Login    Register
User Information
Username:
Password:
We are a free and open
community, all are welcome.
Click here to Register
Sponsored
Who is online

In total there are 72 users online :: 3 registered, 0 hidden and 69 guests


Most users ever online was 155 on Mon Aug 15, 2016 1:40 am

Registered users: Bing [Bot], Google [Bot], Yahoo [Bot] based on users active over the past 5 minutes

The Team
Administrators
Global Moderators
global_moderators.png CS

Im tearing my hair out with this calculation!!!

A place to ask general spud cannon related questions.
Sponsored 
  • Author
    Message

Im tearing my hair out with this calculation!!!

Unread postAuthor: Dave_424 » Tue Jun 02, 2009 3:54 pm

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 semi-automatic that uses a large tank to re-fill 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
  • 0


Dave_424
2nd Lieutenant
2nd Lieutenant
 
Posts: 220
Joined: Thu Aug 02, 2007 7:45 am
Reputation: 0

Unread postAuthor: spot » Tue Jun 02, 2009 5:15 pm

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
  • 0


spot
Private First Class
Private First Class
 
Posts: 27
Joined: Wed Feb 14, 2007 6:52 am
Reputation: 0

Unread postAuthor: mark.f » Tue Jun 02, 2009 5:25 pm

.. or Boyle's law. :wink:
  • 0

User avatar
mark.f
Donating Member
Donating Member
 
Posts: 3464
Joined: Sat May 06, 2006 11:18 am
Country: United States (us)
Reputation: 21

Unread postAuthor: Ragnarok » Tue Jun 02, 2009 5:46 pm

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>
  • 0

Does that thing kinda look like a big cat to you?
User avatar
Ragnarok
Chief of Staff
Chief of Staff
 
Posts: 5339
Joined: Tue Dec 19, 2006 8:23 am
Location: The UK
Reputation: 8

Unread postAuthor: Dave_424 » Tue Jun 02, 2009 5:59 pm

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
  • 0


Dave_424
2nd Lieutenant
2nd Lieutenant
 
Posts: 220
Joined: Thu Aug 02, 2007 7:45 am
Reputation: 0

Unread postAuthor: Gippeto » Tue Jun 02, 2009 11:04 pm

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
  • 0

"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.
User avatar
Gippeto
Donating Member
Donating Member
 
Posts: 2393
Joined: Sat Jan 19, 2008 10:14 am
Location: The Great White North...Canada eh!
Reputation: 11

Unread postAuthor: Dave_424 » Wed Jun 03, 2009 2:03 am

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
  • 0


Dave_424
2nd Lieutenant
2nd Lieutenant
 
Posts: 220
Joined: Thu Aug 02, 2007 7:45 am
Reputation: 0

Sponsored

Sponsor
 


Unread postAuthor: Gippeto » Wed Jun 03, 2009 9:14 am

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?
  • 0

"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.
User avatar
Gippeto
Donating Member
Donating Member
 
Posts: 2393
Joined: Sat Jan 19, 2008 10:14 am
Location: The Great White North...Canada eh!
Reputation: 11

Unread postAuthor: Dave_424 » Wed Jun 03, 2009 9:32 am

Im hopefully getting a 0-200 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 300-400 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
  • 0


Dave_424
2nd Lieutenant
2nd Lieutenant
 
Posts: 220
Joined: Thu Aug 02, 2007 7:45 am
Reputation: 0

Unread postAuthor: spot » Wed Jun 03, 2009 10:31 am

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.
  • 0


spot
Private First Class
Private First Class
 
Posts: 27
Joined: Wed Feb 14, 2007 6:52 am
Reputation: 0

Unread postAuthor: thedeathofall » Thu Jun 04, 2009 11:01 am

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?
  • 0

"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

thedeathofall
2nd Lieutenant
2nd Lieutenant
 
Posts: 213
Joined: Tue May 06, 2008 10:43 pm
Reputation: 0

Unread postAuthor: spot » Thu Jun 04, 2009 11:30 am

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 :)
  • 0

Last edited by spot on Thu Jun 04, 2009 11:48 am, edited 1 time in total.

spot
Private First Class
Private First Class
 
Posts: 27
Joined: Wed Feb 14, 2007 6:52 am
Reputation: 0

Unread postAuthor: Technician1002 » Thu Jun 04, 2009 11:45 am

thedeathofall wrote: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?


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.
  • 0

User avatar
Technician1002
Chief of Staff
Chief of Staff
 
Posts: 5190
Joined: Sat Apr 04, 2009 11:10 am
Reputation: 14

Unread postAuthor: jimmy101 » Thu Jun 04, 2009 1:14 pm

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?
  • 0

Image

jimmy101
Lieutenant General
Lieutenant General
 
Posts: 3128
Joined: Wed Mar 28, 2007 9:48 am
Location: Greenwood, Indiana
Country: United States (us)
Reputation: 7

Unread postAuthor: thedeathofall » Thu Jun 04, 2009 2:05 pm

wait, so its the same formula?

I still don't understand, i may not be reading this correctly.
  • 0

"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

thedeathofall
2nd Lieutenant
2nd Lieutenant
 
Posts: 213
Joined: Tue May 06, 2008 10:43 pm
Reputation: 0

Next

Return to General Spud Cannon Related

Who is online

Registered users: Bing [Bot], Google [Bot], Yahoo [Bot]

Reputation System ©'