A reversed Brayton refrigeration cycle uses air as its working fluid in a closed

Question

A reversed Brayton refrigeration cycle uses air as its working fluid in a closedloop.The air enters the compressor at 100 kPa and -10oC, and enters the turbine at 1000kPa and 30oC. The mass flow rate of the air through the cycle is 1.5 kg/s. Determine the rate at which heat is removed from the cold space and the coefficient of performance of the cycle if (a) the turbine and compressor are isentropic and (b) the turbine and compressor both have isentropic efficiencies of 0.80. Assume that the specific heats of the air are constant.

Explanation / Answer

In the below diagram, the rate at which heat is removed from the cold space is represented by qe.

qe = m Cp (T2-T6)-----------------(1)

T2 is given, to find T6, we use isentropic expansion process 5-6, in which T6/T5 = (p6/p5)[(gamma-1)/gamma]

Since, p6 = p2, we have

T6 = 303 x (100/1000)(1.4-1)/1.4 = 156.93oK

Putting these values in eq.1

qe = 1.5 x 1 x (263-156.93) = 159 kJ/s

Coefficient of performance of the cycle

COP = (T2-T6)/[(T3-T2)-(T5-T6)], here we know all temperatures except, T3.

To find T3, we use isentropic compression process 2-3, in which T3/T2 = (p3/p2)[(gamma-1)/gamma]

So, T3 = 263 x (1000/100)(1.4-1)/1.4 = 507.77oK

Hence, COP = (263-156.93)/[(507.77-263)-(303-156.93)] = 1.074

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