Question

A siphon tube is discharging a liquid of specific gravity $0.9$ from a reservoir as shown in the figure find the velocity of the liquid through the siphon.

Answer 1

Using the Bernoulli theorem,

$P_A-\frac{1}{2}\rho v^2+\rho gh=P_D$

But $P_A=P_D=P_0$

$\therefore$ $\frac{1}{2}\rho v^2=\rho gh$

$\therefore$ $v=\sqrt{2gh}$

Here, $h$ = (4+1)= 5 m

$\therefore$ $v=\sqrt{2\times 9.8\times 5}$

$=9.9$ m/s