The power consumption of a refrigerating machine is determined in terms of kW. However the power consumption of the driving motor is sometimes rated in horsepower (H.P). we have

where Wref is in kW and HP is in metric units,


where Q1 = refrigerating capacity in kW
Since 1 TR = 3.5167 kW, therefore

Example 18.4
The ambient air temperature during summer and winter in a particular locality are 45°C and 15°C respectively. Find the value of Carnot COP for an air-conditioner for cooling and heating, corresponding to refrigeration temperatures of 5°C for summer and heating temperature of 55°C for winter. Also find the theoretical power consumption per ton of refrigeration in each case.
Solution
For summer: T1 = 273 + 5 = 278 K, T2 = 45 + 273 = 318 K
For cooling, Carnot ![]()
Power consumption, ![]()
For winter: T1 = 273 + 15 = 288 K, T2 = 273 + 55 = 328 K
For heating, ![]()
Power consumption, ![]()
Example 18.5
A refrigerator has working temperature of −30°C and 35°C. The actual COP is 0.75 of the maximum. Calculate the power consumption, and heat rejected to the surroundings per ton of refrigeration.
Solution

Power consumption per ton ![]()
Heat rejected per ton = 3.5167 + 1.256 = 4.7727 kW
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