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1 MAE 5310: COMBUSTION FUNDAMENTALS Chemical Time Scales and Partial Equilibrium October 3, 2012 Mechanical and Aerospace Engineering Department Florida.

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Presentation on theme: "1 MAE 5310: COMBUSTION FUNDAMENTALS Chemical Time Scales and Partial Equilibrium October 3, 2012 Mechanical and Aerospace Engineering Department Florida."— Presentation transcript:

1 1 MAE 5310: COMBUSTION FUNDAMENTALS Chemical Time Scales and Partial Equilibrium October 3, 2012 Mechanical and Aerospace Engineering Department Florida Institute of Technology D. R. Kirk

2 2 CHEMICAL TIME SCALES Insight gained from knowledge of chemical time scale,  chem What is approximate magnitude of time scale? Characteristic time is defined as time required for concentration of A to fall from its initial value to a value equal to 1/e times the initial value –[A](t=  chem )/[A] 0 =1/e –RC circuit analog Unimolecular reactions Bimolecular reactions Termolecular reactions Bimolecular reactions case where [B] 0 >> [A] 0 Termolecular reactions case where [B] 0 >> [A] 0

3 3 EXAMPLE: CHEMICAL TIME SCALE CALCULATIONS Reaction 1 is an important step in oxidation of CH 4 Reaction 2 is key step in CO oxidation Reaction 3 is a rate-limiting step in prompt-NO mechanism Reaction 4 is a typical radical recombination reaction Estimate  chem associated with least abundant reactant in each reaction for following 2 conditions –Assume that each of 4 reactions is uncoupled from all others and that third-body collision partner concentration is the sum of the N 2 and H 2 O concentrations) k=1.0x10 8 T 1.6 exp(-1570/T) k=4.76x10 7 T 1.23 exp(-35.2/T) k=2.86x10 8 T 1.1 exp(-10,267/T) k=2.2x10 22 T -2.0 Condition I: Low Temperature T=1344 K, P=1 atm  CH 4 =2.012x10 -4  N 2 =  CO=4.083x10 -3  OH=1.818x10 -4  H=1.418x10 -4  CH=2.082x10 -9  H 2 O= Condition II: High Temperature T=2199 K, P=1 atm  CH 4 =3.773x10 -6  N 2 =  CO=1.106x10 -2  OH=3.678x10 -3  H=6.634x10 -4  CH=9.148x10 -9  H 2 O=

4 4 EXAMPLE SOLUTION Increasing T shortens chemical time scales for all species, but most dramatically for CH 4 + OH and CH + N 2 reactions –For the CH 4 + OH reaction, the dominant factor is the decrease in the given CH 4 concentration –For the CH + N 2 reaction, the large increase in the rate coefficient dominates Characteristic times at either condition vary widely among the various reactions –Notice long characteristic times associated with recombination reaction H + OH + M → H 2 O + M, compared with the bimolecular reactions. –That recombination reactions are relatively slow plays a key role in using partial equilibrium assumptions to simplify complex chemical mechanisms (see next slide) Note that  chem =1/[B] 0 k bimolec could have been used with good accuracy for reaction 3 at both low and high temperature conditions, as well as reaction 1 at high temperature because in these cases one of reactants was much more abundant than other ConditionReactionSpeciesk(T)  chem (ms) I1OH3.15x I2OH3.27x I3CH3.81x I4H1.22x II1CH x II2OH6.05x II3CH1.27x II4H4.55x

5 5 PARTIAL EQUILIBRIUM Many combustion process simultaneously involve both fast and slow reactions such that the fast reactions are rapid in both the forward and reverse directions –These fast reactions are usually chain-propagating or chain-branching steps –The slow reactions are usually the termolecular recombination reactions Treating fast reactions as if they were equilibrated simplifies chemical kinetics by eliminating need to write rate equations for radical species involved, which is called the partial-equilibrium approximation Consider the following hypothetical mechanism:

6 6 PARTIAL EQUILIBRIUM VS. STEADY-STATE Net result of invoking either partial equilibrium assumption or steady-state approximation is the same: –A radical concentration is determined by an algebraic equation rather than through the integration of an ordinary differential equation Keep in mind that physically the two approximations are quite different: 1.Partial equilibrium approximation forcing a reaction, or sets of reactions, to be essentially equilibrated 2.Steady-state approximation forces the individual net production rate of one or more species to be essentially zero There are many examples in combustion literature where partial equilibrium approximation is invoked to simplify a problem Examples: 1.Calculation of CO concentrations during the expansion stroke of a spark-ignition engine 2.Calculation of NO emissions in turbulent jet flames –In both examples, slow recombination reactions cause radical concentrations to build up to levels in excess of their values for full equilibrium


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