Question

A rectangular wing of 1.2 m chord length and aspect ratio 5 is tested in a wind tunnel at an air speed of 60 m/s. The density and the dynamic viscosity of air are 1.3 kg/m\(^3\) and \( 1.8 \times 10^{-5} \) kg/m-s, respectively. A second rectangular wing of the same span, but with an aspect ratio of 6, is to be tested in the same tunnel at the same Reynolds number. The air speed at which the second test should be performed is _________ m/s (answer in integer).

Hint:

For ensuring the same Reynolds number, the airspeed and chord length must be adjusted according to the changes in the aspect ratio. The formula \( V_2 = \frac{V_1 L_1}{L_2} \) helps maintain the same Reynolds number.

Answer 1

Correct Answer: 72

Given Parameters \begin{table}[h] \centering \begin{tabular}{lll} \toprule Parameter & First Wing & Second Wing
\midrule Chord length & 1.2 m & \(c_2\)
Aspect ratio & 5 & 6
Air speed & 60 m/s & \(V_2\)
Span & \(b\) & Same as first wing
\bottomrule \end{tabular} \end{table} Common properties: Air density (\(\rho\)) = 1.3 kg/m\(^3\) Dynamic viscosity (\(\mu\)) = \(1.8 \times 10^{-5}\) kg/m/s
Step 1: Calculate Span of First Wing
The aspect ratio (\(AR\)) is defined as: \[ AR = \frac{{Span}}{{Chord}} = \frac{b}{c} \] For the first wing: \[ 5 = \frac{b}{1.2} \implies b = 5 \times 1.2 = 6 \, {m} \] Step 2: Calculate Chord of Second Wing
For the second wing (same span): \[ 6 = \frac{6}{c_2} \implies c_2 = \frac{6}{6} = 1 \, {m} \] Step 3: Compute Reynolds Number for First Wing
\[ Re = \frac{\rho V c}{\mu} \] \[ Re_1 = \frac{1.3 \times 60 \times 1.2}{1.8 \times 10^{-5}} = \frac{93.6}{1.8 \times 10^{-5}} = 5.2 \times 10^6 \] Step 4: Set Equal Reynolds Numbers
For the second wing: \[ Re_2 = \frac{1.3 \times V_2 \times 1}{1.8 \times 10^{-5}} = 5.2 \times 10^6 \] \[ V_2 = \frac{5.2 \times 10^6 \times 1.8 \times 10^{-5}}{1.3} = \frac{93.6}{1.3} = 72 \, {m/s} \] Final Answer The required air speed for the second test is \(\boxed{72}\) m/s.