Part 7 · Chapter 23

Switching Surges and Travelling Waves

Lightning comes from the sky, but a power system also manufactures its own overvoltages every time a switch opens or closes. To understand both, we need the physics of how a surge actually moves: a voltage disturbance does not appear everywhere at once but races along the line as a travelling wave, reflecting and refracting wherever the line changes character. This chapter builds that wave machinery — velocity, surge impedance, reflection and refraction, the lattice diagram — and then turns it on the switching operations that, on the longest and highest-voltage lines, set the insulation level more than lightning ever does.

High-Voltage Engineering Prof. Mithun Mondal Reading time ≈ 50 min
i What you'll learn
  • How a surge propagates as a travelling wave with velocity \(v = 1/\sqrt{LC}\) and surge impedance \(Z=\sqrt{L/C}\), the two linked by \(V = ZI\).
  • Reflection and refraction at a junction: \(\Gamma=\dfrac{Z_2-Z_1}{Z_2+Z_1}\) and \(\tau=\dfrac{2Z_2}{Z_2+Z_1}\).
  • The special terminations — open (voltage doubling), short, and matched — and what each does to a wave.
  • The Bewley lattice diagram for tracking successive reflections in space and time.
  • The causes of switching overvoltages — energizing, reclosing, fault clearing, capacitor and reactor switching, current chopping and restrikes.
  • The standard \(250/2500~\mu\mathrm{s}\) switching impulse, why switching governs EHV/UHV insulation, and how closing resistors, controlled switching and arresters tame it.
Section 23-1

Surges From Within

Chapter 22 traced overvoltages to an external cause — the lightning flash. But the system is also its own troublemaker. Every time a circuit breaker opens or closes, it suddenly rearranges where energy is stored in the network's inductances and capacitances, and the network responds with a transient: a switching surge. Energizing a long line, reclosing onto a trapped charge, clearing a fault, or switching a capacitor bank all launch such transients, and on the longest extra-high-voltage lines these internally generated surges, not lightning, decide how much insulation the line needs.

Whatever its origin, a surge obeys the same travel law. A disturbance applied at one point on a line does not raise the whole line at once; it sets off a wave of voltage and current that moves along the conductors at close to the speed of light, and behaves at every change of line — a junction, an open end, a transformer — like a wave meeting a boundary. Mastering that wave behaviour is the key that unlocks both switching and lightning surge analysis, so we build it first.

Section 23-2

The Travelling Wave

Model the line as a chain of series inductance \(L\) and shunt capacitance \(C\) per unit length. A voltage step applied at one end charges the first capacitor through the first inductor, which then charges the next, and so on — the disturbance is handed down the line at a finite speed. Solving the resulting wave equation gives two fundamental quantities. The velocity of propagation is

Propagation velocity
\[ v = \frac{1}{\sqrt{LC}} \]
V (voltage wave) v → I = V / Z Z = √(L/C)
A travelling wave — a voltage disturbance moves along the line at velocity \(v=1/\sqrt{LC}\), carrying a current wave of identical shape, the two tied together by the surge impedance through \(V = ZI\)

For an overhead line in air this works out to almost exactly the speed of light, \(3\times10^{8}~\mathrm{m/s}\); in a cable, where the dielectric slows it, it falls to roughly half that. The second quantity is the surge (characteristic) impedance, the fixed ratio of voltage to current within a single travelling wave:

Surge impedance
\[ Z = \sqrt{\frac{L}{C}} \qquad\Longrightarrow\qquad V = Z\,I \;\;(\text{forward wave}) \]

Typical values are \(300\)–\(500~\Omega\) for an overhead line and only \(20\)–\(60~\Omega\) for a cable, whose large capacitance lowers \(Z\). The surge impedance is not a resistance that dissipates energy; it is the ratio the line demands between the voltage and current of a wave passing along it. Every wave carries this fixed partnership — and that is exactly why a wave is disturbed when it reaches a place where the line wants a different ratio.

Section 23-3

Reflection and Refraction

Suppose a wave travelling on a line of surge impedance \(Z_1\) reaches a junction with a second line of impedance \(Z_2\) — an overhead line meeting a cable, say, or a line meeting a transformer. The incoming wave cannot simply continue, because the two sides demand different voltage-to-current ratios. Nature reconciles them by splitting the wave: part refracts (transmits) onward into \(Z_2\), and part reflects back along \(Z_1\), the two adjusting so that voltage and current are continuous across the junction. Enforcing that continuity gives the two governing coefficients:

Reflection and refraction (voltage) coefficients
\[ \Gamma = \frac{Z_2 - Z_1}{Z_2 + Z_1} \qquad\qquad \tau = 1 + \Gamma = \frac{2Z_2}{Z_2 + Z_1} \]
Z₁ Z₂ junction incident V Γ·V (reflected) τ·V (transmitted)
Reflection and refraction at a junction — the incident wave on \(Z_1\) splits into a reflected wave \(\Gamma V\) returning on \(Z_1\) and a transmitted wave \(\tau V\) continuing on \(Z_2\), chosen to keep voltage and current continuous across the boundary

The reflection coefficient \(\Gamma\) lies between \(-1\) and \(+1\). When the wave passes into a higher impedance (\(Z_2>Z_1\)), \(\Gamma\) is positive and the transmitted voltage rises above the incident — it can approach double. When it passes into a lower impedance, \(\Gamma\) is negative and the transmitted voltage falls. The current coefficients are the mirror image (the current reflection coefficient is \(-\Gamma\)), because a voltage that doubles at a high-impedance end must be met by a current that nearly vanishes there. This single pair of formulas governs every junction a surge can meet.

Section 23-4

Open, Short and Matched Ends

Three terminations are worth committing to memory, because the whole network is built from them. Setting \(Z_2\) to its extreme values in the coefficient formulas gives each one immediately.

open Γ = +1 V → 2V short Γ = −1 V → 0 matched (Z) Γ = 0 absorbed
The three key terminations — an open end (\(\Gamma=+1\)) doubles the voltage, a short (\(\Gamma=-1\)) drives it to zero, and a matched end equal to \(Z\) (\(\Gamma=0\)) absorbs the wave with no reflection

At an open circuit (\(Z_2\to\infty\)) the reflection coefficient is \(+1\): the reflected wave adds to the incident, and the voltage at the open end doubles to \(2V\). This is the most important single fact in surge protection — a surge running into an open breaker, an unloaded transformer terminal or a line end is amplified twofold, which is why those points are so vulnerable. At a short circuit (\(Z_2=0\)) the coefficient is \(-1\): the reflected wave inverts and cancels the incident, holding the voltage at zero (while the current doubles). At a matched end (\(Z_2=Z_1\)) the coefficient is \(0\): nothing reflects, and the wave is absorbed as if the line continued forever — the ideal that surge-absorbing terminations strive for.

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Voltage doubling at an open end
\[ V_{\text{open}} = 2V \]

An incident surge arriving at an open-circuited end is reflected with \(\Gamma=+1\), so the total voltage there reaches twice the incoming wave. Transformer terminals, open switches and line ends therefore see up to double the travelling-wave voltage — the reason these are the points most in need of arrester protection.

Section 23-5

The Lattice Diagram

A real surge does not reflect once and stop; it bounces between the two ends of a line, reflecting at each with that end's coefficient, the voltage at any point building up from the overlap of all the waves present at that instant. Keeping track of this by hand is the job of the Bewley lattice diagram — a simple but powerful bookkeeping chart. Distance along the line runs across the top; time runs downward. Each travelling wave is a sloped line crossing the chart in one transit time \(T=\ell/v\); at each end it turns back, multiplied by that end's reflection coefficient. Reading across at any chosen time and place, and summing the waves that have arrived, gives the voltage there and then.

sending (x=0) receiving (x=ℓ) time ↓ T 2T 3T ×Γ_R ×Γ_S ×Γ_R
A Bewley lattice diagram — the incident wave crosses in one transit time \(T\); at the receiving end it is multiplied by \(\Gamma_R\) and returns, at the sending end by \(\Gamma_S\), and so on. Summing the waves present at a point and time gives the voltage there

The lattice makes visible why surges are dangerous and why timing matters. Successive reflections can add, pushing the voltage at a vulnerable point well above the original wave before the transients finally damp out through line losses. It also shows that the worst stress often appears not at the instant the surge arrives but a transit time or two later, once a reflection has returned — a subtlety that pure peak-value thinking misses.

Section 23-6

Switching Overvoltages

With the wave tools in hand, return to the surges the system makes itself. Switching overvoltages arise whenever a breaker abruptly changes the network, and a few mechanisms dominate:

OperationMechanismConcern
Line energizingstep injected at sending end, doubling at the open far endup to \(\sim2~\mathrm{p.u.}\), more with trapped charge
Reclosingclosing onto a line still holding trapped chargecan reach \(3~\mathrm{p.u.}\) or beyond
Fault initiation / clearingsudden collapse and recovery of voltagesteep transients, recovery voltage
Capacitor switchinginrush on closing; restrikes on openingvoltage escalation on restrike
Inductive switchingcurrent chopping before natural zero\(L\,\mathrm{d}i/\mathrm{d}t\) voltage spikes

Two features set switching surges apart from lightning. First, their magnitude depends on the instant in the power-frequency cycle at which the contacts close — the point on the wave — so the surge is essentially random, varying from operation to operation, and must be described by a statistical distribution rather than a single value. Second, the most violent cases involve energy already trapped in the system: reclosing onto a line that still holds charge from a previous opening can superimpose the new surge on the old, and capacitor or reactor switching can pump energy up through repeated restrikes, each one ratcheting the voltage higher. Magnitudes commonly reach \(2\)–\(3~\mathrm{p.u.}\) — lower in crest than a direct lightning stroke, but applied to the whole line at once and lasting far longer.

Section 23-7

The Standard Switching Impulse

Because switching surges are slow-front and statistical, they are tested with their own standard waveform: the \(250/2500~\mu\mathrm{s}\) switching impulse, with a time-to-crest of \(250~\mu\mathrm{s}\) and a time-to-half-value of \(2500~\mu\mathrm{s}\) — hundreds of times slower on the front than the \(1.2/50\) lightning impulse.

V t → 1.2/50 lightning 250/2500 switching 250 µs
Switching versus lightning impulse — the \(250/2500\) switching impulse is a slow, broad hump beside the narrow \(1.2/50\) lightning spike; the slow front is what makes it the most onerous stress for long air gaps

Why a separate waveform? Because long air gaps are weakest against slow-front stress. The breakdown strength of a large air gap, plotted against the front time of the applied impulse, sags to a minimum near the switching-impulse range — a gap can withstand a fast lightning spike better than a slow switching hump of the same crest. As a result, on systems at \(300~\mathrm{kV}\) and above, and decisively at UHV, it is the switching impulse that sets the insulation level, while lightning is held in check by arresters. The withstand itself is statistical: the air gap has a \(50\%\) flashover voltage \(U_{50}\) with a spread, so insulation coordination works in probabilities — a theme taken up fully in Chapter 24.

Section 23-8

Controlling Switching Surges

Unlike lightning, switching surges are made by equipment the utility controls, so they can be attacked at the source. The classic remedy is the pre-insertion closing resistor: a resistor briefly inserted in series as the breaker closes, damping the energizing transient before it is shorted out, which can cut the surge from around \(2.5\) toward \(1.5~\mathrm{p.u.}\). The modern alternative is controlled (point-on-wave) switching, in which the breaker is commanded to close each pole at the instant its voltage is near zero, so almost no transient is launched in the first place. Shunt reactors help bleed away trapped charge so that reclosing does not find a charged line, and over everything sits the surge arrester, which clips whatever overvoltage survives to a safe ceiling.

These measures, together with the travelling-wave understanding of how surges build and where they double, are precisely the inputs the next chapter needs. Knowing the size and shape of both lightning and switching surges, and how arresters and resistors limit them, is what allows the insulation of a whole system to be coordinated — strong enough to survive, economical enough to build.

Section 23-9

Worked Examples

1 Surge impedance and velocity

Problem. An overhead line has \(L = 1.0~\mu\mathrm{H/m}\) and \(C = 11.1~\mathrm{pF/m}\). Find its surge impedance and propagation velocity.

Solution. Use \(Z=\sqrt{L/C}\) and \(v=1/\sqrt{LC}\):

Working
\[ Z = \sqrt{\frac{1.0\times10^{-6}}{11.1\times10^{-12}}} \approx 300~\Omega, \qquad v = \frac{1}{\sqrt{(1.0\times10^{-6})(11.1\times10^{-12})}} \approx 3\times10^{8}~\mathrm{m/s} \]

A surge impedance of about 300 Ω and a velocity of \(3\times10^{8}~\mathrm{m/s}\) — the speed of light, as expected for a line in air.

2 Line meets cable

Problem. A \(1.0~\mathrm{MV}\) wave travels on an overhead line of \(Z_1 = 400~\Omega\) and reaches a cable of \(Z_2 = 50~\Omega\). Find the reflected and transmitted voltages.

Solution. Compute \(\Gamma\) and \(\tau\):

Working
\[ \Gamma = \frac{50-400}{50+400} = -0.78,\quad \tau = \frac{2(50)}{450} = 0.22 \;\Rightarrow\; V_{\text{refl}} = -0.78~\mathrm{MV},\; V_{\text{trans}} = 0.22~\mathrm{MV} \]

Entering the low-impedance cable, the transmitted voltage falls to about 0.22 MV while a large inverted wave reflects back — which is exactly why a short cable section is sometimes used to protect a station by knocking down an incoming surge.

3 Voltage doubling at an open end

Problem. A \(600~\mathrm{kV}\) surge travels toward the open end of a line (an unloaded transformer terminal). What voltage appears at the open end?

Solution. An open end has \(\Gamma=+1\), so the reflected wave equals the incident and they add:

Working
\[ V_{\text{open}} = (1+\Gamma)\,V = 2 \times 600~\mathrm{kV} = 1.2~\mathrm{MV} \]

The open terminal sees 1.2 MV — twice the travelling wave. This doubling is why open points and transformer terminals are the first to be given arrester protection.

4 Travel time and the returning reflection

Problem. A surge is launched at the sending end of a \(150~\mathrm{km}\) open-ended line with \(v=3\times10^{8}~\mathrm{m/s}\). When does the doubled wave from the far end return to the sending end?

Solution. The one-way transit time is \(T=\ell/v\); the reflection returns after a round trip \(2T\):

Working
\[ T = \frac{150\times10^{3}}{3\times10^{8}} = 0.5\times10^{-3}~\mathrm{s} = 500~\mu\mathrm{s}, \qquad 2T = 1000~\mu\mathrm{s} = 1~\mathrm{ms} \]

The reflected wave arrives back after 1 ms. Note this round-trip time is of the same order as the switching-impulse front — a reminder that for switching surges the line's length and timing genuinely shape the stress.

5 A switching surge in real volts

Problem. A \(2.5~\mathrm{p.u.}\) switching surge occurs on a \(420~\mathrm{kV}\) (\(U_m\)) system. Find its crest in kV. (Base = peak phase-to-earth voltage.)

Solution. The per-unit base is \(\sqrt{2}\,U_m/\sqrt{3}\):

Working
\[ 1~\mathrm{p.u.} = \frac{\sqrt{2}\times420}{\sqrt{3}} \approx 343~\mathrm{kV}, \qquad 2.5~\mathrm{p.u.} \approx 857~\mathrm{kV} \]

About 857 kV crest. The line's switching-impulse withstand must clear this with statistical margin — and an arrester is set to clip anything that would exceed it.

Review

Chapter Summary

The travelling wave

A surge moves at \(v=1/\sqrt{LC}\) (\(\approx c\) on a line) with surge impedance \(Z=\sqrt{L/C}\), voltage and current tied by \(V=ZI\).

Reflection & refraction

At a junction, \(\Gamma=\dfrac{Z_2-Z_1}{Z_2+Z_1}\) and \(\tau=1+\Gamma\); into higher \(Z\) the voltage rises, into lower \(Z\) it falls.

Key terminations

Open doubles the voltage (\(\Gamma=+1\)), short cancels it (\(\Gamma=-1\)), matched absorbs it (\(\Gamma=0\)).

Lattice diagram

Tracks successive reflections in space and time; overlapping reflections can build the voltage above the original wave.

Switching surges

Energizing, reclosing onto trapped charge, fault clearing, capacitor restrikes and current chopping; statistical, \(2\)–\(3~\mathrm{p.u.}\)

250/2500 & control

The slow switching impulse governs EHV/UHV air insulation; closing resistors, controlled switching and arresters tame it.

Practice

Problems

For each item, first identify what it tests — the wave parameters, the reflection/refraction coefficients, a special termination, the lattice idea, switching-surge causes, or the switching impulse — then apply it. Difficulty rises down the list.

  1. Define surge impedance and propagation velocity, and explain why \(Z\) is not a power-dissipating resistance.
  2. Write the voltage reflection and refraction coefficients at a junction and state the sign of \(\Gamma\) for a step into higher and into lower impedance.
  3. State what happens to the voltage at an open, a short and a matched end, with the value of \(\Gamma\) for each.
  4. A line has \(L=0.9~\mu\mathrm{H/m}\) and \(C=12.8~\mathrm{pF/m}\). Find \(Z\) and \(v\).
  5. A \(500~\mathrm{kV}\) wave on \(Z_1=350~\Omega\) reaches a line of \(Z_2=450~\Omega\). Find the reflected and transmitted voltages, and comment on the rise.
  6. A \(400~\mathrm{kV}\) surge reaches an open transformer terminal. Find the terminal voltage and explain the consequence.
  7. Explain, using a lattice diagram, why the worst voltage at a point can occur a transit time after the surge first arrives.
  8. List four switching operations that cause overvoltages and explain why their magnitudes are statistical rather than fixed.
  9. Explain why a separate \(250/2500\) switching impulse is needed and why switching governs EHV insulation while lightning governs lower voltages.
  10. A \(3.0~\mathrm{p.u.}\) reclosing surge occurs on a \(765~\mathrm{kV}\) (\(U_m=800~\mathrm{kV}\)) system. Find its crest in kV and name two measures that would reduce it.
Tip: this chapter has one engine and one warning. The engine is the pair \(v=1/\sqrt{LC}\), \(Z=\sqrt{L/C}\) with \(V=ZI\): every surge is a wave carrying a fixed voltage-to-current ratio at near light speed. The warning is what happens at boundaries — \(\Gamma=\dfrac{Z_2-Z_1}{Z_2+Z_1}\) decides how much reflects, and the open-end \(\Gamma=+1\) doubles the voltage at exactly the transformer terminals and line ends you most want to protect. Add the lattice diagram to track the bounces and the statistical, slow-front nature of switching surges, and you have both halves of the system's self-made overvoltages — ready to be coordinated and clipped in Chapter 24.