0
$\begingroup$

I'm having some trouble understanding the phenomenological analysis used to calculate the thickness of a 1D deflagration wave (Combustion Theory by Williams, page 135). I quote the text below:

Consider a stationary plane deflagration wave of thickness $\delta$, into which a combustible gas mixture flows at velocity $v_0$ (the laminar burning velocity). If $q$ is the heat of reaction (energy released per unit mass of the reactant mixture) and $w$ is the reaction rate (mass of the reactant mixture converted per unit volume per second), then $qw\delta$ is the energy released per unit area of the wave per second. The cold incoming gas is heated to a temperature at which it reacts at an appreciable rate, by means of heat conduction from the reaction zone. The rate at which heat is conducted upstream is roughly $\lambda dT/dx \approx \lambda(T_\infty— T_0)/\delta$ (energy per unit area per second), where $\lambda$ is the mean thermal conductivity of the gas, $T$ is the temperature, $x$ is the distance normal to the wave, and the subscripts 0 and $\infty$ identify conditions upstream and downstream of the wave, respectively. If the wave is adiabatic so that no energy is lost downstream or from the sides of the wave, then energy conservation implies that $q = c_p(T_\infty — T_0)$ (where $c_p$ is an average specific heat of the mixture) and that the entire heat released must be conducted upstream, that is, $qw\delta \approx \lambda(T_\infty — T_0)/\delta$. These results lead to the equation $c_p(T_\infty — To)w\delta \approx \lambda(T_\infty — T_0)\delta$, which implies that the thickness of the wave is $$\delta \approx \sqrt{\lambda/c_pw}.$$

There are a couple of points above which I can't wrap my head around. First, when we say that heat is conducted upstream at a rate $\lambda dT/dx \approx \lambda(T_\infty— T_0)/\delta$, are we basically ignoring the role of convection in the heat transfer? Isn't the rate of heat transfer affected by the upstream flow velocity (which is in the opposite direction to the temperature gradient)? Also, what justifies the assumption that all heat released by the flame is to be conducted upstream? Isn't this heat used to increase the temperature of the products themselves (and which are convected downstream of the wave)?

$\endgroup$

1 Answer 1

1
$\begingroup$

Both of your points are actually correct and relevant in an absolute sense, but the approach used to derive these equations uses an asymptotic analysis to neglect small terms. I think answering your questions without providing a more overarching answer would be not be too helpful, so I'll attempt to do both. Before answering those questions, I'll point you to look up "thermal flame theory" or "Mallard and Le Chatelier's 2-zone theory" which will lead to the same derivations seen in Williams (although I generally have not seen the derivations explained the way Williams does).

First, when we say that heat is conducted upstream at a rate λdT/dx≈λ(T∞—T0)/δ, are we basically ignoring the role of convection in the heat transfer? Isn't the rate of heat transfer affected by the upstream flow velocity (which is in the opposite direction to the temperature gradient)?

Let's clarify "convection" here as you stated. Convection by definition is the movement caused within a fluid by a gradient, whereas advection is the transport of some property (i.e. energy) by the motion of the fluid. In this case, thermal gradients are not going to drive any fluid motion (that only happens because of buoyancy in a gravitational field), but the motion of the fluid (aka the velocity) does transport thermal energy in/out of the reaction zone via advection. In the full energy balance of the reaction zone, there are three terms (all on per-area basis):

  • Source term (energy released by the reaction): $qw\delta = C_p(\Delta T)w$
  • Conduction term: $\frac{-k\Delta T}{\delta}$
  • Advection term (assuming average $C_p$): $\dot{m}\Delta h = \rho v_0C_p\Delta T$

Now if we take the ratio of advection rate to thermal diffusion rate (conduction), which is also known as the Peclet number, we get: $$\frac{\rho v_0C_p\Delta T}{\frac{-k\Delta T}{\delta}} = \frac{\delta \rho v_0C_p} {k}$$

Applying an asymptotic analysis based on a very thin reaction zone ($\delta << 1$), we can see that the ratio becomes extremely small, leading to the conclusion that for a thin reaction zone, the heat transfer is dominated by conduction over advection. This assumption is part of the underlying theory, meaning that the equations are based on the assumption of conduction dominating the heat transfer being valid.

Also, what justifies the assumption that all heat released by the flame is to be conducted upstream? Isn't this heat used to increase the temperature of the products themselves (and which are convected downstream of the wave)?

This is also correct. The advection term above shows that some thermal energy will be transported out of the system by fluid motion ($\rho v_0 C_p (T_0-T_{\infty})$). The energy released by the reaction also does go towards increasing the product temperatures up to the adiabatic flame temperature $T_{ad}$, but to understand why these are neglected in the derivations, it's useful to actually understand the 2-zone model.

Explanation

2-zone model

Mallard and Le Chatelier's flame theory basically says that there are 2 zones of interest:

  • the pre-heat zone, where the incoming upstream fuel, at temperature $T_{\infty}$ is heated up via thermal conduction from the reaction zone to the ignition temperature of the fuel, let's call it $T_i$
  • the reaction zone, where the reaction occurs, energy is released as described by the source term, and the gas is raised from temperature $T_i$ up to $T_{ad}$ or $T_f$ (adiabatic flame temp), also termed $T_0$ by Williams

So, since the analysis of the reaction zone itself would show a temperature change from $T_i$ to $T_0$, not the full range of $T_{\infty}$ to $T_0$, assuming that the actual temperature change in the thin reaction zone is relatively small would allow us to ignore the energy that leaves the system downstream (this isn't really a perfect assumption, but we are considering an idealized system to derive these equations). Also, the same assumption about the scale of temperature change in a very thin zone leads to the assumption that all energy is conducted back to the preheat zone. At the same time, we aren't assuming the $\Delta T$ in the reaction zone is 0, otherwise there would be no conduction. If this isn't an accurate or satisfactory description, maybe someone can edit in a better explanation.

As mentioned previously, the thickness of the reaction zone, $\delta$, is very small and allows conduction to dominate. Now if we conduct an energy balance in the preheat zone, where there is no reaction, the change in energy of the gas must equal the conducted heat from the reaction zone. Without going into the derivation here, this gives the relation between reaction zone thickness (aka laminar flame thickness) and flow velocity (which by transforming to a ground reference frame is the laminar flame velocity, $S_L$): $$\delta = \frac{\alpha}{S_L}$$ where thermal diffusivity $\alpha = \frac{k}{\rho C_p}$ If you do the actual derivation based on the energy balance, you would also find a term $\frac{T_i - T_{\infty}}{T_0 - T_i}$, which is O(1) and is neglected as part of the approximation (I honestly don't have a better explanation for this).

The next step of the model connects flame speed to reaction rate, but that's outside the scope of your question (and you can read about it in the provided link or any other search result).

$\endgroup$
2
  • $\begingroup$ That's an amazing answer, thank you very much! I definitely meant to say "advection" rather than "convection". Also, I think you mixed up $T_0$ and $T_\infty$ in Williams' notation ($T_\infty$ is the temperature downstream of the wave). Also, perhaps the ratio in your second-to-last paragraph has to be $(T_i-T_0)/(T_\infty-T_0)$. $\endgroup$
    – Tofi
    Feb 6, 2022 at 10:11
  • $\begingroup$ Yeah you're right, I mixed up the notations. If I get a chance I'll edit later. $\endgroup$ Feb 6, 2022 at 20:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.