Question

Problem-3 (200) In a vapor-compression refrigeration cycle, R134a exits the evaporator as saturated vapor at -32°C. The refrigerant enters the condenser at 14 bar and 170°C, and saturated liquid exits at 14 bar. There is no significant heat transfer between the compressor and its surroundings, and the refrigerant passes through the evaporator with a negligible change in pressure. If the mass flow rate is 2.987 kg/s, determine (a) Refrigeration capacity in KW (100) (b) The power input to the compressor, in kW. (75) (c) The coefficient of performance. (25)

Answer 1

T-s diagram 3 1 1-2 → Isenteopic compression 2-3 → Condensation 3-4 [senthalpic expansion 4-1 Heat addition in eraporator . Criven data: R-134 a 555.097 Ti = -32°C P2 = 14 bar Ta = 170°C W = 2.987 kgls Find specific anthalpies of all states using R-134a tables. hi = hg at - 32°C = 379.06 kJ/kg. ha kilkg (from superheated table at 14 bar & 170°c) h₂ = h4 = hf at (4 bar 275.336 kilkg. ha= ha (: 3-4 is constant enthalpy expansion process) (a) Refrigeration Capacity (RC): RC = m Chicha) 2.987 (379.06-275.336) . RC = 309.82 kW (6) Power input to the compressos, Pe Pe = in Charhi)

Pe = 2.987 (555:097-379.06). .: Pc = 525.82 kW (e) Coefficient of performance, cop Cop= Refrigeration Capacity RC Pc. RE Power input in Compressor Cop="309.82 525.82 Cop=0.589
At times cop may become less than one as in this question where there is a large temperature difference.