Lecture 3. Fuel Cell Thermodynamics

1. Lecture 3. Fuel Cell Thermodynamics • Thermodynamics Review • Heat Potential of a Fuel: Enthalpy of Reaction • Work Potential of a Fuel: Gibbs Free Energy • Reversible Voltage of a Fuel Cell • Fuel Cell Efficiency • Thermal and Mass Balances in Fuel Cells

2. Thermodynamics Review What Is Thermodynamics? The study of the transformation of energy from one form to the other. Thermodynamics can predict upper bound limits on the maximum electrical potential that can be generated in a reaction, or the ideal performance of a fuel cell under ideal conditions.

3. Thermodynamics Review Internal Energy The total intrinsic energy of a fuel is quantified by a property known as internal energy (U). It is associated with microscopic movement and interaction between particles at atomic and molecular levels.

4. Thermodynamics Review First Law The law of conservation of energy: energy can never be created or destroyed, it can only be transformed from one form to the other. dU = dQ – dW = dQ - pdV (closed system)

5. Thermodynamics Review Heat and Work Both are not properties of matter. They represent energy in transit. Work is often called the most “noble” form of energy; it is the a universal donor. Heat is the most “ignoble” form of energy; it is the most universal acceptor. Any form of energy cn eventually be 100% dissipated to the environment as heat, but heat can never be 100% converted to work.

6. Thermodynamics Review Second Law Entropy (S) is a measure of “disorder” of a system. dS = dQrev/ T  0 In any cyclic process the entropy will either increase or remain the same.

7. Thermodynamics Review Second Law It is impossible to extract an amount of heat QH from a hot reservoir and use it all to do work W. Some amount of heat QC must be exhausted to a cold reservoir. It is not possible for heat to flow from a colder body to a warmer body without any work having been done to accomplish this flow.

8. Thermodynamics Review Thermodynamic Potentials Internal Energy (U): Energy needed to create a system in the absence of changes in T or volume. Enthalpy (H): Energy needed to create a system plus the work needed to make room for it (from zero volume). Helmholtz Free Energy (F): Energy needed to create a system minus the energy gain from environment due to spontaneous heat transfer (at constant T). Gibbs Free Energy (G): Energy needed to create a system and make room for it minus energy gain from environment due to heat transfer.

9. Thermodynamics Review Thermodynamic Potentials H = U + PV F = U - TS G = F + PV = U - TS + PV G = H - TS

10. Thermodynamics Review Molar Quantities Intrinsic (size or volume independent) and extrinsic (volume dependent) properties. Molar quantities are intrinsic properties that are independent of system size. Internal energy per mole of gas (kJ/mol) is intrinsic property. It is often convenient to calculate the energy changes due to a reaction one a per-mole basis.

11. Thermodynamics Review Standard State Standard conditions are usually defined as 298.15 K and 1 atm (101.325 Pa), or STP. The standard-state conditions specify that all reactant and product species are present at unit activity.

12. Thermodynamics Review Reversibility Reversible implies thermodynamic equilibrium. In thermodynamics, a reversible process, or reversible cycle if the process is cyclic, is a process that can be "reversed" by means of infinitesimal changes in some property of the system without loss or dissipation of energy.

13. Enthalpy of Reaction The maximum heat energy can be extracted from a fuel is given by the fuel’s enthalpy of a reaction (for a constant pressure process). dH = TdS + VdP = TdS = dU + dW The enthalpy change associated with a combustion reaction is called the heat of combustion. So the enthalpy of a reaction is also called the heat of a reaction (Hrxn).

14. Enthalpy of Reaction The heat of a reaction () can be calculated from the formation enthalpy difference between reactants and products. Usually we use the standard formation enthalpy for each compounds for these calculations. aA + bBmM + nN where A, B are reactants, M, N are products, and a, b, m, and n represent the number of moles.

15. Enthalpy of Reaction =- The enthalpy of reaction is computed from the difference between the molar weighted reactant and product formation enthalpies.

16. Enthalpy of Reaction Heating value or calorific value of a substance, usually a fuel or food, is the amount of heat released during the combustion of a specified amount of it (kJ/mol, kJ/kg, kcal/kg, kcal/mol, BTU/m3) HHV (higher heating value, or gross calorific value, gross energy or upper heating value or higher calorific value HCV): is the heat of combustion by bringing all the products of combustion back to the original pre-combustion temperature, and in particular condensing any vapor produced.

17. Enthalpy of Reaction LHV (lower heating value, net calorific value or lower calorific value LCV): is heating value by subtracting the heat of vaporization of the water vapor from the higher heating value. GHV (gross heating value): accounts for water in the exhaust leaving as vapor, and includes liquid water in the fuel prior to combustion. This value is important for fuels like wood or coal, which will usually contain some amount of water prior to burning.

18. Heating Values of Fuels

19. Enthalpy of Reaction Example 2.1 A direct methanol fuel cell uses methanol as fuel instead of hydrogen. Calculate the and for the combustion reaction: CH3OH + 1.5 O2 CO2 + 2H2O(liq)

20. Enthalpy of Reaction =- = [-393.51+2(-285.83)] – [-245.98+1.5(0)] = -719.19 (kJ/mol) =- = [213.8+269.95)] – [239.83 +1.5(205.14)] = -193.84 (J/mol K)

21. Enthalpy of Reaction The amount of heat energy that a substance can adsorb changes with T, so does the formation enthalpy of a substance. The variation of enthalpy with T is given by: = where is the average specific heat at the average temperature {½(T+T0) }.

22. Enthalpy of Reaction Example 2.2 Determine the enthalpy of formation of water at P= 1atm and temperature of (a) 80 C; (b) 200 C; (c) 650 C; and (d) 1000 C. Notice that for the conditions (a), water can be liquid or vapor, calculation for both cases. [=0 for both H2 and O2; =-285.826 (kJ/mol), -241.826 (kJ/mol) for H2O(l) and H2O(v), respectively.] H2 + 0.5 O2 H2O

23. Enthalpy of Reaction Solution: For T= 80 ºC = 353 K: the average specific heat at 325.5K for H2, O2 , H2O(v),H2O(l) are: 28.964 (J/mol K) 29.535 (J/mol K) 33.860 (J/mol K) 75.339 (J/molK)

24. Enthalpy of Reaction Therefore, the absolute enthalpy of each molecule at 80 ºC can be calculated as: hH2 =,H2+(T-T0)=0+28.964(353-298) = 1,593.0 (J/mol) hO2=,O2+(T-T0)=0+ 29.535(353-298) = 1,624.4 (J/mol)hH2O(l)=,H2O(l)+(T-T0)=-285,826+75.339(353-298) =-281,682 (J/mol) hH2O(v)=,H2O(v)+(T-T0)=-241,826+33.860(353-298) =-239,964 (J/mol)

25. Enthalpy of Reaction The enthalpy of formation of H2O(l) and H2O(v) at 1 atm and 80 ºC can be calculated as: ,H2O(l)(T, P) = hH2O(l)(T,P) – [hH2(T,P)+0.5*hO2(T,P)] = -281,682 – (1,593.0 +0.5*1,624.4) = -248,087 (J/mol) ,H2O(v)(T, P) = hH2O(v)(T,P) – [hH2(T,P)+0.5* hO2(T,P)] = -239,964 – (1,593.0 +0.5*1,624.4) = -242,369 (J/mol)

26. Enthalpy of Reaction Similarly, we can calculate the enthalpy of formation of H2O(v) at 1 atm and 200 ºC, 650 ºC and 1000 ºC: T= 200 ºC, hH2=5,102.1 (J/mol), hO2=5,247.01 (J/mol), hH2O= -235,816 (J/mol), ,H2O(v)(T, P) = hH2O(v)(T,P) – [hH2(T,P)+0.5* hO2(T,P)] = -235,816 – (5,102.1 +0.5*5,247.0) = -243,542 (J/mol)

27. Enthalpy of Reaction T= 650 ºC, hH2=18,334 (J/mol), hO2=20,119 (J/mol), hH2O = -218,996 (J/mol), ,H2O(v) (T, P) = hH2O(v)(T,P) – [hH2(T,P)+0.5* hO2(T,P)] = -218,996 – (18,334 +0.5*20,119) = -247,390 (J/mol) T= 1000 ºC, hH2=28,857 (J/mol), hO2=32,790 (J/mol), hH2O = -204,271 (J/mol), ,H2O(v) (T, P) = hH2O(v)(T,P) – [hH2(T,P)+0.5* hO2(T,P)] = -204,271 – (28,857 +0.5*32,790) = -249,523 (J/mol)

28. Gibbs Free Energy The maximum work can be extracted from a fuel is given by the fuel’s Gibbs free energy change of a reaction. G = H –TS; dG = dH –TdS For constant T, the above equation can be written in terms of molar quantities: - T

29. Reversible Cell Potential Gibbs free energy and electric work dG = dU –TdS + pdV + Vdp dU = TdS –dW = TdS – (pdV + dWelec) dG = [TdS – (pdV + dWelec)] – TdS + pdV + VdP For constant T and P process (dT = dP =0): dG = - dWelec Welec= -grxn

30. Reversible Cell Potential The constant T and P assumption is valid for most fuel cells because the T and P do not change during the fuel cell reaction.

31. Gibbs Free Energy Gibbs free energy and reaction spontaneity G > 0 nonspontaneous (energetically unfavorable) G = 0 equilibrium G < 0 spontaneous (energetically favorable)

32. Reversible Cell Potential Gibbs free energy and Voltage Welec= EQ =E(zF) = - E = - /(zF) For H2-O2 fuel cell, the Gibbs free energy change is of -237 (kJ/mol) under standard conditions for liquid water product, which generates a reversible voltage of 1.23 V.

33. Reversible Cell Potential Example 2.3. Determine the standard reversible cell potential for H2-O2 fuel cell with the following reaction. Obtain the result by considering the product water is (a) liquid (b) vapor, respectively. H2O + 0.5 O2 H2O

34. Reversible Cell Potential Solution: We need to calculate the Gibbs free energy change for the reaction at the standard conditions, then determine the reversible cell potential using the relationship between the Gibbs free energy and the reversible cell potential. There are two ways to calculate the Gibbs free energy change: 1). Using the Gibbs free energy of formation; 2) Using enthalpy and entropy of formation of reactant and product.

35. Reversible Cell Potential Solution: (1). Using Gibbs free energy of formation: • = ,H2O(l)]–[,H2 + 0.5*,H2 ] = -237,180 (J/mol) (Liquid H2O product) • = ,H2O(v)]–[,H2 + 0.5* ,H2 ] = -228,590 (J/mol) (Vapor H2O product)

36. Reversible Cell Potential Solution: The reversible cell potential is: For liquid H2O product: E= -/(zF) =-(-237,180)/(2*96485) = 1.229 (V) For vapor H2O product: E= -/(zF) =-(-228,590)/(2*96485) = 1.185 (V)

37. Open Cell Voltage a A + b B  c C + d D where represents standard Gibbs free energy of formation per mole of molecules, z is the number of electrons transferred for each fuel molecule, ΔE0 is the open cell voltage (OCV) at zero current.

38. Open Cell Voltage for H2 FC represents change in Gibbs free energy of formation per mole of molecules, or molar Gibbs free energy change

39. OCV for H2-O2 Fuel Cells

40. OCV for H2-O2 Fuel Cells OCV (V)

41. OCV for H2-O2 Fuel Cells

42. Reversible Cell Potential Effect of T on reversible cell potential dG = -SdT + VdP E = - /(zF),

43. Reversible Cell Potential For H2-O2 fuel cell,=-44.43 (J/mol K), the variation of reversible cell potential with T is approximated as: =

44. Reversible Cell Potential Effect of P on reversible cell potential dG = -SdT + VdP E = - /(zF), For ideal gas,

45. Reversible Cell Potential Effect of concentration on reversible cell potential Nernst Equation For fuel cell based on H2 + 0.5 O2H2O reaction,

46. Carnot Cycle The efficiency of a heat engine—that is, a device for converting the thermal energy of a system at a high temperature into mechanical work—is best assessed in terms of the Carnot cycle. This cycle is a sequence of thermodynamically reversible operations carried out on a gas between two temperatures, the final state being identical with the initial.

47. Q1 Q=0 Q=0 Q2 Carnot Cycle • A to B: Isothermal expansion • B to C: Adiabatic expansion • C to D: Isothermal compression • D to A: Adiabatic compression

48. Carnot Cycle Assumptions: ideal gas, reversible process The heat absorbed at the higher temperature (T1) is and the heat lost at the lower temperature (T2) is from thermodynamics of adiabatic process, we can prove that for adiabatic process

49. Carnot Cycle : the net work output from the Carnot cycle is:

50. Fuel Cell Efficiency For any energy conversion device, efficiency is of great importance. We define the efficiency  of a conversion process as the amount of useful work extracted from the process relative to the total energy involved by the process.