## passive pressure compensation: time to equalise

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mavpro
Posts: 3
Joined: Thu May 07, 2015 2:13 am

### passive pressure compensation: time to equalise

Bit of a tricky one.....
I have a pressure sensitive underwater vessel (+/- 50kPa operating pressure), say 10L fixed air volume, connected to a pressure compensating bladder via a 2m long, 8mm dia hose (same depth as vessel). The bladder balances the internal pressure of the vessel to avoid it being crushed as it dives. Lets say the bladder has infinite volume & can adjust to a change in pressure instantaneously (take the bladder out of the equation, the vessel needs to equalise with the water depth via the hose). Standard smooth hose, no other obstructions. The depth of the vessel can be controlled, from 2m down to 100m.

I need to be able to perform two calculations;
a) if there was a sudden change of depth (eg 10m down to 15m), how long would it take to equalise
b) what is the constant rate of descent (or ascent) I can maintain without exceeding the limit of the vessel. I realise this could also change with depth, ie 10m to 15m would be a different velocity than 95m to 100m.

I see the application is similar to the 'discharge' calculator (fantastic web site by the way, please keep it up). The mass flow of air would degrade as the pressure gets closer to equalising (flow approaches zero as P1/P2 approaches 1). As its a differential relationship, I've tried iterating the results in excel, but without success. Please help!

Posts: 358
Joined: Mon Feb 08, 2010 7:47 pm

### Re: passive pressure compensation: time to equalise

Hello and thanks for nice words about web site, as indeed it will be up for long time, I can guarantee that.

Is the bladder connected to vessel all of the time?

You are right that flow is not at constant rate and flow rate will be changing during pressure change. With discharge calculator, you can check that for flow through 8 mm, 2 m long hose and 5 m of water head, flow is not choked and flow rate of air is almost 3l/s, and for 10 L of vessel it means that in 1 s pressure will increase by 30%.

I have in the past tried to make relationship for this kind of problems so please take a look here.

That is approach that seems right for me, but never had confirmation if it is so.
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mavpro
Posts: 3
Joined: Thu May 07, 2015 2:13 am

### Re: passive pressure compensation: time to equalise

To answer your first question, the bladder is attached all the time (part of the same equipment).

I think I've got the answer, working back from your earlier post. I used the equation Dt = sqrt [(f*L/2.D.g)*(Dh^2/H-Dh)].
Where Dt = time step
Dh = change in height, m
f = friction factor (I used f = 0.0424)
L = hose length (L = 2m)
D = hose diameter (D = 0.008m)
H = total height difference (H = 5m). H is known because my 'tank B' is infinite, the ocean.

For simplicity, I iterated using excel, so for H = 5m & 100 steps; Dh = 0.05m. Careful to note; the Dh^2 is always 0.05^2, but the H-Dh is the cumulative step, ie at the halfway point it is 5-(50*0.05) = 2.5m. Also, seeing as t approaches infinity as the depths equalise, I 'stopped the clock' when my answer was within 0.2m. The time I calculated (sum of all the time steps, Dt) was 2.58 sec, which sounds plausible.

The second part of my original question was to work out a constant rate of descent. Working backward up the table, I found it takes 1 second to go from 1.35m differential to something approaching zero, therefore I see my answer being 1.35m/sec descent (or ascent).

It all sounds reasonable, can you please verify if my approach is correct?

Posts: 358
Joined: Mon Feb 08, 2010 7:47 pm

### Re: passive pressure compensation: time to equalise

I think so. Yet, as I said I haven't got back confirmation of the derivation that you used also, but all calculation results are very close to the conclusion that you made.

1.35 m is about 13.5 kPa, so as you have +/- 50 kPa that should be with solid reserve, which I think is ok and should be accepted, unless that speed of 1.35 m/s is unacceptably slow.

My thinking, again, is that the % of change in depth should be equal to % of change in L of injected air in the same period of time. For 10 L tank, 1 L of air injected is making 10% of pressure change. So for 1 L/s, depth should change for 10% or 1 m. Using calculator I get that flow is really around 1 L/s for 1 m head difference. So it is again close to conclusion that you made.

In reality I think that it will be even better (safer) situation, but that needs to be confirmed with more sophisticated methods.
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mavpro
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Joined: Thu May 07, 2015 2:13 am

### Re: passive pressure compensation: time to equalise

I realise after I posted there was an error in my logic; it did not take into account the volume of the vessel, in this case 10L. I need to delve a little deeper into your 'gas discharge flow calculator', if I may.

I can see how you derive the mass flow rate (w) using the modified Darcy formula, if the 'enter pipe length' box is not checked. However, how are the formulas modified when;
- pipe length is entered; length (L), kinematic viscoscity (v) & pipe roughness (kr) become relevant
- the modified Darcy formula uses density, whereas I'm assuming the temperature will remain constant (ie use sea water temp, air density will increase as I go deeper). What is the relationship between temperature & density?
- how do you calculate the expansion factor, Y

As I am iterating using a spreadsheet, I need the formulas to drive the results. I'll let you know how I go. Thank you.

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