75 kW Motor Starting: Cable, MCCB or Starter?
An industry has a 500 kVA, 415 V transformer with 5% impedance. Its existing load is 250 kW at 0.85 PF. A new 75 kW, 415 V, 4-pole induction motor (FLC 135 A, starting current 6.5×FLC, acceleration time 9 s) is started DOL from a 95 mm² aluminium XLPE cable, 120 m long, clipped direct. The cable's tabulated rating is 210 A and its voltage drop is 0.41 mV/A/m. On start, the terminal voltage dips about 15% and the 160 A MCCB trips in roughly one out of five starts. The thermal overload relay is set at 135 A and trips after 20 minutes when the motor runs at 80% load. Ambient in the MCC room is 45°C. The plant engineer proposes two fixes: replace the cable with 150 mm², or raise the MCCB rating to 250 A. The motor drives a centrifugal pump.
1. Calculate the continuous and starting voltage drop in the 95 mm² cable. Is the cable the main cause of the 15% dip? Show your working.
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Continuous: ΔU = 0.41 mV/A/m × 135 A × 120 m / 1000 = 6.6 V = 1.6% of 415 V - acceptable. Starting: I_start = 6.5 × 135 = 878 A. ΔU = 0.41 × 878 × 120 / 1000 = 43.2 V = 10.4%. Transformer contribution: starting kVA = √3 × 415 × 878 / 1000 = 631 kVA. Dip ≈ 5% × 631/500 = 6.3%. Adding arithmetically gives 16.7%; the observed 15% is consistent with vector addition and the motor terminal not being at the transformer terminals. So the cable is a major contributor - about 10 points of the 15 - but not the sole cause: even a perfect cable would leave the transformer contributing about 6%. Replacing the cable alone cannot guarantee a compliant start.
2. Evaluate the two proposed fixes - 150 mm² cable or a 250 A MCCB - and recommend a solution, including the effect on starting current and torque.
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150 mm² Al: mV/A/m ≈ 0.28. Starting drop = 0.28 × 878 × 120 / 1000 = 29.5 V = 7.1%; plus transformer 6.3% ≈ 13.4% arithmetic. It helps but is expensive and still marginal; continuous drop is already fine, so the cable is being replaced to fix a starting problem. 250 A MCCB: does not reduce the dip at all, and a 250 A device may not protect the 95 mm² cable (210 A tabulated, less after derating) or the motor. Reject. Recommend a soft starter (or VFD if pump control is needed) with current limit about 3×FLC = 405 A. Cable drop falls to 0.41 × 405 × 120 / 1000 = 19.9 V = 4.8%; transformer dip ≈ 5% × (√3×415×405/1000)/500 = 2.9%; total ≈ 7.7%. Starting torque falls roughly as (I/Is)²: if DOL starting torque is 150% FL, at 3×FLC it is about (3/6.5)² × 150 ≈ 32% FL. A centrifugal pump needs low torque at low speed, so this is usually enough; verify the pump curve. Keep the 160 A MCCB, set the overload per motor FLC, and check that the soft starter is bypassed after acceleration.
3. The overload relay trips after 20 minutes at 80% load. What would you check before changing the setting?
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First measure the actual line current with a clamp meter at 80% load; at 80% of 135 A it should be about 108 A. If it is much higher, the problem is mechanical (pump rubbing, valve throttling) or electrical (single-phasing, unbalanced supply, high harmonics). Check phase imbalance, terminal tightness, contactor and cable insulation, and the relay's CT ratio and setting. The relay is in a 45°C MCC room; a non-compensated thermal relay derates, so a 135 A setting may effectively be about 120 A at 45°C. That is still above 108 A, so a trip suggests either a higher current than assumed or a relay/CT fault. Do not raise the setting to 150 A to stop nuisance trips; that can cook the motor. Replace with an ambient-compensated overload or a motor protection relay, set to the motor nameplate FLC and service factor, and confirm the trip class (Class 10 for a centrifugal pump).