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An 18,000 Btu/h split air conditioner is running at full load to keep a room at 25°C in an environment at 45°C. The power input to the air conditioner compressor is 2.5 kW.

Determine:

1. The COP (Coefficient of Performance) of the air conditioning unit.
2. The rate at which heat is rejected to the ambient from the air conditioner condenser.

[1 Btu = 1,055 J]

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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