Impossible refrigerator (thermodynamics and PV diagrams)

[Avatrin]

It is probably impossible to make this cycle reversible. But the question says it is a reversible heat engine. That means that the working substance is in equilibrium with the surroundings at all times. If so, the cycle can be reversed with an infinitesimal change in conditions. If it is reversed, work will be applied and the heat flow will go in the reverse direction. How is that not a refrigerator or heat pump?

AM

The question doesn't say it is a reversible heat engine. All it says is that we have a rectangular cycle on a PV diagram, and we are trying run it counterclockwise to make a refrigerator (unlike the image where it's running in clockwise direction).

 

[Andrew Mason]

The question doesn't say it is a reversible heat engine. All it says is that we have a rectangular cycle on a PV diagram, and we are trying run it counterclockwise to make a refrigerator (unlike the image where it's running in clockwise direction).

Sorry. I was confusing it with this thread that I am also posting on: https://www.physicsforums.com/threads/carnot-cycle.784322/

The question does not say that the surroundings are at constant temperature. A stirling cycle has constant volume expansion and compression parts and can be made, at least in theory, reversible. It uses a 'regenerator' to change the temperature of the surroundings that the working substance is in thermal contact with. I don't see a material difference between the Stirling cycle and this rectangular one. So, if it could be a reversible heat engine cycle, it has to be able to operate in the reverse direction and use work to cause heat flow in the opposite direction.

AM

 

[DrClaude]

I can use dT ∝Δ(PV)... So, since the system is expanding in volume while pressure remains the same, temperature goes up. So, in legs A and C the opposite is happening of what I wrote should be happening. After coming into contact with the hot reservoir, the temperature of the working substance keep rising.

We're getting confused here. Let's take a step back. Could you please summarize the results you have by answering:

What is the sign of Q in each leg?

What is the relation between the sign of Q and whether it is in contact with the hot or cold reservoir?

What is the relation between temperature of the working substance and the hot or cold reservoir during these exchanges?

 

[Avatrin]

We're getting confused here. Let's take a step back. Could you please summarize the results you have by answering:

What is the sign of Q in each leg?

What is the relation between the sign of Q and whether it is in contact with the hot or cold reservoir?

What is the relation between temperature of the working substance and the hot or cold reservoir during these exchanges?

Well, I do stand by my calculations in post #20 (I am quite sure my calculations for A and C in post #11 were wrong). So,
Q_A = dU_A + P_A(V_D - V_B)<0
Q_B = dU_B = \frac{f}{2} V_B (P_A - P_C)>0
Q_C = dU_C + P_C(V_B - V_D)>0
Q_D = dU_D = \frac{f}{2} V_D (P_C - P_A)<0

If Q is positive, the internal energy of the working substance is rising. So, it must be in contact with the cold reservoir when Q is positive since this is supposed to be a refrigerator. The system is in contact with the hot reservoir when it Q is negative.

If heat is going from the cold reservoir to the system, the system must be colder than the cold reservoir... ... Not necessarily...? But, I did assume that in my previous answer.. C = Q/dT.. So Q = CdT. Since we are not working with something like black holes, C must be positive. A positive Q for the working substance means a negative Q for the reservoir; Temperature must fall there (although, since it is a reservoir, it is negligible). Since they are approaching equilibrium by coming into contact, the first sentence in this paragraph must be correct.

 

[DrClaude]

Well, I do stand by my calculations in post #20 (I am quite sure my calculations for A and C in post #11 were wrong). So,
Q_A = dU_A + P_A(V_D - V_B)<0
Q_B = dU_B = \frac{f}{2} V_B (P_A - P_C)>0
Q_C = dU_C + P_C(V_B - V_D)>0
Q_D = dU_D = \frac{f}{2} V_D (P_C - P_A)<0

If Q is positive, the internal energy of the working substance is rising. So, it must be in contact with the cold reservoir when Q is positive since this is supposed to be a refrigerator. The system is in contact with the hot reservoir when it Q is negative.

If heat is going from the cold reservoir to the system, the system must be colder than the cold reservoir... ... Not necessarily...? But, I did assume that in my previous answer.. C = Q/dT.. So Q = CdT. Since we are not working with something like black holes, C must be positive. A positive Q for the working substance means a negative Q for the reservoir; Temperature must fall there (although, since it is a reservoir, it is negligible). Since they are approaching equilibrium by coming into contact, the first sentence in this paragraph must be correct.

That's pretty much correct. Then take for instance the junction between legs D and C. In leg D, as you calculated, Q < 0, so heat is going from the working substance to the hot reservoir, so T_ws > T_h. Then in leg C Q > 0, so T_ws < T_c. What happens then at point 4 on the diagram?

 

[Avatrin]

That's pretty much correct. Then take for instance the junction between legs D and C. In leg D, as you calculated, Q < 0, so heat is going from the working substance to the hot reservoir, so T_ws > T_h. Then in leg C Q > 0, so T_ws < T_c. What happens then at point 4 on the diagram?

Well, if we graph T_ws, there will either be an impossible discontinuity, or T_h<T_ws<T_c, which is a contradiction.

 

[DrClaude]

Well, if we graph T_ws, there will either be an impossible discontinuity,

But T_ws is a state variable, so it has a defined value at each point in the cycle.

or T_h<T_ws<T_c, which is a contradiction.

And that's the solution.

To summarize, any cycle can be made into a heat engine, as we only need to use a hot reservoir with T_h > max(T_ws) and a cold reservoir with T_c < min(T_ws). The problem with a refrigerator is that it has to cycle between T_ws < T_c and T_ws > T_h, which is not possible for all cycles.