Question

A tank having a small circular hole contains oil on top of water. It is immersed in a large tank of the same oil. Water flows through the hole. What is the velocity of this flow initially? When the flow stops, what would be the position of the oil-water interface in the tank from the bottom. The specific gravity of oil is $\mathrm{0.5.}$

Answer 1

(a) $\Delta p=h({\rho }_w-{\rho }_0)g=\left(10\right)(1000-500)9.8$
$=49000$ N/m${}^2$
Now, $\Delta P=\frac{1}{2}{\rho }_w{v}_2$
$\therefore$ $v=\sqrt{\frac{2\Delta P}{{\rho }_w}}$
$=\sqrt{\frac{2\times 49000}{1000}}$
$=9.8$ m/s
The flow will stop when,
(b) $(10+5){\rho }_{0}g=5{\rho }_{0}g+h{\rho }_{w}g$
$10{\rho }_0=h{\rho }_w$
$\therefore$ $h=\frac{10\times 500}{1000}$
$=5$ m
i.e., flow will stop when the water-oil interface is at a height of 5.0 m.