Wind Tunnel Testing Questions and Answers

Answer: B

The full-scale drag is about 644 N.

Flow similarity for a submerged body means equal Reynolds numbers, and then the drag coefficient is the same. With L the length scale (model half size):

\[ \begin{aligned} V_{m} &= V_{p}\frac{\rho_{p}}{\rho_{m}}\frac{\mu_{m}}{\mu_{p}}\frac{L_{p}}{L_{m}} \\ &= 60 \times \frac{1.2}{1000} \times \frac{1.01 \times 10^{-3}}{1.86 \times 10^{-5}} \times 2 \\ &= 7.82\ \text{m/s} \end{aligned} \]

\[ \begin{aligned} \frac{F_{p}}{F_{m}} &= \frac{\rho_{p}V_{p}^{2}L_{p}^{2}}{\rho_{m}V_{m}^{2}L_{m}^{2}} = \frac{1.2 \times 3600 \times 4}{1000 \times 61.1} \\ &= 0.283 \\ F_{p} &= 2280 \times 0.283 \approx 644\ \text{N} \end{aligned} \]

The same ratio follows directly from \(F \propto \mu^{2}/\rho\) at equal Reynolds number. 322, 1288 and 2576 are this answer times 1/2, 2 and 4, the kind of error a mishandled length-scale factor produces.

Answer: C

The shadowgraph depends on the second derivative of density with respect to spatial coordinate.

Light is bent by refractive index gradients, and refractive index varies linearly with density. In a shadowgraph the light rays are deflected and then allowed to spread before reaching the screen, so the change in illumination depends on how the deflection varies across the field, which is the second derivative \(\partial^{2}\rho/\partial x^{2}\).

The schlieren method responds to the first derivative of density, and interferometry measures density itself. No common optical method responds to the third derivative.

Answer: B

The diffuser throat must be larger than the nozzle throat, with the shock located downstream of the diffuser throat.

During starting, a normal shock passes through the test section at the test Mach number. The total pressure loss across it means the diffuser throat must pass the same mass flow at a lower stagnation pressure, so it needs a larger area: \(A_{t2}/A_{t1} = p_{01}/p_{02} \gt 1\). A diffuser throat the same size as the nozzle throat would choke and the shock could not be swallowed.

Once swallowed, the shock passes through the diffuser throat and settles in the diverging part just downstream of it, where it is weak and its position is stable. A shock held exactly at the throat is unstable: a small disturbance pushes it upstream and the tunnel unstarts. That rules out the two 'just at the diffuser throat' options.

Answer: B

The minimum number of runs is 100.

The aerodynamic coefficients depend on five quantities: speed \(V\), density \(\rho\), viscosity \(\mu\), chord \(c\) and angle of attack \(\alpha\). These involve three base dimensions (mass, length, time), so by the Buckingham pi theorem the five reduce to \(5 - 3 = 2\) independent dimensionless groups:

\[ Re = \frac{\rho V c}{\mu}, \qquad \alpha \]

(the angle is already dimensionless). Taking 10 values of each of the two independent groups needs \(10 \times 10 = 100\) runs.

1,00,000 is \(10^{5}\), what testing all five dimensional parameters separately would cost. 10,000 treats four parameters as independent, and 10 forgets that the angle of attack varies independently of the Reynolds number.