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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.Quote:
Originally Posted by mandy123
$\displaystyle v = \sqrt{2gh}$
further, the rate of change of volume w/r to time has the relationship
$\displaystyle \frac{dV}{dt} = -av$ , where $\displaystyle a$ is the area of the opening.
let $\displaystyle y = 0$ be at the tank's center. the volume of fluid in the tank as a function of the fluid's height above the opening ...
$\displaystyle V = \int_{-3}^{-3+h} 10\sqrt{9 - y^2} \, dy$
$\displaystyle \frac{dV}{dt} = 10\sqrt{9 - (-3+h)^2} \cdot \frac{dh}{dt}$
$\displaystyle -a\sqrt{2gh} = 10\sqrt{6h-h^2} \cdot \frac{dh}{dt}$
$\displaystyle -\frac{a\sqrt{2g}}{10} \, dt = \sqrt{6-h} \, dh$
$\displaystyle -\frac{a\sqrt{2g}}{10} t + C = -\frac{2}{3}(6-h)^{\frac{3}{2}}$
at $\displaystyle t = 0$, $\displaystyle h = 3$ ... $\displaystyle C = -\frac{2}{3}(3)^{\frac{3}{2}}$
when $\displaystyle h = 0$ ...
$\displaystyle t = \frac{20}{3a\sqrt{2g}}\left[(6)^{\frac{3}{2}} - (3)^{\frac{3}{2}}\right]$
$\displaystyle a = \frac{\pi}{144} \, ft^2$
$\displaystyle g = 32 \, \frac{ft}{s^2}$
$\displaystyle t \approx 363 \, s$ 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)
