Impossible refrigerator (thermodynamics and PV diagrams)

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Homework Statement


I have a rectangular PV diagram of a heat engine. Like this one:
Cyclic_process.PNG

I have to show why this cannot work in reverse as a refrigerator.

Homework Equations


U = 0.5fnRT (U = internal energy, f = degrees of freedom, n = number of moles of gas, R = ideal gas constant, T = temperature)
dU = Q + W = TdS - PdV (S = entropy, V = volume, Q = heat, W = work, P = pressure)
PV = nRT


The Attempt at a Solution


According to a website this wouldn't work because the refrigerator would get warmer and warmer until it got warmer than the cold reservoir. However, temperature is a state variable, and when I calculated it, dT = 0 as expected. So, while it may get warmer during the cycle, it will return to its original temperature.

The answer may lie in the entropy, but I am quite sure entropy is a state variable in this case.
 

Answers and Replies

  • #2
Matterwave
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A refrigerator works by using work to pump heat from a colder reservoir to a hotter one. Look at the processes D and B, what processes are those? What work is being done during those two legs of the cycle? What does that tell you about the heat transfer between the engine and the reservoir with which it is in contact? What has to happen if I wanted to turn those legs around to go the other way?
 
  • #3
Matterwave
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Or: you know that a refrigeration cycle requires positive work done by the system to transfer heat from a colder to a hotter environment. What's the sign of the work done per cycle by the system here? (A quick look at your p-V diagram gives the answer).
P.S. : physicists & engineers use dW = pdV whereas you define dW = - pdV.
But you can change that just by running the cycle backwards...?
 
  • #4
rude man
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The answer may lie in the entropy, but I am quite sure entropy is a state variable in this case.
Entropy is always a state variable.
 
  • #5
rude man
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But you can change that just by running the cycle backwards...?
If it's a refrigeration cycle it MUST run counterclockwise, doesn't it?
This is the only part I got right. :oops: (PLus post #4).
 
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  • #6
Matterwave
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If it's a refrigeration cycle it MUST run counterclockwise, doesn't it?
I guess I'm not seeing your point...o.o

But I am not so super well versed in these problems, so I'll defer to your better judgement haha.
 
  • #7
Matterwave
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Well, your diagram shows it to go counterclockwise! :)
Do you see what the sign of W is over a cycle? where W = ∫pdV?

Think: if W could be negative then you would have a machine that transfers heat from a cold place to a hot place without inputing work. That violates the 2nd law (Clausius or Kelvin, I forget which).
I'm not the OP, so I don't want to reveal too much by discussing too much. But the diagram in OP's post is going clockwise...o.o
 
  • #8
rude man
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Matterwave said:
I'm not the OP, so I don't want to reveal too much by discussing too much. But the diagram in OP's post is going clockwise...o.o
You're right, I screwed this up badly. My apologies to the OP and to you. o:)

The question as I see it is why the cycle couldn't be run counterclockwise to provide refrigeration.

OK, here's what I suggest: in order for refrigeration to take place, the net heat Q per cycle removed from the system (the gas) has to be positive.
Determine Q for each of the four processes in the reversed (counterclockwise) cycle to see if this obtains.
 
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  • #9
rude man
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But you can change that just by running the cycle backwards...?
Yeah, dumb statement. See my post #8.
 
  • #10
rude man
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I guess I'm not seeing your point...o.o

But I am not so super well versed in these problems, so I'll defer to your better judgement haha.
Big mistake! See my post #8.
 
  • #11
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Okay, lets start with what I know. I'll assume that I have monoatomic gas (f= 3)
[tex] dU_A = Q_A - P_A (V_B - V_D) [/tex]
[tex] dU_B = Q_B [/tex]
[tex] dU_C = Q_C - P_C (V_D - V_B) [/tex]
[tex] dU_D = Q_D [/tex]
Also: [tex] dU = 1.5NkdT = 1.5d(PV) [/tex]
[tex] dU_A = 1.5P_A (V_B - V_D) = Q_A - P_A (V_B - V_D) [/tex]
[tex] dU_B = Q_B = 1.5V_B (P_C - P_A) [/tex]
[tex] dU_C = Q_C - P_C (V_D - V_B) = 1.5P_C (V_D - V_B) [/tex]
[tex] dU_D = Q_D = 1.5V_D (P_A - P_C) [/tex]
[tex] Q_A = 2.5P_A (V_B - V_D) [/tex]
[tex] Q_C = 2.5P_C (V_D - V_B) [/tex]
I also know that:
[tex] P_A >P_C [/tex] [tex] V_B > V_D [/tex]
So, Q_C and Q_B are negative while the other two are positive. However, I dont see how this is my answer. The space inside the refrigerator in this case is the cold reservoir (the book I am using is "An Introduction to Thermal Physics" by Daniel Schroeder). So, all that this calculation shows is that heat flows from some place and onto some other.

EDIT:
I am not quite sure this is the answer. However, the heat throughout the process is:
[tex] P_A (V_B - V_D)- P_C (V_B - V_D) = (P_A-P_C) (V_B - V_D) > 0[/tex]
This is due to the inequalities I wrote above. So, each cycle, more heat is added to the refrigerator. Maybe the website that claimed that the temperature of the machine rises with each cycle is correct? How can this be?
 
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  • #12
rude man
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Maybe the website that claimed that the temperature of the machine rises with each cycle is correct? How can this be?
In a standard refrigeration cycle the assumption is that you have a hot reservoir with essentially infinite heat capacity (the room air temperature is not significantly affected by the refrigerator). Also, a practical refrigerator either has to shut off some of the time or the heat leakage into the cold side has to offset the heat removed by the system.

The "machine", aka the "system" or in this case the gas, will not heat up indefinitely but moves continuously within well-defined temperatures. If for simplicity you assume an ideal gas, then the temperatures are going to be T1 = pAVD/nR, T2 = pAVB/nR, T3 = pCVB/nR and T4 = pCVD/nR at the corners, and something inbetween everywhere else.

Assuming fixed or slowly varying cold and hot side temperatures, there would have to be continuous heat flow in and out of both the cold side and the hot side in each cycle, with net heat outflow from the cold side and net inflow to the hot side. So the cold & hot reservoirs would act as buffers, regulating the mean temperature of the cold side as desired. If the heat capacity of the cold side were very small then its temperature would vary quite significantly each cycle.

Heat removed from the cold side per cycle = Cp(T3 - T4) + CV(T2 - T3).
Heat added to the hot side per cycle = Cp(T2 - T1) + CV(T1 - T4).
Net heat removed per cycle = |area| of the p-V diagram loop = work done on the system.

Comments welcomed, everyone!
 
  • #13
Andrew Mason
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The Attempt at a Solution


According to a website this wouldn't work because the refrigerator would get warmer and warmer until it got warmer than the cold reservoir. However, temperature is a state variable, and when I calculated it, dT = 0 as expected. So, while it may get warmer during the cycle, it will return to its original temperature.

The answer may lie in the entropy, but I am quite sure entropy is a state variable in this case.
The website answer is wrong. The working substance just has to absorb heat flow from the cold reservoir. It does not have to be an ideal gas. The working substance does not necessarily have to get hotter than the cold reservoir. Even if it did, that does not mean that it would not absorb net heat flow from the cold reservoir.

AM
 
  • #14
rude man
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The website answer is wrong.
Agreed.
The working substance does not necessarily have to get hotter than the cold reservoir.
??? It does some of the time!
 
  • #15
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I noticed that the exercise doesn't really mention ideal gasses either (I assumed I had to use ideal gas equations because the exercise refers back to earlier exercises where rectangular PV cycles were used where we did work with ideal gasses). That makes two of the equations I used irrelevant. I am still not any closer to understanding why this system cannot work in reverse as a refrigerator.
 
  • #16
DrClaude
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So, Q_C and Q_B are negative while the other two are positive. However, I dont see how this is my answer.
That's part of the answer. Think about what the sign of Q mean with respect to the different parts of the refrigeration cycle, namely the exchanges between the working substance and the hot and cold reservoirs.
 
  • #17
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That's part of the answer. Think about what the sign of Q mean with respect to the different parts of the refrigeration cycle, namely the exchanges between the working substance and the hot and cold reservoirs.
But, I no longer know that, do I? Since my system doesn't have to be an ideal gas, is the equipartition theorem necessarily valid? All I know is that dU_B = Q_B and Q_C = dU_C + P_C(V_D- V_B)..

If I, for simplicity, do assume my system comprises of ideal gasses... dU = Q + W.. I know that dU is the change of the internal energy of the system. So, a positive Q means heat is transferred to the system. So, Q_C and Q_B are the parts of the cycle where heat is transferred to the hot reservoir. I also know that net heat over the entire cycle is positive. So, since temperature is a state variable, I suppose the system has to do work to offset it. The work has to be negative.... .... Are you telling me that the system will have to become larger and larger with each cycle?
 
  • #18
DrClaude
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So, a positive Q means heat is transferred to the system. So, Q_C and Q_B are the parts of the cycle where heat is transferred to the hot reservoir.
And conversely, legs A and D are where heat is absorbed from the cold reservoir. What does that say about the different temperatures?

So, since temperature is a state variable
Yes, that's important.


Are you telling me that the system will have to become larger and larger with each cycle?
No. But putting together the above pieces of the puzzle, you should find something weird at a couple of points of the cycle.
 
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  • #19
rude man
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I noticed that the exercise doesn't really mention ideal gasses either (I assumed I had to use ideal gas equations because the exercise refers back to earlier exercises where rectangular PV cycles were used where we did work with ideal gasses). That makes two of the equations I used irrelevant. I am still not any closer to understanding why this system cannot work in reverse as a refrigerator.
Right, you don't need to assume an ideal gas. I merely used that assumption to show how well-controlled the system temperatures are, and that a temperature buildup as your text claims is not going to happen. To the extent that the gas is not ideal there will be small deviations from the ideal gas relation T = pV/nR, of course.
 
  • #20
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I think a few of my equations above may be wrong. The diagram I found above is confusing because the cycle is going in the wrong direction. I cannot find a rectangular diagram for refrigeration, but I'll try again with the equations:

[tex] Δ(PV)_A = P_A(V_D - V_B)[/tex]
[tex] Δ(PV)_B = V_B(P_A - P_C)[/tex]
[tex] Δ(PV)_C = P_C(V_B - V_D)[/tex]
[tex] Δ(PV)_D = V_D(P_C - V_A)[/tex]

And conversely, legs A and D are where heat is absorbed from the cold reservoir. What does that say about the different temperatures?
In the following paragraph I assume my older calculations were correct...

Lets see.... Legs A and D are the parts of the cycle where temperature is rising. However, in both cases, Δ(PV) is negative (so, per the ideal gas law the temperature change must be negative as well). The opposite is the case in B and C. For A and C, this discrepancy can be countered by work. That does not work for B and D... On the other hand, this is only a problem if a refrigerator must be a closed system. If not, this can work by letting ΔN be negative.
 
  • #21
rude man
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I think a few of my equations above may be wrong. The diagram I found above is confusing because the cycle is going in the wrong direction. I cannot find a rectangular diagram for refrigeration, but I'll try again with the equations:

[tex] Δ(PV)_A = P_A(V_D - V_B)[/tex]
[tex] Δ(PV)_B = V_B(P_A - P_C)[/tex]
[tex] Δ(PV)_C = P_C(V_B - V_D)[/tex]
[tex] Δ(PV)_D = V_D(P_C - V_A)[/tex]



In the following paragraph I assume my older calculations were correct...

Lets see.... Legs A and D are the parts of the cycle where temperature is rising.
Not if it's a refrigeration cycle where the arrows go counterclockwise. Are you doing an engine or a refrigerator?
However, in both cases, Δ(PV) is negative (so, per the ideal gas law the temperature change must be negative as well). The opposite is the case in B and C. For A and C, this discrepancy can be countered by work. That does not work for B and D... On the other hand, this is only a problem if a refrigerator must be a closed system. If not, this can work by letting ΔN be negative.
At this point I'm pretty confused. I'll just opine that n is constant and it IS a closed system. Never heard of any other kind ...
 
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  • #22
DrClaude
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I did not check your calculations in detail, because the only point of importance to answer the question is for which legs is Q positive and for which legs it is negative. Now, you need to consider it in terms of Tc, Th, and Tws, where the latter is the temperature of the working substance.
 
  • #23
Andrew Mason
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I noticed that the exercise doesn't really mention ideal gasses either (I assumed I had to use ideal gas equations because the exercise refers back to earlier exercises where rectangular PV cycles were used where we did work with ideal gasses). That makes two of the equations I used irrelevant. I am still not any closer to understanding why this system cannot work in reverse as a refrigerator.
The system DOES work in reverse as a refrigerator. At least it uses work to cause heat flow from the cold reservoir to the hot reservoir. How effective it would be in doing this is another matter. The website is wrong in saying that it cannot cause heat flow from the cold to the hot reservoir.

AM
 
  • #24
DrClaude
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The system DOES work in reverse as a refrigerator. At least it uses work to cause heat flow from the cold reservoir to the hot reservoir. How effective it would be in doing this is another matter. The website is wrong in saying that it cannot cause heat flow from the cold to the hot reservoir.
Without going into the details until the OP figures it out, there is an internal inconsistensy in running the cycle in reverse, such that it is impossible to implement a refrigerator on a square cycle like that.
 
  • #25
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Not if it's a refrigeration cycle where the arrows go counterclockwise. Are you doing an engine or a refrigerator?

At this point I'm pretty confused. I'll just opine that n is constant and it IS a closed system. Never heard of any other kind ...
I am using a refrigeration cycle. Since the cycle in the picture in the original post is moving in the wrong direction, I redrew it on paper by hand, and I am quite sure my new calculations are the right ones (but, I cannot upload it here although I have scanned in the correct cycle).

The reason I am not certain if a heat engine has to be closed or not, is these exercises by the University of Oslo (the part about a diesel engine):
https://www.uio.no/studier/emner/matnat/fys/FYS2160/h14/book/2014-oblig-05.pdf
Here gas is sucked into and ejected from the engine at different parts of the cycle.

I did not check your calculations in detail, because the only point of importance to answer the question is for which legs is Q positive and for which legs it is negative. Now, you need to consider it in terms of Tc, Th, and Tws, where the latter is the temperature of the working substance.
Well, what else can I use...? I know due to the second law of thermodynamics that:
[tex] \frac{T_c}{T_h} \leq \frac{Q_c}{Q_h} [/tex]
Since, [tex] dS = \frac{Q}{T} \geq 0 [/tex]
Also, the maximum efficiency (COP) of the refrigerator:
[tex] COP = \frac{Q_c}{Q_h - Q_c} \leq \frac{T_c}{T_h - T_c} [/tex]

What else do I know? Hmm... T_c<T_h... I also know (assuming a closed system):
[tex] ΔT = \frac{Δ(PV)}{Nk} [/tex]
[tex] ΔT = \frac{2Δ(U)}{fNk} [/tex]

I still do not know where to go on from here.

EDIT:
Just to add some more..The problem cannot lie in one of the legs A, B, C or D alone since all those parts also exist in a rectangular heat engine (which supposedly does not have this problem). So, it must be a combination of the legs causing the problem.

The system DOES work in reverse as a refrigerator. At least it uses work to cause heat flow from the cold reservoir to the hot reservoir. How effective it would be in doing this is another matter. The website is wrong in saying that it cannot cause heat flow from the cold to the hot reservoir.

AM
Well, the website is not alone in claiming it is not possible. The textbook I am using is specifically asking me to explain why it is impossible to use the cycle in a refrigerator.
 
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