Combustion kinetics I¶

Equilibrium is approached in combustion, but often kinetics are limiting and required for an accurate representation of temperature and composition.
Kinetics are important for soot and NOx
Flame extinction and ignition phenomena
Combustion chemistry is complex: hundreds (thousands) of species and reactions.

Elementary reactions¶

$$\frac{d[A]}{dt} = -k[A][B]$$

In general: $$ aA + bB \rightleftharpoons cC + dD$$ $$\frac{d[A]}{dt} = -ak_f[A]^a[B]^b + ak_r[C]^c[D]^d$$
For elementary reactions the powers on [A], [B], etc. are integers, and correspond to the coefficients in the reaction.

$$k_f = \alpha T^{\beta}\exp(-E_a/RT)$$

$\alpha$ is a pre-exponential rate factor.
$\beta$ is a temperature coefficient.
$E_a$ is the activation energy.
Unimolecular reaction: $A\rightarrow B$
Termolecular reaction: $A+B+M\rightarrow \text{products}$
In combustion, we often see species M, which is some generic species.
- Normally occurs in radical reactions.

Lump many elementary steps into one (or more) overall reaction. $$\text{Fuel} + a\text{Oxidizer} \rightarrow b\text{Products}$$
For methane $$\frac{d[CH_4]}{dt} = -k[CH_4]^n[O_2]^m$$
Generally $n$, and $m$ are not integers.
- Can be greater than or less than 1.
The reaction rate is empirical and only valid within the range of conditions used to fit the data.
Use with caution
Can be very convenient.
See Turns Table 5.1
See also Westbrook 1981 simplified reaction mechanisms

Consider reaction $i$, made up of all species $j$ $$\sum_{j=1}^{N_{sp}} \nu_{i,j}^\prime X_j\rightleftharpoons \sum_{j=1}^{N_{sp}} \nu_{i,j}^{\prime\prime}X_j$$
$\nu_{i,j}^{\prime}$ is the reactant coefficient of species $j$ in reaction $i$.
$\nu_{i,j}^{\prime\prime}$ is the product coefficient of species $j$ in reaction $i$.
$X_j$ simplly denotes species $j$ in the reaction.
Most of the $\nu$ are zero, since most reactions only involve a very few species.

$$[V^{\prime}]\mathbf{X} \rightleftharpoons [V^{\prime\prime}]\mathbf{X}$$

$[V^{\prime}]$ has $N_{rxns}$ rows, and $N_{sp}$ columns. $[V^{\prime\prime}]$ also.
$\mathbf{X}$ is the vector of all species.

Species: $O_2$, $H_2$, $H_2O$, $HO_2$, $O$, $H$, $OH$, $M$
Reactions: $$H_2 + O_2 \rightleftharpoons HO_2 + H$$ $$H + O_2 \rightleftharpoons OH + O$$ $$OH + H_2 \rightleftharpoons H_2O + H$$ $$H + O_2 + M \rightleftharpoons HO_2 + M$$

$[V^{\prime}] = $

Rxn	$O_2$	$H_2$	$H$	$OH$	$M$
1	1	1	0	0	0
2	1	0	1	0	0
3	0	1	0	1	0
4	1	0	1	0	1

$[V^{\prime\prime}] = $

Rxn	$H_2O$	$HO_2$	$O$	$H$	$OH$	$M$
1	0	1	0	1	0	0
2	0	0	1	0	1	0
3	1	0	0	1	0	0
4	0	1	0	0	0	1

$$R_{f,i} = k_{f,i}\prod_{j=1}^{N_{sp}}[X_j]^{\nu_{i,j}^{\prime}}$$ $$R_{r,i} = k_{r,i}\prod_{j=1}^{N_{sp}}[X_j]^{\nu_{i,j}^{\prime\prime}}$$

For the reaction $$ aA + bB \rightleftharpoons cC + dD,$$ we have $$R_f = k_f[A]^a[B]^b[C]^0[D]^0 = k_f[A]^a[B]^b,$$ $$R_r = k_r[A]^0[B]^0[C]^c[D]^d = k_r[C]^c[D]^d.$$
The net rate for each reaction $i$ is called the rate of progress variable, and is given by $$q_i = R_{f,i} - R_{r,i}$$
The net reaction rate for species $j$ is then $$\dot{\omega}_j = \sum_{i=1}^{N_{rxns}}\underbrace{(\nu_{i,j}^{\prime\prime}-\nu_{i,j}^{\prime})}_{\nu_{i,j}}q_i$$ Or,

$$\boldsymbol{\omega} = [V]^T\mathbf{q},$$

where the elements of $[V]$ are $\nu_{i,j}^{\prime\prime}-\nu_{i,j}^{\prime}$.

Reaction rate constants are independent of composition.
Hence, for a reaction at equilibrium, $q_i=R_{f,i}-R_{r,i}=0$.
This gives $R_{f,i}=R_{r,i}$, or $$k_{f,i}\prod_j[X_j]^{\nu_{i,j}^\prime} = k_{r,i}\prod_j[X_j]^{\nu_{i,j}^{\prime\prime}}$$ $$\frac{k_{f,i}}{k_{r,i}} = \prod_j[X_j]^{\nu_{i,j}} = \prod_j\left[\frac{P_i}{RT}\right]^{\nu_{i,j}} = \prod_j\left[\frac{P_i}{P_o}\right]^{\nu_{i,j}}\left[\frac{P_o}{RT}\right]^{\nu_{i,j}} = K_{eq}\prod_j\left[\frac{P_o}{RT}\right]^{\nu_{i,j}} = K_c$$

$$\frac{k_{f,i}}{k_{r,i}} = K_c = K_{eq}\prod_j\left[\frac{P_o}{RT}\right]^{\nu_{i,j}}$$

Fast reacting species, like radical intermediates may be formed and destroyed very fast relative to other species.
Assume that the fast species are in steady state relative to the slower species.

$$O + N_2 \rightarrow NO + N\phantom{xxxxx}\text{slow}$$ $$N + O_2 \rightarrow NO + O\phantom{xxxxx}\text{fast}$$

$$\frac{d[N]}{dt} = k_1[O][N_2] - k_2[N][O_2] = 0$$ Solve this for $[N]$

$$[N]_{ss} = \frac{k_1}{k_2}\frac{[O][N_2]}{[O_2]}.$$

We have replaced an ODE with an algebraic expression for $[N]$ in terms of other species.
$[N]_{ss}$ is a function of time though, through the time dependence of $[O]$, $[N_2]$, and $[O_2]$.

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