Air flow and fuel flow calculations 2

[gaber26111]

Hello everybody here,

First, that is not a student work, that is a work i do as freelancing task, if anybody hears about, well, here is what i need help about,

Take a look at that website, which offer engine simulator software, gives accurate calculations for the whole vehicle, with some input data, here is it:

http://www.speed-wiz.com/calculations/engine/engine-flow.htm

I made the calculations for the compression ratio, and displacement, and got very close, may be exact as the software output, and i am stuck with the air flow and fuel flow calculations using the input data which are there, so anyone can help with this?, and which formulas used?

Will appreciate your reply, and again not student work, as i know and respect the forum legacy,
Thanks

 

[Lou Scannon]

I see a problem already on this screen, where the intake air density is higher than the outside air density, even through the intake air temperature is higher than the outside air temperature.
Even the outside air density value seems high to me. At sea level it should be around 0.0742 lb/cu.ft., lower still obviously at 1000 ft altitude.
For the fuel flow calculation, are you already correcting for humidity?
I would suggest making the excess air ratio (lambda) or its inverse, equivalence ratio (phi) an input, as mixture enrichment is commonly used at higher loads in liquid fueled otto engines.

"Schiefgehen will, was schiefgehen kann" - das Murphygesetz

 

[tbuelna]

That calculator is very crude, and small changes in the inputs can have a large effect on the results. For example, the input for VE can have a huge effect on fuel mass flow requirements.

 

[gaber26111]

Thanks hemi and tbeulna for the reply,

well, hemi, at the screen i sent there is no air density at all, not in the inputs neither at the outputs, here is the air density calculations, take a look, as it gives 0.0742 Ib/cu.ft, or very close to be 0.076 Ib/cu.ft FOR "outside air density" at sea level

http://www.speed-wiz.com/calculations/engine-simulation/engine-simulation-conditions.htm

and about the fuel flow, here is a factor for correcting, which is the fuel flow factor, take a look:

http://www.speed-wiz.com/calculations/general/fuel.htm

i tried this stoichemtric air/fuel ratio, which is given multiplied by the fuel factor to give actual air/fuel ratio?, if that is correct, then got fuel flow different than the software output, so my question is how to get the fuel flow using those two screens in same value of the software?

and tbuelna, that software is very accurate in my point of view, very close to actual values, if you calculated in a right and accurate way, you should get same results of the software, and that means it's accurate

any help with the fuel flow calculations?

Note: it's volumetric fuel flow, as the outputs in gallons, and liters.

Thanks again

 

[tbuelna]

What input value did you use for "efficiency"? The VE value can vary widely (from say 70% to 110%) for any particular engine at any given operating point. So unless you have an accurate value for VE over the entire operating range of an engine, how can you claim the results of this analysis tool will be accurate?

 

[gaber26111]

tueblna, it giving more than one value of the VE "volumetric efficeiny", look at this:

http://www.speed-wiz.com/calculations/engine/engine-efficiency.htm

as at that screen are the inputs i used for calculating the VE, and here is how i calcualted based on air flow and displacement:

1- VE based on air flow and displacement
VE = (3456 * air flow)/ (displacement* engine speed
= (3456 * 500)/ (351.859 * 5000) = 0.9822*100= 98.22 %

which is the same as the software output.

by the way, if you look carefully at the website, you would see that every screen, have the inputs, which is from you calculate the outputs, which is given in the bottom of the screen

what i want, is how to calculate the fuel flow per minute, using the givens in the screen i sent its link and here it is again:

http://www.speed-wiz.com/calculations/engine/engine-flow.htm

so do you how we calculate that?, to get same results by the software, or very close to it?

Thanks

 

[BrianPetersen]

If you know the displacement
and you know the revs
and you know that it is a 4 stroke naturally aspirated engine
and you know the volumetric efficiency (but do you, really? I'm not so sure, but I digress)
and you know the local air temperature and presssure a.k.a. density
and you know the air/fuel ratio

... what is so hard about calculating the mass flow rates of both??

I do these calculations in SI units because they make more logical sense to me. You are on your own with the units conversions.

 

[BrianPetersen]

Also, the whole point of an engine simulation is to estimate the volumetric efficiency based on cam timing, intake and exhaust configuration, etc., so I'm not entirely sure what the original poster is missing.

 

[gaber26111]

You don't understand Brain,

it's like a problem to solve, each screen is separate that the other, in each the top values is given, and you calculate to get the outputs.

the original post is about how to calculate the fuel flow per minute, using the inputs of this screen, take a look:

http://www.speed-wiz.com/calculations/engine/engine-flow.htm

which are:

Displacement= 351.859 cubic.inches, Efficiency = 95%, Engine speed= 5000 RPM, No. of cylinders = 8

and required:
Fuel flow per minute in "gallons, Liters and pounds"

if you can get it in SI units, there is no problem, i will convert to English units

hope you understand me now,
Thanks

 

[BrianPetersen]

So what is the problem? Is the calculation sheet on that website wrong?

 

[gaber26111]

No, not wrong, i just don't know how to get the fuel flow, by using those inputs, can you help with calculating it?, to give same results as the software results, or very close to it?

it's a work i am doing, to get the formulas used by that software, while keeping same results, to use in ECU chip programming, so just that fuel flow i can't get, that's what i need the help for

to get the fuel flow per minute, you should use the software inputs, which I've mentioned previously and air mass flow per minute"which already calculated", that's what i am sure about, but which formula and how to get the 0.51 gallons, i don't know

any help?

 

[BrianPetersen]

A simple division of the air mass flow rate (35.3 lb/min) by the fuel mass flow rate (3.06 pounds per minute) gives the air/fuel ratio being used in whatever calculation underlies this spreadsheet, 11.5:1

Where the 11.5:1 comes from, I have no idea, but it's a plausible number, if perhaps rather on the rich side if the fuel involved is gasoline.

 

[Komodo86]

OP this one does fuel flow rate amongst other things:

http://blackartdynamics.com/Engine/EngineThermodynamics.php

I don't know if it is any more useful to you really as the inputs may be limited for what you want, but might be worth a look.

 

[Lou Scannon]

well, hemi, at the screen i sent there is no air density at all, not in the inputs neither at the outputs, here is the air density calculations, take a look, as it gives 0.0742 Ib/cu.ft, or very close to be 0.076 Ib/cu.ft FOR "outside air density" at sea level

http://www.speed-wiz.com/calculations/engine-simul...

That's exactly what I'm trying to point out. For the conditions given, the outside air density should be pretty much spot on 0.742 lb/cu.ft., and, be that as it may, the intake air density, with its higher temperature, needs to be lower than the outside air density, at any rate.

Not sure what you're trying to achieve with these calculators. Some of your terminology is unconventional (at least it seems to me), e.g. "fuel flow factor", "specific energy".
I'm sure we could be more helpful if you would revert to conventional terms, or at least define your terms that are non-conventional.

"Schiefgehen will, was schiefgehen kann" - das Murphygesetz

 

[gaber26111]

ok hemi, here is what i need again

first take a look at that link,

http://www.speed-wiz.com/calculations/engine/engine-flow.htm

Do you see the line of "Fuel flow per minute"?, giving values of:

" 0.51 gallons, 1.93 Liters, 3.06 pounds"?

what i need help with is the how the software got that value of 0.51 gallons?, and other values " same line", it's conversion issue, which i know that already, but it's the "0.51 gallons", which i can't get

Do you see the givens at the top of the screen?, which are :

Displacement, Efficiency, Boost, etc?, you only use it and the "air mass per minute", which already calculated previously by the software to get other results such as the fuel flow per munite, fuel flow per hour, etc

i used that formula for calculating the air mass, and it's straight forward substitution gives very close value of the software result, here is it:

Vaf = (displacement/1728)*(Engine speed (RPM)/2)*VE

where:" Vaf: is the volumetric air flow, and VE: is the volumetric efficiency, which is given already as 95 %

my question is: " how the fuel flow calculated to give 0.51 gallons?", or very close to it?

as it's a task i am working in, using that software to check my calculations results, so i get the equations used, make conversions, etc, and that's for the whole vehicle, including engine, brakes, tires, suspension, engine simulation, etc

any help?

 

[BrianPetersen]

The software calculated the air mass flow rate. By what means, we do not know. (Hint, the tricky bit is the volumetric efficiency - but it's not hard to establish some reasonable bounds on it, only hard to get it accurately)

The software divided the mass air flow rate by the target air/fuel ratio. Where that air/fuel ratio came from, I do not know. But if the fuel in question is gasoline, then the stoichiometric ratio is around 14.6:1 and the max-power ratio is around 12-something:1, so the apparent 11.5:1 is a little rich of the norm, but not completely out of plausibility, particularly if it is a forced-induction engine. By the way, if the fuel in question happens to be ethanol, the 11.5:1 is quite close to stoichiometric.

Dividing the air mass flow rate by the target air/fuel ratio gives the fuel mass flow rate.

Maybe I'm missing something but I don't understand why this concept is so hard.

Every real engine operates not entirely unlike this. The air flow rate "governs". The amount of air going in, sets the bounds on the amount of fuel. The ECU is calibrated with the characteristics of the injectors etc and fires the injectors accordingly. But the amount of air going in "governs" and the amount of fuel "follows".

 

[BrianPetersen]

To clarify even further ... The air mass flow rate (35.3 lb/min) was calculated first.

The fuel mass flow rate was calculated by dividing the air mass flow rate by the target air/fuel ratio.

 

[gaber26111]

Ok Brian, thanks for your reply,

and sorry to forget to mention where the air/fuel ratio came from, here is where we calculate it, take a look,

http://www.speed-wiz.com/calculations/general/fuel.htm

can you get the 0.51 gallons for me?, using that screen and the one i sent before?

i know that the A/F= (Ma/Mf), we use it to get the Mf, and multiply by fuel density, to get the volume of fuel

where: Ma: is the mass of air, and Mf: is the mass of fuel

and just those fuel factor, specific energy factor, and specific energy, confuse me in getting the right result as the software

it's not straight forward calculation, so any help?

 

[Lou Scannon]

The software calculated the air mass flow rate. By what means, we do not know. (Hint, the tricky bit is the volumetric efficiency - but it's not hard to establish some reasonable bounds on it, only hard to get it accurately)

The software divided the mass air flow rate by the target air/fuel ratio. Where that air/fuel ratio came from, I do not know. But if the fuel in question is gasoline, then the stoichiometric ratio is around 14.6:1 and the max-power ratio is around 12-something:1, so the apparent 11.5:1 is a little rich of the norm, but not completely out of plausibility, particularly if it is a forced-induction engine. By the way, if the fuel in question happens to be ethanol, the 11.5:1 is quite close to stoichiometric.

Dividing the air mass flow rate by the target air/fuel ratio gives the fuel mass flow rate.

Maybe I'm missing something but I don't understand why this concept is so hard.

Every real engine operates not entirely unlike this. The air flow rate "governs". The amount of air going in, sets the bounds on the amount of fuel. The ECU is calibrated with the characteristics of the injectors etc and fires the injectors accordingly. But the amount of air going in "governs" and the amount of fuel "follows".


To clarify even further ... The air mass flow rate (35.3 lb/min) was calculated first.

The fuel mass flow rate was calculated by dividing the air mass flow rate by the target air/fuel ratio.

I don't have anything intelligent to add to that.

"Schiefgehen will, was schiefgehen kann" - das Murphygesetz

 

[BrianPetersen]

http://www.speed-wiz.com/calculations/general/fuel.htm

is incorrect.

The screen shot (I am not able to edit anything) suggests E10 gasoline on the example sheet.

Stoich for gasoline is around 14.6 - 14.7:1, stoich for ethanol is around 11:1, stoich for a blend of 90% gasoline and 10% ethanol should be around 14.3:1 not the 13:1 shown on that screen. It's almost as if the ethanol is not being accounted for at all.

Your term "fuel flow factor" - is this intended to be the inverse of "lambda" which is the technically correct term?

Be that as it may, it is not rocket science to get from 3.06 pounds to 1.38 kg and then divide by the density (approx 0.75 kg/litre) to get 1.85 litres (close enough) to 0.5 US gallons.

I really do not understand why simple basic math and units conversions are being so difficult.