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AME 513 Principles of Combustion

AME 513 Principles of Combustion. Lecture 3 Chemical thermodynamics I – 2 nd Law. Outline. Why do we need to invoke chemical equilibrium? Degrees Of Reaction Freedom (DORFs) Conservation of atoms Second Law of Thermodynamics for reactive systems Entropy of an ideal gas mixture

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AME 513 Principles of Combustion

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  1. AME 513Principles of Combustion Lecture 3 Chemical thermodynamics I – 2nd Law

  2. Outline • Why do we need to invoke chemical equilibrium? • Degrees Of Reaction Freedom (DORFs) • Conservation of atoms • Second Law of Thermodynamics for reactive systems • Entropy of an ideal gas mixture • Equilibrium constants • Application of chemical equilibrium to hydrocarbon-air combustion • Application of chemical equilibrium to compression/expansion AME 513 - Fall 2012 - Lecture 3 - Chemical Thermodynamics 2

  3. Why do we need chemical equilibrium? • (From lecture 2) What if we assume more products, e.g. 1CH4 + 2(O2 + 3.77N2)  ? CO2 + ? H2O + ? N2+ ? CO In this case how do we know the amount of CO vs. CO2? • And if if we assume only 3 products, how do we know that the “preferred” products are CO2, H2O and N2 rather than (for example) CO, H2O2 and N2O? • Need chemical equilibrium to decide - use 2nd Law to determine the worst possible end state of the mixture AME 513 - Fall 2012 - Lecture 3 - Chemical Thermodynamics 2

  4. Degrees of reaction freedom (DoRFs) • If we have a reacting “soup” of CO, O2, CO2, H2O, H2 and OH, can we specify changes in the amount of each molecule independently? No, we must conserve each type of atom (3 in this case) • Conservation of C atoms: nCO + nCO2 = constant O atoms: nCO + 2nCO2 + 2nO2 + nH2O + nOH = constant H atoms: 2nH2O + 2nH2 + nOH = constant • 3 equations, 6 unknown ni’s 3 degrees of reaction freedom (DoRFs) • # of DoRFs = # of different molecules (n) - # of different elements • Each DoRF will result in the requirement for one equilibrium constraint AME 513 - Fall 2012 - Lecture 3 - Chemical Thermodynamics 2

  5. Conservation of atoms • Typically we apply conservation of atoms by requiring that the ratios of atoms are constant, i.e. the same in the reactants as the products C atoms: nCO + nCO2 = constant O atoms: nCO + 2nCO2 + 2nO2 + nH2O + nOH = constant H atoms: 2nH2O + 2nH2 + nOH = constant • Specifying nO/nH also would be redundant, so the number of atom ratio constraints = # of atoms - 1 • What are these “constants?” Depends on initial mixture, e.g. for stoichiometric CH4 in O2, nC/nO = 1/4,nC/nH = 1/4 AME 513 - Fall 2012 - Lecture 3 - Chemical Thermodynamics 2

  6. 2nd law of thermo for reacting systems • Constraints for reacting system • First law: dE = Q - W = Q - PdV • Second law: dS ≥ Q/T • Combine: TdS - dU - PdV ≥ 0 for any allowable change in the state of the system • For a system at fixed T and P (e.g. material in a piston-cylinder with fixed weight on top of piston, all placed in isothermal bath: d(TS-U-PV) ≥ 0, or per unit mass d(Ts-u-Pv) ≥ 0 • Define “Gibbs function” g  h - Ts = u + Pv - Ts • Thus for system at fixed T and P: d(-g) ≥ 0 or dg ≤ 0 • Thus at equilibrium, dg = 0or dG = 0 (g or G is minimum) and for each species i µi = chemical potential of species i • Similarly, for system at fixed U and V (e.g. insulated chamber of fixed volume): ds ≥ 0, at equilibrium ds = 0or dS = 0 (s is maximum) AME 513 - Fall 2012 - Lecture 3 - Chemical Thermodynamics 2

  7. Entropy of an ideal gas mixture • Depends on P AND T (unlike h and u, which depend ONLY on T) •   (1) (2) (3) • (1) Start with gases A, B and C at temperature T and Po = 1 atm • (2) Raise/lower each gas to pressure P ≠ 1 atm (same P for all) A P = Pref A P≠Pref C P = Pref C P ≠Pref B P = Pref B P ≠ Pref A, B, C P ≠ Pref AME 513 - Fall 2012 - Lecture 3 - Chemical Thermodynamics 2

  8. Entropy of an ideal gas mixture • (3) Now remove dividers, VA VA + VB + VC (V = volume) • Can also combine Xi and P/Pref terms using partial pressures Pi; for an ideal gas mixture Xi = Pi/P, where (as above) P without subscripts is the total pressure = SPi AME 513 - Fall 2012 - Lecture 3 - Chemical Thermodynamics 2

  9. Entropy of an ideal gas mixture • Most useful form - entropy per unit mass (m), since mass is constant: ln(P/Pref) - entropy associated with pressure different from 1 atm - P > Pref leads to decrease in s Xiln(Xi) - entropy associated with mixing (Xi < 1 means more than 1 specie is present - always leads to increase in entropy since -Xiln(Xi) > 0) • Denominator is just the average molecular weight • Units of s are J/kgK AME 513 - Fall 2012 - Lecture 3 - Chemical Thermodynamics 2

  10. Equilibrium of an ideal gas mixture • How to use this result? dG = 0 means there are n equations of the form ∂G/∂ni = 0 for n unknown species concentrations! • Break system into multiple sub-systems, each with 1 DoRF, e.g.. for a chemical reaction of the form AA + BB  CC + DD e.g. 1 H2 +1 I2 2 HI A = H2, A = -1, B = I2, B = -1, C = HI, C = 2, D = nothing, D = -0 conservation of atoms requires that • G = H – TS, where AME 513 - Fall 2012 - Lecture 3 - Chemical Thermodynamics 2

  11. Equilibrium of an ideal gas mixture AME 513 - Fall 2012 - Lecture 3 - Chemical Thermodynamics 2

  12. Equilibrium of an ideal gas mixture It is fairly inconvenient to work with the exp[-g/RT] terms, but we can tabulate for every species its equilibrium constant Ki, i.e. the value of this term for the equilibrium of the species with its forming elements in their standard states, e.g. So in general for a reaction of the form AA + BB  CC + DD AME 513 - Fall 2012 - Lecture 3 - Chemical Thermodynamics 2

  13. Equilibrium constants • Examples of tabulated data on K - (double-click table to open Excel spreadsheet with all data for CO, O, CO2, C, O2, H, OH, H2O, H2, N2, NO at 200K - 6000K) • Note K = 1 at all T for elements in their standard state (e.g. O2) AME 513 - Fall 2012 - Lecture 3 - Chemical Thermodynamics 2

  14. Chemical equlibrium - example • For a mixture of CO, O2 and CO2 (and nothing else, note 1 DoRF for this case) at 10 atm and 2500K with C:O = 1:3, what are the mole fractions of CO, O2 and CO2? 1 CO2  1 CO + .5 O2 • 3 equations, 3 unknowns: XCO2 = 0.6495, XO2 = 0.3376, XCO = 0.0129 • At 10 atm, 1000K: XCO2 = .6667, XO2 = 0.3333, XCO = 2.19 x 10-11 • At 1 atm, 2500K: XCO2 = 0.6160, XO2 = 0.3460, XCO = 0.0380 • With N2 addition, C:O:N = 1:3:6, 10 atm, 2500K: XCO2 = 0.2141, XO2 = 0.1143, XCO = 0.0073, XN2 = 0.6643 (XCO2/XCO = 29.3 vs. 50.3 without N2 dilution) (note still 1 DoRF in this case) • Note high T, low P and dilution favor dissociation AME 513 - Fall 2012 - Lecture 3 - Chemical Thermodynamics 2

  15. Chemical equlibrium • How do I know to write the equilibrium as 1 CO2  1 CO + .5 O2 and not (for example) 2 CO + 1 O2 2 CO2 For the first form which is the appropriate expression of equilibrium for the 2nd form - so the answer is, it doesn’t matter as long as you’re consistent AME 513 - Fall 2012 - Lecture 3 - Chemical Thermodynamics 2

  16. Chemical equlibrium - hydrocarbons • Reactants: CxHy + rO2 + sN2 (not necessarily stoichiometric) • Assumed products: CO2, CO, O2, O, H2O, OH, H, H2, N2, NO • How many equations? • 10 species, 4 elements  6 DoRFs  6 equil. constraints • 4 types of atoms  3 atom ratio constraints • Conservation of energy (hreactants = hproducts) (constant pressure reaction) or ureactants = uproducts (constant volume reaction) • Pressure = constant or (for const. vol.) • 6 + 3 + 1 + 1 + 1 = 12 equations • How many unknowns? • 10 species  10 mole fractions (Xi) • Temperature • Pressure • 10 + 1 + 1 = 12 equations AME 513 - Fall 2012 - Lecture 3 - Chemical Thermodynamics 2

  17. Chemical equlibrium - hydrocarbons • Equilibrium constraints - not a unique set, but for any set • Each species appear in at least one constraint • Each constraint must have exactly 1 DoRF • Note that the initial fuel molecule does not necessarily appear in the set of products! If the fuel is a large molecule, e.g. C8H18, its entropy is so low compared to other species that the probability of finding it in the equilibrium products is negligible! • Example set (not unique) AME 513 - Fall 2012 - Lecture 3 - Chemical Thermodynamics 2

  18. Chemical equlibrium - hydrocarbons • Atom ratios • Sum of all mole fractions = 1 • Conservation of energy (constant P shown) AME 513 - Fall 2012 - Lecture 3 - Chemical Thermodynamics 2

  19. Chemical equlibrium - hydrocarbons • This set of 12 simultaneous nonlinear algebraic equations looks hopeless, but computer programs (using slightly different methods more amenable to automation) (e.g. GASEQ) exist • Typical result, for stoichiometric CH4-air, 1 atm, constant P AME 513 - Fall 2012 - Lecture 3 - Chemical Thermodynamics 2

  20. Chemical equlibrium - hydrocarbons • Most of products are CO2, H2O and N2 but some dissociation (into the other 7 species) occurs • Product  is much lower than reactants - affects estimation of compression / expansion processes using Pv relations • Bad things like NO and CO appear in relatively high concentrations in products, but will recombine to some extent during expansion • By the time the expansion is completed, according to equilibrium calculations, practically all of the NO and CO should disappear, but in reality they don’t - as T and P decrease during expansion, reaction rates decrease, thus at some point the reaction becomes “frozen”, leaving NO and CO “stuck” at concentrations MUCH higher than equilibrium AME 513 - Fall 2012 - Lecture 3 - Chemical Thermodynamics 2

  21. Adiabatic flame temp. - hydrocarbons AME 513 - Fall 2012 - Lecture 3 - Chemical Thermodynamics 2

  22. Adiabatic flame temp - hydrocarbons • Adiabatic flame temperature (Tad) peaks slightly rich of stoichiometric - since O2 is highly diluted with N2, burning slightly rich ensures all of O2 is consumed without adding a lot of extra unburnable molecules • Tad peaks at ≈ 2200K for CH4, slightly higher for C3H8, iso-octane (C8H18) practically indistinguishable from C3H8 • H2 has far heating value per unit fuel mass, but only slightly higher per unit total mass (due to “heavy” air), so Tad not that much higher • Also - massive dissociation as T increases above ≈ 2400K, keeps peak temperature down near stoichiometric • Also - since stoichiometric is already 29.6% H2 in air (vs. 9.52% for CH4, 4.03% for C3H8), so going rich does not add as many excess fuel molecules • CH4 - O2 MUCH higher - no N2 to soak up thermal energy without contributing enthalpy release • Constant volume - same trends but higher Tad AME 513 - Fall 2012 - Lecture 3 - Chemical Thermodynamics 2

  23. Compression / expansion • Compression / expansion processes are typically assumed to occur at constant entropy (not constant h or u) • Use sreactants = sproducts constant for compression / expansion instead of hreactants = hproducts or ureactants = uproducts • All other relations (atom ratios, Xi = 1, equilibrium constraints) still apply AME 513 - Fall 2012 - Lecture 3 - Chemical Thermodynamics 2

  24. Compression / expansion • Three levels of approximation • Frozen composition (no change in Xi’s), corresponds to infinitely slow reaction • Equilibrium composition (Xi’s change to new equilibrium), corresponds to infinitely fast reaction (since once we get to equilibrium, no further change in composition can occur) • Reacting composition (finite reaction rate, not infinitely fast or slow) - more like reality but MUCH more difficult to analyze since rate equations for 100’s or 1000’s of reactions are involved • Which gives most work output? • Equilibrium - you’re getting everything the gas has to offer; recombination (e.g. H + OH  H2O) gives extra enthalpy release, thus more push on piston or more kinetic energy of exhaust • Frozen - no recombination, no extra heat release • Reacting - somewhere between, most realistic AME 513 - Fall 2012 - Lecture 3 - Chemical Thermodynamics 2

  25. Compression / expansion • Example - expansion of CO2-O2-CO mixture from 10 atm, 2500K to 1 atm in steady-flow control volume (e.g. nozzle) or control mass (e.g piston/cylinder) • Initial state (mixture from lecture 2 where h was calculated): XCO = 0.0129, XO2 = 0.3376, XCO2 = 0.6495, T = 2500 , P = 10 atm h = -3784 kJ/kg, u = -4397 kJ/kg AME 513 - Fall 2012 - Lecture 3 - Chemical Thermodynamics 2

  26. Compression / expansion • Expand at constant entropy to 1 atm, frozen composition: T = 1738K, XCO = 0.0129, XO2 = 0.3376, XCO2 = 0.6495, , h = -4795 kJ/kg, u = -5159 kJ/kg, s = 7381J/kgK • Work done (control volume, steady flow) = hbefore - hafter = +1011 kJ/kg • Work done (control mass) = ubefore - uafter = +762 kJ/kg • Expand at constant entropy to 1 atm, equilibrium composition: T = 1794K, XCO = 0.00022,XO2 = 0.3334, XCO2 = 0.6664 (significant recombination) h = -4811 kJ/kg, u = -5184 kJ/kg, s = 7382 J/kgK • Work done (control volume, steady flow) = +1027 kJ/kg (1.6% higher) • Work done (control mass) = 787 kJ/kg (3.3% higher) • Moral: let your molecules recombine! AME 513 - Fall 2012 - Lecture 3 - Chemical Thermodynamics 2

  27. Summary - Lecture 3 • In order to understand what happens to a chemically reacting mixture if we wait a very long time, we need to apply • 1st Law of Thermodynamics (conservation of energy) - but this doesn’t tell us what the allowable direction of the reaction is; A  B or B  A are equally valid according to the 1st Law • 2nd Law of Thermodynamics (increasing entropy) - invokes restrictions on the direction of allowable processes (if A  B is allowed then B  A isn’t) • Equilibrium occurs when the worst possible end state is reached; once this point is reached, no further chemical reaction is possible unless something is changed, e.g. T, P, V, etc. • The statements of the 2nd law are • ds ≥ 0 (constant u & v, e.g. a rigid, insulated box) • dg = d(h - Ts) ≤ 0 (constant T and P, e.g. isothermal piston/cylinder) AME 513 - Fall 2012 - Lecture 3 - Chemical Thermodynamics 2

  28. Summary - Lecture 3 • The application of the 2nd law leads to complicated looking expressions called equilibrium constraintsthat involve the concentrations of each type of molecule present in the mixture • These equilibrium constraints are coupled with conservation of energy, conservation of each type of atom, and the ideal gas law (or other equations of state, not discussed in this class) to obtain a complete set of equations that determine the end state of the system, e.g. T, P and composition of the combustion products AME 513 - Fall 2012 - Lecture 3 - Chemical Thermodynamics 2

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