Capacitance of Bundled Conductors

Set 7 — Capacitance of Bundled Conductors

Twenty worked problems on the single most consequential piece of EHV line design. Sets 5 and 6 introduced bundling twice, once for each parameter; this set treats it properly — as one geometric change that lowers inductance and raises capacitance simultaneously, and therefore attacks the surge impedance \(\sqrt{L/C}\) from both sides. The result is a line that carries substantially more power on barely more metal, and the arithmetic that establishes it is worth doing carefully.

Textbook Chapter 8 · 20 solved · 12 practice · 3 challenge · 12 MCQs

i Method Recap

Two bundle radii, not one. For inductance, \(D_{sb}\) is built from the sub-conductor's GMR; for capacitance, \(r_b\) is built from its actual radius. The construction is identical — only the starting quantity differs — and the two differ by roughly \(\sqrt{0.7788} = 0.88\) for a two-conductor bundle.
The constructions. Two sub-conductors: \(\sqrt{D_sd}\). Three at the corners of a triangle: \(\sqrt[3]{D_sd^2}\). Four on a square: \(1.09\sqrt[4]{D_sd^3}\), the 1.09 being \(2^{1/8}\) from the diagonals. Replace \(D_s\) by \(r\) throughout for capacitance.
Both parameters move the helpful way. A bigger effective radius shrinks \(\ln(D_m/D_s)\), which lowers \(L\) because the logarithm is in the numerator, and raises \(C\) because it is in the denominator. There is no trade-off to balance here.
Surge impedance. \(Z_c = \sqrt{L/C}\), around 400 \(\Omega\) for a single-conductor line and 250–300 \(\Omega\) when bundled. It is a pure resistance for a lossless line and depends only on geometry, never on length.
Surge impedance loading. \(\text{SIL} = V_L^2/Z_c\) — the three-phase power at which the line's reactive generation exactly balances its reactive absorption, so the voltage profile is flat. Bundling raises SIL in proportion to \(1/Z_c\).
Bundle spacing is not a free parameter. It is set by the spacer hardware, by short-circuit forces between sub-conductors, and by wind response — typically 0.3 to 0.5 m. The electrical gain from wider spacing is logarithmic and the mechanical cost is not.
Check the answer against the bands. Bundled EHV lines land near \(L \approx 0.9\)–\(1.05\) mH/km, \(C \approx 0.011\)–\(0.013\) µF/km, \(Z_c \approx 260\)–\(300\ \Omega\). A surge impedance outside 200–450 \(\Omega\) means an error in one of the radii.

VideoWalkthrough

Problem 1Warm-upTwo Bundle Radii

A two-conductor bundle has sub-conductors of outside radius 1.5 cm and GMR 1.22 cm, spaced 0.45 m apart. Find the bundle radius to be used for inductance and the one to be used for capacitance, and explain why they differ.

Solution

For inductance, the construction starts from the sub-conductor's GMR:

\[ D_{sb} = \sqrt{D_s\,d} = \sqrt{(0.0122)(0.45)} = \sqrt{0.005490} = 0.0741\ \text{m} \]

For capacitance, the identical construction starts from the actual radius:

\[ r_b = \sqrt{r\,d} = \sqrt{(0.015)(0.45)} = \sqrt{0.006750} = 0.0822\ \text{m} \]

Why they differ. The formulas are the same; only the seed quantity differs, for exactly the reason established in Set 6 Problem 2 — magnetic flux penetrates the conductor and electric flux does not. The ratio between them follows the square root of the ratio of the seeds:

\[ \frac{D_{sb}}{r_b} = \sqrt{\frac{D_s}{r}} = \sqrt{\frac{0.0122}{0.015}} = \sqrt{0.813} = 0.902 \]

For a solid round sub-conductor, where \(D_s = 0.7788r\) exactly, the ratio would be \(\sqrt{0.7788} = 0.883\). Here it is 0.902 because a stranded ACSR sub-conductor has a GMR slightly above \(0.7788r\).

The square root halves the discrepancy, which is why the error survives so long undetected. Using the wrong seed for a single conductor gives a 22% error in the radius; for a two-conductor bundle it gives only 10%, and inside a logarithm that becomes about 2% in the parameter. Small enough to look like rounding, large enough to matter in a surge impedance. The discipline is simply to carry two columns — GMR for \(L\), radius for \(C\) — from the data sheet onwards, and never to compute one from the other mid-calculation.

Answer\(D_{sb} = 0.0741\) m for inductance, \(r_b = 0.0822\) m for capacitance

Problem 2Exam level345 kV Line

A 345 kV line uses the bundle of Problem 1, the bundle centres arranged horizontally with 8 m between adjacent phases and the line transposed. Find the equivalent spacing, the inductance per phase per km and the inductive reactance at 50 Hz.

Solution

Horizontal arrangement at 8 m adjacent makes the outer pair 16 m apart:

\[ D_{eq} = \sqrt[3]{(8)(8)(16)} = \sqrt[3]{1024} = 10.08\ \text{m} \]

Note that \(D_{eq}\) is the geometric mean of the distances between bundle centres. The sub-conductors within a bundle are already accounted for in \(D_{sb}\) and must not appear again here.

Inductance per phase, using the inductive bundle radius:

\[ L = 0.2\ln\frac{D_{eq}}{D_{sb}} = 0.2\ln\frac{10.08}{0.0741} = 0.2\ln(136.0) \]

\[ = 0.2(4.9127) = 0.983\ \text{mH/km} \]

Reactance at 50 Hz:

\[ X_L = 2\pi fL = 314.16(0.983\times10^{-3}) = 0.309\ \Omega/\text{km} \]

The value sits at the low end of the 0.3–0.45 \(\Omega\)/km band, which is exactly what a bundled line should do.

Bundle centres for \(D_{eq}\), sub-conductors for \(D_{sb}\) — the separation of scales is what makes the calculation tractable. Because 8 m is so much larger than 0.45 m, treating each bundle as a single object at its centre introduces a negligible error. Were the bundle spacing comparable with the phase spacing the approximation would fail and every sub-conductor would have to be handled individually, as in the composite-conductor calculation of Set 5 Problem 6.

Answer\(D_{eq} = 10.08\) m, \(L = 0.983\) mH/km, \(X_L = 0.309\ \Omega\)/km

Problem 3Warm-upPer-Unit Reactance

The line of Problem 2 is 160 km long. Find its total series reactance and express it in per-unit on a base of 100 MVA, 345 kV.

Solution

Total reactance:

\[ X = 0.309 \times 160 = 49.4\ \Omega \]

Base impedance of the 345 kV zone:

\[ Z_B = \frac{(345)^{2}}{100} = \frac{119\,025}{100} = 1190\ \Omega \]

Per-unit reactance:

\[ X_{pu} = \frac{49.4}{1190} = 0.0415\ \text{p.u.} \]

Roughly 0.026 p.u. per hundred kilometres at 345 kV — a useful figure to carry, and about a third of the 0.083 p.u. per hundred kilometres that Set 5 Problem 19 found at 220 kV.

Two separate effects made this line electrically short, and it is worth keeping them apart. Bundling cut the ohms per kilometre by 27%; raising the voltage from 220 to 345 kV cut the per-unit value by a further factor of \((345/220)^2 = 2.46\) through the base impedance. The second effect is much the larger, and it is the same \(1/V^2\) that governed conductor volume in Set 2. Bundling is what makes EHV construction possible; the voltage is what makes it worthwhile.

Answer\(X = 49.4\ \Omega = 0.0415\) p.u. on 100 MVA, 345 kV

Problem 4Exam levelCapacitance

Find the capacitance to neutral per km and the susceptance of the line of Problem 2, and compare with the same line using a single conductor of radius 1.5 cm.

Solution

Bundled, using the capacitive bundle radius \(r_b = 0.0822\) m:

\[ C_n = \frac{2\pi\varepsilon_0}{\ln(D_{eq}/r_b)} = \frac{5.5632\times10^{-11}}{\ln(10.08/0.0822)} = \frac{5.5632\times10^{-11}}{\ln(122.6)} \]

\[ = \frac{5.5632\times10^{-11}}{4.8090} = 1.157\times10^{-11}\ \text{F/m} = 0.01157\ \mu\text{F/km} \]

Susceptance:

\[ B = 2\pi fC_n = 314.16(1.157\times10^{-8}) = 3.634\times10^{-6}\ \text{S/km} \]

Single conductor, same geometry, \(r = 0.015\) m:

\[ C_n = \frac{5.5632\times10^{-11}}{\ln(10.08/0.015)} = \frac{5.5632\times10^{-11}}{\ln(672.0)} = \frac{5.5632\times10^{-11}}{6.5103} = 8.545\times10^{-12}\ \text{F/m} \]

The increase from bundling:

\[ \frac{11.57 - 8.545}{8.545}\times100 = 35.4\% \]

Set the two results side by side: 27% less inductance and 35% more capacitance from one change. Nothing in line design normally moves two parameters in helpful directions at once — raising the voltage helps the per-unit reactance but worsens corona; widening the spacing helps corona but worsens inductance. Bundling is the exception, and Problems 7 to 9 show that its effect on the surge impedance is therefore the product of both gains rather than either alone.

Answer\(C_n = 0.01157\ \mu\text{F}\)/km, \(B = 3.634\times10^{-6}\) S/km — 35.4% above the single conductor

Problem 5Challenge-liteChoosing the Spacing

Explain what determines the bundle spacing \(d\) in practice, given that the electrical benefit increases without limit as \(d\) grows.

Solution

The electrical gain is logarithmic and therefore weak. For a two-conductor bundle:

\[ L = 0.2\ln\frac{D_{eq}}{\sqrt{D_sd}} = 0.2\left[\ln\frac{D_{eq}}{\sqrt{D_s}} - \tfrac12\ln d\right] \]

Doubling \(d\) removes only \(0.1\ln2 = 0.069\) mH/km — about 7% — as Problem 13 confirms numerically.

What limits it:

\[ \begin{array}{ll} \text{Spacer hardware} & \text{Mass and cost rise with } d; \text{ spacers every 50–70 m} \\ \text{Short-circuit forces} & \text{Attraction between sub-conductors, } \propto 1/d \\ \text{Wind and galloping} & \text{Aerodynamic coupling worsens with wide bundles} \\ \text{Ice and wind loading} & \text{Effective frontal area grows} \\ \text{Corona} & \text{Wider is better — the one force pushing } d \text{ up} \end{array} \]

Where the balance falls. Practical bundle spacings lie between 0.30 and 0.50 m almost universally, with 0.45 m the most common choice. The narrowness of that range across voltages from 220 kV to 765 kV is itself informative: the mechanical constraints scale with the hardware rather than with the voltage, while the electrical benefit is too weak a function of \(d\) to argue with them.

Note that the short-circuit force works against wide spacing only in the sense that a wider bundle needs stronger spacers to resist collapse; the force itself falls as \(1/d\), so a wider bundle experiences less attraction but has further to travel before the sub-conductors clash. Problem 17 quantifies it.

When a design variable enters through a logarithm, something else always ends up deciding it. This is the same pattern as the phase spacing in Set 5 Problem 14, which is set by insulation clearance rather than by inductance, and the conductor size in Set 6 Problem 18, which is set by current rating rather than by capacitance. In line design the electrical parameters are remarkably often the passengers rather than the drivers — which is precisely why the corona limit of Set 8, being a hard physical threshold rather than a logarithm, has so much more influence over EHV geometry than any of them.

AnswerSpacer mechanics, short-circuit forces and wind loading — not the electrical gain, which is only logarithmic. Practical range 0.30–0.50 m

Problem 6Exam levelBundle Number

For the geometry of Problem 2 — \(D_{eq} = 10.08\) m, sub-conductor GMR 1.22 cm and radius 1.5 cm, bundle spacing 0.45 m — tabulate \(D_{sb}\), \(r_b\), \(L\) and \(C_n\) for one, two, three and four sub-conductors per phase.

Solution

Applying the four constructions to each seed. For inductance:

\[ D_{sb} = D_s;\ \sqrt{D_sd};\ \sqrt[3]{D_sd^{2}};\ 1.09\sqrt[4]{D_sd^{3}} \]

\[ = 0.0122;\ 0.0741;\ 0.1352;\ 0.1990\ \text{m} \]

And for capacitance, the same with \(r\) in place of \(D_s\):

\[ r_b = 0.0150;\ 0.0822;\ 0.1449;\ 0.2096\ \text{m} \]

The resulting parameters:

\[ \begin{array}{ccccc} n & D_{sb}\ (\text{m}) & L\ (\text{mH/km}) & r_b\ (\text{m}) & C_n\ (\text{pF/m}) \\ \hline 1 & 0.0122 & 1.343 & 0.0150 & 8.55 \\ 2 & 0.0741 & 0.983 & 0.0822 & 11.57 \\ 3 & 0.1352 & 0.862 & 0.1449 & 13.11 \\ 4 & 0.1990 & 0.785 & 0.2096 & 14.37 \end{array} \]

The increments tell the story better than the values. Going from one to two sub-conductors cuts \(L\) by 26.8%; two to three by a further 12.3%; three to four by 8.9%.

Diminishing returns, and the first step is worth more than the next two together. The reason is structural: \(D_{sb}\) grows as \(d^{(n-1)/n}\), which approaches \(d\) asymptotically, so no amount of sub-conductors can push the effective radius beyond the bundle spacing itself. That ceiling is why bundle numbers above four are essentially never chosen for reactance reasons — where six appear, at 765 kV and above, corona is the reason, as Set 8 explains.

Answer\(L\): 1.343, 0.983, 0.862, 0.785 mH/km · \(C_n\): 8.55, 11.57, 13.11, 14.37 pF/m

Problem 7Exam levelSurge Impedance

Find the surge impedance of the bundled line of Problems 2 and 4, and of the single-conductor line of the same geometry. Explain why it is independent of the line's length.

Solution

The surge impedance of a lossless line:

\[ Z_c = \sqrt{\frac{L}{C}} \]

Both must be in the same length unit, and the units then cancel — which is the first hint that length will not appear in the answer.

Bundled: \(L = 0.983\times10^{-3}\) H/km, \(C = 1.157\times10^{-8}\) F/km:

\[ Z_c = \sqrt{\frac{0.983\times10^{-3}}{1.157\times10^{-8}}} = \sqrt{84\,960} = 291\ \Omega \]

Single conductor: \(L = 1.343\times10^{-3}\), \(C = 8.545\times10^{-9}\):

\[ Z_c = \sqrt{\frac{1.343\times10^{-3}}{8.545\times10^{-9}}} = \sqrt{157\,180} = 396\ \Omega \]

Why length does not enter. \(L\) and \(C\) are both per unit length, and both scale linearly with it. Their ratio therefore has the length divided out before the square root is taken. A 10 km line and a 1000 km line of identical construction have identical surge impedance.

Both values sit in the expected bands — around 400 \(\Omega\) single, around 290 \(\Omega\) bundled.

Surge impedance is the one line parameter that is a property of the cross-section alone. Resistance, reactance and susceptance all scale with length and are therefore properties of a particular line; \(Z_c\) is a property of the design. That is what makes it the natural currency for comparing constructions, and what makes surge impedance loading — the subject of Problem 8 — a meaningful rating for a line before its length is even decided.

AnswerBundled \(Z_c = 291\ \Omega\); single conductor \(396\ \Omega\) — independent of length in both cases

Problem 8Exam levelSurge Impedance Loading

Find the surge impedance loading of the 345 kV bundled line of Problem 7, and explain what is physically special about operating a line at that loading.

Solution

Surge impedance loading is the three-phase power delivered when the line is terminated in its own surge impedance:

\[ \text{SIL} = \frac{V_L^{2}}{Z_c} = \frac{(345\times10^{3})^{2}}{291} = \frac{1.190\times10^{11}}{291} = 4.09\times10^{8}\ \text{W} \]

\[ \text{SIL} = 409\ \text{MW} \]

Note that the line voltage is used directly and no \(\sqrt3\) appears — the three factors of \(1/\sqrt3\) from the phase voltages and the three-phase summation cancel exactly, as they did for charging MVA in Set 6 Problem 10.

What is special about it. At this loading the reactive power absorbed by the series inductance exactly equals the reactive power generated by the shunt capacitance, at every point along the line:

\[ I^{2}X_L = V^{2}B \quad\text{everywhere} \]

The consequences follow immediately. The line neither absorbs nor generates net vars; the voltage magnitude is constant from end to end — a perfectly flat profile; and the line looks, to the system, like a pure resistance.

Below SIL the capacitance dominates and the line is a var source, raising the far-end voltage — the Ferranti effect. Above SIL the inductance dominates and the line is a var sink, depressing it.

SIL is the loading a line "wants" to carry, and real lines rarely oblige. A 345 kV line with a thermal rating around 1000 MW has an SIL of only 409 MW, so it spends most of its life above SIL and needs reactive support; at night it falls below and needs reactors. The whole apparatus of Set 34 — shunt reactors, capacitor banks, SVCs, tap changers — exists to manage the gap between the loading a line actually carries and the one at which it would be reactively self-sufficient.

Answer\(\text{SIL} = 409\) MW; at this loading the voltage profile is flat and net var flow is zero

Problem 9Exam levelBundling and SIL

Using the table of Problem 6, compute the surge impedance and the surge impedance loading at 345 kV for one, two, three and four sub-conductors per phase, and comment.

Solution

Taking \(Z_c = \sqrt{L/C}\) with \(L\) in H/km and \(C\) in F/km for each row:

\[ \begin{array}{ccccc} n & L\ (\text{mH/km}) & C_n\ (\text{pF/m}) & Z_c\ (\Omega) & \text{SIL}\ (\text{MW}) \\ \hline 1 & 1.343 & 8.55 & 396 & 300 \\ 2 & 0.983 & 11.57 & 291 & 409 \\ 3 & 0.862 & 13.11 & 256 & 464 \\ 4 & 0.785 & 14.37 & 234 & 509 \end{array} \]

Checking one row explicitly, \(n = 4\):

\[ Z_c = \sqrt{\frac{0.785\times10^{-3}}{1.437\times10^{-8}}} = \sqrt{54\,630} = 234\ \Omega, \qquad \text{SIL} = \frac{1.190\times10^{11}}{234} = 509\ \text{MW} \]

The gain from bundling, measured on SIL:

\[ \frac{409 - 300}{300} = 36\%\ (n=2), \qquad \frac{509 - 300}{300} = 70\%\ (n=4) \]

A 36% rise in natural loading for twice the metal is the real economic case for bundling. The reason it beats either parameter alone is that \(Z_c = \sqrt{L/C}\) compounds both gains: \(L\) fell by 27% and \(C\) rose by 35%, and \(\sqrt{0.73/1.35} = 0.735\) — a 26.5% fall in \(Z_c\), hence a 36% rise in SIL. This is why the surge impedance, rather than the reactance, is the figure of merit an EHV line designer actually optimises.

Answer\(Z_c\): 396, 291, 256, 234 \(\Omega\); SIL: 300, 409, 464, 509 MW

Problem 10Warm-up400 kV Line

A 400 kV line uses a two-conductor bundle at 0.45 m with sub-conductor GMR 1.22 cm and radius 1.5 cm, the bundles arranged horizontally at 10 m between adjacent phases. Find \(L\), \(C_n\), \(Z_c\) and SIL.

Solution

Equivalent spacing:

\[ D_{eq} = \sqrt[3]{(10)(10)(20)} = \sqrt[3]{2000} = 12.60\ \text{m} \]

Bundle radii are unchanged from Problem 1: \(D_{sb} = 0.0741\) m, \(r_b = 0.0822\) m.

\[ L = 0.2\ln\frac{12.60}{0.0741} = 0.2\ln(170.0) = 0.2(5.1358) = 1.027\ \text{mH/km} \]

\[ C_n = \frac{5.5632\times10^{-11}}{\ln(12.60/0.0822)} = \frac{5.5632\times10^{-11}}{\ln(153.3)} = \frac{5.5632\times10^{-11}}{5.0324} = 1.106\times10^{-11}\ \text{F/m} \]

Surge impedance:

\[ Z_c = \sqrt{\frac{1.027\times10^{-3}}{1.106\times10^{-8}}} = \sqrt{92\,860} = 305\ \Omega \]

Surge impedance loading:

\[ \text{SIL} = \frac{(400\times10^{3})^{2}}{305} = \frac{1.600\times10^{11}}{305} = 525\ \text{MW} \]

Wider phase spacing raised \(Z_c\) from 291 to 305 \(\Omega\), and the higher voltage still won handsomely. Going from 345 to 400 kV forced the phases apart for clearance, which worsened both parameters slightly — but SIL depends on \(V^2\) and rose from 409 to 525 MW regardless. The lesson generalises: at EHV the voltage dominates every geometric consideration, which is exactly why the industry climbed the voltage ladder rather than optimising conductor arrangements at any one rung.

Answer\(L = 1.027\) mH/km, \(C_n = 0.01106\ \mu\text{F}\)/km, \(Z_c = 305\ \Omega\), SIL \(= 525\) MW

Problem 11Exam level765 kV Line

A 765 kV line uses a four-conductor bundle on a 0.45 m square, with the same sub-conductors as before, and the bundles arranged horizontally at 14 m between adjacent phases. Find \(L\), \(C_n\), \(Z_c\) and SIL.

Solution

Equivalent spacing:

\[ D_{eq} = \sqrt[3]{(14)(14)(28)} = \sqrt[3]{5488} = 17.64\ \text{m} \]

Four-conductor bundle radii, from Problem 6:

\[ D_{sb} = 0.1990\ \text{m}, \qquad r_b = 0.2096\ \text{m} \]

Parameters:

\[ L = 0.2\ln\frac{17.64}{0.1990} = 0.2\ln(88.64) = 0.2(4.4846) = 0.897\ \text{mH/km} \]

\[ C_n = \frac{5.5632\times10^{-11}}{\ln(17.64/0.2096)} = \frac{5.5632\times10^{-11}}{\ln(84.16)} = \frac{5.5632\times10^{-11}}{4.4327} = 1.255\times10^{-11}\ \text{F/m} \]

Surge impedance:

\[ Z_c = \sqrt{\frac{0.897\times10^{-3}}{1.255\times10^{-8}}} = \sqrt{71\,470} = 267\ \Omega \]

Surge impedance loading:

\[ \text{SIL} = \frac{(765\times10^{3})^{2}}{267} = \frac{5.852\times10^{11}}{267} = 2192\ \text{MW} \]

Over two gigawatts of natural loading from one circuit — the reason 765 kV exists at all. Against the 409 MW of the 345 kV line of Problem 8, this is a factor of 5.4, obtained from a voltage ratio of 2.2 squared (4.9) and a surge impedance ratio of 291/267 (1.09). A single 765 kV circuit replaces five 345 kV circuits and one right of way replaces five, which is an argument that wins over almost any construction cost.

Answer\(L = 0.897\) mH/km, \(C_n = 0.01255\ \mu\text{F}\)/km, \(Z_c = 267\ \Omega\), SIL \(= 2192\) MW

Problem 12Challenge-liteSize vs Number

For a fixed total quantity of aluminium per phase, is it better to use one large conductor or two smaller ones in a bundle? Compare a single conductor of radius 2.12 cm with a two-conductor bundle of radius 1.5 cm each, both containing the same metal, at \(D_{eq} = 10.08\) m and \(d = 0.45\) m.

Solution

Confirming the metal is equal. Two conductors of 1.5 cm radius against one of \(1.5\sqrt2 = 2.12\) cm:

\[ 2\pi(0.015)^{2} = 1.414\times10^{-3}\ \text{m}^{2}, \qquad \pi(0.0212)^{2} = 1.412\times10^{-3}\ \text{m}^{2}\ \checkmark \]

Single large conductor. Taking \(D_s = 0.813r\) as before, \(D_s = 0.01724\) m:

\[ L = 0.2\ln\frac{10.08}{0.01724} = 0.2\ln(584.7) = 0.2(6.3712) = 1.274\ \text{mH/km} \]

\[ C_n = \frac{5.5632\times10^{-11}}{\ln(10.08/0.0212)} = \frac{5.5632\times10^{-11}}{6.1651} = 9.024\times10^{-12}\ \text{F/m} \]

\[ Z_c = \sqrt{\frac{1.274\times10^{-3}}{9.024\times10^{-9}}} = \sqrt{141\,180} = 376\ \Omega \]

Two-conductor bundle, from Problems 2, 4 and 7:

\[ L = 0.983\ \text{mH/km}, \qquad C_n = 1.157\times10^{-11}\ \text{F/m}, \qquad Z_c = 291\ \Omega \]

Comparing at 345 kV:

\[ \text{SIL}_{\text{single}} = \frac{1.190\times10^{11}}{376} = 317\ \text{MW}, \qquad \text{SIL}_{\text{bundle}} = 409\ \text{MW} \]

\[ \text{Advantage to the bundle} = 29\% \]

Same metal, same towers, 29% more natural loading — the bundle wins outright. The mechanism is the one identified in Set 5 Problem 17: a bundle's effective radius is set by the spacing, which costs nothing, while a solid conductor's is set by its radius, which costs the square of itself in metal. Splitting a conductor in two and holding the halves apart is therefore free money, and the only price is the spacer hardware of Problem 5 and the short-circuit forces of Problem 17. That is why no EHV line has ever been built with a single conductor per phase.

AnswerBundle: \(Z_c = 291\ \Omega\), SIL 409 MW. Single: \(376\ \Omega\), 317 MW — the bundle is 29% better on the same metal

Problem 13Warm-upSpacing Sensitivity

For the two-conductor bundle of Problem 2, compute the inductance at bundle spacings of 0.30 m, 0.45 m and 0.60 m, and quantify the sensitivity.

Solution

Bundle radius at each spacing, with \(D_s = 0.0122\) m:

\[ d = 0.30: \ \sqrt{0.00366} = 0.0605; \qquad d = 0.45: \ 0.0741; \qquad d = 0.60: \ \sqrt{0.00732} = 0.0856\ \text{m} \]

Inductance at each, with \(D_{eq} = 10.08\) m:

\[ \begin{array}{ccc} d\ (\text{m}) & D_{sb}\ (\text{m}) & L\ (\text{mH/km}) \\ \hline 0.30 & 0.0605 & 1.023 \\ 0.45 & 0.0741 & 0.983 \\ 0.60 & 0.0856 & 0.954 \end{array} \]

Doubling the spacing from 0.30 to 0.60 m reduces \(L\) by

\[ \frac{1.023 - 0.954}{1.023}\times100 = 6.7\% \]

The general result, from the derivation in Problem 5:

\[ \Delta L = -0.1\ln 2 = -0.069\ \text{mH/km per doubling of } d \]

Half the sensitivity of the phase spacing, because \(d\) enters under a square root before entering the logarithm.

Doubling the bundle spacing buys 7%; doubling the number of sub-conductors bought 27%. That comparison settles the design question of Problem 5 conclusively. Bundle number is the effective lever and bundle spacing is not, which is fortunate, because number is a free choice while spacing is constrained by hardware, wind and short-circuit forces. Where wider spacing is chosen, the reason is almost always corona rather than reactance.

Answer1.023, 0.983 and 0.954 mH/km — only 6.7% for a doubling of bundle spacing

Problem 14Exam levelCharging MVA

The 400 kV bundled line of Problem 10 is 250 km long. Find its total charging MVA, and express it as a fraction of its surge impedance loading.

Solution

Susceptance per km and for the whole line:

\[ B = 314.16(1.106\times10^{-8}) = 3.475\times10^{-6}\ \text{S/km} \]

\[ B_{\text{total}} = 3.475\times10^{-6}(250) = 8.688\times10^{-4}\ \text{S} \]

Charging MVA:

\[ Q_c = V_L^{2}B_{\text{total}} = (400\times10^{3})^{2}(8.688\times10^{-4}) = 139\ \text{MVAr} \]

Against the SIL of 525 MW from Problem 10:

\[ \frac{139}{525} = 0.265 \]

There is a general relation hiding here. Since \(\text{SIL} = V_L^2/Z_c\) and \(Q_c = V_L^2 B l\):

\[ \frac{Q_c}{\text{SIL}} = BlZ_c = \omega C l \sqrt{L/C} = \omega l\sqrt{LC} = \frac{\omega l}{v} = \beta l \]

where \(\beta l\) is the line's electrical length in radians. Checking: \(\beta = \omega/v = 314.16/2.97\times10^5 = 1.058\times10^{-3}\) rad/km, so \(\beta l = 0.265\) rad \(\checkmark\)

The ratio of charging to natural loading is the electrical length, and that is not a coincidence. It means a line's reactive behaviour can be read off from one number: 0.265 rad is about 15°, or 4% of a wavelength. Lines below roughly 0.1 rad behave as lumped series impedances; above about 0.5 rad the distributed treatment of Set 12 becomes unavoidable. This single dimensionless quantity is what the short, medium and long line classifications of Sets 9 to 12 are really measuring.

Answer\(Q_c = 139\) MVAr \(= 0.265 \times\) SIL, which equals the electrical length \(\beta l\) in radians

Problem 15Exam levelPower Transfer

Compare the maximum steady-state power transfer of the 160 km, 345 kV line of Problem 3 with the same line built using a single conductor, taking \(P_{max} = V_SV_R/X\) with both terminal voltages at 345 kV.

Solution

Bundled, from Problem 3:

\[ X = 49.4\ \Omega, \qquad P_{max} = \frac{(345\times10^{3})^{2}}{49.4} = \frac{1.190\times10^{11}}{49.4} = 2409\ \text{MW} \]

Single conductor, whose inductance was 1.343 mH/km:

\[ X_{\text{per km}} = 314.16(1.343\times10^{-3}) = 0.422\ \Omega/\text{km}, \qquad X = 0.422(160) = 67.5\ \Omega \]

\[ P_{max} = \frac{1.190\times10^{11}}{67.5} = 1763\ \text{MW} \]

The improvement:

\[ \frac{2409 - 1763}{1763}\times100 = 36.6\% \]

Exactly the reciprocal of the 26.8% reactance reduction, since \(P \propto 1/X\): \(1/0.732 = 1.366\).

Both figures are theoretical maxima at \(\delta = 90^\circ\), which no line is ever operated near. A practical limit at \(\delta = 30^\circ\) gives half these values, and the thermal rating usually binds first on a line this short.

Notice that the stability limit and the SIL improved by almost the same 36%, from quite different mechanisms. The stability limit improved because \(L\) alone fell; SIL improved because \(\sqrt{L/C}\) fell. That they agree so closely is a numerical accident of this geometry — but it is a useful one, because it means a single figure of merit describes a bundled line's advantage over an unbundled one, whichever limit happens to bind.

AnswerBundled 2409 MW against 1763 MW single — a 36.6% improvement

Problem 16Challenge-liteEquivalent Conductor

What single solid conductor would have the same inductance as the two-conductor bundle of Problem 1? Find its radius and the metal it would require, and comment.

Solution

The equivalent conductor must have \(D_s = D_{sb} = 0.0741\) m. For a solid round conductor, \(D_s = 0.7788r\):

\[ r_{eq} = \frac{0.0741}{0.7788} = 0.0951\ \text{m} = 9.51\ \text{cm} \]

A conductor 19 cm in diameter — roughly the size of a dinner plate in cross-section.

Metal required by the bundle:

\[ A_{\text{bundle}} = 2\pi(0.015)^{2} = 1.414\times10^{-3}\ \text{m}^{2} = 14.1\ \text{cm}^{2} \]

Metal required by the equivalent single conductor:

\[ A_{eq} = \pi(0.0951)^{2} = 2.841\times10^{-2}\ \text{m}^{2} = 284\ \text{cm}^{2} \]

The ratio:

\[ \frac{284}{14.1} = 20.1 \]

Beyond the cost, such a conductor would be unmanufacturable in continuous lengths, impossible to spool or string, and would weigh twenty times as much — requiring towers designed for a load that no transmission structure has ever been built to carry.

Twenty times the metal, and the bundle achieves it with air. This is the cleanest possible statement of what bundling does: it manufactures the electrical properties of an enormous conductor out of two ordinary ones and the space between them. A hollow conductor of the same outside diameter would work equally well electrically — and hollow conductors were indeed tried in the 1920s — but a bundle achieves the same end with standard hardware, standard conductors and a spacer, which is why it won.

Answer\(r_{eq} = 9.51\) cm, needing \(20.1\times\) the metal of the bundle

Problem 17Challenge-liteShort-Circuit Forces

A two-conductor bundle at 0.45 m spacing carries a symmetrical fault current of 40 kA in its phase. Find the force between the sub-conductors, state its direction, and comment on its significance for spacer design.

Solution

The phase current divides equally between the two sub-conductors:

\[ I_1 = I_2 = \frac{40}{2} = 20\ \text{kA} \]

The force per unit length between two parallel currents:

\[ \frac{F}{l} = \frac{\mu_0 I_1I_2}{2\pi d} = \frac{2\times10^{-7}I_1I_2}{d} \]

Substituting:

\[ \frac{F}{l} = \frac{2\times10^{-7}(2\times10^{4})^{2}}{0.45} = \frac{2\times10^{-7}(4\times10^{8})}{0.45} = \frac{80}{0.45} = 178\ \text{N/m} \]

Direction: attraction. The two sub-conductors carry current in the same direction, and parallel currents in the same direction attract. The bundle tries to collapse on itself.

To appreciate the magnitude: 178 N/m is about 18 kgf per metre of span, applied to a conductor whose own weight is perhaps 1.5 kgf/m. The electromagnetic force during a fault is an order of magnitude above gravity.

Note also that the force is proportional to \(I^2\) and therefore pulsates at twice power frequency, and that the first peak of an asymmetrical fault can reach 2.5 times the symmetrical value — giving a transient force above 1100 N/m.

Spacers exist to resist this force, and it is why they appear every 50 to 70 metres rather than once per span. Between spacers the sub-conductors can pinch together, and if they clash the arc damages both. The design case is not the steady state but the first half-cycle of a close-in fault. This is also the constraint that argues against the wide bundle spacing that corona would prefer: the force falls as \(1/d\), but the distance the conductors must be prevented from travelling grows as \(d\), and the spacer must be stiffer in absolute terms.

Answer\(F/l = 178\) N/m, attractive — about 18 kgf/m, over ten times the conductor's own weight

Problem 18Exam levelDouble Circuit

Two of the 400 kV bundled circuits of Problem 10 are strung on a common tower, the two bundles of each phase being 12 m apart, with an equivalent GMD between phases of 9 m. Find \(L\), \(C_n\), \(Z_c\) and SIL per phase.

Solution

Each phase now consists of two bundles 12 m apart. Treating the pair as a group of two, whose members are themselves bundles of effective radius \(D_{sb} = 0.0741\) m:

\[ D_{sb}^{\text{dc}} = \sqrt{D_{sb}\,D_{aa'}} = \sqrt{(0.0741)(12)} = \sqrt{0.8892} = 0.943\ \text{m} \]

Similarly for capacitance, from \(r_b = 0.0822\) m:

\[ r_b^{\text{dc}} = \sqrt{(0.0822)(12)} = \sqrt{0.9864} = 0.993\ \text{m} \]

Parameters with \(D_m = 9\) m:

\[ L = 0.2\ln\frac{9}{0.943} = 0.2\ln(9.544) = 0.2(2.2559) = 0.451\ \text{mH/km} \]

\[ C_n = \frac{5.5632\times10^{-11}}{\ln(9/0.993)} = \frac{5.5632\times10^{-11}}{\ln(9.063)} = \frac{5.5632\times10^{-11}}{2.2042} = 2.524\times10^{-11}\ \text{F/m} \]

Surge impedance and SIL:

\[ Z_c = \sqrt{\frac{0.451\times10^{-3}}{2.524\times10^{-8}}} = \sqrt{17\,870} = 134\ \Omega \]

\[ \text{SIL} = \frac{1.600\times10^{11}}{134} = 1194\ \text{MW} \]

Against the single circuit's 525 MW, the double circuit carries 2.27 times as much — more than double, for the reason found in Set 5 Problem 13.

A surge impedance of 134 \(\Omega\) is remarkably low, and it comes from the twelve-metre "bundle" that each phase has become. The two circuits are not independent lines that happen to share a tower; electromagnetically they are one line whose phases are enormously wide conductors. This is why a double-circuit outage is so much worse than losing one of two separate lines — and why the two circuits' fault currents, thermal ratings and stability limits cannot be treated as simply additive.

Answer\(L = 0.451\) mH/km, \(C_n = 0.0252\ \mu\text{F}\)/km, \(Z_c = 134\ \Omega\), SIL \(= 1194\) MW

Problem 19Warm-upVelocity and Wavelength

For the bundled 345 kV line of Problems 2 and 4, find the velocity of propagation, the wavelength at 50 Hz, and the phase constant \(\beta\).

Solution

With \(L = 0.983\times10^{-3}\) H/km and \(C = 1.157\times10^{-8}\) F/km:

\[ LC = (0.983\times10^{-3})(1.157\times10^{-8}) = 1.137\times10^{-11}\ \text{s}^{2}/\text{km}^{2} \]

\[ \sqrt{LC} = 3.372\times10^{-6}\ \text{s/km} \]

Velocity of propagation:

\[ v = \frac{1}{\sqrt{LC}} = \frac{1}{3.372\times10^{-6}} = 2.966\times10^{5}\ \text{km/s} = 2.966\times10^{8}\ \text{m/s} \]

That is \(0.989c\) — the shortfall being the internal inductance, exactly as Set 6 Problem 16 predicted.

Wavelength at 50 Hz:

\[ \lambda = \frac{v}{f} = \frac{2.966\times10^{5}}{50} = 5932\ \text{km} \]

Phase constant:

\[ \beta = \omega\sqrt{LC} = 314.16(3.372\times10^{-6}) = 1.059\times10^{-3}\ \text{rad/km} \]

Equivalently \(\beta = 2\pi/\lambda = 6.2832/5932 = 1.059\times10^{-3}\) rad/km \(\checkmark\)

The 160 km line of Problem 3 is therefore \(\beta l = 0.169\) rad, or 9.7° — under 3% of a wavelength.

Bundling changed \(L\) and \(C\) substantially and barely touched the velocity. That follows directly from Set 6 Problem 16: the logarithms nearly cancel in the product \(LC\), so any change that raises \(C\) while lowering \(L\) leaves the product almost fixed. The velocity is a property of the medium — air — and no rearrangement of conductors within it will move it far from \(c\). Surge impedance, being the ratio rather than the product, is entirely at the designer's disposal; velocity is not.

Answer\(v = 2.966\times10^8\) m/s \(= 0.989c\), \(\lambda = 5932\) km, \(\beta = 1.059\times10^{-3}\) rad/km

Problem 20Exam levelComplete Parameters

Assemble the complete parameter set for a 400 kV, 300 km line using the twin bundle of Problem 10: series impedance and shunt admittance in ohms and siemens, both in per-unit on 100 MVA, together with \(Z_c\), SIL and electrical length. Take the conductor resistance as 0.032 \(\Omega\)/km per sub-conductor.

Solution

Resistance. Two sub-conductors in parallel per phase halve it:

\[ R = \frac{0.032}{2} = 0.016\ \Omega/\text{km}, \qquad R_{\text{total}} = 0.016(300) = 4.8\ \Omega \]

Series reactance, from Problem 10's \(L = 1.027\) mH/km:

\[ X = 314.16(1.027\times10^{-3}) = 0.323\ \Omega/\text{km}, \qquad X_{\text{total}} = 96.8\ \Omega \]

\[ Z = 4.8 + j96.8\ \Omega, \qquad \frac{X}{R} = 20.2 \]

Shunt admittance, from \(B = 3.475\times10^{-6}\) S/km:

\[ Y = j(3.475\times10^{-6})(300) = j1.043\times10^{-3}\ \text{S} \]

Per-unit on 100 MVA, 400 kV, where \(Z_B = 1600\ \Omega\) and \(Y_B = 6.25\times10^{-4}\) S:

\[ Z_{pu} = \frac{4.8 + j96.8}{1600} = 0.0030 + j0.0605\ \text{p.u.} \]

\[ Y_{pu} = \frac{j1.043\times10^{-3}}{6.25\times10^{-4}} = j1.669\ \text{p.u.} \]

Derived quantities, from Problem 10 and the method of Problem 14:

\[ Z_c = 305\ \Omega, \qquad \text{SIL} = 525\ \text{MW}, \qquad \beta l = \frac{Q_c}{\text{SIL}} = \frac{(400\times10^3)^2(1.043\times10^{-3})}{525\times10^6} = 0.318\ \text{rad} \]

0.318 rad is 18.2°, or about 5% of a wavelength — squarely in medium-line territory.

Every number the rest of this book needs about this line is now on one page. \(Z_{pu}\) and \(Y_{pu}\) go into the \(Y\)-bus of Set 16 and the load flows of Sets 19 and 20; \(Z_c\) and \(\beta l\) into the wave equation of Set 12; SIL into the stability and compensation studies of Sets 24 and 34; and \(X/R = 20\) into the fault calculations of Set 21, where it justifies discarding the resistance. All of it descends from a conductor radius, a bundle spacing and three phase spacings.

Answer\(Z = 4.8 + j96.8\ \Omega = 0.0030 + j0.0605\) p.u.; \(Y = j1.043\times10^{-3}\) S \(= j1.669\) p.u.; \(Z_c = 305\ \Omega\), SIL 525 MW, \(\beta l = 0.318\) rad

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. A two-conductor bundle has sub-conductor GMR 1.4 cm and radius 1.7 cm at 0.40 m spacing. Find both bundle radii.

Show answer
\(D_{sb} = \sqrt{(0.014)(0.40)} = \mathbf{0.0748}\) m; \(r_b = \sqrt{(0.017)(0.40)} = \mathbf{0.0825}\) m.
P2. That bundle sits at \(D_{eq} = 11\) m. Find \(L\) and \(C_n\).

Show answer
\(L = 0.2\ln(147.1) = \mathbf{0.999}\) mH/km; \(C_n = 5.5632\times10^{-11}/\ln(133.3) = \mathbf{11.35}\) pF/m.
P3. Find the surge impedance of that line.

Show answer
\(\sqrt{(0.999\times10^{-3})/(1.135\times10^{-8})} = \sqrt{88\,020} = \mathbf{297}\ \Omega\).
P4. Find its SIL at 400 kV.

Show answer
\((400\times10^3)^2/297 = \mathbf{539}\) MW.
P5. A three-conductor bundle has sub-conductor GMR 1.2 cm at 0.45 m spacing. Find \(D_{sb}\).

Show answer
\(\sqrt[3]{(0.012)(0.45)^2} = \sqrt[3]{0.00243} = \mathbf{0.1345}\) m.
P6. Why does the four-conductor bundle formula carry a factor 1.09 when the three-conductor one carries none?

Show answer
The square has four diagonals of length \(\sqrt2 d\) among its sixteen distances, contributing \(2^{4/16} \cdot\) — that is \(2^{1/8} = \mathbf{1.09}\). An equilateral triangle has all mutual distances equal, so nothing extra appears.
P7. A line has \(Z_c = 260\ \Omega\). Find its SIL at 765 kV.

Show answer
\((765\times10^3)^2/260 = \mathbf{2251}\) MW.
P8. A line operates at half its SIL. Is it absorbing or generating vars, and does the far-end voltage rise or fall?

Show answer
Generating — capacitance dominates below SIL — so the far-end voltage rises. This is the Ferranti effect.
P9. A two-conductor bundle at 0.5 m carries 50 kA of fault current in its phase. Find the force between sub-conductors.

Show answer
Each carries 25 kA. \(F/l = 2\times10^{-7}(25\,000)^2/0.5 = \mathbf{250}\) N/m, attractive.
P10. A 300 km line has \(\beta = 1.06\times10^{-3}\) rad/km. Find its electrical length in degrees and as a fraction of a wavelength.

Show answer
\(\beta l = 0.318\) rad \(= \mathbf{18.2^\circ}\), or \(0.318/2\pi = \mathbf{5.1\%}\) of a wavelength.
P11. Why does bundling barely change the velocity of propagation?

Show answer
\(v = 1/\sqrt{LC}\) and the logarithms nearly cancel in the product, so \(LC \approx \mu_0\varepsilon_0\) regardless. Only the ratio \(L/C\) — the surge impedance — is at the designer's disposal.
P12. A calculation gives \(Z_c = 850\ \Omega\) for an overhead line. What is wrong?

Show answer
Far above the 200–450 \(\Omega\) band. Almost certainly the bundle radius was omitted — the sub-conductor's own radius used in place of \(D_{sb}\) and \(r_b\), raising \(L\) and lowering \(C\) together.

Challenge

Challenge Problems

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

C1. Show that the ratio of a line's charging MVA to its surge impedance loading equals its electrical length \(\beta l\) in radians, and explain why this makes \(\beta l\) the correct criterion for classifying lines as short, medium or long.

Show answer
The derivation (also given in Problem 14):
\[ \frac{Q_c}{\text{SIL}} = \frac{V_L^2 Bl}{V_L^2/Z_c} = BlZ_c = \omega Cl\sqrt{\frac{L}{C}} = \omega l\sqrt{LC} = \frac{\omega l}{v} = \beta l \]
Why it is the right criterion. Every voltage or length that could have made the comparison arbitrary has cancelled. What remains is a pure ratio of the line's shunt effect to its natural throughput — precisely the quantity that decides whether the shunt branch can be ignored, lumped, or must be distributed.

The classification in these terms:
— Short, \(\beta l \lesssim 0.05\) rad (under ~50 km at 50 Hz): charging under 5% of SIL, shunt branch discarded entirely — Set 9.
— Medium, \(0.05 \lesssim \beta l \lesssim 0.5\) rad (~50–250 km): charging significant, lumped at the ends as a nominal \(\pi\) — Set 11.
— Long, \(\beta l \gtrsim 0.5\) rad (beyond ~250 km): the lumping error itself becomes significant and the distributed solution is needed — Set 12.

Why the usual km-based rule works at all. Since \(v \approx c\) for every overhead line regardless of construction (Problem 19), \(\beta\) is very nearly \(1.05\times10^{-3}\) rad/km on all of them, and the radian criterion translates into a kilometre criterion with little loss. It stops working for cables, where \(v \approx 0.66c\) makes \(\beta\) half as large again — so a cable is "long" at two thirds the length of an overhead line, quite apart from the charging-current problem of Set 6 Problem 14.
C2. Bundling improves both inductance and capacitance. Identify what it costs, and construct the argument an EHV line designer would actually make when choosing between two, three and four sub-conductors.

Show answer
What it costs:
— Metal, in direct proportion to \(n\). This is usually the smallest cost, since the sub-conductors are smaller.
— Spacers, every 50–70 m per span, sized for the short-circuit forces of Problem 17. Cost scales with \(n\) and with \(d\).
— Wind and ice loading, which grows with the bundle's frontal area and drives tower and foundation cost.
— Aeolian vibration and galloping, worse for bundles, requiring dampers.
— Installation, more complex stringing and tensioning.

The designer's argument, in order:
1. Start from corona, not reactance. At 400 kV and above, the surface gradient sets a minimum total conductor surface — Set 8. This usually fixes a floor on \(n\) before any electrical optimisation begins: two at 400 kV, four at 765 kV.
2. Check the thermal rating. The required current-carrying capacity sets a minimum total cross-section, which interacts with \(n\) through the sub-conductor size.
3. Then look at SIL. Problem 9 gives 300, 409, 464, 509 MW for \(n = 1\) to 4 — increments of 36%, 13%, 10%. The first step is decisive, the rest marginal.
4. Weigh against tower cost. The wind loading of a four-bundle can push the tower into a heavier class, and a tower series change costs more than the conductor saving.

The conclusion in practice: \(n\) is nearly always decided by corona and voltage class, with SIL as a welcome by-product rather than the driver. 220 kV single or twin; 400 kV twin or quad; 765 kV quad or hex. A designer who optimised \(n\) on reactance alone would reach roughly the same answer, which is why the two criteria are seldom in conflict.
C3. A double-circuit line was found in Problem 18 to have a surge impedance of 134 \(\Omega\) — less than half that of one circuit. Examine whether this means it can carry more than twice the power, and identify what actually limits it.

Show answer
What the numbers say. Single circuit: \(Z_c = 305\ \Omega\), SIL 525 MW. Double: \(Z_c = 134\ \Omega\), SIL 1194 MW — a factor of 2.27, indeed more than double.

Why more than double. The two circuits are not independent. Each phase has become a "bundle" of two conductors 12 m apart, with an effective radius of 0.94 m against 0.074 m for one circuit's bundle. That mutual coupling lowers \(L\) and raises \(C\) beyond what simple paralleling would give — the same effect Set 5 Problem 13 found for inductance alone.

But SIL is not a rating. Three other limits apply, and any may bind first:
— Thermal. Each conductor still carries only its own rated current. Total thermal capacity is exactly double, not 2.27 times, so above a certain length-independent loading the thermal limit binds and the coupling advantage is unavailable.
— Stability. \(P_{max} = V_SV_R/X\) improves by the reactance ratio, which here is \(1.027/0.451 = 2.28\) — consistent with the SIL gain, so on a long line the coupling advantage is realisable.
— Contingency. This is the binding constraint in practice. A double-circuit tower can lose both circuits to a single event — a tower collapse, a broad fire, a lightning strike causing a double-circuit fault. Planning criteria therefore usually require the system to survive the loss of both, which means the pair cannot be loaded above what one corridor's worth of alternative capacity can replace.

Conclusion. The 2.27 factor is electrically real and matters for stability-limited long lines. It is unavailable thermally, and it is usually irrelevant in planning, because the double-circuit tower's shared failure mode caps the useful loading well below either electrical limit. This is a recurring pattern: the physics gives a bonus, and the reliability criterion declines to spend it.

Self-Test

Multiple-Choice Questions

MCQ 1. The bundle radius used for capacitance is built from the sub-conductor's:
(a) GMR (b) actual radius (c) diameter (d) cross-sectional area

Show answer
(b). The GMR is for inductance only. Two columns must be carried from the data sheet onwards — Problem 1.
MCQ 2. Bundling a line:
(a) raises \(L\) and \(C\) (b) lowers both (c) lowers \(L\), raises \(C\) (d) raises \(L\), lowers \(C\)

Show answer
(c). The logarithm sits in the numerator for \(L\) and the denominator for \(C\), so a larger effective radius helps both.
MCQ 3. The surge impedance of a line depends on:
(a) its length (b) its voltage (c) its cross-sectional geometry only (d) its loading

Show answer
(c). \(L\) and \(C\) are both per unit length, so length divides out of the ratio — Problem 7.
MCQ 4. A typical bundled EHV line has a surge impedance of about:
(a) 50 \(\Omega\) (b) 130 \(\Omega\) (c) 290 \(\Omega\) (d) 800 \(\Omega\)

Show answer
(c). Single-conductor lines run near 400 \(\Omega\); option (b) is a double-circuit value.
MCQ 5. At surge impedance loading a line:
(a) carries its thermal maximum (b) has a flat voltage profile (c) has zero losses (d) is at its stability limit

Show answer
(b). Reactive generation equals reactive absorption at every point, so \(|V|\) is constant end to end — Problem 8.
MCQ 6. SIL is computed as:
(a) \(\sqrt3 V_L^2/Z_c\) (b) \(V_L^2/Z_c\) (c) \(V_{ph}^2/Z_c\) (d) \(3V_L^2/Z_c\)

Show answer
(b). The \(\sqrt3\) factors cancel between the phase voltages and the three-phase summation, exactly as for charging MVA.
MCQ 7. Going from one to two sub-conductors per phase raises SIL by about:
(a) 10% (b) 36% (c) 70% (d) 100%

Show answer
(b) 36%. Option (c) is the gain for four sub-conductors — Problem 9.
MCQ 8. Doubling the bundle spacing reduces inductance by about:
(a) 50% (b) 27% (c) 7% (d) 0%

Show answer
(c) 7%. The spacing enters under a square root before entering the logarithm — Problem 13.
MCQ 9. The force between the sub-conductors of a bundle during a fault is:
(a) repulsive (b) attractive (c) zero (d) alternating in direction

Show answer
(b) attractive — parallel currents in the same direction. It pulsates in magnitude at twice power frequency but never reverses — Problem 17.
MCQ 10. A line operating below SIL has a receiving-end voltage that is:
(a) lower than the sending end (b) higher than the sending end (c) equal (d) unpredictable

Show answer
(b) higher — the Ferranti effect, because capacitance dominates below SIL.
MCQ 11. Bundling changes the velocity of propagation:
(a) substantially upward (b) substantially downward (c) hardly at all (d) to exactly \(c\)

Show answer
(c). \(v = 1/\sqrt{LC}\) and the logarithms cancel in the product, so \(v\) stays near \(0.99c\) — Problem 19.
MCQ 12. The ratio of a line's charging MVA to its SIL equals:
(a) its length in km (b) its electrical length \(\beta l\) in radians (c) its \(X/R\) ratio (d) unity

Show answer
(b). This is the correct criterion for the short/medium/long classification — Problem 14 and Challenge C1.

Reference

Key Formulas

Quantity	Relation	Notes
Bundle radius, inductance	\(D_{sb}\) from sub-conductor \(D_s\)	Two columns must be kept separate
Bundle radius, capacitance	\(r_b\) from sub-conductor \(r\)	\(D_{sb}/r_b = \sqrt{D_s/r}\)
Two sub-conductors	\(\sqrt{D_sd}\)
Three sub-conductors	\(\sqrt[3]{D_sd^2}\)	Equilateral
Four sub-conductors	\(1.09\sqrt[4]{D_sd^3}\)	Square; \(1.09 = 2^{1/8}\)
Inductance	\(L = 0.2\ln(D_{eq}/D_{sb})\) mH/km	\(D_{eq}\) between bundle centres
Capacitance	\(C_n = 2\pi\varepsilon_0/\ln(D_{eq}/r_b)\) F/m
Surge impedance	\(Z_c = \sqrt{L/C}\)	Independent of length; 200–450 \(\Omega\)
Surge impedance loading	\(\text{SIL} = V_L^2/Z_c\)	No \(\sqrt3\); flat voltage profile
Phase constant	\(\beta = \omega\sqrt{LC} = 2\pi/\lambda\)	\(\approx 1.05\times10^{-3}\) rad/km at 50 Hz
Velocity	\(v = 1/\sqrt{LC} \approx 0.99c\)	Barely affected by bundling
Electrical length	\(\beta l = Q_c/\text{SIL}\)	The short/medium/long criterion
Power transfer	\(P_{max} = V_SV_R/X\)	Improves as \(1/X\)
Sub-conductor force	\(F/l = 2\times10^{-7}I_1I_2/d\) N/m	Attractive; sizes the spacers
Bundle resistance	\(R = R_{\text{sub}}/n\)	Sub-conductors in parallel
Typical bundled EHV	\(L \approx 0.9\)–\(1.05\) mH/km, \(C \approx 0.011\)–\(0.013\ \mu\text{F}\)/km	\(Z_c \approx 260\)–\(300\ \Omega\)

Diagnostics

Common Mistakes

Using one bundle radius for both parameters. \(D_{sb}\) is seeded by the GMR and \(r_b\) by the actual radius — Problem 1.
Measuring \(D_{eq}\) to sub-conductors rather than bundle centres. The intra-bundle geometry is already inside \(D_{sb}\) and must not be counted twice — Problem 2.
Omitting the 1.09 in the four-conductor formula. It is \(2^{1/8}\) from the square's diagonals, not a typographical error.
Forgetting that a bundle's resistance is that of one sub-conductor divided by \(n\). They are in parallel — Problem 20.
Inserting a \(\sqrt3\) into the SIL formula. \(\text{SIL} = V_L^2/Z_c\) uses the line voltage directly.
Treating surge impedance as length-dependent. It is a property of the cross-section alone — Problem 7.
Treating SIL as a thermal rating. It is the reactively self-sufficient loading, typically well below the thermal limit — Problem 8.
Mixing length units inside \(Z_c = \sqrt{L/C}\). Both must be per the same unit; H/km with F/m gives an answer wrong by \(\sqrt{1000}\).
Expecting wider bundle spacing to help much. Doubling \(d\) buys 7%; doubling \(n\) buys 27% — Problems 6 and 13.
Assuming the sub-conductors repel during a fault. They attract, and the spacers exist to prevent collapse — Problem 17.
Adding two circuits' parameters as if independent. They are mutually coupled and behave as one very wide bundle — Problem 18.
Accepting \(Z_c\) outside 200–450 \(\Omega\). Almost always a bundle radius omitted or a unit error.

Looking Ahead

Bundling has been justified here entirely on reactance, capacitance and surge impedance — and that is not how it was invented. Every gain in this set is logarithmic and therefore modest; none of them would on its own have persuaded an industry to hang four conductors where one would do.

Set 8 supplies the reason that did. Above about 220 kV the electric field at the conductor surface approaches the breakdown strength of air, and the line begins to lose power continuously into a glow discharge that also generates radio interference and audible noise. The remedy is more conductor surface, distributed rather than concentrated — which is a bundle. The parameters computed in this set are, historically, the by-product; Set 8 computes the driver.