Hmt Bs Me 2014 Introduction Plus Conduction

HEAT AND MASS TRANSFER

1.Heat and Mass Transfer a Practical Approach by Cengel2.Fundamentals of Heat and Mass Transfer by Incropera3.Heat Transfer by Holman4.Transport Phenomenon by Bird

Course Out Line1. BASICS OF HEAT TRANSFER

1. Relationship between Thermodynamics and Heat Transfer2. Heat Transfer Mechanisms

1. Conduction1. Thermal Conductivity2. Thermal Diffusivity

2. Convection3. Radiation

3. Simultaneous Heat Transfer Mechanisms4. The conduction equation in different coordinates with and

without heat generation1. The various forms of the conduction equation2. Heat generation effects and its calculations3. Boundary conditions required for the solution of

conduction equation4. Case studies of various 1D steady state conduction

equation with and without heat generation5. The effect of variable thermal conductivity on conduction

heat transfer

Course Out Line, contd..2. STEADY HEAT CONDUCTION

1. Steady Heat Conduction in Plane Walls2. The Thermal Resistance Concept3. Thermal Resistance Network4. Multilayer Plane Walls5. Thermal Contact Resistance6. Generalized Thermal Resistance Networks7. Heat Conduction in Cylinders and Spheres8. Multilayered Cylinders and Spheres

1. Critical Radius of Insulation9. Enhancement in Conduction Heat Transfer

1. Heat Transfer from Finned Surface2. Fin Equation3. Fin Efficiency4. Fin Effectiveness5. Proper Length of a Fin

Course Out Line, contd..3. TRANSIENT HEAT CONDUCTION

1. Lumped System Analysis2. Criteria for Lumped System Analysis3. Transient Heat Conduction in Large Plane Walls, Long

Cylinders, and Spheres with Spatial Effects4. Transient Heat Conduction in Semi-Infinite Solids5. Transient Heat Conduction in Multidimensional Systems

4. FUNDAMENTALS OF CONVECTION1. Physical Mechanism on Convection2. Velocity Boundary Layer3. Thermal Boundary Layer4. Laminar and Turbulent Flows5. Heat and Momentum Transfer in Turbulent Flow6. Relationship between Fluid friction and heat transfer7. Differential Convection Equations8. Dimensionless form of the convection equation9. Nusselt, Prandtle & Reynolds Numbers & their importance

Course Out Line, contd..4. FUNDAMENTALS OF CONVECTION, contd..

10. General Thermal Analysis11. Constant Surface Heat Flux (qs constant)12. Constant Surface Temperature (Ts _ constant)13. Laminar Flow in Tubes14. Pressure Drop, Temperature Profile and the Nusselt

Number15. Laminar Flow in Noncircular Tubes16. Developing Laminar Flow in the Entrance Region17. Turbulent Flow in Tubes

1. Rough Surfaces2. Developing Turbulent Flow in the Entrance Region3. Turbulent Flow in Noncircular Tubes

18. Flow through Tube Annulus19. Heat Transfer Enhancement

Course Out Line, contd..5. EXTERNAL AND INTERNAL FORCED CONVECTION

1. Empirical and Practical Correlations for External and internal Forced Convection heat transfer for1. Parallel Flow over Flat Plates2. Flow across Cylinders and Spheres3. Flow across Tube Banks

6. NATURAL CONVECTION1. Physical Mechanism of Natural Convection2. Equation of Motion and the Grashof Number3. Empirical correlations for free convection over

1. Vertical / Horizontal Plates and cylinders2. Spheres3. Specific empirical correlations for Air

Course Out Line, contd..7. BOILING AND CONDENSATION

1. Boiling Heat Transfer1. Pool Boiling2. Boiling Regimes and the Boiling Curve3. Heat Transfer Correlations in Pool Boiling4. Flow Boiling

2. Introductory Condensation Heat Transfer3. Film-wise and Drop-wise Condensation

8. RADIATION HEAT TRANSFER1. Introduction to radiation heat transfer2. Thermal Radiation3. Blackbody Radiation4. Radiation Intensity5. Radiative Properties

1. Emissivity2. Absorptivity, Reflectivity, and Transmissivity

6. Kirchhoff’s Law

Course Out Line, contd..8. RADIATION HEAT TRANSFER, contd..

7. The View Factor8. View Factor Relations9. Radiation Heat Transfer: Black Surfaces10. Radiation Heat Transfer: Gray Surfaces11. Radiation Shields and the Radiation Effect12. Radiation Effect on Temperature Measurements

9. HEAT EXCHANGERS1. Types of Heat Exchangers2. The Overall Heat Transfer Coefficient3. Fouling Factor4. Analysis of Heat Exchangers5. The Log Mean Temperature Difference Method6. Types of flows in Heat Exchangers7. Design Methods of Heat Exchangers8. Selection of Heat Exchangers

Course Out Line, contd..10. MASS TRANSFER

1. Introduction2. Analogy between Heat and Mass Transfer3. Mass Diffusion4. Fick’s Law of Diffusion:

1. Stationary Medium Consisting of 2 Species2. Boundary Conditions3. Case Studies

5. Diffusion in a Moving Medium6. Equi-molar Counter diffusion7. Mass Convection8. Analogy between Friction, Heat Transfer & Mass Transfer

Coefficients9. Mass Convection Relations10. Simultaneous Heat and Mass Transfer

MODES OF HEAT TRANSFER Three modes of heat transfer

Conduction Convection Radiation

Forced Natural

Heat is transferred due to temperature

gradient

Heat is transferred by the motion of the

fluid

If the fluid is forced, forced convection

If the motion of the fluid is induced by the temperature, natural convection

Only occur at high temperature

HEAT CONDUCTION Points to Remember

1. Transfer occurs due to the pure random motion of the molecules at molecular level only

2. Simplest among the three modes of heat transfer. 3. It occurs in all three phases of matter (gas, liquid and

solid). 4. In solids, it is the only mechanism of heat transfer.5. The driving force for the heat conduction is the

temperature gradient, that is whenever a gradient in temperature exists there exists a conductive heat flow.

The heat flow is measured as the amount of energy transferred through any given plane per unit area per unit time, The HEAT FLUX.

Note:

Surface Energy BalanceConsider heat flow from hot wall of a house in winter

By conduction through the wall

to the outer surface of wall

Conducted heat is Convected away

Conducted heat is also

radiated awayForced Natural

Heat is transferred due to temperature

gradientHeat is transferred

by the motion of the fluid

If wind is blowing If no wind is blowing

Only significant at high

temperature

Surface Energy BalanceOVER ALL STEADY STATE HEAT TRANSFER

(Law of Conservation of Energy)Heat in = Heat out : In steady state conditions

Heat conducted through wall = heat convected + Heat radiatedQcond = Qcov + Qrad = Qgained/lost (very very important)

OVER ALL STEADY STATE HEAT TRANSFERHeat generated = Heat in = Heat out : In steady state conditions

Heat generated =Heat conducted through wall = heat convected + Heat radiated

Qgen = Qcond = Qcov + Qrad = Qgained/lost (very very important)

OVER ALL TRANSIENT STATE HEAT TRANSFERHeat in Heat out : In transient conditions

Heat in = Heat stored heat conducted and/or convected and/or radiated

HEAT CONDUCTION, contd. -2

From our experience we know the following facts

Direction of heat transfer

To

T1

Cross sectional area

z, thickness

1 oT Tq kz

HEAT CONDUCTION, contd. -3Experience shows that Conductive heat flux

heat flux: energy/area/time Property of material through which the heat is conducted:Thermal conductivity

Temperature gradient

Point wise:

Negative sign:Heat is moving in the direction of negative temperature gradient

Fourier’s lawdTq kdz

The fundamental equation

in heat conduction

z

q

Plane A

HEAT CONDUCTION, contd. -4

dTdz

Thermal conductivity at the plane A, k

Heat flux as a vector

Temperature is a scalar, while heat flux is a vector

HEAT CONDUCTION, contd. -4Cartesian Components of the Heat Flux Vector

z zdTq kdz

y ydTq kdy

x x

dTq kdx

For isotropic materials,kx = ky = kz = k

the heat flux vector will become:

q k T

Where x y zT T Tq q i q j q k k i j kx y z

and

i + j + kx y z

Grad operator

MOMENTUM vs HEAT , Similarity & differences

y ydTq kdy

xvyx

ddy

Heat / momentum flux

Driving Force

Property of Material

REMEMBERThere are three components of the HEAT FLUX VECTOR

What about the HEAT FLUX VECTOR accompanied by the three dimensional VELOCITY VECTOR

THERMAL CONDUCTIVITY

In general, ksolid > kliquid > kgas

Property of material through which the heat is conducted

Units: W/m/K

Materials Thermal conductivityHydrogen @ 100K 0.067Water @ 293K 0.600Liquid sodium @373K 83.70Aluminum 206.0

Insulating materials have low k

Materials Thermal conductivityGlass @ 298K 1.00

Brick @ 293K 0.6 – 1.00Glass wool @298K 0.07

THERMAL CONDUCTIVITY

THERMAL CONDUCTIVITY, contd. -2

Temperature dependence:

Gas Liquid Solid

Temperature

k

Gas: k increases with temperature Liquid: no general rule; e.g. k of alcohols, benzene, toluene decrease with T, while k of glycerin, ethylene glycol increase with T Solid: also no general rule; e.g. k of polyethylene, polypropylene, PTFE decrease with T, while those of insulation materials increase with T

Fairly independent of pressurePressure dependence:

CONVECTION HEAT TRANSFERA mode of energy/heat transfer between 1. A solid surface and the adjacent fluid that is in motion2. It involves the combined effects of conduction and fluid motion

1. Faster the fluid motion, the greater the convection heat transfer.2. No bulk fluid motion, heat transfer between a solid surface and

the adjacent fluid is by pure conduction. 3. Bulk motion of the fluid enhances the heat transfer between the

solid surface and the fluid but the bulk motion complicates the determination of heat transfer rates

Forced Convection: When the fluid is forced to flow over the surface by external means such as a fan, pump, or the windNatural (or Free) Convection: When the fluid motion is caused by buoyancy forces that are induced by density differences due to the variation of temperature in the fluid

Convection heat transfer is observed to be proportional to the temperature difference, and is given by Newton’s law of cooling as

Qconv = (k/x) Awall (Twall –Tbulk) or qconv = h (Twall –Tbulk)

CONVECTION HEAT TRANSFER

THE CONDUCTION EQUATIONThe conduction equation

describes the thermal energy transport by conduction

through an homogeneous solid.

Net Rate of Net Rate of internal and Net Rate of thermalaccumulation = addition by molecular motion energy generation of internal Energy (conduction) with in the CV

A thermal energy balance without any source term can be

written as

Internal energy: is energy associated with the random translational and internal motions of the molecules plus the

energy of interaction between the molecules

THE ENERGY EQUATION, -3

Accumulation

x y z Ut

Volume of element

Internal Energy per unit mass

Net Energy Addition by Conduction

x x y y z zz z zx x x y yy z q q x z q q x y q q

, , are components of the heat flux vector

Putting all those terms into the shell energy balance equation, then dividing by ∆x∆y∆z, we get the following energy balance equation per unit volume:

The case of No thermal energy generation

THE ENERGY EQUATION, -6

Ut

yx zqq qx y z

ConductionAccumulation

U qt

rate of energy input per unit volume by conduction

rate of gain of internal energy per unit volume

q k T

2U kTt

THE ENERGY EQUATION -71. Equations written in terms of internal energy (often difficult to

measure in an experiment) are not directly too useful.2. One prefers equations written in terms of temperature and heat

capacities (usually easier to measure in an experiment.From thermodynamics:

vT v v

U U pdU dv dT p T dv C dTv T T

In solids

1. pressure remains constant2. specific volume remains constant3. Cv Cp

pdU C dT

For solid, Liquid and gases if pressure is constant

2pTC k Tt

21

p

T Tk tC

21 T T

t

SOURCE TERMSThere is thermal energy generation with in the solid

Energy production can be due to:1. Chemical reaction2. Electrical heat3. Viscous dissipation (degradation of mechanical energy)4. Nuclear Reactions

Each energy production will be calculated as per its own physicsHowever, it must have the same consistent dimensions as that of the

original equation i.e energy per unit volume :

.energy production per unit volume g

2.1 T g T

t k

The Conduction Equation with Thermal Energy Generation

Energy Equation In Different Coordinate

2 2 22

2 2 2

T T TTx y z

2 2 22

2 2 2 2

1 1T T T TTr r r r z

2 2 22

2 2 2 2 2 2 2

2 1 1 1tan sin

T T T T TTr r r r r r

2 0gT Poisson Equationk

2 1g TT Diffusion or Fourier Biot Equationk t

2 0T Laplace Equation

BOUNDARY CONDITIONS1. Temperature is specified. This is BC of first kind (Dirichlet BC)2. At the bounding surface, the heat flux equals to the flux supplied.

Boundary condition of the second kind (Neumann BC).

1. Insulated Boundary

2. Thermal Symmetry

3. At the bounding surface, heat flux entering the domain equals the heat flux across the thin film surrounding the object. BC of the 3rd kind (Robin BC).

Concept of heat transfer coefficient

4. At the interface between the two domains, the temperatures and the heat fluxes of those domains are continuous. BC of the fourth kind.

0, 0,0 0

T t T tk

x x

2, 2,0 0

T L t T L tk

x x

1 1

0,0,

T tk h T T t

x

2 2

,,

T L tk h T L t T

x

441 ,1

0,0,surr

T tk T T t

x

4 4

2 ,2

,, surr

T L tk T L t T

x

Special Forms of Conduction Equation

Steady state conduction without any thermal energy generation

2 0T

Steady state conduction with thermal energy generation

2. 1g TTk t

Transient conduction without thermal energy generation

in 3D in 1D (Cartesian)2

2 0d Tdx

2.

0gTk


2

.0d T g

dx k

2 1 TTt

in 3D in 1D (Cartesian)

2

2

1T Tx t

Transient conduction with thermal energy generation


2

. 1T g Tx k t



2 0T


2. 1g TTk t


in 3D in 1D (cylindrical)2

2

1 1 0d dT d T dTrr dr dr dr r dr

2.

0gTk

in 3D in 1D (cylindrical).1 0d dT gr

r dr dr k

2 1 TTt

in 3D in 1D (cylindrical)

1 1d dT Trr dr dr t


in 3D in 1D (cylindrical).1 1d dT g Tr

r dr dr k t



2 0T


2. 1g TTk t


in 3D in 1D (spherical)2

22 2


2.

0gTk

in 3D in 1D (spherical)2

2

.1 0d dT grr dr dr k

2 1 TTt

in 3D in 1D (spherical) 2

2

1 1d dT Trr dr dr t


in 3D in 1D (spherical) 22

.1 1d dT g Trr dr dr k t

Steady state heat Conduction in slab

energy in energy out

constantcross sectionarea, A

thin shell

z z+zno heat production

Steady state

k is constant k is temperaturedependent

2

2 0d Tkdz

0d dTk Tdz dz

2 0T in 1D (Cartesian)2

2 0d Tdz

Steady state heat Conduction in slab, contd. -3

Physical constraint on the boundary

Boundary conditions of the first kind.

Temperature distribution

Temperature distribution is linear (only valid for constant k

and slab geometry).

0

o

L

T z T zT T L


Heat flux

Apply the Fourier’s law;

Be Very Careful and Remember1. that the heat flux in general is a function of z2. So to calculate the heat flux, you first decide

1. where do we evaluate the heat flux at? 2. At z=0 or z=L ?

Fourier lawdTq kdz

o LT Tq kL

0

o

L

T z T zT T L

Linear temperature

distribution

Heat flux proportionalto temperature

difference


Would the heat flux be proportional to temperature difference and temperature distribution linear if k=k(T)

Look at the case when “k” takes the following functional form:

1o ok k T T

Step 6′: Heat balance equation

0 1 0od dT d dTk k T Tdz dz dz dz

Step 7′: Boundary conditionssame as before, that is:z = 0; T = T0

z = L; T = TL

β

β


Step 8′: Temperature distribution

2

20

2

2

o o

L L o

T z T T z T zLT T T T

Thus the temperature distribution across the slab is not linear.

β > 0

β = 0

β < 0


Can you explain the shape of the temperature distribution?

z = 0 z = L

T0β > 0

Lower k, highertemperature gradient

Higher k, lowertemperature gradient

β < 0

Lower k, highertemperature gradient

Higher k, lowertemperature gradient

TL


Heat flux

Apply the Fourier’s law

2

20

2

2

o o

L L o

T z T T z T zLT T T T

1o odTq k T Tdz

k

2

2o L o L

o

T T T Tq k

L

independent of z,as one would

expect physically

Would this minus sign cause a concern to you as it could give zero flux?

Answer: Of course not, see the constraint on β


Note: If we define an average thermal conductivity across the slab as:

( ) 1

12

L L

o o

L L

o o

T T

o oT T

avg o o LT T

T T

k T dT k T T dT

k k T TdT dT

The heat flux equation is:

So far so goodBut are there other things that need to be addressed?

1. Existence of gas film surrounding the slab faces.2. Slab is made of many different materials.

LET’S CONSIDER THESE ONE BY ONE.

o Lavg

T Tq kL

which is an interesting equation

Heat conduction in slab: Gas film resistance

Physical Definition of the Problem

Conduction Heat Eq.Let’s study the constant k first, and then deal with temperature dependent k.

Heat conduction in slab: Gas film resistance, -2

Physical Constraints / Boundary ConditionsWith the existence of the gas film, we talk of BC of the third kind.

0

0; (0)o oz

dTz h T T kdz

; ( )L Lz L

dTz L h T L T kdz

Local Heat Transfer Coefficients = f(Re, Pr, Gr)


Note: T(0) and T(L) are yet known, but we proceed anyway.

( ) (0)( ) (0)

T z T zT L T L

Heat flux

this is the fluxthrough the slab

(0) ( )T T Lq kL


For steady state operation

Rearrange the above equation as:

Final heat flux equation

gas film slabq q

(0) ( )(0) ( )o o L L

T T Lq h T T k h T L T

L

(0) (0) ( ) ( )11

o L

Lo

T T T T L T L Tq

Lk hh

1 1

o L

o L

T Tq

Lh k h

written in terms of known parameters andoperating conditions.


write the heat flux equation as:

1 101

o L

L

T TkqLBi Bi

Where Bi0 and BiL are dimensionless and are defined as:

00 ; L

Lh L h LBi Bik k

which are called the Biot numbers

Physical significance of Biot number

heat transfer through the filmheat transfer through the solid

Bi

1. Bi >>1: Heat transfer through the object is limiting.2. Bi << 1: Heat transfer through the fluid film is limiting.


Avoid the confusion Bi and Nu

Their definitions:

(heat transfer coeff. through the film) (length)thermal conductivity of solid

Bi

(heat transfer coeff through the film) (length)thermal conductivity of fluid

Bi

Nu is seen in many correlations of heat transfer coefficients; hence it involves only film properties.

Bi is seen in the analysis of heat transfer to or from an object.

Heat conduction in slab: Gas film resistance, -6How about T(0) and T(L)?

Recall the following equations:

(0) (0) ( ) ( )11

o L

Lo

T T T T L T L Tq

Lk hh

1 1

o L

o L

T Tq

Lh k h

You get

1(0)

1 1o

o o L

o L

hT T T T

Lh k h

1( )

1 1L

L o L

o L

hT L T T T

Lh k h

Heat conduction in slab: Gas film & k(T)

The conduction heat balance equation is:

0 1 0od dT d dTk k T Tdz dz dz dz

Boundary conditions

0

0; (0)o oz

dTz h T T kdz

; ( )L Lz L

dTz L h T L T kdz

with T(0) being the solid object temperature at z=0, and T(L) being that at z=L.

Temperature distribution & heat flux

(0) ( )avg

T T Lq kL

1

2avg o o Lk k T T

Heat conduction in slab: Gas film & k(T), -2

This flux must be the same as fluxes through the film at both sides of the solid object:

(0) ( )(0) ( )o o avg L L

T T Lq h T T k h T L T

L

Thus, we get

1 1

o L

o avg L

T Tq

Lh k h

1

(0)1 1

oo o L

o avg L

hT T T T

Lh k h

1

( )1 1

LL o L

o avg L

hT L T T T

Lh k h

Two nonlinear algebraic equations in terms of T(0) and T(L) as kavg is afunction of T(0) and T(L).

So farHeat conduction1. Thermal conductivity, k2. First principles of solving heat conduction

problems3. Heat conduction in slab object:4. Heat conduction with gas film in slab object

1. BC of the first kind2. BC of the third kind

5. Constant thermal conductivity6. T-dependent thermal conductivity

Now let us turn to 1. composite objects2. objects of different geometries than slab, 3. Cylinder4. sphere

Composite slab objects

Let deal with constant k, and N layers.

Composite slab objectsUtilizing results obtained earlier for a single slab object,

one can write:

1 2 2 3 11 1 2 1 1 2

1 2

.. N No o N N N N

N

T T T T T Tq h T T k k k h T T

L L L

In the ratio form;

1 1 2 2 3 1 1 2

1 2

1 2 1

...1 1

o N N N N

N

o N N

T T T T T T T T T Tq

L L Lk kh k h

2

1 1

1 1

o N

Nj

jo j N

T Tq

Lh k h

The interfacial temperatures can be found by equating the above two equations

U-1

Heat conduction in cylinderArea is changing along the heat flow direction.

2

2


( ) 0d rq rdr

1 0d dTrr dr dr

( ) dTq r kdr

1 0d dTr kr dr dr

Heat conduction in cylinder, contd. -2

The final heat balance equation valid at a point is:

( ) 0d rq rdr

First-order ODE w.r.t. q

Apply the Fourier law of heat conduction

( ) dTq r kdr

0d dTr kdr dr

k is constant k =k(T)

2

2 0d Tkdr

0d dTk Tdr dr


Physical constraints

At r= R1: T = T1

At r= R2: T = T2

BC of first kind

Case of constant thermal conductivity


22

1 2 1

2

ln( )

ln

rRT r T

T T RR

always associated withcylindrical geometry


Heat flux

22

1 2 1

2

ln( )

ln

rRT r T

T T RR

( ) dTq r kdr

1 2

2

1

( )ln

k T Tq r

RrR

Unlike the slab case, this heat flux is not a constant, but rather decreases as the heat moves away from the center.

This is not entirely unexpected.

r heat flux area

Heat flow =

flux × area

small large smal

l the same

large small larg

e the same


Although the heat flux changes with distance r, the heat flow (energy/time) must be a constant, a requirement of steady state.

Total rate of heat flow out of the cylinder:

2 2

1 2 1 22

2 22

1 1

2 ( ) 2 2ln ln

r R r R

k T T T TQ rL kq r R L Lk

R RRR R

constant as expected

Compare this heat flow with that for the slab geometry.

Slab 1 2x L

AQ k T TL

Cylinder

21 2

2

1

2

lnr R

LQ T Tk

RR

Irrespective of the geometry, the heat flow is proportional to1. thermal conductivity2. temperature differenceOnly difference is geometrical factor


When the annulus is very thin, that is no curvature

R2 - R1 << R1

Would you expect the solution obtained for the cylindrical geometry to reduce to that of slab.

Now let’s consider the case of k=k(T)

1 such that 1o o o Lk k T T T T

Doing all the necessary mathematics and using same BC’s


2

1 11

22

2 1 2 1

1

ln2

ln2

rT r T T r T R

RT T T TR


Heat flux

211 2 1 2

2

1

2ln

kq T T T TRrR

Heat flow

211 2 1 2

2

1

22

ln

Lkq T T T TRR

which is independent of r.

Heat conduction in composite cylinders walls

Exactly the same analysis as that of composite wall slabs

Heat flow

2 1 3 2

1 2

2ln ln

....

in outL T Tq

R R R Rk k

Similarly the analysis of composite cylindrical walls with fluidsat the two surfaces, leads to

1 2

1

21 1

2

ln1 1

f f

ni i

iinner i n outer

L T Tq

R RR h k R h

Heat conduction in cylinder with source -1

Area is changing along the heat flow direction.

Heat balance around the annulus shell

2.

0gTk


�̇�=𝒉𝒆𝒂𝒕 𝒈𝒆𝒏𝒆𝒓𝒂𝒕𝒊𝒐𝒏𝒑𝒆𝒓 𝒖𝒏𝒊𝒕 𝒗𝒐𝒍𝒖𝒎𝒆If heat generated is by electricity then

Solve to get


In particular for solids the Thermal Equation is:

At r= 0: T = finite

At r= 0: T = Tm

Hence C1 = 0


2

1

.

2dT g rr cdr k

1.

2cdT g r

dr k r

2

1 2

.ln

4g rT r c r ck

2 mc T

2.

4mg rT r Tk

OR

At r= R: T = Ts

2

2 4sg Rc Tk

2 2 2

22

. .1

4 4 4s sg r g R g rT r T T r T Rk k k R


Finally the temperature distribution

-0.006 -0.004 -0.002 0.000 0.002 0.004 0.0060

50

100

150

200

250

300

350

400

450

500

550

600

650

Rad

ial T

empe

ratu

re

Radial Distance

To=50 To=100 To=150 To=200 To=250

The influence of shape on heat balanceSphereSlab Cylinder

0d dTkdr dr

0d dTr k

dr dr

2 0d dTr kdr dr

The general form for the three shapes is:

0sd dTr kdr dr

S (the shape factor) = 0, 1, 2

Hmt Bs Me 2014 Introduction Plus Conduction

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Transcript of Hmt Bs Me 2014 Introduction Plus Conduction