A question directly about fuel consumption

[elemein]

Originally Posted by redyaris

As far as i know you can use pounds of air and ponds of fuel to do the calculations so you will have to convert volume of air to weight and volume of fuel to weight. at which point you can do the calculation to find mile per gallon.

Oh so... Let's see... Is this correct?:

1 litre of air weighs about 1.29 grams.

So;

1 L = 1000 cc
2000 RPM = 1000 RPM of combustions

1.29 (grams) / 15.5 = .08322 grams of fuel (I am using 1.29 g as that is a litre of air, and the maximum displacement at 100% VE of the engine is 1 L; so it works out.)

Fuel Consumption = 0.08322 * 0.20 (throttle) * 0.80 (VE) * 1000 (RPM of combustions)
= 0.01664 * 0.80 * 1000
= 0.01331 * 1000
= 13.31 grams of fuel burned per minute

1 litre of gasoline weighs ~737.22 grams
So, 13.31 / 737.22 = 0.018; or 1.8 %

1.8 % of a litre is 18cc

So 18 cc of fuel is burned in this situation?

Am I correct?

[mort]

Originally Posted by elemein

Let's say I am driving a car with a 1.0 L engine (quite small; a Geo Metro perhaps?) I am cruising at 2000 RPM and have the throttle down at 20%. Let's say that the engine is currently running at 80% Volumetric Efficiency, and, it is also running a bit lean at a 15.5:1 A/F ratio.

How much fuel would I use?

Hello elemein,
First using 20% throttle and 80% VE is a bit unclear. VE generally means at full throttle. Assuming you mean to find the amount of fuel used at 20% throttle, and also assuming that 20% throttle means that the amount of air (mass of air) being consumed is 20% as much as would be consumed at full throttle, then your math is fine. Assuming 20% as much mass of air makes everything easy, you consider that the air entering is about STP and the density is about 1.2 gm/l So 200 cc (20% of 1l) is 0.24 gm of air.
0.24 * 1000 (rpm/2) = 240 gm per minute. at 15.5:1 A/F gets you 15.4 gm/minute, or 929 gm/hr of fuel. What speed is this Metro making at 2000 rpm? Is that likely 50 km/hr? 929/50=18 gm/km. The density of gasoline is about 750 gm/l so 0.025 l/km or 40 km/l (about 100 mpg)

Or is that 2000 rpm and 20% throttle closer 10 km/hr? Or 100?

-mort

[elemein]

Originally Posted by mort

Hello elemein,
First using 20% throttle and 80% VE is a bit unclear. VE generally means at full throttle. Assuming you mean to find the amount of fuel used at 20% throttle, and also assuming that 20% throttle means that the amount of air (mass of air) being consumed is 20% as much as would be consumed at full throttle, then your math is fine. Assuming 20% as much mass of air makes everything easy, you consider that the air entering is about STP and the density is about 1.2 gm/l So 200 cc (20% of 1l) is 0.24 gm of air.
0.24 * 1000 (rpm/2) = 240 gm per minute. at 15.5:1 A/F gets you 15.4 gm/minute, or 929 gm/hr of fuel. What speed is this Metro making at 2000 rpm? Is that likely 50 km/hr? 929/50=18 gm/km. The density of gasoline is about 750 gm/l so 0.025 l/km or 40 km/l (about 100 mpg)

Or is that 2000 rpm and 20% throttle closer 10 km/hr? Or 100?

-mort

This math sounds similar to the math made in my previous post; is my previous math correct?

I am not sure what speed the Metro is making at those conditions; I am just sortof making up conditions and it just so happens that the Metro happens to have a 1.0L engine and fits this situation oddly well...

Nonetheless; thank you very much! Your posts have made complete sense and are quite concise. I understand this subject much better! Thanks again!

[mort]

Hi elemein,
I posted before I saw your replies.
Yes, except for spurious VE, quite right.
-m

Originally Posted by elemein

Oh so... Let's see... Is this correct?:

1 litre of air weighs about 1.29 grams.

So;

1 L = 1000 cc
2000 RPM = 1000 RPM of combustions

1.29 (grams) / 15.5 = .08322 grams of fuel (I am using 1.29 g as that is a litre of air, and the maximum displacement at 100% VE of the engine is 1 L; so it works out.)

Fuel Consumption = 0.08322 * 0.20 (throttle) * 0.80 (VE) * 1000 (RPM of combustions)
= 0.01664 * 0.80 * 1000
= 0.01331 * 1000
= 13.31 grams of fuel burned per minute

1 litre of gasoline weighs ~737.22 grams
So, 13.31 / 737.22 = 0.018; or 1.8 %

1.8 % of a litre is 18cc

So 18 cc of fuel is burned in this situation?

Am I correct?

[elemein]

Originally Posted by mort

Hi elemein,
I posted before I saw your replies.
Yes, except for spurious VE, quite right.
-m

Ah thanks mort! It looks like this thread is answered.

[redyaris]

Originally Posted by elemein

Oh so... Let's see... Is this correct?:

1 litre of air weighs about 1.29 grams.

So;

1 L = 1000 cc
2000 RPM = 1000 RPM of combustions

1.29 (grams) / 15.5 = .08322 grams of fuel (I am using 1.29 g as that is a litre of air, and the maximum displacement at 100% VE of the engine is 1 L; so it works out.)

Fuel Consumption = 0.08322 * 0.20 (throttle) * 0.80 (VE) * 1000 (RPM of combustions)
= 0.01664 * 0.80 * 1000
= 0.01331 * 1000
= 13.31 grams of fuel burned per minute

1 litre of gasoline weighs ~737.22 grams
So, 13.31 / 737.22 = 0.018; or 1.8 %

1.8 % of a litre is 18cc

So 18 cc of fuel is burned in this situation?

Am I correct?

0.20 ( throatle ) is a problem because it does not tell you the mass of air getting into the engine. The volume will always be 1000cc regardless of throatle position, what changes is the mass/density of air in the cylinder when the valve closed.
The 0.8 volumetric efficiancy is at optimum engine speed, probably at max torque at wide open throatle.

[elemein]

Originally Posted by redyaris

0.20 ( throatle ) is a problem because it does not tell you the mass of air getting into the engine. The volume will always be 1000cc regardless of throatle position, what changes is the mass/density of air in the cylinder when the valve closed.
The 0.8 volumetric efficiancy is at optimum engine speed, probably at max torque at wide open throatle.

You are a little more precise.

20% throttle does not let by 200 cc of air; it still lets in 1000 cc, but the 20% throttle makes that particular volume of air have the density and pressure of 200 cc at atmospheric.

So yes, you are correct, but the math still works out the same.

[user removed]

Throttle position does not consider the load on the engine. You can have 0 vacuum, and 100% of the engines displacement inducted into each cylinder, per combustion event, if you have a high enough load and low enough RPM.

Higher RPM and lower load, versus lower RPM and higher load can both be achieved at 20% throttle, but the RPM would differ even with an identical throttle position.

regards
mech

[redyaris]

Originally Posted by elemein

You are a little more precise.

20% throttle does not let by 200 cc of air; it still lets in 1000 cc, but the 20% throttle makes that particular volume of air have the density and pressure of 200 cc at atmospheric.

So yes, you are correct, but the math still works out the same.

That is not corect. You would have to do some testing or find some sort of chart, to see what the mass of air traped in the cyclinder is at % throatle. There is no liniar relationship between % throatle position and mass of air traped in the cylinder under variouse % throatle.

[elemein]

Originally Posted by redyaris

That is not corect. You would have to do some testing or find some sort of chart, to see what the mass of air traped in the cyclinder is at % throatle. There is no liniar relationship between % throatle position and mass of air traped in the cylinder under variouse % throatle.

Volumetric efficiency also plays a role in here too. There's a lot of variables, but the math works out to be roughly the same. Volumetric efficiency is something you can measure at EVERY RPM and throttle position; not just at max throttle and most efficient RPM.