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CxCxT += where C1 and C2 are arbitrary constants. Applying the boundary conditions give Heat flux at x = 0: k q CqkC 0101 && \u2212=\u2192=\u2212 Temperature at x = 0: 12121 0)0( TCTCCT =\u2192=+×= Substituting C1 and C2 into the general solution, the variation of temperature is determined to be 90420C90 C W/m5.2 W/m1050)( 2 1 0 +\u2212=°+°\u22c5\u2212=+\u2212= xxTxk q xT & (c) The temperature at x = L (the right surface of the wall) is C-36°=+×\u2212= 90m) 3.0(420)(LT Note that the right surface temperature is lower 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. 2-31 2-68E A large plate is subjected to convection, radiation, and specified temperature on the top surface and no conditions on the bottom surface. The mathematical formulation, the variation of temperature in the plate, and the bottom surface temperature are to be determined for steady one-dimensional heat transfer. Assumptions 1 Heat conduction is steady and one-dimensional since the plate is large relative to its thickness, and the thermal conditions on both sides of the plate are uniform. 2 Thermal conductivity is constant. 3 There is no heat generation in the plate. Properties The thermal conductivity and emissivity are given to be k =7.2 Btu/h\u22c5ft\u22c5°F and \u3b5 = 0.7. x T\u221e h Tsky L 75°F \u3b5 Analysis (a) Taking the direction normal to the surface of the plate to be the x direction with x = 0 at the bottom surface, and the mathematical formulation of this problem can be expressed as 02 2 = dx Td and ])460[(][])([])([)( 4sky 4 22 4 sky 4 TTTThTLTTLTh dx LdTk \u2212++\u2212=\u2212+\u2212=\u2212 \u221e\u221e \u3b5\u3c3\u3b5\u3c3 F75)( 2 °== TLT (b) Integrating the differential equation twice with respect to x yields 1C dx dT = 21)( CxCxT += where C1 and C2 are arbitrary constants. Applying the boundary conditions give Convection at x = L: kTTTThC TTTThkC /]})460[(][{ ])460[(][ 4 sky 4 221 4 sky 4 221 \u2212++\u2212\u2212=\u2192 \u2212++\u2212=\u2212 \u221e \u221e \u3b5\u3c3 \u3b5\u3c3 Temperature at x = L: LCTCTCLCLT 122221 )( \u2212=\u2192=+×= Substituting C1 and C2 into the general solution, the variation of temperature is determined to be )3/1(2.2075 ft )12/4( FftBtu/h 2.7 ]R) 480()R 535)[(RftBtu/h 100.7(0.1714+F)9075)(FftBtu/h 12(F75 )( ])460[(][ )()()( 44428-2 4 sky 4 22 212121 x x xL k TTTTh TCxLTLCTxCxT \u2212\u2212= \u2212°\u22c5\u22c5 \u2212\u22c5\u22c5×°\u2212°\u22c5\u22c5+°= \u2212\u2212++\u2212+=\u2212\u2212=\u2212+= \u221e \u3b5\u3c3 (c) The temperature at x = 0 (the bottom surface of the plate) is F68.3°=\u2212×\u2212= )03/1(2.2075)0(T 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. 2-32 2-69E A large plate is subjected to convection and specified temperature on the top surface and no conditions on the bottom surface. The mathematical formulation, the variation of temperature in the plate, and the bottom surface temperature are to be determined for steady one-dimensional heat transfer. Assumptions 1 Heat conduction is steady and one-dimensional since the plate is large relative to its thickness, and the thermal conditions on both sides of the plate are uniform. 2 Thermal conductivity is constant. 3 There is no heat generation in the plate. Properties The thermal conductivity is given to be k =7.2 Btu/h\u22c5ft\u22c5°F. Analysis (a) Taking the direction normal to the surface of the plate to be the x direction with x = 0 at the bottom surface, the mathematical formulation of this problem can be expressed as 0 2 2 = dx Td x T\u221e h L 75°F and )(])([)( 2 \u221e\u221e \u2212=\u2212=\u2212 TThTLThdx LdTk F75)( 2 °== TLT (b) Integrating the differential equation twice with respect to x yields 1Cdx dT = 21)( CxCxT += where C1 and C2 are arbitrary constants. Applying the boundary conditions give Convection at x = L: kTThCTThkC /)( )( 2121 \u221e\u221e \u2212\u2212=\u2192\u2212=\u2212 Temperature at x = L: LCTCTCLCLT 122221 )( \u2212=\u2192=+×= Substituting C1 and C2 into the general solution, the variation of temperature is determined to be )3/1(2575 ft )12/4( FftBtu/h 2.7 F)9075)(FftBtu/h 12(F75 )()()()()( 2 2 212121 x x xL k TThTCxLTLCTxCxT \u2212\u2212= \u2212°\u22c5\u22c5 °\u2212°\u22c5\u22c5+°= \u2212\u2212+=\u2212\u2212=\u2212+= \u221e (c) The temperature at x = 0 (the bottom surface of the plate) is F66.7°=\u2212×\u2212= )03/1(2575)0(T 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. 2-63 2-129 A spherical liquid nitrogen container is subjected to specified temperature on the inner surface and convection on the outer surface. The mathematical formulation, the variation of temperature, and the rate of evaporation of nitrogen are to be determined for steady one-dimensional heat transfer. Assumptions 1 Heat conduction is steady and one-dimensional since there is no change with time and there is thermal symmetry about the midpoint. 2 Thermal conductivity is constant. 3 There is no heat generation. Properties The thermal conductivity of the tank is given to be k = 18 W/m\u22c5°C. Also, hfg = 198 kJ/kg for nitrogen. Analysis (a) Noting that heat transfer is one-dimensional in the radial r direction, the mathematical formulation of this problem can be expressed as 02 =\u239f\u23a0 \u239e\u239c\u239d \u239b dr dTr dr d and C196)( 11 °\u2212== TrT ])([)( 22 \u221e\u2212=\u2212 TrThdr rdTk (b) Integrating the differential equation once with respect to r gives r1 r2 h T\u221e r -196°C N2 1 2 C dr dTr = Dividing both sides of the equation above by r to bring it to a readily integrable form and then integrating, 2 1 r C dr dT = \u2192 21)( Cr C rT +\u2212= where C1 and C2 are arbitrary constants. Applying the boundary conditions give r = r1: 12 1 1 1 )( TCr C rT =+\u2212= r = r2: \u239f\u239f\u23a0 \u239e \u239c\u239c\u239d \u239b \u2212+\u2212=\u2212 \u221eTCr Ch r Ck 2 2 1 2 2 1 Solving for C1 and C2 simultaneously gives 1 2 21 2 1 1 1 1 12 21 2 12 1 1 and 1 )( r r hr k r r TT T r C TC hr k r r TTr C \u2212\u2212 \u2212+=+= \u2212\u2212 \u2212= \u221e\u221e Substituting C1 and C2 into the general solution, the variation of temperature is determined to be 196)/1.205.1(8.549C)196(1.2 2 1.2 )m 1.2)(C W/m25( C W/m18 2 1.21 C)20196( 1 11)( 2 1 2 1 2 21 2 1 1 1 1 1 1 1 1 \u2212\u2212=°\u2212+\u239f\u23a0 \u239e\u239c\u239d \u239b \u2212 °\u22c5 °\u22c5\u2212\u2212 °\u2212\u2212= +\u239f\u239f\u23a0 \u239e \u239c\u239c\u239d \u239b \u2212 \u2212\u2212 \u2212=+\u239f\u239f\u23a0 \u239e \u239c\u239c\u239d \u239b \u2212=++\u2212= \u221e r r T r r r r hr k r r TT T rr C r C T r C rT (c) The rate of heat transfer through the wall and the rate of evaporation of nitrogen are determined from negative) since tank the(to W 261,200 )m 1.2)(C W/m25( C W/m18 2 1.21 C)20196(m) 1.2()C W/m18(4 1 )( 44)4( 2 21 2 12 12 12 \u2212= °\u22c5 °\u22c5\u2212\u2212 °\u2212\u2212°\u22c5\u2212= \u2212\u2212 \u2212\u2212=\u2212=\u2212=\u2212= \u221e \u3c0 \u3c0\u3c0\u3c0 hr k r r TTr kkC r C rk dx dTkAQ& kg/s 1.32=== J/kg 000,198 J/s 200,261 fgh Qm & & 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. 2-64 2-130 A spherical liquid oxygen container is subjected to specified temperature on the inner surface and convection on the outer surface. The mathematical formulation, the variation of temperature, and the rate of evaporation of oxygen are to be determined for steady one-dimensional heat transfer. Assumptions 1 Heat conduction is steady and one-dimensional since there is no change with time and there is thermal symmetry about the midpoint. 2 Thermal conductivity is constant. 3 There is no heat generation. Properties The thermal conductivity of the tank is given to be k = 18 W/m\u22c5°C. Also, hfg = 213 kJ/kg for