TITLE.PM5

REFRIGERATION CYCLES 747

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\M-therm\Th14-3.pm5

Actual refrigerating effect per kg

= C.O.P.(actual) × work done

= 5.52 × (h 3 – h 2 ) = 5.52 × (1319.22 – 1196.23)

= 678.9 kJ/kg

Heat to be extracted per hour

=

28 1000 335

24

××

= 390833.33 kJ

Heat to be extracted per second =

390833.33

3600

= 108.56 kJ/s.

∴ Mass of refrigerant circulated per second = 108 56

678 9

.

.

= 0.1599 kg

Total work done by the compressor per second

= 0.1599 × (h 3 – h 2 ) = 0.1599 (1319.22 – 1196.23)

= 19.67 kJ/s

i.e., Power required to drive the compressor = 19.67 kW. (Ans.)

Example 14.14. A refrigerating plant works between temperature limits of – 5°C and

25 °C. The working fluid ammonia has a dryness fraction of 0.62 at entry to compressor. If the

machine has a relative efficiency of 55%, calculate the amount of ice formed during a period of

24 hours. The ice is to be formed at 0°C from water at 15°C and 6.4 kg of ammonia is circulated

per minute. Specific heat of water is 4.187 kJ/kg and latent heat of ice is 335 kJ/kg.

Properties of NH 3 (datum – 40°C).

Temp. Liquid heat Latent heat Entropy of liquid

°C kJ/kg kJ/kg kJ/kg K

25 298.9 1167.1 1.124

• 5 158.2 1280.8 0.630
  Solution. Fig. 14.25 shows the T-s diagram of the cycle.
  Enthalpy at point ‘2’, h 2 = hf 2 + x 2 hfg 2 = 158.2 + 0.62 × 1280.8 = 952.3 kJ/kg

Enthalpy at point ‘1’, h 1 = hf 4 = 298.9 kJ/kg
Also, entropy at point ‘2’ = entropy at point ‘3’
i.e., s 2 = s 3

sf 2 + x 2 sfg 2 = sf 3 + x 3 sfg 3

0.630 + 0.62 ×

1280.8

( 5 273)−+ = 1.124 + x^2 ×

1167.1

(25 273)+

i.e., x 3 = 0.63

∴ Enthalpy at point ‘3’, h 3 = hf 3 + x 3 hfg 3

= 298.9 + 0.63 × 1167.1 = 1034.17 kJ/kg

C.O.P.(theoretical) =

hh

hh

21

32

−

−

= −

−

952.3 298.9 =

1034.17 952.3

653.4

81.87

= 7.98.

C.O.P.(actual) = 0.55 × 7.98 = 4.39