TITLE.PM5

244 ENGINEERING THERMODYNAMICS

dharm

/M-therm/th5-2.pm5

Since, W 1 – W 2 = W = 400 kJ

W 2 = W 1 – W = 1670 – 400 = 1270 kJ

∴ Q 4 = 3.306 × 1270 = 4198.6 kJ

Q 3 = Q 4 + W 2 = 4198.6 + 1270 = 5468.6 kJ

Q 2 = Q 1 – W 1 = 2500 – 1670 = 830 kJ.

Heat rejection to the 50°C reservoir

= Q 2 + Q 3 = 830 + 5468.6 = 6298.6 kJ. (Ans.)

(ii) Efficiency of actual heat engine cycle,

η = 0.45 ηmax = 0.45 × 0.668 = 0.3

∴ W 1 = η × Q 1 = 0.3 × 2500 = 750 kJ

∴ W 2 = 750 – 400 = 350 kJ

C.O.P. of the actual refrigerator cycle,

C.O.P. = Q

W

4

2

= 0.45 × 3.306 = 1.48
∴ Q 4 = 350 × 1.48 = 518 kJ. (Ans.)
Q 3 = 518 + 350 = 868 kJ
Q 2 = 2500 – 750 = 1750 kJ
Heat rejected to 50°C reservoir
= Q 2 + Q 3 = 1750 + 868 = 2618 kJ. (Ans.)
+Example 5.13. (i) A reversible heat pump is used to maintain a temperature of 0°C in a
refrigerator when it rejects the heat to the surroundings at 25°C. If the heat removal rate from
the refrigerator is 1440 kJ/min, determine the C.O.P. of the machine and work input required.
(ii)If the required input to run the pump is developed by a reversible engine which receives
heat at 380°C and rejects heat to atmosphere, then determine the overall C.O.P. of the system.
Solution. Refer Fig. 5.15 (a).
(i) Temperature, T 1 = 25 + 273 = 298 K
Temperature, T 2 = 0 + 273 = 273 K

Source

25°C

Sink

0°C

Heat

pump

Q 2

Q 1

W

Source

380°C

Source

0°C

Sink (Atmosphere)

25°C

Heat

engine

Heat

pump

Q 3 Q 1

Q 4 Q 2

(a) Single system (b) Combined system

W

Fig. 5.15