Capacitance of Transmission Lines

Chapter 6 computed the series impedance of a line and, in doing so, answered the question of what happens to a current that flows along it. This chapter answers a different question: what happens because a voltage is applied to it, whether or not any current is drawn at the far end.

Two conductors held at different potentials carry equal and opposite charges on their surfaces. The ratio of that charge to the potential difference is the capacitance, and it is fixed by geometry alone:

On a d.c. line this is the end of the story: the charge is placed once and stays there. On an a.c. line the applied voltage alternates, so the charge alternates, and an alternating charge is a current. That current — the charging current — flows into the line from the sending end and out of it into the conductors' own capacitance to earth and to one another. It has nothing to do with the load. Open-circuit the far end of a long line and the sending-end ammeter still reads.

Four consequences follow, and they are the reason this chapter exists rather than being a footnote to Chapter 6. The charging current adds vectorially to the load current, so the current is no longer the same at both ends of a line and the voltage drop calculation of Chapter 10 must account for it. Being very nearly a leading reactive current, it supplies reactive power to the system — a lightly loaded line is a capacitor bank, and a heavily loaded one is not, which is the whole subject of Chapter 34. On a long open-ended line it raises the receiving-end voltage above the sending-end voltage, the Ferranti effect of Chapter 14. And the ratio of series inductance to shunt capacitance fixes the surge impedance \(Z_c=\sqrt{L/C}\), from which Chapter 12's travelling waves and Chapter 14's surge impedance loading are built.

How large the effect is depends on length. For lines shorter than about 80 km the shunt admittance is small enough that dropping it changes the answer by less than the uncertainty in the data, and Chapter 10's short-line model does exactly that. Between 80 and 250 km the capacitance must be included but may be lumped, as Chapter 11 does. Beyond that it must be treated as distributed, which is Chapter 12.

Take a long straight cylindrical conductor of radius \(r\), isolated in a uniform medium of permittivity \(\varepsilon\), carrying a charge of \(q\) coulombs per metre of length. Because the conductor is a conductor, the charge cannot sit inside it: any interior charge would produce an interior field, which would drive the charge until the field vanished. All of it resides on the surface. Because the conductor is isolated and remote from other charges, the surface charge is uniformly distributed around the periphery, and the field it produces is purely radial.

Draw a cylindrical Gaussian surface of radius \(x \ge r\) and unit axial length, concentric with the conductor. Gauss's law states that the electric flux leaving a closed surface equals the charge enclosed. The flux leaves only through the curved wall — the end faces are parallel to the field — and that wall has area \(2\pi x\) per metre of length. So the flux density is

and the electric field intensity, \(E = D_f/\varepsilon\), is

Compare this with Chapter 6. There, Ampère's law around a circle of radius \(x\) gave \(H = I/2\pi x\), and the whole inductance calculation followed from integrating it. Here Gauss's law around a cylinder of radius \(x\) gives \(E = q/2\pi\varepsilon x\), and the whole capacitance calculation will follow from integrating that. The two laws have produced fields of identical shape.

They have not, however, produced identical situations inside the conductor. Current is distributed through the metal, so a flux tube inside the conductor links part of the current and contributes to the inductance — the internal inductance of Section 6-2, worth \(\tfrac12\times10^{-7}\) H/m, which was absorbed into the fictitious radius \(r' = 0.7788r\). Charge is distributed on the surface, so there is no field inside the conductor at all and no internal contribution to capacitance. This single fact is the reason every capacitance formula in this chapter carries the actual radius \(r\) where the corresponding inductance formula carried \(D_s\).

Potential difference is work per unit charge, \(V = W/q\), and force per unit charge is the field, \(E = F/q\). Moving a test charge from a point \(P_2\) to a point \(P_1\) therefore requires work \(\int E\cdot dx\) taken along the path between them. A positive charge on the conductor repels a positive test charge, so work must be done to push the test charge inward; the near point is at the higher potential.

Let \(P_1\) and \(P_2\) lie at radial distances \(D_1\) and \(D_2\) from the conductor's axis, both outside it. Since the field is conservative, only the radial parts of the path contribute and any circumferential detour is free. Integrating the field of Section 7-2:

Two sign conventions are folded into that single line, and confusing them is the commonest source of error in the problems that follow. The result is positive — \(P_1\) genuinely above \(P_2\) — when \(q\) is positive and \(D_2 > D_1\), that is, when we move outward away from a positive charge. It reverses if the charge is negative, and it reverses again if the two points are taken in the other order. Keep the rule as a sentence: the logarithm is of the far distance over the near distance, and the answer is the potential of the near point above the far one.

The expression has an awkward property. If \(P_2\) is allowed to recede to infinity, \(v_{12}\) grows without bound: the potential of a single infinite line charge relative to infinity does not exist. This is precisely the divergence met in Section 6-3, where the flux linkage of a single conductor out to a remote point also diverged. There it was cured by the fact that the currents in a real circuit sum to zero. Here it will be cured by the fact that the charges do.

Take two parallel conductors of radii \(r_a\) and \(r_b\), their centres a distance \(D\) apart, with \(D\) much larger than either radius. Conductor \(a\) carries \(q_a\) coulombs per metre and conductor \(b\) carries \(q_b = -q_a\), since whatever charge leaves one must arrive on the other.

We want \(v_{ab}\), the potential of \(a\) above \(b\). Consider the charges one at a time. Because of \(q_a\), point \(a\) sits at distance \(r_a\) from the charge and point \(b\) at distance \(D\); by Section 7-3 the contribution is \((q_a/2\pi\varepsilon)\ln(D/r_a)\). Because of \(q_b\), point \(a\) sits at distance \(D\) from that charge and point \(b\) at distance \(r_b\); the contribution is \((q_b/2\pi\varepsilon)\ln(r_b/D)\). Adding:

The capacitance between the conductors follows straight from the definition:

Power system analysis does not want a line-to-line quantity. Chapter 3 established that every three-phase calculation is done per phase against a neutral, and Chapter 4's per-unit machinery assumes the same. So the two-conductor capacitance must be re-expressed as two equal capacitances in series between the conductors, joined at a point which — by symmetry, for equal radii and equal and opposite charges — sits at zero potential. That point is the electrical neutral of the line. Two equal capacitors \(C_n\) in series give \(C_n/2\), so

The capacitive reactance to neutral follows at once. Writing \(\varepsilon = \varepsilon_0 = 8.854\times10^{-12}\) F/m,

The units deserve a moment's attention because they are a persistent trap. \(C_n\) is a capacitance per metre, so it grows in proportion to line length and the total capacitance of a 200 km line is 200 000 times the per-metre value. Reactance, being the reciprocal, falls with length: \(X_c\) in \(\Omega\)·m divided by the length in metres gives the reactance of the whole line in ohms. A shunt element is always quoted per unit length as an admittance or a susceptance for exactly this reason, and Chapters 11 and 12 will use \(y = j\omega C_n\) S/km throughout.

Section 6-5 obtained the flux linkage of one conductor in a group of \(n\) whose currents sum to zero, and everything after it was an application of that one line. The electrostatic case runs in parallel, and it is worth carrying out in full so that the two results can be set side by side.

Let \(n\) parallel conductors carry charges \(q_1, q_2,\dots,q_n\) per metre with \(\sum q_k = 0\), the centres separated by distances \(D_{jk}\). Take a remote reference point \(P\) and compute the potential of conductor 1 above it. The charge \(q_1\) contributes \((q_1/2\pi\varepsilon)\ln(D_{1P}/r_1)\), since conductor 1's own surface is at radius \(r_1\) from its own charge. The charge \(q_k\) contributes \((q_k/2\pi\varepsilon)\ln(D_{kP}/D_{1k})\), because conductor 1's surface is at distance \(D_{1k}\) from that charge. Summing,

The first bracket holds only the geometry of the group. The second holds the remote point, and it must be made to vanish. Use \(q_n = -(q_1+q_2+\cdots+q_{n-1})\):

Every surviving term is the logarithm of a ratio of two distances to the same remote point. The conductors are metres apart and \(P\) is arbitrarily far away, so each ratio tends to unity and each logarithm to zero. The divergence noted at the end of Section 7-3 has cancelled, and it cancelled because the charges sum to zero.

The zero of potential implied here is the point at infinity, which for a charge-balanced group is well defined. It is also, for any symmetric balanced arrangement, the potential of the neutral, so \(V_1\) computed this way is the line-to-neutral voltage of phase 1 — which is exactly the quantity per-phase analysis wants. Potential differences such as \(v_{ab} = V_a - V_b\) are recovered by subtraction whenever they are needed, and the classical textbook derivations that work directly with \(v_{ab}\) and \(v_{ac}\) reach the same answers by a longer road. Both routes appear below, because examinations set both.

Place three identical conductors of radius \(r\) at the corners of an equilateral triangle of side \(D\), carrying balanced charges \(q_a+q_b+q_c=0\). Section 7-5 applied to conductor \(a\) gives

The two neighbouring charges entered with the same coefficient because they are equidistant from \(a\), and their sum was replaced by \(-q_a\). One line of algebra, and the answer contains only \(q_a\).

The classical derivation, which the examination papers still favour, avoids the group formula and works with line voltages. Superposing the three charges on the pair \((a,b)\) and then on the pair \((a,c)\):

The term in \(q_c\) vanished from \(V_{ab}\) because \(c\) is equidistant from \(a\) and \(b\), so it raises both by the same amount and changes their difference not at all. The last step used \(q_b+q_c=-q_a\), which turns \(2\ln(D/r) + \ln(r/D)\) into \(3\ln(D/r)\). The final step is a fact about balanced phasors rather than about capacitance: for a balanced set, \(V_{ab}+V_{ac} = 3V_{an}\), because \(V_{ab}=V_{an}-V_{bn}\) and \(V_{ac}=V_{an}-V_{cn}\), so their sum is \(3V_{an}-(V_{an}+V_{bn}+V_{cn}) = 3V_{an}\). Dividing by three recovers \(V_{an} = (q_a/2\pi\varepsilon)\ln(D/r)\), as before.

Nobody builds equilateral lines, for the reasons Chapter 5 gave and Section 6-8 repeated: flat and vertical tower outlines are cheaper, easier to string and easier to maintain. So the practical arrangement has three unequal spacings \(D_{12}\), \(D_{23}\) and \(D_{31}\), and the three phases then have three different capacitances. A balanced set of applied voltages would produce unbalanced charges, hence unbalanced charging currents, hence a residual current in earth even with a perfectly balanced load.

The remedy is the same as it was for inductance, and it is applied at the same time on the same towers. Transposition rotates the conductors through the three physical positions so that each phase occupies each position for one third of the route. Chapter 6 established the machinery; here it pays off a second time at no extra cost.

Assume, as Chapter 6 did, that the charge per unit length of each phase is the same in all three sections — a very good approximation for a line whose transposition intervals are short compared with a wavelength. Write the potential of conductor \(a\) from Section 7-5 in each of the three sections. In the first, \(a\) is in position 1 with \(b\) in position 2 and \(c\) in position 3; in the second, \(a\) has moved to position 2; in the third, to position 3:

Average the three, which is what the line as a whole presents:

Both \(q_b\) and \(q_c\) collected the same three logarithms, one from each section, which is the whole point of transposing; their sum was then replaced by \(-q_a\); and the average of three logarithms is the logarithm of a cube root.

A practical caution about transposition, carried over from Section 6-8: modern practice on shorter lines often omits it, accepting a small unbalance rather than paying for the transposition towers. The formulas above are then used as approximations, with \(D_{eq}\) computed as though the line were transposed. The error is of the order of one per cent in the phase capacitances of a normal flat-configuration line, which is well inside the uncertainty in the conductor sag and hence in the true spacings.

Everything so far has treated the conductors as isolated in an infinite uniform medium. They are not. They run a few metres above a conducting earth, and the earth alters the electric field. It cannot alter the magnetic field of Chapter 6 to any comparable degree — the earth's permeability is that of free space, so it is magnetically almost invisible — but electrically it is a conductor, and a conductor placed in an electric field is never invisible.

Model the earth as a perfect conductor occupying a horizontal plane of infinite extent. Two facts follow immediately. Charge flows up from the earth to the conductor's underside region and arranges itself so that the earth's surface is an equipotential. And, since field lines meet a perfect conductor at right angles, every line of flux from the overhead conductor arrives at the earth perpendicular to it.

Now the trick. Delete the earth. In its place put a fictitious conductor of the same size, the same distance below the plane as the real conductor is above it, carrying an equal and opposite charge. The field of that pair, by symmetry, has the plane midway between them as an equipotential surface which every field line crosses at right angles. Those are exactly the two conditions the real earth imposed, and the uniqueness theorem of electrostatics says that a field satisfying the boundary conditions is the field. So in the region above the plane the two problems have the same answer, and the second one is a problem in free space of the kind Section 7-5 already solves.

Apply this to the two-wire line of Section 7-4 at height \(h\) above earth. The group now contains four charges: \(q_a\) and \(q_b=-q_a\) above, their images \(-q_a\) and \(+q_a\) below. The sum is zero, so Section 7-5 applies directly. Conductor \(a\) is at distance \(2h\) from its own image and at distance \(H=\sqrt{D^2+4h^2}\) from the image of \(b\), by Pythagoras on the horizontal separation \(D\) and the vertical separation \(2h\). Writing \(v_{ab}\) as the difference of the two potentials, and grouping the free-space terms and the image terms separately:

The three-phase transposed line is handled the same way and is the result worth keeping. Six charges are present: \(q_a,q_b,q_c\) above and their negatives below. Let \(H_1, H_2, H_3\) be the distances from each conductor to its own image — these are simply twice the heights — and \(H_{12}, H_{23}, H_{31}\) the distances from each conductor to the images of the other two. Writing the potential of \(a\) in the first section of the transposition cycle,

Averaging over the three sections, exactly as in Section 7-7, replaces each set of three spacings by its geometric mean and each set of three image distances by its geometric mean; then \(q_b+q_c=-q_a\) closes the algebra.

How large is the correction in practice? For a 132 kV flat line 8 m above ground the answer computed in Example 4 is about six parts in a thousand. That is why the effect of earth is routinely neglected in capacitance calculations for overhead lines: it is smaller than the change caused by conductor sag between towers. Two situations reverse the verdict. Distribution lines and railway feeders run low, where \(D/2h\) is no longer small and the correction reaches a few per cent. And whenever the three charges do not sum to zero — during an earth fault, or in the zero-sequence network of Chapter 23 — the earth is not a correction but the entire return path, and the image method becomes the only way to compute the capacitance at all.

With \(C_n\) known, the current the line draws follows from the definition of capacitance differentiated in time. For a sinusoidal line-to-neutral voltage \(V_{an}\), the charging current per phase is

The two forms agree for the two-wire line, where \(C_{ab}=C_n/2\) and \(V_{ab}=2V_{an}\), so the factors of two cancel. They do not agree if \(C_{ab}\) is carelessly carried into a three-phase problem: there the line-to-line capacitance is a delta-connected element and \(C_\Delta = C_n/3\), not \(C_n/2\), because a star of three \(C_n\) converts to a delta of three \(C_n/3\). The reliable habit is to work per phase to neutral and never to leave it.

Multiplying by three times the phase voltage gives the reactive power the line generates:

What do the formulas actually produce? As with inductance, remarkably little variety, and for the same reason: the logarithm compresses everything. The table below uses the same five lines as the table in Section 6-9 so that the two sets of parameters can be read together.

*The starred entries are equivalent radii of composite arrangements rather than physical conductor radii: \(\sqrt{r\,d}\) for the twin bundle, obtained in Chapter 8, and \(\sqrt{2Rr}\) for the hexagonal double circuit, obtained in Example 6. The single-circuit capacitances vary by only twelve per cent across a fifteen-fold change in spacing and a seven-fold change in voltage. As with reactance, only the devices that raise the effective radius substantially — bundling and paralleling a second circuit — move the number appreciably, and both raise \(C\) while lowering \(L\).

The last quantity to name is the one Chapters 11 and 12 will use. The shunt admittance per unit length is purely capacitive for an overhead line — leakage over the insulator strings is negligible in dry weather and is never modelled — so

and the line's complete per-phase description is the pair \((z, y)\): \(z = R + j\omega L\) from Chapters 5 and 6, \(y = j\omega C_n\) from this chapter. Part 3 does nothing but arrange those two numbers into circuits of increasing fidelity.

Take \(\varepsilon_0=8.854\times10^{-12}\) F/m and \(f=50\) Hz unless a problem says otherwise, and neglect the effect of earth except where it is explicitly asked for. Quote capacitances in μF/km and state clearly whether an answer is line-to-line or line-to-neutral.