Corona and Radio Interference

Set 8 — Corona and Radio Interference

Twenty worked problems on the phenomenon that decided the shape of every EHV line ever built. Corona is not a parameter to be computed but a threshold to be avoided: below the critical disruptive voltage nothing happens at all, and above it the line loses power continuously, radiates interference and audibly hisses. The remedy is more conductor surface — which is why the bundles of Set 7 exist, and why their real justification appears only now.

Textbook Chapter 15 · 20 solved · 12 practice · 3 challenge · 12 MCQs

i Method Recap

Critical disruptive voltage. \(V_c = m_og_o\delta r\ln(d/r)\) kV per phase, r.m.s. Below it there is no corona whatever; the phenomenon has a hard threshold, unlike every other loss mechanism in the subject.
The breakdown strength of air is 30 kV/cm at its peak, so the r.m.s. figure used in these formulas is \(g_o = 30/\sqrt2 = 21.2\) kV/cm. Using 30 where 21.2 is meant overstates \(V_c\) by 41% and is the commonest error in the topic.
Air density factor. \(\delta = \dfrac{3.92b}{273 + t}\) with \(b\) in cm of mercury and \(t\) in °C. It equals 1 at 76 cm and 25 °C. Thin air breaks down more easily, so high altitude and high temperature both lower the corona threshold.
Irregularity factor. \(m_o = 1\) for a polished conductor, 0.98–0.92 for a rough or weathered one, and 0.87–0.80 for a stranded one. Dirt, raindrops and insects all concentrate the field locally, and every one of them lowers the threshold.
Visual critical voltage is higher than the disruptive one: \(V_v = m_vg_o\delta r\big(1 + \tfrac{0.3}{\sqrt{\delta r}}\big)\ln(d/r)\). Corona begins electrically before it can be seen, so a line showing no visible glow may still be losing power.
Peek's loss formula. \(P = 242.2\left(\dfrac{f+25}{\delta}\right)\sqrt{\dfrac{r}{d}}\,(V - V_c)^2\times10^{-5}\) kW/km/phase. The square of the excess is what makes corona so sharply nonlinear — a 10% overvoltage on a line just at threshold can multiply the loss several times.
The surface gradient is the real quantity. \(g = \dfrac{V_{ph}}{r\ln(D/r)}\) kV/cm. Corona occurs when \(g\) exceeds \(m_og_o\delta\); everything else in this set is bookkeeping around that single comparison.

VideoWalkthrough

Problem 1Exam levelDisruptive Voltage

A three-phase line has conductors 2 cm in diameter spaced equilaterally 1 m apart. If the dielectric strength of air is 30 kV/cm (maximum), find the disruptive critical voltage for the line. Take the air density factor \(\delta = 0.952\) and the irregularity factor \(m_o = 0.9\).

Solution

Conductor radius and spacing, in consistent units:

\[ r = \frac{2}{2} = 1\ \text{cm}, \qquad d = 1\ \text{m} = 100\ \text{cm} \]

Convert the dielectric strength to r.m.s. The quoted 30 kV/cm is a peak value; the formula works in r.m.s.:

\[ g_o = \frac{30}{\sqrt2} = 21.2\ \text{kV(r.m.s.)/cm} \]

This single conversion is the most frequently omitted step in the whole topic, and omitting it overstates every answer by 41%.

Applying the disruptive critical voltage formula:

\[ V_c = m_og_o\delta r\ln\frac{d}{r} = 0.9(21.2)(0.952)(1)\ln\frac{100}{1} \]

Evaluating step by step:

\[ 0.9 \times 21.2 = 19.08, \qquad 19.08 \times 0.952 = 18.16, \qquad \ln(100) = 4.605 \]

\[ V_c = 18.16(4.605) = 83.6\ \text{kV per phase (r.m.s.)} \]

As a line voltage:

\[ V_{c,\text{line}} = \sqrt3(83.6) = 144.8\ \text{kV} \]

Corona has a hard threshold, and that makes it unlike every other loss in this book. Resistance dissipates at any current; hysteresis at any flux. Corona does nothing at all until the surface gradient reaches the breakdown strength of air, and then rises as the square of the excess. A line designed to sit 10% below \(V_c\) in fair weather has no corona loss whatever — which is why corona design is about staying under a limit rather than minimising a quantity.

Answer\(V_c = 83.6\) kV/phase, equivalent to \(144.8\) kV between lines

Problem 2Exam levelInverse Problem

A 132 kV line with conductors 1.956 cm in diameter is built so that corona takes place if the line voltage exceeds 210 kV (r.m.s.). Taking the potential gradient at which ionisation occurs as 30 kV/cm, find the spacing between the conductors. Assume smooth conductors and standard atmospheric conditions.

Solution

Gathering the data. Smooth conductors give \(m_o = 1\); standard conditions give \(\delta = 1\):

\[ r = \frac{1.956}{2} = 0.978\ \text{cm}, \qquad g_o = \frac{30}{\sqrt2} = 21.21\ \text{kV/cm} \]

The stated 210 kV is a line voltage; the formula needs the phase value:

\[ V_c = \frac{210}{\sqrt3} = 121.24\ \text{kV per phase} \]

Substituting into the disruptive voltage formula and solving for the logarithm:

\[ 121.24 = (1)(21.21)(1)(0.978)\ln\frac{d}{r} = 20.75\ln\frac{d}{r} \]

\[ \ln\frac{d}{r} = \frac{121.24}{20.75} = 5.844 \]

Taking the exponential:

\[ \frac{d}{r} = e^{5.844} = 345.2 \]

\[ d = 345.2(0.978) = 337.6\ \text{cm} = 3.38\ \text{m} \]

A sanity check: the line operates at 132 kV and coronas only at 210 kV, a margin of 59%. A spacing of 3.4 m is entirely ordinary for 132 kV, so the design is consistent.

The spacing appears only inside a logarithm, which makes this inverse problem badly conditioned — and that is useful to know. A 5% error in \(V_c\) becomes a 30% error in \(d\), because the exponential amplifies whatever the logarithm compressed. In the forward direction this insensitivity is a blessing: the designer can move the phases about almost freely without disturbing the corona margin. Solving backwards for a dimension, as here, is the one situation in which it becomes a trap.

Answer\(d = 337.6\) cm \(\approx 3.38\) m

Problem 3Warm-upAir Density Factor

Find the air density factor at a barometric pressure of 73.6 cm of mercury and a temperature of 40 °C, and verify that the formula gives unity at standard conditions of 76 cm and 25 °C.

Solution

The air density factor:

\[ \delta = \frac{3.92b}{273 + t} \]

with \(b\) in centimetres of mercury and \(t\) in degrees Celsius, so that \(273 + t\) is the absolute temperature.

At the stated conditions:

\[ \delta = \frac{3.92(73.6)}{273 + 40} = \frac{288.5}{313} = 0.922 \]

Verifying at standard conditions:

\[ \delta = \frac{3.92(76)}{273 + 25} = \frac{297.9}{298} = 1.000\ \checkmark \]

The constant 3.92 exists precisely to normalise the expression this way — it is \(298/76\).

At \(\delta = 0.922\) every corona voltage falls by 7.8% relative to standard conditions, since \(\delta\) multiplies \(V_c\) directly.

Thin air breaks down more easily, and both hot weather and high altitude thin it. The physical mechanism is mean free path: in less dense air an electron accelerated by the field travels further before colliding, gains more energy, and is likelier to ionise the molecule it eventually strikes. This is why corona is a summer problem more than a winter one, and why a line designed for the plains must be re-checked before it is built across a plateau — as Problem 4 shows quantitatively.

Answer\(\delta = 0.922\); the formula returns 1.000 at 76 cm and 25 °C

Problem 4Exam levelAltitude

A line has a critical disruptive voltage of 150 kV per phase at sea level and 25 °C. Find its critical voltage at an altitude where the barometric pressure is 60 cm of mercury and the temperature is 35 °C, and state the design implication.

Solution

Air density factor at altitude:

\[ \delta = \frac{3.92(60)}{273 + 35} = \frac{235.2}{308} = 0.764 \]

Since \(V_c \propto \delta\) with all geometry unchanged:

\[ V_c' = 150 \times \frac{0.764}{1.000} = 114.6\ \text{kV per phase} \]

The reduction:

\[ \frac{150 - 114.6}{150}\times100 = 23.6\% \]

The design implication. A line that had a comfortable margin at sea level may be in continuous corona at altitude. The remedies are the ones available throughout this set: a larger conductor, a bundle, or a wider spacing — in that order of effectiveness.

Note that both effects worked the same way here. Lower pressure and higher temperature each reduce air density, so a hot day at altitude is the worst case and must be the design case.

A quarter of the corona margin lost to the atmosphere alone. This is why transmission standards specify corona performance at a stated altitude and why lines crossing high plateaux — the Andes, the Tibetan plateau, the Ethiopian highlands — use conductors visibly larger than their voltage class would suggest elsewhere. It is also why the same line design cannot simply be exported between climates, however identical the electrical duty.

Answer\(\delta = 0.764\), so \(V_c\) falls from 150 to \(114.6\) kV/phase — a 23.6% loss of margin

Problem 5Exam levelVisual Voltage

A three-phase 220 kV line has conductors of radius 1.5 cm spaced 2 m apart in equilateral formation. At a temperature of 40 °C and a barometric pressure of 73.6 cm, with \(m_v = 0.72\), find the visual critical voltage.

Solution

From Problem 3, \(\delta = 0.922\). The geometry:

\[ r = 1.5\ \text{cm}, \qquad d = 200\ \text{cm}, \qquad \ln\frac{d}{r} = \ln(133.3) = 4.893 \]

The visual critical voltage carries an extra bracket relative to the disruptive one:

\[ V_v = m_vg_o\delta r\left(1 + \frac{0.3}{\sqrt{\delta r}}\right)\ln\frac{d}{r} \]

Evaluating the bracket first:

\[ \sqrt{\delta r} = \sqrt{(0.922)(1.5)} = \sqrt{1.383} = 1.176 \]

\[ 1 + \frac{0.3}{1.176} = 1 + 0.255 = 1.255 \]

Now the whole expression:

\[ V_v = 0.72(21.2)(0.922)(1.5)(1.255)(4.893) \]

\[ = 15.26 \times 0.922 \times 1.5 \times 1.255 \times 4.893 = 129.6\ \text{kV per phase} \]

As a line voltage:

\[ V_{v,\text{line}} = \sqrt3(129.6) = 224.5\ \text{kV} \]

The line operates at 220 kV, marginally below the visual threshold — so no glow would be seen, though Problem 6 shows the line is nevertheless in electrical corona.

Corona begins before it can be seen, and the gap between the two thresholds is where lines quietly lose money. The extra bracket \(1 + 0.3/\sqrt{\delta r}\) exists because a visible glow needs not merely ionisation but enough of it to produce light, which requires the field to exceed breakdown over a finite thickness of air rather than at the surface alone. An inspection at night that reveals nothing is therefore no evidence that a line is corona-free — the loss measurement of Problem 7 is the only reliable test.

Answer\(V_v = 129.6\) kV/phase, equivalent to \(224.5\) kV between lines

Problem 6Exam levelThreshold Test

For the line of Problem 5, taking the irregularity factor for the stranded conductor as \(m_o = 0.85\), find the critical disruptive voltage and determine whether the line is in corona at its rated 220 kV.

Solution

Critical disruptive voltage, with \(\delta = 0.922\) and \(\ln(d/r) = 4.893\):

\[ V_c = m_og_o\delta r\ln\frac{d}{r} = 0.85(21.2)(0.922)(1.5)(4.893) \]

Evaluating in stages:

\[ 0.85 \times 21.2 = 18.02, \qquad \times 0.922 = 16.61, \qquad \times 1.5 = 24.92, \qquad \times 4.893 = 121.9 \]

\[ V_c = 121.9\ \text{kV per phase} \]

The operating phase voltage:

\[ V_{ph} = \frac{220}{\sqrt3} = 127.0\ \text{kV} \]

The comparison:

\[ V_{ph} = 127.0 > V_c = 121.9 \quad\Rightarrow\quad \textbf{the line is in corona} \]

\[ V_{ph} - V_c = 5.1\ \text{kV per phase} \]

Yet Problem 5 found the visual threshold at 129.6 kV per phase, above the operating voltage. So the line is losing power to corona while showing no visible glow — the gap identified at the end of Problem 5, here made concrete.

A margin of 5 kV in 127 is 4%, which is far too fine for a design. Every quantity in \(V_c\) is uncertain: \(m_o\) depends on how weathered the conductor is, \(\delta\) on the day's weather, and the surface condition changes entirely in rain. Design practice is to keep the operating gradient at least 10–15% below the fair-weather threshold, precisely so that the ordinary variation in these factors does not push the line over. This line has no such margin and would be redesigned — as Problems 11 and 12 do.

Answer\(V_c = 121.9\) kV/phase against an operating \(127.0\) — in corona, by 5.1 kV

Problem 7Exam levelPeek's Formula

Find the corona power loss per km per phase for the 50 Hz line of Problem 6, using Peek's formula.

Solution

Peek's empirical formula:

\[ P = 242.2\left(\frac{f + 25}{\delta}\right)\sqrt{\frac{r}{d}}\,(V - V_c)^{2}\times10^{-5}\ \text{kW/km/phase} \]

with \(V\) and \(V_c\) both phase voltages in kV r.m.s., and \(r\) and \(d\) in the same length unit.

Evaluating each factor:

\[ \frac{f+25}{\delta} = \frac{75}{0.922} = 81.35 \]

\[ \sqrt{\frac{r}{d}} = \sqrt{\frac{1.5}{200}} = \sqrt{0.0075} = 0.0866 \]

\[ (V - V_c)^{2} = (127.0 - 121.9)^{2} = (5.1)^{2} = 26.0 \]

Multiplying through:

\[ P = 242.2(81.35)(0.0866)(26.0)\times10^{-5} \]

\[ = 19\,703 \times 0.0866 \times 26.0 \times 10^{-5} = 44\,360\times10^{-5} = 0.444\ \text{kW/km/phase} \]

For all three phases:

\[ P_{3\phi} = 3(0.444) = 1.33\ \text{kW/km} \]

Peek's formula is empirical, and its constants carry no derivation whatever. The 242.2, the 25 added to the frequency, the square root of \(r/d\) — all were fitted to laboratory measurements around 1911 and none follows from theory. It is accurate to perhaps \(\pm 20\%\) near the threshold and worse well above it, where Peterson's formula is preferred. What it captures correctly, and what matters, is the \((V - V_c)^2\) dependence — the reason corona is negligible below threshold and punishing above it.

Answer\(P = 0.444\) kW/km/phase, or \(1.33\) kW/km for the three-phase line

Problem 8Warm-upWhole-Line Loss

The line of Problem 7 is 150 km long. Find the total corona loss and its annual energy cost at \(\text{₹}5\) per kWh, assuming these conditions persist throughout the year.

Solution

Total corona loss:

\[ P_{\text{total}} = 1.33 \times 150 = 200\ \text{kW} \]

Annual energy, since corona depends on voltage rather than load and therefore runs continuously:

\[ E = 200 \times 8760 = 1.752\times10^{6}\ \text{kWh} \]

Annual cost:

\[ \text{Cost} = 5 \times 1.752\times10^{6} = \text{₹}8.76\ \text{million per year} \]

The assumption of constant conditions is generous to the line: the calculation used fair-weather figures, and in rain the loss can rise by an order of magnitude, as Problem 14 shows.

Corona loss is a base load, not a peak load, and that is what makes it expensive. The \(I^2R\) loss of Set 2 followed the load and so cost money only in proportion to revenue earned; corona depends on the voltage, which is held constant day and night. A line just over its threshold burns money at three in the morning with nothing connected. Nearly ₹9 million a year buys a great deal of conductor, which is precisely the trade-off Problem 12 evaluates.

Answer\(200\) kW total, \(1.75\) GWh a year, costing about \(\text{₹}8.8\) million

Problem 9Exam levelSurface Gradient

Derive the expression for the voltage gradient at the surface of a conductor in a three-phase line, and evaluate it for the line of Problem 6.

Solution

The potential of a conductor carrying charge \(q\) per unit length, relative to neutral, is

\[ V_{ph} = \frac{q}{2\pi\varepsilon_0}\ln\frac{D}{r} \]

which is the capacitance relation of Set 6 rearranged.

The field at the conductor surface, from Gauss's law:

\[ g = \frac{q}{2\pi\varepsilon_0 r} \]

Eliminating \(q\) between the two:

\[ g = \frac{V_{ph}}{r\ln(D/r)} \]

Every corona criterion in this set is a comparison between this \(g\) and the breakdown strength \(m_og_o\delta\). The \(V_c\) formula is simply that comparison rearranged to give a voltage.

Evaluating for the line of Problem 6, with \(V_{ph} = 127.0\) kV, \(r = 1.5\) cm and \(\ln(D/r) = 4.893\):

\[ g = \frac{127.0}{1.5(4.893)} = \frac{127.0}{7.34} = 17.3\ \text{kV/cm (r.m.s.)} \]

Equivalently as a peak value:

\[ g_{\text{peak}} = \sqrt2(17.3) = 24.5\ \text{kV/cm} \]

The gradient is the physical quantity; the critical voltage is a convenience. Air does not know what voltage a line operates at — it responds to the field at the metal surface, and that field depends on the charge and the radius, not on the distance to the next phase except through the logarithm. Working in gradients rather than voltages makes the bundle calculation of Problem 11 straightforward, and it is how EHV designers actually work, because the gradient can be compared directly against a single material property.

Answer\(g = V_{ph}/[r\ln(D/r)] = 17.3\) kV/cm r.m.s., or 24.5 kV/cm peak

Problem 10Warm-upGradient Criterion

Show that comparing the surface gradient of Problem 9 with the breakdown strength gives the same corona verdict as the voltage comparison of Problem 6.

Solution

The breakdown strength for this conductor and this weather:

\[ g_{\text{crit}} = m_og_o\delta = 0.85(21.2)(0.922) = 16.6\ \text{kV/cm (r.m.s.)} \]

The operating gradient from Problem 9:

\[ g = 17.3\ \text{kV/cm} \]

The verdict:

\[ g = 17.3 > g_{\text{crit}} = 16.6 \quad\Rightarrow\quad \textbf{corona} \]

Agreeing with Problem 6, as it must.

The two criteria are algebraically identical. Dividing the voltage comparison by \(r\ln(D/r)\):

\[ V_{ph} > V_c = m_og_o\delta r\ln\frac{D}{r} \quad\Longleftrightarrow\quad \frac{V_{ph}}{r\ln(D/r)} > m_og_o\delta \quad\Longleftrightarrow\quad g > g_{\text{crit}} \]

The margin, expressed as a ratio:

\[ \frac{g}{g_{\text{crit}}} = \frac{17.3}{16.6} = 1.042 \]

The same 4% excess found in Problem 6, since the two comparisons differ only by a common factor.

Two formulations, one physical statement — but the gradient form generalises and the voltage form does not. For a bundle, or for a conductor near an earthed structure, or for a stranded surface with local field concentration, there is no clean \(V_c\) expression to write down; there is always a gradient to compute. This is why every modern corona study works in kV/cm at the surface, and why the \(V_c\) formulas of Problems 1 to 6 survive mainly as an examination convenience.

Answer\(g_{\text{crit}} = 16.6\) against \(g = 17.3\) kV/cm — corona, in agreement with Problem 6

Problem 11Exam levelBundling

The line of Problem 6 is rebuilt with a two-conductor bundle at 45 cm spacing, each sub-conductor of radius 1.5 cm, keeping the same 2 m phase spacing. Find the new surface gradient and determine whether corona is eliminated.

Solution

The capacitive bundle radius, from Set 7:

\[ r_b = \sqrt{r\,d_b} = \sqrt{(1.5)(45)} = \sqrt{67.5} = 8.22\ \text{cm} \]

The charge on the phase is now shared between two sub-conductors, and the effective radius in the logarithm is \(r_b\). The average surface gradient becomes

\[ g = \frac{V_{ph}}{n\,r\ln(D/r_b)} \]

The \(n r\) in the denominator is the total surface available to carry the charge; the \(r_b\) inside the logarithm is the geometry the rest of the system sees.

Evaluating:

\[ \ln\frac{D}{r_b} = \ln\frac{200}{8.22} = \ln(24.3) = 3.19 \]

\[ g = \frac{127.0}{2(1.5)(3.19)} = \frac{127.0}{9.57} = 13.3\ \text{kV/cm} \]

Against the threshold of 16.6 kV/cm from Problem 10:

\[ g = 13.3 < g_{\text{crit}} = 16.6 \quad\Rightarrow\quad \textbf{no corona} \]

\[ \text{Margin} = \frac{16.6 - 13.3}{16.6}\times100 = 20\% \]

The gradient fell from 17.3 to 13.3 kV/cm, a reduction of 23%, and the line moved from 4% over threshold to 20% under it.

This is what bundling is actually for, and it is why the Set 7 reactance benefits are a by-product. Two mechanisms combine: the charge is shared between two surfaces, halving the charge each must carry, and the effective radius rises from 1.5 to 8.22 cm, which reduces the logarithm. Neither could have been achieved by a single conductor without the twenty-fold metal penalty of Set 7 Problem 16. Every EHV line is bundled because of this calculation, and gets a lower surge impedance for nothing.

Answer\(g = 13.3\) kV/cm against a threshold of 16.6 — corona eliminated, with a 20% margin

Problem 12Challenge-liteConductor Size

As an alternative to bundling, what single-conductor radius would give the line of Problem 6 a 15% margin below the corona threshold? Compare the metal required with the two-conductor bundle of Problem 11.

Solution

The target gradient, 15% below the 16.6 kV/cm threshold:

\[ g_{\text{target}} = 0.85(16.6) = 14.1\ \text{kV/cm} \]

Setting the single-conductor gradient equal to it:

\[ 14.1 = \frac{127.0}{r\ln(200/r)} \quad\Rightarrow\quad r\ln\frac{200}{r} = 9.01 \]

A transcendental equation, solved by trial.

Trying successive values:

\[ \begin{array}{ccc} r\ (\text{cm}) & \ln(200/r) & r\ln(200/r) \\ \hline 1.5 & 4.893 & 7.34 \\ 2.0 & 4.605 & 9.21 \\ 1.95 & 4.631 & 9.03 \\ 1.94 & 4.636 & 8.99 \end{array} \]

\[ r \approx 1.95\ \text{cm} \]

Comparing the metal. The single conductor:

\[ A = \pi(1.95)^{2} = 11.9\ \text{cm}^{2} \]

\[ \text{Bundle: } A = 2\pi(1.5)^{2} = 14.1\ \text{cm}^{2} \]

The single conductor uses less metal — 11.9 against 14.1 cm² — and achieves a comparable margin.

At 220 kV enlarging the conductor is still competitive, and that is exactly why 220 kV lines are often unbundled. The reason is the modest 4% excess this line had to overcome; a small increase in radius sufficed. The picture reverses completely at 400 kV and above, where the required gradient reduction is far larger and the transcendental equation drives \(r\) into the range where a single conductor becomes unmanageable — Problem 19 works that case. The bundle also brings the 27% reactance reduction of Set 7, which this comparison ignores entirely.

Answer\(r \approx 1.95\) cm, using 11.9 cm² against the bundle's 14.1 cm² — enlargement wins at this voltage

Problem 13Warm-upIrregularity Factor

For the line of Problem 6, compute the critical disruptive voltage for a polished conductor (\(m_o = 1.0\)), a weathered one (\(m_o = 0.93\)) and a stranded one (\(m_o = 0.85\)), and state which the designer must use.

Solution

Since \(V_c \propto m_o\) with everything else fixed, and Problem 6 gave 121.9 kV at \(m_o = 0.85\):

\[ V_c = \frac{121.9}{0.85}m_o = 143.4\,m_o \]

Evaluating:

\[ \begin{array}{lcc} \text{Surface} & m_o & V_c\ (\text{kV/phase}) \\ \hline \text{Polished} & 1.00 & 143.4 \\ \text{Weathered} & 0.93 & 133.4 \\ \text{Stranded} & 0.85 & 121.9 \end{array} \]

Against the operating 127.0 kV per phase, the polished and weathered cases are clear of corona and the stranded case is not.

Which to use. The stranded value, 0.85 — and arguably lower. Real conductors are stranded, and they weather, collect dust, attract insects and carry water droplets. The design must assume the worst surface the conductor will ever present, not the best.

A 15% swing in the threshold from surface condition alone, and it is entirely outside the designer's control after commissioning. The irregularity factor is the least defensible number in the whole calculation — it is a fudge covering everything from strand geometry to bird droppings — and it moves the answer more than the atmospheric factor does. This is the deepest reason corona design works to a margin rather than a computed optimum: the input uncertainty is larger than any refinement of the arithmetic could justify.

Answer143.4, 133.4 and 121.9 kV/phase — the stranded value must be used for design

Problem 14Challenge-liteFoul Weather

Explain why corona loss rises so sharply in rain, snow and fog, estimate the effect for the line of Problem 7 taking the foul-weather irregularity factor as 0.6, and comment on how this is handled in practice.

Solution

The mechanism. Water droplets hanging from the underside of a conductor are conducting protrusions with very small radii of curvature. The field at a droplet tip is many times the field at the smooth conductor surface, so ionisation begins there at a fraction of the nominal threshold. Snow and frost do the same with ice crystals; fog wets the surface uniformly and raises the effective roughness.

Quantifying it. With \(m_o = 0.6\) instead of 0.85, and using the proportionality of Problem 13:

\[ V_c = 143.4(0.6) = 86.0\ \text{kV per phase} \]

The excess over threshold grows dramatically:

\[ V - V_c = 127.0 - 86.0 = 41.0\ \text{kV} \quad\text{against}\quad 5.1\ \text{kV in fair weather} \]

Since Peek's loss goes as the square of the excess:

\[ \frac{P_{\text{wet}}}{P_{\text{fair}}} = \left(\frac{41.0}{5.1}\right)^{2} = 64.6 \]

\[ P_{\text{wet}} = 64.6(1.33) = 86\ \text{kW/km} \]

In practice the annual energy is computed by weighting fair and foul weather by their durations. If rain occupies 5% of the year:

\[ \bar{P} = 0.95(1.33) + 0.05(86) = 1.26 + 4.30 = 5.6\ \text{kW/km} \]

Foul weather contributes 77% of the annual corona energy while occupying 5% of the time.

A sixty-fold rise from rain is not an exaggeration — it is the standard EHV experience. Utilities quote fair-weather and foul-weather corona losses separately for exactly this reason, and the annual figure is dominated by a small fraction of the hours. It also explains the audible noise complaints that accompany EHV lines: the crackling heard beneath a 400 kV line in drizzle is corona at its droplet-enhanced threshold, and it is loudest precisely when the loss is greatest.

Answer\(V_c\) falls to 86 kV/phase and the loss rises about 65-fold to 86 kW/km; foul weather dominates the annual energy

Problem 15Exam levelOvervoltage

The line of Problem 7 experiences a 10% overvoltage. Find the new corona loss and the factor by which it has increased.

Solution

The new phase voltage:

\[ V = 1.10(127.0) = 139.7\ \text{kV} \]

\(V_c\) is unchanged at 121.9 kV — it depends on geometry and weather, not on the applied voltage.

The new excess:

\[ V - V_c = 139.7 - 121.9 = 17.8\ \text{kV} \]

Since the loss goes as the square of the excess:

\[ \frac{P_{\text{new}}}{P_{\text{old}}} = \left(\frac{17.8}{5.1}\right)^{2} = (3.49)^{2} = 12.2 \]

\[ P_{\text{new}} = 12.2(1.33) = 16.2\ \text{kW/km} \]

A 10% overvoltage multiplied the corona loss twelvefold.

The amplification comes from the difference, not the voltage. Because the line sat only 4% above threshold, a 10% rise in \(V\) was a 249% rise in \((V - V_c)\), and squaring made it twelvefold. The closer a line operates to its threshold, the more violently its corona loss responds to any disturbance — which is a strong argument for the 15% design margin of Problem 12, quite apart from the fair-weather loss itself. It also links directly to Set 14: the Ferranti effect raises the far-end voltage of a lightly loaded long line, and it is precisely there that corona is worst.

AnswerLoss rises from 1.33 to \(16.2\) kW/km — a factor of 12.2 for a 10% overvoltage

Problem 16Warm-upSpacing

For the line of Problem 6, find the critical disruptive voltage if the phase spacing is increased from 2 m to 4 m, and compare the effectiveness of widening the spacing with that of enlarging the conductor.

Solution

At 4 m spacing:

\[ \ln\frac{d}{r} = \ln\frac{400}{1.5} = \ln(266.7) = 5.586 \]

\[ V_c = 0.85(21.2)(0.922)(1.5)(5.586) = 24.92(5.586) = 139.2\ \text{kV per phase} \]

Against 121.9 kV at 2 m — a gain of 14.2%, and enough to clear the 127.0 kV operating voltage with a 9% margin.

Comparing the two levers. Doubling the spacing multiplied \(V_c\) by \(5.586/4.893 = 1.142\). From Problem 12, raising the radius from 1.5 to 1.95 cm — a 30% increase — gave

\[ \frac{r'\ln(200/r')}{r\ln(200/r)} = \frac{9.01}{7.34} = 1.228 \]

a 22.8% gain, from a much smaller physical change.

The reason is structural: \(r\) appears both as a multiplier and inside the logarithm, while \(d\) appears only inside it.

Radius is a strong lever and spacing is a weak one, which is the reverse of what intuition suggests. Doubling the phase spacing means taller, wider, far more expensive towers and a wider right of way, for a 14% gain; a 30% larger conductor gives 23% for a modest increase in weight. Add that widening the spacing also raises the line's inductance (Set 5) and lowers its capacitance (Set 6), both unhelpfully, and the case is closed: corona is fixed at the conductor, never at the tower.

Answer\(V_c = 139.2\) kV/phase — 14% gain for doubled spacing, against 23% for a 30% larger radius

Problem 17Challenge-liteRadio Interference

Explain the mechanism by which corona produces radio interference and audible noise, why these often govern EHV line design more tightly than the power loss does, and what the design response is.

Solution

The mechanism. Corona is not a steady glow but a sequence of discrete discharge pulses — streamers — each lasting tens of nanoseconds. A pulse that fast has Fourier components extending well into the megahertz range, and the conductor radiates them as an antenna.

The consequences divide by frequency:

\[ \begin{array}{ll} 0.5\text{–}1.6\ \text{MHz} & \text{AM broadcast interference} \\ \text{up to } 30\ \text{MHz} & \text{Radio interference (RI), falling with frequency} \\ \text{above } 50\ \text{MHz} & \text{Television interference, usually minor} \\ \text{audio} & \text{Crackle and hum from the acoustic pressure pulse} \end{array} \]

Why they govern. Three reasons:

\[ \begin{array}{ll} \text{Onset} & \text{RI and noise appear at gradients below the loss threshold} \\ \text{Regulation} & \text{Limits are legally binding; power loss is merely costly} \\ \text{Perception} & \text{Complaints arrive at any audible level, not a threshold} \end{array} \]

The design response is the same as for loss but applied more stringently: keep the maximum surface gradient below roughly 16–17 kV/cm r.m.s. at EHV, which for a given voltage means more sub-conductors and larger ones. It also means attention to hardware — corona rings on insulator strings, smooth fittings, no sharp edges anywhere on the live assembly — since a single protruding bolt can produce more RI than kilometres of conductor.

Audible noise is specified separately, typically below about 50 dB(A) at the edge of the right of way in wet weather, which is the condition of Problem 14.

The binding constraint on a 765 kV line is usually a noise complaint, not an energy bill. The corona loss computed in Problem 8 is a real cost but a manageable one; a line that violates a radio-interference limit cannot be operated at all, and one that generates audible noise at a nearby house creates a permanent liability. This is why EHV bundles are chosen with more sub-conductors than the loss calculation alone would justify, and it is the final reason that Set 7's reactance benefits are best understood as a bonus rather than a purpose.

AnswerNanosecond streamer pulses radiate to 30 MHz; RI and noise onset below the loss threshold and are legally limited, so they set the bundle size

Problem 18Exam levelCorona vs I²R

The 220 kV line of Problem 7 carries 400 A and has a resistance of 0.08 \(\Omega\)/km per phase. Compare its corona loss with its \(I^2R\) loss at full load and at no load.

Solution

At full load. The resistive loss for three phases:

\[ P_{I^2R} = 3I^{2}R = 3(400)^{2}(0.08) = 38\,400\ \text{W} = 38.4\ \text{kW/km} \]

Against the fair-weather corona loss of 1.33 kW/km from Problem 7:

\[ \frac{1.33}{38.4} = 3.5\% \]

Corona is a minor addition at full load.

At no load. The resistive loss falls to essentially zero — only the charging current flows — while the corona loss is unchanged:

\[ P_{I^2R} \approx 0, \qquad P_{\text{corona}} = 1.33\ \text{kW/km} \]

In foul weather at no load, using the 86 kW/km of Problem 14:

\[ P_{\text{corona}} = 86\ \text{kW/km} \quad\text{against}\quad P_{I^2R} \approx 0 \]

More than twice the full-load resistive loss, with no power being delivered at all.

The two losses have opposite load characteristics, and that is what makes corona insidious. \(I^2R\) scales with the square of the current and therefore with the revenue the line earns; corona scales with voltage and is entirely independent of it. A line spends its nights at light load with the resistive loss near zero and the corona loss at its full value — and if it happens to be raining, losing more than it would at full load. Set 2's loss load factor treats the first mechanism; corona simply does not fit that framework at all.

AnswerCorona is 3.5% of \(I^2R\) at full load, but is the only loss at no load — and in rain exceeds the full-load resistive loss

Problem 19Challenge-lite400 kV Design

A 400 kV line has an equivalent phase spacing of 12 m. Taking \(m_o = 0.85\) and \(\delta = 1\), find the single-conductor radius needed to keep the surface gradient 15% below threshold, and hence show why 400 kV lines are bundled.

Solution

Threshold gradient and target:

\[ g_{\text{crit}} = m_og_o\delta = 0.85(21.2)(1) = 18.0\ \text{kV/cm}, \qquad g_{\text{target}} = 0.85(18.0) = 15.3\ \text{kV/cm} \]

Phase voltage:

\[ V_{ph} = \frac{400}{\sqrt3} = 231\ \text{kV} \]

Setting the gradient to the target, with \(D = 1200\) cm:

\[ 15.3 = \frac{231}{r\ln(1200/r)} \quad\Rightarrow\quad r\ln\frac{1200}{r} = 15.1 \]

Solving by trial:

\[ \begin{array}{ccc} r\ (\text{cm}) & \ln(1200/r) & r\ln(1200/r) \\ \hline 2.5 & 6.175 & 15.44 \\ 2.4 & 6.215 & 14.92 \\ 2.45 & 6.195 & 15.18 \end{array} \]

\[ r \approx 2.45\ \text{cm}, \quad\text{a conductor 4.9 cm in diameter} \]

Why this is impractical. The cross-section would be \(\pi(2.45)^2 = 18.9\) cm², weighing roughly 5 kg per metre — well beyond standard ACSR sizes, difficult to string, and imposing tower loads that would drive the whole structure design. And it would be sized for corona alone, carrying far more metal than the current rating requires.

The bundled alternative. Two sub-conductors of 1.5 cm radius at 45 cm, giving \(r_b = \sqrt{(1.5)(45)} = 8.22\) cm:

\[ g = \frac{231}{2(1.5)\ln(1200/8.22)} = \frac{231}{3(4.982)} = \frac{231}{14.95} = 15.5\ \text{kV/cm} \]

Essentially the same gradient, on \(2\pi(1.5)^2 = 14.1\) cm² of metal against 18.9 — and with the 27% reactance reduction of Set 7 thrown in.

At 220 kV enlargement won (Problem 12); at 400 kV bundling wins, and at 765 kV enlargement is not even conceivable. The crossover happens because the required conductor radius grows almost linearly with voltage while the metal grows as its square, whereas a bundle's effective radius grows with the spacing at no metal cost at all. Somewhere around 300 kV the two curves cross, and every line above that voltage in the world is bundled.

Answer\(r \approx 2.45\) cm (18.9 cm²) as a single conductor, against 14.1 cm² bundled for the same gradient

Problem 20Exam levelFull Assessment

Carry out a complete corona assessment of a 400 kV, 250 km line with two-conductor bundles at 45 cm, sub-conductor radius 1.5 cm, equivalent phase spacing 12 m, at 35 °C and 74 cm of mercury, with \(m_o = 0.85\). Find the air density factor, the surface gradient, the margin, and the fair-weather loss if any.

Solution

Step 1 — air density factor:

\[ \delta = \frac{3.92(74)}{273 + 35} = \frac{290.1}{308} = 0.942 \]

Step 2 — threshold gradient:

\[ g_{\text{crit}} = m_og_o\delta = 0.85(21.2)(0.942) = 17.0\ \text{kV/cm} \]

Step 3 — bundle radius and surface gradient:

\[ r_b = \sqrt{(1.5)(45)} = 8.22\ \text{cm}, \qquad \ln\frac{1200}{8.22} = \ln(146.0) = 4.984 \]

\[ g = \frac{231}{2(1.5)(4.984)} = \frac{231}{14.95} = 15.5\ \text{kV/cm} \]

Step 4 — the margin:

\[ \frac{17.0 - 15.5}{17.0}\times100 = 8.8\%\ \text{below threshold} \]

\[ g < g_{\text{crit}} \quad\Rightarrow\quad \textbf{no fair-weather corona, loss} = 0 \]

Step 5 — the caveats. An 8.8% margin is below the 15% normally sought. Three things would erode it:

\[ \begin{array}{ll} \text{Rain } (m_o \to 0.6) & g_{\text{crit}} \to 12.0\ \text{kV/cm — corona, and heavy} \\ \text{A hotter day } (45\,^\circ\text{C}) & \delta \to 0.912,\ g_{\text{crit}} \to 16.5 — \text{margin 6\%} \\ \text{A 5\% overvoltage} & g \to 16.3 — \text{margin 4\%} \end{array} \]

The design is acceptable in fair weather and marginal against the combination. A three-conductor bundle, or 1.8 cm sub-conductors, would restore the margin.

Zero fair-weather loss is the correct design outcome, and it is why corona rarely appears in a loss budget. A properly designed EHV line has no corona at all in fair weather — the phenomenon exists in its life only during rain, during switching overvoltages, and wherever hardware presents a sharp edge. That is a very different kind of engineering from minimising \(I^2R\): the target is not a small number but a threshold never crossed, and every part of the line's geometry that Sets 5 to 7 computed exists partly to keep it uncrossed.

Answer\(\delta = 0.942\), \(g = 15.5\) against \(g_{\text{crit}} = 17.0\) kV/cm — no fair-weather corona, but only an 8.8% margin

Practice

Practice Problems

Work each on paper before opening the answer. The answer is given so you can check yourself; the method is deliberately not.

P1. Find \(\delta\) at 70 cm Hg and 30 °C.

Show answer
\(3.92(70)/303 = 274.4/303 = \mathbf{0.906}\).
P2. A conductor of radius 1 cm sits 150 cm from its neighbours, \(m_o = 0.9\), \(\delta = 1\). Find \(V_c\) per phase.

Show answer
\(0.9(21.2)(1)(1)\ln(150) = 19.08(5.011) = \mathbf{95.6}\) kV/phase.
P3. Why is \(g_o\) taken as 21.2 and not 30 kV/cm?

Show answer
30 kV/cm is the peak breakdown strength; the formulas work in r.m.s., so \(30/\sqrt2 = 21.2\). Using 30 overstates \(V_c\) by 41%.
P4. A line has \(V_c = 110\) kV/phase and operates at 132 kV line voltage. Is it in corona?

Show answer
\(V_{ph} = 132/\sqrt3 = 76.2\) kV \(< 110\). No corona — with a comfortable 31% margin.
P5. Find the surface gradient for \(V_{ph} = 127\) kV, \(r = 2\) cm, \(D = 300\) cm.

Show answer
\(g = 127/[2\ln(150)] = 127/(2\times5.011) = \mathbf{12.7}\) kV/cm.
P6. A line just at threshold suffers a 20% overvoltage. If it was 2 kV over \(V_c\) at 130 kV, by what factor does the loss rise?

Show answer
New excess \(= 156 - 128 = 28\) kV against 2, so \((28/2)^2 = \mathbf{196}\) times. Near threshold the loss is violently nonlinear.
P7. A two-conductor bundle has \(r = 1.6\) cm at 40 cm. Find \(r_b\).

Show answer
\(\sqrt{(1.6)(40)} = \sqrt{64} = \mathbf{8.0}\) cm — five times the sub-conductor radius.
P8. Why does corona get worse at high altitude?

Show answer
Lower pressure lowers \(\delta\), and \(V_c \propto \delta\). Physically, longer mean free paths let electrons gain more energy before colliding, so thin air ionises more easily.
P9. A line loses 0.5 kW/km/phase to corona. Find the annual energy for 200 km, three phases.

Show answer
\(3(0.5)(200) = 300\) kW; \(\times 8760 = \mathbf{2.63}\) GWh a year. Corona runs continuously, independent of load.
P10. Which reduces corona more — doubling the phase spacing or increasing the radius by 30%?

Show answer
The radius: about 23% against 14%, because \(r\) appears both as a multiplier and inside the logarithm while \(d\) appears only inside it — Problem 16.
P11. Why is radio interference often a tighter constraint than corona loss?

Show answer
It begins at a lower gradient than measurable loss, and its limits are legally binding rather than merely expensive — Problem 17.
P12. A student computes \(V_c = 210\) kV/phase for a conductor of 1 cm radius at 1 m spacing. What went wrong?

Show answer
Roughly \(\sqrt2\) times the correct 145 kV (at \(m_o = \delta = 1\)): 30 kV/cm was used instead of 21.2.

Challenge

Challenge Problems

Each of these needs an idea rather than a formula. Decide what the governing principle is before opening the answer.

C1. Corona is the only loss mechanism in this book with a hard threshold. Trace the consequences of that fact for design method, for economics and for the way corona is specified, and contrast it throughout with \(I^2R\) loss.

Show answer
Design method. \(I^2R\) is minimised — Kelvin's law of Set 2 balances it against capital cost and finds an optimum. Corona is avoided — there is no optimum, only a threshold, and the design target is a margin below it. No Kelvin-type trade-off exists because the loss is zero on one side of the boundary and unbounded on the other.

Economics. \(I^2R\) scales with \(I^2\) and therefore with delivered energy, so it is a cost proportional to revenue. Corona scales with \((V - V_c)^2\) and \(V\) is held constant, so it is a fixed cost incurred whether or not the line is earning — Problem 18 found it the only loss at no load. In loss-load-factor terms corona has a loss load factor of 1.0, which no other mechanism approaches.

Specification. \(I^2R\) is specified as an efficiency or a resistance. Corona is specified as a gradient limit (16–17 kV/cm), an RI limit and an audible noise limit — three pass/fail criteria rather than a quantity to report.

Sensitivity. A 10% current error changes \(I^2R\) by 21%. A 10% voltage error changed the corona loss by a factor of 12 in Problem 15, because the difference rather than the voltage is squared. This is why corona calculations demand margins that resistive calculations do not.

Uncertainty. Resistance is known to a fraction of a per cent. \(V_c\) depends on \(m_o\), which swings 15% with surface condition (Problem 13) and collapses to 0.6 in rain (Problem 14). The input uncertainty exceeds any refinement of the arithmetic — which is the deepest reason corona is handled by margin rather than by calculation.

The unifying point: a threshold phenomenon with large parameter uncertainty and violent super-threshold behaviour can only be engineered by staying well clear of it. Everything about EHV conductor selection follows from that.
C2. Problem 12 found enlargement cheaper than bundling at 220 kV, and Problem 19 found the reverse at 400 kV. Derive the scaling that explains the crossover and locate it.

Show answer
The requirement. To hold \(g = V_{ph}/[nr\ln(D/r_{\text{eff}})]\) at a fixed target while \(V_{ph}\) rises, the product \(nr\ln(D/r_{\text{eff}})\) must rise in proportion to \(V_{ph}\).

Route A — enlarge the single conductor. Here \(n = 1\) and \(r_{\text{eff}} = r\), so we need \(r\ln(D/r) \propto V\). The logarithm varies slowly, so to leading order \(r \propto V\) and the metal
\[ A = \pi r^2 \propto V^2 \]
Route B — bundle. Fix the sub-conductor at radius \(r_0\) and vary \(n\). Then \(nr_0\ln(D/r_b) \propto V\), and since \(r_b\) grows with \(n\) the logarithm shrinks — so \(n\) must grow somewhat faster than \(V\). Metal is
\[ A = n\pi r_0^2 \propto V^{1+\epsilon}, \qquad \epsilon \approx 0.2\text{–}0.3 \]
The crossover. Route A costs \(V^2\) and Route B costs roughly \(V^{1.25}\). Bundling must eventually win, and the crossover is where the two curves meet. From the two worked cases:
— 220 kV: 11.9 cm² single against 14.1 cm² bundled — enlargement wins.
— 400 kV: 18.9 cm² single against 14.1 cm² bundled — bundling wins.

Interpolating on the \(V^2\) versus \(V^{1.25}\) laws puts the crossover near \(\mathbf{280\text{–}300\ \text{kV}}\), which matches practice: 220 kV lines are commonly single-conductor, 400 kV lines never are.

Two effects the metal comparison misses, both favouring bundling further: the reactance and SIL gains of Set 7, worth 36% in transfer capability; and manufacturability, since Route A at 400 kV demands a 4.9 cm-diameter conductor that no standard ACSR range offers. Include those and the practical crossover falls below 250 kV.
C3. Corona sets a minimum conductor surface at high voltage while conductor volume economics (Set 2) argues for high voltage. Examine how these two forces interact, and explain what actually terminates the voltage ladder.

Show answer
The two forces. Set 2 Problem 1 found conductor volume \(\propto 1/V^2\) — a powerful argument for raising the voltage. This set finds that the surface gradient \(g \propto V\) for fixed geometry, so corona worsens linearly with voltage and the required conductor surface must grow to match.

How they interact. They act on different quantities, and this is the key point:
— Set 2's argument concerns the conductor's cross-sectional area, set by the current the line must carry with acceptable loss.
— This set's constraint concerns the conductor's surface, set by the field the air must withstand.

Below about 220 kV the current requirement dominates and the conductor is sized for area; corona is satisfied incidentally. Above roughly 300 kV the surface requirement dominates, and lines carry more metal than the current needs purely to present enough surface. A 765 kV line's four-conductor bundle is thermally over-provisioned by a wide margin.

Why bundling resolves the conflict. A bundle decouples surface from cross-section: four sub-conductors of area \(A/4\) present twice the surface of one conductor of area \(A\), since surface goes as \(r\) and area as \(r^2\). This is why the \(1/V^2\) economy of Set 2 could be pursued to 765 kV at all — without bundling it would have stalled around 250 kV against a wall of unmanufacturable conductors.

What actually terminates the ladder. Not corona, which bundling defeats, and not conductor economics, which continues to improve. The binding limits are:
— Insulation and clearances, which grow with \(V\) and drive tower size, right-of-way width and cost — the \(BV\) term of Set 2 Challenge C1.
— Switchgear, whose cost rises steeply and in discrete steps per voltage class.
— Switching overvoltages, which scale with \(V\) and eventually govern insulation rather than lightning does.
— Demand. Above 765 kV a single circuit carries over 2 GW (Set 7 Problem 11), more than most corridors need, so the incremental case weakens.
— HVDC, which above roughly 600 km becomes cheaper and has no corona-driven bundling requirement of the same severity, no charging current, and no stability limit.

The ladder stopped at 765–1150 kV not because a.c. physics forbade more, but because HVDC overtook it for the long distances that alone would justify more.

Self-Test

Multiple-Choice Questions

MCQ 1. The r.m.s. dielectric strength of air used in corona formulas is:
(a) 30 kV/cm (b) 21.2 kV/cm (c) 42.4 kV/cm (d) 15 kV/cm

Show answer
(b). 30 kV/cm is the peak value; \(30/\sqrt2 = 21.2\).
MCQ 2. The air density factor at 76 cm Hg and 25 °C is:
(a) 0.5 (b) 1.0 (c) 1.5 (d) 3.92

Show answer
(b) 1.0. The constant 3.92 is \(298/76\), chosen to normalise exactly this way.
MCQ 3. At high altitude the critical disruptive voltage:
(a) rises (b) falls (c) is unchanged (d) depends on the conductor

Show answer
(b) falls. Thin air has longer mean free paths, so it ionises more easily — Problem 4.
MCQ 4. The visual critical voltage is:
(a) below \(V_c\) (b) equal to \(V_c\) (c) above \(V_c\) (d) unrelated to \(V_c\)

Show answer
(c) above. Corona begins electrically before it can be seen — a line showing no glow may still be losing power.
MCQ 5. The irregularity factor for a stranded conductor is about:
(a) 1.0 (b) 0.85 (c) 1.15 (d) 0.3

Show answer
(b) 0.85. Polished is 1.0; in rain it falls to about 0.6 — Problems 13 and 14.
MCQ 6. Corona loss varies with the excess voltage as:
(a) \((V-V_c)\) (b) \((V-V_c)^2\) (c) \(V^2\) (d) \(\sqrt{V-V_c}\)

Show answer
(b). The square of the excess, not of the voltage — which is why a line near threshold is so sensitive to overvoltage.
MCQ 7. The surface voltage gradient of a conductor is:
(a) \(V_{ph}/r\) (b) \(V_{ph}\ln(D/r)\) (c) \(V_{ph}/[r\ln(D/r)]\) (d) \(V_{ph}r/D\)

Show answer
(c). This is the physical quantity; \(V_c\) is simply the same criterion rearranged — Problems 9 and 10.
MCQ 8. Bundling reduces the surface gradient chiefly by:
(a) reducing the voltage (b) sharing the charge over more surface and raising the effective radius (c) increasing the spacing (d) reducing the current

Show answer
(b). Both mechanisms act together — Problem 11 found a 23% reduction.
MCQ 9. In rain, corona loss typically:
(a) falls (b) is unchanged (c) rises by tens of times (d) rises by about 10%

Show answer
(c). Droplets are sharp conducting protrusions; Problem 14 found a factor of about 65.
MCQ 10. At no load, the corona loss of a line compared with its \(I^2R\) loss is:
(a) smaller (b) equal (c) the only loss present (d) zero

Show answer
(c). Corona depends on voltage, not current, so it runs unchanged with the line unloaded — Problem 18.
MCQ 11. To reduce corona, the most effective single measure is to:
(a) widen the phase spacing (b) increase the conductor radius or bundle (c) lower the frequency (d) raise the towers

Show answer
(b). \(r\) acts both as a multiplier and inside the logarithm; \(d\) acts only inside it — Problem 16.
MCQ 12. Radio interference from a line originates in:
(a) the fundamental current (b) nanosecond corona streamer pulses (c) transformer harmonics (d) conductor vibration

Show answer
(b). Their fast rise time gives components up to 30 MHz, radiated by the conductor — Problem 17.

Reference

Key Formulas

Quantity	Relation	Notes
Breakdown strength of air	\(g_o = 30/\sqrt2 = 21.2\) kV/cm	r.m.s.; 30 kV/cm is the peak
Air density factor	\(\delta = 3.92b/(273+t)\)	\(b\) in cm Hg, \(t\) in °C; unity at 76, 25
Irregularity factor	\(m_o\): 1.0 polished, 0.93 weathered, 0.85 stranded, 0.6 wet	The largest single uncertainty
Critical disruptive voltage	\(V_c = m_og_o\delta r\ln(d/r)\) kV/phase	r.m.s.; hard threshold
Visual critical voltage	\(V_v = m_vg_o\delta r\big(1+\tfrac{0.3}{\sqrt{\delta r}}\big)\ln(d/r)\)	Always above \(V_c\)
Surface gradient	\(g = V_{ph}/[r\ln(D/r)]\) kV/cm	The physical criterion
Bundled gradient	\(g = V_{ph}/[n\,r\ln(D/r_b)]\)	\(n\) sub-conductors, bundle radius \(r_b\)
Corona criterion	\(g > m_og_o\delta\)	Equivalent to \(V_{ph} > V_c\)
Peek's loss formula	\(P = 242.2\left(\dfrac{f+25}{\delta}\right)\sqrt{\dfrac{r}{d}}(V-V_c)^2\times10^{-5}\)	kW/km/phase; empirical, \(\pm20\%\)
Three-phase loss	\(P_{3\phi} = 3P\)	Runs continuously, load-independent
Design gradient limit	\(g \le 16\)–\(17\) kV/cm at EHV	Set by RI and noise, not loss
Design margin	10–15% below threshold	Covers \(m_o\), \(\delta\) and overvoltage
Overvoltage sensitivity	\(P \propto (V-V_c)^2\)	10% overvoltage gave 12× in Problem 15
Foul weather	\(m_o \to 0.6\), loss \(\times\) tens	Dominates the annual energy

Diagnostics

Common Mistakes

Using 30 kV/cm instead of 21.2. The most frequent error in the topic — a 41% overstatement of every voltage — Problem 1.
Comparing \(V_c\) with the line voltage. \(V_c\) is per phase; the operating voltage must be divided by \(\sqrt3\) before comparison — Problem 6.
Mixing units between \(r\) and \(d\). They appear as a ratio, so any unit will do — but the same one, and \(r\) alone also appears as a multiplier in centimetres.
Using the diameter where the radius is meant. A factor of two, both as multiplier and inside the logarithm.
Taking \(m_o = 1\) for a real line. Real conductors are stranded and weather; 0.85 is the design value, 0.6 in rain — Problem 13.
Forgetting that \(\delta\) falls with both altitude and temperature. The worst case is a hot day at altitude, and it must be the design case — Problem 4.
Squaring the voltage rather than the excess in Peek's formula. It is \((V-V_c)^2\), and near threshold the difference is enormous — Problem 15.
Assuming no visible glow means no corona. The visual threshold is above the disruptive one — Problems 5 and 6.
Quoting Peek's formula per phase as though it were the three-phase total. Multiply by three — Problem 7.
Treating corona loss as load-dependent. It depends on voltage alone and is the only loss present at no load — Problem 18.
Widening the spacing to fix a corona problem. Weak, expensive, and it worsens both line parameters — Problem 16.
Designing to zero margin. With \(m_o\) uncertain by 15% and \(\delta\) by 8%, a margin below 10% is not a design — Problems 6 and 20.

Looking Ahead

The line's parameters are now complete and their limits understood. A conductor radius, a bundle arrangement and three spacings give a series impedance, a shunt admittance, a surge impedance and a corona margin — everything that distinguishes one line from another.

From Set 9 the parameters stop being derived and start being used. The question becomes what the line does with them: how much voltage it drops, how its regulation and efficiency behave, and at what length the lumped model breaks down. Set 9 begins with the short line, which discards the capacitance of Sets 6 and 7 altogether; Set 11 restores it as a lumped \(\pi\); and Set 12 abandons lumping entirely for the wave equation, where the 6000 km wavelength of Set 6 Problem 16 finally governs.