Finding the Equivalence Ratio of Combustion (Methane + Air)

[agm]

I'd like to do some experiments with flames at different Equivalence Ratios - but I'm confused as to how I can find the Equivalence Ratio for different conditions. Wiki-article

So the stoichiometric balance for Methane + Air is
CH4 + 2(O2 + 3.76N2) -> CO2 + 2H2O + 7.52N2

Referencing equations online, it looks like the Stoichiometric Fuel to Air ratio for Methane + Air is 1/2 (One mole of methane over two moles of air) - is this correct?

 

[BvU]

Hello @agm ,

$\qquad$!

​

is this correct?

No. As you indicate, the ratio Methane/ oxygen is 1/2 (on a molecular base)

Referencing equations online

Don't post something like that without adding the links.

$\$

 

[agm]

Don't post something like that without adding the links.

So I was mostly referencing this Section 3.2 book summary on Science Direct

No. As you indicate, the ratio Methane/ oxygen is 1/2 (on a molecular base)

So it would not be correct to say the Fuel to AIR ratio is 1/2. Instead the Fuel to Oxidizer (oxygen) is 1/2?

 

[agm]

So apparently you can also find the equivalence ratio on a volumetric flow rate basis - paper

This should work out well, as I'd like to know what the equivalence ratio is for the flowing gasses in a methane + air flame.

If I were to add an inert gas into the flow like Nitrogen, how can I factor in this added volumetric flow rate without affecting some known equivalence ratio. Like if I were to add 1 LPM of Methane with 2 LPM of air (equivalence ratio should be 1), would adding 1 LPM of Nitrogen turn the mixture from stoichiometric to fuel lean? I don't think so since there's still enough Methane and Air to stoichiometrically combust, but intuitively adding in more volumetric flow should disrupt a flame at the end of some tube.

 

[bigfooted]

If I were to add an inert gas into the flow like Nitrogen, how can I factor in this added volumetric flow rate without affecting some known equivalence ratio. Like if I were to add 1 LPM of Methane with 2 LPM of air (equivalence ratio should be 1), would adding 1 LPM of Nitrogen turn the mixture from stoichiometric to fuel lean? I don't think so since there's still enough Methane and Air to stoichiometrically combust, but intuitively adding in more volumetric flow should disrupt a flame at the end of some tube.

You have 1LPM of methane and 2 LPM of air. Air consists of 1 part oxygen and 3.76 part nitrogen (or: 0.21 O2 and 0.79 N2). We assume that volume fractions and mole fractions are the same. That means that for every mole of ch4, you have 2*0.21=0.42 moles of oxygen. The stoichiometric ratio of methane and oxygen according to your first reaction equation is 1:2. So in your example with 1 LPM methane and 2 LPM air, the equivalence ratio is 0.42/2 = 0.21.

 

[agm]

We assume that volume fractions and mole fractions are the same. That means that for every mole of ch4, you have 2*0.21=0.42 moles of oxygen.

Okay so I think I see what the crux of my understanding.

There seem to be different ways to represent equivalence ratio, it seems like one way is a definition of Fuel to Oxidizer (the oxidizer part in air is oxygen, so out of the total volumetric flow of air, I have to extract the oxygen part of it to use in the equivalence ratio)

$$ \phi = \frac{(F/O)}{(F/O\vert_{\text{stoich}})} \hspace{1in} \phi = \frac{(O/F\vert_{\text{stoich}})}{(O/F)}$$

So I've gathered there's one way to represent the stoichiometric fuel to oxidizer ratio on a molecular basis $\frac{CH_4}{O_2}\vert_{\text{stoich}} = \frac{1}{2}$.

I'd like to to convert this stoichiometric ratio from a molecular basis to a volume basis, I might be able to with the following (at STP). Assuming 1LPM of CH4 and 2LPM of O2 (not air) is correct volumetric flow rates for a stoichiometric combustion.

$$ 1[LPM]\vert_{CH_4} \times \frac{1[{\text{mol}}]}{22.4[L]} \approx 0.0446 \left[\frac{\text{mol}}{\text{min}}\right]\vert_{CH_4} \hspace{1in} 2[LPM]\vert_{O_2} \times \frac{1[{\text{mol}}]}{22.4[L]} \approx 0.0893\left[\frac{\text{mol}}{\text{min}}\right]\vert_{O_2}$$

Since I'd be using air, I'm more interested in the volumetric flow of air that corresponds to 2LMP of Oxygen, so I can use the approximation of 21% O2 and 79% N2 in air to get the molar flow rate of air from 2LPM of Oxygen.

$$ \frac{100}{21} \times 0.0893\left[\frac{\text{mol}}{\text{min}}\right]\vert_{O_2} = 0.425 \left[\frac{\text{mol}}{\text{min}}\right]\vert_{AIR}$$

Now since I have both the methane and air in molar flow rates, I can reference online ratios to check my math (remembering air-fuel and oxidizer-fuel ratios are different)
For methane:
$$ \frac{\text{Air}}{\text{Fuel}}\vert_{\text{molar}}^{\text{stoich}} = 9.52 \hspace{1in} \frac{0.425\left[\frac{\text{mol}}{\text{min}}\right]\vert_{AIR}}{0.0446 \left[\frac{\text{mol}}{\text{min}}\right]\vert_{CH_4}} \approx 9.53 $$

If what I've done is correct, then for my experiment I can determine what the volumetric flow rate of methane and air should be to correspond to some equivalence ratio I'd like to study.

$$ \phi = \frac{9.52}{\left[\frac{\text{mol}}{\text{min}}\right]\vert_{AIR}/ \left[\frac{\text{mol}}{\text{min}}\right]\vert_{CH_4}} $$

 

[agm]

So I believe I've managed to find a solution to my problem. There's some nuance we dig through before agreeing on what "equivalence ratio" means. It seems like air is a super-common oxidizer in the combustion-industry, so it makes sense for some people/researchers to use an equivalence ratio based on air. On the other hand, for the same stoichiometric equation, where air is your oxidizer, you can define a separate equivalence ratio based on the amount of oxygen. Either way can be used, and will give you a hint as to how much oxygen or air you need for a certain combustion process.

Example:
$$
\ce{CH4 + 2(O2 + 3.76N2) -> CO2 + 2H2O + 7.52N2}
$$
My question is this - if we have a volume of methane, how much air do we need to combust stoichiometrically (all the air/methane is consumed in the combustion). You could define equivalence ratio in one of these two ways, for simplicity I'll define equivalence ratio on a mole basis (mass defined equivalence ratio is a bit messier).
$$
\phi = \frac{(F/O)}{(F/O\vert_{\text{stoich}})} \hspace{1in} \phi = \frac{(F/A)}{(F/A\vert_{\text{stoich}})}
$$
The moles of air being $2(1+3.76) = 9.52$ and oxygen being $2$ (in each case there is always 1 mole of $\ce{CH4}$ present, so $F\vert_{\text{stoich}} = 1$)
$$
\phi = \frac{(F/O)}{(1/2)} \hspace{1in} \phi = \frac{(F/A)}{(1/9.52)}
$$
Now both of the above equations for the equivalence ratio will tell you how much air you need for a stoichiometric combustion, let's see where $\phi = 1$ (volume/volume flow rate will be directly proportional to moles, so can be used interchangeably here)

$$
\phi = \frac{1[LPM]\vert_{F}/2[LPM]\vert_{O}}{1/2} \hspace{1in} \phi = \frac{1[LPM]\vert_{F}/9.52[LPM]\vert_{A}}{1/9.52}
$$

In either case it's clear the equivalence ratio is 1, further we can find the volumetric flow of air needed to produce 2LPM of oxygen using an estimation of the percent of oxygen in air.

(or: 0.21 O2 and 0.79 N2).

$$
\dot{V}\vert_{\ce{O2}} = 2[LPM] \hspace{1in} (2[LPM]) \left( \frac{100}{21} \right) \approx 9.52[LPM]
$$
This result which may seem incredibly simple; but was escaping me - a few other things I've found out about equivalence ratio which may be worth noting, it's typically easier to interchange moles and volume for your fuel-to-oxidizer ratios. If you do want to use mass or mass flow, you'll need to introduce the molecular weight (atomic mass) of the fuel and oxidizer into your fuel-to-oxidizer ratio. In other words 1 [kg/m] of $\ce{CH4}$ and 2 [kg/m] of $\ce{O2}$ will NOT produce a stoichiometric flame.

 

[bigfooted]

for extra insight: suppose you have a ratio of F/O= 1/4
what is the ratio F/A, and what are the equivalence ratios when using the definitions based on F/A or F/O?

 

[agm]

for extra insight: suppose you have a ratio of F/O= 1/4
what is the ratio F/A, and what are the equivalence ratios when using the definitions based on F/A or F/O?

For this example, we have $\frac{F}{O} = \frac{1}{4}$, assume we're using $\ce{CH4}$. If we need to find the amount of air that corresponds to $4$ units of oxygen, we perform the following based on gas composition of air
$$
(4)_{\ce{O2}} \times \frac{(100)_{\text{Air}}}{(21)_{\ce{O2}}} \approx (19.048)_{\text{Air}}
$$
So for our scenario, we can say we have a fuel-to-oxidizer and fuel-to-air ratios of $\frac{F}{O} = \frac{1}{4}$ and $\frac{F}{A} = \frac{1}{19.048}$. The equivalence ratios using either definition are then
$$
\phi_{\ce{O2}} = \frac{\frac{1}{4}}{\frac{1}{2}} = 0.5 \hspace{1in} \phi_{\text{Air}} = \frac{\frac{1}{19.048}}{\frac{1}{9.52}} \approx 0.5
$$
Equivalence ratio (E. ratio) tells us information about a combustion process, if the E. ratio is less than one, that means we'll have more oxidizer than we need. We can define the amount of oxidizer or air we need for a stoichiometric combustion by looking at a chemical balance for the oxidizer or air. We've demonstrated we can convert oxygen to air and vice versa, so in reality in this case we can define E. ratio in the following ways
$$
\phi_{\ce{O2}} = \frac{\frac{F}{O}}{\frac{1}{2}} \hspace{1in} \phi_{\text{Air}} = \frac{\frac{F}{O\times\ \frac{100}{21}}}{\frac{1}{2\times \frac{100}{21}}}
$$
We can see that this conversion factor cancels, therefore it makes sense that the E. ratio is the same in either case. The choice for using either definition will come down to whatever system you're designing. For instance, if I were to use some esoteric oxidizer, I would never use fuel-to-air ratios as they wouldn't be applicable or useful.

 

[bigfooted]

air or oxygen, it does not matter, you get the same value for the equivalence ratio.