Question

NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. Refrigerant-134a enters an adiabatic compressor as saturated vapor at 100 kPa at a rate of 1.3 m3/min and exits at 1-MPa pressure. The isentropic efficiency of the compressor is 87 percent. 1 MPa R-134a Compressor 100 kPa sat. vapor Determine the temperature of the refrigerant at the exit of the compressor. Use the tables for R-134a. (You must provide an answer before moving on to the next part.) The temperature of the refrigerant at the exit of the compressor is °C.

Answer 1

Solution - Given : P = 100 kPa, Pz = Impa, How zate = 1.3 m3 m 1 min Misenbrapie = 87 l. * from Refrigecont table CA-12) at P, = 100 kPa i (sp. volume, enthalpy, entropy.) is : h, = 234.46 kJkg, s, = 0.95191 kg]K, 2, = 0.19255 m Ikg. Ù m = Now, mass flow rate is : 41 ☆ As 460 1:3160 = m = 0.1125 kg/ see 0.1925S # room Refrigerant table CA-13) at P₂ = 1 MPa (exit enthalpy, ) has 282.76 kg) kg, Now, Actual enthalpy is given by from the efficiency; = ho hi + has-hi 282:76 - 234.46 = 234:46 + ht 0.87 : haal 289.98 kJ I Kg. pressure uue con above value & final determine Sinat temp. (from Table A-13) using interpolations ; .. T2 = 56.25°C