The Theory of a Local Ignition

Abstract

It is shown that the problem on a local ignition comes down to the analysis of the dynamics of a reaction zone under condition of cooling of the ignition center with the inert environment; at the same time the power of a chemical heat source during the induction period can be considered approximately constant. The approximate analytical method is applied for analysis of the problem on a local chain-thermal explosion in the reaction of hydrogen oxidation in the presence of chemically active additive. The concept of an intermediate combustion wave with the maximum temperature equal to the initial temperature of the hot spot is introduced. It is shown that key parameters defining the critical size of a local source of ignition, are the temperature in the center of a local ignition zone, the quantity of the active centers of combustion created with the local source, and presence of active chemical additives in a combustible gaseous mixture. Comparison to experimental data has shown the applicability of the developed approach for the analysis of critical conditions of a local ignition in combustible gas mixtures.

Appendix

Analytical determination of τ _del of hydrogen–air mix at atmospheric pressure.

We consider the branched chain mechanism of hydrogen oxidation, described above. We neglect the consumption of initial reagents during τ _del. Then we have:

$$ \begin{aligned} & \frac{{{\text{d}}H(t)}}{{{\text{d}}t}} = l_{0} + l_{1} H(t) - 2k_{11} MH(t)^{2} \\ & C_{\text{p}} \rho \frac{{{\text{d}}T(t)}}{{{\text{d}}t}} = Q_{1} k_{6} \text{O}_{20} MH(t) + Q_{2} k_{11} MH(t)^{2} \\ \end{aligned} $$

(3.7)

where l ₀ = k _i H₂₀O₂₀ и l ₁ = 2k ₂O₂₀ − k ₅In₀ − k ₆O₂₀ M, H₂₀, O₂₀ и In₀—initial concentration of initial reagents and an additive, C _p is molar thermal capacity at a constant pressure, ρ-density. In the first equation of the system (3.7) we put $H\left( t \right) = \frac{1}{{k_{11} My(t)^{2} }}\frac{{{\text{d}}y(t)}}{{{\text{d}}t}}$, then we get:

$$ \frac{{{\text{d}}^{2} y(t)}}{{{\text{d}}t^{2} }} = l_{0} k_{11} My(t) + l_{1} \frac{{{\text{d}}y(t)}}{{{\text{d}}t}} $$

Its solution under initial condition H(0) = H ₀ (the local source generates only hydrogen atoms) is: $H(t) = \frac{{ - \xi_{1} \,\exp \,(\xi_{1} t)m + n\xi_{2} \,\exp \,(\xi_{2} t)}}{{k_{11} M( - \exp \,(\xi_{1} t)m + n\,\exp \,(\xi_{2} t))}}$ where $\xi_{1} = \frac{1}{2}l_{1} + \sqrt {l_{1}^{2} + 4k_{11} Ml_{0} }$, $\xi_{2} = \frac{1}{2}l_{1} - \sqrt {l_{1}^{2} + 4k_{11} Ml_{0} } ,\quad m = \xi_{1} - k_{11} H_{0}M,\quad n = \xi_{2} - k_{11} H_{0}M$.

Integration of the second equation of the set for T(0) = T ₀ gives:

$$ \begin{aligned} T(t) & = T_{0} + \left( {\frac{{(\xi_{1} + \xi_{2} )\beta }}{{2\xi_{1}^{2} \xi_{2}^{2} }} + \frac{{Q_{1} k_{6} O_{2} }}{{k_{11} M\rho C_{p} }}} \right)\ln \left( {\frac{{ - \exp (\xi_{1} t)m + n\exp (\xi_{2} t)}}{n - m}} \right) \\ & \quad - \frac{\beta t}{{2\xi_{1} \xi_{2} }} - \frac{{\beta mn(\xi_{1} - \xi_{2} )(\exp (\xi_{1} t) + \exp (\xi_{2} t))}}{{2(n - m)\xi_{1}^{2} \xi_{2}^{2} ( - \exp (\xi_{1} t)m + n\exp (\xi_{2} t))}}\quad {\text{where}}\quad \beta = \frac{{Q_{2} }}{{C_{p} \rho k_{11} M}} \\ \end{aligned} $$

(3.8)

According to [23] we will consider that the delay period expires when self-heating exceeds one characteristic interval, namely $\Delta T = T(t) - T_{0} = \frac{{ {RT}_{0}^{2} }}{E}$. We consider the activation energy of the linear branching reaction (k ₂, a limiting stage) as the activation energy.

During the delay period, it is possible to neglect concentration of accumulated hydrogen atoms. In addition, direct calculation shows that it is possible to ignore also the term $\frac{\beta t}{{2\xi_{1} \xi_{2} }}$. The Eq. (3.5) after substituting the values ξ ₁ and ξ ₂ takes a form:

$$ \begin{aligned} \frac{{ {RT}_{0}^{2} }}{E} = \Delta T & = \left( {\frac{{\beta l_{1} }}{{2k_{11}^{2} M^{2} l_{0}^{2} }} + \frac{{Q_{1} k_{6} O_{2} }}{{k_{11} M\rho C_{p} }}} \right) \\ & \quad \ln \left( { - \frac{{ - \exp \left( {\left( {l_{1} + \frac{{k_{11} Ml_{0} }}{{l_{1} }}} \right)t} \right)\left( { - \xi_{2} + k_{11} H_{0}M} \right) + \left( { - \xi_{1} + k_{11} H_{0}M} \right)\exp (\xi_{2} t)}}{{\left( {l_{1} + \frac{{2k_{11} Ml_{0} }}{{l_{1} }}} \right)}}} \right) \\ \end{aligned} $$

(3.9)

We simplify the Eq. (3.9) to get the equation for the delay period in an explicit form:

$$ \begin{aligned} \frac{{ {RT}_{0}^{2} }}{E} = \Delta T & = \left( {\frac{{\beta l_{1} }}{{2k_{11}^{2} M^{2} l_{0}^{2} }} + \frac{{Q_{1} k_{6} O_{2} }}{{k_{11} M\rho C_{p} }}} \right) \\ & \quad \ln \left( {\frac{{\exp (l_{1} t)k_{11} M(H_{0}l_{1} + l_{0} )}}{{l_{1}^{2} }} + 1 - \frac{{k_{11} MH_{0} }}{{l_{1} }} - \frac{{k_{11} Ml_{0} t}}{{l_{1} }}} \right) \\ \end{aligned} $$

(3.10)

Further, we will calculate the value ${RT}_{0}^{2} /E$ for the given conditions (T ₀ = 1000 K, E = 16.7 kcal/mol): ${RT}_{0}^{2} /E \approx 120$. The point of intersection of the dependence (3.11) with a line y = 120 also will give the required value of the delay period τ _del (Fig. 10a, b). As is seen the value of the delay period depends both on the amount of the active centers introduced into a gas mixture at initiation and on the concentration of additive in the gas mixture. We solve the Eq. (3.10) for t, substituting $\frac{{ {RT}_{0}^{2} }}{E}$ instead of ΔT. Then by definition t = τ _del the value of the delay period. We get:

$$ \begin{aligned} \tau_{\text{del}} = & \frac{1}{{k_{11} Ml_{0} l_{1} }}\left( {{\text{Lambert}}W\left( { - \frac{{H_{0}l_{1} + l_{0} }}{{l_{0} }}\exp \left( {\frac{{ - l_{1} \left( {l_{1} \exp \left( {\frac{{ {RT}_{0}^{2} }}{E\alpha }} \right) - l_{1} - k_{11} MH_{0}} \right)}}{{k_{11} Ml_{0} }}} \right)} \right)} \right. \\ \times & \quad \left. {k_{11} Ml_{0} + l_{1}^{2} (\exp \left( {\frac{{ {RT}_{0}^{2} }}{E\alpha }} \right) - 1) + k_{11} MH_{0}l_{1} } \right) \\ \end{aligned} $$

(3.11)

In the Eq. (3.11) LambertW(x) + exp(LambertW(x)) = x by definition,

$$ \alpha = \left( {\frac{{\beta l_{1} }}{{2k_{11}^{2} M^{2} l_{0}^{2} }} + \frac{{Q_{1} k_{6} O_{2} }}{{k_{11} M\rho C_{p} }}} \right) $$

It is easy to show that the results of the calculation by Eqs. (3.9)–(3.11) practically coincide.