correcton
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.
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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.
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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
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