# Thread: Diff Eq for a conical tank

1. ## Diff Eq for a conical tank

Hello -

I'm trying to come up with a differential equation to model a conical tank with a flow in (Fo) and a flow out (F=K(sqrt(h)))which is a function of the valve coefficient and the height of the liquid in the tank. The differential equation has to be in terms of the height of the liquid in the tank which is what I am unsure about. Should the differential equation be in terms of dh/dt or just h itself?

I am given placeholder parameters like Rmax=max radius of cone, Hmax = max height of cone etc. I am not looking for this diff eq to be solved...just looking for a little assistance coming up with the correct equation. Would the volume forumula for a cone come into play here?

Thanks for any help!

2. Yes, it would.

$Volume = \dfrac13 \pi r^2h$

You are looking for how the height of liquid changes with time, so you are looking for $\dfrac{dh}{dt}$

Overall rate of flow is $F_o - F = F_o- k\sqrt{h}$

Then you'll need to use the chain rule:

$\dfrac{dV}{dt} = \dfrac{dV}{dh} \times \dfrac{dh}{dt}$

You'll therefore need to differentiate V with respect to h. This also means that you'll need the relationship between h and r to solve this.

3. In this case r=1/2h as Hmax=10 and Rmax=5 so I'm assuming I can use that as the relationship between h and r.

4. Right!

Use this to express V in terms of h only.

Then, you apply the chain rule.

5. I was never very good at these types of problems so I have a stupid question. How do the two equations come together?

I would say:
$\dfrac{dV}{dt} = \dfrac{dV}{dh} \times \dfrac{dh}{dt}=F_o- k\sqrt{h}$

But I am unsure where to substitute the volume equation I have:
$V=\dfrac{1}{3}\pi(\dfrac{1}{2}h)^2h$

6. In the equation:

$\dfrac{dV}{dt} = \dfrac{dV}{dh} \times \dfrac{dh}{dt}$

You already have $\dfrac{dV}{dt} = F_o- k\sqrt{h}$

Ok, that's one.

You are looking for $\dfrac{dh}{dt}$

So, you must look for some way of getting $\dfrac{dV}{dh}$

This is obtained by differentiating the volume.

$V=\dfrac{1}{3}\pi(\dfrac{1}{2}h)^2h = \dfrac{1}{12}\pi h^3$

Can you find $\dfrac{dV}{dh}$ now?