# Impulse and Momentum (another stupid question)

• posted

I'm struggling with a simple calculation, been through the books and can't get an answer I believe.

A cylinder of internal diameter one inch is closed at one end (ie a can with one end open) is plugged with a bung that weighs say 20grams.

If I create an impulse by suddenly releasing compressed air into the cylinder which causes the bung to be released from the cyclinder at 150PSI and assuming the impulse is of a duration of 50mS how do I calculate the change in momentum of the bung?

I know that momentum is mass x velocity and I know that the impulse is Pressure x time, but the dimensions don't match so there is a bit of extra math to do.

TIA

Steve

PS: I wouldn't have asked this question if that chap in Cherokee County hadn't blown up his cannon!

• posted

You can use Force = Pressure X Area to get the force on the bung.

BUT ...

There are (at least) two complications here. The first is that the force is applied only as long as the bung remains inside the cylinder. You can compute the initial portion of this trajectory with Acceleration = Force / Mass (1) and then "integrating" the acceleration along the length of the cylinder until the bung reaches the limit of the cylinder and the pressure starts to drop rapidly. You can assume that the pressure goes to zero instantaneously and then the projectile has whatever velocity it has when it reached the end of the cylinder. Some formulas of interest are velocity = Acceleration X time (2) distance = 1/2 x acceleration x time x time (3)

You can use equation 1 to compute the acceleration then equation 3 to compute the time the force acts on the projectile and finally equation 2 to compute its velocity at the end of the acceleration phase of the trajectory.

BUT ...

As soon as the bung starts to move there's an aerodynamic drag force impeding that motion and this is usually a function of not only its surface area but also its shape AND its velocity. It's a very complicated thing to even estimate this.

Norm

• posted

I may be wrong, but the chap that blew up his cannon was likely using smokeless powder, not black powder. That's a *huge* mistake, if so. Sorry, can't help you with your math problem. Way over my head.

Harold

• posted

Interesting question.

Well, actually, the bung will start moving long before the full pressure is reached (or not. A Quantum thing? or just not enough constraints?)

When the bung starts moving, the volumn behind it (the charge area) changes (increases), dropping the pressure behind it.

A better way to say it is the _rate_ of pressure increase drops?

ie: slower pressure rise because the charged volumn is increasing.

150 psi for 60 ms - but at what current (cubic somethings per time unit)? How big is the orifice?

If it is big enough to completely pressurize the entire length of the tube, (in 50 ms? way cool!) the bung will probably accellerate the full length of the tube.

Take friction losses into account and you can calculate the exit velocity.

If the orifice won't pass enough (CFM) to fully pressurize the entire tube, the bung, at some point, may have moved far enough (thus increasing the vol behind the bung) that the pressure behind it is actually BELOW ambient pressure - causing the bung to begin DE-cellerating!

Depending on gobs of factors, such as pressure, flow rate, impulse time, friction, mass of the bung, etc. there may be a range of solutions where the bung decelerates to a stop and rebounds back the way it came from, just _before_ it pops out of the open end of the tube.

• posted

On Thu, 3 Jun 2004 18:53:38 +0100, "Steve" vaguely proposed a theory ......and in reply I say!: remove ns from my header address to reply via email\

Almost impossible as described.

If you plug the bung into the open end so it just goes in, and hope that it seals well enough to let air pressure build up, then there is all sorts of trouble. Friction against the tube walls would be the biggest problem. The bung would receive force only as long as it stayed in the tube. So bung length would also affect the result.

Norm Dresner has given you the formulae.But that assumes a relatively loose fit in the tube, and that the bung is far rnough down the tube to have time to accelerate.

*******************************************************

Sometimes in a workplace you find snot on the wall of the toilet cubicles. You feel "What sort of twisted child would do this?"....the internet seems full of them. It's very sad

• posted

Another question is what is the mass of the cylinder (can)? You have given us the mass of the bung, but for all practical purposes, you will be accelerating the can as well as the bung -- in proportion to their masses -- unless you have the can anchored to something immovable. This complicates your calculations even more.

Good Luck, DoN.

• posted

The dimensions don't match because you need force (pressure * area) rather than pressure.

Other "yeah-buts" from other posters are valid, but in response to your question: 150 PSI on 1" dia is 117.8 lbf, times 50 mS is 5.89 lbf-sec or (Where did the 50 mS come from? Well, nevermind....)

The resulting velocity (with MathCAD doing the unit conversions I'd surely screw up) is 1310 meters/sec. Wow! That's right up there with a .30-06 round and well in excess of sonic velocity, so that air would have to be quite hot to maintain pressure over 50 mS of acceleration. One hell of a popgun!

Perhaps your 50 mS assumption merits review?

Why don't you just try it and see how far the bung goes? Even plastic water pipe will withstand 150 PSI. It takes about 150 PSI to blow a disposable PET Coke bottle, this from experimental data.

• posted

For simplicity, I shall assume no losses and constant pressure in the cylinder. Reality would be different but the principles remain the same.

Part of your problem is that impusle is force x time not pressure x time.

1" diameter cylinder has area 0.785 sq.in. 150 psi in that are gives 118 lbs force. 20 grams is .0441 lbs. Ft(Impulse) = dMV(change in momentum). So Vf = 118*.05/(.0441÷32.2) note: mass = weight/g Vf = 4308 feet per second 118 lbs acting on a mass of .0441/32.2 (= .00137) gives an acceleration of 86131 feet/sec^2. Using V^2 = 2aS, S = V^2/2A = 108 feet. Pretty long barrel! Of course, if you tried it, you could hardly ignore aerodynamic effects and mechanical friction not to mention pressure drop 'cause the air couldn't fill the tube fast enough. i.e. Simplifying assumptions would be nonsense.

Now, in a (say) 6"(0.5') barrel, it might be easier to use energy rather than momentum. work(W) = FS(force x distance) = .5*118 = 59 foot pounds. Since kinetic energy is 0.5MV^2, 59 = (0.5)(.0441)V^2. Solving for V gives V = Sqrt(2W/M) = 51.7fps. Since V=At, t = 51.7/86131 = 0.6 msec. As a check on arithmetic, I plugged 0.6msec into the above equation involving impulse and get a final velocity of (118 x 0.0006)÷(.0441÷32.2) = 51.7fps

Hope that helps.

Ted

• posted

The subject is called "internal ballistics" and a Google search on the phrase will bring up about 1500 hits, most of which seem to be on the subject. You will have to figure how a jolt of compressed gas introduced via a port differs from the burning of propellant, but either you know what your system can provide in the way of pressure over time, or you are asking an unanswerable question.

If you can't quantify your internal pressure, you still may have a way to measure the results. Take a page from the same people (firearms experimenters) and fire your bung into a block of modeling clay (compliant and sticky) suspended from strings as a pendulum. Measure how far the pendulum swings (have it drag a thread through a hole and measure the slack after the pendulum stops swinging). Now do a search on the phrase "ballistic pendulum" (2800 hits) and get the math needed to calculate either the momentum or the kinetic energy (you may find that one or the other suits your needs better).

-- --Pete "Peter W. Meek"

• posted

Thanks for the help. I've started plug all of this into excel to make it easy to play with the numbers and can now see what is going on. I can't estimate resistance either from the cylinder walls or from the air so I've ignored them. What I do know is the size of the inlet port and the time it takes to open and close.

Next step is to try this out and measure the momentum of the bung and see how far of course my simple math is.

Best Regards

Steve

• posted

You are getting into the design of air cannon, a difficult subject involving shock waves traveling back and forth inside the cannon and the velocity of sound in your working fluid, air in this case, enters into the problem. Bob Meade

• posted

Would that make the cylinder a bunghole?

• posted

That's a dull knife but if it's the best you have, ...

Keep your final velocity below the speed of sound (about 1000 ft/sec) or all bets are off. Things change pretty drastically at that point.

Ted

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