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What's the formula for determining nitrous volume used per sec?

Old Mar 6, 2008 | 02:31 PM
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Default What's the formula for determining nitrous volume used per sec?

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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Old Mar 6, 2008 | 04:23 PM
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tt...............
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Old Mar 7, 2008 | 05:45 PM
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Anyone? I thought for sure that someone would know this right of the top of their head.
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Old Mar 7, 2008 | 05:47 PM
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1 pound per 100 horse power every 10 seconds
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Old Mar 8, 2008 | 03:23 AM
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Originally Posted by Nitro Dave's Nitrous Outlet
1 pound per 100 horse power every 10 seconds
at what pressure? Kinda looking for a formula to calculate it more exact w/ orifice size vs pressure.
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Old Mar 8, 2008 | 02:14 PM
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its in the sticky
I always figure about 4 lbs of N2O per minute for every 100 hp as a starting point.
http://home.ican.net/~jsetter/nitrous.html
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Old Mar 8, 2008 | 04:16 PM
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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.
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Old Mar 9, 2008 | 01:10 PM
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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
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Old Mar 9, 2008 | 09:55 PM
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Originally Posted by Robert56
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
Perhaps I should explain what I am trying to accomplish with this formula. It doesnt need to be exact to the point of figuring out the pressure loss. The reason that I would like to know this formula is because people rarely tell you what pressure they are running but will usually tell you the "size" of the shot. If you are trying to gage your performance vs that of other similarly setup cars in order to know if the combo is working well, it would be nice to now how to figure out a close estimate of nitrous volume per second to make sure that you are comparing apples to apples. When I say comparing, I dont mean that I beat such and such and he runs low 11's so I must run 10's. No, I am talking about comparing time slips.
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Old Mar 9, 2008 | 11:08 PM
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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
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Old Mar 11, 2008 | 01:00 AM
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Originally Posted by Hennytime
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
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
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Old Mar 11, 2008 | 08:00 AM
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Originally Posted by Robert56
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
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.
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Old Mar 11, 2008 | 08:06 PM
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Originally Posted by 98Z28CobraKiller
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.
Here's what it's really Dependant on, and the basics are in fact in the above equation, we need to know what brand kit, what pressure they run at and jet sizes for any given HP rating. All companies use a n2o to fuel ratio for sizing jets, and this is also based on nitrous pressure. I do have the ratios somewhere in my notes. Anyway, the equation is simple if we knew what kit and jet sizes used in Hennytimes offered up info. We will still end up with a common denominator that we can just plug numbers into and figure things out down to the seconds based on HP of given jet sizes. So, trying to compare John doe's 100 hit to John q's 100hit will only be apples to apples if they run the same brand of kit. I'll try and find the info later and see what i can come up with.
Robert

Edit: from the above,
So a 11.5 run would be:
11.5 / 15 = .766lbs of n2o @ a 100shot at 1050psi
or,
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; Mar 11, 2008 at 08:14 PM.
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