Velocity of a cap on a pressurized chamber 1

[BlaineW]

My inability to google the missing pieces has left me seeking help from the community.

I'm working on a concept project and unfortunately I can't get into too many details.

Imagine a scenario of a pressurized chamber that has a cap of some sort. Imagine it is suddenly released and allowed to shoot off under high pressure. Is there a non experimental way to determine what sort of velocity the cap would fly off at?

The initial force applied is easy enough to figure out, but it would drop very quickly as the pressure dies off, so how does one evaluate the dynamics?

What are some assumptions that should be made?
What equation(s) can be used to predict either the force on the cap as it relates to time/pressure drop, how does the pressure drop nover time?

I would appreciate any insight,

Thanks.

 

[btrueblood]

Equations would be those of fluid dynamics and/or gas dynamics. Grab your old fluids textbook and look up the equations for a fluid jet. You could assume (worst case) that the cap will achieve a velocity roughly equal to that of the fluid jet pushing it.

 

[drawoh]

BlaineW,

Are you trying to accurately model the dynamics of this thing, or are you doing enough work to show someone they have a bad idea?

JHG

 

[MiketheEngineer]

F = mv^2 at least for a few seconds. OR V^2 = F/m

 

[BlaineW]

Ugh...Fluid Dynamics, my worst subject. And that was 10 years ago, I think I may have burned that book. Looks like a trip to the library is in order.

drawoh,

No i'm not looking to get too detailed...it's more of an assessment of feasibility sort of thing (positive spin) but I would like to have some science behind it.

 

[dvd]

Force is usually mass x acceleration not mass x velocity x velocity. The units are wrong, kinetic energy has those units.

The cap sounds uncontrollable. Hard to reliably release. Lots of versions of pressure release actually restrain the cap so that it won't fly off and strike someone. If you are designing a launching system of some sort, you might be able to get a rough estimate by equating the potential energy of the cap with the kinetic energy released.

 

[drawoh]

BlaineW,

Will the cap be released deliberately in a controlled manner, or will this be an accident?

When a gun fires, the bullet accelerates down a barrel. There usually is rifling in the barrel to spin and stabilize the bullet. You should be able to find literature on this.

If the cap is released when something fractures, the fracture must be considered unpredictable. The mass, acceleration and direction will depend on how it fractured and on what remains attached to the cap, I should think.

I must admit, I am not the expert on this.

JHG

 

[btrueblood]

Blaine,

What is in the tank, i.e. gas or liquid? If liquid, will it flash (create vapor when allowed to expand to atmospheric pressure). Of the 3 cases, the liquid one is easy to figure out, gas less so, and flashing liquid toughest.

Can you give us an idea of what you are trying to prove, i.e. that the cap won't exceed some velocity, or that it will, or ...?

 

[berkshire]

one thing you have not described is whether or not the cap is flat, or does it have a flange like a screw cap. If it is like a screw cap, then you would have scenario like a spigot fired weapon as the cap departs.
B.E.

The good engineer does not need to memorize every formula; he just needs to know where he can find them when he needs them. Old professor

 

[rb1957]

the force decays with time, F(t), when the cap is released initially the pressure is acting of the cap ... F(0) = p*A. but it decays quickly as the pressure vents.

and of course, F(t) = m*a(t). if the acceleration was constant then a = ((v-0)*t)/2, you can integrate other forms.

then you have to worry about drag ...

 

[Cockroach]

This is basically the principle of a bomb, except tou have a directed pressure outlet. Force is therefore the pressure times the seal diameter times pi over four.

In theory you need to use gas dynamics, your situation falls under isenthalpic throttling. This situation is extremely common with valves, you are blowing down from one state to ambient using a known working fluid. So the boundary control is the wetted perimeter of the vessel and outlet. It is not steady state, you have mass loss, I.e. The pressurized working medium is allowed to escape to ambient state.

But you can use general dynamics a split second after rupture to obtain the required "muzzle velocity". There are some general equations in Ballistics that are handy, but thermodynamics of the working fluid is far superior. To answer your question, calculate the blow down period using the orifice equation and fluid properties state one to two. Then apply the impulse equation noting the assumption that force times blowdown time is 100% efficiently transferred to your cap of known mass. Once you Haney that initial velocity, the rest is just first year dynamics.

Cute problem, have fun with it!

 

[racookpe1978]

Well, you don't have a "gun barrel" effect going: Once the cap has left the immediate end of its "pipe" (or threaded connection) whatever gas pressure is left in the PV is still present inside the PV (but is bleeding off and will continue to bleed off through the hole) but it will NOT be able to continue to accelerate the cap.

So, break your problem into pieces:
1) Immediately before the hypothetical break, cap speed = 0, t=0, Pressure = something.
2) Acceleration of period of the cap, from t=0 to t=accel
3) Deceleration time (flight time) of the cap, from t=accel to t=impact

1) Your biggest unknown is making all the assumptions about how the cap is going to be "broken" loose from the PV
cleanly and instantaneously with no resistance to movement: champagne bottle "pops" won't happen in the real world of pressure walls tearing irregularly and caps being attached by welds and threaded connections.

2) Vol, Temp, Pressure of the gas from t=0 through t=accel will not vary. (Yes, they will change by a trivial amount since the cap has moved a few fractions of an inch -> ignore that little bitty change in volume.) Air resistance during this period can also be neglected, compared to the acceleration forces,because the cap is starting at zero velocity. You (I assume) know the mass of the cap, and so initial force = initial pressure x inside area of the cap. From that, figure out initial acceleration, then initial velocity.

This is because, to a very good assumption, in the very few microseconds of initial acceleration of the cap, speed is near zero, distance of travel is near zero, and so air resistance is negligible.

3) Once the cap has left the vessel, the internal pressure can't continue to push on the cap, so its speed (from (t=acell to t=impact) will slow down until it reaches terminal velocity (if falling down) or the ground or wall (if "aimed" towards some solid object.

 

[btrueblood]

"3) Once the cap has left the vessel, the internal pressure can't continue to push on the cap, "

I'd modify that statement slightly. If the PV contains blanket-pressurized liquid, the liquid jet could continue pushing the cap for many pipe diameters downstream, until the cap tumbles out of the path of the jet. For a gas, the jet from the broken pipe will continue pushing with very nearly the full stagnation pressure of the tank for at least 1/3 the pipe diameter, and likely up to a full pipe diameter of distance from the cap to the nozzle, and then decay to near zero several pipe diameters away (how many depends on the type of gas and upstream pressure).

 

[btrueblood]

Another case to consider is when the tank hold flashing liquid, and the pipe cap/nipple can conceivably break with some amount of held volume within the broken stub cap/pipe. Here the thrust generated by the held liquid flashing to vapor can be considered using the equations for rocket engines, F = mdot*Vexit, in addition to any push from the gas/liquid two-phase jet pushing from behind...

Did I mention it can get complicated?

 

[drawoh]

...

Did I mention it can get complicated?

I don't think you have accounted for the possible orientations of the cap as it flies away. Are the gasses acting normal to the face of a round disc, or are they acting on the edge of it? Perhaps the cap is spinning.

This is why I brought up guns. A lot of effort is made to control the orientation of the bullet as it leave the barrel. I don't see any control of the cap, unless they built it to work like a gun.

JHG

 

[SNORGY]

Based on what I see, dvd's approach would be the one I would take.

PE (chamber) = KE (cap)

assuming no "tear away" energy losses due to fracture (which will not be a valid assumption), and no velocity imparted to what the cap has broken away from.

Probably gets close enough.

Regards,

SNORGY.

 

[btrueblood]

Drawoh,

Yes, exactly. You would of course include the various breakaway models in the dynamics calculation (an exercize left to the reader). Assume a jet impinges on the cap, it almost doesn't matter (to within the precision of this discussion, which is order-of-magnitude only) if the cap tumbles, there will be a significant area of the cap seeing nearly complete recovery pressure up to a full pipe diameter away from the broken end of the pipe, assuming the break is more or less normal to the pipe axis. For a hinged failure, with the cap rotating before the pipe fully breaks, you have a whole 'nother set of calculations to run.

My exception is to the statement "Once the cap has left the vessel, the internal pressure can't continue to push on the cap", which as you have pointed out is somewhat in error. External and transition ballistics show that the base pressure on a bullet leaving a gun muzzle can have significant impact on its flight path.

 

[btrueblood]

"PE (chamber) = KE (cap)"

Snorgy, that's probably appropriate for a full vessel rupture (like the bottom of a compressed air tank falling out), but might be a tad overestimating the result for a smaller element like a pipe cap. Still, as an upper bound to the problem, it can't be beat.

 

[77JQX]

Like cockroach said, this is basically a bomb problem, only instead of a spherical pressure wave, we have a directed pressure wave. As an approximation, I think we can say the upper bound on the cap's velocity is the speed of sound of the escaping gas (if it is a gas in the vessel). And that would happen only if there was a "gun barrel" effect. Bomb shrapnel doesn't travel as far as bullets, does it?

I think the PE(chamber) = KE(cap) could has the potential to overstate the energy of the cap, since I can imagine the vessel still venting long after the cap is gone. I think cockroach has the answer you're looking for.

BTW, what keeps the rest of the vessel from becoming a rocket if the cap does let loose?

 

[SNORGY]

Thanks btrueblood.

The calculation of trajectory is far easier: simply find the most expensive thing in the area to hit and plot a parabola that intersects with precisely that point in space.

Regards,

SNORGY.