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Thread: Hemispherical water tank

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    MHF Contributor alexmahone's Avatar
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    Hemispherical water tank

    Consider an initially full hemispherical water tank of radius 4 ft, where the radius r of its circular bottom hole is now unknown. At 1 P.M. the bottom hole is opened and at 1:30 P.M. the depth of water in the tank is 2ft.

    (a) Use Torricelli's law in the form $\displaystyle \frac{dV}{dt}=-(0.6)\pi r^2\sqrt{2gy}$ (taking constriction into account) to determine when the tank will be empty.

    (b) What is the radius of the bottom hole?

    My working:

    $\displaystyle \pi(8y-y^2)\frac{dy}{dt}=-(0.6)\pi r^2\sqrt{2gy}$

    $\displaystyle \int(8y^{1/2}-y^{3/2})dy=-\int{0.6*\sqrt{2g}r^2}dt$

    $\displaystyle \frac{16}{3}y^{3/2}-\frac{2}{5}y^{5/2}=-0.6*\sqrt{2g}r^2t+C$

    $\displaystyle y(0)=4$

    $\displaystyle C=\frac{16}{3}*8-\frac{2}{5}*32=\frac{448}{15}$

    $\displaystyle \frac{16}{3}y^{3/2}-\frac{2}{5}y^{5/2}=-0.6*\sqrt{2g}r^2t+\frac{448}{15}$

    $\displaystyle y(1800)=2$

    $\displaystyle \frac{16}{3}*2^{3/2}-\frac{2}{5}*2^{5/2}=-0.6*8*r^2*1800+\frac{448}{15}$

    $\displaystyle 8640r^2=17.0445$

    $\displaystyle r\approx 0.04442\ ft$
    Last edited by alexmahone; Sep 5th 2011 at 09:36 PM. Reason: Figured it out!
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