how much air, how much fuel?

with all this talk of EPA and clean diesel, I was wondering if thier was a way to calculate the amount of air that it takes to to burn "x" amount of fuel.

I am really in the dark on this. I know that the flow rate of a turbo is measuered on cubic feet per minute (CFM) and I am not sure how the flow of the injectors is measured. So if you have a turbo that flows 320 CFM how much fuel could you effiently burn? or if you had a set of injectors that flowed "x"amount how much air would you need to effectively atommize the fuel?

I figure someone on here will know what i am talking about. I think what iam asking is clear but if not I will try to explain it better. any insight to this would be great, because I think it would help people make good decisions on building a DD that is efficeint. I know it amy sound good in thoery but is it really going to work if it is put in to action?

Add more air till it cleans up??

I dont know if there is a formula, but I would just come up with some combos that you are interested in running, and just post em up on here to get more info. Im sure there is somebody that either knows or is running that combo on here. Just a thought to make it easier on ya bud.

I aint changing what I got cause it actually does clean up the smoke really good, I was just wondering if thier was a formula for it. Its a lot of thinking for a easy to answer problem by asking some body that knows, but thier has to be a formula I would think.

well if you can find out the air fuel ratio of diesel it would be easy the air fuel ratio of gas is 12.5/1 and alky is 5/1

I dont know bout diesel's but a gas engine is tuned in GOOD at 12 to 1 air to fuel ratio. but that may not be what your talkn/asking about??? As you can tell im in the dark on this to!!! LOL

I found this discusion , it was wild so I paste it .

The original question mentions 14.7 so it's pretty sure the intention was to find out the stoichiometric air fuel ratio.

CxHy + a(O2+0.79/0.21N2) => xCO2 + y/2H2O + a*0.79/0.21N2
where a=x+y/4 since a C can use up a whole O2 and an H can only use 1/4 of an O2. The 0.79 and 0.21 are the proportions of N2 and O2 in air.

If you take the formula for a hydrocarbon reacting with air you can easily work out how much of each is required.

molecular weight of O2 is 32 and N2 is 28.
Octane C8H18 has a=12.5 and molecular weight 114 ( 8 12's are 96 plus another 18).
Dodecane C12H26 has a=18.5 and molecular weight 170.

So if you crunch the numbers to get the AFR
AFR = a(MWO2+0.79/0.21*MWN2)/MWfuel
you get 15.06 for petrol and 14.95 diesel.

Since hydrocarbons are mainly H-C-H, two to one, and the odd hydrogens at the ends making the difference are light, all the hydrocarbons will have similar stoichiometric air fuel ratios, with the exception of the very light ones where the two extra hydrogen makes up a significant proportion of the weight. Take CH4, the H-C-H weighs 14 and the two H's weigh 2. That's a large portion compared to 2 in 114 or 2 in 170.

So AFR for methane calculates to 17.2 and Ethane to 16.1.
Then propane 15.6 and Butane 15.4. You can see the figures closing in on the figure 15 for petrol and diesel.

Now if you add oxygen things change drastically since oxygen in the fuel means you need less oxygen from the air. And since air is 4/5 nitrogen, if you can cut down on a certain amount of oxygen then you reduce the amount of air you need by a factor of about 5, (drop 4 N2s for each O2 dropped). So the AFR for a fuel like ethanol works out to be about 9.

Now if you say that US gasoline is like petrol but cut with about 5% ethanol you can calculate 5%*9 + 95%*15 = 0.45 + 14.25 = 14.7 and get anAFR for gasoline.

So providing the diesel is all hydrocarbon based the AFR will be around 15. Diesel of plant origin (biodiesel) has oxygen in it and the AFR will be lower and mixtures would be in between, in the same way that the AFR for gasoline is between that for petrol and ethanol, heavily weighted towards the petrol figure. I don't think the figures change much if you take air as being 78% Nitrogen and 1% Argon instead of 79% Nitrogen and as I said before, changing one heavy hydrocarbon for another won't make much difference, ie calculating diesel as a coctail of hydrocarbons would only complicate the calculations and not significantly change the result.

How bout a little english, Greg? LOL

Thanks greg but talk about more than I want less than I need.

lets break this down.

turbo flow is measured in cubic feet per minute

Injector flow is measured How?

How bout a round of applause for com461!!!! That was impresive, didnt follow a thing he said....but impresive!!! :clap: :bow:

That was some excellent info! Now my Brain hurts!!!!

I don't think you will find a clear cut formula, though it would be nice. To many variables when talking injector flow and turbo volume. Here is what works for me.

1. Add fuel until EGT's are to high.
2. Add air until EGT's drop to acceptable area.
3. Repeat steps 1 and 2 until broke

Paul

I'm only an automotive enthusiast, so this statement should be taken as an opinion -

Calculations can be performed all day long, but without proper data acquisition it's not going to get you where you want. This can also get expensive to research, and if you are just some dude that wants to clean up his ride, then tuning/trial and error are going to be the way to go... see paulb's post above.

http://www.turbobygarrett.com/turbobygarrett/tech_center/diesel_tech.html

This will give a little insight, it wont answer your question exactly unless you have specific numbers for your application.

ok that gives me an idea of how that works. Thanks for the input guys!

I think the smoke free AFR in a diesel is around 20:1.

AND I think a common multiplier to get the airflow (in CFM) is multiply 1.75 times the HP goal.

I think that number presumes 100 percent Volumetric Efficiency.

I have never measured it or got as into the calculations as Comp's post above, but I read it in a book or something somewhere once, and I did stay at a Holiday Inn Express once.

The smoke limit is about 18:1 air by mass to diesel fuel by mass. Twoforme2 stated 20:1 which would be better for emmissions.

Now your air volume in CFM needs to be converted to lb/sec (or kg/sec) likewise the deisel injection rate needs to be converted to the same units.

To convert air CFM to mass flow, you need to know the air temp, pressure and density.

In my experience the injection pump output is usually given in cubic centimeters per 1000 strokes (2 revs of the engine to 1 stroke of I.P.).

Sorry I've had too much to drink to delve further into converting volumetric flow of air and diesel to mass flow.:Cheer: