In Electrical Calculations By Admiralty Pdf | Examples
From the Admiralty tables, he knew copper’s resistivity at 20°C: (or 0.0175 Ω·mm²/m). The manual demanded voltage drop not exceed 3% for power circuits.
Load current: (I = P/V = 3000/110 \approx 27.3\ \text{A}). The fuse was rated 40 A — fine for overload. But for short-circuit, the prospective fault current matters.
The Admiralty tables listed nearest standard: copper cable. Installing that solved the tripping. Gibbs noted: “Always account for temperature rise — use 0.0204 Ω·mm²/m at 45°C for safety.” Example 2: Short-Circuit Calculation for a Searchlight A 3 kW searchlight (110 V) suddenly failed. A cable chafed against a bulkhead, causing a dead short. Gibbs needed to prove the protective fuse was correct.
At 440 V, 60 Hz: Capacitance (C = \frac{Q_c}{2\pi f V^2} = \frac{3560}{2\pi \times 60 \times 440^2} \approx 48.7\ \mu\text{F}) per phase. examples in electrical calculations by admiralty pdf
Required correction: (Q_c = Q_1 - Q_2 \approx 3.56\ \text{kVAR}) (capacitive).
Cable data: 16 mm² copper, length 30 m round trip. Resistance: [ R_{cable} = \rho \times \frac{L}{A} = 0.0175 \times \frac{60}{16} \approx 0.0656\ \Omega ]
Gibbs calculated required capacitive reactive power to raise PF to 0.90. From the Admiralty tables, he knew copper’s resistivity
Using the formula: [ R = \frac{V_{drop}}{I} = \frac{1.65}{85} \approx 0.0194\ \Omega ]
Battery internal resistance (from Admiralty battery tables for that bank): ~0.02 Ω. Total resistance ~0.0856 Ω.
Maximum allowable drop per core: 1.65 V (two cores in series). The fuse was rated 40 A — fine for overload
Initial reactive power (Q_1 = \sqrt{S^2 - P^2} = \sqrt{8^2 - 5.2^2} \approx 6.08\ \text{kVAR})
Chief Electrician Arthur Gibbs wiped salt spray from his spectacles. Below decks, the newly installed gyrocompass was humming erratically. The Captain wanted answers. Gibbs reached for the worn, blue-covered manual: — his bible for shipboard power systems. Example 1: Cable Sizing for a Deck Winch The forward mooring winch had been tripping its breaker. Gibbs suspected voltage drop. The winch motor drew 85 A at 110 V DC (common on older naval vessels). The cable run from the main switchboard to the winch was 45 meters of two-core armored cable.
Checking the fuse’s time-current curve (Admiralty Handbook, Plate 12), a 40 A fuse would clear 1285 A in ~0.01 seconds — safe. But the mechanical switch arced badly. Gibbs recommended adding a high-speed circuit breaker. Post-war, HMS Vigilant got new radar. The induction motor load (radar rotating aerial) had a power factor of 0.65 lagging . Apparent power S = 8 kVA, true power P = 5.2 kW. The generator ran hot.
What I can do is provide an based on the type of electrical calculation examples typically found in such Admiralty or naval engineering manuals. This will illustrate the principles, context, and practical application. Story: The Chief Electrician’s Logbook HM Destroyer Vigilant , North Atlantic, 1943
