Part 8 · Chapter 37

Substation Layout, Earthing and Insulation Coordination

A substation is where the network is deliberately made breakable, deliberately made earthable and deliberately made the weakest point of its own insulation — and every one of those three decisions is an engineering trade between what an outage costs, what a shock costs and what a flashover costs.

Electric Power Systems Prof. Mithun Mondal Reading time ≈ 50 min
i What you'll learn
  • What a substation actually does, and how the six standard busbar arrangements trade capital cost against the consequences of a bus fault or a breaker outage.
  • Why \(\mathrm{SF_6}\) lets a gas-insulated substation occupy a tenth of the land of an air-insulated one, and what is given up in return.
  • How the choice of neutral earthing — solid, resistance, reactance or resonant — sets the earth-fault current, the healthy-phase voltage rise and the relaying strategy of Chapter 36.
  • The Peterson coil tuning condition \(L = 1/(3\omega^2 C)\), derived from the requirement that the coil current cancel the line's charging current.
  • How an earthing grid is sized so that ground potential rise never produces a lethal step or touch voltage.
  • Where lightning and switching overvoltages come from, and what a metal-oxide surge arrester does about them.
  • The BIL concept, and how the protective margin \(\mathrm{PM}=(\mathrm{BIL}-V_p)/V_p\) closes the loop between the arrester and the transformer it protects.
Section 37-1

The Substation as a Node of the Network

Every diagram we have drawn since Chapter 3 has treated a bus as a dimensionless point. Chapter 16 gave that point an admittance, Chapter 18 gave it a specification, Chapter 25 gave it a fault level, and Chapter 35 gave it a circuit breaker. On the ground, that point is a fenced rectangle of land carrying steel structures, insulator stacks, current and voltage transformers, disconnectors, breakers, transformers, a control building and a buried copper mesh. The substation is where the abstraction is cashed in.

Four jobs are performed there, and a substation is usually classified by which of them dominates. It transforms voltage — a step-up substation at a generating station lifts \(11\)–\(21\;\mathrm{kV}\) to \(220\) or \(400\;\mathrm{kV}\); a grid substation steps \(400\) down to \(220\), a secondary one \(220\) to \(132\) or \(66\), and a distribution substation \(33\;\mathrm{kV}\) down to \(11\;\mathrm{kV}\) and finally to \(415\;\mathrm{V}\). It switches — a switching substation contains no transformer at all and exists purely so that lines can be connected, separated and isolated. It measures and protects — every instrument transformer feeding the relays of Chapter 36 lives here. And it controls reactive power — the capacitor banks, shunt reactors and static compensators of Chapter 34 are installed on substation buses because that is where the network is accessible.

A fifth category, the converter substation, terminates an HVDC link; Chapter 38 is devoted to what happens inside it.

Why the layout question is not cosmetic. The single-line diagram of a substation determines what happens when one element is lost. A bus fault in one arrangement disconnects the whole station; in another it disconnects nothing. A breaker taken out for maintenance in one scheme forces its circuit out with it; in another the circuit stays energised. Since a large transmission substation may carry \(2000\;\mathrm{MW}\) of transfer, the difference between those outcomes is measured in millions of rupees per event — and that is the currency in which the extra breakers are paid for.

Physically the equipment falls into a short list. Busbars — rigid aluminium tubes or strung ACSR — are the common connection points. Circuit breakers interrupt load and fault current; disconnectors (isolators) carry no interrupting duty at all and are operated only on a dead circuit, to establish a visible open point for safety. Earthing switches bond an isolated section to the grid before anybody touches it. Current and voltage transformers scale the measurands. Surge arresters cap the overvoltage. Wave traps block carrier signals from leaving the protected zone. Above everything run the earth wires that shield the yard from direct lightning strokes, and beneath everything lies the earthing grid of Section 37-6.

Section 37-2

Busbar Arrangements and the Reliability They Buy

The design question is narrow enough to be answered systematically. Given \(n\) circuits — lines and transformers — that must meet at a station, how many buses and how many breakers do we install, and how are they interconnected? Every scheme in use is one of six answers, and they form an almost perfect ladder of increasing cost and increasing security.

Single bus. One busbar; each circuit taps it through one breaker flanked by two disconnectors. It is the cheapest arrangement possible — one breaker per circuit — and the least secure. A fault on the bus, or on any breaker, trips the entire station. Maintenance of a breaker requires its circuit to be de-energised. It is used at distribution voltages and in small industrial stations where a total outage is tolerable.

Sectionalised single bus. Insert a bus-coupler breaker in the middle of the bus, and feed alternate circuits from the two halves. A bus fault now costs half the station instead of all of it, and half the station can be taken out for maintenance while the other half runs. The cost is one extra breaker for the whole station — an outstanding return, which is why almost no genuinely single bus is built above \(11\;\mathrm{kV}\).

Main and transfer bus. A second bus, the transfer bus, runs parallel to the main bus, and a single bus-coupler breaker joins the two. Every circuit can be switched onto the transfer bus and picked up by the coupler breaker, freeing its own breaker for maintenance while the circuit stays in service. The scheme therefore solves the breaker maintenance problem with one extra breaker, but not the bus fault problem: a main-bus fault still trips everything. During a transfer, the circuit is protected by the coupler's relays rather than its own, which complicates the settings of Chapter 36.

Double bus, single breaker. Two full-capacity buses, each circuit selectable onto either by a pair of disconnectors, with a coupler breaker between them. Circuits may be split into two groups, and a bus fault costs only the group on that bus, since the healthy group is switched across. Flexibility is high; the breaker count remains \(n+1\); but a breaker fault still removes its circuit, and the changeover involves disconnector operations.

Double bus, double breaker. Each circuit connects to both buses through its own breaker: two breakers per circuit. Any bus can be lost, any breaker can be lost, and every circuit stays in service, switched by the other breaker. It is the most secure arrangement known and, at \(2n\) breakers, easily the most expensive. It appears only where the load is critical and the voltage is high enough that an outage is intolerable.

Ring bus. Close the bus on itself so that the breakers form a ring and each circuit is tapped between two adjacent breakers. There are \(n\) breakers for \(n\) circuits — no more than a single bus — yet every circuit is fed from two directions. A fault on one circuit opens the two breakers flanking it and the ring becomes an open chain; all other circuits stay energised. The weakness is what happens next: while the ring is open, a second fault splits the station, and a breaker taken out for maintenance opens the ring in the same way. Rings are therefore restricted in practice to about four to six circuits.

Breaker-and-a-half. Between the two buses, string three breakers in series to form a diameter; tap one circuit between the first and second breakers and another circuit between the second and third. Two circuits share three breakers, so the count is \(1.5\) breakers per circuit — the name. The centre breaker is shared. A bus fault trips only the breakers on that bus and every circuit continues to be fed from the other bus through its centre breaker. Any breaker may be removed for maintenance without disconnecting a circuit. It gives very nearly the security of double-bus-double-breaker at three-quarters of the breaker count, and it is the standard arrangement of large \(400\;\mathrm{kV}\) and \(765\;\mathrm{kV}\) substations worldwide.

RING BUS (n breakers) L1 L2 T1 T2 a circuit fault opens two breakers and opens the ring BREAKER-AND-A-HALF (1.5n) BUS A BUS B Line 1 Line 2 Tr 1 Tr 2 shared centre breakers a bus fault trips only that bus; every circuit survives
Two economical answers to the same question — the ring and the breaker-and-a-half diameter
ArrangementBreakers for \(n\) circuitsBus fault costsBreaker maintenanceTypical use
Single bus\(n\)Whole stationCircuit outSmall \(11\;\mathrm{kV}\) stations
Sectionalised bus\(n+1\)Half the stationCircuit out\(11\)–\(33\;\mathrm{kV}\) distribution
Main and transfer\(n+1\)Whole stationCircuit stays in\(66\)–\(132\;\mathrm{kV}\)
Double bus, single breaker\(n+1\)One bus groupCircuit out\(132\)–\(220\;\mathrm{kV}\)
Ring bus\(n\)Nothing (ring opens)Ring opens\(\le 6\) circuits, \(220\;\mathrm{kV}\)
Breaker-and-a-half\(1.5n\)NothingCircuit stays in\(400\)–\(765\;\mathrm{kV}\) grid
Double bus, double breaker\(2n\)NothingCircuit stays inCritical EHV nodes
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The design rule
Security is bought in breakers. Going from \(n\) to \(1.5n\) breakers removes both the bus-fault outage and the maintenance outage; going on to \(2n\) buys almost nothing more.

This is why the breaker-and-a-half scheme dominates EHV practice: it sits exactly at the knee of the cost–security curve. Note also that the two circuits sharing a diameter must be chosen so that losing the whole diameter is survivable — a line and a transformer, never the two halves of a double-circuit line.

Section 37-3

Gas-Insulated Substations

Everything above assumed the insulating medium is atmospheric air, which breaks down at roughly \(30\;\mathrm{kV/cm}\) in a uniform field and considerably less in the divergent fields around real hardware. Clearances are therefore large: a \(400\;\mathrm{kV}\) air-insulated yard needs phase-to-earth clearances of about \(3.5\;\mathrm{m}\) and occupies several hectares. In a city centre that land does not exist, or costs more than the whole substation.

Sulphur hexafluoride, \(\mathrm{SF_6}\), changes the arithmetic. It is an electronegative gas: its molecules capture free electrons and remove them from the avalanche that would otherwise develop into a breakdown. At atmospheric pressure its dielectric strength is about \(2.5\) to \(3\) times that of air, and because strength continues to rise with pressure, at the \(3\)–\(5\;\mathrm{bar}\) used in practice it exceeds air by roughly an order of magnitude. The same gas is an excellent arc-quenching medium, which is why Chapter 35 gave \(\mathrm{SF_6}\) breakers their own section.

In a gas-insulated substation the busbars, disconnectors, earthing switches, current and voltage transformers and the breakers themselves are enclosed in earthed aluminium tubes filled with pressurised \(\mathrm{SF_6}\). The live conductor runs along the axis, supported on cast-resin spacers that also divide the enclosure into separately monitored gas compartments. The result is a coaxial system: fields are almost uniform, so the clearance can be cut to the theoretical minimum, and the whole \(400\;\mathrm{kV}\) station shrinks to perhaps \(10\%\) of the equivalent air-insulated footprint.

PropertyAir-insulated (AIS)Gas-insulated (GIS)
Footprint at \(400\;\mathrm{kV}\)Reference (100%)\(\approx 10\%\)
Exposure to pollution, salt, dustFullNone — sealed enclosure
Capital costLower\(2\)–\(4\times\) higher
Erection timeLong, weather-dependentShort, factory-tested modules
Maintenance intervalYears\(\approx 10\)–\(20\) years
Fault location and repairVisible, quickDifficult; compartment must be evacuated
Environmental issueLand use, audible noise\(\mathrm{SF_6}\) global-warming potential \(\approx 23\,500\)
A different overvoltage problem. Because a GIS is a coaxial transmission line of a few tens of ohms surge impedance, a disconnector operating on a short bus section produces very fast transient overvoltages with rise times of nanoseconds — far faster than the \(1.2\;\mu\mathrm{s}\) lightning impulse the insulation is tested against. These VFTOs are the reason GIS design pays close attention to enclosure bonding and to the transition at the SF₆-to-air bushing, where a fast transient can appear on the outdoor equipment.
Section 37-4

Neutral Earthing: Solid, Resistance, Reactance

Chapter 23 built the zero-sequence network and Chapter 24 showed that the single line-to-ground fault current is

Single line-to-ground fault, from Chapter 24
\[ I_f = 3I_{a0} = \frac{3E_a}{Z_1 + Z_2 + Z_0 + 3Z_n} \]

The impedance \(Z_n\) deliberately inserted between the transformer or generator star point and earth appears tripled in the zero-sequence path and nowhere else. That single algebraic fact is the entire theory of neutral earthing: it is the one knob that changes earth-fault current without touching anything else in the network.

Why not simply leave the neutral unearthed? An isolated neutral system does have one attraction — a single ground fault draws only the small capacitive charging current of the healthy phases, so the system can keep running with the fault on it. But the price is severe. The neutral point shifts by the full phase voltage, so the two healthy phases are stressed to \(\sqrt3\) times normal for as long as the fault lasts, and every piece of insulation must be rated accordingly. Worse, the fault current is small enough to extinguish at a current zero and restrike on the next voltage peak, trapping charge on the line capacitance and escalating in steps — an arcing ground, which can build transient overvoltages to \(5\)–\(6\) times normal and destroy insulation elsewhere in the system.

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Definition — effective earthing
A system is effectively earthed when \(X_0/X_1 \le 3\) and \(R_0/X_1 \le 1\). The healthy-phase voltage during an earth fault then does not exceed \(0.8\) times the line voltage, i.e. an earth-fault factor of about \(1.4\).

The earth-fault factor is the ratio of the highest healthy-phase power-frequency voltage during a fault to the normal phase voltage. It is the number that later fixes the surge arrester rating in Section 37-8, so the earthing decision and the insulation decision are not independent choices.

Solid earthing sets \(Z_n = 0\). The earth-fault factor falls to about \(1.4\) or below, insulation can be graded — the winding near the neutral of a large transformer is wound with less insulation than the line end, a real saving — and the fault current is large, which makes it easy for the earth-fault relays of Chapter 36 to see. The cost is exactly that large current: the single line-to-ground fault current can exceed the three-phase value when \(Z_0 < Z_1\), setting the breaker duty of Chapter 25 and imposing severe mechanical and thermal stress. Solid earthing is universal at \(66\;\mathrm{kV}\) and above.

Resistance earthing inserts \(R_n\). It limits the fault current, and because \(R_n\) is resistive it also damps the transient oscillation between the line capacitance and the system inductance, which suppresses arcing-ground escalation. The current is normally limited to somewhere between full-load current and about \(400\;\mathrm{A}\); enough for selective relaying, small enough that the damage at the fault point is limited. The resistor dissipates \(I_f^2 R_n\), which for a few hundred amperes and tens of ohms is megawatts — it is rated for a short time only, typically \(10\;\mathrm{s}\), consistent with the fault clearing time. Generator neutrals are almost always resistance-earthed, often through a distribution transformer with a low-ohmic secondary resistor, to keep stator iron damage below the threshold at which a rewind becomes a re-core.

Reactance earthing inserts \(X_n\). It limits the current with negligible loss, and it is the usual way to trim a generator or transformer neutral so that the single line-to-ground current does not exceed the three-phase current. The reactance must be kept small — \(X_0 \le 3X_1\) overall — because a large \(X_n\) pushes the system toward resonance with the line capacitance and toward the very restriking problem earthing was meant to prevent. That resonance, however, is not always an enemy. Deliberately tuned, it becomes the Peterson coil.

Section 37-5

The Peterson Coil and the Tuning Condition

Consider a three-phase line with capacitance \(C\) from each phase to earth, and let phase \(a\) develop a solid ground fault. Phase \(a\) is now at earth potential, so its own capacitance is short-circuited and carries nothing. The two healthy phases, however, are raised to the full line voltage with respect to earth: \(V_b\) and \(V_c\) become \(\sqrt3\,V_{ph}\) in magnitude, \(60^\circ\) apart in phase, and each drives a charging current through its own capacitance.

Capacitive current returning through the fault
\[ I_b = \sqrt3\,V_{ph}\,\omega C, \qquad I_c = \sqrt3\,V_{ph}\,\omega C, \qquad \text{angle between them } 60^\circ \]
\[ I_C = 2\,\big(\sqrt3\,V_{ph}\,\omega C\big)\cos 30^\circ = 2\sqrt3\,V_{ph}\,\omega C \times \frac{\sqrt3}{2} = 3\,V_{ph}\,\omega C \]

The capacitive current arriving at the fault is therefore \(3\omega C V_{ph}\) — three times what a single phase would contribute, which is the same factor of three that appears in the zero-sequence algebra of Chapter 22, and for the same reason: all three phase-to-earth capacitances are in parallel as seen from the neutral.

Waldemar Petersen's idea, published in 1916, was to connect an inductor between the neutral and earth so that its current at the fault is equal and opposite. The coil sits at phase-voltage \(V_{ph}\) across it once the neutral has shifted, so it draws

Coil current and the tuning condition
\[ I_L = \frac{V_{ph}}{\omega L} \]
\[ \text{Complete compensation:}\qquad I_L = I_C \quad\Longrightarrow\quad \frac{V_{ph}}{\omega L} = 3\,\omega C\,V_{ph} \]
\[ \boxed{\;L = \frac{1}{3\omega^{2} C}\;} \]

The coil current lags the neutral voltage by \(90^\circ\) and the capacitive current leads it by \(90^\circ\); they are in exact antiphase at the fault point. When the tuning condition holds, the current in the arc is reduced to the small residual made up of the coil's copper loss, the line's leakage conductance and the harmonic content — typically a few percent of \(I_C\). That residual is far too small to sustain an arc, so the fault self-extinguishes and the system continues to run with one phase earthed.

N a b c fault C C L coil I_L → ← I_C at the fault I_L = V/ωL cancels I_C = 3ωCV ⇔ L = 1/(3ω²C) I_C I_L
The arc-suppression coil supplies an inductive current that annuls the line's charging current at the fault

Two practical points follow immediately. First, \(C\) depends on how much line is connected, so a coil tuned for the full network is detuned the moment a feeder is switched out. Real arc-suppression coils are therefore built with a movable plunger or a tapped winding and an automatic controller that measures the neutral displacement voltage and re-tunes continuously; the residual is held to a few percent of \(I_C\). Second, resonant earthing is only worthwhile where the charging current is modest — up to a few hundred amperes — which in practice means \(33\;\mathrm{kV}\) and below on predominantly overhead networks. Cable systems have capacitances one to two orders of magnitude larger (Chapter 9), and the coil would be impossibly big.

MethodEarth-fault currentHealthy-phase voltageArcing ground riskWhere used
Isolated neutralCharging current only\(\sqrt3 V_{ph}\)SevereObsolete; small LV nets
SolidVery large\(\le 1.4 V_{ph}\)None\(66\;\mathrm{kV}\) and above
ResistanceLimited, e.g. \(200\)–\(400\;\mathrm{A}\)Up to \(\sqrt3 V_{ph}\)DampedGenerators, \(3.3\)–\(11\;\mathrm{kV}\) industry
ReactanceTrimmed to \(\le\) 3-phase valueModerate risePresent if \(X_n\) largeGenerator and transformer neutrals
Resonant (Peterson coil)Residual only, a few A\(\sqrt3 V_{ph}\)Self-extinguishing\(\le 33\;\mathrm{kV}\) overhead networks
Earthing and protection are one decision. Chapter 36 built earth-fault relays that respond to \(3I_0\). A resonant-earthed system deliberately removes the very quantity those relays measure, so it must be protected by wattmetric or admittance methods that detect the small resistive residual instead. Choose the earthing first, and the earth-fault protection philosophy is already chosen with it.
Section 37-6

The Earthing Grid: Step and Touch Potential

Everything so far concerned the neutral connection at a single point. The other earthing problem is different in kind: when fault current flows into the soil, the soil is not an equipotential, and people standing in the yard are in the resulting field.

Bury a horizontal mesh of copper conductors — typically \(7\;\mathrm{m}\) to \(15\;\mathrm{m}\) spacing at \(0.5\;\mathrm{m}\) depth, supplemented by driven rods — and bond every structure, fence, enclosure and neutral to it. When a ground fault injects current \(I_G\) into that grid, the whole grid rises to a potential above remote earth called the ground potential rise:

Ground potential rise
\[ \mathrm{GPR} = I_G R_g \]

For a grid of buried length \(L\) covering area \(A\) in soil of resistivity \(\rho\), the resistance to remote earth follows Laurent and Niemann's approximation — a hemispherical-plate term plus a conductor-length term:

Grid resistance
\[ R_g \;=\; \frac{\rho}{4}\sqrt{\frac{\pi}{A}} \;+\; \frac{\rho}{L} \]
\[ \text{Sverak's refinement, including burial depth } h:\quad R_g = \rho\left[\frac{1}{L} + \frac{1}{\sqrt{20A}}\left(1 + \frac{1}{1 + h\sqrt{20/A}}\right)\right] \]

The first term dominates for a large grid: resistance falls as \(1/\sqrt A\), so doubling the area buys only a \(30\%\) reduction. Adding more conductor inside a fixed area helps the second term only, and quickly saturates. This is the fundamental frustration of earthing design — you cannot simply drive \(R_g\) to zero, and a \(1\;\Omega\) grid carrying \(20\;\mathrm{kA}\) is at \(20\;\mathrm{kV}\) above remote earth.

Since the GPR cannot be eliminated, safety is defined not by the grid's absolute potential but by the differences a human body can be exposed to.

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The two dangerous voltages
Step voltage is the potential difference between two feet \(1\;\mathrm{m}\) apart on the surface. Touch voltage is the difference between a hand on an earthed structure and the feet standing \(1\;\mathrm{m}\) away.

A third case, transferred potential, arises when a conductor at remote-earth potential — a telephone pair, a pipe, a fence extending outside the yard — is brought into a station at GPR; the exposure is then the whole GPR, and it is dealt with by isolation, never by grid design.

What matters physiologically is the current through the heart, not the voltage. Dalziel's work fixes the tolerable body current for a shock of duration \(t_s\) seconds as \(I_B = k/\sqrt{t_s}\), with \(k = 0.116\) for a \(50\;\mathrm{kg}\) person and \(0.157\) for a \(70\;\mathrm{kg}\) person. Taking the body resistance as \(1000\;\Omega\) and each foot as a \(0.08\;\mathrm{m}\)-radius plate of resistance \(3\rho_s\), the two feet appear in parallel for touch (\(1.5\rho_s\)) and in series for step (\(6\rho_s\)). Multiplying the tolerable current by the total circuit resistance gives the tolerable voltages.

Tolerable step and touch voltages (IEEE Std 80), \(50\;\mathrm{kg}\) body
\[ E_{\text{step},50} = \big(1000 + 6\,C_s\rho_s\big)\frac{0.116}{\sqrt{t_s}}, \qquad E_{\text{touch},50} = \big(1000 + 1.5\,C_s\rho_s\big)\frac{0.116}{\sqrt{t_s}} \]
\[ C_s = 1 - \frac{0.09\left(1 - \dfrac{\rho}{\rho_s}\right)}{2h_s + 0.09} \]

Here \(\rho_s\) is the resistivity of a surface layer of thickness \(h_s\) — usually \(0.10\)–\(0.15\;\mathrm{m}\) of crushed rock, at \(2000\)–\(5000\;\Omega\mathrm{\cdot m}\) when wet — and \(C_s\) is the derating factor that accounts for the layer being thin enough that the foot still "sees" the native soil beneath. The crushed rock is not decoration: as Example 3 shows, it multiplies the tolerable touch voltage by roughly three, and it is the cheapest safety measure available in a substation yard.

The actual voltages the grid produces are the mesh voltage \(E_m\) — the worst touch voltage, found at the centre of a corner mesh where the surface potential dips furthest below the grid — and the step voltage \(E_s\), largest just outside the perimeter. IEEE 80 gives them as

Attained mesh and step voltages
\[ E_m = \frac{\rho\,K_m K_i I_G}{L_M}, \qquad E_s = \frac{\rho\,K_s K_i I_G}{L_S} \]

with \(K_m\) and \(K_s\) geometric spacing factors, \(K_i\) an irregularity factor, and \(L_M, L_S\) effective buried lengths. The design is safe when \(E_m \le E_{\text{touch}}\) and \(E_s \le E_{\text{step}}\). If it is not, the remedies are ordered by cost: add crushed rock, reduce the mesh spacing, add ground rods around the perimeter, extend the grid area, and — last — reduce \(I_G\) by changing the earthing method of Section 37-4.

surface potential (GPR at the grid) crushed rock ρs grid, depth h soil ρ E_touch E_step over 1 m safety is a difference, not an absolute: GPR may be kilovolts if E_m and E_s stay below tolerable
Ground potential rise, and the step and touch differences that actually endanger people
Section 37-7

Overvoltages: Lightning and Switching

The insulation of a substation is never designed for the working voltage. It is designed for the worst transient that will reach it, and those transients come in three families distinguished by their duration.

Temporary overvoltages last cycles to seconds and are at or near power frequency: the healthy-phase rise during an earth fault (Section 37-4), the Ferranti rise on a lightly loaded long line (Chapter 14), load rejection, and ferroresonance. They are modest in amplitude — \(1.2\) to \(1.5\;\mathrm{pu}\) — but they last long enough to matter for arrester energy, and they set the arrester's continuous operating voltage.

Switching overvoltages last hundreds of microseconds. Their origin is the travelling-wave behaviour of Chapter 12. Energising an unloaded line launches a step into a line whose far end is effectively open; the reflection coefficient there is \(+1\), so the wave doubles, and if the line already carries trapped charge from a previous opening — as it does on a rapid autoreclose — the step is measured from that trapped voltage and the arriving peak can approach \(3\;\mathrm{pu}\). Current chopping in the interruption of small inductive currents, discussed in Chapter 35, is another source. The standard test waveshape is \(250/2500\;\mu\mathrm{s}\).

Lightning overvoltages last microseconds. A return stroke injects a current of typically \(10\)–\(50\;\mathrm{kA}\) (with a median near \(30\;\mathrm{kA}\)) with a front of about \(1\)–\(2\;\mu\mathrm{s}\). If it strikes a phase conductor directly, the current divides in the two directions and produces a travelling wave of amplitude \(\tfrac12 I Z_0\); with \(Z_0 \approx 400\;\Omega\) even a modest \(10\;\mathrm{kA}\) stroke gives \(2\;\mathrm{MV}\), far above any insulation. The standard waveshape is \(1.2/50\;\mu\mathrm{s}\).

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Why direct strokes must be prevented, not withstood
\[ V_{\text{stroke}} = \tfrac{1}{2} I\,Z_0 \;\approx\; \tfrac12 (30\;\mathrm{kA})(400\;\Omega) = 6\;\mathrm{MV} \]

No practical insulation withstands this, so the line and yard are shielded by earth wires strung above the phase conductors and bonded to every tower. A shielding angle of about \(20^\circ\)–\(30^\circ\) (less at greater tower heights) intercepts essentially all strokes; in a substation the equivalent function is performed by earth wires and lightning masts over the yard.

Shielding transfers the problem rather than removing it. The stroke current now flows down the tower and out through its footing resistance \(R_f\), lifting the tower top to \(I R_f\) plus the inductive drop \(L\,\mathrm{d}i/\mathrm{d}t\) along the tower steel. If that potential exceeds the insulator string's critical flashover voltage, the string flashes over from the earthed tower to the live conductor — a back-flashover. The remedy is to reduce \(R_f\): driven rods where the soil allows, or a buried counterpoise wire radiating from each footing where it does not. Example 6 works this through.

The overvoltage that finally reaches a transformer inside the substation is therefore not the stroke voltage but a travelling wave that has already been limited by shielding, by back-flashover, by corona attenuation on the span (Chapter 15) and, last of all, by the surge arrester.

Section 37-8

Surge Arresters

A surge arrester is a device that is an insulator at power frequency and a conductor at surge voltage. Placed from line to earth beside the equipment it protects, it diverts the surge current to the earthing grid and clamps the terminal voltage to a value the insulation can withstand.

The crudest form is a rod gap: two electrodes separated by air. It is cheap and indestructible, but its breakdown voltage scatters widely, it does not clear the power-frequency follow current — the arc, once struck, is fed by the system until a breaker trips — and its volt–time characteristic rises steeply for fast fronts, so it may not operate before the protected insulation does. It survives only as a backup on distribution lines.

The valve-type arrester improved on this by putting a stack of series spark gaps in series with discs of silicon carbide. The gaps hold off the power-frequency voltage; when they break down, the SiC discs conduct, and because SiC is non-linear (\(I \propto V^{\,\alpha}\) with \(\alpha \approx 4\)–\(6\)) their resistance falls sharply during the surge and rises again afterwards, allowing the gaps to interrupt the follow current at the first current zero.

Modern practice is the gapless metal-oxide arrester. Zinc-oxide discs doped with bismuth, antimony and cobalt oxides have an exponent \(\alpha\) of \(25\) to \(50\): the resistance changes by six or more orders of magnitude for a factor-of-two change in voltage. That non-linearity is so extreme that no series gap is needed at all. At continuous operating voltage the arrester draws a leakage current of well under a milliampere, mostly capacitive; at a \(10\;\mathrm{kA}\) surge it conducts freely, and when the surge passes it returns to its insulating state on its own — there is no follow current to interrupt.

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The four ratings that define an arrester
\(U_c\) (MCOV) · \(U_r\) (rated voltage) · \(U_{res}\) (residual or discharge voltage) · energy class

\(U_c\) is the power-frequency voltage it may carry indefinitely. \(U_r\) is the temporary overvoltage it withstands for \(10\;\mathrm{s}\); typically \(U_c \approx 0.8\,U_r\). \(U_{res}\) is the voltage across it at the nominal discharge current (usually \(10\;\mathrm{kA}\), \(8/20\;\mu\mathrm{s}\)) — this is the protective level the insulation actually sees. The energy class states the joules per kilovolt of \(U_r\) it can absorb in one operation.

Selection starts from the temporary overvoltage, because that is what the arrester must survive without conducting. On an effectively earthed system the earth-fault factor is about \(1.4\) (Section 37-4), so with \(U_m\) the highest system voltage,

Rated voltage from the earth-fault factor
\[ U_r \;\ge\; k_{\text{eff}}\,\frac{U_m}{\sqrt3}, \qquad k_{\text{eff}} \approx 1.4 \text{ (effectively earthed)},\quad \approx 1.73 \text{ (isolated or resonant)} \]

A non-effectively earthed system therefore needs an arrester rated roughly \(25\%\) higher, and since \(U_{res}\) scales with \(U_r\), it also gets a worse protective level for the same insulation. The earthing decision of Section 37-4 has come back, exactly as promised.

Placement is as important as rating. The arrester should stand as close to the protected apparatus as the layout allows, because two effects raise the voltage at the equipment above \(U_{res}\). The connecting leads carry the discharge current at high \(\mathrm{d}i/\mathrm{d}t\) and contribute \(L\,\mathrm{d}i/\mathrm{d}t\) at roughly \(1\;\mu\mathrm{H}\) per metre. And the separation distance \(S\) between arrester and transformer allows the incoming wave to keep rising for the round-trip travel time before the arrester's clamping action arrives back:

Voltage at the protected equipment
\[ V_{\text{eq}} \;=\; U_{res} \;+\; L\frac{\mathrm{d}i}{\mathrm{d}t} \;+\; \frac{2S}{v}\,\frac{\mathrm{d}v}{\mathrm{d}t} \]

with \(v \approx 300\;\mathrm{m/\mu s}\) on an overhead line. A steepness of \(1000\;\mathrm{kV/\mu s}\) and a separation of \(20\;\mathrm{m}\) add \(133\;\mathrm{kV}\) — significant at \(220\;\mathrm{kV}\) and decisive at \(66\;\mathrm{kV}\).

Section 37-9

BIL and Insulation Coordination

Insulation coordination is the process of choosing the insulation strength of every item in a substation, and the protective level of the arresters, so that the two are correctly ordered: the arrester always operates first, and whatever flashover does occur happens where it does least harm.

The insulation strength of an item is stated as its Basic Insulation Level — strictly, the basic lightning impulse insulation level, BIL — which is the crest value of a standard \(1.2/50\;\mu\mathrm{s}\) impulse the item withstands. Above \(300\;\mathrm{kV}\) a second number, the switching impulse level (BSL, tested at \(250/2500\;\mu\mathrm{s}\)), becomes the governing figure, because long air gaps are relatively weaker against slow fronts than against fast ones.

🔑
Self-restoring and non-self-restoring insulation
Air gaps and porcelain surfaces recover completely after a flashover; transformer oil-paper and cable dielectric do not.

Coordination therefore deliberately makes the self-restoring insulation the weakest link. Bus support insulators and line entrance gaps are given a lower withstand than the transformer, so that when something must flash over, it is a gap that recovers in milliseconds rather than a winding that must be rewound. For the same reason self-restoring insulation is specified statistically (a \(10\%\) flashover probability, \(U_{50}\) minus \(1.3\sigma\)) while non-self-restoring insulation is specified deterministically.

The coordination itself reduces to one inequality. Let \(V_p\) be the protective level actually appearing at the equipment terminals — the \(V_{\text{eq}}\) of Section 37-8, not the bare \(U_{res}\). Then define

Protective ratio and protective margin
\[ \text{Protective ratio} \;=\; \frac{\mathrm{BIL}}{V_p}, \qquad \mathrm{PM}\,[\%] \;=\; \frac{\mathrm{BIL} - V_p}{V_p}\times 100 \]
🔑
The coordination criterion
\[ \mathrm{PM} \;\ge\; 20\% \quad\Longleftrightarrow\quad \mathrm{BIL} \ge 1.2\,V_p \]

The \(20\%\) allows for the scatter of the arrester's residual voltage, the ageing of the insulation, and the fact that the real surge is not the standard waveshape. Common practice specifies at least \(20\%\) against lightning impulse and \(15\%\) against switching impulse; margins of \(40\%\) or more are routinely achieved at EHV because the standard BIL values are set by air-clearance requirements rather than by the arrester.

kV 0200400 6008001000 peak V_ph 200 U_c 158 U_r 198 U_res 470 V_p at Tr 633 BIL 1050 PM 66%
The insulation coordination ladder for a 245 kV transformer — every level below the BIL, with margin

Standardised insulation levels are tabulated against the highest voltage for equipment \(U_m\), and normally two BIL values are offered at each \(U_m\) — a higher one for exposed installations and a lower one where arrester protection is close and the station is well shielded. Choosing the lower level is a legitimate economy, but only after the margin has been checked.

Nominal system, kV\(U_m\), kVStandard BIL, kV peakSwitching impulse (BSL), kVTypical arrester \(U_r\), kV
1112759–12
333617030–36
6672.532560
132145650120
220245950 / 1050192 / 198
4004201300 / 14251050 / 1175336–360
76580021001550624
Where the whole chapter converges. The earthing method fixes the earth-fault factor; the earth-fault factor fixes the arrester's rated voltage; the rated voltage fixes its residual voltage; the residual voltage plus lead and separation effects fixes \(V_p\); and \(V_p\) times \(1.2\) is the minimum BIL that may be bought. A change made anywhere in Section 37-4 propagates all the way to the transformer specification — which is why substation design is a single coordinated exercise and not a sequence of independent purchases.
Section 37-10

Worked Examples

1 Tuning a Peterson coil

Problem. A \(33\;\mathrm{kV}\), \(50\;\mathrm{Hz}\) overhead network consists of \(100\;\mathrm{km}\) of line with a capacitance to earth of \(0.0045\;\mu\mathrm{F}\) per kilometre per phase. Find the inductance of an arc-suppression coil that gives complete compensation, the current it carries, and its volt-ampere rating.

Solution. The capacitance per phase to earth is \(C = 100 \times 0.0045 = 0.45\;\mu\mathrm{F}\), and \(\omega = 2\pi(50) = 314.16\;\mathrm{rad/s}\).

Coil inductance
\[ L = \frac{1}{3\omega^2 C} = \frac{1}{3\,(314.16)^2\,(0.45\times10^{-6})} = \frac{1}{3\,(98\,696)(0.45\times10^{-6})} \]
\[ = \frac{1}{0.13324} = 7.51\;\mathrm{H} \]

The phase voltage is \(V_{ph} = 33\,000/\sqrt3 = 19\,053\;\mathrm{V}\), and once the fault has displaced the neutral this full voltage appears across the coil. Its reactance is \(\omega L = 314.16 \times 7.51 = 2360\;\Omega\), so

Coil current, checked against the charging current
\[ I_L = \frac{V_{ph}}{\omega L} = \frac{19\,053}{2360} = 8.07\;\mathrm{A} \]
\[ I_C = 3\,\omega C\,V_{ph} = 3(314.16)(0.45\times10^{-6})(19\,053) = 8.08\;\mathrm{A}\quad\text{✓} \]

The two agree to rounding, as the tuning condition demands. The coil rating is

Rating
\[ S = V_{ph} I_L = 19\,053 \times 8.07 = 154 \times 10^{3}\;\mathrm{VA} = 154\;\mathrm{kVA} \]

Rated for a short time only, since the coil carries current only while a ground fault persists. If a further \(40\;\mathrm{km}\) of feeder is switched in, \(C\) rises to \(0.63\;\mu\mathrm{F}\), the required \(L\) falls to \(5.36\;\mathrm{H}\), and a fixed coil would leave a residual capacitive current of \(3\omega V_{ph}\,\Delta C = 3(314.16)(0.18\times10^{-6})(19\,053) = 3.23\;\mathrm{A}\) — which is why the coil is made adjustable.

2 What the earthing method does to the fault current

Problem. At a \(33\;\mathrm{kV}\) bus the sequence reactances seen from the fault point are \(X_1 = X_2 = j6\;\Omega\) and \(X_0 = j4\;\Omega\). Find the single line-to-ground fault current when the transformer neutral is (a) solidly earthed, (b) earthed through \(10\;\Omega\), and (c) find the resistance needed to limit the fault current to \(500\;\mathrm{A}\).

Solution. With \(E_a = 33\,000/\sqrt3 = 19\,053\;\mathrm{V}\), Chapter 24 gives \(I_f = 3E_a/(Z_1+Z_2+Z_0+3Z_n)\), and \(3E_a = 57\,158\;\mathrm{V}\).

(a) Solid earthing
\[ I_f = \frac{57\,158}{|j6 + j6 + j4|} = \frac{57\,158}{16} = 3572\;\mathrm{A} \]
(b) Neutral resistor of \(10\;\Omega\)
\[ I_f = \frac{57\,158}{|30 + j16|} = \frac{57\,158}{\sqrt{900+256}} = \frac{57\,158}{34.0} = 1681\;\mathrm{A} \]

A \(10\;\Omega\) resistor has cut the fault current to \(47\%\) of the solid-earthed value.

(c) Sizing for \(500\;\mathrm{A}\)
\[ |3R_n + j16| = \frac{57\,158}{500} = 114.3\;\Omega \;\Longrightarrow\; 3R_n = \sqrt{114.3^2 - 16^2} = \sqrt{12\,808} = 113.2 \]
\[ R_n = 37.7\;\Omega \]

The resistor dissipates \(I_f^2 R_n = (500)^2(37.7) = 9.4\;\mathrm{MW}\) while the fault lasts — trivially rated for \(10\;\mathrm{s}\), impossible continuously. Note also the price paid: the earth-fault relay now sees \(500\;\mathrm{A}\) instead of \(3572\;\mathrm{A}\), so the sensitivity of the Chapter 36 settings must be revisited, and the healthy-phase voltage rise is no longer held to \(1.4\;\mathrm{pu}\).

3 Tolerable step and touch voltages

Problem. A substation is built on soil of resistivity \(\rho = 100\;\Omega\mathrm{\cdot m}\). The fault clearing time is \(t_s = 0.5\;\mathrm{s}\). Find the tolerable step and touch voltages for a \(50\;\mathrm{kg}\) person (a) on bare soil and (b) with a \(0.10\;\mathrm{m}\) layer of crushed rock of \(\rho_s = 2500\;\Omega\mathrm{\cdot m}\).

Solution — (a) bare soil. With no surface layer, \(\rho_s = \rho\) and \(C_s = 1\). Also \(0.116/\sqrt{0.5} = 0.116/0.7071 = 0.1640\).

Bare soil
\[ E_{\text{touch}} = \big(1000 + 1.5\times100\big)(0.1640) = 1150 \times 0.1640 = 189\;\mathrm{V} \]
\[ E_{\text{step}} = \big(1000 + 6\times100\big)(0.1640) = 1600 \times 0.1640 = 262\;\mathrm{V} \]

(b) With crushed rock. First the derating factor:

Surface-layer derating
\[ C_s = 1 - \frac{0.09\left(1 - \frac{100}{2500}\right)}{2(0.10)+0.09} = 1 - \frac{0.09(0.96)}{0.29} = 1 - \frac{0.0864}{0.29} = 1 - 0.298 = 0.702 \]
\[ C_s\rho_s = 0.702 \times 2500 = 1755\;\Omega\mathrm{\cdot m} \]
Tolerable voltages with the rock layer
\[ E_{\text{touch},50} = \big(1000 + 1.5\times1755\big)(0.1640) = 3633 \times 0.1640 = 596\;\mathrm{V} \]
\[ E_{\text{step},50} = \big(1000 + 6\times1755\big)(0.1640) = 11\,531 \times 0.1640 = 1892\;\mathrm{V} \]

Ten centimetres of stone has raised the tolerable touch voltage from \(189\;\mathrm{V}\) to \(596\;\mathrm{V}\) — a factor of \(3.2\) — and the step voltage by a factor of \(7.2\). Two further observations: the step limit is always the looser of the two, because the feet are in series with the body rather than in parallel; and halving the clearing time to \(0.25\;\mathrm{s}\) would multiply both limits by \(\sqrt2\), so faster protection is itself an earthing measure.

4 Checking an earthing grid

Problem. The station of Example 3 has a square grid \(70\;\mathrm{m} \times 70\;\mathrm{m}\) buried at \(h = 0.5\;\mathrm{m}\), with a total buried conductor length (including rods) of \(L = 1400\;\mathrm{m}\). The grid current is \(I_G = 4000\;\mathrm{A}\). For this geometry \(K_m = 0.89\), \(K_s = 0.41\), \(K_i = 2.17\), \(L_M = 1400\;\mathrm{m}\) and \(L_S = 1130\;\mathrm{m}\). Find the grid resistance, the ground potential rise, and verify the design.

Solution. The area is \(A = 4900\;\mathrm{m^2}\). By the Laurent–Niemann expression,

Grid resistance
\[ R_g = \frac{\rho}{4}\sqrt{\frac{\pi}{A}} + \frac{\rho}{L} = \frac{100}{4}\sqrt{\frac{3.1416}{4900}} + \frac{100}{1400} \]
\[ = 25(0.02532) + 0.0714 = 0.633 + 0.071 = 0.704\;\Omega \]

Sverak's refinement, with \(\sqrt{20A} = \sqrt{98\,000} = 313.0\) and \(h\sqrt{20/A} = 0.5(0.06389) = 0.0319\), gives

Refined value
\[ R_g = 100\left[\frac{1}{1400} + \frac{1}{313.0}\left(1 + \frac{1}{1.0319}\right)\right] = 100\big[7.14\times10^{-4} + 3.194\times10^{-3}(1.969)\big] = 0.700\;\Omega \]
Ground potential rise
\[ \mathrm{GPR} = I_G R_g = 4000 \times 0.700 = 2800\;\mathrm{V} \]

The GPR is far above the \(596\;\mathrm{V}\) tolerable touch voltage, so the design is not settled by the GPR alone; the actual mesh and step voltages must be computed.

Attained mesh and step voltages
\[ E_m = \frac{\rho K_m K_i I_G}{L_M} = \frac{100(0.89)(2.17)(4000)}{1400} = \frac{772\,520}{1400} = 552\;\mathrm{V} \]
\[ E_s = \frac{\rho K_s K_i I_G}{L_S} = \frac{100(0.41)(2.17)(4000)}{1130} = \frac{355\,880}{1130} = 315\;\mathrm{V} \]

Comparing: \(E_m = 552\;\mathrm{V} < E_{\text{touch}} = 596\;\mathrm{V}\) and \(E_s = 315\;\mathrm{V} \ll E_{\text{step}} = 1892\;\mathrm{V}\). The grid is safe, but only just in touch, with a margin of \(7\%\) — and comfortably in step, with a margin of \(500\%\). This is entirely typical: touch voltage governs earthing-grid design. If the margin were negative the first remedy would be to reduce the mesh spacing (raising \(L_M\) and lowering \(K_m\)) rather than to enlarge the grid, since \(R_g\) barely responds to area.

5 Arrester selection and the protective margin

Problem. A \(220\;\mathrm{kV}\) effectively earthed system has \(U_m = 245\;\mathrm{kV}\) and an earth-fault factor of \(1.4\). The transformer BIL is \(1050\;\mathrm{kV}\). A gapless metal-oxide arrester whose residual voltage at \(10\;\mathrm{kA}\) is \(2.37\,U_r\) is to be selected. The arrester is connected by \(3\;\mathrm{m}\) of lead in total and stands \(20\;\mathrm{m}\) from the transformer; the incoming surge has a steepness of \(1000\;\mathrm{kV/\mu s}\) and the discharge current rises at \(10\;\mathrm{kA/\mu s}\). Find the protective margin.

Solution. The rated voltage must cover the temporary overvoltage:

Rated and continuous voltages
\[ U_r \ge 1.4 \times \frac{245}{\sqrt3} = 1.4 \times 141.5 = 198.0\;\mathrm{kV} \;\Longrightarrow\; \text{select } U_r = 198\;\mathrm{kV} \]
\[ U_c \approx 0.8\,U_r = 158\;\mathrm{kV}, \qquad \text{peak phase voltage} = \sqrt2\,(141.5) = 200\;\mathrm{kV} \]
Residual voltage
\[ U_{res} = 2.37 \times 198 = 469 \approx 470\;\mathrm{kV} \]

Now add the two effects that raise the voltage at the transformer. The lead inductance at \(1\;\mu\mathrm{H/m}\) gives

Lead and separation contributions
\[ L\frac{\mathrm{d}i}{\mathrm{d}t} = (3\;\mu\mathrm{H})\left(10\;\frac{\mathrm{kA}}{\mu\mathrm{s}}\right) = 30\;\mathrm{kV} \]
\[ \frac{2S}{v}\frac{\mathrm{d}v}{\mathrm{d}t} = \frac{2(20)}{300}\times 1000 = 0.1333 \times 1000 = 133\;\mathrm{kV} \]
\[ V_p = 470 + 30 + 133 = 633\;\mathrm{kV} \]
Protective margin
\[ \mathrm{PM} = \frac{\mathrm{BIL} - V_p}{V_p}\times100 = \frac{1050 - 633}{633}\times 100 = \frac{417}{633}\times100 = 65.9\% \]

Comfortably above the \(20\%\) requirement. It is instructive to see how the margin erodes. Against the bare residual voltage the margin would be \((1050-470)/470 = 123\%\); the leads cost \(13\) points and the \(20\;\mathrm{m}\) of separation a further \(44\). At \(66\;\mathrm{kV}\), where the BIL is \(325\;\mathrm{kV}\) and the residual perhaps \(160\;\mathrm{kV}\), that same \(163\;\mathrm{kV}\) of lead-plus-separation would take \(V_p\) to \(323\;\mathrm{kV}\) and the margin to \(0.6\%\) — which is precisely why arresters at lower voltages must be mounted on the transformer tank itself.

6 Back-flashover and the value of a counterpoise

Problem. A \(30\;\mathrm{kA}\) stroke with a \(2\;\mu\mathrm{s}\) front terminates on the shield wire at a \(132\;\mathrm{kV}\) tower \(30\;\mathrm{m}\) high with a footing resistance of \(20\;\Omega\). Take the tower inductance as \(0.5\;\mu\mathrm{H/m}\), the coupling factor between shield wire and phase conductor as \(0.25\), and the insulator string CFO as \(660\;\mathrm{kV}\). Determine whether a back-flashover occurs, and repeat with the footing resistance reduced to \(10\;\Omega\) by a counterpoise.

Solution. The tower-top potential is the resistive drop through the footing plus the inductive drop along the tower steel:

Tower-top potential, \(R_f = 20\;\Omega\)
\[ V_R = I R_f = 30\;\mathrm{kA} \times 20\;\Omega = 600\;\mathrm{kV} \]
\[ L_{\text{tower}} = 30\;\mathrm{m} \times 0.5\;\mu\mathrm{H/m} = 15\;\mu\mathrm{H}, \qquad \frac{\mathrm{d}i}{\mathrm{d}t} = \frac{30\;\mathrm{kA}}{2\;\mu\mathrm{s}} = 15\;\frac{\mathrm{kA}}{\mu\mathrm{s}} \]
\[ V_L = 15 \times 15 = 225\;\mathrm{kV}, \qquad V_{\text{top}} = 600 + 225 = 825\;\mathrm{kV} \]

The shield wire carries the tower-top potential and couples a fraction \(k = 0.25\) of it onto the phase conductor, so only \((1-k)\) appears across the string. To that must be added the power-frequency phase voltage at its worst instant, of peak value \(\sqrt2(132/\sqrt3) = 107.8\;\mathrm{kV}\) and opposite polarity:

Stress across the insulator string
\[ V_{\text{string}} = (1-0.25)(825) + 107.8 = 618.8 + 107.8 = 727\;\mathrm{kV} \;>\; 660\;\mathrm{kV} \]

The string flashes over: a back-flashover, and the \(132\;\mathrm{kV}\) line trips on a stroke that never touched a phase conductor.

With a counterpoise, \(R_f = 10\;\Omega\)
\[ V_R = 30 \times 10 = 300\;\mathrm{kV}, \qquad V_{\text{top}} = 300 + 225 = 525\;\mathrm{kV} \]
\[ V_{\text{string}} = 0.75(525) + 107.8 = 393.8 + 107.8 = 502\;\mathrm{kV} \;<\; 660\;\mathrm{kV} \]

No flashover. Halving the footing resistance has removed \(225\;\mathrm{kV}\) of stress at a cost of a few hundred metres of buried wire — the single most cost-effective lightning-performance measure available on a transmission line. Rearranging for the critical stroke current at \(20\;\Omega\): \(0.75\,(20I + 7.5I) + 107.8 = 660\) gives \(20.625 I = 552.2\), so \(I_c = 26.8\;\mathrm{kA}\); with the counterpoise the same algebra gives \(0.75(10I+7.5I) = 552.2\), so \(I_c = 42.1\;\mathrm{kA}\). Since stroke currents above \(42\;\mathrm{kA}\) are far rarer than those above \(27\;\mathrm{kA}\), the outage rate falls by roughly a factor of four.

Review

Chapter Summary

Security costs breakers

\(n\) breakers give a single bus; \(1.5n\) give the breaker-and-a-half scheme, which survives any bus fault or breaker outage.

GIS

Pressurised \(\mathrm{SF_6}\) cuts the \(400\;\mathrm{kV}\) footprint to about \(10\%\), at two to four times the cost.

Neutral earthing

\(Z_n\) appears as \(3Z_n\) in the zero-sequence path — the one knob that sets earth-fault current alone.

Peterson coil

\(L = 1/(3\omega^2C)\) makes \(I_L\) cancel \(I_C = 3\omega C V_{ph}\); the arc self-extinguishes.

GPR

\(\mathrm{GPR}=I_G R_g\) with \(R_g \approx \frac{\rho}{4}\sqrt{\pi/A}+\rho/L\) — falls only as \(1/\sqrt A\).

Touch governs

\(E_{\text{touch}} = (1000+1.5C_s\rho_s)\,0.116/\sqrt{t_s}\); crushed rock triples it, and \(E_m\) is the binding constraint.

Overvoltages

Temporary \(\sim\!1.4\;\mathrm{pu}\); switching to \(3\;\mathrm{pu}\) at \(250/2500\;\mu\mathrm{s}\); lightning \(\tfrac12 IZ_0\) at \(1.2/50\;\mu\mathrm{s}\).

Metal-oxide arresters

\(\alpha \approx 25\)–\(50\) removes the need for a series gap; \(U_r \ge k_{\text{eff}}U_m/\sqrt3\).

Protective margin

\(\mathrm{PM}=(\mathrm{BIL}-V_p)/V_p \ge 20\%\), with \(V_p\) including lead and separation effects.

Practice

Practice Problems

Take \(f = 50\;\mathrm{Hz}\) throughout, and quote the standard formulae you use. Problems 5 to 8 require the results of more than one section.

  1. A \(220\;\mathrm{kV}\) substation has six circuits: two incoming lines, two outgoing lines and two transformers. Count the breakers needed for a sectionalised single bus, a ring bus, a breaker-and-a-half scheme and a double-bus-double-breaker scheme, and state for each what a bus fault costs.
  2. A \(66\;\mathrm{kV}\), \(50\;\mathrm{Hz}\) network has a capacitance to earth of \(0.008\;\mu\mathrm{F}\) per kilometre per phase and a total route length of \(60\;\mathrm{km}\). Find the inductance and current rating of a Peterson coil for complete compensation, and state what happens if \(20\;\mathrm{km}\) of that network is switched out.
  3. A generator has \(X_1 = j0.20\), \(X_2 = j0.18\) and \(X_0 = j0.06\) per unit on its own base of \(50\;\mathrm{MVA}\), \(11\;\mathrm{kV}\). Find the neutral reactance in ohms that makes the single line-to-ground fault current equal to the three-phase fault current.
  4. Repeat Example 3 for a \(70\;\mathrm{kg}\) person, a clearing time of \(0.3\;\mathrm{s}\), and a \(0.15\;\mathrm{m}\) layer of crushed rock at \(3000\;\Omega\mathrm{\cdot m}\) over soil of \(150\;\Omega\mathrm{\cdot m}\). Compare with the values obtained in the worked example and comment on which change contributed most.
  5. A square earthing grid of side \(60\;\mathrm{m}\) uses \(600\;\mathrm{m}\) of buried conductor in soil of \(200\;\Omega\mathrm{\cdot m}\), buried at \(0.5\;\mathrm{m}\). Find \(R_g\) by both expressions given in Section 37-6. If the grid current is \(6\;\mathrm{kA}\), find the GPR, and determine the extra conductor length that would be needed to bring \(R_g\) below \(1\;\Omega\). Comment on whether this is the right remedy.
  6. A \(400\;\mathrm{kV}\) transformer has a BIL of \(1425\;\mathrm{kV}\) and is protected by an arrester of rated voltage \(360\;\mathrm{kV}\) whose residual voltage at \(10\;\mathrm{kA}\) is \(2.3\,U_r\). The arrester is mounted \(15\;\mathrm{m}\) away with \(4\;\mathrm{m}\) of lead. For an incoming surge of steepness \(800\;\mathrm{kV/\mu s}\) and a discharge current rising at \(10\;\mathrm{kA/\mu s}\), find the protective margin. What is the greatest separation distance for which the margin remains at least \(20\%\)?
  7. A \(30\;\mathrm{kA}\) stroke hits a tower of a \(220\;\mathrm{kV}\) line whose insulator string has a CFO of \(1100\;\mathrm{kV}\). The tower is \(35\;\mathrm{m}\) high with an inductance of \(0.5\;\mu\mathrm{H/m}\), the stroke front is \(2\;\mu\mathrm{s}\), and the coupling factor is \(0.3\). Find the largest footing resistance for which no back-flashover occurs, ignoring the power-frequency voltage.
  8. An \(11\;\mathrm{kV}\) system is to be changed from resistance earthing (\(400\;\mathrm{A}\) limit) to solid earthing. List, with reasons, every consequence for: the breaker rating of Chapter 25, the earth-fault relay settings of Chapter 36, the surge arrester rating of Section 37-8, and the transformer insulation specification. State which of these changes save money and which cost it.
Tip: when a substation problem looks like a collection of unrelated calculations, look for the chain. The earthing method sets the earth-fault current, which sets the grid current in the step-and-touch check and the earth-fault factor in the arrester selection; the arrester rating sets the residual voltage, which sets the BIL; and the BIL sets the air clearances, which set the size of the yard, which comes back to the grid area. Almost every examination question on this chapter is one link of that chain in isolation, and every real design is the whole chain iterated until it closes.