Propane Injection-Math

[FeLeX]

If you are planning on making a new combustion and you want to be able to fire it every time, this is for you. If you are planning on putting a propane injection system on your new gun but you don't know what dimensions it should be, this is for you.
If you are trying to figure out what fittings to buy and how to put it all together this ISN'T for you. If you want to find out what to buy and how to put it together go to http://www.bcvids.com and find a video tutorial on how to make a meter. This is just the MATH part.

Ok now, since I got that out of the way lets start. Lets pretend we have a cannon with a chamber that is 4"x5". Lets pretend we have a meter pipe that is 1/2"x6". Lets get their volumes before we do anything else:
Volume Formula: r*r*pi*l, where r is radius (diameter divided by 2), pi is pi (3.14), and l is the length.
Chamber Volume-2*2*3.14*5=62.8 cu.in.
Meter Pipe Volume-.25*.25*3.14*6=1.1775 cu.in.
Another way, and more accurate way, of getting the volume is to pour water in your chamber and in the meter and then using some kind of beaker that has ml measuring on it measure the volume in ml and convert it to cu.in.

Now we need to figure out how much propane by volume your chamber needs to have a combustion. We will assume that 4% propane to air is the best combustion and we get:
Propane by Volume-62.8*.04=2.512 cu.in.

Now we can start figuring out how much psi is needed in the pipe. We are going to use this formula:
p1*v1=p2*v2, where p1 is the psi of propane in the pipe (the value we are looking for), v1 is the meters volume, p2 is the atmospheric pressure in your region, and v2 is required propane by volume.
Substitute the values in and you will get:
p1*1.1775=14.7*2.512
PSI of propane required in the meter: p1=31.36
So there you go, you would be fine with either charging it to 30 psi or 35 wouldn't be that big of a difference in the performance.

The reason I made this how-to is because I couldn't find anything anywhere about how to calculate the PSI of propane for the meter. All I could find is to figure out how big to make the meter itself so I decided why not just use 6" pipe but use different PSI depending on the volume of the chamber. I know this isn't rocket science and that many of you guys know this and all but this is for noobs like me that are trying to make their next gun better than previous and don't know how to do this.

Hope you liked it and I hope the Spud Veterans wont be too judgemental of this. Good Luck, FeLeX

[super spuder]

that should help alot of people

[hi]

I made a really cool self-explanitory method called guess and check, it works just as well....

just kidding, yea, that will help alot, sometimes it is troublesome to get it right.

[FeLeX]

Hehe yea Hi thats how I had to figure it out by myself at first and then I thought about the pressure*volume gas law.
Thanks for the positive feedback guys!

EDIT: Hmmm if you guys like it alot and you think it should get stickied then say so in your post. Thanks again.

[smokestream888]

sweet man thanks. im building a sniper cannon at the moment and by this i can see ill need 13psi of propane aprox. this will be my first propane injected cannon so thanks again man, this helped alot!

vote 1 for teh sticky!

[Modderxtrordanare]

I use this and this because I'm lazy on the math sometimes.

[CpTn_lAw]

i have a little question here...i'm planning on building a small caliber (16 mm) hybrid...i'm planning on using 2x --> 4x mixes, but i'm using butane instead of propane. Burns hotter and louder. The problem i have is that i don't know the exact amount of butane i need to get a perfect mix.
My ignition is the high voltage output of a plasma globe (much more cheap than stungun and very reliable, 10,000 V at very high frequency spark.)

Any help is apreciated...i have some experience in the spudgun aera lol, old members will know, so don't explain things to me like a noob

[jimmy101]

Pretty good post.

The text and formatting could be cleaned up a bit.

There are a few errors in math and logic ...

For example,

Now we need to figure out how much propane by volume your chamber needs to have a combustion. We will assume that 4% propane to air is the best combustion and we get:
Propane by Volume-62.8*.04=2.512 cu.in.

That doesn't give you 4% by volume, it gives 2.512/(62.8+2.512) = 3.85%

If this is going to be stickied it really should have the math etc. correct.

[Kufive]

Jimmy the formula above does give 4%....and the equation you used
"2.512/(62.8+2.512) = 3.85% " is wrong i think, why did you add the 2.512 to 62.8 ?

[FeLeX]

Jimmy I am sorry to say this but I have no clue what you are talking about. And please do tell me whats wrong with my formating?
CpTn you need to find out the perfect % for butane.

[jimmy101]

Lets see, you need 5 molecules of oxygen for every molecule of propane based on the balanced chemical equation;
1C3H8 + 5O2 = 3CO2 + 4H2O
Air is 21% oxygen. For 100in³ of air you need;
(100in³)(0.21)(1/5)=4.2in³ of propane.

Yep, you were right, except the percentage is 4.2 instead of 4. I was confusing myself thinking about syringe injection. With a meter no air is lost from the chamber when the fuel is injected, with a syringe a volume of air equal to the volume of fuel is lost. With a syringe the percentage is 4.02 instead of 4.2.

So, your equation was correct except for the percentage;

Propane by Volume-62.8*.04=2.512 cu.in.

should be;

Propane by Volume-62.8*.042=2.638 cu.in.

/\/\/\/\

The formating of the equations is not very clear.

Chamber Volume-2*2*3.14*5=62.8 cu.in.

Is that a negative sign or a dash? Much clearer as;

Chamber Volume: 2*2*3.14*5=62.8 cu.in.

or;

Chamber Volume = 2*2*3.14*5 = 62.8 cu.in.

[mark.f]

Basically we're just hitting the old displacement wall.

If you inject 4.2% of the chamber volume worth of propane, and the chamber is completely closed off, the volume will remain static and the pressure will rise the slightest bit. The mixture inside will be stoichiometric in it's proportions, just at a slight pressure.

If you inject propane into a chamber where air can escape, then the propane displaces the air. If you choose 4.2% of the chamber volume in this situation, then the mixture will be a tad rich, though at ambient pressure. To make up for this, simply inject 4.03% propane to chamber volume.

Hope that cleared it up, though I believe Jimmy knew it already.

BUT, also, I don't think it really matters. Will 0.17% error in fueling really make a difference?

[FeLeX]

Jimmy why the heck would that be a negative sign? Just use some common sense. And the reason I used 4% was because it isn't that big of a deal and would have probably boosted the pressure in the pipe by .5 tops. You are going to round it off anyways so why go through all the nasty numbers?

[jimmy101]

Felex, if it ain't a negative sign then why use it? It's a mathematical formula for god's sake, a "-" has a meaning.

[jimmy101]

Besides, if you are going to do a post on "Propane Injection-Math"

1. It really should get everything correct.
2. You should never round things off until the end of the calculation.