Air Data Instruments Questions and Answers

Answer: C

Correct answer: 206 knots

\[ V_T = \frac{150}{\sqrt{0.53}} = \frac{150}{0.728} = 206\ \text{knots.} \] The instrument reads low because it was calibrated for sea level density and the air at 20 000 ft delivers less dynamic pressure at any given speed. The error is in the direction that matters for navigation: the aircraft is travelling faster than the dial suggests.

Answer: D

Correct answer: 32 ft/hPa

\[ \frac{10\,000\ \text{ft}}{1013 - 697\ \text{hPa}} = \frac{10\,000}{316} = 31.6 \approx 32\ \text{ft/hPa.} \] The familiar rule of thumb of 30 ft per hectopascal is the near sea level figure; the ratio grows with height as the air thins, reaching about 50 ft/hPa at 20 000 ft, which is why altimeter setting errors matter more at low level.

Answer: A

Correct answer: 12.5 K

\[ \Delta T = 0.2M^2T_s = 0.2 \times 0.5^2 \times 250 = 0.2 \times 0.25 \times 250 = 12.5\ \text{K.} \] Note the square: doubling the Mach number quadruples the rise, so at Mach 1 in the same air the rise would be 50 K. This is why the ram correction is negligible at low speed and dominant at high speed.

Answer: B

Correct answer: 157 knots

Calibrated airspeed is indicated airspeed with both errors applied, each with its own sign: \[ 160 + 1 - 4 = 157\ \text{knots.} \] Instrument error is a property of the individual instrument, position error a property of where the static ports sit in the flow field of that airframe. Applying both with the wrong signs gives 163, which is the distractor.

Answer: C

Correct answer: Both Statement I and Statement II are correct

Both are true and the second explains the first. The instrument measures the difference between total and static pressure, which is dynamic pressure, and dynamic pressure depends on the product of density and the square of the speed. Since the dial is graduated assuming sea level standard density, the reading equals true airspeed only when the air actually is at that density.

Answer: D

Correct answer: Both Statement I and Statement II are correct

Both are true and the second gives the reason. Mach number can be written entirely in terms of the ratio of total to static pressure, and in forming that ratio the density and temperature terms cancel. The machmeter therefore needs only the two pressures already available from the pitot-static system. A true airspeed computation is a different matter and does need temperature.

Answer: A

Both statements are true. With static trapped, the differential grows as pitot pressure rises with speed but shrinks relative to the trapped reference during a climb, and the standard result is an under-reading airspeed in the climb.

Answer: B

Correct answer: Both Statement I and Statement II are correct

Both are true and the second explains the first. The pressure at a static port is the local pressure in the flow around the fuselage, not the free stream pressure, and the difference depends on the flow field. At low speed and high angle of attack the flow over the fuselage is most distorted, so the discrepancy is largest - which is exactly the regime in which an accurate airspeed matters most.

Answer: C

Correct answer: Both Statement I and Statement II are correct

Both are true and the second is the mechanism. Bringing moving air to rest at the probe compresses it, and the compression is close to adiabatic, so the air heats. Total temperature therefore always exceeds static temperature in flight, and the excess grows with the square of the Mach number - about 30 K at Mach 0.8, which is far too large to ignore.

Answer: D

Correct answer: Both Statement I and Statement II are correct

Both are true and the second explains the first. An altimeter converts pressure to height on a fixed assumption about sea level pressure. Fly into a region where that pressure is genuinely lower and the instrument sees the reduced pressure and reports it as extra height, so the aircraft is lower than indicated. The mnemonic is high to low, look out below.