Fuel Ratios and what they actually mean...

[Bleh]

Several times I have read threads on this forum mentioning fuel ratios and on a few occasions, stoichiometric combustion.

But what does this mean?

Simply put, stoichiometric combustion describes the ideal fuel ratio at which the least amount of harmful by-products are produced, mainly in the form of CO2 (a bit more in depth below). With the blend of gasoline more commonly found in our little island and many other countries, this has been derived from the molecular build up of the fuel mix of gasoline and air (bare with it, example to follow).

Now, as many of you may have noticed, when you put juice in the beast, you may have seen on the pump 95 RON (or similar), this describes the octane rating of that particular gasoline blend where RON stands for 'Researched Octane Number'. All octane molecular formations I have seen (bare in mind, I've far from seen them all!!) have a significant carbon (C) and hydrogen (H) content in their make up and it's this carbon content that results in the CO2 production.

The simplest form of octane to carry on with this explanation is isooctane in the form C8H18 (little 8 and 18, and yes, isooctane derived and defined gets a lot more complicated).

Now, many of you may say 'I know how to calculate fuel ratio, it's mass of air/ mass of fuel!' and you'd be right, it is that. But it can be explained in much greater detail with the following equation (here's where the example starts);

Gasoline + x(O2+3.76N2) ---> aCO2 + bH20 + cN2

Where gasoline + x(O2+3.76N2) is the fuel mix and CO2+H2O+N2 is the by-product.

The (O2 + 3.76N2) is derived from the air we breath. Our air is made up of just over 20% oxygen, 78% nitrogen, 1% noble gases and 1% H2O and CO2 - grouping these accordingly, it is said that air is simply 21% oxygen and 79% nitrogen.
Using these values we get the following 0.21 O2/0.21 = 1xO2 (or just O2) and 0.79N2/0.21 = 3.76N2 (explaining the amount of air to nitrogen) and therefore (O2+3.76N2) where x just determines the amount of air required (hope that clears up any questions about that part you were thinking of?!).

Before we delve into solving the equation, I will firstly list some molecular weights (to the nearest whole number for simplicity);

C = 12 g/mole
O = 16 g/mole (therefore, O2 = 32g/mole)
H = 1 g/mole
N = 14 g/mole (therefore N2 = 28 g/mole)

Now we have all the tools we need to complete the job in 2 steps;

Step 1: Solve above equation.

Hopefully, you'll save me another thousand words by just accepting the following method of determination rather than me explaining each time about individual chemical reactions during combustion;

So; Isooctane + x(O2+3.76N2) ---> aCO2 + bH20 + cN2

Therefore; C8H18 + x(O2+3.76N2) ---> aCO2 + bH20 + cN2

Where a = C, b = H/2, x = (2a+b)/2 and c = x(3.76N2), however, as nitrogen appears to be of little concern as out air is mainly made up of the stuff, we'll forget about it as a by-product from this stage on!

Using the conditions above, we now have;

C8H18+ 12.5(O2+3.76N2) ----> 8CO2 + 9H2O (+ 47N2 if you're really interested!?)

Now we have the ideal blend of gasoline and air mix for stoichiometric combustion (or as close as best as possible for this gasoline).

Step 2: Fuel Ratio

Like I said earlier, someone who says fuel ratio is calculated by mass of air/ mass of fuel would be correct and the principle remains the same for this except we're looking at molecular weight to determine the true potential fuel ratio of your blend. For this, we refer back to the molecular weights I listed earlier (and we can now forget about the by-product).

So mass of air takes the form 12.5(O2+3.76N2) as it is the amount of air required for gasoline to be stoichiometric and the isooctane takes the form C8H18 to determine the weight (simply put, 8xC + 18xH), putting this into numerical format we have:

12.5((32)+3.76(28)) / (8*12)+(18*1)
This is equal to:
12.5(137.28) / 114
Which has the final form:
1716 / 114

Which gives a final fuel ratio of 15.05:1 for Isooctane!

So there we have it, fuel ratios and how to determine them explained. Hopefully, what you'll see form this is that you have very little control over fuel ratios (bar buying some octane booster crap) and that your fuel ratio will vary considerably dependent on the blend of gasoline put into your tank.

If you're interested, from the original formula listed above, you can gain so much more valuable information about your fuel and it's potential to be ideal or an absolute nightmare, for example, the air mix can be used to determine the amount of port space taken up by the fuel (meaning less air in ports), as well as NET heat energy releases and calorific values etc.

Interestingly (for me at least), I have come to the conclusion that the mass of fuel must always be 6.8% the mass of air to achieve 14.7:1 (simply derived from (1/14.7)*100=6. and that there appears to be the very thing a majority of fuels achieve, isooctane for example has a value of 6.6% which is very near but not 6.8% (plus it produces quite a bit of CO2, even at 6.6%).

Hope some of you boffins now have a better understanding of fuel ratio and what it is telling you?!

Please, by all means correct me if anything appears inaccurate but I am fairly confident all appears justifiable and correct.

NEEEEEXXXT......

[darkember]

All that said and done what would happen if one were to add a wee bit of browns gas to the mixture via a Jcell??

[Bleh]

Simple,

gasoline + wee bit of browns gas + x(O2+3.76N2) ----> aCO2 +bH2O + cN2

The principle remains the same. A certain amount of air is always required with the mix to produce stoichiometric combustion.

The only thing that really changes on road going vehicles is when it is direct injected or port fed which changes the evaluation for port space required and generally leads to better heat energy releases (in Di engines)!

[tony.mon]

The problem is that the theory assumes complete combustion- the flame path and speed, valve overlap, intake port (including stack) length and port dimensions/shape all have a part to play, as does the crossover and design of exhaust pipe (scavenging)
A low-revving, long stroke (undersquare) and zero- valve overlap engine should be pretty efficient, but doesn't have the most power.

These engines seem to run best with a fairly rich fuel air ratio, partly, I think, because the richer mixture keeps the engine running cooler.
The carbs are pretty big, and aren't very efficient- hence the poor fuel economy- a fair bit of it is shoved out of the exhaust unburnt- or at least, not burnt where it should be; in the head.

Emissions are probably quite poor- luckily bike MOT's are a lot more forgiving than car ones....

[Bleh]

The problem is that the theory assumes complete combustion- the flame path and speed, valve overlap, intake port (including stack) length and port dimensions/shape all have a part to play, as does the crossover and design of exhaust pipe (scavenging)

All irrelevant to determining the ideal fuel ratio for stoichiometric combustion but more so relevant for 'actual' fuel ratios, which when running rich, wouldn't be stoichiometric! All I am trying to do here is offer a better understanding of what fuel ratio is and means when everyone throws the terms about 14.7:1 as 'ideal'. Hopefully it's clear in the explanation?

I think, because the richer mixture keeps the engine running cooler.
The carbs are pretty big, and aren't very efficient- hence the poor fuel economy- a fair bit of it is shoved out of the exhaust unburnt- or at least, not burnt where it should be; in the head.

Emissions are probably quite poor- luckily bike MOT's are a lot more forgiving than car ones....

I fully understand what you are saying in all of the above and agree with it all 100% when considering running conditions. There are several other factor that could also be added such as valve spring rates, valve area (inc. stem) and lift (maybe obvious?!), port surface type (rough ports increase friction on mix traveling through hence polishing ports which also increases overall area (ever so slightly)!), piston shape and crown clearences etc, all have their part to play on gas velocities and therefore performance of engine.

What else is an interesting thought is what fuel is wasted that gets tucked away in the unburned zone (and down the side of the pistons crown clearances) is likely contributing to the already rich mixture and keeping the cylinder cooler before being exhausted. What would also be interesting to see is if this was the case, what potential could running the bike a little leaner have at higher RPM where there's more opportunity to make use of the unburned fuel?
Just a thought and one that may be interesting to look into in the future but I just do not have the time at the moment to build a model to even theorize if this has any potential!

Has anyone altered their compression ratio and noticed and improvement that way by any chance?
From a number of posts I've come across of yours (Tony), you seem to have a lot of mechanical experience (oodles more than me at least) with some of the things you've done, you ever adjusted CR? What extreme have you tried if you have?

[gl_s_r]

park life!

[tony.mon]

Yes, high comps are a must.
I use JE pistons, 11.6:1 I think compared to the 10.5 of the standard ones, but I also normally gas flow my own heads- I have standard heads on at present and it's down about 5Bhp from where it should be at present, so it's interesting to quantify that....

I'll get some gas flowed this winter, but I also have some 1mm oversize inlet valves to fit, so that'll be interesting.

However, when used with an ignition advancer (I generally run a four degree advance) you will need to either use an octane booster or run it on super unleaded.

Interesting stuff, I fitted Lambda sensors on each downpipe so that I could monitor what emissions I was getting on each cylinder separately, rather than having to tune the carbs as a pair.
Even "proper" dynos use a sniffer probe up the exhaust, and that's after the exhaust gases are combined and then split again- so you can't tell which one runs lean and which one runs rich, if there's a difference.

Not finished playing yet, but i currently have a big difference on main jet sizes front and rear, but then I have a very not-standard airbox, which may be having an effect.

[grumpyfrog]

And far too much fiberglass skill!!!!
When you get it done properly in CF then I'll rob it off you. saying that I'll just have to get you to fix it once I've broken it.. thats assuming I know what it is in the first place.




Too bloody clever for me..

[Stephan]

Yes, high comps are a must.
I use JE pistons, 11.6:1 I think compared to the 10.5 of the standard ones ...

Do you mean stock pistons as standard ones? I thought they are 9.4

[StuartWags]

Some excellent stuff here it is a very interesting subject. It is getting towards the research / masters / PhD end of things though. If you can get your head around it then you should stick at it because there aint many who can.

Sadly most of the brain cells i used for this sort of stuff are long gone. You may have or find interesting Internal Combustion Engine Fundamentals by John Heywood (it is not fundamentals it is very heavy duty technical stuff)

Generally, the next step is to add in the NOx, unburned CH but you have to draw the line somewhere (SOx, CO? etc, etc).

As already pointed out, the equations assume complete reaction. In reality there is and equilibrium and then there is the reaction rate or kinetics. i.e mainly time and temperature dependant.

It does not get that much more difficult than the stuff you already done but needs more crayons and paper.

What you end up with is a model that considers the temperature, pressure at fractions of a crank-angle and resolves the combustion equation for that time period and then gives you the output you need.

However, at the end of all that it is lucky if you get a trend that matches the experimental data

There are many CFD packages that can do this much better nowadays but it is not the sort of stuff that runs on your tablet

At least petrol engines have an homogeneous air fuel mixture to start with. I worked on diesel engines where there rich and lean zones in the combustion chamber due to the direct injection and these all have the own chemical kinetics

[Bleh]

Sadly most of the brain cells i used for this sort of stuff are long gone. You may have or find interesting Internal Combustion Engine Fundamentals by John Heywood (it is not fundamentals it is very heavy duty technical stuff)

The book is my bible, I'm in it very frequently. In my final year of BEng (motorcycle) study and the main man for lecturing us this sort of stuff used to be chief engineer for lotus back in the day, also has his name on 16 patents and has had a wonderful career within the trade - needless to say, we hang on his every word. The guy is nothing short of impressive, the only lecturer who doesn't compile lecture notes and pulls it all out of his head... Makes deriving look as easy as making coffee (although I do struggle with differentiation a tad!)

With the CR, a larger ratio on bigger cylinders is generally better. As to the proclaimed ratio within specs, there should always be an element of caution with what's specified and what's real. For example, if each cylinder has a volume of 499cc's (or cm^3) and a claimed ratio of 9.4:1;
Therefore, CR = Volume 1 (V1)/ Volume 2 (V2) = V1/V2 where V1 will be the larger value, (can be explained in other forms also)

So, we are saying 9.4 = 499/V2

Rearranging for V2 we have, V2 = 499/9.4

Which means the smaller volume is equal to 53.1 cc's (combustion chamber volume). Not that I've ever measured the combustion chamber of the VTR but what I have measured, I have found the combustion chamber to rarely (can't recall when it has) exceed 10% of cylinder volume.
(Interestingly, your total volume (Vt) = V1+V2 = 552.1cc's per cylinder)!

So, for tony's mention of 10.5:1, that would indicate combustion chamber size of 47.5cc's which looks a more sensible value to me (personally). (Vt = 546.5 cc's)

So, from 9.4 to 10.5, there is only a potential 5.6cc's difference (or 5.6ml if it makes it easier to envisage?!).

So why the differing possibilities? Not that I've seen the piston in one of these either (only got the one engine and haven't been had time to get the heads off for port polishing) but I would take a stab in the dark that it's not a true flat top?! If it is the minimum volume can be greater than if the piston has cut-outs to encourage swirl etc!
So, from 9.4 to 10.5, there is only a potential 5.6cc's difference (or 5.6ml if it makes it easier to envisage?!).

Best way I have found to measure it all (which many may be aware of) is to measure bore and stroke at BDC with vernier and with the head flat and chamber exposed, clamp a flat plate over the top with a little hole in and use WD-40 (or similar) in liquid form and a pippet to to fill the chamber 1ml at a time. No. of ml's = No. of cc's (as 1ml = 1cc), to the same for the piston (if not a true flat top) at TDC. Calculate volume of gasket to and add to combustion chamber and you'll acquire the more accurate CR for your machine.

Piston 'upgrades' will be manufactured to reduce V2 as much as possible without collision (i.e. piston crown with valves), top this with head skimming and head gasket upgrade and you will definitely improved CR!

[Stephan]

Bleh: VTR1000F has cylinder volume 498 cc, and calculated combustion chamber volume (including head gasket height and cylinder volume above piston at TDC) is 59.26 cc. This is 100%, I did measurement yesterday, and got 59.9 cc (with +/- 0.5 cc accuracy). This is 35cc in head, 21.5cc cylinder, 3.4 head gasket (height 0.45 mm). So you can trust 9.4 compression as well as calculated volumes.

[Bleh]

Nice to know some 'actual' measurements.
Them measurement suggest (V1/V2 = 498/59.26) CR is 8.4 though, somethings amiss (or I ain't seen it!).

[Stephan]

CR from measurement is 9.3:1 as CR = (Combustion chamber vol + cylinder vol) / combustion chamber vol, in values (59.9+498)/59.9

for me it means that I can trust measured volume of head combustion chamber, as difference is minimal. I did this exercise for custom pistons layout ...

[Bleh]

My bad... I've been forgetting all about adding combustion chamber et al. I'll consider myself