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Switching overvoltage with trapped charge and Rs

Short answer

Use series RLC step response with trapped charge -1 pu; maximum is 1+2 exp(-alpha pi/omega). Without Rs it is 2.90 pu; to limit to 1.8 pu, total R must be 177.1 ohm, so Rs is about 167 ohm.

Using the single-phase equivalent lumped parameter model circuit of a 3 phase, 400 kV, 50 Hz, 400 km long EHV transmission line shown below in figure, compute: a) The maximum per unit switching over voltage while closing the main CB without external switching resistance Rs considering the trap charge across the capacitor as −1 pu. b) Switching resistance Rs to be inserted during the closing of the CB so that the maximum switching over voltage in part (a) will be limited to 1.8 pu. [Figure: AC source feeding the line through the main circuit breaker (CB); a resistor Rs in series with an auxiliary CB is connected in parallel with the main CB; series line impedance 10 Ω and 0.5 H to the receiving end Vo; shunt capacitor 5 μF from Vo to ground] Switching over voltage is given as (E − V0)[1 − e^(−αt) (√(α² + ω²)/ω) cos(ωt − φ)] + V0, where tan φ = α/ω, φ = tan⁻¹(α/ω).

When the breaker closes, the lumped circuit is a series R–L–C circuit driven by a step EE (closing at the source peak, E=1E = 1 pu). The capacitor starts at the trapped-charge voltage V0V_0. The given response is

v(t)=(E−V0)[1−e−αtα2+ω2ωcos⁡(ωt−ϕ)]+V0v(t) = (E - V_0)\left[1 - e^{-\alpha t}\frac{\sqrt{\alpha^2 + \omega^2}}{\omega}\cos(\omega t - \phi)\right] + V_0

with α=R/2L\alpha = R/2L and ω=1/LC−α2\omega = \sqrt{1/LC - \alpha^2}.

Setting dv/dt=0dv/dt = 0 gives sin⁡ωt=0\sin\omega t = 0, so the first and largest peak is at t=π/ωt = \pi/\omega. There, cos⁡(π−ϕ)=−cos⁡ϕ\cos(\pi - \phi) = -\cos\phi and α2+ω2cos⁡ϕ/ω=1\sqrt{\alpha^2 + \omega^2}\cos\phi/\omega = 1. Hence

Vmax=V0+(E−V0)(1+e−απ/ω)V_{max} = V_0 + (E - V_0)\left(1 + e^{-\alpha\pi/\omega}\right)

With E=1E = 1 pu and V0=−1V_0 = -1 pu:

Vmax=−1+2(1+e−απ/ω)=1+2e−απ/ωV_{max} = -1 + 2\left(1 + e^{-\alpha\pi/\omega}\right) = 1 + 2e^{-\alpha\pi/\omega}

Circuit data: R=10 ΩR = 10\ \Omega, L=0.5L = 0.5 H, C=5 μC = 5\ \muF (source resistance neglected).

ω0=1LC=10.5×5×10−6=632.46 rad/s\omega_0 = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{0.5 \times 5\times10^{-6}}} = 632.46\ \text{rad/s}

(a) Without RsR_s

α=R2L=102×0.5=10 s−1ω=632.462−102=632.38 rad/sϕ=tan⁡−1(10/632.38)=0.906∘e−απ/ω=e−10π/632.38=e−0.04968=0.9515Vmax=1+2×0.9515=2.903 pu\begin{aligned} \alpha &= \frac{R}{2L} = \frac{10}{2 \times 0.5} = 10\ \text{s}^{-1} \\ \omega &= \sqrt{632.46^2 - 10^2} = 632.38\ \text{rad/s} \\ \phi &= \tan^{-1}(10/632.38) = 0.906^\circ \\ e^{-\alpha\pi/\omega} &= e^{-10\pi/632.38} = e^{-0.04968} = 0.9515 \\ V_{max} &= 1 + 2 \times 0.9515 = 2.903\ \text{pu} \end{aligned}

The peak occurs at t=π/ω=4.97t = \pi/\omega = 4.97 ms.

(b) RsR_s to limit the overvoltage to 1.8 pu

1+2e−απ/ω=1.8e−απ/ω=0.4αω=ln⁡2.5π=0.9163π=0.29166\begin{aligned} 1 + 2e^{-\alpha\pi/\omega} &= 1.8 \\ e^{-\alpha\pi/\omega} &= 0.4 \\ \frac{\alpha}{\omega} &= \frac{\ln 2.5}{\pi} = \frac{0.9163}{\pi} = 0.29166 \end{aligned}

Using ω=ω0/1+(α/ω)2\omega = \omega_0/\sqrt{1 + (\alpha/\omega)^2}:

ω=632.461+0.291662=632.461.04170=607.16 rad/sα=0.29166×607.16=177.09 s−1Rtotal=2Lα=2×0.5×177.09=177.09 ΩRs=177.09−10=167.09 Ω\begin{aligned} \omega &= \frac{632.46}{\sqrt{1 + 0.29166^2}} = \frac{632.46}{1.04170} = 607.16\ \text{rad/s} \\ \alpha &= 0.29166 \times 607.16 = 177.09\ \text{s}^{-1} \\ R_{total} &= 2L\alpha = 2 \times 0.5 \times 177.09 = 177.09\ \Omega \\ R_s &= 177.09 - 10 = 167.09\ \Omega \end{aligned}

Check: e−177.09π/607.16=e−0.9163=0.400e^{-177.09\pi/607.16} = e^{-0.9163} = 0.400, so Vmax=1+0.8=1.80V_{max} = 1 + 0.8 = 1.80 pu ✓

 E ~---+--- main CB ---+--10 ohm--0.5 H--+-- Vo
       |               |                 |
       +--Rs--aux CB---+               5 uF (V0=-1 pu)
                                         |
 ----------------------------------------+-- gnd
CaseTotal R (Ω\Omega)VmaxV_{max} (pu)
(a) no RsR_s102.90
(b) with RsR_s177.091.80

Answer: (a) 2.90 pu; (b) Rs≈R_s \approx 167 Ω. In practice the pre-insertion resistor is bypassed by the main contacts after about 8–10 ms.

✓ Answer checked by Claude Opus 5.5 (high effort)

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