A tank has a small hole in its side at a heighty ${\mathrm{y}}_1$. It is filled with a liquid (density $\mathrm{\rho }$) to a height ${\mathrm{y}}_2$. If the absolute pressure at the top of the fluid is ${\mathrm{P}}_{\mathrm{t}}$, find the velocity with which it leaves the tank. Assume that the cross- sectional area of the tank is larger as compared to that of the hole.

 

From equation of continuity, ${\mathrm{v}}_2 = \frac{{\mathrm{A}}_1}{{\mathrm{A}}_2}{\mathrm{v}}_1$

Since, ${\mathrm{A}}_1<<{\mathrm{A}}_2, {\mathrm{v}}_2 \mathrm{must} \mathrm{be} \mathrm{ver} \mathrm{small} \mathrm{compared} \mathrm{to} \mathrm{velocity} \mathrm{of} \mathrm{efflux} \mathrm{at} \mathrm{the} \mathrm{hole}, \mathrm{therefore} \mathrm{we} \mathrm{can} \mathrm{take} {\mathrm{v}}_2 =0$

Fluid emerging from the hole is open to atmospheric pressure ${\mathrm{P}}_0$, We take two points A and B at the top of the fluid and at the hole respectively. From Bernoulli's principle,

${\mathrm{P}}_0+\frac{1}{2}{\mathrm{\rho v}}_1^2+{\mathrm{\rho gv}}_1 = {\mathrm{P}}_1+{\mathrm{\rho gy}}_2$

Solving for ${\mathrm{v}}_1, \mathrm{we} \mathrm{obtain} {\mathrm{v}}_1= \sqrt{\frac{2({\mathrm{p}}_{\mathrm{t}}-{\mathrm{p}}_0)}{\mathrm{\rho }}}+2\mathrm{gh}$