TITLE.PM5

SECOND LAW OF THERMODYNAMICS AND ENTROPY 243

dharm

/M-therm/th5-2.pm5

+Example 5.12. A reversible heat engine operates between two reservoirs at tempera-
tures 700°C and 50°C. The engine drives a reversible refrigerator which operates between reser-
voirs at temperatures of 50°C and – 25°C. The heat transfer to the engine is 2500 kJ and the net
work output of the combined engine refrigerator plant is 400 kJ.
(i)Determine the heat transfer to the refrigerant and the net heat transfer to the reservoir
at 50°C ;
(ii)Reconsider (i) given that the efficiency of the heat engine and the C.O.P. of the refrig-
erator are each 45 per cent of their maximum possible values.
Solution. Refer Fig. 5.14.

Fig. 5.14

Temperature, T 1 = 700 + 273 = 973 K

Temperature, T 2 = 50 + 273 = 323 K

Temperature, T 3 = – 25 + 273 = 248 K

The heat transfer to the heat engine, Q 1 = 2500 kJ

The network output of the combined engine refrigerator plant,

W = W 1 – W 2 = 400 kJ.

(i) Maximum efficiency of the heat engine cycle is given by

ηmax = 1 –

T

T

2

1

= 1 –

323

973

= 0.668

Again,

W

Q

1

1

= 0.668

∴ W 1 = 0.668 × 2500 = 1670 kJ

(C.O.P.)max =

T

TT

3

23 −

=

248

323 248−

= 3.306

Also, C.O.P. =

Q

W

4

2

= 3.306