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Thermodynamics: Is Stirling engine reversible or irreversible? 
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#1
Jan1813, 08:48 PM

P: 103

Stirling engine: the cycle is composed by two isothermals and 2 isometrica and there are just two heat reservoirs.
By the isometrics I would say that that it is irreversible since you exchanging heat with a reservoir at a different temperature than your gas. However my notes say the contrary.. I guess it refers to the case where there are a infinity of heat reservoirs 


#2
Jan1813, 11:12 PM

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In the first case, you are correct. It is irreversible because heat does not flow isothermally. In the second case, heat flow would occur isothermally but I'll buy a steak dinner to anyone who can explain to me how you actually would build such an engine. AM 


#3
Jan2013, 09:58 AM

P: 35

Your 1st and last sentences as also your guess are correct.
There are infinity of heat reservoirs (HRs) with which the system interacts during a cycle. But, of them only two HRs suffer a change of heat  the two whose Ts correspond to the temperatures of the isothermals. The other HRs suffer equal amounts of gain and loss of heat during each cycle and therefore don't suffer any change. The set of the HRs that do interact with the system but don't suffer a change are known by the name 'regenerator'. Stirling cycle is a reversible cycle and there is no irreversible heat transfer at any temperature during a cycle or during any portion of the cycle. 


#4
Jan2013, 02:00 PM

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Thermodynamics: Is Stirling engine reversible or irreversible?
This is important because if the regenerator is not at the same temperature as the gas, the process is not reversible. In the Carnot cycle this is not a problem because as the temperature of the gas changes (adiabatic compression or expansion) there is no heat flow. So all you need are two reservoirs. But to make the Stirling cycle reversible you need some way to make the temperatures of the regenerator always the same as the gas temperature as the gas increases or decreases its temperature. You would need an infinite number of regenerators all at different temperatures between Th and Tc. AM 


#5
Jan2113, 07:20 AM

P: 35

Perfectly right, except that you mixed up theory and practice.
Your statement about the reversible adiabatics in Carnot cycle is quite right. In reversible isochoric processes, the system suffers heat interactions which give rise to continuous change of temperature of the system. In order that the process be reversible, we need to provide HRs which are at the temperatures as the system is. This, however, is possible only conceptually. Even the isothermal processes in Carnot cycle are conceptual only  for that matter, any and every reversible process is conceptual only  not possible in practice! Failing to provide an infinity of HRs with a range of continuous temperatures, the best we do in practice is to use a regenerator. All processes we carry out in practice are necessarily irreversible  use of regenertor is no exception. 'You would need an infinite number of regenerators all at different temperatures between Th and Tc.' No, we don't need infinite number of regenerators, we just need one  because, an ideal regenerator acts as a set of infinity number of HRs between two temeratures: Th and Tc. P. Radhakrishnamurty 


#6
Jan2113, 08:04 AM

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I don't see a problem to realize a heat bath of varying temperature, e.g. a large amount of gas being compressed adiabatically.



#7
Jan2113, 09:16 AM

P: 35

As long as you don't see a problem to realize a heat bath (better use heat reservoir) of varying temperature, you may go ahead and use it in the isochoric steps of Stirling cycle, which will then work as a reversible heat engine!
P. Radhakrishnamurty 


#8
Jan2113, 09:34 AM

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AM 


#9
Jan2113, 10:33 AM

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For sure this then resembles suspiciously a Carnot cycle.
But at least it does not nearly sound as strange as that of coupling to an infinity of heat baths with infinitesimally small temperature difference. 


#10
Jan2113, 12:45 PM

P: 5,462

Here is a short description of the regenerator used in the 1845 original engine in the foundry in Dundee. Note the significance of the installation in a foundry.
During the ischores in the cycle the energy rejected is the same as the energy required thus if a device could be found to store the energy during the rejection process and yield it up again during the acceptance process, once the engine had started and warmed up, the cycle would be self perpetuating and no further heat energy would be required through the system boundary. The device used to carrry out this process was called a regenerator and in the case of the foundry engine consisted of a matrix of sheet iron plates maintained at a high temperature by the furnace at one end and at a low temperature by a water cooler at the other. Thus the necessary temperature gradient was maintained through the matrix, which was of such a bulk that the necessary heat transfers through the processes did not substantially modify the temperatures. Since, with a regenerator installed, external heat energy is not required to carry out the constant volume processes then the thermal efficiency relies only on the two isothermal processes and is the same as a Carnot engine. 


#11
Jan2113, 08:57 PM

P: 35

You are right.
In fact, perhaps as you would be aware, the efficiency of Stirling cycle is the same as that of Carnot cycle operating with same values of Th and Tc. After all, they bring about the same changes in the surroundings during each cycle of operation (assuming the same capacity), making them indistinguishable from the point of view of the surroundings, and, that is what matters for efficiency calculation. P. Radhakrishnamurty 


#12
Jan2113, 09:01 PM

P: 35

Thanks for the information.
I too read a similar description about a regenerator, Though, I don't recall the source now. P. Radhakrishnamurty 


#13
Jan2213, 10:05 AM

P: 103

I think the efficiency depends on how you "count" the energies. If you are using a regenerator, you should take into account the heat given by the motor in the right isochoric, that heat cancels the heat given in the left isochoric and the efficiency turns out the same as Carnot's and therefore the engine is reversible. However, if you don't take into account that heat (don't use a regenerator) the efficiency is given by:
R(T2T1)*ln(r)/(cv*(T2T1)+R*T1*ln(r)) r=volume ratio cv=heat capacity at constant volume T2=temp of the hot resevoir T1=temp of the cold resevoir 


#14
Jan2213, 11:04 AM

P: 35

You are right. But what is the point you want to make?
It may be of interest to you to note that, there are some, who argue that the efficiency of Stirling cycle is less than the efficiency of Carnot cycle. They don't bother where the rejected heat goes! the area enclosed by the PV diagram of the cycle gives the work and the sum of the Qs (heat supplied at constant volume plus the heat supplied during the isothermal expansion) is taken as the heat input to the system. This issue has lot of consequences on the second law! 


#15
Jan2213, 11:40 AM

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There is necessarily a thermal gradient through the regenerator. One end is thermally connected to the hot reservoir and the other end is thermally connected to the cold reservoir. So heat flows through the regenerator from the hot reservoir to the cold reservoir without doing any work. A heat gradient like that is always nonreversible: i.e. it takes more than an infinitesimal change in conditions to reverse the gradient. AM 


#16
Jan2213, 02:44 PM

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In reviewing some of the literature on the Stirling engine, it appears that the term "reversible" when applied to the Stirling cycle did not mean reversible in the sense of ΔS = 0 but reversible in the sense that its direction can be reversed so that by adding work it operates as a refrigerator. This appears to be how Lord Kelvin used the term.
The cycle requires heat flow either into and out of the gas during all four parts of the cycle. Work is done only on two parts of the cycle, 1 and 3 (with 1 being the isothermal expansion). The efficiency is: [tex]\eta = out/in = (W_{12}  W_{34})/Q_h[/tex] Now the numerator is: [tex] nRT_h\ln\frac{V_2}{V_1}  nRT_c\ln\frac{V_3}{V_4}[/tex] Since V4 = V1 and V2 = V3 (isochoric parts) this is just: [tex] nR\ln\frac{V_2}{V_1}(T_h  T_c)[/tex] Since heat flows into the (ideal) gas during 4 and 1, Qh is: [tex]Q_h = Q_4 + Q_1 = \Delta U_{41} + W_{12} = nC_v(T_hT_c) + nRT_h\ln\frac{V_2}{V_1}[/tex] So the efficiency is: [tex]\eta = W/Q_h = \frac{nR\ln\frac{V_2}{V_1}(T_h  T_c)}{nC_v(T_hT_c) + nRT_h\ln\frac{V_2}{V_1}}[/tex] This reduces to: [tex]\eta = W/Q_h = \frac{(T_h  T_c)}{\frac{C_v(T_hT_c)}{R\ln\frac{V_2}{V_1}} + T_h}[/tex] And this is the problem. For a reversible cycle we know that Qc/Tc = Qh/Th, so efficiency is just: [tex]\eta = W/Q_h = \frac{(T_h  T_c)}{T_h}[/tex] Two reversible engines have to have the same efficiency. So if the Stirling cycle is "reversible" in the modern thermodynamic sense, that Q4 has to disappear. As I said, if you can show me how it disappears, I'll buy you a steak dinner. AM 


#18
Jan2213, 06:06 PM

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[tex]\Delta S_{sys} + \Delta S_{surr} = 0 + \int_c^h dS_{surr} = \int_{T_c}^{T_h} dQ/T = Q_h/T_h + Q_c/T_c = 0[/tex] Qh is negative since it is a flow out of the reservoir. Qc is positive. And there is no change in the system entropy since there is no change in state in a complete cycle. AM 


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