Skip to main content

Balanced circulating current: 10 kV alternator

Short answer

Primary operating current = 1.8×200 = 360 A; Vph = 5774 V. Unprotected = (360×10)/5774 = 0.6235 (62.35%). To protect 80% of winding, maximum earthing resistance = 0.2×5774/360 = 3.21 Ω.

An alternator rated at 10 kV protected by the balanced circulating current system has its neutral grounded through a resistance of 10 ohms. The protective relay is set to operate when there is out of balance current of 1.8 A in the pilot wires, which are connected to the secondary windings of 1000/5 ratio current transformers. Determine: (i) the percentage winding which remain unprotected (ii) The minimum value of earthing resistance required to protect 80% 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 CT secondary out-of-balance current reaches 1.8 A.

Data: 10 kV, star; R_n = 10 Ω; setting 1.8 A; CT 1000/5 (ratio 200).

Step 1: Primary operating current

I_op = 1.8 × 200 = 360 A
V_ph = 10 000 / √3 = 5773.50 V

(i) Percentage unprotected

x × V_ph / R_n = I_op
x = 360 × 10 / 5773.50 = 0.6235

So 62.35% of the winding (from the neutral) is unprotected and only 37.65% is protected. The large earthing resistance limits the fault current too much.

(ii) Earthing resistance to protect 80%

Unprotected part allowed = 20% (x = 0.2):

0.2 × V_ph / R = 360
R = 0.2 × 5773.50 / 360 = 3.208 Ω

Answer: (i) 62.35% of the winding is unprotected; (ii) the earthing resistance must be reduced to 3.21 Ω (or less) to protect 80% of the winding.

✓ Answer checked by Claude Opus 5.5 (high effort)

Related questions

All Switchgear and Protection chapter 7 questions →