Question

A cold storage is to be maintained at 5c while the surroundings are at 35c. The heat leakage from the surroundings into the cold storage is estimated to be 29 kw. The actual

c.O.P. Of the refrigeration plant used is one third that of an ideal plant working between the same temperatures. Find the power required to drive the plant

Answer 1

Answer:

W= 9.39 KW

Explanation:

T1 = 35C = 273+35 = 308K

T2 = 5C= 273+5= 278K

C.O.P (ideal) = T2/ (T1- T2);

Putting values of T1 and T2;

C.O.P = 278/ (308-278)  

C.O.P (ideal) = 9.266

Actual C.OP = 1/3 COP ideal ( as per given in question statement)

Actual COP = 1/3 * 9.266= 3.088

The heat removed from low temperature reservoir (cold storage) must be equal to heat leakage from surroundings to the cold storage(which is 29kw)

Q2= 29KW

Actual COP = Q2/W

W= Q2/COP

W= 29/3.088

W= 9.39 KW