Differential Equations for Water Flow

[coolxal]

Homework Statement

Water flows from a conical tank with circular orifice at the rate

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

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.

Homework Equations

Modified Euler's Method:

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

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, w_{0} = \alpha = y(0) 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)?

 

[lanedance]

first you need to find A(x) as function of x, not just evaulate it at a point

 

[coolxal]

V = (1/3)*A(x)*x
A(x) = (512*(pi/3))/(x/3)

Is that right?

 

[lanedance]

V = (1/3)*A(x)*x
A(x) = (512*(pi/3))/(x/3)

Is that right?

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
\frac{r(x)}{x} = \frac{r(8)}{8}
to find r(x) and so A(x)

it probably helpful to know the volume of a cone
V(x) = \frac{\pi}{3} r^2 x
x = height
r = radius of cone base

 

[coolxal]

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
\frac{r(x)}{x} = \frac{r(8)}{8}
to find r(x) and so A(x)

it probably helpful to know the volume of a cone
V(x) = \frac{\pi}{3} r^2 x
x = height
r = radius of cone base

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 r(x) = \frac{r(8)}{8} x and

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

V(x) = \frac{\pi}{3} (\frac{r(8)}{8} x)^2 x

V(8) = \frac{\pi}{3} (\frac{r(8)}{8} 8)^2 8

512\frac{\pi}{3} = \frac{\pi}{3} (\frac{r(8)}{8} 8)^2 8

\frac{512}{8} = (r(8))^2

8 = r(8)

What do I do with it?