Explanation Given: The volume of the pressure cooker ( V ) is 4 L. The operating pressure of cooker ( P ) is 175 kPa. The volume of liquid in the cooker ( V f ) is 2L. The volume of vapor in the cooker ( V g ) is 2L. The time for which pressure cooker is not to run out of liquid water is ( Δ t ) 1 hr. Calculation: Refer Table A-4, “Saturated water—Temperature table”, obtain the properties of water corresponding to water inlet pressure ( P 1 ) of 175 kPa . The specific volume of saturated liquid water ( v f ) is 0.001057 m 3 / kg . The specific volume of saturated vapor ( v g ) is 1.0037 m 3 / kg . The specific internal energy of liquid water ( v f ) is 486.82 kJ / kg . The specific internal energy of vapor ( v g ) is 2 , 524.5 kJ / kg . Refer Table A-4 of “Saturated water—Pressure table”, obtain the properties of water corresponding to saturated vapor at saturated pressure ( P 2 ) of 175 kPa . The specific volume of saturated vapor ( v 2 = v g ) is 1.0037 m 3 / kg . The specific internal energy of liquid water ( u 2 = u g ) is 2 , 524.5 kJ / kg . Refer Table A-4, “Saturated water—Pressure table”, obtain the properties of water corresponding to saturated vapor at saturated pressure ( P e ) at exit of 175 kPa . The enthalpy of saturated vapor at saturated pressure ( h e = h g ) is 2 , 700.2 kJ / kg Write the mass balance for the uniform-flow system as follows: m in − m out = Δ m system m e = m 1 − m 2 (I) Here, mass of liquid water entering into the cooker is m in , mass of liquid water coming out from the cooker is m out , rate of change in mass of liquid water within system is Δ m system , exit mass of liquid water from cooker is m e , initial mass of liquid water is m 1 and final mass of liquid water is m 2 . Initially, one-half of the volume is filled with liquid and the other half with vapor. V f = V g = 0.002 m 3 . Calculate the initial mass ( m 1 ) of liquid water in the cooker. m 1 = m f + m g m 1 = V f v f + V g v g m 1 = 0 .002 m 3 0.001057 m 3 / kg + 0 .002 m 3 1.0037 m 3 / kg = 1.893 kg + 0.002 kg = 1.895 kg Calculate the final mass ( m 2 ) of liquid water coming out of the cooker. m 2 = V v 2 m 2 = 0.004 m 3 1.0036 m 3 / kg = 0.004 kg Write the formula to calculate the initial total internal energy of the water ( U 1 ) . U 1 = m 1 u 1 U 1 = m f u f + m g u g Here, specific enthalpy of saturated liquid is u f , internal energy of the liquid water at cooker inlet is u 1 , and specific enthalpy of saturated steam is u g

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ISBN: 9780078027680

Author: Yunus A. Cengel Dr., Robert H. Turner, John M. Cimbala

Publisher: McGraw-Hill Education

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Chapter 6, Problem 107P

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Chapter 6, Problem 106P

Chapter 6, Problem 108P

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Fundamentals of Thermal-Fluid Sciences

Ch. 6 - Prob. 1P Ch. 6 - Define mass and volume flow rates. How are they...Ch. 6 - Does the amount of mass entering a control volume...Ch. 6 - Consider a device with one inlet and one outlet....Ch. 6 - The ventilating fan of the bathroom of a building...Ch. 6 - Air whose density is 0.078 lbm/ft3 enters the duct...Ch. 6 - Air enters a 28-cm diameter pipe steadily at 200...Ch. 6 - A steady-flow compressor is used to compress...Ch. 6 - Prob. 9P Ch. 6 - A desktop computer is to be cooled by a fan whose...

Ch. 6 - Prob. 11P Ch. 6 - Prob. 12P Ch. 6 - Prob. 13P Ch. 6 - Prob. 14P Ch. 6 - Prob. 15P Ch. 6 - Prob. 16P Ch. 6 - A house is maintained at 1 atm and 24°C, and warm...Ch. 6 - Prob. 18P Ch. 6 - Prob. 19P Ch. 6 - Prob. 20P Ch. 6 - Prob. 21P Ch. 6 - The kinetic energy of a fluid increases as it is...Ch. 6 - Prob. 23P Ch. 6 - Air enters a nozzle steadily at 50 psia, 140°F,...Ch. 6 - Prob. 25P Ch. 6 - Prob. 26P Ch. 6 - Air at 600 kPa and 500 K enters an adiabatic...Ch. 6 - Prob. 28P Ch. 6 - Prob. 29P Ch. 6 - Air at 13 psia and 65°F enters an adiabatic...Ch. 6 - Prob. 31P Ch. 6 - Prob. 32P Ch. 6 - Prob. 33P Ch. 6 - Steam at 4 MPa and 400°C enters a nozzle steadily...Ch. 6 - Prob. 35P Ch. 6 - Prob. 36P Ch. 6 - Prob. 37P Ch. 6 - Prob. 38P Ch. 6 - Prob. 39P Ch. 6 - Prob. 40P Ch. 6 - Prob. 41P Ch. 6 - Prob. 42P Ch. 6 - Prob. 43P Ch. 6 - Helium is to be compressed from 105 kPa and 295 K...Ch. 6 - Carbon dioxide enters an adiabatic compressor at...Ch. 6 - Air is compressed from 14.7 psia and 60°F to a...Ch. 6 - Prob. 47P Ch. 6 - An adiabatic gas turbine expands air at 1300 kPa...Ch. 6 - Steam flows steadily into a turbine with a mass...Ch. 6 - Prob. 50P Ch. 6 - Prob. 51P Ch. 6 - Prob. 52P Ch. 6 - Prob. 53P Ch. 6 - Prob. 54P Ch. 6 - Refrigerant-134a is throttled from the saturated...Ch. 6 - Prob. 56P Ch. 6 - Prob. 57P Ch. 6 - Prob. 58P Ch. 6 - Prob. 59P Ch. 6 - Prob. 60P Ch. 6 - Prob. 61P Ch. 6 - Prob. 62P Ch. 6 - Prob. 63P Ch. 6 - Prob. 64P Ch. 6 - Prob. 65P Ch. 6 - Prob. 66P Ch. 6 - Prob. 67P Ch. 6 - Prob. 68P Ch. 6 - Prob. 69P Ch. 6 - Prob. 70P Ch. 6 - A thin-walled double-pipe counter-flow heat...Ch. 6 - Prob. 72P Ch. 6 - Prob. 73P Ch. 6 - Prob. 74P Ch. 6 - Prob. 75P Ch. 6 - Prob. 77P Ch. 6 - Prob. 78P Ch. 6 - Prob. 79P Ch. 6 - Prob. 80P Ch. 6 - Prob. 81P Ch. 6 - Prob. 82P Ch. 6 - Prob. 83P Ch. 6 - Prob. 84P Ch. 6 - Prob. 85P Ch. 6 - The components of an electronic system dissipating...Ch. 6 - Prob. 87P Ch. 6 - Prob. 88P Ch. 6 - Prob. 89P Ch. 6 - Prob. 90P Ch. 6 - Prob. 91P Ch. 6 - Prob. 92P Ch. 6 - Prob. 93P Ch. 6 - A house has an electric heating system that...Ch. 6 - Prob. 95P Ch. 6 - Refrigerant-134a enters the condenser of a...Ch. 6 - Prob. 97P Ch. 6 - Prob. 98P Ch. 6 - Prob. 99P Ch. 6 - Prob. 100P Ch. 6 - Air enters the duct of an air-conditioning system...Ch. 6 - Prob. 102P Ch. 6 - A rigid, insulated tank that is initially...Ch. 6 - Prob. 105P Ch. 6 - Prob. 106P Ch. 6 - Prob. 107P Ch. 6 - Prob. 108P Ch. 6 - Prob. 109P Ch. 6 - An air-conditioning system is to be filled from a...Ch. 6 - Prob. 111P Ch. 6 - A 0.06-m3 rigid tank initially contains...Ch. 6 - A 0.3-m3 rigid tank is filled with saturated...Ch. 6 - Prob. 114P Ch. 6 - A 0.3-m3 rigid tank initially contains...Ch. 6 - Prob. 116P Ch. 6 - Prob. 117P Ch. 6 - An insulated 40-ft3 rigid tank contains air at 50...Ch. 6 - A vertical piston–cylinder device initially...Ch. 6 - A vertical piston–cylinder device initially...Ch. 6 - The air in a 6-m × 5-m × 4-m hospital room is to...Ch. 6 - Prob. 124RQ Ch. 6 - Prob. 125RQ Ch. 6 - Prob. 126RQ Ch. 6 - Prob. 127RQ Ch. 6 - Prob. 128RQ Ch. 6 - Prob. 129RQ Ch. 6 - Prob. 130RQ Ch. 6 - Prob. 131RQ Ch. 6 - Prob. 132RQ Ch. 6 - Steam enters a nozzle with a low velocity at 150°C...Ch. 6 - Prob. 134RQ Ch. 6 - Prob. 135RQ Ch. 6 - Prob. 136RQ Ch. 6 - In large steam power plants, the feedwater is...Ch. 6 - Prob. 138RQ Ch. 6 - Prob. 139RQ Ch. 6 - Prob. 140RQ Ch. 6 - Prob. 141RQ Ch. 6 - Prob. 142RQ Ch. 6 - Prob. 143RQ Ch. 6 - Prob. 144RQ Ch. 6 - Prob. 145RQ Ch. 6 - Prob. 146RQ Ch. 6 - Repeat Prob. 6–146 for a copper wire ( = 8950...Ch. 6 - Prob. 148RQ Ch. 6 - Prob. 149RQ Ch. 6 - Prob. 150RQ Ch. 6 - Prob. 151RQ Ch. 6 - Prob. 152RQ Ch. 6 - Prob. 153RQ Ch. 6 - An adiabatic air compressor is to be powered by a...Ch. 6 - Prob. 156RQ Ch. 6 - Prob. 157RQ Ch. 6 - Prob. 158RQ Ch. 6 - Prob. 159RQ Ch. 6 - Prob. 160RQ Ch. 6 - Prob. 161RQ Ch. 6 - Prob. 162RQ Ch. 6 - Prob. 163RQ Ch. 6 - Prob. 164RQ Ch. 6 - Prob. 166RQ Ch. 6 - Prob. 167RQ Ch. 6 - Prob. 168RQ Ch. 6 - Prob. 169RQ

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