Question about fuel consumtion at idle.

[iamkungfool]

I have absolutely no idea how to calculate this or where to begin to look. That being said, I'm working on an essay about ways to make minute alterations to basic driving patterns that will add up to noticeable large scale fuel usage cutbacks.

Basically I need to know what our Hemi's use at idle (I assume they idle 99% of the time in MDS mode). From that I'll extrapolate to some best estimate of "average" idle use per person. When I finish i'll be sure to post.

[MagnumFreak]

It actually idles in 8 cyl mode. Fuel consumption at idle is so low that switching to 4 cyl would provide little benefit. Not sure how to calculate the fuel consumption at idle maybe fnkychkn will chime in with some data.

[LP81]

Here is a potential...

not the most fun way, but just run it out of gas... fill it with a specific amount and then run it dry again... repeat at a higher RPM and see how it comapres to your driving... and once more at higher and compare the time it takes to drain each time with different RPMS...

[iamkungfool]

there is potential there. how about this way:

when you fill it all the way it cuts off automatically. how does it decide this point? is there some kind of switch inside? what i'm asking is how consistant is it? i know it's probably not going to be as consistant as empty lol but i figure i can probably sit at the gas station fill up all the way not move the pump, run it for 20-30 min, pump in more see how much i used and go from there?

[LP81]

I guess depending on how much you wannt pay out and how accurate of results you want.. you could just fill up a 1-2 gallon can and fill it from that each time to be consistant or some other way to measure out exact amounts... the auto shutoff on the pump it's really accurate.. sicne you can still fill up more after it auto shuts off before it will spurt out...

[iamkungfool]

haha considering it takes how many hours to burn up a tank of gas running at 65mph i'd hate to see how long it'd have to idle. i think i'll see next time i go to the gas station...unless fnky can chime in

[LP81]

lol that's why you only do a gallon or 2 at a time... it will go faster then you think...

[magnuman]

My ScanGauge indicates that fuel usage at idle w/o A/C on is .3 GPH, and with the A/C on .4 GPH. I have no way of knowing if this is accurate though. I know some of the fuel functions on the ScanGauge are inaccurate due to the MDS, but 8 cylinder functions may be as correct as the EVIC.

[quarky42]

I would think fnky or another engineer could tell you what the min flow rate of the fuel injection system is. The service manual might have something to say too. If they can't tell you how much fuel you are using in flow rate terms perhaps someone can give us enough information to extrapolate or better yet calculate the fuel usage:

Does the displacement of the engine combined with a fuel to air ratio tell us how much fuel is in the engine on any given "stroke" or revolution?

Engine Displacement: 5654cc or 5.654 Liters. Would a fuel to air ratio tell us how much of that is gasoline? If we can figure out how much fuel is in one revolution of the engine then we can multiply it by what ever your engine idles at. Mine seems to idle around 700 ish.

I'm thinking displacement is the wrong term to use, on the other hand if we knew the volume of the compressed gasoline on the compression stroke, we could still use the fuel to air ratio. (displacement was just an example of a volume term we could find out easily enough.)

[quarky42]

Compression Ratio 9.6:1
Combustion chamber volume 84.9cc


Found this on the web... dunno how accurate it is. (I'd rather use something like the service manual or a better website... but this might be worth considering.)
http://www.allpar.com/mopar/new-mopar-hemi.html

Okay, so I found out that the air to fuel ratio is based on mass. What is the mass of 1cc air or mass of 1cc gasoline. If I knew this I could figure out the volume of air and get the volume of gasoline (or visa versa). With the volume of gasoline in one cylinder you can get the volume for all 8 cylinders (a full revolution) and multiple that by 600 or 700.

I guess I'd need to know what the mass of gasoline is and the mass of air is when compressed 9.6 times. I'm assuming gasoline itself doesn't compress much compared to the air. It is the compressed gas and air that explodes in the combustion chamber. With the mass of gass and mass of air compressed 9.6 to 1, then I can get the volumes that I'm interested in.

[quarky42]

Air (compressed 9.6:1): 12.384 kg/m^3 is 0.012384 g / cc
Gas 737.22 kg/m^3 is 0.73722 g / cc

Correct me where I'm wrong knowledgable people...I'm just playing with the math to see if I can get an estimate.

[quarky42]

I think that leads me to:
1.0514016 grams air in 84.99cc assuming it was all air
But it isn't all air so I divide by 16 and multiple by 15 because it is a 15:1 ratio (assuming the car is trying to maintain a 15:1 a-f ratio)
0.985689 in 15/16ths of the 84.9 cc

I divide by 15 to get the mass of fuel in 1/16th of 84.9cc
0.0657126 grams fuel in 1/16th of 84.9cc


After plugging in the numbers, I can tell I'm off by an order of magnitude here somewhere. I think the method is okay comparing grams of fuel to grams of air. There is a lot more air than fuel.

Maybe there is a more simple way to do this: Somewhere on the forum is our car's air flow at idle. The air coming in the intake is being compressed down 9.6:1 and that mass of air is being combined at a ratio of 15 parts air to 1 part fuel. Get the mass of air consumed at idle, that will give you the mass of fuel consumed at idle... convert that using the mass of fuel to volume given. You'll have the volume of gas consumed at idle at the rate of the air going in (per second or per what ever the rate of air is)

[iamkungfool]

I like where this is going, but I think the vaporized/atomized gasoline might have a different density, well not really because it's not in gaseous form. Ok so all we need is the air/gasoline ratio at idle now right? We'd also have to know what that ratio means right? Is it mass to mass or volume to volume? Because obviously 10 mass units of air versus 10 mass units of gasoline is going to be a different ratio to 10 volume units of air versus 10 volume units of gasoline.

If it's 9-1 and the Combustion chamber volume 84.9cc (is that pre or post compression? i'm going with post, 85x8=680ccs). so that would mean 90% is air and 10% is gasoline, or 68ccs of gasoline per revolution. I'm really very very very very very sure that the ratio has to be different otherwise we'd be getting rediculously goofy mpg or inches per gallon.

i do remember some chemistry so to balance the combustion equation...bah nevermind i found a site that's right: http://www.madsci.org/posts/archives...4809.Ch.r.html the important part is that 25 molecules of O2 react with 2 molecules of C8H18 (octane) . but that still seems odd and it's in a perfect world.

Ugh...this is why i loved and hated chemistry...