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Arc suppression coil for 66 kV 90 km line

Short answer

Total capacitance per phase is 1.8 μF (0.02 μF/km × 90 km). Inductance L = 1/(3ω²C) = 1.876 H at 50 Hz. Phase voltage 38.1 kV; inductive current I_L = 3V_phωC = 64.64 A, giving kVA rating = V_ph × I_L ≈ 2463 kVA (2.46 MVA).

A 90 km long 66 kV, 50 Hz, 3 phase overhead line has capacitance to earth equal to 0.02 μF per km. What should be the inductance and kVA rating of the arc suppression coil?

The arc suppression (Peterson) coil in the neutral must supply an inductive fault current equal to the capacitive current fed by the two healthy phases during a line-to-ground fault.

Principle: during an earth fault on one phase, the healthy phases send a capacitive current I_C = 3 V_ph ωC into the fault. The Peterson coil in the neutral sends an equal inductive current I_L = V_ph / (ωL) through the fault, which cancels I_C, so the arc goes out. For resonance:

I_L = I_C
V_ph/(ωL) = 3 V_ph ωC
        L = 1 / (3 ω² C)

Rating of the coil = V_ph × I_L (the coil sees phase voltage during the fault).

Data

  • V_L = 66 kV, f = 50 Hz, ω = 314.16 rad/s
  • C = 0.02 μF/km × 90 km = 1.8 μF per phase

Inductance

L = 1 / (3 ω² C)
  = 1 / (3 × 314.16² × 1.8 × 10⁻⁶)
  = 1.876 H

kVA rating

V_ph = 66000/√3            = 38105.1 V
I_L  = 3 V_ph ω C
     = 3 × 38105.1 × 314.16 × 1.8×10⁻⁶
     = 64.64 A
Rating = V_ph × I_L = 38105.1 × 64.64
       = 2463.3 kVA ≈ 2.46 MVA

Answer: Inductance = 1.876 H; kVA rating ≈ 2463 kVA.

✓ Answer checked by Claude Opus 5.5 (high effort)

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