Examples In Electrical Calculations By Admiralty Pdf (2025)

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

Gibbs calculated required capacitive reactive power to raise PF to 0.90.

Then cable cross-section area (A): [ A = \frac{\rho \times L}{R} = \frac{0.0175 \times 45}{0.0194} \approx 40.6\ \text{mm}^2 ]

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. examples in electrical calculations by admiralty pdf

For PF=0.90, new apparent power (S_2 = P / 0.90 = 5.2 / 0.90 \approx 5.78\ \text{kVA}) New reactive power (Q_2 = \sqrt{5.78^2 - 5.2^2} \approx 2.52\ \text{kVAR})

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.

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 ] What I can do is provide an based

Maximum allowable drop per core: 1.65 V (two cores in series).

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.

Initial reactive power (Q_1 = \sqrt{S^2 - P^2} = \sqrt{8^2 - 5.2^2} \approx 6.08\ \text{kVAR}) Then cable cross-section area (A): [ A =

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.

Fault current: (I_{short} = 110 / 0.0856 \approx 1285\ \text{A}).

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.