A refrigerant plant of 100 TR capacity uses R-22 as refrigerant. The condensing and evaporating pressure are 11.82 bar and 1.64 bar, respectively. The refrigerant enters the compressor at dry saturated state, whereas it leaves the condenser, subcooled by 10°C. Actual COP is 70% of its theoretical value. Find the following:
a) Theoretical and actual COP
b) Mass flow in kg/s
c) Compressor power.
The cooling capacity=100TR=100x3.516=351.6 kW
The condenser pressure(Pcon)=11.82 bar=11.82x105 Pa
The evaporator pressure(Peva)= 1.64 bar=1.64x105 Pa
From Property Table of R-22:
At point 2: Refrigerant is dry saturated at
P=1.64 bar, from Table h2=238kJ/kg
T2=-30oC, entropy(s2)=979 J/KgK
At point 3: Refrigerant is superheated having same entropy as that of point 2, therefore entropy(s3)=s3=979 J/kgK
P=11.82 bar, from Table for the entropy value entropy(s3) =979 J/KgK. Temperature T3=63.7oC.
At pressure P=11.82 bar and Temp(T)= 63.7oC
At point 4: Refrigerant is subcooled by10 oC , P=11.82 bar, ant T=Tcond-10 oC=29.7 oC-10 oC=19.7 oC
From Table: Enthalpy(h4) =68.8kJ At pressure P=11.82 bar.
At point 1: Refrigerant undergoes constant enthalpy process in expansion valve.
a) Theoretical COP=(h2-h1)/(h3-h2)=(238-68.8)/(288-238)=3.384
Actual COP=0.70x Theoretical COP=0.7x3.384=2.368
b) Mass flow rate (m) is given as,
TR=mass flow rate(m)x(h2-h1)
Mass flow rate(m)= TR /(h2-h1)=351.6/169.2=2.078 kg/s
c) Compressor power=mass flow rate (m)x(h3-h2)=2.078x(288-238)=103.9kW