REFRIGERATION CYCLES 745
dharm
\M-therm\Th14-3.pm5
0.2232 + x 3 ×
138
298
= 0.1251 + 0.6 ×
154
268
= 0.4698∴ x 3 = (0.4698 – 0.2232) ×^298
138
= 0.5325T(K)298268s4 Condensation^3EvaporationCompression
Throttling12Fig. 14.23
Now, h 3 = hf 3 + x 3 hfg 3 = 59.7 + 0.5325 × 138 = 133.2 kJ/kg
Also, h 1 = hf 4 = 59.7 kJ/kgTheoretical C.O.P. = R
Wn = hh
hh21
32−
−
= −
−123.8 59.7
133.2 123.8
= 6.82Actual C.O.P. = ηrelative × (C.O.P.)theoretical = 0.5 × 6.82 = 3.41
Heat extracted from 1 kg of water at 20°C for the formation of 1 kg of ice at 0°C
= 1 × 4.187 × (20 – 0) + 335 = 418.74 kJ/kg
Let mice = Mass of ice formed in kg/min.(C.O.P.)actual = 3.41 =
R
Wn(actual) = m
mh hice. mice
().(kJ/min)
()(kJ/min)×
−
= ×
−418 74 418 74
32 6 133.2 123.8∴ mice = 6 133 2 123 8 3 41
418 74(. .).
.−× = 0.459 kg/min=
0 459 60 24
1000. ××
tonnes (in 24 hours) = 0.661 tonne. (Ans.)
Example 14.13. 28 tonnes of ice from and at 0°C is produced per day in an ammonia
refrigerator. The temperature range in the compressor is from 25°C to – 15°C. The vapour is dry
and saturated at the end of compression and an expansion valve is used. Assuming a co-efficient
of performance of 62% of the theoretical, calculate the power required to drive the compressor.