➢➣An ice plant produces 100 tons of ice in 24 hrs. If the plant uses water at
200 C, find its cooling capacity. Specific heat and latent heat of
solidification of water are 4.2 kJ/kg/K and 334 kJ/kg respectively.
Solution:
Given,
m = 100 tons = 100000 Kg
T = 200C = 293 K
Specific heat, cf = 4.2 kJ/kg/K
Latent heat of solidification, hfg = 334 kJ/kg
Cooling Capacity =?
We know that,
Heat removed from water to ice at 200C,
Q1 = m cf T
= 100000 x 4.2 x 293
=1.23 x108 KJ
Total latent heat of freezing,
Q2 = m hfg
= 100000 x 334
= 3.34
x107 KJ
Total heat removed in 24 hours,
Q = Q1 +
Q2
= 1.23 x108+3.34
x107
= 156.46 x106 KJ
So, Heat removed in 1 minutes
= 156.46 x106/(24
x 60)
= 108.62 x103 KJ/min
Cooling Capacity,
= 108.62 x103/210
=
517.39 TR
So, Cooling Capacity of ice plant is = 517.39 TR
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