AnalysisThe volumes of the tanks are10-19 A simple ideal Rankine cycle with water as the working fluid operates between the specified pressurelimits. The power produced by the turbine, the heat added in the boiler, and the thermal efficiency of the cycle are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis From the steam tables (Tables A-4, A-5, and A-6),kJ/kg 05.43354.1551.417kJ/kg 54.15mkPa 1kJ 1 kPa)100000,15)(/kgm 001043.0()(/kgm 001043.0kJ/kg 51.417inp,1233121inp,3kPa 100 @1kPa 100 @1whhPPwhhffvvvkJ/kg 5.1911)5.2257)(6618.0(51.4176618.00562.63028.13108.5 kPa 100KkJ/kg 3108.5kJ/kg 8.2610 1kPa 000,1544443443333fgffgfhxhhsssxssPshxPThus,kJ/kg 0.149451.4175.191105.4338.26105.19118.261014out23in43outT,hhqhhqhhwkJ/kg 2177.8kJ/kg 699.3The thermal efficiency of the cycle is0.3148.21770.149411inoutthqq10-20 A simple Rankine cycle with water as the working fluid operates between the specified pressure limits. The isentropic efficiency of the turbine, and the thermal efficiency of the cycle are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis From the steam tables (Tables A-4, A-5, and A-6),kJ/kg 05.43354.1551.417kJ/kg 54.15mkPa 1kJ 1 kPa)100000,15)(/kgm 001043.0()(/kgm 001043.0kJ/kg 51.417inp,1233121inp,3kPa 100 @1kPa 100 @1whhPPwhhffvvvkJ/kg 8.1997)5.2257)(70.0(51.417 70.0kPa 100kJ/kg 5.1911)5.2257)(6618.0(51.4176618.00562.63028.13108.5 kPa 100KkJ/kg 3108.5kJ/kg 8.2610 1kPa 000,15444444443443333fgffgsfsfgfshxhhxPhxhhsssxssPshxPThe isentropic efficiency of the turbine is qinqout100 kPa132415 MPasTqinqout100 kPa1324s15 MPasT40.8775.19118.26108.19978.26104343TshhhhThus,kJ/kg 3.158051.4178.1997kJ/kg 2177.805.4338.261014out23inhhqhhqThe thermal efficiency of the cycle is0.2748.21773.158011inoutthqq11-16 An ideal vapor-compression refrigeration cycle with refrigerant-134a as the working fluid is considered. The rate of heat removal from the refrigerated space, the power input to the compressor, the rate of heat rejection to the environment, and the COP are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) In an ideal vapor-compression refrigeration cycle, the compression process is isentropic, the refrigerant enters the compressor as a saturated vapor at the evaporator pressure, and leaves the condenser as saturated liquid at the condenser pressure. From the refrigerant tables (Tables A-12 and A-13),   throttlingkJ/kg 82.88kJ/kg 82.88liquid sat.MPa 7.0C95.34kJ/kg 50.273MPa 7.0KkJ/kg 94779.0kJ/kg 97.236 vaporsat.kPa 12034MPa 7.0 @ 3322122kPa 120 @ 1kPa 120 @ 11hhhhPThssPsshhPfggThen the rate of heat removal from the refrigerated space and the power input to the compressor are determined fromand        kW 1.83kW 7.41kJ/kg 236.97273.50kg/s 0.05kJ/kg 82.8897.236kg/s 0.0512in41hhmWhhmQL(b) The rate of heat rejection to the environment is determined fromkW 9.23 83.141.7inWQQLH(c) The COP of the refrigerator is determined from its definition,4.06kW 1.83kW 7.41COPinRWQLQHQL0.12 MPa12340.7 MPasT·Win··4s13-16 The mass fractions of the constituents of a gas mixture are given. The volumetric analysis of the mixture and the apparent gas constant are to be determined. Properties The molar masses of O2, N2 and CO2 are 32.0, 28, and 44.0 kg/kmol, respectively (Table A-1)Analysis For convenience, consider 100 kg of the mixture. Then the number of moles of each component and the total number of moles arekmol 136.1kg/kmol 44kg 50kg 50kmol 071.1kg/kmol 28kg 30kg 20kmol 625.0kg/kmol 32kg 20kg 20222222222222COCOCOCONNNNOOOOMmNmMmNmMmNmkmol 832.2136.1071.1625.0222CONO NNNNmNoting that the volume fractions are same as the mole fractions, the volume fraction of each component becomes40.1%37.8%22.1%or0.401kmol 2.832kmol 1.136or0.378kmol 2.832kmol 1.071or0.221kmol 2.832kmol 0.625222222COCONNOOmmmNNyNNyNNyThe molar mass and the gas constant of the mixture are determined from their definitions,kmolkg 31.35kmol 2.832kg 100/mmmNmMandKkg/kJ 0.235 kg/kmol 35.31KkJ/kmol 8.314mumMRR 13-33 The masses, temperatures, and pressures of two gases contained in two tanks connected to each otherare given. The valve connecting the tanks is opened and the final temperature is measured. The volume of each tank and the final pressure are to be determined. Assumptions Under specified conditions both N2 and O2 can be treated as ideal gases, and the mixture as anideal gas mixtureProperties The molar masses of N2 and O2 are 28.0 and 32.0 kg/kmol, respectively (Table A-1)mass20% O230% N250% CO2Analysis The volumes of the tanks are 33m 0.465m 0.295kPa 500K) K)(298/kgmkPa kg)(0.2598 (3kPa 300K) K)(298/kgmkPa kg)(0.2968 (13OO3NN2222PmRTPmRTVV333ONtotalm .760m 0.465m 295.022 VVVAlso,kmol 0.09375kg/kmol 32kg 3kg 3kmol 0.03571kg/kmol 28kg 1kg 122222222OOOONNNN MmNmMmNmkmol 0.1295kmol 0.09375kmol 03571.022ON NNNmThus,kPa 422.233m 0.76K) K)(298/kmolmkPa 4kmol)(8.31 (0.1295mumTNRPV13-67 The volume fractions of components of a gas mixture passing through the turbine of a simple ideal Brayton cycle are given. The thermal efficiency of this cycle is to be determined.Assumptions All gases will be modeled as ideal gases with constant specific heats. Properties The molar masses of N2, O2, H2O, and CO2 are 28.0, 32.0, 18.0, and 44.0 kg/kmol, respectively (Table A-1). The constant-pressure specific heats of these gases at room temperature are 1.039, 0.918, 1.8723, and 0.846