Question

A reversible heat engine operates between two reservoirs at temperatures of 600c and 40c. The engine drives a reversible refrigerator which operates between reservoirs at temperatures of 40c and 20c. The heat transfer to the heat engine is 2000 kj and the net work output for the combined engine refrigerator is 360 kj. (i) calculate the heat transfer to the refrigerant and the net heat transfer to the reservoir at 40c. (ii) reconsider (i) given that the efficiency of the heat engine and the
c.O.P. Of the refrigerator are each 40 per cent of their maximum possible values.

Answer 1

Answer:

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 the heat engine cycles,

n = 1 - T2/T1 = 1 - 313/873 = 0.642

W1/Q1 = 0.642W1 = 0.642 * 2000m = 1284kJ...(i)

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

Q4/W2 = 4.216...(ii)

W1 = 1284 -360 = 924KJ

From equation (ii)

Q4 = 4.216 * 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

Explanation:

Answer 2

T1 = 600 + 273 = 873K T2 = 600 + 273 = 873K T3 = 600 + 273 T2 = 40 + 273 = 313K T2 = 40 + 273 = 313K T2 = 40 + 273 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

Explanation: