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Archived Fluid dynamics, water emptying out of cylinder

  1. Apr 14, 2007 #1
    1. The problem statement, all variables and given/known data
    Tank is filled with water to a height h_0 = 1.00 meters
    Cross section area of tank (A_1 = .785 m^2)
    Hole at bottom of tank with an area, A_2 = .002 m^2

    2. Relevant equations
    How much time does it tank to half empty the tank? find t_(1/2).

    Question also provides "a useful antiderivative."
    / = integral sign
    / x^(-1/2) dx = 2x^(1/2)

    3. The attempt at a solution
    V = A_2 * v_hole
    dV/dt = A_2 * v_hole

    2. getting an equation for velocity coming out of hole
    v_hole = [2gh + (v_surf)^2] ; got that from using Bernoulli's principle.

    3. equation of continuity to get v_surf
    v_surf = v_hole * A_hole / A_surf


    v_hole = (2gh)^(1/2) , since [1 - (A_2/A_1)^2] = ~1

    4. rate of change of the level of water in the tank
    dh/dt = (-v_hole * A_2)/(A_1)
    dh/dt = -.0113

    h_2 - h_0 = -.0113 * t_(1/2)
    a half empty tank means, h goes down 1/2, 1/2 of h = 1, h=-.5
    t_(1/2) = -.5m / -.0013 m/s = 44.3 s

    44.3 s is wrong though. I don't get how come. I went through the hints for this problem. It mentions

    dy/dx = f(x) * g(y) ... f(x) and g(y) are known functions

    / dy/g(y) = / f(x)dx

    I don't get which known functions they are referring to. And it's also been probably 2 years since I ever used calculus, so those equations don't make much sense to me.

    Any help would be appreciated.
    Last edited: Apr 14, 2007
  2. jcsd
  3. Jul 16, 2016 #2
    First, we establish an unsteady state mass balance on the tank
    [tex]\frac{dm}{dt} = -w[/tex]
    We can express mass inside the tank as [itex]m = \rho V = \rho A_1 h[/itex], where density and cross section area are constant, so the only variable changing in time is the height of liquid inside the tank. We can also express the mass flow rate out of the tank as [itex]w = \rho A_2 v[/itex]. Now we need to relate the exiting velocity with the height of the liquid. We can go ahead and use Torricelli's law, [itex]v = v = \sqrt{2gh}[/itex], turning our unsteady state model into a quasi-steady state one.
    [tex]A_1 \frac{dh}{dt} = - A_2 \sqrt{2gh}[/tex]
    We want to know the time it takes to empty half of the tank, so we can separate variables and integrate the differential equation from 0 to [itex]t_{\frac{1}{2}}[/itex] and [itex]h_0[/itex] to [itex]\frac{h_0}{2}[/itex]
    [tex]\int_0^{t_{\frac{1}{2}}} dt = - \frac{1}{\sqrt{2g}} \frac{A_1}{A_2} \int_{h_0}^{\frac{h_0}{2}} \frac{dh}{\sqrt{h}}[/tex]
    [tex]t_{\frac{1}{2}} = \frac{2}{\sqrt{2g}} \frac{A_1}{A_2} \left(\sqrt{h_0} - \sqrt{\frac{h_0}{2}} \right)[/tex]
    Finally, we just plug in the values and solve
    [tex]t_{\frac{1}{2}} = \frac{2}{\sqrt{2 \left(9.8 \ \frac{m}{s^2} \right)}} \frac{0.785 \ m^2}{0.002 \ m^2} \left(\sqrt{1 \ m} - \sqrt{\frac{1 \ m}{2}} \right) = 51.93 \ s[/tex]
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