1

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TRANS HEAT TRANSFERHEAT TRANSFERSFER

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HEAT TRANSFERHEAT TRANSFERSME 4453SME 4453

LECTURER: PM DR MAZLAN ABDUL WAHIDhttp://www.fkm.utm.my/~mazlan

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Dr Mazlan

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CChapter hapter 33SteadySteady --State State S

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SteadySteady --State State One Dimensional One Dimensional Heat ConductionHeat Conduction

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Dr Mazlan

PM Dr Mazlan Abdul Wahid

UTM Faculty of Mechanical EngineeringUniversiti Teknologi Malaysiawww.fkm.utm.my/~mazlan

2

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One-Dimensional Steady-State Conduction

• Conduction problems may involve multiple directions and time-

dependent conditions

• Inherently complex – Difficult to determine temperature distributionsSFER

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• Inherently complex – Difficult to determine temperature distributions

• One-dimensional steady-state models can represent accurately

numerous engineering systems

• In this chapter we will:

� Learn how to obtain temperature profiles for common geometries

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Dr Mazlan

� Learn how to obtain temperature profiles for common geometries

with and without heat generation.

� Introduce the concept of thermal resistance and thermal circuits

� Introduce to the analysis of one dimensional conduction analysis

on extended surfaces

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Assumptions

- 1 dimensional heat transfer

- Isothermal surfaces

- Steady stateSFER

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- Steady state

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3

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Fourier’s law

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Thermal Resistance concept

- Conduction

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Electrical world

4

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Thermal Resistance concept- Convection

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The Plane Wall

Consider a simple case of one-dimensional conduction in a plane wall, separating two fluids of different temperature, without energy

1,∞T22, ,hT∞

Cold fluid

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temperature, without energy generation

• Temperature is a function of x

• Heat is transferred in the x-direction

Must consider– Convection from hot fluid to wall

– Conduction through wall

qx

1,sT

2,sT

2,∞T11, ,hT∞

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Dr Mazlan

– Conduction through wall

– Convection from wall to cold fluid

� Begin by determining temperature distribution within the wall x

x=0 x=LHot fluid

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Temperature Distribution

0= dT

kd

t

Tcq

z

Tk

yy

Tk

yx

Tk

x p ∂∂=+

∂∂

∂∂+

∂∂

∂∂+

∂∂

∂∂ •

ρ

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• Heat diffusion equation in the x-direction for steady-state conditions, with no energy generation:

0=

dx

kdx

• Boundary Conditions:2,1, )(,)0( ss TLTTT ==

� qx is constant

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Dr Mazlan

• Temperature profile, assuming constant k:

1,1,2, )()( sss TL

xTTxT +−=

� Temperature varies linearly with x

(3.1)

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Thermal Resistance

Based on the previous solution, the conduction heat transfer rate can be calculated:

( ) ( )kAL

TTTT

L

kA

dx

dTkAq ss

ssx /2,1,

2,1,

−=−=−= (3.2a)

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( )kALLdx ssx /2,1,

Similarly for heat convection, Newton’s law of cooling applies:

hA

TTTThAq S

Sx /1

)()( ∞

∞−=−=

And for radiation heat transfer:TT

TTAhq surs )()(

−=−=

(3.2b)

(3.2c)

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� Recall electric circuit theory - Ohm’s law for elect rical resistance:

Resistance

e DifferencPotentialcurrent Electric =

Ah

TTTTAhq

r

surssursrrad /1

)()(

−=−= (3.2c)

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Thermal Resistance

• We can use this electrical analogy to represent heat transfer problems using the concept of a thermal circuit (equivalent to an electrical circuit).

∆TForce Driving OverallSFER

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� Compare with equations 3.2a-3.2c� The temperature difference is the “potential” or driving force for the

heat flow and the combinations of thermal conductivity, convection coefficient, thickness and area of material act as a resistance to this

∑∆==

R

Tq overall

Resistance

Force Driving Overall

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Dr Mazlan

coefficient, thickness and area of material act as a resistance to this flow:

AhR

hAR

kA

LR

rradtconvtcondt

1,

1, ,,, ===

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Thermal Resistance for Plane Wall

1,∞T

1,sT

22, ,hT∞

Cold fluid

TTTTTTq ssss 2,2,2,1,1,1, ∞∞ −

=−

=−

=SFER

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In terms of overall temperature difference:qx

2,sT

2,∞T

x=0 x=L

11, ,hT∞

Hot fluidR

TTqx

2,1, −= ∞∞

Ah

TT

kAL

TT

Ah

TTq ssss

x2

2,2,2,1,

1

1,1,

/1//1∞∞ −

=−

=−

=

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Dr Mazlan

xx=0 x=L

AhkA

L

AhR

R

tot

totx

21

11 ++=

7

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Composite Walls

Express the following geometry in terms of a an equivalent thermal circuit.

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Composite Walls

What is the heat transfer rate for this system?

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AlternativelyUAq

TRR

TUAq

ttot

x

1=∆==

∆=

∑

where U is the overall heat transfer coefficient and ∆T the overall

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Dr Mazlan

where U is the overall heat transfer coefficient and ∆T the overall temperature difference.

)]/1()/()/()/()/1[(

11

41 hkLkLkLhARU

CCBBAAtot ++++==

8

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Thermal Resistance concept- Convection & Radiation

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9

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Example – single layer wall

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Example – multi layer wall

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Example – single layer window

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Example – two layer window

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Composite Walls – with parallel resistances

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(a) Surfaces normal to the x-direction are isothermal

� For resistances in series: R =R +R +…+R

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(b) Surfaces parallel to x-direction are adiabatic

Rtot=R1+R2+…+Rn

� For resistances in parallel:1/Rtot=1/R1+1/R2+…+1/Rn

12

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Composite Walls – with parallel resistances

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� For resistances in series: Rtot=R1+R2+…+Rn

� For resistances in parallel:1/Rtot=1/R1+1/R2+…+1/Rn

Composite Walls – with parallel resistances

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Parallel heat conduction

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AttentionThe result obtained will be somewhat approximate, since the surfaces of thethird layer will probably not be isothermal, and heat transfer between the firsttwo layers is likely to occur.

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two layers is likely to occur.

14

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Heat loss through a composite wall

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Contact Resistance

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15

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Contact Resistance

The temperature drop across the interface between materials may be appreciable, due to surface roughness S

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surface roughness effects, leading to air pockets. We can define thermal contact resistance:

intTAhq c ∆=33333333

Dr Mazlan

AhTR cc /int∆=

intTAhq cx ∆=

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16

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Thermal contact resistance is observed to decrease with decreasing surface roughness and increasing interface pressure, as expected.

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expected.

Most experimentally determined values of the thermal contact resistance fall between 0.000005 and 0.0005 m2 · °C/W (the corresponding range of thermal contact conductance is 2000 to 200,000 W/m2 · °C).

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17

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Contact resistance of transistors

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18

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Composite Walls – with contact resistances

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Radial Systems-Cylindrical Coordinates

Consider a hollow cylinder, whose inner and outer surfaces are exposed to fluids at different temperatures

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Temperature distribution

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Temperature distribution

19

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• Heat diffusion equation (eq. 2.5) in the r-direction for steady-state conditions, with no energy

t

Tcq

z

Tk

z

Tk

rr

Tkr

rr p ∂∂=+

∂∂

∂∂+

∂∂

∂∂+

∂∂

∂∂ •

ρφφ2

11

01 =

dr

dTkr

dr

d

r 0

1 =

dr

dTr

dr

d

r

for constant k

Temperature Distribution - Thermal resistance for cylinder

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state conditions, with no energy generation:

drdrr

• Boundary Conditions: 2,21,1 )(,)( ss TrTTrT ==

• Fourier’s law: constdr

dTrLk

dr

dTkAqr =−=−= )2( π

0=

drr

drr

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Dr Mazlan

• Temperature profile, assuming constant k:

2,221

2,1, ln)/ln(

)()( s

ss Tr

r

rr

TTrT +

−= � Logarithmic temperature

distribution

HEAT

TRANS

2,221

2,1, ln)/ln(

)()( s

ss Tr

r

rr

TTrT +

−=

Temperature Distribution - thermal resistance for cylinder

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dr

dTrLk

dr

dTkAqr −=−= )2( π

221

)/ln(

)(2

12

21

rr

TTkLq r

−= π

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Dr Mazlan

12

kLrr

TTq r

π2/)/ln( 12

21 −=

kLrrR cyl π2/)/ln( 12=therefore

20

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Temperature Distribution - thermal resistance for cylinder

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Thermal resistance for cylinder

Based on the previous solution, the conduction hear transfer rate can be calculated:

• Fourier’s law: constdT

rLkdT

kAq =π−=−= )2(SFER

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( ) ( ) ( )condt

ssssssx R

TT

Lkrr

TT

rr

TTLkq

,

2,1,

12

2,1,

12

2,1,

)2/()/ln()/ln(

2 −=

−=

−=

ππ

� In terms of equivalent thermal circuit:

2,1, TTq

−= ∞∞

• Fourier’s law: constdr

rLkdr

kAqr =π−=−= )2(

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Dr Mazlan

)2(

1

2

)/ln(

)2(

1

22

12

11

2,1,

LrhkL

rr

LrhR

R

TTq

tot

totx

πππ++=

−= ∞∞

21

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Composite Walls

Express the following geometry in terms of a an equivalent S

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a an equivalent thermal circuit.

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Composite Walls

What is the heat transfer rate?

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where U is the overall heat transfer coefficient. If A=A1=2πr1L:

44

1

3

41

2

31

1

21

1

1lnlnln

11

hr

r

r

r

k

r

r

r

k

r

r

r

k

r

h

U

CBA

++++=

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Dr Mazlan

alternatively we can use A2=2πr2L, A3=2πr3L etc. In all cases:

∑====

tRAUAUAUAU

144332211

22

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Spherical Coordinates

• Fourier’s law:

dTkAqsph −=S

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RRRRRRRR• Starting from Fourier’s law, acknowledging that q r is constant,

independent of r, and assuming that k is constant, derive the

dr

dTrk

drkAqsph

)4( 2π−=

−=

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Dr Mazlan

independent of r, and assuming that k is constant, derive the equation describing the conduction heat transfer ra te. What is the thermal resistance?

HEAT

TRANS dTkrdrq ∫∫ −=2/4/ π

Temperature Distribution - thermal resistance for sphere

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dTkrdrqsph ∫∫ −=2/4/ π

)/1()/1(

)(4

21

21

rr

TsTskq sph −

−= π

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Dr Mazlan

)11

(4

1

21 rrkR sph −=

π

23

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Conduction with Generation

Thermal energy may be generated or consumed due to conversion from some other energy form.

� If thermal energy is generated in the material at the expense of some other energy form, we have a source: is +veq

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other energy form, we have a source: is +ve– Deceleration and absorption of neutrons in a nuclear reactor– Exothermic reactions – Conversion of electrical to thermal energy:

V

RI

V

Eq eg

2

==&

where I is the current, Re the electrical resistance, V the volume of the medium

q

.

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Dr Mazlan

� If thermal energy is consumed we have a sink: is -ve– Endothermic reactions

q

24

.

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The Plane WallConsider one-dimensional, steady-state conduction in a plane wall ofconstant k, with uniform generation, and asymmetric surface conditions:

• Heat diffusion equation (eq. 2.3) :2 qTd .

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02

2

=+k

q

dx

Td

• General Solution:

212

2CxCx

k

qT ++−=

.

.

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Dr Mazlan

• Boundary Conditions:

2,1, )(,)( ss TLTTLT ==−

2k

25

24

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Temperature Profile

22

)(1

2)( 2,1,1,2,

2

22ssss TT

L

xTT

L

x

k

qLxT

++

−+

−= (3.3)

.

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� Profile is parabolic. � Heat flux not independent of x

What happens when:

?0 increases, ,0 <= qqq. . .

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Dr Mazlan 26

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Symmetrical Distribution• When both surfaces are maintained at a

common temperature, Ts,1= Ts,2 = Ts

TxqL

xT +

−=22

1)( (3.4a).

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sTLk

xT +

−=2

12

)(

What is the location of the maximum temperature?

(3.4a)

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Dr Mazlan

2

max

max)(

=−

−∴L

x

TT

TxT

s

(3.4b)

27

25

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Symmetrical Distribution

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� Note that at the plane of symmetry:

0q" 00

0

=⇒=

=

=x

xdx

dT

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Dr Mazlan

� Equivalent to adiabatic surface

28

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Calculation of surface temperature T s

In equations (3.4a) and (3.4b) the surface temperature, Ts is needed.� Boundary condition at the wall:

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)( ∞=

−=− TThdx

dTk s

Lx

Substituting (dT/dx)x=L from equation(3.4a):

qL.

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Dr Mazlan

h

qLTTs += ∞

(3.5)

29

.

26

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ExampleThe steady-state temperature distribution in a composite plane wall of three different materials, each of constant thermal conductivity, is shown in the schematic below.

a) Does heat generation occur in any of sections A,B, or C?SFER

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a) Does heat generation occur in any of sections A,B, or C?b) Based on the schematic, what is the boundary condition at location (4)?c) Comment on the relative magnitudes of q2” and q3” .

d) Comment on the relative magnitudes of kA and kB.

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Dr Mazlan 30

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Example

A plane wall of thickness 0.1 m and thermal conductivity 25 W/m.K having uniform volumetric heat generation of 0.3 MW/m3 is insulated on one side, while the other side is exposed to a fluid at 92°C. The convection heat transfer coefficient between the wall and the fluid is S

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convection heat transfer coefficient between the wall and the fluid is 500 W/m2.K. Determine the maximum temperature in the wall.

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Dr Mazlan 31

27

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Radial Systems

Cylindrical (Tube) Wall Spherical Wall (Shell)

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Solid Cylinder (Circular Rod) Solid Sphere

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Dr Mazlan 32

HEAT

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Radial Systems• Heat diffusion equation in

the r-direction for steady-state conditions:

01 =+

k

q

dr

dTkr

dr

d

r

2 ln CrCrq

T ++−=

.

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• Boundary Conditions: sor

TrTdr

dT ===

)( ,00

• Temperature profile:

o Trqr

rT +

−=22

1)(

hT ,∞

• General Solution: 212 ln

2CrCr

k

qT ++−=

.

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Dr Mazlan

so

o Tr

r

k

qrrT +

−=

21

4)(

L

))(2()( 2∞−= TTLrhLrq soo ππ

h

qrTT o

s 2+= ∞

• Calculation of surface temperature:

and

(3.53)

33

28

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Example

A cylindrical shell of inner and outer radii ri and ro, respectively, is filled with a heat-generating material that provides a uniform volumetric generation rate. The inner surface is insulated, while the outer surface of the shell is exposed to a fluid with a convection coefficient h.

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of the shell is exposed to a fluid with a convection coefficient h.a) Obtain an expression for the steady-state temperature distribution T(r)

in the shell.b) Determine an expression for the heat rate q’ (ro) at the outer radius of

the shell in terms of the heat generation rate and the shell dimensions

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Dr Mazlan 34

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29

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Critical radius of insulation

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Optimum thickness of insulation

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30

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Extended Surfaces (Fins)

An extended surface (also know as a combined conduction-convection system or a fin) is a solid within which heat transfer by conduction is assumed to be one dimensional, while heat is also transferred by convection (and/or radiation) from the surface in a direction transverse S

FER

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convection (and/or radiation) from the surface in a direction transverse to that of conduction

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Dr Mazlan 35

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Extended Surfaces (Fins)

Extended surfaces may exist in many situations but are commonly used as fins to enhance heat transfer by increasing the surface area available for convection (and/or radiation). They are particularly beneficial when h is small, as for a gas and natural convection.

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beneficial when h is small, as for a gas and natural convection.

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Dr Mazlan

� Solutions for various fin geometries can be found in the literature (see for example in textbook).

36

31

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TRANS

Extended surfacesExtended surfaces-- FinsFins

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There are two waysto increase the rate of heat transfer: to increase the convection heat transfer coefficient h or to increase the surface area A.

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Dr Mazlan

to increase the surface area A.

Fins enhance heat transfer from a surface by exposing a larger surface area toconvection and radiation.

HEAT

TRANS

Fin equation

SFER

CCCCCCCCHHHHHHHH

AAAAAAAA

PPPPPPPP

TTTTTTTT

EEEEEEEE

RRRRRRRR

33333333

Dr Mazlan

or

32

HEAT

TRANS

Infinitely long fin

SFER

CCCCCCCCHHHHHHHH

AAAAAAAA

PPPPPPPP

TTTTTTTT

EEEEEEEE

RRRRRRRR

33333333

Dr Mazlan

HEAT

TRANS

Negligible heat loss from the fin tip

SFER

CCCCCCCCHHHHHHHH

AAAAAAAA

PPPPPPPP

TTTTTTTT

EEEEEEEE

RRRRRRRR

33333333

Dr Mazlan

33

HEAT

TRANS

Fin efficiency

SFER

CCCCCCCCHHHHHHHH

AAAAAAAA

PPPPPPPP

TTTTTTTT

EEEEEEEE

RRRRRRRR

33333333

Dr Mazlan

HEAT

TRANS

Efficiency of fins (I)

SFER

CCCCCCCCHHHHHHHH

AAAAAAAA

PPPPPPPP

TTTTTTTT

EEEEEEEE

RRRRRRRR

33333333

Dr Mazlan

34

HEAT

TRANS

Efficiency of fins (II)

SFER

CCCCCCCCHHHHHHHH

AAAAAAAA

PPPPPPPP

TTTTTTTT

EEEEEEEE

RRRRRRRR

33333333

Dr Mazlan

HEAT

TRANS

Fin effectiveness (I)

SFER

CCCCCCCCHHHHHHHH

AAAAAAAA

PPPPPPPP

TTTTTTTT

EEEEEEEE

RRRRRRRR

33333333

Dr Mazlan

35

HEAT

TRANS

Fin effectiveness (II)

SFER

CCCCCCCCHHHHHHHH

AAAAAAAA

PPPPPPPP

TTTTTTTT

EEEEEEEE

RRRRRRRR

33333333

Dr Mazlan

HEAT

TRANS

Proper length of a fin

SFER

CCCCCCCCHHHHHHHH

AAAAAAAA

PPPPPPPP

TTTTTTTT

EEEEEEEE

RRRRRRRR

33333333

Dr Mazlan

36

HEAT

TRANS

Thermal resistance

SFER

CCCCCCCCHHHHHHHH

AAAAAAAA

PPPPPPPP

TTTTTTTT

EEEEEEEE

RRRRRRRR

33333333

Dr Mazlan

HEAT

TRANSSFER

CCCCCCCCHHHHHHHH

AAAAAAAA

PPPPPPPP

TTTTTTTT

EEEEEEEE

RRRRRRRR

33333333

Dr Mazlan