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Quick advanced question about a psi question

Post questions and info about pneumatic (compressed gas) powered cannons here. This includes discussion about valves, pipe types, compressors, alternate gas setups, and anything else relevant.
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Quick advanced question about a psi question

Unread postAuthor: dapallox1 » Tue Mar 18, 2008 11:20 pm

If I had a 22 gallon tank, and hydrogen compressed to 60 psi, how long would 5 liters/min flow out of the tank at 10psi? Sorry if you can't understand the question, if not, just ask about it and I'll try to expain :-D. Thank you everyone who helps me with this.
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Unread postAuthor: STHORNE » Tue Mar 18, 2008 11:26 pm

yeah, not really understanding the question dap. mind explaining a little betta?
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Unread postAuthor: Hubb » Tue Mar 18, 2008 11:40 pm

The time would be dependent on the type of valve (flow coefficient) you are using to dump the big tank.

Are you asking how many times you can fill a 5L tank to 10 psi using a 22gal tank at 60psi? If so, then the answer is just within reach.
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Unread postAuthor: dapallox1 » Tue Mar 18, 2008 11:48 pm

Okay, I'm working on running a car on nothing but HHO gas (hydrogen2 and 0xygen1). It takes 5 liters per min of HHO gas to run a 220hp motor at full speed. My setup I have written down, takes HHO gas, formed from electrolysis, and compresses it into a 22 gallon tank, at 30-60 psi. What I really need to know is (1) what psi I need to be tanking out of the tank to be producing 5 liters of HHO gas, (2) how long will that motor run at full speed on a 22 gallon tank of gas compressed at 30-60 psi, lets just say 60psi. For information, the 22 gallon tank will hold 83.38 liters of gas. And, I think I just answered one of my questions, lol... At 5 liters per min, it would run about 16mins at full speed...damn. But what psi would have to be coming out to produce 5 liters a min, yeah, that is what I need to know. Sorry about everything else, lol. Just one question, "what psi is needed to produce 5 liters a min? Thanks, and I hope that explains everything.
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Unread postAuthor: Hubb » Tue Mar 18, 2008 11:55 pm

Again, that would depend on the valve. Any psi, given enough initial volume, will be able to produce 5 liters a minute.
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Unread postAuthor: Spedy » Wed Mar 19, 2008 12:03 am

Isn't that basically water? I mean.. HHO... H2O... seriously. Do you mean it just runs on steam? The scientist in me has never heard of HHO gas if it isn't steam of some sort, so I'm kind of skeptical...
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Unread postAuthor: Lentamentalisk » Wed Mar 19, 2008 12:10 am

Your question doesnt really work... A given amount of gas will take up different volumes at different pressures. The basic gas equation is PV=nRT, where P is pressure, V is volume, n is moles, R is the universal gas constant (.08206), and T is temperature in Kelvin.

As is the general point of spudguns, a tank, no matter how huge, can be dumped of its pressurized gases in almost no time what so ever. The rate at which it empties is in relation to the flow through your valve.

From what I managed to decipher from your question, You want to have a flow of 5L at 10psi per minute coming out of a 22L 60psi tank. For starters you will need a regulator that regulates the gas down to 10psi, otherwise your flow will decrease as you empty the tank. Doing the calculations, you will get about 22min of run time, due to the fact that once your tank gets below 10psi, you will no longer get the flow you need.

If I misinterpreted, please tell me and I will recalculate.

One really important note, is that it is highly unsafe to store a tank full of hydrogen and oxygen, especially in their photometrically perfect mix, for large periods of time, especially near a source of heat, due to its great capability of exploding your face. I would suggest capturing your hydrogen and oxygen in two different tanks to avoid "facial explosion".

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No, he is talking about H<sub>2</sub> gas and O<sub>2</sub> together in a tank, that he separated using hydrolysis. It is a proven method. http://en.wikipedia.org/wiki/Hydrolysis
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Unread postAuthor: Eddbot » Wed Mar 19, 2008 1:46 am

then why didn't he say H2O2, or HHOO, lol...
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Unread postAuthor: Hubb » Wed Mar 19, 2008 1:55 am

H2O2 is hydrogen peroxide. :wink:
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Unread postAuthor: Daegurth » Wed Mar 19, 2008 4:42 am

It's HHO and not HHOO because the ratio between them is 2:1. This means that when it's converted back into water, there's nothing left over.

Separating them out is a good idea, they are pretty good at the facial explosion thing. Pressurised 100% hydrogen sounds dangerous, but it's almost nonflammable. I presume you already know that O2 is an oxidiser and incombustible in and of itself. I would use their relative densities to separate them- hydrogen will rise fery quickly, oxygen fairly slowly.

Just remember to make the inside of the oxygen chamber out of something very inert.
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Unread postAuthor: dapallox1 » Wed Mar 19, 2008 7:23 am

Thanks guys, Lentamentalisk, you answered my question almost perfectly. Except, I would use a 22 'Gallon' tank not liter, what would it be then? And I didn't think about storing the gases in two different tanks, that would be a lot safer, lol. Oh, and while driving, a series of HHO Cells would be creating gas and being pumped back into the tank(s), to keep the psi up. Eventually, you would have to pull over, and give it time to fill back up, but it would increase your distance.



Spedy wrote:Isn't that basically water? I mean.. HHO... H2O... seriously. Do you mean it just runs on steam? The scientist in me has never heard of HHO gas if it isn't steam of some sort, so I'm kind of skeptical...


Just go to google, and type in "HHO Cell" or "HHO gas". You'll find a lot.
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Unread postAuthor: psycix » Wed Mar 19, 2008 8:28 am

HHO is a name for a 2H2+ O2 mixture
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Unread postAuthor: Antonio » Wed Mar 19, 2008 12:02 pm

dapallox1 wrote: Just one question, "what psi is needed to produce 5 liters a min? .


mass flow=density*v*Area(crosssectional)=5*10^-3/60 [m^3/s]

for an incommpressible flow(well assume the speed is below the speed of sounds to keep it simple):

V=((2(pressure-atmospheric pressure))/(1.229(1-(cross. Area of tank/cross. area of tube)^2))^0.5

>>time it takes =5*10^-3/((1.229(density air)*v*Area)*60)

I know its complicated, so just reply if you dont get something.

This calculation assumes incompressible, frictionless flow
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Unread postAuthor: Ragnarok » Wed Mar 19, 2008 1:07 pm

dapallox1 wrote:Thanks guys, Lentamentalisk, you answered my question almost perfectly. Except, I would use a 22 'Gallon' tank not liter, what would it be then? And I didn't think about storing the gases in two different tanks, that would be a lot safer, lol. Oh, and while driving, a series of HHO Cells would be creating gas and being pumped back into the tank(s), to keep the psi up. Eventually, you would have to pull over, and give it time to fill back up, but it would increase your distance.

A gallon is 4.5 litres, so it would last 4.5 times as long.

However, I do have to express my doubts that an engine could be run on only 5 litres of HHO gas a minute.

Thinking it through 5 litres of Hydrogen gas at 10 psig will only release 30 kJ of energy when burnt.
A 220 hp engine has a power output of 3.6 times that a second.
Input requirements are going to be at least twice that.

Judging by that, your consumption is going to be at least about 450 times your prediction. This means that to run a 220 hp engine, the tanks will last about 14 seconds (assuming lentamentalisk's estimates were correct)

Sorry to tell you this, but your idea really isn't a viable option. I've worked on lots of weird engine designs, and other than the ones which use liquid fuels with low fuel/air ratios, the consumption is a killer.
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Unread postAuthor: Antonio » Wed Mar 19, 2008 2:15 pm

[quote="Ragnarok"][/quote]
Who says that a car with a 220hp engine uses its full capacity? I mean when you are driving with a constant speed you dont need all the power.
I am not sure about ur other assumption about the 5x..... When you have a higher pressure your air flow is going to be faster as well. So you will need to limit your mass flow rate. But k respect that you work on engines.
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