Efficiency of Stirling Cycle - Is it the same as the Carnot Engine?

[laser1]

TL;DR Summary: Efficiency Of Stirling Cycle - Is it the same as the Carnot Engine? Why is my described engine not the same as Stirling engine?

Ok, the problem I was given is this:
One mole of a perfect gas goes through a quasistatic cycle consisting of the following four stages:
1. isothermal expansion from V1 to V2 at temperature TH,
2. cooling at constant volume V2 from TH to TC,
3. isothermal compression from V2 to V1 at TC,
4. heating at constant volume until the system is back in its initial state

Wikipedia states this for a Stirling cycle:

Ok, so it seems to me they are the same. (I am just including this part because my flaw might be in the assumption that they are the same cycle).

This guy (https://www.youtube.com/watch?v=sklO_qZTBeY&t=371s) explains how to calculate the efficiency of the Stirling engine. His logic all looks fine. His final answer is given by η=1−TLTH Now I have a problem because for the original question, I worked out $$\eta = \frac{W}{Q_H} = \frac{RT_C \ln\left(\frac{V_f}{V_i}\right) (T_H - T_C)}{RT_H \ln\left(\frac{V_f}{V_i}\right) + C_V (T_H - T_C)}$$ which is definitely not the same as the efficiency of the Stirling cycle. Explaining in a bit more detail how I achieved the above expression: η is simply work you get over work you pay for.

 

[Andrew Mason]

TL;DR Summary: Efficiency Of Stirling Cycle - Is it the same as the Carnot Engine? Why is my described engine not the same as Stirling engine?

Ok, the problem I was given is this:
One mole of a perfect gas goes through a quasistatic cycle consisting of the following four stages:
1. isothermal expansion from V1 to V2 at temperature TH,
2. cooling at constant volume V2 from TH to TC,
3. isothermal compression from V2 to V1 at TC,
4. heating at constant volume until the system is back in its initial state

Wikipedia states this for a Stirling cycle:

View attachment 352044


View attachment 352045

An engine will be a Carnot engine if all processes occur with the system and surroundings arbitrarily close to equilibrium. A Stirling engine cycle is not such an engine.

The work done by any engine over a complete cycle is always $|Q_h|-|Q_c|$. This is because the system returns to its initial state after a complete cycle so $\Delta U=0$. So:

$$\eta=\frac{W}{Q_h}=\frac{|Q_h|-|Q_c|}{Q_h}$$

For the Sterling engine heat flow occurs on all four parts of the cycle so you have to add the heat flow on parts 1 and 4 for Qh and heat flow on parts 2 and 3 for Qc.

$Q_h=Q1+Q4=nRT_h\ln\left(\frac{V_2}{V_1}\right) + nC_v(T_h-T_c)$

$Q_c=Q2+Q3=nC_v(T_c-T_h) + nRT_c\ln\left(\frac{V_4}{V_3}\right)$

So: $W=Q_h-|Q_c|=nR[T_h\ln\left(\frac{V_2}{V_1}\right)-|T_c\ln\left(\frac{V_4}{V_3}\right)|]$
and, since $V_4=V_1 \text{ and }V_3=V_2$:
$$\eta=\frac{R[T_h\ln\left(\frac{V_2}{V_1}\right)-T_c\ln\left(\frac{V_2}{V_1}\right)]}{RT_h\ln\left(\frac{V_2}{V_1}\right) + C_v(T_h-T_c)}=\frac{R[\ln\left(\frac{V_2}{V_1}\right)](T_h-T_c)}{RT_h\ln\left(\frac{V_2}{V_1}\right) + C_v(T_h-T_c)}$$
which can be rewritten as:
$$\frac{1}{\eta}=\frac{1}{T_h-T_c} +\frac{C_v}{R\ln\left(\frac{V_2}{V_1}\right)}$$

 

[laser1]

In this video then: it is not the Stirling cycle? I noticed he didn't include the 2->3 term for heat added...

 

[Andrew Mason]

In this video then: it is not the Stirling cycle? I noticed he didn't include the 2->3 term for heat added...

He is describing the Stirling cycle but he is not calculating the efficiency properly. He is not taking into account $Q_{2-3}$ in $Q_h$ which is part of the denominator:
$$\eta=\frac{Q_{2-3}+Q_{3-4}-|Q_{4-1}+Q_{3-2}|}{Q_{2-3}+Q_{3-4}}=\frac{Q_{3-4}-|Q_{4-1}|}{Q_{2-3}+Q_{3-4}}$$

 

[TSny]

The video's result for the efficiency is correct under the assumption that the Sterling engine uses a regenerative heat exchanger. This allows the heat expelled in step 2 to be temporarily stored and then returned to the engine in step 4. Ideally, in this case, we don't have to pay for the heat input in step 4. The efficiency in this case is defined as the ratio of the net work done in the cycle to the heat that we had to supply only in step 1.

See this video where the discussion of heat regeneration starts at time 4:33.

 

[Andrew Mason]

There is no way to make a Stirling cycle into a Carnot cycle even theoretically. A theoretical Carnot cycle operates between two reservoirs of unlimited heat capacity at constant temperatures $T_h$ and $T_c$ respectively. There is no way to have the Stirling cycle with a regenerative heat exchanger operate between those reservoirs in a reversible manner.

With reference to this PV diagram:

one could have two systems that are out of step by half a cycle. One could then put either of the systems at point 4 in thermal contact with the other system at point 2 and have some heat from the system flow into the system at 2 instead of into the cold reservoir. But that heat flow will stop when the system temperature over the 4' path drops below the system temperature over the 3' path.

To calculate the efficiency of such a heat engine, you would still have to factor in $Q_{2-3}$ from the hot reservoir and $Q_{4-1}$ to the cold reservoir and they could not be zero.

AM

 

[Philip Koeck]

There is no way to make a Stirling cycle into a Carnot cycle even theoretically. A theoretical Carnot cycle operates between two reservoirs of unlimited heat capacity at constant temperatures $T_h$ and $T_c$ respectively. There is no way to have the Stirling cycle with a regenerative heat exchanger operate between those reservoirs in a reversible manner.

With reference to this PV diagram:
View attachment 352153
one could have two systems that are out of step by half a cycle. One could then put either of the systems at point 4 in thermal contact with the other system at point 2 and have some heat from the system flow into the system at 2 instead of into the cold reservoir. But that heat flow will stop when the system temperature over the 4' path drops below the system temperature over the 3' path.

To calculate the efficiency of such a heat engine, you would still have to factor in $Q_{2-3}$ from the hot reservoir and $Q_{4-1}$ to the cold reservoir and they could not be zero.

AM

Interesting! As you clearly know, Q_2-3 = - Q_4-1 for reversible isochoric processes and an ideal gas system.
This means that if you can provide a regenerator which acts like a separate part of the environment with a continuum of temperatures between T_c and T_h then you would actually get the same efficiency as for the Carnot engine.
The heat going into the regenerator is not lost and it's completely reused (in theory) so you don't need to count it when you calculate efficiency. If you want to see it that way you can include the regenerator into the system.
I don't see why such a regenerator is not possible in theory.
Obviously in practice it's never perfect. So much is clear.

Without such a regenerator Q_4-1 is a part of heat out and Q_2-3 is a part of heat in and you get an efficiency that's lower than Carnot at the same temperatures.

 

[Philip Koeck]

TL;DR Summary: Efficiency Of Stirling Cycle - Is it the same as the Carnot Engine? Why is my described engine not the same as Stirling engine?

Ok, the problem I was given is this:
One mole of a perfect gas goes through a quasistatic cycle consisting of the following four stages:
1. isothermal expansion from V1 to V2 at temperature TH,
2. cooling at constant volume V2 from TH to TC,
3. isothermal compression from V2 to V1 at TC,
4. heating at constant volume until the system is back in its initial state

Wikipedia states this for a Stirling cycle:

View attachment 352044

Ok, so it seems to me they are the same. (I am just including this part because my flaw might be in the assumption that they are the same cycle).

This guy (https://www.youtube.com/watch?v=sklO_qZTBeY&t=371s) explains how to calculate the efficiency of the Stirling engine. His logic all looks fine. His final answer is given by η=1−TLTH Now I have a problem because for the original question, I worked out $$\eta = \frac{W}{Q_H} = \frac{RT_C \ln\left(\frac{V_f}{V_i}\right) (T_H - T_C)}{RT_H \ln\left(\frac{V_f}{V_i}\right) + C_V (T_H - T_C)}$$ which is definitely not the same as the efficiency of the Stirling cycle. Explaining in a bit more detail how I achieved the above expression: η is simply work you get over work you pay for.

View attachment 352045

You might want to look at @Chestermiller's proof in post 2 of this thread:
https://www.physicsforums.com/threa...ounded-by-two-isotherms.1062470/#post-7100817

It turns out that all cycles bounded by two isotherms have efficiencies smaller than Carnot if there is no heat recycling.

 

[Andrew Mason]

Interesting! As you clearly know, Q_2-3 = - Q_4-1 for reversible isochoric processes and an ideal gas system.
This means that if you can provide a regenerator which acts like a separate part of the environment with a continuum of temperatures between T_c and T_h then you would actually get the same efficiency as for the Carnot engine.
The heat going into the regenerator is not lost and it's completely reused (in theory) so you don't need to count it when you calculate efficiency. If you want to see it that way you can include the regenerator into the system.
I don't see why such a regenerator is not possible in theory.
Obviously in practice it's never perfect. So much is clear.

The second law makes it impossible. Heat is energy transfer so it can only be “saved” as internal energy during the constant volume cooling step 2-3. In order to use that saved energy again in the constant volume 4-1 process you need it to be at a higher temperature than the system during that entire process. That is not possible even in theory.

In the Carnot cycle this is not a problem (in theory) because all heat flow occurs while the system is arbitrarily close to equilibrium with the hot and cold reservoirs. Work is recycled rather than heat. Work can be recycled (in theory) without loss of energy (e.g. by lifting a weight and then letting it drop; or by compressing and later expanding a spring). Heat cannot.

AM

 

[TSny]

The second law makes it impossible. Heat is energy transfer so it can only be “saved” as internal energy during the constant volume cooling step 2-3. In order to use that saved energy again in the constant volume 4-1 process you need it to be at a higher temperature than the system during that entire process. That is not possible even in theory.

I believe the two constant-volume processes may be considered reversible (in theory). As @Philip Koeck mentioned, we can invoke a "continuum" of heat reservoirs with temperatures ranging from the cold temperature $T_C$ to the hot temperature $T_H$. The temperature difference between one reservoir and the next is infinitesimal ($dT$).

Thus, in the constant-volume process where the engine cools from $T_H$ to $T_C$, we imagine something like the sketch below. The engine is placed consecutively in contact with the reservoirs.

All heat transfers are between objects which differ only infinitesimally in temperature. So, the heat transfers are essentially reversible. The vertical arrows indicate the direction of heat flow between the engine and the reservoirs. At the end of this process, the reservoirs have gained heat which can be used later in the constant-volume heating process that takes the engine from $T_C$ back to $T_H$, as shown below:

 

[Andrew Mason]

The problem, even in theory, is that you are effectively trying to avoid the second law of thermodynamics.

Although you could conceive of an arbitrarily large number of reservoirs each with infinite heat capacity and each with an arbitrarily small difference in temperature, dT, when you want to reverse the heat flow, you need to add 2dT to each of those arbitrarily large number of reservoirs to make heat flow in the reverse direction.

The first problem is that, since they must have infinite heat capacity, you cannot change the temperature by any amount by any means. So you could not use all the reservoirs.

Another problem is that you are not operating between a hot and cold reservoir at temperatures Th and Tc respectively. You operating between an arbitrarily large number reservoirs of infinite heat capacity, the reservoirs at Th and Tc being just two of them.

AM

 

[TSny]

The problem, even in theory, is that you are effectively trying to avoid the second law of thermodynamics.

Although you could conceive of an arbitrarily large number of reservoirs each with infinite heat capacity and each with an arbitrarily small difference in temperature, dT, when you want to reverse the heat flow, you need to add 2dT to each of those arbitrarily large number of reservoirs to make heat flow in the reverse direction.

The first problem is that, since they must have infinite heat capacity, you cannot change the temperature by any amount by any means. So you could not use all the reservoirs.

I don't see this. It appears to me that there is no need to change the temperature of any of the reservoirs.

Another problem is that you are not operating between a hot and cold reservoir at temperatures Th and Tc respectively. You operating between an arbitrarily large number reservoirs of infinite heat capacity, the reservoirs at Th and Tc being just two of them.

The $T_H$ reservoir does not absorb any heat in the cooling process shown in post #10 but it does give up heat $dQ$ in the heating process. The $T_C$ reservoir absorbs heat $dQ$ in the cooling process but does not give up heat in the heating process. All other reservoirs absorb $dQ$ in the cooling process and give up the same heat $dQ$ in the heating process.

$dQ = C_V dT$ where $C_V$ is the heat capacity at constant volume of the working substance of the engine. We can take the working substance to be an ideal gas, so $C_V$ does not depend on temperature. The ideal gas decreases its entropy during the cooling process by the same amount that it increases its entropy during the heating process.

So, for the cooling and heating processes taken together, only the $T_H$ and $T_C$ reservoirs have any net entropy change. Thus, the net entropy change of the universe for the cooling and heating processes together is $\Delta S_{\rm univ} = dQ(1/T_C - 1/T_H)$. Since $dQ$ goes to zero as the number of reservoirs goes to infinity, we can theoretically make $\Delta S_{\rm univ}$ as small as we like.

 

[Philip Koeck]

The problem, even in theory, is that you are effectively trying to avoid the second law of thermodynamics.

Although you could conceive of an arbitrarily large number of reservoirs each with infinite heat capacity and each with an arbitrarily small difference in temperature, dT, when you want to reverse the heat flow, you need to add 2dT to each of those arbitrarily large number of reservoirs to make heat flow in the reverse direction.

The first problem is that, since they must have infinite heat capacity, you cannot change the temperature by any amount by any means. So you could not use all the reservoirs.

Another problem is that you are not operating between a hot and cold reservoir at temperatures Th and Tc respectively. You operating between an arbitrarily large number reservoirs of infinite heat capacity, the reservoirs at Th and Tc being just two of them.

AM

I was wondering about another type of theoretical regenerator that's finite.
If the regenerator has exactly the same heat capacity as the system, then you would only need to add or remove an infinitesimal amount of heat to the regenerator to keep the cycle going in the right direction. The cycle would be infinitely slow of course.
The infinitesimal amounts of heat that you add to or remove from the regenerator to get the isochoric heating resp. cooling going could come from the infinitely large reservoirs used in the isothermal processes and they wouldn't change the reservoirs.
Does that make sense (in theory)?

 

[Andrew Mason]

I was wondering about another type of theoretical regenerator that's finite.
If the regenerator has exactly the same heat capacity as the system, then you would only need to add or remove an infinitesimal amount of heat to the regenerator to keep the cycle going in the right direction. The cycle would be infinitely slow of course.
The infinitesimal amounts of heat that you add to or remove from the regenerator to get the isochoric heating resp. cooling going could come from the infinitely large reservoirs used in the isothermal processes and they wouldn't change the reservoirs.
Does that make sense (in theory)?

That would seem to be just an indirect way of using the hot and cold reservoirs to provide the isochoric heat transfers. Each transfer from the hot reservoir to the regenerator would be at a finite temperature difference as the temperature of the regenerator increased from Tc to Th. Similarly, each transfer to the cold reservoir from the regenerator would be at a finite temperature difference as the regenerator temperature decreased from Th to Tc. So entropy of the cold reservoir would increase more than the entropy of the hot reservoir would decrease.

I note that the Stirling engine with the regenerator was first patented in 1816 and was touted as a reversible engine, in the sense that the cycle direction can be reversed to provide cooling. Carnot developed his theorem and concept for maximum efficiency of a heat engine in 1824. The concept of entropy took another 30 years or so. It may be that the notion of the Stirling engine being "reversible" stuck even though the concept of "reversibility" changed: ie. zero change in entropy of system and surroundings.

AM

 

[Andrew Mason]

I don't see this. It appears to me that there is no need to change the temperature of any of the reservoirs.



The $T_H$ reservoir does not absorb any heat in the cooling process shown in post #10 but it does give up heat $dQ$ in the heating process. The $T_C$ reservoir absorbs heat $dQ$ in the cooling process but does not give up heat in the heating process. All other reservoirs absorb $dQ$ in the cooling process and give up the same heat $dQ$ in the heating process.

$dQ = C_V dT$ where $C_V$ is the heat capacity at constant volume of the working substance of the engine. We can take the working substance to be an ideal gas, so $C_V$ does not depend on temperature. The ideal gas decreases its entropy during the cooling process by the same amount that it increases its entropy during the heating process.

So, for the cooling and heating processes taken together, only the $T_H$ and $T_C$ reservoirs have any net entropy change. Thus, the net entropy change of the universe for the cooling and heating processes together is $\Delta S_{\rm univ} = dQ(1/T_C - 1/T_H)$. Since $dQ$ goes to zero as the number of reservoirs goes to infinity, we can theoretically make $\Delta S_{\rm univ}$ as small as we like.

One way to do the isochoric processes reversibly would be to:

1. have the regenerator consist of a large volume of ideal gas and,
2. by a quasi-static expansion/compression of the regenerator (using stored mechanical energy from the work output in the isothermal expansion and then adding it back during expansion) allow the heat flow to/from the system gas to occur at an arbitrarily small temperature difference.

But I am not sure that this doesn't amount to just using the regenerator as an intermediate step in what is otherwise essentially a Carnot engine.

AM