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Frank-Kamenetskii theory

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Through Combustion, Arrhenius equation and Mathematics

Problem description

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Consider a vessel maintained at a constant temperature To, containing a homogeneous reacting mixture. Let the characteristic size of the vessel be a. Since the mixture is homogeneous, the density ρ is constant. During the initial period of ignition, the consumption of reactant concentration is negligible (see tf and te below), thus the explosion is governed only by the energy equation. Assuming a one-step global reaction fuel+oxidizer→products+q, where q is the amount of heat released per unit mass of fuel consumed, and a reaction rate governed by Arrhenius law, the energy equation becomes

ρcv∂T∂t=λ∇2T+qρBYFoe−E/(RT)

where

Non-dimensionalization

An increment in temperature of order T−To∼RTo2/E, where RTo2/E is the Frank-Kamenetskii temperature is enough to increase the chemical reaction by amount e, as is evident from the ratio

e−E/RTe−E/RTo=exp⁡[ERTo(1−T0T)].

Non-dimensional scales of time, temperature, length, and heat transfer may be defined as

τ=tte,θ=β(T−To)To,ηj=rja,δ=tcte

where

Note
In a typical combustion process, γ≈6−8,β≈30−100 so that βγ≫1.
Therefore, tf=βγte≫1. That is, fuel consumption time is much longer than ignition time, so fuel consumption is essentially negligible in the study of ignition.
This is why the fuel concentration is assumed to remain the initial fuel concentration YFo.

Substituting the non-dimensional variables in the energy equation from the introduction

∂θ∂τ=1δ1ηj∂∂η(ηj∂θ∂η)+eθ/(1+θ/β)

Since β≫1, the exponential term can be linearized eθ/(1+θ/β)≈eθ, hence

∂θ∂τ=1δ1ηj∂∂η(ηj∂θ∂η)+eθ

At τ=0, we have θ(η,0)=0 and for τ>0, θ needs to satisfy θ(1,τ)=0 and ∂θ/∂η|η=0=0.

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