TRV numerical: 1.7 mH, 0.0025 μF, 7500 A

The transient recovery voltage (TRV) is the oscillatory voltage that appears across the CB contacts just after the arc is extinguished at current zero.

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 = 1.7 mH, C = 0.0025 μF, I = 7500 A rms, f = 50 Hz, neutral grounded (first-pole-to-clear factor ignored).

X_L = 2π × 50 × 1.7×10⁻³ = 0.5341 Ω
V_m = √2 × 7500 × 0.5341 = 5665 V (peak)
√(LC) = √(1.7×10⁻³ × 0.0025×10⁻⁶) = 2.0616×10⁻⁶ s

(i) Natural frequency

f_n = 1 / (2π × 2.0616×10⁻⁶)
    = 77 201 Hz ≈ 77.2 kHz
ω_n = 4.8507 × 10⁵ rad/s

(ii) Peak value of TRV

V_peak = 2 V_m = 2 × 5665 = 11 329 V ≈ 11.33 kV

(iii) Time of peak TRV

t = π / ω_n = 1 / (2 f_n)
  = 1 / (2 × 77 201) = 6.477 × 10⁻⁶ s = 6.48 μs

(iv) Maximum rate of rise of TRV

RRRV_max = V_m ω_n
         = 5665 × 4.8507×10⁵
         = 2.748 × 10⁹ V/s ≈ 2.75 kV/μs

(v) Time of maximum RRRV

t = π / (2 ω_n) = 1 / (4 f_n)
  = 3.238 × 10⁻⁶ s = 3.24 μs

Answer: (i) 77.2 kHz; (ii) 11.33 kV; (iii) 6.48 μs; (iv) 2.75 kV/μs; (v) 3.24 μs.