Problem 7-1 - Example 7-1 In a vapor-compression...

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Example 7-1 In a vapor-compression refrigeration cycle, R12 evaporates at 40°F and condenses at 110°F. The compressor has an efficiency of 82%. Air flows over the evaporator coils at 4 lbm/s. The air enters at 70°F and exits at 55°F. a) Find the power input to the compressor in kW b) Calculate COP for the cycle Solution: The first law for the evaporator is () 14 56 Ra mh h h −= ±± 1 Btu enthalpy(R12, 40, 1) 81.4 lbm hT x == = =
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43 Btu = enthalpy(R12, 110, 0) 33.5 lbm hh T x == = = 5 Btu enthalpy(air, 70) 127 lbm hT = 6 Btu enthalpy(air, 55) 123 lbm = () 56 14 lbm Btu 4 127 123 sl b m Btu 81.4-33.5 lbm a R mh h m ⎛⎞ ⎜⎟ ⎝⎠ ± ± lbm 0.301 s R m = ± To find compressor power, h 2 will be needed. From the second law 12 s s s = 1 Btu entropy(R12, 40, 1) 0.166 lbm R sT x = =
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Problem 7-1 - Example 7-1 In a vapor-compression...

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