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Efficiency and Open Circuit Voltage (OCV) Why we have to study efficiency of fuel cell? From the definition and calculation of efficiency of fuel cell, the limitation of Fuel Cell can be discussed. Also, it can give us • OCV • Pressure effect • Gas concentration effect • Temperature effect The electrical power/energy can be defined as following: Power = VI (w) or Energy = VIt (J) However, the energy of chemical input and output is not so easily defined. It can be said as “chemical energy”, but the “chemical energy” is NOT simple defined. There are several terms to describe the “chemical energy”: • Internal energy • Gibbs free energy • Enthalpy • Helmhotz function These terms are defined in the thermodynamics. So, let’s review it before discuss more further. Thermodynamics Review Internal energy A fuel cell converts energy stored within a fuel into other, more useful forms of energy. The total intrinsic energy of a fuel is quantified by a property known as internal energy (U). First Law The first law of thermodynamics is also known as the law of conservation of energy. That means the energy can never be created or destroyed. It can be expressed as following equation: dEuniv = dEsystem + dEsurroundings = 0 or dEsystem = -dEsurroundings The equation states the change of energy of a (closed) system must be equal to the energy transfer to the surroundings. There are two ways that energy can be transferred between a (closed) system and its surroundings: via heat (Q) or work (W). So, the above equation can be rewritten as: dU = dQ - dW 19 dWmech = pdV and dW = dWmech + dWelectrical The expression for the internal energy change of a system is: dU = dQ – pdV – dWelectrical Second Law The second law of thermodynamics introduces the concept of entropy. Entropy is determined by the number of possible ways of configuring a system. Entropy can be thought of as a measure of “disorder,” since an increasing entropy indicates an increasing number of ways of configuring a system. Except for extremely simple system, it is impossible to calculate entropy exactly. Instead, a system’s entropy is usually inferred based on how heat transfer causes the entropy of the system to change. Q dS = T rev and Q dS > T irr so, dS Q T From above equation, it can be rewrite as following: dS = Q + S gen T It can be concluded that Sgen 0 Enthalpy The enthalpy (H) is a description of thermodynamic potential of a system, which can be used to calculate the "useful" work obtainable from a closed thermodynamic system under constant pressure. The energy needed to create a system plus the work needed to make room for it. H = U + pV dH = dU + Vdp + pdV and dU = TdS – pdV dH = TdS + Vdp 20 Thermodynamic Potentials The thermodynamic potentials indicate how energy can be transferred from one form to another based on the first and second laws of thermodynamics. From the first and second law of thermodynamics and ignore the electrical work, dU = dQ – pdV and Q dS = T rev the equation can be combined as dU = TdS – pdV Two useful relations are driven which show the dependent variables (T and P) are related to variables in the independent variables (S and V): dU =T dS V and dU = p dV S Unfortunately, S and V are not easily measurable in most experiments (there is no such thing as an “entropy meter.”). Therefore, a new thermodynamic potential is needed equivalent to U but depending on quantities that are more readily measured than S and V. Temperature T and pressure P fall into this category. Helmhotz function Helmholtz energy (A) is a thermodynamic potential which measures the “useful” work obtainable from a closed thermodynamic system at a constant temperature. The energy needed to created a system and make room for it minus the energy that you can get from system’s environment due to heat transfer. dF = dU – SdT – TdS and dU = TdS – pdV 21 dF = -SdT – pdV Gibbs free energy The Gibbs free energy (G) can be considered to be the net energy required to created a system and make room for it minus the energy received from the environment due to heat transfer. In the other words, G represents the net energy cost for a system created at a constant environmental temperature T from a negligible initial volume after subtracting what the environment automatically supplied. and dG = dU – SdT – TdS + Vdp + pdV where dU = TdS – pdV dG = -SdT + Vdp If G represents the net energy you had to transfer to create system, then G should also represent the maximum energy that you ever get back out of the system. For Fuel cell, the Gibbs free energy is important. 22 In fuel cell, the “Gibbs free energy of formation”, Gf, is used instead of the “Gibbs free energy” since Gf gives us the energy released. This change is the different between the Gibbs free energy of products and the Gibbs free energy of the inputs or reactants. Gf = G of products – G of reactants Standard State Because most thermodynamic quantities depend on temperature and pressure, it is convenient to reference everything to a standard set of conditions. This set of conditions is called the standard-state. Standard-state conditions are taken as room temperature and atmospheric pressure. (298.15K and 1atm = 101.325kPa ~ 1bar) The standard-state conditions are also known as standard temperature and pressure, or STP. Standard-state condition are designated by a superscript zero, i.e., Po, To, Ho, Go, or So . Molar Quantities Typical notation distinguishes between intensive and extensive variables. Intensive quantities such as temperature and pressure are independent of system size. In the other hand, the extensive quantities, i.e. internal energy or entropy, are dependent of system size. It is conventional to denote intensive quantities with a lowercase letter (p, pressure) and extensive quantities with an uppercase letter (U, internal energy). Molar quantities are intensive. It can be defined as physical quantities per mole and denoted as lowercase letter with bar. For instant, • • • • Internal energy, u (kJ/mol) Enthalpy, h (kJ/mol) Entropy, s (kJ/mol) Gibb free energy, g (kJ/mol) For the quantities at STP, • • • • Internal energy, u o (kJ/mol) Enthalpy, h o (kJ/mol) Entropy, s o (kJ/mol) Gibb free energy, g o (kJ/mol) For Gibbs free energy of formation, the denotation is g f . Reversibility From the below figure, it shows a ball has potential energy (PE=mgh) but no kinetic energy (KE) in position A. When this ball is released at point A, the PE starts to convert to the KE. Until the ball reaches point B, the PE has been converted to KE completely if there is no friction loss, wind resistance. This process without energy loss is reversible which means the ball can roll up the other side and recover its PE. 23 Simple Reversible Process The energy loss involved in the process is called irreversible process. Once the irreversible process happens, the system cannot get back to the original status. Calculating Reaction Enthalpies For a reaction aA + bB mM + nN, the formation of enthalpy (reaction of enthalpy) also can be defined at STP. h of = mh o M + nh o N a h o A + b h o B An expression analogous to above equation may be written for the standard-state entropy of a reaction. s of = ms o M + ns o N a s o A + b s o B Ex: Calculate the h of and s of for the methanol combustion. CH3OH + 3/2 O2 CO2 + 2H2O (liquid) Solution: Chemical Species h of (kJ/mole) s of (J/mole-K) CH3OH -200.95 239.83 O2 0 205.14 CO2 -393.51 213.80 H2O (liquid) -285.83 69.95 h o f = h o CO2 + 2 h o H 2O h o CH 3OH + 1.5 h o O2 =-719.19 kJ/mole o o o o o s f = s CO2 + 2 s H 2O s CH 3OH + 1.5 s O2 =-193.84 J/mole-K Temperature dependence of enthalpy hf = ho f + T To C p (T )dT where T0 = 298.15 K; Cp is the constant-pressure heat capacity of the substance. Temperature dependence of entropy 24 s = so + T C p (T ) To T dT where T0 = 298.15 K; Cp is the constant-pressure heat capacity of the substance. Calculating Gibbs Free Energy of Formation Recall the equation of Gibbs free energy of formation: Gf = G of products – G of reactants It can be re-written in molar quantities form as following: g f = g prod g react Consider a reaction aA + bB mM + nN The Gibbs free energy of formation can be calculated as [ ] [ g f = mg M + ng N a g A + b g B ] Ex: Calculate the Gibbs free energy of formation of the reaction H2 + O2 H2O at STP. Solution: From the reaction equation, the Gibbs free energy can be expressed as: g f = g H 2O g H 2 0.5 g O2 From thermodynamics properties table, the Gibbs free energy at STP (1atm, 25oC) can be found. g H 2O = 306.69 kJ mole g H 2 = 38.96 kJ mole = 61.12 kJ mole g O2 So, g f = (-306.69)-(-38.96+0.5*(-61.12)) = -237.13 (kJ/mole) The other way to determined the Gibbs free energy when the thermodynamic properties of the Gibbs free energy are lacked: Recalling the definition of G and H, where G = U TS + PV and H = U + PV It can be concluded as: G = H - TS 25 Differentiating the expression gives dG = dH – TdS – SdT At the isothermal process (constant temperature), this relationship in terms of molar quantities gives g f = h f T s f Ex: Determine the Gibbs free energy of formation of the reaction H2 + O2 H2O at STP. Solution: The enthalpy and entropy properties can be found from the thermodynamic table as following: h H 2O = 285.83 kJ mole h H 2 = 0 kJ mole s H 2O = 69.95 J mole K and h O2 = 0 kJ mole s H 2 = 130.68 J mole K sO2 = 205.0 J mole K h f = (-285.83)-(0+0.5*0) = -285.83 (kJ/mole) s f = (69.95)-(130.68+0.5*205.0) = -163.23 (J/mole-K) g f = h f T s f =-285830 – (163.23 * 298.15) = -237.16 kJ/mole 2.1 Reversible Open Circuit Voltage (Reversible OCV) Again, from the definition of Gibbs free energy, dG = dU – SdT – TdS + Vdp + pdV and, the internal energy is dU = dQ - dW dU = TdS – pdV – dWelectrical which yields dG as dG = – SdT + Vdp – dWelectrical For a constant-temperature, constant-pressure process (dT, dP = 0) , the equation reduces to dG = – dWelectrical For a reaction using molar quantities, the equation can be written as: Welectrical = g f 26 Remember that the constant temperature, constant pressure assumption used here is not really as restrictive as it seems. The only limitation is that the temperature and pressure do not vary during the reaction process. Since fuel cells usually operate at constant temperature and pressure, this assumption is reasonable. It is important to realize that the expression derived above is valid for different values of temperature and pressure as long as these values are not changing during the reaction. We could apply this equation for T = 200 K and p = 1 atm, or just as validly apply it for T = 400 K and p = 5 atm. For the electrical work, Welectrical = charge (C) * voltage (J/C) = zFE (J) Where F is Faraday constant which means the charge of a mole of electrons. z is the number of electrons produced per mole of the fuel. N is Avogadro’s number, e is the charge on one electron. F = Ne = 6.022 10231.602 10-19 = 96485 C So, zFE = g f The reversible OCV of a fuel cell is, E= g f zF Ex: For PEMFC, please determine its reversible open circuit voltage at STP. Solution: The PEMFC’s overall reaction is H2 + O2 H2O which produced 2 mole electrons per mole fuels. So that, z = 2. From the reaction equation, the Gibbs free energy can be expressed as: g f = g H 2O g H 2 0.5 g O2 From thermodynamics properties table, the Gibbs free energy at STP (1atm, 25oC) can be found. g H 2O = 306.69 kJ mole g H2 = 38.96 kJ mole g O2 = 61.12 kJ mole So, g f = (-306.69)-(-38.96+0.5*(-61.12)) = -237.13 (kJ/mole) The reversible open circuit voltage can be calculated: 27 E= g f zF = - (-273130) / (2 * 96485) = 1.23 (V) The Effect of Temperature, Pressure, and Gas Concentration The reversible open circuit voltage of a fuel cell is determined at the standard-state. However, a fuel cell usually operates at the conditions that vary greatly from the STP. For instant, the SOFC operate at 700~1000 oC. In that case, the Gibbs free energy changes in a chemical reaction vary with temperature, Pressure, and gas concentration. Thus, the reversible OCV will be changed. Reversible OCV Variation with Temperature To understand how the reversible voltage varies with temperature, we need to go back to the original differential expression for Gibbs free energy: dG = -SdT + VdP At the constant pressure, dG = S dT p Use the molar quantities of Gibbs free energy of formation, the equation becomes, d g dT ( ) p = s from the equation, zFE = g f The variation of reversible OCV with temperature can be expressed as following. dE s = dT p nF Therefore, the voltage varies as a function of temperature can be derived. E = s (T T o ) zF So, the reversible OCV varied with temperature (ET) are ET = E o + E = E o + 28 s (T T o ) zF s is assumed to be independent of temperature. If more accurate value of ET is needed, it may be calculated by integrating the heat capacity related temperature dependence of s . From the equation, • If s for a chemical reaction is positive, then ET will increase with temperature. • If s for a chemical reaction is negative, then ET will decrease with temperature. For most fuel cell reactions, s is negative; therefore reversible fuel cell voltages tend to decrease with increasing temperature. For a PEMFC, s =-44.43 J/mole-K. The variation of cell voltage with temperature is approximated as ET = E o + s 44.43 (T T o ) = E o + (T T o ) = E o 2.3 104 (T T o ) zF 2 96485 Thus, for every 100 degrees increase in cell temperature, there is an approximate 23 mV decrease in cell voltage. Should we operate the fuel cell at low temperature since the reversible OCV decreases by increasing the operation temperature? NO! Activation losses tend to decrease with increasing temperature. Therefore real fuel cell performance typically increases with increasing temperature even though the thermodynamically reversible OCV decreases. The other way to determine the reversible OCV reversible OCV varied with temperature (ET) is, [ ] [ g f (T ) = mg M (T ) + ng N (T ) a g A (T ) + b g B (T ) E= ] g f (T ) zF For PEMFC at 400 K, [ ] [ ] g f (T ) = g H 2O (T ) g H 2 (T ) + 0.5 g O2 (T ) =(-317.89)-(-52.73)-0.5(82.51) = 221.405 kJ/mol Thus, E= g f (T ) zF = 221405 = 1.147V 2 96485 Reversible OCV Variation with Pressure and Gas Concentration Consider a general reaction such as jJ + kK mM In the case of gases behaving as “ideal gases”, the activity, a, can be defined as 29 ai = Pi Po where P is the pressure or partial pressure of the gas PO is standard pressure, 0.1 MPa. The activity is proportional to partial pressure. For above reaction equation, the activities can be shown. PJ P P a = Ko ; a M = Mo o ; K P P P aJ = The activities of the reactants and products modify the Gibbs free energy change of a reaction. It can be expressed that in above chemical reaction. a j ak o g f = g f RT ln J m K aM For PEMFC, H2 + O2 H2O g f = o g f 1 2 a H 2 aO2 RT ln a H 2O If the activity of reactant increases, g f becomes more negative, that means more energy is released. Plug into the equation zFE = g f The equation of reversible OCV becomes E= g of zF + k RT a Jj a K ln m zF a M k RT a Jj a K = E o + ln m zF a M The equation in terms of product and reactant activity is called Nernst equations. From the definition of activity, aJ = PJ P P a = Ko ; a M = Mo o ; K P P P and Po = 1 bar with all pressure are given in bar, the equation is simplifies to E = Eo + RT PJ j PKk ln zF PMm 30 All pressures in the equation are partial pressure. If the system pressure is P, and , , are constants depending on the molar masses and concentrations of the species. PJ = P; PK = P; PM = P The equation becomes E = Eo + k k RT Jj K ( j +km) RT Jj K = E o + ln m P ln zF M zF Mm RT + ( j + k m) ln( P) zF The second term of equation can be seen as the effect is occurred by the concentration of fuels. The third is the effect of the system pressure. For PEMFC, H2 + O2 H2O 1 2 RT PH 2 PO2 E=E + ln zF PH 2O o 1 2 RT RT E = Eo + ln ln( P) + zF 4F Partial Pressure In a mixture of gases, the total pressure is the sum of all the “partial pressure” of the components of the mixtures. Partial Pressures of Atmospheric Gases Gas Partial Pressure (MPa) Nitrogen 0.07809 Oxygen 0.02095 Argon 0.00093 Others (including CO2) 0.00003 Total 0.10000 System pressure The system pressure effect can be described by the last term of Nerst equation. For a PEMFC, RT ( j + k m) ln( P) zF RT ln( P) 4F If the pressure changes from P1 to P2, the voltage difference is, 31 V = P RT ln 2 4F P1 A similar effect occurs when studying the change from air to oxygen. Recall the Nerst equation of a PEMFC, 1 2 RT RT E = Eo + ln ln( P) + 2F 4F E = Eo + RT 1 RT RT ln + ln( ) + ln( P) 2F 2 2F 4F So, the voltage change caused by using oxygen instead of air is, V = RT 1 8.314 (80 + 273) 1 RT ln 2 = ln ln = = 0.012V 0.21 0.21 4 96485 4F 1 4F In fact, there are many reports show the change of voltage will more significant than the prediction of Nerst equation that is because the pressure effect will reduce the losses. Efficiency and Efficiency Limit The efficiency of a fuel cell is not straightforward to be defined. From the previous section in this chapter, the formation of Gibbs free energy is used to define the useful energy in a fuel cell. If it were not for irreversibilities, all this energy would be converted into electrical energy, and the efficiency could be said to be 100%. So, the fuel cell efficiency definition can be expressed as: electrical energy produced Gibbs free energy change However, it is not very useful, and is rarely done, as whatever conditions are used the efficiency limit is always 100%. In contrast to a fuel cell, the maximum theoretical efficiency of a conventional heat/expansion engine is described by the Carnot cycle. This efficiency may be derived from classical thermodynamics. It can be expressed as: Carnot = T H Tc T = 1 C TH TH In the expression, TH is the maximum temperature of heat engine and TL is the rejection temperature of the heat engine. For a heat engine that operates at 400 oC (673 K) and reject heat at 50 oC (323 K), the efficiency is 52%. Let’s consider a hydrogen-oxygen reaction, no wonder that occurs in a heat engine or a fuel cell, the total input energy, the energy to produce water (said formation of enthalpy, h f ), is the same. So, the suitable definition of a fuel cell is the ration of useful energy to total energy. 32 However, there are two different values that we can use for h f . For the reaction which produces steam, h f = -243.83 kJ/mole. For the product water is condensed back to liquid, h f =-285.84 kJ/mole. The higher figure is called the higher heating value (HHV) and the lower one is called the lower heating value (LHV). Which of these values should be used in computing a fuel cell’s efficiency? The most equitable calculations of fuel cell efficiency use the HHV. Using the HHV instead of the LHV is appropriate because it acknowledge the true total heat that could theoretically be recovered from hydrogen combustion reaction. Use of the LHV will result in higher, but perhaps misleading, efficiency numbers. From the above discussion, it can be concluded that the efficiency limit (or reversible efficiency, maximum efficiency) can be written as: = g f h f 100% If the equation of efficiency can be expressed in voltage, the calculation can be more convenient to perform. Eh = h f zF =1.48 V if using the HHV So that the efficiency becomes, = E g f V = = 100% Eh h f 1.48 33