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 \(10\).

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.

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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, \[V_{2}=\sqrt{\frac{2\gamma \Delta h}{\rho \left [ 1-\left ( \frac{A_{2}}{A_{1}} \right )^{2} \right ]}}\]\[\gamma = \left ( 1.36 \times 10^{4} \right )\left ( 9.8 \right )=1.33\times 10^{5}N/m^{2}\]\[h = 10.5\,cm = 0.105\,m\]\[\Rightarrow  V_{2} = \sqrt{\frac{2\times \left ( 1.33\times 10^{5} \right )\times 0.105}{1.225\left [ 1-\left ( \frac{1}{10} \right )^{2} \right ]}}=151.8\,m/s\]

 

Answered on 25th June 2021.
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