Underpressure of Submerged Tank

[Ally0138]

Hi,

I have tank which consists of an inner flexible bladder contained within an outer metal shell. The whole tank is submerged in water. The inner bladder contains oil which is transferred in / out via a pump located above the waterline, circa 3 m above the base of the tank. The cavity between the inner bladder and the outer shell is filled with the same water that surrounds the tank. The outer shell is completely watertight, with the exception of a single water intake / outfall line, which rises from the base of the tank to an isolation valve just above the waterline and then drops back down into the middle of the water.

Diagram:

Under normal operation, the isolation valve will be open, allowing water to flow freely in and out of the tank outer shell / cavity as the oil bladder is emptied / filled.


Questions:

1. What I'm trying to understand is what will happen if the isolation valve is left in the closed position and an attempt is made to pump oil out of the bladder. Intuitively, I expect that as the volume of the oil in the bladder decreases, since there's no water flowing into the cavity to fill the volume vacated by the water, the pressure in the tank will drop, which will in turn cause a drop in NPSHa to the oil pump. How could I go about estimating the extent of the pressure drop in this case?

Here's where I'm confused: There's normally no vapour in the tank, it's all liquid phase (at least to start with) and pressure in a liquid is proportional only to depth / height of liquid column. But if we have a fixed volume of liquid and pump some of it out, with nothing flowing in to take its place, shouldn't the pressure in the tank drop?


2. Is there a potential issue with positioning the water isolation valve above the waterline? I.e. the water is effectively being siphoned out of the main tank / reservoir into the tank cavity; could we lose the siphon and suck a load of air into the cavity? (Presumably this won't be an issue if the water line is initially filled [I'm not sure how this will be achieved] and the intake is positioned sufficiently below the waterline?)


Thank you in advance,
Ally

 

[MJCronin]

Homework problem ???/

MJCronin
Sr. Process Engineer

 

[LittleInch]

1) The pressure in the tank will drop, but the key point is finding the lowest pressure in the system.

This will be basically at the pump inlet - where the pressure will fall until you reach the vapour pressure of the oil. Then your pump is trying to pump gas instead of liquid and it won't work very well....

2) The key is what is the highest point of the syphon above water level.

If this gets close to 10 m then you'll vapourise the water and loose syphon. 4-5 m no problem.

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

 

[itsmoked]

3) Establishing the siphon will be a problem as diagrammed because the gas initially in it can possibly end up in the inner box with the bladder.

4) With all the other detail of the drawing you don't show that the big tank is actually vented to atmosphere. Is it?

5) Depending on what the vapor pressure is before the pump just cavitates you could collapse the the outer tank if it's not vented or the inner tank if the valve is closed.

Keith Cress
kcress -

http://www.flaminsystems.com

 

[georgeverghese]

The oil inflow line should be fitted with a top vent at the high point of this fill line if you want to prevent trapped air from getting into the bladder during the initial priming - use of this line. If there is air in the bladder to start with, it can affect how the pressure drops in the bladder as the evacuation pump draws oil out with the external equalisation valve closed.

 

[LittleInch]

Agree that the key issue is collapse of the metal tank.

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

 

[Ally0138]

Homework problem ???/

No. Real life.

 

[itsmoked]

Homework problem ?

No way! How could anything this bizarre be a homework problem.

Keith Cress
kcress -

http://www.flaminsystems.com

 

[Ally0138]

First, thanks for the responses LittleInch, itsmoked, and georgeverghese. I appreciate your input.


LittleInch:

1. I agree. The lowest pressure will be at the pump suction, where there could be a risk of cavitation. But how can I go about estimating the minimum pressure in the tank? I'm looking for a method or a technique that will allow me to estimate this. Can I work backwards from the condition of pump cavitation? i.e. assume that we've reached the point where NPSHa = NPSHr, then just add the frictional and hydrostatic losses/gains at the rated pump flow to determine what the pressure will be in the tank at this point.

2. That's what I thought. This won't be an issue - the valve is positioned approx. 0.4 m above the waterline. Thanks for the confirmation.


itsmoked:

3. I agree. I'm not sure how this will be achieved in practice.

4. Sorry. Yes, the big tank is actually open to atmosphere at the top (not a sealed box as it may appear from my original diagram). I have updated the diagram to reflect this.

5. Yes. Collapse of the inner tank when the valve is closed is exactly what I'm concerned about. That's what I'm trying to work out. How can I work out what is the minimum pressure in the tank if the valve is closed and the pump is running? I'm looking for a method or a technique that will allow me to estimate this.


georgeverghese:

Good point and one that I agree with. This is not my design, but one belonging to a client of mine.


Thanks again,
Ally

 

[LittleInch]

Well this is not easy to get, because if the pump really starts cavitating madly as it will then flow drops to virtually zero so you loose the frictional losses part of the equation.

Just go with NPSHR at the inlet based on no or very little flow ( the lowest number) and work it backwards from there. It might actually be a few metres lower before you get into 100% cavitation / loss of suction as your oil vapourises. As a worst case look up the vapour pressure of the oil at the temperature you're pumping at. The negative pressure at the pump inlet can't be less than that.

With any decent pump you're probably at around 0.5 to 0.7 bar difference from outside to in.

That will collapse a lot of thin square tanks.

A thick domed pressure vessel - no.

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

 

[itsmoked]

Why use the siphon? Can't the inner box just be open to the outer tank?

This removes the "how to start the siphon".
Removes the air bubble delivered to the small tank by the siphon.
Removes the potential for collapsing the small tank.

Short of closing the valve there would be no difference.

It's not clear what the siphon provides to the system.

Keith Cress
kcress -

http://www.flaminsystems.com

 

[LittleInch]

You can only assume that there are vents and drains on this system to completely fill the inner tank before the first fill in the bladder which will then drive the water out through the syphon and fill it.

This really is quite odd though.

why is it like this?

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

 

[georgeverghese]

Suspect the inner tank acts as secondary containment in case the bladder leaks or blows out. Agreed, the tap location for the equalisation line should be at the top of the inner tank, not at the bottom - unless there is something we are missing here.