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Hydraulics exercise - Buoyant force

Hi
I'm first year Civil Engineering student, taking a resit in Fluid Mechanics. Our lecturer left us a bunch of revising exercises. I have a problem with one, can anyone help? This lecturer is currently away and doesn't reply to emails.
Here's the exercise:

A steel pipeline designed to convey oil of relative density 0.8 has an internal diameter of 90cm and an external diameter of 95cm. It is lying on the sea bed, completely immersed and it is anchored at intervals of 8m along its length. Sea water density is 1020kg/m^3 while the density of steel is 7900 kg/m^3. Calculate the largest (the worst case) upward force on each anchor.The given answer is 11.70kN, but I get 28.45kN every time socould someone explain it step by step? Thank you
Original post by MattSz
The given answer is 11.70kN, but I get 28.45kN every time socould someone explain it step by step? Thank you

Can you please post your working?
Reply 2
Original post by RogerOxon
Can you please post your working?


I start by calculating the weight of the displaced water by the pipe, as the buoyant force equals the weight of displaced water.
Area of the end of pipe: A= pi x 0.95^2/4 = 3.15x0.26 = 0.71 m^2
Volume: V=area x lenght= 8m x 0.71m^2 = 5.69 m^3
Mass of displaced water: M=rho x volume = 1020kg/m^3 x 5.69m^3 = 5799.5 kg
Weight of the displaced water: W=m x g = 5799.5kg x 9.81 N/kg = 56893.1 N
And this is in total so one anchor is half of it so 28446.3 N

I assume I forgot about something as there's loads of unused data left, but my lecturer used to give us a bad answer once in a while as well so don't know haha
Original post by MattSz
I start by calculating the weight of the displaced water by the pipe, as the buoyant force equals the weight of displaced water.
Area of the end of pipe: A= pi x 0.95^2/4 = 3.15x0.26 = 0.71 m^2
Volume: V=area x lenght= 8m x 0.71m^2 = 5.69 m^3
Mass of displaced water: M=rho x volume = 1020kg/m^3 x 5.69m^3 = 5799.5 kg
Weight of the displaced water: W=m x g = 5799.5kg x 9.81 N/kg = 56893.1 N
And this is in total so one anchor is half of it so 28446.3 N

I assume I forgot about something as there's loads of unused data left, but my lecturer used to give us a bad answer once in a while as well so don't know haha

I've not checked your numbers (π\pi is not 3.15 though), but there are couple of obvious issues:
- You have omitted to subtract the weight of the pipe and its fluid;
- Each support should be assumed to support 4m either side, i.e. 8m of pipe.

However, the question does miss lots of things that you may want to design for, e.g. the pipe being empty / cleaned, the failure / replacement of a support, etc. You probably don't need to worry about this yet ..
(edited 6 years ago)
Reply 4
If I substract the weight of the pipe with oil in it (35071.1kg that is 344041.5N) then the result is negative. And it's logical as it's fully submerged and it's supposed to remain on the sea bed?
Original post by MattSz
If I substract the weight of the pipe with oil in it (35071.1kg that is 344041.5N) then the result is negative. And it's logical as it's fully submerged and it's supposed to remain on the sea bed?

It looks like they've calculated the upward force with the pipe empty, which isn't unreasonable.
Reply 6
Yeah it seems correct. So "The worst case" is when the pipe is empty. Thank you so much
Original post by MattSz
If I substract the weight of the pipe with oil in it (35071.1kg that is 344041.5N) then the result is negative. And it's logical as it's fully submerged and it's supposed to remain on the sea bed?

I get the weight of the steel at 5660N/m of pipe and the buoyancy at 7090N/m. That'd give a weight of 45.3kN for the pipe and a buoyancy of 56.7kN per 8m section (4m either side of the support). That gives a nett of 11.4kN upwards.
Reply 8
Yeah got the same results thanks :smile:

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