an extension to endoreversible thermodynamics for multi

An Extension toEndoreversible Thermodynamicsfor Multi-Extensity Fluxes andChemical Reaction Processes

Von der Fakultat fur Naturwissenschaften der Technischen Universitat Chemnitzgenehmigte Dissertation zur Erlangung des akademischen Grades

doctor rerum naturalium(Dr. rer. nat.)

vorgelegt von Katharina Wagner, M. Sc.geboren am 25.10.1984 in Frankenberg

eingereicht am 30. April 2014

Gutachter: Prof. Dr. Karl Heinz HoffmannProf. Dr. Bjarne Andresen

Tag der Verteidigung: 20. Juni 2014

Bibliographische Beschreibung

Wagner, KatharinaExtensions to Endoreversible Thermodynamics for Multi-Extensity Fluxesand Chemical Reaction ProcessesTechnische Universitat Chemnitz, Fakultat fur NaturwissenschaftenDissertation (in englischer Sprache), 2014102 Seiten, 28 Abbildungen, 82 Literaturzitate

Referat

In dieser Arbeit erweitere ich den Formalismus der endoreversiblen Thermodynamik,um Flusse mit mehr als einer extensiven Große sowie chemische Reaktionsprozessemodellieren zu konnen. Mit Hilfe dieser Erweiterungen eroffnen sich zahlreiche neueAnwendungsmoglichkeiten fur endoreversible Modelle. Flusse mit mehreren extensi-ven Großen sind fur die Betrachtung von Massestromen ebenso notig wie fur Prozesse,bei denen sowohl Volumen als auch Entropie zwischen zwei Teilsystem ausgetauschtwerden. Fur sowohl reversibel wie auch irreversibel gefuhrte chemische Reaktions-prozesse wird ein neues Teilsystem – der

”Reaktor“ – vorgestellt, welches sich ahnlich

wie endoreversible Maschinen durch Bilanzgleichungen auszeichnet. Der Unterschiedzu den Maschinen besteht in den Produktions- bzw. Vernichtungstermen in den Teil-chenzahlbilanzen sowie der moglichen Entropieproduktion innerhalb des Reaktors.Beide Erweiterungen finden dann in einem endoreversiblen Modell einer Brennstoff-zelle Anwendung. Dabei werden Flusse mehrerer gekoppelter Extensitaten fur denZustrom von Wasserstoff und Sauerstoff sowie fur den Protonentransport durch dieElektrolytmembran benotigt. Chemische Reaktionen treten in der Anode und Katho-de der Brennstoffzelle auf. Diese werden mit dem neu eingefuhrten Teilsystem, demReaktor, eingebunden. Mit Hilfe des Modells werden dann Wirkungsgrad, Zellspan-nung und Leistung einer Brennstoffzelle unter Berucksichtigung der Partialdruckeder Substanzen, der Temperatur sowie der Dissipation beim Protonentransport be-rechnet. Dabei zeigt sich, dass experimentelle Daten fur die Zellspannung sowohlqualitativ als auch naherungsweise quantitativ durch das Modell abgebildet werdenkonnen. Der Vorteil des endoreversiblen Modells liegt dabei in der Moglichkeit, mitnur einem Modell neben den genannten Kenngroßen auch die abgegebene Warmesowie die Entropieproduktion zu quantifizieren und den einzelnen Teilprozessen zu-zuordnen.

Schlagworte

Endoreversible Thermodynamik, Nichtgleichgewichtsthermodynamik, Ideales Gas,van der Waals Gas, Brennstoffzelle, Modellierung, Kreisprozesse, Chemische Reak-tionen, Wirkungsgrad

Abstract

In this thesis extensions to the formalism of endoreversible thermodynamics for multi-extensity fluxes and chemical reactions are introduced. These extensions make itpossible to model a great variety of systems which could not be investigated withstandard endoreversible thermodynamics. Multi-extensity fluxes are important whenstudying processes with matter fluxes or processes in which volume and entropyare exchanged between subsystems. For including reversible as well as irreversiblechemical reaction processes a new type of subsystems is introduced – the so calledreactor. It is similar to endoreversible engines, because the fluxes connected to it arebalanced. The difference appears in the balance equations for particle numbers, whichcontain production or destruction terms, and in the possible entropy production inthe reactor.Both extensions are then applied to an endoreversible fuel cell model. The chemicalreactions in the anode and cathode of the fuel cell are included with the newlyintroduced subsystem – the reactor. For the transport of the reactants and productsas well as the proton transport through the electrolyte membrane, the multi-extensityfluxes are used. This fuel cell model is then used to calculate power output, efficiencyand cell voltage of a fuel cell with irreversibilities in the proton and electron transport.It directly connects the pressure and temperature dependencies of the cell voltagewith the dissipation due to membrane resistance. Additionally, beside the listedperformance measures it is possible to quantify and localize the entropy productionand dissipated heat with only this one model.

Contents

1 Introduction 91.1 Document Structure . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

2 Endoreversible Thermodynamics 132.1 General Formalism . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

2.1.1 Subsystems . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14Reservoirs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15Engines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

2.1.2 Interactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 162.2 Extension for Multi-Extensity Fluxes . . . . . . . . . . . . . . . . . . 172.3 Ideal Gas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

2.3.1 Gibbs Equation and Principal Equation of State . . . . . . . . 212.3.2 Infinite Reservoirs . . . . . . . . . . . . . . . . . . . . . . . . . 232.3.3 Finite Reservoirs . . . . . . . . . . . . . . . . . . . . . . . . . 24

Finite Reservoirs with Constant Pressure . . . . . . . . . . . . 24Finite Reservoirs with Constant Volume . . . . . . . . . . . . 27

2.3.4 Reversible Engines . . . . . . . . . . . . . . . . . . . . . . . . 30Combining Gas Fluxes . . . . . . . . . . . . . . . . . . . . . . 33

2.3.5 Interactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35Interactions with Volume, Particle and Entropy Flux . . . . . 35Interactions with Volume and Entropy Flux . . . . . . . . . . 36Interactions with Particle and Entropy Flux . . . . . . . . . . 40

2.3.6 Example: Pressure Regulator . . . . . . . . . . . . . . . . . . 422.4 Van der Waals Gas . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

2.4.1 Gibbs Equation and Principal Equation of State . . . . . . . . 482.4.2 Interactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49

Interactions with Volume and Entropy Flux . . . . . . . . . . 492.4.3 Example: Pressure Regulator . . . . . . . . . . . . . . . . . . 51

2.5 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53

3 Chemical Reactions 553.1 Chemical Engines Without Reactions . . . . . . . . . . . . . . . . . . 563.2 Chemical Engines with Reactions . . . . . . . . . . . . . . . . . . . . 58

3.2.1 Reversible Reactions vs. Reversible Thermodynamic Processes 59

7

Contents

3.2.2 A New Subsystem – the Reactor . . . . . . . . . . . . . . . . . 603.2.3 Reversible Reaction Processes . . . . . . . . . . . . . . . . . . 63

Reaction at Constant Volume . . . . . . . . . . . . . . . . . . 64Reaction at Constant Pressure . . . . . . . . . . . . . . . . . . 65

3.2.4 Irreversible Reaction Processes . . . . . . . . . . . . . . . . . 67Reaction at Constant Volume . . . . . . . . . . . . . . . . . . 68Reaction at Constant Pressure . . . . . . . . . . . . . . . . . . 69

3.3 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70

4 Fuel Cells 734.1 Structure of a PEM Fuel Cell . . . . . . . . . . . . . . . . . . . . . . 734.2 Cell Voltage and Fuel Cell Efficiency . . . . . . . . . . . . . . . . . . 744.3 Endoreversible Modeling of PEM Fuel Cells . . . . . . . . . . . . . . 76

4.3.1 Reversible Fuel Cell Model . . . . . . . . . . . . . . . . . . . . 764.3.2 Endoreversible Fuel Cell Model . . . . . . . . . . . . . . . . . 79

4.4 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87

5 Conclusion 89

Nomenclature 91

Bibliography 93

8

1 Introduction

Classical equilibrium thermodynamics has been developed in the 19th century. Itoffers a macroscopic theory for the description of thermodynamic systems in equi-librium. It concentrates on states of thermodynamic systems rather than processes.Thermodynamic processes are usually understood as a quasi-static sequences of equi-libria. Within the classical theory, processes are completely reversible, which for realsystems would require infinitely slow execution. This leads to infinitely small rates,e.g. for energy conversion. The Carnot process [17] is an example for a completelyreversible process. The Carnot efficiency ηC = 1 − TL/TH gives the fraction of theheat, which is taken from a high temperature reservoir, that is converted into work.The rest of the heat is released to a lower temperature reservoir. Since the Carnotprocess is completely reversible and there does not occur any dissipation, this is thehighest efficiency reachable in a heat engine which operates between temperaturesTL and TH.

In real processes, however, finite rates are desired. Therefore, finite and thus ir-reversible fluxes are required. Over the years there have been various attempts inengineering and physics to include irreversibilities in thermodynamic descriptions,for instance under labels like finite-time thermodynamics, Onsager theory, extendedthermodynamics and so on. A great number of publications was focused on irre-versibilities in heat engines (e.g. [6, 7]). In [57], Muller et al. give an extensiveoverview of “Thermodynamics of irreversible processes – past and present”, includingthe work of Cattaneo, Prigogine and Onsager.

In 1979 one of the fields in finite-time thermodynamics was labeled “endoreversible”thermodynamics by Rubin [68, 69]. He defined an “endoreversible engine to be anengine such that during its operation its working fluid undergoes reversible transfor-mations.” This definition includes all the reversible engines in classical equilibriumthermodynamics, such as the Carnot engine. “More importantly [. . . ], the subclassof engines which are coupled to the external world via irreversible processes” [68]is included in the definition of endoreversible engines. The most important andmost investigated examples of such endoreversible engines are the Novikov heat en-gine [59, 60] and the Curzon-Ahlborn (CA) heat engine [25]. Fischer et al. [33] foundthat a simple Novikov model can accurately render an Otto engine if the parametersare adjusted appropriately.

The CA engine has been discussed in great detail in literature. In its original form,

9

1 Introduction

it consists of a Carnot heat engine that is coupled irreversibly to the high and lowtemperature reservoirs. The operating point of maximum efficiency for such an en-doreversible heat engine is not desirable, since the power output at this point goes tozero. However, the power over efficiency curve shows an operating point of maximumpower at the CA efficiency, ηCA = 1−

√TL/TH. This efficiency is a good approxima-

tion of real power plant efficiencies (comparisons can be found in [14] and [25]).

Various heat transfer laws and their effect on power output and efficiency of a CAheat engine have been investigated in [39, 46]. Chen et al. [24], Gordon [38] andNulton et al. [61] discussed a generalized heat transfer law of the form q ∝ ∆(T n),which includes the Newtonian heat transfer for n = 1, the Fourier heat transfer forn = −1 and radiation for n = 4. Huleihil et al. [48] as well as Angulo-Brown etal. [10] also proposed a slightly different generalized heat transfer law of the formq ∝ (∆T )n. Also, the effect of staging of two engines on the efficiency and powerwas investigated by Chen and Wu [20] and Rubin and Andresen [70]. Furthermore,Chen et al. [19] discussed staging more than two endoreversible CA engines. Anothervariation in the structure of a CA engine was the incorporation of a heat leak betweenthe hot and cold reservoir, which leads to a loop in the power efficiency diagram (seefor instance the work of De Vos [26], Huleihil and Andresen [49], Aragon-Gonzalezet al. [12], Chen et al. [18] or Gorden et al. [39]). Amelkin et al. [3] studied systems,which are connected to several heat baths with constant temperatures.

Some authors introduced new performance measures for the CA engine to include notonly power and efficiency but also the total entropy production in the optimizationof heat engines. The so called ecological function was investigated, for instance, byAngulo-Brown [8] and Aragon-Gonzales et al. [12]. De Vos [30] optimized a heatengine with respect to economic performance, i.e. he minimized the relative costs ofinvestments and fuels for the achieved power.

Endoreversible thermodynamics can not only be used for research but is also useful ineducation. Agrawal [2] discussed a simplified version of the CA engine for educationalpurpose. When using Newtonian heat transfer, i.e. q ∝ ∆T , with the same constantheat conductance and the same temperature difference between the reservoirs andthe engine, the calculations for power and efficiency become very easy. Still, theentropy production in the interactions and the effect of vanishing power at maximumefficiency can be investigated by the students. A software application for the creationand investigation of simple endoreversible models has been developed by me [78]. Itcan be used to easily compare different heat transfer laws or investigate the influenceof heat leaks on the performance of heat engines.

Another aspect in the investigation of endoreversible systems, especially in moreengineering related publications, is the exergy of systems. Sorin et al. [72] discussedexergy destruction in heat engines at maximum power.

10

1.1 Document Structure

Based on the CA engine, Paez-Hernandez et al. introduced a number of variations inthe design of the heat engine. In [65], they exchanged the working medium of the CAengine, which was originally an ideal gas, with a van der Waals gas. In [64] they in-corporated internal irreversibilities in the Carnot engine, which, however, contradictsthe initial idea of keeping the internal process of the heat engine reversible.

As one leaves the well investigated field of heat engines, the number of publicationsin the field of endoreversible thermodynamics gets very small. The aim of this workis thus to provide an endoreversible formalism that can be used to model a widerange of thermodynamic systems. When real systems are modeled in detail, oftenfluxes of gases or liquids have to be considered. Thus, additional to the well describedendoreversible theory [45, 47] for modeling heat engines, I want to present extensionsto incorporate multi-extensity fluxes. These can be used to include matter fluxes inthe models.

Another aspect, that has been rarely discussed in the scope of endoreversible thermo-dynamics, is modeling chemical reactions. There have already been some attempts inthis area, but usually only for examples with special restrictions. In [40, 47] and [27],for instance, chemical engines without reaction processes are presented. De Vos [27]also investigated reversible reaction processes. In this thesis, however, I want to in-troduce a more general treatment of chemical engines and reactions including both,reversible and irreversible reaction processes.

A further remark has to be made about the results, which can be obtained fromendoreversible thermodynamics. Usually, the models are based on phenomenologicaltransport laws, e.g. for heat or particle transfer. Even very detailed endoreversiblemodels are only an approximation of the real system, because it is generally not possi-ble to include each and every source of dissipation. Hence, endoreversible modeling isused to find realistic bounds for processes, e.g. maximum efficiency or power output,rather than exact descriptions of the state of a system. In providing realistic boundsto thermodynamical processes, it is a powerful tool, as the obtained performancemeasures are often very close to experimental data (cf. e.g. [25] and [33]).

1.1 Document Structure

In chapter 2, first I recall the formalism of endoreversible thermodynamics followingthe review article of Hoffmann et al. [47]. Based on this structure of endoreversiblemodels with reservoirs and engines that are connected with (irreversible) interac-tions, I present an extension to endoreversible thermodynamics for interactions withmultiple extensities. This extension is useful in the treatment of matter flows, whichis demonstrated with some examples for ideal gases and van der Waals gases. In

11

1 Introduction

the examples I comment on the special attention, which is needed in the consider-ation of volume fluxes. Usually, volume and particle fluxes are separated. Finally,the Joule-Thomson effect is modeled using the presented endoreversible formalism.This example demonstrates, how the features of this processes, like constant total en-thalpy, are elegantly obtained from the endoreversible model. It is an example of howendoreversible thermodynamics may help understanding thermodynamic processes.Furthermore, such gas relaxation processes can be included in the fuel cell model inchapter 4.

In chapter 3 I explain how to include chemical engines with and without reactionprocesses in endoreversible models. I review the few attempts of modeling chemi-cal engines in endoreversible thermodynamics and their limitations. Additionally, Ipresent my own, more general approach to (electro-)chemical processes, which coversreversible as well as irreversible reaction processes. Therefore, a new subsystem forendoreversible models is introduced – the so called reactor.

In chapter 4 I use the extensions for multi-extensity fluxes and chemical reactions,which are presented in chapters 2 and 3, to develop an endoreversible model for poly-mer electrolyte membrane fuel cells. Starting from a very basic reversible model, amore detailed model is developed, which includes internal irreversibilities of a fuelcell. The resistance of the membrane, through which protons are exchanged betweenanode and cathode of the fuel cell, is considered as well as activation losses at the elec-trodes. With this model the power output, efficiency and cell voltage are calculated.It is shown, that the results obtained with the endoreversible model qualitatively andquantitatively resemble experimental data.

Finally, in chapter 5 I summarize the results of this thesis and give an outlook onfurther applications of the presented extensions to the endoreversible formalism.

12

2 Endoreversible Thermodynamics

Endoreversible thermodynamics is a formalism to model irreversible systems and pro-cesses. The systems are separated into subsystems and interactions between them.The subsystems exchange energy and extensive quantities (or short extensities). Ex-tensities are thermodynamic quantities that are proportional to the system size, e.g.entropy, volume, particle number and so on. They are carriers for the energy. Thismeans, energy cannot be transferred without any extensity. If there is, for instance,a heat flux, there has to be a flux of entropy as well. Thus, each interaction containsat least two fluxes – an energy flux and an extensity flux. Energy also scales with thesystem size and could be called an extensity, too. However, it is treated separatelyin endoreversible thermodynamics as we will see in the description below.

All the irreversibilities and therefore the entropy production are limited to the in-teractions. The subsystems are all systems in equilibrium or systems with reversibleprocesses only. Hence the name – endoreversible, which means reversible inside.

In this chapter I will briefly explain the endoreversible formalism as it has been usedin the past decades. Then, I will present the extensions of the formalism to includemulti-extensity fluxes and discuss a number of examples, in which ideal gases andvan der Waals gases are transferred. These extensions are needed, e.g. when systemswith matter fluxes are modeled. The matter fluxes usually contain more than oneextensity and the fluxes are coupled through the state of the gas or fluid, which ismodeled. Including matter fluxes in endoreversible models is also creating a linkbetween a basic physical description and models in engineering, which are usuallycloser to real applications. Førland et al. [34], for instance, discussed heat, mass andcharge fluxes in the scope of Onsager’s theory of fluxes and forces.

2.1 General Formalism

A detailed description of endoreversible thermodynamics with a number of exampleswas given by Hoffmann et al. [47]. Following this publication, I will give a briefmathematical description of the classical endoreversible formalism.

In the following, I will denote extensive quantities of subsystem i with Xαi , where α

indicates the different extensities. The energy of subsystem i is Ei(Xαi ), i.e. it can

13


finite infinite steady state cyclic reversible irreversible

EnginesReservoirs Interactions

Endoreversible Thermodynamics

Figure 2.1: Elements of endoreversible models

be expressed as function of its extensities.

Each extensity has a conjugate intensity (or intensive quantity), which can be calcu-lated as the partial derivative of the energy,

Y αi =

∂Ei(Xαi )

∂Xαi

. (2.1)

If, for example, the entropy S is the extensity, the corresponding intensity is thetemperature T = ∂E(S,... )

∂S. Using the notation from equation 2.1 the Gibbs equation

becomes [71]

dEi =∑α

Y αi dXα

i . (2.2)

Thus, each flux of extensity, Jαi = Xαi , carries an accompanying flux of energy

Iαi = Y αi J

αi . (2.3)

Figure 2.1 shows a scheme of the elements of endoreversible models, all of which Iwill discuss in detail in the following sections.

2.1.1 Subsystems

Subsystems are reversible systems that exchange energy and extensities through con-tact points. The contact points are characterized by intensive quantities, e.g. tem-perature, pressure or chemical potential.

There are two types of subsystems – reservoirs and engines. Reservoirs are energyand extensity storage elements. Engines, however, transfer energy between differentcarriers, i.e. extensities.

14

2.1 General Formalism

Reservoirs

Reservoirs have at least one contact point to exchange energy and an extensity. Theycan be divided into finite and infinite reservoirs. Infinite reservoirs are characterizedby their constant intensities. A heat bath with constant temperature, for instance,would be represented by an infinite reservoir. The temperature of such a bath doesnot change when energy and entropy are transferred from or to the reservoir.

Finite reservoirs behave different. Their state is characterized by their energy, whichis a function of all its extensities:

Ei = Ei(Xαi ). (2.4)

Every flux to or from the reservoir changes its extensity and energy as well as theintensity. In the heat bath example, the energy of the reservoir and its temperaturewould increase with an influx of heat and entropy and decrease with an efflux.

Engines

While reservoirs, usually, have only one contact point, engines need at least two,but normally have more than two. Within engines energy and extensities cannot bestored permanently as in reservoirs. Also, they cannot be produced or destroyed.That means, all the energy and extensities entering the engine have to leave it atother contact points. The different contact points with influxes of extensities arenumbered by k and those with effluxes are numbered by l.

The balance equations for extensities and energy for a steady state engine are

0 =∑k

Jαi,k −∑l

Jαi,l for all α and (2.5)

0 =∑α,k

Iαi,k −∑α,l

Iαi,l =∑α,k

Y αi,kJ

αi,k −

∑α,l

Y αi,lJ

αi,l. (2.6)

In case of cyclic engines, the energy and extensity fluxes have to add up to zero in acomplete cycle. Thus, the balance equations are

0 =

tcycle∫0

(∑k

Jαi,k −∑l

Jαi,l

)dt for all α and (2.7)

0 =

tcycle∫0

(∑α,k

Iαi,k −∑α,l

Iαi,l

)dt =

tcycle∫0

(∑α,k

Y αi,kJ

αi,k −

∑α,l

Y αi,lJ

αi,l

)dt. (2.8)

15


Often the balance equations are written as sums over all contact points and it canbe seen from its sign whether it is an influx or efflux. In this work, however, thefluxes shall all be defined as positive. In the balance equations the effluxes arethen subtracted from the influxes. The reason for this will be discussed further indescription of interactions below.

2.1.2 Interactions

Interactions are connections between contact points of two subsystems, through whichenergy and extensities are exchanged. Each contact point belongs to exactly oneinteraction. Interactions are characterized by the extensities, which are transferredand the intensities at the connected contact points. They may either be reversibleor irreversible. In the reversible case only contact points for the exchanged extensityare needed. In the irreversible case, however, entropy is produced. To deposit thisentropy and the energy it carries, an additional contact is needed. An exception areinteractions, where entropy is the transferred extensity. In this case, the irreversiblyproduced entropy can be disposed through the interaction itself and no additionalcontact is needed.

Interactions are either reversible or irreversible and can be one of the following cases:

• The intensities at both contact points i and j are equal, i.e. Y αi,r = Y α

j,r. In thiscase the fluxes of the interaction are not limited, i.e. the energy and extensitiesare completely and instantaneously transferred.

• The transfer is defined by a transport law either for the energy

Iαi,r = Iαi,r({Y αω }, {Xα

ω}, am) (2.9)

or for the extensity

Jαi,r = Jαi,r({Y αω }, {Xα

ω}, am) (2.10)

as functions of the intensities, extensities (in case of interactions at finite reser-voirs) and additional external parameters am.

An example for a transport law would be a heat flux proportional to a temperaturedifference between the connected subsystems,

I = Q = k(Tj − Ti), (2.11)

where k is the heat conductance. The corresponding extensity fluxes, i.e. in this case

16

2.2 Extension for Multi-Extensity Fluxes

entropy fluxes, would be

JS,i =k(Tj − Ti)

Tiand (2.12)

JS,j =k(Tj − Ti)

Tj(2.13)

The distinction between in- and effluxes in the description of engines above wasmade to obtain a consistent formalism in the combination of engines and interactions.When setting up a model, we give each interaction a direction, i.e. we decide thata flux goes from subsystem i to subsystem j. Then the fluxes have the same signthroughout the interaction but it is an efflux at subsystem i while it is an influx atsubsystem j. One may argue that the direction of a flux may not be clear whencreating the model. However, this is not a problem, as the solution of the balanceequations will result in negative fluxes, if the direction of an interaction is differentfrom that of the actual flux.

If the fluxes at an engine are not separated into influxes and effluxes, we encountera change in sign of the flux within the interaction. The flux would be negative atthe subsystem, where it is an efflux, and positive at the subsystem, where it is aninflux. In many publications this version is still used, as often only single enginesare investigated or the convention that fluxes leaving a system are negative is notapplied to reservoirs.

However, in this thesis all the interactions are assigned a direction, such that alsocomplex models with a number of interactions and subsystems can be set up consis-tently.

Figure 2.2 shows, how the reservoirs, engines and interactions are drawn in the fol-lowing. Reservoirs are represented by squares or rectangles, engines are representedby circles and the interactions between them are represented by arrows. Straightarrows are used for reversible interactions while curly arrows are used for irreversibleinteractions.


Usually, endoreversible models contain only interactions with one extensity flux andthe corresponding energy flux. In many real systems, however, we do not find singleextensity fluxes as for instance in heat engines like the Novikov engine [59, 60] orCurzon-Ahlborn engine [25]. Often fluxes can be found that contain more than oneextensity. We might, for instance, observe a tube with a flux of some kind of gasor liquid. The flux may consist of particles at a certain temperature and pressure

17


(a) Reservoir (b) Engine (c) Interaction (d) Contact point

Figure 2.2: Graphical representation of the elements of endoreversible models. Reser-voirs are drawn as rectangles and engines as circles. Straight arrows areused for reversible interactions while curly arrows are used for irreversibleinteractions. The color of the arrow indicates the transferred flux. Redis used for the energy flux while all other colors are used for extensities.Green, for instance, represents entropy fluxes and orange represents vol-ume fluxes. Contact points are the connections between subsystems andinteractions.

and thus combines a number of extensities, namely particle number, entropy andvolume. The particles may even be charged, which would add another extensity tothe flux. The extensities are usually coupled. If we assume, for instance, a flux ofidentically charged particles, i.e. all particles have the same mass, charge, and so on,than particle number and charge flux (electric current) are connected by

Iel = z F J, (2.14)

where Iel is the electric current, z is the charge number, F is the Farraday constantand J is the particle flux. It is usually not convenient to treat those fluxes separatelywith chemical potential µ and electrical potential ϕ as the corresponding intensivequantities. Instead the contact points at the subsystems can be characterized by anelectrochemical potential µ = µ + ϕ z F. With this, the energy flux carried by thecharged particle flux is

µ J = (µ + ϕ z F) J = µJ + ϕ Iel. (2.15)

As we will see below, volume and entropy flux can often be written as functions ofthe particle flux. For ideal and van der Waals gases the formulas are given in sections2.3.1 and 2.4.

Note, however, that not necessarily all extensities are transferred in a matter flux. Itmay, for instance, be the case that particles and entropy are transferred but volumeis not. For a detailed discussion of different combinations of extensities in a flux ofan ideal gas see section 2.3.5.

From equation 2.3 we know that the energy flux can be calculated as the product ofthe extensity fluxes with its intensities. If we have, for instance, an interaction with

18


12 14 16 18 20 22 24 260.00

0.05

0.10

0.15

0.20

0.25

0.30

V�@10-3m

3D

p�@M

PaD 1

2

3

4

(a) pV diagram

199 200 201 202 203 204 205 2060

100

200

300

400

500

S�@J�KD

T�@K

D 1 2

3

4

(b) TS diagram

Figure 2.3: (a) pV and (b) TS diagram for different processes between two states(1 and 2) of an ideal gas. The blue curves (state 1 to state 2) showan isothermal expansion process. The red curves show a process withan isochoric increase in entropy (state 1 to state 3) that is followed byan isenthalpic expansion (state 3 to state 2). The green curves show aprocess with an isenthalpic expansion (state 1 to state 4) that is followedby an isochoric increase in entropy (state 4 to state 2).

entropy and volume flux that enters subsystem i at temperature Ti and pressure pi,we can write for the carried energy flux

Ii = Ti(Si, Vi)JS,i − pi(Si, Vi)JV,i. (2.16)

Note that temperature and pressure at the contact point, i.e. of the subsystem, arefunctions of the extensities and may change with the flux.

Often, we are interested in the total energy, which is transferred through an inter-action during a specified time or till a specified state of the system is reached. In acyclic process, for instance, we may be interested in the total energy, which is trans-ferred at a contact point in one cycle. Therefore, we need to integrate the flux atthis contact point over the cycle time.

If we are only interested in the total change in energy, it is sufficient to know theamounts of entropy, ∆S, and volume, ∆V , which are exchanged in a cycle. Then,we can integrate first over the entropy flux, while the volume is kept constant, andafterwards over the volume flux, while the entropy is kept constant. During thisprocess the temperature and pressure change as they are functions of S and V . ThepV and TS diagram of this process is shown in figure 2.3 (red line). We can alsointegrate over the volume first and over the entropy afterwards (see the green curvesin figure 2.3). Since dEi according to the Gibbs equation 2.2 is a total differentialand therefore does not depend on the path over which we integrate, the total energy

19


carried by the flux is the same in both cases, i.e.

∆E =

tcycle∫0

Ii dt =

∆E∫0

dEi (2.17)

=

∆S∫0

Ti(Si, Vi) dSi −∆V∫0

pi(Si + ∆S, Vi) dVi (2.18)

= −∆V∫0

pi(Si, Vi) dVi +

∆S∫0

Ti(Si, Vi + ∆V ) dSi. (2.19)

If we are, however, interested in the amounts of energy, which are carried by a singleextensity in the flux, e.g. the heat or work, the integration path is important. Thearea under the temperature curve in the TS diagram equals the heat that is exchangedduring a process, i.e. Q =

∫T dS. Accordingly, the work equals the negative area

under the pressure curve in the pV diagram, i.e. W = −∫p dV . Thus, the amount

of heat and work, which is exchanged during a process depends on the temperatureand pressure profile.

Often, one can find isothermal processes (see the blue curves in figure 2.3). In sucha process the temperature at the contact point does not depend on the change ofextensities in the subsystem. This can only be achieved, if entropy and volume arechanged simultaneously in the subsystem. It is then necessary to express the pressureas a function of the constant temperature and the volume instead of entropy andvolume, such that the integration path is an isotherm in the pV diagram. Then, wecan write for the energy

∆E =

∆S∫0

Ti dSi −∆V∫0

pi(Ti, Vi) dVi (2.20)

= Ti ∆S −∆V∫0

pi(Ti, Vi) dVi. (2.21)

In the following sections I will discuss some common subsystems and processes forideal and van der Waals gases. We will see, that depending on the systems and pro-cesses which are modeled, the intensities are needed as functions of different quanti-ties. The ideal gas was chosen, because it is very simple and the calculations of thepresented processes and systems can be done analytically. The van der Waals gas,as an example of a more realistic gas model, is used to demonstrate that the modelscan easily be exchanged. In addition, some effects can only be investigated with agas model which includes particle volumes and interactions.

20

2.3 Ideal Gas

2.3 Ideal Gas

Ideal gases consist of non-interacting particles, which have no volume, i.e. they arepoint masses. They are described using the equation of state

p V = nRT (2.22)

and the internal energy

E(T, n) = cV nRT, (2.23)

which is a function of the amount of substance n and temperature T only and doesnot depend on the volume of the gas. CV = cV nR is the heat capacity of the gaswith cV = f/2. f is the number of degrees of freedom of the molecules in the gas.Particles in mono-atomic gases have only three translational degrees of freedom, i.e.cV = 3/2. Diatomic gases have two additional degrees of freedom for rotation aroundaxes perpendicular to the connection line of the atoms. They also can vibrate alongthe connection line of the atoms, but around standard temperature these degrees offreedom can be neglected. Thus, cV = 5/2 is often used for diatomic gases at lowertemperatures. At high temperatures, the two degrees of freedom for vibration haveto be added resulting in cV = 7/2 [71].

2.3.1 Gibbs Equation and Principal Equation of State

For our endoreversible models we often need a description of the ideal gas withfunctions which depend on the extensities only. It is especially useful to have anexpression for the energy of an reservoir in this form, as it changes according to in-and effluxes of extensities.

If we consider an ideal gas with one type of particles, the extensive quantities areentropy S, volume V and amount of substance n (or alternatively the particle num-ber N = NA n with NA being Avogadro’s constant). The corresponding intensivequantities are temperature T , pressure p and chemical potential µ. Thus, accordingto equation 2.2 the Gibbs equation becomes

dE =∑α

Y α dXα = T dS − p dV + µ dn. (2.24)

The internal energy as a function of the extensities only can be derived from theentropy of an ideal gas, which is (see e.g. [37])

S(T, V, n) = cV nR lnT + nR lnV

n+ nRσ, (2.25)

21


where σ is a constant with respect to T , V and n. If we know the entropy of areference state S0(T0, V0, n0), we can calculate the difference S/n−S0/n0, which doesnot contain σ. Thus, we can write for the entropy

S(T, V, n) = cV nR lnT

T0

+ nR lnV

V0

− nR lnn

n0

+ nS0

n0

. (2.26)

Data for the entropy at standard temperature and pressure, S0, can for instancebe obtained from the online database of the National Institute of Standards andTechnology of the U.S. Department of Commerce [1].

If we solve this equation for T , we get

T (S, V, n) =

(V0 T

cV0

n0

n

Ve

(SnR− S0n0 R

)) 1cV

. (2.27)

Inserting this into equation 2.23, we obtain the internal energy as a function of S, Vand n:

E(S, V, n) = cV nR

(V0 T

cV0

n0

n

Ve

(SnR− S0n0 R

)) 1cV

. (2.28)

Essex and Andresen [31] labeled this equation the principle equation of state.

The intensities as functions of S, V and n are then

T (S, V, n) =

(∂E

∂S

)V,n

=

(V0 T

cV0

n0

n

Ve

(SnR− S0n0 R

)) 1cV

(2.29)

p(S, V, n) = −(∂E

∂V

)S,n

=nR

V

(V0 T

cV0

n0

n

Ve

(SnR− S0n0 R

)) 1cV

(2.30)

µ(S, V, n) =

(∂E

∂n

)S,V

=

(cV R + R− S

n

)(V0 T

cV0

n0

n

Ve

(SnR− S0n0 R

)) 1cV

(2.31)

Let us consider an endoreversible subsystem with an ideal gas. The flux at a contactpoint r shall contain dn moles of particles at chemical potential µr, temperature Trand pressure pr. These particles occupy a volume

dV =RTrpr

dn (2.32)

according to the equation of state 2.22. If we substitute V and V0 in equation 2.26by nRT

pand n0 RT0

p0, respectively, the entropy connected to the flux of dn particles

becomes

dS =

((cV + 1) R ln

TrT0

− R lnprp0

+S0

n0

)dn. (2.33)

22

2.3 Ideal Gas

In the entropy equation S0 is the entropy of n0 moles at temperature T0 and pressurep0 (such that V0 = n0 RT0

p0). Note that not all three extensities of a gas (S, V and

n) have to be exchanged through an interaction. As we will see in the following,there are interactions with only one or two extensities. We may, for instance, findinteractions with particles and entropy but no volume exchange.

For the sake of completeness, let us have a look at mixtures of ideal gases. Abovewe only considered gases with one type of particles. If we have an ideal gas withmore than one type of particles, we need to distinguish between the amounts of thedifferent substances. nl shall be the particle number of substance l. Each of thegases has its own chemical potential µl and partial pressure pl. p would be the totalpressure of the system which can also be written as the sum of the partial pressures,p =

∑l pl. Since the particles of ideal gases do not interact, they all occupy the same

volume, V = Vl. The temperature is the same for all particle types and the entropy ofthe system is the sum over the individual entropies, S =

∑l Sl. The internal energy

of the whole system is the sum of the internal energies of the different particle types.Thus, the Gibbs equation reads

dE =∑l

dEl (2.34)

=∑l

T dSl −∑l

pl dV +∑l

µl dnl (2.35)

= T dS − p dV +∑l

µl dnl. (2.36)

2.3.2 Infinite Reservoirs

In this section I will discuss mass fluxes entering or leaving infinite reservoirs. Thesubstance, which is transferred to or from the reservoir, shall be an ideal gas with onetype of particles. Per definition, in infinite reservoirs the temperature, pressure andchemical potential stay constant regardless of any energy and extensity exchange.

At the contact point of the reservoir, the intensities of a mass flux must be equal tothose of the reservoir, i.e. the flux must have the same temperature, pressure andchemical potential.

The surrounding air of a system could, for instance, be modeled as an infinite reservoirwith fixed temperature and pressure. A heat bath with constant temperature wouldalso be an example of an infinite reservoir.

23


(a)

VR = nR RTR

pR

(b)

Isur

J surV J in

S , J in

I in

Figure 2.4: Finite reservoir with constant pressure. (a) Structure of the reservoir.(b) Endoreversible model with a rectangle representing the reservoir and(curly) arrows to represent (irreversible) interactions.

2.3.3 Finite Reservoirs

When modeling finite reservoirs of ideal gases in endoreversible thermodynamics, onehas to consider a number of aspects. The design of the reservoir influences its behaviorwhen exchanging energy and extensities. It is important whether the reservoir canexchange heat with the surroundings or whether it is insulated. One also needs toknow whether it is a container with fixed walls or whether it can expand when gasis added.

In the following, I will discuss the general cases of finite reservoirs, which have eitherconstant volume (e.g. a pressure cylinder) or constant pressure (e.g. a perfectly elasticballoon). As we will see, for instance in section 2.3.5, finite reservoirs with changingvolume and pressure are possible as well. The energy of a finite reservoir is a functionof its extensities and changes due the fluxes of the connected interactions.

The reservoirs described in the following sections contain nR moles of an ideal gas attemperature TR and pressure pR. The gas shall consist of only one type of atoms ormolecules and only these particles are added or removed. Entropy and volume of thereservoir are SR and VR. Thus the internal energy of the reservoir is ER(SR, VR, nR).It changes according to the flux of gas leaving or entering the reservoir. The extensi-ties SR, VR and nR as well as the intensities TR, pR and µR may also change. Whethersome of the quantities stay constant depends on the type of reservoir and will bediscussed in the detailed description below.

Finite Reservoirs with Constant Pressure

First, I want to discuss a finite reservoir with constant pressure pR and constanttemperature TR. The example system is a cylinder with a movable piston as depictedin figure 2.4a. The piston movement shall be frictionless and the pressure inside the

24

2.3 Ideal Gas

cylinder is the same as the surrounding pressure. The same applies for the temper-ature inside and outside the cylinder. The cylinder is connected to a pipe throughwhich gas is exchanged. In the endoreversible representation (figure 2.4b), the reser-voir needs one contact point through which the gas is exchanged. The extensities atthis contact point are entropy and particles. This interaction is irreversible in mostcases, because the flux of particles is usually caused by differences in the intensities. Iwill discuss this in detail in section 2.3.5, which deals with interactions. Since the vol-ume of the cylinder can change due to the piston movement, a second contact point isneeded at the reservoir for the exchange of volume work with the surroundings. Sincethe pressures inside and outside the cylinder are the same, the volume interaction isa reversible interaction. Note, that only the change in entropy and particle numbercan be realized as a gas flux through a pipe while the volume exchange takes placewith the surroundings. We will see below, that we do not need a heat exchange withthe surroundings to keep the temperature in the reservoir constant.

The process I want to discuss is a flux of particles into the cylinder. The total amountof particles, which is transferred to the reservoir in the considered process, shall be∆nin. As the system is designed to have constant pressure and temperature, theparticle influx causes an increase in the cylinder volume. Furthermore, the particleflux is accompanied by a flux of entropy. The total change in internal energy of thereservoir is derived by integrating over the fluxes at both contact points:

∆ER =

∆Sin∫0

TR dS +

∆nin∫0

µR dn−∆V sur∫0

pR dV (2.37)

= TR ∆Sin + µR ∆nin︸︷︷︸Ein

− pR ∆V sur︸︷︷︸W sur

, (2.38)

where the superscript ‘in’ marks quantities of the gas flux and the superscript ‘sur’marks quantities, which are connected to the volume exchange with the surroundings.The chemical potential of the reservoir has to be constant, when temperature andpressure are constant, since it can be expressed as

µR(TR, pR) = −RTR log

(TcpR p0

Tcp0 pR

). (2.39)

Because the pressure in the reservoir is the same before (quantities marked with ◦)and after the influx, with the equation of state 2.22 we obtain

p◦R = pR (2.40)

n◦R RTR

V ◦R=

(n◦R + ∆nin) RTR

V ◦R + ∆V sur. (2.41)

25


Solving this for ∆V sur, we get

∆V sur = ∆ninV◦

R

n◦R. (2.42)

From constant temperature of the reservoir, i.e.

T ◦R(S◦R, V◦

R , n◦R) = TR(S◦R + ∆SR, V

◦R + ∆V sur, n◦R + ∆nin), (2.43)

we obtain with equation 2.29(V0 T

cV0

n0

n◦RV ◦R

e

(S◦Rn◦

RR− S0n0 R

)) 1cV

=

(V0 T

cV0

n0

n◦R + ∆nin

V ◦R + ∆V sure

(S◦R+∆SR

(n◦R

+∆nin) R− S0n0 R

)) 1cV

. (2.44)

Inserting equation 2.42 and solving for ∆SR gives

∆SR = ∆ninS◦R

n◦R= ∆Sin. (2.45)

This means that the change in entropy in the reservoir equals the entropy influx ofthe gas. There is no need for a heat exchange with the surroundings to keep thetemperature of the reservoir constant.

From equation 2.42 we see, that the change in volume equals the volume that isoccupied by ∆nin particles in the gas flux at the contact point with the reservoir.Thus, the total change in internal energy is the same as if all extensities of the gasflux were added at the gas contact instead of entropy and particles only. However, ifnot only the reservoir is considered but also the interaction, it is necessary to separatethe volume flux from the other fluxes. At the second contact point of the interactionthe pressure could be different. Thus, the particles would occupy a different volume.However, since volume is a conserved quantity, it cannot change within an interaction(see section 2.3.5 for a more detailed discussion of such an interaction).

The internal energy after the influx is

ER = E◦R + ∆Ein +W sur. (2.46)

The extensities of the reservoir after the influx are given by

nR = n◦R + ∆nin, (2.47)

SR = S◦R + ∆Sin and (2.48)

VR = V ◦R + ∆V sur. (2.49)

Likewise, the internal energy and extensities are reduced at constant intensities by amass flux out of the reservoir (i.e. ∆Ein < 0 in equation 2.46).

26

2.3 Ideal Gas

(a)

δQ

(b)

Ibath

JbathS J in

S , J inn

I in

Figure 2.5: Finite reservoir with constant volume and surrounding bath. (a) Struc-ture of the reservoir. (b) Endoreversible model with a rectangle repre-senting the reservoir and (curly) arrows representing (irreversible) inter-actions.

Finite Reservoirs with Constant Volume

In case of constant volume instead of constant pressure of the reservoir the situationis different. We need to distinguish systems according to their ability to exchangeheat with the surroundings.

First, I want to discuss a system, which can exchange heat with a surrounding bathwithout further limitations (i.e. unlimited heat transfer rates), such that it has aconstant temperature TR = T bath. A pressure cylinder to store gas would be anexample system. It shall not be thermally insulated. The general structure of theexample system and its representation as an endoreversible model are shown in figure2.5. Any change in temperature due to a flux of gas into or out of the cylinder iscompensated by a heat exchange with the bath.

Again, I want to discuss the process of an influx of gas into the cylinder. The reservoir,which represents the cylinder, needs two contact points. One for the gas exchange,where particles and entropy are the extensities, and one for the heat exchange with thesurroundings, where entropy is the only extensity. A single volume flux cannot occurat a reservoir with fixed volume. Quantities at the gas contact have the superscript‘in’ and quantities at the contact for the bath have the superscript ‘bath’. Thus, thechange in energy due to the gas flux is

dEin = TR dS + µR(SR, nR) dn. (2.50)

Additionally, the reservoir exchanges the heat δQbath = T bath dSbath with the bathwhen particles and entropy of the influx are transferred to the reservoir. The totalchange in internal energy then becomes

dER = dEin + δQbath. (2.51)

27


The energy and extensities of the reservoir after the influx are obtained by integratingover the fluxes, i.e.

ER = E◦R +

∫dER (2.52)

SR = S◦R +

∆Sin∫0

dS +

∆Sbath∫0

dS, (2.53)

VR = V ◦R = constant and (2.54)

nR = n◦R +

∆nin∫0

dn. (2.55)

The temperature of the reservoir stays constant due to the interaction with thesurrounding bath. Pressure and chemical potential, however, change with the influxof particles:

pR(nR) =nR RTR

VR

and (2.56)

µR(nR) = −RTR ln

(VR T

cVR n0

nR V0 TcV0

), (2.57)

where V0 is the volume of n0 moles of the gas at standard temperature T0 and standardpressure p0.

To calculate the heat exchange with the bath, we will have a look at the entropy ofthe reservoir before and after the influx and the entropy that is carried by the influxitself. Since temperature and volume of the reservoir are constant, the entropy canbe expressed as a function of the amount of substance only. Before the influx, it isn◦R and after the influx it is nR = n◦R + ∆nin. Thus, the entropies are

S◦R = cV n◦R R ln

TR

T0

+ n◦R R lnVR

V0

− n◦R R lnn◦Rn0

+ n◦RS0

n0

, (2.58)

SR = S◦R + ∆Sin + ∆Sbath (2.59)

= cV nR R lnTR

T0

+ nR R lnVR

V0

− nR R lnnR

n0

+ nRS0

n0

. (2.60)

The influx of gas has to have the same intensities as the reservoir at the contact pointduring the whole process. This means, the pressure of the influx has to change withthe changing reservoir’s pressure. With equation 2.33 we get for the entropy of the

28

2.3 Ideal Gas

influx

∆Sin =

∆nin∫0

((cV + 1) R ln

TR

T0

− R lnpR(n◦R + n)

p0

+S0

n0

)dn (2.61)

=cV ∆nin R lnTR

T0

+ ∆nin R lnVR

V0

−∆nin R lnnR

n0

+ ∆nin S0

n0

− n◦R R lnnR

n◦R+ ∆nin R. (2.62)

From this and equations 2.58 till 2.60, we obtain for the entropy and heat exchangewith the bath

∆Sbath =−∆nin R and thus (2.63)

Qbath =−∆nin RTR. (2.64)

If the reservoir is completely insulated, i.e. it cannot exchange heat with the surround-ings, its temperature is not constant anymore. In this case, temperature, pressureand chemical potential change according to mass fluxes into and out of the reservoir.As before, only the part of energy of the mass flux that is carried by entropy andamount of substance can be transferred to the reservoir. Therefore, such a reservoirhas only one contact point for the gas flux. When gas is added to the reservoir, theinternal energy changes according to

ER = E◦R +

∫dEin with (2.65)

dEin = TR(SR, nR) dS + µR(SR, nR) dn. (2.66)

The corresponding changes in extensities are

SR = S◦R +

∆Sin∫0

dS, (2.67)

VR = V ◦R = constant and (2.68)

nR = n◦R +

∆nin∫0

dn, (2.69)

i.e. the entropy and particles of the flux are completely added, while the volume doesnot change. The intensities of the reservoir change and can be determined usingequations 2.29 to 2.31. Since the mass flux has to have the intensities of the reservoirat the contact point, the entropy flux accompanying the particle flux changes as well.It can be calculated as

dSin =

((cV + 1) R ln

TR(SR, nR)

T0

− R lnpR(SR, nR)

p0

+S0

n0

)dn (2.70)

29


Energy fluxEntropy fluxEntropy fluxVolume flux

43

2

1Qcomp

3

Scomp

W4

V

Ein1

Sin

nin

Eout2

Sout

nout

Figure 2.6: Reversible engine to model an isothermal compression or expansion of anideal gas. Red arrows are energy fluxes, green arrows are entropy fluxes,orange arrows are volume fluxes and blue arrows are particle fluxes. Thecontact points are labeled with the numbers inside the circle.

2.3.4 Reversible Engines

Engines in endoreversible thermodynamics are defined as reversible subsystems withinput and output streams. For each extensity and the energy the sum over all fluxeshas to equal zero. This means, there is no entropy production within the engine.

Engines may, for instance, be used to mix different ideal gases or to expand orcompress them.

Let us have a look at the (quasi) reversible, isothermal compression of a gas withone type of particles. It shall be realized in a cylinder with a piston where the gasis sucked in at constant pressure pin through an opening while the piston moves toincrease the volume inside the cylinder. Afterwards, the inlet is closed and the gasis compressed due to the piston movement till the pressure in the gas equals pout.Finally, an outlet is opened and the gas is pushed out of the cylinder at constantpressure pout while the piston moves to decrease the volume again. The process shallbe sufficiently slow, such that the gas has time to equilibrate and the process can beseen as a series of equilibria.

The process shall be represented by a cyclic operating engine. The cycle has threebranches – inlet, compression and outlet of the gas. In the following, quantitiesare assigned to branches with the superscripts ‘in’, ‘out’ and ‘comp’. The engineneeds contacts for the flux of the gas into (subscript 1) and out (subscript 2) of it.Additionally, a contact for the heat exchange (subscript 3) and a contact for thevolume exchange (subscript 4) are needed. The endoreversible model is depicted infigure 2.6.

30

2.3 Ideal Gas

As in the system described in section 2.3.3, through the gas contacts only particles andentropy can be exchanged. The volume exchange takes place at a different contact,which represents the piston movement and the connected volume work. Therefore,the energies, which are exchanged in one cycle through the contacts for the gas inletand outlet are

Ein1 =

Sin1∫

0

T in1 dS +

nin1∫

0

µin1 dn and (2.71)

Eout2 =

Sout2∫

0

T out2 dS +

nout2∫

0

µout2 dn. (2.72)

Since the whole process shall be isothermal, the temperatures of influx and efflux areequal, i.e. T in

1 = T out2 = T . In the same branches of the cycle, in which the in- and

efflux of the gas take place, the piston moves as well such that the pressure is constantduring these branches. The volume work during the inlet and outlet process, whichis exchanged with the surroundings at contact 4 becomes

W in4 = −

V in4∫

0

pin4 dV = −pin

4 Vin

4 = −nin1 RT and (2.73)

W out4 = −

0∫V out

4

pout4 dV = pout

4 V out4 = nout

2 RT. (2.74)

During the compression, there are no fluxes at contacts 1 and 2, i.e. there is no gasexchange. However, volume work

W comp4 = −

V out4∫

V in4

p4(V ) dV (2.75)

and heat

Qcomp3 =

Sout2∫

Sin1

T dS (2.76)

are exchanged. In the integrals above the absolute values of V out and Sout have to beused, since these are quantities leaving the engine, which are negative by definition.V in and Sin, however, are defined as positive, because these are quantities enteringthe engine.

31


After a whole cycle, all the extensities and the energy have to be balanced within theengine. Thus, the balance equations are

0 = V in4 − V out

4 + V comp4 , (2.77)

0 = nin1 − nout

2 , (2.78)

0 = Sin1 − Sout

2 + Scomp3 and (2.79)

0 = Ein1 − Eout

2 +Qcomp3 +W in

4 +W out4 +W comp

4 . (2.80)

The energies of the ideal gas input and output can also be written as Ein = cV RT nin

and Eout = cV RT nout. From equation 2.78 we get nin

1 = nout2 and thus

Ein1 = Eout

2 and (2.81)

W out4 = nin

1 RT = −W in4 . (2.82)

Inserting this into equation 2.80 we obtain

−Qcomp3 = W comp

4 = −V out

4∫V in

4

p4(V ) dV (2.83)

= −V out

4∫V in

4

nin1 RT

VdV (2.84)

= −nin1 RT ln

(V out

4

V in4

). (2.85)

It means that the internal energy of the gas is constant. The work done to the gasduring the compression is compensated by the exchanged heat.

Since the (very slow) isothermal compression of an ideal gas can be treated as acompletely reversible process, it is a standard example in classical thermodynamics(see for instance [41, p. 30], [56, p. 108] or [58, p. 180]). Since the change in internalenergy of an ideal gas is zero for constant temperature and particle number, from thefirst law of thermodynamics,

∆E = Q+W, (2.86)

one gets

0 = ∆E = Q+W and therefore (2.87)

−Q = W = −V out∫V in

p dV = −nin1 RT ln

(V out

4

V in4

). (2.88)

32

2.3 Ideal Gas

This is the same result as derived from the endoreversible engine above.

The classical calculations for this process may seem easier than setting up an en-doreversible model with an engine and all the interactions and balance equations. Inthe classical calculations, the transfer of gas into and out of the cylinder is usuallynot considered. However, when modeling real systems, a mere compression processis not of interest but it may occur as a part in a more complex process. Then, theinlet and outlet of the gas is necessary to connect the engine to other subsystems inthe model.

Combining Gas Fluxes

Combining two or more gas fluxes is a process that often occurs in real systems. Inan endoreversible model engines can be used to represent mixing. For the discussionof such processes we need to distinguish engines where fluxes of different gases aremixed and engines where fluxes of the same gas are mixed. The mixing behavior andthe conditions for reversibility are different in these two cases.

The mixing process shall take place at constant temperature. All the gas streamshave to enter the engine at this temperature in order to avoid entropy production. Ifthe streams would enter the engine at different temperatures, heat exchange betweenthe gases would occur and therefore entropy production.

Such a mixing process could be accomplished as a cyclic process in cylinder with amovable piston. In the first branch the different gases are sucked into cylinder, thentheir volume may be changed, i.e. they may be compressed or expanded, and finallythe mixed gas is let out of the engine.

At the contact points of the engine, the pressures of the incoming gas fluxes need tobe equal to the partial pressures of that gases, which are already in the engine. Thepressures of different particle types can differ. The gases in the output of the engineshall be all the gases of the input. The total pressure in the engine shall be pin

total inthe first branch of the cycle and pout

total in the last branch. In both branches it is thesum over the partial pressures, i.e.

pintotal =

∑l

pinl and pout

total =∑l

poutl , (2.89)

where l indicates the different substances.

The general balance equations for a reversible engine, which is used to mix different

33


gases, are

Volume: 0 =

∫J inV dt−

∫JoutV dt+

∫JchangeV dt, (2.90)

Particles: 0 =

∫J inl dt−

∫Joutl dt, (2.91)

Entropy: 0 =

∫ ∑l

J inS,l dt−

∫ ∑l

JoutS,l dt+

∫q

Tdt, (2.92)

Energy: 0 =

∫ ∑l

I inl dt−

∫ ∑l

Ioutl dt+

∫q dt−

∫p Jchange

V dt, (2.93)

where q is a heat flux. While the energies and entropies of all fluxes are summed upin one equation (see equations 2.93 and 2.92), the particles of each of the differentgases have to be balanced individually in separate balance equations. Whether thereis a heat flux and a change in volume between the in- and efflux of the gas, dependson the actual process. If, for instance, gases are mixed in a vessel with constantvolume, there are no volume fluxes at all. If the mixing process is combined with anisothermal expansion or compression, the term Jchange

V for the volume change and acontact for the heat exchange is needed.

When we have an engine which combines several influxes of the same type of particlesinto one efflux, all the influxes have to be in the same state, i.e. with the sametemperature, pressure and chemical potential. If they were combined in differentstates, an irreversible process would take place until the gas is in equilibrium. Incontrast to distinguishable particles, which can occupy the same volume withoutinteraction, for one type of particles the volumes of the influxes have to be added.The balance equations are then

Volume: 0 =

∫ ∑r

J inV,r dt−

∫JoutV dt+

∫JchangeV dt, (2.94)

Particles: 0 =

∫ ∑r

J inr dt−

∫Jout dt, (2.95)

Entropy: 0 =

∫ ∑r

J inS,r dt−

∫JoutS dt+

∫q

Tdt, (2.96)

Energy: 0 =

∫ ∑r

I inr dt−

∫Iout dt+

∫q dt−

∫p Jchange

V dt, (2.97)

where r indicates the different contact points of the influxes and q is the heat fluxagain.

34

2.3 Ideal Gas

(a)

V1 V2

(b)

1JS, JV , J

I

2

Figure 2.7: (a) A box with an ideal gas at temperature T , pressure p and chemicalpotential µ. Volumes V1 and V2 are separated by a permeable membrane(black dashed line). This separation is moved to a different position(gray dashed line). (b) Endoreversible representation of the system. Therectangles represent the reservoirs, which are connected by straight arrowsrepresenting the reversible interaction.

2.3.5 Interactions

Now, let us have a closer look at interactions with coupled fluxes. For an ideal gasthe considered extensities are entropy, particles and volume. As in the case of singleextensity fluxes, they can either be reversible or irreversible. In the following, threedifferent scenarios will be discussed. In the first scenario all three extensities aretransferred through the interaction. In the second scenario only volume and entropyare transferred but no particles and in the last scenario particles and entropy aretransferred but no volume.

In all the scenarios the interaction connects subsystem 1 with subsystem 2. Theintensities at the two contact points are the pressures p1 and p2, the temperatures T1

and T2 and the chemical potentials µ1 and µ2.

Interactions with Volume, Particle and Entropy Flux

Interactions with volume, particle and entropy flux are rarely found in reality but wecan set up a thought experiment. Let us assume we have a closed box filled with anideal gas as shown in figure 2.7a. Inside the box there is a permeable membrane thatseparates volume 1 from volume 2. In both regions the intensive quantities are thesame, i.e. T1 = T2, p1 = p2 and µ1 = µ2. Now the permeable wall is moved such thatvolume 1 is increased while volume 2 is decreased. This process shall be frictionlessand the intensities in both subsystems shall not change. An analog process wouldbe to remove the separation between the subsystems and to insert it again at adifferent position. Such a process has no effect on the total energy and extensities of

35


(a)

V1 V2

(b)

1

JV , JS

I2

T1 = T2 = constant

Figure 2.8: (a) A box with an ideal gas which is separated into volumes V1 and V2.The separation is moved, e.g. due to pressure differences in the two re-gions. (b) The endoreversible model of this system contains two finitereservoirs which irreversibly exchange volume, entropy and the corre-sponding energy with each other. The arrow representing the energyhas two arrow heads, because the heat and volume work flux have dif-ferent directions. The sum over both, i.e. the net energy flux, is zero forthis process.

the whole system but in the two subsystems volume, entropy, particle number andinternal energy have changed.

In an endoreversible model (see figure 2.7b) this process can be represented by twofinite reservoirs with constant pressure. The reservoirs are connected with an inter-action through which entropy, volume, particle number and energy are exchanged.The whole process is reversible. There is no entropy production and the extensityfluxes leaving subsystem 1 are completely transferred to subsystem 2, i.e.

J1 = J2 = J, (2.98)

JV,1 = JV,2 = JV , (2.99)

JS,1 = JS,2 = JS. (2.100)

Thus, the energy flux leaving subsystem 1 is completely transferred to subsystem 2:

I1 = T JS − p JV + µJ = I2. (2.101)

Interactions with Volume and Entropy Flux

Interactions, where volume and entropy are the exchanged extensities, occur whentwo subsystems are directly connected by a movable, non-insulating wall or mem-brane. An example system is a balloon with its surroundings. As the balloon ex-pands, i.e. its volume increases, the surrounding’s volume decreases and the gas insideand outside of the balloon exchange heat such that the temperature is the same at alltimes. Another example is shown in figure 2.8a. It is a thermally insulated box with

36

2.3 Ideal Gas

constant total volume that is separated into two volumes V1 and V2 by a movablewall. The separating wall is not insulating, i.e. heat can be exchanged between thesubsystems. In this system the temperature shall be the same in both volumes butthe initial pressure shall be bigger in subsystem 1, i.e. T1 = T2 and p◦1 > p◦2. Dueto the insulation of the box and its fixed total volume, there is no heat and workexchange with the surroundings. The total internal energy,

Etotal = E1 + E2 (2.102)

= cV n1 RT + cV n1 RT (2.103)

= cV (n1 + n2) RT, (2.104)

is constant. Since the particle number in both subsystems is constant as well, fromequation 2.104 we obtain that the temperature in the system stays constant. Thus,the internal energy of each of the subsystems is also constant.

The endoreversible model of this system is depicted in figure 2.8b. It consists oftwo finite reservoirs representing the two volumes and an interaction between them,through which volume, entropy and the corresponding energy are exchanged. Bothsubsystems are in equilibrium but the total system is not because of the differencein intensities.

The pressure difference causes an expansion of subsystem 1 and a compression ofsubsystem 2. At the same time heat is exchanged between the subsystems, such thatthe temperature stays constant. Both, the volume and entropy flux of the interactionare directed from subsystem 2 to subsystem 1, when p1 > p2. Particles cannotbe transferred because of the separating wall. Since the volume flux is conservedwithin the interaction, we obtain JV,1 = JV,2 = JV . This is quite intuitive, since thetotal volume is constant and therefore, any volume change of subsystem 1 has to becompensated by a volume change of subsystem 2. Because the total energy of thesystem is constant, the energy flux is conserved within the interaction, i.e.

I1 = I2 (2.105)

T JS,1 − p1 JV,1 = T JS,2 − p2 JV,2 (2.106)

(p1 − p2) JV = T (JS,1 − JS,2). (2.107)

Solving for JS,1 gives

JS,1 = JS,2 +p1 − p2

TJV = JS,2 + σ, (2.108)

where σ is the entropy production.

Since the change of internal energy in each of the subsystems is zero, the fluxes I1

and I2 equal zero. This means, that the work flux, which is exchanged between the

37


subsystems is compensated by a heat flux in the opposite direction. From this, weobtain for the entropy fluxes:

0 = TJS,1 − p1(V1)JV → JS,1 =p1(V1)

TJV , (2.109)

0 = TJS,2 − p2(V2)JV → JS,2 =p2(V2)

TJV . (2.110)

The pressures in subsystems 1 and 2 can be expressed as functions of the volumes,i.e.

p1(V1) =n1 RT

V1

and p2(V2) =n2 RT

V2

. (2.111)

As the process takes place, the pressure difference gets smaller till it reaches zeroat equilibrium. If V ◦1 and V ◦2 are the initial volumes, the total volume exchange∆V between the subsystems till equal pressures are reached, i.e. p1(V ◦1 + ∆V ) =p2(V ◦2 −∆V ), is

∆V =n1 V

◦2 − n2 V

◦1

n1 + n2

. (2.112)

Thus, we can calculate the total entropy that is transferred at the two contact points,i.e. the change in entropy in subsystems 1 and 2, by integrating over the fluxes.

∆S1 =

∫JS,1 dt =

∫p1(V1)

TJV dt

=

V ◦1 +∆V∫V ◦

1

n1 R

V1

dV1

= n1 R lnV ◦1 + ∆V

V ◦1

= n1 R ln

(n1

n1 + n2

V ◦1 + V ◦2V ◦1

)and (2.113)

∆S2 = n2 R lnV ◦2 −∆V

V ◦2

= n2 R ln

(n2

n1 + n2

V ◦1 + V ◦2V ◦2

). (2.114)

If entropy is transferred from subsystem 2 to subsystem 1, ∆S1 > 0 and ∆S2 < 0.The total entropy production within the interaction then is

σtot = ∆S1 + ∆S2 (2.115)

= (n1 + n2) R ln

(V ◦1 + V ◦2n1 + n2

)+ n1 R ln

(n1

V ◦1

)+ n2 R ln

(n2

V ◦2

). (2.116)

38

2.3 Ideal Gas

0

5

10

15

20

25

t

V�@1

0-

3m

3D

V1

V2

(a) Volume

0

50

100

150

200

t

p�@k

PaD

p1

p2

(b) Pressure

0.0

0.5

1.0

1.5

t

Σto

t�@J�KD

(c) Total entropy production

Figure 2.9: (a) Volume and (b) pressure of the two subsystems, which exchange vol-ume according to the transport law in equation 2.117. (c) Accumulatedentropy generation in the interaction between the two subsystems.

If we are not only interested in the final state, when the pressures p1 and p2 are equalbut in the process, we need a transport law for the volume flux. The volume fluxcould, for instance, be proportional to the pressure difference in the subsystems, i.e.

JV = h(p1 − p2), (2.117)

where h is a time-independent parameter. Figures 2.9a and 2.9b show how thevolumes and pressures of subsystems 1 and 2 change during this isothermal process.For the numerical calculations of the process, the system was initialized with V ◦1 =1/2V ◦2 = 8.2 × 10−3 m3 and n1 = 2n2 = 2/3 mol. Thus, for the initial pressures weget p◦1 = 4 p◦2 = 204 kPa. As we are not interested in the speed of the process and it iseasily scalable with the parameter h in equation 2.117, the time axes show no values.Figure 2.9c shows the accumulated entropy, which is irreversibly produced during theprocess. It approaches the result of equation 2.116. The entropy production rate ishigher at the beginning of the process, when the pressure difference is bigger, andgoes to zero as the pressure difference vanishes.

39


(a)

nV1 V2

(b)

1

JS, J

I

2

Figure 2.10: (a) A box with an ideal gas is separated into two regions by a permeablemembrane. Particles and heat can be exchanged through the membranebut the volumes V1 and V2 stay constant. (b) The endoreversible modelof the system consists of two finite reservoirs that exchange entropy,particles and the corresponding energy.

Interactions with Particle and Entropy Flux

Interactions with entropy and particle flux can be observed in tubes that connecttwo subsystems or in systems where two subsystems are separated by a permeablemembrane as in figure 2.10. Particles and entropy can be exchanged through themembrane but the volume of the two regions stays constant. The whole systemshall be insulated and have a fixed volume. Thus, there is no exchange of energy orextensities with the surroundings. As before, we assume that the initial pressure insubsystem 1 is bigger than that in subsystem 2 and that the temperatures are thesame, i.e. p1 > p2 and T1 = T2.

Due to the pressure difference in the subsystems, particles are transferred throughthe membrane. Additionally, entropy is transferred, such that the temperature issame in both subsystems during the whole process. Thus, the extensities of theinteraction are particle number and entropy. The particle number is conserved withinthe interaction, i.e. J1 = J2 = J .

The total internal energy, Etotal = E1 + E2, cannot change, because there is only aninteraction between the two subsystems but none with the surroundings. Since weassumed that the initial temperatures in the subsystem are equal, the total energycan be written as

Etotal = E◦1(n◦1, T◦) + E◦2(n◦2, T

◦) (2.118)

= cV n◦1 RT ◦ + cV n

◦2 RT ◦ (2.119)

= cV (n◦1 + n◦2) RT ◦, (2.120)

where T ◦ is the initial temperature and n◦1 and n◦2 are the initial particle numbers.After ∆n moles of particles are transferred from subsystem 1 to subsystem 2, theparticle numbers in the subsystems are n◦1 − ∆n and n◦2 + ∆n. The total internal

40

2.3 Ideal Gas

energy can than be written as

Etotal = E1(n◦1 −∆n, T ) + E2(n◦2 + ∆n, T ) (2.121)

= cV (n◦1 −∆n) RT + cV (n◦2 + ∆n) RT (2.122)

= cV (n◦1 + n◦2) RT. (2.123)

Comparing equations 2.120 and 2.123, we obtain that T ◦ = T . This means, thetemperature does not change during this process. Note, however, that this is aproperty of the ideal gas, because its internal energy does not depend on the volumeof the gas but on the particle number and temperature only.

Constant total internal energy means that the sum over the changes in the two sub-systems has to equal zero or that the energy flux is conserved within the interaction,i.e.

I1 = I2 (2.124)

TJS,1 + µ1J1 = TJS,2 + µ2J2 (2.125)

T (JS,2 − JS,1) = (µ1 − µ2)J. (2.126)


JS,2 = JS,1 +µ1 − µ2

TJ = JS,1 + σ. (2.127)

The difference in chemical potentials causes the entropy production in this irreversibleprocess. The system reaches equilibrium when the pressures and chemical potentialsin both subsystems are equal. From the pressure equality, p1(n◦1−∆n) = p2(n◦2+∆n),the total particle exchange between the subsystems can be obtained as

∆n =V2 n

◦1 − V1 n

◦2

V1 + V2

. (2.128)

For the particle flux between the subsystems we need a (phenomenological) transportlaw. Let us assume, the particle flux is a function of the difference in chemicalpotentials, i.e.

J = h(µ1 − µ2), (2.129)

where h shall be a constant. The entropy flux at the two contact points equals theentropy change in the corresponding reservoir, since both reservoirs have only onecontact point. In addition, both reservoirs have constant volume and temperature.Thus, the entropy flux at subsystem i can be expressed as a function of the particleflux, i.e.

JS,i =dSidt

=

(∂S(ni, Vi, T )

∂ni

)Vi,T

dnidt

(2.130)

=

(cV R ln

T

T0

+ R lnViV0

− R lnnin0

− R +S0

n0

)J. (2.131)

41


The total change in entropy in subsystem 1, when the particle number is decreasedby ∆n, can be calculated by integrating over the entropy flux or as

∆S1 = S(T, V1, n◦1 −∆n)− S(T, V1, n

◦1) (2.132)

=−∆n

(cV R ln

T

T0

+ R lnV1

V0

+S0

n0

)− (n◦1 −∆n) R ln

n◦1 −∆n

n0

+ n◦1 R lnn◦1n0

. (2.133)

For subsystem 2 the total change in entropy is

∆S2 = S(T, V2, n◦2 + ∆n)− S(T, V2, n

◦2) (2.134)

= ∆n

(cV R ln

T

T0

+ R lnV2

V0

+S0

n0

)− (n◦2 + ∆n) R ln

n◦2 + ∆n

n0

+ n◦2 R lnn◦2n0

(2.135)

Thus, the total gain in entropy, i.e. the irreversibly produced entropy σtot =∫σ dt,

reads

σtot = ∆S1 + ∆S2 (2.136)

= ∆nR lnV2

V1

+ ∆nR lnn◦1 −∆n

n◦2 + ∆n

− n◦1 R lnn◦1 −∆n

n◦1− n◦2 R ln

n◦2 + ∆n

n◦2. (2.137)

With equation 2.128 this becomes

σ =(n◦1 + n◦2) R ln

(V1 + V2

n◦1 + n◦2

)+ n◦1 R ln

(n◦1V1

)+ n◦2 R ln

(n◦2V2

), (2.138)

which is the same as for the process with volume instead of particle exchange, if theinitial states are the same (cf. equation 2.116). This means, in terms of entropyproduction it does not matter whether the pressure difference in the initial state isreduced by particle or volume exchange between the subsystems. This is reasonable,since the total system starts in the same initial state and reaches the same equilib-rium state, in which no extensities or energy are exchanged anymore between thesubsystems. Only the process to reach equilibrium is different.

2.3.6 Example: Pressure Regulator

I will now discuss a more realistic system where the above described formalism is used.Let us have a look at how to model a pressure regulator. A scheme of the pressure

42

2.3 Ideal Gas

V1 V2

p1 p2

(a) Schematic pressure regulator

11 2I1,1

JV,1 JS, J

I

21 2I2,2

JV,2

(b) Endoreversible model

Figure 2.11: (a) A pressure regulator with two volumes V1 and V2, which are separatedby a membrane (blue). The pressures in both regions are constant andp1 > p2. Because of the pressure difference, particles are pressed throughthe membrane and V1 decreases while V2 increases. (b) Endoreversiblemodel of a pressure regulator. Systems 1 and 2 are finite reservoirswith constant pressure. Particles and entropy are exchanged betweensystem 1 and 2. The volume exchange, however, takes place with thesurroundings.

regulator is shown in figure 2.11a. It consists of a porous membrane that separatesa region with high pressure p1 from a region with lower pressure p2. Particles arepressed through the membrane because of the pressure difference and there is noheat exchange with the surroundings as the system shall be insulated. This processis isenthalpic, i.e. the enthalpy H = E + p V does not change. Due to the pressuredrop, a change in temperature of the gas can be observed. This effect is called Joule-Thomson effect. The temperature change can only be found for real gases. For idealgases, however, the process is isothermal since the internal energy of an ideal gasdoes not depend on its volume.

Such a process can be found in many system, which operate with gases. In the fuelcell, which will be discussed in chapter 4, the hydrogen is usually supplied from apressure cylinder. Before it enters the fuel cell, the pressure needs to be regulated.A similar process occurs in a fuel cell, when protons diffuse through the electrolytemembrane. This will be discussed in detail in chapter 4.

An endoreversible model of this system is shown in figure 2.11b. For the two re-gions, which are separated by the porous membrane, two connected reservoirs areneeded. The extensities, which are exchanged between the reservoirs, are particles

43


(i.e. amount of substance) and entropy. Both reservoirs shall have constant pressure.To accomplish this they both need another interaction for volume exchange with thesurroundings, since the reservoirs shall be finite in the described system. Dependingon the system, which is modeled, it may also be useful to replace the reservoirs with aconstant volume reservoir or an infinite reservoir. If the pressure regulator is locatedat a high pressure cylinder, for instance, the cylinder has a constant volume and thepressure in the cylinder decreases when gas is released. On the other side of theregulator the gas may be released to the surroundings at standard pressure, whichwould be modeled best as an infinite reservoir.

Because p1 > p2, particles are transferred from reservoir 1 to reservoir 2. Since thepressures shall be constant, the particle flux out of subsystem 1 causes a volume fluxat contact point 1 with JV,1 = RT1

p1J1. The corresponding work flux to subsystem 1

is

I1,1 = −p1 JV,1 = −RT1 J1. (2.139)

In reservoir 2 the particle number is increased due to the particle influx J2 andtherefore the volume has to increase as well to obtain constant pressure, i. e. JV,2 =RT2

p2J2. The work flux from subsystem 2 is

I2,2 = −p2 JV,2 = −RT2 J2. (2.140)

Since there is no heat exchange with the surroundings, the energy flux leaving sub-system 1 with the particles and entropy is completely added to subsystem 2, i.e.

I = I1,2 = T1 JS,1 + µ1 J1 = I2,1 = T2 JS,2 + µ2 J2. (2.141)

As mentioned above, the process is isenthalpic. This means, the total enthalpy ofthe system, H = E1 + p1 V1 + E2 + p2 V2, is constant. This can also be expressedas dH = dH1 + dH2 = 0, i.e. the total change in enthalpy is zero. In general, thechange in enthalpy is dH = T dS + V dp+ µ dn. For constant pressure systems, thechange in enthalpy reduces to dH = T dS + µ dn. In our example the enthalpy fluxout of reservoir 1 is IH,1 = I1,1. Accordingly, for reservoir 2 we get IH,2 = I2,1. Thus,from the energy conservation within the interaction, equation 2.141, we obtain thatthe enthalpy is conserved as well during the process.

The transferred particles are conserved within the interaction as well, i.e. all theparticles leaving subsystem 1 are entering subsystem 2. Thus we can use J = J1 = J2

and

µi =

((cV + 1) R− Si

ni

)Ti (2.142)

44

2.4 Van der Waals Gas

(cf. equations 2.29 and 2.31) to rewrite equation 2.141:

T1 JS,1 +

((cV + 1) R− S1

n1

)T1 J = T2 JS,2 +

((cV + 1) R− S2

n2

)T2 J. (2.143)

With equation 2.33 we can reduce it to

(cV + 1) RT1 J = (cV + 1) RT2 J (2.144)

T1 = T2. (2.145)

Thus, from the properties of the interaction between reservoirs 1 and 2 we obtain,that the process is isothermal. Using T1 = T2 = T and solving equation 2.141 forJS,2, we get

JS,2 = JS,1 +µ1 − µ2

TJ (2.146)

= JS,1 + σ. (2.147)

The transport of particles through the membrane is an irreversible process and σis the entropy production when a flux of J particles is transferred. Using equation2.142 and

Sini

= (cV + 1) R lnT

T0

− R lnpip0

+S0

n0

, (2.148)

we get for the irreversibly produced entropy

σ = R lnp1

p2

J. (2.149)

With J2 = J1 = J and T1 = T2 = T , we also get I1,1 = I2,2. This means the workflux into subsystem 1 equals the work flux out of subsystem 2.


A van der Waals gas is a model of a real, non-ideal gas. In contrast to an ideal gas itconsists of interacting particles which have a specific volume. The equation of statefor a van der Waals gas is (

p+a n2

V 2

)(V − n b) = nRT, (2.150)

where a is a measure of the attraction between the particles and b is the excludedvolume for one mole of particles [41]. Both are specific quantities of a gas.

45


0.0 0.1 0.2 0.3 0.4 0.5

0

2

4

6

8

10

12

14

V�@10-3m

3D

p�@M

PaD

T = 130

T = 140

T = 150

T = 160

T = 170

T = 180

T = 190

T = 200

Figure 2.12: Pressure over volume according to the equation of state of a van derWaals gas. For low temperatures the pressure has a minimum which isnot realistic. In this region the liquid and gaseous phase coexist and thepressure is constant. The value of this constant pressure can be foundwith the Maxwell construction as described in the text and in figure 2.13.Parameters for oxygen, i.e. a = 0.13983Pa m6

mol2and b = 0.032 × 10−3 m3

mol,

have been taken from [41].

If particles are modeled as hard spheres, which occupy a volume of 4/3π r3. Theminimum distance of the centers of two particles is twice the radius r, i.e. the diameterd. Thus, the excluded volume per particle is 4/3π d3 = 8× 4/3π r3. The parameterb is this volume multiplied with NA/2, i.e. half the number of particles in a mole.The factor 1/2 is included to prevent over counting. For a more detailed discussionof the parameters see, for instance, [56].

The internal energy of a van der Waals gas is

E(T, V, n) = cV nRT − a n2

V, (2.151)

which is a function of the temperature T , the amount of substance n and the volumeV of the gas [58]. The internal energy of an ideal gas does not depend on the volumebut for real gases it does. Thus, the van der Waals gas is an improvement in thedescription of gases. This plays an important role, e.g. in the Joule-Thomson process(see section 2.4.3).

Figure 2.12 shows the pressure over volume using the van der Waals equation for

46


V

p

V1V2

pg

Figure 2.13: Maxwell construction for a van der Waals gas. The graph shows anisotherm (blue) and the corresponding isobar (green) in the phase tran-sition region. The area below (light blue) and above (light red) theisobar are equal, which is a geometric interpretation of equation 2.154.

oxygen. It can be seen that for higher temperatures the graphs are similar to thoseof an ideal gas. The pressure decreases with increasing volume. For lower tem-peratures, however, there are regions where the pressure increases with increasingvolume. This is a non-physical behavior for a gas but can be explained consideringphase changes [56]. In this region the liquid phase and gas phase are both present.In the process of isothermal compression at a temperature below the critical tem-perature, the gas starts to condense at some point. In the equilibrium of gas andliquid, their pressures are equal, i.e. pg = pl. The gas pressure pg is a function ofthe temperature only and does not depend on the volume of the gas in this regionof coexistence of the two phases. Therefore, it is a straight line in the pV diagram.To find the volumes V1 and V2, between which the pressure is constant, the Maxwellconstruction can be used [56].

Because the internal energy has a total differential, the change of energy between V1

and V2 must be independent of the path between those two points. Therefore, it canbe calculated as the integral either over the volume dependent pressure of the vander Waals equation or over the constant pressure pg. Thus, the change in internalenergy during the isothermal compression from V1 to V2 reads

∆E =−V2∫V1

p(V ) dV = −V2∫V1

(nRT

V − n b −a n2

V 2

)dV. (2.152)

47


and

∆E = −V2∫V1

pg dV. (2.153)

By solving the integrals in equations 2.152 and 2.153 and comparing them, we obtain

−nRT lnV2 − n bV1 − n b

− a n2

(1

V2

− 1

V1

)= −pg(V2 − V1). (2.154)

Using this equation and the equation of state 2.150 at pressure pg for V1 and V2, thegas pressure and the two volumes can be calculated.

Equation 2.154 can be easily understood geometrically. If we have a look at the areasbetween the isobar at pg and the isotherm of the van der Waals gas in the regionfrom V1 to V2, the area below the isobar has to equal the area above (see figure 2.13).

2.4.1 Gibbs Equation and Principal Equation of State

As for the ideal gas, for the endoreversible models we need a description of the vander Waals gas with functions which depend on extensities only. Again, the extensivequantities are entropy S, volume V and amount of substance n (or alternatively theparticle number N = NA n). The corresponding intensive quantities are temperatureT , pressure p and chemical potential µ. Thus, according to equation 2.2, the Gibbsequation is

dE =∑α

Y α dXα = T dS − p dV + µ dn. (2.155)

The internal energy as function of the extensities only, i.e. the principle equation ofstate, reads

E(S, V, n) = cV nR

((V0 − n0 b)T

cV0

n0

n

V − n b e

(SnR− S0n0 R

)) 1cV

− a n2

V. (2.156)

As for the ideal gas (cf. section 2.3.1), it can be obtained using equation 2.151 andthe entropy as a function of temperature, volume and amount of substance [41]:

S(T, V, n) = cV nR lnT

T0

+ nR lnV − n bV0 − n0 b

− nR lnn

n0

+ nS0

n0

. (2.157)

If we compare this equation to the entropy of an ideal gas (equation 2.26), we seethat the only difference is, that for the van der Waals gas the volumes are replacedby the reduced volumes.

48


The intensities as functions of S, V and n for a van der Waals are

T (S, V, n) =

((V0 − n0 b)T

cV0

n0

n

V − n b e

(SnR− S0n0 R

)) 1cV

, (2.158)

p(S, V, n) =nR

V − n b

((V0 − n0 b)T

cV0

n0

n

V − n b e

(SnR− S0n0 R

)) 1cV

− a n2

V 2, (2.159)

µ(S, V, n) =

(cV R +

RV

V − b n −S

n

)((V0 − n0 b)T

cV0

n0

n

V − n b e

(SnR− S0n0 R

)) 1cV

− 2 a n

V. (2.160)

2.4.2 Interactions

Interactions with Volume and Entropy Flux

Let us have a look again at the example system of section 2.3.5. It is shown again infigure 2.14. Now, the box shall be filled with a van der Waals gas. For the ideal gasthe process of volume and entropy exchange in such an insulated system is isothermal,because the internal energy does not depend on the volume (cf. section 2.3.5). Whena van der Waals gas is used, the internal energy depends on the volume and thereforethe temperature changes during the process of volume and entropy exchange betweenthe reservoirs (see equation 2.162 below).

The initial temperature shall be the same in both reservoirs and the initial pressureshall be bigger in subsystem 1, i.e. T ◦1 = T ◦2 = T ◦ and p◦1 > p◦2. The volume flux shallbe a function of the pressure difference again, i.e. JV = h(p1 − p2). Since reservoir 1expands while reservoir 2 is compressed, the volume flux is directed from subsystem 2to 1.

Since the box shall be insulated and have a constant total volume, there is no heatand work exchange with the surroundings. Thus, the total internal energy, Etotal =E◦1 +E◦2 = E1 +E2, is constant. According to equation 2.151, it can be expressed as

Etotal = cV (n1 + n2) RT − a n21

V ◦1 + ∆V− a n2

2

V ◦2 −∆V(2.161)

with V1 = V ◦1 + ∆V and V2 = V ◦2 −∆V . Solving for the temperature gives

T (∆V ) =1

cV (n1 + n2) R

(Etotal +

a n21

V ◦1 + ∆V+

a n22

V ◦2 −∆V

). (2.162)

49


(a)

V1 V2

T1 = T2

(b)

1

JV , JS

I

2

(c)

0 2 4 6 8299.85

299.9

299.95

300

DV�@10-3m

3D

T�@K

DFigure 2.14: (a) A box with a van der Waals gas which is divided into two volumes

with different pressures but the same temperature. (b) Endoreversiblemodel of the system with two reservoirs for the volumes 1 and 2 andinteractions for the volume and entropy flux as well as the correspondingenergy flux. (c) Temperature over volume change of a van der Waalsgas in the system on the left. As the volume of subsystem 1 is increasedby ∆V , the volume of subsystem 2 is decreased by the same amount.The temperature decreases according to equation 2.162 till the pressuresare equal in both subsystems. For an ideal gas the temperature staysconstant during the same process (see section 2.3.5).

Figure 2.14 shows the system temperature over ∆V for a system with T ◦ = 300 K,V ◦1 = 1/2V ◦2 = 1/3V0 = 8.2× 10−3 m3 and n1 = 2n2 = 2/3 mol. It can be seen thatthe temperature slightly decreases in this process when a van der Waals gas is used.For an ideal gas the temperature in the same process was constant (cf. section 2.3.5).

Van der Waals parameters for oxygen, i.e. a = 0.13983Pa m6

mol2and b = 0.032×10−3 m3

mol,

from [41] have been used for the calculations. Figure 2.15 shows the volume andpressure development in the two reservoirs.

From the energy conservation in the interaction, we obtain

I1 = I2 (2.163)

T JS,1 − p1 JV = T JS,2 − p2 JV (2.164)


JS,1 = JS,2 +p1 − p2

TJV (2.165)

= JS,2 + σ. (2.166)

50


10

15

20

t

V�@1

0-

3m

3D

V1

V2

(a) Volume

6080

100120140160180200

t

p�@k

PaD

p1

p2

(b) Pressure

Figure 2.15: (a) Volumes and (b) pressures of the two reservoirs in figure 2.8 for avan der Waals gas.

As the process takes place, the pressure difference gets smaller till equilibrium of thetotal system is reached and there is no exchange of energy and extensities betweenthe subsystems, i.e. till p1 = p2.

2.4.3 Example: Pressure Regulator

We will now have a look at the pressure regulator in section 2.3.6 again. The structureof the pressure regulator shall be the same as in fig. 2.11a, i.e. it consists of a porousmembrane that separates a region with high pressure p1 from a region with lowerpressure p2. Particles are pressed through the membrane because of the pressuredifference and insulation of the system prevents heat exchange with the surroundings.While in section 2.3.6 an ideal gas was used, here we consider a van der Waals gas.As for the ideal gas, the process is isenthalpic. However, the difference is that achange in temperature occurs for the van der Waals gas (Joule-Thomson effect) .

The general structure of the endoreversible model of this system is the same as for theideal gas shown in figure 2.11b. The only difference is, that the temperatures in thetwo regions differ for the van der Waals gas, i.e. T1 6= T2. As before, the two regionswith constant pressures p1 and p2 are represented by two connected reservoirs. Theextensities, which are exchanged between the reservoirs, are particles and entropy.Additionally, there is volume exchange with the surroundings at both reservoirs.

Let p1 be larger than p2, such that particles are transferred from reservoir 1 toreservoir 2. To obtain constant pressure, the volume of reservoir 1 needs to bereduced by JV,1 and the volume work flux into subsystem 1 reads

I1,1 = −p1JV,1. (2.167)

51


In reservoir 2 the particle number increases and, therefore, the volume flux JV,2occurs. The corresponding work flux out of the subsystem is

I2,2 = −p2JV,2. (2.168)

Now let us have a look at the interaction between reservoirs 1 and 2. Since there isno heat exchange with the surroundings, the energy flux leaving system 1 (I1,2) withthe particle flux (J) and entropy flux (JS,1) is completely added to system 2 (I2,1),i.e.

I1,2 = T1JS,1 + µ1J = I2,1 = T2JS,2 + µ2J, (2.169)

where J1 = J2 = J was used. Since the pressures p1 and p2 are constant, this equalsthe enthalpy change in the two reservoirs, i.e. I1,2 = IH,1 and I2,1 = IH,2. Fromequation 2.169 we obtain that the enthalpy in the whole system is constant, becausethe total change in enthalpy is IH,2 − IH,1 = 0.

From constant total enthalpy and the equation of state of a van der Waals gas, thetemperature T2 can be calculated. Since the structure of the equation of state ismore complex than that of an ideal gas, using numerical methods for calculations ishelpful.

Solving equation 2.169 for JS,2 we get

JS,2 =T1

T2

JS,1 +µ1 − µ2

T2

J (2.170)

= JS,1 + σ, (2.171)

where σ = µ1−µ2

T2J + (T1

T2− 1)JS,1. As for the ideal gas in section 2.3.6, the transport

of particles through the membrane is an irreversible process. For the ideal gas theirreversibly produced entropy was only due to the difference in chemical potentials,since the temperature was the same in both reservoirs. Here, part of σ is due to theirreversible change in temperature between the reservoirs, because the interactioncontains heat conduction.

Note, that the gas does not necessarily cool down when it expands into the lowerpressure regions. The sign of the temperature change depends on the type of gas andits initial temperature T1. The partial derivative of the temperature with respect tothe pressure at constant enthalpy is called Joule-Thompson coefficient,

δJT =

(∂T

∂p

)H

. (2.172)

In figure 2.16 the Joule-Thomson coefficient for hydrogen and oxygen is shown fora pressure p1 = 101.3 bar before the expansion. For hydrogen at temperatures be-low 179.4 K the coefficient is positive, i.e. with decreasing pressure the temperature

52

2.5 Summary

150 200 250 300

-0.02

-0.01

0.00

0.01

0.02

T1�@KD

∆JT

�@K�ba

rD

hydrogen

(a) a = 0.01945 Pa m6/mol2,b = 0.022× 10−3 m3/mol [41]

180 200 220 240 260 280 300

0.3

0.4

0.5

0.6

0.7

0.8

T1�@KD

∆JT

�@K�ba

rD

oxygen

(b) a = 0.13983 Pa m6/mol2,b = 0.032× 10−3 m3/mol [41]

Figure 2.16: Joule-Thompson coefficient δJT for (a) hydrogen and (b) oxygen as afunction of the temperature. The gases are modelled as van der Waalsgases at p1 = 101.3 bar. The van der Waals parameters a and b aregiven in the captions.

decreases as well. For temperatures above the inversion point the temperature in-creases with decreasing pressure. For oxygen the Joule-Thomson coefficient is alwayspositive in the considered temperature range.

2.5 Summary

Above it was shown how fluxes with multiple extensities can be included in endore-versible models. From the examples we see that it largely depends on the describedprocess, how the necessary quantities are best expressed. The entropy flux at acontact point, for instance, can often be expressed as a function of the connectedparticle or volume flux (cf. equations 2.109, 2.110 or 2.131) and some carefully cho-sen other quantities. For instance, the entropy flux should be expressed as a functionof the temperature in isothermal processes or as a function of the pressure in isobaricprocesses. Also, it is reasonable to express the entropy flux as a function of the sub-systems volume or particle number, in case these are constant. For ideal gases that isalways possible, because the equation of state 2.22 is very simple. For more complexequations of state, it might be difficult to analytically find the required functions. Inthese cases numerical methods may be used as in section 2.4.3.

The example of the pressure regulator shows, that some features of a process can beelegantly obtained from an endoreversible model. The conservation of enthalpy in theJoule-Thomson process is a result of the transport laws in the interaction betweenthe two reservoirs (cf. sections 2.3.6 and 2.4.3). This example also shows that the

53


ideal gas model is not suitable to capture the main effect, i.e. the temperature changedue to a pressure drop. Therefore, in some cases a more realistic gas model, e.g. thevan der Waals gas, has to be used.

54

3 Endoreversible Modeling of ChemicalReactions

A topic, which has been rarely discussed in the field of endoreversible thermody-namics, is modeling chemical reactions. General thermodynamic analysis of chemicalreactions can be found in many textbooks and articles. In their article “On the Ir-reversible Production of Entropy”, Tolman et al. [74] included a section about theentropy production in chemical reactions. Prigogine [67], for instance, discussed theentropy production in irreversible chemical processes. Ondrechen et al. [63, 62] inves-tigated chemically driven engines with a finite-time thermodynamics approach. Theymodeled the reaction with two connected engines – an irreversible chemical reactionengine and a heat engine that is driven by the heat output of the reaction. Mironovaet al. [55] also used finite-time thermodynamics to optimize a chemical reaction withrespect to entropy production by controlling the concentrations. Tsirlin et al. [75]also investigated the thermodynamic efficiency of a chemical reactor.

In the standard endoreversible formalism (see section 2.1) chemical reactions cannotbe treated straightforward. On the one hand, we have reservoirs, which are systemsin equilibrium. This is generally not the case for chemical reactions. They may bein equilibrium in a way that forward and backward reaction are taking place at thesame velocity. This, however, is usually not a process of interest. In most cases,reactions with a preferred direction need to be investigated. On the other hand, wehave engines, which are defined as reversible subsystems without entropy productionbut balanced extensities. In chemical reactions, however, entropy production mayoccur. In addition, the particle numbers change. Thus, neither reservoirs nor enginescan be used as they have been defined in section 2.1 to model chemical reactions.

In the following I will first discuss a type of endoreversible chemical engine thathas been widely discussed in literature. It does not include chemical reactions butis a good starting point for the treatment of reactions in the scope EndoreversibleThermodynamics. Afterwards, approaches to model both reversible and irreversiblereaction processes are discussed.

55

3 Chemical Reactions

(a)TH

qH

TiH

TiL

qL

TL

P

(b)µH

JH

µiH

µiL

JL

µL

P

Energy fluxEntropy fluxParticle flux

Figure 3.1: Comparison of (a) a Curzon-Ahlborn heat engine and (b) a chemicalengine. In the irreversible interactions for the heat engine the heat (orenergy) flux is conserved while for the chemical engine the particle (orextensity) flux is conserved.

3.1 Chemical Engines Without Reactions

First, I want to have a look at a type of endoreversible chemical engines that hasbeen discussed, for instance, in [27, 28, 40, 47]. In these engines power is generated ina reversible engine due to a difference in chemical potentials. The structure of sucha system is depicted in figure 3.1b. It is similar to that of a Curzon-Ahlborn heatengine [25], where heat is transferred irreversibly due to a difference in temperaturesas depicted in figure 3.1a. In both engines two reservoirs are irreversibly connected toan engine. However, in the heat engine the heat flux, i.e. the energy flux, is conservedwithin the interactions while in the chemical engine the particle flux, i.e. the extensityflux, is conserved.

In the chemical engine, the reversible engine is connected to two particle reservoirs– one with high chemical potential µH and one with low chemical potential µL. Theirreversible particle transfer between the reservoirs and the engine is proportional tothe difference in chemical potentials. The transport laws are

JH = hH(µH − µiH) and (3.1)

JL = hL(µiL − µL), (3.2)

where JH and JL are the particle fluxes from the high chemical potential reservoir andto the low chemical potential reservoir, respectively. µiH and µiL are the chemical po-tentials at the contact points of the engine and hH and hL are non-negative constants.The chemical potentials are ordered as µH > µiH > µiL > µL. The corresponding

56

3.1 Chemical Engines Without Reactions

energy fluxes at the contact points of the reservoirs and engine are

IH = µH JH IiH = µiH JH (3.3)

IL = µL JL IiL = µiL JL. (3.4)

The energy flux IH at the reservoir with high chemical potential is bigger than theenergy flux IiH at the engine, since µH > µiH. This means energy is dissipated in thetransport process while the particle flux JH is conserved. The same applies for thefluxes between the engine and the low chemical potential reservoir, i.e. IiL > IL andJL is the same at the connected contact points ‘iL’ and ‘L’.

There are two balance equations for the engine – one for the particle fluxes and onefor the energy fluxes. They are

0 = JH − JL and (3.5)

0 = IiH − IiL − P. (3.6)

Since both particle fluxes have been defined as positive, a minus sign had to beincluded before the effluxes in both balance equations. From the balance of theparticle fluxes (equation 3.5) we get JH = JL = J , i.e. the particle flux is the samebefore and after the engine. The power output can be obtained from the balance ofthe energy fluxes (equation 3.6) using the previous result as

P = IiH − IiL = (µiH − µiL)J. (3.7)

With equations 3.1 and 3.2, we can substitute µiL and J in the power equation andobtain

P = hH(µH − µiH)

(µiH −

hH(µH − µiH)

hL

− µL

). (3.8)

For the discussion of the engine it is not relevant in which form the output poweris used, i.e. which extensities are carrying it. Lin and Chen et al. [53, 81], however,studied such a chemical engine where the output power was used to pump particlesto a higher chemical potential reservoir.

In analogy to heat engines a number of variations of these chemical engines havebeen studied. Chen et al. [22] investigated the influence of a mass leak in a chemicalengine. Staging of two chemical engines was discussed by Chen et al. [21].

Such a chemical engine can be used to describe electrochemical processes, which can,for instance, be found in fuel cells (see section 4.3.2) or batteries. Then, the chemicalpotentials µl have to be replaced by electrochemical potentials µl = µl + z F ϕl.These chemical engines do, however, not include chemical reactions and thereforeonly represent part of the processes in electrochemical cells. In the following sectionI will discuss the incorporation of chemical reactions into endoreversible models.

57


3.2 Chemical Engines with Reactions

Chemical reactions can be written in the form

∑j

νjRj ⇀↽∑k

νkPk, (3.9)

where Rj are the reactants, Pk are the products and νi is the stoichiometric coefficientof substance i. The amount of substance that is processed in a chemical reaction, i.e.the change in particle number of this substance, is

∆nl = νl ξ. (3.10)

ξ is the so called extent of reaction [67]. It is a measure for the total amount ofreactants that is consumed and the amount of products that is produced in a reaction.If we solve equation 3.10 for ξ, we see that the fraction ξ = ∆nl/νl is the same forall reactants and products.

Generally, in chemical engineering one distinguishes between continuous and batchprocessing. In continuous processing, a constant flow of reactants enters the reactor, isprocessed and the products are continuously removed. This is the desired processingin chemical engineering, since it can be automated and is cheaper compared to batchprocessing. Continuous processing is intuitively compatible with the endoreversibleformalism of fluxes and steady state engines. In batch processing, a predeterminedamount of substances is mixed in a vessel, heated, cooled or pressurized and then theproducts are isolated. This is labor intensive but still used when only small amountsare needed or higher value substances are processed. A possibility to include batchprocesses in the formalism of endorversible thermodynamics is to model the reactionas a cyclic process. In one branch the reactants are filled into the vessel, in anotherbranch the reaction takes place and in a last branch the products are removed fromthe vessel. As long as one is not interested in the times of the different branchesand the total reaction time, it is sufficient to consider the total amounts of particles,entropy and so on, instead of the fluxes. In this chapter, I will discuss batch processeswithout considering reaction times. Cyclic processes would be similar, as one cyclecould be treated like a batch process. In cyclic processes, however, the time to fillthe reaction vessel, the reaction time and the time to empty the vessel could beconsidered. Thus, additional measures like a ‘time efficiency’ could be optimized.The balance equations are also equivalent for continuous processes as can be seen insection 4.3, where I discuss endoreversible fuel cell models.

58


3.2.1 Reversible Reactions vs. Reversible ThermodynamicProcesses

When talking about reversible reactions and reversible processes, one has to be care-ful not to confuse the conceptual differences in the meaning and usage of these terms.In chemistry textbooks, reversible reactions are often defined as reactions which cantake place in both forward and backward direction and which reach an equilibrium,where both, reactants and products are present in a not negligible concentration[44, 79]. At the equilibrium point forward and backward reaction take place at thesame rate. Freiser et al. [36], for instance, state that “a reversible reaction is one thatcan be made to proceed in either the forward or the reverse direction by appropriatelyadjusting reaction conditions [. . . ]”. According to Le Chatelier’s principle, dependingon the state of the reacting substances, i.e. their temperature, partial pressures orconcentrations, the ratio of the concentrations of reactants and products at equilib-rium changes. Thus, if a chemical system is in equilibrium and then the temperatureis changed, the reaction rates of the forward and backward reaction change as welland the system approaches a new equilibrium.

At the equilibrium point of a reversible reaction, the process of simultaneous forwardand backward reaction is also a reversible process in the thermodynamic sense, i.e.without entropy production. If the system is not in equilibrium, the reactions aregenerally taking place with finite rates and possibly entropy production. Thus, theymay be irreversible processes. There are, however, (theoretical) possibilities to con-duct a reaction approximately reversible. If a reaction process shall be reversible,there must be a way to use the released reaction energy.

One of them is van’t Hoff’s equilibrium box for reactions of gases (see e.g. the textbookof Becker [13]). It is a box containing all reactants and products of an reaction inequilibrium at constant temperature T . If we consider a reaction

A + B ⇀↽ AB, (3.11)

van’t Hoff’s equilibrium box contains the substances A, B and AB at the equilibriumconcentrations cA, cB and cAB. Additionally, we need two boxes for the reactantsA and B at their initial concentrations c◦A and c◦B. These boxes have a movablepiston, such that the gases can be isothermally expanded or compressed till they reachconcentrations cA and cB. Then, the box containing A is connected to the equilibriumbox but separated with a membrane, which is only permeable for substance A. Sincethe concentrations of A in the equilibrium box and in the connected box are thesame, no particles are exchanged through the membrane. In the same way, the boxcontaining substance B is connected and another box for substance AB. Then, thepistons at all three boxes are moved, such that substances A and B are slowly pushedinto the equilibrium box while substance AB is sucked out of it. The situation inside

59


the box does not change during this process, if it takes place infinitely slowly. Thereactants A and B, which enter the box, react and form the product AB, whichis removed from the equilibrium box, such that the concentrations inside the boxremain the same for all substances. Finally, the box containing the product AB isremoved from the equilibrium box and the gas AB can be expanded or compressedisothermally till the concentration c◦AB is reached. All the expansion or compressionprocesses as well as the reaction process are isothermal and therefore reversible. Thereaction in the equilibrium box can also take place in the opposite direction. Thereverse reaction may even take place at different concentrations c′A, c′B and c′AB butat the same temperature. The equilibrium constant at this temperature T has to bethe same, i.e.

Kc =cA cB

cAB

=c′A c

′B

c′AB

. (3.12)

Then, the initial state of reactants and products as well as its surroundings can bereached again. For a more detailed discussion of van’t Hoff’s equilibrium box and theconnected heat and work exchange during the expansion and compression processesas well as the reaction process see e.g. [13]. Van’t Hoff’s equilibrium box can notonly be used for reaction process but also for separation processes as described byAmelkin et al. [4].

Another way of achieving an (almost) reversible reaction process are electrochemicalcells. In electrolytic cells electric power is used to decompose chemical compounds.The reverse reaction can be found in a galvanic cell, where a reaction causes an electriccurrent. The voltage in such cells slows down the reaction process and thereforereduces dissipation. In reality we will still find dissipation in these processes, e.g. dueto Ohmic resistance in the conductor.

In the following I will use the term reversible only in the thermodynamic sense, i.e.to describe processes without dissipation and hence without entropy production.

3.2.2 A New Subsystem – the Reactor

When modeling chemical reactions, changing particle numbers and in some cases irre-versibility have to be considered. To do so, I want to introduce a new endoreversiblesubsystem, which I will call ‘reactor’. It shall have the following properties. The reac-tants and products enter and leave the reactor as multi-extensity fluxes as describedin chapter 2. The extensities of these fluxes are amount of substance, entropy andpossibly volume. Additionally, a contact with entropy as the extensity is needed atthe reactor for the released (exothermic reaction) or absorbed (endothermic reaction)heat. In case the reaction is not totally irreversible, another contact is needed for

60


the ‘chemical’ work, i.e. usable work generated during the reaction process withoutvolume work. Finally, in case the volume of the reaction vessel is not constant, acontact for the volume work is needed. This may be the same as that for the heatexchange.

The reaction process in the reactor shall take place at constant temperature T . Also,the substances occurring in the reaction equation shall enter the reactor alreadymixed, such that the mixing process with the connected entropy change does nothave to be considered in the treatment of the reactor. They also leave the reactor asa mixture at reaction temperature. Temperature changes of the reaction products dueto released or consumed heat shall be treated separately after the reaction process.

As in endoreversible engines in a reactor the energy fluxes are redistributed overthe different carriers, i.e. extensities, and balanced. That means, all the energyfluxes have to add up to zero to ensure conservation of energy. For some of theextensities, however, the balance equations differ from those of engines. They containa production (or destruction) term in addition to the fluxes in and out of the reactor.

Let us have a closer look at the balance equations for the amount of substance orparticle number. In general, the particles in chemical reactions are not only atoms butalso molecules, ions and electrons, e.g. H2, O2 or e−. As described in section 2.1.1,all the extensities have to sum up to zero within an engine. In chemical reactions,however, the particle numbers do not fulfill this requirement. New particles arecreated while others vanish when molecules are formed or split up.

Let us, for instance, have a look at the particle numbers in the reaction

H2 +1

2O2 ⇀↽ H2O. (3.13)

For one and a half moles of reactants, i.e. hydrogen and oxygen, we get one mole ofproduct, i.e. water. This means, the numbers of molecules decrease for hydrogen andoxygen, while the number of water molecules increases. The total particle number inthe system decreases, as well. Thus it is necessary to balance the particle numberswith a production (or destruction) term according to the stoichiometric equation.

In our example the stoichiometric coefficients are νH2 = −1, νO2 = −1/2 and νH2O =1. Thus, the changes in particle numbers according to equation 3.10 are

∆nH2 = −ξ, (3.14)

∆nO2 = −1

2ξ =

1

2∆nH2 and (3.15)

∆nH2O = ξ. (3.16)

Note, that in the literature different conventions are used for the sign of stoichiometriccoefficients. In some publications, they are all positive and for calculations, e.g. of

61


reaction enthalpies, the difference between products and reactants is used. In othertexts, they are negative for the reactants and positive for the products, so that sumsover all substances can be used. Here, I use the latter possibility, such that it can beeasily included in the endoreversible formalism. Thus, the balance equation for theamount of substance in a reactor is

0 = ninl + ∆nl − nout

l (3.17)

= ninl + νl ξ − nout

l . (3.18)

The superscripts in and out denote quantities before and after the reaction, respec-tively. The term νl ξ is a production term, which did not occur in the standarddefinition of an endoreversible engine. Hoffmann et al. [47] stated that for each ex-tensity the fluxes of all contact points of an engine have to be balanced (cf. equation2.5 at page 15). Note, however, that the reversibility condition is not violated by this.It is possible to have reversible chemical reactions with changing particle numbers.

The balance equation for entropy is similar to those of the particle numbers. At thecontact points of the reactor, the mass fluxes of the chemical substances and theheat exchange contain entropy fluxes. Additionally, entropy may be produced duringchemical reactions. The entropy balance in a reactor then is

0 =∑l

Sinl + σ +

Q

T−∑l

Soutl , (3.19)

where Sinl and Sout

l are the entropies of the substances before and after the reactionrespectively, σ is the irreversibly produced entropy and Q/T is the entropy leaving thereactor through the heat exchange contact. In case of reversible reaction processes σequals zero.

The volume balance is the same as that of a standard endoreversible engine, i.e. thereis no production term:

0 = V in + ∆V − V out. (3.20)

V in and V out are the volumes occupied by the substances before and after the reaction.∆V is the volume exchange with surroundings, in case the volume of the reactionvessel is not constant.

The general energy balance for a chemical reaction is

0 =∑l

Einl +Q+W −

∑l

Eoutl , (3.21)

where Einl and Eout

l are the internal energies of substance i before and after thereaction. From equation 3.19 we obtain, that the exchanged heat Q can be expressed

62


in terms of the entropy difference of the substances before and after the reaction andthe irreversibly produced entropy. Therefore, it can be divided into a reversible andan irreversible term, Q = Qrev +Qirr, with

Qrev = −T(∑

l

Sinl −

∑l

Soutl

)and (3.22)

Qirr = −T σ. (3.23)

The work W contains the chemical work Wchem, which I will discuss later, and thevolume work Wvol. The latter is defined as

Wvol = −V out∫V in

p dV, (3.24)

where p is the total pressure in the system, which can be written as p =∑

l pl.

I will limit the reaction conditions to constant total pressure in the reaction vesselor constant volume of the reaction vessel. As the volume work is zero if the reactorhas constant volume, it appears only in the constant pressure case and can then besimplified to

Wvol = −pV out∫V in

dV = −p (V out − V in) = −p∆V. (3.25)

Eoutl −Ein

l is the change in internal energy of substance l during the reaction. Usingthe Gibbs equation 2.2, it can be expressed as

∆El = Eoutl − Ein

l =

Soutl∫

Sinl

Tl dSl −V outl∫

V inl

pl dVl +

noutl∫

ninl

µl dnl. (3.26)

In the following, I will first discuss reversible chemical reaction processes, where noentropy production occurs. Afterwards, irreversible reaction processes are considered,where entropy is produced and the chemical energy is completely released as heat.

3.2.3 Reversible Reaction Processes

Angulo-Brown et al. [9] discussed an endoreversible reaction using the principle ofvan’t Hoff’s equilibrium box and the so called method of Carnot cycles. The forward

63


reaction takes place at temperature T while the backward reaction takes place at aninfinitesimal lower temperature T − dT .

De Vos [27, 29] investigated chemical reactions in endoreversible thermodynamics.His first approach was an analogy between heat engines and chemical engines likethe one presented in section 3.1. He assumes that the change in internal energy of thechemical substances is completely released as work. There is no entropy productionduring the reaction, i.e. it is reversible. Only in the transport processes before andafter the reaction, entropy is produced due to differences in chemical potentials.Additionally, he states an axiom about conservation of matter. It says that theparticle number n (different notation is used in the article) is conserved within anendoreversible engine. In the treatment of chemical reactions (section 5 of [27]) hechooses an example reaction, A1 ⇀↽ A2, where the axiom is still fulfilled. This,however, is only true for the total particle number but not for the particle numbersof the single substances A1 and A2. Only in the appendix, the axiom is changed toinclude general chemical reactions. There, n is not the particle number anymore butthe reaction constant ξ, which is defined as ξ = nl

νl(also called extend of reaction,

cf. 3.10). So the particle numbers are not balanced within the engine but niνi

has tobe the same for each of the reactants and products.

De Vos’ [27, 29] approach, that the reaction constant is the same for each particleflux can be used as long as in the reaction the reactants are completely processed.This means, before the reaction there are only reactants and after the reaction thereare only products. However, often we find that there are already reaction products inthe vessel before the reaction or that not all of the reactants are processed such thatwe find a mixture of reactants and products. In those cases the balance equations3.18 are a better choice, because they allow a more general description. They can beused for reactions, where all reactants are processed, as well.

Reaction at Constant Volume

First, I want to discuss reversible reaction processes in a reaction vessel with constantvolume V , i.e. the substances occupy the same volume before and after the reaction.The system of balance equations for this reaction is

0 = V in − V out, (3.27)

0 = ninl + νl ξ − nout

l , (3.28)

0 =∑l

Sinl +

Qrev

T−∑l

Soutl , (3.29)

0 =∑l

Einl +Qrev +Wchem −

∑l

Eoutl . (3.30)

64


From the Gibbs equation we know that the change in internal energy of all thechemical substances in the reaction is

dEV=const.

= T dS − p dV +∑l

µl dnl (3.31)

V=const.= T dS +

∑l

µl dnl (3.32)

V=const.= δQrev + δWchem. (3.33)

The reaction shall take place at constant temperature T . Since there is no entropyproduction during a reversible reaction process, the exchanged heat can be calculatedfrom the difference of the entropies before and after the reaction. Thus, we obtainthe reversible heat from equation 3.29 as

Qrev = T∑l

(Soutl − Sin

l

). (3.34)

The chemical work can then be calculated from equation 3.30 as

Wchem =∑l

Eoutl −

∑l

Einl − T

∑l

(Soutl − Sin

l

)(3.35)

=∑l

(Eoutl − T Sout

l

)−∑l

(Einl − T Sin

l

)(3.36)

=∑l

F outl −

∑l

F inl . (3.37)

This means the chemical work equals the change in free energy, F = E − T S. Thechange in free energy dF = −S dT − p dV +

∑l µl dnl reduces to dF =

∑l µl dnl

for constant temperature and volume. Thus, the work can also be expressed as

Wchem =

∫dF =

∑l

noutl∫

ninl

µl(T, V, nl) dnl, (3.38)

where the chemical potentials µl(T, V, nl) =(∂F∂nl

)T,V

= µl(nl) are only functions of

nl, since T and V are constant.

Reaction at Constant Pressure

If a reaction takes place at constant pressure, we have not only chemical but alsovolume work. This means the change in internal energies of the chemical substances

65


EnergyEntropyParticlesVolume

{Rj}

Wvol

Qrev

{Pk}

Wchem

Figure 3.2: Endoreversible representation of a chemical reactor with a reversible re-action process. The input of reactants and the output of products arelabeled with {Rj} and {Pk} respectively. They contain particles, en-tropy, volume and energy. Additionally, there are contact points for thereversible heat exchange, the volume work exchange and the chemicalwork exchange.

is divided into reversible heat and the two work terms. Figure 3.2 shows such anendoreversible reactor. The balance equations for this reaction are then

0 = V in + ∆V − V out, (3.39)


l , (3.40)

0 =∑l

Sinl +

Qrev

T−∑l

Soutl and (3.41)

0 =∑l

Einl +Qrev +Wvol +Wchem −

∑l

Eoutl . (3.42)

The reversible heat can be calculated following equation 3.34.

As stated in equation 3.24, the volume work can be expressed as an integral over−p dV , where p is the total system pressure. Since the system pressure p is constant,we can easily solve this integral and get

Wvol = −V out∫V in

p dV = −p (V out − V in) = −p∆V. (3.43)

66


The chemical work can then be calculated from equation 3.42 as

Wchem =∑l

Eoutl −

∑l

Einl − T

∑l

(Soutl − Sin

l

)+ p(V out − V in) (3.44)

=∑l

(Eoutl − T Sout

l + poutl V out

)−∑l

(Einl − T Sin

l + pinl V

in)

(3.45)

=∑l

Goutl −

∑l

Ginl . (3.46)

This means, we can use the Gibbs free energy, G = E − T S + p V , to calculate thechemical work. The change in the Gibbs free energy dG = −S dT +V dp+

∑l µl dnl

reduces to dG =∑

l µl dnl for constant temperature and pressure. Thus the chemicalwork can also be expressed as

Wchem =

∫dG =

∑l

noutl∫

ninl

µl(T, pl, nl) dnl, (3.47)

where the chemical potentials µl(T, pl, nl) =(∂G∂nl

)T,p

= µl(pl, nl) are functions of pl

and nl, since the temperature and total pressure are constant but not the partialpressures pl. However, substituting nl with nin

l + νl ξ′, the partial pressures can be

expressed as functions of the total system pressure and the extent of reaction, i.e.

pl = pl(p, ξ′) =

nl∑j nj

p =ninl + νl ξ

′∑j(n

inj + νj ξ′)

p. (3.48)

Using this, dnl = νl dξ′ and ξ = (noutl − nin

l )/νl, we can rewrite equation 3.47 as

Wchem =

ξ∫0

∑l

µl(T, p, ξ′) νl dξ′, (3.49)

which can then be easily solved.

3.2.4 Irreversible Reaction Processes

Combustion of complex fuels, like diesel, is a irreversible process. During the reactionheat is released and it is not possible to use that amount of heat to get to theinitial state of the reactants and its surroundings again. Let us consider our examplereaction 3.13 again. If the reaction does not take place in an electrochemical cell or atequilibrium conditions, but instead hydrogen is burned, it is an irreversible process.

67


The difference in the internal energies of the reactants and products is released asheat and, depending on the reaction vessel, as volume work.

Depending on the modeled system, there are two possibilities to include irreversiblechemical reaction processes. The first possibility is to use a reactor with its balanceequations as described in the following sections. This is reasonable, when the reactionis part of a complex system with many interacting subsystems. In this case the fluxof reactants usually comes from another subsystem, is processed in the reactor andthe reaction products and heat are then used in further processes. If, however, thereaction is only the heat source for further processes, it can as well be modeled as aheat reservoir. In this case, the reaction process itself is not of any interest and thereaction products are not used for further processes. Only the released heat is usedto operate a heat engine.

If we assume that a reaction takes place at constant temperature and that all reac-tants and products are ideal gases, their partial pressures are functions of the particlenumber and volume only:

pin/outl =

nin/outl RT

Vin/outl

. (3.50)

In the example reaction 3.13, for hydrogen and oxygen the assumption of an idealgas is reasonable even at standard temperature and high pressures. For water vapor,the equation of state for an ideal gas is only a good estimate for high temperaturesand low pressures. At low temperatures or high pressures another gas model wouldbe more suitable, e.g. the van der Waals gas (cf. 2.4).

Reaction at Constant Volume

If a reaction takes place at constant volume, the reactants and products all have thesame volume and the partial pressures are functions of the particle numbers only.Thus, for ideal gases the change in system pressure during the reaction is

pout − pin =∑l

(noutl − nin

l

) RT

V. (3.51)

With equation 3.18 this can be written as

pout − pin =∑l

νl ξRT

V. (3.52)

68


The balance equations for a reaction at constant volume are

0 = V in − V out, (3.53)


l , (3.54)

0 =∑l

Sinl + σ +

Q

T−∑l

Soutl , (3.55)

0 =∑l

Einl +Q−

∑l

Eoutl . (3.56)

From the last equation we see, that in an irreversible reaction process at constantvolume the heat exchange equals the change in internal energy of the substances. Theexchanged heat can be divided into reversible and irreversible heat, Q = Qrev + Qirr

with

Qrev = T

(∑l

Soutl −

∑l

Sinl

)and (3.57)

Qirr = −T σ. (3.58)

Combining equations 3.55 and 3.56 we obtain

σ =− 1

T

(∑l

(Eoutl − T Sout

l )−∑l

(Einl − T Sin

l )

)(3.59)

=− 1

T

∑l

noutl∫

ninl

µl dnl, (3.60)

where Eoutl − Ein

l =∫

dEl = T∫

dSl −∫pl dV +

∫µl dnl was used.

∫pl dV is zero

as the volume does not change.

Reaction at Constant Pressure

If a reaction takes place at constant pressure p, the volume changes during the reac-tion. The balance equations for such a reaction are then

0 = V in + ∆V − V out, (3.61)


l , (3.62)

0 =∑l

Sinl + σ +

Q

T−∑l

Soutl and (3.63)

0 =∑l

Einl +Q+Wvol −

∑l

Eoutl . (3.64)

69


If we put Wvol = −p∆V = −p (V out − V in) into equation 3.64 and solve for Q, weobtain

Q =∑l

Eoutl −

∑l

Einl + p (V out − V in) (3.65)

=∑l

(Eoutl + pout

l V out)−∑l

(Einl + pin

l Vin)

(3.66)

=∑l

Houtl −

∑l

H inl . (3.67)

This means, the change in enthalpy, H = E + p V , equals the heat exchange of anirreversible process at constant pressure.

3.3 Summary

In this chapter we saw that there are two types of “chemical engines”. The first typegenerates power from a difference in electrochemical potentials but does not includechemical reactions. It can be included in endoreversible models as an engine, sinceall the balance equations for energy and extensities described in section 2.1.1 arefulfilled. The second type is used for chemical reactions. For chemical reactions,engines cannot be used, as the balance equations from section 2.1.1 are generallynot fulfilled for particle number and entropy. Since the particle numbers usuallychange during chemical reactions, i.e. the number of reactants decreases while thenumber of products increases, a destruction or production term is needed in theparticle balance. This can be accomplished, as the change in particle number is welldefined by the stoichiometric equation of the reaction (cf. equation 3.18). In case ofan irreversible reaction process, the entropy balance also contains a production term.Thus, I introduced a new subsystem, the reactor, which includes the production termsfor particle number and entropy but has the same balance equations as conventionalengines for all other extensities and energy.

Note that the introduction of the reactor contradicts the idea of endoreversibility, asit is a subsystem which may include entropy production. However, it seems to bea necessary compromise to accept that a reactor is not always a reversible subsys-tem. It offers the possibility to model a great number of processes, which can notbe considered with the conventional endoreversible formalism (cf. section 2.1). Fur-thermore, the entropy production in the presented reactor can be quantified, sincethe reaction conditions and balance equations are known. This is an advantage overformer attempts of including irreversibilities in endoreversible heat engines. Chen etal. [23] and Angulo-Brown et al. [11] included the losses due to irreversibilities as afixed percentage of the energy influx.

70

3.3 Summary

Finally, I briefly want to discuss the calculation of the reaction heat for ideal gases.The internal energy of substance l reads

El = clV nl RT. (3.68)

If we consider the reaction of equation 3.13 as an irreversible combustion process ina fixed volume, the energy balance with the above equation for the energy becomes

Q =∑l

(Eoutl − Ein

l

)(3.69)

=∑l

clV RT(noutl − nin

l

)(3.70)

= RT∑l

clV (νl ξ) (3.71)

= RT ξ

(−5

2· 1− 5

2· 1

2+ 6 · 1

)(3.72)

=9

4RT ξ, (3.73)

where the particle balance of equation 3.54, cH2V = cO2

V = 5/2 and cH2OV = 6 was used.

From the last equation we see that the exchanged heat Q is positive, i.e. heat inputis needed in this reaction. In reality, however, this is an exothermic reaction. Thisindicates, that one has to be careful using the ideal gas model in the description of theenergy balance of chemical reactions, when realistic values are desired. The reasonfor the difference is that the theoretical values for cV differ from experimental values.For water, the value of cH2O

V at standard temperature T = 298.15 K is 3.037 [56]instead of 6. The experimental values for hydrogen and oxygen are approximately5/2 at standard temperature, i.e. they are almost the same as the theoretical values.Thus, using experimental data for cV , the exchanged heat becomes negative as onewould expect.

Note, that it is still reasonable to use the equation of state of the ideal gas model todescribe the pressure volume dependence of gases at high temperatures. For realisticcalculations of internal energies, however, it is better to use experimental data, e.g.from [1], for the heat capacities, entropies, chemical potentials, etc.

71

4 Fuel Cells

In this chapter I will use the extensions to the formalism of endoreversible thermo-dynamics, which have been introduced in the previous chapters, to model a fuel cell.For a reasonable fuel cell model multi-extensity fluxes will be needed for the fuelsupply of the cell and chemical reactions occur in the cell.

Fuel cells are a well investigated subject in engineering as well as in physics. The fieldsof interest range from geometrical flow simulations [15, 16] as well as temperature andconcentration gradients [51] to simplified finite-time thermodynamics models [76].

There are various types of fuel cells which differ, for instance, in the electrolyte, theworking temperature or the fuel that is used. Some use pure hydrogen, others arefueled with methanol. For an extensive overview on fuel cell types see, for example,the book “Fuel Cell Systems Explained” by Larminie and Dicks [52].

In the following I will concentrate on PEM fuel cells, when it is necessary to limitthe description to special types of fuel cells (e.g. when considering irreversibilities).PEM stands for proton exchange membrane or sometimes also for polymer elec-trolyte membrane. Both names describe the same technique of separating anode andcathode. PEM fuel cells are operated with (pure) hydrogen and have an operatingtemperature in the range from 30°C to 100°C [52]. The reactants and products are allmodeled as ideal gases but using experimental data for the heat capacities, entropiesand chemical potentials from [1].

4.1 Structure of a PEM Fuel Cell

The basic elements of a fuel cell are the anode, the cathode and the electrolyte thatseparates anode and cathode (see figure 4.1). At the anode hydrogen is split intoprotons and electrons, i.e.

H2 → 2 H+ + 2 e−. (4.1)

Only the protons can pass the electrolyte while the electrons cannot. Thus, theelectrons are lead through an outer circuit with a load, where the electric poweris used. At the cathode the protons and electrons react with oxygen and water is

73

4 Fuel Cells

H2

O2

H2O

e− load e−

H+

H+

H+

membraneanode cathode

Figure 4.1: Basic structure of a PEM fuel cell. Anode and cathode are separated bythe polymer electrolyte membrane, which is permeable for protons butnot for electrons. Thus, the electrons move through the outer circuit.Hydrogen is supplied at the anode and oxygen at the cathode. Usually,oxygen is supplied with air and the water vapor is removed by the sameair flow.

produced according to the reaction equation

2 H+ + 2 e− +1

2O2 → H2O. (4.2)

In PEM fuel cells the electrolyte is a solid, porous polymer in which protons aremobile. The electrodes in the anode and cathode contain Platinum which acts as acatalyst for the reaction.

4.2 Cell Voltage and Fuel Cell Efficiency

Typical performance measures for fuel cells are the cell voltage and the efficiency [52].The cell voltage Vfc is usually shown as a function of the current density iel (cf. e.g.[35, 50, 54]). Electrical power of the fuel cell and current density are linked by

Pel = Vfc iel. (4.3)

74

4.2 Cell Voltage and Fuel Cell Efficiency

0 2000 4000 6000-0.4

0.

0.4

0.8

T�@KD

DRG

�D RH

Figure 4.2: Maximum efficiency of a fuel cell over temperature is shown for the reac-tion in equation 4.6. The efficiency gets negative beyond the equilibriumtemperature. This means above this temperature the reaction does notrelease energy. Data at standard pressure was taken from [1]. Gaseouswater was assumed as reaction product.

There can be found some discussion on fuel cell efficiencies in literature [43, 52, 80].A common definition of the fuel cell efficiency ηfc is the ratio of the electrical work,which is produced, and the change in enthalpy of the reaction ∆RH, i.e.

ηfc =Wel

∆RH. (4.4)

Larminie et al. [52] pointed out that the change in enthalpy for the reaction inequation 4.6 can be calculated for liquid water or water vapor as reaction product.Since the difference is not negligible, it is important to know which value was used,when efficiencies are published. However, they state that in most publications thelower value, i.e. that for water vapor, is used, because then higher efficiencies can bereached.

If there were no irreversibilities in a fuel cell, the electrical power output would equalthe change in Gibbs free energy of the reaction (cf. section 4.3.1). Thus, the ideal ormaximum efficiency is [52]

ηmaxfc =

∆RG

∆RH. (4.5)

Since both ∆RG and ∆RH are temperature dependent, the maximum efficiency istemperature dependent as well. Figure 4.2 shows the maximum efficiency over tem-perature. For this diagram water vapor was assumed as reaction product.

Wright [80] and Haynes [43] discussed the efficiency of fuel cells in comparison withthe Carnot efficiency. Fuel cell efficiencies are often said to be not bounded by theCarnot efficiency. Therefor, the Carnot efficiency is calculated using the combustion

75

4 Fuel Cells

Energy fluxEntropy fluxParticle fluxVolume flux

H2

q

Pvol

O2

H2O

Pel

Figure 4.3: Reversible model of a PEM fuel cell. An influx of hydrogen and oxygen isneeded to operate the fuel cell. The products of the reaction are an effluxof water, a heat flux q and power exchange Pvol with the surroundingsdue to the change in volume and the electrical power Pel.

heat of hydrogen and oxygen as the high temperature heat input of the Carnotprocess. Both, Wright and Haynes discuss the misconceptions in this comparison andshow that the same ideal efficiency can be obtained, when the combustion process isincluded in the calculation of the Carnot efficiency as an ideal process.

4.3 Endoreversible Modeling of PEM Fuel Cells

In endoreversible modeling the amount of details that are implemented in the systemdescription is variable. On the one hand, it is possible to consider only the initialenergy input and the final output without taking into account the inner processesof the system. On the other hand, it is possible and sometimes necessary to includea more detailed description of the system’s geometry and processes to capture therelevant dissipation processes.

4.3.1 Reversible Fuel Cell Model

The simplest model of a PEM fuel cell is depicted in figure 4.3. It consists only of areversible chemical engine with contacts for

• the in- and efflux of reactants and products, i.e. hydrogen, oxygen and water,

• the heat and volume work flux to the surroundings and

76


• the electrical work flux.

The latter is the useful work flux, i.e. the electrical power output of the fuel cell. In-stead of considering the two reactions 4.1 and 4.2 of the anode and cathode separately,the combined reaction

H2 +1

2O2 → H2O (4.6)

is investigated in such a simplified model.

Vaudrey et al. investigated a reversible fuel cell model in the first part of [76] and [77].They compared a fuel cell, which operates completely reversible, to a Carnot heatengine. To do so, they divided the fuel cell model into a reversible reaction anda Carnot heat engine. The authors assume, that during the reaction only heat isreleased but no work. To achieve a reversible reaction with this assumption, the heatis released at temperature T ∗ such that the entropy balance

0 = ∆RS(T, p) +Q

T ∗(4.7)

is fulfilled, where ∆RS(T, p) is the entropy difference between the reactants and prod-ucts of the chemical reaction. The temperature T ∗ is different from the temperatureT of the reactants and products. The released heat of the exothermic reaction is thenused as the high temperature heat input for a Carnot heat engine. In the second partof [77] and [76] the released heat of the Carnot heat engine is irreversibly transferredto a reservoir at a lower temperature Tc. Thus, the model is not totally reversibleanymore, but endoreversible, since the irreversibilities are limited to an interaction.The finite rate heat transfer between the engine and the cold reservoir is the limitingfactor of the total fuel cell model of Vaudrey et al. They used different heat transferlaws for this interaction and calculated the efficiency of the endoreversible fuel cellmodel.

Since the assumption of a reversible reaction with heat output only is generally notvery accurate, I will use a reversible engine as discussed in section 3.2.3 instead fora reversible fuel cell model. The balance equations for such a fuel cell model areequations 3.39 to 3.42 or if we assume that the fuel cell operates continuously, webalance the fluxes according to

Volume: 0 = J inV + Jchange

V − JoutV , (4.8)

Particles: 0 = J inH2

+ νH2 ξ − JoutH2, (4.9)

0 = J inO2

+ νO2 ξ − JoutO2, (4.10)

0 = J inH2O + νH2O ξ − Jout

H2O, (4.11)

Entropy: 0 = J inS,H2

+ J inS,O2

+ J inS,H2O +

q

T− Jout

S,H2− Jout

S,O2− Jout

S,H2O (4.12)

Energy: 0 = I inH2

+ I inO2

+ I inH2O + q − p Jchange

V + Pel − IoutH2− Iout

O2− Iout

H2O. (4.13)

77

4 Fuel Cells

0.0 0.5 1.0 1.5 2.00.0

0.5

1.0

1.5

2.0

2.5

iel�@A�cm2D

Pel

�@WD

300K

1000K

2000K

Figure 4.4: Power output Pel over electric current density iel of the reversible fuelcell model at different temperatures. For higher temperatures the poweroutput decreases slightly.

Here, constant total pressure in the fuel cell was assumed [52]. The volume balancein this model does not accurately resemble reality, because all the substances arebalanced with one equation. As described above, the reaction in a fuel cell is actuallydivided into two reactions in the anode and cathode, which are physically separatedby the electrolyte. Thus, the supply and removal of hydrogen on the one hand and thesupply and removal of oxygen and water on the other hand are physically separatedand their volumes are balanced separately. Thus, if we balance all the substances inone equation, the partial pressures of each of the substances are different than thoseresulting from two separate volume balances.

Solving the equation system above, we obtain that the electrical power output is thedifference in Gibbs free energy fluxes of the reactants and products, i.e.

Pel =µoutH2

(J in

H2+ νH2 ξ

)− µin

H2J in

H2

+ µoutO2

(J in

O2+ νO2 ξ

)− µin

O2J in

O2

+ µoutH2O

(J in

H2O + νH2O ξ)− µin

H2O JinH2O. (4.14)

This is the reversible and, therefore, maximum power output from this reaction.Beside the explicit dependence on the extent of reaction rate ξ, the expression forPel also implicitly depends on ξ, because the chemical potentials with the superscript‘out’ change according to the partial pressures in the effluxes. When more of the influxof hydrogen and oxygen is processed, i.e. when ξ gets larger, the partial pressures ofhydrogen and oxygen in the efflux get smaller and that of water gets bigger. However,the explicit linear dependence is dominant, as can be seen in figure 4.4. The diagramshows the power output over the current density iel = 2F ξ.

78


For higher temperatures the power output gets smaller, because the reaction getscloser to its equilibrium (cf. section 3.2.1). At chemical equilibrium the differencein Gibbs free energy between the products and reactants and therefore the poweroutput of the fuel cell would be zero.

With such a simplified model, however, it is not possible to include the dissipationlosses that occur inside a fuel cell. It would be possible to include irreversibilities inthe transport equations. However, this would only effect the supply of hydrogen andoxygen or the removal of water and waste heat. The inner structure of the fuel cell,e.g. the polymer electrolyte membrane, as well as voltage drops and ohmic resistanceare neglected. Therefore, a more detailed model is necessary for an appropriaterepresentation with realistic results regarding the power output and efficiency of afuel cell. Since most of the losses occur inside the fuel cell, the approach of Vaudreyet al. [77, 76], who limited the transfer of waste heat, seems not very accurate.

4.3.2 Endoreversible Fuel Cell Model

Based on the reversible fuel cell model above, I will now develop a more detailedendoreversible model. Dissipation due to the limited electron and proton transferwithin the fuel cell will be considered. Therefore, the model has to be extended inthe following way. Since there are two chemical reactions taking place in a PEMfuel cell (see equations 4.1 and 4.2) it is reasonable to model these with two separatereactors. Between the reactors interactions for the electron and proton transport areneeded. Additionally, a chemical engine without a reaction as discussed in section3.1 needs to be included between the reactors to extract the electrical power of thefuel cell. The irreversibilities, which occur in the fuel cell, shall be limited to theinteractions for the proton and electron transport in this model.

Both, the anode and cathode reaction are modeled as reversible processes, such thatwe only have reversible heat and work exchange at the reactors. The reactor repre-senting the anode reaction shall be subsystem 1, the reactor representing the cathodereaction shall be subsystem 2 and the engine between those shall be subsystem 3. Theproton exchange through the polymer electrolyte membrane is modeled as an irre-versible interaction between subsystems 1 and 2. For the electron flux, i.e. the electriccurrent, we need an interaction between subsystems 1 and 3 and another interactionbetween subsystems 3 and 2. Figure 4.5 shows the structure of this endoreversiblemodel of a PEM fuel cell. The fluxes of hydrogen, oxygen and water are assumed tobe reversible. The figure shows a fuel cell, where total consumption of hydrogen andoxygen is assumed. Usually, that is not the case. Part of the hydrogen is not usedin reactor 1 and oxygen is not supplied in the stoichiometric ratio in reactor 2, sinceit comes from the surrounding air, which is led through the fuel cell [52]. Thus, it is

79

4 Fuel Cells

Energy fluxEntropy fluxParticle fluxVolume flux

H2

q1

Pvol,1

1

H+

2q2

Pvol,2

O2

H2O

3Pel

e−

e−

Figure 4.5: Endoreversible model of a PEM fuel cell. The irreversibilities are limitedto the interactions for the proton and electron transport. Reactors 1 and2 represent the anode and cathode reaction and engine 3 is used for theextraction of electrical power. The reservoirs for the surroundings andthe power output have not been drawn to keep the scheme concise.

necessary to add another interaction for the efflux of the remaining particles. Theremay also be water before the reaction, e.g. from the wet polymer membrane, suchthat an influx of water into reactor 2 could be added. To make the scheme in figure4.5 understandable, I concentrated on the most important parts and did not drawthese interactions. I left out some of the reservoirs, as well, since I am interested inthe process inside the fuel cell rather than the supply and removal of the reactantsand products. In the balance equations below, however, all these interactions will beincluded as they influence the balance of the energy fluxes and therefore the poweroutput, efficiency and cell voltage.

For simplicity, I assume that the fuel cell operates in a steady state. This means, thefluxes and intensities are not time dependent. The supply of hydrogen and oxygenare constant as well as the temperature in the fuel cell.

To calculate the power output in the model, we need to set up the balance equationsfor each of the three subsystems and define the transport laws for the proton and

80


electron interactions. In the following the volume fluxes are denoted by JV = dVdt

, theparticle fluxes of substance l are denoted by Jl, the entropy fluxes of substance l aredenoted by JS,l and the energy fluxes of substance l are denoted by Il. The balanceequations for reactor 1 are

Volume: 0 = J inV,1 + Jchange

V,1 − JoutV,1 , (4.15)

Particles: 0 = J inH2

+ νH2 ξ − JoutH2, (4.16)

0 = νH+,1 ξ − JoutH+,1, (4.17)

0 = νe−,1 ξ − Joute−,1, (4.18)

Entropy: 0 = J inS,H2

+q1

T− Jout

S,H2− Jout

S,H+ (4.19)

Energy: 0 = I inH2

+ q1 − p JchangeV,1 − Iout

H2− Iout

H+ − Ioute−,1. (4.20)

For protons and electrons I assume that all of them are immediately transferredthrough the membrane or outer circle, such that there are no influxes, J in

H+ and J ine− ,

in the particle balances. Accordingly, all protons and electrons are assumed to reactin reactor 2, such that there are no fluxes out of it.

For the second reactor, the balance equations are

Volume: 0 = J inV,2 + Jchange

V,2 − JoutV,2 , (4.21)

Particles: 0 = J inO2

+ νO2 ξ − JoutO2, (4.22)

0 = J inH2O + νH2O ξ − Jout

H2O, (4.23)

0 = J inH+,2 + νH+,2 ξ, (4.24)

0 = J ine−,2 + νe−,2 ξ, (4.25)

Entropy: 0 = J inS,O2

+ J inS,H+ + J in

S,H2O +q2

T− Jout

S,O2− Jout

S,H2O (4.26)

Energy: 0 = I inO2

+ I inH2O + I in

H+ + I ine−,2 + q2 − p Jchange

V,2 − IoutO2− Iout

H2O. (4.27)

For engine 3, there are only two balance equations, since there is only a single exten-sity, the particle flux, carrying the energy:

Particles: 0 = J ine−,3 − Jout

e−,3 (4.28)

Energy: 0 = I ine−,3 + Pel − Iout

e−,3. (4.29)

The energy flux of the electrons at engine k is given by

Ie−,k = µe−,k Je−,k, (4.30)

where µe−,k = µe−,k + z F ϕe−,k is the electrochemical potential. The energy flux ofthe protons through the membrane is carried by particles and entropy, i.e.

IH+,k = T JS,H+,k + µH+,k JH+,k. (4.31)

81

4 Fuel Cells

The proton interaction can be compared with that of the pressure regulator in section2.3.6. The energy fluxes of all the other substances are carried by volume, particleand entropy fluxes, i.e. the energy flux for hydrogen, oxygen and water can be writtenas

Il = T JS,l − pl JV + µl Jl. (4.32)

The total pressure p shall be the same in reactor 1 and 2 and is the sum over thepartial pressures. It can, thus, be written as the sum over all influxes or effluxes ofsubsystems 1 or 2:

p = pinH2

(4.33)

= poutH2

+ poutH+ (4.34)

= pinO2

+ pinH2O + pin

H+ (4.35)

= poutO2

+ poutH2O. (4.36)

Through the interactions, the quantities at all three subsystems are coupled. Sincethe particle fluxes are conserved within interactions, the particle fluxes of electronsand protons are the same at all contact points. This means, the number of electronsleaving reactor 1 is the same as those entering and leaving engine 3 and those enteringreactor 2, i.e. Je− = Jout

e−,1 = J ine−,3 = Jout

e−,3 = J ine−,2. Also, the number of protons leaving

reactor 1 is the same as those entering reactor 2, i.e. JH+ = JoutH+,1 = J in

H+,2. Usingthis, the pressure equations 4.33 to 4.36 and the remaining balance equations, theenergy balance equations of subsystems 1 and 2 can be simplified to

0 = µinH2J in

H2− µout

H2

(J in

H2+ νH2 ξ

)− µH+,1 JH+ − µout

e−,1 Je− (4.37)

0 = µinO2J in

O2− µout

O2

(J in

O2+ νO2 ξ

)+ µin

H2O JinH2O − µout

H2O

(J in

H2O + νH2O ξ)

+ µH+,2 JH+ + µine−,2 Je− . (4.38)

From equations 4.28 and 4.29, we obtain the electrical power output as

Pel = −(µin

e−,3 − µoute−,3

)Je− . (4.39)

Let us check, if the endoreversible model contains the reversible model of section4.3.1. If all the interactions between the subsystems were reversible, the chemicalpotentials were the same at the connected contact points for each of the interactions,i.e. µH+,1 = µH+,2, µout

e−,1 = µine−,3 and µout

e−,3 = µine−,2. Using this, we can combine the

82


three energy balances 4.20, 4.27 and 4.29 and get for the power output

P revel =−

(µout

e−,1 − µine−,2

)Je− (4.40)

= µoutH2

(J in

H2+ νH2 ξ

)− µin

H2J in

H2

+ µoutO2

(J in

O2+ νO2 ξ

)− µin

O2J in

O2

+ µoutH2O

(J in

H2O + νH2O ξ)− µin

H2O JinH2O. (4.41)

As expected, this is the same result as for the reversible model above (see equation4.14).

In case of irreversible interactions, the transport laws for electrons and protons be-tween the subsystems have to be defined. The protons diffuse through the polymerelectrolyte membrane [82]. Thus, according to Fick’s law of diffusion [32] the particleflux for protons shall be proportional to the difference in electrochemical potentialsat the connected contact points, i.e.

JH+ ∝ 1

rm

(µH+,1 − µH+,2) , (4.42)

where rm is the area-specific membrane resistance. The irreversibilities due to sucha transport law are usually called ohmic losses [52], because the voltage drop of thefuel cell due to the membrane resistance, ∆Vm follows approximately Ohm’s law:

∆Vm = iel rm. (4.43)

Assuming, that JH+ is the proton flux per cm2, the current density is

iel = F JH+ . (4.44)

Using this and ∆Vm = 1/F (µH+,1 − µH+,2), we obtain the transport law for theprotons as

JH+ =1

F2 rm

(µH+,1 − µH+,2) . (4.45)

The membrane resistance usually is usually not constant but depends on the op-erating temperature of the fuel cell, the electric current density and the thickness,material and humidity of the membrane. Since there is no accurate theoretical de-scription of these dependencies, we will use an empirical expression. An often citedempirical expression for the resistance of membranes, rm, was given by Mann et al.[54]. They combined a number of previous attempts to obtain:

rm =181.6

(1 + 0.03 iel + 0.062 (T/303)2 i2.5el

)(λ− 0.634− 3 iel) e4.18 ((T−303)/T )

d. (4.46)

83

4 Fuel Cells

0.0 0.2 0.4 0.6 0.8 1.00.0

0.2

0.4

0.6

0.8

1.0

1.2

iel�@A cm-2D

Vfc

�@VD

(a) λ from 8 to 20 in steps of 2

0.0 0.2 0.4 0.6 0.8 1.00.0

0.2

0.4

0.6

0.8

1.0

1.2

iel�@A cm-2D

Vfc

�@VD

(b) d from 250 to 50 µm in steps of 50 µm

0.0 0.2 0.4 0.6 0.8 1.00.0

0.2

0.4

0.6

0.8

1.0

1.2

iel�@A cm-2D

Vfc

�@VD

(c) α from 0.2 to 0.6 in steps of 0.1

0.0 0.2 0.4 0.6 0.8 1.00.0

0.2

0.4

0.6

0.8

1.0

1.2

iel�@A cm-2D

Vfc

�@VD

(d) i0 from 5× 10−4 to 5× 10−9 A cm−2

Figure 4.6: Fuel cell voltage over current density for different values of the parametersλ and d (cf. equation 4.46) and α and i0 (cf. equation 4.48). It can beseen that the activation losses at the electrode (parameters α and i0)cause a lower cell voltage for all current densities. The dissipation due tothe membrane resistance (parameters λ and d) increases with increasingcurrent densities. For all plots the parameters, which are held constant,are λ = 12, d = 200µm, α = 0.4 and i0 = 5×10−4 A cm−2. The ranges ofthe varied parameters are given in the caption of the respective diagram.The first value belongs to the lowest and the second to the highest curve.For i0 the exponent has been changed from -4 to -9 in steps of 1.

It depends on the thickness of the membrane d (in cm), the temperature T (in K)and the current density iel (in A/cm−2). The dimensionless parameter λ is a fittingparameter to capture changes in the resistance due to the humidity and the materialof the membrane. Mann et al. [54] state that it has a maximum value of 23 and thatλ increases with higher amounts of water in the membrane. This expression for the

84


membrane resistance was achieved for Nafion (a trademark of Dupont) membraneswhich are widely used in PEM fuel cells and for which a lot of experimental cellvoltage data was published [54].

Another source of irreversibility are activation losses [52]. In 1905 Tafel observedin experiments with electrochemical cells, that the voltage drop at the surface of anelectrode follows

∆Vact =RT

2αFln

(iel

i0

), (4.47)

which is known as the Tafel equation. The dimensionless parameter α is the so calledcharge transfer coefficient, which depends on the electrode material and the reaction,which takes place at the electrode. It is“the proportion of the electrical energy appliedthat is harnessed in changing the rate of an electrochemical reaction” [52, page 49].For most of the hydrogen electrodes it lies between 0.4 and 0.6 [42]. i0 is the exchangecurrent density, which is a measure for the equilibrium current at the electrode. Inequilibrium the activity at the electrode is not zero, but the forward and backwardelectrode reaction are taking place at the same rate (cf. section 3.2.1). This rate isthe exchange current density i0. Higher exchange current densities are desirable infuel cells, as the electrode is then more ‘active’ [52]. Note that equation 4.47 is onlyvalid for current densities greater than the exchange current density, i.e. iel > i0.

Solving the Tafel equation for iel (it is then called Butler-Vollmer equation [52]) andreplacing iel with F Je− (Je− = JH+ , because νe−,1 = νH+,1), we obtain

Je− =i0F

e2αRT (µe−,1−µe−,3). (4.48)

Here we included the activation losses between subsystems 1 and 3, i.e. at the anodeof the fuel cell.

The transport laws in equations 4.45 and 4.48 contain system specific parameters,namely λ, d, α and i0. The influence of these parameters on the cell voltage Vfc

is shown in figure 4.6. The voltage drop due to the membrane resistance dependshighly on the current density. For low current densities it is negligible while it getssignificant for higher current densities. As one would expect, thicker membranes(higher d) cause bigger voltage drops due to higher resistance. Higher values forλ cause smaller voltage drops, because a higher λ corresponds to a more humidmembrane and thus to a lower resistance. The activation losses are already significantfor small current densities.

Figures 4.7 and 4.8 show the cell voltage over current density for different PEM fuelcells (for values of the parameters see the figure captions). The comparison withexperimental data (dots in the diagrams) shows that qualitatively and quantitatively

85

4 Fuel Cells

0.0 0.5 1.0 1.5 2.00.0

0.2

0.4

0.6

0.8

1.0

iel�@A cm-2D

Vfc

�@VD

Figure 4.7: Cell voltage over current density for a fuel cell operating at tempera-ture T = 298.15 K. The dots show experimental cell voltages of a PEMfuel cell, which were taken from [54, figure 2] (originally published byPrater [66]). The parameters λ and d of equation 4.46 as well as α andi0 of equation 4.48 have been used for fitting. (λ = 17, d = 116µm,α = 0.32, i0 = 2.2× 10−4 A cm−2)

realistic cell voltages can be obtained from the model by adjusting only few param-eters. Since the membrane thickness and electrode material were not given in thepublications, from which the experimental data was taken, d, α and i0 had to befitted. If the fuel cell is better known, less parameters have to be adjusted.

Figure 4.9 shows the power output Pel = iel Vfc and efficiency ηfc (see section 4.2 fordetails) of a PEM fuel cell as a function of the current density. They were calculatedwith the endoreversible fuel cell model with irreversibilities due to membrane resis-tance (cf. equation 4.45) and activation losses (cf. equation 4.48). For these plotsthe parameters from the fit of figure 4.7 have been used. The power output shows amaximum, because the cell voltage decreases while the current density increases.

The entropy production due to the irreversible proton transfer through the membraneand the activation losses at the anode are shown in figure 4.10 for a fuel cell with thesame parameters as in figure 4.7. The entropy production in the proton transfer, i.e.due to the membrane resistance, can be calculated from the difference in the energyfluxes at the connected reactors as

σm =(µH+,1 − µH+,2)

TJH+ . (4.49)

Accordingly, the entropy production due to activation losses, which is located at theelectron transport in our model, reads

σact =(µe−,1 − µe−,3)

TJe− . (4.50)

Both increase with increasing current density.

86

4.4 Summary

0.0 0.2 0.4 0.6 0.80.0

0.2

0.4

0.6

0.8

iel�@A cm-2D

Vfc

�@VD

(a) λ = 6.1, d = 103µm,α = 0.26, i0 = 1.1× 10−4 A cm−2

0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.70.0

0.2

0.4

0.6

0.8

iel�@A cm-2D

Vfc

�@VD

(b) λ = 4.3, d = 75µm,α = 0.44, i0 = 1.3× 10−4 A cm−2

Figure 4.8: Cell voltage over current density for fuel cells operating at temperatureT = 298.15 K. The dots in (a) show experimental cell voltages of a PEMfuel cell from [5, figure 4]. The dots in (b) show experimental cell voltagesof a PEM fuel cell from [73, figure 10B]. The parameters λ and d ofequation 4.46 and α and i0 of equation 4.48 have been used for fitting.

4.4 Summary

In this chapter the methods for incorporating multi-extensity fluxes and chemicalreactions in the endoreversible formalism were used to develop a fuel cell model.First, a basic model was presented, which can be used to calculate the electric poweroutput for given temperatures and pressures of the reactants and products. However,it cannot be used to include the internal irreversibilities of fuel cells. Thus, a complexfuel cell model was developed, which contains irreversibilities due to the resistance inthe proton exchange membrane and activation losses at the electrodes. This modelconnects the influence of the partial pressures of the reactants and products withthe internal irreversibilities. It is thus possible to simultaneously investigate andoptimize the efficiency, power output and cell voltage under different conditions.Furthermore, the entropy production and heat output can be calculated and locatedin different parts of the fuel cell as well. The results presented in section 4.3.2 resembleexperimental data for the cell voltage at different current densities of three differentPEM fuel cells. With the help of few fitting parameters the properties of differentmembranes and electrode materials can be adjusted in the model.

87

4 Fuel Cells

0.0 0.5 1.0 1.5 2.00.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

iel�@A cm-2D

Pel

�@WD

(a) Power output

0.0 0.5 1.0 1.5 2.00.0

0.2

0.4

0.6

0.8

iel�@A cm-2DΗ

fc

(b) Efficiency

Figure 4.9: Electrical power output and efficiency over current density for a fuel celloperating at temperature T = 298.15 K. The same parameters as for thediagram in figure 4.7 have been used.

0.0 0.5 1.0 1.5 2.00.000

0.001

0.002

0.003

0.004

0.005

0.006

iel�@A cm-2D

Σm

(a) σm due to membrane resistance

0.0 0.5 1.0 1.5 2.00.0000

0.0005

0.0010

0.0015

0.0020

0.0025

iel�@A cm-2D

Σac

t

(b) σact due to activation losses

Figure 4.10: Entropy production over current density. (a) σm is the entropy produc-tion due to the membrane resistance. (b) σact is the entropy productiondue to activation losses at the electrode. The same parameters as forthe diagram in figure 4.7 have been used.

88

5 Conclusion

In this thesis two extensions to endoreversible thermodynamics were introduced andapplied to a model of a fuel cell. These extensions open up a wide field of newapplications for endoreversible modeling.

The first extension concerns the interactions in endoreversible models. It was shown,how fluxes with multiple extensities can be included in the formalism. These fluxes oc-cur in many real systems, e.g. in pipes through which gases or liquids are transportedor in combined volume and heat exchanges between subsystems. The extensity fluxesare usually coupled, so that, for instance, the entropy flux can often be expressed asa function of the particle or volume flux. Depending on the constant quantities inthe subsystems and interactions, functions of useful variables for the extensity andenergy fluxes have to be found. The internal energy of gases as a function of theextensities only, i.e. as a function of its natural variables, was presented for an idealgas and a van der Waals gas. It is a reasonable energy function for gas reservoirs,because the energy of reservoirs changes according to the extensity fluxes connectedto it.

For the description of interactions, the intensities and fluxes at the connected contactpoints are needed. The intensities can be expressed as functions of the subsystem’sextensities. However, in the discussed examples it was shown, that depending onthe considered process, functions of other quantities are often more practical. Inisothermal processes, for instance, expressions for all quantities as functions of theconstant temperature simplify the calculations.

When including volume fluxes in endoreversible models, one has to be careful inlocating them. Almost always, volume fluxes and matter fluxes appear at differentcontact points of a subsystem. An influx of particles in a system through a pipe,for instance, may cause a volume exchange of the system with its surroundings butnot with the pipe. The pipe usually has a fixed volume and thus cannot exchangevolume. Note, however, that the volume flux depends on the particle flux in such aprocess.

The second extension, the introduction of a new subsystem – the reactor – enables usto include processes with chemical reactions in endoreversible models. The peculiar-ity of the reactor is that it contains production and destruction terms in the balanceequations for particles. These production terms, which do not appear in conventional

89

5 Conclusion

endoreversible engines, where the fluxes have to be balanced (cf. section 2.1.1), arewell defined by the stoichiometric equation of the considered chemical reaction. Theproduction and destruction of particles does not contradict the reversibility of thesubsystems. However, depending on the considered reaction process, entropy pro-duction may occur in the reactor. If this is the case, the subsystem is no longer areversible system. In my opinion, it seems to be a necessary compromise to acceptthat a reactor is not always a reversible subsystem, as it offers the possibility to modela great number of processes, which can not be considered with the conventional en-doreversible formalism. Furthermore, in many cases irreversible reaction processescan be ignored in endoreversible models. If only the released heat is of interest forfurther processes but not the actual reaction process and the reactants and products,it is sufficient to include the heat as a finite reservoir.

In chapter 4, the extensions have been applied to an endoreversible fuel cell model.First, a very coarse, reversible model was presented, which only consisted of a singlereactor with the in- and effluxes of reactants and products and the heat and workexchange. With this model, the power output of a fuel cell without irreversibilitiescan be calculated, but it is not possible to include the internal irreversibilities ofa fuel cell. Thus, another, more detailed model was developed, which consistedof two reactors for the anode and cathode reaction and an engine for the electricalpower generation. With this model the internal irreversibilities, e.g. due to membraneresistance and activation losses at the electrodes, could be included. The advantageof such an endoreversible model lies in the possibility to simultaneously investigateperformance measures, like cell voltage, efficiency and power output, as well as theentropy production and heat output of the fuel cell. The influence of pressure andtemperature changes as well as different sources of dissipation in the fuel cell canbe investigated with this model. The calculated cell voltages qualitatively and afterfitting only few parameters also quantitatively resemble experimental data.

For further investigations the presented methods can be used to model and optimizea great variety of chemical and electrochemical systems. The discussed fuel cellmodel can be altered to analyze fuel cells with different types of electrolytes, fuelsor operating temperatures. Further sources of dissipation could also be included.Additionally, a more detailed description of the hydrogen supply could be added,since hydrogen is usually stored in high pressure cylinders and then lead through apressure regulator before entering the fuel cell. The presented methods can also beused to find realistic bounds for efficiencies and power output of different kinds ofbatteries. Furthermore, the multi-extensity fluxes can be helpful in studying variouskinds of hydraulic and pneumatic systems.

90

Nomenclature

Greek symbols

µ Chemical potential

µ Electrochemical potential

ν Stoichiometric coefficient

σ Entropy production

ξ Extent of reaction

Latin symbols

Cp Heat capacity at constant pressure

cp = CpnR

CV Heat capacity at constant volume

cV = CVnR

E Energy

F Free energy

F Farraday constant (= 96485 Cmol

)

G Gibbs free energy

H Enthalpy

I Energy flux

iel Electric current density

Iel Electric current

IH Enthalpy flux

J Particle flux

JV Volume flux

JS Entropy flux

N Particle number

n Amount of substance

NA Avogadro constant (= 6.022× 1023 1mol

)

P Power

p Pressure

91

Nomenclature

Q Heat

q Heat flux

R Gas constant (= 8.314 JK mol

)

S Entropy

T Temperature

t Time

V Volume

Vfc Cell voltage

∆Vm Voltage drop in a fuel cell due to membrane resistance

∆Vact Voltage drop in a fuel cell due to activation losses

W Work

z Charge number, e.g. -1 for an electron or +1 for a proton

92

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Selbstandigkeitserklarung gemaß§6 Promotionsordnung

Ich erklare, dass ich die vorliegende Arbeit selbststandig und nur unter Verwendungder angegebenen Literatur und Hilfsmittel angefertigt habe.

Ich erklare, nicht bereits fruher oder gleichzeitig bei anderen Hochschulen oder andieser Universitat ein Promotionsverfahren beantragt zu haben.

Ich erklare, obige Angaben wahrheitsgemaß gemacht zu haben und erkenne die Pro-motionsordnung der Fakultat fur Naturwissenschaften der Technischen UniversitatChemnitz vom 31. Januar 2011 an.

Chemnitz, den 30. April 2014

Katharina Wagner

99

Lebenslauf

Personliche Daten

Name: Katharina Wagner

Geburtsdatum: 25.10.1984

Geburtsort: Frankenberg

Schulischer Werdegang und Studium

1991 – 1995 Grundschule Furth-Glosa, Chemnitz

1995 – 2003 Gymnasium Hohe Straße/Karl-Schmidt-Rottluff-Gymnasium, Chemnitz

Abschluss: Abitur

10/2003 - 08/2008 Studium an der Technischen Universitat Chemnitz

Studiengang: Computational Science

Berufserfahrung, Praktika, Auslandsaufenthalte, etc.

03/2004 Praktikum bei der Chemnitzer Werkstoffmechanik(CWM) GmbH, Chemnitz

09/2004 – 01/2005 Studentische Hilfskraft bei der CWM GmbH, Chemnitz

02/2005 – 06/2006 Hilfswiss. Mitarbeiterin beim Fraunhofer Institut fur Zu-verlassigkeit und Mikrointegration, Chemnitz

09/2005 Praktikum bei der CWM GmbH, Chemnitz

09/2006 – 12/2006 Auslandssemester, University of Helsinki, Finnland

10/2007 – 01/2009 Studentische/wissenschaftliche Hilfskraft am Institut furPhysik, Professur Computerphysik an der TU Chemnitz

101

Lebenslauf

01/2010 – 12/2012 Stipendiatin im Studienforderwerk Klaus Murmann derStiftung der Deutschen Wirtschaft

08/2011 – 09/2011 Forschungsaufenthalt am Institut fur Systemanalyse derRussischen Akademie der Wissenschaften in Pereslawl-Salesski, Russland

09/2011 Studentische Gutachterin in einem Akkreditierungsver-fahren an der Staatl. Universitat Karaganda, Kasachs-tan

04/2012 – 03/2014 Gleichstellungsbeauftragte der Fakultat fur Naturwis-senschaften, TU Chemnitz

07/2013 Forschungsaufenthalt an der University of Oregon in Eu-gene, Oregon, USA

seit 01/2014 Assistant Managing Editor des Journal of Non-Equilibrium Thermodynamics

Veroffentlichungen, Vortrage, Konferenzen

06/2006 Bachelorarbeit”Image Alignment

08/2008 Masterarbeit”A graphic based interface to Endorever-

sible Thermodynamics

06/2009 Joint European Thermodynamics Conference 2009 inKopenhagen, Danemark (Vortrag)

06/2010 Energy Landscapes 2010 in Chemnitz (Teilnahme)

03/2011 DPG Fruhjahrstagung in Dresden (Vortrag)

06/2011 Joint European Thermodynamics Conference 2011 inChemnitz (Organisation)

07/2013 Energy Landscapes 2013 & Optimizing ThermodynamicSystems in Telluride, Colorado, USA (Vortrag)

10/2013 ASME Internal Combustion Engine Fall Technical Con-ference in Dearborn, Michigan, USA (Teilnahme)

03/2014 DPG Fruhjahrstagung in Dresden (Vortrag)

102

an extension to endoreversible thermodynamics for multi

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