Part 3 · Chapter 15

Corona and Radio Interference

Air is the insulation of every overhead line, and it fails locally long before it fails completely — so the ceiling on transmission voltage is not set by the conductor or the tower but by the electric field at the conductor's own surface.

Electric Power Systems Prof. Mithun Mondal Reading time ≈ 45 min
i What you'll learn
  • Why a non-uniform field allows a partial discharge that stops short of flashover, and why the strength of air is quoted as \(30\) kV/cm peak or \(21.1\) kV/cm rms.
  • How the critical disruptive voltage \(V_c = m_o g_o \delta r \ln(d/r)\) follows in three lines from the surface gradient of Chapter 8.
  • Where the air density factor \(\delta = 3.92b/(273+t)\) comes from, and why the constant is \(3.92\) and not something else.
  • Why the glow appears at a higher voltage than the loss does, and what the \(\big(1+0.3/\sqrt{\delta r}\big)\) correction is telling you about thin conductors.
  • Why corona loss obeys \(P \propto f\,(V-V_c)^2\), and the conditions under which Peek's formula may be trusted.
  • How discrete streamer pulses become radio interference, how it is measured in dB above \(1\,\mu\)V/m, and why ultra-high-voltage design is often limited by audible noise instead.
  • Why bundling (Chapter 8) is the only lever strong enough to build a \(400\) kV line at all.
Section 15-1

The Ceiling That Air Puts on Voltage

Everything in Part 2 and Part 3 has treated the line as a set of circuit parameters \(R\), \(L\), \(C\) and \(G\), with the air around the conductors doing nothing but hold the charge in place. Chapter 2 argued that transmission voltage should be as high as the design will allow, because the power a line can carry rises as \(V^2\) while its \(I^2R\) loss falls as \(1/V^2\). Chapter 14 pushed the same argument to its conclusion with surge impedance loading. Nothing so far has said where the argument stops.

It stops at the conductor surface. Chapter 8 computed the quantity that matters, the surface voltage gradient of a single conductor carrying phase-to-neutral voltage \(V_{an}\):

Surface gradient, from Chapter 8
\[ E_r = \frac{V_{an}}{r\,\ln\!\big(D_{eq}/r\big)} \]

Raise \(V_{an}\) and \(E_r\) rises in exact proportion. The strength of air, however, does not rise with anything. It is fixed at roughly \(30\) kilovolts per centimetre by the physics of nitrogen and oxygen molecules, and once the gradient at the conductor surface reaches that value the air immediately around the conductor stops being an insulator. It becomes a thin, luminous, crackling shell of ionised gas that consumes power, radiates radio noise, generates ozone and hisses audibly. That is corona.

The phenomenon is not flashover. A line in corona keeps operating; the discharge is confined to a sheath a few millimetres thick and never bridges the gap to the next phase or to the tower. This chapter is about why the discharge stops where it does, at what voltage it starts, what it costs, and what a designer does about it. The answer to the last question was given in Chapter 8 before its justification: split the phase into a bundle. Here we finally compute the threshold that made bundling unavoidable.

The real design variable is not voltage — it is gradient. A \(400\) kV line and a \(132\) kV line are built to the same surface gradient, somewhere near \(16\) kV/cm rms. Everything that changes between them — bigger conductors, sub-conductor bundles, wider spacing, taller towers — exists to hold that one number down while the voltage in the numerator goes up by a factor of three.
Section 15-2

Ionisation by Collision and the Confined Avalanche

Air is never perfectly neutral. Cosmic rays, ultraviolet light and the natural radioactivity of the ground keep a few hundred ion pairs per cubic centimetre in existence at all times, continuously created and continuously recombining. Those stray free electrons are the seed of everything that follows.

Put a free electron in a field \(E\). It is accelerated, travels a mean free path, and collides with a neutral molecule. Because the molecule is some tens of thousands of times heavier, an elastic collision returns only a tiny fraction of the electron's kinetic energy — the electron therefore keeps most of what the field gave it and is accelerated again. Over a succession of free paths its energy climbs until, at a collision, it exceeds the ionisation energy of the molecule (about \(15.6\) eV for nitrogen, \(12.1\) eV for oxygen). That collision knocks an electron loose. Where there was one free electron there are now two, and a positive ion left behind.

Both electrons repeat the process. The population grows geometrically along the direction of the field, which is Townsend's description of an avalanche: if \(\alpha\) is the number of ionising collisions an electron makes per centimetre of drift, then a single electron starting at \(x=0\) has become

Townsend avalanche growth
\[ n(x) = n_0\,e^{\alpha x}, \qquad \alpha = \alpha\!\left(\frac{E}{p}\right) \]

The coefficient \(\alpha\) depends on the field only through the ratio \(E/p\), because the field sets the energy gained per free path and the pressure sets the length of that free path. This single fact is the origin of the air density correction of Section 15-4, and it is worth carrying forward.

Now the crucial question. If an avalanche multiplies without limit, why does the line not simply flash over? Because the field around a cylindrical conductor is not uniform. At a radial distance \(x\) from the conductor axis the field is

Radial field outside the conductor
\[ E(x) = \frac{V}{x\,\ln(d/r)}, \qquad r \le x \ll d \]

and it falls as \(1/x\). Only in the region where \(E(x)\) exceeds the threshold can \(\alpha\) outrun the rate at which electrons are lost to attachment on oxygen molecules. Beyond that radius the avalanche starves. Denote the threshold gradient by \(g_o\) and find the radius at which the field has decayed to it:

Radius of the ionised envelope
\[ \frac{V}{x_0\,\ln(d/r)} = g_o \quad\Longrightarrow\quad x_0 = \frac{V}{g_o \ln(d/r)} = r\,\frac{V}{V_c} \]

where \(V_c = g_o\, r \ln(d/r)\) is the voltage at which the surface itself has just reached the threshold — the critical disruptive voltage derived properly in the next section. The result is exact and unusually informative: the ionised shell extends to \(r\) times the per-unit overvoltage. At \(V = 1.3\,V_c\) the glowing envelope reaches \(1.3r\); at twice the critical voltage it reaches \(2r\). The discharge is self-limiting because it must live inside a field that it cannot itself extend.

The same relation quietly explains a claim made in every list of corona's "advantages": that corona increases the effective diameter of the conductor. It does, and by exactly the factor \(V/V_c\), because the conducting plasma sheath behaves electrically as part of the conductor.

r ionised envelope, radius x₀ = r·V/V_c conductor cross-section x E g₀ (threshold) r x₀ E(x) = V / [x ln(d/r)] E_r
The avalanche survives only inside the shell where the 1/x field still exceeds the threshold
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Breakdown strength of air
At \(76\) cm of mercury and \(25^\circ\mathrm{C}\), air ionises at a gradient of \(g_o = 30\) kV/cm peak, equivalently \(21.1\) kV/cm rms.

The rms figure is the peak divided by \(\sqrt2\). Since the surface gradient is normally computed from an rms phase voltage, the rms threshold \(21.1\) kV/cm is the one that must be used — mixing an rms gradient with the peak threshold overstates the margin by \(41\%\), and is the single most common error in corona problems.

One asymmetry is worth recording because it shows up on a direct-current line. On a negative conductor the avalanche is launched from the conductor surface into a falling field and terminates quickly, producing a rapid, highly regular train of small pulses (Trichel pulses) and a spotty, mobile glow. On a positive conductor the electrons move toward the conductor, into a rising field, so the discharge develops as long filamentary streamers that push outward — larger pulses, less regular, and a smooth uniform glow. Under alternating voltage both occur, one per half cycle, and the positive streamers dominate everything that follows in Section 15-7.

Section 15-3

The Critical Disruptive Voltage

The critical disruptive voltage \(V_c\) is defined as the lowest phase-to-neutral voltage at which the air at the conductor surface starts to ionise. Its derivation is now a single substitution.

Take two parallel conductors of radius \(r\) whose axes are \(d\) apart, with \(d \gg r\), and let \(V\) be the phase-to-neutral voltage. Chapter 7 gives the charge per unit length as \(q = 2\pi\varepsilon V/\ln(d/r)\), and the field at the surface of a line charge is \(q/2\pi\varepsilon r\), so the surface gradient is

Surface gradient in terms of the phase voltage
\[ g = \frac{q}{2\pi\varepsilon r} = \frac{V}{r\,\ln\!\left(\dfrac{d}{r}\right)}\;\;\text{volts/cm} \]

Corona begins the instant \(g\) reaches the breakdown strength \(g_o\). Setting \(g = g_o\) and solving for the voltage that does it:

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Critical disruptive voltage — ideal conductor, standard air
\[ V_c = g_o\, r \ln\!\left(\frac{d}{r}\right)\;\;\text{kV per phase (rms)}, \qquad g_o = 21.1\;\text{kV/cm (rms)} \]

\(r\) and \(d\) must be in the same units, conventionally centimetres, so that \(d/r\) is dimensionless and \(r\) carries the centimetre that cancels the one in \(g_o\). The answer is a phase-to-neutral voltage; multiply by \(\sqrt3\) for the line value that appears on a nameplate.

Two extensions make this usable. For a three-phase line the spacing \(d\) is replaced by the equivalent equilateral spacing — the geometric mean distance \(D_{eq} = \sqrt[3]{D_{12}D_{23}D_{31}}\) introduced in Chapter 6 and used again for capacitance in Chapter 7. The logarithm is the same logarithm that appeared in both of those chapters, which is not a coincidence: corona, inductance and capacitance are all statements about the same field pattern.

The second extension is that \(V_c\) as written applies to a mirror-smooth conductor in laboratory air. Real conductors are stranded and real air is not at \(76\) cm of mercury. Both corrections multiply the expression, and both reduce it.

Note where \(r\) sits. It appears once as a multiplier and once inside a logarithm, and the two work in opposite directions — increasing \(r\) raises the prefactor but shrinks \(\ln(d/r)\). Since a logarithm changes slowly, the prefactor wins: doubling the conductor radius raises \(V_c\) by roughly \(1.7\) to \(1.8\) times, not by two. Widening the spacing \(d\), by contrast, only moves the logarithm, so doubling \(d\) buys about \(13\%\). This is the whole reason lines are built with fat conductors rather than wide spacings.
Section 15-4

Air Density and Surface Condition

Section 15-2 established that the ionisation coefficient depends on \(E/p\). Turn that around: the field required to sustain ionisation is proportional to the pressure, or more precisely to the number density of molecules, because what really matters is the energy an electron picks up between collisions and that is set by how far apart the molecules are. By the ideal gas law the number density is proportional to \(p/T\). Hence

Derivation of the air density factor
\[ \frac{g_o(b,t)}{g_o(76,\,25^\circ)} \;=\; \frac{b/(273+t)}{76/(273+25)} \;=\; \frac{298}{76}\cdot\frac{b}{273+t} \]
\[ \Longrightarrow\quad \delta \;=\; \frac{3.92\,b}{273+t} \]

The famous constant is nothing more than \(298/76 = 3.921\), with \(b\) the barometric pressure in centimetres of mercury and \(t\) the ambient temperature in degrees Celsius. At \(b = 76\) and \(t = 25^\circ\)C the factor is exactly unity, as it must be. Everywhere else it is less than one: hotter air is thinner, and higher ground has less air above it.

AltitudeBarometric \(b\) (cm Hg)\(\delta\) at \(25^\circ\)C\(\delta\) at \(45^\circ\)C
Sea level76.01.0000.937
500 m71.60.9420.883
1000 m67.40.8870.831
1500 m63.40.8340.782
2000 m59.60.7840.735
3000 m52.60.6920.648

A line at \(2000\) m on a summer afternoon has lost a quarter of its corona margin relative to the same line at sea level in the morning. Substation clearances and insulator creepage in mountainous regions are derated for the same reason.

The second correction is geometric rather than atmospheric. The derivation of \(V_c\) assumed a perfectly cylindrical, perfectly smooth conductor, so that the field on the surface is the same all the way round. A stranded conductor is scalloped: each outer strand is a small cylinder of its own, and the field at the crown of a strand is appreciably higher than the field a smooth cylinder of the same overall diameter would produce. Dirt, scratches from stringing, insect bodies and water droplets do the same thing locally and much more violently. All of it is lumped into an empirical irregularity factor \(m_o \le 1\) multiplying \(V_c\).

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Critical disruptive voltage — working form
\[ \boxed{\;V_c = m_o\,g_o\,\delta\,r\,\ln\!\left(\frac{D_{eq}}{r}\right)\;}\;\;\text{kV per phase (rms)} \]

with \(g_o = 21.1\) kV/cm rms, \(\delta = 3.92b/(273+t)\), and \(m_o\) taken as \(1.0\) for a polished conductor, \(0.98\)–\(0.92\) for a smooth conductor that is weathered or dirty, and \(0.87\)–\(0.80\) for stranded conductors, the lower end applying to few-strand constructions with deeper scalloping.

The equivalent statement in terms of gradient is often more convenient in design: corona begins when the computed rms surface gradient reaches \(m_o g_o \delta\). For a stranded conductor at sea level that threshold is about \(0.85 \times 21.1 = 17.9\) kV/cm, which is why Chapter 8 quoted a design target of \(15\) to \(17\) kV/cm — a margin of roughly ten to fifteen per cent against fair-weather corona, and no margin at all against rain.

Section 15-5

The Visual Critical Voltage

Raise a line slowly past \(V_c\) at night and nothing is seen. Ionisation is occurring — a sensitive instrument registers loss and radio noise — but the eye detects nothing until the voltage is appreciably higher. The voltage at which a continuous glow appears along the whole conductor is the visual critical voltage \(V_v\), and it always exceeds \(V_c\).

The reason is the one already established in Section 15-2. Visible light comes from excited molecules relaxing, and enough of them must be excited within a shell thick enough to register. At \(V\) barely above \(V_c\) the envelope radius \(x_0 = r\,V/V_c\) is barely above \(r\): the ionised shell is a few tens of micrometres thick and produces no perceptible light. The shell must reach an appreciable absolute thickness before it glows, and on a thin conductor, where the \(1/x\) field collapses quickly, that requires a much larger overvoltage than on a fat one.

Peek measured the effect and fitted it. The extra thickness needed turns out to scale as \(\sqrt{\delta r}\), so the correction enters as a term \(0.3/\sqrt{\delta r}\) added to unity:

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Visual critical voltage (Peek)
\[ \boxed{\;V_v = m_v\,g_o\,\delta\,r\left(1 + \frac{0.3}{\sqrt{\delta r}}\right)\ln\!\left(\frac{D_{eq}}{r}\right)\;}\;\;\text{kV per phase (rms)} \]

with \(r\) in centimetres. The irregularity factor \(m_v\) is a different number from \(m_o\): \(1.0\) for a polished conductor, \(0.93\)–\(0.98\) for a locally roughened or weathered surface, and \(0.72\)–\(0.82\) for stranded conductors, seven-strand constructions sitting at the bottom of the range and nineteen-strand and above at the top.

The bracket alone shows how the effect fades with size. At \(\delta = 1\) it is \(1.42\) for a \(0.5\) cm conductor, \(1.30\) at \(1\) cm, \(1.21\) at \(2\) cm and \(1.17\) at \(3\) cm: on a very thin wire the glow is delayed by more than \(40\%\) in voltage, on a heavy transmission conductor by less than \(20\%\). Because \(m_v\) is well below \(m_o\) for stranded conductors, the two effects partly cancel, and the ratio \(V_v/V_c\) for a real stranded line is typically only \(1.05\) to \(1.15\).

Do not design to the visual voltage. \(V_v\) tells you when a night inspection will see something; \(V_c\) tells you when the line begins to lose power and to make radio noise. A line operated between \(V_c\) and \(V_v\) looks perfectly quiet and is nevertheless in corona. Every economic and interference criterion in this chapter is written against \(V_c\).
Section 15-6

Power Loss: Peek's Formula and Its Limits

Every ionising collision costs energy, and the energy comes from the line. It leaves as light, heat, sound, and the chemical energy locked up in ozone and oxides of nitrogen. Before quoting the standard formula it is worth seeing why it has the shape it has.

Consider one half cycle. As the instantaneous voltage climbs past \(V_c\), streamers form and inject a space charge into the sheath. That charge is proportional to how far the voltage has exceeded the critical value, so \(q_{corona} \propto (V - V_c)\). Sweeping it out through the potential difference that drove it costs energy \(\propto (V-V_c) \times q_{corona} \propto (V-V_c)^2\). This happens twice per cycle, so the power is proportional to frequency:

Why the loss has this form
\[ P \;\propto\; f\,(V - V_c)^2 \]

Peek's experiments on test lines supplied the constant and a weak geometric term \(\sqrt{r/d}\), which accounts for how quickly the field falls away from conductors of different slenderness. He also found that the loss did not extrapolate to zero at zero frequency — a direct-voltage conductor in corona still loses power — which is why the frequency appears as \(f + 25\) rather than \(f\).

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Peek's corona loss formula (fair weather)
\[ \boxed{\;P = \frac{242.2}{\delta}\left(f+25\right)\sqrt{\frac{r}{d}}\,\big(V - V_c\big)^2 \times 10^{-5}\;}\;\;\text{kW/km/phase} \]

\(f\) in hertz; \(V\) and \(V_c\) both rms phase-to-neutral voltages in kV; \(r\) and \(d\) in the same units so the ratio is dimensionless. The whole expression is zero for \(V \le V_c\) — the formula describes what happens after the threshold, and returns nonsense if applied below it.

The formula is empirical and its range of validity is narrow. It should be applied only when the supply frequency lies between \(25\) and \(120\) Hz, when the ratio of operating to critical voltage is above about \(0.8\) — that is, when the line is anywhere near the threshold — and when the conductor radius exceeds \(0.25\) cm. Outside those bounds, and particularly for \(V\) very close to \(V_c\), Peek's expression overestimates the loss, sometimes by a large factor, because the quadratic law is a fit to the well-developed discharge and not to its onset.

For the region close to and below the threshold, Peterson's formula is preferred. It replaces the quadratic in \((V-V_c)\) with a term in \(V^2\) multiplied by an experimentally determined function \(F\) of the ratio \(V/V_c\):

Peterson's formula
\[ P = \frac{21\times10^{-6}\,f\,V^2\,F}{\big[\log_{10}(d/r)\big]^{2}}\;\;\text{kW/km/phase} \]

Here \(F\) is read from a measured curve; it is very small below \(V/V_c \approx 0.8\), passes through a knee near unity, and climbs steeply beyond it. The form is instructive even without the curve: the loss is governed by the ratio \(V/V_c\), so a line designed with a comfortable margin loses essentially nothing and a line designed on the margin loses a great deal.

V (phase) P 0.8V_c V_c foul weather fair weather operating V no loss P ∝ (V − V_c)²
Corona loss is zero below threshold and quadratic above it — rain moves the threshold, not the shape

Rain, snow, fog and hoar frost transform the picture. Water hanging from the underside of a conductor forms drops that are drawn into sharp points by the field, and each point is a local field concentrator far worse than any stranding. The universal engineering allowance is to take the critical disruptive voltage in foul weather as

Foul-weather threshold
\[ V_c^{\,\text{foul}} = 0.8\,V_c \]

Because the loss is quadratic in the excess, a \(20\%\) reduction in threshold does far more than a \(20\%\) increase in loss. On a line operating modestly above \(V_c\) the excess \((V-V_c)\) can triple, and the loss therefore rises by nearly an order of magnitude — which is exactly what Example 3 finds. Annual corona energy on a well-designed extra-high-voltage line is dominated almost entirely by the small fraction of hours in which it is raining.

Section 15-7

Radio Interference and Audible Noise

Corona loss is a slow, averaged quantity. The discharge that produces it is anything but slow. Each streamer is a current pulse of a few tens of milliamperes lasting on the order of a hundred nanoseconds, and a conductor in corona emits a great many of them per half cycle at randomly scattered points along its length.

A pulse that short has a very wide spectrum. When it injects charge onto the conductor, the resulting current step splits and travels both ways along the line as a travelling wave at the surge impedance of Chapter 12, and the line — a horizontal wire kilometres long, a few tens of metres above a conducting plane — radiates. That radiation is radio interference, and it lands squarely in the medium-wave broadcast band.

The two polarities behave differently, and the difference matters. Negative Trichel pulses are small, short and extremely regular; positive streamer pulses are perhaps an order of magnitude larger in amplitude and much less regular. On an ac line, radio interference is therefore produced almost entirely during the positive half cycle of each phase.

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How radio interference is quoted
As a field strength in decibels above \(1\,\mu\)V/m, measured with a quasi-peak detector in a \(9\) kHz bandwidth at \(0.5\) or \(1\) MHz, at a stated lateral distance — usually \(15\) or \(20\) m from the outermost phase, or at the edge of the right of way.

\(0\) dB is \(1\,\mu\)V/m and every \(20\) dB is a factor of ten in field strength. A reading of \(40\) dB(\(\mu\)V/m) therefore means \(100\,\mu\)V/m.

Because the mechanism is governed by the surface gradient, so is the interference. A widely used empirical expression, fitted to fair-weather measurements at \(1\) MHz, is

Empirical fair-weather radio interference level
\[ \mathrm{RI} = 3.5\,E_m \;+\; 6\,d_c \;-\; 33\log_{10}\!\left(\frac{D}{20}\right) \;-\; 30 \;\;\;\text{dB above }1\,\mu\text{V/m} \]

with \(E_m\) the maximum conductor surface gradient in kV/cm rms, \(d_c\) the conductor (or sub-conductor) diameter in cm, and \(D\) the radial distance from the conductor in metres. Three features of it deserve attention.

First, the gradient enters linearly in decibels, so every extra kV/cm of surface gradient costs \(3.5\) dB — a factor of \(1.5\) in field strength. Interference is an exponential function of gradient, and this is why a design that is marginal on corona loss is hopeless on interference.

Second, the lateral decay is \(33\log_{10}D\), roughly \(10\) dB per doubling of distance, which is faster than the \(6\) dB per doubling of a single radiating wire. The extra attenuation comes from cancellation between the three phases, whose corona sources are in different places and at different instants of the cycle.

Third, the frequency dependence is steep: above \(1\) MHz the level falls by \(20\) to \(30\) dB per decade. By the time one reaches the VHF television band the contribution of conductor corona is negligible. Television interference near a line is almost always traced to gap-type discharges — a loose split pin, a corroded joint in an insulator string, a cracked ceramic — and not to the conductors themselves, which is a useful diagnostic rule in the field.

lateral distance D (m) dB(µV/m) 102050100 406020 typical 40 dB criterion edge of right of way single conductor, E_m = 19.8 kV/cm twin bundle, E_m = 15.4 kV/cm 15.5 dB
Interference falls about 10 dB per doubling of distance; bundling shifts the whole profile down

What matters to the neighbour is not the noise but the ratio of the wanted signal to it. The usual acceptability criterion for amplitude-modulated broadcast reception is a signal-to-noise ratio of about \(24\) dB at the edge of the right of way, and typical utility specifications limit fair-weather interference to \(40\) to \(45\) dB(\(\mu\)V/m) there. Rain raises the measured level by \(15\) to \(25\) dB, so foul-weather interference is accepted rather than designed against — on the reasoning that atmospheric noise from the same storm is high anyway.

Alongside the radio noise there is sound. A conductor in corona emits a broadband hiss and crackle from the individual streamers, plus a pure tone at twice power frequency produced by the space charge oscillating in and out of the sheath each half cycle. Audible noise is quoted in A-weighted decibels and, like interference, rises very steeply with surface gradient. It is negligible in fair weather and prominent in rain. Design limits at the edge of the right of way are of the order of \(52\) to \(55\) dBA in rain.

At the top of the voltage range, sound wins. Below about \(500\) kV the binding corona constraint is loss and radio interference. Above it, audible noise becomes the governing criterion — the bundle geometry of a \(765\) kV or \(1200\) kV line is chosen so that people living at the edge of the right of way are not kept awake in the rain, and the resulting gradient is comfortably below what loss and interference alone would permit. Chapter 39 returns to this when it treats power quality and public-interface limits.
Section 15-8

The Corona Balance Sheet

Corona is usually presented as a defect, and mostly it is. It is worth being precise about the two respects in which it genuinely helps, because both follow from results already derived and both matter to later chapters.

The first benefit is the enlarged effective radius. Section 15-2 showed that the conducting sheath extends to \(x_0 = r\,V/V_c\). A conducting shell at the conductor potential is the conductor, electrically, so the gradient at the outer edge of the sheath is lower than it would be at a bare surface of radius \(r\). Corona partly relieves the very stress that created it, which is why the discharge stabilises instead of running away into a flashover.

The second benefit is the effect on travelling waves. A lightning or switching surge arriving on a line has a crest far above the operating voltage, so it drives the conductor deep into corona for the duration of its front. The discharge draws energy out of the wave, and the extra charge stored in the sheath acts as additional shunt capacitance which slows the crest of the wave relative to its toe. The surge therefore arrives at the substation both attenuated and with a gentler front than it left with, which relaxes the insulation coordination problem of Chapter 37 considerably. Corona is, in this sense, a distributed surge absorber that costs nothing to install.

Against that stand the costs.

EffectMechanismConsequence
Energy lossIonisation energy drawn from the line every half cycleReduced transmission efficiency; annual energy cost dominated by rainy hours
Non-sinusoidal charging currentCharge injected only above \(V_c\), i.e. near the voltage peaksThird-harmonic component added to the charging current of Chapter 7; harmonic voltage drop along the line
Radio and audible noiseStreamer pulses radiating from the conductorInterference with broadcast reception; noise complaints; right-of-way width driven up
Chemical attackOzone and oxides of nitrogen formed in the discharge; nitric acid in the presence of moistureCorrosion of conductor, fittings and, in enclosed equipment, of insulation
Induced interferenceHarmonic and pulse currents coupling to parallel circuitsNoise on adjacent open-wire communication lines and on line-carrier protection channels
Insulation riskHeavily ionised air near bus-bars and insulator stringsReduced flashover margin at substations rated \(33\) kV and above

Set against the benefits, the balance is decisively negative for steady-state operation and mildly positive for surges. The engineering conclusion is therefore not "eliminate corona" but "keep the line out of corona at operating voltage, and be glad of the corona that appears when a surge passes".

Section 15-9

Designing Corona Out

Every remedy is an attempt to satisfy one inequality:

The design condition
\[ E_{max} \;\le\; m_o\,g_o\,\delta \qquad\text{with margin, at the operating phase voltage} \]

Only the left-hand side is under the designer's control, and Chapter 8 already catalogued what can be done to it. Ranked by how much they actually buy:

Increase the conductor radius. Since \(E \propto 1/[r\ln(D_{eq}/r)]\), this is the strongest single lever, and it is the reason ACSR exists: the steel core lets a conductor have a large diameter without the aluminium cross-section — and therefore the cost and weight — that a solid conductor of the same diameter would need. Where an even larger diameter was wanted without more metal, expanded ACSR with a fibre or paper filler between core and outer layers was used.

Split the phase into a bundle. The gradient of an \(n\)-conductor bundle carries the factor \(n\) in its denominator, as Chapter 8 derived, and the equivalent bundle radius \(r_b\) inside the logarithm helps a little more. This is the only lever that scales, and above about \(220\) kV no line is built without it.

Widen the spacing. \(D_{eq}\) sits inside a logarithm, so the return is poor — doubling the spacing improves the gradient by perhaps \(13\%\) — while tower cross-arms, right-of-way width and line inductance all grow with it. Spacing is set by clearance and swing requirements, not by corona.

Control the surface and the hardware. The factor \(m_o\) is not a constant of nature; it is a description of how carefully the line was built and maintained. Conductors are strung over rollers to avoid scoring, joints are compressed and smoothed, and every fitting that would otherwise present an edge is shrouded. Corona rings — toroids fitted at the line end of an insulator string and around the terminals of high-voltage bushings, arresters and instrument transformers — exist purely to replace a sharp geometry with a large radius of curvature and so cap the local gradient. Inside substations the same reasoning applies to bus-bar corners and clamp designs at \(33\) kV and above.

System voltageTypical conductor arrangementSub-conductor \(r\) (cm)Bundle spacing (cm)Typical \(E_{max}\) (kV/cm rms)
132 kVsingle ACSR (Panther)1.05≈ 14
220 kVsingle ACSR (Zebra) or twin1.43— / 45≈ 16
400 kVtwin ACSR (Moose)1.5945≈ 16
765 kVquad ACSR (Bersimis)1.7545≈ 17
1200 kVoctal bundle1.6–1.8~60≈ 15

The last column is the point of the table. Across a factor of nine in system voltage the surface gradient barely moves. That constancy is not a coincidence or a convention — it is the design condition above, enforced at every voltage level, with the bundle count doing the work that the voltage would otherwise force onto the air.

Section 15-10

Worked Examples

Examples 1 to 6 follow a single line so that each result can be checked against the last: a three-phase \(220\) kV, \(50\) Hz overhead line with conductors of radius \(1.2\) cm arranged at an equivalent spacing of \(250\) cm, with \(m_o = 0.85\) and \(m_v = 0.72\).

1 Critical disruptive voltage at standard conditions

Problem. Find the critical disruptive voltage of the line, per phase and line to line, at \(76\) cm of mercury and \(25^\circ\)C. Is the line in corona at \(220\) kV?

Solution. First the air density factor, which should come out to unity:

Air density factor
\[ \delta = \frac{3.92 \times 76}{273+25} = \frac{297.92}{298} = 1.000 \]

Then the geometry term and the voltage:

Critical disruptive voltage
\[ \ln\!\left(\frac{D_{eq}}{r}\right) = \ln\!\left(\frac{250}{1.2}\right) = \ln(208.33) = 5.3391 \]
\[ V_c = m_o g_o \delta r \ln\!\left(\frac{D_{eq}}{r}\right) = 0.85 \times 21.1 \times 1.000 \times 1.2 \times 5.3391 = 114.9\;\text{kV/phase} \]
\[ V_{c,\text{line}} = \sqrt3 \times 114.9 = 199.0\;\text{kV} \]

The operating phase voltage is \(220/\sqrt3 = 127.0\) kV, which exceeds \(114.9\) kV. Equivalently the line voltage \(220\) kV exceeds \(199.0\) kV. The line is in corona in fair weather, at a ratio \(V/V_c = 1.105\).

2 Visual critical voltage, and what a hot day at altitude does

Problem. Find the visual critical voltage of the same line at standard conditions. Then recompute both \(V_c\) and \(V_v\) for a barometric pressure of \(74\) cm of mercury and a temperature of \(40^\circ\)C.

Solution. At \(\delta = 1\) the Peek bracket is

Visual critical voltage, standard conditions
\[ 1 + \frac{0.3}{\sqrt{\delta r}} = 1 + \frac{0.3}{\sqrt{1.2}} = 1 + 0.2739 = 1.2739 \]
\[ V_v = 0.72 \times 21.1 \times 1.000 \times 1.2 \times 1.2739 \times 5.3391 = 124.0\;\text{kV/phase} \]
\[ V_{v,\text{line}} = \sqrt3 \times 124.0 = 214.8\;\text{kV} \]

The line operates at \(220\) kV, above the visual threshold of \(214.8\) kV as well as the disruptive threshold of \(199.0\) kV: a night inspection would find a continuous glow. The ratio \(V_v/V_c = 1.079\) — the \(27\%\) delay from the Peek bracket is largely undone by \(m_v\) being smaller than \(m_o\).

At \(b=74\) cm Hg and \(t = 40^\circ\)C:

Reduced air density
\[ \delta = \frac{3.92 \times 74}{273+40} = \frac{290.08}{313} = 0.9268 \]
\[ V_c' = 114.9 \times 0.9268 = 106.5\;\text{kV/phase} \;\;\Longrightarrow\;\; 184.4\;\text{kV line} \]
\[ 1 + \frac{0.3}{\sqrt{0.9268 \times 1.2}} = 1 + \frac{0.3}{1.0546} = 1.2845 \]
\[ V_v' = 0.72 \times 21.1 \times 0.9268 \times 1.2 \times 1.2845 \times 5.3391 = 115.9\;\text{kV/phase} \;\;\Longrightarrow\;\; 200.7\;\text{kV line} \]

A drop of \(7.3\%\) in air density has cost \(7.3\%\) of the disruptive voltage but only \(6.5\%\) of the visual voltage, because \(\delta\) appears inside the square root as well and partly compensates. The gap between the two voltages has widened, which is precisely the behaviour of a thinner atmosphere: ionisation starts earlier, but the sheath must grow thicker before it can be seen.

3 Corona loss in fair and foul weather

Problem. Using Peek's formula at standard conditions, find the corona loss of the line in kW per km per phase and for all three phases, in fair weather and in rain.

Solution. Collect the terms once, since only \((V-V_c)\) changes between the two cases:

Constant part of Peek's formula
\[ \frac{242.2}{\delta}(f+25) = 242.2 \times 75 = 18165, \qquad \sqrt{\frac{r}{d}} = \sqrt{\frac{1.2}{250}} = \sqrt{0.0048} = 0.06928 \]
\[ 18165 \times 0.06928 = 1258.5 \]

Fair weather, with \(V = 127.0\) kV and \(V_c = 114.9\) kV:

Fair-weather loss
\[ V - V_c = 12.11\;\text{kV}, \qquad (V-V_c)^2 = 146.6 \]
\[ P = 1258.5 \times 146.6 \times 10^{-5} = 1.845\;\text{kW/km/phase} \;\;\Longrightarrow\;\; 5.54\;\text{kW/km for the line} \]

In rain the threshold falls to \(0.8V_c = 91.93\) kV:

Foul-weather loss
\[ V - 0.8V_c = 127.02 - 91.93 = 35.09\;\text{kV}, \qquad (35.09)^2 = 1231 \]
\[ P = 1258.5 \times 1231 \times 10^{-5} = 15.50\;\text{kW/km/phase} \;\;\Longrightarrow\;\; 46.5\;\text{kW/km for the line} \]

A \(20\%\) reduction in threshold has multiplied the loss by \(8.4\). Squaring an excess that was small to begin with is a brutal amplifier, and it is the reason corona is an economic problem in wet climates and an academic one in dry ones.

4 The gradient, and the bundle that fixes it

Problem. Compute the surface gradient of the single conductor and confirm it agrees with Example 1. Then replace each phase by a twin bundle of the same conductor with \(45\) cm sub-conductor spacing, keeping \(D_{eq} = 250\) cm, and find the new maximum gradient and the new critical disruptive voltage.

Solution. For the single conductor, using the Chapter 8 result:

Single-conductor gradient
\[ E_r = \frac{V_{an}}{r\ln(D_{eq}/r)} = \frac{127.02}{1.2 \times 5.3391} = \frac{127.02}{6.4069} = 19.82\;\text{kV/cm rms} \]

The threshold for this conductor is \(m_o g_o \delta = 0.85 \times 21.1 = 17.94\) kV/cm, so the gradient exceeds it by a factor \(19.82/17.94 = 1.105\) — exactly the ratio \(V/V_c\) found in Example 1. The voltage criterion and the gradient criterion are the same statement written twice.

For the twin bundle, the equivalent radius and the average gradient are

Twin bundle gradient
\[ r_b = \sqrt{r\,d} = \sqrt{1.2 \times 45} = \sqrt{54} = 7.348\;\text{cm}, \qquad \ln\!\left(\frac{250}{7.348}\right) = \ln(34.02) = 3.5270 \]
\[ E_{av} = \frac{V_{an}}{n\,r\,\ln(D_{eq}/r_b)} = \frac{127.02}{2 \times 1.2 \times 3.5270} = \frac{127.02}{8.4648} = 15.01\;\text{kV/cm} \]
\[ E_{max} = E_{av}\left[1 + (n-1)\frac{r}{d}\right] = 15.01\left(1 + \frac{1.2}{45}\right) = 15.01 \times 1.0267 = 15.41\;\text{kV/cm} \]

That is \(14\%\) below the threshold of \(17.94\) kV/cm, so the bundled line is corona-free in fair weather. Inverting the gradient expression gives the corresponding critical disruptive voltage:

Critical disruptive voltage of the bundle
\[ V_c = m_o g_o \delta \cdot \frac{n\,r\,\ln(D_{eq}/r_b)}{1+(n-1)r/d} = 17.94 \times \frac{8.4648}{1.0267} = 147.9\;\text{kV/phase} \]
\[ V_{c,\text{line}} = \sqrt3 \times 147.9 = 256.1\;\text{kV} \]

Splitting the phase has raised the threshold from \(199\) kV to \(256\) kV line-to-line without adding a single kilogram of aluminium per phase beyond the second conductor — and the second conductor was worth having anyway, since it halves the resistance.

5 Radio interference before and after bundling

Problem. Using the empirical formula of Section 15-7, find the fair-weather interference level at \(20\) m from the outer phase for both arrangements of Example 4, and at \(50\) m for the bundled line. A local broadcast station delivers \(5\) mV/m at the site; does either arrangement meet a \(24\) dB signal-to-noise requirement?

Solution. At \(D = 20\) m the distance term vanishes, since \(\log_{10}(20/20) = 0\).

Single conductor, \(E_m = 19.82\) kV/cm, \(d_c = 2.4\) cm
\[ \mathrm{RI} = 3.5(19.82) + 6(2.4) - 0 - 30 = 69.4 + 14.4 - 30 = 53.8\;\text{dB}(\mu\text{V/m}) \]
Twin bundle, \(E_m = 15.41\) kV/cm, sub-conductor \(d_c = 2.4\) cm
\[ \mathrm{RI} = 3.5(15.41) + 14.4 - 30 = 53.9 + 14.4 - 30 = 38.3\;\text{dB}(\mu\text{V/m}) \]

The bundle is quieter by \(15.5\) dB, a factor of \(10^{15.5/20} = 5.93\) in field strength, achieved by a gradient reduction of only \(22\%\). At \(50\) m the bundled line falls further:

Lateral attenuation
\[ \mathrm{RI}(50) = 38.3 - 33\log_{10}\!\left(\frac{50}{20}\right) = 38.3 - 33(0.3979) = 38.3 - 13.1 = 25.2\;\text{dB}(\mu\text{V/m}) \]

The broadcast signal is \(5\) mV/m \(=5000\;\mu\)V/m, which is \(20\log_{10}(5000) = 74.0\) dB above \(1\,\mu\)V/m. At \(20\) m:

Signal-to-noise ratio
\[ \text{single: } 74.0 - 53.8 = 20.2\;\text{dB}, \qquad \text{bundle: } 74.0 - 38.3 = 35.7\;\text{dB} \]

The single-conductor line fails the \(24\) dB criterion by nearly \(4\) dB; the bundled line passes it with a comfortable margin, and would still pass at \(2.5\) times closer. Corona loss and interference are decided by the same gradient, so a design fixed for one is almost always fixed for the other.

6 What the corona costs over a year

Problem. The line is \(200\) km long and it rains for \(20\%\) of the year. Using the losses of Examples 3 and 4, compare the annual corona energy of the single-conductor and bundled designs at a cost of ₹6 per kWh.

Solution. Fair weather occupies \(0.8 \times 8760 = 7008\) h and foul weather \(1752\) h.

Single conductor
\[ E_{\text{fair}} = 5.54\,\tfrac{\text{kW}}{\text{km}} \times 200\,\text{km} \times 7008\,\text{h} = 7.76\;\text{GWh} \]
\[ E_{\text{foul}} = 46.5\,\tfrac{\text{kW}}{\text{km}} \times 200\,\text{km} \times 1752\,\text{h} = 16.29\;\text{GWh} \]
\[ E_{\text{total}} = 24.05\;\text{GWh} \;\;\Longrightarrow\;\; \text{₹}144\;\text{million per year} \]

For the bundled line \(V = 127.0\) kV is below \(V_c = 147.9\) kV, so fair-weather loss is zero. In rain the threshold falls to \(0.8 \times 147.9 = 118.3\) kV, and applying Peek's formula to each sub-conductor with \(r = 1.2\) cm, \(d = D_{eq} = 250\) cm and then doubling for the two sub-conductors:

Bundled line, foul weather
\[ V - 0.8V_c = 127.02 - 118.3 = 8.72\;\text{kV}, \qquad (8.72)^2 = 76.0 \]
\[ P = 2 \times 1258.5 \times 76.0 \times 10^{-5} = 1.91\;\text{kW/km/phase} \;\;\Longrightarrow\;\; 5.74\;\text{kW/km} \]
\[ E_{\text{total}} = 5.74 \times 200 \times 1752 = 2.01\;\text{GWh} \;\;\Longrightarrow\;\; \text{₹}12.1\;\text{million per year} \]

The bundle removes about \(22\) GWh of loss and ₹132 million of cost every year, and it does so on a line whose conductor material has merely doubled. No other corona remedy comes close. Two cautions belong with the number: Peek's formula overestimates near the threshold, so both figures should be read as upper bounds, and applying it sub-conductor by sub-conductor is an approximation that ignores the interaction between them.

Review

Chapter Summary

Threshold

Air ionises at \(30\) kV/cm peak, \(21.1\) kV/cm rms, at \(76\) cm Hg and \(25^\circ\)C.

Confinement

The \(1/x\) field limits the sheath to \(x_0 = r\,V/V_c\), so corona is partial, not flashover.

Disruptive voltage

\(V_c = m_o g_o \delta r \ln(D_{eq}/r)\) kV per phase — multiply by \(\sqrt3\) for the line value.

Air density

\(\delta = 3.92b/(273+t)\), the constant being \(298/76\); altitude and heat both cut the margin.

Visual voltage

\(V_v\) adds the factor \(\big(1+0.3/\sqrt{\delta r}\big)\) and uses \(m_v\), not \(m_o\). Always above \(V_c\).

Loss

Peek: \(P \propto (f+25)\sqrt{r/d}\,(V-V_c)^2/\delta\); zero below \(V_c\), and \(V_c \to 0.8V_c\) in rain.

Interference

Streamer pulses radiate; \(3.5\) dB per kV/cm of gradient, falling \(33\log_{10}D\) laterally.

The fix

Bundling carries \(n\) in the denominator of the gradient — the only lever that scales with voltage.

Practice

Practice Problems

Take \(g_o = 21.1\) kV/cm rms unless a problem states otherwise, and be explicit about whether each voltage you compute is per phase or line to line — most marks lost on corona questions are lost to that one confusion.

  1. A three-phase \(132\) kV line has conductors of \(1.05\) cm radius spaced \(400\) cm apart in an equilateral formation. With \(m_o = 0.85\) and standard atmospheric conditions, find the critical disruptive voltage per phase and line to line, and state whether the line is in corona.
  2. For the line of Problem 1, find the visual critical voltage taking \(m_v = 0.78\). By what percentage does it exceed the disruptive voltage, and how much of that is due to the Peek bracket alone?
  3. The same line is re-sited at an altitude where the barometric pressure is \(63.4\) cm of mercury, with an ambient temperature of \(35^\circ\)C. Compute \(\delta\) and the new critical disruptive voltage, and determine the conductor radius that would restore the original margin at the original spacing.
  4. A three-phase \(220\) kV, \(50\) Hz line has conductors of radius \(1.0\) cm at \(D_{eq} = 300\) cm, with \(m_o = 0.82\) and \(\delta = 0.95\). Find the corona loss per kilometre for the whole line in fair weather and in rain, and the ratio between them.
  5. Show, starting from \(V_c = g_o r \ln(d/r)\), that the conductor radius which maximises \(V_c\) for a fixed spacing \(d\) is \(r = d/e\). Explain why no transmission line is ever built anywhere near that radius.
  6. A \(400\) kV line uses a twin bundle of \(1.59\) cm sub-conductors at \(45\) cm spacing with \(D_{eq} = 1000\) cm. Compute \(r_b\), the average and maximum surface gradients, and the fair-weather margin against a threshold of \(m_o g_o \delta\) with \(m_o = 0.85\) and \(\delta = 1\).
  7. For the bundle of Problem 6, estimate the fair-weather radio interference level at \(20\) m and at \(40\) m using the empirical formula of Section 15-7. If the same line were built with a single conductor of \(1.59\) cm radius, by how many decibels would the level at \(20\) m rise?
  8. A \(275\) kV line loses \(1.6\) kW/km/phase to corona in fair weather. Peek's formula is to be used to estimate the loss after the conductor radius is increased by \(25\%\) at unchanged spacing. Set up the calculation symbolically, identify every term that changes, and explain why the answer is far more sensitive to \(V_c\) than to the \(\sqrt{r/d}\) factor.
Tip: before touching a corona problem, write down the phase voltage, the threshold gradient \(m_o g_o \delta\), and the actual surface gradient. Those three numbers settle everything — whether the line is in corona, by what ratio, and therefore roughly what Peek's formula must return. An answer in which the loss is large while \(V\) barely exceeds \(V_c\), or zero while the gradient is above threshold, is wrong before the arithmetic is checked.