SPSolution30-32

SPSolution30-32 - Naik, Sameer V. Solution for Special...

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Naik, Sameer V. Fall 2009 1000 Solution for Special Problem 30 Given: - Internally reversible, isothermal compression of R134a in a piston-cylinder device Find: - Heat transfer and work (kJ/kg) during the process - Entropy production (kJ/kg-K) during the process - P-v and T-s diagrams - Work on P-v diagram and heat transfer on T-s diagram System: Assumptions: - Negligible leakage in piston-cylinder device, i.e. closed system (mass constant) - Stationary system, i.e., 0 PE = and 0 KE = - Quasistatic process Basic Equation(s): ( 29 2 1 internally reversible q ds q T s s T δ = = - q w u ke pe - = ∆ + ∆ + ∆ j system j j q s T σ = + Solution: (a) Using Table A-11, for P 1 = 2.8 bar, T sat = -1.23 ° C T 1 = 60 ° C > T sat SHV From Table A-12, we get: v 1 = 0.09319 m 3 /kg, u 1 = 277.23 kJ/kg, s 1 = 1.1079 kJ/kg-K Using Table A-11, for P 2 = 14 bar, T sat = 52.43 ° C T 2 = T 1 = 60 ° C > T sat SHV
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Naik, Sameer V. Fall 2009 1000 From Table A-12, we get: v 2 = 0.01495 m 3 /kg, u 2 = 262.17 kJ/kg, s 2 = 0.9297 kJ/kg-K For an internally reversible isothermal process: ( 29 ( 29 ( 29 2 1 kJ 60 273 K 0.9297 1.1079 kg-K q T s s = - = + × - i.e. heat transfer during the compression process is: kJ 59.35 kg q = - = - ; negative sign indicates heat loss from the system. Applying energy balance, we get:
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SPSolution30-32 - Naik, Sameer V. Solution for Special...

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