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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?

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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.

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  • $\begingroup$ Thank you, sir. What a delight. I am still new to COP and the reversible Rankine refrigeration cycle, it's frustrating to not be able to understand this fully. $\endgroup$
    – Qwin
    Commented Jun 13, 2020 at 15:31
  • $\begingroup$ @Qwin So, you found the answer to be acceptable? $\endgroup$
    – Bob D
    Commented Jun 13, 2020 at 16:37
  • $\begingroup$ Yes, sir. I upvoted with great admiration of your knowledge. Thank you. $\endgroup$
    – Qwin
    Commented Jun 14, 2020 at 10:43
  • $\begingroup$ @Qwin I only asked because you didn’t press the “accept” button. New contributors are often unaware of it $\endgroup$
    – Bob D
    Commented Jun 14, 2020 at 10:52
  • $\begingroup$ Where can I find it, sir? $\endgroup$
    – Qwin
    Commented Jun 14, 2020 at 10:53

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