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Thermodynamic Study of a Low Temperature Difference Stirling Engine at Steady State Operation

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Thermodynamic Study of a Low Temperature Difference Stirling Engine at Steady State Operation 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 TDC sink of the engine. Such an engine can run simply 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. T c' 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 : Q= UA(T − T)t ∆ (32) c p 1 Trh c p Trh c cc wc c δQh = Vh − dp + p 1− dVh 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’).
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