Cable and starter selection for a 30 kW motor at the end of a long feeder
A 30 kW, 415 V, three-phase, four-pole induction motor drives a conveyor. The motor nameplate gives 56 A at 0.86 power factor and 91% efficiency. The MCC is 110 m from the motor, and the proposed cable is a 4-core 25 mm² copper XLPE armoured cable clipped to a tray. The cable data give R = 0.83 ohm/km and X = 0.08 ohm/km at operating temperature; the supply transformer is 1000 kVA with 5% impedance. The conveyor needs a high starting torque, and the project engineer proposes a star-delta starter because the motor has six terminals. The plant requires motor terminal voltage during starting to remain above 85% of nominal and running drop below 3%. The existing MCC voltage is 415 V.
1. Calculate the motor full-load current from the data and the running voltage drop. Is 25 mm² cable acceptable for running?
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From output: I = P/(sqrt(3) x V x cos phi x eta) = 30000/(1.732 x 415 x 0.86 x 0.91) = 53.3 A. The nameplate 56 A is higher and should be used for protection and cable checks because it reflects the actual machine. Voltage drop = sqrt(3) x I x L x (R cos phi + X sin phi). R cos phi = 0.83 x 0.86 = 0.714 ohm/km; X sin phi = 0.08 x 0.51 = 0.041 ohm/km; sum = 0.755 ohm/km. dV = 1.732 x 56 x 0.110 x 0.755 = 8.05 V = 1.94% of 415 V. This is below the 3% running limit. The cable is acceptable thermally if its clipped-direct ampacity exceeds the derated 56 A (25 mm² XLPE is typically well above 100 A), but verify grouping and ambient.
2. Estimate the voltage dip during star-delta starting. Does it meet the 85% criterion? What about DOL?
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Take DOL starting current as 6 x FLC = 336 A. Star-delta reduces the line current to one-third: 112 A, and starting torque to one-third of DOL. Cable drop at 112 A = 1.94% x (112/56) = 3.88%. Transformer dip: LV short-circuit current = 1000/(1.732 x 0.415 x 0.05) = 27.8 kA. A 112 A start is 0.4% of this, so transformer regulation is small; the impedance drop is roughly 0.4% using a linear approximation. Total dip is about 4.3%, terminal voltage about 95.7%, which passes the 85% criterion. DOL at 336 A would give cable drop 11.6% plus transformer drop about 1.2%, total about 12.8%, terminal voltage 87.2%, which is marginal, and the supply authority may limit starting current. But the real issue is torque, not voltage.
3. The conveyor needs high starting torque. Is star-delta the right choice? Recommend a starter and give the overload relay setting.
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Star-delta gives only one-third of DOL starting torque, so it is suitable for high-inertia or high-breakaway loads only if the load torque at low speed is well below one-third of motor starting torque. A loaded conveyor often needs more; check the load torque curve. If it cannot accelerate, use a soft starter with current limit (still limited torque) or a VFD, which gives rated torque at low speed and controlled current. For a high-breakaway conveyor, a VFD or a slip-ring motor is the robust choice. Overload relay: if the relay is in the line contactor, set to the motor FLC, 56 A. If it is in the delta contactor phase, it carries 56/sqrt(3) = 32.3 A, so set 32 A. Use Class 10 or 20 depending on starting time, and coordinate with the fuse/MCB.