How to calculate a coefficient of performance (COP)

  • #1
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Moved from a technical forum, so homework template missing

Homework Statement



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Homework Equations



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3. The Attempt at a Solution


Here's my solution attempt:

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NOTE: Evolution 2 to 3 is really an isothermic process, but in the diagram shows as an politropic process.

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4. Further Questions

But it isn't giving me logical values of COP when I compare it with a reverse cycle of Carnot.

Question: How could I calculate the COP and COPmax of this cycle?
Not sure how to arrange heats and work on COPs formula for a triangular cycle like this one.
 

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Answers and Replies

  • #2
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In this cycle, you are removing heat from one part of the surroundings at a low temperature (7088 J) and discharging heat to another part of the surroundings at a higher (range of) temperature (2809 J + 5063 J = 7871 J). To accomplish this pumping of heat, you are supplying work to the system (782.63 J). So, viewed as a heater, its COP is the heat delivered divided by the work done (7871/783). Viewed as a cooler, its COP is the heat removed divided by the work done (7088/783). So basically, except for the sign on cooler performance, both your results are correct.
 
  • #3
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@Chestermiller Thanks! But ...

Can the COP of my cycle be greater than the COP of an equivalent carnot cycle?
I thought that it should always be smaller, even in reversible cycles.
 
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  • #4
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@Chestermiller Thanks! But ...

Can the COP of my cycle be greater than the COP of an equivalent carnot cycle?
I thought that it should always be smaller, even in reversible cycles.
Can you please define the "equivalent Carnot cycle," given that the temperature is changing in steps 1-2 and in 3-1?
 
  • #5
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Can you please define the "equivalent Carnot cycle," given that the temperature is changing in steps 1-2 and in 3-1?
Well... I guess that would be a reverse carnot cycle that works between the same two temperature sinks (243 k and 487k).
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  • #6
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Well... I guess that would be a reverse carnot cycle that works between the same two temperature sinks (243 k and 487k).
View attachment 231938
But this system is not operating between 2 constant temperature sinks at 243 and 487. In step 1-2, the system is being heated along the process path from 243 to 487, so it is being brought into contact with a sequence of constant temperature reservoirs running from the lower temperature to the higher temperature. And, in step 3-1, the system is being cooled along the process path from 487 to 243, so it is being brought into contact with a sequence of constant temperature reservoirs running from the higher temperature to the lower temperature. The only part of the process where a single constant temperature reservoir is being used is step 2-3.
 
  • #7
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So, you are saying that since my cycle has an isothermal process, and for it to be reversible this must be achieved with infinite temperature sources ... then the theoretical COP of a carnot cycle does not have to be greater than the COP of my cycle because it does not really have only 2 sources ...

Sorry, I'm still confused, the carnot cycle does not also have reversible isothermal passes?

I always thought that Efficience and COP from Carnot Cycles where always greater than those of other cycles that were not carnot, reversible or irreversible.

I am very grateful for your answers, but I am a bit harsh with this topic.
 
  • #8
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So, you are saying that since my cycle has an isothermal process, and for it to be reversible this must be achieved with infinite temperature sources ... then the theoretical COP of a carnot cycle does not have to be greater than the COP of my cycle because it does not really have only 2 sources ...
No. That's not what I'm saying. The isothermal (487) reversible step 2-3 is OK to allow a comparison with a Carnot cycle. But, what I'm saying is that that steps 1-2 and 3-1 of your process, although being reversible, are not isothermal. So there is no one temperature along these paths that you can use to compare directly with an isothermal leg of a Carnot cycle. The most you can say is that some temperature between 243 and 487 might be appropriate for such a comparison, but it is not obvious what the specific temperature should be.
 
  • #9
kuruman
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For whatever it's worth, I have encountered problems like this before and the expected answer for the Carnot efficiency assumes a cycle the isothermals of which are at the highest and lowest possible temperatures at which the system may find itself.
 
  • #10
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So, s
No. That's not what I'm saying. The isothermal (487) reversible step 2-3 is OK to allow a comparison with a Carnot cycle. But, what I'm saying is that that steps 1-2 and 3-1 of your process, although being reversible, are not isothermal. So there is no one temperature along these paths that you can use to compare directly with an isothermal leg of a Carnot cycle. The most you can say is that some temperature between 243 and 487 might be appropriate for such a comparison, but it is not obvious what the specific temperature should be.

In short, if I have an isobaric or isochoric process in a cycle, then I can not compare to the COP of a carnot cycle because the sinks are not well defined?
 
  • #11
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For whatever it's worth, I have encountered problems like this before and the expected answer for the Carnot efficiency assumes a cycle the isothermals of which are at the highest and lowest possible temperatures at which the system may find itself.
Do you have an example of that, that would be cool.
 
  • #12
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4,607
So, s


In short, if I have an isobaric or isochoric process in a cycle, then I can not compare to the COP of a carnot cycle because the sinks are not well defined?
Your reversible process is well defined. I just do see how to compare this to a Carnot cycle on a rational basis.
 
  • #13
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Here's an approach to an equivalent Carnot cycle that might make sense.

The transfer of heat to the hot reservoir(s) is 7871 J, and the transfer of entropy to the hot reservoir(s) is 20.16 J/K. So the equivalent Carnot hot reservoir temperature is 7871/20.16 = 390.43 K.

The transfer of heat from the cold reservoir(s) is 7088.62 J, and the transfer of entropy from the cold reservoir(s) is 20.16. So the equivalent Carnot cold reservoir temperature is 7088.62/20.16 = 351.61 K.

Both these values lie between the extreme temperatures of 243K and 487 K. Anyway, use of these two temperatures in a Carnot cycle along with the known heats give exactly the same values for the COPs that we have already calculated.
 

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