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View Full Version : How much energy is stored in HPA tank?



AcemanPB
03-10-2003, 07:04 PM
I was wondering if any knows how much energy is stored in a 68/4500 tank

And if anyone feels like explaining to me how and what formulas to use to get the number I would be very thankful :)

FalconGuy016
03-10-2003, 07:49 PM
I can find this out... hold on (ill be back to edit)

])arKNe$$
03-10-2003, 08:57 PM
68x4500=306000psi.?

AcemanPB
03-10-2003, 08:59 PM
No.

Paintballer86
03-10-2003, 09:19 PM
Originally posted by AcemanPB
No.
I burst out laughing when I read this. (what Aceman wrote)

And lets just say that it is enough energy to blow your arm off if you dis-respect it for even a split second or mistreat the fiber-wrapping

AcemanPB
03-10-2003, 10:01 PM
Well I do know

Work = Pressure(Pa) x Change in volume(m^3)


4500psi = 31026410 Pa

I'm not to sure about change in volume I'm going from .00111432 cubic meters (or 68cu inches) to what? This formula also doesn't take into account the heat or any other factors that would play a role

Kevmaster
03-10-2003, 10:19 PM
Originally posted by Paintballer86

I burst out laughing when I read this. (what Aceman wrote)

And lets just say that it is enough energy to blow your arm off if you dis-respect it for even a split second or mistreat the fiber-wrapping


I think he said it best

AcemanPB
03-10-2003, 10:21 PM
Yes I know this but I'm curious and want to know exactly how much energy we are talking here

rifleman
03-11-2003, 01:07 AM
Well, I've kinda wondered about this myself and I'm not sure if energy is even the right term to measure it.

E = force * distance
which is:
Energy = mass * distance/time/time * distance

Well your mass is easy enough to figure out. I 'think' you'd have to find the velocity and acceleration of the molecules moving within the tank--which is directly proportional to the volume and temperature of the tank.

so in other words... hell if i know :)


Adam

rifleman
03-11-2003, 01:09 AM
^ That equation is for potential energy.. but it may be kinetic energy we're measuring since the molecules 'are' moving inside the tank.

So relate to this ke = .5 * mass * velocity^2

But still, you'd have to find the velocity of the molecules.

askman
03-11-2003, 02:02 PM
in 68ci at 4500psi, multiply the two, you have 306000 lb-in or 25500 foot-lb. One Horse power is 33,000 foot lb per min. same energy as running 1hp motor for 46.4sec.
that is lot of energy. When tank bursts, it is releasing this instantly.(fraction of sec)

25500 foot lb, assuming that you are 255lb wil lift you up 100ft or 10 stories.


One horse is lot of energy. it is 746watt of energy. 746joule/sec

it is same as 34587 joule of energy in that tank.

1 joule is 1newton moving 1 meter.

1 newton is 1kg, 1m/sec^2

xen_100
03-11-2003, 02:19 PM
umm.......I like pie!


I understood very little of that, although it is complettely fasinating.

Hunter
03-11-2003, 02:52 PM
you people have to to much on your mind.

2k1angel
03-11-2003, 03:03 PM
In short it could rip your arm off.

Doc Nickel
03-11-2003, 03:40 PM
Originally posted by Hunter
you people have to to much on your mind.

-It could be argued instead, that you don't have enough.

Doc.

askman
03-11-2003, 04:56 PM
I have to admit that It is bit confusing the way I wrote it.

pound of force is one pound of mass multiplied by gravity.

PSI is pound force per square inch. multiply it by volume (cubic inch), you end up with inch-pound. You can covert the inch-pound to foot-pound by dividing it by 12 (12inch per foot)


thus 68 cubic tank with 4500 psi have 25500 foot pound of energy. foot-pound-force is equal to 1.35582 joule. Thus 25500 foot pound is equavalent to 34573 joules of energy.

1 mole(227gm) of nitroglycerine have about 300k joule of energy, thus our tank has same energy as about 25gm of nitroglycerine. enough to blow your arm off.

hope this clears it up...

hitech
03-11-2003, 05:41 PM
Originally posted by askman
...thus our tank has same energy as about 25gm of nitroglycerine.

Any idea how much nitroglycerine is in an average stick of dynamite?

Rynoboy06
03-11-2003, 05:42 PM
Originally posted by Doc Nickel


-It could be argued instead, that you don't have enough.

Doc.

Go Doc!

FutureMagOwner
03-11-2003, 05:44 PM
Originally posted by askman
25500 foot lb, assuming that you are 255lb wil lift you up 100ft or 10 stories.

so the moral of this story is treat tanks with respect or eat alot of fastfood and only get shot up 10 feet instead lol

xmetal2001
03-11-2003, 11:10 PM
Originally posted by 2k1angel
In short it could rip your arm off.

It would rip off alot more than your arm.

Lets put it this way -- a exploding scuba tank can send an 18 wheeler into the air...they arent THAT much smaller.

Dont expect to be able to get away with losing an arm if your tank actually explodes.

But the idea behind fiber wrapping is basically to keep your tank from exploding and instead it would hopefully stay in one piece while the pressure escapes through the rupture in the tank.

Edit: WOW, weird, i just posted this a few seconds ago and according to the AO clock the post after mine hasn't even happened yet. Its 12:10 and it was made at 12:30 so it shows up after mine. Nifty yet weird.

AcemanPB
03-11-2003, 11:36 PM
34573 joules? I wasn't expecting THAT much energy, wow. Well I was wondering if it's possible to find the least amount of energy required to get a paintball at 300fps (assuming no friction, etc). Now I'm curious how much of that energy is actually "wasted" :) .

askman
03-11-2003, 11:41 PM
34573 joules per tank, assuming that we get 1000-1700 round per tank it takes about 20-35 joule per shot. In reality, we can only use the pressure down to 600psi for mags and 2-300 for even the low pressure markers, so potential energy is less, meaning it probably is more like 20-30 joule.

In 100% efficient condition, assuming 3gm paintball, 30cm barrel, 300fps(84m/s),

d=.5*a*t*t
v=at
.3=.5a*t*t
84=a*t


.3=.5(84)*t :t=.00714sec, a=11765m/s^2 (about 1200g, wow, more than I thought)

f=m*a =.003*11765=35.3 newtons
e=f*d= 35.3*.3= 10.6 Joules per shot

Of course, simpler way is if we use Kinetic energy formula: ke=.5*m*v*v

ke=.5(.003)84*84=10.58 joules

either way, same answer.
So our marker is about 30-50% efficient. so maximum we can get is about 3000 shots even in perfect condition

pbzmag
03-12-2003, 12:55 AM
After reading all those formulas, my head started to hurt! Even Einstien is going "WHAT"!!!

pbzmag