Explanation Express the energy balance equation for closed system. E i n − E o u t = Δ E s y s t e m Here, energy transfer into the control volume is E i n , energy transfer exit from the control volume is E o u t and change in internal energy of system is Δ E s y s t e m . Substitute 0 for E i n , Q o u t for E o u t , and m ( u 3 − u 1 ) for Δ E s y s t e m . Q o u t = m ( u 2 − u 1 ) (I) Here, amount of heat transfer exit from the system is Q o u t , mass of the system is m , internal energy at final state is u 2 and internal energy at initial state is u 1 . Write the formula to calculate the mass ( m ) of refrigerant inside the tank. m = V 1 v 1 (II) Here, initial volume of tank is V 1 and initial specific volume is v 1 . Write the formula to calculate the vapor quality of final state ( x 2 ) . x 2 = v 2 − v f v f − v g (III) Here, final specific volume is v 2 , specific volume of saturated liquid is v f and specific volume of saturated vapor is v g . Calculate the internal energy ( u 2 ) at final state. u 2 = u f + x 2 u f g (IV) Here, vapor quality of final state is x 2 , internal energy of saturated liquid is u f and internal energy of evaporation is u f g . Calculate the piston force ( F ) act at piston cylinder interface. F = A ( P 1 − P 2 ) (V) Here, initial pressure inside the cylinder is P 1 , final pressure inside the cylinder is P 2 and area of cylinder is A . Conclusion: From the Table A-5 of “Saturated water: Pressure”, select the specific volume of saturated liquid ( v f ) , saturated vapor ( v g ) , internal energy of saturated liquid ( u f ) , and evaporation ( u f g ) at saturated pressure of 100 kPa as 0.001043 m 3 / kg , 1.6941 m 3 / kg , 417.40 kJ / kg and 2 , 505.6 kJ / kg respectively. From the Table A-4 of “Saturated water: Temperature”, select the specific volume of saturated liquid ( v f ) , saturated vapor ( v g ) , internal energy of saturated liquid ( u f ) and evaporation ( u f g ) , saturated pressure at temperature of 30 ° C as 0.001004 m 3 / kg , 32.879 m 3 / kg , 125.73 kJ / kg , 2 , 290.2 kJ / kg and 4.2469 kPa respectively. From the saturated water pressure table, select the specific volume of saturated vapor ( v g ) and internal energy of saturated vapor ( u g ) at initial pressure of 100 kPa as 1

8th Edition

ISBN: 9780073398174

Author: Yunus A. Cengel Dr., Michael A. Boles

Publisher: McGraw-Hill Education

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Chapter 4.5, Problem 143RP

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The friction force developed at the piston cylinder interface.

The amount of heat transfer from the piston cylinder device.

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Chapter 4.5, Problem 142RP

Chapter 4.5, Problem 144FEP

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The friction force developed at the piston cylinder interface. The amount of heat transfer from the piston cylinder device.