Boost Pump Feeder Cable: Voltage Drop and Reselection
A 28 V DC essential bus supplies a fuel boost pump through a cable run of 15 m, which means 30 m of conductor (out and back). The pump draws 18 A. The installed cable is 4 mm² copper; take ρ = 1.72 × 10⁻⁸ Ω·m at operating temperature. The AMM requires the voltage at the pump terminals to stay at or above 26.5 V with the bus at 28.0 V. During a wiring modification the stores hold only 6 mm² and 8 mm² copper cable, plus 4 mm² aluminium cable (ρ = 2.82 × 10⁻⁸ Ω·m). The circuit is protected by a 25 A fuse. The aircraft operates in a hot climate and the cable runs in a loom with six other wires.
1. Is the installed 4 mm² copper cable compliant? Show the calculation and state the pump terminal voltage and the power wasted in the cable.
Read the worked answer
Resistance of the run
R = ρL/A = (1.72 × 10⁻⁸ × 30) / (4 × 10⁻⁶) = 0.129 Ω
Voltage drop
ΔV = I·R = 18 × 0.129 = 2.32 V
Terminal voltage = 28.0 − 2.32 = 25.68 V
This is below the 26.5 V minimum, so the installation is not compliant. The permitted drop is only 28.0 − 26.5 = 1.5 V.
Power dissipated in the cable
P = I²R = 18² × 0.129 = 41.8 W
That heat is produced inside a loom of seven wires, so it is also a bundling/derating concern, not just a voltage problem. The fix is to increase the cross-section, not to raise the bus voltage.
2. What is the minimum cross-section that satisfies the AMM limit, and which available cable would you fit? Justify the choice, including why the nearest size above the minimum may still not be the right answer.
Read the worked answer
Required area
A = ρLI/ΔV = (1.72 × 10⁻⁸ × 30 × 18) / 1.5 = 6.19 × 10⁻⁶ m² = 6.19 mm²
Check the nearest standard size, 6 mm²
R = (1.72 × 10⁻⁸ × 30) / (6 × 10⁻⁶) = 0.086 Ω → ΔV = 18 × 0.086 = 1.55 V → terminals at 26.45 V. Still non-compliant, even though 6 mm² is above the calculated minimum.
Check 8 mm²
R = (1.72 × 10⁻⁸ × 30) / (8 × 10⁻⁶) = 0.0617 Ω → ΔV = 1.11 V → terminals at 26.89 V. Compliant, with margin for a cold pump surge current and for the higher resistance of a hot loom.
Decision: fit the 8 mm² copper cable. Lessons: (1) round up to a standard size and then re-verify, never assume the next size works; (2) check the wire's current rating with the applicable derating factors for loom bundling, altitude and ambient temperature; (3) confirm the 25 A fuse still protects the smaller... - actually the larger conductor only improves protection, so the fuse remains correct.
3. A colleague proposes using 4 mm² aluminium cable instead, 'because it is the same size and we have it in stock'. Evaluate this proposal quantitatively and practically.
Read the worked answer
Quantitatively
R(Al) = (2.82 × 10⁻⁸ × 30) / (4 × 10⁻⁶) = 0.212 Ω → ΔV = 18 × 0.212 = 3.81 V → terminals at 24.2 V. Badly non-compliant, and 1.6 times worse than the copper cable that already failed.
To match the 1.5 V requirement in aluminium: A = (2.82 × 10⁻⁸ × 30 × 18) / 1.5 = 10.2 mm² - not a stock item here.
Practically
- Equal cross-section does not mean equal performance: aluminium resistivity is about 64 % higher than copper, and it also needs roughly 1.6× the area for the same current rating.
- Aluminium cold-flows and oxidises, needs special terminals and bimetallic interfaces, and has poorer fatigue behaviour under vibration - this is why it is rare in aircraft looms.
- Fitting it would be an unapproved deviation from the approved wire type in the maintenance manual or standard practices manual.
Answer: reject the proposal. Aluminium is only acceptable if the approved data permits it, at a correctly recalculated cross-section, with approved terminals and a documented engineering basis.