SpudFiles

What pressure would I have to pump a 1 inch by 5 inch chamber to get 110 psi in a 2 inch by 36 inch chamber? I have no idea what eqaution I would use. Thanks.

Oh jeez, well you would need to find the volume of the two, then the compression ratio of pressure, which I'm pretty sure is 1 liter of 100 psi air is equivalent to 6 liters of air at atmospheric pressure, so with that you could probably figure it out with a bit of work...

Any clues as to what this is for? I'm intrigued...

Your question and information provided is very unclear. Do you mean a chamber with a 1 inch diameter that is 5 inches long and a second chamber with a 2 inch diameter that is 36 inches long? And do you mean what pressure you will need in the smaller chamber to get the second chamber to the goal pressure when both are connected?

Exactly what I mean. Well I am building a burst disk cannon and dont want to pump it till it bursts so instead I will make a small chamber with a ball valve so all I have to do is flip the valve open and BOOM. What do you think?

jon_89 wrote:Exactly what I mean. Well I am building a burst disk cannon and dont want to pump it till it bursts so instead I will make a small chamber with a ball valve so all I have to do is flip the valve open and BOOM. What do you think?

the second chamber can be removed and simply replaced with a burst disk right in front of the valve. Having the secondary chamber only decreases firing pressure.

I dont think I quite understand what you mean. I do get the pressure decreasing though.

the burst disk bursts at a set pressure. If you put a Ballvalve behind the burst disk, And open it, The disk will burst at its Set burst pressure, And by that time proboably have the Ballvalve mostly open, Yeilding a Higer powered shot that if you just used a Ballvalve.

Ok I get what you are saying, dont use the secondary chamber then?

The easiest way to do this would use a mixture of the ideal gas law, polytropic process relationships, and constant specific heats. It's a little more complicated than I'd like to explain so I'll just give you the answer. And I'm not even sure if the assumptions made are appropriate so take it for what it's worth.

Assuming no heat transfer you're looking at about 12000 psi of pressure. Assuming an isothermal process you're looking at about 2900 psi of pressure. Reality should be somewhere between the two but much closer to the no heat transfer assumption. This seems extremely high so maybe my math's wrong (or what VH_man said is very very true). Again, take it for what it's worth.

Edit: Let me state that I think my temperature drop assumption is incorrect because this happens too fast for everything to mix well. Consider the no heat transfer number to be extremely approximate.

of course, We can always just try and see what happens

Testing the unkown is how good things get started. So go ahead. Build it. If it doesnt work, Just make a new one untill it does!

But does putting the ball valve before the burst disk still have the same performance as just the burst disk? I remember there was once a topic on this and some said it worked some said it didnt.