Answer:
Explanation:
The pressure at which the saturated liquid-vapor mixture enters the evaporator is, P1=120kPa
P1=120 kPa
Superheated vapor leaves the compressor at temperature
T2=70oC
T2=60oC
To calculate Quality of the refrigerant at the inlet of evaporator,
x=0.3
Power consumed by the compressor,
P=600W
Using the steam table of the refrigerant at P = 120 kPa, the specific enthalpy of the refrigerant at the inlet of evaporator,
h4=hf+xhfgh4=23.32kJ/kg+0.3∗215.54kJ/kgh4=87.982kJ/kg
h4=hf+xhfgh4=21.32 kJ/kg+0.3∗212.54 kJ/kgh4=85.082 kJ/kg
From steam table of the refrigerant at P = 120 kPa,
specific enthalpy of the saturated vapor refrigerant at the inlet of the compressor,
=h1=233.85kJ/kg
s1=0.9355 kJ/kg
For isentropic process, the specific entropy of the refrigerant at the outlet of the compressor S2=S1=0.92545kJ/kg
checking the superheated table of the refrigerant when T=70C
S2=0.9354kJ/kg
The specific enthalpy at the exit of the compressor
h2=284.17kJ/kg
pressure of the refrigerant at the exit
P2=1340.43kPa
a) The mass flow rate of the refrigerant,
P=˙m(h2−h1)600W=˙m(285.17kJ/kg−23.86kJ/kg)˙m=11.92kg/s
b) The condenser pressure:
Pressure of the condenser=P2=1340.43kPa
c) The COP of the refrigerator can be calculated using this formula
COP=h1−h4h1−h2COP=233.86kJ/kg−
85.082
kJ/kg(284.17 kJ/kg−233.86kJ/kg)COP=2.98
