The Stirling Cycle and Heat Load on the Regenerator

Sections on the page :

Description of Cycle
The Regenerator and Efficency of Stirling Cycle
Influence of working fluid on work and heat transfer rates


This page contains a short rehash of the Stirling cycle for heat engines. Some math and knowledge of thermodynamics ( conservation of energy , internal energy , specific heat capacities at constant pressure and constant volume ) is required for understanding. Major finding, beyond what can practically be found in any text book, is that the heat transfer load on the regenerator is influenced by the gas being used through the value of the gas' ratio of specific heats. The heat load on the regenerator increases as we go from monatomic to diatomic gases by a factor of 1.67 and by a factor of more than 2 as the gas molecules contain 3 or more atoms.
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The ideal Stirling Cycle consists of 4 individual processes with end points as marked in the p-V diagram ( p=pressure, V=volume). Its analysis is a standard item in all text books of engineering thermodynamics dealing with gas powered heat engines. So why revisited ? Apart from summarizing the underlying thermodynamic laws and the resulting equations we look at the heat transfer needed for the two isochoric processes of the cycle and the amount of heat needed to facilitate them. Surprisingly, the amount of heat needed for these processes depends strongly on the ratio of the specific heats of the working fluid favoring monatomic ideal gases to be used as working fluid. The term "fluid" is used here in the same way physicists use it : it can mean either a gas or a liquid.
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Description of Cycle

Figure 1 : Ideal Stirling cycle in
pressure-volume diagram
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, TC. 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, TH. 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.

Point in
Figure 1
Table 1 : Volume of Compression
and Expansion Space
If you assume to have an alpha-type Stirlign engine with a compression and an expansion space then Table 1 shows how you can realize the four corner points shown in Figure 1.

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 :

Qij - Wij = Uj - Ui

with :

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

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

Ui = 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 TH ). 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 TH and rejecting heat at temperature TC. With that its thermodynamic efficiency ( = ratio of work gained to heat provided ) must be equal to that of the Carnot cycle :

η = 1 - TC/TH

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Influence of working fluid on work and heat transfer rates

We assume that our working fluid is an ideal gas with constant specific heat capacities.

Hence for the process 3→4 the internal energy of the gas remains constant and :

Q34=W34= ∫ p dV = ∫ n Ru TH /V dV = n Ru TH ln ( V4/V3 )

In the same fashion for the process 1→2 :

Q12=W12= ∫ p dV = ∫ n Ru TC /V dV = n Ru TC ln ( V2/V1 )

And the network per cycle :

Wnet = W34 + W12 = n Ru ( TH - TC ) ln ( V4/V3 ) ;

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

η = 1 - TC/TH

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 TC and TH 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)/(Ru 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 cv ( TH - TC ) = 1/(κ-1) n Ru ( TH - TC )

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.

Influence of ratio κ on heat load in regenerator
GasFormula κ = cp/cv 1/(κ-1)Factor*
HeliumHe 1.6671.51.0
ArgonAr 1.6671.51.0
NitrogenN2 1.4002.51.67
OxygenO2 1.3952.531.69
Air  1.4002.51.67
HydrogenH2 1.4052.4691.65
Carbon DioxideC O2 1.2893.462.31
WaterH2O 1.3273.062.04
MethaneC H4 1.2993.342.23
PropaneC3H8 1.1267.945.29
OctaneC8H18 1.04422.7215.15
* Factor = factor by which the heat load on the regenerator increases in comparison to the case of having a monatomic gas.
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Zig Herzog © 2014
Last revised: 12/05/14