A Refrigeration cycle receives 135 kJmin of heat transfer at

A Refrigeration cycle receives 135 kJ/min of heat transfer at a boundary temperature of -30degree C and rejects heat at a boundary temperature of 27 degree C. If the actual COP is 60 percent of the maximum possible value, (a) determine the power required, in kilowatts, (b) If the low temperature is raised to -10degree C,  find the percent change in the work required in %.

Solution

a)

T_high = 273 + 27 = 300 K

T_low = 273 - 30 = 243 K

Max COP = 1 / (T_high / T_low - 1)

= 1 / (300/243 - 1)

= 4.26

Actual COP = 0.6*4.26 = 2.56

COP = Q_low / W

2.56 = 135 / W

W = 52.77 kJ/min

= 52.77/60 kW

= 0.88 kW

b)

Repeating the above calculations with T_low = 273-10 = 263 K we get

Ideal COP = 1 / (300/263 - 1) = 7.108

Actual COP = 0.6*7.108 = 4.265

W = 135/4.265 = 31.65 kJ/min = 0.53 kW

% change in power input = (0.53 - 0.88) / 0.88 *100 = -40.04% or 40.04% lower