Cable Sizing for a Motor Feeder With Voltage-Drop Constraint
A workshop adds a 45 kW, 400 V, three-phase induction motor at the far end of a 120 m cable run from the main switchboard. The motor has a full-load current of 80 A, a starting current of 6 times full load, and a power factor of 0.86 at full load. The cable is to be run in a tray with three other loaded circuits, giving a grouping derating factor of 0.80. Ambient temperature is 40 degrees C, giving a temperature derating factor of 0.87. The designer proposes a 35 mm2 copper XLPE four-core cable with a tabulated current rating of 129 A (in free air, 30 degrees C, single circuit). The client requires voltage drop at full load to be no more than 4% of 400 V. Cable resistance at operating temperature is 0.65 ohm/km and reactance 0.08 ohm/km.
1. Check whether the proposed cable satisfies the current-carrying capacity requirement, including derating.
Read the worked answer
Apply both derating factors to the tabulated rating:
Derated capacity = 129 x 0.80 x 0.87 = 89.8 A.
The motor full-load current is 80 A, so the cable is adequate for continuous duty with a margin of about 9.8 A (12%).
However, the starting current is 6 x 80 = 480 A. The cable does not need to carry this continuously, but the designer should confirm that the protective device tolerates the starting surge and that the cable's short-time rating is not exceeded during the start. The continuous check passes; the starting check is a separate protection-coordination question, not a cable ampacity failure. The margin is thin enough that any additional grouping or a higher ambient temperature would push the derated capacity below 80 A, so the design has little headroom.
2. Calculate the full-load voltage drop and determine whether the 4% limit is met.
Read the worked answer
Use the three-phase voltage-drop expression:
Vdrop = sqrt(3) x I x L x (R cos(phi) + X sin(phi))
With I = 80 A, L = 0.120 km, R = 0.65 ohm/km, X = 0.08 ohm/km, cos(phi) = 0.86, sin(phi) = sqrt(1 - 0.86^2) = 0.510.
R cos(phi) = 0.65 x 0.86 = 0.559 ohm/km X sin(phi) = 0.08 x 0.510 = 0.0408 ohm/km Sum = 0.600 ohm/km
Vdrop = 1.732 x 80 x 0.120 x 0.600 = 9.98 V.
As a percentage of 400 V: 9.98 / 400 = 2.49%.
The 4% limit (16 V) is met with margin. The binding constraint is therefore ampacity, not voltage drop. If the run were longer than about 190 m at the same current, voltage drop would become the limiting criterion.
3. The client asks whether the cable can be reduced to 25 mm2 to save cost. Evaluate this request.
Read the worked answer
A 25 mm2 cable has a lower tabulated rating and higher resistance, so both checks must be redone.
Ampacity: a typical 25 mm2 XLPE four-core rating in free air is around 100 A. Derated: 100 x 0.80 x 0.87 = 69.6 A, which is below the 80 A full-load current. The cable would be overloaded in continuous duty. This alone rejects the proposal.
Voltage drop: resistance rises to roughly 0.90 ohm/km. Then R cos(phi) = 0.774, plus X sin(phi) 0.041, sum 0.815. Vdrop = 1.732 x 80 x 0.120 x 0.815 = 13.6 V, or 3.4% - still within 4% but with less margin.
The ampacity failure is decisive: reducing to 25 mm2 would leave the cable undersized for continuous current after derating. The correct response is to retain 35 mm2, or consider a 50 mm2 if the client wants headroom for future load or higher ambient temperature. Cost saving cannot justify a cable that fails the derated ampacity check.