Across-the-Line Motor Start on an Islanded Diesel Generator
A 480 V, 60 Hz, 1500 kVA standby diesel generator (0.8 pf, subtransient reactance X″d = 0.12 per unit on its own base) will carry a plant whose largest load is a 400 hp, 480 V induction motor driving a loaded belt conveyor. The motor has a rated full-load current of 477 A and a locked-rotor current of 6.0 times full-load current. Its design-B starting torque is 150% of rated, and the loaded conveyor needs 100% of rated motor torque to break away and accelerate. The plant engineer wants to start the conveyor across the line with the utility breaker open, relying on the generator alone. Other connected loads require the sustained voltage dip to stay at or below 15%. The voltage regulator is fast; the engine is not oversized.
1. Estimate the voltage dip at the 480 V bus during the across-the-line start, the resulting motor torque, and state whether the start will succeed.
Read the worked answer
Locked-rotor current: I_LR = 6.0 × 477 = 2862 A
Locked-rotor kVA: S_LR = √3 × 0.480 kV × 2.862 kA = 2379 kVA
Generator rated current: I_gen = 1500 / (√3 × 0.480) = 1804 A
Per-unit locked-rotor current on the generator base: 2862 / 1804 = 1.586 pu
Voltage dip: ΔV ≈ X″d × I_pu = 0.12 × 1.586 = 0.190 pu → 19.0% dip Terminal voltage during start ≈ 0.81 pu ≈ 389 V
Torque: starting torque scales with V², T = 150% × (0.81)² = 150% × 0.656 = 98% of rated
Verdict: the start fails on two counts.
- The 19% dip exceeds the 15% limit for the other connected loads.
- The motor produces only 98% of rated torque against a 100% breakaway requirement - it will stall or accelerate so slowly that the generator's engine and the motor overload protection become involved. The 2% torque deficit is deceptive: voltage dip is calculated with the motor at standstill, and the generator's own speed/field recovery during the 1–3 s acceleration typically makes the real dip deeper.
Note the dip is driven by the motor's kVA, not the generator's kW rating - the engine has little to do with this calculation.
2. The contractor proposes a 65%-tap autotransformer (or an equivalent soft starter) to fix the problem because it 'cuts the starting current by more than half.' Evaluate this proposal quantitatively.
Read the worked answer
Current from the source: at a 0.65 tap, line current is reduced to 0.65² = 0.42 of the DOL value: I = 0.42 × 2862 = 1202 A → 0.666 pu on the generator base ΔV = 0.12 × 0.666 = 0.080 pu → 8.0% dip ✓ (meets the 15% limit)
But the motor torque falls with the square of the motor voltage: T = 150% × (0.65)² = 150% × 0.4225 = 63% of rated
63% is well below the 100% breakaway requirement. The conveyor will not move; the motor sits at locked rotor on the reduced tap, heating, until the overload trips - and a soft starter produces the same V² torque penalty during the ramp.
Conclusion: reject the proposal. Reduced-voltage starting helps only current-limited (friction/inertia) loads. This is a torque-limited load, so any method that reduces motor terminal voltage during acceleration is self-defeating. The only correct options are:
- a VFD, which delivers full torque at reduced line current (best answer, also gives controlled acceleration), or
- mechanical unloading / a no-load or fluid-coupling start, so the 100% breakaway requirement disappears, or
- a larger generator set.
3. If a larger generator is the chosen path (VFD rejected for cost), size it, keeping X″d = 0.12 pu and assuming the dip limit governs.
Read the worked answer
Which limit governs?
- Other-load limit: dip ≤ 15% → S_gen ≥ 0.12 × 2379 / 0.15 = 1903 kVA
- Torque limit: need V ≥ √(100/150) = 0.816 pu → dip ≤ 18.4% → S_gen ≥ 0.12 × 2379 / 0.184 = 1551 kVA
The 15% other-load limit governs.
Selection: the next standard frame above 1903 kVA is 2000 kVA at 480 V (2500 kVA if a standard step is required and no intermediate size is offered). Check the resulting dip: 0.12 × 2379 / 2000 = 14.3%, and torque = 150% × (0.857)² = 110% of rated, which clears the 100% breakaway with about 10% margin.
Cautions to state in the recommendation:
- The estimate assumes the generator is at rated voltage before the start. If the generator is already carrying load, the pre-start voltage is depressed and the actual dip is worse; re-run with the pre-start load kVA added to the source drop.
- Verify the engine can accept the kW step (motor at 0.81 pu voltage draws roughly 2379 × 0.81 ≈ 1930 kVA at a low power factor); the dip calculation says nothing about frequency dip.
- Confirm X″d is the saturated (rated-voltage) value from the manufacturer; using the unsaturated value underpredicts the dip.