A cold storage plant is required to store 20 tonnes of fish. The tempe

1.	A cold storage plant is required to store 20 tonnes of fish. The temperature of the fish when supplied = 25°C ; storage temperature of fish required = – 8°C ; specific heat of fish above freezing point = 2.93 kJ/kg-°C ; specific heat of fish below freezing point = 1.25 kJ/kg- °C ; freezing point of fish = – 3°C. Latent heat of fish = 232 kJ/kg. If the cooling is achieved within 8 hours ; find out : (i) Capacity of the refrigerating plant. (ii) Carnot cycle C.O.P. between this temperature range. (iii) If the actual C.O.P. is 1/3rd of the Carnot C.O.P. find out the power required to run the plant
Answer» Heat removed in 8 hours from each kg of fish = 1 × 2.93 × [25 – (– 3)] + 232 + 1 × 1.25 [– 3 – ( – 8)] = 82.04 + 232 + 6.25 = 320.29 kJ/kg Heat removed by the plant /min = \(\cfrac{320.29\times20\times1000}{8}\) = 800725 kJ/h (i) Capacity of the refrigerating plant = \(\cfrac{800725}{14000}\) = 57.19 tonnes. T₁ = 25 + 273 = 298 K T₂ = – 8 + 273 = 265 K ∴ C.O.P. of reversed Carnot cycle = \(\cfrac{T_2}{T_1-T_2}\) = \(\cfrac{265}{298-265}\) = 8.03. (iii) Power required : Actual C.O.P. = \(\cfrac13\) x Carnot C.O.P. = \(\cfrac13\) x 8.03 = 2.67 But actual C.O.P. = \(\cfrac{Net \,refrigerating \,effect/min}{Work \,done /min}\) = \(\cfrac{R_n}W\) 2.67 = \(\cfrac{800725}W\) kJ/h W = \(\cfrac{800725}W\) = 299897 kJ/h = 83.3 kJ/s ∴ Power required to run the plant = 83.3 kW.

A cold storage plant is required to store 20 tonnes of fish. The temperature of the fish when supplied = 25°C ; storage temperature of fish required = – 8°C ; specific heat of fish above freezing point = 2.93 kJ/kg-°C ; specific heat of fish below freezing point = 1.25 kJ/kg- °C ; freezing point of fish = – 3°C. Latent heat of fish = 232 kJ/kg. If the cooling is achieved within 8 hours ; find out : (i) Capacity of the refrigerating plant. (ii) Carnot cycle C.O.P. between this temperature range. (iii) If the actual C.O.P. is 1/3rd of the Carnot C.O.P. find out the power required to run the plant

Answer»

Heat removed in 8 hours from each kg of fish

= 1 × 2.93 × [25 – (– 3)] + 232 + 1 × 1.25 [– 3 – ( – 8)]

= 82.04 + 232 + 6.25 = 320.29 kJ/kg

Heat removed by the plant /min

= \(\cfrac{320.29\times20\times1000}{8}\) = 800725 kJ/h

(i) Capacity of the refrigerating plant

= \(\cfrac{800725}{14000}\) = 57.19 tonnes.

T₁ = 25 + 273 = 298 K

T₂ = – 8 + 273 = 265 K

∴ C.O.P. of reversed Carnot cycle

= \(\cfrac{T_2}{T_1-T_2}\) = \(\cfrac{265}{298-265}\) = 8.03.

(iii) Power required :

Actual C.O.P. = \(\cfrac13\) x Carnot C.O.P. = \(\cfrac13\) x 8.03 = 2.67

But actual C.O.P. = \(\cfrac{Net \,refrigerating \,effect/min}{Work \,done /min}\) = \(\cfrac{R_n}W\)

2.67 = \(\cfrac{800725}W\) kJ/h

W = \(\cfrac{800725}W\) = 299897 kJ/h = 83.3 kJ/s

∴ Power required to run the plant = 83.3 kW.