Motor Starting Voltage Dip and Starter Selection for a Loaded Conveyor
A 2 MW, 6.6 kV, 4-pole induction motor is to be started direct-on-line from a 6.6 kV switchboard fed by a 5 MVA, 11/6.6 kV transformer with 6.5% impedance. The 11 kV source fault level is 250 MVA. Motor data: full-load current 210 A, locked-rotor current 6.0 per unit, starting power factor 0.30, and starting torque 100% of full-load torque. The driven load is a loaded conveyor requiring at least 35% of motor full-load torque at standstill, and its torque rises linearly with speed. The plant standard limits the voltage dip at the 6.6 kV bus to 10% and requires running motors to ride through; one 400 kW motor with a contactor drops out below 80% terminal voltage. A soft starter rated 300% current for 30 s and a 65% autotransformer starter are both available. The transformer is already loaded to 3.2 MVA.
1. Estimate the direct-on-line voltage dip and motor terminal voltage, and check compliance with the plant standard.
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Refer the source impedance to 6.6 kV: Zs = 11²/250 = 0.484 Ω at 11 kV; referred to 6.6 kV, Zs = 0.484 × (6.6/11)² = 0.174 Ω. Transformer impedance: Zt = 0.065 × 6.6²/5 = 0.566 Ω. Total source-plus-transformer impedance is 0.740 Ω. Motor starting kVA = √3 × 6.6 kV × (6 × 210 A) = 14.4 MVA; starting impedance = 6.6²/14.4 = 3.03 Ω. Dip = 0.740/(0.740 + 3.03) = 19.6%, so the motor terminal voltage is about 80.4%. This exceeds the 10% plant limit and leaves the 400 kW contactor at its 80% dropout threshold. Starting torque at 80.4% voltage is about 0.647 × 100% = 64.7% of full-load torque, so the conveyor itself would accelerate, but the system limit is violated and the running contactor is at risk.
2. Evaluate the 300% soft starter and the 65% autotransformer against both the dip limit and the conveyor breakaway torque.
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Soft starter at 300% current limit: the supply sees 3 × 210 A = 630 A, so starting kVA = √3 × 6.6 × 0.630 = 7.2 MVA; effective impedance = 6.6²/7.2 = 6.05 Ω; dip = 0.740/(6.05 + 0.740) = 10.9%, slightly above 10%. Motor terminal voltage is about 300/600 = 50% of rated, so torque is about 0.25 × 100% = 25% of full-load torque, below the 35% breakaway requirement; the conveyor stalls. Autotransformer at 65%: motor terminal voltage ≈ 0.65 × 90.6% = 58.9%, torque ≈ 0.347 × 100% = 34.7% of full-load torque, also just below 35% at standstill. Supply dip ≈ 0.65² × 14.4 MVA = 6.08 MVA; effective impedance = 6.6²/6.08 = 7.16 Ω; dip = 0.740/(7.16 + 0.740) = 9.4%, so voltage is acceptable but torque is not. A 70% tap gives about 40% torque but a dip near 10.7%, so it also fails the dip limit.
3. Recommend a practical solution and list the additional checks you would require before committing to it.
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Recommend a VFD (or a slip-recovery/fluid coupling if the conveyor allows) because it can produce full starting torque at about 1.0–1.2 pu motor current, giving a supply dip of roughly 3–4% and no contactor dropout. If a VFD is not available, the only reliable options are to reduce the breakaway load (start the conveyor empty), reinforce the source with a larger transformer or upstream capacity, or use a 70–80% autotransformer and accept a dip slightly above 10% under a formal deviation - not recommended. Check the 400 kW contactor: at 80.4% it is marginal, so use a ride-through/restart scheme or a contactor coil that holds at 70%. Check transformer loading after start (3.2 MVA plus 2 MW) and motor thermal capability for the longer start. Confirm the conveyor breakaway torque by measurement, because the margin between 34.7% and 35% is within data uncertainty.