Question

A manometer connected to a Pitot-static tube has a difference in the height of the two columns of water of 10 cm when the Pitot-static tube is placed in a flow of air at standard sea level conditions. What is the velocity of the airflow

Answer 1

Answer:

v = 40 m/s

Explanation:

Difference in the height of the two columns:

$h_m = \triangle h = 10 cm \\h_m = 0.1 m$

Air density, $\rho_{air} = 1.225 kg/m^3$

Water density, $\rho_{w} = 1000 kg/m^3$

The velocity of air flow is given by the equation:

$v = \sqrt{2gh_m (\frac{p_w}{p_{air}} - 1) } \\\\v = \sqrt{2*9.81*0.1 (\frac{1000}{1.225} - 1) }\\\\v = 40 m/s$