IEC 60364-5-52 / IEC 60364-4-4110 min read

Worked Example: Sizing a 75 kW DOL Motor Cable — Derating, Voltage Drop, Motor Start and Earth-Fault Loop

Circuit C04 of the suite's worked example project. Current capacity alone would allow 70 mm²; cumulative voltage drop and the earth-fault loop push the cable to 95 mm².

Scenario

LoadProcess pump P-101, 75 kW DOL motor, η 0.95, cos φ 0.86, starting current 7 × FLC at cos φ 0.3
Supply400 V TN-S, from MCC-01 (Ik3 30 kA, Ik1 min 12 kA); upstream feeder C03 already drops 2.46 %
CableCu/XLPE/SWA multicore, 150 m, on ladder (method E), 1 tier, 6 circuits touching
SiteOutdoor, shaded, air 50 °C
ProtectionMotor starter: magnetic-only MCCB + overload relay, let-through I²t 1 × 10⁶ A²s
LimitsCircuit VD 3 %, total LV VD 5 %, starting VD 15 % (including 3 % bus dip)

Open the Sizing Engine, click Load worked example project and select C04 to follow along.

Step-by-step calculation

Step 1: Design current

Ib = P / (√3 · U · η · cos φ)
Ib = 75 000 / (√3 × 400 × 0.95 × 0.86)
Ib = 132.5 A

With a motor starter the overload relay is set to Ib, so the cable must carry Iz ≥ 132.5 A.

Step 2: Correction factors

kT = √((90 − θa) / (90 − 30)); kG from IEC 60364-5-52 Table B.52.20
kT = √(40/60) = 0.816; ladder, 1 tier, 6 cables: kG = 0.79
k = 0.816 × 0.79 = 0.645

Step 3: Current capacity

IEC 60364-5-52 Table B.52.5, method E, three loaded conductors, Cu/XLPE:

SizeItabIz = 0.645 × Itab≥ 132.5 A?
50 mm²192 A123.8 ANo
70 mm²246 A158.7 AYes
95 mm²298 A192.2 AYes

Step 4: Running voltage drop

ΔU% = √3 · Ib · L · (R cos φ + X sin φ) / U × 100
95 mm² at 90 °C: R = 0.2475 Ω/km, X = 0.082 Ω/km → √3 × 132.5 × 0.150 × (0.2475 × 0.86 + 0.082 × 0.510) / 400
Circuit 2.19 %; total 2.46 + 2.19 = 4.65 % ≤ 5 %

70 mm² gives 2.90 % for the circuit but 5.35 % in total — over the 5 % limit. The upstream drop matters.

Step 5: Motor starting

ΔU% = √3 · Ist · L · (R cos φst + X sin φst) / U × 100 + bus dip
Ist = 7 × 132.5 = 927.5 A at cos φ 0.3 → 9.19 % + 3 %
12.19 % ≤ 15 %

Step 6: Short-circuit withstand

S ≥ √(I²t) / k
√(1 × 10⁶) / 143
S ≥ 7.0 mm² — not limiting

Step 7: Earth-fault loop impedance

CPC to IEC 60364-5-54 Table 54.2: S/2 = 47.5 → 50 mm² (R2 = 0.4935 Ω/km at 90 °C). Ze from the bus: cmin·U0/Ik1 min = 0.95 × 230.9 / 12 000 = 0.0183 Ω. The MCCB magnetic pick-up Ia = 12 × 132.5 = 1 590 A.

Zs = Ze + |(R1 + R2) + j·2X| · L ≤ cmin · U0 / Ia
Zs = 0.0183 + |(0.2475 + 0.4935) + j0.164| × 0.150 = 0.0183 + 0.1138
Zs = 0.1321 Ω ≤ Zs,max = 0.95 × 230.9 / 1 590 = 0.1380 Ω

70 mm² gives Zs = 0.172 Ω, which fails. The resistive part alone (0.1294 Ω) matches the hand calculation most designers do; the suite adds the loop reactance, which is safe side.

Result summary

CheckRequirementActualStatus
Current capacityIz ≥ 132.5 A192.2 A (95 mm²)✓ PASS
Total voltage drop≤ 5 %4.65 %✓ PASS
Motor starting drop≤ 15 %12.19 %✓ PASS
Short circuitS ≥ 7.0 mm²95 mm²✓ PASS
Earth-fault loopZs ≤ 0.1380 Ω0.1321 Ω✓ PASS
4-core 95 mm² Cu/XLPE/SWA (3C + E, 50 mm² CPC). The suite reports the governing criteria as voltage drop and earth-fault loop: 70 mm² carries the current but gives 5.35 % total voltage drop and Zs = 0.172 Ω.

Key insight: On long motor circuits the cable is rarely sized by current. Always carry the upstream voltage drop into the check, and check the earth-fault loop against the magnetic pick-up of a motor-starter MCCB, which is high.

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Frequently asked questions

Why is Ia 12 × In for this circuit?

A motor-starter MCCB is magnetic-only; its instantaneous pick-up is typically 10–14 × the overload setting. The suite uses 12 × by default — enter the actual setting in the Ia override field.

Where does the upstream 2.46 % come from?

From circuit C03 (main switchboard to MCC-01), which the engine evaluates first because C04 starts at the bus C03 feeds.

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