Consider a low-speed open-circuit subsonic wind tunnel with an inlet-to-throat area ratio of \(12\). The tunnel is turned on, and the pressure difference between the inlet (the settling chamber) and the test section is read as a height difference of \(10 \;cm\) on a U-tube mercury manometer. Calculate the velocity of the air in the test section.
Consider a low-speed open-circuit subsonic wind tunnel with an inlet-to-throat area ratio of \(12\). The tunnel is turned on, and the pressure difference between the inlet (the settling chamber) and the test section is read as a height difference of \(10 \;cm\) on a U-tube mercury manometer.(The density of liquid mercury is \(1.36 × 104 \;kg/m^{3}\).) Calculate the velocity of the air in the test section.
Here inlet to throat area ratio = 12.We need to find the velocity of air in the test section \(V_{2}\).
\[\rho =1.36\times 10^{4}\;kg/m^{3}
\\V_{2}=\sqrt{\frac{2W\Delta h}{\rho \left [ 1-\left ( A_{2}/A_{1} \right )^{2} \right ]}}
\\W=\rho mg=\left ( 1.36\times 10^{4} \right )\left ( 9.8\;m/s^{2} \right )
\\=1.33\times10^{5}\;N/m^{2}
\\\Delta h=10\;cm=0.1\;m
\\\rho =1.23\;kg/m^{3}
\\\frac{A_{2}}{A_{1}}=\frac{1}{12}\]
Therefore
\[V_{2}=\sqrt{\frac{2\left ( 1.33\times 10^{5} \right )\left ( 0.1 \right )}{\left ( 1.23 \right )\left [ 1-\left ( \frac{1}{12} \right )^{2} \right ]}}
\\=147\;m/s\]
