submerged pipe

[markham123]

Guys I need some help on a 24 inch od pipe (23 inch id ) steel pipe. It is 50 ft in length. It is laying ( horizontally) on the bottom of a old river bed in about 30ft of water. I'm assuming the bottom half of the pipe is filled with silt and the other half with water. I'm trying to calculate the total force that will be needed to lift this pipe. I included the weight of the steel pipe, the weight of the silt in the bottom half of the pipe, the weight of the water in the top half of the pipe, and the pressure exerted on the pipe at a depth of 30ft. Did I miss something or am I not looking at this the correct way? The pipe is of course open on both ends. Thanks.

 

[SJones]

Friction of the pipe with the river bed?

Steve Jones
Materials & Corrosion Engineer

http://www.linkedin.com/pub/8/83b/b04

 

[katmar]

I think that what SJones has pointed out will probably be the most important factor, especially if it has become embedded in mud or rocks. Unfortunately it is probably the most difficult factor to assess.

The pressure exerted by the 30 ft of water above the pipe does not come into it. This same pressure is exerted all around the pipe so there is no net downward force.

The water inside the pipe, while it is still submerged, is totally offset by the bouyancy effect. You could take this effect on the steel and silt into account in making the pipe seem a bit lighter while it is submerged - but it will be a small effect.

It may be an idea to lift the pipe from one end only so that as it comes clear of the water the contained silt and water will hopefully drain out the lower end, making the pipe lighter once it is clear of the water.

Katmar Software - Engineering & Risk Analysis Software

http://katmarsoftware.com

"An undefined problem has an infinite number of solutions"

 

[BigInch]

Cohesion of the river bed mud, not friction.

There could be a suction-breaking effect you might need to account for as well. A surface pressed into a river mud, or "muck" could act similar to a suction cup on a glass window, effectively prohibiting hydrostatic balance with the pressure from the water above. A suction cup does not allow atmospheric pressue under the cup to contact that surface to balance with atmospheric pressure on the other side of the cup's membrane. If the bottom is sandy, you will not get the suction cup effect because water pressure from above will infiltrate through the pores to reach the undersurface of the pipe and balance the hydrostatic pressure that is being applied above the pipe. If that is the case, you can subtract the buoyant force.

If the river bottom is muck, then hydrostatic pressures may not fully balance and you will have to temporarily supply that hydrostatic force with the pull load until the "suction is broken" - hydrostatic pressure equalizes. When suction equalizes, you can subtract the buoyant force from the pull load.

If you do not have to remove the pipe in one unbent piece, you could conceivably hook one end only and pull, perhaps bending the pipe, but breaking any "suction" under the bent end with much less force than would be required to break suction all across the pipe if it remained in one straight piece.

Only put off until tomorrow what you are willing to die having left undone. - Pablo Picasso

 

[markham123]

Thanks for the replies. I figured there would be no buoyant force to subtract since there is no air in the pipe. Is this correct?

 

[BigInch]

No. The pipe itself weighs less by the amount of water it displaces; a buoyant force equal to the volume of the steel * density of water. Volume of steel, V_steel = [&pi;]*(R_od-R<sub>id)²/4 * Length
Net weight of pipe in water = Vsteel* (Density of Steel - Density of water)

Density of steel is 490 pcf, density of water is 62.4 if fresh water, or 64 pcf if sea water.

If there was any air displacing water inside the pipe, then you would add that volume * water density to the above.
Divided by 2, in your case, since you assumed mud was filling half the pipe. The mud volume would be subject to buoyant force, so you'd take the same 1/2 pipe volume, Vmud * (density of soil in air - density of water) to get the net weight of mud in water.

Only put off until tomorrow what you are willing to die having left undone. - Pablo Picasso</sub>

 

[dhengr]

That piece of pipe weighs about 6500 lbs. High pressure water jetting around and under the pipe might help loosen it, and brake any suction, particularly if the river flow helps carry debris away. Jetting inside the pipe might help clean it out, any mud inside would be a weight you have to lift. A bladder inside the pipe and filled with air would help in the lifting operation, as would lifting bags and slings under the pipe. Some under water investigation might be in order to assist in determining conditions and equipment which might be needed. You might consider just capping one end, with a sliding shoe as part of the cap, and dragging it out longitudinally to the nearest bank/shore. A Cat, winch and cable would do this, as opposed to larger floating equipment.

 

[MiketheEngineer]

Dynamite - everyone eels has good suggestions

 

[BigInch]

The eels for sure wouldn't like it.

Only put off until tomorrow what you are willing to die having left undone. - Pablo Picasso