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Apparently I have a stupid question.

I got curious and crunched some numbers to determine approximately how much air an engine consumes in total to burn a full tank of gas. Something went horribly wrong and I feel stupid.

So theoretically, I can get this number by multiplying the total amount of fuel in the tank, let's say 15 gallons (56.78 liters), by the air part of the air-fuel ratio, which I read is 15:1. So the engine would consume about 225 gallons (851.7 liters) of air per tank of gas?

But that number makes no sense if you take into account the other way to calculate this: engine size. Let's say we have a four cylinder, two liter engine. Since it's four stroke, the air it will consume should be equivalent to its rpm. So at 1,000 rpm it will consume 1,000 liters of air per minute. Right? That's a very long way from a total of 850 liters for the whole tank of gas.

So what did I do wrong?

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  • $\begingroup$ Did you consider that the fuel goes in as a vapour not a liquid? $\endgroup$ – Solar Mike Feb 23 '18 at 17:56
  • $\begingroup$ strictly following the 15:1 ratio, you could only drive for less than a minute with a full tank according to your second approach $\endgroup$ – OpticalResonator Feb 23 '18 at 18:08
  • $\begingroup$ notes on the volume method. 1) a 4 stroke pulls air every other revolution, so use half your rpm. 2) the air pumped per stroke is substantially less than the geometric volume because the throttle restriction produces a significant vacuum. If you consider running the engine wide open throttle you might get in the ball park that way (but there are always pumping losses) $\endgroup$ – agentp Feb 25 '18 at 20:05
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You probably want to look at stochiometric ratios by mass.

However this is a difficult problem to solve from first principals as for an IC engine air is essentially free and monitoring its consumption in bulk is not a major priority.

I suspect at the mistake you have made is confusing mass ratios with volume ratios. As air is a gas and as such has very low density this make a very big difference.

If we assume that 15:1 air:fuel ratio by mass then we will need 15kg of air for every kg of fuel burned.

Air has a density of about 1.2 kg per cubic metre so that's 18 cubic metes of air per kg of fuel (at atmospheric pressure)

An alternative approach is to consider the swept volume per cylinder and compare with rpm and fuel economy

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