Review -1 School of Aerospace Engineering Copyright © by Jerry M. Seitzman. All rights reserved. AE/ME 6766 Combustion AE/ME 6766 Combustion: Review of Chemical Thermodynamics AE/ME 6766 Combustion: Review of Chemical Thermodynamics

Review -2 School of Aerospace Engineering Copyright © by Jerry M. Seitzman. All rights reserved. AE/ME 6766 Combustion Chemical Thermodynamics Calculation of final state based on some information about initial state –assuming thermodynamic equilibrium achieved Final state properties –temperature (adiabatic flame temperature) conservation of energy, 1 st law –equilibrium composition mass (atom) conservation entropy, 2 nd law Reactants Products

Review -3 School of Aerospace Engineering Copyright © by Jerry M. Seitzman. All rights reserved. AE/ME 6766 Combustion Why Study Thermodynamic Equilibrium? Equilibrium considers where reaction is heading to if given enough time –i.e., what levels are species concentrations, temperatures being “pulled” toward Kinetics (later) considers chemical rates –i.e., how fast a reaction occurs as it tends toward equilibrium x or t T eq Y eq  chem Y, T Initial

Review -4 School of Aerospace Engineering Copyright © by Jerry M. Seitzman. All rights reserved. AE/ME 6766 Combustion Refresher Specific heats: –u=internal energy (kJ/kg) –H=enthalpy (kJ/kg) Enthalpy: Perfect gas mixtures: –Y i =mass fraction=m i /m –  i =mole fraction=N i /N Enthalpy of formation (energy in chemical bonds) Sensible enthalpy

Review -5 School of Aerospace Engineering Copyright © by Jerry M. Seitzman. All rights reserved. AE/ME 6766 Combustion Specific Heat Characteristics Specific heats of non-monotonic gases increase with temperature due to excitation of internal energy modes 10’s K Few 1000K 3/2 5/2 7/2 C v /R Monatomics (He, Ar,..) Diatomics (N 2,O 2 ) Dissociation, ionization

Review -6 School of Aerospace Engineering Copyright © by Jerry M. Seitzman. All rights reserved. AE/ME 6766 Combustion Specific Heat Characteristics (2) Reproduced from Turns, An Introduction to Combustion, 2000

Review -7 School of Aerospace Engineering Copyright © by Jerry M. Seitzman. All rights reserved. AE/ME 6766 Combustion Stoichiometry Stoichiometric quantity of oxidizer is amount required to completely burn a quantity of fuel Fuel-oxidizer ratio, f –sometimes mass fuel/mass oxidizer –or moles fuel/moles oxidizer –Equivalence ratio  (or  ) = f actual /f stoichiometric  = 1; stoichiometric –just enough oxidizer to completely consume fuel  < 1; fuel lean (excess ox.)  > 1; fuel rich (excess fuel)

Review -8 School of Aerospace Engineering Copyright © by Jerry M. Seitzman. All rights reserved. AE/ME 6766 Combustion Adiabatic Combustion Temperature Equilibrium temperature that would be achieved if reactants were converted to equilibrium products without heat addition or loss –energy conservation (1 st Law) provides one equation –need 2 nd condition to fix state 2 Adiabatic Flame Temperature (T ad ) –p constant Constant Volume Reactants Products Reactants (1) Products (2) 0

Review -9 School of Aerospace Engineering Copyright © by Jerry M. Seitzman. All rights reserved. AE/ME 6766 Combustion Calculation First law: Define: Final Result: If product and reactant C p ’s are equal:

Review -10 School of Aerospace Engineering Copyright © by Jerry M. Seitzman. All rights reserved. AE/ME 6766 Combustion Heating Value (HV) Heating value represents the amount of chemical to thermal energy conversion from burning one unit of fuel (in defined oxidizer) –HHV – water in products is liquid –LHV – water in products is gas –Typical LHV in air CH 4 = 50,016 kJ/kg C 3 H 8 =50,368 kJ/kg H 2 = 120,923 kJ/kg CO = 10,107 kJ/kg

Review -11 School of Aerospace Engineering Copyright © by Jerry M. Seitzman. All rights reserved. AE/ME 6766 Combustion Typical Stoichiometric T ad Methane = 2226 K Acetylene = 2539 K Propane= 2267 K Hydrogen=2390 K Carbon Monoxide= 2275 K

Review -12 School of Aerospace Engineering Copyright © by Jerry M. Seitzman. All rights reserved. AE/ME 6766 Combustion  dependence of T ad Heat release/(mass oxidizer +mass fuel) peaks near  =1 –No extra fuel or air mass to heat up –T ad usually peaks slightly rich CH 4 :  =1.04 H 2 :  =1.07

Review -13 School of Aerospace Engineering Copyright © by Jerry M. Seitzman. All rights reserved. AE/ME 6766 Combustion Equilibrium To know T ad, you need to know product species –to this point, we’ve assumed that they are known –However, determining these is actually a part of the solution Typical products: CO 2, H2O, H 2, CO, O, H, OH Problem is coupled –Temperature affects species –Species affects temperature As we’ll discuss, reactants never completely go to products A+B C Equilibrium is reached when the forward and backward rates balance

Review -14 School of Aerospace Engineering Copyright © by Jerry M. Seitzman. All rights reserved. AE/ME 6766 Combustion How to Determine Equilibrium? Given reaction: Equilibrium concentrations given by:

Review -15 School of Aerospace Engineering Copyright © by Jerry M. Seitzman. All rights reserved. AE/ME 6766 Combustion Observations on  G o For endothermic reaction:  G o > 0 For exothermic reaction:  G o < 0 –Typically strongly negative, indicating that reaction tends toward products –e.g. CO oxidation at atmospheric pressure, T 2 =2000 K conclusions: –Mostly CO 2 at equilibrium –However, impossible to get 100% combustion efficiency, even if reaction has infinite time!

Review -16 School of Aerospace Engineering Copyright © by Jerry M. Seitzman. All rights reserved. AE/ME 6766 Combustion Equilibrium Hydrocarbon/Air Combustion Products Major products: –Lean: CO 2, H 2 O, O 2 –Rich: CO 2, CO, H 2 O, H 2, O 2

Review -17 School of Aerospace Engineering Copyright © by Jerry M. Seitzman. All rights reserved. AE/ME 6766 Combustion Equilibrium Hydrocarbon/Air Combustion Products (2) Minor Products: –NO, OH, O, H, H 2 (  <1), CO (  <1)

Review -18 School of Aerospace Engineering Copyright © by Jerry M. Seitzman. All rights reserved. AE/ME 6766 Combustion Temperature Effects on Equilibrium Returning to CO oxidation example, note variation in K p with temperature: –K p = 235 (T=1500 K) = 63 (T=2000 K) = 26 (T=2500 K) –i.e., shift toward reactants with increasing temperature This is a general rule for exothermic reactions –Reaction shifts toward products (reactants) with decreases (increases) in temperature Thus, increasing the preheat temperature will cause equilibrium to shift so that the product temperature doesn’t increase by the same amount Opposite result is true for endothermic reactions –e.g. O 2 dissociation: increasing temperature favors O 2 dissociation

Review -19 School of Aerospace Engineering Copyright © by Jerry M. Seitzman. All rights reserved. AE/ME 6766 Combustion Pressure Effects on Equilibrium Note from K p equation that the pressure term is raised to the power: # product moles - # reactant moles Equilibrium shifts to less (more) moles as pressure increases (decreases) –N prod =N reac : pressure independent –N prod >N reac p increases (decreases) cause shift toward reactants (products) –N prod< N reac p increases (decreases) cause shift toward products (reactants)

Review -20 School of Aerospace Engineering Copyright © by Jerry M. Seitzman. All rights reserved. AE/ME 6766 Combustion Pressure Effects on Dissociation N prod >N reac – –p increases reduce dissociation –T increases increase dissociation

Review -21 School of Aerospace Engineering Copyright © by Jerry M. Seitzman. All rights reserved. AE/ME 6766 Combustion Pressure Effects on Equilibrium Methane Example N prod =N reac – –Thus, above rxn is pressure independent –Result shows that flame temperature increases slightly with pressure, why?

Review -22 School of Aerospace Engineering Copyright © by Jerry M. Seitzman. All rights reserved. AE/ME 6766 Combustion Le Chatelier’s Principle Provides convenient way to remember effects of pressure, temperature on equilibrium –“Given a change in conditions, reaction will shift in such a way as to minimize the effect of this change”