# Draining a conical tank

1. Dec 14, 2012

### Maag

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 * r2 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 * r2 = -(dh/dt) * pi * h2 * tan(a)2

What do I do now? I really need help with this.

Thanks in advance, Thure.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Dec 14, 2012

### Staff: Mentor

The smaller the cross sectional area, the higher the velocity (not the other way around). Also, v1 = -dh/dt. Also, v2 = A1v1/A2.

What is your equation for the volume of fluid in the conical container at any arbitrary time t in terms of h and tan(a) ? If fluid is leaving the bottom of tank at a volumetric flow rate of v2A2, what is your equation for the rate of change of fluid volume in the tank with respect to time? Hint: do a mass balance.

3. Dec 15, 2012

### Maag

If doing a mass balance I get that the amount of fluid draining from the container is equal to the amount of fluid running through the outflow, therefore:

dVoutflow = dVcontainer

Where:

dVcontainer = 1/3 * pi * h * (h*tan(a))

dVoutflow = v2*A2 <=> (A1*(-dh/dt)) / (pi*r2), where r is the radius of the outflow?

4. Dec 15, 2012

### Staff: Mentor

This is not dV of liquid in the container. This is the total volume of liquid in the container:

Vliquid = 1/3 * pi * h * (h*tan(a))

Also:

v2 = -dh/dt
A2 = pi * h * (h*tan(a))

dVliquid/dt = (2/3) * pi * h * (dh/dt) * tan(a)

In your problem statement, A1 is not specified. So you just need to leave it arbitrary. Also, in your problem, h1 = 0, and h2 = h. So the outlet flow rate from the tank is equal to

F = A1 * v1 = A1(2g*h)(1/2)

So, the rate of change of liquid volume in the tank is equal to minus the outlet flow rate:

(2/3) * pi * h * (dh/dt) * tan(a) = - A1(2g*h)(1/2)

5. Dec 15, 2012

### Maag

I have done sorta the same for a container where the cross-sectional area of the container is a constant. However, I ended up with the differential equation:

dh(t)/dt = -(A/B)*√(2g*h(t)), where A is the cross sectional area of the outflow and B is the cross sectional area of the container. Furthermore I reached the solution:

h(t) = (√(h0)-(k*t)/2)2, where k = (A/B)*√2g

So my question is, how can I end up with a result which is similar to that?

6. Dec 15, 2012

### Staff: Mentor

There was an error in my previous post. Here is the corrected version:

Vliquid = 1/3 * pi * h * (h*tan(a))2

Also:

v2 = -dh/dt
A2 = pi * (h*tan(a))2

dVliquid/dt = pi * h2 * (dh/dt) * tan2(a)

In your problem statement, A1 is not specified. So you just need to leave it arbitrary. Also, in your problem, h1 = 0, and h2 = h. So the outlet flow rate from the tank is equal to

F = A1 * v1 = A1(2g*h)(1/2)

So, the rate of change of liquid volume in the tank is equal to minus the outlet flow rate:

pi * h2 * (dh/dt) * tan2(a) = - A1(2g*h)(1/2)

This differential equation has the form:

h2 * (dh/dt) = -k √h

So, just integrate it.