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Aluminium cable for 25 kW single-phase load

Short answer

Using voltage drop formula ΔV = I × ρ × 2L / A with 3% limit (6.6 V), load current 162.34 A, loop length 80 m, aluminium resistivity 0.0283 Ω·mm²/m, minimum area is 55.69 mm²; select 70 mm² aluminium cable (2.39% drop).

Calculate the cross-sectional area of Aluminum cable require to supply a single phase load of 25 KW at power factor of 0.7 lagging which is 40 m away from the voltage source so that voltage drop in the cable does not exceed 3% of the standard supply voltage of 220 V.

The cross-section is found from the permissible voltage drop of the go-and-return (2-wire) circuit.

Constants: resistivity of aluminium ρ = 0.0283 Ω·mm²/m (2.83 × 10⁻⁸ Ω·m); the whole I·R drop is taken (conservative, reactance neglected).

Load current I = P / (V × cosφ)
               = 25000 / (220 × 0.7) = 162.34 A
Allowed drop   = 3 % of 220 = 6.60 V
Loop length    = 2 × 40 = 80 m
ΔV = I × R = I × ρ × 2L / A ≤ 6.60
A  = ρ × 2L × I / ΔV
   = 0.0283 × 80 × 162.34 / 6.60
   = 55.69 mm²

Next standard size: 70 mm².

Check: ΔV = 0.0283 × 80 × 162.34 / 70
          = 5.25 V (2.39 %) < 6.60 V  OK

Answer: minimum area = 55.69 mm²; use a 70 mm² aluminium cable (drop 5.25 V). If only the in-phase component of drop (I·R·cosφ) is considered, A = 38.98 mm² and a 50 mm² cable would just satisfy the drop limit; the current rating of the chosen size must also be checked from the manufacturer's table for 162 A.

✓ Answer checked by Claude Opus 5.5 (high effort)

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