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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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 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 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 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 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 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 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 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 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 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
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
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
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
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
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
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
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
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
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
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
)( 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
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
)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
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
)( 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
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
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
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
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
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
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
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
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
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
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
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
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
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