Assume the hole in the valve body under the burst disk is a 1/8" hole. That works out to a surface are of about 1/100 square inches.
3000 Lbs per square inch divided by 100 only exerts a force of 30 Lbs for a fraction of a second before the pressure drops. Since the tank weighs about 30lbs I find it hard to imagine that it went airborn without sliding and impacting something else before it was up in the air losing energy the whole time.
Put the tank on a table, have it fall off because its spinning and when the valve hits a sharp corner it could fracture. The hole through the now missing top part of the valve is less than 1/4" and the gas now flowing through the 1/4" hole could cause 120lbs of force, which is quite a bit - I'm not sure that I can punch a sandbag with 120 lbs of force.
That kind of force can destroy a dryboard wall, cause very litte damage to a hollow cinderblock wall but appearantly enough force to kill an unprepared firefighter.
Sad, not sure what we'll learn from this as divers, but may he RIP.
Michael
Hmmm.
Pi (R**2) yields the area of a circle. 1/8" is .125".
3.14(.125 * .125) = .049 sq. in.
.049 sq in * 3000 PSI yields 147 lbs of thrust*. When the tank has emptied by 1/3, there is still nearly 100 lbs of thrust. When there's 1000 PSI left, there's still nearly 50 lbs of thrust. That takes at least several seconds.
147 lbs of thrust, at the end of a tank, perpendicular to its axis, *will* make the tank fall over, where if unrestrained it will roll until the thrust is about parallel with the floor, upon which it will spin up *very* quickly on a smooth surface. Like, say, smooth concrete. Or, as I think I posted earlier in the thread, a tiled floor. Clearly, it is likely to stay on the floor unless it hits something, but if it does hit something it is not at all hard for me, at least, to imagine it bouncing into the air with lots and lots of stored rotational energy just waiting for whatever will happen when it hits the next solid object.
Am I missing something somewhere?
* It's actually less than this, most likely, because the flow through the hole is imperfect, with edge effects that reduce the effective "nozzle" size, but this seems like a reasonable first approximation. Also, if the thrust is reduced, it's because the flow is reduced, which means the thrust continues for a longer time. Either way, that sucker is spinning fast on the floor.