REFRIGERATION CYCLES 717

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(ii)For a Carnot cycle ‘heat pump’ :

C.O.P.(heat pump) =

Heat rejected to the hot body/cycle

Work done per cycle

=

Tlm

TT pn

1

12

×

()−×

=

Tpn

TT pn

1

12

×

()−× ()Q lm=pn

=

T

TT

1

12 −

...(14.3)

= 1 + T

TT

2

12 −

...(14.4)
This indicates that C.O.P. of heat pump is greater than that of a refrigerator working on
reversed Carnot cycle between the same temperature limits T 1 and T 2 by unity.
(iii)For a Carnot cycle ‘heat engine’ :
C.O.P.(heat engine) = Work obtained/cycle
Heat supplied/cycle

= −×

×

= −×

×

() ()TT pn

Tlm

TT pn

Tpn

12

1

12

1

()Q lm=pn

= TT

T

12

1

− ...(14.5)

Example 14.1. A Carnot refrigerator requires 1.3 kW per tonne of refrigeration to main-
tain a region at low temperature of – 38°C. Determine :
(i)C.O.P. of Carnot refrigerator
(ii)Higher temperature of the cycle
(iii)The heat delivered and C.O.P. when this device is used as heat pump.
Solution. T 2 = 273 – 38 = 235 K
Power required per tonne of refrigeration = 1.3 kW
(i)C.O.P. of Carnot refrigerator :
C.O.P.(Carnot ref.) =
Heat absorbed
Work done

=

1

3

14000

36060

tonne kJ/h

1. 
   
   
   
   1. 
      

      kJ/h

      
      

      ×× = 2.99. (Ans.)
      (ii)Higher temperature of the cycle, T 1 :

      
      
   
   
   
   

Fig. 14.2