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# HW7s - Spring 2006 Thermodynamics Homework#7 Solutions 1...

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Spring 2006 1 ME 201 Thermodynamics Homework #7 Solutions 1. Refrigerant -134a as saturated vapor at 0.5 MPa is isentropically compressed by a compressor in a refrigeration plant to 1.2 MPa. Determine the enthalpy change for the process and the final fluid phase. Solution: Substance Type: Compressible (R-134a) Problem Type: Process(Isentropic) State 1 State 2 T 1 = 15.71 ° C T 2 = 49.3 ° C P 1 = 0.5 MPa P 2 = 1.2 MPa h 1 = 259.30 kJ/kg h 2 = 277.4 kJ/kg s 1 = 0.9240 kJ/(kg K) s 2 = 0.9240 kJ/(kg K) phase: sat.vap. phase: sup.vap. Italicized values are from R-134a tables, bold values are calculated. At state 1 we know the pressure and that we have saturated liquid, so that the state is fixed. Going to the saturation pressure table, Table A-12, we find kJ/kg 259.30 = h K), kJ/(kg 40 2 9 0. = s C, 5.71 1 = T 1 1 1 ° At state 2 we know the pressure and that we have an isentropic process or K) kJ/(kg .9240 0 = s = s 1 2 To determine the fluid phase, we go to the saturation pressure table at 1.2 MPa and find K) kJ/(kg 0.9130 = s and K) kJ/(kg 0.4244 = s g f Since s 2 > s g , we have a superheated vapor at state 2. Going to the superheat tables,

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HW7s - Spring 2006 Thermodynamics Homework#7 Solutions 1...

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