DFQ: "A vertical conical tank..."

**Noxe** · November 14th 2008, 04:45 PM

Hey guys.

I've spent probably 6 hours today trying to solve a basic differential equation, without success, and I'm about to go crazy.

The problem:

"A vertical conical tank, point down, with a radius of 0.5 meters, and a height of 3 meters, is filled with water. How long does it take to drain the tank through a small hole in the bottom of the tank, when the water level has sunk by 1.5 meters after 45 minutes?"

The answer is supposed to be ~55 minutes.

I'm not gonna do the entire thing here, but here is general way of how I tried solving it:

1) Volume of a cone is $V = \frac{\pi * r^2 * h}{3}$ => $V(t) = \frac{\pi * r^2}{3} * h(t)$
2) General solution to a "draining tank through a small hole" is $\frac{dV}{dt} = -k * \sqrt{h}$
3) Rearranging and finding the derivative, getting $\frac{\pi * r^2}{3} * \frac{dh}{\sqrt{h}}= -k dt$
4) Integrating both sides = $\frac{2\pi * r^2}{3} * \sqrt{h} = -kt+C$
5) Finding an expression for h(t): $h(t) = ((-kt+C) * \frac{3}{2\pi*r^2})^2$
6) Using t = 0 to find C, and then the information about the water level after 45 minutes, to find k.
7) Finding the time when h(t) = 0 (tank is empty). Standard second degree equation.
8) Getting the wrong answer. All but one time, I've gotten t = 2.56 hours, which is completely wrong.

Anyone able to see if there's anything wrong with that? Or am I just making some sort of fluke error while calculating? Anyone able to get it right (~55 minutes)?

Thanks!

**skeeter** · November 14th 2008, 06:44 PM

your mistake is treating r as a constant ... it's not.

I arrived at 54.663 min, but hear this ... this problem has some ugly constants.

$r = \frac{h}{6}$

$V = \frac{\pi}{3}\left(\frac{h}{6}\right)^2 h$

$V = \frac{\pi}{108} h^3$

$\frac{dV}{dt} = \frac{\pi}{36} h^2 \frac{dh}{dt}$

$-k\sqrt{h} = \frac{\pi}{36} h^2 \frac{dh}{dt}$

$-\frac{36k}{\pi} \, dt = h^{\frac{3}{2}} \, dh$

$-\frac{36k}{\pi} t + C = \frac{2}{5} h^{\frac{5}{2}}$

when t = 0 , h = 3 ...

$C = \frac{2}{5} 3^{\frac{5}{2}}$

when t = 45 , h = 1.5

solve for k

I'll leave you to work out the rest ... like I said C and k are two ugly constants. I stored their values in my calculator and found that h = 0 at about t = 55 min.

**Noxe** · November 15th 2008, 02:10 AM

Ah...! Thanks a lot man, you just saved my hair from being viciously ripped out.

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