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?

Please help me, I am lost???(Headbang)

Quote:

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?

Please help me, I am lost???(Headbang)

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.

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!!!(Nod)