# Equivalence ratio explained

To let you know where I'm coming from: I haven't studied chemistry since high school. I am now studying something else, but I have a dataset coming from your field and I would like some help in understanding just the very very basics of what it is.

What I have is $\ce{NOx}$, nitric oxide and nitrogen dioxide in an engine exhaust and the equivalence ratio. It's from the dataset ethanol in the package SemiPar in R, and the original reference is Brinkman (1981). You'll find a figure at the bottom.

What I want: I want to be able to, extremely briefly, explain what this is describing.

My understanding of the equivalence ratio is that if it is below 1, then there is too little air (compared to the stoichometric mix). If it is above 1, that means that there is too much fuel, and not all of it can be used leading to what is called incomplete combustion.

Is this correct? Would you be able to provide a better explanation? Imagine you're trying to explain it to a five-year-old, that's about how much of this I know!

If you've read this far, please also help me with appropriate tags for the question. I'm not sure what to choose.

Reference:

Brinkman, N.D. (1981). Ethanol fuel – a single-cylinder engine study of efficiency and exhaust emissions. SAE transactions Vol. 90, No 810345, 1410–1424.

You are correct. There is a theoretical ratio of air-to-fuel that provides exactly enough air to completely burn the fuel, converting it to $\ce{CO2}$ and $\ce{H2O}$. This theoretical ratio can be calculated knowing (a) the exact composition of the fuel and (b) how much oxygen is in the air (generally ~21%), and (c) the formula weight(s) of the molecules that are in the fuel. For fuel that are pure octane, the combustion reaction is:

$$\ce{\frac{25}{2}O2 + C8H18 -> 8CO2 + 9H2O}$$

Which means 12.5 moles of oxygen are required for every mole of octane. For 12.5 moles of oxygen you need about 60 moles of air because air is only 21% oxygen. That works out to 1740 grams of air (~29 grams per mole). One mole of octane is 114 grams. So the theoretical, or stoichiometric ratio of air to fuel is 1740/114 or about 15.

Most motor fuels are not pure octane. The additives complicate the math a bit, but the stoichiometric ratio is around 14 for many fuels.

The equivalence ratio is just the actual air-to-fuel ratio divided by the theoretical.

My understanding of the equivalence ratio is that if it is below 1, then there is too little air (compared to the stoichometric mix). If it is above 1, that means that there is too much fuel, and not all of it can be used leading to what is called incomplete combustion.

You said the second part backwards. If the ratio is below one, there is too little air (i.e. too much fuel), and if it above one, there is too much air (not enough fuel). So incomplete combustion should happen more at lower equivalence ratios.

Also, the exhaust temperature will be maximized for equivalence ratios close to 1. Otherwise, there are extra reactants, either fuel or air, which remain after combustion and absorb some of the combustion energy, resulting in a lower temperature increase. Since temperature is important for NOx formation, I figured it was worth mentioning.

• Thanks for your answer! I think I'll be able to give just a brief description of the meaning of the data I'm using now. Regarding the incomplete combustion, I more or less just took that from here: engr.colostate.edu/~allan/thermo/page9/page9.html where it says "If it is < 1, the combustion is lean with excess air, and if it is >1, the combustion is rich with incomplete combustion." But not mentioning incomplete combustion and simply saying that <1 means too little air and >1 too much air is probably enough for me. Commented Feb 27, 2015 at 6:25
• That page seems to be speaking of "fuel-air ratios" instead of "air-fuel ratios". People sometimes use one and sometimes the other. It just flips things around. So their "equivalence ratio" is the opposite of ours. If you check the wikipedia page, their (fuel-to-air based ratio) is phi or $\phi$, and the one I used in my answer (air-to-fuel based ratio) is lambda or $\lambda$. Commented Feb 27, 2015 at 7:46

There are roughly 2 definitions of equivalence ratio (ER). φ=AFR(stoich)/AFR(actual) and λ=1/φ

I personally don't understand why they are both referred to as ER but they are so be careful. In the graph that you've shown, ER>1 refers to a rich mixture (not enough air to completely burn). I can say that due to the fact that NOx formation is reduced since there is no oxygen available for oxidation of nitrogen at high temp and pressures. ER<1 is a lean mixture meaning there is more air than needed. This means N2 gets oxidised.

AFR(stoich) is a property of your fuel but AFR(actual) is the condition under which you do the combustion. These are usually for internal combustion engines and by the looks of your graph, it's a compression ignition engine.

feel free to drop me an email if you don't understand what I just wrote! :D It's quite a lot of stuff to get your head around. [email protected]