Direct-on-line start of a 1,250 kW motor on a 6.6 kV bus with a 10% dip limit
A 6.6 kV, 50 Hz plant is fed from an 11/6.6 kV, 5 MVA transformer of 7.5% impedance; the 11 kV bus has a utility fault level of 250 MVA. A new 1,250 kW, 6.6 kV, four-pole motor driving a centrifugal pump is to be started direct-on-line. Its full-load current is 132 A, locked-rotor current 6.0 × FLC and locked-rotor torque 90% of full-load torque; the pump's breakaway torque is 25% of the motor's full-load torque and rises with the square of speed. Existing plant on the 6.6 kV bus draws 3.0 MW at 0.90 pf. The utility's connection agreement limits the voltage dip to 6% at the 11 kV point of common coupling, and the plant standard limits dips on the 6.6 kV bus to 10% so that running contactors do not drop out. A 1,200 kVAR capacitor bank is connected to the 6.6 kV bus. The process allows the pump to be started at reduced speed, but production staff oppose any change that shortens pump life.
1. Calculate the voltage dip at the 6.6 kV bus and at the 11 kV point of common coupling for a direct-on-line start, and state whether each limit is met.
Read the worked answer
Starting demand. Locked-rotor current = 6.0 × 132 = 792 A. Starting apparent power = √3 × 6.6 kV × 0.792 kA = 9.05 MVA.
Source strength at 6.6 kV.
- Transformer alone: 5 MVA / 0.075 = 66.7 MVA
- Utility: 250 MVA (unchanged through the transformer ratio)
- Series combination: 1/(1/250 + 1/66.7) = 52.6 MVA
Dips.
- At the 6.6 kV bus: 9.05/52.6 = 17.2%, so the bus falls to 82.8% for the duration of the start.
- At the 11 kV PCC: 9.05/250 = 3.6%.
Verdict. The utility limit of 6% is met with room to spare. The plant limit of 10% on the 6.6 kV bus is exceeded by a wide margin - the direct-on-line start is not acceptable.
Consequential checks. Running contactors typically hold above 70–75% of coil voltage, so at 82.8% they should stay in - but that is a 17% step repeated at every start, and flicker at the PCC (Pst) must be assessed if the pump starts more than a couple of times an hour, because a 3.6% step repeated frequently breaches planning limits even though a single step does not. The 1,200 kVAR bank must be switched out before the start: a capacitor bank energised with a starting motor can self-excite and produce overvoltage at the end of the start, and it also changes the resonant frequency seen by the VFDs on the bus.
2. Assess whether the motor will actually accelerate the pump at the reduced voltage, and explain how the answer would change for a different driven load.
Read the worked answer
Torque at reduced voltage. Motor torque varies with the square of terminal voltage: at 82.8% the available locked-rotor torque is 0.828² × 90% = 61.7% of full-load torque. The pump needs 25% at standstill, so there is a clear margin at breakaway.
Through the acceleration. As speed rises the motor torque falls from the locked-rotor value towards the breakdown point and then to full-load torque at rated speed, while the pump torque rises as n². The two curves must be compared point by point; for a centrifugal pump with low inertia the margin stays positive throughout and the start is short, of the order of a few seconds. Thermal duty is also satisfied: with 6 × FLC for, say, 8 s the I²t is about 2,300 (pu²·s), far inside the hot stall withstand of a machine this size.
Why the load matters more than the dip. If the driven machine were a positive-displacement compressor or a loaded conveyor with a breakaway torque of 70–80% of motor full-load torque, the available 61.7% would be insufficient. The motor would sit at standstill drawing 792 A, the bus would stay depressed, and the machine would either trip on overload or be damaged - with no indication other than a slowly rising winding temperature. A dip standard expressed only as a percentage is therefore incomplete: the plant standard should also require a torque-margin check using the actual driven-equipment curve.
Cascading risk. If other large motors on the 6.6 kV bus are near their breakdown torque at the moment of the start, they will draw more current, deepen the dip and can stall, turning a 17% dip into a plant trip. That is the reason for limiting the bus dip to 10% in the first place.
3. Recommend a starting method with quantitative justification, considering the plant and utility limits and the process constraint.
Read the worked answer
Options, quantified against the 52.6 MVA source.
Soft starter. Dip = 9.05 × (I_limit/6)/52.6.
- At 3.5 × FLC = 462 A: 5.28 MVA → 10.0% dip; torque = (3.5/6)² × 90% = 30.6%, above the 25% breakaway. Meets torque but sits exactly on the plant limit with no margin.
- At 3.0 × FLC = 396 A: 4.53 MVA → 8.6% dip; torque = (3.0/6)² × 90% = 22.5%, below the pump's 25% breakaway. The motor may fail to break away and sit at locked-rotor current - the worst possible outcome.
Autotransformer at 65% tap. Supply current = 0.65² × 792 = 335 A → 3.83 MVA → 7.3% dip; motor torque = 0.65² × 90% = 38%, comfortably above breakaway. A closed-transition arrangement avoids a second current surge on changeover.
Variable-frequency drive. Current limited to 100–150% of FLC, so the dip is under 3%, torque is available from zero speed, and the process gains soft starting and speed control. Costs more, and needs harmonic and motor-insulation review, but the process explicitly allows reduced-speed starting.
Recommendation. Because the pump cannot be assumed to break away reliably at the current limit a soft starter would need to meet the 10% limit, and because the process permits variable speed, specify a 6.6 kV VFD. If capital cost rules that out, an autotransformer on the 65% tap is the lowest-dip conventional solution and is thermally kinder to the motor and pump than a soft starter running at the torque limit. Whichever is chosen: interlock the start with the capacitor bank, record the actual dip with a disturbance recorder on first start, and re-check the PCC flicker limit for the expected number of starts per hour.