Motor Starting Voltage Dip on a 500 kVA Transformer
A 200 kW, 400 V, three-phase induction motor is supplied from a 500 kVA, 400 V transformer with 5.5% impedance. The motor has full-load efficiency 0.94 and power factor 0.90. Its locked-rotor current is 6 times full-load current, and the plant rule limits starting voltage dip at the motor terminals to 15%. The source upstream of the transformer is stiff enough to be neglected. The driven load is a centrifugal pump that requires very little torque at standstill. A reduced-voltage autotransformer starter with a 65% tap is available if the direct-on-line start fails the limit.
1. Compute the full-load current, the starting current and the resulting voltage dip for a direct-on-line start.
Read the worked answer
Input kVA at full load = 200/(0.94 x 0.90) = 236.4 kVA.
Full-load current = 236,400/(sqrt3 x 400) = 341 A.
Starting current = 6 x 341 = 2047 A.
Starting kVA = sqrt3 x 400 x 2047 = 1.418 MVA.
Per-unit starting kVA on the 500 kVA base = 1.418/0.500 = 2.836 pu.
Dip = I_pu x Z_pu = 2.836 x 0.055 = 0.156 pu = 15.6%.
This exceeds the 15% limit, so direct-on-line starting is not acceptable.
2. Determine the minimum transformer rating that would keep the dip at or below 15%.
Read the worked answer
Required condition: S_start x Z / S_tx <= 0.15.
S_tx >= S_start x Z / 0.15 = 1.418 x 0.055 / 0.15 = 0.520 MVA.
A 500 kVA unit is therefore just short; the next standard size, 630 kVA, gives dip = 1.418 x 0.055/0.630 = 12.4%, which complies. Replacing the transformer is technically effective but expensive compared with a starting method change.
3. Evaluate the 65% autotransformer tap: compute the new dip and comment on torque suitability for the pump.
Read the worked answer
On the 65% tap, motor terminal voltage is 0.65 pu, so motor current is 0.65 x 2047 = 1331 A and line current is 0.65 x 1331 = 865 A (equivalently 0.65^2 x 2047).
Starting kVA = 0.65^2 x 1.418 = 0.599 MVA.
Dip = 0.599 x 0.055/0.500 = 6.6%, well within the 15% limit.
Starting torque falls to 0.65^2 = 42% of direct-on-line torque. For a centrifugal pump, which needs little torque at standstill, this is acceptable; for a loaded conveyor or crusher it could stall. The judgment call is to accept the autotransformer starter because the load is a pump and the dip is comfortably compliant.