Ways of calculating coefficient of performance (COP) Calculating COP with the help of:

*

*Power (i.e. a whole plant)

*Enthalpies (with refrigerant)

*Using Carnot's COP as a comparison (this is mainly theoretical).

The refrigerant could be R134a as a processing medium, where it acts as different temperatures on the source. The source in this case is an immersion heater, even though realistically, we'd use seawater.
I accidentally pressed enter, which published the question. However, I meant to say that, how can I calculate the COP with the help of those three?
 A: I'm not sure about an actual power plant as a whole (I am not conversant in power plant engineering). But I would guess you would need to apply the general definition of COP. The COP is defined as the ratio of the desired heat transfer to the work required to make the transfer. What is considered to be the "desired heat transfer" depends on whether you are looking at it as a heat pump or as an air conditioner/refrigerator.
$$COP_{HP}=\frac{Q_H}{W}$$
$$COP_{AC}=\frac{Q_L}{W}$$
and since $W=Q_{H}-Q_{L}$
$$COP_{HP}=\frac{Q_H}{Q_{H}-Q_L}$$
$$COP_{HP}=\frac{Q_L}{Q_{H}-Q_L}$$
Where $Q_H$ is the heat transfer to the higher temperature environment and $Q_L$ is the heat transfer out of the low temperature environment.
For the Carnot heat pump/refrigerator the COP only depends on the high and low temperature environment temperatures, $T_H$ and $T_L$, or for Carnot
$$COP_{HP}=\frac{T_H}{(T_{H}-T_{L})}$$
$$COP_{AC}=\frac{T_L}{(T_{H}-T_{L})}$$
An example of using enthalpies is the reversible Rankine refrigeration cycle. In this case
$$COP_{HP}=\frac{h_{1}-h_4}{h_{2}-h_1}$$
$$COP_{AC}=\frac{h_{2}-h_3}{h_{2}-h_1}$$
Where
$h_1$ is the enthalpy at the evaporator output
$h_2$ is the enthalpy at the compressor input
$h_3$ is the enthalpy at the condenser output
$h_4$ is the enthalpy at the evaporator input.
Hope this helps.
