Direct-On-Line Starting of a 1250 kW Boiler Feed Pump Motor on a Weak 6.6 kV Board
A 1250 kW, 6.6 kV, 6-pole, 50 Hz squirrel-cage motor drives the boiler feed pump of a 2 x 210 MW station. It is started direct-on-line from the 6.6 kV unit board, which is fed by a 6.3 MVA, 33/6.6 kV transformer of 8% impedance; the 33 kV bus behind it has a fault level of 80 MVA. Nameplate data: full-load current 120 A, locked-rotor current 6.0 pu, locked-rotor torque 100% of rated, breakdown torque 220%. The 33 kV incoming feeder relay has a definite-time overcurrent element set at 400 A with a 0.5 s delay, and the normal station load on that feeder is about 250 A. On every start the board voltage dips visibly, a neighbouring 400 kW auxiliary contactor drops out, and the 33 kV feeder relay has tripped twice this month. The pump is centrifugal, with shut-off torque of 25% of rated and rated torque at rated speed; the start currently takes 22 s.
1. Compute the starting current, the 6.6 kV bus voltage dip and the motor torque available at the dipped voltage. Then explain the 33 kV feeder relay trips.
Read the worked answer
Starting current. I_LR = 6 x 120 = 720 A, giving S_start = sqrt3 x 6.6 x 0.72 = 8.23 MVA.
Source strength. Transformer short-circuit capacity = 6.3/0.08 = 78.75 MVA. In series with 80 MVA upstream: S = (80 x 78.75)/(80 + 78.75) = 39.7 MVA. Ignoring the negligible motor contribution during acceleration, dip = 8.23/(39.7 + 8.23) = 17.2%, so the motor terminal sits at 82.8% of nominal.
Torque. Torque scales with voltage squared: 0.828^2 = 0.686. With 100% locked-rotor torque the available starting torque is about 68.6% of rated, far above the 25% shut-off torque, so the pump does accelerate - slowly, because the excess accelerating torque is modest.
Contactors. 82.8% is below the typical 85% hold-in of an AC contactor, which is exactly why the neighbouring auxiliary drops out. The fix is a contactor with a delayed undervoltage release or a latched starter, not a bigger transformer.
Relay trips. Reflecting 720 A at 6.6 kV to 33 kV gives 720 x 6.6/33 = 144 A. Adding the 250 A station load gives about 394 A, which is at the 400 A setting; the 0.5 s delay is far shorter than the 22 s start, so the element times out on marginal pickup. Check the relay target flags before changing anything - if the flag is on this element, raise the setting with a proper coordination study against the transformer damage curve rather than simply increasing the delay.
2. Compare a 65% autotransformer starter, a soft starter limited to 350% current and a variable-frequency drive on supply current, bus dip and motor torque. Which would you specify for this pump, and why?
Read the worked answer
65% autotransformer. Supply current = 0.65^2 x 6.0 = 2.54 pu = 3.47 MVA, so dip = 3.47/(39.7 + 3.47) = 8.0%. Motor torque falls to 0.65^2 = 42% of DOL, and with the bus at 92% the effective torque is 0.42 x 0.846 = about 36% of rated - still above the 25% shut-off torque, but the transition to full voltage produces a second current surge and a further dip.
Soft starter at 350%. Supply current = (3.5/6) x 6.0 = 3.5 pu = 4.80 MVA, dip = 4.80/(39.7 + 4.80) = 10.8%. Torque = (3.5/6)^2 = 34% of DOL, so about 34% of rated at the dipped voltage. It removes the transition surge if set to a smooth ramp, and it is the cheapest option.
VFD. Input current is roughly 1.0-1.1 pu at near-unity power factor, about 1.4 MVA, so the dip is under 3%. Torque is controlled and the pump can be run at a lower speed instead of throttled.
Judgement. If the pump is currently throttled for flow control, the VFD pays for itself in energy and solves the starting problem completely - specify it. If the pump runs at full speed and throttling is not used, a soft starter with a 350% limit and a 30 s ramp is adequate and much cheaper, provided the feeder relay setting is corrected. The autotransformer gives the least supply current of the two starters but has the worst transition behaviour.