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Question please give answer

The capacity of a refrigerator is 300 TR when working between – 8°C and 25℃. Determine the mass of ice produced per day from water at 25°C. Also find the power required to drive the unit. Assume that the cycle operates on reversed Carnot cycle and latent heat of ice is 335 kJ/kg.

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Given:

Q= 200 TR

T1= -6 ° C = -6+273=267 K

T2= 25° C


Mass of ice produced per day


We know that heat extraction capacity of the refrigerator

= 200 x 210 = 42000 kI/min

Heat removed from 1 kg of water at 25°C to form ice at 0°C (1 TR=210 kJ/min)

= Mass x Sp. heat x Rise in temperature + hfg (ice)

= 1 x 4.187 (25-0) +335 = 439.7 ki/kg

Mass of ice produced per min

= $\frac{42000 }{95.52}$

= 97 7552 Kg/min

and mass of ice produced per day = 95.52 x 60 x 24 = 137550 kg = 137.55 tonnes


Power required to drive the unit

We know that C.O.P. of the reversed Carnot cycle

$\frac {T_1 }{T_2-T_1} $ =${267}{ 298-267}=8.6$

COP. = $ \cfrac {Heat extraction capacity}{Work done per mm} $

Work done per min = 42 000/8.6 = 4884 kJ/min


Power required to drive the unit

$= \frac{4884}{60}$ =81.4 kW

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