Things to ponder with how much pressure you need.

[CHK6]

"I'm not dead yet!" Monty Python

Wow, I read all the above posts. Interesting.

I think I'm on the right track in figuring out pressure loss in piping, but I need to figure out how to calculate SCFM.

Most problems explain what SCFM is and give it a value to calculate ICFM, but never states how one derives SCFM.

Anyone know how to do this?

UPDATE: Never mind I figured it out and have a completed formula. All I need is a couple of reference tables for relating density to temperature and some others.

[pbjosh]

okay everybody,

I did a quick survey on pbnation, and i guess I will post one here soon.

Here is an energy used chart, per gun.

Now, these are not tested numbers, but from a list of peoples postings about which gun, how many shots, and what tank. So, error is included. Just for now, this is just a beginning. I want to fill this up to the point that we can figure out valves and barrel ratios also.

So, if you read this, post something like....

I have a SFL Snagler ACE with an 68/4500 tank and I get 900 shots per tank @ 225 psi

Thanks,

Josh

[CHK6]

To: PBJosh

PBJosh,

The energy the tank holds also heavily depends on local barometric pressure, temperature, and altitude above sea level.

I have also found some other interesting tidbit relating to PSI levels. When talking "shop talk" I found that you either have to say PSIA or PSIG and just saying PSI will get you funny looks from individuals that work with compressed air. And if you say what the PSIG is you have to give the local barometric pressure.

I also found out that the barrel id has an effect of how many shots you get with the energy in a bottle.

Also when calculating pressure temperature you have to convert Farenheit to Rankine. Something I wasn't doing and really scewed my results.

Just sharing the information I have found in better hopes to truely understand it all myself.

Later gator.

[pbjosh]

Well, it all gets down to averages. A person can tell me how much pressure is in their tank, but the gauges on a tank is not to accurate. Also I will get basic averages for the number a shots per tank, generally rounded off to 50 or 100.

Barrel has alot to do with it, but the general average for the gun will be a bigger factor. And part of this is to educate. The amount of energy used per gun on average is a number nobdoy uses to figure out efficinct, but should be incorporated in everybodies ideals for how a gun works. How many people think that low pressure means high efficiency? With a basic graph, this should help people realize what gun they want, but also what tank. If a person thinks that one valve is better just because of its working pressure, then they are not taking into account the amount of energy used per shot.

A different bolt, or even operating pressure can affect all of this also.

Again with the accuratecy of a post to a survey, and the gauges, and everything else, the difference for the local barometric pressure temp, and PSIG and all that, it doesn't really matter. I am dealing with larger numbers and averages than that. And the reason for a large # of averages is to sort that information outta the mix. I don't want an effecientcy number based on a certain air pressure, certain paint, certain barrel. I want an average for that gun, in any local barometric pressure, at any temp, and etc.

So, tell us, how many shots from what gun and the tank you use at what pressure and what velocity.?

Josh

[AGD]

Pbj,

Nice data table!! Glad to see someone wants the facts. I agree with you that a large sample size will average out the variables. I would like to see the formulas you used to get the energy in the tank and the per shot energy.

Thanks

AGD

[CHK6]

An inward view

To answer the question I get 250 to 300 shots on any given day with 3.5 oz CO2 on my Phantom. Pressure ranges from 800 to 1000 on the guage.

[pbjosh]

AGD! proud to have you on this!

Okay, here is all I did to figure out the numbers.

I originally posted over a 2 years ago on the tinker's guild to a question about the pressure it takes to get the ball to 300 fps, like on this post. I was posting on how the number that they are looking for is a value of energy, not pressure.

Then I collected the numbers for the volume of a couple of guns with fixed chamber dump valves. Since energy is a measure of Volume x Pressure, I figured I could get the right numbers from multiplying the Volume of the dump valve with the pressure the gun ran. And, then I smacked my head and said "Hey, all the guns have a fixed amount of energy attached to them! The HPA tank!"

So in the basic 68/3000 tank, multiplied, you get 204,000 ci*lbs of energy in the tank! And if a gun runs till 400 psi, a person would take the volume of the tank, minus the amount of air unused, like 68 ci x 400 psi = 27,200. So, 204,000 - 27,200 = 176,800. And if a gun gets 800 shots, then divide the energy that can be used, 176,800, buy the number of shots, 800, you get the energy used, so:

176,800 / 800 = 221 ci*lbs of energy on each shot.

I just asked about the gun, the tank size and the number of shots @ which pressure. That gave me the total energy used per system. So, even if it doesn't have the exact amount of air used down the barrel, it does give a relative number for the energy used per gun. And I must say, the 'Mags are the best dump shamber out there right now, with 200-225 ci*lbs of energy used. A gun like the Shocker is horrid, with 365-400 ci*lbs common.

Well, that about it, just whiped up an Excel file and that was that.

Now go my children, and spread the word.

Josh

[athomas]

Lets get more elaborate here. I'm no expert at math anymore, its been a few years since I've picked up a math book, but I am interested in some of the dynamics here.

for ease of calculation:

Mag chamber pressure = 400 psi
translates to 220 in-lb of work force

Actual force on the paintball at start is 150lbs when properly vectored, as it is the cross sectional area of the ball that is important, not half of the surface area. 150lbs represents the pressure in one direction(out the barrel).

Chamber = .55 ci
cu area of 1 inch of barrel (id = .690) = .3739 ci

Lets plot the acceleration force of the expanded gas.

distance = force
start(0) = 150 lbs
1 in = 89 lbs
2 in = 63.4 lbs
3 in = 49.2 lbs
4 in = 40.2 lbs
5 in = 34 lbs
6 in = 29.4 lbs
7 in = 26 lbs
8 in = 23.2 lbs
9 in = 21 lbs
10 in = 19.2 lbs

As you can see, the amount of force on the ball decreases as the gas expands and the ball accelerates down the barrel. Therefore the rate of acceleration is not going to be standard. Most acceleration is calculated using a constant force.


Now, I can't find my books with all my formulas, so I need your help here. Can anyone figure out a formula for calculating the actual rate of acceleration of a paintball down the barrel of a paintball gun. If we can come up with a standard formula taking into account the changing volume of the chamber-barrel as the ball accelerates it will help us realize the benefits of changing pressures, volumes and barrel lengths.

I did some quick calculations and noted by eye that the effect of doubling the chamber size and decreasing the chamber pressure doesn't work. It is more like double the chamber size, decrease the pressure by 25%. The amount of in-lbs of useable gas increased. In the other direction we could greatly increase the efficiency by increasing the pressure and decreasing the chamber size. The starting force would be increased and would probably cause the balls to explode in the gun though.

I know my data has gaps in it but the fundamental concept is basically sound.

Hope this helps.

We can do it for the betterment of mankind on the painball field.

[bjjb99]

Warning, a whole lot of theory and physics going on...

Fair warning, I'm going metric on you guys. That being said, here
we go...

First off, some basic formulas.

1: F = m a
2: F = P A
3: P1 * V1 = P0 * V0

Notes:

Formula 1 is your basic "force equals mass times acceleration"
equation. Formula 2 relates force, pressure, and area. Formula 3
describes how pressure and volume interrelate. The "*" symbols in
Formula 3 are multiplication symbols, for those who are used to
seeing "x" instead.

Since we're interested in the acceleration of the ball at any
point in a barrel, we really want to solve Formula 1 in terms
of the variable "a".

a = F / m

Now, we have a pretty good idea what the mass of a paintball is
(around 0.116 ounces, or around 0.0033 kilograms). What we need
to know is how the force on the ball varies as the ball travels
down the barrel. This is where the second formula comes into
play.

The force on the ball as a result of the gas behind it is
dependent on the pressure of that gas and the cross-sectional
area of the paintball.

F = P A

We can calculate the cross-sectional area easily enough. A
paintball is right around 0.68 inches in diameter (0.017 meters).
Area is simply Pi R squared, which gives a final area of
0.00023 square meters. Why am I using square meters for something
so tiny? Because the pressure values I'm going to be using are
expressed in units of force per square meter of area. So now
we have the following for Formula 2:

F = 0.00023 P

We can plug this into our acceleration solution for Formula 1 to
get the following:

a = 0.00023 P / 0.0033 (I plugged in the mass as well)

a = 0.0697 P (simply divided 0.00023 by 0.0033)

Ok. Now we have acceleration expressed as a function of pressure.
Now we have to answer how the pressure changes as the ball travels
down the barrel. I'm going to assume that the barrel we're using
has zero porting in it, just to make my life easier. :)

This brings us to Formula 3:

P1 * V1 = P0 * V0

This basically says that the energy held in a gas essentially
remains constant. Of course, this also assumes that the gas
temperature does not change. We'll make that assumption for
this exercise.

Our initial pressure (P0) is going to be 400 psi
(2.757 megapascals). The initial volume (V0) is 0.55 ci
(0.00000901 cubic meters). We can plug these two values into
Formula 3 and get the following:

P1 * V1 = 24.85 Newton-meters (i.e. Joules)

The volume of space behind the ball as it travels down the barrel
is going to be the volume of the dump chamber (0.55 ci) plus the
area of the barrel times the distance the ball has traveled.

V1 = V0 + A * x,

where x is the distance the ball has traveled, measured in meters.

We can then get the following expression for our pressure:

P1 = 24.85 / (V0 + A * x).

P = 24.85 / (0.00000901 + 0.00023 * x)

Ok. We now know how the pressure should change as a function of
ball position. We can plug this back into our acceleration
formula:

a = 0.0697 P

a = 0.0697 * 24.85 / (0.00000901 + 0.00023 * x)

a = 1.732 / (0.00000901 + 0.00023 * x)

or, if you want to keep all the variables as variables rather than
having the numbers plugged in...

a = P0 V0 A / (m * (V0 + A x) )

P0 measured in pascals (newtons per square meter)
V0 measured in cubic meters
m measured in kilograms
A measured in square meters
x measured in meters

Our acceleration should be in meters per second squared.

And there you have it... one acceleration formula. Keep in mind
that this does not take into account frictional forces experienced
by the ball as it slides against the barrel wall. These forces
are not always insignificant, though if you have a good paint to
barrel match and can blow a paintball out the barrel with ease,
you may well be able to ignore frictional effects.

I've got a semi-handy Microsoft Excel spreadsheet which calculates
a paintball's acceleration down a nonported barrel and determines
what the theoretical exit velocity should be. I'm working on
tuning this equation so that the ball exits the barrel at the
desired velocity and when the pressure in front and behind it are
equal. I can vary the pressure, initial volume, barrel length,
and frictional forces so far.

BJJB

[athomas]

Excellent. That's what we're talking about.

If it isn't too much trouble, could I have a copy of the excel worksheet. It would save me doing the same work all over.

thanks

[bjjb99]

athomas wrote:
>
> If it isn't too much trouble, could I
> have a copy of the excel worksheet. It
> would save me doing the same work all
> over.

I'll have to make sure it's reasonably presentable... after all, I created it to be "me friendly" rather than "user friendly". When I get it ready I'll post an attachment in .zip format, since it doesn't look like this forum supports .xls or .wbk. It's going to be an ancient version of excel (4.0), but the recent versions should still be able to read it.

BJJB

[bjjb99]

Excel Spreadsheet

Well, I've made the spreadsheet as friendly as I could, and have zipped it up so I can attach it. Man, it's just under the filesize limit. I hope this works...

There are a couple of things to keep in mind when playing around with this spreadsheet. First, it does not necessarily represent reality, as it is only a theoretical treatment of the problem at hand. Second, it handles things a bit differently than what has been discussed... basing everything on very small increments of time (calculus a-la excel) rather than ball position.

I have yet to figure out why, but the spreadsheet fails to give even remotely close velocity numbers for an automag's dump chamber design (500+ fps from a 10 inch barrel at 400 psi input?!). I think this has much to do with my inaccurate guesses for the volumes of the power tube and bolt, both of which will cause an immediate pressure drop from the dump chamber initial pressure at the moment of gas release. Perhaps more accurate numbers for these two volumes would correct the spreadsheet results. There are also the matters of non-instantaneous gas relase from the dump chamber and gas leakage around the bolt and power tube (likely small?) to contend with.

Have fun playing around with it. Maybe you can improve some things here and there, though you may want to fiddle around with a copy in case things get out of hand.

BJJB

[CHK6]

Baby steps taken a bit further....

Let me first say I have really enjoyed this thread, I have learned a lot. I am 99.999% almost done in figuring out pressure drop in a pipe with compressible gasses. The following is a massive culmination of research I have gathered from the last three weeks and was able to cross reference verify with other sources.

Here is the generic formula that needs to be solved.

delta(P) = F (L/D) * [(p * u^2)/(2 * g)]

delta(P) is the absolute pressure drop, not guaged.
F is friction factor of material.
L is the length of the pipe.
D is the pipe diameter
p is avg. density at flow conditions.
u is mean velocity
g is gravity

To find (u):
u = Q/A
A = cross section of a circle.
Q = volume flow in entrance conditions. (ICFM)

To find (Q):
Q = (Z * R * T) / (60 * M * Ps)
Z is the compressibility of the gas.
R is universal gas constant
T is pressure temperature
M is molecular weight of the gas
Ps is pressure (PSIA)

To find (p):
p = (P1 / 14.7) * d
d is the density of air at amospheric temperature.

To find (F):
You need a materials chart

To find out if the flow of the gas behind the ball is turbulent calculate the Reynold's Number:
Re = (p * u * D) / y
y is absolute viscosity of air at atmosphereic pressure.
If the result of Re is greater than 4000 the flow of the gas behind the ball is considered turbulent.

Now all I have left is to plug in the numbers. Which I'll do this weekend when I have some time.

[CHK6]

Grrrr....

This doesn't sound right. After some calculations I figured that with a 10 inch SS pipe with a mean velocity of 140 fps using Nitrogen with 80 degree weather the pressure drop is 2.142318 PSIG.

Now that's with a constant flow of gas for the length of the pipe. So if we started with ~28 PSIG we would end up with a pressure guage reading of ~26 PSIG at the end of the pipe. Giving use slightly less than 280 fps. This gives us the solution of show much max volume you need at the lowest pressure possible. Volume being the volume of gas the 10 inch barrel consumes.

So the flip side is how little volume can we get away with at the highest pressure?

Since we can't provide a constant flow of gas for the entire length of the barrel, (well we can, but why stop there and be in-efficient; aka gas piggy), we need to figure out what fixed amount of volume of Nitrogen is needed at under what preesure will accelerate the ball.

[steveg]

Looking at pbjosh's table and concluding that 200in3*lb
is a close enough for our purposes number, and that the
typical maximum output pressure for a fixed air tank is
850psi

200in3*lb/850psi=.235 in3 not suprisingly about half
the volume of the mag at 400psi.

Given the anecdotal nature of pb's table, it is remarkably
constant from marker to marker

so what makes a shocker so wasteful and an impulse so
efficient ?