I am working on a paper at the moment which has to do with draining tanks. I have already set up a differential equation which explains the drain from a tank where the cross-sectional areas of the container and of the outflow are constants. But now I have to set up one which explains the drain from a conical tank where the cross-sectional area of the container varies. And that is my problem. I am finding it difficult to make this particular differential equation. I need it to be a dh/dt equation. I have got the following set up so far: (I got no clue if I am on the right track or not).
From Bernoullis equation I got:
(p1/Pg) + (v12/2g) + h1 = (p2/Pg) + (v22/2g) + h2, where P is the density of the fluid, v is the velocity and g is the gravitational constant, p is the pressure and h is the height. One must assume that the pressure is the same throughout the flow, therefore p1 = p2 = 0 and the parts of the equation containing p is removed:
(v12/2g) + h1 = (v22/2g) + h2
The velocity at the outflow (v2) can be defined as the change in height to the time dt:
Furthermore v2<< v1. Then v1 can be defined as follows:
v1 = (2g*(h2-h1))^(1/2)
From the continuity equation we have that: v1A1 = v2A2. Which means that the velocity at cross-sectional area 1 is different from the velocity at cross-sectional area 2. It also means that the smaller the cross-sectional area is the fluid has a lower velocity and vice versa.
The cross-sectional area of a circle is: A = pi * r^2 but then the radius has to be known. The radius of the circle can be defined as: r = h*tan(a), where a is the angle of the cone. This radius is valid for all heights h. Then by inserting in the continuity I get:
(2g*(h1-h2))^(1/2) * pi * r^2 = -(dh/dt) * pi * h^2 * tan(a)^2
What do I do now?
Thanks in advance, Thure.