Transferência de Calor e Evaporação de Nitrogênio Líquido em Tanque Esférico - Análise de Isolamento

386 A spherical tank filled with liquid nitrogen at 1 atm and 196C is exposed to convection and radiation with the surrounding air and surfaces The rate of evaporation of liquid nitrogen in the tank as a result of the heat gain from the surroundings for the cases of no insulation 5cm thick fiberglass insulation and 2cm thick superinsulation are to be determined Assumptions 1 Heat transfer is steady since the specified thermal conditions at the boundaries do not change with time 2 Heat transfer is onedimensional since there is thermal symmetry about the midpoint 3 The combined heat transfer coefficient is constant and uniform over the entire surface 4 The temperature of the thinshelled spherical tank is said to be nearly equal to the temperature of the nitrogen inside and thus thermal resistance of the tank and the internal convection resistance are negligible Properties The heat of vaporization and density of liquid nitrogen at 1 atm are given to be 198 kJkg and 810 kgm³ respectively The thermal conductivities are given to be k 0035 WmC for fiberglass insulation and k 000005 WmC for super insulation Analysis a The heat transfer rate and the rate of evaporation of the liquid without insulation are A πD² π3 m² 2827 m² Ro 1 hoA 1 35 Wm²C2827 m² 000101 CW Q Ts1 T2 Ro 15 196C 000101 CW 208910 W Q mhfg m Q hfg 208910 kJs 198 kJkg 1055 kgs b The heat transfer rate and the rate of evaporation of the liquid with a 5cm thick layer of fiberglass insulation are A πD² π31 m² 3019 m² Ro 1 hoA 1 35 Wm²C3019 m² 0000946 CW Rinsulation r2 r1 4πk r1 r2 155 15 m 4π0035 WmC155 m15 m 00489 CW Rtotal Ro Rinsulation 0000946 00489 00498 CW Q Ts1 T2 Rtotal 15 196C 00498 CW 4233 W Q mhfg m Q hfg 4233 kJs 198 kJkg 00214 kgs c The heat transfer rate and the rate of evaporation of the liquid with 2cm thick layer of superinsulation is A πD² π304 m² 2903 m² Ro 1 hoA 1 35 Wm²C2903 m² 0000984 CW Rinsulation r2 r1 4πk r1 r2 152 15 m 4π000005 WmC152 m15 m 1396 CW Rtotal Ro Rinsulation 0000984 1396 1396 CW Q Ts1 T2 Rtotal 15 196C 1396 CW 1511 W Q mhfg m Q hfg 001511 kJs 198 kJkg 0000076 kgs 387 A spherical tank filled with liquid oxygen at 1 atm and 183C is exposed to convection and radiation with the surrounding air and surfaces The rate of evaporation of liquid oxygen in the tank as a result of the heat gain from the surroundings for the cases of no insulation 5cm thick fiberglass insulation and 2cm thick superinsulation are to be determined Assumptions 1 Heat transfer is steady since the specified thermal conditions at the boundaries do not change with time 2 Heat transfer is onedimensional since there is thermal symmetry about the midpoint 3 The combined heat transfer coefficient is constant and uniform over the entire surface 4 The temperature of the thinshelled spherical tank is said to be nearly equal to the temperature of the oxygen inside and thus thermal resistance of the tank and the internal convection resistance are negligible Properties The heat of vaporization and density of liquid oxygen at 1 atm are given to be 213 kJkg and 1140 kgm³ respectively The thermal conductivities are given to be k 0035 WmC for fiberglass insulation and k 000005 WmC for super insulation Analysis a The heat transfer rate and the rate of evaporation of the liquid without insulation are A πD² π3 m² 2827 m² Ro 1 hoA 1 35 Wm²C2827 m² 000101 CW Q Ts1 T2 Ro 15 183C 000101 CW 196040 W Q mhfg m Q hfg 196040 kJs 213 kJkg 0920 kgs b The heat transfer rate and the rate of evaporation of the liquid with a 5cm thick layer of fiberglass insulation are A πD² π31 m² 3019 m² Ro 1 hoA 1 35 Wm²C3019 m² 0000946 CW Rinsulation r2 r1 4πk r1 r2 155 15 m 4π0035 WmC155 m15 m 00489 CW Rtotal Ro Rinsulation 0000946 00489 00498 CW Q Ts1 T2 Rtotal 15 183C 00498 CW 3976 W Q mhfg m Q hfg 3976 kJs 213 kJkg 00187 kgs c The heat transfer rate and the rate of evaporation of the liquid with a 2cm superinsulation is A πD² π304 m² 2903 m² Ro 1 hoA 1 35 Wm²C2903 m² 0000984 CW Rinsulation r2 r1 4πk r1 r2 152 15 m 4π000005 WmC152 m15 m 1396 CW Rtotal Ro Rinsulation 0000984 1396 1396 CW Q Ts1 T2 Rtotal 15 183C 1396 CW 1418 W Q mhfg m Q hfg 001418 kJs 213 kJkg 0000067 kgs 388 An electric wire is tightly wrapped with a 1mm thick plastic cover The interface temperature and the effect of doubling the thickness of the plastic cover on the interface temperature are to be determined Assumptions 1 Heat transfer is steady since there is no indication of any change with time 2 Heat transfer is onedimensional since there is thermal symmetry about the centerline and no variation in the axial direction 3 Thermal properties are constant 4 The thermal contact resistance at the interface is negligible 5 Heat transfer coefficient accounts for the radiation effects if any Properties The thermal conductivity of plastic cover is given to be k 015 WmC Analysis In steady operation the rate of heat transfer from the wire is equal to the heat generated within the wire Q Wg VI 8 V13 A 104 W The total thermal resistance is Rconv 1 hoAo 1 24 Wm²Cπ00042 m14 m 02256 CW Rplastic lnr2 r1 2πkL ln21 11 2π015 WmC14 m 00490 CW Rtotal Rconv Rplastic 02256 00490 02746 CW Then the interface temperature becomes Q T1 T2 Rtotal T1 T QRtotal 30C 104 W02746 CW 586C The critical radius of plastic insulation is rcr k h 015 WmC 24 Wm²C 000625 m 625 mm Doubling the thickness of the plastic cover will increase the outer radius of the wire to 3 mm which is less than the critical radius of insulation Therefore doubling the thickness of plastic cover will increase the rate of heat loss and decrease the interface temperature 389 To avoid condensation on the outer surface the necessary thickness of the insulation around a copper pipe that carries liquid oxygen is to be determined Assumptions 1 Heat conduction is steady and onedimensional 2 Thermal properties are constant 3 Thermal contact resistance is negligible Properties The thermal conductivities of the copper pipe and the insulation are given to be 400 Wm C and 005 Wm C respectively Analysis From energy balance and using the thermal resistance concept the following equation is expressed Copy the following line and paste on a blank EES screen to solve the above equation 202002010120D3lnD325e32005ln25202400112020e3 Solving by EES software the outer diameter of the insulation is D3 00839 m The thickness of the insulation necessary to avoid condensation on the outer surface is t D3 D22 00839 m 0025 m2 00295 m Discussion If the insulation thickness is less than 295 mm the outer surface temperature would decrease to the dew point at 10 C where condensation would occur b The total amount of heat transfer during a 24hour period and the amount of ice that will melt during this period are Q Q Δt 64600 kJs24 3600 s 5581 106 kJ mice Qhlf 5581 106 kJ3337 kJkg 16730 kg Check The outer surface temperature of the tank is Q hconvrad Ao T1 Ts Ts T1 Q hconvrad Ao 25C 64600 W 10 5424 Wm2 C20257 m2 43C which is very close to the assumed temperature of 5C for the outer surface temperature used in the evaluation of the radiation heat transfer coefficient Therefore there is no need to repeat the calculations 373 A steam pipe covered with 3cm 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 Heat transfer is steady since there is no indication of any change with time 2 Heat transfer is onedimensional 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 k 15 WmC for steel and k 0038 WmC for glass wool insulation Analysis The inner and the outer surface areas of the insulated pipe per unit length are The individual thermal resistances are Then the steady rate of heat loss from the steam per m pipe length becomes The temperature drops across the pipe and the insulation are 378 Chilled water is flowing inside a pipe The thickness of the insulation needed to reduce the temperature rise of water to onefourth of the original value is to be determined Assumptions 1 Heat transfer is steady since there is no indication of any change with time 2 Heat transfer is onedimensional since there is thermal symmetry about the centerline 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 conductivity is given to be k 005 WmC for insulation Analysis The rate of heat transfer without the insulation is Qold mcpΔT 098 kgs4180 Jkg C8 7C 4096 W The total resistance in this case is Qold T TwRtotal 4096 W 30 75C Rtotal Rtotal 0005493CW The convection resistance on the outer surface is Ro 1 ho Ao 1 9 Wm2 Cπ004 m200 m 0004421 CW The rest of thermal resistances are due to convection resistance on the inner surface and the resistance of the pipe and it is determined from R1 Rtotal Ro 0005493 0004421 0001072 CW The rate of heat transfer with the insulation is Qnew mcp ΔT 098 kgs4180 Jkg C025C 1024 W The total thermal resistance with the insulation is Qnew T TwRtotalnew 1024 W 30 7 7252C Rtotalnew Rtotalnew 002234CW It is expressed by Rtotalnew R1 Ronew Rins R1 1ho Ao lnD2 D1 2πkins L 002234CW 0001072 19 Wm2 CπD2 200 m lnD2 004 2π005 Wm C200 m Solving this equation by trialerror or by using an equation solver such as EES we obtain D2 01406 m The following line in EES is used 002234 0001072 19piD2200 lnD2004 2pi005200 Then the required thickness of the insulation becomes tins D2 D12 01406 0042 00503 m 503 cm 379 Steam flows in a steel pipe which is insulated by gypsum plaster The rate of heat transfer from the steam and the temperature on the outside surface of the insulation are be determined Assumptions 1 Heat transfer is steady since there is no indication of any change with time 2 Heat transfer is onedimensional since there is thermal symmetry about the centerline and no variation in the axial direction 3 Thermal conductivities are constant 4 The thermal contact resistance at the interface is negligible Properties a The thermal conductivities of steel and gypsum plaster are given to be 50 and 05 WmC respectively Analysis The thermal resistances are Ri 1 hi Ai 1 800 Wm2 Cπ006 m20 m 00003316 CW Rsteel lnD2 D1 2πksteel L ln8 6 2π50 Wm C20 m 00000458 CW Rins lnD3 D2 2πkins L ln16 8 2π05 Wm C20 m 0011032 CW Ro 1 ho Ao 1 200 Wm2 Cπ016 m20 m 00004974 CW The total thermal resistance and the rate of heat transfer are Rtotal Ri Rsteel Rins Ro 00003316 00000458 0011032 00004974 0011907 CW Q Ti To Rtotal 200 10C 0011907 m2 CW 15957 W b The temperature at the outer surface of the insulation is determined from Q Ts To Ro 15957 W Ts 10C 00004974 m2 CW Ts 179C 380E A steam pipe covered with 2in thick fiberglass insulation is subjected to convection on its surfaces The rate of heat loss from the steam per unit length and the error involved in neglecting the thermal resistance of the steel pipe in calculations are to be determined Assumptions 1 Heat transfer is steady since there is no indication of any change with time 2 Heat transfer is onedimensional 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 k 87 BtuhftF for steel and k 0020 BtuhftF for fiberglass insulation Analysis The inner and outer surface areas of the insulated pipe are Ai πDi L π35 12 ft1 ft 0916 ft2 Ao πDo L π8 12 ft1 ft 2094 ft2 The individual resistances are Ri 1 hi Ai 1 30 Btuhft2 F0916 ft2 0036 h FBtu R1 Rpipe lnr2 r1 2πk1 L ln2 175 2π87 BtuhftF1 ft 0002 h FBtu R2 Rinsulation lnr3 r2 2πk2 L ln4 2 2π0020 BtuhftF1 ft 5516 h FBtu Ro 1 ho Ao 1 5 Btuhft2 F2094 ft2 0096 h FBtu Rtotal Ri R1 R2 Ro 0036 0002 5516 0096 565 h FBtu Then the steady rate of heat loss from the steam per ft pipe length becomes Q T1 T2 Rtotal 450 55F 565 h FBtu 6991 Btuh If the thermal resistance of the steel pipe is neglected the new value of total thermal resistance will be Rtotal R1 R2 Ro 0036 5516 0096 5648 hFBtu Then the percentage error involved in calculations becomes error 565 5648 hFBtu 565 hFBtu 100 0035 which is insignificant 381 Hot water is flowing through a 15m section of a cast iron pipe The pipe is exposed to cold air and surfaces in the basement The rate of heat loss from the hot water and the average velocity of the water in the pipe as it passes through the basement are to be determined Assumptions 1 Heat transfer is steady since there is no indication of any change with time 2 Heat transfer is onedimensional since there is thermal symmetry about the center line and no variation in the axial direction 3 Thermal properties are constant Properties The thermal conductivity and emissivity of cast iron are given to be k 52 WmC and ε 07 Analysis The individual resistances are Ai πDiL π004 m15 m 1885 m2 Ao πDoL π0046 m15 m 2168 m2 Ri 1hiAi 1120 Wm2C1885 m2 000442 CW Rpipe lnr2 r1 2πk1L ln23 2 2π52 WmC15 m 000003 CW The outer surface temperature of the pipe will be somewhat below the water temperature Assuming the outer surface temperature of the pipe to be 80C we will check this assumption later the radiation heat transfer coefficient is determined to be hrad εσT2o2 T2surrT2 Tsurr 07567 108 Wm2K4353 K2 283 K2353 283 5167 Wm2K Since the surrounding medium and surfaces are at the same temperature the radiation and convection heat transfer coefficients can be added and the result can be taken as the combined heat transfer coefficient Then hcombined hrad hconv2 5167 15 2017 Wm2C Ro 1hcombinedAo 12017 Wm2C2168 m2 002287 CW Rtotal Ri Rpipe Ro 000442 000003 002287 002732 CW The rate of heat loss from the hot water pipe then becomes Q To1 To2 Rtotal 90 10C 002732 CW 2928 W For a temperature drop of 3C the mass flow rate of water and the average velocity of water must be Q m cp ΔT m Q cp ΔT 2928 Js 4180 JkgC3 C 02335 kgs m ρVAc V m ρAc 02335 kgs 1000 kgm3π004 m24 0186 ms Discussion The outer surface temperature of the pipe is Q To1 Ts Ri Rpipe 2928 W 90 TsC 000442 000003CW Ts 770C which is close to the value assumed for the surface temperature in the evaluation of the radiation resistance Therefore there is no need to repeat the calculations 382 Hot water is flowing through a 15 m section of a copper pipe The pipe is exposed to cold air and surfaces in the basement The rate of heat loss from the hot water and the average velocity of the water in the pipe as it passes through the basement are to be determined Assumptions 1 Heat transfer is steady since there is no indication of any change with time 2 Heat transfer is onedimensional since there is thermal symmetry about the centerline and no variation in the axial direction 3 Thermal properties are constant Properties The thermal conductivity and emissivity of copper are given to be k 386 WmC and ε 07 Analysis The individual resistances are Ai πDiL π004 m15 m 1885 m2 Ao πDoL π0046 m15 m 2168 m2 Ri 1hiAi 1120 Wm2C1885 m2 000442 CW Rpipe lnr2 r1 2πkL ln23 2 2π386 WmC15 m 00000038 CW The outer surface temperature of the pipe will be somewhat below the water temperature Assuming the outer surface temperature of the pipe to be 80C we will check this assumption later the radiation heat transfer coefficient is determined to be hrad εσT2o2 Tsurr2T2 Tsurr 07567 108 Wm2K4353 K2 283 K2353 283 5167 Wm2K Since the surrounding medium and surfaces are at the same temperature the radiation and convection heat transfer coefficients can be added and the result can be taken as the combined heat transfer coefficient Then hcombined hrad hconv2 5167 15 2017 Wm2C Ro 1hcombinedAo 12017 Wm2C2168 m2 002287 CW Rtotal Ri Rpipe Ro 000442 00000038 002287 002733 CW The rate of heat loss from the hot water pipe then becomes Q To1 To2 Rtotal 90 10C 002733 CW 2927 W For a temperature drop of 3C the mass flow rate of water and the average velocity of water must be Q m cp ΔT m Q cp ΔT 2927 Js 4180 JkgC3 C 02334 kgs m ρVAc V m ρAc 02334 kgs 1000 kgm3π004 m24 0186 ms Discussion The outer surface temperature of the pipe is Q To1 Ts Ri Rpipe 2927 W 90 TsC 000442 00000038CW Ts 771C 384E Steam exiting the turbine of a steam power plant at 100F is to be condensed in a large condenser by cooling water flowing through copper tubes For specified heat transfer coefficients and 001in thick scale build up on the inner surface the length of the tube required to condense steam at a rate of 120 lbmh is to be determined Assumptions 1 Heat transfer is steady since there is no indication of any change with time 2 Heat transfer is onedimensional since there is thermal symmetry about the centerline and no variation in the axial direction 3 Thermal properties are constant 4 Heat transfer coefficients are constant and uniform over the surfaces Properties The thermal conductivities are given to be k 223 BtuhftF for copper tube and be k 05 BtuhftF for the mineral deposit The heat of vaporization of water at 100F is given to be 1037 Btulbm Analysis When a 001in thick layer of deposit forms on the inner surface of the pipe the inner diameter of the pipe will reduce from 04 in to 038 in The individual thermal resistances are Ai πDiL π038 12 ft1 ft 0099 ft2 Ao πDoL π06 12 ft1 ft 0157 ft2 Ri 1hiAi 135 Btuhft2F0099 ft2 02886 hFBtu Rpipe lnr2 r1 2πkL ln3 2 2π223 BtuhftF1 ft 000029 hFBtu Rdeposit lnr1 rdep 2πk2L ln02 019 2π05 BtuhftF1 ft 001633 hFBtu Ro 1hoAo 11500 Btuhft2F0157 ft2 000425 hFBtu Rtotal Ri Rpipe Rdeposit Ro 02886 000029 001633 000425 03095 hFBtu The heat transfer rate per ft length of the tube is Q To1 To2 Rtotal 100 70F 03095 FBtu 969 Btuh The total rate of heat transfer required to condense steam at a rate of 120 lbmh and the length of the tube required can be determined to be Qtotal m hg 120 lbmh1037 Btulbm 124440 Btuh Tube length Qtotal Q 124440 969 1284 ft 365 In an experiment the convection heat transfer coefficients of a air and b water flowing over the metal foil are to be determined Assumptions 1 Steady operating conditions exist 2 Heat transfer is onedimensional 3 Thermal properties are constant 4 Thermal resistance of the thin metal foil is negligible Properties Thermal conductivity of the slab is given to be k 0023 Wm K and the emissivity of the metal foil is 002 Analysis The thermal resistances are Rcond LkA Rconv 1hA and Rrad 1hradA From energy balance and using the thermal resistance concept the following equation is expressed T T1Rconv Tsurr T1Rrad qelecA T1 T2Rcond or 1Rconv T1 T2Rcond Tsurr T1Rrad qelecA 1T T1 h T1 T2Lk Tsurr T11hrad qelec 1T T1 a For air flowing over the metal foil the radiation heat transfer coefficient is hrad eσTs² Tsurr²Ts Tsurr 00256710⁸ Wm² K⁴423² 293²K²423 293K 0215 Wm² K The convection heat transfer coefficient for air flowing over the metal foil is h 150 20 K0025 m0023 Wm K 20 150 K10215 Wm² K 5000 Wm² 120 150 K 373 Wm² K b For water flowing over the metal foil the radiation heat transfer coefficient is hrad eσTs² Tsurr²Ts Tsurr 00256710⁸ Wm² K⁴303² 293²K²303 293K 01201 Wm² K The convection heat transfer coefficient for water flowing over the metal foil is h 30 20 K0025 m0023 Wm K 20 30 K101201 Wm² K 5000 Wm² 120 30 K 499 Wm² K Discussion If heat transfer by conduction through the slab and radiation on the metal foil surface is neglected the convection heat transfer coefficient for the case with air flow would deviate by 32 from the result in part a while the convection heat transfer coefficient for the case with water flow would deviate by 02 from the result in part b 366 A kiln is made of 20 cm thick concrete walls and ceiling The two ends of the kiln are made of thin sheet metal covered with 2cm thick styrofoam For specified indoor and outdoor temperatures the rate of heat transfer from the kiln is to be determined Assumptions 1 Heat transfer is steady since there is no indication of change with time 2 Heat transfer through the walls and ceiling is onedimensional 3 Thermal conductivities are constant 4 Heat transfer coefficients account for the radiation heat transfer 5 Heat loss through the floor is negligible 6 Thermal resistance of sheet metal is negligible Properties The thermal conductivities are given to be k 09 Wm C for concrete and k 0033 Wm C for styrofoam insulation Analysis In this problem there is a question of which surface area to use We will use the outer surface area for outer convection resistance the inner surface area for inner convection resistance and the average area for the conduction resistance Or we could use the inner or the outer surface areas in the calculation of all thermal resistances with little loss in accuracy For top and the two side surfaces Ri 1hiAi 13000 Wm² C40 m13 12 m 0007110⁴ CW Rconcrete LkAave 02 m09 Wm C40 m13 06 m 448010⁴ CW Ro 1hoAo 125 Wm² C40 m13 m 076910⁴ CW Rtotal Ri Rconcrete Ro 00071 4480 076910⁴ 525610⁴ CW and Qtopsides Tin ToutRtotal 40 4C 525610⁴ CW 83700 W Heat loss through the end surface of the kiln with styrofoam Ri 1hiAi 13000 Wm² C4 045 04 m² 020110⁴ CW Rstyrofoam LkAave 002 m0033 Wm C4 025 02 m² 00332 CW Ro 1hoAo 125 Wm² C4 5 m² 00020 CW Rtotal Ri Rstyrofoam Ro 020110⁴ 00332 00020 00352 CW and Qend surface Tin ToutRtotal 40 4C 00352 CW 1250 W Then the total rate of heat transfer from the kiln becomes Qtotal Qtopsides 2Qside 83700 2 1250 86200 W 372 A spherical container filled with iced water is subjected to convection and radiation heat transfer at its outer surface The rate of heat transfer and the amount of ice that melts per day are to be determined Assumptions 1 Heat transfer is steady since the specified thermal conditions at the boundaries do not change with time 2 Heat transfer is onedimensional since there is thermal symmetry about the midpoint 3 Thermal conductivity is constant Properties The thermal conductivity of steel is given to be k 15 Wm C The heat of fusion of water at 1 atm is hif 3337 kJkg The outer surface of the tank is black and thus its emissivity is e 1 Analysis a The inner and the outer surface areas of sphere are Ai πDi² π8 m² 20106 m² Ao πDo² π803 m² 20257 m² We assume the outer surface temperature T2 to be 5C after comparing convection heat transfer coefficients at the inner and the outer surfaces of the tank With this assumption the radiation heat transfer coefficient can be determined from hrad eσT2² Tsurr²T2 Tsurr 156710⁸ Wm² K⁴273 5 K² 273 25 K²273 25 K273 5 K 5424 Wm² K The individual thermal resistances are Rconvi 1hiA 180 Wm² C20106 m² 0000062 CW R1 Rsphere r2 r14πk r1 r2 4015 40 m 4π15 Wm C4015 m40 m 0000005 CW Rconvo 1hoA 110 Wm² C20257 m² 0000494 CW Rrad 1hradA 15424 Wm² C20257 m² 0000910 CW 1Reqv 1Rconvo 1Rrad 10000494 10000910 Reqv 0000320 CW Rtotal Rconvi R1 Reqv 0000062 0000005 0000320 0000387 CW Then the steady rate of heat transfer to the iced water becomes Q T1 T2 Rtotal 25 0C 0000387 CW 64600 W 4th wall with double pane windows Rglass Rair Rglass Ri Rwall Ro Rwall Lwall kA R value A 231 m²CW 20 4 512 18m² 0033382 CW Rglass Lglass kA 0005 m 078 Wm²C12 18m² 0002968 CW Rair Lair kA 0015 m 0026 Wm²C12 18m² 0267094 CW Rwindow 2Rglass Rair 2 0002968 0267094 027303 CW 1 Reqv 1 Rwall 5 1 Rwindow 1 0033382 5 027303 Reqv 0020717 CW Rtotal Ri Reqv Ro 0001786 0020717 0000694 0023197 CW Then Q T1 T2 Rtotal 24 8C 0023197 CW 690 W The rate of heat transfer which will be saved if the single pane windows are converted to double pane windows is Qsave Qsinglepane Qdoublepane 5224 690 4534 W The amount of energy and money saved during a 7month long heating season by switching from single pane to double pane windows become Qsave Qsave Δt 4534 kW7 30 24 h 22851 kWh Money savings Energy savedUnit cost of energy 22851 kWh008kWh 1828 337 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 is steady since the temperatures of the food compartment and the kitchen air remain constant at the specified values 2 Heat transfer is onedimensional 3 Thermal conductivities are constant 4 Heat transfer coefficients account for the radiation effects Properties The thermal conductivities are given to be k 151 WmC for sheet metal and 0035 WmC for fiberglass insulation Analysis The minimum thickness of insulation can be determined by assuming the outer surface temperature of the refrigerator to be 20C 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 Q hoATroom Ts out 9 Wm²C1 m²24 20C 36 W Using the thermal resistance network heat transfer between the room and the refrigerated space can be expressed as Q Troom Trefrig Rtotal QA Troom Trefrig 1ho 2Lkmetal Lkinsulation 1hi Substituting 36 Wm² 24 2C 19 Wm²C 20001 m 151 Wm²C L 0035 Wm²C 1 4 Wm²C Solving for L the minimum thickness of insulation is determined to be L 000875 m 0875 cm 341 Warm air blowing over the inner surface of an automobile windshield is used for defrosting ice accumulated on the outer surface The convection heat transfer coefficient for the warm air blowing over the inner surface of the windshield necessary to cause the accumulated ice to begin melting is to be determined Assumptions 1 Steady operating conditions exist 2 Heat transfer through the windshield is onedimensional 3 Thermal properties are constant 4 Heat transfer by radiation is negligible 5 The automobile is operating at 1 atm Properties Thermal conductivity of the windshield is given to be k 14 WmC Analysis The thermal resistances are Ri 1 hiA Ro 1 hoA and Rwin L kA From energy balance and using the thermal resistance concept the following equation is expressed T o T1 Ro T1 T i Rwin Ri Ri T1 T i T o T1 Ro Rwin or 1 hi T1 T i T o T1 1 ho L k For the ice to begin melting the outer surface temperature of the windshield T1 should be at least 0 C The convection heat transfer coefficient for the warm air is hi T1 T i T o T1 1 ho L k¹ 0 25 C 10 0 C 1 200 Wm²C 0005 m 14 WmC¹ 112 Wm²C Discussion In practical situations the ambient temperature and the convective heat transfer coefficient outside the automobile vary with weather conditions and the automobile speed Therefore the convection heat transfer coefficient of the warm air necessary to melt the ice should be varied as well This is done by adjusting the warm air flow rate and temperature 342 The thermal contact conductance for an aluminum plate attached on a copper plate that is heated electrically is to be determined Assumptions 1 Steady operating conditions exist 2 Heat transfer is onedimensional 3 Thermal properties are constant 4 Heat transfer by radiation is negligible Properties The thermal conductivity of the aluminum plate is given to be 235 WmC Analysis The thermal resistances are Rcond L kA and Rconv 1 hA From energy balance and using the thermal resistance concept the following equation is expressed qelec A T1 T Rc A Rcond Rconv or qelec A T1 T Rc A LkA 1hA Rearranging the equation and solving for the contact resistance yields Rc T1 T qelec Lk 1h 100 20C 5300 Wm² 0025 m 235 WmC 1 67 Wm²C 625810⁵ m²CW The thermal contact conductance is hc 1 Rc 16000 Wm²C Discussion By comparing the value of the thermal contact conductance with the values listed in Table 32 the surface conditions of the plates appear to be milled 350 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 is onedimensional since the plate is large 3 Thermal conductivities are constant Properties The thermal conductivities are given to be k 386 WmC for copper plates and k 026 WmC for epoxy boards The contact conductance at the interface of copperepoxy layers is given to be hc 6000 Wm²C Analysis The thermal resistances of different layers for unit surface area of 1 m² are Rcontact 1hc Ac 16000 Wm²C1 m² 000017 CW Rplate LkA 0001 m 386 WmC1 m² 2610⁶ CW Repoxy LkA 0007 m 026 WmC1 m² 002692 CW The total thermal resistance is Rtotal 2Rcontact Rplate 2Repoxy 2000017 2610⁶ 2002692 005419 CW Then the percent error involved in the total thermal resistance of the plate if the thermal contact resistances are ignored is determined to be Error 2Rcontact Rtotal 100 2000017 005419 100 063 which is negligible 356 A wall consists of horizontal bricks separated by plaster layers There are also plaster layers on each side of the wall and a rigid foam on the inner side of the wall The rate of heat transfer through the wall is to be determined Assumptions 1 Heat transfer is steady since there is no indication of change with time 2 Heat transfer through the wall is onedimensional 3 Thermal conductivities are constant 4 Heat transfer by radiation is disregarded Properties The thermal conductivities are given to be k 072 WmC for bricks k 022 WmC for plaster layers and 0026 WmC for the rigid foam Analysis We consider 1 m deep and 028 m high portion of wall which is representative of the entire wall The thermal resistance network and individual resistances are Ri Rconv1 1h1 A 110 Wm²C0281 m² 0357 CW R1 Rfoam LkA 002 m 0026 WmC0281 m² 2747 CW R2 R6 Rplaster side LkA 002 m 022 WmC0281 m² 0325 CW R3 R5 Rplaster center Lho A 015 m 022 WmC00151 m² 4545 CW R4 Rbrick LkA 015 m 072 WmC0251 m² 0833 CW Ro Rconv2 1h2 A 120 Wm²C0281 m² 0179 CW 1Rmid 1R3 1R4 1R5 14545 10833 14545 Rmid 0804 CW Rtotal Ri R1 2R2 Rmid Ro 0357 2747 20325 0804 0179 4737 CW The steady rate of heat transfer through the wall per 028 m² is Q T1 T2 Rtotal 22 4C 4737 CW 549 W Then steady rate of heat transfer through the entire wall becomes Qtotal 549 W 46 m² 028 m² 470 W 328 A circuit board houses 100 chips each dissipating 006 W The surface heat flux the surface temperature of the chips and the thermal resistance between the surface of the board and the cooling medium are to be determined Assumptions 1 Steady operating conditions exist 2 Heat transfer from the back surface of the board is negligible 2 Heat is transferred uniformly from the entire front surface Analysis a The heat flux on the surface of the circuit board is As 012 m018 m 00216 m² q Q As 100 x 006 W 00216 m² 278 Wm² b The surface temperature of the chips is Q hAsTs T Ts T Q hAs 40C 100 x 006 W 10 Wm²C00216 m² 678C c The thermal resistance is Rconv 1 hAs 1 10 Wm²C00216 m² 463CW 329 A person is dissipating heat at a rate of 150 W by natural convection and radiation to the surrounding air and surfaces For a given deep body temperature the outer skin temperature is to be determined Assumptions 1 Steady operating conditions exist 2 The heat transfer coefficient is constant and uniform over the entire exposed surface of the person 3 The surrounding surfaces are at the same temperature as the indoor air temperature 4 Heat generation within the 05cm thick outer layer of the tissue is negligible Properties The thermal conductivity of the tissue near the skin is given to be k 03 WmC Analysis The skin temperature can be determined directly from Q kA T1 Tskin L Tskin T1 QL kA 37C 150 W0005 m 03 WmC17 m² 355C 330 A doublepane window is considered The rate of heat loss through the window and the temperature difference across the largest thermal resistance are to be determined Assumptions 1 Steady operating conditions exist 2 Heat transfer coefficients are constant Properties The thermal conductivities of glass and air are given to be 078 WmK and 0025 Wm K respectively Analysis a The rate of heat transfer through the window is determined to be Q AΔT 1hi Lskg Laka Lskg 1ho 1 x 15 m²20 20C 140 Wm²C 0004 m078 WmC 0005 m0025 WmC 0004 m078 WmC 120 Wm²C 1 x 15 m²20 20C 0025 0000513 02 0000513 005 210 W b Noting that the largest resistance is through the air gap the temperature difference across the air gap is determined from ΔTa Q Ra Q Laka A 210 W0005 m0025 WmC1 x 15 m² 28C 336 Two of the walls of a house have no windows while the other two walls have single or doublepane windows The average rate of heat transfer through each wall and the amount of money this household will save per heating season by converting the single pane windows to double pane windows are to be determined Assumptions 1 Heat transfer through the window is steady since the indoor and outdoor temperatures remain constant at the specified values 2 Heat transfer is onedimensional since any significant temperature gradients will exist in the direction from the indoors to the outdoors 3 Thermal conductivities of the glass and air are constant 4 Heat transfer by radiation is disregarded Properties The thermal conductivities are given to be k 0026 Wm C for air and 078 WmC for glass Analysis The rate of heat transfer through each wall can be determined by applying thermal resistance network The convection resistances at the inner and outer surfaces are common in all cases Walls without windows Ri 1h1A 17 Wm2 C10 4 m2 0003571 CW Rwall LwallkA RvalueA 231 m2 CW10 4 m2 005775 CW Ro 1h0A 118 Wm2 C10 4 m2 0001389 CW Rtotal Ri Rwall Ro 0003571 005775 0001389 006271 CW Then Q T1 T2Rtotal 24 8C006271 CW 2551 W Wall with single pane windows Ri 1h1A 17 Wm2 C20 4 m2 0001786 CW Rwall LwallkA RvalueA 231 m2 CW20 4 512 18 m2 0033382 CW Rglass LglasskA 0005 m078 Wm2 C12 18 m2 0002968 CW 1Requiv 1Rwall 5 1Rglass 10033382 50002968 Requiv 0000583 CW Ro 1h0A 118 Wm2 C20 4 m2 0000694 CW Rtotal Ri Requiv Ro 0001786 0000583 0000694 0003063 CW Then Q T1 T2Rtotal 24 8C0003063 CW 5224 W 320 A doublepane window consists of two layers of glass separated by a stagnant air space For specified indoors and outdoors temperatures the rate of heat loss through the window and the inner surface temperature of the window are to be determined Assumptions 1 Heat transfer through the window is steady since the indoor and outdoor temperatures remain constant at the specified values 2 Heat transfer is onedimensional since any significant temperature gradients will exist in the direction from the indoors to the outdoors 3 Thermal conductivities of the glass and air are constant 4 Heat transfer by radiation is negligible Properties The thermal conductivity of the glass and air are given to be kglass 078 Wm C and kair 0026 Wm C Analysis The area of the window and the individual resistances are A 15 m 24 m 36 m2 Ri Rconv 1 1h1A 110 Wm2 C36 m2 002778 CW R1 R3 Rglass L1k1A 0003 m078 WmC36 m2 000107 CW R2 Rair L2k2A 0012 m0026 WmC36 m2 012821 CW Ro Rconv 2 1h2A 125 Wm2 C36 m2 001111 CW Rtotal Rconv 1 2R1 R2 Rconv 2 002778 2000107 012821 001111 016924 CW The steady rate of heat transfer through window glass then becomes Q T1 T2Rtotal 21 5C016924 CW 154 W The inner surface temperature of the window glass can be determined from Q T1 T1Rconv 1 T1 T1 Q Rconv 1 21C 154 W002778 CW 167C 321 A doublepane window consists of two layers of glass separated by an evacuated space For specified indoors and outdoors temperatures the rate of heat loss through the window and the inner surface temperature of the window are to be determined Assumptions 1 Heat transfer through the window is steady since the indoor and outdoor temperatures remain constant at the specified values 2 Heat transfer is onedimensional since any significant temperature gradients will exist in the direction from the indoors to the outdoors 3 Thermal conductivity of the glass is constant 4 Heat transfer by radiation is negligible Properties The thermal conductivity of the glass is given to be kglass 078 Wm C Analysis Heat cannot be conducted through an evacuated space since the thermal conductivity of vacuum is zero no medium to conduct heat and thus its thermal resistance is zero Therefore if radiation is disregarded the heat transfer through the window will be zero Then the answer of this problem is zero since the problem states to disregard radiation Discussion In reality heat will be transferred between the glasses by radiation We do not know the inner surface temperatures of windows In order to determine radiation heat resistance we assume them to be 5C and 15C respectively and take the emissivity to be 1 Then individual resistances are A 15 m 24 m 36 m2 Ri Rconv 1 1h1A 110 Wm2 C36 m2 002778 CW R1 R3 Rglass L1k1A 0003 m078 WmC36 m2 000107 CW Rrad 1εσAT52 Tsur2T5 Tsur 1567 108 Wm2 K436 m22882 2782288 278K3 005402 CW Ro Rconv 2 1h2A 125 Wm2 C36 m2 001111 CW Rtotal Rconv 1 2R1 Rrad Rconv 2 002778 2000107 005402 001111 009505 CW The steady rate of heat transfer through window glass then becomes Q T1 T2Rtotal 21 5C009505 CW 274 W The inner surface temperature of the window glass can be determined from Q T1 T1Rconv 1 T1 T1 Q Rconv 1 21C 274 W002778 CW 134C Similarly the inner surface temperatures of the glasses are calculated to be 131 and 17 C we had assumed them to be 15 and 5 C when determining the radiation resistance We can improve the result obtained by reevaluating the radiation resistance and repeating the calculations 324 A cylindrical resistor on a circuit board dissipates 015 W of power steadily in a specified environment The amount of heat dissipated in 24 h the surface heat flux and the surface temperature of the resistor are to be determined Assumptions 1 Steady operating conditions exist 2 Heat is transferred uniformly from all surfaces of the resistor Analysis a The amount of heat this resistor dissipates during a 24hour period is Q Q dot Δ t 015 W24 h 36 Wh b The heat flux on the surface of the resistor is As 2 πD² 4 πDL 2 π0003 m² 4 π 0003 m0012 m 0000127 m² q dot Q dot As 015 W 0000127 m² 1179 Wm² c The surface temperature of the resistor can be determined from Q dot hAs Ts T Ts T Q dot hAs 35C 015 W 9 Wm²C0000127 m² 166C 325 A very thin transparent heating element is attached to the inner surface of an automobile window for defogging purposes the inside surface temperature of the window is to be determined Assumptions 1 Steady operating conditions exist 2 Heat transfer through the window is onedimensional 3 Thermal properties are constant 4 Heat transfer by radiation is negligible 5 Thermal resistance of the thin heating element is negligible Properties Thermal conductivity of the window is given to be k 12 Wm C Analysis The thermal resistances are Ri 1 hi A Ro 1 ho A and Rwin L kA From energy balance and using the thermal resistance concept the following equation is expressed Ti T1 Ri q doth A T1 To Rwin Ro or Ti T1 1hi A q doth A T1 To LkA 1ho A Ti T11hi q doth T1 To Lk 1ho 22C T1 115 Wm² C 1300 Wm² T1 5C 0005 m 12 Wm C 1100 Wm² C Copy the following line and paste on a blank EES screen to solve the above equation 22T1115 1300 T15000512 1100 Solving by EES software the inside surface temperature of the window is T1 149 C Discussion In actuality the ambient temperature and the convective heat transfer coefficient outside the automobile vary with weather conditions and the automobile speed To maintain the inner surface temperature of the window it is necessary to vary the heat flux to the heating element according to the outside condition 326 A process of bonding a transparent film onto a solid plate is taking place inside a heated chamber The temperatures inside the heated chamber and on the transparent film surface are to be determined Assumptions 1 Steady operating conditions exist 2 Heat transfer is onedimensional 3 Thermal properties are constant 4 Heat transfer by radiation is negligible 5 Thermal contact resistance is negligible Properties The thermal conductivities of the transparent film and the solid plate are given to be 005 Wm C and 12 Wm C respectively Analysis The thermal resistances are Rconv 1hA Rf Lf kf A and Rs Ls ks A Using the thermal resistance concept the following equation is expressed T Tb Rconv Rf Tb T2 Rs Rearranging and solving for the temperature inside the chamber yields T Tb T2 Rs Rconv Rf Tb 70 52C 0013 m 12 Wm C 1 70 Wm² C 0001 m 005 Wm C 70C 127C The surface temperature of the transparent film is T1 Tb Rf Tb T2 Rs T1 Tb T2 Rs Rf Tb Tb T2 Ls ks Lf kf Tb T1 70 52C 0013 m 12 Wm C 0001 m 005 Wm C 70C 103C Discussion If a thicker transparent film were to be bonded on the solid plate then the inside temperature of the heated chamber would have to be higher to maintain the temperature of the bond at 70C 327 A power transistor dissipates 015 W of power steadily in a specified environment The amount of heat dissipated in 24 h the surface heat flux and the surface temperature of the resistor are to be determined Assumptions 1 Steady operating conditions exist 2 Heat is transferred uniformly from all surfaces of the transistor Analysis a The amount of heat this transistor dissipates during a 24hour period is Q Q Δt 015 W24 h 36 Wh 00036 kWh b The heat flux on the surface of the transistor is As 2 πD²4 πDL 2 π 0005 m²4 π0005 m0004 m 00001021 m² q QAs 015 W00001021 m² 1469 Wm² c The surface temperature of the transistor can be determined from Q hAsTs T Ts T QhAs 30C 015 W 18 Wm² C00001021 m² 1116C 319 The two surfaces of a window are maintained at specified temperatures The rate of heat loss through the window and the inner surface temperature are to be determined Assumptions 1 Heat transfer through the window is steady since the surface temperatures remain constant at the specified values 2 Heat transfer is onedimensional since any significant temperature gradients will exist in the direction from the indoors to the outdoors 3 Thermal conductivity is constant 4 Heat transfer by radiation is negligible Properties The thermal conductivity of the glass is given to be k 078 Wm C Analysis The area of the window and the individual resistances are A 15 m 24 m 36 m² Ri Rconv1 1h1A 110 Wm² C36 m² 002778 CW Rglass Lk1A 0006 m 078 Wm C36 m² 000214 CW Ro Rconv2 1h2A 125 Wm² C36 m² 001111 CW Rtotal Rconv1 Rglass Rconv2 002778 000214 001111 004103 CW The steady rate of heat transfer through window glass is then Q T1 T2Rtotal 24 5C 004103 CW 707 W The inner surface temperature of the window glass can be determined from Q T1 T1Rconv1 T1 T1 QRconv1 24C 707 W002778 CW 44C 317 The two surfaces of a wall are maintained at specified temperatures The rate of heat loss through the wall is to be determined Assumptions 1 Heat transfer through the wall is steady since the surface temperatures remain constant at the specified values 2 Heat transfer is onedimensional since any significant temperature gradients will exist in the direction from the indoors to the outdoors 3 Thermal conductivity is constant Properties The thermal conductivity is given to be k 08 Wm C Analysis The surface area of the wall and the rate of heat loss through the wall are A 3 m 6 m 18 m² Q kA T1 T2L 08 Wm C18 m²14 5C 025 m 518 W 318 Heat is transferred steadily to the boiling water in an aluminum pan The inner surface temperature of the bottom of the pan is given The boiling heat transfer coefficient and the outer surface temperature of the bottom of the pan are to be determined Assumptions 1 Steady operating conditions exist 2 Heat transfer is onedimensional since the thickness of the bottom of the pan is small relative to its diameter 3 The thermal conductivity of the pan is constant Properties The thermal conductivity of the aluminum pan is given to be k 237 WmC Analysis a The boiling heat transfer coefficient is As πD24 π025 m24 00491 m2 Q hAsTs T h QAsTs T 800 W00491 m2108 95C 1254 Wm2C b The outer surface temperature of the bottom of the pan is Q kA Tsouter TsinnerL Tsouter Tsinner QLkA 108C 800 W0005 m237 WmC00491 m2 1083C