# Math Help - liquid draining from tank

1. ## liquid draining from tank

A cylindrical tank with length of 5 ft. and radius 3 ft. is situated with its axis horizontal. If a circular bottom hole with a radius of 1 in. is opened and the tank is initially half full of xylene, how long will it take for the liquid to drain completely?

2. Originally Posted by mandy123
A cylindrical tank with length of 5 ft. and radius 3 ft. is situated with its axis horizontal. If a circular bottom hole with a radius of 1 in. is opened and the tank is initially half full of xylene, how long will it take for the liquid to drain completely?

this is an application of Toricelli's principle ... the speed at which a fluid exits a hole is a function of the height of the fluid above the opening.

$v = \sqrt{2gh}$

further, the rate of change of volume w/r to time has the relationship

$\frac{dV}{dt} = -av$ , where $a$ is the area of the opening.

let $y = 0$ be at the tank's center. the volume of fluid in the tank as a function of the fluid's height above the opening ...

$V = \int_{-3}^{-3+h} 10\sqrt{9 - y^2} \, dy$

$\frac{dV}{dt} = 10\sqrt{9 - (-3+h)^2} \cdot \frac{dh}{dt}$

$-a\sqrt{2gh} = 10\sqrt{6h-h^2} \cdot \frac{dh}{dt}$

$-\frac{a\sqrt{2g}}{10} \, dt = \sqrt{6-h} \, dh$

$-\frac{a\sqrt{2g}}{10} t + C = -\frac{2}{3}(6-h)^{\frac{3}{2}}$

at $t = 0$, $h = 3$ ... $C = -\frac{2}{3}(3)^{\frac{3}{2}}$

when $h = 0$ ...

$t = \frac{20}{3a\sqrt{2g}}\left[(6)^{\frac{3}{2}} - (3)^{\frac{3}{2}}\right]$

$a = \frac{\pi}{144} \, ft^2$

$g = 32 \, \frac{ft}{s^2}$

$t \approx 363 \, s$ or a little over 6 min.

hope I didn't commit any egregious errors ... I'm sure someone will let us both know if I did.

3. Thank you so much, I thought it had something to do with torricelli's law, but I had no clue as where to start, but you explained it very well, and it makes more sense!!! Thanks!!!