Adding a 250 kW Motor to an Existing 1 MVA Board: DOL, Star-Delta or VFD
An 11 kV/415 V, 1 MVA, 6% transformer feeds a 250 kW, 415 V, 4-pole, 50 Hz cage induction motor through 25 m of 4-core 185 mm² cable (r = 0.10 Ω/km, x = 0.08 Ω/km per conductor). Motor data: full-load efficiency 0.94, power factor 0.88, full-load slip 1.33%, direct-on-line starting current 6.5 times full-load current. The driven conveyor needs at least 40% of rated torque to break away from rest and about 55% to accelerate to rated speed. The utility limits the voltage dip at the 415 V busbar to 8% during motor starting. The plant wants to add this motor to an existing board without changing the transformer. Two options are on the table: a star-delta starter already in stores, and a new 250 kW variable-frequency drive. The transformer's off-load +2.5% tap is currently set at nominal.
1. Calculate the full-load and DOL starting currents, the source impedance referred to 415 V, and the resulting starting dip. Is the 8% limit satisfied?
Read the worked answer
Full-load current: In = 250 000/(√3 × 415 × 0.94 × 0.88) = 250 000/594.6 = 420 A Ist = 6.5 × 420 = 2733 A
Impedances referred to 415 V:
- Transformer: Z = 0.06 × 415²/1×10⁶ = 0.01033 Ω (predominantly reactance, so treat as X)
- Cable: R = 0.10 × 0.025 = 0.0025 Ω; X = 0.08 × 0.025 = 0.0020 Ω
- |Z| = √(0.0025² + 0.01233²) = 0.01259 Ω
Dip = 2733 × 0.01259 = 34.4 V = 8.3% of 415 V. Using the starting power factor (cos φ ≈ 0.25) in ΔV ≈ I(R cos φ + X sin φ) gives the same 8.3%.
So the DOL start marginally fails the 8% limit. The +2.5% tap raises the bus to about 425 V and shaves a fraction off the percentage, but that is not a defensible fix - it stresses the motor insulation and other loads. The starting method must change.
2. Evaluate the star-delta starter against both the voltage-dip limit and the conveyor's torque requirement. Would you approve it?
Read the worked answer
Star-delta reduces the line current to one third of DOL: 2733/3 ≈ 911 A, giving a dip of roughly 8.3/3 ≈ 2.8%, comfortably inside the 8% limit.
The torque, however, falls as the square of the applied voltage, i.e. to one third:
- DOL starting torque = (Ist/In)² × s_fl = 6.5² × 0.0133 = 0.56 per unit (56% of rated)
- Star-delta: 0.56/3 = 0.19 per unit (19% of rated)
The conveyor needs 40% to break away and 55% to accelerate. At 19% the motor will simply sit stalled in star with locked-rotor current flowing until the starter times out or the protection trips. The open-transition changeover also produces a second current and torque transient that can damage the gearbox.
Verdict: reject star-delta for this duty. It is only suitable for loads that start unloaded, such as centrifugal fans or unloaded compressors, where the breakaway torque is a small fraction of rated.
3. Size up the VFD option: starting dip, torque capability, transformer loading and the additional checks you would insist on before approving it.
Read the worked answer
With a 150% current limit the drive draws about 630 A at near-unity displacement factor (diode front end, say 0.98): ΔV ≈ 630(0.0025 × 0.98 + 0.01233 × 0.2) = 630 × 0.00492 = 3.1 V ≈ 0.75% - far inside the 8% limit, and full torque is available from zero speed, so the 55% acceleration requirement is easily met.
Transformer loading: the motor draws 250/(0.94 × 0.88) = 302 kVA, only 30% of the 1 MVA unit, so no transformer change is needed.
Checks before approval:
- Harmonics - a 250 kW 6-pulse drive injects roughly 30–40% current THD. Fit a 4% line reactor or DC choke and verify the board's IEEE 519 TDD limit at the PCC with a harmonic study.
- Motor suitability - confirm the winding is inverter-duty rated for the cable length (25 m is undemanding) and that the shaft-voltage/bearing-current risk is handled.
- Protection and earthing - the drive limits fault current, so the existing overcurrent grading and earth-fault settings must be re-checked.
- An autotransformer at a 65% tap would give only 0.65² × 56% = 24% torque - also insufficient, so it is not an alternative.