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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.

1 Answer

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Answer:

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).

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