Question

Determine the COP of a heat pump that supplies energy to a house at a rate of 8200 kJ/h for each kW of electric power it draws. Also, determine the rate of energy absorption from the outdoor air.

Answer 1

`COP_(HP) = 2.278`,
`\dot Q_(L) = 1.278\,kW`

Step-by-step explanation:

The heat supply is:

`Q_(H) = (8200\,(kJ)/(h))\cdot ((1\,h)/(3600\,s) )`

`Q_(H) = 2.278\,kW`

The Coefficient of Performance of a Heat Pump is:

`COP_(HP) = (\dot Q_(H))/(\dot W)`

`COP_(HP) = (2.278\,kW)/(1\,kW)`

`COP_(HP) = 2.278`

The rate of heat absorption from the outdoor air is:

`\dot Q_(L) = \dot Q_(H) - \dot W`

`\dot Q_(L) = 2.278\,kW-1\,kW`

`\dot Q_(L) = 1.278\,kW`