Ideal Isothermal AnalysisTheory

### 1. Introduction and Definitions

Gustav Schmidt (Sep 16,1826 - Jan 27, 1883) published the isothermal analysis of the Stirling cycle in 1871 and provided a closed form solution for the case of sinusoidally varying volumes of the expansion and the compression space.

We use here very much the same notation as used by Urieli.

The following assumptions are basis of the subsequent analysis :

• An ideal gas is used as working fluid and the ideal gas law connecting pressure p, volume V, mass m, and temperature T to each other via the specific gas constant R :

(1)     p V = m R T

• The pressure, p , is the same everywhere inside the engine and varies only with time.

• The heat transfer conditions are sufficient to keep the gas inside the compression space, volume Vc, and inside the kooler space, volume Vk, at constant temperature Tc at all times. Therefore, the masses of gas inside the compression and the kooler space, respectively, are given by :

(2)     mc = p Vc / ( R Tc )

(3)     mk = p Vk / ( R Tc )

• The heat transfer conditions are sufficient to keep the gas inside the expansion space, volume Ve, and heater space, volume Vh, at constant temperature Th at all times. Therefore, the masses of gas inside the expansion and the heater space, respectively, are given by :

(4)     me = p Ve / ( R Th )

(5)     mh = p Vh / ( R Th )

• The heat transfer conditions are sufficient to keep the temperature distribution inside the regenerator, volume Vr , linear, varying from Tc where the regenerator is connected to the kooler to Th at the heater side. Therefore the mass of gas inside the regenerator space is given by :

(6)     mr = p Vr ln(Th/Tc) / ( R ( Th - Tc ) )

• The volume of the compression space varies in sinusoidal fashion :

(7)     Vc = Vclc + 0.5 Vswc( 1 + cos(Θ) )

Θ = crank angle, proportional to time
Vclc = clearance volume = minimum of Vc
Vswc = swept volume
Vc has a maximum of Vclc + Vswc at Θ = 0° and a minimum of Vclc at Θ = 180°
• The volume of the expansion space varies in sinusoidal fashion :

(8)     Ve = Vcle + 0.5 Vswe( 1 + cos(Θ + δ) )

Θ = crank angle, proportional to time
Vcle = clearance volume = minimum of Ve
Vswe = swept volume , max{Ve} = Vcle + Vswe
δ = volume phase lag angle, usually around 90°.
At Θ = 0° the compression space is its maximum while the expansion space is already "half-way" down to its minimum which it reaches when Θ = 180 - δ ≈ 90°. This is followed by the compression space reaching its minimum at Θ = 180°
• The total mass of gas inside the engine, mtot, does not vary in time.

(9)     mtot = mc + mk + mr + mh + me

### 2. Pressure

#### 2.1 Pressure as Function of Crank Angle

An equation for pressure p as function of angle Θ can now be found by substituting Eq.(2) to (8) into Eq.(9) and solving for p. It is of advantage to operate as much as possible with dimensionless variables which can be achieved by simple algebraic rearrangement. appropriate reference variables. As a results units do not matter anymore (but temperature must always measured in absolute degrees) because they naturally cancel out. We define 3 dimensionless variables e, c, and d :

(10)   e  =  0.5

(11)   f  =  0.5

 Vswc Vswe
 Th Tc

(12)   d  =

 Vh + Vcle Vswe
+
 Th Tc
 Vk + Vclc Vswe
+
 Th Th - Tc
 Vr Vswe
ln(Th/Tc)

With that :

(13)   p  =

 mtot R Th Vswe
 1 (e+f+d) + c cos(Θ) + e cos(Θ+δ)

Eq.(13) allows us to determine the pressure p for any value of the crank angle Θ provided we know all volumes, temperatures, the phase angle, and the total mass of working gas. The latter can be determined by knowing the average pressure inside the engine. To this end we need to integrate Eq.(13). It is of advantage to bring Eq.(13) into a different form using the geometric identity :

cos(Θ+δ)   =   cos(Θ) cos(δ) - sin(Θ) sin(δ)

(14a)   p  =

 mtot R Th Vswe
 1 B + C cos(Θ + β)

where :

(14b)    B = e + f + d

(14c)    C = √ e² + 2 e f cos(δ) + f²

and β can be obtained from :

(14d) sin(β) = f sin(δ) / C    cos(β) = (c+f cos(δ)) / C

#### 2.2 Average Pressure

The average, over angle Θ from zero to 2π, of the pressure can be calculated from

pave   =

 1 2π
0 p dΘ  =
 mtot R Th Vswe
 1 2π
0
 dΘ B + Ccos(Θ + β)

The solution to the integral is explained in Integrals in support of Schmidt Analysis (look for integral I1) and we arrive at :

(15)   pave  =

 mtot R Th Vswe
 1 √B² -C²

It is interesting to note that the average pressure is exactly equal to the geometric mean of the minimum and maximum pressure.

#### 2.3 Minimum and maximum Pressure

According to Eq.(14a) the minimum pressure is reached when cos(Θ+β) = +1 and the maximum is reached when this term is equal to -1 :

(16)   pmin  =

 mtot R Th Vswe
 1 B + C

(17) (17)  pmax  =

 mtot R Th Vswe
 1 B - C

It takes a just little algebra to show that

B > C

under all circumstances and of course physics requires the same ( that is pmax > 0).

#### 3.1 Work done by Expansion Space

By definition :

We = p dVe

in which the integration has to be performed over a complete cycle. Differentiating Eq.(8) with respect to Θ :

dVe = - 0.5 Vswe sin(Θ+δ) dΘ

and using Eq.(14a) for the pressure we obtain :

We = - mtot R Th e 0

 sin(θ+δ) dΘ B + Ccos(Θ + β)

The solution to the integral is explained in Integrals in support of Schmidt Analysis (look for integral I3) with which we get :

We = - mtot R Th

 2 π e C
(
 B √B² - C²
- 1) sin(β-δ)

Using the definition for 'e', sin(β) and cos(β) from Eq.(14d) and some trigonometry we finally get :

(18a)    We = mtot R Th

 π f C2
(
 B √B² - C²
- 1) sin(δ)

Or alternatively,

(18b)    We = pave Vswe

 π f C2
( B - √B² - C² ) sin(δ)

#### 3.2 Work done by Compression Space

By definition :

Wc = p dVc

in which the integration has to be performed over a complete cycle. Differentiating Eq.(7) with respect to Θ :

dVc = - e Vswc sin( Θ ) dΘ

and using Eq.(14a) for the pressure we obtain :

Wc = - mtot R Th e (Vswc/Vswe) 0

 sin(θ) dΘ B + Ccos(Θ + β)

The solution to the integral is explained in Integrals in support of Schmidt Analysis (look for integral I2) with which we get after some algebra and trigonometry :

(19a)    Wc = - mtot R Tc

 π f C2
(
 B √B²-C²
- 1) sin(δ)

Or alternatively :

(19b)    Wc = - pave Vswe

 Tc Th
 π f C2
( B - √B² - C² ) sin(δ)

The minus sign indicates that work has to be done on this space to perform the required piston motion.

#### 3.3 Net Work done by Engine

Wnet = We + Wc

With the results from Eq.(17) and (18) we get :

(20a)    Wnet = mtot R ( Th - Tc )

 π f C2
(
 B √B² - C²
- 1 ) sin(δ)

Or alternatively :

(20b)    Wnet = pave Vswe ( 1 -

 Tc Th
)
 π f C2
( B - √B² - C² ) sin(δ)

An important insight from Eq.(20b) is that the net work is independent of the type of working gas being used. The gas constant does not appear in any of the appearing coefficients. The same holds true for the work of the compression space, Wc Eq.(19b), and the work of the expansion space, We Eq.(18b).

### 4. Thermodynamic Efficiency

The thermodynamic efficiency for heat engines is generally defined as the ratio of benefit ( Wnet ) to costs ( Qh + Qe = heat supplied per cycle ).

η = Wnet/(Qh + Qe)

To determine Qh and Qe we look at the law of conservation of energy for each of the subspaces involved and obtain :

Qc = Wc
Qk = Wk = 0
Qr = Wr = 0
Qh = Wh = 0
Qe = We

It is the assumption of isothermal behaviour of all subspaces which simplifies the law of conservation of mass so drastically because the gas masses flowing in and out of each subspace have at all times a temperature identical to that of the subspace itself and therefore the net energy transport per cycle of these mass flows is zero for each subspace. In addition, the works for the kooler, regenerator, and heater are zero because the volumes of these spaces do not change in time ( dV = 0 in ∫ p dV ).

With that :

(21)   η =

 Wnet We
= 1 -
 Tc Th

using Eq.(18) and (19) for We and Wnet, respectively. (Those familiar with the Carnot efficiency and its underlying physical principle shouldn't be surprised at all).

### 5. Optimization

#### 5.1 Some general Remarks

Within the confines of the Schmidt analysis we have available for modification the temperatures Tc and Th, the volumes of the individual subspaces, the total mass mtot, the type of gas, and the volume phase lag δ.

There is no need to discuss the temperatures Tc and Th much, their ratio directly affects the efficiency in an obvious way and their difference the net work per cycle, Wnet, in an equally obvious way. Hence, in the following we assume Tc and Th to be given.

Because for given temperatures Tc and Th the efficiency is identical for all Schmidt engines we can only optimize Wnet as given in Eq.(20b). The influence of pave and Vswe is quite obvious.

#### 5.2 Influence of Vclc, Vk, Vr, Vh, and Vcle

It is generally accepted in the Stirling community that the volumes of these sub spaces are to be kept at a minimum as much as heat transfer and manufacturing considerations allow. It is interesting to note that the Schmidt analysis points in the same direction.

The only influence the volumes Vclc, Vk, Vr, Vh, and Vcle have on Eq.(20b) is through the term B which decreases as any of these volumes decreases. Any decrease in B leads to an increase in Wnet according to Eq.(20b) !!! ( Take the derivative of Wnet with respect to B, it'll be negative.)

From Eq.(12) and (14b) we also see that B decreases more dramatically with a decrease in (Vclc + Vk) than with a decrease in (Vh+Vcle), namely by a factor of (Th/Tc). The affect of the regenerator volume Vr is similary enhanced by the factor Th/(Th -Tc) ln(Th/Tc), the value of which lies somewhere between 1 and Th/Tc.

#### 5.3 Optimal volume phase lag, δ

The optimum for the volume phase lag angle δ is achieved for example by keeping all parameters in Eq.(20b) constant but varying the volume phase angle δ systematically until the largest value for the network Wnet is achieved. δ influences Wnet mostly through the sin(δ)-term at the right end of Eq.(20b) but also through the term C², see Eq.(14c). This suggest that the optimum for the volume phase angle is close to 90°, a value commonly suggested.

Mathematically speaking, we have to take the derivative of Eq.(20b) with respect to δ and setting it to zero. After some algebra an equation containing the optimum volume phase lag can be obtained :

(22)   cos(δ) = √1 + μ² - μ

(23)   μ =

 1 f
(B + √B² - C² )B² - C²

For the definitions of B and C² see Eq.(14b) and (14c). Eq.(22) can not be solved explicitly for δ because δ appears to the left of the equal sign and on the right through the term C² and there doesn't seem to be a way to solve for δ or its cosine or sine explicitly. But it renders itself nicely to an iterative solution.
This means that you can take an estimate for δ ( for example 90° ) to determine the term C², then determine μ from Eq.(23) and finally find an improved value for δ via Eq.(22) which you can use as a starting value for another go-around. Usually, after 2 iterations you obtain changes for angle δ only of a thousands of a degree or less. For cases I have investigated the optimum value for δ is always close to 90° which is an indication that the behavior of Wnet as given by Eq.(20b) is dominated by the sine-term. That in turns tells us that the volume phase lag is not an extremely critical parameter, for example sin(60°)=sin(120°)=0.866 which does not compare badly with sin(90°) = 1.

#### 5.4 Optimization for Compression and Expansion Space

Although it seem to be impossible to provide an exxact proof experience with the computer based determination of Wnet indicates that the net work per cycle will increase as either the swept volume of the expansion space, Vswe, or that of the compression space, Vswc is increased, albeit with deminishig return. On the other hand, increasing these volumes increases the size and construction cost of an engine.

Based on the previous findings that Wnet continuously increases as the compression and/or expansion space are increased, we must employ an additional restriction on these volumes.

For example, we could demand that the sum of these two volumes is a constant and then ask for how best to divide up the available space.

Because we could not find an analytical solution which optimizes the division of a given volume into compression and expansion space a computer program is provided.

### 6. Optimization Program

#### 6.1 What it needs and does

A sample case demonstrates the capabilities of this program.

As input to the program the user provides his/her engine specifications, for example :

 Average pressure pave 200.0 kPa Working gas Air Heater temperature Th 923.0 °K Kooler temperature Tc 300.0 °K Volume phase lag δ 95.569 ° Clearance volume, compressions space Vclc 8.0 cm³ Swept volume, compression space Vswc 61.045 cm³ Volume of kooler space Vk 31.21 cm³ Volume of regenerator Vr 34.89 cm³ Volume of heater space Vh 28.51 cm³ Clearance volume of expansion space Vcle 10.0 cm³ Swept volume of expansion space Vswe 61.045 cm³
Some of you might recognize these data as ross90.dat

The program in turn provides three sets of responses :

One pertaining exactly to the user's configuration, a second which keeps all of user's input data except optimizes the volume phase lag, and thirdly a set of results which optimizes volume phase lag and Vswc/Vswe keeping the value of Vswc+Vswe at its original value.
The program also produces a contour plot of the network produced as function of volume phase lag and swept-volume ratio. Typically, such graphs show that the optimum is fairly flat, meaning that even larger changes deviations from the optimal values for δ or Vswc/Vswe will not reduce the network per cycle dramatically.

### 7. Heat Transfer Calculations

#### 7.1 Introduction

The derivation of equation and associated discussions in this Section are not directly linked to optimization of the geometric layout of a Stirling engine but are related to the discussion of what working gases are preferrable.

It is commonly argued that the choice of working gas of a Stirling engine has an influence on the amount of net work produced due to different properties of gases. In particular, it is thought that flow friction losses and heat transfer rates can be positively influenced. Helium and Hydrogen are commonly mentioned as preferable over air.

In addition to these arguments, we propose that the amount of heat which has to be transferred in the different subspaces of a Stirling engine ( compression space , kooler , regenerator , heater , expansion space ) is influenced quite dramatically by the value of the ratio of specific heats, κ=cp/cv, of the gas. Roughly speaking the value of κ is 1.667 for monatomic, 1.4 for diatomic and less than 1.3 for molecules with a higher number of atoms ( A short note on ideal gases). This influence was demonstrated during a more in depth analysis of The ideal Stirling Cycle and Heat Load on the Regenerator which showed that the heat load per cycle is proportional to the value of 1/(1-κ). Hence, the amount of heat to be removed from/ added to the gas by the regenerator matrix increases by a factor of over 2 (two) when switching from Helium to CO2 or methane. The basic reason for this is that the energy balance on each individual subspace is affected by the specific heat at constant volume, cv, and - in a different way - by the specific heat at constant pressure, cp. The program, Schmidt , now calculates various heat transfer rates based on the equations derived below.

In the equations below, the letter "d" in front of a quantity as in dVe denotes a (infinitesimal) small change of the quantity itself, in this example Ve. Such equations can be obtained by differentiation

#### 7.2 Basic Equations used on all subspaces

Change of volume of compression space versus change of crank angle Θ. Obtained by differentiating Eq.(7) :

dVc = -0.5*Vswcsin(Θ) dΘ

Change of volume of expansion space versus change of crank angle Θ. Obtained by differentiating Eq.(8) :

dVe = -0.5*Vswesin(Θ+δ) dΘ

Change in pressure p as function of changes in volume of compression and expansion space. Obtained by differentiating Eq.(13) :

dp = - p²/(mtot R Th) (dVe + (Th/Tc) dVc)

Part of the assumptions underlying the Schmidt analysis is to assume the working medium to be an ideal gas with constant specific heat capacities. As a result the specific internal energy, u , of an ideal gas at temperature T is :

u = cv ( T - T0 ) [kJ/kg]

And then by definition (h=u+p*v) the specific enthalpy becomes :

h = cp T - cv T0 {kJ/kg]

The temperature T0 is inserted to provide the best fit between the real behavior of u and the linear relationship. For monatomic gases like He, Ne, Ar etc. T0=0 over a temperature range of a few thousand degrees. We will later see that T0 itself will drop out of the important equations to follow.

Also know that for ideal gases the following relationship holds exactly :

cp - cv = R

#### 7.3 Heat transfer inside compression space

Because the temperature is assumed to stay constant, the specific internal energy for this space is :

u = cv ( Tc - T0)

and the specific enthalpy of the gas leaving is :

h = cp Tc - cv T0

Because the kooler - exchanging mass with the compression space - is also at temperature Tc the equation for the enthalpy is also correct when mass is flowing into the compression space.

Let dmc be a small amount of gas entering the compression space while the crank rotates by a small angle dΘ and let dQc be a small amount of heat flowing into the gas. Conservation of energy then becomes :

dmc cv ( Tc - T0) = dQc - p dVc + dmc ( cp Tc - cv T0)

The temperature T0 drops out and rearranging for dQc which is the quantity we wish to calculate :

dQc = p dVc - ( cp - cv ) Tc dmc = p dVc - R Tc dmc

The small volume change dVc is known and dmc can be obtained by differentiating the ideal gas law in which for the compression space the pressure p, the volume Vc, and the mas mc change with time :

p Vc = mc R Tc

p dVc + Vc dp = dmc R Tc

Using this to eliminate dmc from the equation for dQc we arrive at :

dQc = - Vc dp

In the computer program "Schmidt" we determine dQc incrementing dΘ be small amounts and summing up ( basically integrating ) over a complete cycle. Because during a cycle the pressure increases (dp>0) and decreases (dp<0) heat is transferred in and out of the gas during a complete cycle with more heat removed than gained. The program keeps track separately of the sum of all the positve dQc and the negatives.

#### 7.4 Heat transfer inside expansion space

The analysis is exactly identical to that of the compression space in section 7.3.

dQe = - Ve dp

Integrating over a complete cycle produces a positive Qe representing the net heat flowing into the gas in the expansion space.

#### 7.5 Heat transfer inside kooler space

Because the temperature is assumed to stay constant, the specific internal energy for this space is :

u = cv ( Tc - T0)

and the specific enthalpy of the gas leaving is :

h = cp Tc - cv T0

Because the kooler receives gas from the compression space and the regenerator with a temperature Tc , the equation for the enthalpy is also correct when mass is flowing into the kooler space regardless as to whether it is coming from the compression space or the regenerator. Because the volume of the kooler space is constant the energy balance becomes :

dmk cv ( Tc - T0) = dQk + dmk ( cp Tc - cv T0)

Re-arranging :

dQk = - ( cp - cv ) Tc dmk = - R Tc dmk

The small change of mass inside the kooler space, dmk, can be obtained by differentiating the ideal gas law in which for the kooler space only the pressure p and the mass mk change with time :

p Vk = mk R Tc

Vk dp = dmk R Tc

Using this to eliminate dmk from the equation for dQk we arrive at :

dQk = - Vk dp

Because during a cycle the pressure in the engine arrives back at its starting value Qk = ∫ dQk = 0 which is the well-known artifact of the Schmidt analysis. The heat transferred inside the kooler space into the gas during that part of the cycle while the pressure is decreasing :

Qk = Vk ( pmax - pmin )

is exactly transferred out of the gas while the pressure is increasing.

#### 7.6 Heat transfer inside heater space

The analysis for the heater space is exactly like that for the kooler space with the results :

dQh = - Vh dp

and the same conclusion are to be drawn.

#### 7.7 Heat transfer inside the regenerator

Things are a bit more complicated for the regenerator because on its inside the temperature varies (linear variation is the classical assumption) from Tc on the kooler side to Th on the heater side. In addition the enthalphy of the gas exchanged between kooler and regenerator has an enthalphy corresponding to Tc while for the gas exchange between heater and regenerator Th determines the enthalpy.

We irst look at the internal energy of the regenerator. Let "A" be its free cross-section and "L" be its length and let "dx" be a small slice thereof. In the following all integrals go from x=0 ( kooler side ) to x=L ( heater side ) and the temperature varies linearly :

(7.7.1)    T = Tc + x/L ( Th - Tc )

The mass inside the regenerator with ρ=density becomes now :

(7.7.2)    mr = ∫ ρ A dx

and with ρ = p / ( R T ) (from ideal gas law) we get

(7.7.3)    mr = ∫ p / ( R T ) A dx = p A / R &int dx/T = p Vr / ( R Tr )

Here we used Vr = A L = volume of regenerator and Tr = effective temperature of regenerator defined by :

(7.7.4)    Tr = ( Th - Tc ) / ln(Th/Tc)

The equation to keep for later use is that for small changes of mr :

(7.7.5)    dmr = Vr / ( R Tr ) dp

In similar fashion we can derive an equation for the internal energy of the gas inside the regenerator.

(7.7.6)    U = ∫ cv ( T - T0 ) ρ A dx = cv ∫ T ρ A dx - cv T0 ∫ ρ A dx = cv ∫ p/R A dx - cv T0 mr

Finally :

(7.7.7)    U = Vr (cv/R) p - cv T0 mr

(7.7.8)    dU = Vr (cv/R) dp - cv T0 dmr

We are now in the position to set up the energy balance for the regenerator but still need "dm1" and "dm2" the masses leaving the regenerator to the kooler and the heater space, respectively.

(7.7.9)    dU = Vr (cv/R) dp - cv T0 dmr = dQr - dm1 ( cp Tc - cv T0 ) - dm2 ( cp Th - cv T0 )

Because of conservation of mass : dmr = - (dm1+dm2) and all terms containing T0 drop out again to give :

(7.7.10)    dQr = Vr (cv/R) dp + cp Tc dm1 + cp Th dm2

We have already an expression for "dp" but don't know dm1 and dm2 yet. But by virtue of conservation of mass and previous expressions for dmk and dmc :

(7.7.11)    dm1 = dmk + dmc = Vk/(R Tc) dp + Vc/(R Tc) dp + p/(R Tc) dVc

In the same fashion :

(7.7.12)    dm2 = dmh + dme = Vh/(R Th) dp + Ve/(R Th) dp + p/(R Th) dVe

Substituting back :

(7.7.13)    dQr = [Vr cv/R + (Vc+Vk+Vh+Ve) cp/R) ] dp + p cp/R ( dVe + dVc )

With &kappa = cp/cv we get :

(7.7.14)    dQr = { [ Vr + κ(Vc+Vk+Vh+Ve) ] dp + κ p ( dVe + dVc ) } / { κ-1 }

Again, program "Schmidt" sums up separately all positive dQr ( heat transferred from the matrix into the gas ) and all negatives and verifies that the net heat transfer during a complete cycle is zero as indicated by above equation.

#### 7.8 Summary

For each value of the crank angle Θ from 0 to 2π in steps of some small increment dΘ the above equations are evaluated. First Vc and Ve and their increments dVc and dVe and from that the change in pressure dp. With that all heat flow increments dQc etc. can be evaluated and appropriately summed to the contributions from the previous Θ's.

The program and above equations have been triple checked.

We performed calculations on what is referred to as the ross90 data which are presented in Section 6.1 of this webpage. The Input-webpage for the program "Schmidt" offers an easy option to input these data into the program with the additional option to choose the working gas ( air , helium , hydrogen , carbon dioxide , methane ).

As predicted theoretically, a change in working gas has no affect on the efficiency, net work and heat transfer in the compression, kooler, heater, and expansion space. The mass of working gas is proportional to the inverse of the specific gas constant. The heat transferred inside the regenerator into the gas while it is being moved from the heater to the kooler (Qr1 in the table below) is strongly affected by κ = ratio of specific heats. ( the same amount of heat is transferred out of the gas when the gas streams in the reverse direction). The table below shows the results for the ross90 geometry when volume phase lag and swept-volume ratio are not optimized.

 Rspec [J/(kg k)] k=cp/cv mgas [g] Qr1 [J] Air 287.0 1.400 0.2476 31.7887 Hydrogen 4124.0 1.405 0.0172 31.4917 Helium 2076.9 1.667 0.0342 22.1790 Carbon Dioxide 188.9 1.289 0.3762 41.0438 Methane 518.2 1.299 0.3762 39.9276

When comparing - for example - the two gases Helium and Carbon Dioxide the analysis of the ideal Stirling cycle predicted that the heat transfer rates would change by a factor of (1.667-1)/(1.289-1) = 2.308 while the Schmidt analysis produces a ratio of 41.0438/22.1790 = 1.851 for the ross90 geometry. Although not quite as bad as predicted, the increase in heat transfer load on the regenerator with decreasing value of ratio of specific heats, κ, is quite significant. Note , that there is no significant difference between air and hydrogen.

For the above tabulated cases the net work per cycle was a constant 3.6136 [J] which is significantly lower than the heat load on the regenerator. This emphasizes once more the importance of the regenerator.

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