1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Ordinary Differential Equations

  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

    lanedance

    User Avatar
    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

    lanedance

    User Avatar
    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
    r(8)

    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
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Ordinary Differential Equations
Loading...