# Draining a conical tank

• Dec 14th 2012, 03:59 AM
Thure
Draining a conical tank
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:
v2= -(dh/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?

• Dec 14th 2012, 06:45 AM
ebaines
Re: Draining a conical tank
First I would make the variables consistent. On the left hand side you use the term $h_1-h_2$, but on the right hand side you just use $h$. I would stick with just $h$, so you have:

$\pi R^2 \sqrt {2gh} = - \pi h^2 \tan^2(a) \frac {dh}{dt}$

Note that 'R' here is a constant - namely the radius of the outlet hole.

You can rearrange to get a differential equation:

$\frac {dh}{dt} = - \sqrt {2g} R^2 \tan^2(a) h^{-3/2}$

If we set the cionstant $K = \sqrt {2g} R^2 \tan^2(a)$ this can be written as $\frac {dh}{dt} = - K h^{-3/2}$. Rearrange:

$h^{3/2} dh = -Kdt$, and integrate to get $\frac 2 5 h^{(5/2)}= Kt +C$, or $h(t) = [\frac 5 2 (-Kt+C)]^{(2/5)}$ The value for C can be found from the initial conditions: $C = \frac 2 5 h_0 ^{5/2}$