Week_2_2

Objectives:

1. 1-D Heat Conduction in a Plane Wall

2. Electrical Resistance Analogy

3. Interface Resistance

4. 1-D Heat Flow in Cylindrical/ Spherical Coordinates

5. Electrical Resistance Analogy in Cylindrical Coords.

ME 3345 Heat Transfer

Reading Assignment: 3.1 3.4

1.With Uniform Volumetric Energy Generation; Assumptions ?

Solution:

2. What if there is no heat generation?

Steady State Temperature Distribution in a Plane Wall System

2 2 2

2 2 2

1T T T q T

k tx y z

21 2( )

2

qT x x C x C

k

1 2( )T x C x C

02

2

k

q

x

T

x

qq

Note: The form of the equation giving the steady state temperature

distribution is independent of boundary conditions. However, the

Temperature magnitude does depend on B.C.s.

1. Prescribed Temperature at Boundary: (e.g. T = To)

2. Uniform Heat Flux at boundary: (q = qo)

3. Adiabatic at Boundary: dT/dx = 0

4. Convection and Radiation at Boundary:

44

surTTTThdx

dTk

Example: What is the 1-D temperature distribution in a Plane Wall with

no energy generation subjected convection on both surfaces?

Assume that the thermal conductivity, k, is constant, no heat generation

uniform, fluid temperature and convective heat transfer coefficient ,

T and h, respectively.

K

x

0q

0 L

h, T h, T

T1 T2

R

I

V

I = V/R

We may also use an electrical resistance analogy to

help solve heat conduction problems.

Electrical Analog of Heat Conduction

T1 T2

Thermal resistance:

th

LR

kA

LR

A

Electrical resistance:

xth

kA Tq T

L R

xq

V= T

I = qx

In order to develop the electrical resistance for each

mode of heat transfer, we must look at the linear

heat transport coefficients:

1. Conduction:

2. Convection:

3. Radiation:

th

kA Lq T R

L kA

hARThAq

th

1

AhRTAhq

r

thr

1

(no energy generation!!!)

Heat Transfer Through a Plane Wall

L

A

,1sT

,2 2,T h

,2sT

,1 1,T h

Hot

Fluid

Cold

Fluid

,1 ,2

1 2 3

( )

( )x

T Tq

R R R

AhR

2

1

3AhR

1

1

1 kA

LR

2

qx

T,1 T,2 Ts,1 Ts,2

Thermal Circuits

1 2

1 1, where is the .x tot

tot

T Lq R total thermal resist

R h A kA h A

1, where ( ) is the .x totq UA T U R A overall heat transfer coeff

Thermal Contact Resistance

(Very Small)

TA TB

Unit Area

q"x

2, [m K/W]

A Bt c

x

T TR

q

Contact resistance values range from

10 6 10 3 m2K/W, depending on

(1) materials, (2) surface roughness,

and (3) contact pressure.

Effective thermal conductivity: ,/c t ck R

, , /t c t cR R A

A B

A B

A B

L L

k k

1T

2T

Example: A composite wall with contact resistance .

What is the heat flux if the temperatures are known. cR

1 2 and T T

1 2

1 2

/ /

xtot

A A c B B

T Tq

R

T T

L k R L k

Answer:

xq1T AT BT 2

T

/A AL k /B BL kcR

How to improve Contact Resistance ??

Apply Pressure

How to improve Contact Resistance ??

Use filler material

Soft metal (indium, lead, tin, silver)

Thermal grease

Epoxy materials

Soldering

Very active research area

Tables 3-1 and 3-2 give values for some solid-solid contact

resistance.

Types of contact: metal-metal, metal-insulators, etc.

Interfacial materials: vacuum, air, soft metal, solder, grease,

plastic, etc.

Contact resistance is often important in practice but only

empirical theories exist.

Some Comments Regarding Contact Resistance

Thermal Contact Resistance

(Very Small)

TA TB

Unit Area

q"x

Thermal contact resistance gives rise

to a temperature discontinuity at the

interface. A surface energy balance

would be:

qx is continuous =

dx

dTk

dx

dTk ba

Temperature is discontinuous =

BActx TTRq"

,

"

Radial Heat Transfer

For 1-D, steady state

constant k k(T)

w/o heat generation, we have in the cylindrical coordinates,

0d dT

rdr dr

1

dTr C

dr

1 2( ) lnT r C r C

(2 )r cdT dT

q kA k rLdr dr r r+dr

r

qr

qr+dr

/r r cdT

q q A kdr

For a cylindrical shell with known surface temperatures

T2T1

r2

r1

1 2( ) lnT r C r C

1 1 1 2ln( )T C r C

2 1 2 2ln( )T C r C

(1)

(2)

Solving for C1 and C2 yields the following temperature distribution:

2

2

2

1

21 ln

ln

)( Tr

r

rr

TTrT

T2T1

r2

r1

T

r

T1

T2

r1 r2

T1 > T2 T

r

T1

T2

r1 r2

T1 < T2

The temperature distributions

What about temperature gradient?

q= k = constant

A and dT/dr not constant

Note that / is getting smaller as the radius is getting larger.dT dr

How to get resistance ??

T2T1

r2

r1

Heat transfer rate and thermal resistance

12

21

/ln22

rr

TTLk

dr

dTrLk

dr

dTkAqr

Rate of heat transfer:

Heat flux:

12

21

/ln rr

TT

r

k

dr

dTk

A

qr

2

2

2

1

21 ln

ln

)( Tr

r

rr

TTrT

Constant

T2T1

r2

r1

Thermal Resistance:

12

21

/ln2

rr

TTLkqr

Lk

rrT

Lk

rr

TTqr

2

/ln

1

2

/ln 1212

12

Lk

rrR condt

2

/ln 12,

The cross-sectional areas are different (inner radius vs. outer radius)

Composite Cylindrical Walls

1 1 2 2(2 ) (2 ) ...

rtot

Tq UA

R

UA U r L U r L

Different overall heat

transfer coefficients.

Composite Cylindrical Walls

Spherical shell

T2T1

r2

r1

24rdT dT

q kA krdr dr

2

4

rq dr dTk r

2 2

1 1

1 22

1 2

4 ( )

4 1/ 1/

r Trrr T

t

q k T Tdr TdT q

k r r Rr

1 1

12

1

4 ( )

4 1/ 1/

r Trrr T

q k T TdrdT q

k r rr

11 1 2

1 2

1/ 1/( ) ( )

1/ 1/

r rT r T T T

r r