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Please help - hybrid calculations

Post questions and info about hybrid (compressed gas with fuel) powered cannons here. This includes discussion about fuels, ratios, ignition systems, build types, safety, and anything else relevant.
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Please help - hybrid calculations

Unread postAuthor: Jolly Roger » Mon Aug 28, 2006 8:40 am

Okay guys this might take a while to explain because its one of those scenarios where its hard to describe what your actually asking, so bear with me.. And please don't send in replies not intended on answering the questions, please... Say I had a chamber of hmm.. 5000cc, and my meter pipe was hmm.. 1000cc I am using OXYGEN, not air, and the stoichiometric ratio of oxygen to a type of fuel is 3.7:1. Now what I want to find out (without using the burnt lake fuel tool) is how to work out the supply pressure of the fuel meter pipe for different mixes. On burntlake fuel tool I've put in the calculation for the chamber (5000cc) the atmoshperic pressure (14.7psia) the fuel mix (0.270) which is 3.7:1 oxygen to fuel ratio, and I played around with the supply pressure until the result for the meter pipe volume was at 1000cc. The supply pressure recorded 19.8psi. This is I know a 1x mix (fuel tool is only for combustion) Do I just double that pressure to get a 2x mix? ie. 39.6psi. Then would I just add oxygen until the chamber pressure reached:

f(mix) = mix(14.7)-14.7 psi
f(2) =2(14.7)-14.7 psi
= 14.7 psi

I hope thats right. If it is then how do I project what the chamber pressure should be before adding the oxygen? Thats the real question...
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Last edited by Jolly Roger on Mon Aug 28, 2006 10:20 am, edited 2 times in total.
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Unread postAuthor: Shrimphead » Mon Aug 28, 2006 5:20 pm

Don't use pure oxygen in a hybrid. By doing so you take out the "buffer" gases that prevent problems. It could easily explode on you.
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Unread postAuthor: drac » Mon Aug 28, 2006 6:35 pm

Exactely what I said a couple posts ago! ;)
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Unread postAuthor: Shrimphead » Mon Aug 28, 2006 6:49 pm

Ya I thought I was kind of quoting someone. But that's just how it popped into my mind to say it.
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Unread postAuthor: Jolly Roger » Mon Aug 28, 2006 10:35 pm

Come on guys. That's what I said in my post, it's only an example, its so I can use with any fuel air mix... If I said propane and air someone would just give me the answer and not explain how to work it out.. So does anyone know how to work it out?
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Unread postAuthor: drac » Tue Aug 29, 2006 9:46 pm

You need our suggestions when you're considering doing something as stupid as this. Oxygen/propane and PVC DO NOT MIX. It will explode and send shards of PVC death at you. i wouldn't even attempt this in a metal cannon!


Don't you dare get pissy at us, we're protecting your stupid ass from getting killed. With our help, you're bound to get suggestion like these, but they're only for your betterment. We don't want anyone getting killed because they didn't know.

EDIT: OK, done ranting. With that formula you have, you don't need to add oxygen. All the oxygen is in the compressed air you're adding, and it's much safer.

Didn't mean to be an asshole, sorry bout that.
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Unread postAuthor: dongfang » Wed Aug 30, 2006 5:53 pm

Hi,

Are you assuming that the gas remaining in the metering pipe is pure fuel, or that there is a homogenous mixture of fuel and oxy in the chamber as well as in the meter, when you close the meter to chamber valve?

Although (or BECUASE) I have done a 40% ogygen shot @ athmospheric pressure before, I agree with those who try to talk you out of actually trying 100% oxygen. It's powerful stuff. At least make a scaled down experiment with 60% oxygen @ 1.5 bar in a 500 cc tank, that you cover with one foot of soil. When that has scared you shitless, just be happy that you didn't try with 100% in a big tank.

But I'd be happy to see if I can help you with the numbers, just because the formula would be useful in other contexts. So, what is your assumption on the remaining gas in the meter pipe?

But think abt what you are doing once more --- if you stuff 5 liters of oxy in a tank at 5 bar, you have the equivalent of 125 liters of air, not considering that the a fuel air mix will just explode at most, while the oxy thing will detonate, and distribute your brains over a couple of acres.

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Unread postAuthor: Jolly Roger » Thu Aug 31, 2006 12:02 am

Yeah guys ok, disregard the oxy example then... But yeah dongfang I'm just trying to find out what the chamber pressure would be when I let the pure fuel from the meter pipe into the chamber (before pressurizing with air). At first I assumed if the chamber was say twice the volume of the meter pipe the chamber pressure would just be half the pressure that the meter pipe was at... If you follow. ie.

Meter pipe volume = 1000cc
Chamber volume = 2000cc
Meter pressure = 20psi
Then chamber pressure = 10psi (Before air pressurization)

It seems simple enough but I didn't account for the fact that the actual volume the fuel would be covering would be that of the chamber AND the meter...
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Unread postAuthor: dongfang » Thu Aug 31, 2006 11:06 am

Well, before opening the meter-chamber valve, you have

Meter has vcolume V_1
Chamber has volume V_2

N_1 moles of gas in the meter
N_2 moles of gas in the chamber

Pressure in meter is P_1 bar
Pressure in chamber is 1 bar

Temperature is constant (we wait for it to settle @ environment temp).

After opening and waiting, we have basically a chamber of volume V_1+
V_2, with N_1+N_2 moles of gas in it.

The ideal gas law is:

P * V = N * R * T

so

P * (V_1 + V_2) = (N_1 + N_2) * R * T

P = (N_1 + N_2) * R * T / (V_1 + V_2)

Rewriting the Ns for each pre-opening chamber, using the ideal gas law again:

N_1 = P_1 * V_1 / (R * T_1)
N_2 = P_2 * V_2 / (R * T_2)

So (using the assumption that T = T_1 = T_2):

P = (P_1 * V_1 / (R * T) + P_2 * V_2 / (R * T)) * R * T / (V_1 + V_2)

Simplifying:

P = (P_1 * V_1 + P_2 * V_2) / (V_1 + V_2)


So, for a meter pipe of 1000 cc and a chamber pipe of 2000 cc, with 2 bar (1 bar above athmospheric) in the meter:

P = (2 * 1000 + 1 * 2000) / (3000) = 1.25


You will also want to know how much of the gas in the chamber is actually fuel. Assume that ALL the gas in the meter is fuel (meter and chamber did not mix perfectly; there was just a hiss of fuel from meter to chamber):

There used to be

N_1 = P_1 * V_1 / (R * T) moles of fuel in the meter. Now, there are

N_r = P * V_1 / (R * T) moles left there. The other

N_c = (P_1 - P) * V_1 / (R * T) moles went into the chamber.

For the example, that would be

(2 - 1.25) * 1000 / (R * T) moles.

R = universal gas constant = 8.3145 J/mol K

and T = 20 deg Celcius = 293 K.


(2 - 1.25) * 1000 / (R * T) moles = 0.308 moles.

If the gas was butane, that would be (wikipedia) 17.86 grams of it (the weight of a small mouse - not little for a gas charge).

To burn it, you need 0.77 moles of oxygen, or 3.66 moles of air.
You have N_2 = P_2 * V_2 / (R * T) = 1 * 2000 / R * 293 = 0.83 moles of
air in there from the before. You need to add another N_d = 2.84 moles, by adding a partial pressure of

P_d = N_d * R * T / V_2

= 2.84 * R * 293 / 2000 = 3.46 bar, ending at 3.46 + 1.25 = near 5 bar.

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Unread postAuthor: Jolly Roger » Thu Aug 31, 2006 12:37 pm

Thanks heaps mate, that's what I was looking for. If I had any spud bux left (donated them all to sniperman) I would show you my gratitude.

Again thanks for the help,
Jolly Roger
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