The COP
If we had no +15°C ground water and we were assumed to use outdoor air at -10°C as the low temperature source (to which remove the heat), we would have, for each kW absorbed, a yield of 3.33 kW.
With a system of electric heating elements, for each kW electric absorbed, we obtain 1 kW heat.
The ratio of heat provided to electrical energy consumed of a heat pump is called COP (Coefficient Of Performance).
With a system of electric heating elements, for each kW electric absorbed, we obtain 1 kW heat.
The ratio of heat provided to electrical energy consumed of a heat pump is called COP (Coefficient Of Performance).
The figure’s example shows that COP is 3.33 / 1 = 3.33 (provided kW / kW absorbed).
In the previous example we had 6.47 kW (kW returns) / 1 (kW absorbed ) = 6.47 COP.
Another example : A heat pump absorbs 4 kW of electricity and makes 18.40 thermal kW. 18.40 / 4 = 4.6 ( COP ) .
The heat pump is nothing more than a refrigerator which uses the heat produced by the high temperature source .
In the previous example we had 6.47 kW (kW returns) / 1 (kW absorbed ) = 6.47 COP.
Another example : A heat pump absorbs 4 kW of electricity and makes 18.40 thermal kW. 18.40 / 4 = 4.6 ( COP ) .
The heat pump is nothing more than a refrigerator which uses the heat produced by the high temperature source .
