Ground-Run Noise and Ultrasonic Coverage
A returning-to-service package contains two independent items. (1) With one engine at idle, the sound pressure level is 90 dB at 1.0 m from the intake lip. Both engines are identical, their noise is incoherent, and the aerodrome limit at the nearest boundary is 80 dB. The boundary point is 3.0 m from the nearer intake and 9.0 m from the other, the second engine being 6 m further away. (2) A 5 MHz probe with a 10 mm crystal is used on a 2.0 mm aluminium wing skin to search for a suspected 0.4 mm inclusion. The set is calibrated for 6300 m/s. A technician proposes to sign off both items from these two readings alone.
1. Estimate the level at the boundary and decide whether the ground run complies.
Read the worked answer
Spherical spreading: L = L₁m − 20 log₁₀ r.
- Near engine at 3 m: 90 − 20 log 3 = 90 − 9.5 = 80.5 dB
- Far engine at 9 m: 90 − 20 log 9 = 90 − 19.1 = 70.9 dB
Incoherent sources add on power: L = 10 log(10^8.05 + 10^7.09) = 10 log(1.24 × 10⁸) = 80.9 dB.
That is about 1 dB above the 80 dB limit, so the run does not comply. Two cautions on the estimate: a hard apron adds a reflecting image of up to +3 dB, the intake is not a point source at 3 m, and atmospheric absorption has been ignored - the true figure could be 2–3 dB worse. Note the far engine contributes only ~0.4 dB; turning the aircraft or screening the near intake is the effective fix, relocating the far one is not.
2. Is the 5 MHz probe capable of detecting the 0.4 mm inclusion? Show the numbers.
Read the worked answer
Wavelength: λ = c/f = 6300/(5 × 10⁶) = 1.26 mm.
The usual detection limit is about λ/2 ≈ 0.63 mm (some texts use λ). A 0.4 mm inclusion is below that, so a 'no defect found' result with this probe is not evidence of absence.
A 10 MHz probe gives λ = 0.63 mm and a limit near 0.32 mm, at the cost of higher attenuation and a shorter useful range. A focused probe or an angle beam may be needed for a volumetric defect of that size.
Judgement: the technician cannot sign off item (2). He can sign off item (1) only as a 'does not comply' finding.
3. The skin is only 2.0 mm thick. What does the near-field length say about this measurement?
Read the worked answer
Near-field length for a circular crystal: N = D²/(4λ)
N = (10 × 10⁻³)² / (4 × 1.26 × 10⁻³) = 1.0 × 10⁻⁴ / 5.04 × 10⁻³ = 19.8 mm.
The first ~20 mm in front of the crystal is in the near field, where the beam has not yet formed and echoes merge with the transmit pulse; the practical dead zone is several millimetres even with good damping. A 2.0 mm skin lies entirely inside that region for this probe.
So neither the thickness nor any defect sizing from this setup is reliable, whatever the screen shows. Correct method: a delay-line or purpose-designed thin-material probe, calibrated on a step wedge of the same alloy, with the gain set from a reference reflector rather than from the back wall.