# Find the velocity of air in the test section of a low speed open circuit subsonic wind tunnel which has an inlet -to-throat area ratio of $$10$$. The U-tube mercury manometer read as a height difference of $$10.5\,cm$$ for the pressure difference between the inlet and the test section of the wind tunnel.

Find the velocity of air in the test section of a low speed open circuit subsonic wind tunnel which has an inlet -to-throat area ratio of $$10$$. The U-tube mercury manometer read as a height difference of $$10.5\,cm$$ for the pressure difference between the inlet and the test section of the wind tunnel.

Kisan Kumar Asked on 15th June 2021 in
Velocity of air in the test section of a low speed open circuit subsonic wind tunnel is given by  $V_{2} = \sqrt{\frac{2\left ( p_{1} – p_{2} \right )}{\rho \left [ 1 – \left ( \frac{A_{2}}{A_{1}} \right )^{2} \right ]}}$Pressure difference, $$p_{1}-p_{2}$$ is measured by means of a manometer.$p_{1}-p_{2}= \gamma \Delta h$Here, $$\gamma$$ = density of mercury * acceleration due to gravity
$$\Delta h$$ = Difference in heights of the liquid in the manometer between its two side.
Therefore, $V_{2}=\sqrt{\frac{2\gamma \Delta h}{\rho \left [ 1-\left ( \frac{A_{2}}{A_{1}} \right )^{2} \right ]}}$$\gamma = \left ( 1.36 \times 10^{4} \right )\left ( 9.8 \right )=1.33\times 10^{5}N/m^{2}$$h = 10.5\,cm = 0.105\,m$$\Rightarrow V_{2} = \sqrt{\frac{2\times \left ( 1.33\times 10^{5} \right )\times 0.105}{1.225\left [ 1-\left ( \frac{1}{10} \right )^{2} \right ]}}=151.8\,m/s$