Skip to main content

Differential protection: 11 kV, 100 MVA generator

Short answer

Primary operating current = 1×200 = 200 A; Vph = 6351 V. Unprotected fraction = (200×6)/6351 = 0.189. Thus 81.1% of the winding is protected.

A 11 kV, 100 MVA generator is grounded through a resistance of 6 Ω. The CTs have a ratio of 1000/5. The relay is set to operate when there is an out of balance current of 1 A. What percentage of the generator winding will be protected by percentage differential scheme of protection?

In a differential scheme, an earth fault at fraction x of the winding from the neutral gives a fault current x·V_ph / R_n. The relay sees this as out-of-balance current; the part of the winding near the neutral, where the fault current is below the setting, is unprotected.

Data: 11 kV, 100 MVA; R_n = 6 Ω; CT 1000/5 (ratio 200); relay setting 1 A; winding impedance neglected.

I_op (primary) = 1 × 200 = 200 A
V_ph = 11 000 / √3 = 6350.85 V

x = I_op × R_n / V_ph
  = 200 × 6 / 6350.85
  = 0.1890   (unprotected)

Protected = 1 − 0.1890 = 0.8110

 Earth ──R_n=6Ω── N ──[ 18.9% ]──[   81.1%   ]── line
                      unprotected   protected

Answer: about 81.1% of the generator winding is protected; the 18.9% nearest the neutral is unprotected.

✓ Answer checked by Claude Opus 5.5 (high effort)

Related questions

All Switchgear and Protection chapter 7 questions →