Final temperature of an adiabatic process between two reservoirs

[Baibhab Bose]

Homework Statement

A heat engine is operated between two bodies that are kept at constant pressure. The constant-
pressure heat capacity Cp of the reservoirs is independent of temperature. Initially the
reservoirs are at temperatures 300 K and 402 K. If, after some time, they come to a common
final temperature Tf , the process remaining adiabatic, what is the value of Tf (in Kelvin) ?

Relevant Equations

from 1st law, dQ=dU+PdV. Not sure the ideal gas equation or the P*V^(gamma)=K equation would be applicable here or not since they are for gases.

*Here, no mention of these reservoirs being a gas, so I'm not sure if I can use the PV=nRT or the P*V^(gamma)=K equation.
SO I am left with only the 1st law.
If I can write dQ1( going out from object 1)= Cp (indep of T)*(Tf-T1)
dQ2(coming into the object 2)= Cp*(T2-Tf), where T2>T1; given.
Then is it right to assume that since the process is adiabatic, no heat is lost to the environment an the heat which goes out from object 1 would reach Object 2? Then dQ1=dQ2 would be logical. But from this equality the solution is not correct.
And what is the significance of the pressure being constant?
I can't get any further.

 

[256bits]

A heat engine
...
dQ1=dQ2

Hi
Is that correct.
where does the work go?

 

[Baibhab Bose]

This is my question too.
From the 2nd law: The the heat cannot be totally converted to work.
but is that the other way around too?
can't there be only flow of heat and no work?

 

[256bits]

This is my question too.
From the 2nd law: The the heat cannot be totally converted to work.
but is that the other way around too?
can't there be only flow of heat and no work?

Sure - if the two resevoirs are in contact, the heat flows from the hot to the cold side.
But that wouldn't be a heat engine, which has work involved.

Production of work, such as a steam power plant, where heat moves from the hot to cold, and some work is extracted, so less heat is transferred to the cold side than injected from the hot.

What does efficiency of the heat engine given the temperatures say?

 

[Chestermiller]

What is the change in entropy of the working fluid used in the engine if the engine operates in a cycle? What is the change in entropy of the combination of reservoirs plus working fluid if the process is carried out reversibly and the combined overall system is adiabatic? What is the total change in entropy of the two reservoirs?

 

[Baibhab Bose]

Sure - if the two resevoirs are in contact, the heat flows from the hot to the cold side.
But that wouldn't be a heat engine, which has work involved.

Production of work, such as a steam power plant, where heat moves from the hot to cold, and some work is extracted, so less heat is transferred to the cold side than injected from the hot.

What does efficiency of the heat engine given the temperatures say?

Definition of efficiency being $\frac {work}{heat input} = 1-\frac {Q2} {Q1}$ when we evaluate this for a Carnot cycle, it becomes $1-\frac {T2}{T1}$.
So, at what point this 'engine'; producing work, can be equivalent to a Carnot engine?

 

[Baibhab Bose]

What is the change in entropy of the working fluid used in the engine if the engine operates in a cycle? What is the change in entropy of the combination of reservoirs plus working fluid if the process is carried out reversibly and the combined overall system is adiabatic? What is the total change in entropy of the two reservoirs?

The total system is adiabatic. SO, no heat is lost to the environment. hence $ΔS_{total}=0, as ΔQ_{total}=0$. And $ΔS_{total}=ΔS_1+ΔS_2 =0$
Where $ΔS_i=\frac{ΔQi}{Ti}$
And $ΔQ1=Cp(T_f-T_1) { }and{ } ΔQ_2=Cp(T_2-T_f)$
Is this right?
And will there be another $ΔS_3= \frac{ΔQ_1-ΔQ_2}{T_m}$ The entropy change of the fluid,
And in that case what is this $T_m$ ?

 

[Chestermiller]

The total system is adiabatic. SO, no heat is lost to the environment. hence $ΔS_{total}=0, as ΔQ_{total}=0$. And $ΔS_{total}=ΔS_1+ΔS_2 =0$
Where $ΔS_i=\frac{ΔQi}{Ti}$
And $ΔQ1=Cp(T_f-T_1) { }and{ } ΔQ_2=Cp(T_2-T_f)$
Is this right?
And will there be another $ΔS_3= \frac{ΔQ_1-ΔQ_2}{T_m}$ The entropy change of the fluid,
And in that case what is this $T_m$ ?

The entropy change for the working fluid in the engine is zero, because the engine is operating in a cycle, and, since entropy is a function of state, there is no change in entropy over a cycle.

Your equation for the entropy changes of the individual reservoirs is incorrect. To get the entropy change of each, you need to integrate over a reversible path for each (finite) reservoir individually. If the mass of each reservoir is M and its heat capacity is Cp, for a reversible path, $$dS=\frac{dQ}{T}=MC_p\frac{dT}{T}$$Based on this, what do you get for the entropy changes of the two reservoirs? These must add up to zero.

 

[Baibhab Bose]

Okay, that is understandable. But what keeps bugging me is the concept of reversibility.
Over a cycle the change of state function entropy for the fluid is 0 as you said.
The fluid is taking the heat from object1, doing some work on the way and sending the remnant heat to the object2.
A cycle would only happen when a same amount of external work is done ON the system, so the Q2 heat from the colder Object2 can be sent to the hot Object1 with an addition from the Work to make the heat arriving at Object1 Q1.
Thus a cycle is formed and thus the Total entropy is 0.
And in that case Object1, under this cycle it has let out Q1 and got it back at the end of the cycle; so no change of heat and no change in entropy. Same for the Object2.

 

[256bits]

A cycle would only happen when a same amount of external work is done ON the system, so the Q2 heat from the colder Object2 can be sent to the hot Object1 with an addition from the Work to make the heat arriving at Object1 Q1.
Thus a cycle is formed and thus the Total entropy is 0.
And in that case Object1, under this cycle it has let out Q1 and got it back at the end of the cycle; so no change of heat and no change in entropy. Same for the Object2

The problem does not state that a heat pump is operating between the two reservoirs.
You are adding to the problem.
As stated the heat engine operates in a cycle, Rankine or whatever.

 

[Chestermiller]

The problem does not state that a heat pump is operating between the two reservoirs.
You are adding to the problem.
As stated the heat engine operates in a cycle, Rankine or whatever.

What do these words mean to you: "Initially the reservoirs are at temperatures 300 K and 402 K. If, after some time, they come to a common final temperature Tf.."

 

[Chestermiller]

Okay, that is understandable. But what keeps bugging me is the concept of reversibility.
Over a cycle the change of state function entropy for the fluid is 0 as you said.
The fluid is taking the heat from object1, doing some work on the way and sending the remnant heat to the object2.
A cycle would only happen when a same amount of external work is done ON the system, so the Q2 heat from the colder Object2 can be sent to the hot Object1 with an addition from the Work to make the heat arriving at Object1 Q1.
Thus a cycle is formed and thus the Total entropy is 0.
And in that case Object1, under this cycle it has let out Q1 and got it back at the end of the cycle; so no change of heat and no change in entropy. Same for the Object2.

Yes. So?

You are aware that, in this problem, the working fluid must be put through multiple cycles in succession, with the hot reservoir temperature getting just a little lower and the cold reservoir temperature getting just a little higher over each successive cycle, right? During each cycle, only a tiny amount of heat is transferred from the hot reservoir to the working fluid, a tiny amount of work is done by the working fluid, and an even tinier amount of heat is rejected to the cold reservoir. The cumulative net effect of these successive cycles is that a finite amount of heat gets transferred from the hot reservoir to the working fluid, a finite amount of work gets done, and a finite amount of heat gets transferred from the working fluid to the cold reservoir. In the end, the hot reservoir is at Tf, the cold reservoir is at Tf, and the working fluid is at Tf, and no more heat transfer orwork is possible.

 

[256bits]

What do these words mean to you: "Initially the reservoirs are at temperatures 300 K and 402 K. If, after some time, they come to a common final temperature Tf.."

Not sure what you are asking. Terminology for the masses?
In normal terminology,
A heat engine, as stated in the problem, would be extracting work from the difference in temperature.
An input of work, would involve a heat pump and the temperatures would diverge.

Your explanation in the previous post is what I had in mind.

 

[Chestermiller]

Not sure what you are asking. Terminology for the masses?
In normal terminology,
A heat engine, as stated in the problem, would be extracting work from the difference in temperature.
An input of work, would involve a heat pump and the temperatures would diverge.

Your explanation in the previous post is what I had in mind.

What they are describing in the problem statement is an engine operating between two ideal but finite-heat-capacity reservoirs, rather than one operating between two reservoirs of infinite heat capacity (where, in this latter case, their temperatures would remain constant).

 

[256bits]

What they are describing in the problem statement is an engine operating between two ideal but finite-heat-capacity reservoirs, rather than one operating between two reservoirs of infinite heat capacity (where, in this latter case, their temperatures would remain constant).

Agreed.

 

[Baibhab Bose]

Yes. So?

You are aware that, in this problem, the working fluid must be put through multiple cycles in succession, with the hot reservoir temperature getting just a little lower and the cold reservoir temperature getting just a little higher over each successive cycle, right? During each cycle, only a tiny amount of heat is transferred from the hot reservoir to the working fluid, a tiny amount of work is done by the working fluid, and an even tinier amount of heat is rejected to the cold reservoir. The cumulative net effect of these successive cycles is that a finite amount of heat gets transferred from the hot reservoir to the working fluid, a finite amount of work gets done, and a finite amount of heat gets transferred from the working fluid to the cold reservoir. In the end, the hot reservoir is at Tf, the cold reservoir is at Tf, and the working fluid is at Tf, and no more heat transfer orwork is possible.

Yes. I got the difference between two bodies with finite temperatures and two reservoirs with infinite heat capacity. Since here we have bodies with finite temperatures, the temperatures come to an equilibrium by means of infinitesimal exchanges of heat tagged with a tinier amount of lost work.
Question: In these cycles, the fact that the heat is being transported in infinitesimal amounts make this a 'reversible' process, right?
But why a cycle? I mean I can't commensurate the idea of a cycle to this scenario.
Cycle means, returning back to the same state after some activities.
Here, where is the "returning back' happening?, since infinitesimal heats are being transferred to the other body, and not coming back!

 

[Chestermiller]

Yes. I got the difference between two bodies with finite temperatures and two reservoirs with infinite heat capacity. Since here we have bodies with finite temperatures, the temperatures come to an equilibrium by means of infinitesimal exchanges of heat tagged with a tinier amount of lost work.
Question: In these cycles, the fact that the heat is being transported in infinitesimal amounts make this a 'reversible' process, right?

Yes.

But why a cycle? I mean I can't commensurate the idea of a cycle to this scenario.
Cycle means, returning back to the same state after some activities.
Here, where is the "returning back' happening?, since infinitesimal heats are being transferred to the other body, and not coming back!

It isn't the reservoirs that are experiencing the cycle (just as in the Carnot cycle). It is the working fluid that is experiencing cycle(s). Actually in this process, the working fluid experiences multiple cycles in succession.

Here is a typical cycle that the working fluid (assumed to be an ideal gas) experiences: Temporarily assume that, in each cycle, the working fluid starts out at the ultimate final temperature Tf (i.e., the final temperature for the overall process). Then,

1. It is compressed adiabatically and reversibly until it reaches the current temperature of the hotter reservoir.
2. It receives a tiny (differential) amount of heat from the hot reservoir, expanding reversibly while the hot reservoir cools slightly. But, in essence, this takes place virtually at constant temperature.
3. It is expanded adiabatically and reversibly until it reaches the current temperature of the colder reservoir.
4. It rejects a tiny (differential) amount of heat to the cold reservoir, compressing reversibly while the colder reservoir heats slightly. But, in essence, this takes place virtually at constant temperature.
5. It is compressed adiabatically and reversibly until it again reaches Tf. This completes the cycle.

In reality, even if the working fluid initially starts out at a temperature that differs from Tf, this does not matter since we assume that the mass of working fluid is negligible compared to the masses of the two reservoirs, so its ultimate change in entropy is negligible.

 

[256bits]

5. It is compressed adiabatically and reversibly until it again reaches Tf. This completes the cycle

After what @Baibhab Bose wrote in post 16, it finally clicked about his mention of returning to the same state after one cycle.
I had the same idea as you that it would be an iterative process starting out at temperature T_f.
{ PS Rankine was just an example of the working fluid going through a cycle and not an indication what type of heat engine should be used here - ]

After some thought though, would not the working fluid have to return to T_f, and as well as one other state variable, say P_f, to describe the same initial state, and completion of a cycle.

On the TdS diagram, the two adiabatic expansion/compression lines have to meet somewhere, as well as the isotherms, at a common T_f for the final solution and no more work can be done.
Subsequently, on the PV diagram, the cycle has to shrink continuously also at a final P_f and V_f.
In both diagrams, the final state would have to be a state that the engine has not been in before ( subsequent cycles also even if with each cycle to the next the difference in state is infinitesimal ).

Your thought.

 

[Chestermiller]

After what @Baibhab Bose wrote in post 16, it finally clicked about his mention of returning to the same state after one cycle.
I had the same idea as you that it would be an iterative process starting out at temperature T_f.
{ PS Rankine was just an example of the working fluid going through a cycle and not an indication what type of heat engine should be used here - ]

After some thought though, would not the working fluid have to return to T_f, and as well as one other state variable, say P_f, to describe the same initial state, and completion of a cycle.

On the TdS diagram, the two adiabatic expansion/compression lines have to meet somewhere, as well as the isotherms, at a common T_f for the final solution and no more work can be done.
Subsequently, on the PV diagram, the cycle has to shrink continuously also at a final P_f and V_f.
In both diagrams, the final state would have to be a state that the engine has not been in before ( subsequent cycles also even if with each cycle to the next the difference in state is infinitesimal ).

Your thought.

I don't want to think about this too much because, as I said, it really doesn't matter since the mass of the working fluid is assumed to be negligible compared to that of the reservoir fluids. But, I think what you are saying is that, even if we started out at (Pf,Tf), we couldn't execute perfect cycles passing through Pf at Tf at the end of each cycle. I don't know whether this is the case, but we might still be able to figure out how to make it work. But spending more time on this just doesn't seem worth it.

 

[Chestermiller]

@256bits I think I can devise a way of doing it. I would keep the left hand adiabat fixed, and carry out step 1 up to a temperature of $T_h-\Delta T$. The ideal gas in each cycle would be put through a perfect Carnot cycle, although the cycles would be getting smaller, holding the left hand adiabat fixed. So at the high temperature end, on the first cycle, we would have the high temperature reservoir cool down to the working fluid temperature, and thus $$Q_h=MC\Delta T=nR(T_h-\Delta T)\ln{(V_2/V_1)}$$This would determine V2 (which would get closer to V1 as we made $\Delta T$ smaller. At the cold end of the cycle, we would compress the gas, such that $$\frac{Q_h}{(T_h-\Delta T)}=\frac{Q_c}{(T_c+(\Delta T)^*)}$$This would determine the temperature that the working fluid is compressed at, where $$Q_c=MC(\Delta T)^*$$(assuming that the two reservoirs have the same mass and heat capacity).

The cycle would end on the same left hand adiabat the I started on. I would then do the second cycle with another decrease of $\Delta T$ at the hot end.

Thoughts?

 

[256bits]

I don't want to think about this too much because, as I said, it really doesn't matter since the mass of the working fluid is assumed to be negligible compared to that of the reservoir fluids. But, I think what you are saying is that, even if we started out at (Pf,Tf), we couldn't execute perfect cycles passing through Pf at Tf at the end of each cycle. I don't know whether this is the case, but we might still be able to figure out how to make it work. But spending more time on this just doesn't seem worth it.

The small mass of the working fluid should solve the problem for the reasons you stated.
A larger mass of working fluid seems to have the returning to the initial state problem - it could be faulty logic.
As we could replace the Tf point, I think, with an infinite reservoir and have two Carnot engines operating - one between Tf and Tc and the other between Tf and Th. The Tf isotherm should then be fixed for both engines.

 

[256bits]

@256bits I think I can devise a way of doing it. I would keep the left hand adiabat fixed, and carry out step 1 up to a temperature of $T_h-\Delta T$. The ideal gas in each cycle would be put through a perfect Carnot cycle, although the cycles would be getting smaller, holding the left hand adiabat fixed. So at the high temperature end, on the first cycle, we would have the high temperature reservoir cool down to the working fluid temperature, and thus $$Q_h=MC\Delta T=nR(T_h-\Delta T)\ln{(V_2/V_1)}$$This would determine V2 (which would get closer to V1 as we made $\Delta T$ smaller. At the cold end of the cycle, we would compress the gas, such that $$\frac{Q_h}{(T_h-\Delta T)}=\frac{Q_c}{(T_c+(\Delta T)^*)}$$This would determine the temperature that the working fluid is compressed at, where $$Q_c=MC(\Delta T)^*$$(assuming that the two reservoirs have the same mass and heat capacity).

The cycle would end on the same left hand adiabat the I started on. I would then do the second cycle with another decrease of $\Delta T$ at the hot end.

Thoughts?

Also interesting.
I think that should work also considering $\Delta T$ is small.

PS
[ I guess the next step is working it out with a cycle other than Carnot, and compare that Tf to the reversible engine.
Oh, the humanity
]

 

[Baibhab Bose]

Yes.

It isn't the reservoirs that are experiencing the cycle (just as in the Carnot cycle). It is the working fluid that is experiencing cycle(s). Actually in this process, the working fluid experiences multiple cycles in succession.

Here is a typical cycle that the working fluid (assumed to be an ideal gas) experiences: Temporarily assume that, in each cycle, the working fluid starts out at the ultimate final temperature Tf (i.e., the final temperature for the overall process). Then,

1. It is compressed adiabatically and reversibly until it reaches the current temperature of the hotter reservoir.
2. It receives a tiny (differential) amount of heat from the hot reservoir, expanding reversibly while the hot reservoir cools slightly. But, in essence, this takes place virtually at constant temperature.
3. It is expanded adiabatically and reversibly until it reaches the current temperature of the colder reservoir.
4. It rejects a tiny (differential) amount of heat to the cold reservoir, compressing reversibly while the colder reservoir heats slightly. But, in essence, this takes place virtually at constant temperature.
5. It is compressed adiabatically and reversibly until it again reaches Tf. This completes the cycle.

In reality, even if the working fluid initially starts out at a temperature that differs from Tf, this does not matter since we assume that the mass of working fluid is negligible compared to the masses of the two reservoirs, so its ultimate change in entropy is negligible.

So here, assuming the working fluid is in contact with the two objects, after the increase of temperature of the fluid under compression, why sudden it should take heat from the body whose temperature is close to itself and not lose heat to the cold object in contact?
And also for two objects in contact by means of a fluid, would it be illogical to think that heat from the hot body just simply flows through the fluid to the cold one and attains and equilibrium at Tf? Why should be resort to this cyclic process of compression and expansion?

 

[Chestermiller]

The working fluid is not in contact with the hot reservoir and the cold reservoir at the same time. First it is brought in contact contact with the hot reservoir and expands. Then it expands adiabatically and reversibly. Then it is brought in contact with the cold reservoir and compresses. Then it compresses adiabatically and reversibly. This is a process involving four steps in succession.

 

[Baibhab Bose]

Thats very enlightening. I have always thought of these processes as touch-and-flow kind of trivial processes. And have always considered cycles like carnot as a special kind of dynamics between two reservoirs.
Never knew that they are correlated as such.

 

[Chestermiller]

Thats very enlightening. I have always thought of these processes as touch-and-flow kind of trivial processes. And have always considered cycles like carnot as a special kind of dynamics between two reservoirs.
Never knew that they are correlated as such.

It is not surprising that you have been very confused about all this.