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Discussion Starter · #1 ·
Could we have some knoledge on this please

I am told the ideal fuel ratio is 14.7:1 & that at this level LMBDA should be 1.0

On my G-force graph I have 9.6 LMBDA until 3700 RPM after which itb drops to 8.8 at 4000 RPM

When the graph hits 8.8 my BHP apears to drop off or straight line to 6000 RPM

Can any one explain this
 

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Fatman said:
Could we have some knoledge on this please

I am told the ideal fuel ratio is 14.7:1 & that at this level LMBDA should be 1.0
Who told you that?
For reading I'd suggest Internal Combustion Engine Fundamentals by John B. Heywood. A very lovely 900page textbook purely on engine design/theory.

I'm currently on my 2nd reading of it at the moment.

Andy
 

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Discussion Starter · #4 ·
Re: Re: Fuel Ratio

AndyF_RSX said:
Who told you that?
For reading I'd suggest Internal Combustion Engine Fundamentals by John B. Heywood. A very lovely 900page textbook purely on engine design/theory.

I'm currently on my 2nd reading of it at the moment.

Andy
To be fair andy, I was looking at the shortcut route ie getting some insight in the next few days.

Would you care to get me started with some surface knoledge?

Would you like a mature apprentice?

Can you comment on my graph
 

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As a rough guide, a lambda reading >1 = an oxygen rich (or fuel lean) mixture, a reading of <1 = a fuel rich mixture.

It is generally regarded that a mixture of around 0.9 lambda is optimum for producing the highest combustion efficiency (and hence, highest engine torque output). A mixture running at around 1.1 lambda is the best for giving good fuel consumption.

*Most* turbo'd cars tend to run slightly richer than 0.9 lambda to give a bit better protection from detonation. This is because using excess fuel helps by 1) cooling the intake charge and 2) slowing down the flame speed in the burning mixture.

Will the do or do you want me to go back a step and explain where lambda comes from and how it relates to the combustion chemical equations?

Andy
 

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Discussion Starter · #6 ·
AndyF_RSX said:
As a rough guide, a lambda reading >1 = an oxygen rich (or fuel lean) mixture, a reading of <1 = a fuel rich mixture.

It is generally regarded that a mixture of around 0.9 lambda is optimum for producing the highest combustion efficiency (and hence, highest engine torque output). A mixture running at around 1.1 lambda is the best for giving good fuel consumption.

*Most* turbo'd cars tend to run slightly richer than 0.9 lambda to give a bit better protection from detonation. This is because using excess fuel helps by 1) cooling the intake charge and 2) slowing down the flame speed in the burning mixture.

Will the do or do you want me to go back a step and explain where lambda comes from and how it relates to the combustion chemical equations?

Andy
That would be usefull,any knoledge is usefull

Looking at my graph would you say I was barking up the wrog tree then

I have seen a similar curve that was more rounded using a similar turbo that continued to climb beyond 4000RPM, mine reaches a similar top BHP but straight lines between the 4000, 6000 RPM points

Thanks
 

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Lambda is a measurement of the patial pressure of oxygen in the car's exhaust gases compared to the partial pressure of oxygen in the atmosphere.

It gives an indication as to how complete the combustion is in the engine.

For complete combustion, all of the fuel (carbon and hydrogen) reacts with a certain level of oxygen (based on chemical equations) and will produce only carbon dioxide and water.

The chemical formula for petrol is generally regarded as being a 7 chain carbon molecule with a single unsaturated carbon-carbon bond. That is chemistry speak to say that each molecule of petrol contains 7 carbon atoms and 14 hydrogen atoms attached to them.

Each of these reacts with a known quantity of oxygen to produce carbon dioxide and water during the combustion process.

If the incoming air contains the correct amount of oxygen (no more or no less) then the mixture is said to be stoichiometrically correct.

The chemical equation for combustion of the 2 petrol molecules mentioned would then be written like follows:

C7H14 + O2 ------> CO2 + H2O

petrol + oxygen -> carbon dioxide + water

Now, for a chemical equation to work correctly you need to have the same number of atoms on each side of the equation.

Looking at the equation above, you can see that on the left hand side of the equation we have 7 carbon atoms and on the right hand side we only have 1 in the carbon dioxide.
So to balance the equation it means we have to produce 7 molecules of CO2.
Doing this with all of the other atoms as well means that the overall chemical equation is:

C7H14 + 10.5 O2 ---------> 7 CO2 + 7 H2O

Now assuming that air contains approximately 21% oxygen the 10.5 molecules of oxygen translates into 50 molecules of air.

So we have 1 molecule of C7H14 reacts wirh 50 molecules of air for complete combustion.

Converting these numbers to a mass of material requires the molecular weight. Now, 1 mole (which is a specific number of atoms) of carbon weighs 12 grammes, oxygen is 16 grammes, hydrogen is 1 gram and 1 mole of air weighs approximately 28.84 grammes

So 1 mole of C7H14 weighs in at 98 grammes
and 50 moles of air weighs 1442 grammes

Now we are getting somewhere.......

So we now know that for every 1442 grammes of air we put into our engine, we require 98 grammes of petrol to react with it for it to burn completely.

So if we divide the mass of air by the mass of petrol used we get......

1442 / 98 = 14.71 which is the stoichiometric air/fuel ratio.
 

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If we now look at uncomplete combustion......

When you run the engine fuel lean, it means that all of the fuel gets used up and you have excess air that is left unreacted.
This means that you will end up with water, carbon dioxide and air as your products of combustion.

As a result of this, your fuel mixture is no longer stoichiometric because you have more air than you need and so the air/fuel ratio goes up.

e.g. if you were reacting 1 mole of C7H14 with 60 moles of air then you would have an air/fuel ratio of.....

(60*28.84) / (1*98) = 17.66

In the same way, if you run fuel rich, you don't burn all of the petrol and so you are left with unburnt material in the exhaust and you consume all of the available air

e.g. if you were reacting 1 mole of C7H14 with only 40 moles of air then you would have an air/fuel ratio of....

(40*28.84) / (1*98) = 11.77


The actual lambda value on your graph is calculated as to how far away from stoichiometric you are so in the 2 examples above, it is calculated as

17.66 / 14.7 = 11.20

11.77 / 14.7 = 0.80


The way that a lambda probe works is that it compares the pressure of oxygen between the exhaust gases and the outside atmosphere.

If the exhaust gases are rich in fuel then there is no pressure of oxygen on one side of the probe compared to the partial pressure of oxygen from the atmosphere on the other side of the probe and the probe gives a voltage output based on how much the difference.

In the same way, if the exhaust gases are rich in oxygen then there is some oxygen pressure in the exhaust to compare to the pressure in the atmosphere.

Hope this helps a bit

Andy
 

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Discussion Starter · #10 ·
Thanks andy

Chemistary wasnt my strong point but that is a great help

I am sure I will have to read this several times before it soaks in

If the LMBDA figure is 0.8 then some of the fuel is being throw down the exhaust unburt then.

If the perfect mix of 1.0 was obtained, as we travel away from 1.0 how does that affect power generated.

Sorry its been a hard day and I am being lazy.
 

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Discussion Starter · #11 ·
Is there some reason that alot of power graphs straight line between 4000 & 6000 RPM
 

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Fatman,
A lambda figure of 0.8 does indeed mean that there is excess fuel in the cylinders however as I mentioned above, it has been proven that a figure of around 0.9 is the best for power output because you also need to factor in the length of time for the complete combustion to take place and this then gets very technical because you start to look at burning rates and flame propogation rates across the combustion chamber and a whole host of other things.

Generally on turbo'd cars, most will run around 0.8-0.85 because as I said in my first post, it helps to reduce the risk of detonation because the excess fuel is vapourised in the cylinder and this removes heat from the intake air charge and the excess fuel also actually slows down the flame speed during combustion because if you get the flame speed too high then you get uncontrolled burning which is one of the causes of detonation.

Andy

P.S. A lot of power graphs straight line between 4000-6000 rpm because the boost level is fairly constant and so the torque is relatively constant and the power graph is only rising because the revs are rising. During the spool up phase of the turbo, the revs are rising and the torque is rising as the boost builds and so you will see a rapid climb in the power curve.
 

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Discussion Starter · #13 ·
AndyF_RSX said:
Fatman,

P.S. A lot of power graphs straight line between 4000-6000 rpm because the boost level is fairly constant and so the torque is relatively constant and the power graph is only rising because the revs are rising. During the spool up phase of the turbo, the revs are rising and the torque is rising as the boost builds and so you will see a rapid climb in the power curve.
The more I learn, the less I realise I know

1. So does this mean a graph that is curved all the way to maximum revs may still be spooling up or be a oversized turbo exhibiting Lag

2. That at 4000 RPM the curve goes straigh line so the turbo has reached maximum RPM

Thanks again Andy
 

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1. Yes possibly, or it may also be just trying to reach a really high boost level before stabilising out.

2. No, at whenever the line straightens it would mean that you've reached your set boost leve.

Have a look through MLR modified and you will see that every car will have a steep curve up to the point where they reach their set boost level and then the power curve starts to level off. As most peoples boost level drops off as the revs build, the torque starts to fall away and so the power comes to a peak.

Andy
 

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One thing that always makes me sad is to see the drop off in torque as the revs rise (I think my standard car hits 280ftlb @ ~3500rpm and has dropped near to 230ftlb or so @6500 to get the power figure of 280bhp). Is it feasable to tune the boost characteristics to try and keep the torque dead level across most of the rev range (ie increase boost as rpm increases to provide the same torque at the crank), or is the torque drop off more to do with the smaller time constraint for combustion at higher rpm, or perhaps also to do with powertrain losses increasing as the rotational velocity of the gearbox and diffs increases (assuming an RR graph rather than an engine dyno, obviously)?

ie if I keep the td05-16g6, would programming a boost controller so I got 300lbft from 3500-6500 rpm, assuming a boost increasing from ~1.1bar @ 3500 to probably ~1.5bar @ 6500 and hitting the maximum efficient boost level for the turbo.

I ask because this would give me a peak power of around 360bhp @ 6500 (assuming 300lbft) and since the torque curve is nice and flat the whole way, would also make for a very predictable and forgiving car to drive (3000rpm on the standard GSR box means there is plenty of leeway on gearchanges). It would only be gaining top end on a standard car, but it should, in theory, also be fairly easy and cost effective to get a car to that level of tune (boost controller + ECU + fuel pump + maybe bigger injectors).

Or am I just being silly, and haven't taken something important into account?
 

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Numpty,
The graphs from my RSX are on this thread here.

If you look at the tractive effort, you will see that it is pretty level from 4000-5500 rpm but then starts to fall away as the boost level drops from 1.4-1.2 bar.

Holding 1.4 bar all the way to the redline would have probably been sufficient to see my torque curve flatline to more or less 6,000 rpm and I think it may have been due to the lack of loading on the rollers that meant I got a large drop off in boost because on the road, I didn't see the same level of drop off.

Andy
 

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I'm still intrigued why the torque drops off on most cars as you get higher up the revs, especially on sports models. Is it a deliberate thing to relieve mechanical stress on the rods as the revs increase, or is it a function of something else that I'm not seeing?

I need to save up a couple of k and talk to Mr Shead about getting a flat torque curve I think... :cool:
 

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Generally it's to protect from detonation.

High torque is a direct relationship to high cylinder pressures.

High cylinder pressures at high revs is risky

Andy
 
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