How quick will this pressured vessel vent

[bertyboy]

I asked a question recently about how
quick this pressured vessel will vent.
I have attached a sketch :-
Can someone show via some formula how
long it will take to vent, or at lest
drop half its pressure ?

Thanks a lot !

 

[ivymike]

here's an idea... haven't thought about it terribly much.

try to answer the question "how fast do I have to lower this thing to keep the water inside _X_ feet below the surface of the water outside"

Take 2 ft as an example... how fast do you have to lower it to keep the water inside 2ft below the water outside?

* 2ft gives you a pressure (24in)...
* here are some orifice flow calcs - looks like 0.65 might be a good coefficient

http://www.jegasho.net/rotron pages/orifice flow calculation.htm

* pressure is 24in water
* k is 4005 if using in water
* V at the orifice is 213fps or 145mph (sounds high)
* V of the can is V_orifice * (Dorifice/Dcan)^2 or 2.3ft/sec

using 4ft water gives 3.2ft/sec
using 8ft water gives 4.5ft/sec

 

[waross]

Some more interesting questions may be: When the vessel submerges;
1 Will it vent fast enough to avoid implosion.
2 At what depth will it implode.

Based on a real world event. Don't ask.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[Cockroach]

This is a thermodynamic problem, you would need to use either the air tables for given internal state of compressed air and then isenthalpic throttling assuming adiabatic conditions to ambient.

I've actually done this with valves and used the nitrogen tables to tie down the initial and end states of air. Remember that air is 78% nitrogen, so the nitrogen tables lend good support to your calculation.

I went from 1440 psi at +30C shop conditions to 14.7 psi finding heavy frosting of the vent valve and icing at the opening. Temperature drop was -41C, effectively the jet stream was at -11C near the opening thus explaining the frost effect. The jet stream velocity was not quite sonic, somewhere around 75% sonic velocity for ambient thermal conditions.

Give it a try, but be sure to wear ear plugs!

Kenneth J Hueston, PEng
Principal
Sturni-Hueston Engineering Inc
Edmonton, Alberta Canada

 

[gerhardl]

You can make a fair and simple approximation, if you can find the approximate kv- value for your vent.

Looking at table for ball valves I find DN500 (PN10) BV Kv=42500, DN600 Kv=62000, approximate middle for 560mm (your vent opening if figure is correctly read) -> kv 52000. This would be the maximum value. Sharp edges and turbulence would probably cut this down, but let us say not longer down than for a DN600 double eccentric BFL valve with approximate Kv=37.000.

Your driving force at start is the hydraulic pressure from sea below = H = 2,1m = 0,21 bar.

Your volume of air to escape in Norm cubic meter to equalize the pressure inside and out is equal to the volume earlier replaced by water : Height 8,1m - L = (8,1 - 6,69)m = 1,41m Volume to escape: 29m3

Now for supercritical air flow (P2 < P1/2) we have QN = 13,36 x P1 x kv.

13,36 x 0,21 x 52000 = 145 891 m3/h through the escape opening, and 13,36 x 0,21 x 37000 = 103 807 m3/h.

For 29m3 to escape from results above gives then between 0,74 to 1,00 seconds, at an air velocity through the 560mm diameter opening of between 164 and 117 m/s, which seems reasonable as a forced velocity. Speed of sound in air: 343m/s.

Even if we question this approximation and says as an example that we will never reach a higher air velocity than 30m/s (which is less than 10% of speed of sound, and in my opinion far below the true velocity in this case) you will land at under 4 seconds!

(Note: I have not checked your height calculations)

 

[saplanti]

I guess all the approaches above uses the same assumption that the lower part of the cylinder is a lot heavier than the upper part that connected onto the hook and the motion is always vertically.

My guess is this: The upper part of the cylinder is heavier than the lower part and when it is lowered under the air damping by orifice effect the cylinder will go the the horizontal position first and get fully filled with water. Therefore, under the gravity and hook control it will go deeper quickly.

How about that.

Kind regards.

Ibrahim Demir

 

[gerhardl]

Bertyboy:

My assumption was that you already had lowered the cylinder to the given condition in your sketch, and that in this 'static' situation the vent cover suddenly broke, but with cylinder still supported by crane and chain in the given position.

From the others it seems unclear what your 'real' problem is: how to obtain a controlled lowering condition, controlled venting by lowering, or other.

Please comment.

 

[saplanti]

Bertyboy,

The upper hole helps lowering cylinder vertically in the water. While the cylinder is lowered slowly water displaces through the hole. I guess the selection of lowering velocity is the problem and the hole acts as an orifice.

Lowering velocity and area of the cylinder gives the flow rate. However you need to do an orifice calculation to find the chocking flow rate through the orifice for the ultimate flow rate which will give you the maximum lowering velocity.

I guess this is the principle used. To me my previous post is still adequate assumption at the surface of the water to lower into the water fast. Everything still depends on the mass balance of the system and the contact speed to the water. Otherwise the air trapped inside will act like a damper the cylinder will bounce and tipped over.

In case the lower part is the heaviest part of the cylinder and the cylinder is lowered in to the water slower enough you may still do the similar orifice calculation for air and find the lowering velocity for air displacement.

Regards,

Ibrahim Demir