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Factors in choosing a generator

Short answer

Generator selection factors: MVA rating from turbine output and power factor (0.85–0.9), voltage (6.6–11 kV), speed/poles (N=120f/p), shaft arrangement, reactances and SCR, inertia constant, efficiency above 97%, insulation class F/B, air cooling, excitation (brushless or static), high-resistance neutral grounding, grid requirements, transport limits, and cost.

Discuss the various factors which should be considered for the choice of generator during the design of a power plant.

The generator must match the turbine, the grid and the site. The main factors are:

  1. Rating (MVA) and power factor: S=Pturbine ηg/cos⁡ϕS = P_{turbine}\,\eta_g / \cos\phi; power factor usually 0.85–0.9 lagging so the unit can supply reactive power. Some overload margin (e.g. 10%) is often specified.
  2. Rated voltage: usually 6.6 kV or 11 kV for hydro units (higher, e.g. 13.8–15.75 kV, for very large units); chosen with GSU and bus/cable cost.
  3. Speed and number of poles: fixed by the turbine: N=120f/pN = 120f/p. Hydro units are slow, salient-pole machines with many poles.
  4. Shaft arrangement: vertical (large Francis/Kaplan) or horizontal (small Pelton/Francis).
  5. Reactances and SCR: Xd′′X''_d sets fault level and breaker rating; SCR and XdX_d set stability and voltage regulation.
  6. Inertia constant (GD²): larger inertia limits speed rise on load rejection and improves stability, but costs more.
  7. Efficiency and losses: high efficiency (above 97–98%) saves energy over the plant life.
  8. Insulation and temperature rise: class F insulation with class B temperature rise is common; altitude derating above 1000 m.
  9. Cooling: air-cooled (open or closed with air–water coolers) for hydro.
  10. Excitation system: brushless for small/medium units, static for large units.
  11. Neutral grounding: high-resistance grounding via NGT.
  12. Grid requirements: capability curve, reactive power range, fault ride-through, harmonics.
  13. Transport and erection limits: weight and size on hill roads; split stator if needed.
  14. Standards and cost: IEC 60034; capital cost versus loss capitalisation.

Example

For a 10 MW turbine output, ηg=0.97\eta_g = 0.97, pf 0.85: generator rating =10×0.97/0.85≈11.4= 10 \times 0.97/0.85 \approx 11.4 MVA, 11 kV, 50 Hz; if the turbine speed is 600 rpm, poles =120×50/600=10= 120 \times 50/600 = 10.

✓ Answer checked by Claude Opus 5.5 (high effort)

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