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## Thermodynamics: Is Stirling engine reversible or irreversible?

 Quote by tsuwal how do you know that if it is resersible, Qc/Tc = -Qh/Th?
If it is reversible, ΔS = 0:

$$\Delta S_{sys} + \Delta S_{surr} = 0 + \int_c^h dS_{surr} = \int_{T_c}^{T_h} dQ/T = Q_h/T_h + Q_c/T_c = 0$$

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

 Quote by Andrew Mason If heat flow from the gas or from the regenerator to the cold reservoir is not isothermal or if heat flow from the hot reservoir to the gas or to the regenerator is not isothermal, then the engine cannot be reversible. I have yet to see an explanation that this is the case even in theory. 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 non-reversible: i.e. it takes more than an infinitesimal change in conditions to reverse the gradient. AM
I can see the validity of the point you make.

The ideal regenerator must now be understood like this: It is equivalent to a series of HRs with a continuous range of temperatures (in our case Th to Tc). Just as the HRs have no thermal or heat communication from one to the other, heat transfer through the regenerator from one end to the other end is prohibited, by definition. There exists a temperature difference between the two ends of the regenerator ΔT, but spacially there is no connectivity between two points for heat to flow. Heat flows only from the regenerator to the system and vice versa, but no heat transfer tangential to the plane of (or within) the regenerator. This is the reason the regenerator needs no additional support by way of an external source of energy to maintain the temperature difference between its ends. This is the concept of regenerator.
Regenerator is a conceptual device and not a practical device. To the extent the practical regenerator is different from the ideal (or conceptual one) we can take the practical Stirling engine to fall short of efficiency from the corresponding Carnot engine. Similarly, the real Stirling engine will be irreversible to the extent the regenerator falls short of ideality.

 Quote by Andrew Mason 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: $$\eta = out/in = (W_{1-2} - W_{3-4})/Q_h$$ Now the numerator is: $$nRT_h\ln\frac{V_2}{V_1} - nRT_c\ln\frac{V_3}{V_4}$$ Since V4 = V1 and V2 = V3 (isochoric parts) this is just: $$nR\ln\frac{V_2}{V_1}(T_h - T_c)$$ Since heat flows into the (ideal) gas during 4 and 1, Qh is: $$Q_h = Q_4 + Q_1 = \Delta U_{4-1} + W_{1-2} = nC_v(T_h-T_c) + nRT_h\ln\frac{V_2}{V_1}$$ So the efficiency is: $$\eta = W/Q_h = \frac{nR\ln\frac{V_2}{V_1}(T_h - T_c)}{nC_v(T_h-T_c) + nRT_h\ln\frac{V_2}{V_1}}$$ This reduces to: $$\eta = W/Q_h = \frac{(T_h - T_c)}{\frac{C_v(T_h-T_c)}{R\ln\frac{V_2}{V_1}} + T_h}$$ And this is the problem. For a reversible cycle we know that Qc/Tc = -Qh/Th, so efficiency is just: $$\eta = W/Q_h = \frac{(T_h - T_c)}{T_h}$$ 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
Your argument is perfectly valid. No one can find fault with it.

The problem, however is with the interpretation of the terms Q and W for efficiency calculations.

W poses no problem. It is Q in the denominator of efficiency equation that is problematic. One way to see it is as you saw it. But there is a different way of seeing at it! And that is to see the HRs that suffered loss of heat. In the case of Stirling engine it is just one HR that is at temperature Th that suffers heat loss Qh. It is this Qh that must find place in the denominator of efficiency and nothing else. That gives us Stirling efficiency equal to Carnot efficiency.

I may add an interesting experience of mine here.

Some years ago (1998-1999) AJP called for papers for its special theme issue on Thermal and statistical physics. Harvey Gould and Jan Tobochnik were the Editors of the issue. I submitted a paper which discussed the issue about the Q in the denominator of efficiency of Ideal heat engine cycles. It was more with specific reference to the right angled triangle PV cycle of R H Dickerson and J Mottmann that appeared in AJP 62,(1994) 737, where they dealt with calculation of efficiency.

My paper was found suitable by the editors and they asked me to send it by email. I sent it that way. Unfortunately Prof. Tobochnik found it contained virus and asked me to remove the virus and submit. I was new to use computers and didn't know how to remove it. While I was seeking help from others to get the virus removed the last day for acceptance passed.
Prof. J Tobochnik was not prepared to remove the virus at his end and was not prepared to include my paper in the special issue after the dead line of dates passed. He suggested I might submit it to AJP. I submitted it to AJP. AJP promptly rejected the same saying that it was a original paper and not suitable for publication in AJP.

By this time I lost interest in pursuing it further and that paper remained unpublished.

Prof. H S Leff who reviewed my paper found it interesting and offered me lots of help but in vain.

 Recognitions: Science Advisor I had a look at the design of a Stirling motor. Apparently the regenerator is a block with fine pores and a temperature gradient. If the gas is pumped sufficiently slowly through it, its temperature will always be in local equilibrium with that of the block at the current position. Hence irreversibility is due to heat conduction in the material of the block (which can be reduced chosing a material with a low heat conductance, like some ceramic) and the heat conduction of the gas along the pores (which can be reduced by reducing the size of the pores). So theoretically the irreversibility of a Stirling motor can be made small.

 Quote by DrDu I had a look at the design of a Stirling motor. Apparently the regenerator is a block with fine pores and a temperature gradient. If the gas is pumped sufficiently slowly through it, its temperature will always be in local equilibrium with that of the block at the current position. Hence irreversibility is due to heat conduction in the material of the block (which can be reduced chosing a material with a low heat conductance, like some ceramic) and the heat conduction of the gas along the pores (which can be reduced by reducing the size of the pores). So theoretically the irreversibility of a Stirling motor can be made small.
Perfectly right. In the limit no heat conduction within the block ideality is reached and reversibility of the cycle is ensured. If possible you may send me some more information about the design of Stirling motor and the regenerator.

 Both Lewitt (Thermodynamics Applied to Heat Engines) and Robinson & Dickson (Applied Thermodynamics) Contain quite a few pages describing/analysing Stirling and Ericsson engines, including drawings of the original engine and regenerator. Lewitt also has a worked example from past London University exams. PM me if you need more information.

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