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:

(p_{1}/Pg) + (v_{1}^{2}/2g) + h_{1}= (p_{2}/Pg) + (v_{2}^{2}/2g) + h_{2}, 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 p_{1}= p_{2}= 0 and the parts of the equation containing p is removed:

(v_{1}^{2}/2g) + h_{1}= (v_{2}^{2}/2g) + h_{2 }The velocity at the outflow (v_{2}) can be defined as the change in height to the time dt:

v_{2}= -(dh/dt)

Furthermore v_{2}<< v_{1}. Then v_{1}can be defined as follows:

v_{1}= (2g*(h_{2}-h_{1}))^(1/2)

From the continuity equation we have that: v_{1}A_{1}= v_{2}A_{2}. 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*(h_{1}-h_{2}))^(1/2) * pi * r^2 = -(dh/dt) * pi * h^2 * tan(a)^2

What do I do now?

Thanks in advance, Thure.