Explanation Write the expression for the energy balance equation for a closed system. E i n − E o u t = Δ E s y s t e m (I) Here, net energy transfer into the control volume is E i n , net energy transfer exit from the control volume is E o u t and change in internal energy of the system is Δ E s y s t e m Write the expression for the final entropy ( s 2 ) of the system. s 2 = s f + x 2 s f g (II) Here, the final dry fraction is x 2 , the entropy of saturated liquid is s f and entropy of evaporation is s f g . Write the expression for the initial internal energy in the isentropic process ( u 2 s ) . u 2 s = u f + x 2 s u f g (III) Here, the isentropic final dry fraction is x 2 s , the internal energy of saturated liquid is u f and internal energy of evaporation is u f g . Write the expression for the isentropic efficiency. η = w a , o u t w s , o u t (IV) Here, isentropic efficiency is η , actual work output is w a , o u t and isentropic work output is w s , o u t . Conclusion: Substitute Q for E i n , W o u t for E o u t and Δ U for Δ E s y s t e m in Equation (I). Q − W o u t = Δ U (V) Here, heat transfer is Q , work output is W o u t and change in internal energy is Δ U . Substitute 0 for Q and m ( u 2 − u 1 ) for Δ U in Equation (V). − W o u t = m ( u 2 − u 1 ) − W o u t m = ( u 2 − u 1 ) − w o u t = ( u 2 − u 1 ) Here, mass flow is m , initial internal energy is u 1 and final internal energy is u 2 . Write the expression for actual work input w a , o u t . w a , o u t = ( u 1 − u 2 a ) (VI) Here, the actual final internal energy is u 2 a . Write the expression for isentropic work input w s , o u t . w s , o u t = ( u 1 − u 2 s ) Here, the isentropic final internal energy is u 2 s . Refer to Table A-13, “Superheated refrigerant-134a”, obtain the below properties at the pressure of 700 kPa . s 1 = s 2 = 0.9642 kJ / kg ⋅ K u 1 = 256.41 kJ / kg Here, final entropy is s 2 . Refer to Table A-12, “Saturated refrigerant-134a-Pressure table”, obtain the below properties at the pressure of 60 kPa . u f = 3.795 kJ / kg u f g = 205.34 kJ / kg s f = 0.01633 kJ / kg ⋅ K s f g = 0.94812 kJ / kg ⋅ K Substitute 0.01633 kJ / kg ⋅ K for s f , 0.9642 kJ / kg ⋅ K for s 2 and 0.94812 kJ / kg ⋅ K for s f g in Equation (II). 0.9642 kJ / kg ⋅ K = 0.01633 kJ / kg ⋅ K + x 2 ( 0.94812 kJ / kg ⋅ K ) x 2 = 0.9997 Substitute 3.795 kJ / kg for u f , 0

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Chapter 7.13, Problem 178RP

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The actual work output and the final enthalpy for an isentropic expansion efficiency of 80 percent.

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Chapter 7.13, Problem 177RP

Chapter 7.13, Problem 179RP

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