Int. J. of Thermodynamics ISSN 1301-9724 Vol. 10 (No. 4), pp. 165-176, December 2007
Thermodynamic Study of a Low Temperature Difference Stirling Engine at Steady State Operation
Nadia MARTAJ *, Lavinia GROSU 1*, Pierre ROCHELLE *,**
* Laboratoire d’Energétique et d’Economie d’Energie 50, rue de sèvres, 92 410 Ville d’Avray ** Laboratoire de Mécanique Physique 2, av. de la Gare de Ceinture 78310 Saint Cyr l’Ecole [email protected]
Abstract In the current energy economy context, the use of renewable energies and the valuation of lost energies are the subject of many studies. From this point of view, the Stirling engine draws attention of the researchers for its many advantages. This paper presents a thermodynamic analysis of a low temperature Stirling engine at steady state operation; energy, entropy and exergy balances being presented at each main element of the engine. A zero dimensional numerical model describing the variables evolution (pressure, volumes, masses, exchanged energies, irreversibilities...) as function of the crankshaft angle is also presented. The calculated irreversibilities are due to imperfect regeneration and temperature differences between gas and wall in the hot and cold exchangers. A favourable comparison was made with experimental results obtained on an small size engine. Keywords: Stirling engine, numerical model, thermodynamic analysis, imperfect regeneration.
1. Introduction Cold Sink In the past few yars the understanding of the Stirling engine has shown considerable growth. Vc TDC Many new applications were developed, one of y these applications being the low temperature lr y0+l r difference Stirling engine. This new type of Vh Stirling engine is able to operate with very low BDC temperature difference between the source and the sink of the engine. Such an engine can run simply TDC Heat Source x placed on a hot cup of coffee or on the hand. More x0 practical applications (pumps) have also appeared. BDC
This study presents an energetic, entropic and dead volume
exergetic analysis of a gamma type engine which allowed the development of an equation system describing the processes occurring at every Figure.1. Low temperature difference Stirling element of the engine. The numerical model engine allows the evaluation of the processes from the 2. Model of low temperature difference engine energetic, entropic and exergetic point of view as function of the crankshaft angle (kinematics- 2.1 Volume and pressure expressions thermodynamics coupling). For this type of engine, compression and The hot air engine configuration which one expansion spaces are defined by the positions (x proposes to study is presented on Figure 1 . The and y) of the pistons compared to the two pistons (working piston and displacer piston) corresponding top dead centers. Compression and are connected to the same crankshaft with an expansion volumes can be expressed according to appropriate out-of-phase angle ; the displacer the instantaneous pistons positions by using the piston is used jointly as a regenerator. engine geometry.
1 Author to whom correspondence should be a d dressed . Int. J. of Thermodynamics, Vol. 10 (No. 4) 165 dV dV Vh= V mh +−( y0 yA ) d + xA p (1) p (c+ h ) T Tp2 dVdV V= V + yA (2) dp =−ch =− ch + (8) c mc d V V V m T T c+ h + r c h Pistons positions in cylinders Tc T h T r
From a reference state (working piston at 2.2 Energy analysis: BDC), the instantaneous position of the working piston in the small cylinder is written according to The differential form of the energy balance the angle of crankshaft like: for an open system is written generally by :
x δQ+ δ W +∑ hdmi i = cdmT v ( ) (9) x = 0 1( + cosθ ) (3) 2 2.2.1 Regenerator cell: The position of the displacer piston, which The regenerator/displacer reciprocating separates compression and expansion spaces in the movement forces the air of the cooling cell large cylinder, is given by: towards the heating cell and conversely ; it is also y useful to store and release the heat exchanged with y =0 (1 + cos(θ − ϕ )) (4) the regenerator wall during this transfer. It is 2 supposed that the regeneration is imperfect (Figure 2 ). The fluid temperature at the exit of the with ϕ generally approximated by π . 2 regenerator (orifice at abscissa 0) towards the cold
The dead space of the regenerator (annular in cell Tc' is higher than Tc and the fluid this case) is defined by: temperature at the exit of the regenerator (orifice
2 2 at abscissa lr ) towards the hot cell Th' is lower Vr=π.( R cyl − Rl d ). r (5) than Th . It remains constant during the operation of the engine contrary to hot an cold volumes which T change with positions x and y of the pistons. Th In the following, the engine is considered as made up of three cells: compression, regeneration Tr and heating cell. Tc' The instantaneous pressure, considered as Th' uniform in the engine and its variation can be Tr expressed by using the mass balance. T The total gas mass m locked up in the engine, c which is the sum of the gas masses of the three cells, remains constant during the operation of the 0 z engine, therefore, by assuming that the gas is lr perfect: Figure 2. Regenerator T-z diagram Let define the regenerator efficiency by : p VcVr V h mc++== m r m h m ++ (6) rTc T r T h Th −Tc' Th' −Tc Tr ηr = = =1− (10) dmc+dmr +dmh=0 Th −Tc Th −Tc Th −Tc
Tr being the mean temperature in the regenerator (see Equation (12)), then : where Tr represents the temperature pinch in the regenerator, assumed identical at the two p= m (7) extreme orifices (between C and R and between H Vc Vh Vr + + and R): rTc rTh rTr
While differentiating (7) and using (6) one Tr =Tc' −Tc =Th −Th' (11) obtains dp in the following form, assuming that the In this constant-volume cell of heat storage temperatures are constant: and release, the work exchanged is null and the average temperature is supposed to be constant
Tr .
166 Int. J. of Thermodynamics, Vol. 10 (No. 4) T+ T pV = mrT (18) T = c h (12) c c c r 2 and its differential form : In the regenerator cell, the equation (9) can dp be written: + dVc =dmc + dTc (19) p Vc mc Tc
cv δQr= cTdm prc c + cTdm prh h + Vdp r (13) The cooling cell is supposed to be isothermal. r Thus the differential form of the state equation is reduced to: the mass dm i is positive when it enters the volume i. dp + dVc =dmc (20) Two cases arise for each opening: p Vc mc
- flow from (C) towards (R): dmc <0 , then according to (18):
Trc =Tc , Trh= T h' . p Vc mc = (21) r Tc - flow from (H) towards (R): dmh <0 , then and : Trh =Th , Trc= T c' . To determine the mass in this cell and its Vcdp p dV c 1 dmc =+ =() Vc dp + p dV c differential, the equation of state is written. Tcr T c r rT c (22) pV r= m r rT r In this way the equation (16) becomes p V r mr = (14) c c p r T δQ = −c T dm + v V dp + pdV r c p rc c r c r c and the mass transfer balance gives: c p 1 T c p T δQ =V − rc dp + p 1− rc dV V c c c r r γ Tc r Tc dmr=− dm c − dm h = dp (15) rT r (23)
2.2.2 Cooling cell 2.2.3 Heating cell For the cooling cell ( Figure 3 ), which have a If the equation (9) is applied to this cell, the single communication orifice, one can write the result is equation (9) in the following form: cv cp δ Qhprhh=− cTdm + Vdp h + pdV h (24) δQc +δWc +cp Trc dmc =cv.d(mT )c (16) r r as stated before, if dmc >0 ( elementary mass If dm h > 0 then Trh= T h ' else Trh= T h . entering the compression volume) then Trc =Tc' , else Trc =Tc .
δQh δWc =−p dVc (17)
δQc Th (H)
Tc (C) δWc dm dm he hs δWh
dmce dmcs Figure 4. Heating cell Using the same reasoning as for the previous Figure 3. Cooling cell cells, the air mass and its differential can be One determines the air mass and its obtained: differential in the compression volume by using the perfect gas state equation: p V h mh = (25) r T h Int. J. of Thermodynamics, Vol. 10 (No. 4) 167 Vhdp p dV h 1 ηth dmh =+ =() Vh dp + p dV h ηII = (31) Thr T h r rT h ηcarnot (26) The temperatures of the walls considered as In this way the equation (24) becomes sources are given by the following relations :
c 1 T c T Qc= UA(T cc wc − T)t c (32) δQ = V p − rh dp + p p 1− rh dV h h h QQUAT+= − T t (33) r γ Th r Th h r hh( wh h ) (26’) where t is the engine revolution period and Ui 2.2.4 Engine balance is the heat transfer coefficient between the gas and During a crankshaft rotation the quantities of the cell wall i. heat exchanged in the cooling and heating cells 2.3 Entropy analysis are: Q = δQ and Q = δQ obtained by h ∫ h c ∫ c In the same way as previously, the entropic integration of the equations (23) and (26’). balance of an open system is written in the form: With the considered assumptions δQ dS= +se.dme −ss.dms +δπ (34) (Tc'− T c = T h − T h ' ), an additional correcting T quantity of heat Qr , to bring by the hot source, is applied to the three cells. essential. 2.3.1 Regenerator cell This deficit is due to the inequality of the masses transferred at the interfaces cooling cell– The entropy balance of this cell is: regenerator cell and heating cell–regenerator cell (different densities). δ Qr dSr=− sdm rccrhh − sdm + δπ r (35) Tr δQ = c T dm + c T dm ∫ r ∫ p rc c ∫ p rh h src and srh are the specific entropies of the Notice : if ηr =100% then Tri = T i = cst fluid associated with the mass flow through the throughout the cycle and Qr =0 . entries of the cell: The overall engine balance is written : Trc p sscrc=0 + p .ln( ) − r .ln( ) (36) T p W + Qc + Qh + Qr = 0 0 0 W = −(Q + Q + Q ) (27) Trh p c h r sscrh=0 + p .ln( ) − r .ln( ) (37) T p This work is carried out by the working 0 0 piston during compression and expansion. It can s , T , p are reference parameters. be also calculated by the following expression: 0 0 0 Two possible cases arise, again: dW =dWc +dWh - first case, flow between (R) and (C) : if dm c > 0 dWc and dWh having been expressed previously. After integration: then src= s c' (and Trc= T c' ), else src= s c (and
Trc= T c ). W =Wc +Wh (28) - second case : flow between (H) and (R), if This work is also: dm h > 0 then srh= s h' (and Trh= T h' ), else s= s (and ). W =−∫ pdV (29) rh h Trh= T h with p : the internal pressure. 2.3.2 Cooling cell and, dV : the total air volume variation For the cooling cell case, one obtains: The thermal engine efficiency can be δ Qc expressed by: dSc= + sdm rcc. (38) Tc η = W (30) th As it was already noted, the specific Qh +Qr entropies depend on the direction of the flow: and the second low efficiency by : 168 Int. J. of Thermodynamics, Vol. 10 (No. 4) If dm c > 0 then src= s c ' (and Trc= T c' ), else 2.3.4 Engine entropy balance src= s c (and Trc= T c ). The engine entropy balance is written excluding external entropy generation below, The entropy flow follows the heat transfer using (35), (38) and (41) : direction and increases with the temperature difference. δQ δQ δQ dS +dS +dS = c + h + r +δπ =0 (44) external c h r r The entropy generation δπc , due to Tc Th Tr the temperature pinch between the cold source
(wall) and the gas ( Tsc and Tc ), is the difference where δπr is the regenerator entropy generation. between the entropy received by the cold source δQc δQh δQr and the entropy transferred by the cooling cell δπr =−( + + ) (45) (Figure 5 ). Tc Th Tr
δπc =dSsc−dSc (39) 2.4 Exergy analysis 1 1 In order to evaluate the effectiveness of each δπc =δQc − (40) Tsc Tc cell, one now proposes to write the exergetic balance for an open system:
Tc T dEx=−(10 )δδ Q ++ W pdV . +∑ exdmf . − T δπ δQc 0i i 0int dSc T (46) δπ Cooling cell c applied to each one of these cells.
f ex i : represents the specific exergy associated δQc dSsc with the mass flow dm i . Tsc Figure 5. Functional entropy balance diagram of f ex=−− h h Ts − s (47) the cooling cell i( i0) 0( i 0 ) 2.3.3 Heating cell In the following, for our particular application, the external exergetic expenditure is One writes the entropy balance in the heating from the cold sink (T c
Th f f dS else exrc = ex c' . δQh h δπh Second case : for a flow from (H) to (R)
Heating cell f f if dm h < 0 , then exrh= ex h ,
f f else exrh= ex h' . δQh dS sh Tsh Figure 6. Functional entropy diagram of the heating cell Int. J. of Thermodynamics, Vol. 10 (No. 4) 169 2.4.2 Cooling cell T T0 sc c δExδQC = 1( − )δQ represents the exergy of The exergy balance in the compression space Tsc can be expressed by : heat δQc at the temperature Tsc and, T0 f (49) dExc=− (1 )Qδ cc ++ δ W p.dV0 crcc + ex.dm Tc T0 Tc δEx = 1( − )δQc is the exergy of heat δQc δQC c f T exrc=( h rc −− h0) Ts 0( rc − s 0 ) f but at the temperature Tc . exrc= cT p( rc −− T0) Ts 0( rc − s 0 ) (50) Then where, the interface entropy is expressed by c 1 1 δExlost =T0 ( − )δQc =T0 δπc (52) Tsc Tc Trc p sscrc=0 + p .ln( ) − r .ln( ) T0 p 0 2.4.3 Heating cell: As in paragraph 2-3, the specific exergies In a similar way to the cell of compression depend on the interface temperatures and the treatment, one makes the exergy balance in the reference state considered : space of expansion and one obtains: f f f f If dmc >0 , then ex= ex , else ex= ex . rc c' rc c T0 f dExh=− (1 )Qδ hch ++ δ W pdV0 + ex.dmrh h (53) The functional diagram of the cooling cell Th (Figure 7 ) reveals the exergies exchanged by the exf = cT −− T Ts − s (54) gas with the cold sink. rh prh( 0) 0( rh 0 ) T p where ssc= +.ln(rh ) − r .ln( ) . Tsc rh0 p T0 p 0 Tsc δQc δEx f f f f δQc c If dm > 0 then ex= ex else ex= ex . δExlost h rh h ' rh h
Cooling cell The functional diagram of exergy of this cell is presented in Figure 8 .
δExTc δQc δQc Tsh (=T0 ) Tc T δEx sh =0 δQ δQsh h Figure 7. Functional exergetic diagram of the δExh cooling cell lost
Heating cell The direction of transfer of exergy is opposed to the direction of transfer of heat since the level T of the temperatures is lower than the ambient δQh δEx h δQ temperature of reference T . The exergy of a fluid h Th Tsc c Tc δExδQC =δExlost +δExδQC (51) h Th δExlost =δExδQh (55) where, 170 Int. J. of Thermodynamics, Vol. 10 (No. 4) Tsh T0 3. Results and analyses: where δExδQh = 1( − )δQh =0 represents the Tsh The model described previously is used to represent the operation of the engine during one exergy of heat δ Qh at the temperature Tsh =T0 rotation of crankshaft. and The initial data are listed in the following T T0 δEx h = 1( − )δQh represents the exergy of heat table: δQh Th TABLE I. INITIAL DATA δQh at the temperature Th . a) Dimensional and mechanical data of the actual Then engine h 1 1 working piston stroke: δExlost =T0 ( − )δQh =T0 δπh (56) 0,007 x0 (m) Th Tsh working piston diameter: 0,0095 2.4.4 Exergetic characteristics of the Dp (m) engine: working piston swept -7 3 4,9618.10 volume: V p (m ) The exergetic efficiency is defined generally displacer piston stroke : by the ratio: 0,007 Useful effect/Exergetic expenditure, it is y0 (m) written: displacer piston diameter: 0,077 D (m) W W d η = = (57) displacer piston swept -5 ex 3,2596.10 Tsh 3 Ex T0 volume: V d (m ) Q+ Q 1− ()Q + Q h r T h r proportion of cold dead sh 0,1 volume: V mc /V d (-) if T >T and T ≈T . proportion of hot dead sh 0 sc 0 0,1 volume: V mh /V d (-) proportion of regenerator W W 0,2 Or ηex = = (58) dead volume: V r/V d (-) Tsc Ex T0 out-of-phase angle of the Qc 1− Q c pistons: ϕ ( ° ) 90 Tsc number of revolutions of 180 the engine: N(tr/min) if Tsh ≈T0 and Tsc Int. J. of Thermodynamics, Vol. 10 (No. 4) 171 We defined the data of the state of reference (0) as 1,012E+05 it follows: 1,010E+05 1,008E+05 T0 =296 3, K 1,006E+05 5 p0 =10 Pa p[Pa] 1,004E+05 s0 = ,6 858 J /kg K 1,002E+05 1,000E+05 3.1 First simulation results: 9,980E+04 0.008 9,960E+04 4,56E-05 4,57E-05 4,58E-05 4,59E-05 4,60E-05 4,61E-05 4,62E-05 0.006 Vt[m3 ] y[m] 0.004 Figure 12. Engine p-V diagram The out-of-phase movements of the two 0.002 pistons (engine piston and displacer piston) can be followed on Figure 9 . 0 Figure 10 shows the variations of hot and 0 0.002 0.004 0.006 0.008 x[m] cold volumes according to the crankangle. It illustrates well the low variations of the working Figure 9. Working piston (x) and displacer piston volume compared to those of the hot or cold (y) instantaneous positions volumes. Variation of the working volume is not 3 shown because it is nearly one percent of the V [m ] 4,000E-05 (H) displacer swept volume. (C) 3,000E-05 Figures 11 and 12 show the variations of pressure in the cylinder according to the angle of crankshaft or the total volume of the engine (p-V 2,000E-05 diagram). The following table provides the results of 1,000E-05 the simulation: TABLE II. RESULTS FOR A COMPLETE 0,000E+00 CYCLE SIMULATION 0 60 120 180 240 300 -4 cycle work: Wcycle J -1,909.10 crankshaft angle [ ] cold side heat: Qc J -2 Figure 10. Expansion volume, compression [ ] -6,705.10 volume and working volume versus crankshaft hot side heat: Qh[J ] 6,720.10 -2 angle regenerator heat: Qr J -5 [ ] 3,768.10 p[Pa ] 1,015E+05 thermal efficiency: ηth[%] 0,28 hot side temp.: Th[K] 292 1,010E+05 cold side temp.: Tc[K] 290 Carnot efficiency.: ηcarnot[%] 0,7 1,005E+05 hot wall temp.: Twh[K] 296,3 cold wall temp.: Twc K 285,6 [ ] -7 1,000E+05 reg. entropy prod.: π r[J /K ] 9,292.10 -6 cold entropy prod.: π c[J /K ] 3,495.10 -6 9,950E+04 hot entropy prod.: π h[J /K ] 3,363.10 0 100 200 300 r -4 regen. exergy loss: Exlost[J ] 2,753.10 crankshaft angle c -3 cold exergy loss: Exlost[J ] 1,035.10 Figure 11. Pressure versus crankshaft hot exergy loss: Exh [J ] 9,967. 10 -4 angle lost exergetic efficiency: ηex[%] 7,63 max. exergetic eff.: ηexmax [%] 18,7 172 Int. J. of Thermodynamics, Vol. 10 (No. 4) 3.2 Sensitivity study: 3.2.2 Cold side gas temperature: The central parameters of this study of TABLE IV. SENSITIVITY OF THE sensitivity are: EFFICIENCY OUTPUTS COMPARED TO THE The proportion of the regenerator dead COLD GAS TEMPERATURE volume Vmr /Vd = 0,2 Tc[K] Th −Tc ηth [%] ηcarnot[%] ηex [%] ηII[%] The temperature of the cold side gas (air) 291 1 0,20 0,34 8,16 58,62 Tc =290K 290 2 0,28 0,68 7,64 41,41 The regenerator efficiency ηr =50% . 289 3 0,33 1,03 6,59 32,07 3.2.1 Regenerator dead volume: 288 4 0,36 1,37 5,67 26,14 TABLE III. SENSITIVITY OF THE One notices on Figure 14 the evolution of the EFFICIENCY OUTPUTS VERSUS efficiencies according to the temperature of the gas PROPORTION OF REGENERATOR DEAD (air) on the cold side, therefore of the difference of VOLUME 90 9 ηII th ex Vr/ V d [−] η [%] η [%] ηth , ηex 0,1 0,30 7,77 ηex 0,2 0,28 7,64 60 6 0,3 0,27 7,50 ηII 30 3 0,3065 7,8 η ηex th η th 0 0 0,2965 ηex 7,65 1 2 3 4 Th −Tc 0,2865 ηth 7,5 Figure 14. ηth , ηex and ηII versus 0,2765 temperature difference Th −Tc 0,2665 7,35 temperature Th −Tc . It is obvious that for a more 0,1 0,2 0,3 important difference of temperature the engine uses better the heat provided at each cycle what Vr/ V d implies a higher thermal efficiency. On the other hand, since the air will be colder Figure 13. ηth and ηex variations on the cold side, the exergetic expenditure which versus regenerator dead volume ratio represents the exergy Q at temperature T also c SC Vmr /Vd increases, which makes decrease the exergetic Figure 13 presents the influence of efficiency. regenerator dead volume on the thermal and exergetic efficiencies. It is obviously necessary to On Figure 15 , one traces entropy generation limit as far as possible this dead volume since it is according to the difference of the hot and cold a volume of gas which handicaps the engine temperatures of the gas. One realizes that if the performance. difference in temperature increases, entropy generation in the three cells of the engine increase, which cause a drop in the degree of quality of the engine. Int. J. of Thermodynamics, Vol. 10 (No. 4) 173 TABLE V. SENSITIVITY OF ENTROPY As one can see on Figure 16 and TABLE VI, PRODUCTIONS VERSUS TEMPERATURE OF the efficiency of the regenerator has a great THE COLD GAS influence on the thermal and exergetic efficiencies of the engine. A better regeneration increases the Tc[K] Th −Tc π c[J /K ] π h[J /K] π r[J /K ] thermal efficiency, since same work will be -6 -6 -7 291 1 1,718.10 1,677.10 2,306.10 provided for a lower energy expenditure from the 290 2 3,495.10 -6 3,364. 0 -6 9,292.10 -7 hot source. -6 -6 -6 289 3 5,949.10 5,638.10 2,106.10 The exergetic efficiency increase is due to the 288 4 9,119.10 -6 8,507.10 -6 3,772.10 -6 reduction of the exergy expenditure at the cold sink. In fact, the temperature of this sink increases, approaching the temperature of the gas (a constant for this study of sensitivity) which decreases its 0,000012 energy quality, as TSC TABLE VII. SENSITIVITY OF THE 0,000008 πc TEMPERATURE OF THE WALLS TO THE EVOLUTION OF THE REGENERATOR EFFICIENCY (with fixed Tc and T h) π h 0,000004 ηr[%] Twh[K] Twc[K ] Tw 0 298,867 283,144 15,723 π r 25 297,599 284,412 13,187 0 50 296,331 285,680 10,651 1 2 3 4 75 295,063 286,948 8,115 Th −Tc 100 293,795 288,216 5,579 Figure 15. Entropy productions On Figure 17 , one sees that if the regenerator π c, π h and π r versus Th −Tc efficiency increases, the difference of temperatures between the hot and cold walls decreases with a 3.2.3 Regenerator efficiency: 300 TABLE VI. SENSITIVITY OF THE OTHER Twh EFFICIENCY OUTPUTS VERSUS 294 REGENERATOR EFFICIENCY ηr[%] ηth [%] ηex [%] ηII[%] 288 Twc 0 0,18 3,23 26,15 25 0,22 4,75 32,07 282 50 0,28 7,64 41,46 0 0,25 0,5 0,75 1 75 0,40 14,25 58,62 100 0,68 35,26 100 ηr Figure17. Variations of Twc and Twh versus ηr 45,00% 0,90% ηex ηth constant (Tc −Th ). This is due to the reduction of 30,00% 0,60% the transferred heat quantities. TABLE VIII. SENSITIVITY OF ENTROPY GENERATIONS TO THE REGENERATOR ηth EFFICIENCY 15,00% 0,30% ηex ηr[%] π c[J /K ] π h[J /K] π r[J /K ] 0 8,883.10 -6 8,381. 10 -6 1,858. 10 -6 0,00% 0,00% -6 -6 -6 25 5,874. 10 5,60. 10 1,393. 10 0 0,25 0,5 0,75 1 -6 -6 -7 50 3,495. 10 3,36. 10 9,291. 10 -6 -6 -7 η r 75 1,736. 10 1,690. 10 4,645. 10 -7 -7 100 5,907. 10 5,84. 10 0 Figure 16. Variations of ηth and ηex versus ηr 174 Int. J. of Thermodynamics, Vol. 10 (No. 4) 0,0000122 5. Conclusion: In this paper, we present a thermodynamic study of one γ type Stirling engine at low 0,000009 πc temperature difference and steady state operation. Energy, entropy and exergy balances were carried out at each element of the engine, according to the 0,000006 π h kinematics of the pistons (displacer and working pistons). A sensitivity study of the engine efficiencies 0,000003 and creations of entropy with respect to various π r parameters of the model (regenerator efficiency, cold temperature of gas, dead volume of the 0 regenerator) was carried out. 0 0,25 0,5 0,75 1 ηr The results of simulation make it possible to Figure 18. Variations of π c, π h and π r locate the optimum conditions for operation of this engine which give the best efficiencies and the versus ηr smallest productions of entropy, to minimize the operating cost. This model was validated by a Generation of entropy in the three cells of the comparison with experimental results of actual engine decrease with the regenerator efficiency engines. increase : higher is this efficiency, less important are the losses due to the differences of temperature Nomenclature and the losses due to imperfect regeneration. A Section, [m 2] 4. Comparison with experimental results: cp Specific heat at p=cst, [J/kg K] c Specific heat at V= cst, [J/kg K] Tests on a nearly identical engine as ours v Ex Exergy, [J] with low T were carried out by H. Roussel [7]. Specific exergy, [J/kg] For two experimental points, he published the ex following results : h Specific enthalpy, [J/kg] l Length, [m] Th− T c p Pressure, [Pa] Tw N[tr/min] W[J ] Q Heat, [J] 3 21 199 2,89.10 -4 R Radius, [m] 2,6 5 ,5 20 2,46.10 -4 r Specific gas constant, [J/kg K] The results obtained by our model are located S Entropy, [J/K] s Specific entropy, [J/kg K] in the same order of magnitude while taking ηr =0 T Temperature, [K] and Th− T c =3 ° C . Our calculated work in this t Time, [s] 2 −4 U Heat exchange coefficient, [W/m K] case, Wcal = 2,91.10 J , and that measured by H. -4 u Velocity, [m/s] Roussel (2.89.10 J) are very close. We carried V Volume, [m 3] out tests on our engine, very similar to those of H. W Work, [J] Roussel, which confirm his results. x Position of the working piston, [m] A test bench is under development ; pressure y Position of the displacer, [m] and piston position pick-ups will allow the direct θ Angle of crankshaft, [-] calculation of the indicated work provided by gas, π Entropy generation, [J/K] by the layout of the real cycle on a p-V diagram ; ϕ Out-of-phase angle, [-] temperature sensors, a controlled electric heating, η Efficiency, [-] an effective insulation, various mechanical adjustments (strokes, out-of-phase angle) Indices: supplement this equipment and will make it possible to obtain the indicated efficiency and the a ambiant conditions Carnot efficiency of the engine, as well as a c cold modulation of the kinematics of the engine. This cyl cylinder will make it possible to better validate this d displacer thermodynamic model. i inlet, outlet Int. J. of Thermodynamics, Vol. 10 (No. 4) 175 g gas PRIETO J.I., GARCIA D., 2005,”Comparaison h hot between Kolin’s law for power and other criteria int internal for preliminary design of Kinematic Stirling m dead (volume) engines” Thermo- and GFD modelling of Stirling p working piston machines, pp.389-397. r regenerator ROBSON A., 2005, “Development of a computer sc cold source model to simulate a low temperature differential sh hot source Ringbom Stirling engine”, Thermo- and GFD th thermal modelling of Stirling machines, pp.350-357. w wall 0 reference state ROCHELLE P., 2005, ”LDT Stirling engine simulation and optimization using finite dimension References thermodynamics”, Thermo- and GFD modelling of Stirling machines, pp.358-366 FEIDT M., LESAOS K., COSTEA M., PETRESCU S., “Optimal allocation of HEX WAGNER A., SYRED N., ELSNER M., 2005, inventory associated with fixed power output or ”First order calculation of Stirling engines”, fixed heat transfer rate input”, Int. J. Applied Thermo- and GFD modelling of Stirling machines, Thermodtnamics, vol 5, n°1, p25-36, mars 2002 pp.367-379. MARTAJ N., GROSU L., ROCHELLE P., 2006 www.ent.ohiou.edu/~urieli/stirling/me422.html “Exergetical analysis and design optimization of www.photologie.net/indexStirBlue.html the Stirling engine”, Int. J. of Exergy, Vol.3, No. 1, pp.45-67 176 Int. J. of Thermodynamics, Vol. 10 (No. 4)