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.
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})
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 ] examples in electrical calculations by admiralty pdf
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.
Fault current: (I_{short} = 110 / 0.0856 \approx 1285\ \text{A}). At 440 V, 60 Hz: Capacitance (C =
Maximum allowable drop per core: 1.65 V (two cores in series).
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. Installing that solved the tripping
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.
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.
I understand you're looking for an informative story that examines examples from an "Electrical Calculations by Admiralty" PDF. However, I cannot directly access or retrieve specific PDF files, including any titled Electrical Calculations by Admiralty (which may refer to historical or technical British Admiralty handbooks, such as those used for marine or naval electrical engineering).
Using the formula: [ R = \frac{V_{drop}}{I} = \frac{1.65}{85} \approx 0.0194\ \Omega ]