Question

Required Information NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to retum to this part. Refrigerant-134a enters an adiabatic compressor as saturated vapor at 100 kPa at a rate of 0.9 m/min and exits at 1MPa pressure. The Isentropic efficiency of the compressor is 87 percent. 1 MPа R-134a Compressor 100 kPa sal 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 0.0780 O'C.

Answer 1

Solution :- o from saturated Refrigerant 134a Pressure table ; 2 At lookfa : 0.19254 m3/kg g@ kPa hoz ngon z KJ/kg 234.44 100 kPa &i= Sga 0.95183 kJ/rgk 100 kPa Now, we determine mass flow rate; mo n = z Sol 0.9 Go 0.19254 ط M = 0.078 Kg Is T 1 mpa 2 28 100 kPa 3 $-$25 8 for I sentropic process 1-25 we determine the ideal exit Bnthalpy using the initial entropy and the final pressure by looking superheated R-134 a table;

1 MPa : (kJirg) 8 (Kis/lg 1) 21.71 0.9199 he 0.95183 282.74 0.9525 has 282.74 Dag5183 - 0.9525 :) (27.1.7) – 282.74) 09179 -0.9525 .. has 202.526 KI/kg Now I sentropic Efficiency of compressor is given by; has-hi hz-hi 0.84 282.526 234.44 hz 234.44 ha 289.71 Kollcg From the actual exit enthalpy ie; hz and final pressure we can determine the final temperature from superheated R-134a table; At 1MPa : Troc) h CRIM) 50 282.74 T, 289.71 60 293.38 Tz - 60 + 289.7–293.30 :) (so-60) 282.74 - 293-38 Tz = 56.55 °C tory,