Using Bernoulli to determine pressure of equalizing flow 2

[Roylance]

Greetings all,
I wanted to carry out an analysis of the following scenario. A tall cylindrical container is located at depth subsea. This container is filled with water but these contents are maintained at a lesser pressure than the surrounding seawater. The interior of this container has an approximate volume 350,000 cubic feet and also features a port at the top which can be opened and closed by a valve.
The pressure differential between the local exterior & interior hydrostatic pressures is 75psi.
For the situation where the valve is open, I’ve been given a value of 60 fps for the flow velocity.
From the above, I wanted to determine what the pressure would be within the constriction formed by the valve bore. For this I made use of Bernoulli’s Equation, working between two points. The upper point, let’s call it Point A, to be a distance above the container at which the flow speed directed into the valve bore would be negligible i.e. zero. Point B, the lower point, I took this to be located at the base of the valve bore. Calculating to determine the pressure at point B gave a result that was a negative pressure.
I used absolute pressure values throughout rather than gauge so it puzzled me a little that I should end up with a negative value and a substantial one at that. Is this an indication that the flow regime in such a scenario would no longer be best explained by Bernoulli? I had in mind that the pressure drop between the static fluid and the rapid fluid flow through the valve may be resulting in cavitation, hence the calculated negative pressure. Is that a correct interpretation?
Varying both the fluid velocity through the bore and the estimated elevation of Point A above Point B gave more sensible (as in positive pressure values) with reducing the fluid velocity having the biggest effect. The element that particularly troubles me as being imprecise is having to estimate Point A. When I say estimate, I think uneducated guess is a more apt description. Could anyone point me in the direction of a better way to determine even just a very approximate distance?
Thanks in advance.

 

[BigInch]

Not Bernoulli.
Flow through the valve is described by the valve's CV (gpm/psi^0.5) coefficient.
As you can see by the units of CV, a 75psi differential pressure will give you the flow rate thorugh the valve.


Learn from the mistakes of others. You don't have time to make them all yourself.

 

[Roylance]

Thanks for the reply BigInch.
Why not Bernoulli though? The valve is of the gate type so the passage through which equalization flow takes place is essentially constant bore diameter. There isn't a published Cv value for this passage. I could attempt to calculate/estimate one though.

 

[BigInch]

perhaps a velocity of 60 fps is not possible with 75 psi

Learn from the mistakes of others. You don't have time to make them all yourself.

 

[LittleInch]

You need to sketch out what you're thinking as it is not clear. Apart from the difficulty of keeping a liquid filled large container 75psi below the subsea pressure, it is not clear if you are looking at a valve at the end of pipe or essentially an orifice. orifices have entry and exit losses as well so you're probably using the wrong equation or ignoring some impact or losses, but it is difficult to see from words alone, e.g. where is the 60fps - in the valve bore or somewhere else? what else is connected?

My motto: Learn something new every day

Also: There's usually a good reason why everyone does it that way

 

[BigInch]

It seems to be a mathematical problem at this point, and inlet/outlet losses wouldn't affect his calculations, since he's not using them. I think he's starting with 60 fps and finding that there is only 55 something ft of head, which doesn't equal anywhere near 75 psi.

Learn from the mistakes of others. You don't have time to make them all yourself.

 

[Roylance]

Hopefully the attached sketch clarifies matters.

 

[LittleInch]

I just stuck this data into an orifice calculation, because that is basically what you have, and got 0.00186ft/sec(!) with 75 psi differential

Something seriously wrong with your data - either there is a lot more pressure difference or the whole is a lot lot bigger or the velocity figure is wrong.

Figure out which two elements are true (diff pressure or hole size or flow veleocity) then work out the third one...

My motto: Learn something new every day

Also: There's usually a good reason why everyone does it that way

 

[BigInch]

That can't be right. Scaling that up, snails are rocket sleds.

Learn from the mistakes of others. You don't have time to make them all yourself.

 

[LittleInch]

Sorry,

Friday afternoon syndrome.

orifice calc gives about 75 feet/sec, previous one was giving me the velocity in the very big tube I put in - my error.

So in the end it actually looks about right.

How about putting down your calculation and see where it is going wrong?

My motto: Learn something new every day

Also: There's usually a good reason why everyone does it that way

 

[bimr]

You guys are missing something. The poster stated that the container at depth is already full.

Since water is nearly incompressible, it will not take much water to bring water pressure up to test pressure. For example, if you had an 8 inch water main, 1,000 feet long and all of the air was removed or purged from the water line, it would only take about one cup of water to bring the pressure up to 150 PSI. For your example, that would be less than 8.3 cubic feet of water.

This is Pascal's law not Bernoulli's.

http://www.hydraulicsonline.com/hydraulic-principles

This would not be a steady state situation. The water flow will diminish quickly over times. You also have other issues such as the flexure of the container walls and that the water pressure changes with depth, but lets not get into that.

 

[BigInch]

Not missing it. That would fall out in the next step. If we ever get there, that is.

Learn from the mistakes of others. You don't have time to make them all yourself.

 

[77JQX]

Here's my take: At 60 fps, with ordinary water, the velocity head (v^2/2g) works out to 24 psi. That the kinetic part of your conservation of energy in Bernoilli's equation. You'll need to work out the friction and elevation parts for the rest of the potential energy lost in the 75 psi differential. So in the throat of your valve, the pressure will be Outside P - friction - elevation change - 24 psi. I don't see how your answer came out negative.

 

[BigInch]

subtract 75 from 24


Learn from the mistakes of others. You don't have time to make them all yourself.

 

[LittleInch]

bimr, the OP also said "these contents are maintained at a lesser pressure than the surrounding seawater". How (or why) is immaterial to us - this is the statement made so we have to go along with it. I agree if it is not "maintained" then your statements apply and the calculation becomes a transient thing. The cylindrical container is also pretty big - 9,700m3 is not small.

My motto: Learn something new every day

Also: There's usually a good reason why everyone does it that way

 

[bimr]

LittleInch, your comment is understood. It appears that the poster has not completely thought this out. If the contents are maintained at a lower pressure, than the poster must be pumping out of the container at the same time the inflow occurs. But, he does not mention another port on the tank to accomplish this "maintained pressure" comment. If this is in fact happening, his outflow will more or less be the same as the inflow.

Regardless of the "maintained pressure" comment, the inflow and/or the outflow will be very small flow of fluid since water is not compressible; the container is full; and you are only pressurizing/depressurizing the container, not filling it.

 

[Roylance]

Thanks for all the input and comments.
bimr - not quite sure how I've not fully thought the scenario out. I am describing something that exists in reality and not a hypothetical design. As LittleInch says, the how/why is immaterial, hence my omitting the detail from the OP. Pumping would not necessarily be required, just access to a lesser hydrostatic column of fluid would maintain the reduced pressure within the container.

 

[bimr]

I understand now. Bernoulli's equation does not completely describe the scenario.

Your scenario can be simplified to one of continuous flow through a valve with a 75 psi pressure drop across the valve.

It is difficult or often impossible to measure the lowest pressure through a valve, for example in the Vena Contracta. Valve manufactures use a Cavitation Index to predict cavitation, not Bernoulli's equation.

http://neutrium.net/fluid_flow/cavitation-in-restriction-orifices/

It is also difficult to estimate the elevation of Point A (to be a distance above the valve at which the flow speed directed into the valve bore would be negligible). A vortex equation may be used to approximate the distance.

http://www.pumpfundamentals.com/help11.html

http://en.wikipedia.org/wiki/Vortex

http://web.mit.edu/fluids-modules/

http://www/potential_flows/LecturesHTML/lec08/lecture8.html