Mathmatical modeling with Toricelli's law

[Eclair_de_XII]

Homework Statement

"Fluid will leak out of a hole at the base of a cylindrical container at a rate proportional to the square root of the height of the fluid's surface from the base. If a cylindrical container is initially filled to a height of $h(0)=12 in.$, and it takes one minute for the height to reach $h(60)=3 in.$, how long will it take for all of the fluid to leak out?"

Homework Equations

$h(0)=12 in.$
$h(60)=3 in.$
$\frac{dh}{dt}=-u(t)h(t)$
$u(t)=\sqrt{h}$

The Attempt at a Solution

$\frac{dh}{dt}=-u(t)h(t)=h^{\frac{3}{2}}$
$h^{-\frac{3}{2}}dh=-dt$
$\int h^{-\frac{3}{2}}dh=-\int dt$
$-2h^{-\frac{1}{2}}=-t+C$
$h^{-\frac{1}{2}}=2t+C$
$h(t)=\frac{1}{(2t+C)^2}$

I am very completely sure that this is not correct. For one, $h(t)$ can never be zero. I know I went wrong when trying to derive the expression for $\frac{dh}{dt}$. And I have two conditions at $t=0$ and $t=60$ that I don't think will ever be satisfied simultaneously. What am I doing wrong?

 

[fresh_42]

Why don't you use the Volume $V$ of the liquid as function of height, and height as function of time. Then you get a first order non-linear DE with two boundary conditions.

Maybe you tried something similar, but I can't see what $u(t)$ should be or where the proportion factor has gone to.

 

[Eclair_de_XII]

So I should try to find an expression for $\frac{dV}{dh}\frac{dh}{dt}$, then. I guess I'll start by finding $\frac{dV}{dh}$ and $\frac{dh}{dt}$ and multiplying them by each other?

So... $\frac{dV}{dh}=\pi r^2$. I don't think that's right, since I didn't include the factor of $\sqrt{h}$ in it.

 

[fresh_42]

What is the volume of a cylinder? There is a height in the formula. Then $\frac{dV}{dt}$ is given up to a factor. This combines to a formula of $h(t),h(t)'$ which can be solved. I used WolframAlpha, but a polynomial Ansatz will do.

 

[epenguin]

I don't see why you write dh/dt = h^3/2 and not just Kh^½. You may be having some thoughts about the volume decrease, but volume has a constant relation to height in a cylindrical container.

The rate when h is very small Is very small (" tending to 0 as h tends to 0") so so long as this model holds, the cylinder will never completely empty surely? I guess you will just have to fix some practical point like 95% empty or 99% empty and say what the corresponding times are.

 

[Eclair_de_XII]

What is the volume of a cylinder?

It's $V(h)=\pi r^2 h$, I think.

Then $\frac{dV}{dt}$ is given up to a factor.

How would I write volume as a function of time, though?

I don't see why you write $\frac{dh}{dt} = h^{\frac{3}{2}}$ and not just $Kh^½$

I think I misunderstood what was meant by "a rate proportional to the square root of the height". So I should express the time for $V(t_f)=0.01$ as a function of $K$, then, since it's not given (or maybe it's $K=\sqrt{2g}$)?

 

[fresh_42]

It's $V(h)=\pi r^2 h$, I think.
How would I write volume as a function of time, though?

Just be more exact! Write $V(h(t))=\pi r^2 h(t)$ and differentiate. And don't forget the proportion factor in the formula for $\frac{d}{dt} V(h(t))$.

 

[Eclair_de_XII]

Oh, that sounds easier to do. Let's see... is it $\frac{dV}{dh} \frac{dh}{dt}=(\pi r^2)⋅[h'(t)]=(\pi r^2)⋅(K\sqrt{h})=\frac{dV}{dt}$?

 

[fresh_42]

Oh, that sounds easier to do. Let's see... is it $\frac{dV}{dh} \frac{dh}{dt}=(\pi r^2)⋅[h'(t)]=(\pi r^2)⋅(K\sqrt{h})=\frac{dV}{dt}$?

You messed up the factors. Of course you can always write $V'(t) \sim \sqrt{h(t)}$ as $V'(t) = (\pi r^2 K) \sqrt{h(t)}$ but regarding the text given, I would write $V'(t) = K \sqrt{h(t)}$. But anyway, perhaps this is easier to handle. You just have to keep in mind, that you defined the proportion factor to be $\pi r^2 K$ and not simply $K$, if you plug in your boundary conditions.

 

[epenguin]

I think I misunderstood what was meant by "a rate proportional to the square root of the height". So I should express the time for $V(t_f)=0.01$ as a function of $K$, then, since it's not given (or maybe it's $K=\sqrt{2g}$)?

I think you should read the phrase "proportional to..." as meaning you are not required to come up with any fundamental theory explaining the K in terms of something else.

Also you can think of volume change and its relation to height change, and get formulae involving π, but if you have a formula involving π multiplied by another formula involving an arbitrary constant K, then all you've got is the new formula involving another arbitrary constant K' say. So you might as well leave it out these complications and just have a formula involving a constant and √h .
The actual numerical value of this K is something you can determine from the experimental measurements given in the problem.

I'm not saying your other thoughts were valueless. They would come in handy if they gave you a problem changing the diameter. Or the shape of the vessel, give scope for exercises related to this one.

 

[Eclair_de_XII]

Okay. Thanks.