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.

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.

Maximum allowable drop per core: 1.65 V (two cores in series). examples in electrical calculations by admiralty pdf

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

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). The Admiralty tables listed nearest standard: copper cable

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 ]

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}) A cable chafed against a bulkhead, causing a dead short

Required correction: (Q_c = Q_1 - Q_2 \approx 3.56\ \text{kVAR}) (capacitive).

Battery internal resistance (from Admiralty battery tables for that bank): ~0.02 Ω. Total resistance ~0.0856 Ω.

Using the formula: [ R = \frac{V_{drop}}{I} = \frac{1.65}{85} \approx 0.0194\ \Omega ]