Question

A reversible heat engine operates between two reservoirs at temperature of 600°C and 40°C. The Engine drives a reversible refrigerator which operates between reservoirs at temperature of 40°C and – 20°C. The heat transfer to the heat engine is 2000KJ and net work output of combined engine refrigerator plant is 360KJ. Evaluate the heat transfer to the refrigerator and the net heat transfer to the reservoir at 40°C.

Answer 1

T1 = 600 + 273 = 873K

T2 = 40 + 273 = 313K

T3 = – 20 + 273 = 253K

Heat transfer to engine = 200KJ

Net work output of the plant = 360KJ

Efficiency of heat engine cycle,

η = 1 – T2/T1 = 1 – 313/873 = 0.642

W1/Q1 = 0.642W1 = 0.642 x 2000 = 1284KJ ...(i)

C.O.P. = T3/(T2 – T3) = 253/(313 – 253) = 4.216

Q4/W2 = 4.216 ...(ii)

W1 – W2 = 360; W2 = W1 – 360

W2 = 1284 – 360 = 924KJ

From equation (ii)

Q4 = 4.216 x 924 = 3895.6KJ

Q3 = Q4 + W2 = 3895.6 + 924

Q3 = 4819.6KJ

Q2 = Q1 – W1 = 2000 – 1284

Q2 = 716KJ

Heat rejected to reservoir at 40°C = Q2 + Q3 = 716 + 4819.6

Heat rejected to reservoir at 40°C = 5535.6KJ

Heat transfer to refrigerator, Q4 = 3895.6KJ