HW14s - Spring 2006 Thermodynamics Homework 14 Solution 1....

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Spring 2006 1 ME 201 Thermodynamics Homework 14 Solution 1. Consider the Carnot cycle occurring in a piston-cylinder device containing refrigerant-134a with operating conditions given below: Process A: Isothermal heat addition at T H = 30 ° C to convert saturated liquid to saturated vapor Process B: Isentropic and adiabatic expansion to T L = -20 ° C Process C: Isothermal heat removal at -20 ° C Process D: Isentropic and adiabatic compression back to the initial state With R-134a as the working fluid for this cycle calculate the thermal efficiency using η th net in = W Q Compare this to the ideal Carnot cycle efficiency given by η Carnot L H = 1 - T T Solution: We shall make our calculations process by process Process A Working Fluid: R-134a(compressible) System: Closed System Process: Isothermal State 1 State 2 T 1 = 30°C T 2 = 30°C P 1 = 770.64 kPa P 2 = 770.64 kPa v 1 = 0.0008421 m 3 /kg v 2 = 0.026622 m 3 /kg u 1 = 92.93 kJ/kg u 2 = 246.14 kJ/kg s 1 = 0.34789 kJ/(kg K) s 2 = 0.91879 kJ/(kg K) phase: sat.liq . phase: sat.vap. italicized values from tables Initial State: Fixed
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ME 201 Thermodynamics Spring 2006 2 Final State: Fixed W sh = 0 Q = ???? W bnd = ???? 1 st Law: u 2 - u 1 = q - w bnd Pdv: w bnd = P(v 2 -v 1 ) Both states are fixed, so we can go to the R-134a tables and look up the remaining properties. Since our process is both isothermal and isobaric due to the phase change,
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This note was uploaded on 07/25/2008 for the course ME 201 taught by Professor Somerton during the Spring '06 term at Michigan State University.

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HW14s - Spring 2006 Thermodynamics Homework 14 Solution 1....

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