BOOKCOMP, Inc. — John Wiley & Sons / Page 191 / 2nd Proofs / Heat Transfer Handbook / Bejan STEADY ONE-DIMENSIONAL CONDUCTION 191 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 [191], (31) Lines: 1465 to 1501 ——— 0.20709pt PgVar ——— Normal Page * PgEnds: Eject [191], (31) The solution of eq. (3.108) that satisfies eqs. (3.109) is T = T s,2 + ˙qr 2 2 4k 1 − r r 2 2 − ˙qr 2 2 4k 1 − r 1 r 2 2 + T s,2 − T s,1 ln(r 2 /r) ln(r 2 /r 1 ) (3.110) If the inside and outside surfaces are cooled by convection, the inside with fluid at T ∞,1 with heat transfer coefficient h 2 , the overall energy balance gives ˙q r 2 2 − r 2 1 = 2h 1 r 1 T s,1 − T ∞,1 + 2h 2 r 2 T s,2 − T ∞,2 (3.111) and for the case of T s,1 = T s,2 , eq. (3.110) is reduced to T = T s + ˙qr 2 2 4k 1 − r r 2 2 − ˙qr 2 2 4k 1 − r 1 r 2 2 ln(r 2 /r) ln(r 2 /r 1 ) (3.112) Similarly, when T ∞,1 = T ∞,2 = T ∞ and h 1 = h 2 = h, eq. (3.111) is reduced to T s − T ∞ = ˙q(r 2 − r 1 ) 2h (3.113) Solid Cylinder Equation (3.108) also applies to the solid cylinder of Fig. 3.16, but the boundary conditions change to r 0 r k T s T max L q Cold fluid ,Th ϱ . Figure 3.16 Conduction in a solid cylinder with uniform internal energy generation. BOOKCOMP, Inc. — John Wiley & Sons / Page 192 / 2nd Proofs / Heat Transfer Handbook / Bejan 192 CONDUCTION HEAT TRANSFER 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 [192], (32) Lines: 1501 to 1553 ——— 9.7792pt PgVar ——— Normal Page PgEnds: T E X [192], (32) dT dr r=0 = 0 and T(r = r 0 ) = T s (3.114) The temperature distribution is given by T = T s + ˙q 4k r 2 0 − r 2 (3.115) and the maximum temperature occurs along the centerline at r = 0: T max = T s + ˙qr 2 0 4k (3.116) If the outside surface of the cylinder is cooled by convection to a fluid at T ∞ through a heat transfer coefficient h, the overall energy balance gives T s − T ∞ = ˙qr 0 2h (3.117) Hollow Sphere For the hollow sphere shown in Fig. 3.17, the appropriate form of eq. (3.6) with constant thermal conductivity k is 1 r 2 d dr r 2 dT dr + ˙q k = 0 (3.118) and eq. (3.109) provides the boundary conditions T(r = r 1 ) = T s,1 and T(r = r 2 ) = T s,2 (3.109) r 1 r 2 r k T s,2 T s,1 q Th ϱ ,2 2 , Th ϱ,1 1 , . Figure 3.17 Conduction in a hollow sphere with uniform internal energy generation. BOOKCOMP, Inc. — John Wiley & Sons / Page 193 / 2nd Proofs / Heat Transfer Handbook / Bejan STEADY ONE-DIMENSIONAL CONDUCTION 193 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 [193], (33) Lines: 1553 to 1617 ——— 3.10439pt PgVar ——— Normal Page * PgEnds: Eject [193], (33) The temperature distribution is found to be T = T s,2 + ˙qr 2 2 6k 1 − r r 2 2 − ˙qr 2 2 6k 1 − r 1 r 2 2 + T s,2 − T s,1 1/r − 1/r 2 1/r 1 − 1/r 2 (3.119) If the inside and outside surfaces are cooled by convection, the inside with fluid at T ∞,1 with heat transfer coefficient h 1 and the outside with fluid at T ∞,2 with heat transfer coefficient h 2 , the overall energy balance gives ˙q r 3 2 − r 3 1 3 = h 1 r 2 1 T s,1 − T ∞,1 + h 2 r 2 2 T s,2 − T ∞,2 (3.120) and when T s,1 = T s,2 = T s , eq. (3.119) reduces to T = T s + ˙qr 2 2 6k 1 − r r 2 2 − ˙qr 2 2 6k 1 − r 1 r 2 2 1/r − 1/r 2 1/r 1 − 1/r 2 (3.121) When T ∞,1 = T ∞,2 = T ∞ , and h 1 = h 2 = h, eq. (3.120) is reduced to T s − T ∞ = ˙q r 3 2 − r 3 1 3h r 2 1 + r 2 2 (3.122) Solid Sphere For the solid sphere, eq. (3.118), 1 r 2 d dr r 2 dT dr + ˙q k = 0 (3.118) must be solved subject to the boundary conditions of eqs. (3.114): dT dr r=0 = 0 and T(r = r 0 ) = T s (3.114) The solution is T = T s + ˙q 6k r 2 0 − r 2 (3.123) The maximum temperature that occurs at the center of the sphere where r = 0is T max = T s + ˙qr 2 0 6k (3.124) BOOKCOMP, Inc. — John Wiley & Sons / Page 194 / 2nd Proofs / Heat Transfer Handbook / Bejan 194 CONDUCTION HEAT TRANSFER 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 [194], (34) Lines: 1617 to 1665 ——— 6.41006pt PgVar ——— Normal Page PgEnds: T E X [194], (34) and if the cooling at the outside surface is to a fluid at T ∞ via a heat transfer coefficient of h, the overall energy balance gives T s − T ∞ = ˙qr 0 3h (3.125) 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 of the system. The conducting medium may be nonhomogeneous, causing thermal conductivity to vary with location. Similarly, the temperature dependence of thermal conductivity cannot be ignored if the temperature difference driving the conduction process is large and the assumption of uniform heat generation may prove too restrictive. For example, when the shield of a nuclear reactor is irradiated with gamma rays, the resulting release of energy decays exponentially with distance from the irradiated surface, making the heat generation location dependent. A more realistic modeling of heat generation due to the passage of electric current or a chemical reaction requires that ˙q be treated as temperature dependent. Finally, if the 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 of such situations. 3.5.1 Location-Dependent Thermal Conductivity Plane Wall Consider the plane wall of Fig. 3.6 and let the thermal conductivity k increase linearly with x in accordance with k = k 0 (1 + ax) (3.126) where k 0 is the thermal conductivity at x = 0 and a is a measure of the variation of k with x. The equation governing the temperature distribution is d dx k dT dx = 0 (3.127) Solving eq. (3.127) subject to the boundary conditions of eq. (3.66), T(x = 0) = T s,1 and T(x = L) = T s,2 (3.66) BOOKCOMP, Inc. — John Wiley & Sons / Page 195 / 2nd Proofs / Heat Transfer Handbook / Bejan MORE ADVANCED STEADY ONE-DIMENSIONAL CONDUCTION 195 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 [195], (35) Lines: 1665 to 1725 ——— -1.69618pt PgVar ——— Normal Page PgEnds: T E X [195], (35) gives T = T s,1 + T s,2 − T s,1 ln(1 +ax) ln(1 +aL) (3.128) and the rate of heat transfer will be q = k 0 Aa(T s,1 − T s,2 ) ln(1 +aL) (3.129) In the limit, as a → 0, eqs. (3.128) and (3.129) reduce to eqs. (3.67) and (3.68), respectively. Now consider the case where k is of the form k = k 0 (1 + ax 2 ) (3.130) The solutions for T and q are given by T = T s,1 + T s,2 − T s,1 arctan √ ax arctan √ aL (3.131) and the rate of heat transfer will be q = k 0 A √ a(T s,1 − T s,2 ) arctan √ aL (3.132) Hollow Cylinder When modified to allow for the location-dependent thermal conductivity of the form k = a(1 + br) (3.133) 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,1 ln r 2 1 + br 2 1 + br r + T s,2 ln 1 + br 1 r 1 r 1 + br ln 1 + br 1 1 + br 2 r 2 r 1 (3.134) q = 2πaL(T s,1 − T s,2 ) ln 1 + br 1 1 + br 2 r 2 r 1 (3.135) When b = 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. BOOKCOMP, Inc. — John Wiley & Sons / Page 196 / 2nd Proofs / Heat Transfer Handbook / Bejan 196 CONDUCTION HEAT TRANSFER 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 [196], (36) Lines: 1725 to 1790 ——— 0.19121pt PgVar ——— Normal Page * PgEnds: Eject [196], (36) 3.5.2 Temperature-Dependent Thermal Conductivity Plane Wall Let the thermal conductivity k of the plane wall of Fig. 3.6 be a linear function of the temperature T , expressed as k = k 0 (1 + aT ) (3.136) Equation (3.127) will then take the form d dx k 0 (1 + aT ) dT dx = 0 (3.137) which must be integrated using the boundary conditions of eqs. (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 2 aT 2 (3.138) Differentiation of eq. (3.138) with respect to x gives dT ∗ dx = (1 + aT ) dT dx (3.139) which allows eq. (3.137) to be written as d 2 T ∗ dx 2 = 0 (3.140) The boundary conditions of eq. (3.66) in terms of T ∗ become T ∗ s,1 (x = 0) = T s,1 + 1 2 aT 2 s,1 and T ∗ s,2 (x = L) = T s,2 + 1 2 aT 2 s,2 (3.141) The solution for T ∗ is T ∗ = T ∗ s,1 + T ∗ s,2 − T ∗ s,1 x L (3.142) Once T ∗ has been found, T can be reclaimed by solving the quadratic of eq. (3.138), which gives T = 1 a −1 + √ 1 + 2aT ∗ (3.143) and the rate of heat transfer can be shown to be BOOKCOMP, Inc. — John Wiley & Sons / Page 197 / 2nd Proofs / Heat Transfer Handbook / Bejan MORE ADVANCED STEADY ONE-DIMENSIONAL CONDUCTION 197 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 [197], (37) Lines: 1790 to 1858 ——— 1.89345pt PgVar ——— Normal Page PgEnds: T E X [197], (37) q = k m A(T s,1 − T s,2 ) L (3.144) where k m = k 0 (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 of the thermal conductivity with temperature represented by k = k 0 (1 + aT 2 ) (3.145) the temperature distribution in the plane wall (Fig. 3.6) is given by the cubic equation T + 1 3 aT 3 = T s,1 + 1 3 aT 3 s,1 + 1 3 a T 3 s,2 − T 3 s,1 + (T s,2 − T s,1 ) x L (3.146) and the corresponding rate of heat transfer is q = Ak 0 T s,1 − T s,2 1 + (a/3) T 2 s,1 + T s,1 T s,2 + T 2 s,2 L (3.147) Hollow Cylinder For the thermal conductivity temperature relation of eq. (3.136), the Kirchhoff transformation of eq. (3.138) can also be used for the hollow cylinder of Fig. 3.7. The final result for T ∗ is T ∗ = T ∗ s,1 + T ∗ s,1 − T ∗ s,2 ln(r 1 /r 2 ) ln r r 1 (3.148) where T ∗ s,1 and T ∗ s,2 are as given in eq. (3.141). Once T ∗ for any radius r is found from eq. (3.148), eq. (3.143) can be used to find the corresponding value of T . The rate of heat transfer then follows as q = 2πk m L T s,1 − T s,2 ln(r 2 /r 1 ) (3.149) where k m is 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/r 2 − 1/r 1 1 r 1 − 1 r (3.150) BOOKCOMP, Inc. — John Wiley & Sons / Page 198 / 2nd Proofs / Heat Transfer Handbook / Bejan 198 CONDUCTION HEAT TRANSFER 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 [198], (38) Lines: 1858 to 1890 ——— 0.66515pt PgVar ——— Normal Page PgEnds: T E X [198], (38) T = 1 a −1 + √ 1 + 2aT ∗ (3.151) q = 4πk m T s,1 − T s,2 1/r 1 − 1/r 2 (3.152) 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 that k(x, T ) can be expressed as a product of two 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 of the form ˙q =˙q 0 1 − x L (3.153) The temperature distribution in the wall is given by T = T s,1 + ˙q 0 Lx 2k − ˙q 0 2k x 2 − x 3 3L (3.154) Figure 3.18 Plane wall with linearly decaying, location-dependent internal energy gener- ation. BOOKCOMP, Inc. — John Wiley & Sons / Page 199 / 2nd Proofs / Heat Transfer Handbook / Bejan MORE ADVANCED STEADY ONE-DIMENSIONAL CONDUCTION 199 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 [199], (39) Lines: 1890 to 1951 ——— 0.7885pt PgVar ——— Normal Page * PgEnds: Eject [199], (39) and the maximum temperature occurs at x = L for an insulated face: T = T s,1 + ˙q 0 L 2 6k (3.155) Next, assume that the plane wall of Fig. 3.6 represents the shield of a nuclear reactor. The absorption of gamma radiation at the left surface (x = 0) triggers energy release into the shield which decays exponentially with the penetration distance x and can be represented by the relation ˙q = q 0 ae −ax (3.156) where q 0 (W/m 2 ) is the incident radiation heat flux and a m −1 is the absorption coefficient of the shield. The temperature distribution in the shield is T = T s,1 + q 0 ak 1 − e −ax + T s,2 − T s,1 + q 0 ak e −aL − 1 x L (3.157) and the maximum temperature occurs at x = 1 a ln q 0 aL ak(T s,1 − T s,2 ) + q 0 (1 − e −aL ) (3.158) Solid Cylinder Reconsidering the solid cylinder of Fig. 3.16, ˙q will now be assumed to vary linearly with the radial distance r, that is, ˙q = ar (3.159) where a (W/m 4 ) is a constant. The temperature distribution in this case is T = T ∞ + ar 2 0 3h + a 9k r 3 0 − r 3 (3.160) from which the centerline (r = 0) temperature T c and the surface (r = r 0 ) tempera- ture T s follow as T c = T ∞ + ar 2 0 3h + ar 3 0 9k (3.161) T s = T ∞ + ar 2 0 3h (3.162) 3.5.4 Temperature-Dependent Energy Generation In this section we present a collection of results 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 BOOKCOMP, Inc. — John Wiley & Sons / Page 200 / 2nd Proofs / Heat Transfer Handbook / Bejan 200 CONDUCTION HEAT TRANSFER 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 [200], (40) Lines: 1951 to 2014 ——— 0.49226pt PgVar ——— Normal Page PgEnds: T E X [200], (40) ˙q =˙q s [ 1 + a(T − T s ) ] (3.163) where ˙q S is the energy generation at the surface temperature T s and a is a constant. Plane Wall For a plane wall of thickness 2L having identical surface temperatures T s on both faces, the temperature distribution is T = T s + 1 a cos nx cos nL − 1 (3.164) where n = √ a ˙q s /k and nL < π/2 to ensure that the temperatures remain finite. If the convection cooling, characterized by the temperature T ∞ and heat transfer coefficient h, is identical on both faces of the wall, the relationship between T s and T ∞ is given by T s = T ∞ + m h tan nL (3.165) where m = √ ˙q s k/a. Solid Cylinder For a solid cylinder of radius r 0 , the temperature distribution is T = T s + 1 a J 0 (nr) J 0 (nr 0 ) − 1 (3.166) where n = √ a ˙q s /k and J 0 is the Bessel function of the first kind of zero order (see Section 3.3.5). The parallel counterpart of eq. (3.165) is T s = T ∞ + m h J 1 (nr 0 ) J 0 (nr 0 ) (3.167) where m = √ ˙q s k/a,n = √ a ˙q s /k, and J 1 is the Bessel function of the first kind of order 1. In eqs. (3.166) and (3.167), nr 0 < 2.4048 to assure finite temperatures in the cylinder. Solid Sphere For a solid sphere of radius r 0 , the temperature distribution is T = T s + 1 a r 0 r sin nr sin nr 0 − 1 (3.168) where nr 0 < π to assure finite temperatures in the sphere and the relationship between T s and the coolant temperature T ∞ is T = T s + k hr 0 a [ 1 − (nr 0 ) cot nr 0 ] (3.169) where, here too, nr 0 < π to assure finite temperatures in the sphere. . Wiley & Sons / Page 192 / 2nd Proofs / Heat Transfer Handbook / Bejan 192 CONDUCTION HEAT TRANSFER 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 [192],. Wiley & Sons / Page 194 / 2nd Proofs / Heat Transfer Handbook / Bejan 194 CONDUCTION HEAT TRANSFER 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 [194],. Wiley & Sons / Page 196 / 2nd Proofs / Heat Transfer Handbook / Bejan 196 CONDUCTION HEAT TRANSFER 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 [196],