Heat Transfer Handbook part 21 pot

Figure 3.16 Conduction in a solid cylinder with uniform internal energy generation... Figure 3.17 Conduction in a hollow sphere with uniform internal energy generation... 3.5 MORE ADVANC

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The solution ofeq (3.108) that satisfies eqs (3.109) is

T = T s,2+ ˙qr22

4k



1−

r

r2

2

− ˙qr22

4k



1−

r1

r2

2 +T s,2 − T s,1



ln(r2/r)

ln(r2/r1) (3.110)

Ifthe inside and outside surfaces are cooled by convection, the inside with fluid at

T ∞,1with heat transfer coefficienth2, the overall energy balance gives

˙qr2

2− r2 1



= 2h1 r1



T s,1 − T ∞,1

+ 2h2 r2



T s,2 − T ∞,2

(3.111) and for the case ofT s,1 = T s,2, eq (3.110) is reduced to

T = T s+ ˙qr22

4k



1−

r

r2

2

− ˙qr22

4k



1−

r 1

r2

2

ln(r2/r)

ln(r2/r1) (3.112)

Similarly, whenT ∞,1 = T ∞,2 = T∞andh1= h2 = h, eq (3.111) is reduced to

T s − T∞= ˙q(r2 − r1 )

Solid Cylinder Equation (3.108) also applies to the solid cylinder ofFig 3.16, but the boundary conditions change to

r0

r k

T s

Tmax

L

q

Cold fluid ,

T h⬁

.

Figure 3.16 Conduction in a solid cylinder with uniform internal energy generation

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dT dr





r=0 = 0 and T (r = r0 ) = T s (3.114) The temperature distribution is given by

T = T s+ ˙q

4k



r2

0− r2

(3.115) and the maximum temperature occurs along the centerline atr = 0:

Tmax= T s+ ˙qr02

Ifthe outside surface ofthe cylinder is cooled by convection to a fluid atT∞through

a heat transfer coefficienth, the overall energy balance gives

T s − T∞= ˙qr0

Hollow Sphere For the hollow sphere shown in Fig 3.17, the appropriate form ofeq (3.6) with constant thermal conductivityk is

1

r2

d dr

r2dT dr



and eq (3.109) provides the boundary conditions

T (r = r1) = T s,1 and T (r = r2) = T s,2 (3.109)

r1

r2 r k

T s,2

T s,1 q

T⬁,2 2,h

T⬁,1 1,h

Figure 3.17 Conduction in a hollow sphere with uniform internal energy generation

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The temperature distribution is found to be

T = T s,2+ ˙qr22

6k



1−

r

r2

2

− ˙qr22

6k



1−

r1

r2

2 +T s,2 − T s,1



1/r − 1/r2

1/r1− 1/r2 (3.119)

Ifthe inside and outside surfaces are cooled by convection, the inside with fluid at

T ∞,1 with heat transfer coefficienth1 and the outside with fluid atT ∞,2 with heat

transfer coefficienth2, the overall energy balance gives

˙qr3

2 − r3 1



3 = h1 r2

1



T s,1 − T ∞,1+ h2 r2

2



T s,2 − T ∞,2 (3.120) and whenT s,1 = T s,2 = T s, eq (3.119) reduces to

T = T s+ ˙qr22

6k



1−

r

r2

2

− ˙qr22

6k



1−

r1

r2

2

1/r − 1/r2

1/r1− 1/r2 (3.121)

WhenT ∞,1 = T ∞,2 = T∞, andh1= h2 = h, eq (3.120) is reduced to

T s − T∞= ˙q



r3

2− r3 1



3hr2

1 + r2 2

Solid Sphere For the solid sphere, eq (3.118),

1

r2

d dr

r2dT dr

 + k ˙q = 0 (3.118) must be solved subject to the boundary conditions ofeqs (3.114):

dT dr





r=0 = 0 and T (r = r0 ) = T s (3.114) The solution is

T = T s+ ˙q

6k



r2

0 − r2

(3.123) The maximum temperature that occurs at the center ofthe sphere wherer = 0 is

Tmax= T s+ ˙qr02

194 CONDUCTION HEAT TRANSFER

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and ifthe cooling at the outside surface is to a fluid atT∞via a heat transfer coefficient

ofh, the overall energy balance gives

T s − T∞= ˙qr0

For additional analytical results for one-dimensional steady conduction with uni-form internal heat generation, the reader should consult Incropera and DeWitt (1996, App C)

3.5 MORE ADVANCED STEADY ONE-DIMENSIONAL CONDUCTION

The results presented in Section 3.4 have been based on assumptions such as constant thermal conductivity, uniform heat generation, and pure convective cooling or heating

at the boundary In some applications, these assumptions may introduce significant errors in predicting the thermal behavior ofthe system

The conducting medium may be nonhomogeneous, causing thermal conductivity

to vary with location Similarly, the temperature dependence ofthermal conductivity cannot be ignored ifthe temperature difference driving the conduction process is large and the assumption ofuniform heat generation may prove too restrictive For example, when the shield ofa nuclear reactor is irradiated with gamma rays, the resulting release ofenergy decays exponentially with distance from the irradiated surface, making the heat generation location dependent A more realistic modeling ofheat generation due to the passage ofelectric current or a chemical reaction requires that ˙q be treated as temperature dependent Finally, ifthe heat transfer process at a

boundary is driven by natural convection, radiation becomes equally important and must be taken into account This section is devoted to a discussion ofsuch situations

3.5.1 Location-Dependent Thermal Conductivity

Plane Wall Consider the plane wall ofFig 3.6 and let the thermal conductivityk

increase linearly withx in accordance with

wherek0is the thermal conductivity atx = 0 and a is a measure ofthe variation of

k with x The equation governing the temperature distribution is

d dx

k dT dx



Solving eq (3.127) subject to the boundary conditions ofeq (3.66),

T (x = 0) = T s,1 and T (x = L) = T s,2 (3.66)

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gives

T = T s,1+T s,2 − T s,1  ln(1 + ax)

and the rate ofheat transfer will be

q = k0Aa(T s,1 − T s,2 )

In the limit, asa → 0, eqs (3.128) and (3.129) reduce to eqs (3.67) and (3.68),

respectively

Now consider the case wherek is ofthe form

The solutions forT and q are given by

T = T s,1+T s,2 − T s,1 arctan√ax

arctan√aL (3.131) and the rate ofheat transfer will be

q = k0A√a(T s,1 − T s,2 )

Hollow Cylinder When modified to allow for the location-dependent thermal conductivity ofthe form

analysis of Section 3.4.2 for a hollow cylinder (Fig 3.7) gives the following results for the temperature distribution and the rate of heat transfer:

T =

T s,1ln

r2

1+ br2

1+ br r



+ T s,2ln

1+ br1

r1

r

1+ br



ln

1+ br1

1+ br2

r2

r1

q = 2πaL(Ts,1 − T s,2 )

ln

1+ br1

1+ br2

r2

r1

Whenb = 0, k = a and the thermal conductivity is constant In this case, eqs (3.134)

and (3.135) are reduced to eqs (3.71) and (3.72), respectively

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3.5.2 Temperature-Dependent Thermal Conductivity

Plane Wall Let the thermal conductivityk ofthe plane wall ofFig 3.6 be a linear

function of the temperatureT , expressed as

Equation (3.127) will then take the form

d dx



k0(1 + aT ) dT dx



which must be integrated using the boundary conditions ofeqs (3.66):

T (x = 0) = T s,1 and T (x = L) = T s,2 (3.66) The solution is facilitated by the introduction of a new variable,T∗, defined by the

Kirchhoff transformation:

T∗ =

 T 0

(1 + aT ) dT = T +1

2aT2 (3.138) Differentiation of eq (3.138) with respect tox gives

dT∗

dx = (1 + aT )

dT

which allows eq (3.137) to be written as

d2T∗

The boundary conditions ofeq (3.66) in terms ofT∗become

T∗

s,1 (x = 0) = T s,1+1

2aT2

s,1 and T∗

s,2 (x = L) = T s,2+1

2aT2

s,2 (3.141)

The solution forT∗is

T∗= T∗

s,1+T∗

s,2 − T∗

s,1  x

OnceT∗has been found,T can be reclaimed by solving the quadratic ofeq (3.138),

which gives

T = 1 a



−1 +√1+ 2aT∗

(3.143) and the rate ofheat transfer can be shown to be

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q = k m A(T s,1 − T s,2 )

wherek m = k0 (1 + aT s,m ) is the thermal conductivity at the mean temperature,

T s,m= T s,1 + T s,2

2 For a variation ofthe thermal conductivity with temperature represented by

the temperature distribution in the plane wall (Fig 3.6) is given by the cubic equation

T +1

3aT3= T s,1+1

3aT3

s,1+

 1

3aT3

s,2 − T3

s,1



+ (T s,2 − T s,1 )

x

L (3.146)

and the corresponding rate ofheat transfer is

q = Ak0



T s,1 − T s,2

1+ (a/3)T2

s,1 + T s,1 T s,2 + T2

s,2

Hollow Cylinder For the thermal conductivity temperature relation ofeq (3.136), the Kirchhofftransformation ofeq (3.138) can also be used for the hollow cylinder ofFig 3.7 The final result forT∗is

T∗= T∗

s,1+T s,1∗ − T s,2∗

ln(r1/r2) ln

r

r1

(3.148) whereT∗

s,1andT∗

s,2are as given in eq (3.141) OnceT∗for any radiusr is found from

eq (3.148), eq (3.143) can be used to find the corresponding value ofT The rate of

heat transfer then follows as

q = 2πkm L



T s,1 − T s,2

ln(r2/r1) (3.149)

wherek mis the thermal conductivity at the mean temperature,

T s,m= T s,1 + T s,2

2

Hollow Sphere The results for a hollow sphere (Fig 3.8) whose thermal conduc-tivity–temperature variation follows eq (3.136) are

T∗ = T∗

s,1+ T s,1∗ − T s,2∗

1/r2− 1/r1

1

r1

−1

r



(3.150)

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T = 1 a



−1 +√1+ 2aT∗

(3.151)

q = 4πk m



T s,1 − T s,2

Gebhart (1993) provides one-dimensional steady conduction analyses for single and composite solids when the thermal conductivity varies simultaneously with lo-cation and temperature He also gives expressions for the conduction resistances of a plane wall, a hollow cylinder, and a hollow sphere for three cases of variable thermal conductivity:k = k(T ), k = k(x) or k(r), and k = k(x, T ) = k(T )f (x) Note that

the last case assumes thatk(x, T ) can be expressed as a product oftwo functions, k(T ) and f (x), each a function of a single variable.

3.5.3 Location-Dependent Energy Generation

Plane Wall Figure 3.18 presents a plane wall that experiences location-dependent energy generation ofthe form

˙q = ˙q01−L x (3.153) The temperature distribution in the wall is given by

T = T s,1+ ˙q0 Lx

2k −

˙q0

2k

x2− x3

3L



(3.154)

Figure 3.18 Plane wall with linearly decaying, location-dependent internal energy gener-ation

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and the maximum temperature occurs atx = L for an insulated face:

T = T s,1+ ˙q0 L2

Next, assume that the plane wall ofFig 3.6 represents the shield ofa nuclear reactor The absorption ofgamma radiation at the left surface (x = 0) triggers energy

release into the shield which decays exponentially with the penetration distancex

and can be represented by the relation

˙q = q

whereq

0 (W/m2) is the incident radiation heat flux andam−1

is the absorption coefficient of the shield

The temperature distribution in the shield is

T = T s,1+q0

ak



1− e −ax

+

T s,2 − T s,1+q0

ak



e −aL− 1 x

L (3.157)

and the maximum temperature occurs at

x = 1

aln

q

0aL ak(T s,1 − T s,2 ) + q

Solid Cylinder Reconsidering the solid cylinder ofFig 3.16, ˙q will now be

assumed to vary linearly with the radial distancer, that is,

wherea (W/m4) is a constant The temperature distribution in this case is

T = T∞+ar02

3h +

a

9k



r3

0− r3

(3.160)

from which the centerline (r = 0) temperature T cand the surface (r = r0) tempera-tureT sfollow as

T c = T∞+ar02

3h +

ar3 0

T s = T∞+ar02

3.5.4 Temperature-Dependent Energy Generation

In this section we present a collection ofresults for one-dimensional steady conduc-tion in a plane wall, a solid cylinder, and a solid sphere when each experiences energy generation that increases linearly with local temperature in accordance with

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˙q = ˙q s[1+ a(T − T s )] (3.163) where ˙q Sis the energy generation at the surface temperatureT sanda is a constant.

Plane Wall For a plane wall ofthickness 2L having identical surface temperatures

T son both faces, the temperature distribution is

T = T s+1

a

cosnx cosnL− 1



(3.164)

wheren =√a ˙q s /k and nL < π/2 to ensure that the temperatures remain finite.

Ifthe convection cooling, characterized by the temperatureT∞and heat transfer coefficienth, is identical on both faces of the wall, the relationship between T s and

T∞is given by

T s = T∞+m h tannL (3.165) wherem =√˙q s k/a.

Solid Cylinder For a solid cylinder ofradiusr0, the temperature distribution is

T = T s+1a

J

0(nr)

J0(nr0)− 1



(3.166)

wheren =√a ˙q s /k and J0is the Bessel function ofthe first kind ofzero order (see Section 3.3.5) The parallel counterpart ofeq (3.165) is

T s = T∞+m h J1(nr0)

J0(nr0) (3.167)

wherem =√˙q s k/a, n =√a ˙q s /k, and J1is the Bessel function of the first kind of order 1 In eqs (3.166) and (3.167),nr0< 2.4048 to assure finite temperatures in the

cylinder

Solid Sphere For a solid sphere ofradiusr0, the temperature distribution is

T = T s+1

a

r0

r

sinnr

sinnr0

− 1



(3.168)

where nr0 < π to assure finite temperatures in the sphere and the relationship

betweenT sand the coolant temperatureT∞is

T = T s+hr k

0a[1− (nr0 ) cot nr0] (3.169)

where, here too,nr0< π to assure finite temperatures in the sphere.