# Air fuel ratio

107 views
0

Possibly a bit of a smooth brained question. So how exactly does the math of the air to fuel ratio work in a combustion engine. As an example; ethanol, ratio 9:1 (just for simplicity). Let’s say you have a cylinder that’s 500cc. It makes no sense to me that in one 4 stroke cycle, the cylinder will intake 450cc air and 50cc ethanol! That’s an insane amount of fuel regardless of fuel type and obviously is not mechanically possible. Somethings going completely over my head. Or is it simply that a cylinder take in a very small amount of air too? Nowhere even close to its capacity?

In: 2

It’s mass ratio, not volume, it might intake 0.6 grams of air (about 500 ml) and 0.07 grams of ethanol if the ratio is 9:1

And even if it was a volume ratio, you’d still have to calculate the fact that it’s vapor volume, not liquid volume, because stuff expands as it changes from liquid to gas

500cc is the volume of the fuel after burning. Thats a lot more than before because a liquid fuel becomes a hot gas wich expands a lot.

It takes a tiny amount of air and fuel in, wich is automatically controlled by measuring what you have too much of right now with the labda sensor. Too much fuel? Reduce a little. Too much air? Use a little more fuel