Wind Tunnel Testing Questions and Answers

Answer: A

A thin oil film is dragged along by the surface shear, so when it dries it records the skin-friction line pattern. Separation and reattachment lines, transition fronts and vortex footprints all show up clearly, which makes it the standard first look at a new configuration.

Answer: B

Illuminating a seeded plane twice in quick succession and cross-correlating the two images gives the velocity field across the whole plane at once. It is non-intrusive and rich, though seeding a large tunnel uniformly and safely is a substantial practical problem.

Answer: C

The paint's luminescence is quenched by oxygen, so its brightness falls as the local pressure rises. It gives a continuous pressure map rather than the handful of values discrete taps provide, at the cost of careful temperature correction and calibration.

Answer: D

A sting entering through the base changes the base pressure and the near wake, and a strut disturbs the flow around it. The usual remedy is to repeat the run with a dummy support present and subtract the difference, which isolates the interference.

Answer: A

The ratio measures how efficiently the circuit converts shaft power into kinetic energy at the model, and values well above one are normal for a good closed-circuit tunnel because the air is recirculated. It is the figure of merit used when comparing tunnel designs.

Answer: B

Continuity for an incompressible flow gives the velocity ratio as the area ratio, so \( 6 \times 9 = 54 \) m/s. The large contraction is chosen for flow quality rather than speed: it lowers the turbulence intensity relative to the mean velocity as it accelerates the stream.

Answer: C

Reynolds number is proportional to the characteristic length at fixed speed, density and viscosity, so a one-fifth model runs at one fifth the Reynolds number. Transition will sit proportionally further aft than at full scale, which is exactly why a trip strip is fitted.

Answer: D

Correct answer: 1531 Pa

\[ q = \tfrac{1}{2}\rho V^{2} = 0.5 \times 1.225 \times 50^{2} = 0.5 \times 1.225 \times 2500 = 1531\ \mathrm{Pa}. \] Omitting the factor of one half gives the 3063 offered. Dynamic pressure matters because every aerodynamic coefficient is a force divided by \( qS \), so an error in it carries straight through to every result the tunnel produces.

Answer: A

The ratio is \( 0.08/2.0 = 0.04 \), that is four per cent. This sits inside the usual guideline of five per cent or so, above which the blockage corrections become large enough to cast doubt on the data rather than merely adjust it.

Answer: B

Dividing through, \( 240 / (1500 \times 0.4) = 240/600 = 0.40 \). Reducing raw balance readings to coefficients in this way is what makes the result transferable to full scale, provided the similarity conditions have been respected.