Question

A siphon discharges oil of specific gravity 0.8 from a reservoir into open air as shown in figure. The head loss from point 1 to point 2 is $V^2/g$; where $V$ is the velocity of flow in pipe and from point 2 to point 3 is $3V^2/2g$. The volume flow rate in the siphon is _______lps.

• A. 7.9

Answer 1

The correct option is A 7.9
Applying Bernoulli`s equation between 1 and 3
$\frac{p_1}{\rho g}+\frac{V_1^2}{2g}+z_1=\frac{p_3}{\rho g}+\frac{V_3^2}{2g}+z_3+h_{L(1-3)}$
$0+0+5=0+\frac{V^2}{2g}+0+\frac{V^2}{g}+\frac{3V^2}{2g}$ [Assuming $V_1=0$]
$\begin{array}{c}V=4.04m/sec\end{array}$
Volume flow rate, $Q=A\times V$
$=\frac{\pi }{4}\times (0.05{)}^2\times 4.04$
$=7.932\times {10}^{-3}{m}^3/sec$
$=7.93lps$