Induction Motor Starting at Reduced Bus Voltage
A 500 hp, 4,160 V, 60 Hz, three-phase induction motor drives a conveyor that requires 60% of rated torque at standstill and during acceleration. The motor has a locked-rotor torque of 110% of rated torque and a locked-rotor current of 5.5 times full-load current. The motor is started across the line from a 4,160 V bus supplied by a 2,500 kVA transformer with 6.5% impedance. A recent start causes the bus voltage to dip to 88% of nominal, and the conveyor fails to accelerate within the 10 s permitted by the starter. The plant wants to know whether the motor can start at 88% voltage and whether reduced-voltage starting would help or hurt.
1. Calculate the available starting torque at 88% voltage and compare it with the conveyor requirement. What does the comparison mean for the start?
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Available starting torque scales with terminal voltage squared:
Tst = TLR (V/Vn)² = 1.10(0.88)² = 0.852 pu.
That is 85.2% of rated torque, while the conveyor requires 60%. The net accelerating torque at breakaway is about 25.2 percentage points, so the motor should start if the 88% voltage is maintained. The margin is small, however: at 80% voltage the torque falls to 1.10(0.80)² = 0.704 pu, leaving only 10.4 points. Any additional dip, load torque increase, or misalignment could stall the conveyor. The reported failure suggests the 88% dip may be worse at the motor terminals or that the load torque is not constant. A trial start should be monitored for current and terminal voltage.
2. Would a reduced-voltage starter improve the situation? Compare an autotransformer on a 65% tap with an 80% tap.
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Reduced-voltage starting reduces motor torque by the square of the applied voltage, so it usually hurts a high-breakaway load. An autotransformer on a 65% tap gives approximately Tst = 1.10(0.65)² = 0.465 pu, which is below the 60% load requirement even with a full bus. An 80% tap gives 1.10(0.80)² = 0.704 pu at nominal bus, but if the bus is already at 88%, the motor sees 0.80×0.88 = 0.704 pu and torque is only 1.10(0.704)² = 0.545 pu, still below 60%. A reactor starter has the same square-law torque penalty. A soft starter can limit current but cannot create torque. The practical fixes are a stronger supply, a lower-impedance transformer, larger conductors, mechanical unloading, or a motor with higher locked-rotor torque.
3. What additional data would you request before approving a trial start or a starter change?
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Request the motor speed-torque curve and the conveyor load-torque curve, not just the 60% breakaway value. Also obtain the actual transformer impedance and source fault MVA, the motor-terminal voltage during the failed start, protective relay settings, starter type and acceleration-time limit, and the mechanical condition of the conveyor. Confirm whether the 10 s limit is a starter setting or a process requirement. A plant test should record three-phase voltage and current at the motor terminals. If the voltage recovers only after the starter trips, the problem may be protection or the starter, not torque. If voltage stays depressed, the supply impedance is the limiting factor.