# Gasoline stoichiometry calculation

1. Sep 3, 2007

### GPracer2500

I'm trying to figure out where the often cited 14.7:1 stoich a/f ratio for gasoline comes from. I've always thought it was calculated from a primary reference fuel containing some mixture of iso-octane and n-heptane. But I've done the calculations for those hydrocarbons and my numbers don't match up. I did some figuring in THIS THREAD. I did the math again with less rounding and I'm getting:

iso-octane: 15.03:1
n-heptane: 15.08:1

I'm trying to tweak my calculations (concerning how to figure in the atmosphere that is NOT oxygen) to see if that is my problem. If nothing comes of that....

My question is, were is 14.7:1 coming from? Which hydrocarbons and in what mix are "they" using to come up with that figure? And what composition of air would be used? Does anyone know?

note: I'm not looking to figure any real-world situation. Rather, I'm concerned with how the 14.7:1 figure is generated.

2. Sep 3, 2007

### Staff: Mentor

For gasoline it is apparently 14.7/1 air to fuel.

http://en.wikipedia.org/wiki/Air-fuel_ratio

Looking at octane - M ~ 114.23 gm/mol
http://en.wikipedia.org/wiki/Octane

Then air is ~79% N2 (M=28) and 21% O2 (M=32) .

One has to determine the moles of O2 then mass of air for that number of moles of O.

3. Sep 3, 2007

### GPracer2500

Thanks. I think I've already done that though:

1 mole C8H18 and 12.5 moles of O2.
12.5 moles O2 = 400 units mass.
3.76 times as much nitrogen as oxygen in air (79%/21%): (3.76)(12.5 moles) = 47 moles N2. 47 moles N2 = 1316 units mass.
1316 + 400 = 1716
1716.599 / 114.232 = 15.03:1

n-heptane works out to 15.08:1.

That's where I'm stuck. I'm suspicious that I've made a mistake in determining how much "air" it takes to get the required moles of oxygen. Aren't the 79% and 21% values by volume? Do I need the oxygen-to-air relationship in weight instead?

4. Sep 3, 2007

### Staff: Mentor

Yes, but since 1 mole of a gas at STP is 22.4 l, those would also be mole fractions.

Yes, or technically by mass. But that's what one would do if one determines the moles of O2 needed and then added the moles of N2 associated with number of moles of O2.

I think your method is correct, but some how it doesn't yield 14.7 which is 0.98 * 15.

No site I can find explains how A/F = 14.7. They just refer to it as the stoichiometric ratio.

5. Sep 4, 2007

### AlephZero

The "78.084% N2, 20.946% O2, etc" composition is for dry air.

Wikipedia gives the water vapour content as "typically 1% to 4% near the surface". Since practical engines don't run on dry air, does this explain the 2% discrepancy between your calculation and the "standard" value used by engine designers?

6. Sep 4, 2007

### FredGarvin

There is also a very long line of the other constituents of air that are not included in the simple hand calcs we do in thermo classes. We also dismiss the formations of $$NO$$ and $$NO_2$$. I believe the 14.7 number is an experimentally derived number.

Last edited: Sep 4, 2007
7. Sep 4, 2007

### Staff: Mentor

I was wondering about that myself. It could.

Also, as Fred mentioned, there are other constituents to consider. I've seen one formula dealing with CO and NOx in the exhaust, but I thought the ideal A/F assumed ideal combustion - but perhaps not. Certainly NOx would lead to less than ideal combustion.

8. Sep 4, 2007

### brewnog

The formula I've seen used defined stoichiometric air fuel ratio in terms of the relative carbon/hydrogen content of the fuel in question, along with the molecular weights of the constituents.

Unfortunately, it only accounts for the combustion of a single hydrocarbon fuel in an oxygen and nitrogen mix, and using typical fuel data here, the calculated stoichiometric AFR is something like 15.15.

Given that Haywood quotes 14.6 as being the stoichiometric AFR for gasoline, I believe this number is derived purely by experiment.

9. Sep 5, 2007

### TexanJohn

It's been way too long since I tried to balance chemistry equations. I don't know if these guys know their stuff, but their product does seem to work (it's an EFI controller). You might find this article an interesting read. I tried to compute the A/F ratio using their math and still didn't come up with 14.7:1.

10. Sep 5, 2007

### brewnog

The problem with AFR control units is that even with a lambda sensor, the system still runs on a map which is tweaked and scaled up/down to give an optimum AFR, which probably still won't actually equal 14.7:1.

11. Sep 5, 2007

### TexanJohn

I think there are a multitude of reasons 'why' an EFI system won't always have the desired AFR (whether it is target of stoich or something different). There are problems ranging from concept (a feedback system based on a lambda sensor is always going to be at least 1 cycle behind even if operating conditions remain static), processing speed, accuracy, method of determining of the mass of air, ... :)

I was just providing a link to a fairly detailed discussion of the computation of a stoich AFR. Using the formula's on the page I referenced, I get 15.03:1.

12. Dec 22, 2007

### GPracer2500

I found an SAE paper: Stoichiometric Air-Fuel Ratios of Automotive Fuels

http://www.sae.org/technical/standards/J1829_200210

Anyone have access to SAE docs? I can't pay \$50 for that. Plus, I'm suspicious it doesn't actually define a "standard" gasoline but rather guides one through the stoich calculations given an already known gasoline composition. I'd still love to read it though.

13. Dec 23, 2007

### ray b

perfect vs real world
not all the gasoline becomes a vapor
some puddles in the intake track
some burns later in the headers on the way out

so the 14.x vs the 15.03:1. or whatever is real vs ideal

14. Dec 23, 2007

### TVP45

Use the percentage of air by weight (0.239 I think). Although you can switch back and forth between weight and volume when looking at reactions, you can't get away with that with mixtures. The AFR will come out just about 14.7. In real life that changes with all the reformulations, seasonal variations, ethanol additions, etc, but a Lambda sensor handles that just fine.

Generally if you're just looking to talk about gasoline, use 70% heptane and 30% octane and ignore all those iso-thingys - they just make my head hurt.

15. Dec 23, 2007

### GPracer2500

I understand what your saying. But my understanding is the 14.7 a/f ratio is an ideal number (i.e. it assumes perfect combustion). If that's not the case then what defines the variables used to arrive at 14.7 a/f ratio for gasoline? I mean literally--what gasoline, engine, "air", etc. where used to experimentally (or otherwise) determine 14.7 as an a/f ratio for gasoline.

That's the crux of my question. Certainly there must be some specific basis for arriving at 14.7? It's hard for me to imagine that it's just a guess--especially when it's suppose to be an ideal number.

16. Dec 23, 2007

### GPracer2500

Using data from Bruce Hamilton's superbly cited Automotive Gasoline FAQ, I'm getting oxygen as a 23.14% constituent of air, by weight. So using 23.14% instead of the 21% I used before:

1 mole C8H18 and 12.5 moles of O2.
12.5 moles O2 = 400 units mass.
3.32 times as much nitrogen as oxygen in air (76.86%/23.14%): (3.32)(12.5 moles) = 41.5 moles N2. 41.5 moles N2 = 1162 units mass.
1162 + 400 = 1716
1562 / 114= 13.7:1 a/f ratio for iso-octane

Making the same change in my calculation for n-heptane, I get a 13.75:1 a/f ratio.

Different results but the same problem. No combination of iso-octane and n-heptane could possibly be stoich at 14.7:1.

Yes, I understand. I began believing that stoich = 14.7 for gasoline was based on a binary reference fuel containing iso-octane and n-heptane. I'm struggling to get any result in which that could possibly be the case. So I'm still left wondering what specific mix of hydrocarbons was/is used as the reference fuel upon which 14.7 is based.

Unless I'm really confused, all isomers of octane and heptane share the same number molecules (octane = C8H18, heptane = C7H16). So for stoich calculations I don't think it matters which isomers are used. I've been saying iso-octane and n-heptane but I could have just as easily been saying octane and heptane.

17. Dec 23, 2007

### TVP45

14.7 is the standard AFR. It is based on a mixture that is approximately 70% heptane, 30% octane. It is the ratio for complete combustion. It is not a guess, but there are real-world differences.

Gasoline is formulated differently in some states and at certain times of year. Gasoline from different wells has slightly different mixes. Because of this, the actual AFR for any particular gasoline may vary from 14.5 - 14.8.

You normally do not want to run an engine at 14.7 (stoichiometric) because (1) it runs hot; (2) there is almost never complete mixing and burning in a cylinder; and (3) it leads to increased NOx emissions.

The lambda sensor reads excess oxygen in the exhaust stream. That might be set for, say, 1% (I don't remember the standard number but that's close) and will typically vary depending on operating conditions. So, if the AFR is 14 or 14.5 or whatever, it doesn't matter. The engine adjusts to give you the required excess oxygen.

18. Dec 23, 2007

### GPracer2500

That was my original hypothesis (not that precise ratio--but some ratio of those two hydrocarbons). Yet I can't get the chemistry to work. Can you show me the stoichiometry calculations that demonstrate that ratio (or any ratio) of those two hydrocarbons being stoich at 14.7? Or provide some source of info I can use to confirm 70/30 heptane/octane is the reference fuel used arrive at the commonly cited 14.7 figure?

According to my calculations, heptane and octane are stoich at nearly identical a/f ratios (and no result I've come up with comes out to 14.7). How could any combination of the two be anything other than the stoich they both practically share?

I'll accept that 70/30 heptane/octane is stoich at 14.7 but only if I see the stoich calculations that demonstrate it.

Yes, yes--I'm aware of all that. My interest in the precise origins of the 14.7 figure are purely academic.

thanks

19. Dec 24, 2007

### TVP45

My apologies. I didn't understand your question. I think I now see your point. So, to actually calculate that 14.7, you have to use the actual constituents of typical gasoline. Just for a start, try throwing in 10% ethyl benzene (any of the -enes should help your calculations since they are less saturated).

Thanks for a good question that got me thinking about facile assumptions (I've mostly worked with stuff like natural gas or coke gas and just sort of accepted whatever I had heard about gasoline).

20. Dec 24, 2007

### GPracer2500

No worries. I'm away for the holidays but will pick this up again when I can....