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 Lfuel / 14.7 Lair).

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 Lair / 2 rev) * (1 Lfuel / 14.7 Lair)

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

  • AttackBunny@lemmy.world
    3·
    1 year ago

    There are a lot of variables your missing. One thing, afr doesn’t refer to liquid fuel.

    That’s said, you need to know things like (not a totally inclusive list, just first things that come to mind).

    Pulse width of injectors Are injectors batch fire or sequential (assuming fuel injected) Speed density or hot wire flow meter What’s the max output of the pump and injectors Battery voltage Sea level or altitude Ambient temp and humidity Displacement of engine Is the engine new and fresh or old and worn out? Fuel type and rating. Pump has? Methanol? Leaded? n2o?

    To give you a little reference, when we build smaller displacement engines for more power, we frequently use a 340lph pump. In a perfect world that means that pump can move 340 liters per hour. So in a perfect world at max duty cycle that pump (which is bigger than most factory pumps) can move absolutely maximum of 5.666666 liters per minute. There’s no way at 800rpm any engines are moving 4.5L like you’re figuring. But, pumps never run at max (maybe as a lift pump, maybe). Then you have to factor in restriction and what’s the output of the injector. It gets even more complex when you factor in fuel type and the rest of the questions I had above.

    Basically you need a LOT more info to accurately calculate consumption.

    • glibg10b@lemmy.mlOP
      11·
      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 Lair * 1.212 gair/Lair * (1 gfuel / 14.7 gair) / 800 gfuel/Lfuel * (900/2)/min * 60 min/h

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