Question

Pressure limits in a two-stage cooling system are 0.9 MPa and 250kPa. Refrigerant condenser
comes out as saturated liquid and works at 700kPa pressure
reduced to the pressure of the evaporation chamber. Meanwhile, some of the refrigerant evaporates and is mixed with the fluid from the low pressure compressor. The mixture is then
It is compressed to condenser pressure with high pressure compressor. The liquid in the evaporation chamber is reduced to evaporator pressure and draws heat from the cooled environment as it passes through the evaporator. Refrigerant flow rate in the low pressure compressor is 0.18 kg / s. Since the refrigerant from the evaporator is saturated steam and the isentropic efficiency of both compressors is 85%,

Warm environment QH 5 4 Condenser: Expansion valve High-pressure compressor Flash chamber 3 7 2 Expansion valve Low-pressure compressor Evaporator 8 1 cu QL Cold refrigerated space

a) Calculate the refrigerant flow rate in the high pressure compressor
b) Calculate the heat drawn from the cooled environment
c) Calculate the COP value of the cooling system
d) Mass flow, compressor efficiency and heat drawn by a single-stage cycle with the same pressure limits.
and calculate the COP value.

Answer 1

Please refer the following images for the solution.

Given: = 0.9MPa = g bar Pe=250kPa = 2.5bar Pint = 700 kPa = 7 bar m. p=im, ing=ing = n = 0.18 kg/s. Nise 8544 Solution Refrigeration cycle an P-h chart 5 gbar 4 7 7 bar 3/9 6 2.5 bar 8 1 h → From P-h chart, R-134a h = 394.82 KJ/kg the = 416.11 kJ/kg 63-411.99 kJ/kg bs=h6=249.21 kJ/kg hy = hq = 236.84 kJ /kg Now, enthalpy at g and 4 need to calculate first Taking mass balance flash chamber = intima mg = 0.18 ting - ii) balance Taking energy across the nghe i h7+ ghz (0.18+z) hoinghy tinh (from equation () (0.18+ma) *249.21 = (0-18 x 236.84.) + max411.99 44.8578 +249.21mg = 42.6312 +411.99 m2 ..Mg = 0.01368 kg/s aC TOSS flash chamber,

ing=0. 18+ miz 20:18 +0.01368 balance ing=ing = rm4 = mg = 0.19368 kgfs. Now taking energy at mixing chamber righz + m₂ hz = mghg (0.01368463 +0.18x416-))) = 0.19368 xbg ng= 415.82 kJ /kg Considering 9-4 ag an isentropic compression process The = 421.26 kJ/kg a) The refrigerant flow role in the high pressure compressor ng = n = 0.19368 kg/s (calculated above) 6) The heat drawn from the cooled environment തി = C. (2) COX Oto Q = mg Chi-he) = 0.18% ( 394.82 - 236.84) 10 = 28.44 kW c) The COP value of the cooling system. COP= Q Nc where, tica mix (hz-he) + mgx Cha-hg) Nise hise

. We 0.18( 416.11 - 394.82) 0.19368 (421.26 -415.82) + 0.85 0.85 Wc=4.5+ 1.24 : W. = 5.74kW COP= PL 28.44 We 5.74 Cop=4.95 (d) Single stage-cycle on p. h chart 3 pt 4 same Mass h From Ph chart, hi = 394.82 kJ/kg h2=421.60 kJ/kg 63 = 14 -- 429.@gokeolog 249.21 KJ/kg flow for the capacity, 0 = Mref Chicha) 28.44 hinha 394.82-249.21 mirep =0.1953 19/5 Compressor Work, we amefx Cha-hy/-0.1953x (421.60-394.82) 22.00.23 km/ The 6.15 kW In 0.85

COP= QL 28.44 Wc COP = 4.62 6.15