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Im tearing my hair out with this calculation!!!Hey guys, having a bit of a problem about working out the pressure drop in the main tank when it fills a smaller tank
Basically I am designing a semiautomatic that uses a large tank to refill the smaller tank for each shot I have the volumes for both tanks but thy are to about 10000000000 decimal places because I dont want to round it Is there a claculation that will allow me to work out how drastic the pressure drops off. so like starting tank pressure 300psi (The main tank is filled with the secondary tank for the first shot 300 psi 270 psi 250 psi 230 psi etc. etc. All help welcome
When you can give us the measurements, I can send you a table that will describe the pressure left after each shot. If you want to calculate it yourself: You should use the combined gas law. The temperature stays constant (for simple calculations, calculating temperature drop when the gas expands will be way overkill), so you can just drop the T. Thus the formula becomes p1V1 = p2V2
In other words, considering the process diabatic, your equation is:
Pressure left = Big tank initial pressure * Big tank volume / (Big tank volume + Small tank volume)<sup>number of fills taken from big tank</sup>
Does that thing kinda look like a big cat to you?
hey thanks for that, i have gone through 5 diffrent calculators and still can't get the answer.
So here are the details The main tank is 530 ml filled to 30 BAR The tank on the gun is 93 ml
Dave,
Using the formula that Spot provided you, and as Ragnarok states; P1 * V1 = P2 * V2 P1 is 30Bar V1 is 530ml V2 is 530ml + 93ml (total expanded volume) P2 is the new pressure at the expanded volume. So; 30Bar * 530ml = P2 * 623ml Or 30Bar * 530ml / 623ml = P2 P2 = 25.5Bar (the chamber pressure remaining after "charging" your firing chamber once.) After the second "charge", you'll have 21.7Bar remaining in the chamber, and so on. 18.46Bar 15.7 13.37 11.37 9.67 8.23 7 5.96
"It could be that the purpose of your life is to serve as a warning to others" – unknown
Liberalism is a mental disorder, reality is it's cure.
Thanks everyone for the help,
so it looks like i can get about 6 useful shots from a 30 BAR fill. Hmmm..................... It looks like i need to buy an inline regulator. The trouble is finding one with a high enough max input pressure. Could I buy one that is strong like a brass one and then use like 30 BAR input and about 10 BAR output. I think that the reg would hold up okay. What do you think? Dave
Most any welding gas regulator will be rated for 3000psi inlet.
The problem is that the outlet pressures are typically low. But then, 10Bar is only 145psi, you might be able to find something on ebay. Size/weight may also be an issue. Have you considered a regulator designed for paintball?
"It could be that the purpose of your life is to serve as a warning to others" – unknown
Liberalism is a mental disorder, reality is it's cure.
Im hopefully getting a 0200 BAR oxygen regulator. If size and weight seem to be an issue, then I might have the main tank at 150 BAR and the reg to 300400 psi. I have a nice long piece of hydraulic tube i could use.
could a galvanized malleable iron 1/2"  1/4" bell reducer take 150 BAR I think it could be want to make sure
Dave_424 Gippeto tested the iron fittings using his dead weight tester. While he didn't test a bell reducer, I think you would be safe, considering the other fittings withstood more than 8000 PSI (although you would need a safety margin of say 3, the fittings withstood nearly 4 (!) times your pressure.
Okay, i don't mean to hijack, but what if you use a regulator? what formula will tell the pressure drop if the main tank is pressurized to say 3000 (HPA) and then regulated to 150 for the second tank?
"I'm spending time without a gender for tax reasons. It's great if I get hit in the groin, but a total nightmare in the bathroom."
Rag Obsequium parit amicos; veritas parit odium. Cicero
What I would then do (And I'm by no means qualified, haven't even finished high school) is calculate the moles of gas in the chamber. Then subtract the moles of gad that needs to stay in the chamber (so if you want to regulate 50 psi, 50 psi should stay in the chamber). Next you calculate the moles needed in the secondary chamber to get to the pressure you want. Then you divide the first number (moles in chamber  moles that need to stay in chamber) by the second number (moles in secondary chamber) and round the answer DOWN. Hopefully I explained it clear enough
HINT: 1 mole of gas, under ideal circumstances (= 0°C, 1013hPa) has a volume of 22.4 liters. Again, if you post the full dimensions I will make a table with the pressure of the chamber after each refill
Last edited by spot on Thu Jun 04, 2009 11:48 am, edited 1 time in total.
In this instance the volume at STP (Standard temperature and pressure) can be calculated that is needed to change from the original pressure to the final pressure in the chamber. Knowing this mass of gas initially, the amount removed and finding the volume remaining, the pressure changes in the source volume can be calculated and therefore the number of shots that can be delivered before the source is depleted.
Spot and Tech are making it more complex than it needs to be. It is still just P1V1=P2V2, just have to figure out what the correct values are.
The slave tank is at lower pressure. What is the equivalent volume that the big tank would have to be increased by to account for the VolumePressure product of the small tank?
wait, so its the same formula?
I still don't understand, i may not be reading this correctly.
"I'm spending time without a gender for tax reasons. It's great if I get hit in the groin, but a total nightmare in the bathroom."
Rag Obsequium parit amicos; veritas parit odium. Cicero
 
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