Question

A refrigeration cycle absorbs 20,000 kJ/h of heat from a space maintained at -45^oC and rejects heat to the surrounding air at 30^oC.
i. Determine the actual power input to the refrigeration cycle if the coefficient of performance (COP) of the refrigeration cycle is 25% of that of a reversible refrigerating cycle operating between the same two temperature reservoirs (kW).
ii. Determine the minimum work input to the refrigeration cycle (kW).
iii. Will the heat rejected to the surrounding by the actual refrigeration be greater, smaller or the same as the heat rejected to the surrounding by the reversible refrigeration cycle?

Answer 1

Ans

(i) For reversible refrigerating cycle, (COP)_rev. = T2 / (T1 -T2)

= 228 / (303 - 228)

= 3.04

For actual cycle, (COP)_actual = 25% of (COP)_rev.

= 0.25 * 3.04

= 0.76

Also, (COP)_actual = Q2 / W

0.76 = 20000 / W

      W = 26315.79 kJ/h

= ( 26315.79 / 3600) kW

= 7.31 kW (ans)

where, 'W' is the actual power input

(ii) Work input will be minimum in case of refrigeration cycle, therefore,

(COP)_rev. = Q2 / W_min   

3.04 = 20000 / W_min

  W_min = 6578.95 kJ/h

= (6578.95 / 3600) kW

= 1.83 kW (ans)

(iii) Heat rejected by a actual refrigeration cycle is always greater than the reversible refrigeration cycle. We can show this as follows

Heat rejected by actual refrigeration cycle, (Q2)_actual = Q1 + W

= 20000 + 26315.79

= 46315.79 kJ/h

Heat rejected by reversiblel refrigeration cycle, (Q2)rev = Q1 + W_rev

= 20000 + 6578.95

= 26578.95 kJ/h

So, we see that   (Q2)_actual > (Q2)rev