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in a cylindrical uranium fuel rod. The temperature difference between the center and the surface of the fuel rod is to be determined. Assumptions 1 Heat transfer is steady since there is no indication of any change with time. 2 Heat transfer is one-dimensional since there is thermal symmetry about the center line and no change in the axial direction. 3 Thermal conductivity is constant. 4 Heat generation is uniform. Properties The thermal conductivity of uranium at room temperature is k = 27.6 W/m\u22c5°C (Table A-3). Analysis The temperature difference between the center and the surface of the fuel rods is determined from Ts e D C92.8°=° ×==\u2212 C) W/m.6.27(4 m) 016.0)( W/m104( 4 2372 gen k re TT oso & 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. 2-72 2-138 A large plane wall is subjected to convection on the inner and outer surfaces. The mathematical formulation, the variation of temperature, and the temperatures at the inner and outer surfaces to be determined for steady one-dimensional heat transfer. Assumptions 1 Heat conduction is steady and one-dimensional. 2 Thermal conductivity is constant. 3 There is no heat generation. Properties The thermal conductivity is given to be k = 0.77 W/m\u22c5°C. Analysis (a) Taking the direction normal to the surface of the wall to be the x direction with x = 0 at the inner surface, the mathematical formulation of this problem can be expressed as 02 2 = dx Td k h1 T\u221e1 L h2 T\u221e2 and dx dTkTTh )0()]0([ 11 \u2212=\u2212\u221e ])([)( 22 \u221e\u2212=\u2212 TLThdx LdTk (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 x = 0: 12111 )]0([ kCCCTh \u2212=+×\u2212\u221e x = L: ])[( 22121 \u221e\u2212+=\u2212 TCLChkC Substituting the given values, these equations can be written as 12 77.0)27(5 CC \u2212=\u2212 )82.0)(12(77.0 211 \u2212+=\u2212 CCC Solving these equations simultaneously give 20 45.45 21 =\u2212= CC Substituting into the general solution, the variation of temperature is determined to be 21 and CC xxT 45.4520)( \u2212= (c) The temperatures at the inner and outer surfaces are C10.9 C20 °=×\u2212= °=×\u2212= 2.045.4520)( 045.4520)0( LT 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. 2-73 2-139 A hollow pipe is subjected to specified temperatures at the inner and outer surfaces. There is also heat generation in the pipe. The variation of temperature in the pipe and the center surface temperature of the pipe are to be determined for steady one-dimensional heat transfer. Assumptions 1 Heat conduction is steady and one-dimensional since the pipe is long relative to its thickness, and there is thermal symmetry about the centerline. 2 Thermal conductivity is constant. Properties The thermal conductivity is given to be k = 14 W/m\u22c5°C. Analysis The rate of heat generation is determined from [ ] 3222122gen W/m750,264/)m 17(m) 3.0(m) 4.0( W000,254/)( =\u2212=\u2212== \u3c0\u3c0 LDD WWe && & V Noting that heat transfer is one-dimensional in the radial r direction, the mathematical formulation of this problem can be expressed as 01 gen =+\u239f\u23a0 \u239e\u239c\u239d \u239b k e dr dTr dr d r & r2 T2 r r1 T1 egen and C60)( 11 °== TrT C80)( 22 °== TrT Rearranging the differential equation 0gen =\u2212=\u239f\u23a0 \u239e\u239c\u239d \u239b k re dr dTr dr d & and then integrating once with respect to r, 1 2 gen 2 C k re dr dTr +\u2212= & Rearranging the differential equation again r C k re dr dT 1gen 2 +\u2212= & and finally integrating again with respect to r, we obtain 21 2 gen ln 4 )( CrC k re rT ++\u2212= & where C1 and C2 are arbitrary constants. Applying the boundary conditions give r = r1: 211 2 1gen 1 ln4 )( CrC k re rT ++\u2212= & r = r2: 221 2 2gen 2 ln4 )( CrC k re rT ++\u2212= & Substituting the given values, these equations can be written as 21 2 )15.0ln( )14(4 )15.0)(750,26(60 CC ++\u2212= 21 2 )20.0ln( )14(4 )20.0)(750,26(80 CC ++\u2212= Solving for simultaneously gives 21 and CC 8.257 58.98 21 == CC Substituting into the general solution, the variation of temperature is determined to be 21 and CC rrrrrT ln58.987.4778.2578.257ln58.98 )14(4 750,26)( 2 2 +\u2212=++\u2212= The temperature at the center surface of the pipe is determined by setting radius r to be 17.5 cm, which is the average of the inner radius and outer radius. C71.3°=+\u2212= )175.0ln(58.98)175.0(7.4778.257)( 2rT 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. 2-74 2-140 Heat is generated in a plane wall. Heat is supplied from one side which is insulated while the other side is subjected to convection with water. The convection coefficient, the variation of temperature in the wall, and the location and the value of the maximum temperature in the wall are to be determined. Assumptions 1 Heat transfer is steady since there is no indication of any change with time. 2 Heat transfer is one-dimensional since the wall is large relative to its thickness. 3 Thermal conductivity is constant. 4 Heat generation is uniform. Analysis (a) Noting that the heat flux and the heat generated will be transferred to the water, the heat transfer coefficient is determined from the Newton\u2019s law of cooling to be Ts L k x Heater Insulation T\u221e , h sq& gene&C W/m400 2 °\u22c5=°\u2212 += \u2212 += \u221e C40)(90 m) )(0.04 W/m(10) W/m(16,000 352 gen TT Leq h s s && (b) The variation of temperature in the wall is in the form of T(x) = ax2+bx+c. First, the coefficient a is determined as follows k e dT Tdke dx Tdk gen 2 2 gen2 2 0 & & \u2212=\u2192=+ cbxx k e Tbx k e dx dT ++\u2212=+\u2212= 2gengen 2 and && \u2192 2 35 gen C/m2500 )C W/m20(2 W/m10 2 °\u2212=°\u22c5=\u2212= k e a & Applying the first boundary condition: x = 0, T(0) = Ts \u2192 c = Ts = 90ºC As the second boundary condition, we can use either s Lx q dx dTk \u2212=\u2212 = \u2192 sqbk Le k =\u239f\u239f\u23a0 \u239e \u239c\u239c\u239d \u239b +\u2212 gen& \u2192 ( ) ( ) C/m100004.01016000 20 11 5 gen °=×+=+= Leqkb s & or )( 0 \u221e = \u2212\u2212=\u2212 TTh dx dTk s x k(a×0+b) = h(Ts -T\u221e) \u2192 C/m1000)4090(20 400 °=\u2212=b Substituting the coefficients, the variation of temperature becomes 9010002500)( 2 ++\u2212= xxxT (c) The x-coordinate of Tmax is xvertex= -b/(2a) = 1000/(2×2500) = 0.2 m = 20 cm. This is outside of the wall boundary, to the left, so Tmax is at the left surface of the wall. Its value is determined to be C126°=++\u2212=++\u2212== 90)04.0(1000)04.0(25009010002500)( 22max LLLTT The direction of qs(L) (in the negative x direction) indicates that at x = L the temperature increases in the positive x direction. If a is negative, the T plot is like in Fig. 1, which shows Tmax at x=L. If a is positive, the T plot could only be like in Fig. 2, which is incompatible with the direction of heat transfer at the surface in contact with the water. So, temperature distribution can only be like in Fig. 1, where Tmax is at x=L, and this was determined without using numerical values for a, b, or 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. 2-75 Fig. 1 qs(0) qs(L) Slope Fig. 2 qs(L) qs(0) Slope Here, heat transfer and slope are incompatible This part could also