Superoxygenated Water/Methanol Injection

[Richard L]

I did not calculate the power increase due to oxgen, it could be 0.00001 %, but it is still an increase. The combination of cold water may improve the yield.

Sturat was merely trying to replace the oygen displaced by water vapour. As fas as I am concerned, I am interested any any form of experiment, someone wil be injecting pure oxygen, or peroxide soon.

[gaiaresearch]

I'll happily take the extra half cup of chilled O2, for the chill and the O2. Geeeez, I could have pressurised my tanks with carbon dioxide and contributed no extra oxygen and would probably have been shot down for displacing the teaspoon of dissolved O2 in the non superoxygenated water.
I have nothing to lose and something to gain. When the data comes in, if it is positive, it will in all probability be denied, on the basis of positions already taken. Life is full of paradoxes and anomalies. If one does not stick ones's neck out, one might never realise that there is air above the surface of the water. :lol:

Peace

Stuart

[masterp2]

I think it should be tested. Nobody will shoot down charted results in a controlled experiment. You should not let others condemn something you feel so passionate about. Moan as much as you want, I am pretty sure, that there will be a performance decrease. By supercooling the solution, it will inject at a much reduced rate, calculate it yourself.. So unless you compensate for the temp induced viscosity change, by increasing the pressure, or bigger nozzles, less solution is atomized. If you measure flow rate in the controlled experiment, and they are the same, you will know the solution did not stay cold to the nozzle.

Don't believe me? That's ok. Go to the kitchen and turn the water on the hot side. Measure the volume of cold water coming out for 10 seconds, then when it is good and hot, measure for 10 more seconds in another container, and tell me if you don't get 20% more.

[Richard L]

Originally Posted by hotrod

Leaving aside all the other issues lets look at the real world numbers.
people sometimes forget just how small milligrams/liter is.

Assuming you are achieving the 80 milligram / liter of disolved oxygen mentioned in that paper (which is the upper limit claimed).

You have a 400 hp engine, and you run the 1/4 mile in 13 seconds flat. You are injecting pure superoxygenated water at a rate of 500 cc/min.

Your engine will process a maximum of 42 lbs/min of air as it moves down the 1/4 mile. In the 13 seconds of the run that comes out to 42 x .02316 = 9.7272 lb O2, or 4416.1488 grams of O2.

Meanwhile your WI is injecting 500 cc of water / min or 108.33 cc of water. This equals 0.10833 Liters, at 80 mg/liter you have added 8.666 mg of additional oxygen to the 4416.1488 that is in the air.

You have increased the engines oxygen intake by 0.0001962 %.

Since one cubic foot of air holds about 0.016212 lb of O2, or 0.000009382 lb/ cubic inch or .0042594 gm of O2 per cubic inch, your additional .008666 gram of dissolved O2 is approximately equal to the oxygen contained in one half of a cubic inch of air.

And that is a best case scenario.


I think I'll just stick with plain cold water and not worry about dissolved oxygen content.

Larry

I read your calculation with interest, couldn't quite get the correct % of oxygen increase by mass.

4400g of oxygen injected against 80mg/litre/min (40mg/500ml/min). The % of oxygen is 40/4400 = 0.9%, for a 400BHP engine, 1% extra oxygen is 4 BHP - is small but an significant increase?

[hotrod]

Richard, your right I did make a minor mistake in the above calculation, but I think your off by a factor of 1000 on yours. Your dividing milligrams by grams.

Larry

Correction -- It looks like I did drop a step in my first set of calculaions, here is the corrected numbers

Assuming you are achieving the 80 milligram / liter of dissolved oxygen mentioned in that paper (which is the upper limit claimed).

You have a 400 hp engine, and you run the 1/4 mile in 13 seconds flat. You are injecting pure superoxygenated water at a rate of 500 cc/min.

===============
Computed based on air consumed in 13 second quarter mile pass:

Your engine will process a maximum of 42 lbs/min of air as it moves down the 1/4 mile.
In the 13 seconds of the run that comes out to 13/60 = 0.21667 x 42 = 9.1 lb of air

9.1 x .2316 = 2.10756 lb O2, or 956.8 grams of O2. <---- air is 23.16% O2 by mass

Meanwhile your WI is injecting 500 cc of water / min
13/60 x 500 = 108.33 cc of water.
This equals 0.10833 Liters, at 80 mg/liter you have:

0.080 x 0.10833 = 0.008664 gm of additional oxygen to the 956.8 that is in the air.
{ corrected the above had one too many zeros on the milligrams expressed as grams the original was 0.0080 x 0.10833
Boy its hard to keep track of all the zeros when your dealing with quantities that are so small}

You have increased the engines oxygen uptake by .000009058

If hp is directly proportional to O2, then you've increased the power out by
400 x .000009057 = 0.0036339 hp

If the engines power peak of 400 hp is at 6000 rpm, you have increased the engines torque at that rpm by .00318 ft lb or 0.05088 ft ounces.
===============
computed in oxygen uptake / min

42 lbs / min air = 19068 gm/min air
Oxygen content is .2316 by mass.
19068 x .2316 = 4416.1488 gm/min


500 cc/min at 80 mg/liter = 40 mg/min = .004 gm/min

.004 / 4416.1488 = .000009058

[masterp2]

I can't get 1%. This whole thread is bizarre.

I think larry is off by a decimal place, I think because he forgot to factor that air charge is only 21% O2.

I get .001% O2 increase, so I'd say .004 HP, unmeasurable gain.

[masterp2]

Ooops, looks like he beat me to it.

[gaiaresearch]

And for every 1 degree Centigrade drop, you gain 3 hp

[gaiaresearch]

And with every calculation error I lose faith in your numbers.

Stuart

[hotrod]

And for every 1 degree Centigrade drop, you gain 3 hp

Not correct! Power varies approximately by the square root of the change in absolute temperature of the intake air (assuming the ignition timing and such are optimized for the temperature change).

If you measure hp at 20 deg C, ( 293 deg K ) and at 25 deg C ( 298 deg K )
then the power difference will be approximately (sqrt (298/293)) = sqrt (1.01706) = 1.008497.

That makes the power increase approximately equal to .001699 / deg or .1699%.

On a 400 hp engine for each 1 deg C drop in intake temp you would gain about 0.67 hp. The cooling effect of your setup is 1870 times more significant than the alleged increase in oxygen.

Larry