Answer :
The COP of the air conditioning unit is 7.596, and the rate at which heat is rejected to the ambient is 27,990 kJ/h.
How can we calculate COP and rate of heat rejection ?
To determine the COP of an 18,000 Btu/h split air conditioner and the rate at which heat is rejected to the environment at 45°C, follow these steps:
- Calculate the cooling capacity in kilojoules per hour:
18,000 Btu/h × 1,055 kJ/Btu = 18,990 kJ/h - Calculate the COP (Coefficient of Performance) by dividing the cooling capacity by the power input to the compressor:
COP = Cooling Capacity / Power Input
COP = 18,990 kJ/h / 2,500 W
COP = 18,990 kJ/h / 2.5 kW (convert W to kW)
COP = 7.596
The COP of the air conditioning unit is 7.596. - Determine the rate at which heat is rejected to the ambient from the air conditioner condenser:
Heat Rejected = Cooling Capacity + Power Input
Heat Rejected = 18,990 kJ/h + 2.5 kW × 3600 s/h (convert kW to kJ/h)
Heat Rejected = 18,990 kJ/h + 9,000 kJ/h
Heat Rejected = 27,990 kJ/h
The rate at which heat is rejected to the ambient from the air conditioner condenser is 27,990 kJ/h.
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