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Kyoung Hoon Kim 1 - 1

CHAPTER 2

Heat Conduction Equation

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2.1 Introduction

temperature and heat transfer

§ temperature: scalar (magnitude)

heat transfer = vector

(magnitude and direction)

§driving force of heat

transfer: temperaturedifference§ temperature: function of

position and time

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2.1 Introduction

Steady vs transient heat

transfer

1) steady: time independent

transient (unsteady): time

dependent

2) lumped systems: variationwith time, but not with

position

Steady and transient heatconduction in a plane wall

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2.1 Introduction

Multidimensional heat transfer

1) three, two, one dimensional2) Fourier's law of heat conduction:

n

T -kaQ

dxdT -kAQ

¶

¶=

=

:general

:ldimensionaone

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2.1 Introduction

Heat generation

G = gV

G = heat generation rate, Wg = heat generation rate per

unit volume, W/m3

Heat is generatedin the heating coils

of an electric range.

The absorption of solar

radiation by water can be

treated as heat generation.

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2.2 One dimensional heat conduction equation

energy balance

(rate of heat conduction

at x)

-(rate of heat conduction

- at x+Dx)+ ( rate of heat generation

inside the element)

= (rate of change of

energy content of

the element)

[ ] [ ] [ ] ( )[ ]

[ ] [ ]

p

n

n

p

p

p x x x

p x x x

ck/

t

T

k

g

x

T r

xr r

t

T

k

g

x

T r

xr r r

t

T

k

g

x

T

t T

k g

xT r

xr

t

T

k

c

k

g

x

T A

x A x

T k q

t

T c g

x

Aq

A x

t T c g

xqAqA

A x A

T c x At

x gAqAqA

r a

a p

a p

a

a

r

r

r

r

ydiffusivitthermalwhere

11:)2n4Aspherical(

11:)1n2l(Acylindrica

1:0)nzyAcartesian(

11:formgeneral

1:

)(1:0

1:1

2

2

2

2

2

==

¶¶=+÷

ø öç

è æ

¶¶

¶¶=®=

¶

¶=+÷

ø

öçè

æ ¶

¶

¶

¶=®D=

¶

¶=+

¶

¶=®DD=

¶¶=+÷

ø öç

è æ

¶¶

¶¶

¶

¶=+÷

ø

öçè

æ ¶

¶

¶

¶

¶

¶-=

¶

¶=+

¶

¶-®D

¶¶=+D --D´

D¶

¶=D+-

D+

D+

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2.3 General heat conduction equation

rectangular coordinates

t

T

k

g

T ¶

¶

=+Ñ a

12

spherical coordinates

cylindrical coordinates

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Specified temperature BC 1

st

kind)

2.4 Boundary and initial conditions

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Specified heat flux BC 2

nd

kind)


specified

heat fluxinsulation symmetric

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Convection BC 3

rd

kind)


convection boundary conditions on

the two surfaces of a plane wall

the assumed direction of heattransfer at a boundary has no

effect on the boundary

condition expression

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Radiation BC


radiation boundary conditions onboth surfaces of a plane wall

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Interface BC


boundary conditions at the

interface of two bodies inperfect contact

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<Ex 2-11> Heating conduction in a plane wall

Consider a large plane wall of

thickness L, thermal

conductivity k, and surface area

A. The two sides of the wall are

maintained at constant

temperatures T1 and T2,

respectively. Determine a) the

variation of temperature within

the wall and the value of

temperature at x0 and b) the

rate of heat conduction throughthe wall under steady conditions.

Input ata

thickness of the plateL 20 cm×:= left surface temperature T1 120 C×:=surface area A 15 m

2×:= right surface temperature T2 50 C×:=

thermal conductivity k 1.20W

m K ××:= a position x0 10 cm×:=

Solution

1C dxdT = 21 C xC T +=

governing eq: 02

2

=dx

T d

1)0( T T =

221)( T C LC LT =+=

BC’s:

L

T T kAQ 21 -

=

&2)( T LT =

12)0( T C T == &

or L

x

T T

T T =

-

-

12

1 x L

T T T T 12

1

-+=

T0 T1 T2 T1-

Lx0×+:= T0 85C=

Q k A× T1 T2-

L×:= Q 6.3kW=

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<Ex 2-12> A wall with various sets of boundary conditions

Consider steady one-dimensional heat conduction in a large plane wall of thickness L and

constant thermal conductivity k with no heat generation. Obtain expressions for the variation of

temperature within the wall for the following pairs of boundary conditions.

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<Ex 2-13> Heat conduction in the base plate of an iron

Consider the base plate of a Q0 household iron that has a thickness of L, base area A, andthermal conductivity of k. The inner surface of the base plate is subjected to uniform heat fluxgenerated by the resistance heaters inside, and the outer surface loses heat to the

surroundings at Ta by convection. Taking the convection heat transfer coefficient to be h anddisregarding heat loss by radiation, obtain for the variation of temperature in the base plate, andevaluate the temperatures at the inner and the outer surfaces.

A, k

h, Ta

L

heat

flux cond conv

Input ata

thickness of the plate L 0.50cm×:= heat transfer coefficienth 80 W× m 2-

× K 1-

×:=

surface area A 300cm2

×:= temperature of air Ta 20 C×:=

thermal conductivity k 15 W× m 1-

× K 1-

×:= electric power of iron Q 1200W×:=

Solution

1C dx

dT = 21 C xC T +=

01

)0(qkC

dx

dT k =-=-

governing eq: 02

2=

dxT d

0

)0(q

dx

dT k =- [ ]aT LT h

dx

LdT k -=- )(

)(

[ ]a

T C LC hqdx

LdT k -+==- )(

)(210

BC’s:

aT hk

x Lq xT +÷

ø

öçè

æ +-

= 1

)( 0

heat flux q0 Q

A:= q0 40 kW

m2

=

T0 q0 L

k

1

h+æ

è ö ø

× Ta+:= T0 533.333C=

TL q0 1

h

æ è

ö ø

× Ta+:= TL 520C=

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<Ex 2-14> Heat conduction in a solar heated wall

Consider a large plane wall of thickness L and thermal conductivity k in space. The wall iscovered with white porcelain tiles that has an emissivity of e and a solar absorptivity of a. Theinner surface of the wall is maintained at T1 at all times, while the outer surface is exposed to

solar radiation that is incident at a rate of qsolar. The outer surface is also losing heat byradiation to deep space at Tspace. Determine the temperature of the outer surface of the walland the rate of heat transfer through the wall when steady operating conditions are reached.

What would you response be if no solar radiation was incident on the surface?

e

k

Tspace

T1

L

acond

solar

radiation

Input ata

thickness of the plate L 6 cm×:=

thermal conductivity k 1.20 W

m K ××:=

emissivity e 0.85:=

absorptivity a 0.26:=

solar heat flux qsolar 800W

m2

×:=

inner surface temperatureT1 300 K ×:=

deep spacetemperature T2 0 K ×:=

1C dx

dT = 21 C xC T +=

12)0( T C T ==

governing eq: 02

2

=dx

T d

1)0( T T =

[ ] solar space qT LT dx LdT k a es --=- 44)()(

solar space qT T LC kC a es --+=- 44

111 )(

BC’s:

: nonlinear eq for C1

11)( T LC LT T L +==

1kC dx

dT k q -=-=

Solution

solution for C1 C1 0:=

C1 root e s× C1 L× T1+( )4

T24

-× a qsolar ×- k C1×+ C1,:= C1 121.497- K

m=

TL C1 L× T1+:= TL 292.71K =

q k - C1×:= q 145.796W m 2-

×=

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<Ex 2-15> Heat loss through a steam pipe

Consider a steam pipe of length L, inner radius r1, outer radius r2, and thermal conductivity k.The inner and outer surfaces of the pipe are maintained at average temperatures T1 and T2,respectively. Obtain a general relation for the temperature distribution inside the pipe under

steady conditions, and determine the rate of heat loss from the steam through the pipe.

r 1

k

T2

r 2

T1

L

Input ata

inner radius r1 6 cm×:=

outer radius r2 8 cm×:=

length of the pipe L 20 m×:=

thermal conductance of the pipek 20 W× m 1-

× K 1-

×:=

inner surface temperature T1 150 C×:=

outer surface temperature T2 60 K ×:=

Solution

1C

dr

dT r = 21

ln C r C T +=

12111 ln)( T C r C r T =+= 22212 ln)( T C r C r T =+=

&01

=÷ ø

öçè

æ dr

dT r

dr

d

r 11 )( T r T = 22 )( T r T =

)/ln(

)/ln(

12

1

12

1

r r

r r

T T

T T =

-

-

)/ln(2)2(

12

21

r r

T T kL

dr

dT rLk Q

-=-= p p

&

Q 2 p× k L× T1 T2-

lnr2

r1

æ è

ö ø

×:=

Q 786.266kW=

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<Ex 2-16> Heat conduction through a spherical shell

Consider a spherical container of inner radius r1, outer radius r2, and thermal conductivity k.

The inner and outer surfaces of the container are maintained at constant temperatures of

T1 and T2, respectively, as a result of some chemical reactions occurring inside. Obtain a

general relation for the temperature distribution inside the shell under steady conditions,and determine the rate of heat loss from the container.

r 1

kT2

r 2

T1

Input ata

inner radius r1 8 cm×:=

outer radius r2 10 cm×:=

thermal conductance of the shell k 45 W

m K ××:=

inner surface temperature T1 200C×:=

outer surface temperature T2 80 K ×:=

Solution

&01 2

2 =÷

ø

öçè

æ dr

dT r

dr

d

r 11 )( T r T = 22 )( T r T =

1

2 C dr

dT r = 2

1 C r

C T +-=

12

1

11)( T C

r

C r T =+-= 22

2

12)( T C

r

C r T =+-=

21

1

12

1

11

11

r r

r r

T T

T T

-

-=--

12

2121

2 4)4(r r

T T r kr

dr

dT r k Q

-

-=-= p p

&Q 4 p× k r1× r2×

T 1 T2-

r2 r1-×:=

Q 27.143kW=

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