Pressure Vessel Failure

The initial speed is 0 whit F (force) equal to P x area. When the bolts shear, between plate and tank is forming a small gap that is increasing, as the plate is moving up. Because of that, the pressure over the bottom of the plate goes lower, depending of the conformation of the surrounding area, initial pressure, gas viscosity (temperature), and tank volume but is not 0. You calculate P as function of distance from the initial position and integrate from D = 0 (distance where there is no gap) to the position where P = 1 (atmospheric pressure). That multiplied with plate surface minus (Weight x D) will give you the energy that the plate has when only gravity is acting over it from which you determine the speed. The gas flow problem is very tricky because it depends of the surfaces that surround the "discharge".

Thank you for actually addressing the posted question. Do you have any other suggestions on how to calculate Pressure as a function of gap (distance between the top of the cylindrical vessel and the underside of the plate which is moving upward)?

When the plate leaves the top of the vessel, the pressure beneath it is atmosphere. Your pressure energy from inside the vessel has been converted to velocity. You need to find velocity beneath the plate instead of pressure.

I agree with your statement. The unknown is how far the plate moves up before the pressure has dropped to atmospheric. I need to know this distance to calculate velocity. The reason for interest in this calculation is to make sure that there was no other catalyst(s) involved in this failure.

I don't know of any easy hand equation to get what you're looking for. If you use computational fluid dynamics software, you can find the local pressures that are decreasing and changing to velocities in various directions.

As a system, though, you would have to assume it's just like a pipe that vents to atmosphere or a jet engine exit. At the point of the exit, the pressure is all atmosphere. The reason there is still a force on the plate is the pressure has been converted to velocity.

What I'm trying to say is that as soon as you get an air gap between the plate and vessel, you're pressure against the plate is atmosphere.

If you're looking at this from an environmental standpoint, did you still have pressure going into the vessel when the top popped off? If so I guess you could treat this as a hole in a pressurized vessel or something.

All depends of what is around the recipient. Think that if it is like a bullet in a barrel is different then when the volume around the gap is free of obstacles. In the last case, consider that just after all the bolts shears in the same instant, (very unlikely), moment at which the pressure in the recipient have reached 90 psi, the distance between the lead and the cylinder become dD. Now, we apply some engineering simplification and consider that the fluid is perfect, the pressure is the same in every point in the recipient, including the gap surface (both things untrue, as it comes in place the viscosity), and is an isotermic transformation (very untrue). Outside the gap surface, the pressure is atmospheric pressure at all times (untrue). On the surface of the gap, which is s1 = dD x pi x 59.5, the gas has instantly the speed V caused by its former potential energy given by pressure. We have to introduce another variable, time t with the relation between them, dD = a x dt/2, a (acceleration) = [(P x S plate)/plate mass -g]. The pressure in the recipient decreases adiabatic (almost), but to simplify, we consider it of constant temperature. The gas goes out of the volume at a rate of GAS 1 mass = dt x V x s1 x (gas gravity at 1 atm). The pressure that remain in the recipient is P = (GAS mass - GAS1 mass) x H cylinder / (GAS mass x (H + dD), in which GAS mass is determined from initial condition. Put boundary conditions, replace, etc

GA

But can you please answer the OP's request for calculating the P as afunction of the distance. I think I would also like to know.

Thanks

I am also interested to know but seems like the folks are going in another direction. Please some one answer the question seriously so we the unknowing folks can learn from y'all learned ones.

Regards;

Nadeem

<......It is not designed as a pressure vessel........Any advice.....would be greatly appreciated.....>

Get as far away from the remainder of the installation as possible and get others as far away from the remainder of the installation as possible before someone gets hurt/killed.

In the USA it is required, by law, that the owner of a facility perform what is called a HAZOP anaysis.

In this analysis, "what if" scenarios are proposed and evaluated by competent, experienced people.

Sometimes, insurance people also review the HAZOP.

When someone gets hurt or killed, the first thing the govenment investigation team requires is "as-built" PIDs....certified by a PE.

About this time, the MBA plant owner moves all of his money offshore and forms an LLC.

The PE is hunted down ( his professional seal on the PIDs is valid forever) removed from the assisted living home and then sent to jail.

The plant owner then lives in the ropics with his new young girl friend...

Questions:

Was there a relief valve protecting the tank or vessel ?

If not, why not ?????????

You're thinking the same way I am. (why don't we ever get pictures of these things?) I don't care if it's 2 PSI or 100 PSI it should have some type of relief, pressure and possibly temperature. Plumbed to a "tee" so if it goes off it does take off like a rocket corkscrewing all over the place.

I'm not sure why he's focused on calculating how far the lid should have gone? There's very little chance all bolts failed at the exact moment. 1 or 2 had to go first allowing pressure to begin escaping. The actual pressure behind the acting plate would have been less than the highest (90 PSI) reached as some would have been released until all the bolts failed. Again, I'm not sure why the focus on the distance the lid would travel. The answer is "too far" regardless of it's actual distance. Add the proper safety devices and the lid will never become a projectile.

How about a more interesting and less bogus challenge than this? Like how many people you could kill by this criminal design. Hints: you need bolt material stress figure, vassel thickness and material and some calculations on human presence and movement near it. Think many guys in this forum could help you with these calculations. S.M.

The biggest flaw in design is non provision of safety relief valve. Top plate is like flanged dummy of a 59.5" pipe. If you design a flanged dummy for 59.5" pipe for 90 psig pressure will be very high thickness. In your case, it appears plate thickness is less (designed for 2.7 psig), which has bulged at high pressure due to large diameter and caused failure.

The solution to the problem as presented requires some assumptions. However no assumptions are necessary to calculate the stress on each bolt at failure. Each 5/8-11 bolt would be stressed to 97ksi at 90psi internal pressure. This is clearly more than a mild steel bolt can withstand. If you can find the cover, I suggest you double the number of bolts or use high strength bolts. Better yet, install a safety valve or rupture disc set at about 10 psi.

This is a ballistic flight with vertical launch.

You need to decide on the initial velocity which depends on the time the accelerating force works on the lid before the pressure dissipates.

For curiosity, if you consider Vo might be ~100"/sec

And then Time of flight to the highest point = 100/386.32 = 0.26 sec

And then the maximum height = 0.5xVo^2 / 386.32 = 12.94" ~13"

The initial velocity is the matter here and needs to be calculated better than estimated. I have not included any drag.

You know I haven't spent much time thinking about this problem on a calculation level, but I guess you could use the impulse-momentum equation to find the initial velocity.

Ft=mv

F=P/A

t=guess...maybe 0.1 seconds?

m=mass of the lid.

The initial speed can be obtained if we consider that the bursting will be instantenous and therefore, no acceleration takes place after ( approximation that is near enough...?). In this case, the energy produced, will be converted to kinetic and:

E = 1/2 x m x V² = F x V since the distance travel is in inches/sec, therefore, V = Vo here.

from the above, Vo = 2 x F / m and F = P x Area (I think that in your submission you wrote F = P/A ... typing error?)

with the given data: F = 90 x 589530 lbf, and m = 2000 lb, then Vo = 289.5" / sec

With your formulation we would have got Vo = 14.47 "/sec which will produce a lift of 0.27" ---> not feasible(?), while the above Vo will produce 9 feet approx. (of course, this might be excessive because the drag has not been worked in but ??)

Maybe the OP can tell us how far the lid did travel in the accident that he is talking about ??

This needs more imputs from someone better experienced in the matter.

"from the above, Vo = 2 x F / m and F = P x Area (I think that in your submission you wrote F = P/A ... typing error?)"

Too early in the morning error not typing lol.

I think you made a mistake too. Energy is equal to work or F x d not F x V.

Regarding the Vo that you got from my equation, is that based on my guess for t? My t guess was just that (don't know if you could even call it an educated guess). Like I said though, I haven't taken much time to think about the math here. Just throwing out little things to brainstorm.

Yes, I took your t = 0.01sec.

F x d true, but d in a unit of time, will be distance /t which is a speed, or:

V = d/t and replacing, d/t = 2 x F / mass

Since Speed V is distance in Unit of time, which is 1 sec, then for this purpose, we can write F x d = F x V in one sec, even though the action time is unknown.

I hope I am not too wrong.

Ballistic equation of mouvement:

For a vertical Launching: H = -1/2 * g * t² + Vo * t (1)

From this we can obtain the equation of the Speed by getting the derivative:

H' = V =- 2/2 * g * t + Vo ===> V = g * t (2)

By taking V = 0 at the maximum height: Vo = g * t ==> tmax = Vo/g .

Replacing t by tmax in equation (1) we get Hmax.

If we can obtain Vo, then we can get the total time and the maximum height

For Vo:

The total energy available at the bursting moment:

E = 1/2 * m * Vo²

Also, E = F * d were d is the working distance.

But d = Vo here Instantaneous sine V = d/t for t = 1sec

Therefore, F * Vo = 1/2 * m * Vo²

And : Vo = 2 * F / m (3)

For the purpose of the OP Data: Vo = 2 * 289529.2 lbf / 2000lb = 289.5292 Inches/sec

Tmax = 289.5292 / 386.2197 = 0.749649 sec (tmax ~ 0.75 sec)

Hmax = ½ * 386.2197 * 0.749649² + 289.5292 * 0.749649 = 108.5226 inches Or 9.04355 feet (Hmax ~9 ft)