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Homework Help: Differential Equations for Water Flow

  1. Apr 15, 2010 #1
    1. The problem statement, all variables and given/known data
    Water flows from a conical tank with circular orifice at the rate

    [tex]\frac{dx}{dt} = -0.6*\pi*r^2\sqrt{2g}\frac{\sqrt{x}}{A(x)}[/tex]

    r is the radius of the orifice, x is the height of the liquid from the vertex of cone, A(x) is area of the cross section of the tank x units above the orifice. Suppose r = 0.1 ft, g = 32.1 ft/s, tank has initial water level of 8ft and initial volume of 512*(pi/3). Computer water level after 10 min with h = 20s.

    2. Relevant equations

    Modified Euler's Method:

    [tex]w_{0} = \alpha[/tex]
    [tex]w_{i+1} = w_{i} + \frac{h}{2}[f(t_{i}, w_{i}) + f(t_{i+1}, w_{i} + hf(t_{i}, w_{i}))][/tex]

    3. The attempt at a solution
    Is A(8) = (512*(pi/3))/(8/3) = 201.0619?

    What do I set for y and t? If the question was f(y, t) = y' = -y + t + 1, 0 <= t <= 1, y(0) = 1, h is the step size, [tex]w_{0} = \alpha = y(0)[/tex] is the initial condition, t is variable between 0 to 1 with step size h, ...

    But if I input all those numbers into the equation I get dx/dt = -0.6*pi*0.1^2*sqrt(2*32.1)*sqrt(x)/201.0619 which leaves me with just x. I assume x = t in this case but what is y? How do I get it into the form f(y, t)?
  2. jcsd
  3. Apr 15, 2010 #2


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    Homework Helper

    first you need to find A(x) as function of x, not just evaulate it at a point
  4. Apr 15, 2010 #3
    V = (1/3)*A(x)*x
    A(x) = (512*(pi/3))/(x/3)

    Is that right?
  5. Apr 15, 2010 #4


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    Homework Helper

    not quite, the first line is true, but in the 2nd line you actually subsititute for x = 8ft

    so to move forwards you know that at x = 8ft
    V = 512*(pi/3)

    use that to solve for the radius of the cone base at the point x = 8ft

    as the triangles are similar you can then use the fact
    [tex]\frac{r(x)}{x} = \frac{r(8)}{8} [/tex]
    to find r(x) and so A(x)

    it probably helpful to know the volume of a cone
    [tex] V(x) = \frac{\pi}{3} r^2 x [/tex]
    x = height
    r = radius of cone base
  6. Apr 15, 2010 #5
    I used the formula V = (1/3)*B*h where B = A(x) the area and h = x the height so shouldn't A(x) = (512*(pi/3))/(x/3)?

    So if [tex]r(x) = \frac{r(8)}{8} x [/tex] and

    substitute it for r in [tex] V(x) = \frac{\pi}{3} r^2 x [/tex] it's

    [tex] V(x) = \frac{\pi}{3} (\frac{r(8)}{8} x)^2 x [/tex]

    [tex] V(8) = \frac{\pi}{3} (\frac{r(8)}{8} 8)^2 8 [/tex]

    [tex] 512\frac{\pi}{3} = \frac{\pi}{3} (\frac{r(8)}{8} 8)^2 8 [/tex]

    [tex] \frac{512}{8} = (r(8))^2[/tex]

    [tex] 8 = r(8)[/tex]

    What do I do with it?
    Last edited: Apr 16, 2010
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