Matt.Sz

It took me so long to realise that "the worst case" means an empty pipe, thank you all! Got the right result finally

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 mean the buyoant force is the only force acting upwards and the question is about this force...

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

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