WHAT AM I LOOKING FOR ON MY A/F RATIO

[THE ICE MAN]

O.K. I talked to 2 guys on the board that have good knowledge and experience on water injection and will try some of the things we discussed
I am looking to get this bike at least 205 MPH in 9/10's with the motor making a reliable 230 H.P.
Putting the motor in tommorow will tune in next week, and keep you updated.

[Richard L]

There is water injection Hayabusa currently running water inejction on the 200mph.org site.

I will try to invite him over here to help answering some of your more involved questions.

[mx5]

Originally Posted by THE ICE MAN

They say 12.5 to 1 air fuel
and 10.5 -11.1 AIR to (FLUIDS) RATIO water and fuel

Huh?!? You think 20% water to fuel, can move your AFR from 12.5 to 10.5 ..... ? this just doesn't sound right. have you tried calculating whats your Air to Water ratio?

I think you got it the other way around. 10.5-11.1:1 WITHOUT water and 12.5:1 WITH water, since with water you don't need that extra fuel for cooling the cylinder walls ...

[JohnA]

Originally Posted by mx5

Huh?!? You think 20% water to fuel, can move your AFR from 12.5 to 10.5 ..... ? this just doesn't sound right. have you tried calculating whats your Air to Water ratio?

I think you got it the other way around. 10.5-11.1:1 WITHOUT water and 12.5:1 WITH water, since with water you don't need that extra fuel for cooling the cylinder walls ...

Ah, but the second one is air to FLUIDS ratio. So fuel is effectively less than before.
Interesting take on AFR, but it assumes a linear relationship between fuel reduction and water mixtures added. I wouldn't be so sure about this one.

[mx5]

Originally Posted by JohnA

Ah, but the second one is air to FLUIDS ratio. So fuel is effectively less than before.
Interesting take on AFR, but it assumes a linear relationship between fuel reduction and water mixtures added. I wouldn't be so sure about this one.

Yes, and that's what I told to calculate. Now get the calculator and check how much replacing 1.5% of air with fluid (roughly, assuming about 20% of fuel injected water) changes your air to fluid ratio in a 12.5:1 air to fluid mixture ... Changing 1.5% of the air with water doesn't make the air to fluid ratio to 10:1, at least according to my calculator

[hotrod]

Ratios and proportions are always fertile ground for confusion, because there are so many ways to interpret them especially if they are stated as word problems rather than a math formula.

Lets put some numers on this and see what comes out, based on how I am reading this.

Assume:
1300 cc engine displacement, and 6 - 9 psi boost with an assumed VE of about .90. Just to pick an operation point lets say 8000 rpm.


That would give a swept engine volume of about 213 CFM at .90 VE or 192 CFM of air at a manifold absolute pressure of 23.7 psi ( pressure ratio of 1.61) for a true air flow of about 310 CFM or about 21.7 lbs of air/min.
(sorry guys I grew up with imperial units )

At a 12.5:1 air fuel ratio (by mass) you would need 1.736 lb/min of fuel, if the WI rate is at 22% of fuel,(the max recommedation) then you have 0.38 lb/min of injectant.

Your total "fluids to air ratio" becomes 21.7 : (1.736 + .38) or 10.255:1.

At least that is how I am inerpreting the recommendation from the RBR calculator.

Larry

[Richard L]

Water to fluid ratio is quite straight forward to calculate

But if you want specifically on the cooling effect due to evaporation, you need to take the latent heat of each liquid into account and re-calculate to that effect.

Some time ago I have made a chart up to show how much water is require to replace the rich-air/fuel ratio, here it is again - It was done quite quickly and do point out any mistakes plaese:





This one I have included 25% of merthanol:

[Richard L]

If the chart is correct, it appears that all you need is 3% water/fuel will replace two points of afr (10-12.5).

I'd better explain how the chart was calculated:

Without water:
10Kg air + 0.95kg fuel = 10.5 afr
.
.
.
10Kg air + 0.80kg fuel - 12.5 afr



With water
10kg air + 0.95kg fuel + 0.00475kg = 10.5 afr + 5% wfr
.
.
.
10kg air + 0.95kg fuel + 0.095kg = 10.5 afr + 10% wfr

Substitude the latent heat of each liquid and plot the result with excel.

Latent heat of fuel = 350KJ/kg
Latent heat of Methanol = 1109 Kg/kJ
Latent heat of water 2256KJ/kg