Constraints on Onsager coefficients from quasi-static and reversible operations
CConstraints on Onsager coefficients from quasi-static and reversible operations
Ramandeep S. Johal ∗ Department of Physical Sciences,Indian Institute of Science Education and Research Mohali,Sector 81, S.A.S. Nagar,Manauli PO 140306, Punjab, India
The performance of a generic, cyclic heat engine between two heat reservoirs is discussed withina linear-irreversible framework. The Onsager reciprocal relation is derived as a consequence ofthe equivalence between quasi-static and reversible operations, under the tight-coupling condition.When the latter condition is relaxed, it is possible to achieve reversible cycle in a finite duration.Onsager reciprocity must be violated when either the quasi-static cycle is not reversible, or thereversible cycle is not quasi-static.
A quasi-static process, despite being an idealized con-cept, is of great importance in equilibrium thermody-namics [1]. Several textbook models of heat cycles suchas Carnot cycle, Otto cycle and so on, are based on theseprocesses which run infinitely slowly, but may or maynot be reversible. Further, any framework for nonequi-librium processes involving rates and fluxes, quite under-standably, must approach the limiting case of equilibriumtheory when these fluxes become negligible in responseto vanishing thermodynamic forces. In particular, in thelinear response regime [2], the generalized fluxes are as-sumed to be linear functions of the forces. The resultinglinear-irreversible framework is immensely successful inunifying diverse, coupled phenomena such as Seebeck,Peltier, Dufour effects and so on [3, 4].In this letter, we apply the linear-irreversible frame-work to a heat engine operating between two heat reser-voirs. The restriction to near-equilibrium conditions de-mands that the temperature difference between the reser-voirs is small and the duration of the heat cycle ( τ )is sufficiently long. Further, the generalized fluxes arenot defined as instantaneous quantities (time derivativesof macroscopic variables, as in Onsager formalism [2]),but as time-averages over one cycle ( X/τ ). We deter-mine the constraints on the Onsager coefficients in theflux-force relations, taking into account the ideal limitsof quasi-static and reversible cycles. A connection willbe established between the reciprocal property of On-sager coefficients and the equivalence of quasi-static andreversible cycles [5]. It is shown that violation of thisreciprocity implies that quasi-static and reversible cyclesare not equivalent. Further, it is possible to constructa reversible cycle in a finite duration, or in other words,achieve a finite power output at Carnot efficiency.Let Q h denote the heat absorbed in one cycle from thehot reservoir at temperature T h , and Q c be the amountof heat rejected to the cold reservoir at temperature T c .The work performed by the engine is W = Q h − Q c > , (1) ∗ [email protected] while the total change in the entropy of the reservoirs is∆ S = Q c T c − Q h T h . (2)The working medium undergoes a cycle and so does notinvolve a net change in its entropy. Now, if Q c is treatedas a floating variable, it may be eliminated from Eqs.(1) and (2), so that the second law inequality (∆ S ≥ W ≤ Q h η C , where η C = 1 − T c /T h is the Carnotefficiency. Thus, the second law sets an upper boundfor work as W rev = Q h η C , obtained under a reversibleoperation (∆ S = 0). In general, we can write W = W rev − T c ∆ S, (3)The deficit, W rev − W , is referred to as the lost work—the energy that is not available for work during the irre-versible operation [6].Now, let us consider the time-dependence of theenergy-conversion process. The rate of total entropy pro-duction per cycle, ˙ S ≡ ∆ S/τ , can be written as:˙ S = 1 τ (cid:18) − WT c (cid:19) + ˙ Q h (cid:18) T c − T h (cid:19) , (4)where ˙ Q h = Q h /τ . Following the flux-force frameworkof linear-irreversible thermodynamics, we identify twogeneralized fluxes ( J i ) and the corresponding thermody-namic forces ( X i ): J = 1 τ , X = − WT c , (5) J = ˙ Q h , X = 1 T c − T h , (6)which imply ˙ S ≡ J X + J X . J is the number of cy-cles executed per unit time, and so represents the rate atwhich the heat cycle proceeds. Since the forces and thefluxes are assumed to be small, we are dealing here withlong cycle durations close to equilibrium. The process2, satisfying J X >
0, denotes a spontaneous processleading to positive entropy production, while the pro-cess 1 with J X < a r X i v : . [ c ond - m a t . s t a t - m ec h ] S e p
2. The linear regime implies that the flux-force relationsare in the form: J i = (cid:80) j =1 L ij X j , where i = 1 ,
2. Here,the phenomenological coefficients L ij will be referred toas Onsager coefficients. They are assumed fixed withinthe regime of small forces. Then, ˙ S becomes a binaryquadratic form in X and X , and the inequality ˙ S ≥ L , L ≥ , L L ≥ ( L + L ) . (7)There seems no reason, per se, to assume the reciprocalrelation ( L = L ) which was originally derived by On-sager from the principle of microscopic reversibility andthe theory of equilibrium fluctuations [7].Explicitly, the flux-force relations are given by:1 τ = − L WT c + L η C T c , (8) Q h τ = − L WT c + L η C T c . (9)Assuming L >
0, we can eliminate W from the abovepair of equations, to write: Q h τ = L L τ + | L | L η C T c , (10)where the determinant | L | ≡ L L − L L satisfies | L | ≥
0. Let us first assume, | L | = 0, which is also knownas the tight-coupling condition [8]. Under this condition,the fluxes J and J become proportional to each other.Therefore, Eq. (10) yields: Q h = L L > , (11)which requires L >
0. Then, from the | L | = 0 condi-tion, we have L ≥ τ → ∞ ) in Eq. (8). The work output in thislimit is given by: W qs = ( L /L ) η C . Then, Eq. (8)may be rewritten in the form: W = W qs − T c L τ . (12)Thus, the work output may be controlled by varying τ ,while the magnitude of Q h is kept fixed. So, as τ isdecreased, the work output decreases and vanishes at theminimum duration: τ min = T c /L η C .It is well known that a quasi-static operation may ormay not imply a reversible operation [1, 9]. If it does,then as τ → ∞ for our cycle, we also have ∆ S →
0. Thisequivalence has a number of interesting consequences.We can now write: W qs = W rev , which yields Q h = L /L . Upon comparison with Eq. (11), we obtain: L = L , (13)thus obtaining the reciprocal relation for Onsager coef-ficients. Conversely, from the general relations derived above, we can write: W qs = ( L /L ) W rev . Thus, thevalidity of Onsager reciprocity for macroscopic linear-irreversible, cyclic engines implies that quasi-static andreversible operations of the engine are equivalent. Thisis the first main result of this letter.Furthermore, under the condition W qs = W rev , thecomparison of Eqs. (3) and (12) yields:∆ S = 1 L τ . (14)Thus, the total entropy produced varies inversely withthe cycle duration—consistent with the assumption thatthe quasi-static operation implies reversible operation.On the other hand, a quasi-static process may not bereversible, and involve some entropy production ∆ S qs >
0. A familiar situation is the presence of friction betweendifferent parts of the engine, which may not vanish even ifthe cycle is made infinitely slow. Then, from Eq. (3), wecan write: W qs = W rev − T c ∆ S qs . So, in general, we have: W qs ≤ W rev , which implies that L ≤ L . Thus, thecondition L < L represents the fact that the quasi-static work is less than the reversible work—indicatingthe presence of friction or viscous forces. Related to this,we also have the result for quasi-static efficiency: η qs = W qs /Q h = ( L /L ) η C ≤ η C .If | L | >
0, then from Eq. (10), Q h diverges in thequasi-static limit [10]. Since W qs is finite, so the efficiencyvanishes in the quasi-static limit. Clearly, the latter doesnot imply a reversible cycle in the | L | > Q h is finite). In recent years, the possibility of achieving theCarnot efficiency in finite time, or in other words, a finitepower output along with Carnot efficiency, has attractedattention [11–16]. We now approach this issue for thecase of cyclic, linear-irreversible engines.Eliminating τ from Eqs. (8) and (9), and using thereversible work condition, W = Q h η C , we obtain thequadratic equation: L Q h − ( L + L ) Q h + L = 0,whose solutions are: Q h = L + L ± (cid:112) ( L + L ) − L L L . (15)The only real solution in the above is obtained when thethe third condition in Eq. (7) reduces to an equality:( L + L ) = 4 L L , (16)and therefore Q h = L + L L ≥ . (17)The magnitude of the reversible work—performed ina finite duration—is then given by W ( τ )rev = ( L + L ) η C / L . Note that the above expression for Q h holds specifically for the reversible operation. Unlike thetight-coupling case ( | L | = 0) that yields a fixed magni-tude for Q h (Eq. (11)), here we have | L | (cid:54) = 0, and so ingeneral, Q h depends on the cycle duration (Eq. (10)).The duration τ rev of the reversible cycle can be calcu-lated from Eqs. (8) and (17) as:1 τ rev = ( L − L ) η C T c , (18)which requires L ≥ L . Using the positivity conditionon the quasi-static work, W qs = ( L /L ) η C >
0, wecan set L >
0. Then, the inequality L + L ≥ L ≥ − L . These considerationsconstrain the possible values of L as follows: − L ≤ L ≤ L . (19)Since, the minimum allowed value of L is − L , sothe minimum value of τ rev is T c /L η C ≡ τ min at whichthe work output vanishes. On the other hand, for( L − L ) → + , we have τ rev → ∞ , implying thatif Onsager reciprocity holds, then the reversible opera-tion is obtained only in the quasi-static limit. In otherwords, Onsager reciprocity must be violated for the heatcycle undergoing reversible operation in a finite duration.The power output at reversible operation, P rev = W ( τ )rev /τ rev , is given by: P rev = L − L L η T c . (20)Thus, for a given value of L , the power output at re-versible operation vanishes at both the extreme values of L (Eq. (19)); at its minimum value, we have zero workoutput, whereas at the maximum value of L , the On-sager reciprocity holds, leading to a diverging durationof the reversible cycle.Using the second inequality in Eq. (19), we can write( L /L ) η C ≥ ( L + L ) η C / L , or W qs ≥ W ( τ )rev ,where the equality is obtained when τ rev → ∞ (see Eq.(18)). Thus, we have the interesting result that the finite-time reversible work is bounded from above by the quasi-static work. Based on the results obtained so far, we mayorder the different work outputs as follows: W ( τ )rev ≤ W qs ≤ W rev . (21)Fig. 2 clarifies the meaning of the above inequalities interms of relative magnitudes of the Onsager coefficients.Clearly, the equalities are obtained when Onsager reci-procity is satisfied and the reversible work is obtainedjust in the quasi-static limit.Exhibiting a reversible cycle in a finite duration for thelinear-irreversible heat engine is our second main result.For convenience, we denote such a heat engine as R ( τ ) ,which requires the conditions | L | > L >
0, apartfrom Eqs. (16) and (19), to be satisfied. Fig. 1 showsa case study of such an engine, where the Onsager co-efficients can be chosen so as to achieve reversible cycle
100 200 300 400 500 τ η
100 200 300 400 500 τ Δ S
100 200 300 400 500 τ W
100 200 300 400 500 τ W / τ ( d )( c ) W qs ( a ) ( b ) η C Figure 1. An example of the R ( τ ) heat engine, showing re-versible operation in a cycle of finite duration. The param-eters are set at L = 1 , L = 4 , L = 3 and L = 1,consistent with the conditions L >
0, Eqs. (16) and (19).Also, T c = 75 and T h = 100, so that η C = 0 .
25. All quantitiesare plotted versus the cycle duration τ whose minimum valueis τ min = T c /L η C = 100 units. (a) Efficiency η obtainsthe Carnot bound at duration τ rev = 300 (see Eq. (18)). Forlonger durations, the efficiency vanishes since Q h diverges. (b)Total increase in entropy per cycle, ∆ S , vanishes at τ rev . (c)Work output approaches the quasi-static limit W qs for longcycle durations. The work at reversible operation W ( τ )rev ( • )is lower than W qs . (d) Power output becomes maximum at τ ∗ = 2 τ min = 200. The corresponding EMP is given by Eq.(23). • denotes the power output under reversible conditions(Eq. (20)). with finite duration. Further, it can be easily shown that | L | = 0 case does not allow reversible cycle in a finiteduration, since it would violate the second law.Finally, we study the optimization of average poweroutput, ˙ W = W/τ . Setting d ˙ W /dτ = 0, the optimal du-ration of the cycle is: τ ∗ = 2 τ min . We note that themaximum power condition is not affected by whether | L | > | L | = 0. The work performed at maxi-mum power is W ∗ = W qs /
2. So, the maximum poweris ˙ W ∗ = W ∗ /τ ∗ = L η / L T c , and the efficiency atmaximum power (EMP), η ∗ = W ∗ /Q h , is: η ∗ = L | L | + L L η C . (22)The above expression was also derived as EMP for anautonomous engine based on a thermoelectric setup inthe presence of external magnetic field [11] and can beexpressed in terms of two parameters (apart from η C ):the asymmetry ratio ( x ≡ L /L ) and the generalizedfigure of merit ( y ≡ L L / | L | ). For | L | = 0 or thetight-coupling condition, the above formula is simplifiedto η ∗ = η qs /
2. Since L ≤ L for the tightly cou-pled case, so we conclude that η ∗ ≤ η C /
2. The upperbound of half-Carnot value is obtained when quasi-staticand reversible operations are equivalent and so the re-ciprocal relation holds. This result was known for au-tonomous, linear-irreversible engines obeying the tight- L = L L < L L > L Quasi - static Reversible W qs = W rev W qs < W rev W rev ( τ ) < W qs a ) b ) | L | = | L | > Figure 2. a) A cyclic, linear-irreversible engine follows theOnsager reciprocal relation ( L = L ) if it approaches thequasi-static and the reversible operations, simultaneously. Wehave L < L , when the quasi-static cycle is not reversible,whereas L > L holds when the reversible cycle is notquasi-static, or in other words, achieved in a finite duration.b) The corresponding comparison between different magni-tudes of work output per cycle, also illustrating the inequali-ties in Eq. (21). coupling condition along with Onsager reciprocity [8, 17].It was further noted in Ref. [11] that there are no ther-modynamic constraints on the allowed values of the realparameter x . However, for the case of an R ( τ ) engine,there is a constraint on x due to Eq. (19), so that wehave − ≤ /x ≤ +1, or | x | ≥
1. Thus, for such engines,the EMP is simplified to the form: η ∗ = η C /x ) . (23)Notably, the half-Carnot value, established earlier as theupper bound for EMP, is breached here. Moreover, EMPis a function only of the parameter x , and as | x | → ∞ ,the EMP can approach the Carnot bound. Concluding, we have considered the performance of acyclic heat engine within linear-irreversible framework.It is remarkable that the idealized processes of equilib-rium thermodynamics have a bearing on the reciprocalproperties of phenomenological coefficients describing thestrength of couplings in the near-equilibrium regime. Ourmain general conclusions are: Onsager reciprocal relationimplies that the reversible operation is obtained in infi-nite time or quasi-static limit. Since W qs is finite, so inorder to obtain Carnot efficiency, Q h should also be fi-nite in that limit. This necessarily implies | L | = 0. Onthe other hand, even for | L | = 0, we may have a quasi-static operation which is irreversible. This requires vio-lation of Onsager reciprocity—in particular, L < L .When | L | > Q h diverges in the quasi-static limit andso the efficiency vanishes. Interestingly, there is possibil-ity of reversible operation in a finite duration, for which L > L must be obeyed apart from other conditionsmentioned in the text. Thus, violation of the Onsager re-ciprocal relation implies that quasi-static and reversibleoperations are not equivalent (see also Fig. 2). It may beremarked that our treatment of Onsager reciprocity andits violation has been more at an abstract level, withoutgoing into the aspect of concrete, physical realizations.Even so, the consistency of some of the results with thesteady-state thermoelectric setups [11], such as the con-straints for reversible operation in finite duration, indi-cate a wider basis for these results.The model considered here involves only two forces,which is the simplest example of an irreversible systemexhibiting coupled processes. Several lines of inquirymay be visualized in this context. It would be inter-esting to highlight the parallels relating Onsager reci-procity with vanishing fluxes and reversible operations inthe case of autonomous engines. A generalization incor-porating more than two forces and/or larger number ofheat reservoirs is important. Apart from power output,the study of other figures of merit including refrigeratormodels will be desirable. Finally, the study of impli-cations for stochastic thermal machines would foster aninteresting line of inquiry. [1] H. B Callen, Thermodynamics and an introduction tothermostatistics , 2nd ed. (New York : Wiley, 1985).[2] L. Onsager, “Reciprocal relations in irreversible pro-cesses. i.” Phys. Rev. , 405–426 (1931).[3] H. B. Callen, “The application of Onsager’s reciprocal re-lations to thermoelectric, thermomagnetic, and galvano-magnetic effects,” Phys. Rev. , 1349–1358 (1948).[4] G. Lebon and D. Jou, Understanding Non-equilibriumThermodynamics: Foundations, Applications, Frontiers ,SpringerLink: Springer e-Books (Springer Berlin Heidel-berg, 2008).[5] A quasi-static operation and a reversible operation aresaid to be equivalent if one implies the other. [6] M. W. Zemansky,
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