Question

Water flows through a pipe of diameter 0.30 m. What would be the velocity V for the conditions shown in the figure below?

• A. 1.084

Answer 1

The correct option is A 1.084

Manometric reading,
x = 0.3 m
Specific gravity of manometer fluid,
$S_{mano}$=0.8
Density of manometer fluid
${\rho }_{mano}=0.8\times 1000kg/m^3$
$=800kg/m^3$

$\therefore h=x[1-\frac{{\rho }_{mano}}{{\rho }_{pipe}}]=0.3[1-\frac{800}{1000}]$

$=0.06mofwater$

$Velocityofpoint1,$

$V=\sqrt{2gh}=\sqrt{2\times 9.81\times 0.06}$

$=1.084m/s$

Points To Remember
This problem is based on the pitot tube.
Pitot tube: It is a device which is basically used to measure the local velocities in the pipe flow.
Here, velocity at point 1: (V=$\sqrt{2gh}$)is theoretical
velocity

• For actual velocity multiply V by the $C_v$ (coefficient of velocity).
• Value of $C_v$ for pitot tube is 0.97 to 0.99

Hence
$V_{actual}=C_{v}V=C_v\sqrt{2gh}=0.97\times 1.084$
=1.05 m/s