Adrian Bejan(Editor), Allan D. Kraus (Editor). Heat transfer Handbok (776115), страница 20
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A vessel containing nuclear waste experiencesenergy generation as the waste undergoes a slow process of disintegration.BOOKCOMP, Inc. — John Wiley & Sons / Page 188 / 2nd Proofs / Heat Transfer Handbook / Bejan[188], (28)Lines: 1334 to 1377———1.12001pt PgVar———Short PagePgEnds: TEX[188], (28)189STEADY ONE-DIMENSIONAL CONDUCTION123456789101112131415161718192021222324252627282930313233343536373839404142434445Plane Wall Consider first a plane wall of thickness 2L, made of a material havinga thermal conductivity k, as shown in Fig.
3.14. The volumetric rate of energy generation in the wall is q̇ (W/m 3). Let Ts,1 and Ts,2 be the two surface temperatures, andassuming the thermal conductivity of the wall to be constant, the appropriate form ofeq. (3.4) isq̇d 2T+ =0dx 2k(3.100)and the boundary conditions areT (x = −L) = Ts,1andT (x = L) = Ts,2(3.101)The solution for the temperature distribution isT = Ts,2 − Ts,1q̇ 2Ts,2 + Ts,1L − x2 +x+2k2L2[189], (29)(3.102)The maximum temperature occurs at x = k(Ts,2 − Ts,1 )/2Lq̇ and is given byTmaxk(Ts,2 − Ts,1 )2q̇L2Ts,2 + Ts,1+=+2k8q̇L22———(3.103)If the face at x = −L is cooled by convection with a heat transfer coefficient h1 andcoolant temperature T∞,1 , and the face at x = +L is cooled by convection with aheat transfer coefficient h2 and coolant temperature T∞,2 , the overall energy balancegives2q̇L = h1 (Ts,1 − T∞,1 ) + h2 (Ts,2 − T∞,2 )(3.104)x⫺L⫹L.qTs,1Ts,2h1 ,T⬁,1h2 ,T⬁,2Figure 3.14 Conduction in a plane wall with uniform internal energy generation.BOOKCOMP, Inc.
— John Wiley & Sons / Page 189 / 2nd Proofs / Heat Transfer Handbook / BejanLines: 1377 to 14200.18512pt PgVar———Short PagePgEnds: TEX[189], (29)190When Ts,1 = Ts,2 , the temperature distribution is symmetrical about x = 0 and isgiven byT =q̇ 2L − x 2 + Ts2k(3.105)and the maximum temperature occurs at the midplane (x = 0), which can be expressed asTmax =q̇L2+ Ts2k(3.106)When T∞,1 = T∞,2 = T∞ and h1 = h2 = h, eq. (3.104) reduces toTs − T ∞ =q̇Lh(3.107)Hollow Cylinder For the hollow cylinder shown in Fig.
3.15, the appropriate formof eq. (3.5) for constant k is1 ddTq̇r+ =0(3.108)r drdrkand the boundary conditions areT (r = r1 ) = Ts,1andT (r = r2 ) = Ts,2[190], (30)Lines: 1420 to 1465———-1.06888pt PgVar———Normal PagePgEnds: TEX(3.109)[190], (30)Cold fluidT⬁,2, h2Ts,2k.qTs,1 r1r2L123456789101112131415161718192021222324252627282930313233343536373839404142434445CONDUCTION HEAT TRANSFERrCold fluidT⬁,1, h1Figure 3.15 Conduction in a hollow cylinder with uniform internal energy generation.BOOKCOMP, Inc. — John Wiley & Sons / Page 190 / 2nd Proofs / Heat Transfer Handbook / Bejan191STEADY ONE-DIMENSIONAL CONDUCTION123456789101112131415161718192021222324252627282930313233343536373839404142434445The solution of eq.
(3.108) that satisfies eqs. (3.109) is 2 q̇r22rT = Ts,2 +1−4kr2 2 ln(r2 /r)q̇r22r1−+ Ts,2 − Ts,11−4kr2ln(r2 /r1 )(3.110)If the inside and outside surfaces are cooled by convection, the inside with fluid atT∞,1 with heat transfer coefficient h2 , the overall energy balance givesq̇ r22 − r12 = 2h1 r1 Ts,1 − T∞,1 + 2h2 r2 Ts,2 − T∞,2and for the case of Ts,1 = Ts,2 , eq. (3.110) is reduced to 2 2 q̇r22q̇r22ln(r2 /r)rr1T = Ts +−1−1−4kr24kr2ln(r2 /r1 )(3.111)[191], (31)(3.112)———Similarly, when T∞,1 = T∞,2 = T∞ and h1 = h2 = h, eq.
(3.111) is reduced toTs − T∞ =q̇(r2 − r1 )2hCold fluidT⬁, hTmaxTsr0rkLFigure 3.16 Conduction in a solid cylinder with uniform internal energy generation.BOOKCOMP, Inc. — John Wiley & Sons / Page 191 / 2nd Proofs / Heat Transfer Handbook / Bejan0.20709pt PgVar———Normal Page(3.113)* PgEnds: EjectSolid Cylinder Equation (3.108) also applies to the solid cylinder of Fig. 3.16,but the boundary conditions change to.qLines: 1465 to 1501[191], (31)192123456789101112131415161718192021222324252627282930313233343536373839404142434445CONDUCTION HEAT TRANSFERdT =0dr r=0T (r = r0 ) = Tsand(3.114)The temperature distribution is given byT = Ts +q̇ 2r0 − r 24k(3.115)and the maximum temperature occurs along the centerline at r = 0:Tmax = Ts +q̇r024k(3.116)If the outside surface of the cylinder is cooled by convection to a fluid at T∞ througha heat transfer coefficient h, the overall energy balance givesTs − T∞ =q̇r02h(3.117)[192], (32)Lines: 1501 to 1553———Hollow Sphere For the hollow sphere shown in Fig.
3.17, the appropriate formof eq. (3.6) with constant thermal conductivity k is1 dq̇2 dTr+ =0(3.118)r 2 drdrkandT (r = r2 ) = Ts,2(3.109)T⬁ ,2,h2Ts,2Ts,1.qr1T⬁,1, h1rr2kFigure 3.17 Conduction in a hollow sphere with uniform internal energy generation.BOOKCOMP, Inc. — John Wiley & Sons / Page 192 / 2nd Proofs / Heat Transfer Handbook / Bejan———Normal PagePgEnds: TEX[192], (32)and eq. (3.109) provides the boundary conditionsT (r = r1 ) = Ts,19.7792pt PgVarSTEADY ONE-DIMENSIONAL CONDUCTION123456789101112131415161718192021222324252627282930313233343536373839404142434445193The temperature distribution is found to be 2 q̇r22r1−T = Ts,2 +6kr2 2 1/r − 1/r2q̇r22r1−+ Ts,2 − Ts,11−6kr21/r1 − 1/r2(3.119)If the inside and outside surfaces are cooled by convection, the inside with fluid atT∞,1 with heat transfer coefficient h1 and the outside with fluid at T∞,2 with heattransfer coefficient h2 , the overall energy balance givesq̇ r23 − r13= h1 r12 Ts,1 − T∞,1 + h2 r22 Ts,2 − T∞,23(3.120)and when Ts,1 = Ts,2 = Ts , eq.
(3.119) reduces toT = Ts +q̇r226k1−rr22 −q̇r226k1−r1r22 Lines: 1553 to 16171/r − 1/r21/r1 − 1/r2———(3.121)Solid Sphereq̇ r23 − r13=3h r12 + r22(3.122)For the solid sphere, eq. (3.118),1 dr 2 drq̇2 dTr+ =0drkmust be solved subject to the boundary conditions of eqs. (3.114):dT = 0 and T (r = r0 ) = Tsdr r=0(3.118)(3.114)The solution isT = Ts +q̇ 2r0 − r 26k(3.123)The maximum temperature that occurs at the center of the sphere where r = 0 isTmax = Ts +BOOKCOMP, Inc. — John Wiley & Sons / Page 193 / 2nd Proofs / Heat Transfer Handbook / Bejanq̇r026k3.10439pt PgVar———Normal Page* PgEnds: EjectWhen T∞,1 = T∞,2 = T∞ , and h1 = h2 = h, eq. (3.120) is reduced toTs − T∞[193], (33)(3.124)[193], (33)194123456789101112131415161718192021222324252627282930313233343536373839404142434445CONDUCTION HEAT TRANSFERand if the cooling at the outside surface is to a fluid at T∞ via a heat transfer coefficientof h, the overall energy balance givesTs − T∞ =q̇r03h(3.125)For additional analytical results for one-dimensional steady conduction with uniform internal heat generation, the reader should consult Incropera and DeWitt (1996,App.
C).3.5MORE ADVANCED STEADY ONE-DIMENSIONAL CONDUCTIONThe results presented in Section 3.4 have been based on assumptions such as constantthermal conductivity, uniform heat generation, and pure convective cooling or heatingat the boundary. In some applications, these assumptions may introduce significanterrors in predicting the thermal behavior of the system.The conducting medium may be nonhomogeneous, causing thermal conductivityto vary with location. Similarly, the temperature dependence of thermal conductivitycannot be ignored if the temperature difference driving the conduction process islarge and the assumption of uniform heat generation may prove too restrictive.
Forexample, when the shield of a nuclear reactor is irradiated with gamma rays, theresulting release of energy decays exponentially with distance from the irradiatedsurface, making the heat generation location dependent. A more realistic modelingof heat generation due to the passage of electric current or a chemical reaction requiresthat q̇ be treated as temperature dependent.
Finally, if the heat transfer process at aboundary is driven by natural convection, radiation becomes equally important andmust be taken into account. This section is devoted to a discussion of such situations.3.5.1Location-Dependent Thermal ConductivityPlane Wall Consider the plane wall of Fig. 3.6 and let the thermal conductivity kincrease linearly with x in accordance withk = k0 (1 + ax)(3.126)where k0 is the thermal conductivity at x = 0 and a is a measure of the variation ofk with x. The equation governing the temperature distribution isddxkdTdx=0(3.127)Solving eq.
(3.127) subject to the boundary conditions of eq. (3.66),T (x = 0) = Ts,1andBOOKCOMP, Inc. — John Wiley & Sons / Page 194 / 2nd Proofs / Heat Transfer Handbook / BejanT (x = L) = Ts,2(3.66)[194], (34)Lines: 1617 to 1665———6.41006pt PgVar———Normal PagePgEnds: TEX[194], (34)MORE ADVANCED STEADY ONE-DIMENSIONAL CONDUCTION123456789101112131415161718192021222324252627282930313233343536373839404142434445195gives ln(1 + ax)T = Ts,1 + Ts,2 − Ts,1ln(1 + aL)(3.128)and the rate of heat transfer will bek0 Aa(Ts,1 − Ts,2 )ln(1 + aL)q=(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 formk = k0 (1 + ax 2 )(3.130)The solutions for T and q are given by√ arctan ax√ T = Ts,1 + Ts,2 − Ts,1arctan aLLines: 1665 to 1725(3.131)(3.132)[195], (35)Hollow Cylinder When modified to allow for the location-dependent thermalconductivity of the formk = a(1 + br)(3.133)analysis of Section 3.4.2 for a hollow cylinder (Fig.
3.7) gives the following resultsfor the temperature distribution and the rate of heat transfer:r2 1 + br1 + br1 rTs,1 ln+ Ts,2 ln1 + br2 rr1 1 + brT =(3.134)1 + br1 r2ln1 + br2 r1q=2πaL(Ts,1 − Ts,2 )1 + br1 r2ln1 + br2 r1(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 195 / 2nd Proofs / Heat Transfer Handbook / Bejan———-1.69618pt PgVar———Normal PagePgEnds: TEXand the rate of heat transfer will be√k0 A a(Ts,1 − Ts,2 )√ q=arctan aL[195], (35)196123456789101112131415161718192021222324252627282930313233343536373839404142434445CONDUCTION HEAT TRANSFER3.5.2 Temperature-Dependent Thermal ConductivityPlane Wall Let the thermal conductivity k of the plane wall of Fig. 3.6 be a linearfunction of the temperature T , expressed ask = k0 (1 + aT )Equation (3.127) will then take the formdTdk0 (1 + aT )=0dxdx(3.136)(3.137)which must be integrated using the boundary conditions of eqs. (3.66):T (x = 0) = Ts,1andT (x = L) = Ts,2(3.66)The solution is facilitated by the introduction of a new variable, T ∗ , defined by theKirchhoff transformation: T(1 + aT ) dT = T + 21 aT 2(3.138)T∗ =0Differentiation of eq.
(3.138) with respect to x givesdTdT ∗= (1 + aT )dxdx(3.139)which allows eq. (3.137) to be written asd 2T ∗=0dx 2(3.140)The boundary conditions of eq. (3.66) in terms of T ∗ become∗2Ts,1(x = 0) = Ts,1 + 21 aTs,1and∗2Ts,2(x = L) = Ts,2 + 21 aTs,2(3.141)The solution for T ∗ isx ∗∗∗T ∗ = Ts,1+ Ts,2− Ts,1L(3.142)Once T ∗ has been found, T can be reclaimed by solving the quadratic of eq.
(3.138),which givesT =√1−1 + 1 + 2aT ∗aand the rate of heat transfer can be shown to beBOOKCOMP, Inc. — John Wiley & Sons / Page 196 / 2nd Proofs / Heat Transfer Handbook / Bejan(3.143)[196], (36)Lines: 1725 to 1790———0.19121pt PgVar———Normal Page* PgEnds: Eject[196], (36)197MORE ADVANCED STEADY ONE-DIMENSIONAL CONDUCTION123456789101112131415161718192021222324252627282930313233343536373839404142434445q=km A(Ts,1 − Ts,2 )L(3.144)where km = k0 (1 + aTs,m ) is the thermal conductivity at the mean temperature,Ts,m =Ts,1 + Ts,22For a variation of the thermal conductivity with temperature represented byk = k0 (1 + aT 2 )(3.145)the temperature distribution in the plane wall (Fig. 3.6) is given by the cubic equation1 31 3x1 33a Ts,2 − Ts,1 + (Ts,2 − Ts,1 )(3.146)T + aT = Ts,1 + aTs,1 +333Land the corresponding rate of heat transfer is 22+ Ts,1 Ts,2 + Ts,2Ak0 Ts,1 − Ts,2 1 + (a/3) Ts,1q=LLines: 1790 to 1858———(3.147)Hollow Cylinder For the thermal conductivity--temperature relation of eq.
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