HW11s - Spring 2006 Thermodynamics Homework 11 Solution 1...

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Spring 2006 1 ME 201 Thermodynamics Homework 11 Solution 1. One component in a household refrigerator is the compressor where refrigerant 134-a enters as saturated vapor at -24 ° F and is isentropically compressed to 30 psia. Determine the work required in Btu/lb m and the exit temperature of the refrigerant. Solution: System Type: Control Volume (Compressor) Process: Isentropic Working Fluid: Refrigerant -134a (compressible) State 1 (inlet) State 2 (outlet) T 1 = -24 ° F T 2 = 26.7 ° F P 1 = 11.62 psia P 2 = 30 psia s 1 = 0.2286 Btu/(lb m ⋅° F) s 2 = 0.2286 Btu/(lb m ⋅° F) h 1 = 99.51 Btu/lb m h 2 = 107.63 Btu/lb m phase: sat. vap. phase: sup. vap. Italicized values are from R-134a table. Bold values are calculated. Inlet State: Fixed Final State: UNKNOWN Q = 0 sh W = UNKNOWN KE and PE are negligible 1st Law: ( ) W - h - h m = 0 sh 2 1 F Approach: We use our process description to fix the final states. Then the first law can be used to determine the required shaft work.
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HW11s - Spring 2006 Thermodynamics Homework 11 Solution 1...

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