TITLE.PM5

718 ENGINEERING THERMODYNAMICS

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C.O.P.(Carnot ref.) = T

TT

2

12 −

i.e., 2.99 =

235

T 1 − 235

∴ T 1 =^235

299.

+ 235 = 313.6 K

= 313.6 – 273 = 40.6°C. (Ans.)

(iii)Heat delivered as heat pump

= Heat absorbed + Work done

=

14000

60 + 1.3 × 60 = 311.3 kJ/min. (Ans.)

C.O.P.(heat pump) =

Heat delivered

Work done

311.3

1.3 60

=
× = 3.99. (Ans.)
Example 14.2. A refrigerating system operates on the reversed Carnot cycle. The higher
temperature of the refrigerant in the system is 35°C and the lower temperature is – 15°C. The
capacity is to be 12 tonnes. Neglect all losses. Determine :
(i)Co-efficient of performance.
(ii)Heat rejected from the system per hour.
(iii)Power required.
Solution. (i) T 1 = 273 + 35 = 308 K
T 2 = 273 – 15 = 258 K
Capacity = 12 tonne

C.O.P. =

T

TT

2

12

258

− 308 258

=

− = 5.16. (Ans.)

(ii)Heat rejected from the system per hour :

C.O.P. =

Refrigerating effect

Work input

5.16 =

12 14000 kJ/h

Work input

×

∴ Work input =

12 14000

516

×

.

= 32558 kJ/h.

Thus, heat rejected/hour = Refrigerating effect/hour + Work input/hour

= 12 × 14000 + 32558 = 200558 kJ/h. (Ans.)

(iii)Power required :

Power required =

Work input/hour

60 60

32558

× 60 60

=

×

= 9.04 kW. (Ans.)

Example 14.3. A cold storage is to be maintained at – 5°C while the surroundings are at
35 °C. 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. (AMIE)