Motor starting voltage dip and torque constraint
A 6.6 kV bus supplies a 2 MVA, 6.6/3.3 kV transformer with 8% impedance. A 600 kW, 3.3 kV induction motor driving a loaded conveyor is started direct-on-line. The motor draws 6.0 times its full-load current during starting, and the conveyor cannot tolerate a terminal voltage dip greater than 12%. Recent measurements show a 20% dip at the 3.3 kV bus, and the upstream 6.6 kV bus dips by 8%. The transformer already carries 500 kW of other load. The utility limits starting current to 3.5 times full-load current and requires a power-factor correction check. The plant has a 6-week shutdown window and a budget that permits either a soft starter or an autotransformer starter, but not a VFD. The motor accelerates in 8 seconds at rated voltage; the load torque is 60% of rated during start. The protection relay has a 1.1 pu overload setting and a 5-second short-time delay.
1. Estimate the voltage dip for DOL starting and for a soft starter limited to 3.0 times full-load current, stating your assumptions.
Read the worked answer
Use the transformer short-circuit capacity as the dominant source. For 2 MVA and 8% impedance, S_sc = 2/0.08 = 25 MVA. At full load the motor input is about 600/(0.94×0.9) ≈ 710 kVA. DOL at 6× FLC gives a starting demand of about 6×710 ≈ 4.26 MVA. Approximate dip = 4.26/25 ≈ 17%, and with the upstream source impedance the measured 20% is credible. For a soft starter at 3.0× FLC, the starting demand is about 3×710 ≈ 2.13 MVA, giving a dip of about 8.5%; even with the upstream contribution it should stay below 12%. The estimate assumes the motor terminal voltage and current are approximately proportional during the limited-current start, and that other transformer load is steady. The soft starter therefore solves the voltage-dip problem but not necessarily the torque problem.
2. Recommend the starting method, taking account of the conveyor's 60% load torque and the 3.5× current limit.
Read the worked answer
The voltage calculation favours a reduced-current starter, but the torque check is decisive. At 3× current, the motor terminal voltage is roughly half of DOL, so starting torque is roughly 0.25 of the DOL torque. If the motor's DOL starting torque is about 1.8× rated, the soft starter gives only about 0.45× rated torque, which is below the 0.6× load torque; the conveyor may stall or the start may be incomplete.
A 75% tap autotransformer is a better fit: motor voltage is 0.75, starting torque is 0.5625×DOL ≈ 1.0× rated, and line current is 0.5625×6 ≈ 3.4× FLC, within the 3.5× limit. Alternatively, use a soft starter with kick-start or a higher current limit of 3.5× and verify acceleration; that gives about 0.58× voltage and 0.34×DOL torque, still marginal. I would specify a 75% tap autotransformer with transition at about 80% speed, or a soft starter only if a site test proves acceleration. A VFD would be best but is outside the budget.
3. State three commissioning tests and one protection setting change needed before handover.
Read the worked answer
Test: (1) locked-rotor current and starting kVA at the 3.3 kV bus, to confirm the 3.5× limit; (2) acceleration time and minimum terminal voltage at both the 3.3 kV and 6.6 kV buses, to confirm the 12% dip limit and that the conveyor accelerates; (3) thermographic scan of the starter, cable terminations, and motor during and after start. Also check the autotransformer transition current.
Protection: the existing 5-second short-time delay is shorter than the 8-second start, so it may trip during a normal start; increase the short-time delay to about 12-15 seconds while keeping the 1.1 pu overload setting, and confirm the thermal capacity. If a soft starter is used, set the current limit at 3.5× and the ramp time longer than the 8-second acceleration but within the motor's locked-rotor thermal limit.