What's the formula for determining nitrous volume used per sec?
#1
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I'm sure that there are several guys on here that could provide a formula for nitrous consumption per second by using the orifice size vs bottle pressure assuming that the feed lines and noids can flow the volume. Could one of you mathematical types please provide it? Thanks in advance.
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its in the sticky
http://home.ican.net/~jsetter/nitrous.html
I always figure about 4 lbs of N2O per minute for every 100 hp as a starting point.
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I guess that I havent made clear what I am looking for here. What I would like is a formula to which I can plug in the size of the jet (orifice size) and the PRESSURE at which the bottle is running in order to determine the volume of nitrous being used AT THAT PRESSURE. These ballpark guestimates assume a given bottle pressure. I would like to know how (numerical value) that volume increases and decreases based upon PRESSURE.
Thanks.
Thanks.
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#8
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This is really difficult to do, as the dropping n2o pressure will change everything. This psi drop is also hard to calculate as it also varies per size shot. Also, the volumetric efficiency of the motor as well as the n2o system will change the numbers too. Now with a push system we may be able to get a good, or close, formula. I do like this type of thing, though don't have the time right now to think up the formula. The general formulas above will get ya in the ball park though.
Robert
Robert
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This is really difficult to do, as the dropping n2o pressure will change everything. This psi drop is also hard to calculate as it also varies per size shot. Also, the volumetric efficiency of the motor as well as the n2o system will change the numbers too. Now with a push system we may be able to get a good, or close, formula. I do like this type of thing, though don't have the time right now to think up the formula. The general formulas above will get ya in the ball park though.
Robert
Robert
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i *think* the 4lbs per minute at 100 shot is at 1050psi, so with those variables im sure any decent graduate or even undergrad math student could come up with it. ill ask a few friends
#11
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sec / 15 = lbs used in a ¼mile run using a 100 shot and 1050psi
So a 11.5 run would be:
11.5 / 15 = .766lbs of n2o @ a 100shot at 1050psi
The 15 is the common denominator for 100rwhp shot, or real close, at least according to the info supplied by Hennytime.
I think we would need the common denominator for each size shot, as I am not sure if it would be the same, don't have the time to really think this out, and only spent about 3 minutes on the above.
There are more variables per company, as different psi is used and the ratio of spray to fuel to get the same hp gains, so...
Robert
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Here it is:
sec / 15 = lbs used in a ¼mile run using a 100 shot and 1050psi
So a 11.5 run would be:
11.5 / 15 = .766lbs of n2o @ a 100shot at 1050psi
The 15 is the common denominator for 100rwhp shot, or real close, at least according to the info supplied by Hennytime.
I think we would need the common denominator for each size shot, as I am not sure if it would be the same, don't have the time to really think this out, and only spent about 3 minutes on the above.
There are more variables per company, as different psi is used and the ratio of spray to fuel to get the same hp gains, so...
Robert
sec / 15 = lbs used in a ¼mile run using a 100 shot and 1050psi
So a 11.5 run would be:
11.5 / 15 = .766lbs of n2o @ a 100shot at 1050psi
The 15 is the common denominator for 100rwhp shot, or real close, at least according to the info supplied by Hennytime.
I think we would need the common denominator for each size shot, as I am not sure if it would be the same, don't have the time to really think this out, and only spent about 3 minutes on the above.
There are more variables per company, as different psi is used and the ratio of spray to fuel to get the same hp gains, so...
Robert
How much nitrous will flow per second thru a .010" orifice at 1000 psi (assuming that the pressure is constant)? It doesnt seem like it should be too complex an equation. It's a flowrate calculation that I am after.
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You're chasing horsepower numbers that arent important in what I am asking. Here's what I need a formula for with very simplified numbers.
How much nitrous will flow per second thru a .010" orifice at 1000 psi (assuming that the pressure is constant)? It doesnt seem like it should be too complex an equation. It's a flowrate calculation that I am after.
How much nitrous will flow per second thru a .010" orifice at 1000 psi (assuming that the pressure is constant)? It doesnt seem like it should be too complex an equation. It's a flowrate calculation that I am after.
Robert
Edit: from the above,
So a 11.5 run would be:
11.5 / 15 = .766lbs of n2o @ a 100shot at 1050psi
11.5 / 15 = .766lbs of n2o @ a 100shot at 1050psi
1sec = .066lbs n2o running a .XX jet
Then we could take this down to an exact .010 and have a .010 = .0xxlbs per second. Just need the jet sizes and psi per kit/company.
Last edited by Robert56; 03-11-2008 at 08:14 PM.