Example – Heat Pump – Heating and Air Conditioning
- Calculate the amount of heat (Qhot) the heat pump can add to a room?
- If the heat pump were turned to the cooling mode (i.e. to act as an air conditioner in the summer), what would you expect its coefficient of performance to be? Assume all else stays the same and neglect all other losses.
From the COP, which is defined as:
the amount of heat the heat pump can add to a room is equal to:
Qhot = COPheating x W = 3 x 1500 = 4500 W or 4500 J/s
In case of the cooling mode, the heat pump (air conditioner) with 1500 W motor can take heat Qcold from inside the house and then dump Qhot = 4500 W to the hot outside. Using the first law of thermodynamics, which states:
Qcold + W = Qhot,
we obtain the heat, Qcold = 3000 W. From the definition: COPcooling = 3000/1500 = 2.
Note that, in this example we have many assumptions. For example, we assumed that the temperature difference (Thot – Tcold) is the same for both modes. But we have swapped reservoirs, without any impact on COP. It is only an illustrative example.
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