Refrigeration cycle with two compressors

[murrdpirate0]

Homework Statement

Here is the cycle:

I am to assume cold-air standard applies.

I need to:
a)Draw a T-s diagram
b)Find where heat is transferred and work is done
c)Calculate the coefficient of performance

Homework Equations

COP = 1 / (T_H/T_L - 1)

The Attempt at a Solution

For the T-s diagram, as you can see on the image above, I just labeled what happens to entropy on each process (except I forgot to label the turbine, which like the compressor would be constant s ideally). I'm not really sure if this is an ideal case or not, so I guess s=const could be wrong.

Each compressor does work on the fluid, the fluid does work on the turbine, heat is transferred from the fluid to the surroundings in the intercoolers between the compressors, and heat is transferred from the surroundings to the fluid between 6 and 1. So that settles the second part of the hw.

Assuming that the fluid reaches equilibrium with the environment in the intercooling stage and the evaporation stage, T_H would be 293K and T_L would be 293. So COP would be infinity. Which obviously is impossible, but I think that could be the intention of this problem since we are starting to learn about irreversibility (and also because it seems to say that the turbine is driving the compressors, which would be a perpetual motion machine). But is it possible I'm neglecting something? Even if T_H was higher, what would be the point of the second compressor? Like say T₃=T₅=T_H=303. Then COP would calculate the same whether there was one compressor or two. Perhaps if there were only 1 compressor, T₆ would be higher?

 

[Q_Goest]

Hi murrdpirate0,
a) You're correct about dS = 0 for compressors and turbine assuming perfect isentropic performance (isentropic efficiency = 100%). And there is a change in entropy through the heat exchangers, but I'd suggest not showing this entropy change as an arrow. For the case of a fluid that's cooling at constant pressure, such as for the two heat exchangers where heat is rejected by the system, the entropy of the fluid drops, so showing an arrow pointing into the heat exchanger may be misleading. Or did you mean to show that entropy was dropping because the arrow is pointing down?

b) I think you've answered this one correctly.

c) It's been a long time since I did refrigeration cycles, but I believe T_H is the high temperature reservoir and T_L is the low temp reservoir. From Wikipedia:
where T_hot and T_cold are the temperatures of the hot and cold heat reservoirs respectively.
Ref: http://en.wikipedia.org/wiki/Coefficient_of_performance
Read up on COP and try this part again.

 

[Mene]

Hello,

Thank you for sharing your solution attempt. I am a scientist and I would like to provide you with some feedback on your response.

Firstly, your T-s diagram seems to be correct. It shows the different processes involved in the refrigeration cycle and the changes in entropy. However, it is important to note that the compressor and turbine processes are not ideal, as they are not reversible. Therefore, the entropy of the fluid would increase during these processes.

Secondly, you are correct in identifying the heat transfer and work done in the cycle. However, it is important to note that the heat transfer from the surroundings to the fluid at stage 6 is not a part of the refrigeration cycle. This is simply a heat rejection process to maintain the cycle.

Thirdly, your calculation of the coefficient of performance (COP) is incorrect. The COP equation you have used is for a Carnot cycle, which is an idealized cycle. For a refrigeration cycle, the COP is given by COP = QL/W, where QL is the heat absorbed from the cold reservoir (stage 6) and W is the work done by the compressor. In this case, the COP would not be infinity, as you have correctly pointed out, but it would be higher than a single compressor cycle.

Finally, the purpose of the second compressor is to increase the pressure of the fluid, which increases its temperature and makes it easier to reject heat to the surroundings at stage 6. This allows for a higher COP as compared to a single compressor cycle.

I hope this helps clarify some of your doubts. Keep up the good work in learning about thermodynamics and irreversibility!