Inductance of Transmission Lines

Chapter 4 drew every transmission line as a series impedance \(R + jX\) and then, in the reactance diagram, quietly dropped the \(R\). That was not laziness. For a 132 kV line on Panther conductor the resistance is about \(0.14\;\Omega\)/km and the reactance about \(0.39\;\Omega\)/km; at 400 kV on bundled Moose the ratio widens to ten or more. The series reactance is what limits how much power a line can carry, what sets the voltage drop from one end to the other, what determines the current in a fault, and what appears in the swing equation of Chapter 27. Computing it correctly is therefore the most consequential single calculation in Part 2.

Inductance is defined by flux linkage per ampere,

and for a coil the flux linkage is unambiguous: a flux \(\phi\) that threads \(N\) turns links \(N\phi\). A transmission line has no turns. What takes their place is a subtler bookkeeping: a tube of flux at radius \(x\) from the conductor axis links only that part of the current which flows inside radius \(x\), so a flux tube inside the conductor is weighted by the fraction \(x^2/r^2\), while one outside is weighted by unity. Getting that weighting right, and then deciding how far out to integrate, is the whole of the calculation.

The plan of the chapter is to build the answer in layers. Sections 6-2 and 6-3 handle a single isolated conductor, inside and outside. Section 6-4 assembles a two-wire line from them and extracts the trick that removes the internal term from every subsequent formula. Section 6-5 generalises to any number of conductors whose currents sum to zero — the condition that a balanced power circuit always satisfies — and Sections 6-6 to 6-9 are that one result applied to stranded conductors, to three-phase lines, and to double circuits.

Two assumptions run through all of it, and both are good ones for a power line at 50 Hz. The conductor material is non-magnetic, so \(\mu = \mu_0 = 4\pi\times10^{-7}\) H/m — true for copper and aluminium, and the reason the steel core of ACSR is excluded from the calculation. And the current density is uniform over the cross-section, which is the statement that skin effect is being neglected; Chapter 5 measured that error and found it to be a few per cent.

Take a long, straight, solid cylindrical conductor of radius \(r\) carrying a current \(I\), and suppose for the moment that the return path is so far away that it disturbs nothing. By symmetry the magnetic field lines are circles concentric with the conductor, and \(H\) is constant in magnitude along any such circle and tangent to it. Ampère's law applied to a circle of radius \(x\) inside the conductor gives

The second line is where the uniform-current-density assumption enters: only the current enclosed by the circle of radius \(x\) counts, and that is the total current scaled by the ratio of the areas. Multiplying by the permeability gives the flux density, and integrating across an annular element of thickness \(dx\) over one metre of length gives the flux in that element:

Now the weighting. The flux \(d\phi\) forms a closed loop around the axis, but it encircles only the current \(I_x\), not the whole of \(I\). In the language of flux linkage it links a fraction \(x^2/r^2\) of a "turn", so

The cancellation has a physical reading. Making the conductor fatter spreads the current over a larger area, which weakens the internal field at any given fraction of the radius; but it also provides more volume in which that field can store energy. The two effects are exactly equal and opposite.

Outside the conductor the same argument is simpler, because every flux tube encircles the entire current. Take two points \(P_1\) and \(P_2\) at distances \(D_1\) and \(D_2\) from the conductor axis, with \(D_1 \ge r\), and ask for the flux linkage produced between them.

One caution about \(D_2 \to \infty\). Taken alone, the integral diverges: a single isolated conductor carrying a steady current has infinite inductance per metre. That is not a defect of the analysis but a statement that a single conductor carrying current with no return path does not exist. In every real circuit there is a return, its flux opposes this one, and the divergence cancels. Section 6-5 shows the cancellation happening explicitly.

Consider two parallel conductors of radii \(r_1\) and \(r_2\), their axes a distance \(D\) apart, one carrying \(+I\) and the other \(-I\). Ask for the inductance associated with conductor 1.

Flux produced by conductor 1's own current links conductor 1's circuit out to the location of conductor 2, and no further: a flux tube at a distance greater than \(D+r_2\) encircles both \(+I\) and \(-I\), a net zero, and links nothing. A tube between \(D-r_2\) and \(D+r_2\) links a part of conductor 2's current and is awkward to treat exactly, but the conductors of a power line are centimetres thick and metres apart, so \(r_2 \ll D\) and it is entirely legitimate to integrate out to \(D\) and stop. With that, the flux linkage of conductor 1 is the internal term of Section 6-2 plus the external term of Section 6-3 taken from \(r_1\) to \(D\):

The bracket can be tidied by noticing that \(\tfrac14 = \ln e^{1/4}\), and that adding a logarithm is multiplying inside it:

Conductor 2 carries \(-I\), so the flux it produces around itself circulates the opposite way; but conductor 2's circuit is also traversed the opposite way, and the two reversals cancel. Its flux linkage adds to conductor 1's rather than subtracting from it, and the two inductances simply sum. For constant permeability that superposition is exact.

A three-phase line has three conductors, a double circuit has six, a stranded conductor has dozens of filaments. All of them are handled by one general result, and the only condition it needs is the one every balanced power circuit satisfies: the currents sum to zero.

Let \(n\) parallel conductors carry currents \(I_1, I_2, \dots, I_n\) with \(\sum I_k = 0\). Choose a remote point \(P\) and, using Section 6-3, write the flux linkage of conductor 1 caused by each current in turn out to that point. The contribution of \(I_1\) starts at conductor 1's own surface, so it uses \(r_1'\); the contribution of \(I_k\) starts at the distance \(D_{1k}\) between the conductors, because flux from \(I_k\) closer to conductor \(k\) than that does not encircle conductor 1 at all.

Split every logarithm of a quotient into a difference and collect the two kinds of term:

The first bracket contains only the geometry of the conductor group. The second contains the remote point, and it is the one that must be made to disappear. Use the condition \(\sum I_k = 0\) in the form \(I_n = -(I_1+I_2+\cdots+I_{n-1})\) and substitute it into the last term of the second bracket:

Every surviving term is now a logarithm of a ratio of distances to \(P\). As \(P\) recedes, those distances become more and more nearly equal — the conductors are metres apart and \(P\) is kilometres away — so each ratio tends to \(1\) and each logarithm to \(0\). The divergence noted at the end of Section 6-3 has cancelled, exactly as promised, and it did so because the currents sum to zero.

Chapter 5 established that every practical conductor is stranded, and that magnetically it must be treated as a bundle of parallel filaments rather than a solid rod. Section 6-5 now supplies the machinery.

Let a single-phase line consist of two composite conductors. Conductor \(X\) is made of \(n\) identical filaments \(a, b, \dots, n\), each carrying \(I/n\); conductor \(Y\) is the return, made of \(m\) identical filaments \(a', b', \dots, m\), each carrying \(-I/m\). The currents sum to zero, so the group formula applies. Applying it to filament \(a\):

The second line is worth pausing over, because it is where the roots come from. A sum of logarithms divided by \(n\) is the logarithm of an \(n\)-th root of a product; the minus sign turns the \(Y\)-filament product into a numerator. Filament \(a\) carries \(I/n\), so its inductance is

Each filament has a different \(L\), because each sits in a different position. The filaments are in parallel and share the current, so the inductance of conductor \(X\) as a whole is the average of the filament inductances divided by the number in parallel:

Substituting the filament expressions and collecting the \(mn\) distances that appear in every numerator and the \(n^2\) that appear in every denominator gives the result in a form that is nothing but two geometric means.

Three consequences deserve to be drawn out. First, the formula contains the results of Section 6-4 as a special case: for two solid conductors \(n=m=1\), \(D_m = D\), \(D_s = r'\), and \(L_X = 2\times10^{-7}\ln(D/r')\) as before. Second, \(D_s\) is a property of the conductor alone — it can be computed once by the manufacturer and printed in a catalogue, which is exactly what the \(D_s\) column of Chapter 5's ACSR table is. Third, \(D_m\) is a property of the spacing alone, and when the conductors are far apart compared with their own size, \(D_m\) is very nearly the centre-to-centre distance, which is why hand calculations on ordinary lines use the spacing directly and lose nothing.

Put three conductors at the vertices of an equilateral triangle of side \(D\), each with geometric mean radius \(D_s\), carrying balanced currents \(I_a+I_b+I_c=0\). The group formula of Section 6-5 applied to conductor \(a\) gives

The last two terms carry the same coefficient because \(b\) and \(c\) are equidistant from \(a\). Replace \(I_b + I_c\) by \(-I_a\):

Two features of this result are worth noticing. The flux linkage of phase \(a\) came out proportional to \(I_a\) alone, with the other two currents having vanished into the balance condition; that is what allows a single per-phase inductance to be quoted at all. And the answer is the same whether the line has a neutral conductor or not, because a balanced set draws no neutral current and a conductor carrying no current produces no flux.

The trouble is that nobody builds equilateral lines. Three conductors at the corners of a triangle require a tower with an awkward outline; horizontal (flat) and vertical configurations are cheaper to build, easier to string and easier to maintain. So the practical case is the unsymmetrical one.

Let the three conductors sit at three positions with mutual distances \(D_{12}\), \(D_{23}\) and \(D_{31}\), not all equal. Writing out \(\lambda_a\), \(\lambda_b\) and \(\lambda_c\) from the group formula now produces three different expressions, because the middle conductor is closer to both its neighbours than the two outer conductors are to each other. The consequence is real and troublesome: the three phases have different inductances, so equal applied voltages produce unequal currents, and balanced currents produce unequal voltage drops. A line built this way delivers an unbalanced supply even to a perfectly balanced load, and the residual unbalance drives current into the neutral and the earth and induces voltages in nearby telephone circuits.

The remedy is transposition: exchange the physical positions of the conductors at regular intervals along the route, so that each conductor occupies each of the three positions for one third of the length. The line is then geometrically identical as seen from any of the three phases, and the inductance per phase is the average of the three.

Carry out that averaging. In the first section \(a\) occupies position 1, \(b\) position 2 and \(c\) position 3; in the second, \(a\) moves to position 2, \(b\) to 3, \(c\) to 1; in the third, \(a\) is at 3, \(b\) at 1, \(c\) at 2. Writing the group formula for phase \(a\) in each section,

Average the three. The \(I_b\) terms collect the same three distances as the \(I_c\) terms, so both acquire the coefficient \(\tfrac13\ln\big[1/(D_{12}D_{23}D_{31})\big]\), and \(I_b + I_c = -I_a\) once more:

For the common flat configuration with adjacent spacing \(D\) and outer spacing \(2D\), \(D_{eq} = \sqrt[3]{D\cdot D\cdot 2D} = D\sqrt[3]{2} = 1.26\,D\). A flat line therefore has slightly more inductance than an equilateral line of the same adjacent spacing, and slightly less than one of the same overall width — the geometric mean sitting, as always, between the extremes.

Modern practice frequently omits physical transposition altogether. Full transposition requires a special tower and a costly interruption of the conductor, and on a line of moderate length the resulting unbalance is a fraction of a per cent — small enough to be absorbed. The formula is used anyway, because the assumption of a transposed line is what makes a single per-phase impedance meaningful, and every method from Chapter 16 onward depends on there being one. Where the unbalance genuinely matters — very long EHV lines, or lines feeding sensitive loads — transposition is provided, or the phases of a double circuit are arranged so that the two circuits' unbalances oppose.

A double-circuit line carries two three-phase circuits on one set of towers. Electrically the two circuits are in parallel, but they are close enough to be magnetically coupled, so the inductance is not simply half of one circuit's value; the mutual flux must be accounted for. The machinery of Section 6-6 does it without any new ideas: treat the two conductors belonging to phase \(a\) — call them \(a\) and \(a'\) — as the two filaments of a single composite conductor, and likewise for \(b\) and \(c\).

The self GMD of the phase-\(a\) group is the geometric mean of the two conductors' own \(D_s\) values and the distance between them; since \(D_{aa'}\) is metres and \(D_s\) is centimetres, the composite \(D_S^A\) is far larger than \(D_s\), and the inductance correspondingly smaller. That is the whole benefit of a double circuit beyond simple redundancy — and it is maximised by placing the two conductors of a phase as far apart as the tower allows, which means the two circuits should be arranged with their phase sequences reversed relative to one another, \(a\,b\,c\) down one side and \(c\,b\,a\) down the other.

The hexagonal arrangement makes the point cleanly. Six conductors sit at the vertices of a regular hexagon of circumradius \(R\), with the two conductors of a phase at opposite vertices. In a regular hexagon the chord subtending \(60^\circ\) equals \(R\), the chord subtending \(120^\circ\) is \(\sqrt3 R\), and the diameter is \(2R\). Taking phases \(a\) and \(b\), one pair of conductors is separated by \(R\) at each of two vertex pairs and by \(\sqrt3 R\) at the other two, so

Example 5 puts numbers to it and finds that the double circuit gives less than half the inductance of a single circuit of the same \(D_{eq}\) — the mutual coupling, arranged this way, helps rather than hinders.

What do all these formulas actually produce? Remarkably little variety. The logarithm compresses everything, and for overhead lines the answer always lands near the same figure.

Take \(\mu_r = 1\) and 50 Hz throughout, and assume full transposition unless told otherwise. Where a conductor is described by its physical radius, remember to convert to \(r'\).