Attainability of Maximum Work and the Reversible Efficiency of Minimally Nonlinear Irreversible Heat Engines

M. Ponmurugan

doi:10.1515/jnet-2018-0009

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Attainability of Maximum Work and the Reversible Efficiency of Minimally Nonlinear Irreversible Heat Engines

Veröffentlicht/Copyright: 16. Februar 2019

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Aus der Zeitschrift Journal of Non-Equilibrium Thermodynamics Band 44 Heft 2

Abstract

We use the general formulation of irreversible thermodynamics and study the minimally nonlinear irreversible model of heat engines operating between a time-varying hot heat source of finite size and a cold heat reservoir of infinite size. We find the criterion under which the optimized efficiency obtained by this minimally nonlinear irreversible heat engine can reach the reversible efficiency under the tight coupling condition: a condition of no heat leakage between the system and the reservoirs. We assume the rate of heat transfer from the hot to the cold heat reservoir obeys Fourier’s law and discuss physical conditions under which one can obtain the reversible efficiency in a finite time with finite power. We also calculate the efficiency at maximum power for the minimally nonlinear irreversible heat engine under the nontight coupling condition.

Keywords: heat engine; irreversible thermodynamics; exergy; efficiency

Acknowledgment

I thank R. Arun for the critical reading of the manuscript. I also thank the anonymous referees for their critical comments and valuable suggestions.

We have k(T)=β(T)−X2(T)Tca1(T) and p(T,T˙)=1+2a1(T)(g(T)+Cv(T)T˙). The partial differentiation of p with respect to T and T˙ is given by

(49)∂p∂T˙=2a1Cv,

(50)∂p∂T=2∂(a1g)∂T+2T˙∂(a1Cv)∂T.

By using eqs. (23)–(50) one can calculate

(51)∂J3∂T=∂∂T(k[1+p]+βg)

(52)+T˙∂∂T((β−1)Cv),∂J3∂T˙=kpa1Cv+(β−1)Cv,

(53)∂∂T˙(∂J3∂T˙)=−kp3/2a12Cv2,

(54)∂∂T(∂J3∂T˙)=∂∂T(kpa1Cv)+∂∂T((β−1)Cv).

For optimization eq. (28) can be rewritten in terms of T(t) as

(55)T¨∂∂T˙(∂J3∂T˙)+T˙∂∂T(∂J3∂T˙)−∂J3∂T=0.

By using eqs. (51)–(54) in the above equation one can obtain

(56)−T¨ka12Cv2p3/2+T˙∂∂T(kpa1Cv)−∂∂T(k[1+p]+βg)=0.

Since

(57)ddt(kpa1Cv)=−T¨ka12Cv2p3/2+T˙∂∂T(kpa1Cv)

and

(58)∂∂T˙(k[1+p]+βg)=kpa1Cv,

eq. (56) can be rewritten as

(59)ddt(∂Y∂T˙)−∂Y∂T=0,

where Y(T,T˙)=k[1+p]+βg. We get an equation of the same type as eq. (59) for the other value of J1=J1− with Y(T,T˙)=k[1−p]+βg. Therefore, for two different values of J1=J1±, one can get the same equation.

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Received: 2018-03-23

Revised: 2018-11-26

Accepted: 2018-12-12

Published Online: 2019-02-16

Published in Print: 2019-04-26

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Artikel in diesem Heft

https://doi.org/10.1515/jnet-2018-0009

Schlagwörter für diesen Artikel

heat engine; irreversible thermodynamics; exergy; efficiency