SECOND LAW OF THERMODYNAMICS AND ENTROPY 245

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/M-therm/th5-2.pm5

Heat removal rate from the refrigerator,

Q 1 = 1440 kJ/min = 24 kJ/s

Now, co-efficient of performance, for reversible heat pump,

C.O.P. =

T

TT

1

12 −

=

298

()298 273− = 11.92. (Ans.)

∴ (C.O.P.)ref. =

T

TT

2

12

273

− 298 273

=

−

= 10.92

Now, 10.92 = Q

WW

(^1) =^24
∴ W = 2.2 kW
i.e., Work input required = 2.2 kW. (Ans.)
Q 2 = Q 1 + W = 24 + 2.2 = 26.2 kJ/s
(ii) Refer Fig. 5.15 (b).
The overall C.O.P. is given by,
C.O.P. =
Heat removed from the refrigerator
Heat supplied from the source
=^1
3
Q
Q ...(i)
For the reversible engine, we can write
Q
T
Q
T
3
3
4
4

or
QW
T
4 +
3

Q
T
4
4
or QQ^4 +2.

• =


• 
  

  2
  380 273 25 273
  4
  ()()
  or QQ^4 +2.2=
  653 298
  4
  or 298(Q 4 + 2.2) = 653 Q 4
  or Q 4 (653 – 298) = 298 × 2.2
  or Q 4 = 298 2 2
  653 298
  ×
  −
  .
  ()
  = 1.847 kJ/s
  ∴ Q 3 = Q 4 + W = 1.847 + 2.2 = 4.047 kJ/s
  Substituting this value in eqn. (i), we get
  C.O.P. =^24
  4 047.
  = 5.93. (Ans.)
  If the purpose of the system is to supply the heat to the sink at 25°C, then
  Overall C.O.P. =
  QQ
  Q
  24
  3

  
  


• 
  

  = 26.2 1.847
  4.047

  
  


• = 6.93. (Ans.)
  Example 5.14. An ice plant working on a reversed Carnot cycle heat pump produces
  15 tonnes of ice per day. The ice is formed from water at 0°C and the formed ice is maintained at
  0 °C. The heat is rejected to the atmosphere at 25°C. The heat pump used to run the ice plant is