Overvoltages in Power System

For energising a lumped series $R$–$L$–$C$ line model with a step voltage $E$ and initial (trapped) capacitor voltage $V_0$, the capacitor voltage is

$$V_c(t) = (E - V_0)\left[1 - e^{-\alpha t}\frac{\sqrt{\alpha^2+\omega^2}}{\omega}\cos(\omega t - \phi)\right] + V_0, \quad \tan\phi = \frac{\alpha}{\omega}$$

with $\alpha = \frac{R}{2L}$ and $\omega = \sqrt{\frac{1}{LC} - \frac{R^2}{4L^2}}$.

Condition for maximum

Let $K = \sqrt{\alpha^2+\omega^2}$, so $\sin\phi = \alpha/K$ and $\cos\phi = \omega/K$. Differentiate:

$$\begin{aligned} \frac{dV_c}{dt} &= (E-V_0)\frac{K}{\omega}e^{-\alpha t}\left[\alpha\cos(\omega t-\phi) + \omega\sin(\omega t-\phi)\right] \\ &= (E-V_0)\frac{K^2}{\omega}e^{-\alpha t}\left[\sin\phi\cos(\omega t-\phi) + \cos\phi\sin(\omega t-\phi)\right] \\ &= (E-V_0)\frac{\alpha^2+\omega^2}{\omega}e^{-\alpha t}\sin\omega t \end{aligned}$$

Setting $\frac{dV_c}{dt} = 0$ gives $\sin\omega t = 0$, so $\omega t = 0, \pi, 2\pi, \dots$ The first (largest) maximum is at

$$t_m = \frac{\pi}{\omega}$$

Maximum switching overvoltage

At $\omega t = \pi$: $\cos(\pi - \phi) = -\cos\phi = -\frac{\omega}{K}$, so

$$\begin{aligned} V_{max} &= (E-V_0)\left[1 + e^{-\alpha\pi/\omega}\frac{K}{\omega}\cdot\frac{\omega}{K}\right] + V_0 \\ &= (E-V_0)\left(1 + e^{-\alpha\pi/\omega}\right) + V_0 \\ &= E + (E - V_0)\,e^{-\alpha\pi/\omega} \end{aligned}$$

Special cases:

• No trapped charge ($V_0 = 0$): $V_{max} = E(1 + e^{-\alpha\pi/\omega})$, up to $2E$ when $R = 0$.
• Trapped charge of opposite polarity ($V_0 = -E$, as in fast reclosing): $V_{max} = E(1 + 2e^{-\alpha\pi/\omega})$, up to $3E$ when $R = 0$.
• $V_0 = E$: no overvoltage.

So the overvoltage is maximum when (i) the breaker closes at the peak of source voltage (so the step $E$ is the peak), (ii) the line holds a trapped charge of opposite polarity, and (iii) damping is small.

Effect of R, L and C

The damping exponent can be written as

$$\frac{\alpha}{\omega} = \frac{R/2L}{\sqrt{\frac{1}{LC} - \frac{R^2}{4L^2}}} = \frac{R}{\sqrt{4L/C - R^2}}$$

• Increasing R: $\alpha/\omega$ increases, $e^{-\alpha\pi/\omega}$ falls, so $V_{max}$ decreases. For $R \ge 2\sqrt{L/C}$ the circuit is overdamped and there is no overshoot ($V_{max} \to E$). This is why closing (pre-insertion) resistors are used.
• Increasing L: $4L/C$ increases, $\alpha/\omega$ falls (and $\alpha = R/2L$ falls), so damping reduces and $V_{max}$ increases. The oscillation also becomes slower.
• Increasing C: $4L/C$ decreases, $\alpha/\omega$ rises; $\alpha$ is unchanged but $\omega$ falls, so the peak occurs later after more decay, and $V_{max}$ decreases slightly; the period $2\pi/\omega$ becomes longer.

In real lines $R$ is small, so the damping factor $e^{-\alpha\pi/\omega}$ is close to 1 and overvoltages of 2–3 pu occur unless resistors or arresters are used.

Given: $L = 1.3$ mH/km $= 1.3 \times 10^{-3}$ H/km, $C = 8.855 \times 10^{-9}$ F/km, length $l = 200$ km, $f = 50$ Hz. The line is treated as lossless.

(i) Surge impedance

$$Z_c = \sqrt{\frac{L}{C}} = \sqrt{\frac{1.3 \times 10^{-3}}{8.855 \times 10^{-9}}} = \sqrt{146810} = 383.2\ \Omega$$

(ii) Velocity of propagation

$$v = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{1.3\times10^{-3} \times 8.855\times10^{-9}}} = \frac{1}{3.393\times10^{-6}} = 2.947 \times 10^{5}\ \text{km/s}$$

(about 98% of the speed of light).

(iii) Travel time and period of transient

One-way travel time:

$$\tau = \frac{l}{v} = \frac{200}{2.947\times10^5} = 6.786\times10^{-4}\ \text{s} = 0.679\ \text{ms}$$

For an open-ended line energised from a source, the voltage wave reflects back and forth and the transient oscillation has a period of four travel times:

$$T = 4\tau = 2.714\ \text{ms}, \qquad f_{osc} = \frac{1}{T} = 368.4\ \text{Hz}$$

(iv) Propagation constant

For a lossless line $\gamma = \alpha + j\beta$ with $\alpha = 0$ and

$$\beta = \omega\sqrt{LC} = 2\pi \times 50 \times 3.393\times10^{-6} = 1.066\times10^{-3}\ \text{rad/km}$$ $$\gamma = j\,1.066\times10^{-3}\ \text{rad/km}$$

For the whole line, $\beta l = 0.2132$ rad $= 12.2^\circ$. (Wavelength $\lambda = v/f = 5895$ km.)

Answer: $Z_c = 383.2\ \Omega$; $v = 2.947\times10^5$ km/s; $\tau = 0.679$ ms (period $4\tau = 2.71$ ms); $\gamma = j1.066\times10^{-3}$ rad/km.

Consider a lossless line of length $L$ with inductance $l$ and capacitance $c$ per unit length, open at the receiving end (no load). Let $E_s$ be the sending-end phase voltage and $E_r$ the receiving-end voltage.

Line equations

For a lossless line the propagation constant is $\gamma = j\beta$ with

$$\beta = \omega\sqrt{lc} = \frac{2\pi}{\lambda}, \qquad Z_c = \sqrt{\frac{l}{c}}$$

The long-line equations at distance $x$ from the receiving end are

$$\begin{aligned} E(x) &= E_r\cos\beta x + jZ_cI_r\sin\beta x \\ I(x) &= I_r\cos\beta x + j\frac{E_r}{Z_c}\sin\beta x \end{aligned}$$

At no load

$I_r = 0$, so at the sending end ($x = L$):

$$E_s = E_r\cos\beta L, \qquad I_s = j\frac{E_r}{Z_c}\sin\beta L$$

Hence $E_r = \dfrac{E_s}{\cos\beta L}$ (this is the Ferranti rise), and

$$I_s = j\frac{E_s}{Z_c}\frac{\sin\beta L}{\cos\beta L} = j\frac{E_s}{Z_c}\tan\beta L$$

The sending-end current leads $E_s$ by $90^\circ$: it is a pure charging current.

Reactive power supplied by the source

Taking $E_s$ as reference, the complex power per phase is

$$S_s = E_sI_s^* = E_s\left(-j\frac{E_s}{Z_c}\tan\beta L\right) = -j\frac{E_s^2}{Z_c}\tan\beta L$$

The real power is zero (lossless line). The magnitude of reactive power is

$$Q_0 = \frac{E_s^2}{Z_c}\tan\beta L$$

the negative sign meaning the source absorbs lagging VArs, i.e. it supplies leading (capacitive) charging VArs to the line.

Substitution

With $\frac{1}{Z_c} = \sqrt{\frac{c}{l}}$ and $\beta L = \frac{2\pi L}{\lambda}$:

$$\boxed{Q_0 = E_s^2\sqrt{\frac{c}{l}}\tan\frac{2\pi L}{\lambda}}$$

which is the required result.

Remarks

• For a short line, $\tan\beta L \approx \beta L$, giving $Q_0 \approx E_s^2\,\omega cL$, the usual $\omega CV^2$.
• $Q_0$ grows rapidly with length and becomes infinite at $L = \lambda/4$ (1500 km at 50 Hz), so long EHV lines need shunt reactors to absorb charging VArs and limit the Ferranti rise.

Assumptions: the line is represented by the given single-phase lumped series $R$–$L$–$C$ circuit; the breaker closes at the peak of the source voltage $E$ (worst case), the line has no trapped charge, and 1 pu = peak of source phase voltage. When the auxiliary breaker closes first, $R_s$ is in series with the line resistance.

Given: $R = 10\ \Omega$, $L = 0.5$ H, $C = 5\ \mu$F.

Formula

For energisation of a series $RLC$ circuit by a step $E$ with $V_0 = 0$:

$$V_{max} = E\left(1 + e^{-\alpha\pi/\omega}\right), \quad \alpha = \frac{R_T}{2L}, \quad \omega = \sqrt{\frac{1}{LC} - \alpha^2}$$

where $R_T = R_s + 10$. Also

$$\frac{\alpha}{\omega} = \frac{R_T}{\sqrt{4L/C - R_T^2}}$$

Overvoltage without $R_s$ (check)

$\alpha = 10/(2 \times 0.5) = 10\ \text{s}^{-1}$, $\omega = \sqrt{1/(0.5 \times 5\times10^{-6}) - 10^2} = 632.4$ rad/s

$$V_{max} = E(1 + e^{-10\pi/632.4}) = 1.952\ \text{pu}$$

which exceeds 1.8 pu, so a resistor is needed.

Required total resistance

$$\begin{aligned} 1 + e^{-\alpha\pi/\omega} &= 1.8 \\ e^{-\alpha\pi/\omega} &= 0.8 \\ \frac{\alpha}{\omega} &= \frac{-\ln 0.8}{\pi} = \frac{0.22314}{\pi} = 0.07103 \end{aligned}$$

Let $k = 0.07103$. From $\frac{R_T}{\sqrt{4L/C - R_T^2}} = k$:

$$R_T = \frac{k\sqrt{4L/C}}{\sqrt{1+k^2}}$$ $$\sqrt{\frac{4L}{C}} = \sqrt{\frac{4 \times 0.5}{5\times10^{-6}}} = \sqrt{4\times10^5} = 632.46\ \Omega$$ $$R_T = \frac{0.07103 \times 632.46}{\sqrt{1 + 0.07103^2}} = \frac{44.92}{1.00252} = 44.81\ \Omega$$

Switching resistance

$$R_s = R_T - R = 44.81 - 10 = 34.81\ \Omega$$

Check: $\alpha = 44.81/(2\times0.5) = 44.81\ \text{s}^{-1}$, $\omega = \sqrt{4\times10^5 - 44.81^2} = 630.87$ rad/s, $1 + e^{-44.81\pi/630.87} = 1.800$ pu.

Answer: $R_s \approx 34.8\ \Omega$ (total series resistance 44.8 Ω) limits the maximum switching overvoltage to 1.8 pu. After a short insertion time (about 8–10 ms) the main CB closes and shorts $R_s$.

A surge travelling on a cable (low surge impedance) into an overhead line (high surge impedance) is partly reflected and the junction voltage rises above the incident value.

Surge impedances

Cable:

$$Z_1 = \sqrt{\frac{L_1}{C_1}} = \sqrt{\frac{0.3\times10^{-3}}{0.4\times10^{-6}}} = \sqrt{750} = 27.39\ \Omega$$

Overhead line:

$$Z_2 = \sqrt{\frac{L_2}{C_2}} = \sqrt{\frac{1.5\times10^{-3}}{0.012\times10^{-6}}} = \sqrt{125000} = 353.55\ \Omega$$

Transmitted (junction) voltage

Transmission coefficient:

$$\tau = \frac{2Z_2}{Z_1 + Z_2} = \frac{2 \times 353.55}{27.39 + 353.55} = 1.856$$ $$V_t = \tau V = 1.856 \times 15 = 27.84\ \text{kV}$$

Reflected voltage on the cable:

$$V_r = \frac{Z_2 - Z_1}{Z_1 + Z_2}V = \frac{326.17}{380.94} \times 15 = 12.84\ \text{kV}$$

Check: $V + V_r = 15 + 12.84 = 27.84$ kV $= V_t$.

Currents: incident $I = 15000/27.39 = 547.7$ A; transmitted into the line $I_t = 27843/353.55 = 78.75$ A.

Answer: the voltage at the junction rises to about 27.84 kV (a rise of 12.84 kV over the 15 kV surge).

The line is represented by a series $R$–$L$–$C$ circuit energised by a step $E = 343$ kV. Given $R = 12.4\ \Omega$, $L = 0.4$ H, $C = 4\ \mu$F.

(i) Attenuation factor

$$a = \frac{R}{2L} = \frac{12.4}{2 \times 0.4} = 15.5\ \text{s}^{-1}$$

(ii) Natural angular frequency and frequency

Undamped:

$$\omega_0 = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{0.4 \times 4\times10^{-6}}} = 790.57\ \text{rad/s}$$

Damped:

$$\omega = \sqrt{\omega_0^2 - a^2} = \sqrt{790.57^2 - 15.5^2} = 790.42\ \text{rad/s}$$ $$f = \frac{\omega}{2\pi} = \frac{790.42}{2\pi} = 125.8\ \text{Hz}$$

(Since $R \ll 2\sqrt{L/C} = 632.5\ \Omega$, the circuit is oscillatory.)

(iii) Peak voltage across C

The capacitor voltage reaches its first peak at $t = \pi/\omega = 3.97$ ms:

$$V_{max} = E + (E - V_0)e^{-a\pi/\omega}$$

Damping factor:

$$e^{-a\pi/\omega} = e^{-15.5\pi/790.42} = e^{-0.0616} = 0.9403$$

(a) No trapped charge ($V_0 = 0$):

$$V_{max} = 343(1 + 0.9403) = 665.5\ \text{kV} \;(1.94\ \text{pu})$$

(b) Trapped charge of 343 kV. The worst (and usual textbook) case is reclosing when the trapped charge is of opposite polarity to the source, $V_0 = -343$ kV:

$$V_{max} = 343 + (343 + 343)(0.9403) = 343 + 645.0 = 988.0\ \text{kV} \;(2.88\ \text{pu})$$

If the trapped charge had the same polarity ($V_0 = +343$ kV), $E - V_0 = 0$ and there is no oscillation: $V_{max} = 343$ kV.

Answer: $a = 15.5\ \text{s}^{-1}$; $\omega = 790.4$ rad/s, $f = 125.8$ Hz; $V_{max} = 665.5$ kV with no trapped charge and $988.0$ kV with 343 kV trapped charge of opposite polarity.

Instead of an arrester, a short length of cable of low surge impedance $Z_c$ is inserted between the overhead line ($Z_0 = 400\ \Omega$) and the transformer ($Z_t = 1000\ \Omega$). At the line–cable junction most of the surge is reflected back, so only a small voltage enters the cable; it then doubles (nearly) at the transformer. $Z_c$ is chosen so that the transformer voltage is 800 kV.

 Overhead line     Cable        Transformer
 Z0 = 400 ohm ---- Zc ---------- Zt = 1000 ohm
   1500 kV ->   J1           J2

Assumption: only the first transmitted wave is considered (the cable is long enough that successive reflections inside it arrive after the surge crest).

Voltages

At junction J1 (line to cable):

$$V_1 = \frac{2Z_c}{Z_0 + Z_c}\times 1500$$

At junction J2 (cable to transformer):

$$V_2 = \frac{2Z_t}{Z_c + Z_t}V_1 = 1500 \times \frac{2Z_c}{400 + Z_c}\times\frac{2000}{Z_c + 1000}$$

Condition $V_2 = 800$ kV

$$\begin{aligned} 800(400 + Z_c)(1000 + Z_c) &= 1500 \times 4000\,Z_c \\ (400 + Z_c)(1000 + Z_c) &= 7500\,Z_c \\ Z_c^2 + 1400Z_c + 4\times10^5 &= 7500Z_c \\ Z_c^2 - 6100Z_c + 4\times10^5 &= 0 \end{aligned}$$ $$Z_c = \frac{6100 \pm \sqrt{6100^2 - 1.6\times10^6}}{2} = \frac{6100 \pm 5967.4}{2}$$ $$Z_c = 66.3\ \Omega \quad \text{or} \quad 6034\ \Omega$$

A cable has low surge impedance (30–70 Ω typical), so $Z_c = 66.3\ \Omega$ is taken. Any cable with $Z_c \le 66.3\ \Omega$ keeps the transformer at or below 800 kV.

Voltage rating of the cable

$$V_1 = \frac{2 \times 66.3}{400 + 66.3}\times 1500 = 426.5\ \text{kV}$$

Check: $V_2 = \frac{2000}{66.3 + 1000}\times 426.5 = 800$ kV.

Without the cable, the transformer would see $\frac{2 \times 1000}{1400}\times1500 = 2143$ kV.

Answer: use a cable of surge impedance about 66.3 Ω with a surge (impulse) voltage rating of about 427 kV (peak); the transformer voltage is then limited to 800 kV.

When lightning strikes the tower top, the stroke current $I_s$ divides between all the paths that meet at the tower top. These paths are in parallel: the stroke channel itself (surge impedance $Z_s$), the tower footing resistance $R$, and the ground wires (surge impedance $Z_g$ each). So the tower-top potential is

$$V_t = I_s Z_{eq}, \qquad \frac{1}{Z_{eq}} = \frac{1}{Z_s} + \frac{1}{R} + \frac{n}{Z_g}$$

where $n$ is the number of ground wires. (Tower inductance and travel time in the tower are neglected.)

Data: $R = 40\ \Omega$, $Z_g = 500\ \Omega$ (each), $Z_s = 400\ \Omega$, $I_s = 50$ kA.

(i)(a) Considering all impedances (two ground wires)

$$\begin{aligned} \frac{1}{Z_{eq}} &= \frac{1}{400} + \frac{1}{40} + \frac{2}{500} \\ &= 0.0025 + 0.025 + 0.004 = 0.0315\ \text{S} \\ Z_{eq} &= 31.75\ \Omega \\ V_t &= 50 \times 31.75 = 1587.3\ \text{kV} \end{aligned}$$

(i)(b) Neglecting ground wires and stroke impedance

Only the footing resistance carries the current:

$$V_t = I_s R = 50 \times 40 = 2000\ \text{kV}$$

(i)(c) One ground wire and stroke impedance

$$\begin{aligned} \frac{1}{Z_{eq}} &= \frac{1}{400} + \frac{1}{40} + \frac{1}{500} = 0.0295\ \text{S} \\ Z_{eq} &= 33.90\ \Omega \\ V_t &= 50 \times 33.90 = 1694.9\ \text{kV} \end{aligned}$$

The ground wires and the stroke channel take part of the current, so they lower the tower-top potential. The footing resistance is the largest single factor.

(ii) Voltage across the insulator string, case (a)

The surge on the ground wire induces a voltage $K V_t$ on the phase conductor through the coupling factor $K$. The conductor rises with the tower, so the insulator string sees only the difference:

$$\begin{aligned} V_{ins} &= V_t - K V_t = (1 - K) V_t \\ &= (1 - 0.2) \times 1587.3 \\ &= 1269.8\ \text{kV} \approx 1270\ \text{kV} \end{aligned}$$

If this exceeds the flashover voltage of the string, a back-flashover occurs from the tower (at high potential) to the conductor.

Number of discs to prevent back-flashover

Lightning strokes normally come with rain, so the wet (rain) flashover voltage of 80 kV per disc is used (the conservative choice):

$$\begin{aligned} n &= \frac{V_{ins}}{V_{rain}} = \frac{1269.8}{80} = 15.87 \\ n &= 16\ \text{discs (rounded up)} \end{aligned}$$

If dry conditions are assumed: $1269.8/125 = 10.16$, so 11 discs. Since a thunderstorm is wet, the higher number is chosen.

Answer: $V_t$ = 1587.3 kV, 2000 kV and 1694.9 kV for (a), (b), (c); insulator voltage = 1269.8 kV; minimum 16 standard discs (rain FOV basis).

Note: the textbook method ignores the power-frequency voltage of the conductor and assumes the per-disc flashover values add linearly. In practice the string would be checked against its impulse flashover level and the footing resistance would be reduced.

Ferroresonance is a non-linear resonance between a capacitance and an iron-cored (saturable) inductance, such as a transformer magnetising reactance. It produces sustained, distorted overvoltages and overcurrents, often 2–4 pu, with sudden jumps between operating states.

Why it happens

In a series circuit of capacitance $C$ and saturable inductance $L$, resonance needs $\omega L = 1/(\omega C)$. A linear inductor has a fixed $L$, but an iron-core inductance falls sharply when the core saturates. So for a wide range of $C$ there is some flux level at which $\omega L_{sat} \approx 1/(\omega C)$. The circuit can then "jump" into a high-voltage resonant state and stay there. Several stable states can exist for the same source voltage. Which one occurs depends on initial conditions such as residual flux and the switching instant.

  Vs ~---| C |---+
                 |
               L(sat)   (transformer
                 |       magnetising)
  ---------------+------- ground

Causes in power systems

1. Single-phase switching of an unloaded or lightly loaded transformer. One or two phases are open (fuse blown, breaker pole stuck, single-pole switching). The open phase is then fed through line or cable capacitance in series with the magnetising reactance.
2. Transformer fed through a long cable, or through the grading capacitors of open breakers. The cable capacitance or the breaker grading capacitance forms the series $C$.
3. Voltage transformers (PTs) in ungrounded or isolated-neutral systems. The PT magnetising reactance resonates with the phase-to-ground capacitance of the network.
4. Lightly loaded transformers on series-compensated lines.
5. Broken conductor faults, or loss of the ground connection on a wye-connected transformer.

Effects

• High, distorted overvoltages that can damage insulation and surge arresters.
• Overheating, noise and failure of PTs and transformers.

It is avoided by three-pole switching, loading the transformer or adding damping resistors, grounding the neutral, and avoiding long cable–transformer combinations with no load.

When a circuit breaker interrupts a small inductive current, such as the magnetising current of an unloaded transformer or a shunt reactor current, the arc is unstable. The breaker may force the current to zero before its natural zero. This is called current chopping, and it produces a high overvoltage.

How the overvoltage builds up

• Just before chopping, the current in the inductance is $i_0$. The energy stored in the magnetic field is $\tfrac{1}{2} L i_0^2$.
• After the breaker opens, the inductive current cannot stop at once. It has no path except the stray capacitance $C$ of the transformer winding and bushings. The magnetic energy is transferred to this capacitance and oscillates at the frequency

$$f = \frac{1}{2\pi\sqrt{LC}}$$

• At the peak of the oscillation all the energy is in the capacitance:

$$\tfrac{1}{2} C V_m^2 = \tfrac{1}{2} L i_0^2 \;\Rightarrow\; V_m = i_0\sqrt{\frac{L}{C}}$$

Because $L$ is large (henries) and $C$ is small (nanofarads), $\sqrt{L/C}$ is tens of kilo-ohms. So even a few amperes of chopped current gives hundreds of kV. For example, $i_0 = 10$ A, $L = 5$ H and $C = 0.01\ \mu$F give $V_m = 223.6$ kV.

    CB          i0 chopped
 ---/ ---+----------+
         |          |
         C (stray)  L (transformer)
         |          |
 --------+----------+--- ground

Consequences

• The high voltage across the breaker contacts can cause restrikes. Each restrike starts a new oscillation and the voltage may escalate.
• The transformer insulation is stressed.

Remedies: resistance switching (a resistor across the contacts damps the oscillation), surge arresters at the transformer, and breakers that do not chop current (for example SF$_6$ breakers with suitable design).

At a junction with two branch lines, the branches act as two impedances in parallel for the incoming surge. The transmitted (refracted) voltage is the same on both branches.

Data: $V = 200$ kV, $Z_1 = 500\ \Omega$, $Z_2 = 600\ \Omega$, $Z_3 = 200\ \Omega$.

Equivalent impedance of the branches

$$Z_p = \frac{Z_2 Z_3}{Z_2 + Z_3} = \frac{600 \times 200}{800} = 150\ \Omega$$

Transmitted voltage

$$\begin{aligned} V_t &= \frac{2 Z_p}{Z_1 + Z_p} V = \frac{2 \times 150}{500 + 150} \times 200 \\ &= 92.31\ \text{kV} \end{aligned}$$

This same voltage travels into both branch lines.

Transmitted currents

$$\begin{aligned} I_2 &= \frac{V_t}{Z_2} = \frac{92.31}{600} = 0.1538\ \text{kA} = 153.8\ \text{A} \\ I_3 &= \frac{V_t}{Z_3} = \frac{92.31}{200} = 0.4615\ \text{kA} = 461.5\ \text{A} \end{aligned}$$

Check with the reflected wave

• Reflected voltage: $V_r = V_t - V = 92.31 - 200 = -107.69$ kV
• Incident current: $I = 200/500 = 400$ A; reflected current: $I_r = -V_r/Z_1 = +215.4$ A
• Current in line 1 at the junction: $400 + 215.4 = 615.4$ A $= I_2 + I_3 = 153.8 + 461.5$ ✓

 Z1=500 ohm          +--- Z2=600: 92.31 kV, 153.8 A
 200 kV ---->  J ----+
                     +--- Z3=200: 92.31 kV, 461.5 A

Answer: transmitted voltage = 92.31 kV in each branch; currents = 153.8 A (600 Ω line) and 461.5 A (200 Ω line).

When lightning strikes the tower top, the stroke current $I_s$ divides between all the paths that meet at the tower top. These paths are in parallel: the stroke channel itself (surge impedance $Z_s$), the tower footing resistance $R$, and the ground wires (surge impedance $Z_g$ each). So the tower-top potential is

$$V_t = I_s Z_{eq}, \qquad \frac{1}{Z_{eq}} = \frac{1}{Z_s} + \frac{1}{R} + \frac{n}{Z_g}$$

where $n$ is the number of ground wires. (Tower inductance and travel time in the tower are neglected.)

Data: $R = 40\ \Omega$, $Z_g = 500\ \Omega$ (each), $Z_s = 400\ \Omega$, $I_s = 50$ kA.

(i)(a) Considering all impedances (two ground wires)

$$\begin{aligned} \frac{1}{Z_{eq}} &= \frac{1}{400} + \frac{1}{40} + \frac{2}{500} \\ &= 0.0025 + 0.025 + 0.004 = 0.0315\ \text{S} \\ Z_{eq} &= 31.75\ \Omega \\ V_t &= 50 \times 31.75 = 1587.3\ \text{kV} \end{aligned}$$

(i)(b) Neglecting ground wires and stroke impedance

Only the footing resistance carries the current:

$$V_t = I_s R = 50 \times 40 = 2000\ \text{kV}$$

(i)(c) One ground wire and stroke impedance

$$\begin{aligned} \frac{1}{Z_{eq}} &= \frac{1}{400} + \frac{1}{40} + \frac{1}{500} = 0.0295\ \text{S} \\ Z_{eq} &= 33.90\ \Omega \\ V_t &= 50 \times 33.90 = 1694.9\ \text{kV} \end{aligned}$$

The ground wires and the stroke channel take part of the current, so they lower the tower-top potential. The footing resistance is the largest single factor.

(ii) Voltage across the insulator string, case (a)

The surge on the ground wire induces a voltage $K V_t$ on the phase conductor through the coupling factor $K$. The conductor rises with the tower, so the insulator string sees only the difference:

$$\begin{aligned} V_{ins} &= V_t - K V_t = (1 - K) V_t \\ &= (1 - 0.2) \times 1587.3 \\ &= 1269.8\ \text{kV} \approx 1270\ \text{kV} \end{aligned}$$

If this exceeds the flashover voltage of the string, a back-flashover occurs from the tower (at high potential) to the conductor.

Answer: $V_t$ = 1587.3 kV (a), 2000 kV (b), 1694.9 kV (c); insulator string voltage in case (a) = 1269.8 kV.

A travelling wave meeting a change of surge impedance is partly reflected and partly transmitted. The transmitted voltage is

$$V_t = \frac{2 Z_2}{Z_1 + Z_2} V$$

Voltage entering the cable

$Z_1 = Z_0 = 400\ \Omega$ (overhead line), $Z_2 = Z_c = 100\ \Omega$ (cable), $V = 1000$ kV.

$$\begin{aligned} V_{cable} &= \frac{2 \times 100}{400 + 100} \times 1000 \\ &= 0.4 \times 1000 = 400\ \text{kV} \end{aligned}$$

Reflected voltage on the line $= 400 - 1000 = -600$ kV. The low cable impedance greatly reduces the surge entering the cable.

Voltage at the transformer

The transformer has practically infinite surge impedance (bushing capacitance neglected), so the cable end behaves as an open circuit. The reflection coefficient is +1 and the voltage doubles:

$$V_{tr} = 2 \times V_{cable} = 2 \times 400 = 800\ \text{kV}$$

 OH line 400 ohm     cable 100 ohm     transformer
 1000 kV --->  J1 ---- 400 kV ---> J2   (open: 800 kV)

Note: the reflected 400 kV wave from the transformer travels back to J1, where it is partly reflected again. Over successive reflections the transformer voltage settles towards $2 \times 1000 = 2000$ kV (the open-end value for the original line). A short cable therefore only protects the transformer during the first few microseconds, unless an arrester is also used.

Answer: voltage in the cable = 400 kV; transformer voltage (first arrival) = 800 kV.

When lightning strikes the tower top, the stroke current $I_s$ divides between all the paths that meet at the tower top. These paths are in parallel: the stroke channel itself (surge impedance $Z_s$), the tower footing resistance $R$, and the ground wires (surge impedance $Z_g$ each). So the tower-top potential is

$$V_t = I_s Z_{eq}, \qquad \frac{1}{Z_{eq}} = \frac{1}{Z_s} + \frac{1}{R} + \frac{n}{Z_g}$$

where $n$ is the number of ground wires. (Tower inductance and travel time in the tower are neglected.)

Data: $R = 40\ \Omega$, $Z_g = 500\ \Omega$ (each), $Z_s = 400\ \Omega$, $I_s = 50$ kA.

(i)(a) Considering all impedances (two ground wires)

$$\begin{aligned} \frac{1}{Z_{eq}} &= \frac{1}{400} + \frac{1}{40} + \frac{2}{500} \\ &= 0.0025 + 0.025 + 0.004 = 0.0315\ \text{S} \\ Z_{eq} &= 31.75\ \Omega \\ V_t &= 50 \times 31.75 = 1587.3\ \text{kV} \end{aligned}$$

(i)(b) Neglecting ground wires and stroke impedance

Only the footing resistance carries the current:

$$V_t = I_s R = 50 \times 40 = 2000\ \text{kV}$$

(i)(c) One ground wire and stroke impedance

$$\begin{aligned} \frac{1}{Z_{eq}} &= \frac{1}{400} + \frac{1}{40} + \frac{1}{500} = 0.0295\ \text{S} \\ Z_{eq} &= 33.90\ \Omega \\ V_t &= 50 \times 33.90 = 1694.9\ \text{kV} \end{aligned}$$

The ground wires and the stroke channel take part of the current, so they lower the tower-top potential. The footing resistance is the largest single factor.

Answer: (a) 1587.3 kV, (b) 2000 kV, (c) 1694.9 kV.

Switching overvoltages are transient overvoltages produced when breakers or switches change the network configuration. They are the most important overvoltages for EHV and UHV lines (above about 400 kV), where they set the insulation levels.

Major causes

1. Energising (closing) a long unloaded line. The step voltage starts travelling waves that double at the open end. With trapped charge from an earlier opening, the overvoltage can reach about 3 pu.
2. High-speed reclosing on a line with trapped charge.
3. Interruption of small inductive currents. Transformer magnetising current or reactor current is chopped before its natural zero.
4. Interruption of capacitive currents. De-energising unloaded lines, cables or capacitor banks can cause restrikes and voltage escalation.
5. Fault initiation and fault clearing, including out-of-phase switching.
6. Load rejection, together with the Ferranti effect and generator overspeed. (Usually counted as a temporary overvoltage.)
7. Ferroresonance during single-pole switching of unloaded transformers.

Overvoltage due to interruption of inductive current

When a breaker opens an unloaded transformer or reactor, the current is small (a few amperes). The arc becomes unstable and the breaker may cut the current abruptly at a value $i_0$ before its natural zero. This is current chopping.

  Vs ~--- CB ---+-----------+
            i0  |           |
               C stray     L (transformer)
                |           |
  --------------+-----------+--- ground

• At the instant of chopping, the magnetic energy in the transformer inductance is $\tfrac{1}{2} L i_0^2$.
• The current in $L$ cannot change suddenly, so it flows into the stray capacitance $C$ of the winding and bushing. Energy oscillates between $L$ and $C$ at

$$f = \frac{1}{2\pi\sqrt{LC}}$$

• When all the energy is in $C$:

$$\tfrac{1}{2} C V_m^2 = \tfrac{1}{2} L i_0^2 + \tfrac{1}{2} C V_0^2$$

where $V_0$ is the capacitor voltage at the chopping instant. When $V_0$ is small:

$$V_m \approx i_0 \sqrt{\frac{L}{C}}$$

The surge impedance $\sqrt{L/C}$ of a transformer is very high (tens of kΩ). So chopping only 5–10 A can produce more than 200 kV.

Effects: the high voltage appears across the breaker contacts and can cause restrikes, with the voltage escalating, and it stresses the transformer insulation.

Remedies:

• Resistance switching: a resistor across the breaker contacts damps the oscillation and absorbs the energy.
• Surge arresters at the transformer terminals.
• Breakers designed with low chopping levels.

When the breaker closes, the line acts as a series R–L–C circuit excited by a step voltage $E$ (closing at the peak of the source voltage, 1 pu). With no trapped charge, the capacitor voltage is

$$v_C(t) = E\left[1 - e^{-\alpha t}\frac{\sqrt{\alpha^2+\omega^2}}{\omega}\cos(\omega t - \phi)\right], \quad \alpha = \frac{R}{2L},\ \omega = \sqrt{\frac{1}{LC} - \alpha^2}$$

Setting $dv_C/dt = 0$ gives $\sin\omega t = 0$, so the first (largest) peak is at $t = \pi/\omega$:

$$V_{max} = E\left(1 + e^{-\alpha\pi/\omega}\right)$$

Data: $L = 0.5$ H, $C = 5\ \mu$F. Total resistance $R = 0.12 + 10 + R_s$ (source + line + switching resistor in series during closing).

$$\omega_0 = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{0.5 \times 5\times10^{-6}}} = 632.46\ \text{rad/s}$$

Without $R_s$ (for reference)

$R = 10.12\ \Omega$, so $\alpha = 10.12/(2\times0.5) = 10.12\ \text{s}^{-1}$ and $\omega = 632.37$ rad/s.

$$V_{max} = 1 + e^{-10.12\pi/632.37} = 1 + 0.951 = 1.951\ \text{pu}$$

Required $R_s$ for 1.8 pu

$$\begin{aligned} 1 + e^{-\alpha\pi/\omega} &= 1.8 \\ e^{-\alpha\pi/\omega} &= 0.8 \\ \frac{\alpha}{\omega} &= \frac{\ln(1/0.8)}{\pi} = \frac{0.22314}{\pi} = 0.07103 \end{aligned}$$

Using $\omega^2 = \omega_0^2 - \alpha^2$:

$$\begin{aligned} \omega &= \frac{\omega_0}{\sqrt{1 + (\alpha/\omega)^2}} = \frac{632.46}{\sqrt{1 + 0.07103^2}} = 630.87\ \text{rad/s} \\ \alpha &= 0.07103 \times 630.87 = 44.81\ \text{s}^{-1} \\ R &= 2L\alpha = 2 \times 0.5 \times 44.81 = 44.81\ \Omega \\ R_s &= 44.81 - 10 - 0.12 = 34.69\ \Omega \end{aligned}$$

Check: with $R = 44.81\ \Omega$, $e^{-44.81\pi/630.87} = 0.800$, so $V_{max} = 1.80$ pu ✓

 Vs ~--0.12--+--CB--+--10 ohm--0.5 H--+-- Vo
             |      |                 |
             +-Rs-/-+               5 uF
                                      |
 -------------------------------------+-- gnd

The resistor is in circuit only for the first few milliseconds (the pre-insertion time). The main contacts then short it out, which produces a second, smaller transient.

Answer: $R_s \approx$ 34.7 Ω (total circuit resistance 44.81 Ω), assuming no trapped charge and closing at the voltage peak.

A lossless line has series inductance $L$ and shunt capacitance $C$ per unit length. When a source $V$ is switched on, a voltage wave and a current wave travel along the line together.

Impedance offered to the travelling wave

Let the wavefront travel with velocity $v$. In a time $dt$ it covers a length $dx = v\,dt$, and this new length is charged to voltage $V$:

• Charge added: $dq = C\,dx\,V$, so the current is $I = \dfrac{dq}{dt} = CVv$ ... (1)
• Flux set up in the new length: $d\psi = L\,dx\,I$, so the voltage is $V = \dfrac{d\psi}{dt} = LIv$ ... (2)

Dividing (2) by (1):

$$\frac{V}{I} = \frac{LIv}{CVv} \;\Rightarrow\; \left(\frac{V}{I}\right)^2 = \frac{L}{C} \;\Rightarrow\; \frac{V}{I} = \sqrt{\frac{L}{C}} = Z_0$$

Multiplying (1) and (2): $VI = LC\,VI\,v^2$, so $v = 1/\sqrt{LC}$.

For an infinite line (or before the wave reaches the open end) there is no reflected wave. The source therefore sees a pure resistance $Z_0 = \sqrt{L/C}$, the surge impedance, even though the line contains only $L$ and $C$. Typical values are 300–500 Ω for overhead lines and 30–60 Ω for cables.

The waves consist of forward and backward components

For a lossless line the telegraph equations are

$$\frac{\partial v}{\partial x} = -L\frac{\partial i}{\partial t}, \qquad \frac{\partial i}{\partial x} = -C\frac{\partial v}{\partial t}$$

Eliminating $i$ gives the wave equation:

$$\frac{\partial^2 v}{\partial x^2} = LC\frac{\partial^2 v}{\partial t^2} = \frac{1}{v_p^2}\frac{\partial^2 v}{\partial t^2}$$

Its general (d'Alembert) solution is

$$v(x,t) = f_1(x - v_p t) + f_2(x + v_p t)$$

• $f_1(x - v_p t)$ keeps its shape while $x$ increases with $t$. It is the forward wave.
• $f_2(x + v_p t)$ moves towards decreasing $x$. It is the backward wave.

Substituting into the first equation gives the current:

$$i(x,t) = \frac{1}{Z_0}\left[f_1(x - v_p t) - f_2(x + v_p t)\right]$$

So $v_f/i_f = +Z_0$ for the forward wave and $v_b/i_b = -Z_0$ for the backward wave.

With the receiving end open, the current there must be zero. When the forward wave arrives it is reflected with $v_b = +v_f$ and $i_b = -i_f$. The voltage doubles to $2V$ and the current becomes zero. The total voltage and current on the line are then the sums of the forward and backward waves. Only on an infinite line, where no reflection exists, is the ratio $v/i$ always $Z_0$.

The Ferranti effect is the rise of receiving-end voltage above the sending-end voltage on a long line at no load or light load. The line's charging current flows through its series inductance and raises the voltage along the line. It is a main cause of temporary overvoltage.

Mathematical explanation

For a lossless line of length $l$, phase constant $\beta = \omega\sqrt{LC}$ and surge impedance $Z_c = \sqrt{L/C}$:

$$V_s = V_r\cos\beta l + jZ_c I_r\sin\beta l$$

At no load $I_r = 0$:

$$\frac{V_r}{V_s} = \frac{1}{\cos\beta l}$$

At 50 Hz, $\beta \approx 6^\circ$ per 100 km. For example, a 400 km line with $L = 1$ mH/km and $C = 10$ nF/km gives $\beta l = 0.397$ rad and $V_r = 1.085\,V_s$. A 1500 km line ($\beta l = 90^\circ$) gives an infinite voltage.

Reduction by shunt reactor compensation

Connect a shunt reactor $jX_L$ at the receiving end. At no load, the reactor current is

$$I_r = \frac{V_r}{jX_L}$$

Substituting:

$$\begin{aligned} V_s &= V_r\cos\beta l + jZ_c\sin\beta l \cdot \frac{V_r}{jX_L} \\ \frac{V_r}{V_s} &= \frac{1}{\cos\beta l + \dfrac{Z_c}{X_L}\sin\beta l} \end{aligned}$$

The extra positive term $\dfrac{Z_c}{X_L}\sin\beta l$ in the denominator reduces $V_r$. For $V_r = V_s$:

$$\cos\beta l + \frac{Z_c}{X_L}\sin\beta l = 1 \;\Rightarrow\; X_L = \frac{Z_c\sin\beta l}{1 - \cos\beta l} = Z_c\cot\frac{\beta l}{2}$$

Physically, the reactor draws a lagging current that cancels part of the line's leading charging current. A smaller net current flows through the series inductance, so the voltage rise is smaller.

Other methods

1. Shunt reactors at the ends and at intermediate stations; switched or controlled reactors for varying load.
2. Static VAR compensators (SVC) and STATCOMs, which absorb reactive power under light load.
3. Series capacitors are not used for this. They reduce the effective electrical length for power transfer but do not cure the no-load rise.
4. Sectionalising long lines with intermediate substations, so that $\beta l$ of each section is small.
5. Generator under-excitation (absorbing VArs), within stability limits.
6. Proper switching sequence: energise the line from the stronger end and connect the reactors before or with the line.