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Math not working out

Post questions and info about combustion (flammable vapor) powered cannons here. This includes discussion about fuels, ratios, ignition systems, safety, and anything else relevant.
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Math not working out

Unread postAuthor: sandman » Sat Nov 03, 2007 8:34 pm

Ok, im learning Visual Basic.NET now with the help of an awesome teacher :)
So i thought i would start out small with something like a fuel tool, it is supposed to give you the pressure that is supposed to be in your meter. Well, i used the equation shown here and im not getting the same numbers as my stoichiometry. So i come to you with questions hoping to get answers, i will include pics to help you im assisting me.
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Attachments
fuletool.JPG
you can see the code too to see if the problem is in there, if you can understand the code
metermath.jpg
there is my math, the scanner cut off the side of the page so i painted it in, and for you that understand it, yes i know that i didn't have to convert moles of O2 to grams but i wasn't thinking at the time and it doesn't change the answer.

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Unread postAuthor: benstern » Sat Nov 03, 2007 8:54 pm

Thats unpossible math always works!! :o
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Unread postAuthor: sandman » Sat Nov 03, 2007 9:05 pm

i know, so i blame windows

if the answers were close i would not be complaining but they are way off
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Unread postAuthor: clide » Sat Nov 03, 2007 11:56 pm

A few things
That formula you used calculates pressure in ATM, multiply by 14.7 to get psi.
Second, I'm not sure why they subtracted 1 at the end of that thread. I would think Pm = 0.042*Vc/Vm would give you the gauge pressure needed for the meter.

Making those two changes and making the chamber 19 inches like you used in the hand calculation I get the same answer that you got.

A few things you can do to simplify the program too. Since you are doing the same calculations to the chamber and meter diameters to get the chamber and meter volume and then divide the two, those calculations cancel out and you can simply use Pm = (Dc^2*Lc*0.042)/(Dm^2*Lm)

E.G.
<a href="http://www.google.com/search?q=.042*2.835%5E2*19%2F%28.5397%5E2*12%29*14.6959488">Google Calculation</a>
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Last edited by clide on Sun Nov 04, 2007 1:34 pm, edited 1 time in total.

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Unread postAuthor: sandman » Sun Nov 04, 2007 8:56 am

ok, i will fix that when i get back home
thank you very much clide
i would have never thought about it giving me atmospheres
again thank you :D
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Unread postAuthor: mark.f » Sun Nov 04, 2007 9:01 am

sandman, the equation you were looking for is in the first post, which is:

<div align="center"><img src="http://www.markfh11q.net/MATH/derived.gif"></div>

Second, if you're dealing with gauge pressure only, (like in a fuel tool), simply .042(V<sub>c</sub>/V<sub>m</sub>) would work. I'm still fairly certain that nobody even got the point of that thread. The whole point was simply for me to find out why you end up with gauge atmospheres when you input absolute atmospheres. And, I figured out why.

Clide, I believe the whole reason I put -1 at the end of that thread was a simple mistake. For some reason, I must have put that there and said it's the same. I'm editing it now.
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Last edited by mark.f on Sun Nov 04, 2007 9:04 am, edited 1 time in total.
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Unread postAuthor: sandman » Sun Nov 04, 2007 9:03 am

ok, i shall change that too thanks :)
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Unread postAuthor: mark.f » Sun Nov 04, 2007 9:06 am

Wait, change what? Make sure you got what I was saying: .042(V<sub>c</sub>/V<sub>m</sub>) will work for a fuel tool, because you're not interested in absolute pressure of the meter.
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Unread postAuthor: sandman » Sun Nov 04, 2007 12:44 pm

yes so i will take out the -1 that is in there now

Edit: with those changes they come within 1 PSI of each other so im happy :)

thanks to everyone who helped
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