Question

Water is flowing through a horizontal pipe of varying cross section. At any two places, the diameters of the pipe are 4cm and 2cm. If the pressure difference between the two places is 4.5 cm (of water), then determine the rate of flow of water in the pipe.

Answer 1

Suppose the velocity at first place = $v$

Velocity at second place = $v^{\prime }$

By the equation of continuity, $Av=A^{\prime }{v}^{\prime }$

$\frac{v^{\prime }}{v}=\frac{\pi (4/2{)}^2}{\pi (2/2{)}^2}=\frac{\pi \times 4}{\pi \times 1}=4$.

By Bernouli equation, $P+\frac{1}{2}\rho v^2=P^{\prime }+\frac{1}{2}\rho v^{\prime 2}$

or $P-P^{\prime }=\frac{1}{2}\rho v^{\prime 2}-\frac{1}{2}\rho v^2$

or $4.5=\frac{1}{2}\rho \left[\right(4v{)}^2-v^2]$

or $4.5=\frac{1}{2}\rho \times 3v^2$

or $4.5=\frac{3}{2}\times 1\times v^2$

or $v^2=3$

or $v=\sqrt{3}cm/s$

The rate of flow = $Av=4\pi \times \sqrt{3}=21.8c{m}^3/s$.