Question

Two manometric tubes are mounted on a horizontal pipe of varying cross-section at the sections $S_1$ and $S_2$ (figure shown above). The volume of water flowing across the pipe's section per unit time if the difference in water columns is equal to $\Delta h$ is given by $Q=S_1{S}_2\sqrt{\frac{xg\Delta h}{S_2^2-S_1^2}}$. Find the value of $x$.

Answer 1

From the conservation of mass
$v_1{S}_1=v_2{S}_2$ (1)
But $S_1<S_2$ as shown in figure above, therefore
$v_1>v_2$
As every streamline is horizontal between 1 and 2, Bernoulli'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
$p_2-p_1=\rho g\Delta h$ (2)
From Bernoulli's theorem between points 1 and 2 of a 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 $\rho g\Delta h=\frac{1}{2}\rho (v_1^2-v_2^2)$ (3) (using equation (2))
using (1) in (3), 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 sec
$Q=v_1{S}_1=S_1{S}_2\sqrt{\frac{2g\Delta h}{S_2^2-S_1^2}}$