Single-phase Cu cable for three loads

The cable must carry the total current from the supply end and keep the voltage drop at the farthest point within 2 %. Since the loads are distributed along the run, the drop is found by the current-moment method.

Assumptions: 10 HP is the motor output with η taken as 1 (not given); heaters are at unity pf; ρ (Cu) = 0.0178 Ω·mm²/m; drop in both go and return conductors (2-wire circuit), reactance neglected.

Load currents

10 kW heater (75 m): I = 10000 / 230       = 43.48 A
12 kW heater (35 m): I = 12000 / 230       = 52.17 A
10 HP motor  (70 m): P = 10 × 746 = 7460 W
                     I = 7460 / (230 × 0.8) = 40.54 A

Total current at supply end

I = 43.48 + 52.17 + 40.54 (0.8 − j0.6)
  = 128.08 − j24.33
|I| = 130.38 A   (arithmetic sum 136.20 A)

The largest cable in the table (50 mm², 90 A) cannot carry 130.38 A, so one cable from the table is not enough by current rating.

Voltage drop condition

Allowed drop = 2 % of 230 = 4.60 V
Σ I·cosφ·L = 52.17×35 + 40.54×0.8×70 + 43.48×75
           = 1826.09 + 2270.43 + 3260.87
           = 7357.39 A·m
ΔV = 2 ρ Σ(I cosφ L) / A ≤ 4.60
A ≥ 2 × 0.0178 × 7357.39 / 4.60 = 56.94 mm²

Selection

Answer: required area ≥ 56.94 mm² and current ≈ 130 A; using the given table, use two 50 mm² Cu cables in parallel per conductor (100 mm² total, 180 A, drop 2.62 V = 1.14 %). A single 70 mm² or 95 mm² Cu cable could be used if its rating (from a fuller table) exceeds 130 A.