Question

6. Refrigerant-134a enters an adiabatic compressor as saturated vapor at 100 kPa at a rate of 0.7 m3/min and exits at 1 MPa pressure. If the isentropic efficiency of the compressor is 87%, determine (a) the temperature of the refrigerant at the exit of the compressor, (b) the power input (in kW), and (c) the rate of entropy generation during this process.

Answer 1

Obtain the properties of saturated refrigerant 134a from table A-12 (Saturated refrigerant 134a-pressure table) in the text book, at pressure P-100kPa Specific volume of saturated vapour is v v 0.19254 m/kg Specific enthalpy of saturated vapour is h -h 234.44 kJ/kg Specific entropy of saturated vapour is s s0.95183 kJ/kg-K Since, the process is an adiabatic process, entropy will be constant sS0.95183 kJ/kg-K Obtain the enthalpy at state-2s from table A-13 (Superheated refrigerant R-134a) in the text book, at pressure R 1MPa and entropy s2 0.95183 kJ/kg-K by using interpolation h, -271.71 0.95183-0.9179 282.74- 271.71 0.9525-0.9179 hs 271.71 0.9806 11.03 hs (0.9806 (11.03)+271.71 - 282.526 kJ/kg Calculate the actual enthalpy at state-2 using the isentropic efficiency of compressor. Sen 22a Substitute 282.526 kJ/kg for h, 234.44 kJ/kg for h, and 0.87 for 7sen 282.526 234.44 0.87- ha-234.44

48.086 0.87 ha 234.44 48.086 + 234.44 a 0.87 - 289.71 kJ/kg a) Obtain the enthalpy at state-2s from table A-13 (Superheated refrigerant R-134a) in the text book, at pressure R 1MPa and entropy ha -289.71 kJ/kg by using interpolation Ti.-50 289.71-282.74 60-50 293.38- 282.74 T). -50 20.6551 T (0.6551) (10)+50 T 56.551°C b) Calculate the power input to compressor Calculate the mass flow rate Substitute 0.7 m3/min for И and 0.19254 m/kg for v1.

0.7m3/min 0.19254m3/kg 0.01167 m3/s 0.19254 0.0606 kg/s Substitute 0.0606 kg/s for m, 289.71 kJ/kg for ha, and 234.44 kJ/kg for h in equation 0.0606(289.71-234.44) Wt3.349kW act