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expression for the outer surface temperature T2, 0 4 surr 4 22 ])273[()( qTTTTh &=\u2212++\u2212 \u221e \u3b5\u3c3 (c) Substituting the known quantities into the implicit relation above gives 2442 428 2 2 W/m000,100]295)273)[(K W/m1067.5(7.0)26)(C W/m30( =\u2212+\u22c5×+\u2212°\u22c5 \u2212 TT Using an equation solver (or a trial and error approach), the outer surface temperature is determined from the relation above to be T2 = 896°C PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 2-68 2-134E The concrete slab roof of a house is subjected to specified temperature at the bottom surface and convection and radiation at the top surface. The temperature of the top surface of the roof and the rate of heat transfer are to be determined when steady operating conditions are reached. Assumptions 1 Steady operating conditions are reached. 2 Heat transfer is one-dimensional since the roof area is large relative to its thickness, and the thermal conditions on both sides of the roof are uniform. 3 Thermal properties are constant. 4 There is no heat generation in the wall. Properties The thermal conductivity and emissivity are given to be k = 1.1 Btu/h\u22c5ft\u22c5°F and \u3b5 = 0.8. Analysis In steady operation, heat conduction through the roof must be equal to net heat transfer from the outer surface. Therefore, taking the outer surface temperature of the roof to be T2 (in °F), ])460[()( 4sky 4 22 21 TTATThA L TT kA \u2212++\u2212=\u2212 \u221e \u3c3\u3b5 x T\u221e h T1 L Tsky Canceling the area A and substituting the known quantities, 444 2 428 2 22 R]310)460)[(RftBtu/h 101714.0(8.0 F)50)(FftBtu/h 2.3( ft 0.8 F)62( )FftBtu/h 1.1( \u2212+\u22c5\u22c5×+ °\u2212°\u22c5\u22c5=°\u2212°\u22c5\u22c5 \u2212 T T T Using an equation solver (or the trial and error method), the outer surface temperature is determined to be T2 = 38°F Then the rate of heat transfer through the roof becomes Btu/h 28,875=°\u2212×°\u22c5\u22c5=\u2212= ft 0.8 F)3862()ft 3525)(FftBtu/h 1.1( 221 L TT kAQ& Discussion The positive sign indicates that the direction of heat transfer is from the inside to the outside. Therefore, the house is losing heat as expected. PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 2-69 2-135 The surface and interface temperatures of a resistance wire covered with a plastic layer are to be determined. Assumptions 1 Heat transfer is steady since there is no change with time. 2 Heat transfer is one- dimensional since this two-layer heat transfer problem possesses symmetry about the center line and involves no change in the axial direction, and thus T = T(r) . 3 Thermal conductivities are constant. 4 Heat generation in the wire is uniform. Properties It is given that C W/m18wire °\u22c5=k and C W/m8.1plastic °\u22c5=k . Analysis Letting denote the unknown interface temperature, the mathematical formulation of the heat transfer problem in the wire can be expressed as IT 01 gen =+\u239f\u23a0 \u239e\u239c\u239d \u239b k e dr dTr dr d r & r2 r r1 egen T\u221e h with and ITrT =)( 1 0)0( =dr dT Multiplying both sides of the differential equation by r, rearranging, and integrating give r k e dr dTr dr d gen&\u2212=\u239f\u23a0 \u239e\u239c\u239d \u239b \u2192 1 2 gen 2 Cr k e dr dTr +\u2212= & (a) Applying the boundary condition at the center (r = 0) gives B.C. at r = 0: 0 0 2 )0( 0 11 gen =\u2192+×\u2212=× CC k e dr dT & Dividing both sides of Eq. (a) by r to bring it to a readily integrable form and integrating, r k e dr dT 2 gen&\u2212= \u2192 22gen4)( Crk e rT +\u2212= & (b) Applying the other boundary condition at 1rr = , B. C. at : 1rr = 21gen2221gen 4 4 rk e TCCr k e T II && +=\u2192+\u2212= Substituting this relation into Eq. (b) and rearranging give 2C )( 4 )( 221 wire gen wire rrk e TrT I \u2212+= & (c) Plastic layer The mathematical formulation of heat transfer problem in the plastic can be expressed as 0=\u239f\u23a0 \u239e\u239c\u239d \u239b dr dTr dr d with and ITrT =)( 1 ])([)( 22 \u221e\u2212=\u2212 TrThdr rdTk The solution of the differential equation is determined by integration to be 1Cdr dTr = \u2192 r C dr dT 1= \u2192 21 ln)( CrCrT += where C1 and C2 are arbitrary constants. Applying the boundary conditions give r = r1: 112211 ln ln rCTCTCrC II \u2212=\u2192=+ PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 2-70 r = r2: ])ln[( 221 2 1 \u221e\u2212+=\u2212 TCrChr Ck \u2192 21 2 1 ln hr k r r TTC I + \u2212= \u221e Substituting C1 and C2 into the general solution, the variation of temperature in plastic is determined to be 1 2 plastic 1 2 111plastic ln ln lnln)( r r hr k r r TT TrCTrCrT III + \u2212+=\u2212+= \u221e We have already utilized the first interface condition by setting the wire and plastic layer temperatures equal to at the interface . The interface temperature is determined from the second interface condition that the heat flux in the wire and the plastic layer at IT 1rr = IT 1rr = must be the same: 1 2 plastic 1 2 plastic 1gen1plastic plastic 1wire wire 1 ln 2 )()( r hr k r r TT k re dr rdT k dr rdT k I + \u2212\u2212=\u2192\u2212=\u2212 \u221e& Solving for and substituting the given values, the interface temperature is determined to be IT C97.1°°+\u239f\u239f\u23a0 \u239e \u239c\u239c\u239d \u239b °\u22c5 °\u22c5+°\u22c5 ×= +\u239f\u239f\u23a0 \u239e \u239c\u239c\u239d \u239b += \u221e =C25 m) C)(0.007 W/m(14 C W/m1.8 m 0.003 m 007.0ln C) W/m2(1.8 m) 003.0)( W/m105.1( ln 2 2 236 2 plastic 1 2 plastic 2 1gen T hr k r r k re TI & Knowing the interface temperature, the temperature at the center line (r = 0) is obtained by substituting the known quantities into Eq. (c), C97.3°=°\u22c5× ×°=+= C) W/m(184 m) 003.0)( W/m105.1(+C1.97 4 )0( 236 wire 2 1gen wire k re TT I & Thus the temperature of the centerline will be slightly above the interface temperature. PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 2-71 2-136 A cylindrical shell with variable conductivity is subjected to specified temperatures on both sides. The rate of heat transfer through the shell is to be determined. r2 T2 r r1 T1 k(T) Assumptions 1 Heat transfer is given to be steady and one- dimensional. 2 Thermal conductivity varies quadratically. 3 There is no heat generation. Properties The thermal conductivity is given to be . )1()( 20 TkTk \u3b2+= Analysis When the variation of thermal conductivity with temperature k(T) is known, the average value of the thermal conductivity in the temperature range between is determined from 21 and TT ( ) ( ) ( )\u23a5\u23a6\u23a4\u23a2\u23a3\u23a1 +++= \u2212 \u23a5\u23a6 \u23a4\u23a2\u23a3 \u23a1 \u2212+\u2212 =\u2212 \u239f\u23a0 \u239e\u239c\u239d \u239b + =\u2212 + =\u2212= \u222b\u222b 2 121 2 20 12 3 1 3 2120 12 3 0 12 2 0 12 avg 3 1 33 )1()( 2 1 2 1 2 1 TTTTk TT TTTTk TT TTk TT dTTk TT dTTk k T T T T T T \u3b2 \u3b2\u3b2\u3b2 This relation is based on the requirement that the rate of heat transfer through a medium with constant average thermal conductivity equals the rate of heat transfer through the same medium with variable conductivity k(T). avgk Then the rate of heat conduction through the cylindrical shell can be determined from Eq. 2-77 to be ( ) )/ln(3 12 )/ln( 2 12 212 121 2 20 12 21 avgcylinder rr TT LTTTTk rr TT LkQ \u2212 \u23a5\u23a6 \u23a4\u23a2\u23a3 \u23a1 +++=\u2212= \u3b2\u3c0\u3c0& Discussion We would obtain the same result if we substituted the given k(T) relation into the second part of Eq. 2-77, and performed the indicated integration. 2-137 Heat is generated uniformly