Cyclic Machine: Reversible Process?

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The discussion centers on determining whether a cyclic machine, which receives 325 kJ from a 1000 K reservoir, rejects 125 kJ to a 400 K reservoir, and produces 200 kJ of work, is reversible, irreversible, or impossible. Participants analyze the efficiency of the machine compared to the Carnot efficiency, concluding that since the machine's efficiency (0.615) exceeds the Carnot efficiency (0.6), it violates the second law of thermodynamics, making it impossible. The importance of entropy changes in assessing reversibility is emphasized, with a total entropy change indicating that the process cannot be reversible. The conversation also touches on the misconception that processes can be reversible if they approach Carnot efficiency, clarifying that no real process can equal Carnot efficiency. Ultimately, the consensus is that the machine's operation is impossible due to its efficiency exceeding the theoretical limit.
JSBeckton
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A cyclic machine receives 325 kJ from a 1000 K energy reservoir. It rejects 125 kJ to a 400 K energy reservoir, the cycle produces 200 kJ of work as output. Is this cycle reversible, irrerversible, or impossible?

Not exactly sure what he is looking for here. But, since W=QH-QL, does that mean that I can say that the process cold be reversed as a refrigerator so W=QH-QL and is therefore reversible?
 
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Entropy! Entropy! :biggrin:
 
could you elaborate a little?
 
The total entropy change for a reversible process should equal zero. The entropy of the system can change, but the entropy change of the surroundings cancels out this change, so that the overall change is zero. To find out whether the process is reversible or not, you should look at the entropy changes of the system and it's surrondings.
 
JSBeckton said:
A cyclic machine receives 325 kJ from a 1000 K energy reservoir. It rejects 125 kJ to a 400 K energy reservoir, the cycle produces 200 kJ of work as output. Is this cycle reversible, irrerversible, or impossible?

Not exactly sure what he is looking for here. But, since W=QH-QL, does that mean that I can say that the process cold be reversed as a refrigerator so W=QH-QL and is therefore reversible?
As the other posters have said, you could calculate entropy directly and see if it is less than, greater than or equal to 0. Which of the three choices represents entropy = 0; greater than 0; less than 0?

But you could also determine this indirectly (it is still based on entropy) by determining the efficiency of a Carnot engine between these temperatures. The actual efficiency is \eta = W/Q_H. See if the Carnot efficiency is greater, less than or equal to this efficiency and base your answer on that.

AM
 
Andrew, I was headed in that direction, I got thermal efficiency of 0.615 for the cycle, and a thermal efificiency of 0.6 for the carnot. My book is pretty vague on this.

I know that it is impossible to have a greater efficiency than carnot, but its pretty close, how close does it have to be to carnot?
I'm getting only a 2 1/2 percent difference but that means that some energy must be lost to the surroundings. Is there an official "lithmus" test for this? Or since its very close to carnot, can I call it reverseable?
 
JSBeckton said:
Andrew, I was headed in that direction, I got thermal efficiency of 0.615 for the cycle, and a thermal efificiency of 0.6 for the carnot. My book is pretty vague on this.

I know that it is impossible to have a greater efficiency than carnot, but its pretty close, how close does it have to be to carnot?
I'm getting only a 2 1/2 percent difference but that means that some energy must be lost to the surroundings. Is there an official "lithmus" test for this? Or since its very close to carnot, can I call it reverseable?
If the Carnot efficiency is exceeded by any amount at all, you violate the second law of thermodynamics. Carnot is the limit - you can approach it but can never exceed it.

AM
 
Ok, but what does that say about how close to carnot you have to be for it to be considered a reversible process? Is 2.5% close enough?
 
JSBeckton said:
Ok, but what does that say about how close to carnot you have to be for it to be considered a reversible process? Is 2.5% close enough?
?? If the Carnot efficiency cannot be exceeded, why would you think that it could be exceeded by 2.5% and not violate the second law?

AM
 
  • #10
I do not think that it was exceeded by 2.5%, I think that it came to 2.5% less efficient than carnot. I still don't understand what i am supposed to base my decision off of. How do I determine from this information if it is reversible?
 
  • #11
JSBeckton said:
I do not think that it was exceeded by 2.5%, I think that it came to 2.5% less efficient than carnot. I still don't understand what i am supposed to base my decision off of. How do I determine from this information if it is reversible?
You found that the efficiency of the Carnot engine, which is the highest efficiency possible, was 60%. You found that the efficiency of this engine was supposed to be 61.5%.

According to the second law, the maximum efficiency possible is 60%. Anything greater is impossible. Therefore, is an engine that is 61.5% possible?

AM
 
  • #12
If the efficiency is the same as the carnot then it is reversible. If it is more than the carnot it is impossilbe. If it is less than the carnot, then it is nonreversable.
 
  • #13
NotMrX said:
If the efficiency is the same as the carnot then it is reversible. If it is more than the carnot it is impossilbe. If it is less than the carnot, then it is nonreversable.
Yes. So which is it?

AM
 
  • #14
JSBeckton said:
A cyclic machine receives 325 kJ from a 1000 K energy reservoir. It rejects 125 kJ to a 400 K energy reservoir, the cycle produces 200 kJ of work as output. Is this cycle reversible, irrerversible, or impossible?

Not exactly sure what he is looking for here. But, since W=QH-QL, does that mean that I can say that the process cold be reversed as a refrigerator so W=QH-QL and is therefore reversible?

Q_i=Q_h=325KJ

Q_o=Q_c=125KJ

W=Q_h-Q_c=325-125=200KJ

e=\frac{w}{Q_h}=\frac{200}{325}=\frac{8}{13}=.61

e_c=\frac{T_h-T_c}{T-h}=\frac{1000-400}{1000}=0.6

e > e_c
Therefore impossible.
 
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  • #15
JSBeckton said:
A cyclic machine receives 325 kJ from a 1000 K energy reservoir. It rejects 125 kJ to a 400 K energy reservoir, the cycle produces 200 kJ of work as output. Is this cycle reversible, irrerversible, or impossible?

Not exactly sure what he is looking for here. But, since W=QH-QL, does that mean that I can say that the process cold be reversed as a refrigerator so W=QH-QL and is therefore reversible?

I didn't read all the posts. What they are saying is correct of course they probably have more knowledge than me, I don't foramally study physics. The easy way to do this problem is the entropy way.

DeltaS\deltaS=\frac{Q}{T}

DeltaSh\DeltaS_h=\frac{-325000}{1000}=-325J/K

DeltaSc\DeltaS_c=\frac{125000}{400}=312.5J/k

DeltaSu\DeltaS_u=312.5-325=-12.5J/K

This defies the second law since entropy of universe should be increasing. For a refrigerator heat is going towards hot sink, away from cold sink, and you have to put work into it. This problem doesn't follow those parameters. If you work through a few chapters in a general physics book it would probabaly help to get a feel for these problems.
 
  • #16
Thanks everyone,I see what I have failed to notice all along, for some reason in my head I was thinking that the cycle was LESS efficient instead of MORE efficient. So obviously its impossible!

However, if it were switched like I was thinking. say that the carnot was .615 and the real was .60, would this be considered reversible?

I guess that i am thinking that nothing can be as efficient as carnot, therefore technically nothing can = carnot and therefore nothing would be a reversible process, but that's not the case. What am I missing?
 
  • #17
JSBeckton said:
Thanks everyone,I see what I have failed to notice all along, for some reason in my head I was thinking that the cycle was LESS efficient instead of MORE efficient. So obviously its impossible!

However, if it were switched like I was thinking. say that the carnot was .615 and the real was .60, would this be considered reversible?

I guess that i am thinking that nothing can be as efficient as carnot, therefore technically nothing can = carnot and therefore nothing would be a reversible process, but that's not the case. What am I missing?
Nothing can be MORE efficient than a Carnot cycle. One might be able to approach (ie. get arbitrarily close to) that efficiency but one cannot exceed it. I am not aware of any thermodynamic process that actually is able to achieve this efficiency. What do you have in mind when you say "that is not the case"?

AM
 
  • #18
If these two statements are true:

1) A process is reversible only if it's efficiency equals that of carnot.
2) No real process is able to equal carnot.

Than logic would lead one to believe that: No real process is reversible.

If that's the case why do they say that weights put on a piston/cylinder can be removed, making it a reversible process?
 
  • #19
JSBeckton said:
If these two statements are true:

1) A process is reversible only if it's efficiency equals that of carnot.
2) No real process is able to equal carnot.

Than logic would lead one to believe that: No real process is reversible.

If that's the case why do they say that weights put on a piston/cylinder can be removed, making it a reversible process?
Who is "they"? Do you have a cite or quote? They may be talking about something other than thermodynamic reversibility, as in "my car is reversible". A thermodynamically reversible process is one that can be reversed with an infinitessimal change of initial conditions. In real life, this can never be achieved. Its usefulness is in establishing the limit on the amount of useable work that can be extracted from heat.

AM
 

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