Consider two reversible refrigerators operating in series. The first refrigerato

ID: 1862200 • Letter: C

Question

Consider two reversible refrigerators operating in series. The first refrigerator removes

heat at the rate of 250 kW from a space to be maintained at -23C and rejects heat to an

intermediate temperature reservoir. The second refrigerator removes the heat received by

the intermediate temperature reservoir and rejects heat to the ambient reservoir at 27C.

Coefficient of performance of the second refrigerator is 80% of the first refrigerator.

(a) Determine the temperature (C) of the intermediate reservoir.

(b) Calculate the total power (kW) required to operate both the refrigerators.

SOME ANSWERS ARE ATTACHED BUT NEED PROCEDURE

Answer: (a) Not attached; (b) 50 kW; (c) 300 kW, heat transfer into the reservoir

Let T1(cooling space temperature)= -23+273=250K

T2=temperature of second reservoir

T3 (ambient temperature)= 27+273=300K

Heat removal rate by primary refrigerator (Q1)=250kW

COP1=T1/(T2-T1)=250/(T2-250)

and COP2=T2/(T3-T2)=T2/(300-T2)

according to question COP2=0.80*COP1

T2/(300-T2)= 0.80*250/(T2-250)

on solving this equation we get T2=271.22K= -1.77 C

so COP1=250/(271.22-250)=11.78

COP2=271.22/(300-271.22)=9.42

Now for primary Refrigerator

COP1=Q1/W1

11.78=250/W1

W1=21.22kW

Now heat removal rate by secondary refrigerator(Q2)=Q1+W!=250+21.22=271.22 kW

COP2=Q2/W2

9.42=271.22/W2

W2=28.79kW

Hence

(a)Temperature of intermidiate reservoir(T2)= -1.77 C

(b)Total power required to operate both refrigerator= W1+W2=21.22+28.79=50.01kW

(c)rate of heat transfer for the ambient reservoir= Q2+W2=271.22+28.79=300.01 kW

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