[ReverendDexter]

I was postulating this on the porcelain throne, and I'm trying to figure out where my math is failing:

If I have a 5.0 liter 4-stroke engine, that means that I process 2.5L of air-fuel mixture with every revolution of the motor (assuming 100% volumetric efficiency).

That means that if I'm cruising a 2000 rpm, the motor is processing 5000L of air-fuel mixture every minute. At a stochiometric ratio of 14:1 (I think that's actually lean, but I'm not sure what's ideal), that would mean every liter of air-fuel mixture has 15 "parts" - 14 of air, and 1 of fuel.

If we divide 5000L by 15, we get 333.33 L of fuel consumed in that 1 minute. Converted to gallons (at 3.8L/gallon), that's 87.7 gallons of fuel.

87.7 gallons to drive 1 minute at 2k rpm with a 5.0L engine? Nuh-uh. At the most it should be 0.1 gallons. (assuming 2k rpm is ~60mph, 1 minute of operation should travel 1 mile, and assuming 10mpg).

What's missing here? My immediate assumption is there's some conversion between the volume of gasoline as a liquid and the volume of gasoline as a gas (roughly 900:1 based on the above), but that doesn't make sense - gasoline does NOT exist as a gas inside an internal combustion engine, right? Isn't it just aerosolized liquid?

[ReverendDexter]

Seriously? No responses whatsoever?


Does no one know, or does no one care?

[Dansby]

I just don't know...haha. I'm interested to see the correct solution, but I'm an english person, haha.

[Safi]

I may take a stab at it tomorrow....I'm too tired now

[unlicensed to drive]

I don't know, and I'm sure that's why others have failed to respond. I have looked over your calculations a few times and I don't see what you did wrong to result in that way off answer. You may be onto something concerning a liquid gas to vapor gas conversion.

[08blkgt500]

usually im very good with math/conversions and i agree the answer cant be 87.7 gallons. i tried doing it myself, got the same answer. i tracked it backwards on your word problem. all i can say is maybe its not 2.5L of air-fuel mixture per revolution? either that or the stochiometric isnt 14:1?

[cusp]

Your error is using volume instead of mass. Chemical reactions work on mass.

So you need to take 14 Mol of air and 1 mol of gasoline for your reaction. I don't know the Molecular weight off hand of your elements. But thats the key to getting things to calculate out correctly.

[08blkgt500]

^ that could be it

[sikedsyko]

Fuck we just learned about this in Chemistry today... But there's alot of factors involved. and isn't 5.0 the amount of displacement? I would think it would be alot less volume that the mixture is being inserted too

but yeah I'll try again when it's not 5a.m. and I have class in 4 hours

Ideal Chemical equation is something like

2C8H18 + 25O2 -> 18H20 + 16CO2

[~C~]

I haven't done this shit in over 15 years, but it seems to me that the parts are wrong. They should be PPM, or parts per million so i dont think taking 5000 and dividing by 15 is right. If you got 87 gals I think it should be something in the range of .087 or so. This would make more sense.

[Trick Tuners]

you need to use pounds of air and pounds of fuel instead of liters. Stoich is 14.7 not 14.1 (at least for regular gas)

[cusp]

Originally Posted by Trick Tuners you need to use pounds of air and pounds of fuel instead of liters. Stoich is 14.7 not 14.1 (at least for regular gas)

Pounds work.

A mole has 6.0221415×10^23 atoms or molucules and is measured in mass not volume. Boyles law describes the physics of gases. A 5.0L engine is a measure of volume. So depending on things like pressure, (Density Altitude and/or Forced Induction) and Temperature, the mass of air and fuel pumped through that volume will change. Horse power comes from the fuel when it combines with oxygen to release heat. So to get your reaction to come out right, you need to have 14.7 Moles of Oxygen (air is 18 percent O2) and 1 mole of gasoline.

[JASonic]

This is interesting...
Oxygen in the air is what Stoch is based on? Not the entire mass of air in the cylinder before combustion? If its based on weight not volume then you would need a shitload of air to meet the liquid weight of gasoline....and that makes sense when thinking 14:1.


Air weighs 14.7 pounds per square inch at sea level. IDK...Im confused now.

[Trick Tuners]

Originally Posted by JASonic Air weighs 14.7 pounds per square inch at sea level. IDK...Im confused now.

haha, no it doesn't! Each breath i take would weight at least 50 lbs!! Its 14.7psi of pressure from all the air in the atmosphere pushing down. So a 302 would spit out 4439lbs of air for 2 revolutions! From a quick search air at STP is about 1300g/M^3. The ECU deals in # of air and # of fuel. MAF measurements are converted to # of fuel and thats why injectors are rated in #/hr instead of a volume. (ricers like CCs)

[Dennis]

am i the only one who got a headache just reading that shit lol

[sickfox]

Originally Posted by Dennis am i the only one who got a headache just reading that shit lol

no.

[sickfox]

2+2=4 4-2=2 2/2=1 1-1=0

[cusp]

Originally Posted by JASonic This is interesting... Oxygen in the air is what Stoch is based on? Not the entire mass of air in the cylinder before combustion? If its based on weight not volume then you would need a shitload of air to meet the liquid weight of gasoline....and that makes sense when thinking 14:1. Air weighs 14.7 pounds per square inch at sea level. IDK...Im confused now.

The air column above you exerts a pressure 14.7 PSI during "standard" conditions. But the column extends into space. So in effect that 1 Sq inch is very tall. O2 in the air is what the reaction is based on. The other 82 percent of the gasses to not combust or if some trace molecules happen to, they are in amounts to small to make a difference.

[101fng]

so how much does a cubic foot of air weigh?

[Trick Tuners]

Link

About 0.075 lbs. Caculated using the ideal gas law: PV=nRT. Assuming the T=68F (528R) and P=1atm with the MW of air approx. 28.8 lb/lb-mole. The gas constant R, using these units is 0.7302. Rearranging the formula to give: mass = MW*(PV/RT)

the air constitution is aproximatly this: * 75.523% nitrogen * 23.133% oxygen * 1.288% argon * 0.035% carbon dioxide * 0.001267% neon * 0.00029% methane * 0.00033% krypton * 0.000724% helium * 0.0000038 % hydrogen so, the molar mass of the air is the sum of: N2 = 0.75523 * (2 * 14.0067) = 21.15656008 O2 = 0.23133 * (2 * 15.9994) = 7.402282404 Ar = 0.01288 * 39.9481 = 0.514531528 CO2 = 0.00035 * (12.0108 + (2 * 15.9994)) = 0.01540336 Ne = 0.00001267 * 20.1798 = 2.55678066*10^-4 CH4 = 0.0000029 * (12.0108 + (4 * 1.00795)) = 4.652354*10^-5 Kr = 0.0000033 * 83.7982 = 2.7653406*10^-4 He = 0.00000724 * 4.0026 = 2.8978824*10^-5 H2 = 0.000000038 * (2 * 1.00795) = 7.66042*10^-8 21.15656008 + 7.402282404 + 0.514531528 + 0.01540336 + 2.55678066*10^-4 + 4.652354*10^-5 + 2.7653406*10^-4 + 2.8978824*10^-5 + 7.66042*10^-8 = 29.08938516