Question

15% A refrigerator using refrigerant-134a as the working fluid operates on an actual vapor compression ugeration cycle between 0.12 MPa and 1 MPa. The refrigerant leaves the compressor at 60°C and leaves condenser as a saturated liquid. If the mass flow rate is 1 kg/s, determine the COP of the system. If an adiabatic, reversible expansion device were available and used to expand the liquid leaving the condenser, how much would the COP improve by using this device instead of the throttle device?

Answer 1

(C.O.P)throttle = 2.30

and C.O.P increase by 45.21%

(C.O.Pdevice = 3.34

The detailed solution is given below:

Saturated Liquid 330 Throttling » devices a Gardensex y 6oC Evaporator 1 Evaporator Com o Dollaba # State! P,= 120kpa 3 hi= hg@120kpa = 236.97 kJ kg Sato rapor . S = Sg @ Rokpa= 0.94779 kJ / kg ok # State 2 lo- Imba 2 h = 293.38 kilky - To= 60c , S2 = 0.9850 KJ ingk # State as - P2= impa 3 has = 281 45 1kg Sas= Sy = 0.94 779 kJingk # State 3 P= Impa s h = hf@lmpa = 107.32 kJ lkg Sat, liquich I * State 4 Due to throtting her hz = 107.32 kilky

# Cop= A - mich ha) - Win iha-hi) 236.97 - 107.32 293.38 — 236-97 = 2,298 16.0.2 7.30 # If an adiabatic, reversible expansion device is then the process become is en tropic, SA= $z= femp = 0.3918965ingk & Sp = 0,3189 Tigk ? ha= 48.6 45 kg #60.p= hi ha 236.97-48.6 293.38-236.97 COP = 3.34

Hence Cop increased by using expansion devil. # Increase in Cop= 3.34-2.30 xloo į 2.30 = 45.77510