Question

1. An industrial freezer uses a multi-stage vapor-compression cycle with R-134a as the working fluid. The evaporator operates at 140 kPa, the condenser at 1600 kPa, and the flash chamber 600 kPa. The refrigerant leaves the evaporator as a super-heated vapor at -15°C, and leaves the condenser as a sub-cooled liquid at 50°C. Both compressors used in the system have isentropic efficiencies of 85%. For a cooling load of 25 kW,

(a) What is the mass flow rate of refrigerant through the evaporator, and what fraction of the mass flow rate through the condenser does this represent?

(b) What is the COP of the system?

Answer 1

P-1600 otd- 9 P- Goo 7 P = 140 kpe P = 140kpa T,= -15c, from Property table & Chart h = 242 kJ/kg. for R-134 a S, = 0.957 kilugh w = 273 lejling = 268:35 bylke. Also we know Ps = 1600lepa Ts = 50°c us= 123 kglung , SG= 0.441 kg kg k hgan (Iso entalpic expansion) Dryness fraction at point 6 =) ho = hafta (hq) 2 = (ho-h88) -

Saturation properties at 600 kpa That = 2.6'c , " = 81.5 kglug ug = 180.9 ksting, ung: 2624 x= 125 hg- hot 123 - 81.5 uf f 180.9 : 0.23 wp-hof = 81.5 kglang Now cooling load = 25 kw Let m, ma are the mass flow rate through Evaporator & condensor respectively & m₂ is the mass of refrigerent that by passed the evaporator through flash chamber. 0:30letom s m, 3 0.91m)_@ Cooling load - 95 kw = m ( niha) = m = 0.156 kg/sec. from equation 6 m = 0.203 ugls. m = my-m, = 0.047 kg/sec. )

from energy on energy balance at point ③ Oo99 x 262,4 + - S6 * 2-68-35 : 0203 x 3 uz 266.97 leging. Also from Tos Chant of R 134a we get - Sz: 0.937 Kupk Sz= Sy = 0.937 (Isentropic compression). - hú= 288 kglug lcomp haha - - haha 4 - 291.7 kapo kulegt - compressor work - m, (hy-h) & in Inghy) 0156( 268.35-242) + 0.203(291.7.-266.97) 1. = g. 13 kw mass flow through evaporator = m = 0.156 Also mi 0.77 6 cop = evaporater heat loss total work done 9.13 = 2.74