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.

(Note: [1 Btu = 1,055 kJ])

COP of the air conditioning unit is 7.596, and the rate at which heat is rejected to the ambient from the air conditioner condenser is 21.49 kW.

To determine the Coefficient of Performance (COP) of the 18,000 Btu/h split air conditioner and the rate at which heat is rejected to the ambient from the air conditioner condenser, follow these steps:

1. Convert the air conditioner capacity from Btu/h to kW:
18,000 Btu/h * (1,055 kJ / 1 Btu) * (1 kW / 1,000 kJ) = 18,990 W = 18.99 kW

2. Calculate the COP:
COP = Cooling Capacity (kW) / Power Input (kW)
COP = 18.99 kW / 2.5 kW = 7.596

3. Calculate the heat rejected to the ambient:
Heat Rejected = Cooling Capacity + Power Input
Heat Rejected = 18.99 kW + 2.5 kW = 21.49 kW

So, the COP of the air conditioning unit is 7.596, and the rate at which heat is rejected to the ambient from the air conditioner condenser is 21.49 kW.

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