MAE 5310: COMBUSTION FUNDAMENTALS Chemical Equilibrium: Heat of Combustion and Adiabatic Flame Temperature January 18, 2017 Mechanical and Aerospace Engineering Department Florida Institute of Technology D. R. Kirk

ADIABATIC FLAME TEMPERATURE For an adiabatic combustion process, with no change in KE or PE, temperature of products is called Adiabatic Flame Temperature Maximum temperature that can be achieved for given concentrations of reactants Incomplete combustion or heat transfer from reactants act to lower temperature Adiabatic flame temperature is generally a good estimate of actual temperature achieved in a flame, since chemical time scales are often shorter than those associated with transfer of heat and work Most common is constant-pressure adiabatic flame temperature Conceptually simple, but in practice difficult to evaluate because requires detailed knowledge of product composition, which is function of temperature

SUPPLEMENTAL SLIDES (GLASSMAN)

1st LAW FOR COMBUSTION PROBLEMS (GLASSMAN) Most general form (rarely used, but know what each term means) Note on sign of Q is negative. This is consistent with the 1st Law, but you must recall that Qp the is heat evolved, or the heat given off by the system, or the heat of combustion or heat of reaction. Sensible enthalpy change from T=298 to some reference Term is zero if reference T=298 Sensible enthalpy change (kJ/mol) relative to some reference T Enthalpy of formation of products at T=298 K Sensible enthalpy change relative to some reference T Term is zero if reactants enter at some reference T Sensible enthalpy change from T=298 to some reference Term is zero if reference T=298 Enthalpy of formation of reactants at T=298 K

1st LAW FOR COMBUSTION PROBLEMS (GLASSMAN) Much more common, and what we will use in MAE 5310 Sensible enthalpy change relative to T=298 K Enthalpy of formation of products at T=298 K Sensible enthalpy change relative to T=298 K This term is zero if reactants enter the system at T=298 K Enthalpy of formation of products at T=298 K

ADIABATIC FLAME TEMPERATURE (GLASSMAN) For an adiabatic combustion process, with no change in KE or PE, temperature of products is called Adiabatic Flame Temperature, Tad Maximum temperature that can be achieved for given concentrations of reactants Incomplete combustion or heat transfer from the reactants act to lower the temperature The adiabatic flame temperature is generally a good estimate of the actual temperature achieved in a flame, since the chemical time scales are often shorter than those associated with transfer of heat and work

EXAMPLE: FUEL-LEAN OCTANE-AIR COMBUSTION Calculate Tad of normal octane (liquid) burning in air at f= 0.5 Assume no dissociation of stable products formed All reactants are at 298 K and system operates at a pressure of 1 atm Compare results with figure

THERMOCHEMICAL DATA (GLASSMAN): CO2, H2O

THERMOCHEMICAL DATA (GLASSMAN): N2, O2

COMMENT ON NOTATION

CHEMICAL EQUILIBRIUM So far we have calculated adiabatic flame temperature assuming complete combustion All fuel is completely oxidized to form CO2, H2O and excess O2 and N2 are carried through unaffected Assumption reasonable for T < 1250 K, but most combustion systems operate at higher T Species that are normally stable at ambient conditions dissociate Concentration is determined by a balance between oxidation and formation Balance is a function of T, P and concentration Note: The chemical equilibrium relations we will use still only approximate the species concentrations in a combustion process. That is, they rest on the assumption that the conditions are constant for a sufficiently long time for all the reactions to reach equilibrium

ADDITIONAL PRODUCT FORMATION NO Dissociation: Complete Combustion Equivalence ratio less than or equal unity, f ≤ 1 The products formed are: CO2, H2O, O2, and N2 Equivalence ratio greater than unity, f > 1 The products formed are: CO2, CO, H2O, H2, and N2 WITH Dissociation Products formed include: CO2, CO H2O, H2, H, OH, O2, O, NO, N2, and N Concentration is dependent on T, P and f

CHEMICAL EQUILIBRIUM FOR A FIXED-MASS SYSTEM If final temperature of combustion reaction is high enough, CO2 will dissociate Can calculate adiabatic flame temperature as function of a (a = fraction of CO2 dissociated) Must consider second law: dS ≥ 0 Composition of system will shift toward point of maximum entropy when approaching from either side, since dS is positive Once maximum entropy is reached no further changes since would violate second law (dS)U,V,m = 0

PROPANE-AIR COMBUSTION AT 1 ATM Adopted From: Turns, S. R PROPANE-AIR COMBUSTION AT 1 ATM Adopted From: Turns, S.R., Introduction to Combustion

2nd LAW OF THERMODYNAMICS: GIBBS FREE ENERGY To arrive at equilibrium relations, employ 2nd Law of Thermodynamics State 2nd Law in terms of Gibbs Free Energy, G=H-TS For a closed system at constant T and P, the Gibbs free energy is a minimum at thermodynamic equilibrium Basic Thermodynamic Relations Criteria for Equilibrium

MORE USEFUL FORMS Gi=Hi-TSi Separate Gi into a pressure dependent and term and pressure independent term Recall that Cpi is not a function of pressure The more negative DGº is, the larger Kp is, and the more spontaneous the reaction is Another useful form Equation 40

COMMENTS ON KP The Kp of a reaction depends on temperature only Independent of pressure of equilibrium mixture Not affected by presence of inert gases The Kp of the reverse reaction is 1/ Kp The larger the Kp, the more complete the reaction If Kp > 1,000 (or ln Kp > 7) reaction assumed complete If Kp < 0.001 (or ln Kp < -7) reaction assumed not to occur Mixture pressure affects the equilibrium composition (but not Kp) Presence of inert gases affects the equilibrium composition When stoichiometric coefficients are doubled, the value of Kp is squared Free electronic in the equilbrium composition can be treated as an ideal gas Equilibrium calculations provide information on the equilibrium composition of a reaction, not on the reaction rate

ADIABATIC COMBUSTION EQUILIBRIUM Previously we have considered: Known Stoichiometry + 1st Law (Energy Balance) → Adiabatic Flame Temperature Glassman Problems 1-4 Known P and T + 2nd Law (Equilibrium Relations) → Stoichiometry Glassman Problems 5-9 Now we can combine these: 1st Law (Energy Balance) + 2nd Law (Equilibrium Relations) → Adiabatic Flame Temperature + Stoichiometry Glassman Problems 10-14 Solution Scheme Guess a T=Tguess Do equilibrium calculation to solve for species concentrations at Tguess Plug into 1st Law We want F(Tguess)=0 If F(Tguess) > 0, then initial guess was too high If F(Tguess) < 0, then initial guess was too low Increment Tguess

EXAMPLE: MONOPROPELLANT ROCKET PROPULSION Monopropellant rockets are simple propulsion systems that rely on chemicals which, when energized, decompose Decomposition creates both the fuel and an oxidizer (which allows the fuel to burn), which then react with each other Because they only use a single propellant, monopropellant rockets are quite simple and reliable, but not very efficient Mainly used to make small adjustments such as attitude control Typical Specific Impulse: 100-300 sec Typical Thrust: 0.1-100 N Monopropellant hydrazine, N2H4, thrusters that were used for trajectory correction maneuvers (TCMs) during interplanetary cruise, thrust vector control (TVC) during VOI, orbit trim maneuvers during the mapping mission, and attitude control when the action wheels are being desaturated. The rocket motors are clustered in modules located on the end of outrigger booms in order to increase their moment arm and thus decrease attitude control propellant requirements. Twelve 0.9-N (Newton) and four 22-N rocket motors are used for attitude control, with thrust being provided by eight 445-N rocket motors or by the 0.9-N motors for small TCMs.

EXAMPLE: MONOPROPELLANT ROCKET PROPULSION Rocket propellant chemists have proposed a new, high-energy liquid oxidizer, penta-oxygen, O5, which is also a monopropellant. Calculate the monopropellant decomposition temperature at a chamber pressure of 10 atm. If it is assumed the only products are O atoms and O2 molecules. Supplemental Information: Heat of formation of new oxidizer is estimated to be very high: 1,025 kJ/mole O5 enters system at 298 K Amount of O2 and O should be calculated for one mole of O5 decomposing Solution Technique Solution is iterative In order to calculate the final temperature of the mixture, one needs to the composition of the mixture at equilibrium – but – the composition of the mixture can only be found if the final temperature is known. Strategy: Guess final temperature and calculate mixture composition using equilibrium concept Once composition is known at guessed temperature, adiabatic flame temperature is calculated and checked against guessed value Repeat process until guessed and calculated temperatures are same Answer lies somewhere between 4,000 and 5,000 K

PRACTICAL APPLICATION: RECUPERATION A recuperator is a heat exchanger in which energy from a steady flow of hot combustion products, called flue gases, is transferred to the air supplied to the combustion process

SOME COMMON TYPES OF RECUPERATORS http://www. hardtech. es/hgg_tt_hrt Tubes cage radiation recuperator Tubes cage radiation recuperator working at 1,200ºC Double shell radiation recuperator Installation consisting of a tubes cage recuperator and a double shell one, series-connected

EXAMPLE: RECUPERATION (TURNS) A recuperator, as shown in figure, is employed in a natural-gas-fired heating-treating furnace. The furnace operates at atmospheric pressure with an equivalence ratio of 0.9. The fuel gas enters the burner at 298 K, while the air is pre-heated. Determine the effect of air preheat on the adiabatic flame temperature of the flame zone for a range of inlet air temperatures from 298 K to 1,000 K. What fuel savings result from preheating the air from 298 K to 600 K? Assume that temperature of flue gases at furnace exit, prior to entering recuperator, is 1700 K, both with and without preheat. Radiant-tube burner with coupled Recuperator for indirect firing. Note that All flue gases pass through the recuperator Source: Turns, An Introduction to Combustion

NASA CEA PRACTICE PROBLEM Consider combustion of a methane-air mixture at 10 atm (both fuel and air are at 10 atm). Plot mole fractions, ci, for the species CO2, CO, H2O, H2, OH, O2, N2, NO vs. temperature for f=1 for a temperature range from 1000 to 2500 K. Calculate Tflame and mole fractions as a function of f for adiabatic combustion. Plot these results vs. f and discuss at what value the peak flame temperature occurs. Comment on this value in light of the discussion found in Chapter 1, Part 2 of Glassman. Compare Tflame from part (2) to what would be obtained assuming complete oxidation (burning in only oxygen) and what would be obtained assuming complete combustion (burning in air).