第3章习题答案

3-4 Convection heat transfer through the wall is expressed as ()s .Q hA T T ∞=−. In steady heat transfer,heat transfer rate to the wall and from the wall are equal. Therefore at the outer surface which has convection heat transfer coefficient three times that of the inner surface will experience three times smaller temperature drop compared to the inner surface. Therefore, at the outer surface, the temperature will be closer to the surrounding air temperature.3-16 The wall of a refrigerator is constructed of fiberglass insulation sandwiched between two layers of sheet metal. The minimum thickness of insulation that needs to be used in the wall in order to avoid condensation on the outer surfaces is to be determined.Assumptions 1 Heat transfer through the refrigerator walls issteady since the temperatures of the food compartment and the kitchen air remain constant at the specified values. 2 Heat transferin one-dimensional. 3 Thermal conductivities are constant. 4 Heat transfer coefficients account for the radiation effects.Properties The thermal conductivities are given to beC 1.51W/m °⋅=k for sheet metal and C 0.035W/m °⋅ for fiberglass insulation.Analysis The minimum thickness of insulation can be determined by assuming the outer surface temperature of the refrigerator to be 10℃. In steady operation, the rate of heat transfer through the refrigerator wall is constant, and thus heat transfer between the room and the refrigerated space is equal to the heat transfer between the room and the outer surface of the refrigerator. Considering a unit surface area, 45W C 20))(25C)(1m (9W/m )(22,0.=°−°⋅=−=out s room T T A h QUsing the thermal resistance network, heat transfer between the room and the refrigerated space can be expressed as totalrefrigroom R T T Q −=. iinsulation metal refrigroom h k L k L h T T A Q 1)(2(21/0.+++−=Substituting Cm W C m W L C m W m C m W Cm W °⋅+°⋅+°⋅×+°⋅°−=22222/41/035.0/1.15001.02/91)325(/45Solving for L the minimum thickness of insulation is determined to be cm m L 45.00045.0==°C 23-17 A thin copper plate is sandwiched between two epoxy boards. The error involved in the total thermal resistance of the plate if the thermal contact conductances are ignored is to be determined.Assumptions 1 Steady operating conditions exist. 2 Heat transfer isone-dimensional since the plate is large. 3 Thermal conductivities areconstant.Properties The thermal conductivities are given to be C 386W/m °⋅=k for copper plate and C 0.26W/m °⋅=kfor epoxy board. The contact conductance at the interface of copper-epoxy layers is given to be C 6000W/m 2°⋅=c h .Analysis (a) The thermal resistances of different layers for unit surface area of 21m areC/W 1067.1)C)(1m (6000W/m 11422contact °×=°⋅==−c c A h R C/W 102.6)C)(1m (386W/m 0.001m 62plate °×=°⋅==−kA L R C/W 0.01923)C)(1m (0.26W/m 0.005m 2epoxy °=°⋅==kA L R The total thermal resistance isC/W 0.0387970.019232102.61067.122264epoxy plate contact total °=×+×+××=++=−−R R R Rthen the percent error involved in the total thermal resistance of the plate if the thermal contact resistances are ignored is determined to be%86.010*******.01067.121002Error %4total contact =×××=×=−R R Which is negligible.Heat flow3-19 A steam pipe covered with 3-cm thick glass wool insulation is subjected to convection on its surfaces. The rate of heat transfer per unit length and the temperature drops across the pipe and the insulation are to be determined.Assumptions 1 a 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 variation in the axial direction. 3 Thermal conductivities are constant. 4 The thermal contact resistance at the interface is negligible.Properties The thermal conductivities are given to be C 15W/m °⋅=k for steel and C 0.038W/m °⋅=k for glass wool insulation.Analysis (a) The inner and the outer surface areas of the insulated pipe per unit length are 2110.157m m)π(0.05m)(1===L D A π 200361.0)1)(06.0055.0(m m m L D A =+==ππ The individual thermal resistances are C/W 0.08)C)(0.157m (80W/m 112211.°=°⋅==A h R conv=0.00101=3.089C/W 0.1847)C)(0.361m (15W/m 112200.°=°⋅==A h R conv C/W 3.3550.18473.0890.001010.080.211.°=+++=+++=conv conv total R R R R RThen the steady rate of heat loss from the steam per m. pipe length becomes93.9W C/W3.355C5)(32021.=°°−=−=∞∞total R T T Q The temperature drops across the pipe and the insulation areC 0.095C/W)00101(93.9W)(0..°=°==Δpipe pipe R Q T C 290C/W)089(93.9W)(3..°=°==Δinsulation insulation R Q T3-22 A spherical ball is covered with 1-mm thick plastic insulation. It is to be determined if the plastic insulation on the ball will increase or decrease heat transfer from it. Assumptions 1 Heat transfer from the ball is steady since there is no indication of anychange with time. 2 Heat transfer is one-dimensional since there is thermal symmetryabout the midpoint. 3 Thermal properties are constant. 4 The thermal contact resistance at the interface is negligible.Properties The thermal conductivity of plastic cover is given to be C m W k °⋅=/13.0.Analysis The critical radius of plastic insulation for the spherical ball isSince the outer temperature of the ball with insulation is smaller than critical radius of insulation, plasticinsulation will increase heat transfer from the wire.3-25 Two cast iron steam pipes are connected to each other through two 1-cm thick flanges exposed to cold ambient air. The average outer surface temperature of the pipe, the fin efficiency, the rate of heat transfer from the flanges, and the equivalent pipe length of the flange for heat transfer are to be determined.Assumptions 1 steady operating conditions exist. 2 the temperature along the flanges (fins) varies in one directiononly (normal to the pipe). 3 the heat transfer coefficient is constant and uniform over the entire fin surface. 4 the thermal properties of the fins are constant. 5 the heattransfer coefficient accounts for the effect of radiation from the fins.Properties the thermal conductivity of the cast iron is given to be C 50W/m °⋅=k .Analysis (a) we treat the flanges as fins, the individual thermal resistances are21.73m 6m)π(0.092m)(===L D A i i π 200 1.88m )π(0.1m)(6m ===L D A π C/W 0.0032)C)(1.73m (180W/m 1122conv.i °=°⋅==i i A h R°CPlastic21cond ln()ln(5/4.6)0.00004C/W 22(50W/m C)(6πm)r r R kL π−===°⋅°C/W 0.0213)C)(1.88m (25W/m 112200cond.0°=°⋅==A h R C/W 0.02450.02130.000040.0032conv.0cond conv.i total °=++=++=R R R RThe rate of heat transfer and average outer surface temperature of the pipe areW 6737C0.0245C)21(200total 21.=°°−==∞−∞R T T Q C 175.4C/W)0213(7673W)(0.C 21conv.o .22conv.o22.°=°+°=+=⎯→⎯−=∞∞R Q T T R T T Q(b) The fin efficiency can be determined from Fig 3-31 to be :0.29C)(0.02m)(52W/m C25W/m m)00.02(0.05m 2(2=°⋅°⋅+=+=kt h t L ξ22221220.0597m 0.02m)2ππ(0.1m)(](0.05m)2ππ[(0.1m 2)(2=+−=+−=rt r r A fin ππThe heat transfer rate from the flanges is214.6WC )21)(175.4C)(0.0597m 0.88(25W/m )(22b fin fin fin.max .fin inned .=°−°⋅=−==∞T T hA Q Q f ηη(c) A 6-m long section of the steam pipe is losing heat at a rate of 7673 W or 7673/6 =1278.8 W per m length. Then for heat transfer purposes the flanges section is equivalent toEquivalent length=(214.6w)/(1278.8w/m)=0.1678m=16.78cmTherefore, the flanges acts like a fin and increases the heat transfer by 16.78/2=8.39time.88.0fin =η。