Question

C.C. RH. 1111 net 0.7 VO-200 m3/min T=30°C =50% Comp Turbine Comp Ther=1300K TL=300K Q-100 kW Evap. Cond P=1.4 bar P=8 bar ne 0.7 R134-a Tor=?? EX. Valve A mini power station is used to drive a refrigeration cycle to cool the outside air. For the given diagram find power station thermal efficiency, COP of refrigeration cycle and cooling coil efficiency. (Hint: 85% of the power station output power is transfereed to the compressor).

Answer 1

Sol Let us consider the 3 (1300K) Gas power cycle. Given g = 100kw no 0.2 4 **(300K) consider Heat supply process 8 = 10o = 1.005 (1360-12') :. Tz' 12.00.49 K in C (T3-72') m = - 1- T2/T, n = 07 - Tz - To T-300 Tal-Ti 1200.49- 300 Tz =930.34 K n = Wi-z Way 300 930.34 9 What Wzy Wiz = 67,75 kW 67:7 ./. n that 85% of power ation ant * It is given that is supplied to the Refregination plant > 85% of 62.25 57.59 kW 2 3 • 8MP 2 ) GR-134a 0.14mpa 1 ។ 3 from We her The refrigeration table of R-134a. get Tempurator = 2972k 0.14 MPa T condensur = T3 304.33 K ; =0.35 kJ/kg hg, at 0.14 mpa - 210:22 KJ kg 0.9322kJ /kg.k at 0.8 MPa = 264.15 KJ lks 0.9066 kJ/kg .k A-1349 Se= Sg þzg=hy S2g ha= hq ho aut T 1 kg 0.8 MPa 93.42 Kilty hy = bu = Throttling process 93.42 KJ ²8=2 0.9322 0.9066+ 0.85 ln (Tz - Tz = 313.63K Tz-Ti 313.63-257.2 Ti-Ti T2 = 337.81 k GCT 2 -ty) 264.15+ 0.85 (337.3 30433 16=0.7= Т.' - 2522 he's hagt - 304.22 3 he' 292 KJ 1kg :. COP - hahu ha is oh, -> 210.27-93.42 1.429 292 2 10.22 COP = 1.429 * For cooling coil Tc = Teunporates = 257.2K = -15.8°C Ti 30°c To=25°C (Let atmospheric temp inside the room) } 1- 25+15.2 30 +15.8 10-912).