How To Calculate Cable Size Verified -

For a three-phase circuit (line-to-line): [ V_d = \sqrt3 \times I_b \times (R \cos\phi + X \sin\phi) \times L ]

: Voltage drop. For 50 mm², AC resistance at 70°C ≈ 0.494 Ω/km, reactance ≈ 0.088 Ω/km. Three-phase drop: [ V_d = \sqrt3 \times 90.3 \times (0.494 \times 0.85 + 0.088 \times \sin(\cos^-10.85)) \times 0.120 ] sinϕ = 0.527. R term = 0.494×0.85 = 0.4199 X term = 0.088×0.527 = 0.0464 Sum = 0.4663 Ω/km per phase. ( V_d = 1.732 \times 90.3 \times 0.4663 \times 0.120 = 8.74V ) Percent = 8.74/400×100 = 2.18% < 3% OK.

Abstract The correct sizing of electrical cables is a critical task in power system design, directly impacting safety, efficiency, reliability, and economic viability. An undersized cable leads to overheating, insulation breakdown, voltage drops, energy losses, and fire hazards. An oversized cable results in unnecessary material costs, difficult installation, and reduced fault detection sensitivity. This paper provides a rigorous, step-by-step examination of the scientific and regulatory principles governing cable sizing. It explores the four fundamental determinants: current-carrying capacity (ampacity), voltage drop, short-circuit temperature rise, and economic optimization. The paper derives key formulas, explains correction factors for installation conditions, and presents worked examples based on international standards (IEC 60364 and BS 7671, with reference to NEC guidelines). 1. Introduction Cable sizing is not a simple lookup from a table; it is a multi-variable optimization problem. The primary goal is to select a conductor cross-sectional area (usually in mm² or AWG/kcmil) such that under all expected operating conditions, the cable’s temperature remains within the insulation’s rating, the voltage at the load remains within tolerance, and the cable can withstand fault currents without damage.

: ( I_b = \frac50000\sqrt3 \times 400 \times 0.85 \times 0.94 = \frac50000553.6 \approx 90.3 A )

: Check earth fault loop impedance (not shown in detail here) – likely passes with 70 mm².

: Short-circuit withstand. ( k=115 ). ( S_min = \sqrt10000^2 \times 0.4/115 = \sqrt40\times 10^6/115 = 6324/115 = 55 mm^2 ). Required minimum 70 mm². Our 50 mm² fails. Therefore increase to 70 mm².

: Select ( I_n = 100A ) (circuit breaker).

[ I_z = I_ref \times k_temp \times k_group \times k_soil \times k_depth \times k_therm ]