HEAT TRANSFER IN SOLID CYLINDER
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER-COMPOSITE
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER-MULTY LAYER
  • HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER
  • HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER (2)
  • HEAT TRANSFER IN SPHERE
  • HEAT TRANSFER IN SPHERE -MULTY LAYER
  • HEAT TRANSFER IN SPHERE -MULTY LAYER (2)
  • Slide 38
  • Slide 39
  • Slide 40
  • Slide 41
  • CRITICAL RADIUS
  • Plane wall- Insulation thickness -study
  • Problem5
  • Slide 45
  • Slide 46
  • Slide 47
  • GENERAL HEAT CONDUCTION EQUATION IN CYLINDRICAL COORDINATES
  • Slide 49
  • Slide 50
  • Slide 51
  • GENERAL HEAT CONDUCTION EQUATION IN SPHERICAL COORDINATES
  • Slide 53
  • Slide 54
  • Slide 55
  • General Heat Conduction Equation
  • Temperature profile
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Slide 67
Page 34: hmt chap 1

bull Problem 1 The wall in a furnace consists of 125 mm thick refractory bricks and

125 mm thick insulating fire bricks separated by an air gap A 12 mm thick plaster covers the outer wall The inner surface of wall is at 1100oC and the ambient temperature 25oC the heat transfer coefficient on the outer wall to the air is 17 Wm2K and the resistance to heat flow of the air gap is 016 KW the thermal conductivities of refractory brick insulating firebrick and plaster are 16 03 and 014 WmK respectively

Calculatebull (a) the rate of heat loss per unit area of the wall surfacebull (b) the interface temperatures throughout the wallbull (c) the temperature at the outside surface of the wall

bull Problem 2Two large aluminum plates (k=240 WmK) each 2 cm thick with 10 mm surface roughness are placed in contact under 105 Nm2 pressure in air The temperature at the outside surfaces are 390 oC and 406 oCCalculate

A) the heat flux B) the temperature drop due to the contact resistance C) the contact temperatures Thermal contact resistance with air as the interfacial fluid for

10 mm roughness is 27510-4 m2KW

bull Problem 3bull Steam at 350 oC flowing in a pipe (k=80 WmK) of 5 cm

inner diameter and 56 cm outer diameter is covered with 3 cm thick insulation of k=005 WmK Heat is lost to surroundings at 5 oC by natural convection and radiation the combined h being 20 Wm2K Taking the heat transfer coefficient inside the pipe as 60 Wm2K

bull Determinebull A) the rate of heat loss from the steam per unit length of

the pipebull B) the temperature drop across the pipe and the insulation

bull PROBLEM4bull A nuclear reactor has a sperical pressure vessel 750 mm inside

diameter 80 mm wall thickness The temperature at the inner surface is 500 oC If the tempertarue at the outer surface is 495 oC

bull Calculatea) The rate of heat loss through the metal wallb) If the rate of heat loss is to limited to 305Whow much inulation

thickness needs to be applied Thermal conductivity of steel wall=46 WmoC Thermal conductivity of insulation=004 WmoC Film coefficient of heat tranfer at inner and outer surface are 200

Wm2 oC and 15 Wm2 oC Ambient air temperature=28 oC

CRITICAL RADIUS

bull The radius upto which heat flow increases and after decreases At critical radius q becomes qmax

bull The addition of insulation always increases (Rth)condbull Rtotal= (Rth)cond + (Rth)conv

bull The addition of insulation may reduce the (Rth)conv due to increase in surface area

bull r1ltrc -heat transfer increases by adding insulation till r1=rc If

insulation thickness is further increased the rate of heat loss will decrease

r1 gt rc

- heat transfer decreases by adding insulation above the rc

Plane wall- Insulation

thickness -study

Cylinder amp Sphere- Insulation thickness -

study

Problem5bull A 3mm diameter and 5 mm long electric wire is tightly

wrapped with a 2 mm thick plastic cover whose K=015 wmoC Electrical measurements indicate that a current of 10 A passes through the wire and there is a voltage drop of 8 V along the wire If the insulated wire is exposed to a medium at Ta=30oC with a heat transfer coefficient of h=12 wm2oC Determine the temperature at the interface of the wire and the plastic cover in steady operation Also determine whether doubling the thickness of the plastic cover will increase or decrease this interface temperature

dx

TdzdyKQ xx )(

dxQx

QQ xxdxx ][

dxdx

TdzdyK

xQQdQ xdxxxx ])([)(

)1()]([)(

ddzdydx

x

TK

xdQ xx

GENERAL HEAT CONDUCTION EQUATION IN CARTESIAN COORDINATES

Consider an unsteady (transient) three dimensional heat conduction with internal heat generation in heterogeneous amp anisotropic PLANE WALL

Cartesian co-ordinates x y z

Elemental length dxdydz

Consider an elemental volume ABCDEFGH = (dxdydz)

Let Temperature distribution is a function of T=T(xyz)

Let

=internal heat generation per unit volume per unit time (W m3)q= rate of heat flow (W)Q= qdt=total heat flow (J)r=mass density of the material element(kgm3)Cp=specific heat of the material element (Jkg K)

gq

Plane (y - z) lsquoxrsquo direction

Quantity of heat leaving from the face EFGH in lsquoXrsquo direction

Quantity of heat flowing into the face ABCD in lsquoXrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoXrsquo direction

dy

TdzdxKQ yy )(

dyQy

QQ yydyy ][

dydy

TdzdxK

yQQdQ ydyyyy ])([)(

)2()]([)(

ddzdydx

y

TK

ydQ yy

Plane (x - z) lsquoyrsquo direction

Quantity of heat flowing into the face ABFE in lsquoYrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoYrsquo direction

dz

TdydxKQ zz )(

dzQz

QQ zzdzz ][

dzdz

TdydxK

zQQdQ zdzzzz ])([)(

)3()]([)(

ddzdydx

z

TK

zdQ zz

Plane (x - y) lsquozrsquo direction

Quantity of heat flowing into the face ADHE in lsquoZrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoZrsquo direction

Quantity of heat leaving from the face DCGH in lsquoYrsquo direction

Quantity of heat leaving from the face BCGF in lsquoZrsquo direction

)4()()(

ddzdydxqQgg

Total amount of internal heat generation in the element

Ein + Egen = Eout+Est (Ein- Eout)+ Egen=Est

T

Cddzdydxddzdydxqddzdydxz

TK

zy

TK

yx

TK

x pgzyx )()())](()()([

T

pC

gq

z

TzK

zy

Ty

Kyx

TxK

x][][][

stgzyx QQdQdQdQ

)(

)5()()(

dT

CdzdydxQ pst

Total heat energy stored in the element

(ie thermal energy of the element can be increased because of heat accumulation in the element due to heat flow amp internal heat generation)

)( volumem d

TmCQ pst )(

T

CqTK pg)(

KKKyKx z If material is homogeneous amp isotropic

T

pC

gq

z

T

zy

T

yx

T

xK ][][][

T

Kp

C

Kgq

z

T

y

T

x

T2

2

2

2

2

2

T

Kgq

z

T

y

T

x

T 12

2

2

2

2

2

Thermal diffusivity pC

K

T

K

qT g

12

according to thermodynamics first law

)( zr )( dzrddr

)( dzrddr

Cylindrical polar co-ordinates

Elemental length

Consider an elemental volume ABCDEFGH =

GENERAL HEAT CONDUCTION EQUATION IN CYLINDRICAL COORDINATES

gq

gq

Consider an unsteady (transient) three dimensional heat conduction with internal heat generation in heterogeneous amp anisotropic cylindrical geometries (eg rods and pipes)

)( zrTLetTemperature distribution is a function of T=

Let

=internal heat generation per unit volume per unit time (W m3)q= rate of heat flow (W)Q= qdt=total heat flow (J)r=mass density of the material element(kgm3)Cp=specific heat of the material element (Jkg K)

planerdirectionz

planezrdirection

planezdirectionr

)(

)(

)(

LetHeat flow in

drdr

TdzrdK

rdQ

QQdQ

rr

drrrr

])([

)1(][

r

Tr

rddrdzdKdQ rr

Heat accumulation in the element due to heat flow by conduction in lsquorrsquo direction

rddr

TdzdrK

rdQ

QQdQ d

])([

][

2

T

r

ddrdzrdKdQ

)2(][

2

2

2

T

r

ddrdzrdKdQ

Heat accumulation in the element due to heat flow by conduction in lsquoqrsquo direction

Plane (z- q ) lsquorrsquo direction

dr

TdzrdKQ rr

)(

drQr

QQ rrdrr ][

Quantity of heat leaving from the face EFGH in lsquorrsquo direction

Quantity of heat flowing into the face ABCD in lsquorrsquo direction

Plane (r-z) lsquoq lsquodirection

Quantity of heat leaving from the face AEGC in lsquoqrsquo direction

d

r

TdzdrKQ )(

rdQ

rQQ d ][

Quantity of heat flowing into the face BFHD in lsquoqrsquo direction

)3())((2

2

z

TdzdrrdKdQ zz

)5()()(

dT

CdzdrrdQ pst

)4()()(

ddzdrrdqQgg

Total amount of internal heat generation in the element

Total heat energy stored in the element

(ie thermal energy of the element can be increased because of heat accumulation in the element due to heat flow amp internal heat generation)

according to thermodynamics first law

Ein + Egen = Eout + Est (Ein ndash Eout) + Egen = Est

)( volumem

)()(z

T

zdzdrrdKdQ zz

Heat accumulation in the element due to heat flow by conduction in lsquozrsquo direction

dzdz

TdrrdK

zdQ

QQdQ

zz

dzzzz

])([

d

TmCQ pst )(

Plane (r-q) lsquozrsquo direction

dz

TdrrdKQ zz )(

dzQz

QQ zzdzz ][ Quantity of heat leaving from the face CDHG in lsquozrsquo direction

Quantity of heat flowing into the face ABFE in lsquozrsquo direction

stgzr QQdQdQdQ

)(

r

r ][)(rT

rr

ddrdzdKr

][)(

2

2

2

T

r

ddrdzrdK ))((2

2

z

TddzdrrdK z

ddzdrrdq g )(

dT

Cdzdrrd p )(

+ + +

=

T

CqTK pg)(KKKK zr

[gq

)(1

r

Tr

rrK r

)(1

2

2

2 T

rK )(

2

2

z

TK z

)( ddzdrrd )(

ddzdrrd

TC p

+ + + ] =

If material is homogeneous amp isotropic

)(1

r

Tr

rr

)(1

2

2

2 T

r)(

2

2

z

T

K

q g

T

K

C p

)(1

r

Tr

rr

)(1

2

2

2 T

r )(2

2

z

T

K

q g

T

1

+ + + =

)(1

r

Tr

rrK r

)(1

2

2

2 T

rK )(

2

2

z

TK z

gq

T

C p + + + =

+ + + = Thermal diffusivity

T

K

qT g

12

pC

K

(Ein ndash Eout) + Egen = Est

)( rT

)( r

)sin( drrddr)sin( drrddr

planerdirection

planerdirection

planedirectionr

)(

)(

)(

Spherical polar co-ordinates

Elemental length

Consider an elemental volume ABCDEFGH =

GENERAL HEAT CONDUCTION EQUATION IN SPHERICAL COORDINATES

Let

=internal heat generation per unit volume per unit time (W m3)q= rate of heat flow (W)Q= qdt=total heat flow (J)r=mass density of the material element(kgm3)Cp=specific heat of the material element (Jkg K)

gq

Consider an unsteady (transient) three dimensional heat conduction with internal heat generation in heterogeneous amp anisotropic spherical geometries

LetTemperature distribution is a function of T=

X

Y

Z

drT

drrdKQ rr )sin(

drQr

QQ rrdrr ][

drdr

TdrrdK

rdQ

QQdQ

rr

drrrr

])sin([

)1(]([sin 2

rT

rr

ddrddKdQ rr

Plane ( -q f) lsquorrsquo directionQuantity of heat flowing into the face ABCD in lsquorrsquo direction

Quantity of heat leaving from the face EFGH in lsquorrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquorrsquo direction

drT

drdrKQ )sin(

rdQ

rQQ d ][

rddr

TdrdrK

rdQ

QQdQ d

])sin([

])[(sin

2

T

r

drdrdrKdQ

)2(])(sin[

2

T

r

drdrddrKdQ

Quantity of heat leaving from the face BFHD in lsquoqrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoqrsquo direction

X

Y

Z

Plane (r- f ) lsquoq lsquodirection

Quantity of heat flowing into the face AEGC in lsquoqrsquo direction

dr

TdrrdKQ

sin)(

drQ

rQQ d sin][

sin

drdr

TdrrdK

rdQ

QQdQ d

sin]sin

)([sin

][sinsin

)sin(

T

rr

ddrdrrdKdQ

)3()(sin

)sin(2

2

22

2

T

r

ddddrrKdQ

Plane (r- ) q lsquofrsquo directionQuantity of heat flowing into the face ABFE in lsquofrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquozrsquo direction

)4()sin()(

ddrrddrqQgg

)5()sin()(

dT

CdrrddrQ pst

Total amount of internal heat generation in the element

Total heat energy stored in the element

(Ein ndash Eout) + Egen = Est

d

TmCQ pst )(

)( volumem

Quantity of heat leaving from the face CDHG in lsquofrsquo direction

according to thermodynamics first law

Ein + Egen = Eout + Est

X

Y

Z

stgr QQdQdQdQ

)(

]([sin 2

r

Tr

rddrddKr

])(sin[

2

T

rdrdrddrK

)(sin

)sin(2

2

22

2

T

r

ddddrrK

ddrrddrq g )sin(

dT

Cdrrddr p )sin(

2

2

r

r

sin

sin

]([ 22 r

Tr

rr

Kr

])(sin[sin2

TrK

)(sin 2

2

22

T

r

Kgq

T

Cp

]([1

[ 22 r

Tr

rrK

])(sin[sin1

2

T

r)](

sin

12

2

22 T

r gq

T

Cp

]([1

[ 22 r

Tr

rr

])(sin[sin1

2

T

r)](

sin

12

2

22 T

rK

q g

T

K

Cp

]([1

[ 22 r

Tr

rr

])(sin[sin1

2

T

r)](

sin

12

2

22 T

rK

q g

T1

KKKKr If material is homogeneous amp isotropic

TCqTK pg)(

T

K

qT g

12 pC

K

Thermal diffusivity

(Ein ndash Eout) + Egen = Est

General Heat Conduction Equation

PLANE WALL CYLINDER SPHERE

Temperature profile

Plane wall SphereCylinder

bull A copper conductor (k = 380 WmoC resistivity ρ = 2x10-8 Ω m) having inner and outer radii 10 cm and 225 cm respectively is carrying a current density of 4800 amperes cm2 the conductor is internally cooled and a constant temperature of 65oC is maintained at the inner surface and there is no heat transfer through the insulation surrounding the conductor

Determinebull (i) The maximum temperature of the conductor and the radius at

which it occursbull (ii) The internal heat transfer rate

bull A chemical reaction takes place in a packed bed (k=06 WmoC) between two coaxial cylinders with radii 15 mm and 45 mm the inner surface is at 580oC and it is insulated Assuming the reaction rate of 055 MWm3 in the reactor volume find the temperature at the outer surface of the reactor

bull A current of 300 amperes passes through a stainless steel wire of 25 mm diameter and k=20 WmoC The resistivity of the wire is 70x10-8 ohm-m and length of the wire is 2 m if the wire is submerged in a fluid maintained at 50oC and h at the wire surface is 4000 Wm2oC calculate the steady state temperature at the centre and at the surface of the wire

bull An approximately spherical shaped orange ( k=023 WmoC) 90 mm diameter undergoes ripping process and generates 5100 Wm3 of energy If external surface the orange is at 8oC determine

bull (i) temperature at the centre of the orange and

bull (ii) heat flow from the outer surface of the orange

bull FINS External surfaces to used to cooling area and hence convection

heat transfer can be improvedbull APPLICATIONS IC engine cylinder ( air cooling) Heat exchangers automobile

radiators computer chip refrigerators electric motors generators and transformers etc

bull ASSUMPTIONS MADE IN ANALYSIS OF HEAT TRANFER THROUGH FIN

1 Fin material is homogenous and isotropic constthermal conductivity

2 Temperature at any cs of the fin is uniform3 lsquohrsquo is uniform over entire surface4 No internal heat generation5 Thermal contact resistance is negligible6 Radiation heat transfer is negligible7 Heat conduction is steady state and one dimensional

bull Problem 1

bull A stainless steel fin (k=20 WmK) having a diameter of 20 mm and a length of 01 m is attached to a wall at 300oC The ambient temperature is 50oC and the heat transfer coefficient is 10 Wm2K The fin tip is insulated Determine

bull (A) The rate of heat dissipation from the finbull (B) The temperature at the fin tipbull (C) The rate of heat transfer from the wall area covered by the fin if

the fin was not usedbull (D) The heat transfer rate from the same fin geometry if the

stainless steel fin is replaced by a fictitious fin with infinite thermal conductivity

bull Problem2bull One end of a long rod is inserted into a furnace while the other

projects into ambient temperature Under steady state the temperature of the rod is measured at two points 75mm apart and found to be 125oC and 885oC respectively while the ambient temperature is 20oC If the rod is 25mm in diameter and h is 2336 Wm2K

bull Find the thermal conductivity of the rod material

  • HEAT AND MASS TRANSFER
  • Thermodynamics Vs Heat transfer
  • MODE OF HEAT TRANSFER
  • MODE OF HEAT TRANSFER (2)
  • CONDUCTION- MECHANISM
  • CONCECTION-MECHANISM
  • RADIATION-MECHANISM
  • Conduction- Convection- Radiation
  • Slide 9
  • Slide 10
  • Fourierrsquos Law of Heat Conduction
  • Slide 12
  • Newtonrsquos Law of Cooling
  • Laws of Thermal Radiation
  • LAWS FOR HEAT TRANSFER
  • ELECTRICAL ampTHERMAL ANALOGY
  • Slide 17
  • Slide 18
  • PLANE WALL-UNIDIRECTIONAL
  • PLANE WALL -SERIES
  • PLANE WALL IN SERIES
  • PLANE WALL -PARALLEL
  • PLANE WALL-PARALLEL
  • PLANE WALL-COMBINATION
  • PLANE WALL-FILM RESISTANCE
  • PLANE WALL- THERMAL CONTACT RESISTANCE
  • PLANE WALL- THERMAL CONTACT RESISTANCE (2)
  • CYLINDER-SOLID
  • HEAT TRANSFER IN SOLID CYLINDER
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER-COMPOSITE
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER-MULTY LAYER
  • HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER
  • HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER (2)
  • HEAT TRANSFER IN SPHERE
  • HEAT TRANSFER IN SPHERE -MULTY LAYER
  • HEAT TRANSFER IN SPHERE -MULTY LAYER (2)
  • Slide 38
  • Slide 39
  • Slide 40
  • Slide 41
  • CRITICAL RADIUS
  • Plane wall- Insulation thickness -study
  • Problem5
  • Slide 45
  • Slide 46
  • Slide 47
  • GENERAL HEAT CONDUCTION EQUATION IN CYLINDRICAL COORDINATES
  • Slide 49
  • Slide 50
  • Slide 51
  • GENERAL HEAT CONDUCTION EQUATION IN SPHERICAL COORDINATES
  • Slide 53
  • Slide 54
  • Slide 55
  • General Heat Conduction Equation
  • Temperature profile
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Slide 67
Page 35: hmt chap 1

bull Problem 2Two large aluminum plates (k=240 WmK) each 2 cm thick with 10 mm surface roughness are placed in contact under 105 Nm2 pressure in air The temperature at the outside surfaces are 390 oC and 406 oCCalculate

A) the heat flux B) the temperature drop due to the contact resistance C) the contact temperatures Thermal contact resistance with air as the interfacial fluid for

10 mm roughness is 27510-4 m2KW

bull Problem 3bull Steam at 350 oC flowing in a pipe (k=80 WmK) of 5 cm

inner diameter and 56 cm outer diameter is covered with 3 cm thick insulation of k=005 WmK Heat is lost to surroundings at 5 oC by natural convection and radiation the combined h being 20 Wm2K Taking the heat transfer coefficient inside the pipe as 60 Wm2K

bull Determinebull A) the rate of heat loss from the steam per unit length of

the pipebull B) the temperature drop across the pipe and the insulation

bull PROBLEM4bull A nuclear reactor has a sperical pressure vessel 750 mm inside

diameter 80 mm wall thickness The temperature at the inner surface is 500 oC If the tempertarue at the outer surface is 495 oC

bull Calculatea) The rate of heat loss through the metal wallb) If the rate of heat loss is to limited to 305Whow much inulation

thickness needs to be applied Thermal conductivity of steel wall=46 WmoC Thermal conductivity of insulation=004 WmoC Film coefficient of heat tranfer at inner and outer surface are 200

Wm2 oC and 15 Wm2 oC Ambient air temperature=28 oC

CRITICAL RADIUS

bull The radius upto which heat flow increases and after decreases At critical radius q becomes qmax

bull The addition of insulation always increases (Rth)condbull Rtotal= (Rth)cond + (Rth)conv

bull The addition of insulation may reduce the (Rth)conv due to increase in surface area

bull r1ltrc -heat transfer increases by adding insulation till r1=rc If

insulation thickness is further increased the rate of heat loss will decrease

r1 gt rc

- heat transfer decreases by adding insulation above the rc

Plane wall- Insulation

thickness -study

Cylinder amp Sphere- Insulation thickness -

study

Problem5bull A 3mm diameter and 5 mm long electric wire is tightly

wrapped with a 2 mm thick plastic cover whose K=015 wmoC Electrical measurements indicate that a current of 10 A passes through the wire and there is a voltage drop of 8 V along the wire If the insulated wire is exposed to a medium at Ta=30oC with a heat transfer coefficient of h=12 wm2oC Determine the temperature at the interface of the wire and the plastic cover in steady operation Also determine whether doubling the thickness of the plastic cover will increase or decrease this interface temperature

dx

TdzdyKQ xx )(

dxQx

QQ xxdxx ][

dxdx

TdzdyK

xQQdQ xdxxxx ])([)(

)1()]([)(

ddzdydx

x

TK

xdQ xx

GENERAL HEAT CONDUCTION EQUATION IN CARTESIAN COORDINATES

Consider an unsteady (transient) three dimensional heat conduction with internal heat generation in heterogeneous amp anisotropic PLANE WALL

Cartesian co-ordinates x y z

Elemental length dxdydz

Consider an elemental volume ABCDEFGH = (dxdydz)

Let Temperature distribution is a function of T=T(xyz)

Let

=internal heat generation per unit volume per unit time (W m3)q= rate of heat flow (W)Q= qdt=total heat flow (J)r=mass density of the material element(kgm3)Cp=specific heat of the material element (Jkg K)

gq

Plane (y - z) lsquoxrsquo direction

Quantity of heat leaving from the face EFGH in lsquoXrsquo direction

Quantity of heat flowing into the face ABCD in lsquoXrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoXrsquo direction

dy

TdzdxKQ yy )(

dyQy

QQ yydyy ][

dydy

TdzdxK

yQQdQ ydyyyy ])([)(

)2()]([)(

ddzdydx

y

TK

ydQ yy

Plane (x - z) lsquoyrsquo direction

Quantity of heat flowing into the face ABFE in lsquoYrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoYrsquo direction

dz

TdydxKQ zz )(

dzQz

QQ zzdzz ][

dzdz

TdydxK

zQQdQ zdzzzz ])([)(

)3()]([)(

ddzdydx

z

TK

zdQ zz

Plane (x - y) lsquozrsquo direction

Quantity of heat flowing into the face ADHE in lsquoZrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoZrsquo direction

Quantity of heat leaving from the face DCGH in lsquoYrsquo direction

Quantity of heat leaving from the face BCGF in lsquoZrsquo direction

)4()()(

ddzdydxqQgg

Total amount of internal heat generation in the element

Ein + Egen = Eout+Est (Ein- Eout)+ Egen=Est

T

Cddzdydxddzdydxqddzdydxz

TK

zy

TK

yx

TK

x pgzyx )()())](()()([

T

pC

gq

z

TzK

zy

Ty

Kyx

TxK

x][][][

stgzyx QQdQdQdQ

)(

)5()()(

dT

CdzdydxQ pst

Total heat energy stored in the element

(ie thermal energy of the element can be increased because of heat accumulation in the element due to heat flow amp internal heat generation)

)( volumem d

TmCQ pst )(

T

CqTK pg)(

KKKyKx z If material is homogeneous amp isotropic

T

pC

gq

z

T

zy

T

yx

T

xK ][][][

T

Kp

C

Kgq

z

T

y

T

x

T2

2

2

2

2

2

T

Kgq

z

T

y

T

x

T 12

2

2

2

2

2

Thermal diffusivity pC

K

T

K

qT g

12

according to thermodynamics first law

)( zr )( dzrddr

)( dzrddr

Cylindrical polar co-ordinates

Elemental length

Consider an elemental volume ABCDEFGH =

GENERAL HEAT CONDUCTION EQUATION IN CYLINDRICAL COORDINATES

gq

gq

Consider an unsteady (transient) three dimensional heat conduction with internal heat generation in heterogeneous amp anisotropic cylindrical geometries (eg rods and pipes)

)( zrTLetTemperature distribution is a function of T=

Let

=internal heat generation per unit volume per unit time (W m3)q= rate of heat flow (W)Q= qdt=total heat flow (J)r=mass density of the material element(kgm3)Cp=specific heat of the material element (Jkg K)

planerdirectionz

planezrdirection

planezdirectionr

)(

)(

)(

LetHeat flow in

drdr

TdzrdK

rdQ

QQdQ

rr

drrrr

])([

)1(][

r

Tr

rddrdzdKdQ rr

Heat accumulation in the element due to heat flow by conduction in lsquorrsquo direction

rddr

TdzdrK

rdQ

QQdQ d

])([

][

2

T

r

ddrdzrdKdQ

)2(][

2

2

2

T

r

ddrdzrdKdQ

Heat accumulation in the element due to heat flow by conduction in lsquoqrsquo direction

Plane (z- q ) lsquorrsquo direction

dr

TdzrdKQ rr

)(

drQr

QQ rrdrr ][

Quantity of heat leaving from the face EFGH in lsquorrsquo direction

Quantity of heat flowing into the face ABCD in lsquorrsquo direction

Plane (r-z) lsquoq lsquodirection

Quantity of heat leaving from the face AEGC in lsquoqrsquo direction

d

r

TdzdrKQ )(

rdQ

rQQ d ][

Quantity of heat flowing into the face BFHD in lsquoqrsquo direction

)3())((2

2

z

TdzdrrdKdQ zz

)5()()(

dT

CdzdrrdQ pst

)4()()(

ddzdrrdqQgg

Total amount of internal heat generation in the element

Total heat energy stored in the element

(ie thermal energy of the element can be increased because of heat accumulation in the element due to heat flow amp internal heat generation)

according to thermodynamics first law

Ein + Egen = Eout + Est (Ein ndash Eout) + Egen = Est

)( volumem

)()(z

T

zdzdrrdKdQ zz

Heat accumulation in the element due to heat flow by conduction in lsquozrsquo direction

dzdz

TdrrdK

zdQ

QQdQ

zz

dzzzz

])([

d

TmCQ pst )(

Plane (r-q) lsquozrsquo direction

dz

TdrrdKQ zz )(

dzQz

QQ zzdzz ][ Quantity of heat leaving from the face CDHG in lsquozrsquo direction

Quantity of heat flowing into the face ABFE in lsquozrsquo direction

stgzr QQdQdQdQ

)(

r

r ][)(rT

rr

ddrdzdKr

][)(

2

2

2

T

r

ddrdzrdK ))((2

2

z

TddzdrrdK z

ddzdrrdq g )(

dT

Cdzdrrd p )(

+ + +

=

T

CqTK pg)(KKKK zr

[gq

)(1

r

Tr

rrK r

)(1

2

2

2 T

rK )(

2

2

z

TK z

)( ddzdrrd )(

ddzdrrd

TC p

+ + + ] =

If material is homogeneous amp isotropic

)(1

r

Tr

rr

)(1

2

2

2 T

r)(

2

2

z

T

K

q g

T

K

C p

)(1

r

Tr

rr

)(1

2

2

2 T

r )(2

2

z

T

K

q g

T

1

+ + + =

)(1

r

Tr

rrK r

)(1

2

2

2 T

rK )(

2

2

z

TK z

gq

T

C p + + + =

+ + + = Thermal diffusivity

T

K

qT g

12

pC

K

(Ein ndash Eout) + Egen = Est

)( rT

)( r

)sin( drrddr)sin( drrddr

planerdirection

planerdirection

planedirectionr

)(

)(

)(

Spherical polar co-ordinates

Elemental length

Consider an elemental volume ABCDEFGH =

GENERAL HEAT CONDUCTION EQUATION IN SPHERICAL COORDINATES

Let

=internal heat generation per unit volume per unit time (W m3)q= rate of heat flow (W)Q= qdt=total heat flow (J)r=mass density of the material element(kgm3)Cp=specific heat of the material element (Jkg K)

gq

Consider an unsteady (transient) three dimensional heat conduction with internal heat generation in heterogeneous amp anisotropic spherical geometries

LetTemperature distribution is a function of T=

X

Y

Z

drT

drrdKQ rr )sin(

drQr

QQ rrdrr ][

drdr

TdrrdK

rdQ

QQdQ

rr

drrrr

])sin([

)1(]([sin 2

rT

rr

ddrddKdQ rr

Plane ( -q f) lsquorrsquo directionQuantity of heat flowing into the face ABCD in lsquorrsquo direction

Quantity of heat leaving from the face EFGH in lsquorrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquorrsquo direction

drT

drdrKQ )sin(

rdQ

rQQ d ][

rddr

TdrdrK

rdQ

QQdQ d

])sin([

])[(sin

2

T

r

drdrdrKdQ

)2(])(sin[

2

T

r

drdrddrKdQ

Quantity of heat leaving from the face BFHD in lsquoqrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoqrsquo direction

X

Y

Z

Plane (r- f ) lsquoq lsquodirection

Quantity of heat flowing into the face AEGC in lsquoqrsquo direction

dr

TdrrdKQ

sin)(

drQ

rQQ d sin][

sin

drdr

TdrrdK

rdQ

QQdQ d

sin]sin

)([sin

][sinsin

)sin(

T

rr

ddrdrrdKdQ

)3()(sin

)sin(2

2

22

2

T

r

ddddrrKdQ

Plane (r- ) q lsquofrsquo directionQuantity of heat flowing into the face ABFE in lsquofrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquozrsquo direction

)4()sin()(

ddrrddrqQgg

)5()sin()(

dT

CdrrddrQ pst

Total amount of internal heat generation in the element

Total heat energy stored in the element

(Ein ndash Eout) + Egen = Est

d

TmCQ pst )(

)( volumem

Quantity of heat leaving from the face CDHG in lsquofrsquo direction

according to thermodynamics first law

Ein + Egen = Eout + Est

X

Y

Z

stgr QQdQdQdQ

)(

]([sin 2

r

Tr

rddrddKr

])(sin[

2

T

rdrdrddrK

)(sin

)sin(2

2

22

2

T

r

ddddrrK

ddrrddrq g )sin(

dT

Cdrrddr p )sin(

2

2

r

r

sin

sin

]([ 22 r

Tr

rr

Kr

])(sin[sin2

TrK

)(sin 2

2

22

T

r

Kgq

T

Cp

]([1

[ 22 r

Tr

rrK

])(sin[sin1

2

T

r)](

sin

12

2

22 T

r gq

T

Cp

]([1

[ 22 r

Tr

rr

])(sin[sin1

2

T

r)](

sin

12

2

22 T

rK

q g

T

K

Cp

]([1

[ 22 r

Tr

rr

])(sin[sin1

2

T

r)](

sin

12

2

22 T

rK

q g

T1

KKKKr If material is homogeneous amp isotropic

TCqTK pg)(

T

K

qT g

12 pC

K

Thermal diffusivity

(Ein ndash Eout) + Egen = Est

General Heat Conduction Equation

PLANE WALL CYLINDER SPHERE

Temperature profile

Plane wall SphereCylinder

bull A copper conductor (k = 380 WmoC resistivity ρ = 2x10-8 Ω m) having inner and outer radii 10 cm and 225 cm respectively is carrying a current density of 4800 amperes cm2 the conductor is internally cooled and a constant temperature of 65oC is maintained at the inner surface and there is no heat transfer through the insulation surrounding the conductor

Determinebull (i) The maximum temperature of the conductor and the radius at

which it occursbull (ii) The internal heat transfer rate

bull A chemical reaction takes place in a packed bed (k=06 WmoC) between two coaxial cylinders with radii 15 mm and 45 mm the inner surface is at 580oC and it is insulated Assuming the reaction rate of 055 MWm3 in the reactor volume find the temperature at the outer surface of the reactor

bull A current of 300 amperes passes through a stainless steel wire of 25 mm diameter and k=20 WmoC The resistivity of the wire is 70x10-8 ohm-m and length of the wire is 2 m if the wire is submerged in a fluid maintained at 50oC and h at the wire surface is 4000 Wm2oC calculate the steady state temperature at the centre and at the surface of the wire

bull An approximately spherical shaped orange ( k=023 WmoC) 90 mm diameter undergoes ripping process and generates 5100 Wm3 of energy If external surface the orange is at 8oC determine

bull (i) temperature at the centre of the orange and

bull (ii) heat flow from the outer surface of the orange

bull FINS External surfaces to used to cooling area and hence convection

heat transfer can be improvedbull APPLICATIONS IC engine cylinder ( air cooling) Heat exchangers automobile

radiators computer chip refrigerators electric motors generators and transformers etc

bull ASSUMPTIONS MADE IN ANALYSIS OF HEAT TRANFER THROUGH FIN

1 Fin material is homogenous and isotropic constthermal conductivity

2 Temperature at any cs of the fin is uniform3 lsquohrsquo is uniform over entire surface4 No internal heat generation5 Thermal contact resistance is negligible6 Radiation heat transfer is negligible7 Heat conduction is steady state and one dimensional

bull Problem 1

bull A stainless steel fin (k=20 WmK) having a diameter of 20 mm and a length of 01 m is attached to a wall at 300oC The ambient temperature is 50oC and the heat transfer coefficient is 10 Wm2K The fin tip is insulated Determine

bull (A) The rate of heat dissipation from the finbull (B) The temperature at the fin tipbull (C) The rate of heat transfer from the wall area covered by the fin if

the fin was not usedbull (D) The heat transfer rate from the same fin geometry if the

stainless steel fin is replaced by a fictitious fin with infinite thermal conductivity

bull Problem2bull One end of a long rod is inserted into a furnace while the other

projects into ambient temperature Under steady state the temperature of the rod is measured at two points 75mm apart and found to be 125oC and 885oC respectively while the ambient temperature is 20oC If the rod is 25mm in diameter and h is 2336 Wm2K

bull Find the thermal conductivity of the rod material

  • HEAT AND MASS TRANSFER
  • Thermodynamics Vs Heat transfer
  • MODE OF HEAT TRANSFER
  • MODE OF HEAT TRANSFER (2)
  • CONDUCTION- MECHANISM
  • CONCECTION-MECHANISM
  • RADIATION-MECHANISM
  • Conduction- Convection- Radiation
  • Slide 9
  • Slide 10
  • Fourierrsquos Law of Heat Conduction
  • Slide 12
  • Newtonrsquos Law of Cooling
  • Laws of Thermal Radiation
  • LAWS FOR HEAT TRANSFER
  • ELECTRICAL ampTHERMAL ANALOGY
  • Slide 17
  • Slide 18
  • PLANE WALL-UNIDIRECTIONAL
  • PLANE WALL -SERIES
  • PLANE WALL IN SERIES
  • PLANE WALL -PARALLEL
  • PLANE WALL-PARALLEL
  • PLANE WALL-COMBINATION
  • PLANE WALL-FILM RESISTANCE
  • PLANE WALL- THERMAL CONTACT RESISTANCE
  • PLANE WALL- THERMAL CONTACT RESISTANCE (2)
  • CYLINDER-SOLID
  • HEAT TRANSFER IN SOLID CYLINDER
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER-COMPOSITE
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER-MULTY LAYER
  • HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER
  • HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER (2)
  • HEAT TRANSFER IN SPHERE
  • HEAT TRANSFER IN SPHERE -MULTY LAYER
  • HEAT TRANSFER IN SPHERE -MULTY LAYER (2)
  • Slide 38
  • Slide 39
  • Slide 40
  • Slide 41
  • CRITICAL RADIUS
  • Plane wall- Insulation thickness -study
  • Problem5
  • Slide 45
  • Slide 46
  • Slide 47
  • GENERAL HEAT CONDUCTION EQUATION IN CYLINDRICAL COORDINATES
  • Slide 49
  • Slide 50
  • Slide 51
  • GENERAL HEAT CONDUCTION EQUATION IN SPHERICAL COORDINATES
  • Slide 53
  • Slide 54
  • Slide 55
  • General Heat Conduction Equation
  • Temperature profile
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Slide 67
Page 36: hmt chap 1

bull Problem 3bull Steam at 350 oC flowing in a pipe (k=80 WmK) of 5 cm

inner diameter and 56 cm outer diameter is covered with 3 cm thick insulation of k=005 WmK Heat is lost to surroundings at 5 oC by natural convection and radiation the combined h being 20 Wm2K Taking the heat transfer coefficient inside the pipe as 60 Wm2K

bull Determinebull A) the rate of heat loss from the steam per unit length of

the pipebull B) the temperature drop across the pipe and the insulation

bull PROBLEM4bull A nuclear reactor has a sperical pressure vessel 750 mm inside

diameter 80 mm wall thickness The temperature at the inner surface is 500 oC If the tempertarue at the outer surface is 495 oC

bull Calculatea) The rate of heat loss through the metal wallb) If the rate of heat loss is to limited to 305Whow much inulation

thickness needs to be applied Thermal conductivity of steel wall=46 WmoC Thermal conductivity of insulation=004 WmoC Film coefficient of heat tranfer at inner and outer surface are 200

Wm2 oC and 15 Wm2 oC Ambient air temperature=28 oC

CRITICAL RADIUS

bull The radius upto which heat flow increases and after decreases At critical radius q becomes qmax

bull The addition of insulation always increases (Rth)condbull Rtotal= (Rth)cond + (Rth)conv

bull The addition of insulation may reduce the (Rth)conv due to increase in surface area

bull r1ltrc -heat transfer increases by adding insulation till r1=rc If

insulation thickness is further increased the rate of heat loss will decrease

r1 gt rc

- heat transfer decreases by adding insulation above the rc

Plane wall- Insulation

thickness -study

Cylinder amp Sphere- Insulation thickness -

study

Problem5bull A 3mm diameter and 5 mm long electric wire is tightly

wrapped with a 2 mm thick plastic cover whose K=015 wmoC Electrical measurements indicate that a current of 10 A passes through the wire and there is a voltage drop of 8 V along the wire If the insulated wire is exposed to a medium at Ta=30oC with a heat transfer coefficient of h=12 wm2oC Determine the temperature at the interface of the wire and the plastic cover in steady operation Also determine whether doubling the thickness of the plastic cover will increase or decrease this interface temperature

dx

TdzdyKQ xx )(

dxQx

QQ xxdxx ][

dxdx

TdzdyK

xQQdQ xdxxxx ])([)(

)1()]([)(

ddzdydx

x

TK

xdQ xx

GENERAL HEAT CONDUCTION EQUATION IN CARTESIAN COORDINATES

Consider an unsteady (transient) three dimensional heat conduction with internal heat generation in heterogeneous amp anisotropic PLANE WALL

Cartesian co-ordinates x y z

Elemental length dxdydz

Consider an elemental volume ABCDEFGH = (dxdydz)

Let Temperature distribution is a function of T=T(xyz)

Let

=internal heat generation per unit volume per unit time (W m3)q= rate of heat flow (W)Q= qdt=total heat flow (J)r=mass density of the material element(kgm3)Cp=specific heat of the material element (Jkg K)

gq

Plane (y - z) lsquoxrsquo direction

Quantity of heat leaving from the face EFGH in lsquoXrsquo direction

Quantity of heat flowing into the face ABCD in lsquoXrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoXrsquo direction

dy

TdzdxKQ yy )(

dyQy

QQ yydyy ][

dydy

TdzdxK

yQQdQ ydyyyy ])([)(

)2()]([)(

ddzdydx

y

TK

ydQ yy

Plane (x - z) lsquoyrsquo direction

Quantity of heat flowing into the face ABFE in lsquoYrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoYrsquo direction

dz

TdydxKQ zz )(

dzQz

QQ zzdzz ][

dzdz

TdydxK

zQQdQ zdzzzz ])([)(

)3()]([)(

ddzdydx

z

TK

zdQ zz

Plane (x - y) lsquozrsquo direction

Quantity of heat flowing into the face ADHE in lsquoZrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoZrsquo direction

Quantity of heat leaving from the face DCGH in lsquoYrsquo direction

Quantity of heat leaving from the face BCGF in lsquoZrsquo direction

)4()()(

ddzdydxqQgg

Total amount of internal heat generation in the element

Ein + Egen = Eout+Est (Ein- Eout)+ Egen=Est

T

Cddzdydxddzdydxqddzdydxz

TK

zy

TK

yx

TK

x pgzyx )()())](()()([

T

pC

gq

z

TzK

zy

Ty

Kyx

TxK

x][][][

stgzyx QQdQdQdQ

)(

)5()()(

dT

CdzdydxQ pst

Total heat energy stored in the element

(ie thermal energy of the element can be increased because of heat accumulation in the element due to heat flow amp internal heat generation)

)( volumem d

TmCQ pst )(

T

CqTK pg)(

KKKyKx z If material is homogeneous amp isotropic

T

pC

gq

z

T

zy

T

yx

T

xK ][][][

T

Kp

C

Kgq

z

T

y

T

x

T2

2

2

2

2

2

T

Kgq

z

T

y

T

x

T 12

2

2

2

2

2

Thermal diffusivity pC

K

T

K

qT g

12

according to thermodynamics first law

)( zr )( dzrddr

)( dzrddr

Cylindrical polar co-ordinates

Elemental length

Consider an elemental volume ABCDEFGH =

GENERAL HEAT CONDUCTION EQUATION IN CYLINDRICAL COORDINATES

gq

gq

Consider an unsteady (transient) three dimensional heat conduction with internal heat generation in heterogeneous amp anisotropic cylindrical geometries (eg rods and pipes)

)( zrTLetTemperature distribution is a function of T=

Let

=internal heat generation per unit volume per unit time (W m3)q= rate of heat flow (W)Q= qdt=total heat flow (J)r=mass density of the material element(kgm3)Cp=specific heat of the material element (Jkg K)

planerdirectionz

planezrdirection

planezdirectionr

)(

)(

)(

LetHeat flow in

drdr

TdzrdK

rdQ

QQdQ

rr

drrrr

])([

)1(][

r

Tr

rddrdzdKdQ rr

Heat accumulation in the element due to heat flow by conduction in lsquorrsquo direction

rddr

TdzdrK

rdQ

QQdQ d

])([

][

2

T

r

ddrdzrdKdQ

)2(][

2

2

2

T

r

ddrdzrdKdQ

Heat accumulation in the element due to heat flow by conduction in lsquoqrsquo direction

Plane (z- q ) lsquorrsquo direction

dr

TdzrdKQ rr

)(

drQr

QQ rrdrr ][

Quantity of heat leaving from the face EFGH in lsquorrsquo direction

Quantity of heat flowing into the face ABCD in lsquorrsquo direction

Plane (r-z) lsquoq lsquodirection

Quantity of heat leaving from the face AEGC in lsquoqrsquo direction

d

r

TdzdrKQ )(

rdQ

rQQ d ][

Quantity of heat flowing into the face BFHD in lsquoqrsquo direction

)3())((2

2

z

TdzdrrdKdQ zz

)5()()(

dT

CdzdrrdQ pst

)4()()(

ddzdrrdqQgg

Total amount of internal heat generation in the element

Total heat energy stored in the element

(ie thermal energy of the element can be increased because of heat accumulation in the element due to heat flow amp internal heat generation)

according to thermodynamics first law

Ein + Egen = Eout + Est (Ein ndash Eout) + Egen = Est

)( volumem

)()(z

T

zdzdrrdKdQ zz

Heat accumulation in the element due to heat flow by conduction in lsquozrsquo direction

dzdz

TdrrdK

zdQ

QQdQ

zz

dzzzz

])([

d

TmCQ pst )(

Plane (r-q) lsquozrsquo direction

dz

TdrrdKQ zz )(

dzQz

QQ zzdzz ][ Quantity of heat leaving from the face CDHG in lsquozrsquo direction

Quantity of heat flowing into the face ABFE in lsquozrsquo direction

stgzr QQdQdQdQ

)(

r

r ][)(rT

rr

ddrdzdKr

][)(

2

2

2

T

r

ddrdzrdK ))((2

2

z

TddzdrrdK z

ddzdrrdq g )(

dT

Cdzdrrd p )(

+ + +

=

T

CqTK pg)(KKKK zr

[gq

)(1

r

Tr

rrK r

)(1

2

2

2 T

rK )(

2

2

z

TK z

)( ddzdrrd )(

ddzdrrd

TC p

+ + + ] =

If material is homogeneous amp isotropic

)(1

r

Tr

rr

)(1

2

2

2 T

r)(

2

2

z

T

K

q g

T

K

C p

)(1

r

Tr

rr

)(1

2

2

2 T

r )(2

2

z

T

K

q g

T

1

+ + + =

)(1

r

Tr

rrK r

)(1

2

2

2 T

rK )(

2

2

z

TK z

gq

T

C p + + + =

+ + + = Thermal diffusivity

T

K

qT g

12

pC

K

(Ein ndash Eout) + Egen = Est

)( rT

)( r

)sin( drrddr)sin( drrddr

planerdirection

planerdirection

planedirectionr

)(

)(

)(

Spherical polar co-ordinates

Elemental length

Consider an elemental volume ABCDEFGH =

GENERAL HEAT CONDUCTION EQUATION IN SPHERICAL COORDINATES

Let

=internal heat generation per unit volume per unit time (W m3)q= rate of heat flow (W)Q= qdt=total heat flow (J)r=mass density of the material element(kgm3)Cp=specific heat of the material element (Jkg K)

gq

Consider an unsteady (transient) three dimensional heat conduction with internal heat generation in heterogeneous amp anisotropic spherical geometries

LetTemperature distribution is a function of T=

X

Y

Z

drT

drrdKQ rr )sin(

drQr

QQ rrdrr ][

drdr

TdrrdK

rdQ

QQdQ

rr

drrrr

])sin([

)1(]([sin 2

rT

rr

ddrddKdQ rr

Plane ( -q f) lsquorrsquo directionQuantity of heat flowing into the face ABCD in lsquorrsquo direction

Quantity of heat leaving from the face EFGH in lsquorrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquorrsquo direction

drT

drdrKQ )sin(

rdQ

rQQ d ][

rddr

TdrdrK

rdQ

QQdQ d

])sin([

])[(sin

2

T

r

drdrdrKdQ

)2(])(sin[

2

T

r

drdrddrKdQ

Quantity of heat leaving from the face BFHD in lsquoqrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoqrsquo direction

X

Y

Z

Plane (r- f ) lsquoq lsquodirection

Quantity of heat flowing into the face AEGC in lsquoqrsquo direction

dr

TdrrdKQ

sin)(

drQ

rQQ d sin][

sin

drdr

TdrrdK

rdQ

QQdQ d

sin]sin

)([sin

][sinsin

)sin(

T

rr

ddrdrrdKdQ

)3()(sin

)sin(2

2

22

2

T

r

ddddrrKdQ

Plane (r- ) q lsquofrsquo directionQuantity of heat flowing into the face ABFE in lsquofrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquozrsquo direction

)4()sin()(

ddrrddrqQgg

)5()sin()(

dT

CdrrddrQ pst

Total amount of internal heat generation in the element

Total heat energy stored in the element

(Ein ndash Eout) + Egen = Est

d

TmCQ pst )(

)( volumem

Quantity of heat leaving from the face CDHG in lsquofrsquo direction

according to thermodynamics first law

Ein + Egen = Eout + Est

X

Y

Z

stgr QQdQdQdQ

)(

]([sin 2

r

Tr

rddrddKr

])(sin[

2

T

rdrdrddrK

)(sin

)sin(2

2

22

2

T

r

ddddrrK

ddrrddrq g )sin(

dT

Cdrrddr p )sin(

2

2

r

r

sin

sin

]([ 22 r

Tr

rr

Kr

])(sin[sin2

TrK

)(sin 2

2

22

T

r

Kgq

T

Cp

]([1

[ 22 r

Tr

rrK

])(sin[sin1

2

T

r)](

sin

12

2

22 T

r gq

T

Cp

]([1

[ 22 r

Tr

rr

])(sin[sin1

2

T

r)](

sin

12

2

22 T

rK

q g

T

K

Cp

]([1

[ 22 r

Tr

rr

])(sin[sin1

2

T

r)](

sin

12

2

22 T

rK

q g

T1

KKKKr If material is homogeneous amp isotropic

TCqTK pg)(

T

K

qT g

12 pC

K

Thermal diffusivity

(Ein ndash Eout) + Egen = Est

General Heat Conduction Equation

PLANE WALL CYLINDER SPHERE

Temperature profile

Plane wall SphereCylinder

bull A copper conductor (k = 380 WmoC resistivity ρ = 2x10-8 Ω m) having inner and outer radii 10 cm and 225 cm respectively is carrying a current density of 4800 amperes cm2 the conductor is internally cooled and a constant temperature of 65oC is maintained at the inner surface and there is no heat transfer through the insulation surrounding the conductor

Determinebull (i) The maximum temperature of the conductor and the radius at

which it occursbull (ii) The internal heat transfer rate

bull A chemical reaction takes place in a packed bed (k=06 WmoC) between two coaxial cylinders with radii 15 mm and 45 mm the inner surface is at 580oC and it is insulated Assuming the reaction rate of 055 MWm3 in the reactor volume find the temperature at the outer surface of the reactor

bull A current of 300 amperes passes through a stainless steel wire of 25 mm diameter and k=20 WmoC The resistivity of the wire is 70x10-8 ohm-m and length of the wire is 2 m if the wire is submerged in a fluid maintained at 50oC and h at the wire surface is 4000 Wm2oC calculate the steady state temperature at the centre and at the surface of the wire

bull An approximately spherical shaped orange ( k=023 WmoC) 90 mm diameter undergoes ripping process and generates 5100 Wm3 of energy If external surface the orange is at 8oC determine

bull (i) temperature at the centre of the orange and

bull (ii) heat flow from the outer surface of the orange

bull FINS External surfaces to used to cooling area and hence convection

heat transfer can be improvedbull APPLICATIONS IC engine cylinder ( air cooling) Heat exchangers automobile

radiators computer chip refrigerators electric motors generators and transformers etc

bull ASSUMPTIONS MADE IN ANALYSIS OF HEAT TRANFER THROUGH FIN

1 Fin material is homogenous and isotropic constthermal conductivity

2 Temperature at any cs of the fin is uniform3 lsquohrsquo is uniform over entire surface4 No internal heat generation5 Thermal contact resistance is negligible6 Radiation heat transfer is negligible7 Heat conduction is steady state and one dimensional

bull Problem 1

bull A stainless steel fin (k=20 WmK) having a diameter of 20 mm and a length of 01 m is attached to a wall at 300oC The ambient temperature is 50oC and the heat transfer coefficient is 10 Wm2K The fin tip is insulated Determine

bull (A) The rate of heat dissipation from the finbull (B) The temperature at the fin tipbull (C) The rate of heat transfer from the wall area covered by the fin if

the fin was not usedbull (D) The heat transfer rate from the same fin geometry if the

stainless steel fin is replaced by a fictitious fin with infinite thermal conductivity

bull Problem2bull One end of a long rod is inserted into a furnace while the other

projects into ambient temperature Under steady state the temperature of the rod is measured at two points 75mm apart and found to be 125oC and 885oC respectively while the ambient temperature is 20oC If the rod is 25mm in diameter and h is 2336 Wm2K

bull Find the thermal conductivity of the rod material

  • HEAT AND MASS TRANSFER
  • Thermodynamics Vs Heat transfer
  • MODE OF HEAT TRANSFER
  • MODE OF HEAT TRANSFER (2)
  • CONDUCTION- MECHANISM
  • CONCECTION-MECHANISM
  • RADIATION-MECHANISM
  • Conduction- Convection- Radiation
  • Slide 9
  • Slide 10
  • Fourierrsquos Law of Heat Conduction
  • Slide 12
  • Newtonrsquos Law of Cooling
  • Laws of Thermal Radiation
  • LAWS FOR HEAT TRANSFER
  • ELECTRICAL ampTHERMAL ANALOGY
  • Slide 17
  • Slide 18
  • PLANE WALL-UNIDIRECTIONAL
  • PLANE WALL -SERIES
  • PLANE WALL IN SERIES
  • PLANE WALL -PARALLEL
  • PLANE WALL-PARALLEL
  • PLANE WALL-COMBINATION
  • PLANE WALL-FILM RESISTANCE
  • PLANE WALL- THERMAL CONTACT RESISTANCE
  • PLANE WALL- THERMAL CONTACT RESISTANCE (2)
  • CYLINDER-SOLID
  • HEAT TRANSFER IN SOLID CYLINDER
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER-COMPOSITE
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER-MULTY LAYER
  • HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER
  • HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER (2)
  • HEAT TRANSFER IN SPHERE
  • HEAT TRANSFER IN SPHERE -MULTY LAYER
  • HEAT TRANSFER IN SPHERE -MULTY LAYER (2)
  • Slide 38
  • Slide 39
  • Slide 40
  • Slide 41
  • CRITICAL RADIUS
  • Plane wall- Insulation thickness -study
  • Problem5
  • Slide 45
  • Slide 46
  • Slide 47
  • GENERAL HEAT CONDUCTION EQUATION IN CYLINDRICAL COORDINATES
  • Slide 49
  • Slide 50
  • Slide 51
  • GENERAL HEAT CONDUCTION EQUATION IN SPHERICAL COORDINATES
  • Slide 53
  • Slide 54
  • Slide 55
  • General Heat Conduction Equation
  • Temperature profile
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Slide 67
Page 37: hmt chap 1

bull PROBLEM4bull A nuclear reactor has a sperical pressure vessel 750 mm inside

diameter 80 mm wall thickness The temperature at the inner surface is 500 oC If the tempertarue at the outer surface is 495 oC

bull Calculatea) The rate of heat loss through the metal wallb) If the rate of heat loss is to limited to 305Whow much inulation

thickness needs to be applied Thermal conductivity of steel wall=46 WmoC Thermal conductivity of insulation=004 WmoC Film coefficient of heat tranfer at inner and outer surface are 200

Wm2 oC and 15 Wm2 oC Ambient air temperature=28 oC

CRITICAL RADIUS

bull The radius upto which heat flow increases and after decreases At critical radius q becomes qmax

bull The addition of insulation always increases (Rth)condbull Rtotal= (Rth)cond + (Rth)conv

bull The addition of insulation may reduce the (Rth)conv due to increase in surface area

bull r1ltrc -heat transfer increases by adding insulation till r1=rc If

insulation thickness is further increased the rate of heat loss will decrease

r1 gt rc

- heat transfer decreases by adding insulation above the rc

Plane wall- Insulation

thickness -study

Cylinder amp Sphere- Insulation thickness -

study

Problem5bull A 3mm diameter and 5 mm long electric wire is tightly

wrapped with a 2 mm thick plastic cover whose K=015 wmoC Electrical measurements indicate that a current of 10 A passes through the wire and there is a voltage drop of 8 V along the wire If the insulated wire is exposed to a medium at Ta=30oC with a heat transfer coefficient of h=12 wm2oC Determine the temperature at the interface of the wire and the plastic cover in steady operation Also determine whether doubling the thickness of the plastic cover will increase or decrease this interface temperature

dx

TdzdyKQ xx )(

dxQx

QQ xxdxx ][

dxdx

TdzdyK

xQQdQ xdxxxx ])([)(

)1()]([)(

ddzdydx

x

TK

xdQ xx

GENERAL HEAT CONDUCTION EQUATION IN CARTESIAN COORDINATES

Consider an unsteady (transient) three dimensional heat conduction with internal heat generation in heterogeneous amp anisotropic PLANE WALL

Cartesian co-ordinates x y z

Elemental length dxdydz

Consider an elemental volume ABCDEFGH = (dxdydz)

Let Temperature distribution is a function of T=T(xyz)

Let

=internal heat generation per unit volume per unit time (W m3)q= rate of heat flow (W)Q= qdt=total heat flow (J)r=mass density of the material element(kgm3)Cp=specific heat of the material element (Jkg K)

gq

Plane (y - z) lsquoxrsquo direction

Quantity of heat leaving from the face EFGH in lsquoXrsquo direction

Quantity of heat flowing into the face ABCD in lsquoXrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoXrsquo direction

dy

TdzdxKQ yy )(

dyQy

QQ yydyy ][

dydy

TdzdxK

yQQdQ ydyyyy ])([)(

)2()]([)(

ddzdydx

y

TK

ydQ yy

Plane (x - z) lsquoyrsquo direction

Quantity of heat flowing into the face ABFE in lsquoYrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoYrsquo direction

dz

TdydxKQ zz )(

dzQz

QQ zzdzz ][

dzdz

TdydxK

zQQdQ zdzzzz ])([)(

)3()]([)(

ddzdydx

z

TK

zdQ zz

Plane (x - y) lsquozrsquo direction

Quantity of heat flowing into the face ADHE in lsquoZrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoZrsquo direction

Quantity of heat leaving from the face DCGH in lsquoYrsquo direction

Quantity of heat leaving from the face BCGF in lsquoZrsquo direction

)4()()(

ddzdydxqQgg

Total amount of internal heat generation in the element

Ein + Egen = Eout+Est (Ein- Eout)+ Egen=Est

T

Cddzdydxddzdydxqddzdydxz

TK

zy

TK

yx

TK

x pgzyx )()())](()()([

T

pC

gq

z

TzK

zy

Ty

Kyx

TxK

x][][][

stgzyx QQdQdQdQ

)(

)5()()(

dT

CdzdydxQ pst

Total heat energy stored in the element

(ie thermal energy of the element can be increased because of heat accumulation in the element due to heat flow amp internal heat generation)

)( volumem d

TmCQ pst )(

T

CqTK pg)(

KKKyKx z If material is homogeneous amp isotropic

T

pC

gq

z

T

zy

T

yx

T

xK ][][][

T

Kp

C

Kgq

z

T

y

T

x

T2

2

2

2

2

2

T

Kgq

z

T

y

T

x

T 12

2

2

2

2

2

Thermal diffusivity pC

K

T

K

qT g

12

according to thermodynamics first law

)( zr )( dzrddr

)( dzrddr

Cylindrical polar co-ordinates

Elemental length

Consider an elemental volume ABCDEFGH =

GENERAL HEAT CONDUCTION EQUATION IN CYLINDRICAL COORDINATES

gq

gq

Consider an unsteady (transient) three dimensional heat conduction with internal heat generation in heterogeneous amp anisotropic cylindrical geometries (eg rods and pipes)

)( zrTLetTemperature distribution is a function of T=

Let

=internal heat generation per unit volume per unit time (W m3)q= rate of heat flow (W)Q= qdt=total heat flow (J)r=mass density of the material element(kgm3)Cp=specific heat of the material element (Jkg K)

planerdirectionz

planezrdirection

planezdirectionr

)(

)(

)(

LetHeat flow in

drdr

TdzrdK

rdQ

QQdQ

rr

drrrr

])([

)1(][

r

Tr

rddrdzdKdQ rr

Heat accumulation in the element due to heat flow by conduction in lsquorrsquo direction

rddr

TdzdrK

rdQ

QQdQ d

])([

][

2

T

r

ddrdzrdKdQ

)2(][

2

2

2

T

r

ddrdzrdKdQ

Heat accumulation in the element due to heat flow by conduction in lsquoqrsquo direction

Plane (z- q ) lsquorrsquo direction

dr

TdzrdKQ rr

)(

drQr

QQ rrdrr ][

Quantity of heat leaving from the face EFGH in lsquorrsquo direction

Quantity of heat flowing into the face ABCD in lsquorrsquo direction

Plane (r-z) lsquoq lsquodirection

Quantity of heat leaving from the face AEGC in lsquoqrsquo direction

d

r

TdzdrKQ )(

rdQ

rQQ d ][

Quantity of heat flowing into the face BFHD in lsquoqrsquo direction

)3())((2

2

z

TdzdrrdKdQ zz

)5()()(

dT

CdzdrrdQ pst

)4()()(

ddzdrrdqQgg

Total amount of internal heat generation in the element

Total heat energy stored in the element

(ie thermal energy of the element can be increased because of heat accumulation in the element due to heat flow amp internal heat generation)

according to thermodynamics first law

Ein + Egen = Eout + Est (Ein ndash Eout) + Egen = Est

)( volumem

)()(z

T

zdzdrrdKdQ zz

Heat accumulation in the element due to heat flow by conduction in lsquozrsquo direction

dzdz

TdrrdK

zdQ

QQdQ

zz

dzzzz

])([

d

TmCQ pst )(

Plane (r-q) lsquozrsquo direction

dz

TdrrdKQ zz )(

dzQz

QQ zzdzz ][ Quantity of heat leaving from the face CDHG in lsquozrsquo direction

Quantity of heat flowing into the face ABFE in lsquozrsquo direction

stgzr QQdQdQdQ

)(

r

r ][)(rT

rr

ddrdzdKr

][)(

2

2

2

T

r

ddrdzrdK ))((2

2

z

TddzdrrdK z

ddzdrrdq g )(

dT

Cdzdrrd p )(

+ + +

=

T

CqTK pg)(KKKK zr

[gq

)(1

r

Tr

rrK r

)(1

2

2

2 T

rK )(

2

2

z

TK z

)( ddzdrrd )(

ddzdrrd

TC p

+ + + ] =

If material is homogeneous amp isotropic

)(1

r

Tr

rr

)(1

2

2

2 T

r)(

2

2

z

T

K

q g

T

K

C p

)(1

r

Tr

rr

)(1

2

2

2 T

r )(2

2

z

T

K

q g

T

1

+ + + =

)(1

r

Tr

rrK r

)(1

2

2

2 T

rK )(

2

2

z

TK z

gq

T

C p + + + =

+ + + = Thermal diffusivity

T

K

qT g

12

pC

K

(Ein ndash Eout) + Egen = Est

)( rT

)( r

)sin( drrddr)sin( drrddr

planerdirection

planerdirection

planedirectionr

)(

)(

)(

Spherical polar co-ordinates

Elemental length

Consider an elemental volume ABCDEFGH =

GENERAL HEAT CONDUCTION EQUATION IN SPHERICAL COORDINATES

Let

=internal heat generation per unit volume per unit time (W m3)q= rate of heat flow (W)Q= qdt=total heat flow (J)r=mass density of the material element(kgm3)Cp=specific heat of the material element (Jkg K)

gq

Consider an unsteady (transient) three dimensional heat conduction with internal heat generation in heterogeneous amp anisotropic spherical geometries

LetTemperature distribution is a function of T=

X

Y

Z

drT

drrdKQ rr )sin(

drQr

QQ rrdrr ][

drdr

TdrrdK

rdQ

QQdQ

rr

drrrr

])sin([

)1(]([sin 2

rT

rr

ddrddKdQ rr

Plane ( -q f) lsquorrsquo directionQuantity of heat flowing into the face ABCD in lsquorrsquo direction

Quantity of heat leaving from the face EFGH in lsquorrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquorrsquo direction

drT

drdrKQ )sin(

rdQ

rQQ d ][

rddr

TdrdrK

rdQ

QQdQ d

])sin([

])[(sin

2

T

r

drdrdrKdQ

)2(])(sin[

2

T

r

drdrddrKdQ

Quantity of heat leaving from the face BFHD in lsquoqrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoqrsquo direction

X

Y

Z

Plane (r- f ) lsquoq lsquodirection

Quantity of heat flowing into the face AEGC in lsquoqrsquo direction

dr

TdrrdKQ

sin)(

drQ

rQQ d sin][

sin

drdr

TdrrdK

rdQ

QQdQ d

sin]sin

)([sin

][sinsin

)sin(

T

rr

ddrdrrdKdQ

)3()(sin

)sin(2

2

22

2

T

r

ddddrrKdQ

Plane (r- ) q lsquofrsquo directionQuantity of heat flowing into the face ABFE in lsquofrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquozrsquo direction

)4()sin()(

ddrrddrqQgg

)5()sin()(

dT

CdrrddrQ pst

Total amount of internal heat generation in the element

Total heat energy stored in the element

(Ein ndash Eout) + Egen = Est

d

TmCQ pst )(

)( volumem

Quantity of heat leaving from the face CDHG in lsquofrsquo direction

according to thermodynamics first law

Ein + Egen = Eout + Est

X

Y

Z

stgr QQdQdQdQ

)(

]([sin 2

r

Tr

rddrddKr

])(sin[

2

T

rdrdrddrK

)(sin

)sin(2

2

22

2

T

r

ddddrrK

ddrrddrq g )sin(

dT

Cdrrddr p )sin(

2

2

r

r

sin

sin

]([ 22 r

Tr

rr

Kr

])(sin[sin2

TrK

)(sin 2

2

22

T

r

Kgq

T

Cp

]([1

[ 22 r

Tr

rrK

])(sin[sin1

2

T

r)](

sin

12

2

22 T

r gq

T

Cp

]([1

[ 22 r

Tr

rr

])(sin[sin1

2

T

r)](

sin

12

2

22 T

rK

q g

T

K

Cp

]([1

[ 22 r

Tr

rr

])(sin[sin1

2

T

r)](

sin

12

2

22 T

rK

q g

T1

KKKKr If material is homogeneous amp isotropic

TCqTK pg)(

T

K

qT g

12 pC

K

Thermal diffusivity

(Ein ndash Eout) + Egen = Est

General Heat Conduction Equation

PLANE WALL CYLINDER SPHERE

Temperature profile

Plane wall SphereCylinder

bull A copper conductor (k = 380 WmoC resistivity ρ = 2x10-8 Ω m) having inner and outer radii 10 cm and 225 cm respectively is carrying a current density of 4800 amperes cm2 the conductor is internally cooled and a constant temperature of 65oC is maintained at the inner surface and there is no heat transfer through the insulation surrounding the conductor

Determinebull (i) The maximum temperature of the conductor and the radius at

which it occursbull (ii) The internal heat transfer rate

bull A chemical reaction takes place in a packed bed (k=06 WmoC) between two coaxial cylinders with radii 15 mm and 45 mm the inner surface is at 580oC and it is insulated Assuming the reaction rate of 055 MWm3 in the reactor volume find the temperature at the outer surface of the reactor

bull A current of 300 amperes passes through a stainless steel wire of 25 mm diameter and k=20 WmoC The resistivity of the wire is 70x10-8 ohm-m and length of the wire is 2 m if the wire is submerged in a fluid maintained at 50oC and h at the wire surface is 4000 Wm2oC calculate the steady state temperature at the centre and at the surface of the wire

bull An approximately spherical shaped orange ( k=023 WmoC) 90 mm diameter undergoes ripping process and generates 5100 Wm3 of energy If external surface the orange is at 8oC determine

bull (i) temperature at the centre of the orange and

bull (ii) heat flow from the outer surface of the orange

bull FINS External surfaces to used to cooling area and hence convection

heat transfer can be improvedbull APPLICATIONS IC engine cylinder ( air cooling) Heat exchangers automobile

radiators computer chip refrigerators electric motors generators and transformers etc

bull ASSUMPTIONS MADE IN ANALYSIS OF HEAT TRANFER THROUGH FIN

1 Fin material is homogenous and isotropic constthermal conductivity

2 Temperature at any cs of the fin is uniform3 lsquohrsquo is uniform over entire surface4 No internal heat generation5 Thermal contact resistance is negligible6 Radiation heat transfer is negligible7 Heat conduction is steady state and one dimensional

bull Problem 1

bull A stainless steel fin (k=20 WmK) having a diameter of 20 mm and a length of 01 m is attached to a wall at 300oC The ambient temperature is 50oC and the heat transfer coefficient is 10 Wm2K The fin tip is insulated Determine

bull (A) The rate of heat dissipation from the finbull (B) The temperature at the fin tipbull (C) The rate of heat transfer from the wall area covered by the fin if

the fin was not usedbull (D) The heat transfer rate from the same fin geometry if the

stainless steel fin is replaced by a fictitious fin with infinite thermal conductivity

bull Problem2bull One end of a long rod is inserted into a furnace while the other

projects into ambient temperature Under steady state the temperature of the rod is measured at two points 75mm apart and found to be 125oC and 885oC respectively while the ambient temperature is 20oC If the rod is 25mm in diameter and h is 2336 Wm2K

bull Find the thermal conductivity of the rod material

  • HEAT AND MASS TRANSFER
  • Thermodynamics Vs Heat transfer
  • MODE OF HEAT TRANSFER
  • MODE OF HEAT TRANSFER (2)
  • CONDUCTION- MECHANISM
  • CONCECTION-MECHANISM
  • RADIATION-MECHANISM
  • Conduction- Convection- Radiation
  • Slide 9
  • Slide 10
  • Fourierrsquos Law of Heat Conduction
  • Slide 12
  • Newtonrsquos Law of Cooling
  • Laws of Thermal Radiation
  • LAWS FOR HEAT TRANSFER
  • ELECTRICAL ampTHERMAL ANALOGY
  • Slide 17
  • Slide 18
  • PLANE WALL-UNIDIRECTIONAL
  • PLANE WALL -SERIES
  • PLANE WALL IN SERIES
  • PLANE WALL -PARALLEL
  • PLANE WALL-PARALLEL
  • PLANE WALL-COMBINATION
  • PLANE WALL-FILM RESISTANCE
  • PLANE WALL- THERMAL CONTACT RESISTANCE
  • PLANE WALL- THERMAL CONTACT RESISTANCE (2)
  • CYLINDER-SOLID
  • HEAT TRANSFER IN SOLID CYLINDER
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER-COMPOSITE
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER-MULTY LAYER
  • HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER
  • HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER (2)
  • HEAT TRANSFER IN SPHERE
  • HEAT TRANSFER IN SPHERE -MULTY LAYER
  • HEAT TRANSFER IN SPHERE -MULTY LAYER (2)
  • Slide 38
  • Slide 39
  • Slide 40
  • Slide 41
  • CRITICAL RADIUS
  • Plane wall- Insulation thickness -study
  • Problem5
  • Slide 45
  • Slide 46
  • Slide 47
  • GENERAL HEAT CONDUCTION EQUATION IN CYLINDRICAL COORDINATES
  • Slide 49
  • Slide 50
  • Slide 51
  • GENERAL HEAT CONDUCTION EQUATION IN SPHERICAL COORDINATES
  • Slide 53
  • Slide 54
  • Slide 55
  • General Heat Conduction Equation
  • Temperature profile
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Slide 67
Page 38: hmt chap 1

CRITICAL RADIUS

bull The radius upto which heat flow increases and after decreases At critical radius q becomes qmax

bull The addition of insulation always increases (Rth)condbull Rtotal= (Rth)cond + (Rth)conv

bull The addition of insulation may reduce the (Rth)conv due to increase in surface area

bull r1ltrc -heat transfer increases by adding insulation till r1=rc If

insulation thickness is further increased the rate of heat loss will decrease

r1 gt rc

- heat transfer decreases by adding insulation above the rc

Plane wall- Insulation

thickness -study

Cylinder amp Sphere- Insulation thickness -

study

Problem5bull A 3mm diameter and 5 mm long electric wire is tightly

wrapped with a 2 mm thick plastic cover whose K=015 wmoC Electrical measurements indicate that a current of 10 A passes through the wire and there is a voltage drop of 8 V along the wire If the insulated wire is exposed to a medium at Ta=30oC with a heat transfer coefficient of h=12 wm2oC Determine the temperature at the interface of the wire and the plastic cover in steady operation Also determine whether doubling the thickness of the plastic cover will increase or decrease this interface temperature

dx

TdzdyKQ xx )(

dxQx

QQ xxdxx ][

dxdx

TdzdyK

xQQdQ xdxxxx ])([)(

)1()]([)(

ddzdydx

x

TK

xdQ xx

GENERAL HEAT CONDUCTION EQUATION IN CARTESIAN COORDINATES

Consider an unsteady (transient) three dimensional heat conduction with internal heat generation in heterogeneous amp anisotropic PLANE WALL

Cartesian co-ordinates x y z

Elemental length dxdydz

Consider an elemental volume ABCDEFGH = (dxdydz)

Let Temperature distribution is a function of T=T(xyz)

Let

=internal heat generation per unit volume per unit time (W m3)q= rate of heat flow (W)Q= qdt=total heat flow (J)r=mass density of the material element(kgm3)Cp=specific heat of the material element (Jkg K)

gq

Plane (y - z) lsquoxrsquo direction

Quantity of heat leaving from the face EFGH in lsquoXrsquo direction

Quantity of heat flowing into the face ABCD in lsquoXrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoXrsquo direction

dy

TdzdxKQ yy )(

dyQy

QQ yydyy ][

dydy

TdzdxK

yQQdQ ydyyyy ])([)(

)2()]([)(

ddzdydx

y

TK

ydQ yy

Plane (x - z) lsquoyrsquo direction

Quantity of heat flowing into the face ABFE in lsquoYrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoYrsquo direction

dz

TdydxKQ zz )(

dzQz

QQ zzdzz ][

dzdz

TdydxK

zQQdQ zdzzzz ])([)(

)3()]([)(

ddzdydx

z

TK

zdQ zz

Plane (x - y) lsquozrsquo direction

Quantity of heat flowing into the face ADHE in lsquoZrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoZrsquo direction

Quantity of heat leaving from the face DCGH in lsquoYrsquo direction

Quantity of heat leaving from the face BCGF in lsquoZrsquo direction

)4()()(

ddzdydxqQgg

Total amount of internal heat generation in the element

Ein + Egen = Eout+Est (Ein- Eout)+ Egen=Est

T

Cddzdydxddzdydxqddzdydxz

TK

zy

TK

yx

TK

x pgzyx )()())](()()([

T

pC

gq

z

TzK

zy

Ty

Kyx

TxK

x][][][

stgzyx QQdQdQdQ

)(

)5()()(

dT

CdzdydxQ pst

Total heat energy stored in the element

(ie thermal energy of the element can be increased because of heat accumulation in the element due to heat flow amp internal heat generation)

)( volumem d

TmCQ pst )(

T

CqTK pg)(

KKKyKx z If material is homogeneous amp isotropic

T

pC

gq

z

T

zy

T

yx

T

xK ][][][

T

Kp

C

Kgq

z

T

y

T

x

T2

2

2

2

2

2

T

Kgq

z

T

y

T

x

T 12

2

2

2

2

2

Thermal diffusivity pC

K

T

K

qT g

12

according to thermodynamics first law

)( zr )( dzrddr

)( dzrddr

Cylindrical polar co-ordinates

Elemental length

Consider an elemental volume ABCDEFGH =

GENERAL HEAT CONDUCTION EQUATION IN CYLINDRICAL COORDINATES

gq

gq

Consider an unsteady (transient) three dimensional heat conduction with internal heat generation in heterogeneous amp anisotropic cylindrical geometries (eg rods and pipes)

)( zrTLetTemperature distribution is a function of T=

Let

=internal heat generation per unit volume per unit time (W m3)q= rate of heat flow (W)Q= qdt=total heat flow (J)r=mass density of the material element(kgm3)Cp=specific heat of the material element (Jkg K)

planerdirectionz

planezrdirection

planezdirectionr

)(

)(

)(

LetHeat flow in

drdr

TdzrdK

rdQ

QQdQ

rr

drrrr

])([

)1(][

r

Tr

rddrdzdKdQ rr

Heat accumulation in the element due to heat flow by conduction in lsquorrsquo direction

rddr

TdzdrK

rdQ

QQdQ d

])([

][

2

T

r

ddrdzrdKdQ

)2(][

2

2

2

T

r

ddrdzrdKdQ

Heat accumulation in the element due to heat flow by conduction in lsquoqrsquo direction

Plane (z- q ) lsquorrsquo direction

dr

TdzrdKQ rr

)(

drQr

QQ rrdrr ][

Quantity of heat leaving from the face EFGH in lsquorrsquo direction

Quantity of heat flowing into the face ABCD in lsquorrsquo direction

Plane (r-z) lsquoq lsquodirection

Quantity of heat leaving from the face AEGC in lsquoqrsquo direction

d

r

TdzdrKQ )(

rdQ

rQQ d ][

Quantity of heat flowing into the face BFHD in lsquoqrsquo direction

)3())((2

2

z

TdzdrrdKdQ zz

)5()()(

dT

CdzdrrdQ pst

)4()()(

ddzdrrdqQgg

Total amount of internal heat generation in the element

Total heat energy stored in the element

(ie thermal energy of the element can be increased because of heat accumulation in the element due to heat flow amp internal heat generation)

according to thermodynamics first law

Ein + Egen = Eout + Est (Ein ndash Eout) + Egen = Est

)( volumem

)()(z

T

zdzdrrdKdQ zz

Heat accumulation in the element due to heat flow by conduction in lsquozrsquo direction

dzdz

TdrrdK

zdQ

QQdQ

zz

dzzzz

])([

d

TmCQ pst )(

Plane (r-q) lsquozrsquo direction

dz

TdrrdKQ zz )(

dzQz

QQ zzdzz ][ Quantity of heat leaving from the face CDHG in lsquozrsquo direction

Quantity of heat flowing into the face ABFE in lsquozrsquo direction

stgzr QQdQdQdQ

)(

r

r ][)(rT

rr

ddrdzdKr

][)(

2

2

2

T

r

ddrdzrdK ))((2

2

z

TddzdrrdK z

ddzdrrdq g )(

dT

Cdzdrrd p )(

+ + +

=

T

CqTK pg)(KKKK zr

[gq

)(1

r

Tr

rrK r

)(1

2

2

2 T

rK )(

2

2

z

TK z

)( ddzdrrd )(

ddzdrrd

TC p

+ + + ] =

If material is homogeneous amp isotropic

)(1

r

Tr

rr

)(1

2

2

2 T

r)(

2

2

z

T

K

q g

T

K

C p

)(1

r

Tr

rr

)(1

2

2

2 T

r )(2

2

z

T

K

q g

T

1

+ + + =

)(1

r

Tr

rrK r

)(1

2

2

2 T

rK )(

2

2

z

TK z

gq

T

C p + + + =

+ + + = Thermal diffusivity

T

K

qT g

12

pC

K

(Ein ndash Eout) + Egen = Est

)( rT

)( r

)sin( drrddr)sin( drrddr

planerdirection

planerdirection

planedirectionr

)(

)(

)(

Spherical polar co-ordinates

Elemental length

Consider an elemental volume ABCDEFGH =

GENERAL HEAT CONDUCTION EQUATION IN SPHERICAL COORDINATES

Let

=internal heat generation per unit volume per unit time (W m3)q= rate of heat flow (W)Q= qdt=total heat flow (J)r=mass density of the material element(kgm3)Cp=specific heat of the material element (Jkg K)

gq

Consider an unsteady (transient) three dimensional heat conduction with internal heat generation in heterogeneous amp anisotropic spherical geometries

LetTemperature distribution is a function of T=

X

Y

Z

drT

drrdKQ rr )sin(

drQr

QQ rrdrr ][

drdr

TdrrdK

rdQ

QQdQ

rr

drrrr

])sin([

)1(]([sin 2

rT

rr

ddrddKdQ rr

Plane ( -q f) lsquorrsquo directionQuantity of heat flowing into the face ABCD in lsquorrsquo direction

Quantity of heat leaving from the face EFGH in lsquorrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquorrsquo direction

drT

drdrKQ )sin(

rdQ

rQQ d ][

rddr

TdrdrK

rdQ

QQdQ d

])sin([

])[(sin

2

T

r

drdrdrKdQ

)2(])(sin[

2

T

r

drdrddrKdQ

Quantity of heat leaving from the face BFHD in lsquoqrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoqrsquo direction

X

Y

Z

Plane (r- f ) lsquoq lsquodirection

Quantity of heat flowing into the face AEGC in lsquoqrsquo direction

dr

TdrrdKQ

sin)(

drQ

rQQ d sin][

sin

drdr

TdrrdK

rdQ

QQdQ d

sin]sin

)([sin

][sinsin

)sin(

T

rr

ddrdrrdKdQ

)3()(sin

)sin(2

2

22

2

T

r

ddddrrKdQ

Plane (r- ) q lsquofrsquo directionQuantity of heat flowing into the face ABFE in lsquofrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquozrsquo direction

)4()sin()(

ddrrddrqQgg

)5()sin()(

dT

CdrrddrQ pst

Total amount of internal heat generation in the element

Total heat energy stored in the element

(Ein ndash Eout) + Egen = Est

d

TmCQ pst )(

)( volumem

Quantity of heat leaving from the face CDHG in lsquofrsquo direction

according to thermodynamics first law

Ein + Egen = Eout + Est

X

Y

Z

stgr QQdQdQdQ

)(

]([sin 2

r

Tr

rddrddKr

])(sin[

2

T

rdrdrddrK

)(sin

)sin(2

2

22

2

T

r

ddddrrK

ddrrddrq g )sin(

dT

Cdrrddr p )sin(

2

2

r

r

sin

sin

]([ 22 r

Tr

rr

Kr

])(sin[sin2

TrK

)(sin 2

2

22

T

r

Kgq

T

Cp

]([1

[ 22 r

Tr

rrK

])(sin[sin1

2

T

r)](

sin

12

2

22 T

r gq

T

Cp

]([1

[ 22 r

Tr

rr

])(sin[sin1

2

T

r)](

sin

12

2

22 T

rK

q g

T

K

Cp

]([1

[ 22 r

Tr

rr

])(sin[sin1

2

T

r)](

sin

12

2

22 T

rK

q g

T1

KKKKr If material is homogeneous amp isotropic

TCqTK pg)(

T

K

qT g

12 pC

K

Thermal diffusivity

(Ein ndash Eout) + Egen = Est

General Heat Conduction Equation

PLANE WALL CYLINDER SPHERE

Temperature profile

Plane wall SphereCylinder

bull A copper conductor (k = 380 WmoC resistivity ρ = 2x10-8 Ω m) having inner and outer radii 10 cm and 225 cm respectively is carrying a current density of 4800 amperes cm2 the conductor is internally cooled and a constant temperature of 65oC is maintained at the inner surface and there is no heat transfer through the insulation surrounding the conductor

Determinebull (i) The maximum temperature of the conductor and the radius at

which it occursbull (ii) The internal heat transfer rate

bull A chemical reaction takes place in a packed bed (k=06 WmoC) between two coaxial cylinders with radii 15 mm and 45 mm the inner surface is at 580oC and it is insulated Assuming the reaction rate of 055 MWm3 in the reactor volume find the temperature at the outer surface of the reactor

bull A current of 300 amperes passes through a stainless steel wire of 25 mm diameter and k=20 WmoC The resistivity of the wire is 70x10-8 ohm-m and length of the wire is 2 m if the wire is submerged in a fluid maintained at 50oC and h at the wire surface is 4000 Wm2oC calculate the steady state temperature at the centre and at the surface of the wire

bull An approximately spherical shaped orange ( k=023 WmoC) 90 mm diameter undergoes ripping process and generates 5100 Wm3 of energy If external surface the orange is at 8oC determine

bull (i) temperature at the centre of the orange and

bull (ii) heat flow from the outer surface of the orange

bull FINS External surfaces to used to cooling area and hence convection

heat transfer can be improvedbull APPLICATIONS IC engine cylinder ( air cooling) Heat exchangers automobile

radiators computer chip refrigerators electric motors generators and transformers etc

bull ASSUMPTIONS MADE IN ANALYSIS OF HEAT TRANFER THROUGH FIN

1 Fin material is homogenous and isotropic constthermal conductivity

2 Temperature at any cs of the fin is uniform3 lsquohrsquo is uniform over entire surface4 No internal heat generation5 Thermal contact resistance is negligible6 Radiation heat transfer is negligible7 Heat conduction is steady state and one dimensional

bull Problem 1

bull A stainless steel fin (k=20 WmK) having a diameter of 20 mm and a length of 01 m is attached to a wall at 300oC The ambient temperature is 50oC and the heat transfer coefficient is 10 Wm2K The fin tip is insulated Determine

bull (A) The rate of heat dissipation from the finbull (B) The temperature at the fin tipbull (C) The rate of heat transfer from the wall area covered by the fin if

the fin was not usedbull (D) The heat transfer rate from the same fin geometry if the

stainless steel fin is replaced by a fictitious fin with infinite thermal conductivity

bull Problem2bull One end of a long rod is inserted into a furnace while the other

projects into ambient temperature Under steady state the temperature of the rod is measured at two points 75mm apart and found to be 125oC and 885oC respectively while the ambient temperature is 20oC If the rod is 25mm in diameter and h is 2336 Wm2K

bull Find the thermal conductivity of the rod material

  • HEAT AND MASS TRANSFER
  • Thermodynamics Vs Heat transfer
  • MODE OF HEAT TRANSFER
  • MODE OF HEAT TRANSFER (2)
  • CONDUCTION- MECHANISM
  • CONCECTION-MECHANISM
  • RADIATION-MECHANISM
  • Conduction- Convection- Radiation
  • Slide 9
  • Slide 10
  • Fourierrsquos Law of Heat Conduction
  • Slide 12
  • Newtonrsquos Law of Cooling
  • Laws of Thermal Radiation
  • LAWS FOR HEAT TRANSFER
  • ELECTRICAL ampTHERMAL ANALOGY
  • Slide 17
  • Slide 18
  • PLANE WALL-UNIDIRECTIONAL
  • PLANE WALL -SERIES
  • PLANE WALL IN SERIES
  • PLANE WALL -PARALLEL
  • PLANE WALL-PARALLEL
  • PLANE WALL-COMBINATION
  • PLANE WALL-FILM RESISTANCE
  • PLANE WALL- THERMAL CONTACT RESISTANCE
  • PLANE WALL- THERMAL CONTACT RESISTANCE (2)
  • CYLINDER-SOLID
  • HEAT TRANSFER IN SOLID CYLINDER
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER-COMPOSITE
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER-MULTY LAYER
  • HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER
  • HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER (2)
  • HEAT TRANSFER IN SPHERE
  • HEAT TRANSFER IN SPHERE -MULTY LAYER
  • HEAT TRANSFER IN SPHERE -MULTY LAYER (2)
  • Slide 38
  • Slide 39
  • Slide 40
  • Slide 41
  • CRITICAL RADIUS
  • Plane wall- Insulation thickness -study
  • Problem5
  • Slide 45
  • Slide 46
  • Slide 47
  • GENERAL HEAT CONDUCTION EQUATION IN CYLINDRICAL COORDINATES
  • Slide 49
  • Slide 50
  • Slide 51
  • GENERAL HEAT CONDUCTION EQUATION IN SPHERICAL COORDINATES
  • Slide 53
  • Slide 54
  • Slide 55
  • General Heat Conduction Equation
  • Temperature profile
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Slide 67
Page 39: hmt chap 1

Plane wall- Insulation

thickness -study

Cylinder amp Sphere- Insulation thickness -

study

Problem5bull A 3mm diameter and 5 mm long electric wire is tightly

wrapped with a 2 mm thick plastic cover whose K=015 wmoC Electrical measurements indicate that a current of 10 A passes through the wire and there is a voltage drop of 8 V along the wire If the insulated wire is exposed to a medium at Ta=30oC with a heat transfer coefficient of h=12 wm2oC Determine the temperature at the interface of the wire and the plastic cover in steady operation Also determine whether doubling the thickness of the plastic cover will increase or decrease this interface temperature

dx

TdzdyKQ xx )(

dxQx

QQ xxdxx ][

dxdx

TdzdyK

xQQdQ xdxxxx ])([)(

)1()]([)(

ddzdydx

x

TK

xdQ xx

GENERAL HEAT CONDUCTION EQUATION IN CARTESIAN COORDINATES

Consider an unsteady (transient) three dimensional heat conduction with internal heat generation in heterogeneous amp anisotropic PLANE WALL

Cartesian co-ordinates x y z

Elemental length dxdydz

Consider an elemental volume ABCDEFGH = (dxdydz)

Let Temperature distribution is a function of T=T(xyz)

Let

=internal heat generation per unit volume per unit time (W m3)q= rate of heat flow (W)Q= qdt=total heat flow (J)r=mass density of the material element(kgm3)Cp=specific heat of the material element (Jkg K)

gq

Plane (y - z) lsquoxrsquo direction

Quantity of heat leaving from the face EFGH in lsquoXrsquo direction

Quantity of heat flowing into the face ABCD in lsquoXrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoXrsquo direction

dy

TdzdxKQ yy )(

dyQy

QQ yydyy ][

dydy

TdzdxK

yQQdQ ydyyyy ])([)(

)2()]([)(

ddzdydx

y

TK

ydQ yy

Plane (x - z) lsquoyrsquo direction

Quantity of heat flowing into the face ABFE in lsquoYrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoYrsquo direction

dz

TdydxKQ zz )(

dzQz

QQ zzdzz ][

dzdz

TdydxK

zQQdQ zdzzzz ])([)(

)3()]([)(

ddzdydx

z

TK

zdQ zz

Plane (x - y) lsquozrsquo direction

Quantity of heat flowing into the face ADHE in lsquoZrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoZrsquo direction

Quantity of heat leaving from the face DCGH in lsquoYrsquo direction

Quantity of heat leaving from the face BCGF in lsquoZrsquo direction

)4()()(

ddzdydxqQgg

Total amount of internal heat generation in the element

Ein + Egen = Eout+Est (Ein- Eout)+ Egen=Est

T

Cddzdydxddzdydxqddzdydxz

TK

zy

TK

yx

TK

x pgzyx )()())](()()([

T

pC

gq

z

TzK

zy

Ty

Kyx

TxK

x][][][

stgzyx QQdQdQdQ

)(

)5()()(

dT

CdzdydxQ pst

Total heat energy stored in the element

(ie thermal energy of the element can be increased because of heat accumulation in the element due to heat flow amp internal heat generation)

)( volumem d

TmCQ pst )(

T

CqTK pg)(

KKKyKx z If material is homogeneous amp isotropic

T

pC

gq

z

T

zy

T

yx

T

xK ][][][

T

Kp

C

Kgq

z

T

y

T

x

T2

2

2

2

2

2

T

Kgq

z

T

y

T

x

T 12

2

2

2

2

2

Thermal diffusivity pC

K

T

K

qT g

12

according to thermodynamics first law

)( zr )( dzrddr

)( dzrddr

Cylindrical polar co-ordinates

Elemental length

Consider an elemental volume ABCDEFGH =

GENERAL HEAT CONDUCTION EQUATION IN CYLINDRICAL COORDINATES

gq

gq

Consider an unsteady (transient) three dimensional heat conduction with internal heat generation in heterogeneous amp anisotropic cylindrical geometries (eg rods and pipes)

)( zrTLetTemperature distribution is a function of T=

Let

=internal heat generation per unit volume per unit time (W m3)q= rate of heat flow (W)Q= qdt=total heat flow (J)r=mass density of the material element(kgm3)Cp=specific heat of the material element (Jkg K)

planerdirectionz

planezrdirection

planezdirectionr

)(

)(

)(

LetHeat flow in

drdr

TdzrdK

rdQ

QQdQ

rr

drrrr

])([

)1(][

r

Tr

rddrdzdKdQ rr

Heat accumulation in the element due to heat flow by conduction in lsquorrsquo direction

rddr

TdzdrK

rdQ

QQdQ d

])([

][

2

T

r

ddrdzrdKdQ

)2(][

2

2

2

T

r

ddrdzrdKdQ

Heat accumulation in the element due to heat flow by conduction in lsquoqrsquo direction

Plane (z- q ) lsquorrsquo direction

dr

TdzrdKQ rr

)(

drQr

QQ rrdrr ][

Quantity of heat leaving from the face EFGH in lsquorrsquo direction

Quantity of heat flowing into the face ABCD in lsquorrsquo direction

Plane (r-z) lsquoq lsquodirection

Quantity of heat leaving from the face AEGC in lsquoqrsquo direction

d

r

TdzdrKQ )(

rdQ

rQQ d ][

Quantity of heat flowing into the face BFHD in lsquoqrsquo direction

)3())((2

2

z

TdzdrrdKdQ zz

)5()()(

dT

CdzdrrdQ pst

)4()()(

ddzdrrdqQgg

Total amount of internal heat generation in the element

Total heat energy stored in the element

(ie thermal energy of the element can be increased because of heat accumulation in the element due to heat flow amp internal heat generation)

according to thermodynamics first law

Ein + Egen = Eout + Est (Ein ndash Eout) + Egen = Est

)( volumem

)()(z

T

zdzdrrdKdQ zz

Heat accumulation in the element due to heat flow by conduction in lsquozrsquo direction

dzdz

TdrrdK

zdQ

QQdQ

zz

dzzzz

])([

d

TmCQ pst )(

Plane (r-q) lsquozrsquo direction

dz

TdrrdKQ zz )(

dzQz

QQ zzdzz ][ Quantity of heat leaving from the face CDHG in lsquozrsquo direction

Quantity of heat flowing into the face ABFE in lsquozrsquo direction

stgzr QQdQdQdQ

)(

r

r ][)(rT

rr

ddrdzdKr

][)(

2

2

2

T

r

ddrdzrdK ))((2

2

z

TddzdrrdK z

ddzdrrdq g )(

dT

Cdzdrrd p )(

+ + +

=

T

CqTK pg)(KKKK zr

[gq

)(1

r

Tr

rrK r

)(1

2

2

2 T

rK )(

2

2

z

TK z

)( ddzdrrd )(

ddzdrrd

TC p

+ + + ] =

If material is homogeneous amp isotropic

)(1

r

Tr

rr

)(1

2

2

2 T

r)(

2

2

z

T

K

q g

T

K

C p

)(1

r

Tr

rr

)(1

2

2

2 T

r )(2

2

z

T

K

q g

T

1

+ + + =

)(1

r

Tr

rrK r

)(1

2

2

2 T

rK )(

2

2

z

TK z

gq

T

C p + + + =

+ + + = Thermal diffusivity

T

K

qT g

12

pC

K

(Ein ndash Eout) + Egen = Est

)( rT

)( r

)sin( drrddr)sin( drrddr

planerdirection

planerdirection

planedirectionr

)(

)(

)(

Spherical polar co-ordinates

Elemental length

Consider an elemental volume ABCDEFGH =

GENERAL HEAT CONDUCTION EQUATION IN SPHERICAL COORDINATES

Let

=internal heat generation per unit volume per unit time (W m3)q= rate of heat flow (W)Q= qdt=total heat flow (J)r=mass density of the material element(kgm3)Cp=specific heat of the material element (Jkg K)

gq

Consider an unsteady (transient) three dimensional heat conduction with internal heat generation in heterogeneous amp anisotropic spherical geometries

LetTemperature distribution is a function of T=

X

Y

Z

drT

drrdKQ rr )sin(

drQr

QQ rrdrr ][

drdr

TdrrdK

rdQ

QQdQ

rr

drrrr

])sin([

)1(]([sin 2

rT

rr

ddrddKdQ rr

Plane ( -q f) lsquorrsquo directionQuantity of heat flowing into the face ABCD in lsquorrsquo direction

Quantity of heat leaving from the face EFGH in lsquorrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquorrsquo direction

drT

drdrKQ )sin(

rdQ

rQQ d ][

rddr

TdrdrK

rdQ

QQdQ d

])sin([

])[(sin

2

T

r

drdrdrKdQ

)2(])(sin[

2

T

r

drdrddrKdQ

Quantity of heat leaving from the face BFHD in lsquoqrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoqrsquo direction

X

Y

Z

Plane (r- f ) lsquoq lsquodirection

Quantity of heat flowing into the face AEGC in lsquoqrsquo direction

dr

TdrrdKQ

sin)(

drQ

rQQ d sin][

sin

drdr

TdrrdK

rdQ

QQdQ d

sin]sin

)([sin

][sinsin

)sin(

T

rr

ddrdrrdKdQ

)3()(sin

)sin(2

2

22

2

T

r

ddddrrKdQ

Plane (r- ) q lsquofrsquo directionQuantity of heat flowing into the face ABFE in lsquofrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquozrsquo direction

)4()sin()(

ddrrddrqQgg

)5()sin()(

dT

CdrrddrQ pst

Total amount of internal heat generation in the element

Total heat energy stored in the element

(Ein ndash Eout) + Egen = Est

d

TmCQ pst )(

)( volumem

Quantity of heat leaving from the face CDHG in lsquofrsquo direction

according to thermodynamics first law

Ein + Egen = Eout + Est

X

Y

Z

stgr QQdQdQdQ

)(

]([sin 2

r

Tr

rddrddKr

])(sin[

2

T

rdrdrddrK

)(sin

)sin(2

2

22

2

T

r

ddddrrK

ddrrddrq g )sin(

dT

Cdrrddr p )sin(

2

2

r

r

sin

sin

]([ 22 r

Tr

rr

Kr

])(sin[sin2

TrK

)(sin 2

2

22

T

r

Kgq

T

Cp

]([1

[ 22 r

Tr

rrK

])(sin[sin1

2

T

r)](

sin

12

2

22 T

r gq

T

Cp

]([1

[ 22 r

Tr

rr

])(sin[sin1

2

T

r)](

sin

12

2

22 T

rK

q g

T

K

Cp

]([1

[ 22 r

Tr

rr

])(sin[sin1

2

T

r)](

sin

12

2

22 T

rK

q g

T1

KKKKr If material is homogeneous amp isotropic

TCqTK pg)(

T

K

qT g

12 pC

K

Thermal diffusivity

(Ein ndash Eout) + Egen = Est

General Heat Conduction Equation

PLANE WALL CYLINDER SPHERE

Temperature profile

Plane wall SphereCylinder

bull A copper conductor (k = 380 WmoC resistivity ρ = 2x10-8 Ω m) having inner and outer radii 10 cm and 225 cm respectively is carrying a current density of 4800 amperes cm2 the conductor is internally cooled and a constant temperature of 65oC is maintained at the inner surface and there is no heat transfer through the insulation surrounding the conductor

Determinebull (i) The maximum temperature of the conductor and the radius at

which it occursbull (ii) The internal heat transfer rate

bull A chemical reaction takes place in a packed bed (k=06 WmoC) between two coaxial cylinders with radii 15 mm and 45 mm the inner surface is at 580oC and it is insulated Assuming the reaction rate of 055 MWm3 in the reactor volume find the temperature at the outer surface of the reactor

bull A current of 300 amperes passes through a stainless steel wire of 25 mm diameter and k=20 WmoC The resistivity of the wire is 70x10-8 ohm-m and length of the wire is 2 m if the wire is submerged in a fluid maintained at 50oC and h at the wire surface is 4000 Wm2oC calculate the steady state temperature at the centre and at the surface of the wire

bull An approximately spherical shaped orange ( k=023 WmoC) 90 mm diameter undergoes ripping process and generates 5100 Wm3 of energy If external surface the orange is at 8oC determine

bull (i) temperature at the centre of the orange and

bull (ii) heat flow from the outer surface of the orange

bull FINS External surfaces to used to cooling area and hence convection

heat transfer can be improvedbull APPLICATIONS IC engine cylinder ( air cooling) Heat exchangers automobile

radiators computer chip refrigerators electric motors generators and transformers etc

bull ASSUMPTIONS MADE IN ANALYSIS OF HEAT TRANFER THROUGH FIN

1 Fin material is homogenous and isotropic constthermal conductivity

2 Temperature at any cs of the fin is uniform3 lsquohrsquo is uniform over entire surface4 No internal heat generation5 Thermal contact resistance is negligible6 Radiation heat transfer is negligible7 Heat conduction is steady state and one dimensional

bull Problem 1

bull A stainless steel fin (k=20 WmK) having a diameter of 20 mm and a length of 01 m is attached to a wall at 300oC The ambient temperature is 50oC and the heat transfer coefficient is 10 Wm2K The fin tip is insulated Determine

bull (A) The rate of heat dissipation from the finbull (B) The temperature at the fin tipbull (C) The rate of heat transfer from the wall area covered by the fin if

the fin was not usedbull (D) The heat transfer rate from the same fin geometry if the

stainless steel fin is replaced by a fictitious fin with infinite thermal conductivity

bull Problem2bull One end of a long rod is inserted into a furnace while the other

projects into ambient temperature Under steady state the temperature of the rod is measured at two points 75mm apart and found to be 125oC and 885oC respectively while the ambient temperature is 20oC If the rod is 25mm in diameter and h is 2336 Wm2K

bull Find the thermal conductivity of the rod material

  • HEAT AND MASS TRANSFER
  • Thermodynamics Vs Heat transfer
  • MODE OF HEAT TRANSFER
  • MODE OF HEAT TRANSFER (2)
  • CONDUCTION- MECHANISM
  • CONCECTION-MECHANISM
  • RADIATION-MECHANISM
  • Conduction- Convection- Radiation
  • Slide 9
  • Slide 10
  • Fourierrsquos Law of Heat Conduction
  • Slide 12
  • Newtonrsquos Law of Cooling
  • Laws of Thermal Radiation
  • LAWS FOR HEAT TRANSFER
  • ELECTRICAL ampTHERMAL ANALOGY
  • Slide 17
  • Slide 18
  • PLANE WALL-UNIDIRECTIONAL
  • PLANE WALL -SERIES
  • PLANE WALL IN SERIES
  • PLANE WALL -PARALLEL
  • PLANE WALL-PARALLEL
  • PLANE WALL-COMBINATION
  • PLANE WALL-FILM RESISTANCE
  • PLANE WALL- THERMAL CONTACT RESISTANCE
  • PLANE WALL- THERMAL CONTACT RESISTANCE (2)
  • CYLINDER-SOLID
  • HEAT TRANSFER IN SOLID CYLINDER
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER-COMPOSITE
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER-MULTY LAYER
  • HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER
  • HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER (2)
  • HEAT TRANSFER IN SPHERE
  • HEAT TRANSFER IN SPHERE -MULTY LAYER
  • HEAT TRANSFER IN SPHERE -MULTY LAYER (2)
  • Slide 38
  • Slide 39
  • Slide 40
  • Slide 41
  • CRITICAL RADIUS
  • Plane wall- Insulation thickness -study
  • Problem5
  • Slide 45
  • Slide 46
  • Slide 47
  • GENERAL HEAT CONDUCTION EQUATION IN CYLINDRICAL COORDINATES
  • Slide 49
  • Slide 50
  • Slide 51
  • GENERAL HEAT CONDUCTION EQUATION IN SPHERICAL COORDINATES
  • Slide 53
  • Slide 54
  • Slide 55
  • General Heat Conduction Equation
  • Temperature profile
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Slide 67
Page 40: hmt chap 1

Problem5bull A 3mm diameter and 5 mm long electric wire is tightly

wrapped with a 2 mm thick plastic cover whose K=015 wmoC Electrical measurements indicate that a current of 10 A passes through the wire and there is a voltage drop of 8 V along the wire If the insulated wire is exposed to a medium at Ta=30oC with a heat transfer coefficient of h=12 wm2oC Determine the temperature at the interface of the wire and the plastic cover in steady operation Also determine whether doubling the thickness of the plastic cover will increase or decrease this interface temperature

dx

TdzdyKQ xx )(

dxQx

QQ xxdxx ][

dxdx

TdzdyK

xQQdQ xdxxxx ])([)(

)1()]([)(

ddzdydx

x

TK

xdQ xx

GENERAL HEAT CONDUCTION EQUATION IN CARTESIAN COORDINATES

Consider an unsteady (transient) three dimensional heat conduction with internal heat generation in heterogeneous amp anisotropic PLANE WALL

Cartesian co-ordinates x y z

Elemental length dxdydz

Consider an elemental volume ABCDEFGH = (dxdydz)

Let Temperature distribution is a function of T=T(xyz)

Let

=internal heat generation per unit volume per unit time (W m3)q= rate of heat flow (W)Q= qdt=total heat flow (J)r=mass density of the material element(kgm3)Cp=specific heat of the material element (Jkg K)

gq

Plane (y - z) lsquoxrsquo direction

Quantity of heat leaving from the face EFGH in lsquoXrsquo direction

Quantity of heat flowing into the face ABCD in lsquoXrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoXrsquo direction

dy

TdzdxKQ yy )(

dyQy

QQ yydyy ][

dydy

TdzdxK

yQQdQ ydyyyy ])([)(

)2()]([)(

ddzdydx

y

TK

ydQ yy

Plane (x - z) lsquoyrsquo direction

Quantity of heat flowing into the face ABFE in lsquoYrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoYrsquo direction

dz

TdydxKQ zz )(

dzQz

QQ zzdzz ][

dzdz

TdydxK

zQQdQ zdzzzz ])([)(

)3()]([)(

ddzdydx

z

TK

zdQ zz

Plane (x - y) lsquozrsquo direction

Quantity of heat flowing into the face ADHE in lsquoZrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoZrsquo direction

Quantity of heat leaving from the face DCGH in lsquoYrsquo direction

Quantity of heat leaving from the face BCGF in lsquoZrsquo direction

)4()()(

ddzdydxqQgg

Total amount of internal heat generation in the element

Ein + Egen = Eout+Est (Ein- Eout)+ Egen=Est

T

Cddzdydxddzdydxqddzdydxz

TK

zy

TK

yx

TK

x pgzyx )()())](()()([

T

pC

gq

z

TzK

zy

Ty

Kyx

TxK

x][][][

stgzyx QQdQdQdQ

)(

)5()()(

dT

CdzdydxQ pst

Total heat energy stored in the element

(ie thermal energy of the element can be increased because of heat accumulation in the element due to heat flow amp internal heat generation)

)( volumem d

TmCQ pst )(

T

CqTK pg)(

KKKyKx z If material is homogeneous amp isotropic

T

pC

gq

z

T

zy

T

yx

T

xK ][][][

T

Kp

C

Kgq

z

T

y

T

x

T2

2

2

2

2

2

T

Kgq

z

T

y

T

x

T 12

2

2

2

2

2

Thermal diffusivity pC

K

T

K

qT g

12

according to thermodynamics first law

)( zr )( dzrddr

)( dzrddr

Cylindrical polar co-ordinates

Elemental length

Consider an elemental volume ABCDEFGH =

GENERAL HEAT CONDUCTION EQUATION IN CYLINDRICAL COORDINATES

gq

gq

Consider an unsteady (transient) three dimensional heat conduction with internal heat generation in heterogeneous amp anisotropic cylindrical geometries (eg rods and pipes)

)( zrTLetTemperature distribution is a function of T=

Let

=internal heat generation per unit volume per unit time (W m3)q= rate of heat flow (W)Q= qdt=total heat flow (J)r=mass density of the material element(kgm3)Cp=specific heat of the material element (Jkg K)

planerdirectionz

planezrdirection

planezdirectionr

)(

)(

)(

LetHeat flow in

drdr

TdzrdK

rdQ

QQdQ

rr

drrrr

])([

)1(][

r

Tr

rddrdzdKdQ rr

Heat accumulation in the element due to heat flow by conduction in lsquorrsquo direction

rddr

TdzdrK

rdQ

QQdQ d

])([

][

2

T

r

ddrdzrdKdQ

)2(][

2

2

2

T

r

ddrdzrdKdQ

Heat accumulation in the element due to heat flow by conduction in lsquoqrsquo direction

Plane (z- q ) lsquorrsquo direction

dr

TdzrdKQ rr

)(

drQr

QQ rrdrr ][

Quantity of heat leaving from the face EFGH in lsquorrsquo direction

Quantity of heat flowing into the face ABCD in lsquorrsquo direction

Plane (r-z) lsquoq lsquodirection

Quantity of heat leaving from the face AEGC in lsquoqrsquo direction

d

r

TdzdrKQ )(

rdQ

rQQ d ][

Quantity of heat flowing into the face BFHD in lsquoqrsquo direction

)3())((2

2

z

TdzdrrdKdQ zz

)5()()(

dT

CdzdrrdQ pst

)4()()(

ddzdrrdqQgg

Total amount of internal heat generation in the element

Total heat energy stored in the element

(ie thermal energy of the element can be increased because of heat accumulation in the element due to heat flow amp internal heat generation)

according to thermodynamics first law

Ein + Egen = Eout + Est (Ein ndash Eout) + Egen = Est

)( volumem

)()(z

T

zdzdrrdKdQ zz

Heat accumulation in the element due to heat flow by conduction in lsquozrsquo direction

dzdz

TdrrdK

zdQ

QQdQ

zz

dzzzz

])([

d

TmCQ pst )(

Plane (r-q) lsquozrsquo direction

dz

TdrrdKQ zz )(

dzQz

QQ zzdzz ][ Quantity of heat leaving from the face CDHG in lsquozrsquo direction

Quantity of heat flowing into the face ABFE in lsquozrsquo direction

stgzr QQdQdQdQ

)(

r

r ][)(rT

rr

ddrdzdKr

][)(

2

2

2

T

r

ddrdzrdK ))((2

2

z

TddzdrrdK z

ddzdrrdq g )(

dT

Cdzdrrd p )(

+ + +

=

T

CqTK pg)(KKKK zr

[gq

)(1

r

Tr

rrK r

)(1

2

2

2 T

rK )(

2

2

z

TK z

)( ddzdrrd )(

ddzdrrd

TC p

+ + + ] =

If material is homogeneous amp isotropic

)(1

r

Tr

rr

)(1

2

2

2 T

r)(

2

2

z

T

K

q g

T

K

C p

)(1

r

Tr

rr

)(1

2

2

2 T

r )(2

2

z

T

K

q g

T

1

+ + + =

)(1

r

Tr

rrK r

)(1

2

2

2 T

rK )(

2

2

z

TK z

gq

T

C p + + + =

+ + + = Thermal diffusivity

T

K

qT g

12

pC

K

(Ein ndash Eout) + Egen = Est

)( rT

)( r

)sin( drrddr)sin( drrddr

planerdirection

planerdirection

planedirectionr

)(

)(

)(

Spherical polar co-ordinates

Elemental length

Consider an elemental volume ABCDEFGH =

GENERAL HEAT CONDUCTION EQUATION IN SPHERICAL COORDINATES

Let

=internal heat generation per unit volume per unit time (W m3)q= rate of heat flow (W)Q= qdt=total heat flow (J)r=mass density of the material element(kgm3)Cp=specific heat of the material element (Jkg K)

gq

Consider an unsteady (transient) three dimensional heat conduction with internal heat generation in heterogeneous amp anisotropic spherical geometries

LetTemperature distribution is a function of T=

X

Y

Z

drT

drrdKQ rr )sin(

drQr

QQ rrdrr ][

drdr

TdrrdK

rdQ

QQdQ

rr

drrrr

])sin([

)1(]([sin 2

rT

rr

ddrddKdQ rr

Plane ( -q f) lsquorrsquo directionQuantity of heat flowing into the face ABCD in lsquorrsquo direction

Quantity of heat leaving from the face EFGH in lsquorrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquorrsquo direction

drT

drdrKQ )sin(

rdQ

rQQ d ][

rddr

TdrdrK

rdQ

QQdQ d

])sin([

])[(sin

2

T

r

drdrdrKdQ

)2(])(sin[

2

T

r

drdrddrKdQ

Quantity of heat leaving from the face BFHD in lsquoqrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoqrsquo direction

X

Y

Z

Plane (r- f ) lsquoq lsquodirection

Quantity of heat flowing into the face AEGC in lsquoqrsquo direction

dr

TdrrdKQ

sin)(

drQ

rQQ d sin][

sin

drdr

TdrrdK

rdQ

QQdQ d

sin]sin

)([sin

][sinsin

)sin(

T

rr

ddrdrrdKdQ

)3()(sin

)sin(2

2

22

2

T

r

ddddrrKdQ

Plane (r- ) q lsquofrsquo directionQuantity of heat flowing into the face ABFE in lsquofrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquozrsquo direction

)4()sin()(

ddrrddrqQgg

)5()sin()(

dT

CdrrddrQ pst

Total amount of internal heat generation in the element

Total heat energy stored in the element

(Ein ndash Eout) + Egen = Est

d

TmCQ pst )(

)( volumem

Quantity of heat leaving from the face CDHG in lsquofrsquo direction

according to thermodynamics first law

Ein + Egen = Eout + Est

X

Y

Z

stgr QQdQdQdQ

)(

]([sin 2

r

Tr

rddrddKr

])(sin[

2

T

rdrdrddrK

)(sin

)sin(2

2

22

2

T

r

ddddrrK

ddrrddrq g )sin(

dT

Cdrrddr p )sin(

2

2

r

r

sin

sin

]([ 22 r

Tr

rr

Kr

])(sin[sin2

TrK

)(sin 2

2

22

T

r

Kgq

T

Cp

]([1

[ 22 r

Tr

rrK

])(sin[sin1

2

T

r)](

sin

12

2

22 T

r gq

T

Cp

]([1

[ 22 r

Tr

rr

])(sin[sin1

2

T

r)](

sin

12

2

22 T

rK

q g

T

K

Cp

]([1

[ 22 r

Tr

rr

])(sin[sin1

2

T

r)](

sin

12

2

22 T

rK

q g

T1

KKKKr If material is homogeneous amp isotropic

TCqTK pg)(

T

K

qT g

12 pC

K

Thermal diffusivity

(Ein ndash Eout) + Egen = Est

General Heat Conduction Equation

PLANE WALL CYLINDER SPHERE

Temperature profile

Plane wall SphereCylinder

bull A copper conductor (k = 380 WmoC resistivity ρ = 2x10-8 Ω m) having inner and outer radii 10 cm and 225 cm respectively is carrying a current density of 4800 amperes cm2 the conductor is internally cooled and a constant temperature of 65oC is maintained at the inner surface and there is no heat transfer through the insulation surrounding the conductor

Determinebull (i) The maximum temperature of the conductor and the radius at

which it occursbull (ii) The internal heat transfer rate

bull A chemical reaction takes place in a packed bed (k=06 WmoC) between two coaxial cylinders with radii 15 mm and 45 mm the inner surface is at 580oC and it is insulated Assuming the reaction rate of 055 MWm3 in the reactor volume find the temperature at the outer surface of the reactor

bull A current of 300 amperes passes through a stainless steel wire of 25 mm diameter and k=20 WmoC The resistivity of the wire is 70x10-8 ohm-m and length of the wire is 2 m if the wire is submerged in a fluid maintained at 50oC and h at the wire surface is 4000 Wm2oC calculate the steady state temperature at the centre and at the surface of the wire

bull An approximately spherical shaped orange ( k=023 WmoC) 90 mm diameter undergoes ripping process and generates 5100 Wm3 of energy If external surface the orange is at 8oC determine

bull (i) temperature at the centre of the orange and

bull (ii) heat flow from the outer surface of the orange

bull FINS External surfaces to used to cooling area and hence convection

heat transfer can be improvedbull APPLICATIONS IC engine cylinder ( air cooling) Heat exchangers automobile

radiators computer chip refrigerators electric motors generators and transformers etc

bull ASSUMPTIONS MADE IN ANALYSIS OF HEAT TRANFER THROUGH FIN

1 Fin material is homogenous and isotropic constthermal conductivity

2 Temperature at any cs of the fin is uniform3 lsquohrsquo is uniform over entire surface4 No internal heat generation5 Thermal contact resistance is negligible6 Radiation heat transfer is negligible7 Heat conduction is steady state and one dimensional

bull Problem 1

bull A stainless steel fin (k=20 WmK) having a diameter of 20 mm and a length of 01 m is attached to a wall at 300oC The ambient temperature is 50oC and the heat transfer coefficient is 10 Wm2K The fin tip is insulated Determine

bull (A) The rate of heat dissipation from the finbull (B) The temperature at the fin tipbull (C) The rate of heat transfer from the wall area covered by the fin if

the fin was not usedbull (D) The heat transfer rate from the same fin geometry if the

stainless steel fin is replaced by a fictitious fin with infinite thermal conductivity

bull Problem2bull One end of a long rod is inserted into a furnace while the other

projects into ambient temperature Under steady state the temperature of the rod is measured at two points 75mm apart and found to be 125oC and 885oC respectively while the ambient temperature is 20oC If the rod is 25mm in diameter and h is 2336 Wm2K

bull Find the thermal conductivity of the rod material

  • HEAT AND MASS TRANSFER
  • Thermodynamics Vs Heat transfer
  • MODE OF HEAT TRANSFER
  • MODE OF HEAT TRANSFER (2)
  • CONDUCTION- MECHANISM
  • CONCECTION-MECHANISM
  • RADIATION-MECHANISM
  • Conduction- Convection- Radiation
  • Slide 9
  • Slide 10
  • Fourierrsquos Law of Heat Conduction
  • Slide 12
  • Newtonrsquos Law of Cooling
  • Laws of Thermal Radiation
  • LAWS FOR HEAT TRANSFER
  • ELECTRICAL ampTHERMAL ANALOGY
  • Slide 17
  • Slide 18
  • PLANE WALL-UNIDIRECTIONAL
  • PLANE WALL -SERIES
  • PLANE WALL IN SERIES
  • PLANE WALL -PARALLEL
  • PLANE WALL-PARALLEL
  • PLANE WALL-COMBINATION
  • PLANE WALL-FILM RESISTANCE
  • PLANE WALL- THERMAL CONTACT RESISTANCE
  • PLANE WALL- THERMAL CONTACT RESISTANCE (2)
  • CYLINDER-SOLID
  • HEAT TRANSFER IN SOLID CYLINDER
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER-COMPOSITE
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER-MULTY LAYER
  • HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER
  • HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER (2)
  • HEAT TRANSFER IN SPHERE
  • HEAT TRANSFER IN SPHERE -MULTY LAYER
  • HEAT TRANSFER IN SPHERE -MULTY LAYER (2)
  • Slide 38
  • Slide 39
  • Slide 40
  • Slide 41
  • CRITICAL RADIUS
  • Plane wall- Insulation thickness -study
  • Problem5
  • Slide 45
  • Slide 46
  • Slide 47
  • GENERAL HEAT CONDUCTION EQUATION IN CYLINDRICAL COORDINATES
  • Slide 49
  • Slide 50
  • Slide 51
  • GENERAL HEAT CONDUCTION EQUATION IN SPHERICAL COORDINATES
  • Slide 53
  • Slide 54
  • Slide 55
  • General Heat Conduction Equation
  • Temperature profile
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Slide 67
Page 41: hmt chap 1

dx

TdzdyKQ xx )(

dxQx

QQ xxdxx ][

dxdx

TdzdyK

xQQdQ xdxxxx ])([)(

)1()]([)(

ddzdydx

x

TK

xdQ xx

GENERAL HEAT CONDUCTION EQUATION IN CARTESIAN COORDINATES

Consider an unsteady (transient) three dimensional heat conduction with internal heat generation in heterogeneous amp anisotropic PLANE WALL

Cartesian co-ordinates x y z

Elemental length dxdydz

Consider an elemental volume ABCDEFGH = (dxdydz)

Let Temperature distribution is a function of T=T(xyz)

Let

=internal heat generation per unit volume per unit time (W m3)q= rate of heat flow (W)Q= qdt=total heat flow (J)r=mass density of the material element(kgm3)Cp=specific heat of the material element (Jkg K)

gq

Plane (y - z) lsquoxrsquo direction

Quantity of heat leaving from the face EFGH in lsquoXrsquo direction

Quantity of heat flowing into the face ABCD in lsquoXrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoXrsquo direction

dy

TdzdxKQ yy )(

dyQy

QQ yydyy ][

dydy

TdzdxK

yQQdQ ydyyyy ])([)(

)2()]([)(

ddzdydx

y

TK

ydQ yy

Plane (x - z) lsquoyrsquo direction

Quantity of heat flowing into the face ABFE in lsquoYrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoYrsquo direction

dz

TdydxKQ zz )(

dzQz

QQ zzdzz ][

dzdz

TdydxK

zQQdQ zdzzzz ])([)(

)3()]([)(

ddzdydx

z

TK

zdQ zz

Plane (x - y) lsquozrsquo direction

Quantity of heat flowing into the face ADHE in lsquoZrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoZrsquo direction

Quantity of heat leaving from the face DCGH in lsquoYrsquo direction

Quantity of heat leaving from the face BCGF in lsquoZrsquo direction

)4()()(

ddzdydxqQgg

Total amount of internal heat generation in the element

Ein + Egen = Eout+Est (Ein- Eout)+ Egen=Est

T

Cddzdydxddzdydxqddzdydxz

TK

zy

TK

yx

TK

x pgzyx )()())](()()([

T

pC

gq

z

TzK

zy

Ty

Kyx

TxK

x][][][

stgzyx QQdQdQdQ

)(

)5()()(

dT

CdzdydxQ pst

Total heat energy stored in the element

(ie thermal energy of the element can be increased because of heat accumulation in the element due to heat flow amp internal heat generation)

)( volumem d

TmCQ pst )(

T

CqTK pg)(

KKKyKx z If material is homogeneous amp isotropic

T

pC

gq

z

T

zy

T

yx

T

xK ][][][

T

Kp

C

Kgq

z

T

y

T

x

T2

2

2

2

2

2

T

Kgq

z

T

y

T

x

T 12

2

2

2

2

2

Thermal diffusivity pC

K

T

K

qT g

12

according to thermodynamics first law

)( zr )( dzrddr

)( dzrddr

Cylindrical polar co-ordinates

Elemental length

Consider an elemental volume ABCDEFGH =

GENERAL HEAT CONDUCTION EQUATION IN CYLINDRICAL COORDINATES

gq

gq

Consider an unsteady (transient) three dimensional heat conduction with internal heat generation in heterogeneous amp anisotropic cylindrical geometries (eg rods and pipes)

)( zrTLetTemperature distribution is a function of T=

Let

=internal heat generation per unit volume per unit time (W m3)q= rate of heat flow (W)Q= qdt=total heat flow (J)r=mass density of the material element(kgm3)Cp=specific heat of the material element (Jkg K)

planerdirectionz

planezrdirection

planezdirectionr

)(

)(

)(

LetHeat flow in

drdr

TdzrdK

rdQ

QQdQ

rr

drrrr

])([

)1(][

r

Tr

rddrdzdKdQ rr

Heat accumulation in the element due to heat flow by conduction in lsquorrsquo direction

rddr

TdzdrK

rdQ

QQdQ d

])([

][

2

T

r

ddrdzrdKdQ

)2(][

2

2

2

T

r

ddrdzrdKdQ

Heat accumulation in the element due to heat flow by conduction in lsquoqrsquo direction

Plane (z- q ) lsquorrsquo direction

dr

TdzrdKQ rr

)(

drQr

QQ rrdrr ][

Quantity of heat leaving from the face EFGH in lsquorrsquo direction

Quantity of heat flowing into the face ABCD in lsquorrsquo direction

Plane (r-z) lsquoq lsquodirection

Quantity of heat leaving from the face AEGC in lsquoqrsquo direction

d

r

TdzdrKQ )(

rdQ

rQQ d ][

Quantity of heat flowing into the face BFHD in lsquoqrsquo direction

)3())((2

2

z

TdzdrrdKdQ zz

)5()()(

dT

CdzdrrdQ pst

)4()()(

ddzdrrdqQgg

Total amount of internal heat generation in the element

Total heat energy stored in the element

(ie thermal energy of the element can be increased because of heat accumulation in the element due to heat flow amp internal heat generation)

according to thermodynamics first law

Ein + Egen = Eout + Est (Ein ndash Eout) + Egen = Est

)( volumem

)()(z

T

zdzdrrdKdQ zz

Heat accumulation in the element due to heat flow by conduction in lsquozrsquo direction

dzdz

TdrrdK

zdQ

QQdQ

zz

dzzzz

])([

d

TmCQ pst )(

Plane (r-q) lsquozrsquo direction

dz

TdrrdKQ zz )(

dzQz

QQ zzdzz ][ Quantity of heat leaving from the face CDHG in lsquozrsquo direction

Quantity of heat flowing into the face ABFE in lsquozrsquo direction

stgzr QQdQdQdQ

)(

r

r ][)(rT

rr

ddrdzdKr

][)(

2

2

2

T

r

ddrdzrdK ))((2

2

z

TddzdrrdK z

ddzdrrdq g )(

dT

Cdzdrrd p )(

+ + +

=

T

CqTK pg)(KKKK zr

[gq

)(1

r

Tr

rrK r

)(1

2

2

2 T

rK )(

2

2

z

TK z

)( ddzdrrd )(

ddzdrrd

TC p

+ + + ] =

If material is homogeneous amp isotropic

)(1

r

Tr

rr

)(1

2

2

2 T

r)(

2

2

z

T

K

q g

T

K

C p

)(1

r

Tr

rr

)(1

2

2

2 T

r )(2

2

z

T

K

q g

T

1

+ + + =

)(1

r

Tr

rrK r

)(1

2

2

2 T

rK )(

2

2

z

TK z

gq

T

C p + + + =

+ + + = Thermal diffusivity

T

K

qT g

12

pC

K

(Ein ndash Eout) + Egen = Est

)( rT

)( r

)sin( drrddr)sin( drrddr

planerdirection

planerdirection

planedirectionr

)(

)(

)(

Spherical polar co-ordinates

Elemental length

Consider an elemental volume ABCDEFGH =

GENERAL HEAT CONDUCTION EQUATION IN SPHERICAL COORDINATES

Let

=internal heat generation per unit volume per unit time (W m3)q= rate of heat flow (W)Q= qdt=total heat flow (J)r=mass density of the material element(kgm3)Cp=specific heat of the material element (Jkg K)

gq

Consider an unsteady (transient) three dimensional heat conduction with internal heat generation in heterogeneous amp anisotropic spherical geometries

LetTemperature distribution is a function of T=

X

Y

Z

drT

drrdKQ rr )sin(

drQr

QQ rrdrr ][

drdr

TdrrdK

rdQ

QQdQ

rr

drrrr

])sin([

)1(]([sin 2

rT

rr

ddrddKdQ rr

Plane ( -q f) lsquorrsquo directionQuantity of heat flowing into the face ABCD in lsquorrsquo direction

Quantity of heat leaving from the face EFGH in lsquorrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquorrsquo direction

drT

drdrKQ )sin(

rdQ

rQQ d ][

rddr

TdrdrK

rdQ

QQdQ d

])sin([

])[(sin

2

T

r

drdrdrKdQ

)2(])(sin[

2

T

r

drdrddrKdQ

Quantity of heat leaving from the face BFHD in lsquoqrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoqrsquo direction

X

Y

Z

Plane (r- f ) lsquoq lsquodirection

Quantity of heat flowing into the face AEGC in lsquoqrsquo direction

dr

TdrrdKQ

sin)(

drQ

rQQ d sin][

sin

drdr

TdrrdK

rdQ

QQdQ d

sin]sin

)([sin

][sinsin

)sin(

T

rr

ddrdrrdKdQ

)3()(sin

)sin(2

2

22

2

T

r

ddddrrKdQ

Plane (r- ) q lsquofrsquo directionQuantity of heat flowing into the face ABFE in lsquofrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquozrsquo direction

)4()sin()(

ddrrddrqQgg

)5()sin()(

dT

CdrrddrQ pst

Total amount of internal heat generation in the element

Total heat energy stored in the element

(Ein ndash Eout) + Egen = Est

d

TmCQ pst )(

)( volumem

Quantity of heat leaving from the face CDHG in lsquofrsquo direction

according to thermodynamics first law

Ein + Egen = Eout + Est

X

Y

Z

stgr QQdQdQdQ

)(

]([sin 2

r

Tr

rddrddKr

])(sin[

2

T

rdrdrddrK

)(sin

)sin(2

2

22

2

T

r

ddddrrK

ddrrddrq g )sin(

dT

Cdrrddr p )sin(

2

2

r

r

sin

sin

]([ 22 r

Tr

rr

Kr

])(sin[sin2

TrK

)(sin 2

2

22

T

r

Kgq

T

Cp

]([1

[ 22 r

Tr

rrK

])(sin[sin1

2

T

r)](

sin

12

2

22 T

r gq

T

Cp

]([1

[ 22 r

Tr

rr

])(sin[sin1

2

T

r)](

sin

12

2

22 T

rK

q g

T

K

Cp

]([1

[ 22 r

Tr

rr

])(sin[sin1

2

T

r)](

sin

12

2

22 T

rK

q g

T1

KKKKr If material is homogeneous amp isotropic

TCqTK pg)(

T

K

qT g

12 pC

K

Thermal diffusivity

(Ein ndash Eout) + Egen = Est

General Heat Conduction Equation

PLANE WALL CYLINDER SPHERE

Temperature profile

Plane wall SphereCylinder

bull A copper conductor (k = 380 WmoC resistivity ρ = 2x10-8 Ω m) having inner and outer radii 10 cm and 225 cm respectively is carrying a current density of 4800 amperes cm2 the conductor is internally cooled and a constant temperature of 65oC is maintained at the inner surface and there is no heat transfer through the insulation surrounding the conductor

Determinebull (i) The maximum temperature of the conductor and the radius at

which it occursbull (ii) The internal heat transfer rate

bull A chemical reaction takes place in a packed bed (k=06 WmoC) between two coaxial cylinders with radii 15 mm and 45 mm the inner surface is at 580oC and it is insulated Assuming the reaction rate of 055 MWm3 in the reactor volume find the temperature at the outer surface of the reactor

bull A current of 300 amperes passes through a stainless steel wire of 25 mm diameter and k=20 WmoC The resistivity of the wire is 70x10-8 ohm-m and length of the wire is 2 m if the wire is submerged in a fluid maintained at 50oC and h at the wire surface is 4000 Wm2oC calculate the steady state temperature at the centre and at the surface of the wire

bull An approximately spherical shaped orange ( k=023 WmoC) 90 mm diameter undergoes ripping process and generates 5100 Wm3 of energy If external surface the orange is at 8oC determine

bull (i) temperature at the centre of the orange and

bull (ii) heat flow from the outer surface of the orange

bull FINS External surfaces to used to cooling area and hence convection

heat transfer can be improvedbull APPLICATIONS IC engine cylinder ( air cooling) Heat exchangers automobile

radiators computer chip refrigerators electric motors generators and transformers etc

bull ASSUMPTIONS MADE IN ANALYSIS OF HEAT TRANFER THROUGH FIN

1 Fin material is homogenous and isotropic constthermal conductivity

2 Temperature at any cs of the fin is uniform3 lsquohrsquo is uniform over entire surface4 No internal heat generation5 Thermal contact resistance is negligible6 Radiation heat transfer is negligible7 Heat conduction is steady state and one dimensional

bull Problem 1

bull A stainless steel fin (k=20 WmK) having a diameter of 20 mm and a length of 01 m is attached to a wall at 300oC The ambient temperature is 50oC and the heat transfer coefficient is 10 Wm2K The fin tip is insulated Determine

bull (A) The rate of heat dissipation from the finbull (B) The temperature at the fin tipbull (C) The rate of heat transfer from the wall area covered by the fin if

the fin was not usedbull (D) The heat transfer rate from the same fin geometry if the

stainless steel fin is replaced by a fictitious fin with infinite thermal conductivity

bull Problem2bull One end of a long rod is inserted into a furnace while the other

projects into ambient temperature Under steady state the temperature of the rod is measured at two points 75mm apart and found to be 125oC and 885oC respectively while the ambient temperature is 20oC If the rod is 25mm in diameter and h is 2336 Wm2K

bull Find the thermal conductivity of the rod material

  • HEAT AND MASS TRANSFER
  • Thermodynamics Vs Heat transfer
  • MODE OF HEAT TRANSFER
  • MODE OF HEAT TRANSFER (2)
  • CONDUCTION- MECHANISM
  • CONCECTION-MECHANISM
  • RADIATION-MECHANISM
  • Conduction- Convection- Radiation
  • Slide 9
  • Slide 10
  • Fourierrsquos Law of Heat Conduction
  • Slide 12
  • Newtonrsquos Law of Cooling
  • Laws of Thermal Radiation
  • LAWS FOR HEAT TRANSFER
  • ELECTRICAL ampTHERMAL ANALOGY
  • Slide 17
  • Slide 18
  • PLANE WALL-UNIDIRECTIONAL
  • PLANE WALL -SERIES
  • PLANE WALL IN SERIES
  • PLANE WALL -PARALLEL
  • PLANE WALL-PARALLEL
  • PLANE WALL-COMBINATION
  • PLANE WALL-FILM RESISTANCE
  • PLANE WALL- THERMAL CONTACT RESISTANCE
  • PLANE WALL- THERMAL CONTACT RESISTANCE (2)
  • CYLINDER-SOLID
  • HEAT TRANSFER IN SOLID CYLINDER
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER-COMPOSITE
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER-MULTY LAYER
  • HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER
  • HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER (2)
  • HEAT TRANSFER IN SPHERE
  • HEAT TRANSFER IN SPHERE -MULTY LAYER
  • HEAT TRANSFER IN SPHERE -MULTY LAYER (2)
  • Slide 38
  • Slide 39
  • Slide 40
  • Slide 41
  • CRITICAL RADIUS
  • Plane wall- Insulation thickness -study
  • Problem5
  • Slide 45
  • Slide 46
  • Slide 47
  • GENERAL HEAT CONDUCTION EQUATION IN CYLINDRICAL COORDINATES
  • Slide 49
  • Slide 50
  • Slide 51
  • GENERAL HEAT CONDUCTION EQUATION IN SPHERICAL COORDINATES
  • Slide 53
  • Slide 54
  • Slide 55
  • General Heat Conduction Equation
  • Temperature profile
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Slide 67
Page 42: hmt chap 1

dy

TdzdxKQ yy )(

dyQy

QQ yydyy ][

dydy

TdzdxK

yQQdQ ydyyyy ])([)(

)2()]([)(

ddzdydx

y

TK

ydQ yy

Plane (x - z) lsquoyrsquo direction

Quantity of heat flowing into the face ABFE in lsquoYrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoYrsquo direction

dz

TdydxKQ zz )(

dzQz

QQ zzdzz ][

dzdz

TdydxK

zQQdQ zdzzzz ])([)(

)3()]([)(

ddzdydx

z

TK

zdQ zz

Plane (x - y) lsquozrsquo direction

Quantity of heat flowing into the face ADHE in lsquoZrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoZrsquo direction

Quantity of heat leaving from the face DCGH in lsquoYrsquo direction

Quantity of heat leaving from the face BCGF in lsquoZrsquo direction

)4()()(

ddzdydxqQgg

Total amount of internal heat generation in the element

Ein + Egen = Eout+Est (Ein- Eout)+ Egen=Est

T

Cddzdydxddzdydxqddzdydxz

TK

zy

TK

yx

TK

x pgzyx )()())](()()([

T

pC

gq

z

TzK

zy

Ty

Kyx

TxK

x][][][

stgzyx QQdQdQdQ

)(

)5()()(

dT

CdzdydxQ pst

Total heat energy stored in the element

(ie thermal energy of the element can be increased because of heat accumulation in the element due to heat flow amp internal heat generation)

)( volumem d

TmCQ pst )(

T

CqTK pg)(

KKKyKx z If material is homogeneous amp isotropic

T

pC

gq

z

T

zy

T

yx

T

xK ][][][

T

Kp

C

Kgq

z

T

y

T

x

T2

2

2

2

2

2

T

Kgq

z

T

y

T

x

T 12

2

2

2

2

2

Thermal diffusivity pC

K

T

K

qT g

12

according to thermodynamics first law

)( zr )( dzrddr

)( dzrddr

Cylindrical polar co-ordinates

Elemental length

Consider an elemental volume ABCDEFGH =

GENERAL HEAT CONDUCTION EQUATION IN CYLINDRICAL COORDINATES

gq

gq

Consider an unsteady (transient) three dimensional heat conduction with internal heat generation in heterogeneous amp anisotropic cylindrical geometries (eg rods and pipes)

)( zrTLetTemperature distribution is a function of T=

Let

=internal heat generation per unit volume per unit time (W m3)q= rate of heat flow (W)Q= qdt=total heat flow (J)r=mass density of the material element(kgm3)Cp=specific heat of the material element (Jkg K)

planerdirectionz

planezrdirection

planezdirectionr

)(

)(

)(

LetHeat flow in

drdr

TdzrdK

rdQ

QQdQ

rr

drrrr

])([

)1(][

r

Tr

rddrdzdKdQ rr

Heat accumulation in the element due to heat flow by conduction in lsquorrsquo direction

rddr

TdzdrK

rdQ

QQdQ d

])([

][

2

T

r

ddrdzrdKdQ

)2(][

2

2

2

T

r

ddrdzrdKdQ

Heat accumulation in the element due to heat flow by conduction in lsquoqrsquo direction

Plane (z- q ) lsquorrsquo direction

dr

TdzrdKQ rr

)(

drQr

QQ rrdrr ][

Quantity of heat leaving from the face EFGH in lsquorrsquo direction

Quantity of heat flowing into the face ABCD in lsquorrsquo direction

Plane (r-z) lsquoq lsquodirection

Quantity of heat leaving from the face AEGC in lsquoqrsquo direction

d

r

TdzdrKQ )(

rdQ

rQQ d ][

Quantity of heat flowing into the face BFHD in lsquoqrsquo direction

)3())((2

2

z

TdzdrrdKdQ zz

)5()()(

dT

CdzdrrdQ pst

)4()()(

ddzdrrdqQgg

Total amount of internal heat generation in the element

Total heat energy stored in the element

(ie thermal energy of the element can be increased because of heat accumulation in the element due to heat flow amp internal heat generation)

according to thermodynamics first law

Ein + Egen = Eout + Est (Ein ndash Eout) + Egen = Est

)( volumem

)()(z

T

zdzdrrdKdQ zz

Heat accumulation in the element due to heat flow by conduction in lsquozrsquo direction

dzdz

TdrrdK

zdQ

QQdQ

zz

dzzzz

])([

d

TmCQ pst )(

Plane (r-q) lsquozrsquo direction

dz

TdrrdKQ zz )(

dzQz

QQ zzdzz ][ Quantity of heat leaving from the face CDHG in lsquozrsquo direction

Quantity of heat flowing into the face ABFE in lsquozrsquo direction

stgzr QQdQdQdQ

)(

r

r ][)(rT

rr

ddrdzdKr

][)(

2

2

2

T

r

ddrdzrdK ))((2

2

z

TddzdrrdK z

ddzdrrdq g )(

dT

Cdzdrrd p )(

+ + +

=

T

CqTK pg)(KKKK zr

[gq

)(1

r

Tr

rrK r

)(1

2

2

2 T

rK )(

2

2

z

TK z

)( ddzdrrd )(

ddzdrrd

TC p

+ + + ] =

If material is homogeneous amp isotropic

)(1

r

Tr

rr

)(1

2

2

2 T

r)(

2

2

z

T

K

q g

T

K

C p

)(1

r

Tr

rr

)(1

2

2

2 T

r )(2

2

z

T

K

q g

T

1

+ + + =

)(1

r

Tr

rrK r

)(1

2

2

2 T

rK )(

2

2

z

TK z

gq

T

C p + + + =

+ + + = Thermal diffusivity

T

K

qT g

12

pC

K

(Ein ndash Eout) + Egen = Est

)( rT

)( r

)sin( drrddr)sin( drrddr

planerdirection

planerdirection

planedirectionr

)(

)(

)(

Spherical polar co-ordinates

Elemental length

Consider an elemental volume ABCDEFGH =

GENERAL HEAT CONDUCTION EQUATION IN SPHERICAL COORDINATES

Let

=internal heat generation per unit volume per unit time (W m3)q= rate of heat flow (W)Q= qdt=total heat flow (J)r=mass density of the material element(kgm3)Cp=specific heat of the material element (Jkg K)

gq

Consider an unsteady (transient) three dimensional heat conduction with internal heat generation in heterogeneous amp anisotropic spherical geometries

LetTemperature distribution is a function of T=

X

Y

Z

drT

drrdKQ rr )sin(

drQr

QQ rrdrr ][

drdr

TdrrdK

rdQ

QQdQ

rr

drrrr

])sin([

)1(]([sin 2

rT

rr

ddrddKdQ rr

Plane ( -q f) lsquorrsquo directionQuantity of heat flowing into the face ABCD in lsquorrsquo direction

Quantity of heat leaving from the face EFGH in lsquorrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquorrsquo direction

drT

drdrKQ )sin(

rdQ

rQQ d ][

rddr

TdrdrK

rdQ

QQdQ d

])sin([

])[(sin

2

T

r

drdrdrKdQ

)2(])(sin[

2

T

r

drdrddrKdQ

Quantity of heat leaving from the face BFHD in lsquoqrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoqrsquo direction

X

Y

Z

Plane (r- f ) lsquoq lsquodirection

Quantity of heat flowing into the face AEGC in lsquoqrsquo direction

dr

TdrrdKQ

sin)(

drQ

rQQ d sin][

sin

drdr

TdrrdK

rdQ

QQdQ d

sin]sin

)([sin

][sinsin

)sin(

T

rr

ddrdrrdKdQ

)3()(sin

)sin(2

2

22

2

T

r

ddddrrKdQ

Plane (r- ) q lsquofrsquo directionQuantity of heat flowing into the face ABFE in lsquofrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquozrsquo direction

)4()sin()(

ddrrddrqQgg

)5()sin()(

dT

CdrrddrQ pst

Total amount of internal heat generation in the element

Total heat energy stored in the element

(Ein ndash Eout) + Egen = Est

d

TmCQ pst )(

)( volumem

Quantity of heat leaving from the face CDHG in lsquofrsquo direction

according to thermodynamics first law

Ein + Egen = Eout + Est

X

Y

Z

stgr QQdQdQdQ

)(

]([sin 2

r

Tr

rddrddKr

])(sin[

2

T

rdrdrddrK

)(sin

)sin(2

2

22

2

T

r

ddddrrK

ddrrddrq g )sin(

dT

Cdrrddr p )sin(

2

2

r

r

sin

sin

]([ 22 r

Tr

rr

Kr

])(sin[sin2

TrK

)(sin 2

2

22

T

r

Kgq

T

Cp

]([1

[ 22 r

Tr

rrK

])(sin[sin1

2

T

r)](

sin

12

2

22 T

r gq

T

Cp

]([1

[ 22 r

Tr

rr

])(sin[sin1

2

T

r)](

sin

12

2

22 T

rK

q g

T

K

Cp

]([1

[ 22 r

Tr

rr

])(sin[sin1

2

T

r)](

sin

12

2

22 T

rK

q g

T1

KKKKr If material is homogeneous amp isotropic

TCqTK pg)(

T

K

qT g

12 pC

K

Thermal diffusivity

(Ein ndash Eout) + Egen = Est

General Heat Conduction Equation

PLANE WALL CYLINDER SPHERE

Temperature profile

Plane wall SphereCylinder

bull A copper conductor (k = 380 WmoC resistivity ρ = 2x10-8 Ω m) having inner and outer radii 10 cm and 225 cm respectively is carrying a current density of 4800 amperes cm2 the conductor is internally cooled and a constant temperature of 65oC is maintained at the inner surface and there is no heat transfer through the insulation surrounding the conductor

Determinebull (i) The maximum temperature of the conductor and the radius at

which it occursbull (ii) The internal heat transfer rate

bull A chemical reaction takes place in a packed bed (k=06 WmoC) between two coaxial cylinders with radii 15 mm and 45 mm the inner surface is at 580oC and it is insulated Assuming the reaction rate of 055 MWm3 in the reactor volume find the temperature at the outer surface of the reactor

bull A current of 300 amperes passes through a stainless steel wire of 25 mm diameter and k=20 WmoC The resistivity of the wire is 70x10-8 ohm-m and length of the wire is 2 m if the wire is submerged in a fluid maintained at 50oC and h at the wire surface is 4000 Wm2oC calculate the steady state temperature at the centre and at the surface of the wire

bull An approximately spherical shaped orange ( k=023 WmoC) 90 mm diameter undergoes ripping process and generates 5100 Wm3 of energy If external surface the orange is at 8oC determine

bull (i) temperature at the centre of the orange and

bull (ii) heat flow from the outer surface of the orange

bull FINS External surfaces to used to cooling area and hence convection

heat transfer can be improvedbull APPLICATIONS IC engine cylinder ( air cooling) Heat exchangers automobile

radiators computer chip refrigerators electric motors generators and transformers etc

bull ASSUMPTIONS MADE IN ANALYSIS OF HEAT TRANFER THROUGH FIN

1 Fin material is homogenous and isotropic constthermal conductivity

2 Temperature at any cs of the fin is uniform3 lsquohrsquo is uniform over entire surface4 No internal heat generation5 Thermal contact resistance is negligible6 Radiation heat transfer is negligible7 Heat conduction is steady state and one dimensional

bull Problem 1

bull A stainless steel fin (k=20 WmK) having a diameter of 20 mm and a length of 01 m is attached to a wall at 300oC The ambient temperature is 50oC and the heat transfer coefficient is 10 Wm2K The fin tip is insulated Determine

bull (A) The rate of heat dissipation from the finbull (B) The temperature at the fin tipbull (C) The rate of heat transfer from the wall area covered by the fin if

the fin was not usedbull (D) The heat transfer rate from the same fin geometry if the

stainless steel fin is replaced by a fictitious fin with infinite thermal conductivity

bull Problem2bull One end of a long rod is inserted into a furnace while the other

projects into ambient temperature Under steady state the temperature of the rod is measured at two points 75mm apart and found to be 125oC and 885oC respectively while the ambient temperature is 20oC If the rod is 25mm in diameter and h is 2336 Wm2K

bull Find the thermal conductivity of the rod material

  • HEAT AND MASS TRANSFER
  • Thermodynamics Vs Heat transfer
  • MODE OF HEAT TRANSFER
  • MODE OF HEAT TRANSFER (2)
  • CONDUCTION- MECHANISM
  • CONCECTION-MECHANISM
  • RADIATION-MECHANISM
  • Conduction- Convection- Radiation
  • Slide 9
  • Slide 10
  • Fourierrsquos Law of Heat Conduction
  • Slide 12
  • Newtonrsquos Law of Cooling
  • Laws of Thermal Radiation
  • LAWS FOR HEAT TRANSFER
  • ELECTRICAL ampTHERMAL ANALOGY
  • Slide 17
  • Slide 18
  • PLANE WALL-UNIDIRECTIONAL
  • PLANE WALL -SERIES
  • PLANE WALL IN SERIES
  • PLANE WALL -PARALLEL
  • PLANE WALL-PARALLEL
  • PLANE WALL-COMBINATION
  • PLANE WALL-FILM RESISTANCE
  • PLANE WALL- THERMAL CONTACT RESISTANCE
  • PLANE WALL- THERMAL CONTACT RESISTANCE (2)
  • CYLINDER-SOLID
  • HEAT TRANSFER IN SOLID CYLINDER
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER-COMPOSITE
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER-MULTY LAYER
  • HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER
  • HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER (2)
  • HEAT TRANSFER IN SPHERE
  • HEAT TRANSFER IN SPHERE -MULTY LAYER
  • HEAT TRANSFER IN SPHERE -MULTY LAYER (2)
  • Slide 38
  • Slide 39
  • Slide 40
  • Slide 41
  • CRITICAL RADIUS
  • Plane wall- Insulation thickness -study
  • Problem5
  • Slide 45
  • Slide 46
  • Slide 47
  • GENERAL HEAT CONDUCTION EQUATION IN CYLINDRICAL COORDINATES
  • Slide 49
  • Slide 50
  • Slide 51
  • GENERAL HEAT CONDUCTION EQUATION IN SPHERICAL COORDINATES
  • Slide 53
  • Slide 54
  • Slide 55
  • General Heat Conduction Equation
  • Temperature profile
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Slide 67
Page 43: hmt chap 1

Ein + Egen = Eout+Est (Ein- Eout)+ Egen=Est

T

Cddzdydxddzdydxqddzdydxz

TK

zy

TK

yx

TK

x pgzyx )()())](()()([

T

pC

gq

z

TzK

zy

Ty

Kyx

TxK

x][][][

stgzyx QQdQdQdQ

)(

)5()()(

dT

CdzdydxQ pst

Total heat energy stored in the element

(ie thermal energy of the element can be increased because of heat accumulation in the element due to heat flow amp internal heat generation)

)( volumem d

TmCQ pst )(

T

CqTK pg)(

KKKyKx z If material is homogeneous amp isotropic

T

pC

gq

z

T

zy

T

yx

T

xK ][][][

T

Kp

C

Kgq

z

T

y

T

x

T2

2

2

2

2

2

T

Kgq

z

T

y

T

x

T 12

2

2

2

2

2

Thermal diffusivity pC

K

T

K

qT g

12

according to thermodynamics first law

)( zr )( dzrddr

)( dzrddr

Cylindrical polar co-ordinates

Elemental length

Consider an elemental volume ABCDEFGH =

GENERAL HEAT CONDUCTION EQUATION IN CYLINDRICAL COORDINATES

gq

gq

Consider an unsteady (transient) three dimensional heat conduction with internal heat generation in heterogeneous amp anisotropic cylindrical geometries (eg rods and pipes)

)( zrTLetTemperature distribution is a function of T=

Let

=internal heat generation per unit volume per unit time (W m3)q= rate of heat flow (W)Q= qdt=total heat flow (J)r=mass density of the material element(kgm3)Cp=specific heat of the material element (Jkg K)

planerdirectionz

planezrdirection

planezdirectionr

)(

)(

)(

LetHeat flow in

drdr

TdzrdK

rdQ

QQdQ

rr

drrrr

])([

)1(][

r

Tr

rddrdzdKdQ rr

Heat accumulation in the element due to heat flow by conduction in lsquorrsquo direction

rddr

TdzdrK

rdQ

QQdQ d

])([

][

2

T

r

ddrdzrdKdQ

)2(][

2

2

2

T

r

ddrdzrdKdQ

Heat accumulation in the element due to heat flow by conduction in lsquoqrsquo direction

Plane (z- q ) lsquorrsquo direction

dr

TdzrdKQ rr

)(

drQr

QQ rrdrr ][

Quantity of heat leaving from the face EFGH in lsquorrsquo direction

Quantity of heat flowing into the face ABCD in lsquorrsquo direction

Plane (r-z) lsquoq lsquodirection

Quantity of heat leaving from the face AEGC in lsquoqrsquo direction

d

r

TdzdrKQ )(

rdQ

rQQ d ][

Quantity of heat flowing into the face BFHD in lsquoqrsquo direction

)3())((2

2

z

TdzdrrdKdQ zz

)5()()(

dT

CdzdrrdQ pst

)4()()(

ddzdrrdqQgg

Total amount of internal heat generation in the element

Total heat energy stored in the element

(ie thermal energy of the element can be increased because of heat accumulation in the element due to heat flow amp internal heat generation)

according to thermodynamics first law

Ein + Egen = Eout + Est (Ein ndash Eout) + Egen = Est

)( volumem

)()(z

T

zdzdrrdKdQ zz

Heat accumulation in the element due to heat flow by conduction in lsquozrsquo direction

dzdz

TdrrdK

zdQ

QQdQ

zz

dzzzz

])([

d

TmCQ pst )(

Plane (r-q) lsquozrsquo direction

dz

TdrrdKQ zz )(

dzQz

QQ zzdzz ][ Quantity of heat leaving from the face CDHG in lsquozrsquo direction

Quantity of heat flowing into the face ABFE in lsquozrsquo direction

stgzr QQdQdQdQ

)(

r

r ][)(rT

rr

ddrdzdKr

][)(

2

2

2

T

r

ddrdzrdK ))((2

2

z

TddzdrrdK z

ddzdrrdq g )(

dT

Cdzdrrd p )(

+ + +

=

T

CqTK pg)(KKKK zr

[gq

)(1

r

Tr

rrK r

)(1

2

2

2 T

rK )(

2

2

z

TK z

)( ddzdrrd )(

ddzdrrd

TC p

+ + + ] =

If material is homogeneous amp isotropic

)(1

r

Tr

rr

)(1

2

2

2 T

r)(

2

2

z

T

K

q g

T

K

C p

)(1

r

Tr

rr

)(1

2

2

2 T

r )(2

2

z

T

K

q g

T

1

+ + + =

)(1

r

Tr

rrK r

)(1

2

2

2 T

rK )(

2

2

z

TK z

gq

T

C p + + + =

+ + + = Thermal diffusivity

T

K

qT g

12

pC

K

(Ein ndash Eout) + Egen = Est

)( rT

)( r

)sin( drrddr)sin( drrddr

planerdirection

planerdirection

planedirectionr

)(

)(

)(

Spherical polar co-ordinates

Elemental length

Consider an elemental volume ABCDEFGH =

GENERAL HEAT CONDUCTION EQUATION IN SPHERICAL COORDINATES

Let

=internal heat generation per unit volume per unit time (W m3)q= rate of heat flow (W)Q= qdt=total heat flow (J)r=mass density of the material element(kgm3)Cp=specific heat of the material element (Jkg K)

gq

Consider an unsteady (transient) three dimensional heat conduction with internal heat generation in heterogeneous amp anisotropic spherical geometries

LetTemperature distribution is a function of T=

X

Y

Z

drT

drrdKQ rr )sin(

drQr

QQ rrdrr ][

drdr

TdrrdK

rdQ

QQdQ

rr

drrrr

])sin([

)1(]([sin 2

rT

rr

ddrddKdQ rr

Plane ( -q f) lsquorrsquo directionQuantity of heat flowing into the face ABCD in lsquorrsquo direction

Quantity of heat leaving from the face EFGH in lsquorrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquorrsquo direction

drT

drdrKQ )sin(

rdQ

rQQ d ][

rddr

TdrdrK

rdQ

QQdQ d

])sin([

])[(sin

2

T

r

drdrdrKdQ

)2(])(sin[

2

T

r

drdrddrKdQ

Quantity of heat leaving from the face BFHD in lsquoqrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoqrsquo direction

X

Y

Z

Plane (r- f ) lsquoq lsquodirection

Quantity of heat flowing into the face AEGC in lsquoqrsquo direction

dr

TdrrdKQ

sin)(

drQ

rQQ d sin][

sin

drdr

TdrrdK

rdQ

QQdQ d

sin]sin

)([sin

][sinsin

)sin(

T

rr

ddrdrrdKdQ

)3()(sin

)sin(2

2

22

2

T

r

ddddrrKdQ

Plane (r- ) q lsquofrsquo directionQuantity of heat flowing into the face ABFE in lsquofrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquozrsquo direction

)4()sin()(

ddrrddrqQgg

)5()sin()(

dT

CdrrddrQ pst

Total amount of internal heat generation in the element

Total heat energy stored in the element

(Ein ndash Eout) + Egen = Est

d

TmCQ pst )(

)( volumem

Quantity of heat leaving from the face CDHG in lsquofrsquo direction

according to thermodynamics first law

Ein + Egen = Eout + Est

X

Y

Z

stgr QQdQdQdQ

)(

]([sin 2

r

Tr

rddrddKr

])(sin[

2

T

rdrdrddrK

)(sin

)sin(2

2

22

2

T

r

ddddrrK

ddrrddrq g )sin(

dT

Cdrrddr p )sin(

2

2

r

r

sin

sin

]([ 22 r

Tr

rr

Kr

])(sin[sin2

TrK

)(sin 2

2

22

T

r

Kgq

T

Cp

]([1

[ 22 r

Tr

rrK

])(sin[sin1

2

T

r)](

sin

12

2

22 T

r gq

T

Cp

]([1

[ 22 r

Tr

rr

])(sin[sin1

2

T

r)](

sin

12

2

22 T

rK

q g

T

K

Cp

]([1

[ 22 r

Tr

rr

])(sin[sin1

2

T

r)](

sin

12

2

22 T

rK

q g

T1

KKKKr If material is homogeneous amp isotropic

TCqTK pg)(

T

K

qT g

12 pC

K

Thermal diffusivity

(Ein ndash Eout) + Egen = Est

General Heat Conduction Equation

PLANE WALL CYLINDER SPHERE

Temperature profile

Plane wall SphereCylinder

bull A copper conductor (k = 380 WmoC resistivity ρ = 2x10-8 Ω m) having inner and outer radii 10 cm and 225 cm respectively is carrying a current density of 4800 amperes cm2 the conductor is internally cooled and a constant temperature of 65oC is maintained at the inner surface and there is no heat transfer through the insulation surrounding the conductor

Determinebull (i) The maximum temperature of the conductor and the radius at

which it occursbull (ii) The internal heat transfer rate

bull A chemical reaction takes place in a packed bed (k=06 WmoC) between two coaxial cylinders with radii 15 mm and 45 mm the inner surface is at 580oC and it is insulated Assuming the reaction rate of 055 MWm3 in the reactor volume find the temperature at the outer surface of the reactor

bull A current of 300 amperes passes through a stainless steel wire of 25 mm diameter and k=20 WmoC The resistivity of the wire is 70x10-8 ohm-m and length of the wire is 2 m if the wire is submerged in a fluid maintained at 50oC and h at the wire surface is 4000 Wm2oC calculate the steady state temperature at the centre and at the surface of the wire

bull An approximately spherical shaped orange ( k=023 WmoC) 90 mm diameter undergoes ripping process and generates 5100 Wm3 of energy If external surface the orange is at 8oC determine

bull (i) temperature at the centre of the orange and

bull (ii) heat flow from the outer surface of the orange

bull FINS External surfaces to used to cooling area and hence convection

heat transfer can be improvedbull APPLICATIONS IC engine cylinder ( air cooling) Heat exchangers automobile

radiators computer chip refrigerators electric motors generators and transformers etc

bull ASSUMPTIONS MADE IN ANALYSIS OF HEAT TRANFER THROUGH FIN

1 Fin material is homogenous and isotropic constthermal conductivity

2 Temperature at any cs of the fin is uniform3 lsquohrsquo is uniform over entire surface4 No internal heat generation5 Thermal contact resistance is negligible6 Radiation heat transfer is negligible7 Heat conduction is steady state and one dimensional

bull Problem 1

bull A stainless steel fin (k=20 WmK) having a diameter of 20 mm and a length of 01 m is attached to a wall at 300oC The ambient temperature is 50oC and the heat transfer coefficient is 10 Wm2K The fin tip is insulated Determine

bull (A) The rate of heat dissipation from the finbull (B) The temperature at the fin tipbull (C) The rate of heat transfer from the wall area covered by the fin if

the fin was not usedbull (D) The heat transfer rate from the same fin geometry if the

stainless steel fin is replaced by a fictitious fin with infinite thermal conductivity

bull Problem2bull One end of a long rod is inserted into a furnace while the other

projects into ambient temperature Under steady state the temperature of the rod is measured at two points 75mm apart and found to be 125oC and 885oC respectively while the ambient temperature is 20oC If the rod is 25mm in diameter and h is 2336 Wm2K

bull Find the thermal conductivity of the rod material

  • HEAT AND MASS TRANSFER
  • Thermodynamics Vs Heat transfer
  • MODE OF HEAT TRANSFER
  • MODE OF HEAT TRANSFER (2)
  • CONDUCTION- MECHANISM
  • CONCECTION-MECHANISM
  • RADIATION-MECHANISM
  • Conduction- Convection- Radiation
  • Slide 9
  • Slide 10
  • Fourierrsquos Law of Heat Conduction
  • Slide 12
  • Newtonrsquos Law of Cooling
  • Laws of Thermal Radiation
  • LAWS FOR HEAT TRANSFER
  • ELECTRICAL ampTHERMAL ANALOGY
  • Slide 17
  • Slide 18
  • PLANE WALL-UNIDIRECTIONAL
  • PLANE WALL -SERIES
  • PLANE WALL IN SERIES
  • PLANE WALL -PARALLEL
  • PLANE WALL-PARALLEL
  • PLANE WALL-COMBINATION
  • PLANE WALL-FILM RESISTANCE
  • PLANE WALL- THERMAL CONTACT RESISTANCE
  • PLANE WALL- THERMAL CONTACT RESISTANCE (2)
  • CYLINDER-SOLID
  • HEAT TRANSFER IN SOLID CYLINDER
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER-COMPOSITE
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER-MULTY LAYER
  • HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER
  • HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER (2)
  • HEAT TRANSFER IN SPHERE
  • HEAT TRANSFER IN SPHERE -MULTY LAYER
  • HEAT TRANSFER IN SPHERE -MULTY LAYER (2)
  • Slide 38
  • Slide 39
  • Slide 40
  • Slide 41
  • CRITICAL RADIUS
  • Plane wall- Insulation thickness -study
  • Problem5
  • Slide 45
  • Slide 46
  • Slide 47
  • GENERAL HEAT CONDUCTION EQUATION IN CYLINDRICAL COORDINATES
  • Slide 49
  • Slide 50
  • Slide 51
  • GENERAL HEAT CONDUCTION EQUATION IN SPHERICAL COORDINATES
  • Slide 53
  • Slide 54
  • Slide 55
  • General Heat Conduction Equation
  • Temperature profile
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Slide 67
Page 44: hmt chap 1

)( zr )( dzrddr

)( dzrddr

Cylindrical polar co-ordinates

Elemental length

Consider an elemental volume ABCDEFGH =

GENERAL HEAT CONDUCTION EQUATION IN CYLINDRICAL COORDINATES

gq

gq

Consider an unsteady (transient) three dimensional heat conduction with internal heat generation in heterogeneous amp anisotropic cylindrical geometries (eg rods and pipes)

)( zrTLetTemperature distribution is a function of T=

Let

=internal heat generation per unit volume per unit time (W m3)q= rate of heat flow (W)Q= qdt=total heat flow (J)r=mass density of the material element(kgm3)Cp=specific heat of the material element (Jkg K)

planerdirectionz

planezrdirection

planezdirectionr

)(

)(

)(

LetHeat flow in

drdr

TdzrdK

rdQ

QQdQ

rr

drrrr

])([

)1(][

r

Tr

rddrdzdKdQ rr

Heat accumulation in the element due to heat flow by conduction in lsquorrsquo direction

rddr

TdzdrK

rdQ

QQdQ d

])([

][

2

T

r

ddrdzrdKdQ

)2(][

2

2

2

T

r

ddrdzrdKdQ

Heat accumulation in the element due to heat flow by conduction in lsquoqrsquo direction

Plane (z- q ) lsquorrsquo direction

dr

TdzrdKQ rr

)(

drQr

QQ rrdrr ][

Quantity of heat leaving from the face EFGH in lsquorrsquo direction

Quantity of heat flowing into the face ABCD in lsquorrsquo direction

Plane (r-z) lsquoq lsquodirection

Quantity of heat leaving from the face AEGC in lsquoqrsquo direction

d

r

TdzdrKQ )(

rdQ

rQQ d ][

Quantity of heat flowing into the face BFHD in lsquoqrsquo direction

)3())((2

2

z

TdzdrrdKdQ zz

)5()()(

dT

CdzdrrdQ pst

)4()()(

ddzdrrdqQgg

Total amount of internal heat generation in the element

Total heat energy stored in the element

(ie thermal energy of the element can be increased because of heat accumulation in the element due to heat flow amp internal heat generation)

according to thermodynamics first law

Ein + Egen = Eout + Est (Ein ndash Eout) + Egen = Est

)( volumem

)()(z

T

zdzdrrdKdQ zz

Heat accumulation in the element due to heat flow by conduction in lsquozrsquo direction

dzdz

TdrrdK

zdQ

QQdQ

zz

dzzzz

])([

d

TmCQ pst )(

Plane (r-q) lsquozrsquo direction

dz

TdrrdKQ zz )(

dzQz

QQ zzdzz ][ Quantity of heat leaving from the face CDHG in lsquozrsquo direction

Quantity of heat flowing into the face ABFE in lsquozrsquo direction

stgzr QQdQdQdQ

)(

r

r ][)(rT

rr

ddrdzdKr

][)(

2

2

2

T

r

ddrdzrdK ))((2

2

z

TddzdrrdK z

ddzdrrdq g )(

dT

Cdzdrrd p )(

+ + +

=

T

CqTK pg)(KKKK zr

[gq

)(1

r

Tr

rrK r

)(1

2

2

2 T

rK )(

2

2

z

TK z

)( ddzdrrd )(

ddzdrrd

TC p

+ + + ] =

If material is homogeneous amp isotropic

)(1

r

Tr

rr

)(1

2

2

2 T

r)(

2

2

z

T

K

q g

T

K

C p

)(1

r

Tr

rr

)(1

2

2

2 T

r )(2

2

z

T

K

q g

T

1

+ + + =

)(1

r

Tr

rrK r

)(1

2

2

2 T

rK )(

2

2

z

TK z

gq

T

C p + + + =

+ + + = Thermal diffusivity

T

K

qT g

12

pC

K

(Ein ndash Eout) + Egen = Est

)( rT

)( r

)sin( drrddr)sin( drrddr

planerdirection

planerdirection

planedirectionr

)(

)(

)(

Spherical polar co-ordinates

Elemental length

Consider an elemental volume ABCDEFGH =

GENERAL HEAT CONDUCTION EQUATION IN SPHERICAL COORDINATES

Let

=internal heat generation per unit volume per unit time (W m3)q= rate of heat flow (W)Q= qdt=total heat flow (J)r=mass density of the material element(kgm3)Cp=specific heat of the material element (Jkg K)

gq

Consider an unsteady (transient) three dimensional heat conduction with internal heat generation in heterogeneous amp anisotropic spherical geometries

LetTemperature distribution is a function of T=

X

Y

Z

drT

drrdKQ rr )sin(

drQr

QQ rrdrr ][

drdr

TdrrdK

rdQ

QQdQ

rr

drrrr

])sin([

)1(]([sin 2

rT

rr

ddrddKdQ rr

Plane ( -q f) lsquorrsquo directionQuantity of heat flowing into the face ABCD in lsquorrsquo direction

Quantity of heat leaving from the face EFGH in lsquorrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquorrsquo direction

drT

drdrKQ )sin(

rdQ

rQQ d ][

rddr

TdrdrK

rdQ

QQdQ d

])sin([

])[(sin

2

T

r

drdrdrKdQ

)2(])(sin[

2

T

r

drdrddrKdQ

Quantity of heat leaving from the face BFHD in lsquoqrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoqrsquo direction

X

Y

Z

Plane (r- f ) lsquoq lsquodirection

Quantity of heat flowing into the face AEGC in lsquoqrsquo direction

dr

TdrrdKQ

sin)(

drQ

rQQ d sin][

sin

drdr

TdrrdK

rdQ

QQdQ d

sin]sin

)([sin

][sinsin

)sin(

T

rr

ddrdrrdKdQ

)3()(sin

)sin(2

2

22

2

T

r

ddddrrKdQ

Plane (r- ) q lsquofrsquo directionQuantity of heat flowing into the face ABFE in lsquofrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquozrsquo direction

)4()sin()(

ddrrddrqQgg

)5()sin()(

dT

CdrrddrQ pst

Total amount of internal heat generation in the element

Total heat energy stored in the element

(Ein ndash Eout) + Egen = Est

d

TmCQ pst )(

)( volumem

Quantity of heat leaving from the face CDHG in lsquofrsquo direction

according to thermodynamics first law

Ein + Egen = Eout + Est

X

Y

Z

stgr QQdQdQdQ

)(

]([sin 2

r

Tr

rddrddKr

])(sin[

2

T

rdrdrddrK

)(sin

)sin(2

2

22

2

T

r

ddddrrK

ddrrddrq g )sin(

dT

Cdrrddr p )sin(

2

2

r

r

sin

sin

]([ 22 r

Tr

rr

Kr

])(sin[sin2

TrK

)(sin 2

2

22

T

r

Kgq

T

Cp

]([1

[ 22 r

Tr

rrK

])(sin[sin1

2

T

r)](

sin

12

2

22 T

r gq

T

Cp

]([1

[ 22 r

Tr

rr

])(sin[sin1

2

T

r)](

sin

12

2

22 T

rK

q g

T

K

Cp

]([1

[ 22 r

Tr

rr

])(sin[sin1

2

T

r)](

sin

12

2

22 T

rK

q g

T1

KKKKr If material is homogeneous amp isotropic

TCqTK pg)(

T

K

qT g

12 pC

K

Thermal diffusivity

(Ein ndash Eout) + Egen = Est

General Heat Conduction Equation

PLANE WALL CYLINDER SPHERE

Temperature profile

Plane wall SphereCylinder

bull A copper conductor (k = 380 WmoC resistivity ρ = 2x10-8 Ω m) having inner and outer radii 10 cm and 225 cm respectively is carrying a current density of 4800 amperes cm2 the conductor is internally cooled and a constant temperature of 65oC is maintained at the inner surface and there is no heat transfer through the insulation surrounding the conductor

Determinebull (i) The maximum temperature of the conductor and the radius at

which it occursbull (ii) The internal heat transfer rate

bull A chemical reaction takes place in a packed bed (k=06 WmoC) between two coaxial cylinders with radii 15 mm and 45 mm the inner surface is at 580oC and it is insulated Assuming the reaction rate of 055 MWm3 in the reactor volume find the temperature at the outer surface of the reactor

bull A current of 300 amperes passes through a stainless steel wire of 25 mm diameter and k=20 WmoC The resistivity of the wire is 70x10-8 ohm-m and length of the wire is 2 m if the wire is submerged in a fluid maintained at 50oC and h at the wire surface is 4000 Wm2oC calculate the steady state temperature at the centre and at the surface of the wire

bull An approximately spherical shaped orange ( k=023 WmoC) 90 mm diameter undergoes ripping process and generates 5100 Wm3 of energy If external surface the orange is at 8oC determine

bull (i) temperature at the centre of the orange and

bull (ii) heat flow from the outer surface of the orange

bull FINS External surfaces to used to cooling area and hence convection

heat transfer can be improvedbull APPLICATIONS IC engine cylinder ( air cooling) Heat exchangers automobile

radiators computer chip refrigerators electric motors generators and transformers etc

bull ASSUMPTIONS MADE IN ANALYSIS OF HEAT TRANFER THROUGH FIN

1 Fin material is homogenous and isotropic constthermal conductivity

2 Temperature at any cs of the fin is uniform3 lsquohrsquo is uniform over entire surface4 No internal heat generation5 Thermal contact resistance is negligible6 Radiation heat transfer is negligible7 Heat conduction is steady state and one dimensional

bull Problem 1

bull A stainless steel fin (k=20 WmK) having a diameter of 20 mm and a length of 01 m is attached to a wall at 300oC The ambient temperature is 50oC and the heat transfer coefficient is 10 Wm2K The fin tip is insulated Determine

bull (A) The rate of heat dissipation from the finbull (B) The temperature at the fin tipbull (C) The rate of heat transfer from the wall area covered by the fin if

the fin was not usedbull (D) The heat transfer rate from the same fin geometry if the

stainless steel fin is replaced by a fictitious fin with infinite thermal conductivity

bull Problem2bull One end of a long rod is inserted into a furnace while the other

projects into ambient temperature Under steady state the temperature of the rod is measured at two points 75mm apart and found to be 125oC and 885oC respectively while the ambient temperature is 20oC If the rod is 25mm in diameter and h is 2336 Wm2K

bull Find the thermal conductivity of the rod material

  • HEAT AND MASS TRANSFER
  • Thermodynamics Vs Heat transfer
  • MODE OF HEAT TRANSFER
  • MODE OF HEAT TRANSFER (2)
  • CONDUCTION- MECHANISM
  • CONCECTION-MECHANISM
  • RADIATION-MECHANISM
  • Conduction- Convection- Radiation
  • Slide 9
  • Slide 10
  • Fourierrsquos Law of Heat Conduction
  • Slide 12
  • Newtonrsquos Law of Cooling
  • Laws of Thermal Radiation
  • LAWS FOR HEAT TRANSFER
  • ELECTRICAL ampTHERMAL ANALOGY
  • Slide 17
  • Slide 18
  • PLANE WALL-UNIDIRECTIONAL
  • PLANE WALL -SERIES
  • PLANE WALL IN SERIES
  • PLANE WALL -PARALLEL
  • PLANE WALL-PARALLEL
  • PLANE WALL-COMBINATION
  • PLANE WALL-FILM RESISTANCE
  • PLANE WALL- THERMAL CONTACT RESISTANCE
  • PLANE WALL- THERMAL CONTACT RESISTANCE (2)
  • CYLINDER-SOLID
  • HEAT TRANSFER IN SOLID CYLINDER
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER-COMPOSITE
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER-MULTY LAYER
  • HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER
  • HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER (2)
  • HEAT TRANSFER IN SPHERE
  • HEAT TRANSFER IN SPHERE -MULTY LAYER
  • HEAT TRANSFER IN SPHERE -MULTY LAYER (2)
  • Slide 38
  • Slide 39
  • Slide 40
  • Slide 41
  • CRITICAL RADIUS
  • Plane wall- Insulation thickness -study
  • Problem5
  • Slide 45
  • Slide 46
  • Slide 47
  • GENERAL HEAT CONDUCTION EQUATION IN CYLINDRICAL COORDINATES
  • Slide 49
  • Slide 50
  • Slide 51
  • GENERAL HEAT CONDUCTION EQUATION IN SPHERICAL COORDINATES
  • Slide 53
  • Slide 54
  • Slide 55
  • General Heat Conduction Equation
  • Temperature profile
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Slide 67
Page 45: hmt chap 1

drdr

TdzrdK

rdQ

QQdQ

rr

drrrr

])([

)1(][

r

Tr

rddrdzdKdQ rr

Heat accumulation in the element due to heat flow by conduction in lsquorrsquo direction

rddr

TdzdrK

rdQ

QQdQ d

])([

][

2

T

r

ddrdzrdKdQ

)2(][

2

2

2

T

r

ddrdzrdKdQ

Heat accumulation in the element due to heat flow by conduction in lsquoqrsquo direction

Plane (z- q ) lsquorrsquo direction

dr

TdzrdKQ rr

)(

drQr

QQ rrdrr ][

Quantity of heat leaving from the face EFGH in lsquorrsquo direction

Quantity of heat flowing into the face ABCD in lsquorrsquo direction

Plane (r-z) lsquoq lsquodirection

Quantity of heat leaving from the face AEGC in lsquoqrsquo direction

d

r

TdzdrKQ )(

rdQ

rQQ d ][

Quantity of heat flowing into the face BFHD in lsquoqrsquo direction

)3())((2

2

z

TdzdrrdKdQ zz

)5()()(

dT

CdzdrrdQ pst

)4()()(

ddzdrrdqQgg

Total amount of internal heat generation in the element

Total heat energy stored in the element

(ie thermal energy of the element can be increased because of heat accumulation in the element due to heat flow amp internal heat generation)

according to thermodynamics first law

Ein + Egen = Eout + Est (Ein ndash Eout) + Egen = Est

)( volumem

)()(z

T

zdzdrrdKdQ zz

Heat accumulation in the element due to heat flow by conduction in lsquozrsquo direction

dzdz

TdrrdK

zdQ

QQdQ

zz

dzzzz

])([

d

TmCQ pst )(

Plane (r-q) lsquozrsquo direction

dz

TdrrdKQ zz )(

dzQz

QQ zzdzz ][ Quantity of heat leaving from the face CDHG in lsquozrsquo direction

Quantity of heat flowing into the face ABFE in lsquozrsquo direction

stgzr QQdQdQdQ

)(

r

r ][)(rT

rr

ddrdzdKr

][)(

2

2

2

T

r

ddrdzrdK ))((2

2

z

TddzdrrdK z

ddzdrrdq g )(

dT

Cdzdrrd p )(

+ + +

=

T

CqTK pg)(KKKK zr

[gq

)(1

r

Tr

rrK r

)(1

2

2

2 T

rK )(

2

2

z

TK z

)( ddzdrrd )(

ddzdrrd

TC p

+ + + ] =

If material is homogeneous amp isotropic

)(1

r

Tr

rr

)(1

2

2

2 T

r)(

2

2

z

T

K

q g

T

K

C p

)(1

r

Tr

rr

)(1

2

2

2 T

r )(2

2

z

T

K

q g

T

1

+ + + =

)(1

r

Tr

rrK r

)(1

2

2

2 T

rK )(

2

2

z

TK z

gq

T

C p + + + =

+ + + = Thermal diffusivity

T

K

qT g

12

pC

K

(Ein ndash Eout) + Egen = Est

)( rT

)( r

)sin( drrddr)sin( drrddr

planerdirection

planerdirection

planedirectionr

)(

)(

)(

Spherical polar co-ordinates

Elemental length

Consider an elemental volume ABCDEFGH =

GENERAL HEAT CONDUCTION EQUATION IN SPHERICAL COORDINATES

Let

=internal heat generation per unit volume per unit time (W m3)q= rate of heat flow (W)Q= qdt=total heat flow (J)r=mass density of the material element(kgm3)Cp=specific heat of the material element (Jkg K)

gq

Consider an unsteady (transient) three dimensional heat conduction with internal heat generation in heterogeneous amp anisotropic spherical geometries

LetTemperature distribution is a function of T=

X

Y

Z

drT

drrdKQ rr )sin(

drQr

QQ rrdrr ][

drdr

TdrrdK

rdQ

QQdQ

rr

drrrr

])sin([

)1(]([sin 2

rT

rr

ddrddKdQ rr

Plane ( -q f) lsquorrsquo directionQuantity of heat flowing into the face ABCD in lsquorrsquo direction

Quantity of heat leaving from the face EFGH in lsquorrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquorrsquo direction

drT

drdrKQ )sin(

rdQ

rQQ d ][

rddr

TdrdrK

rdQ

QQdQ d

])sin([

])[(sin

2

T

r

drdrdrKdQ

)2(])(sin[

2

T

r

drdrddrKdQ

Quantity of heat leaving from the face BFHD in lsquoqrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoqrsquo direction

X

Y

Z

Plane (r- f ) lsquoq lsquodirection

Quantity of heat flowing into the face AEGC in lsquoqrsquo direction

dr

TdrrdKQ

sin)(

drQ

rQQ d sin][

sin

drdr

TdrrdK

rdQ

QQdQ d

sin]sin

)([sin

][sinsin

)sin(

T

rr

ddrdrrdKdQ

)3()(sin

)sin(2

2

22

2

T

r

ddddrrKdQ

Plane (r- ) q lsquofrsquo directionQuantity of heat flowing into the face ABFE in lsquofrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquozrsquo direction

)4()sin()(

ddrrddrqQgg

)5()sin()(

dT

CdrrddrQ pst

Total amount of internal heat generation in the element

Total heat energy stored in the element

(Ein ndash Eout) + Egen = Est

d

TmCQ pst )(

)( volumem

Quantity of heat leaving from the face CDHG in lsquofrsquo direction

according to thermodynamics first law

Ein + Egen = Eout + Est

X

Y

Z

stgr QQdQdQdQ

)(

]([sin 2

r

Tr

rddrddKr

])(sin[

2

T

rdrdrddrK

)(sin

)sin(2

2

22

2

T

r

ddddrrK

ddrrddrq g )sin(

dT

Cdrrddr p )sin(

2

2

r

r

sin

sin

]([ 22 r

Tr

rr

Kr

])(sin[sin2

TrK

)(sin 2

2

22

T

r

Kgq

T

Cp

]([1

[ 22 r

Tr

rrK

])(sin[sin1

2

T

r)](

sin

12

2

22 T

r gq

T

Cp

]([1

[ 22 r

Tr

rr

])(sin[sin1

2

T

r)](

sin

12

2

22 T

rK

q g

T

K

Cp

]([1

[ 22 r

Tr

rr

])(sin[sin1

2

T

r)](

sin

12

2

22 T

rK

q g

T1

KKKKr If material is homogeneous amp isotropic

TCqTK pg)(

T

K

qT g

12 pC

K

Thermal diffusivity

(Ein ndash Eout) + Egen = Est

General Heat Conduction Equation

PLANE WALL CYLINDER SPHERE

Temperature profile

Plane wall SphereCylinder

bull A copper conductor (k = 380 WmoC resistivity ρ = 2x10-8 Ω m) having inner and outer radii 10 cm and 225 cm respectively is carrying a current density of 4800 amperes cm2 the conductor is internally cooled and a constant temperature of 65oC is maintained at the inner surface and there is no heat transfer through the insulation surrounding the conductor

Determinebull (i) The maximum temperature of the conductor and the radius at

which it occursbull (ii) The internal heat transfer rate

bull A chemical reaction takes place in a packed bed (k=06 WmoC) between two coaxial cylinders with radii 15 mm and 45 mm the inner surface is at 580oC and it is insulated Assuming the reaction rate of 055 MWm3 in the reactor volume find the temperature at the outer surface of the reactor

bull A current of 300 amperes passes through a stainless steel wire of 25 mm diameter and k=20 WmoC The resistivity of the wire is 70x10-8 ohm-m and length of the wire is 2 m if the wire is submerged in a fluid maintained at 50oC and h at the wire surface is 4000 Wm2oC calculate the steady state temperature at the centre and at the surface of the wire

bull An approximately spherical shaped orange ( k=023 WmoC) 90 mm diameter undergoes ripping process and generates 5100 Wm3 of energy If external surface the orange is at 8oC determine

bull (i) temperature at the centre of the orange and

bull (ii) heat flow from the outer surface of the orange

bull FINS External surfaces to used to cooling area and hence convection

heat transfer can be improvedbull APPLICATIONS IC engine cylinder ( air cooling) Heat exchangers automobile

radiators computer chip refrigerators electric motors generators and transformers etc

bull ASSUMPTIONS MADE IN ANALYSIS OF HEAT TRANFER THROUGH FIN

1 Fin material is homogenous and isotropic constthermal conductivity

2 Temperature at any cs of the fin is uniform3 lsquohrsquo is uniform over entire surface4 No internal heat generation5 Thermal contact resistance is negligible6 Radiation heat transfer is negligible7 Heat conduction is steady state and one dimensional

bull Problem 1

bull A stainless steel fin (k=20 WmK) having a diameter of 20 mm and a length of 01 m is attached to a wall at 300oC The ambient temperature is 50oC and the heat transfer coefficient is 10 Wm2K The fin tip is insulated Determine

bull (A) The rate of heat dissipation from the finbull (B) The temperature at the fin tipbull (C) The rate of heat transfer from the wall area covered by the fin if

the fin was not usedbull (D) The heat transfer rate from the same fin geometry if the

stainless steel fin is replaced by a fictitious fin with infinite thermal conductivity

bull Problem2bull One end of a long rod is inserted into a furnace while the other

projects into ambient temperature Under steady state the temperature of the rod is measured at two points 75mm apart and found to be 125oC and 885oC respectively while the ambient temperature is 20oC If the rod is 25mm in diameter and h is 2336 Wm2K

bull Find the thermal conductivity of the rod material

  • HEAT AND MASS TRANSFER
  • Thermodynamics Vs Heat transfer
  • MODE OF HEAT TRANSFER
  • MODE OF HEAT TRANSFER (2)
  • CONDUCTION- MECHANISM
  • CONCECTION-MECHANISM
  • RADIATION-MECHANISM
  • Conduction- Convection- Radiation
  • Slide 9
  • Slide 10
  • Fourierrsquos Law of Heat Conduction
  • Slide 12
  • Newtonrsquos Law of Cooling
  • Laws of Thermal Radiation
  • LAWS FOR HEAT TRANSFER
  • ELECTRICAL ampTHERMAL ANALOGY
  • Slide 17
  • Slide 18
  • PLANE WALL-UNIDIRECTIONAL
  • PLANE WALL -SERIES
  • PLANE WALL IN SERIES
  • PLANE WALL -PARALLEL
  • PLANE WALL-PARALLEL
  • PLANE WALL-COMBINATION
  • PLANE WALL-FILM RESISTANCE
  • PLANE WALL- THERMAL CONTACT RESISTANCE
  • PLANE WALL- THERMAL CONTACT RESISTANCE (2)
  • CYLINDER-SOLID
  • HEAT TRANSFER IN SOLID CYLINDER
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER-COMPOSITE
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER-MULTY LAYER
  • HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER
  • HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER (2)
  • HEAT TRANSFER IN SPHERE
  • HEAT TRANSFER IN SPHERE -MULTY LAYER
  • HEAT TRANSFER IN SPHERE -MULTY LAYER (2)
  • Slide 38
  • Slide 39
  • Slide 40
  • Slide 41
  • CRITICAL RADIUS
  • Plane wall- Insulation thickness -study
  • Problem5
  • Slide 45
  • Slide 46
  • Slide 47
  • GENERAL HEAT CONDUCTION EQUATION IN CYLINDRICAL COORDINATES
  • Slide 49
  • Slide 50
  • Slide 51
  • GENERAL HEAT CONDUCTION EQUATION IN SPHERICAL COORDINATES
  • Slide 53
  • Slide 54
  • Slide 55
  • General Heat Conduction Equation
  • Temperature profile
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Slide 67
Page 46: hmt chap 1

)3())((2

2

z

TdzdrrdKdQ zz

)5()()(

dT

CdzdrrdQ pst

)4()()(

ddzdrrdqQgg

Total amount of internal heat generation in the element

Total heat energy stored in the element

(ie thermal energy of the element can be increased because of heat accumulation in the element due to heat flow amp internal heat generation)

according to thermodynamics first law

Ein + Egen = Eout + Est (Ein ndash Eout) + Egen = Est

)( volumem

)()(z

T

zdzdrrdKdQ zz

Heat accumulation in the element due to heat flow by conduction in lsquozrsquo direction

dzdz

TdrrdK

zdQ

QQdQ

zz

dzzzz

])([

d

TmCQ pst )(

Plane (r-q) lsquozrsquo direction

dz

TdrrdKQ zz )(

dzQz

QQ zzdzz ][ Quantity of heat leaving from the face CDHG in lsquozrsquo direction

Quantity of heat flowing into the face ABFE in lsquozrsquo direction

stgzr QQdQdQdQ

)(

r

r ][)(rT

rr

ddrdzdKr

][)(

2

2

2

T

r

ddrdzrdK ))((2

2

z

TddzdrrdK z

ddzdrrdq g )(

dT

Cdzdrrd p )(

+ + +

=

T

CqTK pg)(KKKK zr

[gq

)(1

r

Tr

rrK r

)(1

2

2

2 T

rK )(

2

2

z

TK z

)( ddzdrrd )(

ddzdrrd

TC p

+ + + ] =

If material is homogeneous amp isotropic

)(1

r

Tr

rr

)(1

2

2

2 T

r)(

2

2

z

T

K

q g

T

K

C p

)(1

r

Tr

rr

)(1

2

2

2 T

r )(2

2

z

T

K

q g

T

1

+ + + =

)(1

r

Tr

rrK r

)(1

2

2

2 T

rK )(

2

2

z

TK z

gq

T

C p + + + =

+ + + = Thermal diffusivity

T

K

qT g

12

pC

K

(Ein ndash Eout) + Egen = Est

)( rT

)( r

)sin( drrddr)sin( drrddr

planerdirection

planerdirection

planedirectionr

)(

)(

)(

Spherical polar co-ordinates

Elemental length

Consider an elemental volume ABCDEFGH =

GENERAL HEAT CONDUCTION EQUATION IN SPHERICAL COORDINATES

Let

=internal heat generation per unit volume per unit time (W m3)q= rate of heat flow (W)Q= qdt=total heat flow (J)r=mass density of the material element(kgm3)Cp=specific heat of the material element (Jkg K)

gq

Consider an unsteady (transient) three dimensional heat conduction with internal heat generation in heterogeneous amp anisotropic spherical geometries

LetTemperature distribution is a function of T=

X

Y

Z

drT

drrdKQ rr )sin(

drQr

QQ rrdrr ][

drdr

TdrrdK

rdQ

QQdQ

rr

drrrr

])sin([

)1(]([sin 2

rT

rr

ddrddKdQ rr

Plane ( -q f) lsquorrsquo directionQuantity of heat flowing into the face ABCD in lsquorrsquo direction

Quantity of heat leaving from the face EFGH in lsquorrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquorrsquo direction

drT

drdrKQ )sin(

rdQ

rQQ d ][

rddr

TdrdrK

rdQ

QQdQ d

])sin([

])[(sin

2

T

r

drdrdrKdQ

)2(])(sin[

2

T

r

drdrddrKdQ

Quantity of heat leaving from the face BFHD in lsquoqrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoqrsquo direction

X

Y

Z

Plane (r- f ) lsquoq lsquodirection

Quantity of heat flowing into the face AEGC in lsquoqrsquo direction

dr

TdrrdKQ

sin)(

drQ

rQQ d sin][

sin

drdr

TdrrdK

rdQ

QQdQ d

sin]sin

)([sin

][sinsin

)sin(

T

rr

ddrdrrdKdQ

)3()(sin

)sin(2

2

22

2

T

r

ddddrrKdQ

Plane (r- ) q lsquofrsquo directionQuantity of heat flowing into the face ABFE in lsquofrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquozrsquo direction

)4()sin()(

ddrrddrqQgg

)5()sin()(

dT

CdrrddrQ pst

Total amount of internal heat generation in the element

Total heat energy stored in the element

(Ein ndash Eout) + Egen = Est

d

TmCQ pst )(

)( volumem

Quantity of heat leaving from the face CDHG in lsquofrsquo direction

according to thermodynamics first law

Ein + Egen = Eout + Est

X

Y

Z

stgr QQdQdQdQ

)(

]([sin 2

r

Tr

rddrddKr

])(sin[

2

T

rdrdrddrK

)(sin

)sin(2

2

22

2

T

r

ddddrrK

ddrrddrq g )sin(

dT

Cdrrddr p )sin(

2

2

r

r

sin

sin

]([ 22 r

Tr

rr

Kr

])(sin[sin2

TrK

)(sin 2

2

22

T

r

Kgq

T

Cp

]([1

[ 22 r

Tr

rrK

])(sin[sin1

2

T

r)](

sin

12

2

22 T

r gq

T

Cp

]([1

[ 22 r

Tr

rr

])(sin[sin1

2

T

r)](

sin

12

2

22 T

rK

q g

T

K

Cp

]([1

[ 22 r

Tr

rr

])(sin[sin1

2

T

r)](

sin

12

2

22 T

rK

q g

T1

KKKKr If material is homogeneous amp isotropic

TCqTK pg)(

T

K

qT g

12 pC

K

Thermal diffusivity

(Ein ndash Eout) + Egen = Est

General Heat Conduction Equation

PLANE WALL CYLINDER SPHERE

Temperature profile

Plane wall SphereCylinder

bull A copper conductor (k = 380 WmoC resistivity ρ = 2x10-8 Ω m) having inner and outer radii 10 cm and 225 cm respectively is carrying a current density of 4800 amperes cm2 the conductor is internally cooled and a constant temperature of 65oC is maintained at the inner surface and there is no heat transfer through the insulation surrounding the conductor

Determinebull (i) The maximum temperature of the conductor and the radius at

which it occursbull (ii) The internal heat transfer rate

bull A chemical reaction takes place in a packed bed (k=06 WmoC) between two coaxial cylinders with radii 15 mm and 45 mm the inner surface is at 580oC and it is insulated Assuming the reaction rate of 055 MWm3 in the reactor volume find the temperature at the outer surface of the reactor

bull A current of 300 amperes passes through a stainless steel wire of 25 mm diameter and k=20 WmoC The resistivity of the wire is 70x10-8 ohm-m and length of the wire is 2 m if the wire is submerged in a fluid maintained at 50oC and h at the wire surface is 4000 Wm2oC calculate the steady state temperature at the centre and at the surface of the wire

bull An approximately spherical shaped orange ( k=023 WmoC) 90 mm diameter undergoes ripping process and generates 5100 Wm3 of energy If external surface the orange is at 8oC determine

bull (i) temperature at the centre of the orange and

bull (ii) heat flow from the outer surface of the orange

bull FINS External surfaces to used to cooling area and hence convection

heat transfer can be improvedbull APPLICATIONS IC engine cylinder ( air cooling) Heat exchangers automobile

radiators computer chip refrigerators electric motors generators and transformers etc

bull ASSUMPTIONS MADE IN ANALYSIS OF HEAT TRANFER THROUGH FIN

1 Fin material is homogenous and isotropic constthermal conductivity

2 Temperature at any cs of the fin is uniform3 lsquohrsquo is uniform over entire surface4 No internal heat generation5 Thermal contact resistance is negligible6 Radiation heat transfer is negligible7 Heat conduction is steady state and one dimensional

bull Problem 1

bull A stainless steel fin (k=20 WmK) having a diameter of 20 mm and a length of 01 m is attached to a wall at 300oC The ambient temperature is 50oC and the heat transfer coefficient is 10 Wm2K The fin tip is insulated Determine

bull (A) The rate of heat dissipation from the finbull (B) The temperature at the fin tipbull (C) The rate of heat transfer from the wall area covered by the fin if

the fin was not usedbull (D) The heat transfer rate from the same fin geometry if the

stainless steel fin is replaced by a fictitious fin with infinite thermal conductivity

bull Problem2bull One end of a long rod is inserted into a furnace while the other

projects into ambient temperature Under steady state the temperature of the rod is measured at two points 75mm apart and found to be 125oC and 885oC respectively while the ambient temperature is 20oC If the rod is 25mm in diameter and h is 2336 Wm2K

bull Find the thermal conductivity of the rod material

  • HEAT AND MASS TRANSFER
  • Thermodynamics Vs Heat transfer
  • MODE OF HEAT TRANSFER
  • MODE OF HEAT TRANSFER (2)
  • CONDUCTION- MECHANISM
  • CONCECTION-MECHANISM
  • RADIATION-MECHANISM
  • Conduction- Convection- Radiation
  • Slide 9
  • Slide 10
  • Fourierrsquos Law of Heat Conduction
  • Slide 12
  • Newtonrsquos Law of Cooling
  • Laws of Thermal Radiation
  • LAWS FOR HEAT TRANSFER
  • ELECTRICAL ampTHERMAL ANALOGY
  • Slide 17
  • Slide 18
  • PLANE WALL-UNIDIRECTIONAL
  • PLANE WALL -SERIES
  • PLANE WALL IN SERIES
  • PLANE WALL -PARALLEL
  • PLANE WALL-PARALLEL
  • PLANE WALL-COMBINATION
  • PLANE WALL-FILM RESISTANCE
  • PLANE WALL- THERMAL CONTACT RESISTANCE
  • PLANE WALL- THERMAL CONTACT RESISTANCE (2)
  • CYLINDER-SOLID
  • HEAT TRANSFER IN SOLID CYLINDER
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER-COMPOSITE
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER-MULTY LAYER
  • HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER
  • HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER (2)
  • HEAT TRANSFER IN SPHERE
  • HEAT TRANSFER IN SPHERE -MULTY LAYER
  • HEAT TRANSFER IN SPHERE -MULTY LAYER (2)
  • Slide 38
  • Slide 39
  • Slide 40
  • Slide 41
  • CRITICAL RADIUS
  • Plane wall- Insulation thickness -study
  • Problem5
  • Slide 45
  • Slide 46
  • Slide 47
  • GENERAL HEAT CONDUCTION EQUATION IN CYLINDRICAL COORDINATES
  • Slide 49
  • Slide 50
  • Slide 51
  • GENERAL HEAT CONDUCTION EQUATION IN SPHERICAL COORDINATES
  • Slide 53
  • Slide 54
  • Slide 55
  • General Heat Conduction Equation
  • Temperature profile
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Slide 67
Page 47: hmt chap 1

stgzr QQdQdQdQ

)(

r

r ][)(rT

rr

ddrdzdKr

][)(

2

2

2

T

r

ddrdzrdK ))((2

2

z

TddzdrrdK z

ddzdrrdq g )(

dT

Cdzdrrd p )(

+ + +

=

T

CqTK pg)(KKKK zr

[gq

)(1

r

Tr

rrK r

)(1

2

2

2 T

rK )(

2

2

z

TK z

)( ddzdrrd )(

ddzdrrd

TC p

+ + + ] =

If material is homogeneous amp isotropic

)(1

r

Tr

rr

)(1

2

2

2 T

r)(

2

2

z

T

K

q g

T

K

C p

)(1

r

Tr

rr

)(1

2

2

2 T

r )(2

2

z

T

K

q g

T

1

+ + + =

)(1

r

Tr

rrK r

)(1

2

2

2 T

rK )(

2

2

z

TK z

gq

T

C p + + + =

+ + + = Thermal diffusivity

T

K

qT g

12

pC

K

(Ein ndash Eout) + Egen = Est

)( rT

)( r

)sin( drrddr)sin( drrddr

planerdirection

planerdirection

planedirectionr

)(

)(

)(

Spherical polar co-ordinates

Elemental length

Consider an elemental volume ABCDEFGH =

GENERAL HEAT CONDUCTION EQUATION IN SPHERICAL COORDINATES

Let

=internal heat generation per unit volume per unit time (W m3)q= rate of heat flow (W)Q= qdt=total heat flow (J)r=mass density of the material element(kgm3)Cp=specific heat of the material element (Jkg K)

gq

Consider an unsteady (transient) three dimensional heat conduction with internal heat generation in heterogeneous amp anisotropic spherical geometries

LetTemperature distribution is a function of T=

X

Y

Z

drT

drrdKQ rr )sin(

drQr

QQ rrdrr ][

drdr

TdrrdK

rdQ

QQdQ

rr

drrrr

])sin([

)1(]([sin 2

rT

rr

ddrddKdQ rr

Plane ( -q f) lsquorrsquo directionQuantity of heat flowing into the face ABCD in lsquorrsquo direction

Quantity of heat leaving from the face EFGH in lsquorrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquorrsquo direction

drT

drdrKQ )sin(

rdQ

rQQ d ][

rddr

TdrdrK

rdQ

QQdQ d

])sin([

])[(sin

2

T

r

drdrdrKdQ

)2(])(sin[

2

T

r

drdrddrKdQ

Quantity of heat leaving from the face BFHD in lsquoqrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoqrsquo direction

X

Y

Z

Plane (r- f ) lsquoq lsquodirection

Quantity of heat flowing into the face AEGC in lsquoqrsquo direction

dr

TdrrdKQ

sin)(

drQ

rQQ d sin][

sin

drdr

TdrrdK

rdQ

QQdQ d

sin]sin

)([sin

][sinsin

)sin(

T

rr

ddrdrrdKdQ

)3()(sin

)sin(2

2

22

2

T

r

ddddrrKdQ

Plane (r- ) q lsquofrsquo directionQuantity of heat flowing into the face ABFE in lsquofrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquozrsquo direction

)4()sin()(

ddrrddrqQgg

)5()sin()(

dT

CdrrddrQ pst

Total amount of internal heat generation in the element

Total heat energy stored in the element

(Ein ndash Eout) + Egen = Est

d

TmCQ pst )(

)( volumem

Quantity of heat leaving from the face CDHG in lsquofrsquo direction

according to thermodynamics first law

Ein + Egen = Eout + Est

X

Y

Z

stgr QQdQdQdQ

)(

]([sin 2

r

Tr

rddrddKr

])(sin[

2

T

rdrdrddrK

)(sin

)sin(2

2

22

2

T

r

ddddrrK

ddrrddrq g )sin(

dT

Cdrrddr p )sin(

2

2

r

r

sin

sin

]([ 22 r

Tr

rr

Kr

])(sin[sin2

TrK

)(sin 2

2

22

T

r

Kgq

T

Cp

]([1

[ 22 r

Tr

rrK

])(sin[sin1

2

T

r)](

sin

12

2

22 T

r gq

T

Cp

]([1

[ 22 r

Tr

rr

])(sin[sin1

2

T

r)](

sin

12

2

22 T

rK

q g

T

K

Cp

]([1

[ 22 r

Tr

rr

])(sin[sin1

2

T

r)](

sin

12

2

22 T

rK

q g

T1

KKKKr If material is homogeneous amp isotropic

TCqTK pg)(

T

K

qT g

12 pC

K

Thermal diffusivity

(Ein ndash Eout) + Egen = Est

General Heat Conduction Equation

PLANE WALL CYLINDER SPHERE

Temperature profile

Plane wall SphereCylinder

bull A copper conductor (k = 380 WmoC resistivity ρ = 2x10-8 Ω m) having inner and outer radii 10 cm and 225 cm respectively is carrying a current density of 4800 amperes cm2 the conductor is internally cooled and a constant temperature of 65oC is maintained at the inner surface and there is no heat transfer through the insulation surrounding the conductor

Determinebull (i) The maximum temperature of the conductor and the radius at

which it occursbull (ii) The internal heat transfer rate

bull A chemical reaction takes place in a packed bed (k=06 WmoC) between two coaxial cylinders with radii 15 mm and 45 mm the inner surface is at 580oC and it is insulated Assuming the reaction rate of 055 MWm3 in the reactor volume find the temperature at the outer surface of the reactor

bull A current of 300 amperes passes through a stainless steel wire of 25 mm diameter and k=20 WmoC The resistivity of the wire is 70x10-8 ohm-m and length of the wire is 2 m if the wire is submerged in a fluid maintained at 50oC and h at the wire surface is 4000 Wm2oC calculate the steady state temperature at the centre and at the surface of the wire

bull An approximately spherical shaped orange ( k=023 WmoC) 90 mm diameter undergoes ripping process and generates 5100 Wm3 of energy If external surface the orange is at 8oC determine

bull (i) temperature at the centre of the orange and

bull (ii) heat flow from the outer surface of the orange

bull FINS External surfaces to used to cooling area and hence convection

heat transfer can be improvedbull APPLICATIONS IC engine cylinder ( air cooling) Heat exchangers automobile

radiators computer chip refrigerators electric motors generators and transformers etc

bull ASSUMPTIONS MADE IN ANALYSIS OF HEAT TRANFER THROUGH FIN

1 Fin material is homogenous and isotropic constthermal conductivity

2 Temperature at any cs of the fin is uniform3 lsquohrsquo is uniform over entire surface4 No internal heat generation5 Thermal contact resistance is negligible6 Radiation heat transfer is negligible7 Heat conduction is steady state and one dimensional

bull Problem 1

bull A stainless steel fin (k=20 WmK) having a diameter of 20 mm and a length of 01 m is attached to a wall at 300oC The ambient temperature is 50oC and the heat transfer coefficient is 10 Wm2K The fin tip is insulated Determine

bull (A) The rate of heat dissipation from the finbull (B) The temperature at the fin tipbull (C) The rate of heat transfer from the wall area covered by the fin if

the fin was not usedbull (D) The heat transfer rate from the same fin geometry if the

stainless steel fin is replaced by a fictitious fin with infinite thermal conductivity

bull Problem2bull One end of a long rod is inserted into a furnace while the other

projects into ambient temperature Under steady state the temperature of the rod is measured at two points 75mm apart and found to be 125oC and 885oC respectively while the ambient temperature is 20oC If the rod is 25mm in diameter and h is 2336 Wm2K

bull Find the thermal conductivity of the rod material

  • HEAT AND MASS TRANSFER
  • Thermodynamics Vs Heat transfer
  • MODE OF HEAT TRANSFER
  • MODE OF HEAT TRANSFER (2)
  • CONDUCTION- MECHANISM
  • CONCECTION-MECHANISM
  • RADIATION-MECHANISM
  • Conduction- Convection- Radiation
  • Slide 9
  • Slide 10
  • Fourierrsquos Law of Heat Conduction
  • Slide 12
  • Newtonrsquos Law of Cooling
  • Laws of Thermal Radiation
  • LAWS FOR HEAT TRANSFER
  • ELECTRICAL ampTHERMAL ANALOGY
  • Slide 17
  • Slide 18
  • PLANE WALL-UNIDIRECTIONAL
  • PLANE WALL -SERIES
  • PLANE WALL IN SERIES
  • PLANE WALL -PARALLEL
  • PLANE WALL-PARALLEL
  • PLANE WALL-COMBINATION
  • PLANE WALL-FILM RESISTANCE
  • PLANE WALL- THERMAL CONTACT RESISTANCE
  • PLANE WALL- THERMAL CONTACT RESISTANCE (2)
  • CYLINDER-SOLID
  • HEAT TRANSFER IN SOLID CYLINDER
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER-COMPOSITE
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER-MULTY LAYER
  • HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER
  • HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER (2)
  • HEAT TRANSFER IN SPHERE
  • HEAT TRANSFER IN SPHERE -MULTY LAYER
  • HEAT TRANSFER IN SPHERE -MULTY LAYER (2)
  • Slide 38
  • Slide 39
  • Slide 40
  • Slide 41
  • CRITICAL RADIUS
  • Plane wall- Insulation thickness -study
  • Problem5
  • Slide 45
  • Slide 46
  • Slide 47
  • GENERAL HEAT CONDUCTION EQUATION IN CYLINDRICAL COORDINATES
  • Slide 49
  • Slide 50
  • Slide 51
  • GENERAL HEAT CONDUCTION EQUATION IN SPHERICAL COORDINATES
  • Slide 53
  • Slide 54
  • Slide 55
  • General Heat Conduction Equation
  • Temperature profile
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Slide 67
Page 48: hmt chap 1

)( rT

)( r

)sin( drrddr)sin( drrddr

planerdirection

planerdirection

planedirectionr

)(

)(

)(

Spherical polar co-ordinates

Elemental length

Consider an elemental volume ABCDEFGH =

GENERAL HEAT CONDUCTION EQUATION IN SPHERICAL COORDINATES

Let

=internal heat generation per unit volume per unit time (W m3)q= rate of heat flow (W)Q= qdt=total heat flow (J)r=mass density of the material element(kgm3)Cp=specific heat of the material element (Jkg K)

gq

Consider an unsteady (transient) three dimensional heat conduction with internal heat generation in heterogeneous amp anisotropic spherical geometries

LetTemperature distribution is a function of T=

X

Y

Z

drT

drrdKQ rr )sin(

drQr

QQ rrdrr ][

drdr

TdrrdK

rdQ

QQdQ

rr

drrrr

])sin([

)1(]([sin 2

rT

rr

ddrddKdQ rr

Plane ( -q f) lsquorrsquo directionQuantity of heat flowing into the face ABCD in lsquorrsquo direction

Quantity of heat leaving from the face EFGH in lsquorrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquorrsquo direction

drT

drdrKQ )sin(

rdQ

rQQ d ][

rddr

TdrdrK

rdQ

QQdQ d

])sin([

])[(sin

2

T

r

drdrdrKdQ

)2(])(sin[

2

T

r

drdrddrKdQ

Quantity of heat leaving from the face BFHD in lsquoqrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoqrsquo direction

X

Y

Z

Plane (r- f ) lsquoq lsquodirection

Quantity of heat flowing into the face AEGC in lsquoqrsquo direction

dr

TdrrdKQ

sin)(

drQ

rQQ d sin][

sin

drdr

TdrrdK

rdQ

QQdQ d

sin]sin

)([sin

][sinsin

)sin(

T

rr

ddrdrrdKdQ

)3()(sin

)sin(2

2

22

2

T

r

ddddrrKdQ

Plane (r- ) q lsquofrsquo directionQuantity of heat flowing into the face ABFE in lsquofrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquozrsquo direction

)4()sin()(

ddrrddrqQgg

)5()sin()(

dT

CdrrddrQ pst

Total amount of internal heat generation in the element

Total heat energy stored in the element

(Ein ndash Eout) + Egen = Est

d

TmCQ pst )(

)( volumem

Quantity of heat leaving from the face CDHG in lsquofrsquo direction

according to thermodynamics first law

Ein + Egen = Eout + Est

X

Y

Z

stgr QQdQdQdQ

)(

]([sin 2

r

Tr

rddrddKr

])(sin[

2

T

rdrdrddrK

)(sin

)sin(2

2

22

2

T

r

ddddrrK

ddrrddrq g )sin(

dT

Cdrrddr p )sin(

2

2

r

r

sin

sin

]([ 22 r

Tr

rr

Kr

])(sin[sin2

TrK

)(sin 2

2

22

T

r

Kgq

T

Cp

]([1

[ 22 r

Tr

rrK

])(sin[sin1

2

T

r)](

sin

12

2

22 T

r gq

T

Cp

]([1

[ 22 r

Tr

rr

])(sin[sin1

2

T

r)](

sin

12

2

22 T

rK

q g

T

K

Cp

]([1

[ 22 r

Tr

rr

])(sin[sin1

2

T

r)](

sin

12

2

22 T

rK

q g

T1

KKKKr If material is homogeneous amp isotropic

TCqTK pg)(

T

K

qT g

12 pC

K

Thermal diffusivity

(Ein ndash Eout) + Egen = Est

General Heat Conduction Equation

PLANE WALL CYLINDER SPHERE

Temperature profile

Plane wall SphereCylinder

bull A copper conductor (k = 380 WmoC resistivity ρ = 2x10-8 Ω m) having inner and outer radii 10 cm and 225 cm respectively is carrying a current density of 4800 amperes cm2 the conductor is internally cooled and a constant temperature of 65oC is maintained at the inner surface and there is no heat transfer through the insulation surrounding the conductor

Determinebull (i) The maximum temperature of the conductor and the radius at

which it occursbull (ii) The internal heat transfer rate

bull A chemical reaction takes place in a packed bed (k=06 WmoC) between two coaxial cylinders with radii 15 mm and 45 mm the inner surface is at 580oC and it is insulated Assuming the reaction rate of 055 MWm3 in the reactor volume find the temperature at the outer surface of the reactor

bull A current of 300 amperes passes through a stainless steel wire of 25 mm diameter and k=20 WmoC The resistivity of the wire is 70x10-8 ohm-m and length of the wire is 2 m if the wire is submerged in a fluid maintained at 50oC and h at the wire surface is 4000 Wm2oC calculate the steady state temperature at the centre and at the surface of the wire

bull An approximately spherical shaped orange ( k=023 WmoC) 90 mm diameter undergoes ripping process and generates 5100 Wm3 of energy If external surface the orange is at 8oC determine

bull (i) temperature at the centre of the orange and

bull (ii) heat flow from the outer surface of the orange

bull FINS External surfaces to used to cooling area and hence convection

heat transfer can be improvedbull APPLICATIONS IC engine cylinder ( air cooling) Heat exchangers automobile

radiators computer chip refrigerators electric motors generators and transformers etc

bull ASSUMPTIONS MADE IN ANALYSIS OF HEAT TRANFER THROUGH FIN

1 Fin material is homogenous and isotropic constthermal conductivity

2 Temperature at any cs of the fin is uniform3 lsquohrsquo is uniform over entire surface4 No internal heat generation5 Thermal contact resistance is negligible6 Radiation heat transfer is negligible7 Heat conduction is steady state and one dimensional

bull Problem 1

bull A stainless steel fin (k=20 WmK) having a diameter of 20 mm and a length of 01 m is attached to a wall at 300oC The ambient temperature is 50oC and the heat transfer coefficient is 10 Wm2K The fin tip is insulated Determine

bull (A) The rate of heat dissipation from the finbull (B) The temperature at the fin tipbull (C) The rate of heat transfer from the wall area covered by the fin if

the fin was not usedbull (D) The heat transfer rate from the same fin geometry if the

stainless steel fin is replaced by a fictitious fin with infinite thermal conductivity

bull Problem2bull One end of a long rod is inserted into a furnace while the other

projects into ambient temperature Under steady state the temperature of the rod is measured at two points 75mm apart and found to be 125oC and 885oC respectively while the ambient temperature is 20oC If the rod is 25mm in diameter and h is 2336 Wm2K

bull Find the thermal conductivity of the rod material

  • HEAT AND MASS TRANSFER
  • Thermodynamics Vs Heat transfer
  • MODE OF HEAT TRANSFER
  • MODE OF HEAT TRANSFER (2)
  • CONDUCTION- MECHANISM
  • CONCECTION-MECHANISM
  • RADIATION-MECHANISM
  • Conduction- Convection- Radiation
  • Slide 9
  • Slide 10
  • Fourierrsquos Law of Heat Conduction
  • Slide 12
  • Newtonrsquos Law of Cooling
  • Laws of Thermal Radiation
  • LAWS FOR HEAT TRANSFER
  • ELECTRICAL ampTHERMAL ANALOGY
  • Slide 17
  • Slide 18
  • PLANE WALL-UNIDIRECTIONAL
  • PLANE WALL -SERIES
  • PLANE WALL IN SERIES
  • PLANE WALL -PARALLEL
  • PLANE WALL-PARALLEL
  • PLANE WALL-COMBINATION
  • PLANE WALL-FILM RESISTANCE
  • PLANE WALL- THERMAL CONTACT RESISTANCE
  • PLANE WALL- THERMAL CONTACT RESISTANCE (2)
  • CYLINDER-SOLID
  • HEAT TRANSFER IN SOLID CYLINDER
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER-COMPOSITE
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER-MULTY LAYER
  • HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER
  • HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER (2)
  • HEAT TRANSFER IN SPHERE
  • HEAT TRANSFER IN SPHERE -MULTY LAYER
  • HEAT TRANSFER IN SPHERE -MULTY LAYER (2)
  • Slide 38
  • Slide 39
  • Slide 40
  • Slide 41
  • CRITICAL RADIUS
  • Plane wall- Insulation thickness -study
  • Problem5
  • Slide 45
  • Slide 46
  • Slide 47
  • GENERAL HEAT CONDUCTION EQUATION IN CYLINDRICAL COORDINATES
  • Slide 49
  • Slide 50
  • Slide 51
  • GENERAL HEAT CONDUCTION EQUATION IN SPHERICAL COORDINATES
  • Slide 53
  • Slide 54
  • Slide 55
  • General Heat Conduction Equation
  • Temperature profile
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Slide 67
Page 49: hmt chap 1

drT

drrdKQ rr )sin(

drQr

QQ rrdrr ][

drdr

TdrrdK

rdQ

QQdQ

rr

drrrr

])sin([

)1(]([sin 2

rT

rr

ddrddKdQ rr

Plane ( -q f) lsquorrsquo directionQuantity of heat flowing into the face ABCD in lsquorrsquo direction

Quantity of heat leaving from the face EFGH in lsquorrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquorrsquo direction

drT

drdrKQ )sin(

rdQ

rQQ d ][

rddr

TdrdrK

rdQ

QQdQ d

])sin([

])[(sin

2

T

r

drdrdrKdQ

)2(])(sin[

2

T

r

drdrddrKdQ

Quantity of heat leaving from the face BFHD in lsquoqrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquoqrsquo direction

X

Y

Z

Plane (r- f ) lsquoq lsquodirection

Quantity of heat flowing into the face AEGC in lsquoqrsquo direction

dr

TdrrdKQ

sin)(

drQ

rQQ d sin][

sin

drdr

TdrrdK

rdQ

QQdQ d

sin]sin

)([sin

][sinsin

)sin(

T

rr

ddrdrrdKdQ

)3()(sin

)sin(2

2

22

2

T

r

ddddrrKdQ

Plane (r- ) q lsquofrsquo directionQuantity of heat flowing into the face ABFE in lsquofrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquozrsquo direction

)4()sin()(

ddrrddrqQgg

)5()sin()(

dT

CdrrddrQ pst

Total amount of internal heat generation in the element

Total heat energy stored in the element

(Ein ndash Eout) + Egen = Est

d

TmCQ pst )(

)( volumem

Quantity of heat leaving from the face CDHG in lsquofrsquo direction

according to thermodynamics first law

Ein + Egen = Eout + Est

X

Y

Z

stgr QQdQdQdQ

)(

]([sin 2

r

Tr

rddrddKr

])(sin[

2

T

rdrdrddrK

)(sin

)sin(2

2

22

2

T

r

ddddrrK

ddrrddrq g )sin(

dT

Cdrrddr p )sin(

2

2

r

r

sin

sin

]([ 22 r

Tr

rr

Kr

])(sin[sin2

TrK

)(sin 2

2

22

T

r

Kgq

T

Cp

]([1

[ 22 r

Tr

rrK

])(sin[sin1

2

T

r)](

sin

12

2

22 T

r gq

T

Cp

]([1

[ 22 r

Tr

rr

])(sin[sin1

2

T

r)](

sin

12

2

22 T

rK

q g

T

K

Cp

]([1

[ 22 r

Tr

rr

])(sin[sin1

2

T

r)](

sin

12

2

22 T

rK

q g

T1

KKKKr If material is homogeneous amp isotropic

TCqTK pg)(

T

K

qT g

12 pC

K

Thermal diffusivity

(Ein ndash Eout) + Egen = Est

General Heat Conduction Equation

PLANE WALL CYLINDER SPHERE

Temperature profile

Plane wall SphereCylinder

bull A copper conductor (k = 380 WmoC resistivity ρ = 2x10-8 Ω m) having inner and outer radii 10 cm and 225 cm respectively is carrying a current density of 4800 amperes cm2 the conductor is internally cooled and a constant temperature of 65oC is maintained at the inner surface and there is no heat transfer through the insulation surrounding the conductor

Determinebull (i) The maximum temperature of the conductor and the radius at

which it occursbull (ii) The internal heat transfer rate

bull A chemical reaction takes place in a packed bed (k=06 WmoC) between two coaxial cylinders with radii 15 mm and 45 mm the inner surface is at 580oC and it is insulated Assuming the reaction rate of 055 MWm3 in the reactor volume find the temperature at the outer surface of the reactor

bull A current of 300 amperes passes through a stainless steel wire of 25 mm diameter and k=20 WmoC The resistivity of the wire is 70x10-8 ohm-m and length of the wire is 2 m if the wire is submerged in a fluid maintained at 50oC and h at the wire surface is 4000 Wm2oC calculate the steady state temperature at the centre and at the surface of the wire

bull An approximately spherical shaped orange ( k=023 WmoC) 90 mm diameter undergoes ripping process and generates 5100 Wm3 of energy If external surface the orange is at 8oC determine

bull (i) temperature at the centre of the orange and

bull (ii) heat flow from the outer surface of the orange

bull FINS External surfaces to used to cooling area and hence convection

heat transfer can be improvedbull APPLICATIONS IC engine cylinder ( air cooling) Heat exchangers automobile

radiators computer chip refrigerators electric motors generators and transformers etc

bull ASSUMPTIONS MADE IN ANALYSIS OF HEAT TRANFER THROUGH FIN

1 Fin material is homogenous and isotropic constthermal conductivity

2 Temperature at any cs of the fin is uniform3 lsquohrsquo is uniform over entire surface4 No internal heat generation5 Thermal contact resistance is negligible6 Radiation heat transfer is negligible7 Heat conduction is steady state and one dimensional

bull Problem 1

bull A stainless steel fin (k=20 WmK) having a diameter of 20 mm and a length of 01 m is attached to a wall at 300oC The ambient temperature is 50oC and the heat transfer coefficient is 10 Wm2K The fin tip is insulated Determine

bull (A) The rate of heat dissipation from the finbull (B) The temperature at the fin tipbull (C) The rate of heat transfer from the wall area covered by the fin if

the fin was not usedbull (D) The heat transfer rate from the same fin geometry if the

stainless steel fin is replaced by a fictitious fin with infinite thermal conductivity

bull Problem2bull One end of a long rod is inserted into a furnace while the other

projects into ambient temperature Under steady state the temperature of the rod is measured at two points 75mm apart and found to be 125oC and 885oC respectively while the ambient temperature is 20oC If the rod is 25mm in diameter and h is 2336 Wm2K

bull Find the thermal conductivity of the rod material

  • HEAT AND MASS TRANSFER
  • Thermodynamics Vs Heat transfer
  • MODE OF HEAT TRANSFER
  • MODE OF HEAT TRANSFER (2)
  • CONDUCTION- MECHANISM
  • CONCECTION-MECHANISM
  • RADIATION-MECHANISM
  • Conduction- Convection- Radiation
  • Slide 9
  • Slide 10
  • Fourierrsquos Law of Heat Conduction
  • Slide 12
  • Newtonrsquos Law of Cooling
  • Laws of Thermal Radiation
  • LAWS FOR HEAT TRANSFER
  • ELECTRICAL ampTHERMAL ANALOGY
  • Slide 17
  • Slide 18
  • PLANE WALL-UNIDIRECTIONAL
  • PLANE WALL -SERIES
  • PLANE WALL IN SERIES
  • PLANE WALL -PARALLEL
  • PLANE WALL-PARALLEL
  • PLANE WALL-COMBINATION
  • PLANE WALL-FILM RESISTANCE
  • PLANE WALL- THERMAL CONTACT RESISTANCE
  • PLANE WALL- THERMAL CONTACT RESISTANCE (2)
  • CYLINDER-SOLID
  • HEAT TRANSFER IN SOLID CYLINDER
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER-COMPOSITE
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER-MULTY LAYER
  • HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER
  • HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER (2)
  • HEAT TRANSFER IN SPHERE
  • HEAT TRANSFER IN SPHERE -MULTY LAYER
  • HEAT TRANSFER IN SPHERE -MULTY LAYER (2)
  • Slide 38
  • Slide 39
  • Slide 40
  • Slide 41
  • CRITICAL RADIUS
  • Plane wall- Insulation thickness -study
  • Problem5
  • Slide 45
  • Slide 46
  • Slide 47
  • GENERAL HEAT CONDUCTION EQUATION IN CYLINDRICAL COORDINATES
  • Slide 49
  • Slide 50
  • Slide 51
  • GENERAL HEAT CONDUCTION EQUATION IN SPHERICAL COORDINATES
  • Slide 53
  • Slide 54
  • Slide 55
  • General Heat Conduction Equation
  • Temperature profile
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Slide 67
Page 50: hmt chap 1

dr

TdrrdKQ

sin)(

drQ

rQQ d sin][

sin

drdr

TdrrdK

rdQ

QQdQ d

sin]sin

)([sin

][sinsin

)sin(

T

rr

ddrdrrdKdQ

)3()(sin

)sin(2

2

22

2

T

r

ddddrrKdQ

Plane (r- ) q lsquofrsquo directionQuantity of heat flowing into the face ABFE in lsquofrsquo direction

Heat accumulation in the element due to heat flow by conduction in lsquozrsquo direction

)4()sin()(

ddrrddrqQgg

)5()sin()(

dT

CdrrddrQ pst

Total amount of internal heat generation in the element

Total heat energy stored in the element

(Ein ndash Eout) + Egen = Est

d

TmCQ pst )(

)( volumem

Quantity of heat leaving from the face CDHG in lsquofrsquo direction

according to thermodynamics first law

Ein + Egen = Eout + Est

X

Y

Z

stgr QQdQdQdQ

)(

]([sin 2

r

Tr

rddrddKr

])(sin[

2

T

rdrdrddrK

)(sin

)sin(2

2

22

2

T

r

ddddrrK

ddrrddrq g )sin(

dT

Cdrrddr p )sin(

2

2

r

r

sin

sin

]([ 22 r

Tr

rr

Kr

])(sin[sin2

TrK

)(sin 2

2

22

T

r

Kgq

T

Cp

]([1

[ 22 r

Tr

rrK

])(sin[sin1

2

T

r)](

sin

12

2

22 T

r gq

T

Cp

]([1

[ 22 r

Tr

rr

])(sin[sin1

2

T

r)](

sin

12

2

22 T

rK

q g

T

K

Cp

]([1

[ 22 r

Tr

rr

])(sin[sin1

2

T

r)](

sin

12

2

22 T

rK

q g

T1

KKKKr If material is homogeneous amp isotropic

TCqTK pg)(

T

K

qT g

12 pC

K

Thermal diffusivity

(Ein ndash Eout) + Egen = Est

General Heat Conduction Equation

PLANE WALL CYLINDER SPHERE

Temperature profile

Plane wall SphereCylinder

bull A copper conductor (k = 380 WmoC resistivity ρ = 2x10-8 Ω m) having inner and outer radii 10 cm and 225 cm respectively is carrying a current density of 4800 amperes cm2 the conductor is internally cooled and a constant temperature of 65oC is maintained at the inner surface and there is no heat transfer through the insulation surrounding the conductor

Determinebull (i) The maximum temperature of the conductor and the radius at

which it occursbull (ii) The internal heat transfer rate

bull A chemical reaction takes place in a packed bed (k=06 WmoC) between two coaxial cylinders with radii 15 mm and 45 mm the inner surface is at 580oC and it is insulated Assuming the reaction rate of 055 MWm3 in the reactor volume find the temperature at the outer surface of the reactor

bull A current of 300 amperes passes through a stainless steel wire of 25 mm diameter and k=20 WmoC The resistivity of the wire is 70x10-8 ohm-m and length of the wire is 2 m if the wire is submerged in a fluid maintained at 50oC and h at the wire surface is 4000 Wm2oC calculate the steady state temperature at the centre and at the surface of the wire

bull An approximately spherical shaped orange ( k=023 WmoC) 90 mm diameter undergoes ripping process and generates 5100 Wm3 of energy If external surface the orange is at 8oC determine

bull (i) temperature at the centre of the orange and

bull (ii) heat flow from the outer surface of the orange

bull FINS External surfaces to used to cooling area and hence convection

heat transfer can be improvedbull APPLICATIONS IC engine cylinder ( air cooling) Heat exchangers automobile

radiators computer chip refrigerators electric motors generators and transformers etc

bull ASSUMPTIONS MADE IN ANALYSIS OF HEAT TRANFER THROUGH FIN

1 Fin material is homogenous and isotropic constthermal conductivity

2 Temperature at any cs of the fin is uniform3 lsquohrsquo is uniform over entire surface4 No internal heat generation5 Thermal contact resistance is negligible6 Radiation heat transfer is negligible7 Heat conduction is steady state and one dimensional

bull Problem 1

bull A stainless steel fin (k=20 WmK) having a diameter of 20 mm and a length of 01 m is attached to a wall at 300oC The ambient temperature is 50oC and the heat transfer coefficient is 10 Wm2K The fin tip is insulated Determine

bull (A) The rate of heat dissipation from the finbull (B) The temperature at the fin tipbull (C) The rate of heat transfer from the wall area covered by the fin if

the fin was not usedbull (D) The heat transfer rate from the same fin geometry if the

stainless steel fin is replaced by a fictitious fin with infinite thermal conductivity

bull Problem2bull One end of a long rod is inserted into a furnace while the other

projects into ambient temperature Under steady state the temperature of the rod is measured at two points 75mm apart and found to be 125oC and 885oC respectively while the ambient temperature is 20oC If the rod is 25mm in diameter and h is 2336 Wm2K

bull Find the thermal conductivity of the rod material

  • HEAT AND MASS TRANSFER
  • Thermodynamics Vs Heat transfer
  • MODE OF HEAT TRANSFER
  • MODE OF HEAT TRANSFER (2)
  • CONDUCTION- MECHANISM
  • CONCECTION-MECHANISM
  • RADIATION-MECHANISM
  • Conduction- Convection- Radiation
  • Slide 9
  • Slide 10
  • Fourierrsquos Law of Heat Conduction
  • Slide 12
  • Newtonrsquos Law of Cooling
  • Laws of Thermal Radiation
  • LAWS FOR HEAT TRANSFER
  • ELECTRICAL ampTHERMAL ANALOGY
  • Slide 17
  • Slide 18
  • PLANE WALL-UNIDIRECTIONAL
  • PLANE WALL -SERIES
  • PLANE WALL IN SERIES
  • PLANE WALL -PARALLEL
  • PLANE WALL-PARALLEL
  • PLANE WALL-COMBINATION
  • PLANE WALL-FILM RESISTANCE
  • PLANE WALL- THERMAL CONTACT RESISTANCE
  • PLANE WALL- THERMAL CONTACT RESISTANCE (2)
  • CYLINDER-SOLID
  • HEAT TRANSFER IN SOLID CYLINDER
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER-COMPOSITE
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER-MULTY LAYER
  • HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER
  • HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER (2)
  • HEAT TRANSFER IN SPHERE
  • HEAT TRANSFER IN SPHERE -MULTY LAYER
  • HEAT TRANSFER IN SPHERE -MULTY LAYER (2)
  • Slide 38
  • Slide 39
  • Slide 40
  • Slide 41
  • CRITICAL RADIUS
  • Plane wall- Insulation thickness -study
  • Problem5
  • Slide 45
  • Slide 46
  • Slide 47
  • GENERAL HEAT CONDUCTION EQUATION IN CYLINDRICAL COORDINATES
  • Slide 49
  • Slide 50
  • Slide 51
  • GENERAL HEAT CONDUCTION EQUATION IN SPHERICAL COORDINATES
  • Slide 53
  • Slide 54
  • Slide 55
  • General Heat Conduction Equation
  • Temperature profile
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Slide 67
Page 51: hmt chap 1

stgr QQdQdQdQ

)(

]([sin 2

r

Tr

rddrddKr

])(sin[

2

T

rdrdrddrK

)(sin

)sin(2

2

22

2

T

r

ddddrrK

ddrrddrq g )sin(

dT

Cdrrddr p )sin(

2

2

r

r

sin

sin

]([ 22 r

Tr

rr

Kr

])(sin[sin2

TrK

)(sin 2

2

22

T

r

Kgq

T

Cp

]([1

[ 22 r

Tr

rrK

])(sin[sin1

2

T

r)](

sin

12

2

22 T

r gq

T

Cp

]([1

[ 22 r

Tr

rr

])(sin[sin1

2

T

r)](

sin

12

2

22 T

rK

q g

T

K

Cp

]([1

[ 22 r

Tr

rr

])(sin[sin1

2

T

r)](

sin

12

2

22 T

rK

q g

T1

KKKKr If material is homogeneous amp isotropic

TCqTK pg)(

T

K

qT g

12 pC

K

Thermal diffusivity

(Ein ndash Eout) + Egen = Est

General Heat Conduction Equation

PLANE WALL CYLINDER SPHERE

Temperature profile

Plane wall SphereCylinder

bull A copper conductor (k = 380 WmoC resistivity ρ = 2x10-8 Ω m) having inner and outer radii 10 cm and 225 cm respectively is carrying a current density of 4800 amperes cm2 the conductor is internally cooled and a constant temperature of 65oC is maintained at the inner surface and there is no heat transfer through the insulation surrounding the conductor

Determinebull (i) The maximum temperature of the conductor and the radius at

which it occursbull (ii) The internal heat transfer rate

bull A chemical reaction takes place in a packed bed (k=06 WmoC) between two coaxial cylinders with radii 15 mm and 45 mm the inner surface is at 580oC and it is insulated Assuming the reaction rate of 055 MWm3 in the reactor volume find the temperature at the outer surface of the reactor

bull A current of 300 amperes passes through a stainless steel wire of 25 mm diameter and k=20 WmoC The resistivity of the wire is 70x10-8 ohm-m and length of the wire is 2 m if the wire is submerged in a fluid maintained at 50oC and h at the wire surface is 4000 Wm2oC calculate the steady state temperature at the centre and at the surface of the wire

bull An approximately spherical shaped orange ( k=023 WmoC) 90 mm diameter undergoes ripping process and generates 5100 Wm3 of energy If external surface the orange is at 8oC determine

bull (i) temperature at the centre of the orange and

bull (ii) heat flow from the outer surface of the orange

bull FINS External surfaces to used to cooling area and hence convection

heat transfer can be improvedbull APPLICATIONS IC engine cylinder ( air cooling) Heat exchangers automobile

radiators computer chip refrigerators electric motors generators and transformers etc

bull ASSUMPTIONS MADE IN ANALYSIS OF HEAT TRANFER THROUGH FIN

1 Fin material is homogenous and isotropic constthermal conductivity

2 Temperature at any cs of the fin is uniform3 lsquohrsquo is uniform over entire surface4 No internal heat generation5 Thermal contact resistance is negligible6 Radiation heat transfer is negligible7 Heat conduction is steady state and one dimensional

bull Problem 1

bull A stainless steel fin (k=20 WmK) having a diameter of 20 mm and a length of 01 m is attached to a wall at 300oC The ambient temperature is 50oC and the heat transfer coefficient is 10 Wm2K The fin tip is insulated Determine

bull (A) The rate of heat dissipation from the finbull (B) The temperature at the fin tipbull (C) The rate of heat transfer from the wall area covered by the fin if

the fin was not usedbull (D) The heat transfer rate from the same fin geometry if the

stainless steel fin is replaced by a fictitious fin with infinite thermal conductivity

bull Problem2bull One end of a long rod is inserted into a furnace while the other

projects into ambient temperature Under steady state the temperature of the rod is measured at two points 75mm apart and found to be 125oC and 885oC respectively while the ambient temperature is 20oC If the rod is 25mm in diameter and h is 2336 Wm2K

bull Find the thermal conductivity of the rod material

  • HEAT AND MASS TRANSFER
  • Thermodynamics Vs Heat transfer
  • MODE OF HEAT TRANSFER
  • MODE OF HEAT TRANSFER (2)
  • CONDUCTION- MECHANISM
  • CONCECTION-MECHANISM
  • RADIATION-MECHANISM
  • Conduction- Convection- Radiation
  • Slide 9
  • Slide 10
  • Fourierrsquos Law of Heat Conduction
  • Slide 12
  • Newtonrsquos Law of Cooling
  • Laws of Thermal Radiation
  • LAWS FOR HEAT TRANSFER
  • ELECTRICAL ampTHERMAL ANALOGY
  • Slide 17
  • Slide 18
  • PLANE WALL-UNIDIRECTIONAL
  • PLANE WALL -SERIES
  • PLANE WALL IN SERIES
  • PLANE WALL -PARALLEL
  • PLANE WALL-PARALLEL
  • PLANE WALL-COMBINATION
  • PLANE WALL-FILM RESISTANCE
  • PLANE WALL- THERMAL CONTACT RESISTANCE
  • PLANE WALL- THERMAL CONTACT RESISTANCE (2)
  • CYLINDER-SOLID
  • HEAT TRANSFER IN SOLID CYLINDER
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER-COMPOSITE
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER-MULTY LAYER
  • HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER
  • HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER (2)
  • HEAT TRANSFER IN SPHERE
  • HEAT TRANSFER IN SPHERE -MULTY LAYER
  • HEAT TRANSFER IN SPHERE -MULTY LAYER (2)
  • Slide 38
  • Slide 39
  • Slide 40
  • Slide 41
  • CRITICAL RADIUS
  • Plane wall- Insulation thickness -study
  • Problem5
  • Slide 45
  • Slide 46
  • Slide 47
  • GENERAL HEAT CONDUCTION EQUATION IN CYLINDRICAL COORDINATES
  • Slide 49
  • Slide 50
  • Slide 51
  • GENERAL HEAT CONDUCTION EQUATION IN SPHERICAL COORDINATES
  • Slide 53
  • Slide 54
  • Slide 55
  • General Heat Conduction Equation
  • Temperature profile
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Slide 67
Page 52: hmt chap 1

General Heat Conduction Equation

PLANE WALL CYLINDER SPHERE

Temperature profile

Plane wall SphereCylinder

bull A copper conductor (k = 380 WmoC resistivity ρ = 2x10-8 Ω m) having inner and outer radii 10 cm and 225 cm respectively is carrying a current density of 4800 amperes cm2 the conductor is internally cooled and a constant temperature of 65oC is maintained at the inner surface and there is no heat transfer through the insulation surrounding the conductor

Determinebull (i) The maximum temperature of the conductor and the radius at

which it occursbull (ii) The internal heat transfer rate

bull A chemical reaction takes place in a packed bed (k=06 WmoC) between two coaxial cylinders with radii 15 mm and 45 mm the inner surface is at 580oC and it is insulated Assuming the reaction rate of 055 MWm3 in the reactor volume find the temperature at the outer surface of the reactor

bull A current of 300 amperes passes through a stainless steel wire of 25 mm diameter and k=20 WmoC The resistivity of the wire is 70x10-8 ohm-m and length of the wire is 2 m if the wire is submerged in a fluid maintained at 50oC and h at the wire surface is 4000 Wm2oC calculate the steady state temperature at the centre and at the surface of the wire

bull An approximately spherical shaped orange ( k=023 WmoC) 90 mm diameter undergoes ripping process and generates 5100 Wm3 of energy If external surface the orange is at 8oC determine

bull (i) temperature at the centre of the orange and

bull (ii) heat flow from the outer surface of the orange

bull FINS External surfaces to used to cooling area and hence convection

heat transfer can be improvedbull APPLICATIONS IC engine cylinder ( air cooling) Heat exchangers automobile

radiators computer chip refrigerators electric motors generators and transformers etc

bull ASSUMPTIONS MADE IN ANALYSIS OF HEAT TRANFER THROUGH FIN

1 Fin material is homogenous and isotropic constthermal conductivity

2 Temperature at any cs of the fin is uniform3 lsquohrsquo is uniform over entire surface4 No internal heat generation5 Thermal contact resistance is negligible6 Radiation heat transfer is negligible7 Heat conduction is steady state and one dimensional

bull Problem 1

bull A stainless steel fin (k=20 WmK) having a diameter of 20 mm and a length of 01 m is attached to a wall at 300oC The ambient temperature is 50oC and the heat transfer coefficient is 10 Wm2K The fin tip is insulated Determine

bull (A) The rate of heat dissipation from the finbull (B) The temperature at the fin tipbull (C) The rate of heat transfer from the wall area covered by the fin if

the fin was not usedbull (D) The heat transfer rate from the same fin geometry if the

stainless steel fin is replaced by a fictitious fin with infinite thermal conductivity

bull Problem2bull One end of a long rod is inserted into a furnace while the other

projects into ambient temperature Under steady state the temperature of the rod is measured at two points 75mm apart and found to be 125oC and 885oC respectively while the ambient temperature is 20oC If the rod is 25mm in diameter and h is 2336 Wm2K

bull Find the thermal conductivity of the rod material

  • HEAT AND MASS TRANSFER
  • Thermodynamics Vs Heat transfer
  • MODE OF HEAT TRANSFER
  • MODE OF HEAT TRANSFER (2)
  • CONDUCTION- MECHANISM
  • CONCECTION-MECHANISM
  • RADIATION-MECHANISM
  • Conduction- Convection- Radiation
  • Slide 9
  • Slide 10
  • Fourierrsquos Law of Heat Conduction
  • Slide 12
  • Newtonrsquos Law of Cooling
  • Laws of Thermal Radiation
  • LAWS FOR HEAT TRANSFER
  • ELECTRICAL ampTHERMAL ANALOGY
  • Slide 17
  • Slide 18
  • PLANE WALL-UNIDIRECTIONAL
  • PLANE WALL -SERIES
  • PLANE WALL IN SERIES
  • PLANE WALL -PARALLEL
  • PLANE WALL-PARALLEL
  • PLANE WALL-COMBINATION
  • PLANE WALL-FILM RESISTANCE
  • PLANE WALL- THERMAL CONTACT RESISTANCE
  • PLANE WALL- THERMAL CONTACT RESISTANCE (2)
  • CYLINDER-SOLID
  • HEAT TRANSFER IN SOLID CYLINDER
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER-COMPOSITE
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER-MULTY LAYER
  • HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER
  • HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER (2)
  • HEAT TRANSFER IN SPHERE
  • HEAT TRANSFER IN SPHERE -MULTY LAYER
  • HEAT TRANSFER IN SPHERE -MULTY LAYER (2)
  • Slide 38
  • Slide 39
  • Slide 40
  • Slide 41
  • CRITICAL RADIUS
  • Plane wall- Insulation thickness -study
  • Problem5
  • Slide 45
  • Slide 46
  • Slide 47
  • GENERAL HEAT CONDUCTION EQUATION IN CYLINDRICAL COORDINATES
  • Slide 49
  • Slide 50
  • Slide 51
  • GENERAL HEAT CONDUCTION EQUATION IN SPHERICAL COORDINATES
  • Slide 53
  • Slide 54
  • Slide 55
  • General Heat Conduction Equation
  • Temperature profile
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Slide 67
Page 53: hmt chap 1

Temperature profile

Plane wall SphereCylinder

bull A copper conductor (k = 380 WmoC resistivity ρ = 2x10-8 Ω m) having inner and outer radii 10 cm and 225 cm respectively is carrying a current density of 4800 amperes cm2 the conductor is internally cooled and a constant temperature of 65oC is maintained at the inner surface and there is no heat transfer through the insulation surrounding the conductor

Determinebull (i) The maximum temperature of the conductor and the radius at

which it occursbull (ii) The internal heat transfer rate

bull A chemical reaction takes place in a packed bed (k=06 WmoC) between two coaxial cylinders with radii 15 mm and 45 mm the inner surface is at 580oC and it is insulated Assuming the reaction rate of 055 MWm3 in the reactor volume find the temperature at the outer surface of the reactor

bull A current of 300 amperes passes through a stainless steel wire of 25 mm diameter and k=20 WmoC The resistivity of the wire is 70x10-8 ohm-m and length of the wire is 2 m if the wire is submerged in a fluid maintained at 50oC and h at the wire surface is 4000 Wm2oC calculate the steady state temperature at the centre and at the surface of the wire

bull An approximately spherical shaped orange ( k=023 WmoC) 90 mm diameter undergoes ripping process and generates 5100 Wm3 of energy If external surface the orange is at 8oC determine

bull (i) temperature at the centre of the orange and

bull (ii) heat flow from the outer surface of the orange

bull FINS External surfaces to used to cooling area and hence convection

heat transfer can be improvedbull APPLICATIONS IC engine cylinder ( air cooling) Heat exchangers automobile

radiators computer chip refrigerators electric motors generators and transformers etc

bull ASSUMPTIONS MADE IN ANALYSIS OF HEAT TRANFER THROUGH FIN

1 Fin material is homogenous and isotropic constthermal conductivity

2 Temperature at any cs of the fin is uniform3 lsquohrsquo is uniform over entire surface4 No internal heat generation5 Thermal contact resistance is negligible6 Radiation heat transfer is negligible7 Heat conduction is steady state and one dimensional

bull Problem 1

bull A stainless steel fin (k=20 WmK) having a diameter of 20 mm and a length of 01 m is attached to a wall at 300oC The ambient temperature is 50oC and the heat transfer coefficient is 10 Wm2K The fin tip is insulated Determine

bull (A) The rate of heat dissipation from the finbull (B) The temperature at the fin tipbull (C) The rate of heat transfer from the wall area covered by the fin if

the fin was not usedbull (D) The heat transfer rate from the same fin geometry if the

stainless steel fin is replaced by a fictitious fin with infinite thermal conductivity

bull Problem2bull One end of a long rod is inserted into a furnace while the other

projects into ambient temperature Under steady state the temperature of the rod is measured at two points 75mm apart and found to be 125oC and 885oC respectively while the ambient temperature is 20oC If the rod is 25mm in diameter and h is 2336 Wm2K

bull Find the thermal conductivity of the rod material

  • HEAT AND MASS TRANSFER
  • Thermodynamics Vs Heat transfer
  • MODE OF HEAT TRANSFER
  • MODE OF HEAT TRANSFER (2)
  • CONDUCTION- MECHANISM
  • CONCECTION-MECHANISM
  • RADIATION-MECHANISM
  • Conduction- Convection- Radiation
  • Slide 9
  • Slide 10
  • Fourierrsquos Law of Heat Conduction
  • Slide 12
  • Newtonrsquos Law of Cooling
  • Laws of Thermal Radiation
  • LAWS FOR HEAT TRANSFER
  • ELECTRICAL ampTHERMAL ANALOGY
  • Slide 17
  • Slide 18
  • PLANE WALL-UNIDIRECTIONAL
  • PLANE WALL -SERIES
  • PLANE WALL IN SERIES
  • PLANE WALL -PARALLEL
  • PLANE WALL-PARALLEL
  • PLANE WALL-COMBINATION
  • PLANE WALL-FILM RESISTANCE
  • PLANE WALL- THERMAL CONTACT RESISTANCE
  • PLANE WALL- THERMAL CONTACT RESISTANCE (2)
  • CYLINDER-SOLID
  • HEAT TRANSFER IN SOLID CYLINDER
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER-COMPOSITE
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER-MULTY LAYER
  • HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER
  • HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER (2)
  • HEAT TRANSFER IN SPHERE
  • HEAT TRANSFER IN SPHERE -MULTY LAYER
  • HEAT TRANSFER IN SPHERE -MULTY LAYER (2)
  • Slide 38
  • Slide 39
  • Slide 40
  • Slide 41
  • CRITICAL RADIUS
  • Plane wall- Insulation thickness -study
  • Problem5
  • Slide 45
  • Slide 46
  • Slide 47
  • GENERAL HEAT CONDUCTION EQUATION IN CYLINDRICAL COORDINATES
  • Slide 49
  • Slide 50
  • Slide 51
  • GENERAL HEAT CONDUCTION EQUATION IN SPHERICAL COORDINATES
  • Slide 53
  • Slide 54
  • Slide 55
  • General Heat Conduction Equation
  • Temperature profile
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Slide 67
Page 54: hmt chap 1

bull A copper conductor (k = 380 WmoC resistivity ρ = 2x10-8 Ω m) having inner and outer radii 10 cm and 225 cm respectively is carrying a current density of 4800 amperes cm2 the conductor is internally cooled and a constant temperature of 65oC is maintained at the inner surface and there is no heat transfer through the insulation surrounding the conductor

Determinebull (i) The maximum temperature of the conductor and the radius at

which it occursbull (ii) The internal heat transfer rate

bull A chemical reaction takes place in a packed bed (k=06 WmoC) between two coaxial cylinders with radii 15 mm and 45 mm the inner surface is at 580oC and it is insulated Assuming the reaction rate of 055 MWm3 in the reactor volume find the temperature at the outer surface of the reactor

bull A current of 300 amperes passes through a stainless steel wire of 25 mm diameter and k=20 WmoC The resistivity of the wire is 70x10-8 ohm-m and length of the wire is 2 m if the wire is submerged in a fluid maintained at 50oC and h at the wire surface is 4000 Wm2oC calculate the steady state temperature at the centre and at the surface of the wire

bull An approximately spherical shaped orange ( k=023 WmoC) 90 mm diameter undergoes ripping process and generates 5100 Wm3 of energy If external surface the orange is at 8oC determine

bull (i) temperature at the centre of the orange and

bull (ii) heat flow from the outer surface of the orange

bull FINS External surfaces to used to cooling area and hence convection

heat transfer can be improvedbull APPLICATIONS IC engine cylinder ( air cooling) Heat exchangers automobile

radiators computer chip refrigerators electric motors generators and transformers etc

bull ASSUMPTIONS MADE IN ANALYSIS OF HEAT TRANFER THROUGH FIN

1 Fin material is homogenous and isotropic constthermal conductivity

2 Temperature at any cs of the fin is uniform3 lsquohrsquo is uniform over entire surface4 No internal heat generation5 Thermal contact resistance is negligible6 Radiation heat transfer is negligible7 Heat conduction is steady state and one dimensional

bull Problem 1

bull A stainless steel fin (k=20 WmK) having a diameter of 20 mm and a length of 01 m is attached to a wall at 300oC The ambient temperature is 50oC and the heat transfer coefficient is 10 Wm2K The fin tip is insulated Determine

bull (A) The rate of heat dissipation from the finbull (B) The temperature at the fin tipbull (C) The rate of heat transfer from the wall area covered by the fin if

the fin was not usedbull (D) The heat transfer rate from the same fin geometry if the

stainless steel fin is replaced by a fictitious fin with infinite thermal conductivity

bull Problem2bull One end of a long rod is inserted into a furnace while the other

projects into ambient temperature Under steady state the temperature of the rod is measured at two points 75mm apart and found to be 125oC and 885oC respectively while the ambient temperature is 20oC If the rod is 25mm in diameter and h is 2336 Wm2K

bull Find the thermal conductivity of the rod material

  • HEAT AND MASS TRANSFER
  • Thermodynamics Vs Heat transfer
  • MODE OF HEAT TRANSFER
  • MODE OF HEAT TRANSFER (2)
  • CONDUCTION- MECHANISM
  • CONCECTION-MECHANISM
  • RADIATION-MECHANISM
  • Conduction- Convection- Radiation
  • Slide 9
  • Slide 10
  • Fourierrsquos Law of Heat Conduction
  • Slide 12
  • Newtonrsquos Law of Cooling
  • Laws of Thermal Radiation
  • LAWS FOR HEAT TRANSFER
  • ELECTRICAL ampTHERMAL ANALOGY
  • Slide 17
  • Slide 18
  • PLANE WALL-UNIDIRECTIONAL
  • PLANE WALL -SERIES
  • PLANE WALL IN SERIES
  • PLANE WALL -PARALLEL
  • PLANE WALL-PARALLEL
  • PLANE WALL-COMBINATION
  • PLANE WALL-FILM RESISTANCE
  • PLANE WALL- THERMAL CONTACT RESISTANCE
  • PLANE WALL- THERMAL CONTACT RESISTANCE (2)
  • CYLINDER-SOLID
  • HEAT TRANSFER IN SOLID CYLINDER
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER-COMPOSITE
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER-MULTY LAYER
  • HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER
  • HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER (2)
  • HEAT TRANSFER IN SPHERE
  • HEAT TRANSFER IN SPHERE -MULTY LAYER
  • HEAT TRANSFER IN SPHERE -MULTY LAYER (2)
  • Slide 38
  • Slide 39
  • Slide 40
  • Slide 41
  • CRITICAL RADIUS
  • Plane wall- Insulation thickness -study
  • Problem5
  • Slide 45
  • Slide 46
  • Slide 47
  • GENERAL HEAT CONDUCTION EQUATION IN CYLINDRICAL COORDINATES
  • Slide 49
  • Slide 50
  • Slide 51
  • GENERAL HEAT CONDUCTION EQUATION IN SPHERICAL COORDINATES
  • Slide 53
  • Slide 54
  • Slide 55
  • General Heat Conduction Equation
  • Temperature profile
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Slide 67
Page 55: hmt chap 1

bull A chemical reaction takes place in a packed bed (k=06 WmoC) between two coaxial cylinders with radii 15 mm and 45 mm the inner surface is at 580oC and it is insulated Assuming the reaction rate of 055 MWm3 in the reactor volume find the temperature at the outer surface of the reactor

bull A current of 300 amperes passes through a stainless steel wire of 25 mm diameter and k=20 WmoC The resistivity of the wire is 70x10-8 ohm-m and length of the wire is 2 m if the wire is submerged in a fluid maintained at 50oC and h at the wire surface is 4000 Wm2oC calculate the steady state temperature at the centre and at the surface of the wire

bull An approximately spherical shaped orange ( k=023 WmoC) 90 mm diameter undergoes ripping process and generates 5100 Wm3 of energy If external surface the orange is at 8oC determine

bull (i) temperature at the centre of the orange and

bull (ii) heat flow from the outer surface of the orange

bull FINS External surfaces to used to cooling area and hence convection

heat transfer can be improvedbull APPLICATIONS IC engine cylinder ( air cooling) Heat exchangers automobile

radiators computer chip refrigerators electric motors generators and transformers etc

bull ASSUMPTIONS MADE IN ANALYSIS OF HEAT TRANFER THROUGH FIN

1 Fin material is homogenous and isotropic constthermal conductivity

2 Temperature at any cs of the fin is uniform3 lsquohrsquo is uniform over entire surface4 No internal heat generation5 Thermal contact resistance is negligible6 Radiation heat transfer is negligible7 Heat conduction is steady state and one dimensional

bull Problem 1

bull A stainless steel fin (k=20 WmK) having a diameter of 20 mm and a length of 01 m is attached to a wall at 300oC The ambient temperature is 50oC and the heat transfer coefficient is 10 Wm2K The fin tip is insulated Determine

bull (A) The rate of heat dissipation from the finbull (B) The temperature at the fin tipbull (C) The rate of heat transfer from the wall area covered by the fin if

the fin was not usedbull (D) The heat transfer rate from the same fin geometry if the

stainless steel fin is replaced by a fictitious fin with infinite thermal conductivity

bull Problem2bull One end of a long rod is inserted into a furnace while the other

projects into ambient temperature Under steady state the temperature of the rod is measured at two points 75mm apart and found to be 125oC and 885oC respectively while the ambient temperature is 20oC If the rod is 25mm in diameter and h is 2336 Wm2K

bull Find the thermal conductivity of the rod material

  • HEAT AND MASS TRANSFER
  • Thermodynamics Vs Heat transfer
  • MODE OF HEAT TRANSFER
  • MODE OF HEAT TRANSFER (2)
  • CONDUCTION- MECHANISM
  • CONCECTION-MECHANISM
  • RADIATION-MECHANISM
  • Conduction- Convection- Radiation
  • Slide 9
  • Slide 10
  • Fourierrsquos Law of Heat Conduction
  • Slide 12
  • Newtonrsquos Law of Cooling
  • Laws of Thermal Radiation
  • LAWS FOR HEAT TRANSFER
  • ELECTRICAL ampTHERMAL ANALOGY
  • Slide 17
  • Slide 18
  • PLANE WALL-UNIDIRECTIONAL
  • PLANE WALL -SERIES
  • PLANE WALL IN SERIES
  • PLANE WALL -PARALLEL
  • PLANE WALL-PARALLEL
  • PLANE WALL-COMBINATION
  • PLANE WALL-FILM RESISTANCE
  • PLANE WALL- THERMAL CONTACT RESISTANCE
  • PLANE WALL- THERMAL CONTACT RESISTANCE (2)
  • CYLINDER-SOLID
  • HEAT TRANSFER IN SOLID CYLINDER
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER-COMPOSITE
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER-MULTY LAYER
  • HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER
  • HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER (2)
  • HEAT TRANSFER IN SPHERE
  • HEAT TRANSFER IN SPHERE -MULTY LAYER
  • HEAT TRANSFER IN SPHERE -MULTY LAYER (2)
  • Slide 38
  • Slide 39
  • Slide 40
  • Slide 41
  • CRITICAL RADIUS
  • Plane wall- Insulation thickness -study
  • Problem5
  • Slide 45
  • Slide 46
  • Slide 47
  • GENERAL HEAT CONDUCTION EQUATION IN CYLINDRICAL COORDINATES
  • Slide 49
  • Slide 50
  • Slide 51
  • GENERAL HEAT CONDUCTION EQUATION IN SPHERICAL COORDINATES
  • Slide 53
  • Slide 54
  • Slide 55
  • General Heat Conduction Equation
  • Temperature profile
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Slide 67
Page 56: hmt chap 1

bull A current of 300 amperes passes through a stainless steel wire of 25 mm diameter and k=20 WmoC The resistivity of the wire is 70x10-8 ohm-m and length of the wire is 2 m if the wire is submerged in a fluid maintained at 50oC and h at the wire surface is 4000 Wm2oC calculate the steady state temperature at the centre and at the surface of the wire

bull An approximately spherical shaped orange ( k=023 WmoC) 90 mm diameter undergoes ripping process and generates 5100 Wm3 of energy If external surface the orange is at 8oC determine

bull (i) temperature at the centre of the orange and

bull (ii) heat flow from the outer surface of the orange

bull FINS External surfaces to used to cooling area and hence convection

heat transfer can be improvedbull APPLICATIONS IC engine cylinder ( air cooling) Heat exchangers automobile

radiators computer chip refrigerators electric motors generators and transformers etc

bull ASSUMPTIONS MADE IN ANALYSIS OF HEAT TRANFER THROUGH FIN

1 Fin material is homogenous and isotropic constthermal conductivity

2 Temperature at any cs of the fin is uniform3 lsquohrsquo is uniform over entire surface4 No internal heat generation5 Thermal contact resistance is negligible6 Radiation heat transfer is negligible7 Heat conduction is steady state and one dimensional

bull Problem 1

bull A stainless steel fin (k=20 WmK) having a diameter of 20 mm and a length of 01 m is attached to a wall at 300oC The ambient temperature is 50oC and the heat transfer coefficient is 10 Wm2K The fin tip is insulated Determine

bull (A) The rate of heat dissipation from the finbull (B) The temperature at the fin tipbull (C) The rate of heat transfer from the wall area covered by the fin if

the fin was not usedbull (D) The heat transfer rate from the same fin geometry if the

stainless steel fin is replaced by a fictitious fin with infinite thermal conductivity

bull Problem2bull One end of a long rod is inserted into a furnace while the other

projects into ambient temperature Under steady state the temperature of the rod is measured at two points 75mm apart and found to be 125oC and 885oC respectively while the ambient temperature is 20oC If the rod is 25mm in diameter and h is 2336 Wm2K

bull Find the thermal conductivity of the rod material

  • HEAT AND MASS TRANSFER
  • Thermodynamics Vs Heat transfer
  • MODE OF HEAT TRANSFER
  • MODE OF HEAT TRANSFER (2)
  • CONDUCTION- MECHANISM
  • CONCECTION-MECHANISM
  • RADIATION-MECHANISM
  • Conduction- Convection- Radiation
  • Slide 9
  • Slide 10
  • Fourierrsquos Law of Heat Conduction
  • Slide 12
  • Newtonrsquos Law of Cooling
  • Laws of Thermal Radiation
  • LAWS FOR HEAT TRANSFER
  • ELECTRICAL ampTHERMAL ANALOGY
  • Slide 17
  • Slide 18
  • PLANE WALL-UNIDIRECTIONAL
  • PLANE WALL -SERIES
  • PLANE WALL IN SERIES
  • PLANE WALL -PARALLEL
  • PLANE WALL-PARALLEL
  • PLANE WALL-COMBINATION
  • PLANE WALL-FILM RESISTANCE
  • PLANE WALL- THERMAL CONTACT RESISTANCE
  • PLANE WALL- THERMAL CONTACT RESISTANCE (2)
  • CYLINDER-SOLID
  • HEAT TRANSFER IN SOLID CYLINDER
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER-COMPOSITE
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER-MULTY LAYER
  • HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER
  • HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER (2)
  • HEAT TRANSFER IN SPHERE
  • HEAT TRANSFER IN SPHERE -MULTY LAYER
  • HEAT TRANSFER IN SPHERE -MULTY LAYER (2)
  • Slide 38
  • Slide 39
  • Slide 40
  • Slide 41
  • CRITICAL RADIUS
  • Plane wall- Insulation thickness -study
  • Problem5
  • Slide 45
  • Slide 46
  • Slide 47
  • GENERAL HEAT CONDUCTION EQUATION IN CYLINDRICAL COORDINATES
  • Slide 49
  • Slide 50
  • Slide 51
  • GENERAL HEAT CONDUCTION EQUATION IN SPHERICAL COORDINATES
  • Slide 53
  • Slide 54
  • Slide 55
  • General Heat Conduction Equation
  • Temperature profile
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Slide 67
Page 57: hmt chap 1

bull An approximately spherical shaped orange ( k=023 WmoC) 90 mm diameter undergoes ripping process and generates 5100 Wm3 of energy If external surface the orange is at 8oC determine

bull (i) temperature at the centre of the orange and

bull (ii) heat flow from the outer surface of the orange

bull FINS External surfaces to used to cooling area and hence convection

heat transfer can be improvedbull APPLICATIONS IC engine cylinder ( air cooling) Heat exchangers automobile

radiators computer chip refrigerators electric motors generators and transformers etc

bull ASSUMPTIONS MADE IN ANALYSIS OF HEAT TRANFER THROUGH FIN

1 Fin material is homogenous and isotropic constthermal conductivity

2 Temperature at any cs of the fin is uniform3 lsquohrsquo is uniform over entire surface4 No internal heat generation5 Thermal contact resistance is negligible6 Radiation heat transfer is negligible7 Heat conduction is steady state and one dimensional

bull Problem 1

bull A stainless steel fin (k=20 WmK) having a diameter of 20 mm and a length of 01 m is attached to a wall at 300oC The ambient temperature is 50oC and the heat transfer coefficient is 10 Wm2K The fin tip is insulated Determine

bull (A) The rate of heat dissipation from the finbull (B) The temperature at the fin tipbull (C) The rate of heat transfer from the wall area covered by the fin if

the fin was not usedbull (D) The heat transfer rate from the same fin geometry if the

stainless steel fin is replaced by a fictitious fin with infinite thermal conductivity

bull Problem2bull One end of a long rod is inserted into a furnace while the other

projects into ambient temperature Under steady state the temperature of the rod is measured at two points 75mm apart and found to be 125oC and 885oC respectively while the ambient temperature is 20oC If the rod is 25mm in diameter and h is 2336 Wm2K

bull Find the thermal conductivity of the rod material

  • HEAT AND MASS TRANSFER
  • Thermodynamics Vs Heat transfer
  • MODE OF HEAT TRANSFER
  • MODE OF HEAT TRANSFER (2)
  • CONDUCTION- MECHANISM
  • CONCECTION-MECHANISM
  • RADIATION-MECHANISM
  • Conduction- Convection- Radiation
  • Slide 9
  • Slide 10
  • Fourierrsquos Law of Heat Conduction
  • Slide 12
  • Newtonrsquos Law of Cooling
  • Laws of Thermal Radiation
  • LAWS FOR HEAT TRANSFER
  • ELECTRICAL ampTHERMAL ANALOGY
  • Slide 17
  • Slide 18
  • PLANE WALL-UNIDIRECTIONAL
  • PLANE WALL -SERIES
  • PLANE WALL IN SERIES
  • PLANE WALL -PARALLEL
  • PLANE WALL-PARALLEL
  • PLANE WALL-COMBINATION
  • PLANE WALL-FILM RESISTANCE
  • PLANE WALL- THERMAL CONTACT RESISTANCE
  • PLANE WALL- THERMAL CONTACT RESISTANCE (2)
  • CYLINDER-SOLID
  • HEAT TRANSFER IN SOLID CYLINDER
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER-COMPOSITE
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER-MULTY LAYER
  • HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER
  • HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER (2)
  • HEAT TRANSFER IN SPHERE
  • HEAT TRANSFER IN SPHERE -MULTY LAYER
  • HEAT TRANSFER IN SPHERE -MULTY LAYER (2)
  • Slide 38
  • Slide 39
  • Slide 40
  • Slide 41
  • CRITICAL RADIUS
  • Plane wall- Insulation thickness -study
  • Problem5
  • Slide 45
  • Slide 46
  • Slide 47
  • GENERAL HEAT CONDUCTION EQUATION IN CYLINDRICAL COORDINATES
  • Slide 49
  • Slide 50
  • Slide 51
  • GENERAL HEAT CONDUCTION EQUATION IN SPHERICAL COORDINATES
  • Slide 53
  • Slide 54
  • Slide 55
  • General Heat Conduction Equation
  • Temperature profile
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Slide 67
Page 58: hmt chap 1

bull FINS External surfaces to used to cooling area and hence convection

heat transfer can be improvedbull APPLICATIONS IC engine cylinder ( air cooling) Heat exchangers automobile

radiators computer chip refrigerators electric motors generators and transformers etc

bull ASSUMPTIONS MADE IN ANALYSIS OF HEAT TRANFER THROUGH FIN

1 Fin material is homogenous and isotropic constthermal conductivity

2 Temperature at any cs of the fin is uniform3 lsquohrsquo is uniform over entire surface4 No internal heat generation5 Thermal contact resistance is negligible6 Radiation heat transfer is negligible7 Heat conduction is steady state and one dimensional

bull Problem 1

bull A stainless steel fin (k=20 WmK) having a diameter of 20 mm and a length of 01 m is attached to a wall at 300oC The ambient temperature is 50oC and the heat transfer coefficient is 10 Wm2K The fin tip is insulated Determine

bull (A) The rate of heat dissipation from the finbull (B) The temperature at the fin tipbull (C) The rate of heat transfer from the wall area covered by the fin if

the fin was not usedbull (D) The heat transfer rate from the same fin geometry if the

stainless steel fin is replaced by a fictitious fin with infinite thermal conductivity

bull Problem2bull One end of a long rod is inserted into a furnace while the other

projects into ambient temperature Under steady state the temperature of the rod is measured at two points 75mm apart and found to be 125oC and 885oC respectively while the ambient temperature is 20oC If the rod is 25mm in diameter and h is 2336 Wm2K

bull Find the thermal conductivity of the rod material

  • HEAT AND MASS TRANSFER
  • Thermodynamics Vs Heat transfer
  • MODE OF HEAT TRANSFER
  • MODE OF HEAT TRANSFER (2)
  • CONDUCTION- MECHANISM
  • CONCECTION-MECHANISM
  • RADIATION-MECHANISM
  • Conduction- Convection- Radiation
  • Slide 9
  • Slide 10
  • Fourierrsquos Law of Heat Conduction
  • Slide 12
  • Newtonrsquos Law of Cooling
  • Laws of Thermal Radiation
  • LAWS FOR HEAT TRANSFER
  • ELECTRICAL ampTHERMAL ANALOGY
  • Slide 17
  • Slide 18
  • PLANE WALL-UNIDIRECTIONAL
  • PLANE WALL -SERIES
  • PLANE WALL IN SERIES
  • PLANE WALL -PARALLEL
  • PLANE WALL-PARALLEL
  • PLANE WALL-COMBINATION
  • PLANE WALL-FILM RESISTANCE
  • PLANE WALL- THERMAL CONTACT RESISTANCE
  • PLANE WALL- THERMAL CONTACT RESISTANCE (2)
  • CYLINDER-SOLID
  • HEAT TRANSFER IN SOLID CYLINDER
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER-COMPOSITE
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER-MULTY LAYER
  • HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER
  • HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER (2)
  • HEAT TRANSFER IN SPHERE
  • HEAT TRANSFER IN SPHERE -MULTY LAYER
  • HEAT TRANSFER IN SPHERE -MULTY LAYER (2)
  • Slide 38
  • Slide 39
  • Slide 40
  • Slide 41
  • CRITICAL RADIUS
  • Plane wall- Insulation thickness -study
  • Problem5
  • Slide 45
  • Slide 46
  • Slide 47
  • GENERAL HEAT CONDUCTION EQUATION IN CYLINDRICAL COORDINATES
  • Slide 49
  • Slide 50
  • Slide 51
  • GENERAL HEAT CONDUCTION EQUATION IN SPHERICAL COORDINATES
  • Slide 53
  • Slide 54
  • Slide 55
  • General Heat Conduction Equation
  • Temperature profile
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Slide 67
Page 59: hmt chap 1

bull Problem 1

bull A stainless steel fin (k=20 WmK) having a diameter of 20 mm and a length of 01 m is attached to a wall at 300oC The ambient temperature is 50oC and the heat transfer coefficient is 10 Wm2K The fin tip is insulated Determine

bull (A) The rate of heat dissipation from the finbull (B) The temperature at the fin tipbull (C) The rate of heat transfer from the wall area covered by the fin if

the fin was not usedbull (D) The heat transfer rate from the same fin geometry if the

stainless steel fin is replaced by a fictitious fin with infinite thermal conductivity

bull Problem2bull One end of a long rod is inserted into a furnace while the other

projects into ambient temperature Under steady state the temperature of the rod is measured at two points 75mm apart and found to be 125oC and 885oC respectively while the ambient temperature is 20oC If the rod is 25mm in diameter and h is 2336 Wm2K

bull Find the thermal conductivity of the rod material

  • HEAT AND MASS TRANSFER
  • Thermodynamics Vs Heat transfer
  • MODE OF HEAT TRANSFER
  • MODE OF HEAT TRANSFER (2)
  • CONDUCTION- MECHANISM
  • CONCECTION-MECHANISM
  • RADIATION-MECHANISM
  • Conduction- Convection- Radiation
  • Slide 9
  • Slide 10
  • Fourierrsquos Law of Heat Conduction
  • Slide 12
  • Newtonrsquos Law of Cooling
  • Laws of Thermal Radiation
  • LAWS FOR HEAT TRANSFER
  • ELECTRICAL ampTHERMAL ANALOGY
  • Slide 17
  • Slide 18
  • PLANE WALL-UNIDIRECTIONAL
  • PLANE WALL -SERIES
  • PLANE WALL IN SERIES
  • PLANE WALL -PARALLEL
  • PLANE WALL-PARALLEL
  • PLANE WALL-COMBINATION
  • PLANE WALL-FILM RESISTANCE
  • PLANE WALL- THERMAL CONTACT RESISTANCE
  • PLANE WALL- THERMAL CONTACT RESISTANCE (2)
  • CYLINDER-SOLID
  • HEAT TRANSFER IN SOLID CYLINDER
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER-COMPOSITE
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER-MULTY LAYER
  • HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER
  • HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER (2)
  • HEAT TRANSFER IN SPHERE
  • HEAT TRANSFER IN SPHERE -MULTY LAYER
  • HEAT TRANSFER IN SPHERE -MULTY LAYER (2)
  • Slide 38
  • Slide 39
  • Slide 40
  • Slide 41
  • CRITICAL RADIUS
  • Plane wall- Insulation thickness -study
  • Problem5
  • Slide 45
  • Slide 46
  • Slide 47
  • GENERAL HEAT CONDUCTION EQUATION IN CYLINDRICAL COORDINATES
  • Slide 49
  • Slide 50
  • Slide 51
  • GENERAL HEAT CONDUCTION EQUATION IN SPHERICAL COORDINATES
  • Slide 53
  • Slide 54
  • Slide 55
  • General Heat Conduction Equation
  • Temperature profile
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Slide 67
Page 60: hmt chap 1

bull Problem2bull One end of a long rod is inserted into a furnace while the other

projects into ambient temperature Under steady state the temperature of the rod is measured at two points 75mm apart and found to be 125oC and 885oC respectively while the ambient temperature is 20oC If the rod is 25mm in diameter and h is 2336 Wm2K

bull Find the thermal conductivity of the rod material

  • HEAT AND MASS TRANSFER
  • Thermodynamics Vs Heat transfer
  • MODE OF HEAT TRANSFER
  • MODE OF HEAT TRANSFER (2)
  • CONDUCTION- MECHANISM
  • CONCECTION-MECHANISM
  • RADIATION-MECHANISM
  • Conduction- Convection- Radiation
  • Slide 9
  • Slide 10
  • Fourierrsquos Law of Heat Conduction
  • Slide 12
  • Newtonrsquos Law of Cooling
  • Laws of Thermal Radiation
  • LAWS FOR HEAT TRANSFER
  • ELECTRICAL ampTHERMAL ANALOGY
  • Slide 17
  • Slide 18
  • PLANE WALL-UNIDIRECTIONAL
  • PLANE WALL -SERIES
  • PLANE WALL IN SERIES
  • PLANE WALL -PARALLEL
  • PLANE WALL-PARALLEL
  • PLANE WALL-COMBINATION
  • PLANE WALL-FILM RESISTANCE
  • PLANE WALL- THERMAL CONTACT RESISTANCE
  • PLANE WALL- THERMAL CONTACT RESISTANCE (2)
  • CYLINDER-SOLID
  • HEAT TRANSFER IN SOLID CYLINDER
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER-COMPOSITE
  • HEAT TRANSFER RATE IN HOLLOW CYLINDER-MULTY LAYER
  • HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER
  • HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER (2)
  • HEAT TRANSFER IN SPHERE
  • HEAT TRANSFER IN SPHERE -MULTY LAYER
  • HEAT TRANSFER IN SPHERE -MULTY LAYER (2)
  • Slide 38
  • Slide 39
  • Slide 40
  • Slide 41
  • CRITICAL RADIUS
  • Plane wall- Insulation thickness -study
  • Problem5
  • Slide 45
  • Slide 46
  • Slide 47
  • GENERAL HEAT CONDUCTION EQUATION IN CYLINDRICAL COORDINATES
  • Slide 49
  • Slide 50
  • Slide 51
  • GENERAL HEAT CONDUCTION EQUATION IN SPHERICAL COORDINATES
  • Slide 53
  • Slide 54
  • Slide 55
  • General Heat Conduction Equation
  • Temperature profile
  • Slide 58
  • Slide 59
  • Slide 60
  • Slide 61
  • Slide 62
  • Slide 63
  • Slide 64
  • Slide 65
  • Slide 66
  • Slide 67