The ideal Stirling Cycle and Heat Load on the Regenerator

Content of this page :

Abstract
Introduction
Description of Cycle
The Regenerator and Efficency of Stirling Cycle
A short note on ideal gases
Influence of working fluid on work and heat transfer rates

In the ideal Stirling cycle for a heat engine the isothermal compression stroke ( process 1→2 ) takes place by removing heat, Q₁₂, from the working fluid in order to keep the temperature at a constant low value, T_C. This is followed by an isochoric compression (process 2→3). The latter takes place because heat of the amount Q₂₃ is added to the working fluid. Expansion (process 3→4) takes place at constant temperature, T_H. In order to keep the temperature constant, heat, Q₃₄, has to be added to the working fluid. Finally, the pressure is lowered at constant volume by removing heat, Q₄₁, from the working fluid to close the cycle.

The Regenerator and Efficency of Stirling Cycle

Basis of the analysis of a cycle is the 1. law of thermodynamics of all its sub processes. In this case we have a working fluid of constant mass which for a process with starting point i and end point j is :

Q_ij - W_ij = U_j - U_i

with :

Q_ij = heat transferred into the working fluid when fluid changes from state i to state j

W_ij = ∫ p dV = work gained during the process i→j

U_i = internal energy of working fluid at state i

In particular, for the process 2→3 the volume does not change and therefore W₂₃ is zero and :

Q₂₃ = U₃ - U₂

and similarily for the process 4→1 :

Q₄₁ = U₁ - U₄

If we choose a working fluid for which the internal energy depends only on temperature ( liquids and ideal gases for example ) the internal energies U₃ and U₄ are equal ( same temperature T_H ). Along the same lines : U₂=U₁. Therefore :

Q₄₁ = - Q₂₃

Or in other words, the heat we have to remove from the gas during the process 4→1 can be used to heat the gas during the process 2→3, which is exactly the function of the regenerator.

Once a regenerator is employed, the interaction of a Stirling engine with its surrounding is restricted to receiving heat at temperature T_H and rejecting heat at temperature T_C. With that its thermodynamic efficiency ( = ratio of work gained to heat provided ) must be equal to that of the Carnot cycle :

η = 1 - T_C/T_H

The behavior of ideal gases , that is how temperature, pressure, volume, and other properties are related to each other is governed by 2 laws. The first, the ideal gas law, is well known, but written down in many different forms which , of course, are all related to each other. For example :

p V = m R_s T    or     p V = n R_u T

p [N/m²] = absolute pressure ( p=0 for vaccuum )

V [m³] = volume occupied by ideal gas

m [kg] = mass of ideal gas

R_s [kJ/(kg K)] = specific gas constants, its value depends on the gas at hand.

T [K] = absolute temperature

n [kmol] = number of moles of gas , 1 mol = 6.023×10²³ molecules (Avogadro's number)

R_u = universal gas constant = 8.31434 kJ/(kmol K)

The above two forms of the ideal gas law are equivalent because :

m/n = M [kg/kmol] = molecular mass of gas , often but falsly called the molecular weight

and

R_s = R_u/M

Another property of ideal gases, much less known, is that its specific internal energy u, enthalpy h, and specific heat capacities c_p and c_v depend ONLY on the temperature of the gas and that :

c_p - c_v = R_s

For some ideal gases, specifically all the noble gases He,Ne,Ar,Kr,Xe, the specific heat capacities are completely temperature independent and their ratio is :

κ = c_p/c_v = 1.667     for all monatomic gases

κ = 1.4     for diatomic gases

A last question of course arises : Under which circumstances does a real life gas behave like an ideal gas ? The answer is somewhat ambiguous because the ideal gas is an abstraction which is approached by every gas as the pressure is lowered and the temperature is raised. If a few percent accuracy is sufficient then a good guideline would that if the absolute temperature of interest is more than twice that of the critical temperature ( listed as T_crit in above table ) you are in good shape. If your temperatures are lower than that, your pressure of interest better be below 0.2 times the critical pressure ( listed as p_crit in above table). If you need a clearer answer, then you either have to get the property data of your gas ( the NIST website at http://webbook.nist.gov/chemistry/fluid/ is a very good source ) or google for the term Compressibility Factor= p V/(R T) which is of course 1 for ideal gases but ≠1 for real gases.

A last comment concerning criticial temperature and pressure, just in case you don't know these :

If you have a gas at a temperature above its critical temperature then the gas will occupy less and less volume as the pressure is increased provided you keep the temperature constant by removing heat as you compress. It will never condense as the pressure increases and we have no clear definition for when you would stop calling it a gas versus a liquid.
If your temperature at which you compress and remove heat is below the critical temperature your gas will ultimately condense into a liquid at a particular pressure as you remove heat, i.e 1 atm for water at 100°C.
When condensing/boiling at a particular temperature takes place all property data of the liquid and of the gas phase have values which are unique for the gas and the given temperature, including pressure. As the boiling temperature is increased the values of all properties ( like internal energy, enthalphy, specific volume, index of refraction etc. ) of the liquid approach those of the gas phase. At the critical temperatur/pressure they are equal.
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We assume that our working fluid is an ideal gas with constant specific heat capacities.

Q₃₄=W₃₄= ∫ p dV = &int n R_u T_H /V dV = n R_u T_H ln ( V₄/V₃ )

In the same fashion for the process 1→2 :

Q₁₂=W₁₂= ∫ p dV = &int n R_u T_C /V dV = n R_u T_C ln ( V₂/V₁ )

And the network per cycle :

W_net = W₃₄ + W₁₂ = n R_u ( T_H - T_C ) ln ( V₄/V₃ ) ;

The heat supplied at temperature T_H ( remember we employ a regenerator ) is equal to Q₃₄ which then gives us a thermodynamic efficiency ( ratio of benefit to costs ) of

η = 1 - T_C/T_H

This is standard thermodynamic stuff and nothing new.

We wish to take a look though at the influence of changing the working fluid, for example He vs. H₂ versus air. Of course we have to keep the temperatures T_C and T_H the same. Furthermore, the volume of our engine, V₀, remains constant and we load the engine to the same initial pressure, p₀, at the same initial temperature, T₀. In this case, the number of moles of gas , n , loaded into the engine remains constant :

n = (p₀ V₀)/(R_u T₀)

With that, the numerical values for the above works, heats, and the efficiency are not affected by a change of gas.

The story is different though for the regenerator. The heat transferred from the regenerator material into the gas during the process 2→3 and removal from the gas during 4→1 is equal to the change in internal energy of the gas which for constant specific heat capacities is :

Q₂₃ = m c_v ( T_H - T_C ) = 1/(κ-1) n R_u ( T_H - T_C )

Q₂₃ is influenced by the ratio of specific heats which changes as we go from monatomic, to diatomic and then to gases with even more atoms per molecule.