Question

A Pitot tube (Shown in figure above) is mounted along the axis of a gas pipeline whose cross-sectional area is equal to $S$. Assuming the viscosity to be negligible, the volume of gas flowing across the section of the pipe per unit time, if the difference in the liquid columns is equal to $\Delta h$, and the densities of the liquid and the gas are ${\rho }_0$ and $\rho$ respectively.

Answer 1

Applying Bernoulli's theorem for the point $A$ and $B$,
$P_A=P_B+\frac{1}{2}\rho v^2$ as, $v_A=0$
or, $\frac{1}{2}\rho v^2=P_A-P_B=\Delta h{\rho }_{0}g$
So, $v=\sqrt{\frac{2\Delta h{\rho }_{0}g}{\rho }}$
Thus, rate of flow of gas, $Q=Sv=S\sqrt{\frac{2\Delta h{\rho }_{0}g}{\rho }}$
The gas flows over the tube past it at $B$. But at $A$ the gas becomes stationary as the gas will move into the tube which already contains gas.
In applying Bernoulli's theorem we should remember that:
$\frac{p}{\rho }+\frac{1}{2}{v}^2+gz$ is constant along a streamline.

In the present case, we are really applying Bernoulli's theorem somewhat indirectly. The streamline at $A$ is not the streamline at $B$. Nevertheless the result is correct. To be convinced of this, we need only apply Bernoulli's theorem to the streamline that goes through $A$ by comparing the situation at $A$ with that above $B$ on the same level. In steady conditions, this agrees with the result derived because there cannot be a transverse pressure differential.