Answer
Required power = 30.81kW
COP = 4.87
Work Step by Step
Given values and data-
Refrigerant 134a as a working fluid.
$Q_{L}$ = 150kW = 150kJ/sec
Temperature = $-12^{\circ}C$ (evaporator)
Pressure = 800kPa = 0.8 MPa (Condenser)
An ideal vapour compression refrigeration cycle work in 4 states.
Now from the table of refrigerant 134a , The enthalpies of all four states are:-
$1^{st}$ state :
$T_{1}$= $-12^{\circ}C$
$h_{1}$=243.34 kJ/kg
$2^{nd}$ state :
$P_{2}$=0.8MPa
$h_{2}$=273.71kJ/kg
$3^{rd}$ state :
$P_{3}$ = 0.8MPa
$h_{3}$=95.48kJ/kg
$4^{th}$ state (throttling process-heat remains constant):
$h_{4}$=95.48kJ/kg
Now the mass flow rate of liquid given by formula:
$Q_{L}$ = m'($h_{1}$-$h_{4}$)
here m' is mass flow rate
m' = $Q_{L}$$\div$($h_{1}$-$h_{4}$)
m' = 150kJ/sec$\div$(243.34-95.48)kJ/kg
m'=1.014 kg/sec
Then power required
$W_{in}$ = m'($h_{2}$-$h_{1}$)
$W_{in}$=1.014kg/sec(273.71 -243.34)kJ/kg
$W_{in}$=30.81 kW
CoP of the refrigerator is
CoP$_{r}$ = $Q_{L}$$\div$$W_{in}$
CoP$_{r}$= 150kW$\div$30.81kW
CoP$_{r}$=4.87
