Question

1. Saturated R134a vapor enters the uninsulated compressor of a home central air- conditioning system at 4 °C. The flow rate of R134a through the compressor is 0.1 kg/s, and the electrical power input is 5 kW. The exit state is 65 °C with 1000 kPa. Any heat transfer from the compressor is to the ambient environment at 27 °C. a) Determine the rate of entropy production for the process in kW/K. b) If the change in enthalpy for the isentropic compression is 28.5 kJ/kg, calculate the isentropic efficiency of the compressor. Show the actual and isentropic compression processes in a T-s diagram. (15 points)

Answer 1

enthaley * 1. @ from S.F.EE Compressob inlet en - h, and exit- he and power input = 5 kW . h, t Q - het w s from saturated R134 a table, h, = 249.53 kg/kg at 9, saturated vapour Condition. and 2 = 296.85 KJ/kg at 1000 Kea, 65 € Q = T (296-85 - 249.53) - 5 0.1 47.32 - 5 = -0.268 KW... Cheat rejected to outside) from Chant, s, 0.9169 ks/kJ.K, s 0-99305 ks/Xq.K as Charge in entropy of systems in = 0.1*(0-99305 - 0.9169) KJ/s. K = 7.615 W/k w/k =-0.893 W/k -) entropy Production = 6.72 x103 KW/K A Clarge in en entropy of outside = 268 360 (4.615 - 0.893) xi 73 KW/K b loookpg line 1-2' = is enthopie process 1-2 = Actual Compression Ta 1990 S is entropie OFT efficieney i sentropie work Actual work 28.5 47.32 60.23% A