Bundled Conductors
Above about 230 kV a single conductor per phase cannot be made large enough to hold the electric field at its surface below the value at which air breaks down, and the answer — splitting the phase into two, three or four sub-conductors held apart by spacers — turns out to improve every line parameter at once.
- Why the surface voltage gradient, not the current rating, is what limits a single conductor at EHV, and how to compute it from Chapter 7's capacitance.
- How Chapter 6's composite-conductor formula produces the bundle GMR \(\sqrt{D_sd}\), \(\sqrt[3]{D_sd^2}\) and \(1.09\sqrt[4]{D_sd^3}\) with no new theory at all.
- Why the same three expressions with \(r\) in place of \(D_s\) give the equivalent radius for capacitance — the one place Chapter 7's \(r\)-versus-\(D_s\) distinction pays a dividend.
- How bundling lowers \(L\) by a quarter and raises \(C\) by a third, and why those two changes reinforce each other in \(Z_c=\sqrt{L/C}\).
- Why surge impedance loading — the real reason bundles are used at 400 kV and above — rises by a third for a twin bundle and by two thirds for a quad.
- Why increasing the sub-conductor spacing \(d\) gives sharply diminishing returns, and what actually fixes it in practice.
- What bundling costs: heavier hardware, wider towers, larger short-circuit forces between sub-conductors, and more charging MVAr to absorb.
Where the Single Conductor Runs Out
Chapter 5 selected conductors on current-carrying grounds: an ACSR of a certain cross-section carries a certain current at a certain temperature rise, and if more current is wanted a larger conductor is chosen. That reasoning is complete at 33 kV and at 132 kV. It fails at 400 kV, and it fails for a reason that has nothing to do with heat.
The electric field at the surface of an energised conductor rises as the voltage rises. When that field exceeds roughly 30 kV/cm at the peak of the cycle, the air immediately adjacent to the metal ionises, and the conductor begins to lose power continuously into a faint bluish glow accompanied by a hiss, an ozone smell, and broadband radio noise. This is corona, the subject of Chapter 15. It wastes energy, it corrodes hardware, and its radio interference is a regulatory matter.
The field at a conductor's surface falls as the conductor is made fatter. So the naive remedy is a larger conductor — and up to a point that is what is done, which is why 220 kV lines use Zebra rather than Dog. But the remedy scales badly. Example 4 works the numbers: at 400 kV a single conductor would need a radius of over 2 cm, and at 765 kV over 4.5 cm, meaning a solid conductor 9 cm in diameter. Such a conductor would be enormously heavy, would require towers and hardware sized for that weight, would have most of its aluminium doing nothing useful because skin effect confines the current to the outer layers, and would be very nearly impossible to handle on a stringing drum.
The alternative was recognised in the 1930s and became universal from the 1950s: keep the metal cross-section that the current demands, but distribute it over two, three or four separate conductors held a fixed distance apart. Electrically the group presents a much larger effective surface to the field. Mechanically it is several familiar conductors rather than one unfamiliar one. That group is a bundle, and its sub-conductors are held in formation by spacers installed every 50 to 70 metres along the span.
The two inequalities \(r \ll d \ll D_{eq}\) are what make the bundle behave, seen from another phase, like one large conductor while behaving, seen from close up, like several small ones. Typical figures are \(r \approx 1.6\) cm, \(d = 45\) cm and \(D_{eq} \approx 14\) m, so \(d/r\) is about 28 and \(D_{eq}/d\) about 31.
Corona control was the original motive and remains the binding constraint. But the moment the phase is split, its geometric mean radius rises, and Chapter 6's formula \(L = 2\times10^{-7}\ln(D_{eq}/D_s)\) says the inductance must fall. Simultaneously its equivalent radius for capacitance rises, and Chapter 7's \(C_n = 2\pi\varepsilon/\ln(D_{eq}/r)\) says the capacitance must rise. Falling \(L\) and rising \(C\) both push the surge impedance down and the power-transfer capability up. Bundling is therefore that rare engineering intervention that improves every relevant quantity simultaneously, and this chapter's business is to compute by how much.
The Surface Voltage Gradient
Before anything can be designed, the quantity being controlled must be computable. Chapter 7 supplies it in two lines. The field at radius \(x\) from a conductor carrying \(q\) coulombs per metre is \(E = q/2\pi\varepsilon x\); at the conductor's own surface \(x=r\), so
Every term of the capacitance has cancelled except the geometry, and the result depends only on the phase voltage, the conductor radius and the spacing. It is conventionally quoted in kilovolts per centimetre, with \(V_{an}\) in kV and \(r\) in cm.
The threshold to compare it against comes from Chapter 15 and from Peek's experiments: air at standard temperature and pressure breaks down at about \(30\) kV/cm peak, which for a sinusoid is \(21.1\) kV/cm rms. Since \(E_r\) computed from an rms voltage is itself an rms figure, the comparison must be made consistently, and the usual design target is an rms gradient of \(15\) to \(17\) kV/cm — comfortably below the ideal-conductor threshold, because a real conductor is stranded, its surface is scratched and dirtied in service, and rain drops hanging from it concentrate the field enormously.
Now split the phase into \(n\) sub-conductors of the same radius \(r\), spaced \(d\) apart. The total charge \(q\) on the phase is set by the phase's capacitance and divides equally between the sub-conductors, so each carries \(q/n\). Seen from far away the bundle acts as a single conductor of equivalent radius \(r_b\), so \(q = 2\pi\varepsilon V_{an}/\ln(D_{eq}/r_b)\). The average gradient at the surface of one sub-conductor is that sub-conductor's own charge divided by \(2\pi\varepsilon r\):
Two separate effects reduce the gradient, and both appear in the denominator. The factor \(n\) is the sharing of charge between sub-conductors. The replacement of \(r\) by the much larger \(r_b\) inside the logarithm raises the total charge somewhat and works the other way, but only weakly, because it sits inside a logarithm while \(n\) does not. The net effect is close to a factor of \(n\) divided by a modest correction.
The sub-conductors are not, however, all at the same potential gradient around their circumference. Each one sees the field of its neighbours superposed on its own, and the field is largest on the side facing into the bundle. To first order, with \(d \gg r\), the maximum gradient on a sub-conductor of an \(n\)-bundle is
For a twin bundle with \(r=1.6\) cm and \(d=45\) cm the bracket is \(1.036\), a correction of under four per cent; for a quad it is \(1.11\). This is small enough that the average gradient is a useful design figure and the correction is applied at the end — but it is also the reason \(d\) is never made too small, since the bracket grows as \(d\) shrinks.
Bundle Geometry and Spacers
Three geometries account for essentially every bundled line in service. A twin bundle places two sub-conductors on a horizontal line, \(d\) apart. A triple places three at the vertices of an equilateral triangle of side \(d\). A quad places four at the corners of a square of side \(d\), so the diagonal is \(\sqrt2\,d\). Six- and eight-conductor bundles exist on 1000 kV and above and follow the same pattern of a regular polygon.
The sub-conductor spacing \(d\) is very nearly standardised at 45 cm across the world, for reasons that are partly electrical and partly mechanical. Electrically, Section 8-5 will show that the benefit of increasing \(d\) is logarithmic and therefore weak: going from 45 cm to 60 cm buys under three per cent of reactance. Mechanically, a wider bundle needs a longer, heavier spacer, presents a larger effective area to wind, and forces the whole tower to widen. And in the other direction, sub-conductors that are too close attract each other violently during a short circuit — the force between two parallel currents goes as \(1/d\) — and can clash.
The spacer itself is a small aluminium-alloy frame clamped to each sub-conductor. It does three jobs. It holds the spacing constant along the whole span, so that the geometry assumed in the calculations is the geometry that exists. It prevents clashing under wind, ice and short-circuit forces. And it connects the sub-conductors electrically in parallel, so that the phase behaves as one conductor at the terminals. Modern spacers incorporate elastomeric bushings and are called spacer dampers, because they also absorb the sub-span oscillation that wind flowing off a windward sub-conductor can excite in a leeward one.
The GMR of a Bundle
No new theory is needed here. Section 6-6 established that a composite conductor of \(n\) filaments has a self geometric mean distance
the \(n^2\)-th root of the product of all \(n^2\) distances from each filament to every filament, including itself. A bundle is precisely such a composite conductor, with the sub-conductors as its filaments and each sub-conductor's own tabulated \(D_s\) as its self-distance. Working out the products for the three standard geometries is arithmetic.
For the twin bundle there are \(2^2 = 4\) distances: \(D_{aa}=D_s\), \(D_{ab}=d\), \(D_{ba}=d\), \(D_{bb}=D_s\). Their product is \((D_s d)^2\), so
For the triple there are \(3^2=9\) distances. Each sub-conductor contributes its own \(D_s\) and two distances \(d\) to the others, so each row of the matrix contributes \(D_s d^2\) and the total product is \((D_s d^2)^3\):
For the quad there are \(4^2=16\) distances. From any corner of a square the distances are: \(D_s\) to itself, \(d\) to each of the two adjacent corners, and \(\sqrt2 d\) to the opposite corner. Each row of the matrix therefore contributes \(D_s\cdot d\cdot d\cdot\sqrt2 d = \sqrt2 D_s d^3\), and there are four identical rows:
The coefficient \(1.09\) is exactly \(2^{1/8}\), and it arises solely from the two diagonals of the square being longer than the sides by \(\sqrt2\). No such coefficient appears for \(n=2\) or \(n=3\) because in those geometries every inter-conductor distance is the same \(d\).
The magnitudes are what matter. A Moose conductor has \(D_s = 1.277\) cm; a twin bundle at 45 cm has \(D_s^{\,b} = \sqrt{0.01277\times0.45} = 7.58\) cm, six times larger; a quad has \(20.1\) cm, sixteen times larger. This is the entire mechanism by which bundling changes the line parameters: the group looks, to the other phases, like a conductor an order of magnitude fatter than any of its members.
Inductance of a Bundled Line
The master formula of Chapter 6 is untouched. Substitute the bundle's self-GMD for the conductor's, and the mutual GMD between phases remains \(D_{eq}\), computed from the centre-to-centre distances between bundles because \(d \ll D_{eq}\) makes the difference between "centre of bundle" and "each sub-conductor" negligible under the logarithm.
The spacings \(D_{12}, D_{23}, D_{31}\) are measured between bundle centres. The approximation involved — replacing \(D_{12}\pm d\) by \(D_{12}\) — introduces an error of order \((d/D_{12})^2\), which for \(d=0.45\) m and \(D_{12}=11\) m is under two parts in a thousand.
Take the standard Indian 400 kV line: Moose conductor, \(D_s = 1.277\) cm, horizontal configuration with 11 m between adjacent phases so \(D_{eq}=11\sqrt[3]{2}=13.859\) m. Then
The twin bundle removes \(25.5\%\) of the inductance and the quad \(39.5\%\). At 50 Hz the reactance falls from \(0.439\) to \(0.327\) and to \(0.266\;\Omega\)/km. Section 6-9 observed that every single-circuit overhead line has about \(0.4\;\Omega\)/km of reactance; bundling is one of only two devices that move that number materially, the other being series capacitors (Chapter 38).
Why the benefit saturates is visible in the formula. Doubling \(d\) adds \(\ln 2\) divided by \(n\) to \(\ln D_s^{\,b}\) — for a twin bundle, only \(0.347\) — against a logarithm of about \(5.2\). Example 6 puts numbers to it: raising \(d\) from 45 cm to 60 cm buys \(2.8\%\) of reactance while widening every tower. The sub-conductor count, which enters as a root index, is the lever that works; the sub-conductor spacing is not.
Capacitance of a Bundled Line
Chapter 7 insisted that capacitance calculations use the actual radius \(r\) and never the geometric mean radius, because charge sits on the surface and there is no internal field. That distinction has been a caution until now; here it becomes a working rule with a consequence.
Derive the result directly rather than asserting it, following Section 7-5. Take a twin-bundle three-phase line in the first section of its transposition cycle, with the bundle of phase \(a\) at position 1 comprising sub-conductors \(a\) and \(a'\), and likewise for \(b\) and \(c\). The sub-conductors of a bundle are connected in parallel by the spacers, so they are at the same potential and, being identical and identically placed relative to the distant phases, share the phase charge equally: each carries \(q_a/2\).
Write \(V_{ab}\) by superposition, remembering that \(d \ll D_{12}\) so that a sub-conductor of bundle \(b\) is at distance \(D_{12}\) from either sub-conductor of bundle \(a\) to within a fraction of a per cent:
The sub-conductor \(a\) is at radius \(r\) from its own charge and \(D_{12}\) from bundle \(b\); the sub-conductor \(a'\) is at distance \(d\) from \(a\)'s charge and, again, \(D_{12}\) from bundle \(b\). Half the sum of two logarithms is the logarithm of a square root, so \(\tfrac12\big[\ln(D_{12}/r)+\ln(D_{12}/d)\big] = \ln\big(D_{12}/\sqrt{rd}\big)\), and the same happens in the \(q_b\) term:
This is term-for-term the expression Section 7-7 obtained for a single-conductor line, with \(r\) replaced by \(\sqrt{rd}\) and nothing else changed. The averaging over the transposition cycle and the elimination of \(q_b+q_c\) therefore proceed exactly as before, and the answer follows without further work.
The equivalent radius \(r_b\) is obtained from the bundle-GMR expressions of Section 8-4 by the single substitution \(D_s \to r\). This is the practical dividend of Chapter 7's distinction: a bundle has two effective radii, \(D_s^{\,b}\) for inductance and \(r_b\) for capacitance, and \(r_b\) is always the larger of the two.
For the 400 kV line above, with Moose of \(r=1.5885\) cm:
The twin bundle raises the capacitance by \(32.8\%\) and the quad by \(62.1\%\). Where the inductance change is a benefit without qualification, this one has two faces: it raises the line's ability to carry power, and it raises the charging reactive power that has to be absorbed at light load. Section 8-8 puts both on the same page.
Gradient Reduction: the Original Purpose
Return to the reason bundling exists. Apply the formula of Section 8-2 to the 400 kV line, with \(V_{an}=400/\sqrt3=230.94\) kV and \(r=1.5885\) cm:
The single-conductor figure of \(21.47\) kV/cm rms is \(30.4\) kV/cm at the peak of the cycle. That is above the disruptive gradient of air, which means such a line would be in corona continuously, in fair weather, along its whole length. It is not a marginal case; it is unbuildable. The twin bundle brings the maximum gradient to \(14.76\) kV/cm rms, inside the design target, and the quad to under \(10\).
The comparison also shows why the answer is to add conductors rather than to enlarge one. Going from a twin bundle to a quad divides the gradient by \(1.64\) — nearly the factor of two that the charge-sharing alone would give, reduced only slightly by the logarithm. Achieving the same reduction with a single conductor would require multiplying its radius by a similar factor, hence its weight by roughly four.
Surge Impedance, SIL and Charging
Bundling lowers \(L\) and raises \(C\). The two changes act in the same direction on the ratio that governs a line's power-carrying character:
Chapter 14 will derive both properly and explain why a line loaded at exactly its SIL neither absorbs nor generates reactive power, and why SIL is the natural yardstick for a line's capability. For the present purpose only the arithmetic is needed. Working with the 400 kV Moose line:
| Bundle | \(D_s^{\,b}\) (m) | \(r_b\) (m) | \(L\) (mH/km) | \(X\) (Ω/km) | \(C_n\) (μF/km) | \(Z_c\) (Ω) | SIL (MW) | \(E_{av}\) (kV/cm) |
|---|---|---|---|---|---|---|---|---|
| single | 0.01277 | 0.01589 | 1.3980 | 0.4392 | 0.008216 | 412.5 | 388 | 21.47 |
| twin, \(d=45\) cm | 0.07581 | 0.08455 | 1.0417 | 0.3273 | 0.010909 | 309.0 | 518 | 14.26 |
| triple, \(d=45\) cm | 0.13726 | 0.14762 | 0.9230 | 0.2900 | 0.012248 | 274.5 | 583 | 10.67 |
| quad, \(d=45\) cm | 0.20141 | 0.21271 | 0.8463 | 0.2659 | 0.013319 | 252.1 | 635 | 8.70 |
The surge impedance falls from \(412\;\Omega\) to \(309\;\Omega\) for a twin bundle and to \(252\;\Omega\) for a quad, so the surge impedance loading rises by \(33\%\) and \(64\%\) respectively. That gain — not the reactance reduction alone, and certainly not the ampacity — is the number a transmission planner quotes when justifying a bundle. Applied across a range of voltage levels the pattern is consistent:
| System | Configuration | \(D_{eq}\) | \(X\) (Ω/km) | \(C_n\) (μF/km) | \(Z_c\) (Ω) | SIL (MW) | \(Q_c\) (MVAr/km) |
|---|---|---|---|---|---|---|---|
| 132 kV | single Panther, flat 3 m | 3.78 m | 0.385 | 0.00945 | 360 | 48 | 0.052 |
| 220 kV | single Zebra, flat 6 m | 7.56 m | 0.408 | 0.00887 | 382 | 127 | 0.135 |
| 400 kV | twin Moose, flat 11 m, \(d=45\) cm | 13.86 m | 0.327 | 0.01091 | 309 | 518 | 0.548 |
| 765 kV | quad Bersimis, \(d=45\) cm | 15.0 m | 0.269 | 0.01315 | 255 | 2292 | 2.417 |
The charging reactive power tells the other half of the story. From Section 7-9, \(Q_c = \omega C_n V_{LL}^2\), and the bundled 400 kV line generates \(0.548\) MVAr per kilometre against \(0.413\) for a hypothetical single-conductor line of the same geometry. A 300 km 400 kV line therefore generates about \(165\) MVAr with no load on it whatsoever, which must be absorbed by shunt reactors at light load or the receiving-end voltage will rise unacceptably. Bundling has made that requirement a third larger. Chapter 34 sizes the reactors; Example 5 makes the estimate.
The Full Balance Sheet
Textbook lists of "advantages of bundled conductors" tend to run to eight items without distinguishing between the decisive and the incidental. The honest ordering follows from what has been computed.
The reason it is done. The surface gradient falls by a factor approaching \(n\), which is what makes operation above 230 kV possible at all. Corona loss (Chapter 15) falls with it, as does radio and television interference and audible noise — the last being a genuine constraint on line routing near habitation.
The reason it is welcome. The series reactance falls by a quarter to two fifths, raising the steady-state transfer limit \(P = (V_sV_r/X)\sin\delta\), improving voltage regulation (Chapter 13), and increasing the transient stability margin (Chapters 28 and 29). The surge impedance falls, so SIL rises by a third to two thirds.
Incidental benefits. The sub-conductors are individually smaller, so skin effect is less pronounced and the a.c. resistance is closer to the d.c. value than it would be for a single conductor of the same total section. The bundle presents more surface area to the air, so it cools better and its current rating exceeds that of an equal-section single conductor. And the sub-conductors are ordinary catalogue items, strung with ordinary equipment.
The costs. Every one of them is real and each has caused trouble on some line.
| Penalty | Mechanism | Consequence |
|---|---|---|
| More hardware | Spacers every 50–70 m, multiple clamps and dead-ends per phase | Higher capital cost and more items to inspect |
| Wider towers | Bundle occupies \(d\) plus conductor diameter, and swings as a unit | Larger cross-arms, more right-of-way |
| Greater wind and ice load | \(n\) conductors present \(n\) times the projected area | Heavier towers and foundations |
| Short-circuit pinch forces | Parallel currents in the same direction attract, force \(\propto I^2/d\) | Spacers must resist bundle collapse during faults |
| Sub-span oscillation | Wake of the windward sub-conductor excites the leeward one | Requires spacer dampers, not plain spacers |
| More charging MVAr | \(C_n\) is a third larger, and \(Q_c=\omega C_nV_{LL}^2\) | Larger shunt reactors needed at light load |
None of these overturns the case above 230 kV, where the alternative is simply not available. Below 230 kV they usually do overturn it, which is why 132 kV and 220 kV lines are almost always single-conductor even though a bundle would improve their electrical parameters exactly as computed above. The decision is not "is bundling better?" but "is bundling worth its cost at this voltage?", and the crossover sits where the surface gradient forces the answer.
Worked Examples
Problem. A 400 kV, 50 Hz transposed line has twin bundles of Moose ACSR (\(r=1.5885\) cm, \(D_s=1.277\) cm) with \(d=45\) cm, in a horizontal configuration with 11 m between adjacent phases. Find \(D_{eq}\), \(D_s^{\,b}\), \(r_b\), the inductance and capacitance per kilometre, and the reactances. Compare each with a single Moose conductor on the same towers.
Solution. The equivalent spacing is the same in both cases and comes first.
For the single conductor, \(\ln(13.859/0.01277)=6.9898\) gives \(L=1.3980\) mH/km and \(X=0.4392\;\Omega\)/km; \(\ln(13.859/0.015885)=6.7713\) gives \(C_n=0.008216\;\mu\)F/km. So the bundle removes \(25.5\%\) of the reactance and adds \(32.8\%\) to the capacitance. The two effective radii are related by \(r_b/D_s^{\,b} = \sqrt{r/D_s} = \sqrt{1.2439} = 1.115\): a bundle compresses the gap between the electric and magnetic effective radii, because the \(n\)-th root spreads the single altered factor over \(n\) positions.
Problem. A transposed line uses triple bundles of a conductor with \(r = 1.59\) cm and \(D_s = 1.35\) cm, with \(d = 40\) cm, and \(D_{eq} = 12\) m. Find \(L\), \(C_n\), \(X\), and the surge impedance.
Solution. Both equivalent radii are cube roots for a triple bundle.
A surge impedance of \(270\;\Omega\) is between the twin and quad values of Section 8-8, as it must be. Triple bundles are comparatively rare — the step from two sub-conductors to four is usually taken directly, because a square bundle is mechanically more convenient than a triangular one and its spacers are simpler.
Problem. Derive the quad-bundle GMR from first principles by writing out all sixteen distances, and evaluate it for Moose (\(D_s = 1.277\) cm) at \(d = 45\) cm. Then find the equivalent radius \(r_b\) for capacitance.
Solution. Label the corners of the square \(1,2,3,4\) in order, so that \(1\)–\(2\), \(2\)–\(3\), \(3\)–\(4\) and \(4\)–\(1\) are sides of length \(d\) while \(1\)–\(3\) and \(2\)–\(4\) are diagonals of length \(\sqrt2 d\). The \(4\times4\) matrix of distances is
Every row has the same content: one \(D_s\), two \(d\)'s and one \(\sqrt2 d\). So each row multiplies to \(\sqrt2 D_s d^3\) and the whole \(16\)-fold product is \(\big(\sqrt2 D_s d^3\big)^4\). Taking the sixteenth root takes the fourth root of the row product:
For capacitance the same arithmetic with \(r=0.015885\) m in place of \(D_s\):
A quad bundle of 3.2 cm conductors behaves electrically like a single conductor over 40 cm in diameter — while weighing what four ordinary conductors weigh and being strung the same way.
Problem. For the 400 kV line of Example 1, find the radius a single conductor would need in order to hold the surface gradient at 17 kV/cm rms. Repeat for a 765 kV line with \(D_{eq}=15\) m, and comment.
Solution. The gradient of Section 8-2 must be inverted, and since \(r\) appears both outside and inside the logarithm the solution is found by trial. With \(V_{an}=230.94\) kV and \(D_{eq}=1385.9\) cm:
| \(r\) (cm) | \(\ln(D_{eq}/r)\) | \(E\) (kV/cm rms) | peak (kV/cm) |
|---|---|---|---|
| 1.5885 (Moose) | 6.771 | 21.47 | 30.4 |
| 1.8 | 6.646 | 19.30 | 27.3 |
| 2.0 | 6.541 | 17.65 | 25.0 |
| 2.1 | 6.492 | 16.94 | 24.0 |
| 2.5 | 6.318 | 14.62 | 20.7 |
Interpolating between the last two rows gives \(r \approx 2.09\) cm, a conductor of \(41.8\) mm diameter — roughly \(1.7\) times the cross-section of Moose. Not impossible, but every kilogram of that extra aluminium is carried by the towers and most of it is idle, because skin effect keeps the current in the outer layers.
At 765 kV, \(V_{an}=441.7\) kV and \(D_{eq}=1500\) cm:
A single conductor of \(9\) cm diameter, with about eight times the section of Moose, weighing some \(16\) kg per metre. It would need drums, tensioners and towers that do not exist. Compare the quad Bersimis bundle actually used, which achieves \(E_{av}=14.9\) kV/cm with four ordinary conductors — and simultaneously delivers a surge impedance of \(255\;\Omega\) and a SIL of 2292 MW.
Problem. The 400 kV twin-bundle line of Example 1 is 300 km long. Find its total charging current per phase and reactive generation at rated voltage, and size shunt reactors to absorb 80 per cent of it, split equally between the two ends. Compare the reactive generation with the line's SIL.
Solution. From Example 1, \(C_n = 0.010909\;\mu\)F/km.
In practice this is rounded to a standard rating — 80 MVAr per end is the common Indian choice on 400 kV circuits of this length, giving about 97 per cent compensation. The charging current of 237 A is a substantial fraction of a twin-Moose line's rating of roughly 1700 A, and it flows whether or not any power is being delivered.
Set against the SIL of 518 MW: at that loading the line's series reactance absorbs exactly the reactive power the shunt capacitance generates, and the reactors would be switched out. Below SIL the line is a net reactive source and the reactors are needed; above it the line is a net reactive sink and reactive support is needed instead. Chapter 14 makes this precise and Chapter 34 turns it into a control problem.
Problem. For the twin-bundle line of Example 1, recompute \(L\), \(C_n\) and \(Z_c\) with \(d=30\) cm and with \(d=60\) cm. Comment on what actually fixes the sub-conductor spacing.
Solution. Both equivalent radii scale as \(\sqrt d\), so a change in \(d\) enters the parameters only through half a logarithm.
| \(d\) (cm) | \(D_s^{\,b}\) (m) | \(r_b\) (m) | \(L\) (mH/km) | \(X\) (Ω/km) | \(C_n\) (μF/km) | \(Z_c\) (Ω) | \(E_{max}\) (kV/cm) |
|---|---|---|---|---|---|---|---|
| 30 | 0.06190 | 0.06903 | 1.0823 | 0.3400 | 0.010492 | 321.2 | 14.44 |
| 45 | 0.07581 | 0.08455 | 1.0417 | 0.3273 | 0.010909 | 309.0 | 14.76 |
| 60 | 0.08753 | 0.09763 | 1.0129 | 0.3182 | 0.011226 | 300.4 | 15.06 |
Doubling the spacing from 30 cm to 60 cm buys \(6.4\%\) of reactance and \(7.0\%\) of capacitance, and reduces \(Z_c\) by \(6.5\%\). Going from the standard 45 cm to 60 cm buys only \(2.8\%\) — while requiring longer spacers, a wider bundle swing envelope, and cross-arms extended by 15 cm on every tower of the route.
The gradient hardly moves at all, and for an instructive reason: shrinking \(d\) shrinks \(r_b\), which enlarges \(\ln(D_{eq}/r_b)\) and so lowers \(E_{av}\) — from \(14.67\) kV/cm at 60 cm to \(13.71\) at 30 cm — while the proximity factor \([1+(n-1)r/d]\) rises from \(1.026\) to \(1.053\) over the same range. The two effects very nearly cancel, leaving \(E_{max}\) between \(14.4\) and \(15.1\) kV/cm throughout. Since the electrical case is weak in both directions, the spacing is settled on mechanical grounds: far enough apart that the sub-conductors cannot clash under short-circuit pinch forces or ice-shedding, close enough that the spacer is short and light. Forty-five centimetres, about \(14\) conductor diameters, is where those two arguments meet.
Chapter Summary
Surface gradient \(E = V_{an}/[r\ln(D_{eq}/r)]\) must stay below about 17 kV/cm rms.
\(\sqrt{D_sd}\), \(\sqrt[3]{D_sd^2}\), \(1.09\sqrt[4]{D_sd^3}\) — Chapter 6's composite formula, nothing new.
Same three expressions with \(r\) for \(D_s\), because charge lives on the surface.
\(L=2\times10^{-7}\ln(D_{eq}/D_s^{\,b})\): a quarter lower for twin, two fifths for quad.
\(C_n=2\pi\varepsilon/\ln(D_{eq}/r_b)\): a third higher for twin, two thirds for quad.
\(E_{av}=V_{an}/[nr\ln(D_{eq}/r_b)]\), with \(E_{max}=E_{av}[1+(n-1)r/d]\).
\(Z_c=\sqrt{L/C}\) falls both ways, so SIL \(=V_{LL}^2/Z_c\) gains 33% (twin), 64% (quad).
\(d\) enters as half a logarithm; 45 cm is set by spacers and pinch forces, not by \(X\).
Practice Problems
Take \(\varepsilon_0=8.854\times10^{-12}\) F/m and \(f=50\) Hz throughout, assume the lines are transposed, and measure phase spacings between bundle centres. State clearly which effective radius — \(D_s^{\,b}\) or \(r_b\) — each part of an answer uses.
- A twin bundle uses conductors with \(D_s = 1.10\) cm and \(r = 1.40\) cm at \(d = 40\) cm. Compute \(D_s^{\,b}\) and \(r_b\), and verify that \(r_b/D_s^{\,b} = \sqrt{r/D_s}\). What is the corresponding relation for a quad bundle?
- A 400 kV transposed line has quad bundles of a conductor with \(D_s = 1.20\) cm and \(r = 1.50\) cm, \(d = 45\) cm, and phase spacings of 12 m, 12 m and 24 m. Find \(D_{eq}\), \(L\), \(C_n\), \(X\) and \(X_c\) per kilometre.
- For the line of Problem 2, compute the average and maximum surface voltage gradients at rated voltage, and state whether the design target of 17 kV/cm rms is met.
- Show from \(L = 2\times10^{-7}\ln(D_{eq}/D_s^{\,b})\) that for a twin bundle the fractional reduction in inductance relative to a single conductor is \(\tfrac12\ln(d/D_s)\div\ln(D_{eq}/D_s)\). Evaluate it for \(D_s = 1.277\) cm, \(d = 45\) cm and \(D_{eq}=13.86\) m, and confirm the \(25.5\%\) quoted in Section 8-5.
- A 220 kV line on single Zebra conductor is to be re-strung as a twin bundle at \(d = 40\) cm on the same towers (\(D_{eq}=7.56\) m, \(r = 1.431\) cm, \(D_s = 1.151\) cm). Compute the new \(X\), \(C_n\), \(Z_c\) and SIL, and state the percentage change in each.
- The line of Problem 2 is 400 km long. Find its total charging reactive power at rated voltage and the rating of shunt reactors needed at each end to compensate 70 per cent of it.
- A six-conductor bundle places its sub-conductors at the vertices of a regular hexagon of side \(d\); the five distances from any vertex to the other five are \(d,\ \sqrt3 d,\ 2d,\ \sqrt3 d,\ d\). Show that \(r_b = \sqrt[6]{6\,r\,d^{5}}\), evaluate it for \(r=1.5885\) cm and \(d=45\) cm, and find by what factor the spacing of a quad bundle of the same conductors would have to be increased to match it. Comment on whether that is practical.
- Explain, using the formulas of Sections 8-5 and 8-6, why bundling reduces the surge impedance whereas widening the phase spacing \(D_{eq}\) increases it, and why only one of the two increases the surge impedance loading.