Calculating idle fuel consumption (lemmy.ml)

submitted 1 year ago* (last edited 1 year ago) by glibg10b@lemmy.ml to c/cars@lemmy.ml

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Some research I've done suggests that a small idling engine should consume about 0.2 - 0.3 gal/h, which is about 0.8 - 1.1 L/h. However, the following calculations I've done are off by a mile.

At idle, engines typically operate in a rich-running condition, which implies a best-case air-fuel ratio of roughly 14.7:1 (i.e. 1 L~fuel~ / 14.7 L~air~).

According to this article, engines typically have a volumetric efficiency of 15% at idle. This means the volume of fresh air drawn into each cylinder is about 15% of the cylinder's volume.

A 4-stroke engine sucks air into all cylinders in two revolutions.

Using this information, I've calculated the idle fuel consumption for a 1.1L engine at 800 RPM as follows:

rate = (800 rev / min) * (15% * 1.1 L~air~ / 2 rev) * (1 L~fuel~ / 14.7 L~air~)

This comes out to roughly 4.5 L~fuel~ / min, or 269 L~fuel~ / h. What gives?

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[-] glibg10b@lemmy.ml 0 points 1 year ago

None of this matters because we already know the AFR, which tells us how much fuel is being injected. The ECU uses the O2 sensors and VE map to ensure the measured AFR matches the commanded AFR

One thing, afr doesn’t refer to liquid fuel

You're right, it refers to air and fuel mass regardless of state of matter. Here's the corrected expression using the vapor density of air at 18 °C and the liquid density of petrol:

15% * 1.1 L~air~ * 1.212 g~air~/L~air~ * (1 g~fuel~ / 14.7 g~air~) / 800 g~fuel~/L~fuel~ * (900/2)/min * 60 min/h

This roughly equals 0.46 L~fuel~/h, which is close. I think the difference can be attributed to the volumetric efficiency estimation's accuracy

[-] glibg10b@lemmy.ml 1 points 1 year ago

@AttackBunny@lemmy.world a downvote with no reply is not useful feedback

this post was submitted on 26 Aug 2023

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