Set 29 — Bundled Conductors
A bundle is not a thicker conductor. It is several ordinary conductors held a few hundred millimetres apart by spacers, and everything it does follows from one piece of geometry: the sub-conductors sit on a regular polygon, and the product of the distances from any one of them to the rest is \(nA^{n-1}\). That single result gives the bundle's self geometric mean distance, and from it the inductance, the reactance, the capacitance, the surge impedance loading and the surface voltage gradient all follow in order.
This set works outward from that geometry. Problems 1 to 4 establish the bundle GMR for two, three and four sub-conductors and evaluate it for real ACSR. Problems 5 to 7 take it into inductance, into the reactance a 400 kV line saves, and into the honest comparison with a single conductor carrying the same aluminium. Problems 8 to 12 turn to the electric field at the conductor surface, where bundling matters even more than it does in the reactance — including the spacing that minimises the gradient, and why a single conductor is simply not admissible at 400 kV. Problems 13 to 20 close with capacitance, surge impedance loading, a 765 kV quad line, a broken sub-conductor and a complete line.
The polygon product rule. For \(n\) points equally spaced on a circle of radius \(A\), the product of the distances from one point to the other \(n-1\) is exactly \(nA^{n-1}\). Everything else in this set is arithmetic on top of that one identity, and Challenge C1 proves it.
Bundle self-GMD. Treating the \(n\) sub-conductors as one composite conductor carrying \(I/n\) each, \(D_{sb} = \left(n\,D_s\,A^{n-1}\right)^{1/n}\), where \(D_s\) is the sub-conductor's own GMR. The bundle radius for capacitance is the same expression with \(r\) in place of \(D_s\).
The three standard cases follow by substituting \(A = d/2\), \(d/\sqrt3\), \(d/\sqrt2\): \(D_{sb} = \sqrt{D_sd}\), \(\sqrt[3]{D_sd^{2}}\) and \(1.09\sqrt[4]{D_sd^{3}}\). The 1.09 is \(2^{1/8}\), not a rounded empirical figure.
Inductance and capacitance then read off unchanged from Sets 5 and 6, with \(D_{sb}\) and \(r_b\) replacing \(D_s\) and \(r\): \(L = 0.2\ln(D_{eq}/D_{sb})\) mH/km and \(C_n = 0.0556/\ln(D_{eq}/r_b)\) µF/km. The GMD between phases is unchanged — Problem 19 shows why.
The reactance falls and the capacitance rises, so the surge impedance \(Z_c = \sqrt{L/C}\) falls on both counts and the surge impedance loading \(\mathrm{SIL} = V^{2}/Z_c\) rises. Problem 16 shows why the SIL gain is smaller than the reactance saving.
Surface gradient is the real reason for bundling. The average gradient at a sub-conductor surface is \(E_{av} = V_{ph}/\!\left(n\,r\ln(D_{eq}/r_b)\right)\); the maximum, to first order, is \(E_{max} = E_{av}\left[1 + (n-1)r/A\right]\). Corona begins near 21.1 kV/cm rms.
Bundle spacing has an optimum, and it is flat. Widening the bundle lowers \(E_{av}\) but worsens the non-uniformity factor, so \(E_{max}\) has a minimum — near 0.30 m for a twin at 400 kV. Practical spacings of 0.45 m sit 0.7% above it, which is why mechanical considerations are allowed to choose.
A phase conductor is made up of \(n\) identical sub-conductors, each of geometric mean radius \(D_s\), arranged symmetrically on a circle of radius \(A\) and each carrying \(I/n\). Show that the self geometric mean distance of the bundle is
and reduce it to the standard forms for \(n = 2, 3, 4\) in terms of the spacing \(d\) between adjacent sub-conductors.
Start from the composite-conductor definition of Set 5. For a group of \(n\) filaments sharing the current equally, the self GMD is the \(n^2\)-th root of the product of all \(n^2\) distances, a filament's distance to itself being its own GMR:
This is the same definition used for a stranded conductor in Set 5 Problem 4; nothing about it assumes the filaments are touching. A bundle is a stranded conductor whose strands happen to be half a metre apart.
Separate the self terms from the mutual terms. Each row \(k\) of the double product contributes one \(D_s\) and the \(n-1\) distances from conductor \(k\) to the others:
where \(P_k\) is the product of the distances from sub-conductor \(k\) to all the others.
Symmetry makes every \(P_k\) the same. A regular polygon looks identical from each of its vertices, so \(P_k = P\) for all \(k\) and
Two numbers only: the sub-conductor's own GMR, and the product of its distances to its neighbours. This is the whole of bundle geometry.
The polygon product rule supplies \(P\). For \(n\) points equally spaced on a circle of radius \(A\), the product of the distances from any one of them to the remaining \(n-1\) is exactly
The proof is short — the vertices are the roots of \(z^{n} = A^{n}\), so \(z^{n}-A^{n} = \prod_k (z - z_k)\); divide by \((z-z_1)\), let \(z \to z_1\), and the limit is \(nz_1^{\,n-1}\), of modulus \(nA^{n-1}\). Challenge C1 works it out in full.
Combining the two results gives the general bundle GMR:
Now substitute the circumradius of each standard bundle, which is where the familiar forms come from. For a two-conductor bundle the two sub-conductors are the ends of a diameter, so \(A = d/2\):
For an equilateral triangle of side \(d\) the circumradius is \(A = d/\sqrt3\):
And for a square of side \(d\) the circumradius is \(A = d/\sqrt2\):
The famous 1.09 is \(2^{1/8}\). Textbooks quote it as a decimal and it is often mistaken for a fitted constant:
It is exact, and it arises solely because the square's circumradius is \(d/\sqrt2\) rather than \(d/2\). Writing the general formula rather than memorising three special cases removes the possibility of misremembering it.
The capacitance version is identical in form. Charge resides on the surface, so the outside radius \(r\) replaces the flux-linkage radius \(D_s\):
Since \(r > D_s\) for any real conductor, \(r_b > D_{sb}\) always — but only by the \(n\)-th root of the ratio, so bundling narrows the gap between the two radii. Problem 2 puts numbers on it.
A 400 kV line uses a two-conductor bundle of ACSR Moose at a sub-conductor spacing of 450 mm. Moose has an overall diameter of 31.77 mm and a geometric mean radius of 1.267 cm. Find the bundle's self-GMD and its equivalent radius for capacitance, and state by what factor each exceeds the corresponding single-conductor value.
Collect the sub-conductor data in centimetres, since gradients later in this set want centimetres:
The ratio \(D_s/r = 0.798\) is typical of a 54/7 ACSR, and sits a little above the 0.7788 of a solid round conductor. A stranded conductor with a steel core carries its aluminium in an outer annulus, and metal pushed away from the axis links less internal flux — so the GMR is a larger fraction of the outside radius than a solid rod's.
Self-GMD of the bundle, from Problem 1 with \(n = 2\):
Equivalent radius for capacitance, the same formula with \(r\):
The multiplying factors:
A bundle of two 31.8 mm conductors behaves inductively like a single conductor of GMR 7.55 cm — six times the real value, and equivalent to a solid conductor of radius \(7.551/0.7788 = 9.70\) cm, that is, a rod 194 mm in diameter. No such conductor is made, and none could be strung.
The two radii have moved closer together, which is the point flagged at the end of Problem 1:
The square root halves the discrepancy in logarithmic terms. That is why a bundled line's inductive and capacitive geometries are more nearly reciprocal than a single-conductor line's — a fact Problem 16 turns into a statement about surge impedance.
A three-conductor bundle uses ACSR Zebra — overall diameter 28.62 mm, GMR 1.144 cm — on an equilateral triangle of side 450 mm. Find the circumradius, the bundle self-GMD and the bundle radius for capacitance, and verify the answer by both routes: the general formula of Problem 1 and the reduced form.
Circumradius of the equilateral triangle. The centroid of an equilateral triangle of side \(d\) lies at \(d/\sqrt3\) from each vertex:
Not \(d/2\) — this is the commonest slip when the general formula is used without care. The sub-conductors sit further from the bundle centre than half the spacing.
Route 1 — the general formula with \(n = 3\):
Route 2 — the reduced form, which must agree:
The factor 3 in the general form and the \(1/3\) hidden in \(A^2 = d^2/3\) cancel exactly, which is why the triangular case has no leading constant.
Bundle radius for capacitance, with \(r = 28.62/2 = 1.431\) cm:
The multiplying factors, and the comparison with a twin:
Going from two sub-conductors to three nearly doubles the multiplying factor even though the aluminium has risen by only 50%. That is the geometric leverage bundling exploits, and Problem 16 shows how quickly it saturates.
A 765 kV line uses a four-conductor bundle of ACSR Bersimis — overall diameter 35.04 mm, GMR 1.400 cm — on a square of side 457 mm. Find \(A\), \(D_{sb}\) and \(r_b\), confirm that the leading constant really is \(2^{1/8}\), and state the diameter of the single solid conductor that would have the same GMR.
The square's circumradius is half its diagonal:
The diagonal itself is \(45.7\sqrt2 = 64.63\) cm, so each sub-conductor has two neighbours at 45.7 cm and one at 64.63 cm. The polygon product rule packages exactly this: \(45.7^2 \times 64.63 = 134\,980 = 4A^{3}\).
Self-GMD by the general formula:
And by the reduced form, as a check on the constant:
The two agree to four figures. Had 1.09 been an empirical rounding rather than \(2^{1/8}\), the two routes would disagree in the third figure.
Bundle radius for capacitance, with \(r = 1.752\) cm:
The equivalent solid conductor. A solid round conductor of radius \(R\) has \(D_s = 0.7788R\), so matching the bundle's GMR requires
A half-metre solid aluminium rod, weighing some 600 kg per metre. Four Bersimis conductors weigh about 8.6 kg per metre together. The bundle achieves the same electrical radius for roughly 1.4% of the metal, and that ratio — not any subtlety of field theory — is the reason every EHV line in the world is bundled.
The multiplying factor for the quad:
Against 5.96 for a twin and 11.6 for a triple at the same 45 cm spacing. Note that the factor is set almost entirely by \(n\) and \(d\); the sub-conductor enters only as \(D_s^{1/n}\), which for \(n=4\) is a fourth root.
The twin-Moose bundle of Problem 2 is used on a 400 kV, 50 Hz line whose three bundle centres lie in one horizontal plane, 11 m between adjacent phases. The line is fully transposed. Find the equivalent spacing, the inductance per phase per kilometre, the inductive reactance per kilometre, and the total series reactance of a 400 km circuit.
Equivalent spacing. With the outer phases 22 m apart, the mutual GMD of the transposed line is the geometric mean of the three inter-phase distances, exactly as in Set 5:
The distances are measured between bundle centres. Problem 19 shows that this introduces an error of under 0.05%, which justifies the practice.
Inductance per phase, with \(D_{sb} = 0.07551\) m from Problem 2:
Both lengths must be in the same unit before the ratio is taken. Feeding 13.86 m against 7.551 cm gives \(0.2\ln 1.835 = 0.121\) mH/km, an answer wrong by a factor of nine and the single commonest arithmetic failure in this topic.
Reactance per kilometre at 50 Hz:
This is the number to remember for an Indian 400 kV twin-Moose line: about a third of an ohm per kilometre per phase. Planning studies routinely quote 0.32–0.33.
Total for 400 km:
Sanity check on the logarithm. The whole answer hangs on \(\ln(D_{eq}/D_{sb})\), and for real overhead lines this quantity is remarkably confined:
Every overhead line ever built has \(\ln(D_{eq}/D_{sb})\) between about 4 and 7, so \(L\) lies between 0.8 and 1.4 mH/km and \(x_L\) between 0.25 and 0.45 Ω/km. An answer outside that band is an arithmetic error, not a discovery.
Rebuild the line of Problem 5 with a single Moose conductor per phase, the geometry unchanged. By what percentage does bundling reduce the series reactance, and by what percentage does it raise the steady-state power limit \(P_{max} = V_SV_R/X\) with both terminal voltages held at 400 kV? Explain why the two percentages differ.
Single-conductor inductance, using \(D_s = 1.267\) cm directly:
The reactance saving:
A quarter of the series reactance removed by hanging a second conductor 450 mm from the first. Nothing else available to a line designer — not tower geometry, not conductor size — comes close to that.
The power limit is inversely proportional to \(X\), so:
Why 25.5% becomes 34.2%. The two are reciprocal statements about the same number, and a reduction of fraction \(k\) in the denominator is an increase of \(k/(1-k)\) in the quotient:
The asymmetry grows with \(k\): a 50% reactance saving would double the power limit, not raise it by half. Quoting one figure when the other is meant is a real source of confusion in feasibility reports.
A caution about what this limit means. \(P_{max} = V^2/X\) is the theoretical maximum at a load angle of 90°, which no system operates at. With the usual 30° limit the transfer is \(V^2\sin 30^\circ/X\):
The ratio is unchanged — bundling buys the same 34.2% at any load angle — but the absolute numbers are halved, and it is these that a 400 km line is actually planned around. Compare with the thermal limit: twin Moose at 75 °C carries about 1600 A, or 1109 MVA, so a 400 km line is stability-limited and not thermally limited.
The comparison of Problem 6 is unfair: the bundle carries twice the metal. Repeat it honestly. Moose has an aluminium area of 528.5 mm², so the twin bundle carries 1057 mm² per phase. Compare it against a single conductor of the same 1057 mm² of aluminium and the same construction, on the same 13.86 m geometry, for reactance, capacitance, surge impedance, SIL and maximum surface gradient. State what survives of the case for bundling.
Size the equal-metal single conductor. Doubling the area doubles \(r^2\), and geometrically similar construction scales every linear dimension — including the GMR — by the same \(\sqrt2\):
A conductor 44.9 mm in overall diameter. Such conductors are made — the largest ACSR sizes reach about 46 mm — but they are near the practical limit for stringing, damping and hardware.
Its inductance and reactance:
Against 1.3995 mH/km for one Moose. Doubling the aluminium of a single conductor saved 5.0% of the reactance; bundling the same aluminium saved 25.5%.
Its capacitance, from Set 6:
The twin bundle, for comparison — values from Problems 5, 13 and 14:
The gradient figures deserve their own line. For the single conductor \(n = 1\) and the surface field is uniform:
Comfortably under the 21.1 kV/cm at which corona begins. So the honest verdict on corona is not that a single conductor at 400 kV is impossible — it is that a single conductor of ordinary size is impossible, which is Problem 12.
What survives. At equal aluminium the bundle still wins every category, but the margins are smaller than the naive comparison of Problem 6 suggested:
Roughly six-sevenths of the electrical benefit is genuinely geometric and one-seventh was bought with metal. The gradient advantage, which looked overwhelming, is largely a size effect once the sizes are equalised.
And the arguments that are not electrical at all, which decide the matter in practice:
Derive the average electric field strength at the surface of a sub-conductor of an \(n\)-conductor bundle in terms of the phase voltage and the line geometry, and evaluate it for the 400 kV twin-Moose line of Problem 5.
Start from the charge on the phase. The line-to-neutral capacitance found in Set 6 relates charge per unit length to phase voltage:
The charge \(q\) belongs to the phase as a whole. The bundle is an equipotential — the sub-conductors are bonded at every spacer — so the charge divides equally, \(q/n\) on each.
Gauss's law at one sub-conductor. Take a cylinder just outside one sub-conductor of radius \(r\). The flux leaving it is the charge it encloses, and dividing by the surface area gives the average normal field:
Substitute for \(q\) and the \(2\pi\varepsilon_0\) cancels:
Note which radius appears where. The \(r\) outside the logarithm is the physical sub-conductor radius, because that is the surface the field sits on; the \(r_b\) inside is the bundle radius, because that is what sets the charge. Interchanging them is the classic error and Problem 12's mistake list returns to it.
Evaluate for the 400 kV line. Work throughout in centimetres and kilovolts so the answer emerges in kV/cm:
Interpretation. The critical disruptive gradient of air at standard conditions is 21.1 kV/cm rms (30 kV/cm peak), so the average gradient sits at
of the corona threshold. That looks like a comfortable margin, and it is not — the average is not what starts corona. Problem 9 supplies the maximum.
A word on rms against peak. Every gradient in this set is rms, matching the rms 21.1 kV/cm threshold; some texts work in peak throughout and quote 30 kV/cm. Either convention is consistent, mixing them is not, and a factor \(\sqrt2\) in a corona calculation is the difference between a line that glows and one that does not.
The average gradient of Problem 8 is not what starts corona. Derive the maximum gradient on a sub-conductor surface, show that to first order
and evaluate it for the 400 kV twin-Moose line. Give the minimum gradient too, and say where on the surface each occurs.
Why the field is not uniform. Each sub-conductor carries \(q/n\), and each sits in the field of the other \(n-1\). Like charges repel, so the external field at any sub-conductor points away from the bundle axis. On the outer face it adds to the conductor's own field; on the inner face it opposes it.
The external field at one sub-conductor. Treating the others as line charges \(q/n\) at distances \(D_k\), and resolving each along the outward radial direction:
The tangential components cancel by symmetry, which is why only the radial sum survives.
A polygon identity does the sum. For \(n\) points equally spaced on a circle of radius \(A\),
Check it on the three cases. \(n=2\): one neighbour, \(\theta = 0\), \(D = 2A\), giving \(1/2A\). \(n=3\): two neighbours at \(D = A\sqrt3\) with \(\cos\theta = \sqrt3/2\), giving \(2 \times \frac{\sqrt3/2}{A\sqrt3} = 1/A\). \(n=4\): two at \(D = A\sqrt2\) with \(\cos\theta = 1/\sqrt2\) and one at \(D = 2A\) with \(\cos\theta = 1\), giving \(2\times\frac{1}{2A} + \frac{1}{2A} = 3/2A\). All three are \((n-1)/2A\).
A conducting cylinder in an external field doubles it. The classic two-dimensional result: a cylinder placed in a uniform transverse field \(E_{ext}\) acquires an induced surface field of \(2E_{ext}\) at the two poles lying along the field. So at the outer point
Factor out \(E_{av}\), and the promised result appears:
The bracket is the non-uniformity factor of Markt and Mengele. It is a first-order result: the induced charge redistribution is itself neglected in computing \(E_{ext}\), which is accurate while \(r \ll A\) — for a 400 kV twin, \(r/A = 0.07\), so the error is well under 1%.
Evaluate for the twin bundle, with \(A = d/2 = 22.5\) cm:
A 15% swing from one side of a sub-conductor to the other. Corona, when it starts, starts on the outer faces of the bundle — which is exactly where field photographs of energised EHV bundles show it.
Where the line now stands against the 21.1 kV/cm threshold:
Using \(E_{av}\) as the design figure would overstate the margin by 7% of the threshold — enough to matter, since practical designs aim at 0.75–0.80 and a stranded conductor's irregularity factor \(m_0 \approx 0.85\) reduces the effective threshold to about 17.9 kV/cm anyway. Set 12 on corona works with the same numbers from the voltage side.
Widening a bundle raises \(r_b\), which lowers \(E_{av}\), but it also raises \(A\), which lowers the non-uniformity factor — so the two effects act in the same direction on the factor and in opposite directions overall. Show that \(E_{max}\) nevertheless has a minimum, derive the condition for it, and find the optimum spacing for the 400 kV twin-Moose line.
Write \(E_{max}\) as a function of \(A\) alone. Collecting Problems 8 and 9:
Expand the logarithm so the dependence on \(A\) is explicit:
So \(\Lambda' = -\dfrac{n-1}{nA}\), and everything else in \(E_{max}\) is elementary.
Why a minimum must exist. As \(A \to 0\) the bracket \(1+(n-1)r/A \to \infty\) faster than \(\Lambda\) grows, so \(E_{max}\to\infty\); as \(A\) grows towards \(D_{eq}\), \(\Lambda \to 0\) and \(E_{max}\to\infty\) again. A positive continuous function that diverges at both ends of an interval has an interior minimum.
Differentiate and set to zero. With \(c = (n-1)r\) and \(g = (1+c/A)/\Lambda\):
Implicit, because \(\Lambda\) itself contains \(A\). But \(\Lambda\) is a logarithm and moves slowly, so two or three iterations settle it.
Iterate for the twin-Moose line: \(n = 2\), \(r = 1.5885\) cm, \(D_{eq} = 1386\) cm, so \(A_{opt} = r(2\Lambda - 1)\). Start from the practical value \(A = 22.5\) cm:
The gradient at the optimum:
Against 15.26 kV/cm at the standard 45 cm spacing — an improvement of 0.66%. All that theory for two-thirds of one per cent.
That is the real result, and it is a negative one. The optimum is genuine, and it is genuinely worthless. Since the electrical penalty for departing from \(d_{opt}\) is negligible over a wide range, the spacing is settled on other grounds entirely:
For Moose, \(d/D_{sub} = 450/31.8 = 14.2\), comfortably inside the wake-stability rule. That constraint — not the gradient — is what fixes 450 mm as the industry standard for a 400 kV twin.
The optimum for other bundles follows from the same formula, and it always lands well below the practical spacing:
Tabulate \(r_b\), \(E_{av}\), \(E_{max}\) and the series reactance of the 400 kV twin-Moose line for bundle spacings from 200 mm to 600 mm. Quantify how flat the gradient minimum is, and say what the reactance column implies for the choice of spacing.
The three quantities as functions of \(d\), all from Problems 5, 8 and 9 with \(D_{eq} = 1386\) cm:
with \(d\) in centimetres, and remembering \(A = d/2\) so \((n-1)r/A = 2r/d = 3.177/d\).
The table:
How flat the minimum is. Over the entire band 250–500 mm — a factor of two in spacing — \(E_{max}\) moves from 15.19 to 15.32 kV/cm:
One per cent for a doubling of the spacing. A stationary point is quadratic in its neighbourhood, so this flatness is expected — but the practical consequence is worth stating plainly: the gradient cannot be the criterion that fixes the spacing.
The reactance column tells a different story. It falls monotonically and by rather more:
Ten per cent of reactance against one per cent of gradient. If any electrical quantity is to choose the spacing, it is the reactance, and it argues for the widest bundle the hardware will carry.
Why the two behave so differently. The reactance depends on \(d\) only through \(\ln D_{sb} = \tfrac12\ln(D_s d)\) — one-sided, monotone, no competing term. The gradient has two terms pulling opposite ways, and they cancel to first order at the optimum. A quantity with an interior optimum is always less sensitive near it than a monotone one.
Take the 400 kV geometry of Problem 5 with a single Moose conductor per phase. Find its surface gradient and decide whether the line will corona in fair weather, taking the irregularity factor of a stranded conductor as \(m_0 = 0.85\). Then find the smallest single conductor that would be acceptable, and repeat the exercise at 765 kV.
Gradient of the single conductor. With \(n = 1\) the bundle radius is the conductor radius, and there is no non-uniformity factor:
The threshold to compare it with. Corona starts when the surface gradient reaches \(m_0 g_0 \delta\), with \(g_0 = 21.1\) kV/cm rms and \(\delta = 1\) at standard conditions:
Twenty per cent over the threshold, in fair weather at sea level. In rain, when water drops on the underside of the conductor act as sharp protrusions, the effective \(m_0\) falls towards 0.6–0.7 and the excess becomes 50%. The line would be in continuous heavy corona.
Compare with the twin bundle on the identical geometry:
The bundle turns a 20% violation into a 15% margin. That, not the reactance, is why 400 kV lines were bundled from the first day they were built.
The smallest acceptable single conductor. Set \(E = E_{crit}\) and solve for \(r\):
The left side is not invertible in closed form, so iterate: \(r = 1.95 \Rightarrow 12.80\); \(r = 2.00 \Rightarrow 13.08\); \(r = 1.963 \Rightarrow 12.87\).
Conductors of that size exist — ACSR Chukar is 40.7 mm — so a single-conductor 400 kV line is not electrically impossible, merely bad. Such a conductor has \(D_s \approx 1.56\) cm, giving \(x_L = 0.427\) Ω/km and a SIL of 400 MW — 30% more reactance and 23% less SIL than the twin bundle, on three-quarters of the aluminium. The metal saved is real; the performance given up is much larger.
At 765 kV the argument becomes absolute. With \(D_{eq} = 17.01\) m and \(V_{ph} = 441.7\) kV:
Nothing of the kind is manufactured, and if it were, no tower could be economically designed to carry three of them plus ice and wind. Above about 300 kV a bundle is not a refinement; it is the only way the line exists at all.
Find the line-to-neutral capacitance, the shunt susceptance and the charging current per kilometre of the 400 kV twin-Moose line, and the total charging reactive power of a 400 km circuit. Compare the capacitance with the single-conductor version and note which radius enters where.
Capacitance to neutral, from Set 6 with \(r_b\) in place of \(r\):
The constant 0.05563 µF/km is \(2\pi\varepsilon_0\) expressed per kilometre; many texts write 0.0556 or 0.02412 for the base-10 form. Earth-image effects are neglected, which for a 400 kV line at 12 m mean height costs about 1% — Set 6 Problem 5 quantifies it.
Susceptance and charging current per kilometre:
Over 400 km:
Two hundred megavars generated by an unloaded line, which is why a 400 km EHV circuit is normally fitted with shunt reactors — typically 50–80 MVAr at each end, plus a line reactor. Set 6 Problem 19 treats the compensation calculation.
Against the single conductor:
The capacitance rises by about the same proportion as the reactance falls, which is no coincidence — both are governed by logarithms of the same two radii, and Problem 16 makes the relation exact.
Which radius goes where — the single most-repeated error in this topic:
Using \(D_{sb}\) for the capacitance here would give 0.01067 µF/km, low by 2.2%. Not fatal, but it propagates into \(Z_c\) and SIL and it is entirely avoidable.
Find the surge impedance and the surge impedance loading of the 400 kV twin-Moose line and of the single-conductor version, state the percentage gain, and verify the common approximation \(Z_c \approx 60\ln\!\left(D_{eq}/\sqrt{D_{sb}r_b}\right)\).
Surge impedance of the bundled line, from the values already established:
Both quantities are per kilometre and the kilometres cancel, which is why \(Z_c\) is a property of the cross-section alone and does not depend on the line's length.
Surge impedance loading:
The published SIL of an Indian 400 kV twin-Moose line is about 515 MW, so the calculation is reproducing a real number to within the rounding of the conductor data.
The single-conductor line:
The approximation. Substituting the two formulae into \(\sqrt{L/C}\) and writing \(\Lambda_L = \ln(D_{eq}/D_{sb})\), \(\Lambda_C = \ln(D_{eq}/r_b)\):
So the exact statement uses the geometric mean of the two logarithms. The familiar form \(60\ln(D_{eq}/\sqrt{D_{sb}r_b})\) is \(60 \times \tfrac12(\Lambda_L+\Lambda_C)\) — the arithmetic mean.
Compare the two on this line, with \(\Lambda_L = 5.2124\) and \(\Lambda_C = 5.0994\):
A discrepancy of 0.08%, because the arithmetic and geometric means of two nearly equal numbers agree to second order. The approximation is safe precisely because \(r\) and \(D_s\) are close — Problem 2 showed that bundling brings them closer still.
What SIL means, in one line. At \(P = \text{SIL}\) the line's own charging exactly supplies its own \(I^2X\) demand:
Equal, as they must be. The line is then reactively self-sufficient, the voltage profile is flat from end to end, and the reactive power flow is zero everywhere — the condition Set 9 calls natural loading.
A 765 kV, 50 Hz line uses the quad Bersimis bundle of Problem 4 — square of side 457 mm — with the three bundle centres horizontal at 13.5 m between adjacent phases, fully transposed. Find \(D_{eq}\), \(L\), \(x_L\), \(C_n\), \(Z_c\), SIL and \(E_{max}\), and comment on the corona margin.
Equivalent spacing:
Inductance and reactance, with \(D_{sb} = 0.2085\) m from Problem 4:
Lower than the 400 kV line's 0.3275 Ω/km despite phases 13.5 m apart rather than 11 m — the quad bundle more than pays for the wider geometry.
Capacitance, with \(r_b = 0.2205\) m:
Surge impedance and SIL:
One 765 kV circuit carries at natural loading more than four times what a 400 kV circuit carries, on 1.9 times the voltage and 2.6 times the metal. That leverage is why the 765 kV level exists.
Surface gradient, with \(V_{ph} = 765/\sqrt3 = 441.7\) kV, \(n = 4\), \(r = 1.752\) cm and \(A = 32.32\) cm:
The corona margin is thin. Against the same \(m_0g_0 = 17.94\) kV/cm:
A 765 kV line runs much closer to the corona threshold than a 400 kV one, and its non-uniformity factor is more than twice as costly. That is why 765 kV designs are audited for audible noise and radio interference as a matter of routine, and why some utilities move to a six-conductor bundle rather than accept the margin.
Summary of the two lines side by side:
The last row divides SIL by the aluminium per phase (1057 mm² against 2756 mm²). Moving to 765 kV raises the power carried per unit of metal by two-thirds, which is the whole economic argument for EHV.
Keeping the Moose sub-conductor, the 450 mm spacing and the 13.86 m geometry fixed, tabulate \(x_L\), \(C_n\), \(Z_c\) and SIL for \(n = 1, 2, 3, 4\). Show analytically that the SIL gain is the geometric mean of the inductive and capacitive gains, explain why the SIL gain exceeds the reactance saving, and quantify how fast the returns diminish against the metal added.
The analytic relation first, because it makes the table readable. Writing \(\Lambda_L = \ln(D_{eq}/D_{sb})\) and \(\Lambda_C = \ln(D_{eq}/r_b)\), every quantity in the problem is a simple function of those two logarithms:
Hence the geometric-mean statement. If bundling multiplies \(\Lambda_L\) by \(\alpha\) and \(\Lambda_C\) by \(\beta\), then
The SIL factor is the geometric mean of the reactance factor and the capacitance factor. For the single-to-twin step, \(\alpha = 5.2124/6.9975 = 0.7449\) and \(\beta = 5.0994/6.7715 = 0.7531\), so \(1/\sqrt{\alpha\beta} = 1.335\) — the 33.5% of Problem 14, obtained without touching a square root of an impedance.
Why the SIL gain beats the reactance saving. The reactance saving is \(1-\alpha = 25.5\%\), a fractional reduction; the SIL gain is \(1/\sqrt{\alpha\beta} - 1 = 33.5\%\), a fractional increase, and the two are reciprocal-shaped statements. Comparing them properly:
SIL rises slightly less than the power limit because the capacitance improves a shade less than the inductance does — \(\beta > \alpha\), since \(r_b\) exceeds \(D_{sb}\). Had they improved equally, the two figures would coincide exactly.
The table, all at \(D_{eq} = 13.86\) m and \(d = 45\) cm:
The returns, step by step, against the metal each step costs:
Every step costs exactly one Moose, so the megawatt column compares like with like: the second sub-conductor is worth twice the third and two and a half times the fourth. Carrying the table on, a fifth sub-conductor adds about 7%, and a sixth less than 5% — which is why five- and six-conductor bundles appear only where the surface gradient, not the SIL, is the binding constraint, as at 1000 kV.
The mathematical reason for the saturation is that the benefit enters through \(\ln D_{sb}\), and
As \(n\) grows the second term tends to \(\ln A\) and the first to zero, so \(D_{sb} \to A\) — the bundle GMR can never exceed the circle the sub-conductors sit on. With \(A = 31.8\) cm for a square of side 45 cm, the quad is already at 20.1 cm, or 63% of the ceiling. Adding conductors buys the remaining 37% and no more; widening the bundle is the only way to raise the ceiling itself.
One sub-conductor of the \(a\)-phase bundle on the 400 kV line of Problem 5 fails at a tension clamp and is removed; the remaining conductor and the other two phases are untouched. The line is left in service carrying 1000 A per phase. Quantify every consequence you can — reactance, gradient, losses, thermal state, balance — and say whether the line should be run.
The \(a\)-phase becomes a single conductor. Its self-GMD collapses from the bundle value to the sub-conductor's own:
Reactance of that phase rises to the single-conductor value of Problem 6:
The line is no longer balanced, and transposition cannot fix it. Transposition equalises the mutual geometry between phases; here it is one phase's self impedance that has changed, and no rotation of positions restores symmetry:
So the sequence networks of Set 21 acquire off-diagonal terms: a positive-sequence current now produces negative- and zero-sequence voltage drops. A balanced load draws unbalanced current, and negative-sequence relays on nearby generators will see a standing input.
Surface gradient — the consequence that decides the matter. With \(n = 1\) the whole phase charge now sits on one conductor:
Twenty per cent above the corona threshold, exactly as in Problem 12 — because it is the same conductor in the same geometry. The phase will be in continuous visible corona, audible from the ground, with measurable radio and television interference and an ozone smell under the line.
Current and losses. The surviving conductor carries the entire 1000 A instead of 500 A. With \(R_{sub} = 0.067\) Ω/km at 75 °C, per kilometre of the \(a\)-phase:
Exactly double, and it appears in one conductor rather than two. Over 400 km that is 13.4 MW of extra loss in one phase. At a loss-load factor of 0.7 that is about 82 GWh a year, worth roughly ₹33 crore at ₹4 per kWh.
The thermal state is the immediate danger. A single Moose is rated at about 800 A for a 75 °C conductor temperature, which is why the bundle is rated 1600 A:
Since the temperature rise above ambient is roughly proportional to the heat input at a fixed wind speed, a conductor sitting at 75 °C above a 40 °C ambient — a 35 °C rise — would settle near \(40 + 1.56 \times 35 = 95\) °C. Aluminium begins to anneal and lose tensile strength above about 90 °C, sag increases and clearance to ground falls. And this is at 1000 A; at the bundle's rated 1600 A the surviving conductor would be at twice its own rating and would fail.
The verdict, and the order of the reasoning. Three independent findings, any one of which is sufficient:
The line should be taken out and repaired. If it must be run to keep a load supplied, it can be run derated — at 800 A per phase the thermal problem disappears and only the corona and unbalance remain, both tolerable for a few days. That is the practical answer a control-room engineer gives.
Why the failure happens at all, and what actually breaks. A bundle's sub-conductors are held by spacers every 45–70 m. A spacer that seizes or fatigues allows the sub-conductors to clash in wind, and the resulting arcing damage at the clash point is the usual precursor to a broken strand. The failure mode this problem describes is therefore almost always a spacer failure two steps earlier, and it is the reason spacer-dampers are inspected on EHV lines rather than merely installed.
A 220 kV line uses a single ACSR Zebra per phase — diameter 28.62 mm, GMR 1.144 cm, aluminium 428.9 mm², mass 1.62 kg/m, ac resistance 0.0856 Ω/km at 75 °C — with phases 7 m apart horizontally. Decide whether bundling it as a twin at 400 mm is justified. Compute the gradient, the reactance and the SIL both ways, then settle the question on economic grounds at loadings of 200 A and 400 A.
Geometry:
Test the corona case first, because if it binds, the economics are irrelevant. For the single conductor:
Comfortably inside the limit, with room for the air-density factor to fall to 0.77 — an altitude of around 2000 m — before corona appears. Corona does not require a bundle at 220 kV. That is the whole difference from Problem 12, and it means the decision here is an economic one and not a physical one.
The twin for comparison, with \(d = 40\) cm:
Reactance, capacitance and SIL both ways:
Electrically the twin wins everywhere, by margins as large as at 400 kV. The question is whether the margins are worth paying for.
Note first that SIL is not the binding limit at 220 kV. A single Zebra carries about 700 A thermally, which at 220 kV is
So a 220 kV circuit routinely runs at twice its natural loading, absorbing reactive power rather than generating it. Raising SIL from 123 to 168 MW does not raise what the line can carry; it changes the reactive balance. The genuine benefits are the reactance and the losses.
The cost of the second conductor, per kilometre of line, at indicative Indian prices:
The benefit is the loss saving, since the twin halves the phase resistance from 0.0856 to 0.0428 Ω/km. At a line current \(I\):
Simple payback on ₹21.9 lakh/km:
Seventeen years is longer than most utilities' evaluation horizon and the answer is no; four years and a bit is a clear yes. The decision turns entirely on the loading, exactly as Kelvin's law of economic conductor size predicts — and the correct question to ask the planner is not "should we bundle?" but "what will this corridor carry in ten years?"
The verdict, stated properly. At 220 kV bundling is an economic conductor decision, not an insulation decision:
Twin-conductor 220 kV lines are built in India on exactly this reasoning, on corridors where the alternative is a second circuit or an early conversion to 400 kV. Both of those alternatives should be costed in the same comparison.
Every calculation in this set has used the distance between bundle centres as the inter-phase spacing. Justify it. Compute the exact mutual GMD between two twin bundles whose centres are 11 m apart with the sub-conductors lying along the line of centres at 450 mm, obtain a general expression for the error, and state when the approximation would fail.
Write down the four distances. With the two sub-conductors of each bundle at \(\pm d/2\) along the line of centres, the four conductor-to-conductor distances are
Evaluate:
Where that number comes from. Taking logarithms before evaluating shows the structure:
The first-order terms in \(d/D\) have cancelled: one conductor is nearer by \(d\) and one is farther by \(d\), and in a logarithmic average those offset. Only the second-order term survives, and \((0.45/11)^{2}/4 = 4.18\times10^{-4}\) reproduces the answer.
The general result, obtained the same way for an \(n\)-conductor bundle. The \(n\) sub-conductors are the roots of \(z^{n} = A^{n}\) about their centre, so the log-average of the distances from an external point at distance \(D\) is
Doing it once for each bundle and adding:
Test it on the three standard bundles — and note how violently it collapses with \(n\):
A quad bundle's centre is an exact stand-in for the bundle to better than one part in a million. The reason is that a symmetric group of \(n\) charges has no multipole moment below the \(n\)-th, so the leading error is \((A/D)^n\).
Orientation matters only at the same order. If the twin bundle is rotated to lie perpendicular to the line of centres, the four distances become \(D\), \(D\), \(\sqrt{D^{2}+d^{2}}\), \(\sqrt{D^{2}+d^{2}}\) and
The same magnitude, opposite sign — as the general expression predicts, since rotating by \(\pi/n\) flips the sign of \((A/D)^{n}\). A vertical twin bundle and a horizontal one differ by 0.08% in mutual GMD, which is beneath the accuracy of any conductor catalogue.
Effect on the answer that matters. The inductance depends on \(\ln(D_{eq}/D_{sb}) = 5.2124\), so an error of \(4.19\times10^{-4}\) in \(\ln D_{eq}\) gives
Eight parts in a hundred thousand. The approximation is not merely acceptable; it is far below every other uncertainty in the calculation, including the third figure of the conductor's GMR.
When it would fail. The error scales as \((A/D)^{n}\), so it becomes visible only when the bundle is large compared with the phase spacing:
The last row is the practical case: on a compact or double-circuit tower, conductors of different phases can come within a metre or two, and there the sub-conductor positions must be used individually. Modern high-surge-impedance-loading designs, which deliberately expand the bundle and compact the phases, are precisely the case where the shortcut breaks.
Assemble the complete parameter set for the 400 km, 400 kV twin-Moose line of Problem 5, taking the ac resistance of each sub-conductor as 0.067 Ω/km at 75 °C. Give \(R\), \(X\) and \(B\) in ohms and siemens and in per-unit on 100 MVA, then \(Z_c\), SIL, the electrical length, the exact \(ABCD\) constants, the no-load voltage rise and the loss at natural loading.
Resistance per phase. Two sub-conductors in parallel, each 0.067 Ω/km:
The parallel halving is exact here because the sub-conductors are bonded at every spacer and carry equal currents; a bundle without spacers would not share equally and this step would be wrong.
The per-kilometre set, collecting Problems 5, 13 and 14:
Totals for 400 km:
Per-unit on 100 MVA, 400 kV:
The susceptance in per-unit is numerically the charging MVAr divided by the base MVA — \(219.4/100 = 2.194\) — which is a useful check and a reminder that a large per-unit susceptance is normal, not an error. The \(X/R\) ratio is 9.8, typical of an EHV line.
Surge impedance and natural loading, now taking the resistance into account:
The lossless value of Problem 14 was 309.1 Ω; including \(R\) changes the magnitude by 0.3% and adds a small negative angle. SIL is quoted from the lossless value, 517.6 MW.
Propagation constant and electrical length:
At 24.3° the line is long enough that the nominal-π model is no longer adequate; Set 10's exact equations are required. The velocity being 99% of light is the usual result for an overhead line and is a good check on \(L\) and \(C\) together — if it comes out above \(c\), one of them is wrong.
The exact \(ABCD\) constants:
Check: \(AD - BC = 1\), as it must for a passive reciprocal two-port. Note that \(|B| = 127.8\) Ω is 3% below the lumped \(|Z| = 131.7\) Ω — the distributed capacitance partially compensates the series reactance, which is why a long line's true power limit exceeds the lumped estimate.
The two performance figures that follow:
Against the 1221 MW that Problem 6's lumped estimate gave — 2.5% optimistic in the right direction. And the 9.7% no-load rise on a 400 kV line means 439 kV at the open end, above the 420 kV highest system voltage, so shunt reactors are not optional on a line of this length.
Loss at natural loading:
Four per cent over 400 km at natural loading — the figure that makes EHV transmission worth building. The same power at 220 kV would need two Zebra circuits and would lose about four times as much.
The complete summary card:
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-conductors of GMR 1.10 cm at 400 mm spacing. Find \(D_{sb}\).
Show answer
\(\sqrt{1.10 \times 40} = \mathbf{6.63}\) cm. Problem 1.P2. A four-conductor bundle has sub-conductors of GMR 1.40 cm on a square of side 450 mm. Find \(D_{sb}\).
Show answer
\(1.0905\sqrt[4]{1.40 \times 45^{3}} = \mathbf{20.61}\) cm. Problem 4.P3. A three-conductor bundle uses conductors of outside radius 1.60 cm at 450 mm. Find the bundle radius for capacitance.
Show answer
\(\sqrt[3]{1.60 \times 45^{2}} = \mathbf{14.80}\) cm — note \(r\), not \(D_s\). Problem 3.P4. Where does the constant 1.0905 in the four-conductor formula come from?
Show answer
It is \(\mathbf{2^{1/8}}\) exactly, arising because a square's circumradius is \(d/\sqrt2\). Problem 1.P5. A twin bundle has \(D_{sb} = 7.55\) cm on a line with \(D_{eq} = 13.86\) m. Find \(x_L\) at 50 Hz.
Show answer
\(0.2\ln(183.5) = 1.0425\) mH/km, so \(x_L = \mathbf{0.3275}\) Ω/km. Problem 5.P6. Bundling drops \(\Lambda_L\) from 6.998 to 5.212 and \(\Lambda_C\) from 6.772 to 5.099. Find the percentage gain in SIL without computing any impedance.
Show answer
\(1/\sqrt{0.7449 \times 0.7531} = 1.335\), so \(\mathbf{+33.5\%}\). Problem 16.P7. Find the average surface gradient of the 400 kV twin-Moose line (\(r = 1.5885\) cm, \(r_b = 8.455\) cm, \(D_{eq} = 13.86\) m).
Show answer
\(230.94/(2 \times 1.5885 \times 5.0994) = \mathbf{14.25}\) kV/cm rms. Problem 8.P8. Find the non-uniformity factor for that bundle, and hence \(E_{max}\).
Show answer
\(1 + 1.5885/22.5 = \mathbf{1.071}\), so \(E_{max} = \mathbf{15.26}\) kV/cm. Problem 9.P9. Two twin bundles have their centres 11 m apart, sub-conductors 450 mm apart along the line of centres. Find the mutual GMD and the error in using 11 m.
Show answer
\([121(121-0.2025)]^{1/4} = \mathbf{10.9954}\) m; error \(\mathbf{-0.042\%}\). Problem 19.P10. As the number of sub-conductors grows with the circumradius \(A\) held fixed, what does \(D_{sb}\) approach?
Show answer
\(\mathbf{A}\) itself — the bundle becomes a hollow cylindrical shell. Any computed \(D_{sb} > A\) is an arithmetic error. Problem 16.P11. A single Moose conductor is used at 400 kV with \(D_{eq} = 13.86\) m. Find its surface gradient and say whether it coronas, taking \(m_0 = 0.85\).
Show answer
\(\mathbf{21.47}\) kV/cm against a threshold of 17.94 — yes, by 20%. Problem 12.P12. One sub-conductor of a twin bundle is lost. By what percentage does that phase's series reactance rise?
Show answer
\(0.4397/0.3275 = 1.342\), so \(\mathbf{+34.2\%}\) — and the gradient rises 40.7%. Problem 17.
Challenge Problems
Three problems that want an idea rather than a formula — a proof, a design that pushes against its own limits, and a diagnosis from a single measured number.
C1 — The polygon product rule. Prove that for \(n\) points equally spaced on a circle of radius \(A\), the product of the distances from any one of them to the other \(n-1\) is exactly \(nA^{n-1}\). Then use it to prove that \(D_{sb} < A\) always, with equality only in the limit.
Show answer
Set up in the complex plane. Place the \(n\) points at the \(n\)-th roots of \(A^{n}\):
\[ z_k = A\,e^{\,i2\pi k/n}, \qquad k = 0,1,\dots,n-1 \]These are exactly the roots of the polynomial \(p(z) = z^{n} - A^{n}\), so it factorises as
\[ z^{n} - A^{n} = \prod_{k=0}^{n-1}(z - z_k) \]Isolate the factor belonging to the point of interest. Take \(z_0 = A\) and divide both sides by \((z - z_0)\):
\[ \frac{z^{n}-A^{n}}{z - A} = \prod_{k=1}^{n-1}(z - z_k) \]Let \(z \to A\). The right-hand side becomes the product we want. The left-hand side is a \(0/0\) limit; by L'Hôpital, or by recognising the definition of the derivative,
\[ \lim_{z\to A}\frac{z^{n}-A^{n}}{z-A} = \left.\frac{d}{dz}\left(z^{n}-A^{n}\right)\right|_{z=A} = nA^{\,n-1} \]Taking moduli, \(\prod_{k=1}^{n-1}|A - z_k| = nA^{n-1}\), which is the rule. The same argument at any other vertex gives the same answer, which is the symmetry step Problem 1 assumed.
A direct check for \(n = 6\), since the six-conductor bundle is real. The distances from one vertex of a regular hexagon of circumradius \(A\) to the others are \(A\), \(A\sqrt3\), \(2A\), \(A\sqrt3\), \(A\):
\[ A \cdot A\sqrt3 \cdot 2A \cdot A\sqrt3 \cdot A = 6A^{5} \quad\checkmark \]Now the ceiling. From Problem 1, \(D_{sb} = (nD_sA^{n-1})^{1/n}\), so
\[ \frac{D_{sb}}{A} = \left(\frac{nD_s}{A}\right)^{1/n} \]The ratio is less than 1 whenever \(nD_s < A\). For a 400 kV twin, \(nD_s = 2.53\) cm against \(A = 22.5\) cm; for the quad of Problem 4, \(5.6\) cm against 32.3 cm. In every practical bundle the inequality is comfortable.
And in the limit. Take logarithms:
\[ \ln\frac{D_{sb}}{A} = \frac{1}{n}\ln\frac{nD_s}{A} \;\to\; 0 \quad \text{as } n \to \infty \]because \(\ln n / n \to 0\). So \(D_{sb} \to A\) from below: an infinitely subdivided bundle is a hollow conducting shell of radius \(A\), whose GMR is exactly \(A\) — which is also the standard result for a thin-walled tube, obtained here by a completely different route. The two agreeing is a good sign that neither is wrong.
The engineering content of the proof is the ceiling. No number of sub-conductors can make a bundle behave like a conductor larger than the circle they sit on, so a design that needs more electrical radius must widen the bundle, not subdivide it further. Problem 16's table of diminishing returns is this theorem in numerical form.
C2 — The compact expanded-bundle line. A high-surge-impedance-loading design for 400 kV compacts the phase spacing to 8 m and expands the twin Moose bundle to 1.2 m. Compute \(L\), \(C_n\), \(Z_c\), SIL and \(E_{max}\), decide whether the design is admissible, and identify every assumption made earlier in this set that it invalidates.
Show answer
Geometry. \(D_{eq} = 8\sqrt[3]{2} = 10.08\) m; with \(d = 120\) cm,
\[ D_{sb} = \sqrt{1.267 \times 120} = 12.33\ \text{cm}, \qquad r_b = \sqrt{1.5885 \times 120} = 13.81\ \text{cm} \]Parameters.
\[ L = 0.2\ln\frac{1007.9}{12.33} = 0.8807\ \text{mH/km}, \qquad x_L = 0.2767\ \Omega/\text{km} \]\[ C_n = \frac{0.05563}{\ln(1007.9/13.81)} = 0.01297\ \mu\text{F/km}, \qquad Z_c = 260.6\ \Omega, \qquad \text{SIL} = 614\ \text{MW} \]The gain is real and large: 614 MW against 518 MW for the conventional line of Problem 14, a rise of 18.6%, with the reactance down 15.5% — from one 400 kV circuit, without changing the conductor or the voltage. That is what the technique is for.
Now the gradient, which is where the bill arrives. Compacting the phases reduces \(\ln(D_{eq}/r_b)\), and the gradient is inversely proportional to it:
\[ E_{av} = \frac{230.94}{2(1.5885)(4.2901)} = 16.94\ \text{kV/cm} \]\[ E_{max} = 16.94\left(1 + \frac{1.5885}{60}\right) = 16.94 \times 1.0265 = 17.39\ \text{kV/cm} \]\[ \frac{17.39}{17.94} = 0.969 \]Verdict: electrically admissible, practically not. The design sits at 97% of the fair-weather corona threshold at standard air density. Any of the following removes the margin entirely: an altitude of 300 m (\(\delta = 0.965\)), a 5% overvoltage, rain, or a conductor with a slightly worse irregularity factor. A design office would reject it and either use a larger sub-conductor or go to a triple bundle.
What it invalidates, item by item:
\[ \begin{array}{ll} \text{Problem 19} & A/D = 0.6/8 = 0.075,\ \text{so the bundle-centre GMD is out by } 0.56\% \\ \text{Problem 10} & \text{the spacing is now far past } d_{opt};\ E_{max}\ \text{is rising, not flat} \\ \text{Problem 9} & r/A = 0.026,\ \text{so the first-order factor is now very accurate — this one survives} \\ \text{Problem 6} & \text{clearance and switching-surge withstand, not reactance, now limit the phase spacing} \end{array} \]And two mechanical consequences that the electrical calculation cannot see. A 1.2 m bundle has four times the short-circuit pinch energy per spacer span of a 0.45 m bundle, and the sub-conductors are far enough apart that each behaves aerodynamically as an isolated conductor — so sub-span oscillation is less of a problem, but galloping amplitude and the spacer's bending moment are both worse.
The general lesson. Compacting the phases and expanding the bundle both raise SIL, and both raise the surface gradient. The two levers are not independent, and a design that pulls them together runs out of corona margin long before it runs out of SIL. The correct design variable to add is sub-conductor diameter, which raises SIL a little and lowers the gradient a lot.
C3 — The commissioning measurement. A newly built 400 km, 400 kV twin-Moose line is designed for 0.3275 Ω/km. The commissioning test measures 141.2 Ω total, that is 0.3530 Ω/km. The tower geometry has been surveyed and is correct. Identify the cause quantitatively, list the alternatives you eliminated, and say what else the fault would have changed.
Show answer
Convert the discrepancy into a length ratio. Everything enters the reactance through one logarithm, and at 50 Hz one unit of that logarithm is worth
\[ 2\pi(50)(0.2\times10^{-3}) = 0.06283\ \Omega/\text{km per unit of }\ln \]\[ \Delta\ln = \frac{0.3530-0.3275}{0.06283} = 0.4058 \quad\Rightarrow\quad \frac{(D_{eq}/D_{sb})_{\text{actual}}}{(D_{eq}/D_{sb})_{\text{design}}} = e^{0.4058} = 1.501 \]A factor of exactly 1.50 — and only two quantities can produce it. Either \(D_{eq}\) is 1.5 times larger than designed, or \(D_{sb}\) is 1.5 times smaller.
Eliminate the first. A \(D_{eq}\) of \(1.5 \times 13.86 = 20.8\) m would need adjacent phases 16.5 m apart. The survey excludes it, and so does the tower drawing: no 400 kV crossarm is 33 m wide.
So \(D_{sb}\) is wrong:
\[ D_{sb} = \frac{7.551}{1.501} = 5.032\ \text{cm} \quad\Rightarrow\quad d = \frac{D_{sb}^{2}}{D_s} = \frac{5.032^{2}}{1.267} = 20.0\ \text{cm} \]The bundle has been strung with 200 mm spacers instead of 450 mm. That is a stock item, it is a plausible warehouse error, and 200 mm is a real spacer size — used on 132 kV twin bundles. The number falls out exactly, which is the signature of a discrete error rather than an accumulation of tolerances.
The alternatives, and why each fails:
\[ \begin{array}{ll} \text{Wrong conductor (smaller } D_s) & \text{would need } D_s = 1.267/2.25 = 0.563\ \text{cm, a 132 kV conductor — visible on site} \\ \text{Line not transposed} & \text{changes phase impedances unequally, not all three by the same } 7.8\% \\ \text{One sub-conductor missing} & \text{would give } 0.4397,\ \text{not } 0.3530 \\ \text{Measurement error} & \text{possible, but 7.8\% is far outside test-set accuracy} \\ \text{Temperature or frequency} & \text{neither affects the inductive reactance} \end{array} \]What else the fault changed, and these are the corroborating measurements to call for:
\[ \begin{array}{lcc} & \text{design }(d=450) & \text{actual }(d=200) \\ \hline r_b\ (\text{cm}) & 8.455 & 5.637 \\ C_n\ (\mu\text{F/km}) & 0.01091 & 0.01011 \\ Q_c\ \text{over }400\ \text{km (MVAr)} & 219.4 & 203.2 \\ Z_c\ (\Omega) & 309.1 & 333.4 \\ \text{SIL (MW)} & 517.6 & 479.9 \\ E_{max}\ (\text{kV/cm}) & 15.26 & 15.30 \end{array} \]A capacitance measurement would confirm it independently: the charging current should be 7.4% low. Notice that the gradient is almost unchanged — Problem 11's flat optimum again — so a corona survey would find nothing and would be the wrong test to order.
The cost of leaving it. SIL down 38 MW and the power limit down 7.2% on a circuit built to carry 500 MW over 400 km. Restringing the spacers on 400 km of twin bundle is a large job — roughly 6700 spacers per phase — but the alternative is a permanent 7% derating of a line that cost several hundred crore. It is corrected.
The general lesson: a single well-chosen measurement can identify a discrete construction error, because the logarithmic structure of line parameters turns a percentage discrepancy into a clean ratio. Convert to \(\Delta\ln\) first; the ratio it produces is usually a recognisable number.
Multiple-Choice Questions
MCQ 1. The self-GMD of a two-conductor bundle of spacing \(d\) is:
(a) \(D_sd\) (b) \(\sqrt{D_sd}\) (c) \(2\sqrt{D_sd}\) (d) \(\sqrt{D_sd/2}\)Show answer
(b). Option (d) is what you get by using \(A = d/2\) without the compensating factor \(n = 2\) in the general formula — a very common slip. Problem 1.MCQ 2. The constant 1.09 in the four-conductor bundle formula is:
(a) an empirical fit (b) \(2^{1/8}\) (c) \(4^{1/4}\) (d) \(\sqrt{2}/\sqrt[3]{2}\)Show answer
(b) = 1.0905, exact. It comes from the square's circumradius \(d/\sqrt2\). Option (c) is \(\sqrt2 = 1.414\) and would be right only if the sub-conductors were all at distance \(d\) from each other. Problem 1.MCQ 3. In the bundle radius used for capacitance, the quantity that replaces \(D_s\) is:
(a) the conductor's GMR (b) its outside radius (c) \(0.7788r\) (d) the bundle circumradiusShow answer
(b). Charge sits on the surface, so no internal-flux correction applies; (c) is the reverse substitution and would lower the capacitance by about 2%. Problem 13.MCQ 4. The circumradius of a three-conductor bundle of side \(d\) is:
(a) \(d/2\) (b) \(d/\sqrt2\) (c) \(d/\sqrt3\) (d) \(d\sqrt3\)Show answer
(c). Answer (a) treats the triangle as if the third conductor were absent and gives \(D_{sb}\) 9% low; (b) is the square's value. Problem 3.MCQ 5. Bundling a 400 kV line as a twin reduces its series reactance by roughly:
(a) 5% (b) 25% (c) 50% (d) 75%Show answer
(b) — 25.5% for the line of Problem 6. Answer (a) is what doubling a single conductor's aluminium achieves, which is precisely the comparison that makes bundling look good. Problems 6 and 7.MCQ 6. The maximum surface gradient on a sub-conductor occurs:
(a) on the face towards its neighbour (b) on the face away from the bundle axis (c) uniformly around it (d) at the spacer clampShow answer
(b). Like charges repel, so the neighbours' field points outward and adds there. Answer (a) reverses the sign and would predict corona on the inside of the bundle, which is not what is observed. Problem 9.MCQ 7. The non-uniformity factor of an \(n\)-conductor bundle is:
(a) \(1+(n-1)r/A\) (b) \(1+nr/A\) (c) \(1+r/d\) (d) \(n\)Show answer
(a). It is \(n-1\) because a conductor is not acted on by its own charge. Answer (c) is the twin case with \(A\) mistaken for \(d\) and halves the correction. Problem 9.MCQ 8. The surge impedance loading of a 400 kV twin-Moose line is about:
(a) 130 MW (b) 260 MW (c) 515 MW (d) 2200 MWShow answer
(c). Answer (a) is a 220 kV single-Zebra circuit and (d) is 765 kV quad — both appear in this set, and knowing all three by heart is worth more than any formula. Problems 14, 15 and 18.MCQ 9. As the number of sub-conductors grows with the circumradius fixed, \(D_{sb}\) tends to:
(a) zero (b) \(A\) (c) infinity (d) \(nD_s\)Show answer
(b) — the bundle becomes a hollow cylindrical shell, whose GMR is its radius. This is a free sanity check: any \(D_{sb}\) exceeding \(A\) is arithmetically impossible. Problem 16 and Challenge C1.MCQ 10. Doubling the bundle spacing of a 400 kV twin from 250 to 500 mm changes \(E_{max}\) by about:
(a) 1% (b) 10% (c) 30% (d) 50%Show answer
(a). The gradient has an interior minimum near 300 mm and is flat around it. Answer (b) is roughly what the reactance does over the same range, and confusing the two is what leads people to believe the spacing is chosen for corona. Problem 11.MCQ 11. Using the distance between bundle centres as the inter-phase spacing, for twin bundles 11 m apart at 450 mm, is in error by about:
(a) 4% (b) 0.4% (c) 0.04% (d) exactly zeroShow answer
(c). Not (d): the error vanishes only to first order, the residual being \(-d^{2}/4D^{2}\). For a quad bundle it really is negligible — 2 parts in \(10^{8}\). Problem 19.MCQ 12. A single conductor on the 400 kV geometry of this set would need an overall diameter of at least about:
(a) 20 mm (b) 32 mm (c) 39 mm (d) 82 mmShow answer
(c), to keep \(E\) below \(m_0g_0 = 17.94\) kV/cm. Answer (b) is Moose itself, which fails by 20%; (d) is the 765 kV requirement. Problem 12.
Key Formulas
| Statement | Relation | Notes |
|---|---|---|
| Polygon product rule | \(\prod_{k\neq1}|z_1-z_k| = nA^{\,n-1}\) | Proved in Challenge C1 |
| Bundle circumradius | \(A = d/2,\ d/\sqrt3,\ d/\sqrt2\) | \(n = 2, 3, 4\) |
| Bundle self-GMD | \(D_{sb} = \left(nD_sA^{\,n-1}\right)^{1/n}\) | Ceiling at \(D_{sb} = A\) |
| Standard forms | \(\sqrt{D_sd},\ \sqrt[3]{D_sd^{2}},\ 1.0905\sqrt[4]{D_sd^{3}}\) | 1.0905 is \(2^{1/8}\) |
| Bundle radius for capacitance | \(r_b = \left(nrA^{\,n-1}\right)^{1/n}\) | Outside radius, not GMR |
| Inductance per phase | \(L = 0.2\ln(D_{eq}/D_{sb})\) mH/km | Transposed; \(x_L = 2\pi fL\) |
| Capacitance to neutral | \(C_n = 0.05563/\ln(D_{eq}/r_b)\) µF/km | Earth image neglected |
| Surge impedance | \(Z_c = 59.96\sqrt{\Lambda_L\Lambda_C}\) | \(\approx 60\ln(D_{eq}/\sqrt{D_{sb}r_b})\) to 0.1% |
| Surge impedance loading | \(\mathrm{SIL} = V_{LL}^{2}/Z_c\) | 518 MW at 400 kV twin |
| SIL gain on bundling | \(1/\sqrt{\alpha\beta}\) | Geometric mean of the two log ratios |
| Average surface gradient | \(E_{av} = V_{ph}/\!\left[n\,r\ln(D_{eq}/r_b)\right]\) | \(r\) outside the log, \(r_b\) inside |
| Maximum surface gradient | \(E_{max} = E_{av}\left[1+(n-1)r/A\right]\) | Markt–Mengele, first order |
| Corona threshold | \(E_{crit} = m_0g_0\delta\), \(g_0 = 21.1\) kV/cm rms | 17.9 kV/cm for stranded ACSR |
| Optimum bundle spacing | \(A_{opt} = r\left[n\Lambda - (n-1)\right]\) | Implicit; flat to 1% either side |
| Mutual GMD error | \(\ln(D_m/D) \approx -\tfrac{2}{n}(A/D)^{n}\) | 0.04% for a twin at 11 m |
| Bundle resistance | \(R = R_{sub}/n\) | Requires spacers to share equally |
| Power limit | \(P_{max} = V_SV_R/X\) | Rises as \(k/(1-k)\) for a saving \(k\) |
Common Mistakes
Taking \(A = d/2\) for a three- or four-conductor bundle. The circumradius is \(d/\sqrt3\) and \(d/\sqrt2\); using \(d/2\) for a triple gives \(D_{sb} = 12.02\) cm instead of 13.23 — Problem 3.
Using \(D_{sb}\) for capacitance or \(r_b\) for inductance. Flux linkage wants the GMR, charge wants the surface. The swap costs 2.2% in \(C_n\) and propagates into \(Z_c\) and SIL — Problem 13.
Mixing metres and centimetres inside the logarithm. \(\ln(13.86/7.551)\) instead of \(\ln(13.86/0.07551)\) gives 0.121 mH/km, wrong by a factor of nine — Problem 5. Always check that the answer lies between 0.8 and 1.4 mH/km.
Putting \(r_b\) outside the logarithm in the gradient formula. The charge is set by \(r_b\), but it sits on a surface of radius \(r\). Using \(r_b\) in both places gives 2.7 instead of 14.25 kV/cm and makes every line look corona-free — Problem 8.
Designing to the average gradient. Corona starts at the maximum, which for a twin is 7% higher and for a quad 16% higher. Problem 15's 765 kV line is at 0.94 of the threshold on \(E_{max}\) and a comfortable 0.81 on \(E_{av}\) — the second number would approve a line that glows.
Believing the bundle spacing is chosen to minimise the gradient. The optimum is real but the curve is flat to 1% across a factor of two in spacing; short-circuit forces and sub-span oscillation choose it — Problems 10 and 11.
Comparing a bundle against one sub-conductor rather than against equal metal. The 25.5% reactance saving becomes 21.6% and the 28.9% gradient benefit becomes 4.6% when the aluminium is equalised. Both comparisons are legitimate; quoting the first while implying the second is not — Problem 7.
Quoting a reactance saving as though it were a power gain. A 25.5% reduction in \(X\) is a 34.2% rise in \(V^{2}/X\), and a 33.5% rise in SIL. Three different numbers describing one change — Problems 6 and 16.
Assuming a bundled line degrades gracefully. Losing one sub-conductor of a twin leaves the circuit energised while overloading the survivor by 25%, raising its gradient 41% past the corona threshold and unbalancing the sequence networks — Problem 17.
Applying the bundle-centre spacing on a compact or double-circuit tower. The error scales as \((A/D)^{n}\): negligible at 11 m, but 2.2% when phases come within 1.5 m — Problem 19 and Challenge C2.
The bundle is now a solved object. Two numbers — a self-GMD of 7.551 cm and a radius of 8.455 cm — carried a 400 kV twin-Moose line from a conductor catalogue through to a surge impedance of 309 Ω, a natural loading of 518 MW, a maximum surface gradient of 15.3 kV/cm and a set of \(ABCD\) constants for four hundred kilometres of it.
Set 7 takes the same geometry and works the capacitance side in more detail; Set 12 takes the gradient of Problems 8 to 12 into the corona loss and radio-interference formulae; and Sets 9 and 10 take the \(z\) and \(y\) of Problem 20 into the medium- and long-line models where the 24.3° electrical length of this circuit finally matters. The parameter work of Part 2 ends here; everything after it is what a line does with these numbers.