Air Discharge 1

[Alph]

I have been asked a question by a colleague, it has left me unsure how to answer it.

A vertical cylinder type suction pile(ID=5m, L=6m), enclosed at the top, bottom is open ended (like a upside down bucket. It is lowered onto the seabed (water depth of 450m)at 2 m/s. The top of the the cylinder has a 600mm vent.

Q. At what depth will all the air be displaced or will all the air have been displaced by the time it reaches the 450m seabed depth ?

My thougts:-
At 2 m/s, it will take 225s to reach seabed. Volume of Cylinder is approx 118m³. Therefore Average air discharge is 0.52m³/s.

Assuming small dp of 0.1 bar across the vent and back calculating flow for the given orifice size. If the calculated flow is greater than than 0.52m³/s, then cylinder will be mainly be air free by the time it reaches the seabed.

Is there a more accurate/ detailed approach which can be used to answer the above question.

Regards
Al

 

[StephenA]

At 10m water depth the volume of air will be 59m3 (if none has escaped) and will decrease further as the depth increases.

At 450m the volume will be 45 times smaller than the original volume.

Is it important to know when the air will be all out of the cylinder or only that it will all be out by the time that the cylinder reaches the seabed?

Stephen Argles
Land & Marine

http://www.landandmarine.com

 

[BigInch]

At the drop rate that's a constant pressure change of 0.2 Bar/sec. Maybe the rate of volume lost out the orifice is greater.

But using your rate through the orifice as a constant 0.5 m3/s, plus the rate of volume lost to compression (apx. 1/2 each 10 m), you're out of air by 33 meters, so that would be 16.5 seconds.

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[Alph]

Thanks - It was simpler than I thougt.

Al

 

[BigInch]

Sorry, its 1/2 the first 10 meters, then 1/3 the next 10, 1/4 the next 10....

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"What gets us into trouble is not what we don't know, its what we know for sure" - Mark Twain

 

[sailoday28]

BigInch (Petroleum)
Shouln't the pressure drop across the orifice consider, the difference in elevation of the lower air water interface and the elevation of the water at the outlet of the orifice. Of course the air water interface will rise as the suction pile sinks.
dp=rho^gdz.
I don't doubt that the specified velocity must also be included.

Regards

 

[BigInch]

Depends on the accuracy needed in the answer.
Do you think he cares about whether its 14 or 16 seconds or even 22.5? He's got 225 seconds according to the lowering rate. Will 5% error in the time to zero air make a big difference, especially if its a 5% error in the first 33 meters? I wouldn't think so, but consider the question asked, if you like.

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"What gets us into trouble is not what we don't know, its what we know for sure" - Mark Twain

 

[sailoday28]

Alph (Petroleum)
You have specified the rate of decent. Did this take into account the bouyant force on the cylinder?

Regards

 

[BigInch]

Probably not as it appears to be a constant.

I would have figured that out for him, except I didn't know the wall thickness of the pipe and the end cap to get the volume and weight in water and of course we're also missing the weight of the orifice fitting, or even if there is an orifice fitting, or even if its made of steel for that matter. We'd also have to know the profiled temperature of the water on the way down, and if its fresh or salt, so we could include the exact density and viscosity, so to figure the drag too.

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"What gets us into trouble is not what we don't know, its what we know for sure" - Mark Twain