TRV numerical: 1.6 mH, 0.0032 μF, 8000 A

When the breaker interrupts a bus fault fed by an alternator, the voltage across its contacts rises as a transient oscillation of the system L and C. The fault current lags the voltage by nearly 90°, so at current zero the system voltage is at its peak V_m (the active recovery voltage). The restriking voltage is v = V_m(1 − cos ωₙt), which peaks at 2V_m.

The system voltage is not given, so it is found from the fault current. With only the alternator reactance limiting the current:

V_ph(rms) = I × X_L,   V_m = √2 × I × X_L

Data

L = 1.6 mH, C = 0.0032 μF, I = 8000 A rms, f = 50 Hz.

(i) Frequency of oscillation

√(LC) = √(1.6×10⁻³ × 0.0032×10⁻⁶)
      = 2.2627 × 10⁻⁶ s
f_n   = 1 / (2π √(LC))
      = 1 / (2π × 2.2627×10⁻⁶)
      = 70 337 Hz ≈ 70.34 kHz
ω_n   = 2π f_n = 4.4194 × 10⁵ rad/s

(ii) Active recovery voltage

X_L = 2π × 50 × 1.6×10⁻³ = 0.5027 Ω
V_m = √2 × 8000 × 0.5027
    = 5687 V ≈ 5.69 kV (peak, phase value)

(The maximum restriking voltage would be 2V_m = 11.37 kV.)

(iii) Time for maximum RRRV

v      = V_m (1 − cos ω_n t)
dv/dt  = V_m ω_n sin ω_n t
Maximum when ω_n t = π/2:
t = π / (2 ω_n) = 1 / (4 f_n)
  = 1 / (4 × 70 337)
  = 3.554 × 10⁻⁶ s = 3.55 μs

(iv) Maximum RRRV

RRRV_max = V_m ω_n
         = 5687 × 4.4194×10⁵
         = 2.513 × 10⁹ V/s
         = 2513 V/μs ≈ 2.51 kV/μs

Answer: (i) f_n = 70.34 kHz; (ii) active recovery voltage = 5.69 kV (peak); (iii) t = 3.55 μs; (iv) maximum RRRV = 2.51 kV/μs.