Determine Time taken to gain cooling:

A cold storage plant is needed to store up 20 tonnes of fish. The fish is supplied at a 30°C. The specific heat of fish above freezing point is 2.93 kJ/kg K. The specific heat offish below freezing pt. is 1.26 kJ/kg K. The fish is stored up in cold storage that is maintained at - 8^oC. The freezing pt of fish is - 4°C. The latent heat of fish is 235 kJ/kg. If the plant needs 75 kW to drive it, determine

(1) The capacity of the plant, &

(2) Time taken to gain cooling.

Suppose actual C.O.P. of the plant like 0.3 of the Carnot C.O.P.

Solution

It is Given:   m = 20 t = 20000 kg; T₂ = 30°C = 30 + 273 = 303 K;

cAF = 2.93 kJ/kg K; cBF = 1.26 kJ/kgK; T₁ = - 8°C = -8 + 273 = 265 K;

T₃ = -4°C = -4 + 273 = 269 k; hfg = 235 kJ/kg; P = 75 kW = 75 kJ/s

(1)        Estimate Capacity of the plant

As we know that Carnot C.O.P.

Actual COP. = 0.3 * 6.97 = 2.091 and heat eliminate by the plant

= Actual COP. * Work required

= 2.091 * 75 = 156.8 kJ/s = 156.8 * 60 = 9408 kJ/min

Then Capacity of the plant

= 9408 /210 = 44.8 TR            Ans.

(b)        Time taken to gain cooling

As we know that heat eliminated from the fish above freezing point,

Q₁ = m * cAF (T₂ - T₃)

= 20000 * 2.93 ( 303 - 269)

= 1.992 * 106 kJ

Likewise, heat eliminated from the fish below freezing point,

Q₂ = m x cBF (T3 - T1)

= 20000 * 1.26 (269 - 265)

= 0.101 *106 kJ

and net latent heat of fish,

Q₃ = m * hfg (fish)

= 20000 * 235

= 4.7 * 106 kJ

Net heat eliminated by the plant

= Q₁ + Q₂ + Q₃

= 1.992 * 10⁶ + 0.101 * 10⁶ + 4.7 * 10⁶

= 6.793 * 10⁶ kJ

and time taken to gain cooling

6.793 * 10⁶ = 722 min = 12.03 h   Ans.