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Earth fault protection: 20 MVA alternator

Short answer

Primary operating current = 1.5×200 = 300 A; Vph = 6351 V. Unprotected = (300×5)/6351 = 0.236, so protected = 76.4%. For 90% protection, earthing resistance cannot exceed 0.10×6351/300 = 2.12 Ω.

The neutral point of a three-phase 20 MVA, 11 kV alternator is earthed through a resistance of 5 Ω, the relay is set to operate when there is an out of balance current of 1.5 A. The CTs have a ratio 1000/5. What percentage of winding is protected against an earth fault and what should be the minimum value of earthing resistance to protect 90% of the winding?

For an earth fault at fraction x of the winding from the neutral, the fault current is x·V_ph / R_n (winding impedance neglected). The relay operates when the corresponding CT secondary current reaches the setting.

Data: 20 MVA, 11 kV, star; R_n = 5 Ω; relay setting 1.5 A; CT 1000/5 (ratio 200).

Step 1: Primary operating current

I_op (primary) = 1.5 × 200 = 300 A
V_ph = 11 000 / √3 = 6350.85 V

Step 2: Percentage of winding protected

x × V_ph / R_n = I_op
x = 300 × 5 / 6350.85 = 0.2362   (unprotected)
Protected = 1 − 0.2362 = 0.7638 → 76.38%

Step 3: Earthing resistance to protect 90%

To protect 90%, only x = 0.10 may be unprotected:

0.10 × V_ph / R = 300
R = 0.10 × 6350.85 / 300 = 2.117 Ω

Any resistance larger than this leaves more than 10% unprotected, so 2.117 Ω is the limiting (maximum permissible) value of earthing resistance; a lower resistance protects more winding but gives larger earth fault currents.

Answer: 76.38% of the winding is protected; earthing resistance must be 2.117 Ω (not more) to protect 90% of the winding.

✓ Answer checked by Claude Opus 5.5 (high effort)

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