TRV numerical: 1.5 mH, 0.0025 μF, 7000 A

When the CB interrupts the busbar fault, the voltage across its contacts rises from zero towards the system voltage as an L–C oscillation (the restriking voltage). Its steady value after the transient is the recovery voltage.

  A (alternator)    L=1.5 mH      CB
 (~)---------------^^^^^^----+---o/ o---+ bus
  |                          |          |  fault
 earth                    C=0.0025 μF   X
                             |
                            earth

Method

At fault current zero the system voltage is at its peak V_m (current lags by about 90°). The voltage across the opening contacts then follows the L–C transient (resistance neglected):

v(t) = V_m (1 − cos ω_n t),   ω_n = 1/√(LC)
peak  = 2 V_m      at ω_n t = π     (t = 1/(2 f_n))
dv/dt = V_m ω_n sin ω_n t
max   = V_m ω_n    at ω_n t = π/2   (t = 1/(4 f_n))

The source voltage is not given, so it is obtained from the fault current limited by the generator reactance: V_m = √2 × I × ωL (phase peak).

Data

L = 1.5 mH/phase, C = 0.0025 μF/phase, I = 7000 A rms, f = 50 Hz.

(i) Active recovery voltage

X_L = 2π × 50 × 1.5×10⁻³ = 0.4712 Ω
V_m = √2 × 7000 × 0.4712
    = 4665 V ≈ 4.67 kV (peak, per phase)

(ii) Frequency of oscillation

√(LC) = √(1.5×10⁻³ × 0.0025×10⁻⁶) = 1.9365×10⁻⁶ s
f_n   = 1 / (2π × 1.9365×10⁻⁶)
      = 82 187 Hz ≈ 82.19 kHz
ω_n   = 5.164 × 10⁵ rad/s

(iii) Time to reach peak restriking voltage

t = 1 / (2 f_n) = 1 / (2 × 82 187)
  = 6.084 × 10⁻⁶ s = 6.08 μs
Peak restriking voltage = 2 V_m = 9.33 kV

(iv) Maximum RRRV

RRRV_max = V_m ω_n
         = 4665 × 5.164×10⁵
         = 2.409 × 10⁹ V/s ≈ 2.41 kV/μs
(occurs at t = 1/(4 f_n) = 3.04 μs)

Answer: (i) active recovery voltage = 4.67 kV (peak); (ii) f_n = 82.19 kHz; (iii) t = 6.08 μs; (iv) maximum RRRV = 2.41 kV/μs.