About the Efficiency of the Regenerator in the Stirling Engine

and the Function of the Volume Ratio V_max / V_min

Dipl.-Ing. Peter Fette
Am Schaeferloch 16,
D-75045 Walzbachtal / Germany

ABSTRACT:

Attention: some browsers may not produce correct greek letters but show them as latin equivalents

The twice double acting Stirling Engine introduced at the 6'th International Stirling Engine Conference ISEC at Eindhoven Netherlands in May 1993 (Report No. ISEC-93003) is an engine which can realize realative high volume ratios V_max/ V_min . Whenever a heatsource and a cooling water supply and their corresponding haetexchangers have enough power for the energy transmission then a higher volume ratio of V_max/ V_min will lead to a higher work output per revolution. Theoretical work is done with regard to the efficiency of the regenerator, the volume ratio of the engine and the adiabate exponent k = C_p / C_v of the working fluid. It can be shown, that the efficiency of the regenerator can be lowered for the benefit of a lower loss of pressure in it. For very high volume ratios the regenerator can be a suitable tube only. Formulas and diagrams illustrate the correlation of Carnot efficiency and the thermal efficiency of the process with the parameters: volume ratio V_max/V_min, k and the regenerator efficiency h_reg.
All diagrams and calculationen were made with the program STMOT2 |6|, base of this program is described in |7|

Nonenclature:

C_p Specific heat of working fluid at constant pressure
C_v Specific heat of working fluid at constant volume
k Adiabate exponent; the ratio of the specific heats k = C_p/ C_v
M _GAS Mass of working fluid within the engine.
h_C Carnot efficiency of the ideal process h_C = (TE-TC) / TE
h_reg Regenerator efficiency
RV Loss factor for the regenerator: RV = 1 - h_reg
h_therm Resulting thermal efficiency of the process
Q_1-2 Heat rejected by the cooler at the compression phase of the ideal Stirling engine at "TC". For the indexes 1-2-3-4 look to Fig. 3
Q_2-3 Isochoric heat supplied at V_min
Q_reg external isochoric heat supplied at V_min if the regenerators efficiency is h_reg < 1, then Q_reg = (1- h _reg ) * Q_2-3= RV * Q_2-3
Q_3-4 External heat supplied at the expansion phase of the ideal Stirling engine at "TE"
Q_4-1 Isochoric heat rejected at V_max
R Gas constant: R = C_p - C_v
S Entropie of the working fluid
TC Constant cold Temperature of the working fluid in the compression space of the ideal Stirling engine
TE Constant hot Temperature of the working fluid in the expansion space of the ideal Stirling engine
VE stroke volume of the expansion cylinder
VC stroke volume of the compression cylinder
VG total instantaneous gas volume in the engine inclusive all dead spaces
V_max is the maximum volume of all working spaces in the engine (including heater, cooler and regenerator)
V_min is the minimum volume of all these spaces
V_max/ V_min the volume ratio.
W Usable work output per cycle
WE Work done during the expansion phase
WC Work to be supplied during the compression phase.

The theoretical considerations of Finkelstein |2| were a challenge to make further theoretical and practical investigations for the Stirling engine decribed in |1|. So this theoretical work is specially done for this engine. The results may not be transfered in all consequences to other Stirling engines, but the tendency of these results should be valid for other engines too.

In the engine |1| a change of the volume ratio is easy to be performed by variation of the working liquid within the cylinders below the gas cushion. Insofar this engine is an optimal testengine too. There are some reasons that a regenerator for this Stirling engine cannot be optimally designed in the common way. E.g. because of too much loss of pressure in the regenerator, and, for this engine is able to work with compound fluids, a commonly designed regenerator will not work with condensating steam, it has additional disadvantages too. A regenerator designed for this engine will work at a loss! But the questions to be answered are: What a loss at the regenerator is tolerable, and what can we do to increase the thermal efficiency with others than an optimal commonly designed regenerator?

These are the common questions. There have been done a lot of theoretical invetigations on this territory, some described in |3|. This work will not only add one more, but it makes an attempt to give an other point of view to these questions, naturally with special attention to the engine of |1| for heatsources of low to medium temperatures.

Going out from an ideal isothermal Stirling process we introduce the effectiveness of the regenerator to this process and then call this process the "disidealized" isothermal Stirling process having a variable efficiency of the regenerator |5|.

The thermal efficiency of a process is defined as the ratio between "W" the external work done per cycle and "Q_ext" the external heat supplied per cycle to the engine.

(1)

Concerning all phases the efficiency of the ideal Stirling process is:

(2)

In this ideal Stirling process it is supposed, the regenerator is of 100 % efficiency; h_reg = 1 . Then the isochoric heat supplied by the regenerator
Q_{2 - 3} at constant volume V _min is equivalent to the isochoric heat Q _{4 - 1} that is received by the regenerator at constant volume V_max.
So Q _{2 - 3} + Q_{4 - 1} = 0. and the external supplied heat for the ideal process is: Q_ext = Q_{3 - 4} . Now the ideal efficiency is:

(3)

Imagine a disidealized Stirling process with a regenerator at a loss. h _reg < 1 . We define a lossfactor RV = 1 - h_reg .

Now imagine that the work "W" done per cycle should be the same for the ideal as for the disidealized process. Then for the case of h_reg < 1 additional heat must be supplied by the heater to get the temperature "TE" at the end of the isochoric change of state done by the regenerator and before beginning the isothermal expansion. And at the other isochoric change of state, -cooling done by the regenerator- additional heat must be rejected by the cooler to get the temperature "TC" before beginning the isothermal compression phase at V_max. This part of the energy which the regenerator cannot supply to the process may be defined as Q_reg = RV * Q_2-3

With these considerations we now define a thermal efficiency of the disidealized process:

(4)

We now must get this equation in a suitable form by evaluating the above expressions: Q_{3 - 4} the heat supplied at the isothermal expansion phase is equal to the external work done during isothermal expansion:

(5)

for the compression phase:

(6)

The isochoric heat supplied at V_minis:

(7)

We now get for h _therm

(8)

We make further substitutions:

Carnot efficiency h _C = ( TE - TC) / TE .
The adiabate exponent k = C _p/ C _v and the gas constant R = C _p - C_v
leads to the term C _v / R = 1 / (k - 1).

We now get h _therm in the wanted form for computer calculations:

(9)

The meaning of equation (9) is:

The resulting thermal efficiency of a disidealized Stirling process -that is an isothermal Stirling process with a regenerator at a loss of RV= 1 - h_reg - is a function of:

its Carnot efficiency h_C depending on the maximum "TE" and minimum "TC" temperatures within the process.

its regenerator efficiency h_reg resp. the regenerator loss: RV = 1 - h_reg
its volume ratio V_max / V_min
the adiabate exponent k of the working fluid; where:

k = 1.6 for one-atomic gases like Helium, Argon,

k = 1.4 for two-atomic gases like Hydrogen, Air,

k = 1.3 for three-atomic gases like Carbondioxyd CO₂

Fig. 1 represents the calculation of Equation (9) for 5 different volume ratios:

V_max/ V_min= 1.2, 1.6, 2.5, 5.0 and 10. and for
2 regenerators:

one regenerator at an efficiency of h_reg = 50 %
and one "null" regenerator (h_reg = 0 means: The regenerator is a tube only).

The working fluid is air for all cases k = 1.4 .

The abszissa in this diagram is the Carnot efficiency in the range from 5 <= h_c <= 80%. The ordinate is the resulting thermal efficiency according to equation (9).

The drawn example in Fig. 1 says:

If You know the temperatures "TE" and "TC" of a disidealized Stirling process, which in this example will lead to a Carnot efficiency of 28% and You have a volume ratio of V_max/ V_min= 1.6 in Your engine which is working with a regenerator at a loss of RV = 50 % then You will get a resulting thermal efficiency of this process of h_therm= 16 %.

A second result of this example is: You will get the same efficiency of h _therm= 16 % if You have only a tube, a regenerator with h _reg = 0 , and when You design this engine with a volume ratio of V_max/V_min = 2.5 .

Fig. 2 shows the dependency of the working fluids characteristics k = C _p/ C_v.

The diagram is calculated:

for an one atomic inert gas like Helium or Argon and for a two atomic gas like Hydrogen or Air.
for 5 volume ratios V_max/ V_min= 1.1, 1.6, 2.0, 5.0 and 10. Two regenerator efficiencies are calculated:
h_reg = 20 % for the one atomic fluid and
h_reg = 45 % for the two atomic working fluid.

These regenerator efficiencies are specially choosen to demonstrate, that You will get the same thermal efficiency now with the same volume ratio in the engine but with a worse regenerator if You choose Helium or Argon instead of air as the working fluid.

This result of equation (9) is to be taken with care, because if You change the working fluid, equation (9) does not take into account the thermal conductivity of the working fluid, which is also important for the efficiency of the process. But on the other hand if You design a slow rotating engine as described in |1| with an optimal heattransfer to the working fluid, Argon will be the better working fluid than air inspite of its slightly worse thermal conductivity. (Naturally the more expensive inert gas Helium is the best for its thermal conductivity is atleast 4 times better than that of air !)

For the engine described in |1| a FORTRAN 77 computer program STMOT2 |6| is develloped for the calculation and graphical output of several 100 functions of interrest.

Fig.3 and Fig.4 show two individual T-S diagrams for the same engine:

In Fig.3 the working fluid is Air with k = 1.4 , and

In Fig.4 the working fluid is Argon with k = 1.6 .

The effect of the different k's is best seen in the gradient of the isochoric change of state lines (2-3) and (4-1) in these both figures.
This gradient d T/d S is equal to :

(10)

The "frames" in both T-S diagrams represent the ideal isothermal Stirling process, the graph within the frame is the calculated T-S diagram of the real process of this a - type engine |1| driven by a crankgear. In both cases the included area of the graphs represent the same external work done per cycle. Have a look to the different gradients d T/d S of these "frames" which are borders of the isochoric heat.

The isochoric heat exchanged by the regenerator is much less at the "frame" in Fig.4 (Argon) than in Fig.3 with air as the working fluid.

The thermal efficiency h_therm of the "calculated real process" is improved from h _therm = 20.6 % in the case of the working fluid is Air to h _therm = 23.7 % in the case of Argon.

According to equation (9) the thermal efficiency h _therm of the "disidealized" isothermal process -the "frames" in Fig. 3 and Fig. 4- is improved from h _therm = 25.6 % in the case of the working fluid is Air to h _therm = 29.1 % in the case of Argon.

The calculations |6| of the processes in Fig.3 and Fig.4 are both done with:

Phaseangle d = 90 ø for the real engines process; pure sinusodial piston movement; Dh=10. Dk=10. R=10. cm; hot and cold deadspaces are 30 % of each stroke volume. See input for STMOT2 |6|
Isothermal temperatures TE = 500 K, TC = 300 K.; h_C= 40 % ; P0 = 5 bar
The regenerator efficiency in both cases is h _reg = 50 %; regenerator volume VR=0.15 l,
The volume ratio in both cases is V_max/ V_min= 2.43

There are made 8 additional simulation runs with the computerprogramm STMOT2 to demonstrate the influence oft the idle pressure P0 (this is the gas pressure within the resting and cold engine), the volumeratio V_max / V _min, the adiabatic exponent k and the regenerator efficiency h_reg to the usable power and the usable efficiency h_n of the engine. These results are presented in a table (Some more information to the output of STMOT2 see in |6|. The input datasets for these 8 simulations are described in section "5 of sample calculations" . The reader may copy the input data from here to make his own calculations with STMOT2)

All values of interest in the table for the working fluid Helium are marked red and are marked blue for the fluid air. Some more remarks are to be made:

In column 1 of the table RUN NO. points to the No. of the simulation run.
In column 7 : ETA_R is the regenerator efficiency h_reg
In column 9 : P0 is the idle pressure P0 of the resting and cold engine; P1MEAN is the mean working pressure.
In column 10 : Vmax/Vmin is the volume ratio
In column 13 : ETA_N is the usable efficiency, which is written h_n in the signatures to Fig.(1)(2) and Fig.(5)(6).
In column 15 : POWER USABLE is the mean usable power for the calculated revolution.
In column 16 : DQZU1 MAX is the maximum of the graph for the supplied heat at this revolution.
You will find more information to this table output in chapter I.6.2 of the user guide to STMOT2 (in german only).

The first 4 simulation runs concern the increase of usable power "POWER USABLE" as a reaction of the higher volumen ratio; run No (1) and run No (2) are calculated for air with a k = 1.4 and run No (3) and run No(4) are calculated for Helium with a k = 1.6 . Here a short extract from the table

(1) idle pressure P0 =16 bar, regenerator efficiency h_reg = 80 %, Vmax/Vmin = 2.71 ----> POWER = 16377 Watt, h_n =30.46 %
(2) idle pressure P0 =16 bar, regenerator efficiency h_reg = 80 %, Vmax/Vmin = 1.29 ----> POWER = 343< Watt, h_n =17.99 %

With Helium as the working fluid and equal conditions for the engine:

(3) idle pressure P0 =16 bar, regenerator efficiency h_reg = 80 %, Vmax/Vmin = 2.71 ----> POWER = 16377 Watt, h_n =33.39 %
(4) idle pressure P0 =16 bar, regenerator efficiency h_reg = 80 %, Vmax/Vmin = 1.29 ----> POWER = 343< Watt h_n =22.47 %

The strong increase of power from POWER = 343 Watt in simulation runs No. (2) and (4) to POWER = 1637 Watt in runs No (1) and (3) naturally is only obtained when "heater" and "cooler" can perform the needed high energy transmissions !! The maximum values for the heat supplied "DQZU1 MAX" are in column 16: of the table. For Helium as the working fluid a smaller maximum value "DQZU1 MAX" is calculated as it is to be expeted in accordance with equation (9).

Fig.(1)(2) shows as functions of the crank angle the graphs for: supplied heat "DQZU1", the removed energy "DQAB1", the idle pressure P1 and the mean power for the calculated revolution "POWERN". Here all graphs of runs No (1) and (2) with air as the workind fluid are put together in 1 diagram. The signals (1) and (2) in Fig.(1)(2) are in accordance with these run No's. 1 and 2 of the table. The values for Helium are shown in the table at RUN NO. 3 and 4.

In the next 4 simulation runs (5) and (6) with air and runs (7) and (8) with Helium the usable power of the engine shall be hold constant at 316 to 319 Watt. It shall be shown here, that this power can be reached with a high volume ratio Vmax/Vmin = 2.71 at a significant lower idle pressure "P0"= 3.3 bar (or a significant lower mean pressure "P1MEAN") and at a lower regenerator efficiency h_reg= 40 %, than the same power is obtained with a lower volume ratio Vmax/Vmin = 1.29 but with P0=15 bar and h_reg = 80%. Next to this, the simulation shows the maximum point for the heat supplied "DQZU1_MAX" is significant lower, when the engine works with this high volume ratio at this power of 316 - 319 Watt. Here again a short extract from the table

(5) usable power = 316 Watt, idle pressure P0 =3.3 bar, regenerator efficiency h_reg = 40 %, volume ratio Vmax/Vmin = 2.71
---->h_n =19.59 % , maximal heat supplied "DQZU1_MAX" = 2496 Watt

(6) usable power = 319 Watt, idle pressure P0 =15 bar, regenerator efficiency h_reg = 80 %, volume ratio Vmax/Vmin = 1.29
---->h_n =17.89 % , maximal heat supplied "DQZU1_MAX" = 5266 Watt

With Helium as the working fluid and the same usable power the simulation results are:

(7) usable power = 316 Watt, idle pressure P0 =3.3 bar, regenerator efficiency h_reg = 40 %, volume ratio Vmax/Vmin = 2.71
---->h_n =23.92 % , maximal heat supplied "DQZU1_MAX" = 2135 Watt
(8) usable power = 319 Watt, idle pressure P0 =15 bar, regenerator efficiency h_reg = 80 %, volume ratio Vmax/Vmin = 1.29
---->h_n =22.34 % , maximal heat supplied "DQZU1_MAX" = 4977 Watt

These simulation results seem to be impossible on the first view. A good way to explain this phenomenon of the results of runs no. (5) and (6) for air as the working fluid will give the T-S Diagrams in Fig.(5)(6)TS, and for the runs (7) and (8) with Helium the T-S Diagrams in Fig.(7)(8)TS will give an explanation:
The work done per revolution, is represented by the area included by the graphs in these diagrams. This work minus the friction losses result in the POWER USABLE: 316 resp. 319 Watt. The areas included by the "frames" around the graphs, are equivalent to the work of the ideal Stirling process. If we now look to the ideal Stirling processes for the simulation runs (5) and (6) (or to the runs (7) and (8)), then indeed we find a quite other result. The work done "W_i" of the ideal Stirling process is equal to the numerator in eq.(8) -see obove:

W_i = M_Gas * R * (TE - TC) * ln ( V_max / V_min )

here the gasmass "M_Gas" is proportional to the idle pressure P0. By this the work of the ideal Stirling processes from simulation run (6) with the high idle pressure P0 = 15 bar is significant higher as in run no. (5) where P0 = 3.3 bar. You easy will see, when comparing the "frames" in both diagrams of Fig.(5)(6)TS (and in both diagrams of Fig.(7)(8)TS). The ideal Stirling process does'nt know deadspaces, the real Stirling process, in which acts a continuously working crank gear, must consider the deadspaces. Now in the real Stirling process a high volume ratio -this means small deadspace- results in a high "load exploitation factor" -here 37.11% of the ideal Stirling process is reached. (for the "load exploitation factor" also look to my unshortened report to the 6#th ISEC in Eindhoven |1|.) The large deads pace, which is defined in the input data to simulation runs no. (6) and (8), here now leads to a smaller load exploitation factor -here 8.73% of the maximal possible -the ideal- process, so you will only get nearly the same work done inspite of the higher mean pressure resp. the higher idle pressure P0 as in runs no. (5) and (7). In column 13 of the table values for AUSN_I are these load exploitation factors.

The total gasmass in run no. (5) is GM01= 1.394851 gr air. This is a significant lower gasmass in the engine as in run no. (6) where the total gasmass is GM01= 23.520571 gr air (see below columns 17 to 19 in theTabelle). Therefore heat to be supplied to warm up the gasmass in run no. (5) will be much more lower. The maximal supplied heat you find in column 16 of the table with a value of "DQZU1_MAX" = 2496 Watt, and for the run no. (6) the maximal heat supplied is "DQZU1_MAX" = 5266 Watt.

Like Fig.(1)(2) now the Fig.(5)(6) shows as functions of the crank angle the graphs for: supplied heat "DQZU1", the removed energy "DQAB1", the idle pressure P1 and the mean power for the calculated revolution "POWERN". Here all graphs of runs No (5) and (6) with air as the working fluid are put together in this 1 diagram. The signals (5) and (6) in Fig.(1)(2) are in accordance with these run No's. 5 and 6 of the table. The values for Helium are shown in the table at RUN NO. 7 and 8.

The usable efficiency ETA_N in column 13 of the table is written in Fig.(1)(2) and Fig.(5)(6) with "h_n". These values can also be found for the Fig.(5)(6) at the RUN NO. (5) and (6) in the table. Please notice the usable efficiency h_n = 19.59 % in RUN NO. 5 is a bit higher than in RUN NO. 6 where h_n = 17.89 % inspite of a higher regenerator efficiency in RUN NO 6. Here h_reg= 40 % is half as large as h_reg= 80 % in RUN NO 6.

(The usable efficiency "h_n" and the usable power "POWERN" take into account the mechanical friction losses of the engine. Therefore "h_n" is less than h_therm.)

Conclusion:

Each Stirling engine is wanted to have a high ratio h _therm / h _C ; this is specially wanted for those engines which are working with low to medium temperatures. The regenerator of the engine|1| will work at a loss, when it should be designed for working with compound fluids. One possibility of equalizing the regenerators loss in double acting engines is to design it as a counterflow heatexchanger as described in |1|.

A further improvement for a better h _therm is got by a higher volume ratio V_max/ V_min. This additionally can be achieved by following the rules in |2| with the ratio of the stroke volumes in expansion and compression spaces made equal to their temperature ratio: VE/VC = TE/TC. See the section "4 of sample calculations". (Raising the volume ratio V_max/V_minat least is limited by the effectiveness of the heatexchangers!)

A third but simple improvement will be to choose an one atomic inert gas rather than a two (or three)- atomic gas as the working fluid. (If in this sense Helium will be better than Hydrogen -with its twice better thermal conductivity than Helium- is to be tested; but Argon should be better in this sense than air.)

Equation (9) is derived from a "disidealized" Stirling process with isochoric heat transmission at a constant volume. Concerning h _therm, V_max/V_min and k the calculations of the crankgear driven engine |1| are similar to the results of equation (9), because of the prerequisites coming from this engine are more qualified for a comparisation of this process to the "disidealised" isothermal process than the more adiabatic processes of common Stirling engines:

Heater and cooler are not external, they are within the expansion space resp. within the compression space of this engine.
The heattransfer to the working fluid can be made nearly isothermal by irrigating the working fluid with a hot resp. with a cold working liquid |1|.
The deadspaces can be reduced essentially to the regenerator space only.

A new point of view for the calculation of the heatprocess of this engine shall be discussed here:

An isothermal change of state is defined by :

An adiabate change of state is defined by :

A polytrope change of state is defined by :

with the exponent n < k

The change of state in this process |1| may be defined by

The exponent e is less than the exponent for an adiabate or polytrope change of state. This, because the heattransmission here is extremly forced during the expansion and compression phase; insofar the changes of state in this process are close to isothermal.
So the exponent e will be 1 < e << n. It will be e nearly =1.
As a derivation of the above formula temperatures can be calculated with:

(11)

An other assumption which is made when calculating this process is to have the same instantaneous pressure in all spaces |4|. With these assumptions the above mentioned similar results for the "disidealised" and the nearly isothermal process of the engine |1| are obtained.

Literature:

|1| P.Fette: A Stirling Engine Able to Work with Compound Fluids Using Heat Energy of Low to Medium Temperatures. Presented at the 6'thISEC in Eindhoven 1993; ISEC-93-003. See also the report at this homepage.

|2| T. Finkelstein: Theoretical Considerations: The Concept of the"Cumulative Mass" and "Thermodynamic Mid-Plane" and the Resulting Volume Ratios, Europeen Stirling Forum Osnabrück 1994

|3| G. Walker: Stirling Engines, Clarendon Press Oxford 1980

|4| R.J.Meijers: Der Philips Stirlingmotor, Motoren Technische Zeitschrift MTZ 29(1968)/7, DK 621.41/43 (492) Philips

|5| P.Fette: Neuer Wärmekraftprozess für die Nutzung von Niedertemperatur Wärme, Brennstoff Wärme Kraft Bd.39 1987 Nr.11, VDI Verlag Düsseldorf

|6| P.Fette: STMOT2 ein FORTRAN 77 Computerprogramm für Studien des Stirlingprozesses

|7| P. Fette Grundlagen für die Berechnung des Wärmekraftprozesses und der Dynamik von a -Typ Stirlingmotoren mit dem Programm STMOT2

Fig.1 The Effect of V_max / V _min and the Regenerator efficiency h_reg on the Thermal Efficiency h_therm

The upper black line with 3 blue rectangles stands for h _reg = 100 % for all V_max/ V_min and for all k's