Wind Tunnel Testing Questions and Answers

Answer: C

Reynolds number goes as density over viscosity at a fixed speed and size, so the two effects multiply: \( 2.7 \times 2.2 = 5.9 \), close to six. This is the whole case for cryogenic testing, and it is achieved without touching the Mach number.

Answer: D

Multiplying, \( 1.4 \times 200\,000 = 280\,000 \). The tunnel's own turbulence promotes transition, so the model behaves as it would at a higher Reynolds number in still air. Dividing rather than multiplying is the usual error and gives 143,000.

Answer: A

Matching Reynolds number at the same density and viscosity requires the product of speed and length to be equal, so the model must be run ten times faster, at 700 m/s. That is beyond Mach 2 and would destroy the very similarity being sought, which is exactly why the Reynolds-Mach conflict cannot be escaped by speed alone.

Answer: B

The ratio is \( 900/300 = 3.0 \). Values above one are possible because the air is recirculated and only the circuit losses need to be made up each pass, which is the principal economic argument for a closed-circuit layout.

Answer: C

Dividing, \( 900 / (2000 \times 0.5) = 900/1000 = 0.90 \). A lift coefficient near unity is a typical cruise-to-climb value, and the same figure would apply at full scale provided the Reynolds number effects have been properly accounted for.

Answer: D

Density is proportional to pressure at a fixed temperature, and Reynolds number is proportional to density, so the factor is four. Pressurisation is therefore a direct and linear lever, limited only by the cost of the pressure shell and the loads it puts on the model.

Answer: A

An isentropic area ratio of 1.687 corresponds to Mach 2 on the supersonic branch for air. The area ratio alone fixes the test-section Mach number, which is why a supersonic tunnel changes Mach number by changing nozzle blocks or by using flexible walls.

Answer: B

The stored volume at test conditions is \( 20 \times 20 = 400 \) m\( ^3 \), so at \( 8 \) m\( ^3 \) per second the run lasts about 50 s, less the margin needed to hold conditions steady. Short run times are the defining constraint of the blowdown layout.

Answer: C

Horizontal buoyancy is a spurious rearward force produced by the tunnel's own streamwise pressure gradient, so it is subtracted: \( 30 - 1.5 = 28.5 \) N. Five per cent is a significant share of a drag measurement, which is why many tunnels use slightly diverging walls to remove the gradient at source.

Answer: D

The frontal area is \( \pi \times 0.075^2 = 0.0177 \) m\( ^2 \) against a section area of 1.0 m\( ^2 \), giving 1.8 per cent. Comfortably inside the usual guideline, so the blockage corrections here would be a small adjustment rather than a source of doubt.