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# Efflux Velocity of Fluid through a Small Orifice in a Tube

A horizontal tube of length $$L$$, open at $$A$$ and closed at $$B$$, is filled with an ideal fluid. The end $$B$$ has a small oriffice. The tube is set in rotation in the horizontal plane with angular velocity $$\omega$$ about a vertical axis passing through $$A$$. Show that the efflux velocity of the fluid is given by $$v=\omega l\sqrt{\frac{2L}{l}-1}$$ where $$l$$ is the length of the fluid.

Solution:

Consider a mass element $$dm$$ of the fluid at a distance $$x$$ from the vertical axis. The centrifugal force on $$dm$$ is
$$Df=dm\omega^2x$$ $$=dm\frac{dv}{dt}$$ $$=dm \frac{dv}{dx}v$$
$$vdv=\omega^2 xdx$$
$$\int vdv=\omega^2 \int xdx$$ $$\frac{v^2}{2}=\frac{\omega^2}{2} \int_{L}^{L-l}$$
So,
$$v=\omega l\sqrt{\frac{2L}{l}-1}$$

October 17, 2017