Determine the refrigeration capacity:

Five hundred kgs fruits are supplied to a cold storage at 20^oC. The cold storage is maintained at temp 5°C & the fruits get cooled up to the storage temperature within 10 hours. The latent freezing heat is 105 kJ/kg & specific heat of fruit is 1.256 kJ/kg K. Determine the refrigeration capacity of the plant.

Solution

It is given that: m = 500 kg; T₂ = 20°C = 20 + 273 = 293K;

T₁ = - 5C = - 5 + 273 = 268 K; hfg = 105 kJ/kg, cF = 1.256 kJ/kg K

As we know that heat eliminate from the fruits in 10 hrs is ,

Q₁ = m cF (T₂ - T₁)

= 500 * 1.256(293 - 268) = 15700 kJ

and net latent heat of freezing,

Q₂ = m . hfg = 500 * 105 = 52500 kJ

Net heat eliminate in 10 hrs,

Q = Q_l + Q₂ = 15700 + 52500 = 68200 kJ

and Net heat eliminated in one minute

= 68200/10 * 60 = 113.7 kJ/min

Refrigeration capacity of the plant

= 113.7/210 = 0.541 TR (1 TR = 210 kJ/min)  Ans.