Question

Two manometric tubes are mounted on a horizontal pipe of varying cross-section at the section $S_1$ and $S_2$ (Fig. $1.81$). Find the volume of water flowing across the pipe in unit time if the difference in water columns is equal to $\Delta h$.

Answer 1

From the conservation of mass

$v_1{S}_1=v_2{S}_2$

But $S_1<S_2$ as shown in the figure of the problem, therefore

As every streamlines is horizontal between $1$ & $2$m Bernoull's theorem becomes

$p+\frac{1}{2}\rho v^2=$ constant, which gives

$p_1<p_2$ as $v_1>v_2$

As the difference in height of the water column is $\Delta h$, therefore

$p_2-p_1=\rho g\Delta h$

From Bernoull's theorem between points $1$ and $2$ of streamline

$p_1+\frac{1}{2}\rho v_1^2=p_2+\frac{1}{2}\rho v_2^2$

or, $p_2-p_1=\frac{1}{2}\rho (v_1^2-v_2^2)$

or $\rho g\Delta h=\frac{1}{2}\rho (v_1^2-v_2^2)$

using $\left(1\right)$ in $\left(3\right)$, we get

$v_1=S_2\sqrt{\frac{2g\Delta h}{S_2^2-S_1^2}}$

Hence the sought volume of water flowing per see

$Q=v_1{S}_1=S_1{S}_2\sqrt{\frac{2g\Delta h}{S_2^2-S_1^2}}$