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16 Chapter Mechanisms of Heat Transfer 16-45E A 200-ft long section of a steam pipe passes through an open space at a specified temperature. The rate of heat loss from the steam pipe and the annual cost of this energy lost are to be determined. Assumptions 1 Steady operating conditions exist. 2 Heat transfer by radiation is disregarded. 3 The convection heat transfer coefficient is constant and uniform over the surface. Analysis (a) The rate of heat loss from the steam pipe is As DL (4 / 12 ft)(200 ft) 2 280 F D =4 in 209.4 ft 2 Qpipe hAs (Ts Tair ) = 289,000 Btu/h (6 Btu/h.ft . F)(209.4 ft )(280 50) F 2 L=200 ft Q Air,50 F (b) The amount of heat loss per year is Q Q t (289,000 Btu / h)(365 24 h / yr) 2.532 109 Btu / yr The amount of gas consumption per year in the furnace that has an efficiency of 86% is Annual Energy Loss 2.532 10 9 Btu/yr 1 therm 0.86 100,000 Btu 29,438 therms/yr Then the annual cost of the energy lost becomes Energy cost (Annual energy loss)(Unit cost of energy) = (29,438 therms / yr)($0.58 / therm) $17,074 / yr 16-46 A 4-m diameter spherical tank filled with liquid nitrogen at 1 atm and -196 C is exposed to convection with ambient air. The rate of evaporation of liquid nitrogen in the tank as a result of the heat transfer from the ambient air is to be determined. Assumptions 1 Steady operating conditions exist. 2 Heat transfer by radiation is disregarded. 3 The convection heat transfer coefficient is constant and uniform over the surface. 4 The temperature of the thin-shelled spherical tank is nearly equal to the temperature of the nitrogen inside. Properties The heat of vaporization and density of liquid nitrogen at 1 atm are given to be 198 kJ/kg and 810 kg/m3, respectively. Vapor Analysis The rate of heat transfer to the nitrogen tank is As D2 (4 m) 2 50.27 m 2 Air 20 C 1 atm Liquid N2 Q hAs (Ts Tair ) 271,430 W (25 W/m 2 . C) (50.27 m 2 )[20 ( 196)] C Then the rate of evaporation of liquid nitrogen in the tank is determined to be Q mh fg m Q h fg 271.430 kJ/s 198 kJ/kg 1.37 kg/s Q -196 C 16-20 Chapter 16 Mechanisms of Heat Transfer 16-47 A 4-m diameter spherical tank filled with liquid oxygen at 1 atm and -183 C is exposed to convection with ambient air. The rate of evaporation of liquid oxygen in the tank as a result of the heat transfer from the ambient air is to be determined. Assumptions 1 Steady operating conditions exist. 2 Heat transfer by radiation is disregarded. 3 The convection heat transfer coefficient is constant and uniform over the surface. 4 The temperature of the thin-shelled spherical tank is nearly equal to the temperature of the oxygen inside. Properties The heat of vaporization and density of liquid oxygen at 1 atm are given to be 213 kJ/kg and 1140 kg/m3, respectively. Vapor Analysis The rate of heat transfer to the oxygen tank is As D2 (4 m) 2 50.27 m 2 Air 20 C 1 atm Liquid O2 -183 C Q hAs (Ts Tair ) 255,120 W (25 W/m 2 . C) (50.27 m 2 )[20 ( 183)] C Then the rate of evaporation of liquid oxygen in the tank is determined to be Q mh fg m Q h fg 255120 kJ / s . 213 kJ / kg 1.20 kg / s Q 16-21 Chapter 16 Mechanisms of Heat Transfer 16-48 "GIVEN" D=4 "[m]" T_s=-196 "[C]" "T_air=20 [C], parameter to be varied" h=25 "[W/m^2-C]" "PROPERTIES" h_fg=198 "[kJ/kg]" "ANALYSIS" A=pi*D^2 Q_dot=h*A*(T_air-T_s) m_dot_evap=(Q_dot*Convert(J/s, kJ/s))/h_fg Tair [C] 0 2.5 5 7.5 10 12.5 15 17.5 20 22.5 25 27.5 30 32.5 35 mevap [kg/s] 1.244 1.26 1.276 1.292 1.307 1.323 1.339 1.355 1.371 1.387 1.403 1.418 1.434 1.45 1.466 16-22 Chapter 16 Mechanisms of Heat Transfer 16-49 A person with a specified surface temperature is subjected to radiation heat transfer in a room at specified wall temperatures. The rate of radiation heat loss from the person is to be determined. Assumptions 1 Steady operating conditions exist. 2 Heat transfer by convection is disregarded. 3 The emissivity of the person is constant and uniform over the exposed surface. Properties The average emissivity of the person is given to be 0.7. Analysis Noting that the person is completely enclosed by the surrounding surfaces, the net rates of radiation heat transfer from the body to the surrounding walls, ceiling, and the floor in both cases are (a) Tsurr = 300 K Qrad 4 As (Ts4 Tsurr ) Tsurr (0.7)(5.67 10 = 37.4 W 8 W/m 2 .K 4 )(1.7 m 2 )[(32 + 273) 4 (300 K) 4 ]K 4 (b) Tsurr = 280 K Q rad As (Ts4 = 169 W 4 Tsurr ) 8 Qrad 32 C (0.7)(5.67 10 W/m .K )(1.7 m )[(32 + 273) 2 4 2 4 (280 K) ]K 4 4 Discussion Note that the radiation heat transfer goes up by more than 4 times as the temperature of the surrounding surfaces drops from 300 K to 280 K. 16-23 Chapter 16 Mechanisms of Heat Transfer 16-50 A circuit board houses 80 closely spaced logic chips on one side, each dissipating 0.06 W. All the heat generated in the chips is conducted across the circuit board. The temperature difference between the two sides of the circuit board is to be determined. Assumptions 1 Steady operating conditions exist. 2 Thermal properties of the board are constant. 3 All the heat generated in the chips is conducted across the circuit board. Properties The effective thermal conductivity of the board is given to be k = 16 W/m C. Analysis The total rate of heat dissipated by the chips is Q 80 (0.06 W) 4.8 W Chips Then the temperature difference between the front and back surfaces of the board is A (012 m)(0.18 m) . Q kA T L T 0.0216 m2 QL kA (4.8 W)(0.003 m) (16 W/m. C)(0.0216 m 2 ) 0.042 C Q Discussion Note that the circuit board is nearly isothermal. 16-51 A sealed electronic box dissipating a total of 100 W of power is placed in a vacuum chamber. If this box is to be cooled by radiation alone and the outer surface temperature of the box is not to exceed 55 C, the temperature the surrounding surfaces must be kept is to be determined. Assumptions 1 Steady operating conditions exist. 2 Heat transfer by convection is disregarded. 3 The emissivity of the box is constant and uniform over the exposed surface. 4 Heat transfer from the bottom surface of the box to the stand is negligible. Properties The emissivity of the outer surface of the box is given to be 0.95. Analysis Disregarding the base area, the total heat transfer area of the electronic box is As (0.4 m)(0.4 m) 4 (0.2 m)(0.4 m) 0.48 m 2 The radiation heat transfer from the box can be expressed as Q rad 100 W As (Ts 4 Tsurr ) 8 4 (0.95)(5.67 10 W/m 2 .K 4 )(0.48 m 2 ) (55 273 K ) 4 Tsurr 4 100 W = 0.95 Ts =55 C which gives Tsurr = 296.3 K = 23.3 C. Therefore, the temperature of the surrounding surfaces must be less than 23.3 C. 16-24 Chapter 16 Mechanisms of Heat Transfer 16-52 Using the conversion factors between W and Btu/h, m and ft, and K and R, the Stefan-Boltzmann constant 567 10 8 W / m2 . K4 is to be expressed in the English unit, Btu / h. ft 2 . R 4 . . Analysis The conversion factors for W, m, and K are given in conversion tables to be 1 W = 3.41214 Btu / h 1 m = 3.2808 ft 1 K = 1.8 R Substituting gives the Stefan-Boltzmann constant in the desired units, 5.67 W / m 2 .K 4 = 5.67 3.41214 Btu / h (3.2808 ft) 2 (1.8 R) 4 0.171 Btu / h.ft 2 .R 4 16-53 Using the conversion factors between W and Btu/h, m and ft, and C and F, the convection coefficient in SI units is to be expressed in Btu/h.ft 2. F. Analysis The conversion factors for W and m are straightforward, and are given in conversion tables to be 1 W = 3.41214 Btu / h 1 m = 3.2808 ft The proper conversion factor between C into F in this case is 1 C = 1.8 F since the C in the unit W/m2. C represents per C change in temperature, and 1 C change in temperature corresponds to a change of 1.8 F. Substituting, we get 1 W / m2 . C = 3.41214 Btu / h (3.2808 ft) 2 (1.8 F) 01761 Btu / h.ft 2 . F . which is the desired conversion factor. Therefore, the given convection heat transfer coefficient in English units is h 20 W/m 2 . C = 20 0.1761 Btu/h.ft 2 . F 3.52 Btu/h.ft 2 . F 16-25 Chapter 16 Mechanisms of Heat Transfer Simultaneous Heat Transfer Mechanisms 16-54C All three modes of heat transfer can not occur simultaneously in a medium. A medium may involve two of them simultaneously. 16-55C (a) Conduction and convection: No. (b) Conduction and radiation: Yes. Example: A hot surface on the ceiling. (c) Convection and radiation: Yes. Example: Heat transfer from the human body. 16-56C The human body loses heat by convection, radiation, and evaporation in both summer and winter. In summer, we can keep cool by dressing lightly, staying in cooler environments, turning a fan on, avoiding humid places and direct exposure to the sun. In winter, we can keep warm by dressing heavily, staying in a warmer environment, and avoiding drafts. 16-57C The fan increases the air motion around the body and thus the convection heat transfer coefficient, which increases the rate of heat transfer from the body by convection and evaporation. In rooms with high ceilings, ceiling fans are used in winter to force the warm air at the top downward to increase the air temperature at the body level. This is usually done by forcing the air up which hits the ceiling and moves downward in a gently manner to avoid drafts. 16-58 The total rate of heat transfer from a person by both convection and radiation to the surrounding air and surfaces at specified temperatures is to be determined. Assumptions 1 Steady operating conditions exist. 2 The person is completely surrounded by the interior surfaces of the room. 3 The surrounding surfaces are at the same temperature as the air in the room. 4 Heat conduction to the floor through the feet is negligible. 5 The convection coefficient is constant and uniform over the entire surface of the person. Properties The emissivity of a person is given to be = 0.9. Analysis The person is completely enclosed by the surrounding surfaces, and he or she will lose heat to the surrounding air by convection, and to the surrounding surfaces by radiation. The total rate of heat loss from the person is determined from Qrad 4 As (Ts4 Tsurr ) = 84.8 W (0.90)(5.67 10 8 W/m 2 .K 4 )(1.7 m 2 )[(32 + 273) 4 (23 + 273) 4 ]K 4 Qconv hAs T (5W/m 2 K)(1.7m 2 )(32 23) C 76.5W Tsurr 23 C and Qtotal Qconv Qrad 84.8 765 161.3 W . Qrad 32 C =0.9 Qconv Discussion Note that heat transfer from the person by evaporation, which is of comparable magnitude, is not considered in this problem. 16-26 Chapter 16 Mechanisms of Heat Transfer 16-59 Two large plates at specified temperatures are held parallel to each other. The rate of heat transfer between the plates is to be determined for the cases of still air, regular insulation, and super insulation between the plates. Assumptions 1 Steady operating conditions exist since the plate temperatures remain constant. 2 Heat transfer is one-dimensional since the plates are large. 3 The surfaces are black and thus = 1. 4 There are no convection currents in the air space between the plates. Properties The thermal conductivities are k = 0.00015 W/m C for super insulation, k = 0.01979 W/m C at -50 C (Table A-22) for air, and k = 0.036 W/m C for fiberglass insulation (Table A-28). Analysis (a) Disregarding any natural convection currents, the rates of conduction and radiation heat transfer Qcond Q rad Q total kA T1 T2 (290 150) K (0.01979 W/m 2 . C)(1 m 2 ) L 0.02 m 4 4 As (T1 T2 ) (150 K ) 4 139 W T1 T2 Q 372 W 1(5.67 10 8 W/m 2 .K 4 )(1m 2 ) (290 K ) 4 Qcond Q rad 139 372 511 W (b) When the air space between the plates is evacuated, there will be radiation heat transfer only. Therefore, Qtotal Qrad 372 W 2 cm (c) In this case there will be conduction heat transfer through the fiberglass insulation only, Qtotal Qcond kA T1 T2 L (0.036 W / m.o C)(1 m2 ) (290 150) K 0.02 m 252 W (d) In the case of superinsulation, the rate of heat transfer will be Qtotal Qcond kA T1 T2 L (0.00015 W / m. C)(1 m2 ) (290 150) K 0.02 m 1.05 W Discussion Note that superinsulators are very effective in reducing heat transfer between to surfaces. 16-27 Chapter 16 Mechanisms of Heat Transfer 16-60 The convection heat transfer coefficient for heat transfer from an electrically heated wire to air is to be determined by measuring temperatures when steady operating conditions are reached and the electric power consumed. Assumptions 1 Steady operating conditions exist since the temperature readings do not change with time. 2 Radiation heat transfer is negligible. Analysis In steady operation, the rate of heat loss from the wire equals the rate of heat generation in the wire as a result of resistance heating. That is, Q E generated VI (110 V)(3 A) = 330 W D =0.2 cm 240 C The surface area of the wire is As ( D) L (0.002 m)(1.4 m) = 0.00880 m 2 L = 1.4 m Q Air, 20 C The Newton's law of cooling for convection heat transfer is expressed as Q hAs (Ts T ) Disregarding any heat transfer by radiation , the convection heat transfer coefficient is determined to be h Q As (T1 T ) 330 W (0.00880 m )(240 20) C 2 170.5 W/m 2 . C Discussion If the temperature of the surrounding surfaces is equal to the air temperature in the room, the value obtained above actually represents the combined convection and radiation heat transfer coefficient. 16-28 Chapter 16 Mechanisms of Heat Transfer 16-61 "GIVEN" L=1.4 "[m]" D=0.002 "[m]" T_infinity=20 "[C]" "T_s=240 [C], parameter to be varied" V=110 "[Volt]" I=3 "[Ampere]" "ANALYSIS" Q_dot=V*I A=pi*D*L Q_dot=h*A*(T_s-T_infinity) h [W/m2.C] 468.9 375.2 312.6 268 234.5 208.4 187.6 170.5 156.3 144.3 134 Ts [C] 100 120 140 160 180 200 220 240 260 280 300 16-29 Chapter 16 Mechanisms of Heat Transfer 16-62E A spherical ball whose surface is maintained at a temperature of 170 F is suspended in the middle of a room at 70 F. The total rate of heat transfer from the ball is to be determined. Assumptions 1 Steady operating conditions exist since the ball surface and the surrounding air and surfaces remain at constant temperatures. 2 The thermal properties of the ball and the convection heat transfer coefficient are constant and uniform. Properties The emissivity of the ball surface is given to be = 0.8. Analysis The heat transfer surface area is As = D = (2/12 ft) = 0.08727 ft Under steady conditions, the rates of convection and radiation heat transfer are Qconv Q rad Air 70 F 170 F D = 2 in Q hAs T (12Btu/h.ft 2 F)(0.08727ft 2 )(170 70) F 104.7 Btu/h 8 As (Ts4 To4 ) 0.8(0.08727ft 2 )(0.1714 10 9.4 Btu/h Btu/h.ft 2 R 4 )[(170 + 460R) 4 (70 + 460R) 4 ] Therefore, Qtotal Qconv Qrad 104.7 9.4 114.1 Btu / h Discussion Note that heat loss by convection is several times that of heat loss by radiation. The radiation heat loss can further be reduced by coating the ball with a low-emissivity material. 16-30 Chapter 16 Mechanisms of Heat Transfer 16-63 A 1000-W iron is left on the iron board with its base exposed to the air at 20 C. The temperature of the base of the iron is to be determined in steady operation. Assumptions 1 Steady operating conditions exist. 2 The thermal properties of the iron base and the convection heat transfer coefficient are constant and uniform. 3 The temperature of the surrounding surfaces is the same as the temperature of the surrounding air. Properties The emissivity of the base surface is given to be = 0.6. Analysis At steady conditions, the 1000 W energy supplied to the iron will be dissipated to the surroundings by convection and radiation heat transfer. Therefore, Qtotal where Qconv Qconv Qrad 1000 W hAs T (35 W/m 2 K)(0.02 m 2 )(Ts 293 K) 0.7(Ts 293 K) W and Q rad As (Ts4 To4 ) 0.6(0.02m 2 )(5.67 10 (293K) 4 ] W 8 W/m 2 K 4 )[Ts4 (293K) 4 ] 0.06804 10 8 [Ts4 Substituting, 1000 W 0.7(Ts 293 K) 0.06804 10 8 [Ts4 (293 K) 4 ] Iron 1000 W Solving by trial and error gives Ts 947 K 674 C Discussion We note that the iron will dissipate all the energy it receives by convection and radiation when its surface temperature reaches 947 K. 16-64 A spacecraft in space absorbs radiation solar while losing heat to deep space by thermal radiation. The surface temperature of the spacecraft is to be determined when steady conditions are reached.. Assumptions 1 Steady operating conditions exist since the surface temperatures of the wall remain constant at the specified values. 2 Thermal properties of the wall are constant. Properties The outer surface of a spacecraft has an emissivity of 0.8 and an absorptivity of 0.3. Analysis When the heat loss from the outer surface of the spacecraft by radiation equals the solar radiation absorbed, the surface temperature can be determined from Qsolar absorbed Qsolar 0.3 As (950W/m ) 2 Qrad 4 As (Ts4 Tspace ) 950 W/m2 8 2 0.8 As (5.67 10 W/m K 4 )[Ts4 (0K) ] 4 Canceling the surface area A and solving for Ts gives Ts 281.5 K = 0.3 = 0.8 . Qrad 16-31 Chapter 16 Mechanisms of Heat Transfer 16-65 A spherical tank located outdoors is used to store iced water at 0 C. The rate of heat transfer to the iced water in the tank and the amount of ice at 0 C that melts during a 24-h period are to be determined. Assumptions 1 Steady operating conditions exist since the surface temperatures of the wall remain constant at the specified values. 2 Thermal properties of the tank and the convection heat transfer coefficient is constant and uniform. 3 The average surrounding surface temperature for radiation exchange is 15 C. 4 The thermal resistance of the tank is negligible, and the entire steel tank is at 0 C. Properties The heat of fusion of water at atmospheric pressure is hif outer surface of the tank is 0.6. Analysis (a) The outer surface area of the spherical tank is As D2 (3.02 m) 2 28.65 m 2 333. 7 kJ / kg . The emissivity of the Then the rates of heat transfer to the tank by convection and radiation become Qconv Q rad hAs (T As Q 4 (Tsurr Ts ) Q rad (30 W/m 2 . C)(28.65 m 2 )(25 0) C (0.6)(28.65 m )(5.67 10 22,780 W 2 -8 2 21,488 W (273 K ) 4 ] 1292 W Ts4 ) W/m .K 4 )[(288 K) 4 Q total conv 21,488 1292 (b) The amount of heat transfer during a 24-hour period is Q Q t (22.78 kJ/s)(24 3600 s) 1,968,000 kJ Air 25 C Iced water 0C 0 C 1 cm Then the amount of ice that melts during this period becomes Q mhif m Q hif 1,968,000 kJ 333.7 kJ/kg 5898 kg Q Discussion The amount of ice that melts can be reduced to a small fraction by insulating the tank. 16-32 Chapter 16 Mechanisms of Heat Transfer 16-66 The roof of a house with a gas furnace consists of a 15-cm thick concrete that is losing heat to the outdoors by radiation and convection. The rate of heat transfer through the roof and the money lost through the roof that night during a 14 hour period are to be determined. Assumptions 1 Steady operating conditions exist. 2 The emissivity and thermal conductivity of the roof are constant. Properties The thermal conductivity of the concrete is given to be k = 2 W/m C. The emissivity of the outer surface of the roof is given to be 0.9. Analysis In steady operation, heat transfer from the outer surface of the roof to the surroundings by convection and radiation must be equal to the heat transfer through the roof by conduction. That is, Q Qroof, cond Qroof to surroundings, conv+rad The inner surface temperature of the roof is given to be Ts,in = 15 C. Letting Ts,out denote the outer surface temperatures of the roof, the energy balance above can be expressed as Q kA Ts,in Ts,out L ho A(Ts,out Tsurr ) A (Ts,out 4 Tsurr 4 ) Tsky = 255 K Q Q (2 W / m. C)(300 m2 ) 2 15 C Ts,out 015 m . (15 W / m . C)(300 m )(Ts,out 10) C 2 (0.9)(300 m2 )(5.67 10 8 W / m2 .K 4 ) (Ts,out 273 K) 4 (255 K) 4 Solving the equations above using an equation solver (or by trial and error) gives Q 25,450 W and Ts, out 8.64 C Then the amount of natural gas consumption during a 16-hour period is E gas Qtotal 0.85 Q t 0.85 (25.450 kJ/s)(14 3600 s) 1 therm 0.85 105,500 kJ 14.3 therms Finally, the money lost through the roof during that period is Money lost (14.3 therms)($0.60 / therm) $8.58 16-33 Chapter 16 Mechanisms of Heat Transfer 16-67E A flat plate solar collector is placed horizontally on the roof of a house. The rate of heat loss from the collector by convection and radiation during a calm day are to be determined. Assumptions 1 Steady operating conditions exist. 2 The emissivity and convection heat transfer coefficient are constant and uniform. 3 The exposed surface, ambient, and sky temperatures remain constant. Properties The emissivity of the outer surface of the collector is given to be 0.9. Analysis The exposed surface area of the collector is As (5 ft)(15 ft) 75 ft 2 Tsky = 50 F Air, 70 F Solar collector 5625 Btu/h 2 Q Noting that the exposed surface temperature of the collector is 100 F, the total rate of heat loss from the collector the environment by convection and radiation becomes Qconv Q rad hAs (T As Ts ) (2.5 Btu/h.ft 2 . F)(75 ft 2 )(100 70) F (0.9)(75 ft )(0.1714 10 2 -8 4 (Tsurr Ts4 ) Btu/h.ft .R 4 )[(100 460 R) 4 (50 460 R ) 4 ] 3551 Btu/h and Qtotal Qconv Qrad 5625 3551 9176 Btu / h 16-34 Chapter 16 Mechanisms of Heat Transfer Review Problems 16-68 A standing man is subjected to high winds and thus high convection coefficients. The rate of heat loss from this man by convection in still air at 20 C, in windy air, and the wind-chill factor are to be determined. Assumptions 1 A standing man can be modeled as a 30-cm diameter, 170-cm long vertical cylinder with both the top and bottom surfaces insulated. 2 The exposed surface temperature of the person and the convection heat transfer coefficient is constant and uniform. 3 Heat loss by radiation is negligible. Analysis The heat transfer surface area of the person is As = DL = (0.3 m)(1.70 m) = 1.60 m The rate of heat loss from this man by convection in still air is Qstill air = hAs T = (15 W/m C)(1.60 m )(34 - 20) C = 336 W In windy air it would be Qwindy air = hAs T = (50 W/m C)(1.60 m )(34 - 20) C = 1120 W To lose heat at this rate in still air, the air temperature must be 1120 W = (hAs T)still air = (15 W/m C)(1.60 m )(34 - Teffective) C which gives Teffective = -12.7 C That is, the windy air at 20 C feels as cold as still air at -12.7 C as a result of the wind-chill effect. Therefore, the wind-chill factor in this case is Fwind-chill = 20 - (-12.7) = 32.7 C Windy weather 16-69 The backside of the thin metal plate is insulated and the front side is exposed to solar radiation. The surface temperature of the plate is to be determined when it stabilizes. Assumptions 1 Steady operating conditions exist. 2 Heat transfer through the insulated side of the plate is negligible. 3 The heat transfer coefficient is constant and uniform over the plate. 4 Radiation heat transfer is negligible. Properties The solar absorptivity of the plate is given to be = 0.7. Analysis When the heat loss from the plate by convection equals the solar radiation absorbed, the surface temperature of the plate can be determined from Qsolar absorbed Qsolar 0.7 A 700W/m 2 Qconv hAs (Ts (30W/m To ) 2 700 W/m2 = 0.7 air, 10 C . Qrad C) As (Ts 10) Canceling the surface area As and solving for Ts gives Ts 26.3 C 16-35 Chapter 16 Mechanisms of Heat Transfer 16-70 A room is to be heated by 1 ton of hot water contained in a tank placed in the room. The minimum initial temperature of the water is to be determined if it to meet the heating requirements of this room for a 24-h period. Assumptions 1 Water is an incompressible substance with constant specific heats. 2 Air is an ideal gas with constant specific heats. 3 The energy stored in the container itself is negligible relative to the energy stored in water. 4 The room is maintained at 20 C at all times. 5 The hot water is to meet the heating requirements of this room for a 24-h period. Properties The specific heat of water at room temperature is C = 4.18 kJ/kg C (Table A-15). Analysis Heat loss from the room during a 24-h period is Qloss = (10,000 kJ/h)(24 h) = 240,000 kJ Taking the contents of the room, including the water, as our system, the energy balance can be written as E in E out Net energy transfer by heat, work, and mass E system Change in internal, kinetic, potential, etc. energies Qout U U water U 0 air 10,000 kJ/h or -Qout = [mC(T2 - T1)]water Substituting, -240,000 kJ = (1000 kg)(4.18 kJ/kg C)(20 - T1) It gives T1 = 77.4 C where T1 is the temperature of the water when it is first brought into the room. water 20 C 16-71 The base surface of a cubical furnace is surrounded by black surfaces at a specified temperature. The net rate of radiation heat transfer to the base surface from the top and side surfaces is to be determined. Assumptions 1 Steady operating conditions exist. 2 The top and side surfaces of the furnace closely approximate black surfaces. 3 The properties of the surfaces are constant. Properties The emissivity of the base surface is = 0.7. Black furnace 1200 K Analysis The base surface is completely surrounded by the top and side surfaces. Then using the radiation relation for a surface completely surrounded by another large (or black) surface, the net rate of radiation heat transfer from the top and side surfaces to the base is determined to be Qrad, base 4 4 A (Tbase Tsurr ) Base, 800 K (0.7)(3 3 m 2 )(5.67 10 -8 W/m 2 .K 4 )[(1200 K) 4 594,400 W (800 K) 4 ] 16-36 Chapter 16 Mechanisms of Heat Transfer 16-72 A refrigerator consumes 600 W of power when operating, and its motor remains on for 5 min and then off for 15 min periodically. The average thermal conductivity of the refrigerator walls and the annual cost of operating this refrigerator are to be determined. Assumptions 1 Quasi-steady operating conditions exist. 2 The inner and outer surface temperatures of the refrigerator remain constant. Analysis The total surface area of the refrigerator where heat transfer takes place is Atotal 2 (18 12) (18 0.8) (12 0.8) . . . . 9.12 m2 Since the refrigerator has a COP of 2.5, the rate of heat removal from the refrigerated space, which is equal to the rate of heat gain in steady operation, is Q We COP (600 W) 2.5 1500 W But the refrigerator operates a quarter of the time (5 min on, 15 min off). Therefore, the average rate of heat gain is Qave Q / 4 (1500 W) / 4 = 375 W Then the thermal conductivity of refrigerator walls is determined to be Qave kA Tave L k Qave L A Tave (375 W)(0.03 m) (9.12 m2 )(17 6) C 0.112 W / m. C The total number of hours this refrigerator remains on per year is t 365 24 / 4 2190 h Then the total amount of electricity consumed during a one-year period and the annular cost of operating this refrigerator are Annual Electricit y Usage We t (0.6 kW)(2190 h/yr) 1314 kWh/yr $105.1/yr Annual cost (1314 kWh/yr )($0.08 / kWh ) 16-37 Chapter 16 Mechanisms of Heat Transfer 16-73 Engine valves are to be heated in a heat treatment section. The amount of heat transfer, the average rate of heat transfer, the average heat flux, and the number of valves that can be heat treated daily are to be determined. Assumptions Constant properties given in the problem can be used. Properties The average specific heat and density of valves are given to be Cp = 440 J/kg. C and kg/m3. = 7840 Analysis (a) The amount of heat transferred to the valve is simply the change in its internal energy, and is determined from Q U mC p (T2 T1 ) (0.0788 kg)(0.440 kJ/kg. C)(800 - 40) C = 26.35 kJ (b) The average rate of heat transfer can be determined from Qave Q t 26.35 kJ 5 60 s 0.0878 kW 87.8 W Engine valve T1 = 40 C T2 = 800 C D = 0.8 cm L = 10 cm (c) The average heat flux is determined from q ave Qave As Qave 2 DL 87.8 W 2 (0.008 m)(0.1m) 1.75 10 4 W/m 2 (d) The number of valves that can be heat treated daily is Number of valves (10 60 min)(25 valves) (5 min) 3000 valves 16-38 Chapter 16 Mechanisms of Heat Transfer 16-74 The glass cover of a flat plate solar collector with specified inner and outer surface temperatures is considered. The fraction of heat lost from the glass cover by radiation is to be determined. Assumptions 1 Steady operating conditions exist since the surface temperatures of the glass remain constant at the specified values. 2 Thermal properties of the glass are constant. Properties The thermal conductivity of the glass is given to be k = 0.7 W/m C. Analysis Under steady conditions, the rate of heat transfer through the glass by conduction is Qcond kA T L (0.7 W/m C)(2.2 m 2 ) (28 25) C 0.006 m 770 W The rate of heat transfer from the glass by convection is Qconv hA T (10 W/m 2 C)(2.2 m 2 )(25 15) C 220 W Q Under steady conditions, the heat transferred through the cover by conduction should be transferred from the outer surface by convection and radiation. That is, 28 C L=0.6 cm 25 C Air, 15 C h=10 W/m2. C A = 2.2 m2 Qrad Qcond Qconv 770 220 550 W Then the fraction of heat transferred by radiation becomes f Qrad Q cond 550 770 0.714 (or 71.4%) 16-39 Chapter 16 Mechanisms of Heat Transfer 16-75 The range of U-factors for windows are given. The range for the rate of heat loss through the window of a house is to be determined. Assumptions 1 Steady operating conditions exist. 2 Heat losses associated with the infiltration of air through the cracks/openings are not considered. Analysis The rate of heat transfer through the window can be determined from Qwindow U overall Awindow (Tin Tout ) Window Q where Ti and To are the indoor and outdoor air temperatures, respectively, Uoverall is the U-factor (the overall heat transfer coefficient) of the window, and Awindow is the window area. Substituting, Maximum heat loss: Minimum heat loss: Q window, max Q window, min (6.25 W/m 2 (1.25 W/m 2 C)(1.2 1.8 m 2 )[20 ( 8)] C C)(1.2 1.8 m 2 )[20 ( 8)] C 20 C -8 C 378 W 76 W Discussion Note that the rate of heat loss through windows of identical size may differ by a factor of 5, depending on how the windows are constructed. 16-40 Chapter 16 Mechanisms of Heat Transfer 16-76 "GIVEN" A=1.2*1.8 "[m^2]" T_1=20 "[C]" T_2=-8 "[C]" "U=1.25 [W/m^2-C], parameter to be varied" "ANALYSIS" Q_dot_window=U*A*(T_16-T_2) U [W/m2.C] 1.25 1.75 2.25 2.75 3.25 3.75 4.25 4.75 5.25 5.75 6.25 Q window [W] 75.6 105.8 136.1 166.3 196.6 226.8 257 287.3 317.5 347.8 378 16-41 Chapter 16 Mechanisms of Heat Transfer 16-77 The windows of a house in Atlanta are of double door type with wood frames and metal spacers. The average rate of heat loss through the windows in winter is to be determined. Assumptions 1 Steady operating conditions exist. 2 Heat losses associated with the infiltration of air through the cracks/openings are not considered. Analysis The rate of heat transfer through the window can be determined from Q window, ave Window U overall Awindow (Tin Tout , ave ) Q 22 C 11.3 C where Ti and To are the indoor and outdoor air temperatures, respectively, Uoverall is the U-factor (the overall heat transfer coefficient) of the window, and Awindow is the window area. Substituting, Q window, ave (2.50 W/m 2 C)(20 m 2 )(22 11.3) C 535 W Discussion This is the "average" rate of heat transfer through the window in winter in the absence of any infiltration. 16-78 Boiling experiments are conducted by heating water at 1 atm pressure with an electric resistance wire, and measuring the power consumed by the wire as well as temperatures. The boiling heat transfer coefficient is to be determined. Assumptions 1 Steady operating conditions exist. 2 Heat losses from the water container are negligible. Analysis The heat transfer area of the heater wire is A DL (0.002 m)(0.50 m) 0.003142 m2 Noting that 4100 W of electric power is consumed when the heater surface temperature is 130 C, the boiling heat transfer coefficient is determined from Newton's law of cooling Q hA(Ts Tsat ) to be h Q A(Ts Tsat ) 4100 W (0.003142 m 2 )(130 100) C 43,500 W/m 2 C Water 100 C Heater 130 C 16-79 . . . 16-81 Design and Essay Problems 16-42

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