DOL Starting of a 22 kW Motor on a 250 kVA Transformer
A factory has a 250 kVA, 400 V, 50 Hz transformer with a short-circuit impedance of 5.5%. The existing balanced load is 120 kVA at 0.85 power factor. A new 22 kW, 400 V, three-phase cage motor is to be started direct-on-line. The motor has a full-load efficiency of 91% and power factor 0.86. Its starting current is 6.5 times full-load current and the starting power factor is 0.35. The motor cable is 35 m of 4 mm² Cu, protected by a 63 A motor-rated MCCB. The production manager wants DOL starting because it is cheapest and the motor is only 22 kW. The electrical contractor is asked to check whether the transformer and cable can accept the start and what alternative is needed if the voltage dip is excessive.
1. Calculate the motor full-load current and starting current. Estimate the transformer voltage dip during DOL starting, stating your assumptions.
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The motor full-load current is IFL = P / (√3 × U × η × cos φ) = 22000 / (1.732 × 400 × 0.91 × 0.86) ≈ 40.6 A. The DOL starting current is 6.5 × 40.6 ≈ 264 A. The starting apparent power is approximately √3 × 400 × 264 ≈ 183 kVA. Using the transformer short-circuit impedance, the approximate dip from the motor start alone is (183 / 250) × 5.5% ≈ 4.0%. The existing 120 kVA load adds further drop, and the transformer is briefly loaded to about 120 + 183 = 303 kVA, or 121% of rating. A realistic dip at the motor terminals is therefore higher, perhaps 6–8%, before cable drop is added. These are approximate values; the actual dip depends on source impedance and load power factors.
2. Check the 4 mm² cable for voltage drop during normal running and starting. Use mV/A/m ≈ 9.5 for 4 mm² three-phase.
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The 4 mm² cable is too small for the running current: its typical three-phase ampacity is around 32 A, below the 40.6 A full-load current, so the 63 A MCCB does not protect the cable against overload. The running voltage drop is about 40.6 × 35 × 9.5 / 1000 = 13.5 V, or 3.4% of 400 V, which is acceptable by itself. During DOL starting, however, the drop is about 264 × 35 × 9.5 / 1000 = 87.8 V, or 22% of 400 V, which is unacceptable and will cause a severe dip at the motor terminals. Replace the cable with 10 mm² Cu: its ampacity is about 50 A, and the running drop falls to about 40.6 × 35 × 3.8 / 1000 = 5.4 V.
3. Recommend a starting method and protection changes. Explain why simply increasing the MCCB rating is not the answer.
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The 4 mm² cable must be replaced, and the MCCB should be sized to protect the new cable, for example a 50 A motor-rated device for 10 mm² Cu. Increasing the MCCB rating would worsen cable protection and would not reduce the starting current or voltage dip. For starting, use a soft starter or a star-delta starter if the load torque permits. A soft starter can limit the starting current to about 2.5–3 times full-load current; at 2.5 times, the 10 mm² cable drop is only about 101 × 35 × 3.8 / 1000 = 13.4 V, or 3.4%, which is far better than DOL. Star-delta reduces starting current and torque to one third of DOL values, so the driven load must be able to accelerate on reduced torque. Check the transformer loading, motor protection, and cable ampacity after the change.