Question

An air conditioner removes heat steadily from a house at a rate of 700 kJ/min while drawing electric power at a rate of 5.10 kW. Determine:

a. the COP of this air conditioner
b. the rate of heat transfer to the outside air.

Answer 1

a) The Coefficient of Performance of this air conditioner is 2.288.

b) The rate of heat transfer to the outside air is 16.767 kilowatts (1006.02 kilojoules per minute).

Step-by-step explanation:

a) Air conditioners are applications of refrigeration cycles, whose performance is analyzed by means of the Coefficient of Performance (
`COP_(R)`), dimensionless:

`COP_(R) = (\dot Q_(L))/(\dot W)` (Eq. 1)

Where:

`\dot Q_(L)` - Heat removal rate from the house, measured in kilowatts.

`\dot W`- Electric power, measured in kilowatts.

If we know that
`\dot W = 5.10\,kW` and
`\dot Q_(L) = 11.667\,kW`, the Coefficient of Performance of this air conditioner is:

`COP_(R) = (11.667\,kW)/(5.10\,kW)`

`COP_(R) = 2.288`

The Coefficient of Performance of this air conditioner is 2.288.

b) We find the rate of heat transfer to the outside air (
`\dot Q_(H)`), measured in kilowatts, by applying the First Law of Thermodynamics:

`\dot Q_(H) = \dot Q_(L)+\dot W` (Eq. 2)

If we get that
`\dot W = 5.10\,kW` and
`\dot Q_(L) = 11.667\,kW`, then the rate of heat transfer to the outside air is:

`\dot Q_(H) = 11.667\,kW+5.10\,kW`

`\dot Q_(H) = 16.767\,kW`

The rate of heat transfer to the outside air is 16.767 kilowatts (1006.02 kilojoules per minute).