Fluid Mechanics

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A cylindrical tank 2m diameter and 4m long with its axis horizontal, is half filled with water and half filled with oil of density 800kg/m^3. Determine the magnitude and position of the net hydrostatic force on one end of the tank.

The answers should be 27.9kN, 1.26m from the top.

Can someone please explain how to work it out?

  • Fluid Mechanics -

    For a uniform liquid, the pressure at a depth of h equals ρh.
    For a tank of radius r (length does not matter) containing a mixture of oil and water, the hydrostatic pressure at any point h below the top is
    p1(h)=0.8h for 0<h≤r, and
    p2(h)=0.8h+(h-r)(1-0.8) for r<h≤2r

    The width of the tank at any height h from the top is
    w(h)=sqrt(r²-(r-h)²)

    The magnitude of the hydrostatic force, F, is the sum of the integrals
    F=∫ 0 r p1(h)dh + ∫r 2r p2(h)dh
    =2.646g N

    The total moment about the top:
    Fh=∫ 0 r x*p1(h)dh + ∫r 2r x*p2(h)dh
    =3.353g N-m

    The position can be found by dividing Fh/F = 1.267 m from the top.

    Note that 2.646g gives 26.0 N as opposed to the 27.9 N given in your answer.

    Check my calculations/the answer given in the book.

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