How do you set up a differential equation for draining a conical tank?

[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.

 

[Chestermiller]

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.

 

[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:

dV_outflow = dV_container

Where:

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

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

 

[Chestermiller]

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:

dV_outflow = dV_container

Where:

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

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

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

Also:

v₂ = -dh/dt
A₂ = pi * h * (h*tan(a))

dV_liquid/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)

 

[Maag]

I have done sort of 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?

 

[Chestermiller]

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

V_liquid = 1/3 * pi * h * (h*tan(a))²

Also:

v₂ = -dh/dt
A₂ = pi * (h*tan(a))²

dV_liquid/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.