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2 Thermal conductivity is constant. 3 Heat is generated uniformly. Analysis The differential equation and the boundary conditions for this heat conduction problem can be expressed as 01 gen 2 2 =+\u2202 \u2202+\u239f\u23a0 \u239e\u239c\u239d \u239b \u2202 \u2202 \u2202 \u2202 k e z T r Tr rr & h T\u221e egen qH z ro Hqz HrTk z rT &=\u2202 \u2202 =\u2202 \u2202 ),( 0)0,( ]),([ ),( 0),0( \u221e\u2212=\u2202 \u2202\u2212 =\u2202 \u2202 TzrTh r zrT k r zT o o 2-126E A large plane wall is subjected to a specified temperature on the left (inner) surface and solar radiation and heat loss by radiation to space on the right (outer) surface. The temperature of the right surface of the wall 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 wall is large relative to its thickness, and the thermal conditions on both sides of the wall are uniform. 3 Thermal properties are constant. 4 There is no heat generation in the wall. qsolar x 520 R L T2 Properties The properties of the plate are given to be k = 1.2 Btu/h\u22c5ft\u22c5°F and \u3b5 = 0.80, and 60.0=s\u3b1 . Analysis In steady operation, heat conduction through the wall must be equal to net heat transfer from the outer surface. Therefore, taking the outer surface temperature of the plate to be T2 (absolute, in R), solar 4 2 21 qATA L TTkA ssss &\u3b1\u3b5\u3c3 \u2212=\u2212 Canceling the area A and substituting the known quantities, )ftBtu/h 300(60.0)RftBtu/h 101714.0(8.0 ft 0.8 R) 520( )FftBtu/h 2.1( 242 4282 \u22c5\u2212\u22c5\u22c5×=\u2212°\u22c5\u22c5 \u2212 TT Solving for T2 gives the outer surface temperature to be T2 = 553.9 R Then the rate of heat transfer through the wall becomes 2ftBtu/h 50.9 \u22c5\u2212=\u2212°\u22c5\u22c5=\u2212= ft 0.8 R )9.553520()FftBtu/h 2.1(21 L TT kq& (per unit area) Discussion The negative sign indicates that the direction of heat transfer is from the outside to the inside. Therefore, the structure is gaining heat. 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. 2-61 2-127E A large plane wall is subjected to a specified temperature on the left (inner) surface and heat loss by radiation to space on the right (outer) surface. The temperature of the right surface of the wall 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 wall is large relative to its thickness, and the thermal conditions on both sides of the wall are uniform. 3 Thermal properties are constant. 4 There is no heat generation in the wall. Properties The properties of the plate are given to be k = 1.2 Btu/h\u22c5ft\u22c5°F and \u3b5 = 0.80. Analysis In steady operation, heat conduction through the wall must be equal to net heat transfer from the outer surface. Therefore, taking the outer surface temperature of the plate to be T2 (absolute, in R), 42 21 TA L TT kA ss \u3b5\u3c3=\u2212 Canceling the area A and substituting the known quantities, 42 4282 )RftBtu/h 101714.0(8.0 ft 0.5 R) 520( )FftBtu/h 2.1( T T \u22c5\u22c5×=\u2212°\u22c5\u22c5 \u2212 Solving for T2 gives the outer surface temperature to be T2 = 487.7 R x 520 R L T2 Then the rate of heat transfer through the wall becomes 2ftBtu/h 77.5 \u22c5=\u2212°\u22c5\u22c5=\u2212= ft 0.5 R )7.487520()FftBtu/h 2.1(21 L TT kq& (per unit area) Discussion The positive sign indicates that the direction of heat transfer is from the inside to the outside. Therefore, the structure 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. 2-62 2-128 A steam pipe is subjected to convection on both the inner and outer surfaces. The mathematical formulation of the problem and expressions for the variation of temperature in the pipe and on the outer surface temperature are to be obtained 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 center line. 2 Thermal conductivity is constant. 3 There is no heat generation in the pipe. Analysis (a) Noting that heat transfer is steady and one- dimensional in the radial r direction, the mathematical formulation of this problem can be expressed as r2 r r1Ti hi To ho 0=\u239f\u23a0 \u239e\u239c\u239d \u239b dr dTr dr d and )]([)( 11 rTThdr rdTk ii \u2212=\u2212 ])([)( 22 oo TrThdr rdTk \u2212=\u2212 (b) Integrating the differential equation once with respect to r gives 1Cdr dTr = Dividing both sides of the equation above by r to bring it to a readily integrable form and then integrating, r C dr dT 1= 21 ln)( CrCrT += where C1 and C2 are arbitrary constants. Applying the boundary conditions give r = r1: )]ln([ 211 1 1 CrCTh r Ck ii +\u2212=\u2212 r = r2: ])ln[( 221 2 1 oo TCrChr Ck \u2212+=\u2212 Solving for C1 and C2 simultaneously gives \u239f\u239f\u23a0 \u239e \u239c\u239c\u239d \u239b \u2212 ++ \u2212\u2212=\u239f\u239f\u23a0 \u239e \u239c\u239c\u239d \u239b \u2212\u2212= ++ \u2212= 1 1 211 2 0 1 112 211 2 0 1 ln ln ln and ln rh kr rh k rh k r r TTT rh krCTC rh k rh k r r TTC i oi i i i i oi i Substituting into the general solution and simplifying, we get the variation of temperature to be 21 and CC 211 2 11 1 111 ln ln )(lnln)( rh k rh k r r rh k r r T rh krCTrCrT oi i i i i ++ + +=\u2212\u2212+= (c) The outer surface temperature is determined by simply replacing r in the relation above by r2. We get 211 2 11 2 2 ln ln )( rh k rh k r r rh k r r TrT oi i i ++ + += 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. 3-1 Chapter 3 STEADY HEAT CONDUCTION Steady Heat Conduction in Plane Walls 3-1C (a) If the lateral surfaces of the rod are insulated, the heat transfer surface area of the cylindrical rod is the bottom or the top surface area of the rod, . (b) If the top and the bottom surfaces of the rod are insulated, the heat transfer area of the rod is the lateral surface area of the rod, 4/2DAs \u3c0= DLA \u3c0= . 3-2C In steady heat conduction, the rate of heat transfer into the wall is equal to the rate of heat transfer out of it. Also, the temperature at any point in the wall remains constant. Therefore, the energy content of the wall does not change during steady heat conduction. However, the temperature along the wall and thus the energy content of the wall will change during transient conduction. 3-3C The temperature distribution in a plane wall will be a straight line during steady and one dimensional heat transfer with constant wall thermal conductivity. 3-4C The thermal resistance of a medium represents the resistance of that medium against heat transfer. 3-5C The combined heat transfer coefficient represents the combined effects of radiation and convection heat transfers on a surface, and is defined as hcombined = hconvection + hradiation. It offers the convenience of incorporating the effects of radiation in the convection heat transfer coefficient, and to ignore radiation in heat transfer calculations. 3-6C Yes. The convection resistance can be defined as the inverse of the convection heat transfer coefficient per unit surface area since it is defined as )/(1 hARconv = . 3-7C The convection and the radiation resistances at a surface are parallel since both the convection and radiation heat transfers occur simultaneously. 3-8C For a surface of A at which the convection and radiation heat transfer coefficients