Pressure Vessel getting punctured with small hole - Trying to find rate of pressure release

[gn87man]

I have a pressure vessel with a known pressure.

The pressure vessel is made from 7050 aluminum, 0.125" sheet thickness.

The pressure vessel is suddenly punctured, with a known hole size.

I am trying to determine the rate of pressure decrease after the puncture occurs.

I would like to plot the pressure decrease on one axis and the time on the other axis for a plot diagram.

Could someone please provide me with the equation(s) needed to determine the pressure decrease due to the known hole size in the pressure vessel in relation to the time (sec) lapse from the moment of puncture?

Thank you very much in advance for any help.

 

[zdas04]

The FAQ might be a place to start. You might look at faq378-1864. Don't let the title derail you.

David Simpson, PE
MuleShoe Engineering

In questions of science, the authority of a thousand is not worth the humble reasoning of a single individual. Galileo Galilei, Italian Physicist

 

[gn87man]

zdas04, thanks so much! I will check out the FAQ you posted.

Have a great day!

 

[gn87man]

Hello zdas04,

Thanks so much for posting the FAQ above. I have reviewed the FAQ, but I have a few questions...

The flow rate is given in MSCF (thousand standard cubic feet).

I am trying to find the remaining pressure in the vessel "x" amount of time after the vessel is punctured.

I figure that I need to somehow convert the MSCF (from the FAQ article) into psi/sec.

Any ideas?

Thanks so much!

 

[racookpe1978]

Do you have to ask how to do that conversion? What classes have you finished in the gas laws?

By the way, what is the size of your pressure vessel? What is the initial pressure? Homework assignments are not permitted in this forum.

 

[gn87man]

Hi racookpe1978,

Thanks for the response.

I have finished all the classes needed (including fluid mechanics) for a B.S. in Mechanical Engineering. However, it has been quite some time since I graduated college and I do not normally deal with pressure vessels in my daily work. I am sure if I do enough research, I can determine this conversion. I just thought that some nice, helpful person in these forums might save me several hours of research by providing the quick answer for me.

The pressure vessel is 10 feet in diameter and 40 feet in length. The initial pressure is 10 psi.

Thank you in advance for any help you can provide.

 

[racookpe1978]

So, to continue the instruction.

What is the volume of the PV? (I am assuming that the 10 ft diameter x 40 foot length implies a horizontal tank, but you have not identified what kind of ends it has. hemisphere or flanged and dished? Elliptical?

I will assume the 10 foot x 40 foot is ID, but there is little difference in wall volume at that size for that low a pressure.

What tank and gas temperature are you starting at?

Now, at 10 psi initial from that big a tank, you cannot realistically measure the change in pressure over a short period of time for a very small hole. Equally, after a long enough amount of time for any size hole, the final pressure will be 0.0 psia. So, what size hole over what period of time are you concerned about?

 

[gn87man]

The Pressure Vessel volume = 3140 ft^3

The ends are flat.

Tank and gas temperature is 70 deg F

Outside temperature is 60 deg F.

The hole size area is 3 in^2

I am concerned about the period of time from the moment of puncture until about 5 seconds after puncture occurs.

Thanks so much for your help.

 

[zdas04]

PV stil equalls nRTZ. Determine the mass in the vessel at the start. Flow it for a (short) time, determine how much mass you've removed and determine the new pressure. Repeat until the presssure is zero.

David Simpson, PE
MuleShoe Engineering

In questions of science, the authority of a thousand is not worth the humble reasoning of a single individual. Galileo Galilei, Italian Physicist

 

[gn87man]

Thanks so much zdas04!

 

[Guest102023]

Worried about a possible driveby shooting?
Aaaaaaah, I see, it's a take-home exam question.

"If you don't have time to do the job right the first time, when are you going to find time to repair it?"

 

[LittleInch]

gn87,

This is a crazy question. 5 seconds flow through a minute hole from a HUGE tank (relative to the hole) with a very low pressure difference. If you get an difference in pressure to four decimal places I would be surprised, far less that it would actually be measurable.

Are you sure you have the right data?

Anyone should be able to see that this mass loss in that period of time is minute.

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

 

[racookpe1978]

Yeah. That's what i see as well. If initial pressure was to one decimal point = 10.0 psig from a 3,000 cubic foot tank, then after 5 seconds it is still within one decimal point of 10.0 psig.

Going down, obviously. But the actual "coefficient of flow" from a 1 inch hole depends too much on what you assume for the hole: ragged edges of a busted wall, smooth transition, orifice plate, simple open pipe, quick-turn (90 degree-opening ball-valve) valve into a (very short!) pipe into open air, relief valve opening (at 10 psig ??), etc, etc, etc.

 

[gn87man]

Thank you brimstoner and LittleInch for your replies.

Yes, I believe I have the correct data.

And no, this is not a take home exam.

My recollection of pressure vessel topics is a bit foggy because it's been 15 years since I graduated college and I have not had to use it in my particular line of work, until now.

Thanks again to everyone for the help.

 

[georgeverghese]

This can most likely be worked out by setting up a differential equation.

dm/dt = current rate of depressurisation in kg/sec

where dm/dt = current massrate escaping from this vessel, and rate of depress = f(P, hole area) in the vessel

But dm = dP.V.MW/(ZRT), given that PV = ZnRT, hence dP.V = dn. zRT ; gas mol wt = MW

Hence dP.V.MW/ZRT = f(P)dt ; or dP/f(P,hole area) = (ZRT/V.MW) dt

Assume and use z,T averaged values throughout the depressure.

Integrate this differential between the limits of P1 to P2 for t = 0 to t, using consistent units.

If we have (P2 / P_ext) > critical press ratio of approx 2, with P_ext being pressure outside this vessel, then flow is always critical in this pressure interval, and we can use the flow through a PSV as the flow rate formula in f(P, hole area).

 

[georgeverghese]

If P1 =10psig = 25psia, then flow is subcritical all the time. The flow formula for subcritical flow is not conveniently integrated. Think Mathcad can do this for you.
Else use the differential form to work out the time for each small decrement in pressure.

 

[israelkk]

Assuming orifice discharge coefficient of 0.8, according to my calculation, even if it was a critical flow case (for example discharge to a vacuum) the pressure in the vessel after 5 seconds will still be over 24.1 psia. I used a propriety closed form solution for isothermal flow and adiabatic flow discharge for critical flow and both give more than 24.1 psi. Therefore, since this case is subcritical flow, the pressure inside the vessel will be even higher after 5 seconds.

 

[gn87man]

georgeverghese and israelkk....thank you very much for your replies.