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Good old question about buoyancy and riser in oil industry

  1. May 22, 2007 #1


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    I've got a scenario and would like a confirmation or rejection on it's correctness.

    If i lower a steel riser (for simplicity it is 3000 m and vertical, open ended, cylindrical steel pipe, completely flush) into drilling mud (assume water) the surface weight would be the steel weight in mud. When I set the riser down (say i latch it on to an existing riser that is already in place with top at ~3000m), the weight of the riser at surface immediately becomes the air weight of the riser when it latches.


    In reality the steel would buckle a bit by the upforce before beeing latched and even more so when it is latched and surface pulling reduced to zero so the whole string rests on itself (far fetched, but example purpose). How does this affect the buoyancy force on the steel pipe?? does it contribute to the buoyancy force acting on the riser??
  2. jcsd
  3. May 22, 2007 #2


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    Honestly, I am not following the terminology that you are using. What do you have that is 3 km in length? What do you mean by "latching?" Can you do a diagram?

    Anyways, the apparent weight is going to be the weight out of water - the weight of the displaced water. So, take the amount of the riser that is under water and calculate that volume. Use that volume and the weight density of water to calculate the buoyant force.
  4. May 22, 2007 #3


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    Hi. sorry for the oil terms. here's to simplify:

    3000 m stell jointed tube, flush on outside and inside and open ended (no bottom, and might in real life be 9 5/8 inch in outside diameter, far fethced that it is vertical though) beeing lowered into a drilled hole and latched (screw onto an exciting in place pipe of same type about 3000 m into the hole.

    Becuase buoyancy upforce acts on casing from below, would this not stop to act on this area when it is set down (latched, or screw) on the steel tube already in place??

    hope this is more clear....(?)
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