Pounds per Square Inch

[Walking Stick]

(If this should be posted in The Workshop, please move it there; I just thought I might get more help in this board.)

Here's the situation.
You have 500 pounds per square inch of regulated gas going into a valve.

The valve is opened when a plunger, connected to a seat with an area of 0.25", is depressed.
How many pounds of force is require to depress the plunger?

What is the answer?
Is the 500 pounds per square inch a constant?

Is it a ratio, like 500:1, where it would then be 125:0.25?

Is it an inverse relationship? A 4x smaller area requires 4x greater pressure?

Please help. I do not know fluid dynamics that well. Thank you.

[MANN]

It is alot more complicated than I want to get into right now, but think about a bottle jack picking up a car. You are not exerting 2000 lbs of force to pick up the car, just as the the reg valve is not opened with 500 psi.

Just fyi 500 psi that is forced on .25 diameter square inch would be 24.53125 psi

I can get into more detail if you are really wanting to understand, but right now Im going out to eat.

[echoback]

I'd like to know more. I like messing around with thermodynamics and stuff, so any information you can give would be appreciated.

What other variables are taken into consideration for the above equation? I got 24.5437 psi from a ratio of areas.

[Walking Stick]

If you do find the time, YES, it would be greatly appreciated and invaluable if you could post the details (formulas, calculations, factors, etc.) of how to figure this out and truly understand it. Thank you.

[jenarelJAM]

Alright, let me see if I can take a stab at this.

You've got 500 psi(hereafter referred to as lb/in^2) on a pin that has a .25" diameter. All force is normal(perpendicular) to the surface, so all you need to think about is that one side of the pin.

.25" diameter means a surface area of pi*(.25/2)^2=~ .049 in^2(square inches)

Now, you've got 500 lb/in^2 * .049 in^2 = 24.53 lb.

My answer is slightly different because I used 3.14 for pi and rounded early, but it doesn't make a difference.

That number (24.53 lb) is the constant force the air pressure is putting on that pin. If you want to get really fancy, you can add in spring tension too, which will increase the constant force on that pin. The equation is F=kx (force=constant*distance)

However, that still probably won't help you a ton, since, even though you could calculate how much energy it would take to open the valve and to keep it open a certain amount of time before it closed again, <i>if the pressures remained constant</i>, the problem is that as soon as the valve opens, the pressures change, and you no longer have a constant force pressing on that pin. You've got the hammer spring, you've got the valve spring in the opposite direction, you've got variable air pressure behind the valve, and you've got variable air pressure in front of the valve escaping through the gun at the same time.

If I were to build a paintball gun from scratch, I would do guess and check. But that's just me. I'm 17 and a senior in high school. Maybe when I get to college I'll learn how to deal with varying pressures, but right now, I have no idea...

Anyways, that's where those numbers came from.

If you tell me a bit more about what you're trying to do with that number, I can try to run some more equations for you.

/and could people please use correct units? It's really confusing to see people refer to pounds, a unit of force, as psi, a unit of pressure, if you don't know what they're talking about.

[sniper1rfa]

At first it's simple.


Its pounds per square inch. So if it is acting on one square inch, it exerts 500 pounds.

Acts on 2 square inches, it exerts 1000 pounds. Half a square inch, 250 pounds.


Of course, the second you open the valve, the pressures change and it is a moot point. Mathematically modelling a poppet valve, like a cocker, is stupendously hard. Like, graduate work type hard. Trust me, i tried.

[echoback]

But as far as paintball is concerned, stating that "the input pressure is constant" is perfectly reasonable, and you won't have to consider the changing pressure inside the valve. While the valve may not put out the exact amount of air that you calculated, the velocity adjustments (mainspring, dwell, etc.) can easily compensate for this.

[echoback]

By the way, can anyone verify/disprove my work?

Chamber A and chamber B are connected by an orifice 0.25in in diameter. If the pressure in chamber A is 500psi, the flow will be limited to 24.54psi (assuming chamber A is constantly supplied, i.e. volume doesn't matter).

In practical terms:
If a spool valve has 500psi in it, and the exhaust port is 0.25in wide, the output will be limited to 25.54psi, right?

[Walking Stick]

Hey, thanks for the help! I thought it was that simple, but then I started second guessing myself and making all these variables. I did make a few typing mistakes and should have been more specific. But I just needed the numbers and calculations so I could see where I was in fault.
This is actually for the high pressure Tippmann CVX (poppet) valve. I just roughed an estimate for the valve seat area (which is not nice and flat >>http://www.tippmannparts.com/index.cfm?fuseaction=catalog.prodInfo&productID=43 8&categoryID=11<<) based off the valve plug's diameter of 0.5 in. That I then roughed into an area of 0,25 SQ. in. (which I erroneously told you was 0.25 in.) for simplicity's sake (for whomever calculating and for me following).
The pin I need to depress a distance of 9/32", which is also opposed by the valve spring. I have three valves and have found the pounds required to depress the pin on each vary. They have been measured as 7.8 lbs., 8.6 lbs. and 15.2 lbs.
Thanks again. Now I'll be able to move on in my pneu project! (Yes, pun intended.)

[Pneumagger]

By the way, can anyone verify/disprove my work?

Chamber A and chamber B are connected by an orifice 0.25in in diameter. If the pressure in chamber A is 500psi, the flow will be limited to 24.54psi (assuming chamber A is constantly supplied, i.e. volume doesn't matter).

In practical terms:
If a spool valve has 500psi in it, and the exhaust port is 0.25in wide, the output will be limited to 25.54psi, right?

The main problem is assuming chamber A stays supplied once open. The only way you can assume that is by having 2 very large (read as: infinate) reservoirs of known pressure feeding through a small hole.

And once you get that set up then you calculate the flowrate (volume/second) through a specified opening... not the pressure. Pressure doesn't flow... it pushes.

[Spider-TW]

The main problem is assuming chamber A stays supplied once open. The only way you can assume that is by having 2 very large (read as: infinate) reservoirs of known pressure feeding through a small hole.

And once you get that set up then you calculate the flowrate (volume/second) through a specified opening... not the pressure. Pressure doesn't flow... it pushes.

With a balanced spool valve you can assume it will stay open, but as you said echo, your A chamber is infinite, so the pressure will not change at all. The ~24 pounds of force is what is required to lift a poppet valve off the seat (from the inside), or the force that is needed to seal the same diameter hole from the outside. Once your typical marker poppet is open, the rapid flow of gas around the valve is a major force as it sucks the poppet back down, in addition to the static pressures you have on each side.

Spool valves are a lot like the 1/4 turn plug valves on the flatline regulators in that the only force you have to overcome is the friction from the seals. The problem is that the friction is due to the high pressure pushing the valve plug against the valve body, making it hard to twist (plug) or slide (spool).

You cannot really directly control flow in a compressible substance. The way flow is usually controlled in gases is to put a regulator across a small (specified ) opening. Google "constant flow relay" and you should be able to find a diagram.

If you are tinkering with poppets, keep in mind that the standard blowback design passes gas down the stem against the hammer. This is one of the points of the flats on the sides of the 'plunger', or stem, and the point of the hammer oring. Low pressure blowback valves have larger holes for more air, but they really need a stem that passes less gas back to the hammer because you need to allow the valve to be open for a longer time (push less on the hammer which is holding the valve open).

Considering that most markers get about 1000 rounds out of a 68cuin/4500psi bottle (gauge pressure) and it goes from 0 to 300 fps in one foot (barrel length), a marker valve needs to pass about 20 cu. in. of gas (at atmospheric pressure 14.7, or 15 psi while im handwaving) in about 6 to 7 milliseconds. That's your end result that you probably can't avoid. However, the money is in your new gas system, which you have already figured out paintball players will pay for

I suspect that some of the 'efficiency' of low pressure guns is that they accelerate the ball further down the barrel and you don't loose velocity like you do on a high pressure gun when they both have 12 or 14 inch barrels. I think AGD (Tom K.) said (Historic Posts?) that most guns reach 280 fps in the first 8 inches. This means that to get 280 at the muzzle, you're hitting something higher while the ball is in the barrel (on a high pressure gun).

[echoback]

And once you get that set up then you calculate the flowrate (volume/second) through a specified opening... not the pressure. Pressure doesn't flow... it pushes.

Gas only flows because it pushes. Bring a chamber down to 14.7psia, and you will have zero flow down the barrel, because the chamber is the same pressure as the air outside. You only get flow when you have some pressure difference between two areas.

In order to calculate the flow rate you need to know the velocity of the escaping gas, which is calculated as v = sqrt( 2p / d ), where 'p' equals the pressure and 'd' equals the density. So the flow rate does, in fact, change based on the pressure inside the chamber. A chamber at higher pressure will force more air through the same orifice than a chamber with lower pressure, basically.

What I was trying to find out is if I did my calculations correctly when finding the pressure of the gas escaping from the orifice. With a chamber at 500psi and an orifice of 0.049in^2, the escaping pressure should be 24.5psi through a simple ratio. I don't know if there is a more complex equation for "pressure drop through an orifice", but I've just been using p2 = p1 * A, where 'p2' is the orifice pressure (lbf/in^2), 'p1' is the chamber pressure (lbf/in^2), and 'A' is the orifice area (in^2). I figure it must be close, since it's true in a practical sense: A small hole in a balloon puts out less pressure than a big hole.

By the way, here is my work for the flow rate of "500psi through 0.049in^2". As you can see, pressure is very important to the equation:
Density of air:
1.2kg/m^3 @ 15psia @ 20* C

Gas velocity:
v = sqrt( 2p / d )

v: Velocity (m/s)
p: Pressure (N/m^2)
d: Density (kg/m^3)

1psi = 6900N/m^2

v = sqrt( (2 * 3450000N/m^2) / 40kgm^3 )
:. v = 415.3312m/s

Flow rate:
Q = A * v

Q: Flow (mm^3/s)
A: Area (mm^2)
v: Velocity (mm/s)

1in = 2.54mm

Q = 0.3167mm^2 * 41533.12mm/s
:. Q = 13153.53mm^3/s

[Spider-TW]

Actually the flow through an orifice has much more to do with the ratio of the bore of the pipe or chamber and the diameter of the hole (beta ratio). Check the ASME orifice sizing standards (library). Unfortunately, valves do not qualify as orifi (?) and those calculations are empirical (derived from experimental data). Check the fisher valve website.

[echoback]

... those calculations are empirical (derived from experimental data).

That's what I said. 500psi (lbf/in^2) in a chamber, released through a 0.049in^2 orifice, should drop down to ~25psi. I'm taking this stuff from accepted fluid dynamics equations, and the numbers are from our above example.

Over the past hour or so I've been getting back in touch with the Venturi effect, which governs pressure through an orifice, so I can correctly calculate the orifice pressure... Forgot about that Venturi guy. I'll try to get a more correct calculation for the pressure in a bit.

[edit] We've debated enough that somebody should just go ahead and test this...

[phat4life07]

F=kx (force=constant*distance)

just to make the correction, spring tension is K=(1/2)kx^2 (kinetic = one half * spring tension * compressed distance of spring spared.)



your question is not very hard to answer at all if you understand how to draw a force diagram, and couple force equations.

here are few equaltions that you can use (depends on how much info you know), since there are many ways to do it, and those should cover 90% of the ways to do it:

EF=MA (total force = mass * accelaration) the units of your answer will be in newtons (N)
K=(1/2)kx^2 answer is in jouls
J=deltaP/deltaT (impluse = change in momentum over change in time) answers will be in kilogram meters per second sqared (kgm/s^2)
Pi=Pf (initial momentum = final momentun) units will be in kilogram meters per second (kgm/s)
Ki=Kf (initail kinetic = final kinetic) *this equation only apply to elastic explosions/collision, or you can also use this equation to find ennergy) units will be in jouls

you will also need couple of the velocity/accelaration/position equations, example: Vf=Vi+at

and to find the EXACT force to active the valve pushing the pin you need to know the coefficient of friction of the pin on orings, wall (there are two different kinds, static and kinetic) and then you can use the normal force (MG) times the coefficient of friction and then you will get your answer.