Speed of a cap when released from a pressure vessel

[Benjamin2019]

Hi,
Pressure vessels have caps on top of them. These caps are secured with tri clamps and if you need to remove them you have to make sure that the tank's pressure is released. If the tank is under pressure (say 10psi) and an operator removes the cap (A is known), what is the speed of the cap?

 

[BvU]

Hello Benjamin, ##\quad## ##\quad## !

For starters you can assume an acceleration of ## A\Delta p /m## with ##m## the mass of the cap.
To estimate final speed you have to estimate how long the ##\Delta p## lasts effectively. That isn't so clear cut and requires more detailed knowledge.
Manhole covers https://www.telegraph.co.uk/news/worldnews/northamerica/usa/11534365/Explosion-shoots-manhole-cover-100-feet-into-air.html are a well known source of fatal accidents

 

[Benjamin2019]

Thank you for the quick reply. If Δt is known, what is the formula to find the speed?

 

[profbuxton]

Pressure vessels "cap" are not fitted to the outside, they are usually oval shaped and fitted internally with the clamps holding them up from inside against the vessel opening. So even if you undo the clamps with pressure inside the pressure will hold the "cap' in place till all pressure is released(by valve open etc).

 

[SirLollington]

Like others said, it really depends on the specific system. There are many factors that influence the final velocity of the cap, such as the mass of the cap, the surface area of the cap, how large the pressure vessel is and the overall geometry around the cap. On top of that, if the pressure is allowed to escape past the cap while it's accelerating (which I assume it would be), or the cap reaches the speed of sound in the pressurized gas, the process becomes more difficult to model.

I'm not entirely sure how sharply the pressure decreases as soon as some of the gas can escape, but for a rough estimate, we can assume that the pressure decreases linearly to 0 over Δt, which you've said is known.

The initial force on the cap is based on it's cross-sectional area and the pressure in the vessel. The formula for that is ##F = p * π * r^2##, where p is the pressure and r is the radius of the cap. The force also decreases linearly to 0 over Δt. Because the decrease in force is linear, this is equivalent to half of the initial force being applied the entirety of that time.

To get the velocity, all we have to do then is get the acceleration and multiply it by the amount of time the force is applied for: ##V = Δt * F/M##

Putting all that together, we get:
$$v = Δt * \frac {p * π * r^2} {2M},$$
where v is the final velocity, p is the pressure, r is the radius of the cap, and M is the mass of the cap.

Assuming a pressure of 10 psi (which is roughly 68.95 kPa) and a cap that weighs 10 kg and has a diameter of 50 cm, and assuming the pressure dissipates entirely in 50 ms, we get:

$$v = 0.05 * \frac {68500 * π * 0.5^2} {2 * 10} = 0.05 * \frac { 53772.5 } {2 * 10} = 134.43125 m/s ≅ 441 ft/s$$

Note that this doesn't consider other factors such as that the cap will likely begin spinning immediately, which both releases the pressure faster and decreases the affected surface area. I also don't know how realistic the values I picked are. :^)

To give you a better answer, I'd need to know more about the specific case you're working with.