Chapter 12 — Line and cable parameters

Four chapters of line modelling have quietly assumed that $Z_C$ and $\tau$, or $R^{\prime }$, $L^{\prime }$, and $C^{\prime }$, are supplied from elsewhere. Someone must compute them, and the raw material of that computation is geometry: how high the conductors hang, how far apart they are, how thick they are, and how conductive the soil beneath them is. This chapter closes the loop.

These calculations are rarely implemented by hand, since dedicated line-constants tools exist and are reliable. Their output must nevertheless be read, and a judgement must be made as to whether the numbers on a datasheet are the numbers a given model requires. Understanding where those numbers come from is what makes such a judgement possible.

Not yet in NumaSim

The Model's Data entry list includes a Physical (coming soon) option. It is a placeholder, and selecting it stops the build with a message. NumaSim has no geometry-based parameter calculator and no cable-specific component today. Reduced parameters must be supplied, computed elsewhere or by hand as this chapter demonstrates. A cable is entered like any other line: given the correct $Z_C$ and $\tau$, the travelling-wave model is indifferent to what the conductor looks like.

Learning objectives

By the end of this chapter you should be able to:

• Compute a line's capacitance matrix from Maxwell's potential coefficients and the method of images.
• Explain why the earth return needs Carson's correction, and how the complex depth of penetration replaces it in closed form.
• Handle bundled conductors, either by an equivalent GMR or by explicit subconductors with Kron reduction.
• Eliminate earth wires from the impedance matrix, and say what that does to the zero-sequence parameters.
• Describe how cable parameters are built from coaxial loop equations, and why cables differ so sharply from overhead lines.
• Convert a utility datasheet or a line-constants report into the numbers NumaSim's three-phase entry methods want.

12.1 Capacitance: potential coefficients and images

Start with electrostatics, which is the easier half. A set of conductors carrying charges $q_1,\dots ,q_n$ per unit length produces potentials that are a linear function of those charges:

$$\mathbf{v}={\mathbf{P}}^{\prime }\mathbf{q},\text{so}{\mathbf{C}}^{\prime }=[{\mathbf{P}}^{\prime }{]}^{-1}.$$

${\mathbf{P}}^{\prime }$ is the matrix of potential coefficients, and its entries come straight from geometry. But there is a boundary condition to satisfy: the earth is a conductor, so the electric field must meet it at right angles.

The classical device for this is the method of images. Replace the earth plane by a mirror conductor at depth $-h$ carrying the opposite charge. The two charges together produce exactly zero potential on the plane between them, which is precisely the boundary condition the earth imposes — so the fictitious image reproduces the real earth's effect on everything above it.

The method of images. A conductor at height $h_i$ over a perfectly conducting earth is replaced by the conductor plus a mirror image at depth $-h_i$ carrying the opposite charge, which reproduces the earth's boundary condition exactly. The mutual term uses $D_{ij}$, the distance to the other conductor's image, against $d_{ij}$, the direct distance.

With images in place the entries are elementary. The self term compares the conductor's radius with the distance to its own image, $2h_i$:

$$P_{ii}^{\prime }=\frac{1}{2\pi {\epsilon }_0}\ln \frac{2h_i}{r_i},$$

and the mutual term compares the distance to the other conductor's image with the direct distance:

$$P_{ij}^{\prime }=\frac{1}{2\pi {\epsilon }_0}\ln \frac{D_{ij}}{d_{ij}}.$$

Inverting ${\mathbf{P}}^{\prime }$ gives the capacitance matrix, which is the ${\mathbf{Y}}^{\prime }=j\omega {\mathbf{C}}^{\prime }$ of Chapter 10. Note the pleasant fact that none of this depends on frequency: for overhead lines the capacitance really is constant, which is why Chapter 11's frequency dependence was entirely about ${\mathbf{Z}}^{\prime }$.

Which height? A conductor is not at a constant height — it sags between towers, and it spends more of its length near the low point than near the attachment. The standard correction uses the average height of the span,

$$h=h_{\text{tower}}-\frac{2}{3}\text{sag},$$

the $2/3$ being the average of a parabolic span. For a tower height of 30 m and a sag of 9 m, the effective height is 24 m — a 20 % correction, so it is not optional.

12.2 Series impedance and the earth return

Now the harder half. The same image construction would give the series inductance directly if the earth were a perfect conductor. It is not. Real soil has a resistivity of tens to thousands of ohm-metres, the return current spreads through a large volume rather than hugging the surface, and the depth it reaches depends on frequency.

Carson solved this in 1926, and his result splits the earth-return impedance into the ideal-image term plus a correction:

$$Z_{ij}^{\prime }=\underset{\text{perfect earth}}{\underbrace{\frac{j\omega {\mu }_0}{2\pi }\ln \frac{D_{ij}}{d_{ij}}}}+\underset{\text{Carson correction}}{\underbrace{\frac{j\omega {\mu }_0}{\pi }{\int }_0^{\mathrm{\infty }}(\cdots )d\lambda }}.$$

The correction integral has no closed form. It is evaluated as an infinite series that converges quickly at power frequency and badly at high frequency, which made it awkward exactly where EMT studies need it.

The complex depth of penetration

The modern approach replaces the whole integral with a single, startlingly simple substitution. Keep the image formula, but put the image not at depth $h_j$ but at depth $h_j+2p$, where

$$p=\sqrt{\frac{\rho }{j\omega {\mu }_0}},|p|=\sqrt{\frac{\rho }{\omega {\mu }_0}},$$

$\rho$ being the soil resistivity. The depth is complex, which is what lets one term carry both the inductance change and the resistance the lossy earth contributes.

Physically, $|p|$ is the depth at which the equivalent return current flows, and its behaviour explains Chapter 11 completely. For $\rho =100\mathrm{\Omega }\text{m}$:

| Frequency | $|p|$ | | --- | --- | | 60 Hz | 460 m | | 1 kHz | 113 m | | 100 kHz | 11 m |

At power frequency the return current is nearly half a kilometre underground, enclosing an enormous loop — hence the large $L_0$ of Chapter 10. As frequency rises the return climbs toward the surface, the loop shrinks, $L_0$ falls, and the current is forced through a smaller volume of poorly-conducting soil, so $R_0$ rises. The approximation agrees with Carson's integral to within a fraction of a percent across the whole practical band, and unlike the integral it is a closed form.

The conductor's own impedance

One more piece: the conductor is not a filament. Its internal impedance adds a resistance and a small internal inductance, both frequency dependent, and the exact solution for a solid round conductor involves Bessel functions of complex argument (conventionally written with the Kelvin functions $\text{ber}$ and $\text{bei}$).

Two limiting cases are worth remembering. At DC the resistance is ${\rho }_c/A$ and the internal inductance is ${\mu }_0/8\pi$ per unit length, which is $0.05\mu \text{H/m}$ for any solid round conductor regardless of size. At high frequency the current occupies only a surface skin of depth $\delta =\sqrt{2{\rho }_c/\omega \mu }$, so the resistance rises as $\sqrt{f}$ and the internal inductance falls toward zero.

The internal inductance is seldom written down explicitly, owing to a convenient piece of bookkeeping: the geometric mean radius. For a solid round conductor,

$$\text{GMR}=r{e}^{-1/4}\approx 0.7788r,$$

and substituting GMR for $r$ in the external-inductance formula includes the internal inductance automatically. This is why conductor datasheets quote a GMR rather than an internal inductance, and also why GMR is used for inductance while the actual radius is used for capacitance: the internal-inductance device has no electrostatic counterpart.

12.3 Bundles and earth wires

Two features of a real tower must be dealt with before a $3\times 3$ matrix can be reached.

Bundled conductors. Transmission at EHV uses two, three, or four subconductors per phase, spaced a few tens of centimetres apart, to reduce the surface gradient and hence corona. The bundle may be treated as one equivalent conductor, with

$${\text{GMR}}_{\text{bundle}}=(n{\text{GMR}}_c{A}^{n-1}{)}^{1/n}$$

for $n$ subconductors on a circle of radius $A$, or every subconductor may be entered explicitly and merged afterwards. The second route is more accurate and more laborious.

Earth wires. Overhead ground wires run along the tower tops for lightning shielding and are bonded to earth at every structure. They carry current and couple to the phases, so they must appear in the matrix. Their voltage is essentially zero everywhere, however, which is precisely the condition permitting a variable to be eliminated.

Both cases call for the same operation, Kron reduction. Partitioning the impedance matrix into the conductors retained ($p$) and those eliminated ($g$), the reduced matrix is

$${\mathbf{Z}}_{\text{red}}={\mathbf{Z}}_{pp}-{\mathbf{Z}}_{pg}{\mathbf{Z}}_{gg}^{-1}{\mathbf{Z}}_{gp}.$$

This is not an approximation but exact elimination: the same Gaussian-elimination step performed by the solver of Chapter 3, applied to the parameter matrix instead of the network matrix.

Reducing a tower to the matrix a three-phase model needs. Subconductors within a bundle are merged, then earth wires are eliminated by Kron reduction, leaving a $3\times 3$ matrix. Both steps are exact.

Eliminating the earth wires has a consequence worth anticipating. They provide a return path much closer to the phase conductors than the earth is, so they reduce $L_0$ and raise $R_0$ — a shielded line has a noticeably different zero sequence from an unshielded one, while its positive sequence barely changes. And the reduced matrix is generally not perfectly balanced even when the raw one looked symmetric, because the earth wires are not symmetric with respect to the three phases. Balanced data is what remains after averaging that out, which transposition physically achieves.

12.4 Cables

A cable is a different animal, and the difference is worth stating in one sentence: in an overhead line the conductors are metres apart in air, while in a cable they are millimetres apart across a solid dielectric. Everything follows from that.

Cable layers laid out by distance from the axis, with the three current loops above. Loop 1 is the core returning through the sheath, loop 2 the sheath returning through the armour, loop 3 the armour returning through earth. Loops couple only through the conductor they share.

Because the geometry is coaxial, the natural variables are not conductor currents but loop currents, each confined between two adjacent cylindrical surfaces. Each loop impedance is a sum of three tubular terms — the outer surface of the inner conductor, the insulation in between, and the inner surface of the outer conductor:

$$Z_{\text{loop 1}}=Z_{\text{core, outer}}+Z_{\text{insul 1}}+Z_{\text{sheath, inner}}.$$

The tubular impedances themselves come from modified Bessel functions of the tube's inner and outer radii. Two loops couple only through the conductor they share, and the mutual term is the tube mutual impedance, which has a memorable closed form: it is inversely proportional to the product of the tube's inner and outer radii. That single structural fact — adjacent loops couple, non-adjacent loops do not — is what makes the formulation tractable. Once the loop impedance matrix is assembled it is transformed back to conductor quantities, and if the sheath is grounded it is eliminated by the same Kron reduction as an earth wire.

Capacitance, by contrast, is almost trivial. Each insulation layer is a coaxial capacitor:

$$C=\frac{2\pi {\epsilon }_0{\epsilon }_r}{\ln (r_{\text{out}}/r_{\text{in}})},$$

and there is no mutual capacitance between cables when each has a grounded sheath, because the sheath is a complete electrostatic shield. The shunt admittance matrix is diagonal, which is one respect in which cables are simpler than lines.

The numbers that fall out are strikingly different from an overhead line's. With conductors millimetres apart and a dielectric of ${\epsilon }_r\approx 2.3$ to $4$, the capacitance is one to two orders of magnitude larger. Therefore:

$$Z_C=\sqrt{\frac{L^{\prime }}{C^{\prime }}}\sim 20\text{–}50\mathrm{\Omega },\nu =\frac{1}{\sqrt{L^{\prime }{C}^{\prime }}}\approx \frac{c}{\sqrt{{\epsilon }_r}}\sim 1.5\text{–}2\times {10}^8\text{m/s}.$$

Tens of ohms rather than hundreds, and about half the speed of light rather than nearly all of it. Both matter for Chapter 9's arithmetic: a cable's low surge impedance makes a line-to-cable junction a strong reflection point, and its low velocity means a short cable can still have a travel time worth modelling.

12.5 From a datasheet to the model

Most practical work begins not from geometry but from a table of sequence quantities at power frequency. The conversion is worked through below.

Suppose a 150 km, 230 kV line is reported at 60 Hz as

	$R$ ($\mathrm{\Omega }$/km)	$X$ ($\mathrm{\Omega }$/km)	$B$ ($\mu$S/km)
Positive sequence	0.05	0.40	3.9
Zero sequence	0.30	1.20	2.4

Step 1 — strip the frequency. Reactance and susceptance are $\omega L$ and $\omega C$, so with $\omega =2\pi (60)=377$:

$$L_1=\frac{0.40}{377}=1.061\text{mH/km},C_1=\frac{3.9\times {10}^{-6}}{377}=0.01034\mu \text{F/km},$$$$L_0=\frac{1.20}{377}=3.183\text{mH/km},C_0=\frac{2.4\times {10}^{-6}}{377}=0.006366\mu \text{F/km}.$$

Step 2 — enter it. Those six numbers plus the length are exactly what Bergeron — Sequence RLC asks for, and it is the entry to prefer, because it is the only three-phase form that preserves $R_0\ne R_1$.

Step 3 — convert to surge impedances if required. The modal quantities follow directly:

$$Z_{C1}=\sqrt{\frac{1.061\times {10}^{-3}}{1.034\times {10}^{-8}}}=320\mathrm{\Omega },{\tau }_1=150\sqrt{(1.061\times {10}^{-3})(1.034\times {10}^{-8})}=497\mu \text{s},$$$$Z_{C0}=\sqrt{\frac{3.183\times {10}^{-3}}{6.366\times {10}^{-9}}}=707\mathrm{\Omega },{\tau }_0=150\sqrt{(3.183\times {10}^{-3})(6.366\times {10}^{-9})}=675\mu \text{s}.$$

These results should be checked before being relied upon. The aerial velocity is $150/497\mu \text{s}=3.0\times {10}^5\text{km/s}$, the speed of light, exactly as an aerial mode should be. The ground mode comes out at $2.2\times {10}^5\text{km/s}$, about $0.74c$, which is within the expected range. Finally $Z_{C0}/Z_{C1}=2.2$ and ${\tau }_0/{\tau }_1=1.36$, both typical values. Had any of these emerged wrong — a velocity above $c$, for instance — the fault would lie in the input data or in a unit conversion.

Note what this entry costs: sequence surge impedance describes a lossless line, so the $R_1$ and $R_0$ from the datasheet are discarded.

Two different quantities, one symbol

A datasheet's "$Z_1$" is a series impedance in ohms per kilometre — $R_1+j{X}_1$. The Z1 that the sequence surge-impedance entry asks for is a surge impedance in ohms, $\sqrt{L_1/C_1}$. The two are unrelated, and entering one where the other belongs produces a line wrong by orders of magnitude while appearing perfectly plausible on the form. The quickest safeguard is the velocity test above.

12.6 Summary

• Capacitance comes from Maxwell's potential coefficients with the earth represented by image conductors; ${\mathbf{C}}^{\prime }=[{\mathbf{P}}^{\prime }{]}^{-1}$, and it does not vary with frequency. Use the average height $h_{\text{tower}}-\frac{2}{3}\text{sag}$.
• Series impedance needs the earth's finite resistivity. Carson's integral is the classical answer; placing the image at complex depth $h+2p$ with $p=\sqrt{\rho /j\omega {\mu }_0}$ reproduces it in closed form.
• The return-current depth $|p|$ shrinks from hundreds of metres at power frequency to metres at hundreds of kilohertz, which is the physical origin of the frequency dependence described in Chapter 11.
• Conductor internal impedance is a Bessel-function problem, usually sidestepped by using GMR $\approx 0.7788r$ in place of the radius for inductance — while capacitance still uses the true radius.
• Bundles are merged and earth wires eliminated by Kron reduction, which is exact. Earth wires lower $L_0$ and raise $R_0$.
• Cables are formulated in coaxial loop variables with tubular Bessel impedances, and have no mutual capacitance when sheaths are grounded. Their much larger $C^{\prime }$ gives $Z_C$ of tens of ohms and roughly half the speed of light.
• To go from a power-frequency datasheet to a model, divide $X$ and $B$ by $\omega$; prefer sequence-RLC entry so the resistances survive; and always check the implied propagation velocities.

12.7 Problems

Problem 12.1. A conductor of radius 15 mm hangs from 32 m towers with 9 m of sag. Compute the effective height and the self potential coefficient.

Solution 12.1

Average height:

$$h=32-\frac{2}{3}(9)=32-6=26\text{m}.$$

Self potential coefficient, with $1/2\pi {\epsilon }_0=1.797\times {10}^{10}\text{m/F}$:

$$P_{ii}^{\prime }=\frac{1}{2\pi {\epsilon }_0}\ln \frac{2h}{r}=1.797\times {10}^{10}\ln \frac{52}{0.015}=1.797\times {10}^{10}\ln (3467).$$$$\ln (3467)=8.151\Rightarrow P_{ii}^{\prime }=1.465\times {10}^{11}\text{m/F}.$$

For a single conductor the capacitance is the reciprocal, $6.8\text{pF/m}=6.8\text{nF/km}$ — the right order for an overhead line. Had we used the tower height instead of the average, we would have got 26 m of clearance wrong by 6, changing the answer by about 3 %.

Problem 12.2. Compute the complex-depth magnitude $|p|$ at 60 Hz for soil resistivities of $20$, $100$, and $1000\mathrm{\Omega }\text{m}$. What does the spread imply about how carefully soil resistivity needs to be known?

Solution 12.2

$|p|=\sqrt{\rho /\omega {\mu }_0}$, with $\omega {\mu }_0=377(1.257\times {10}^{-6})=4.738\times {10}^{-4}$:

$$\rho =20:|p|=\sqrt{20/4.738\times {10}^{-4}}=\sqrt{4.22\times {10}^4}=205\text{m},$$$$\rho =100:|p|=\sqrt{2.11\times {10}^5}=459\text{m},$$$$\rho =1000:|p|=\sqrt{2.11\times {10}^6}=1453\text{m}.$$

A fifty-fold spread in resistivity gives only a seven-fold spread in depth, because of the square root — and the depth enters the impedance inside a logarithm, which compresses it further. A factor of two error in assumed soil resistivity moves the zero-sequence inductance by only a few percent.

That is a genuinely useful piece of reassurance: soil resistivity is the hardest input to know and one of the least critical. It is not unimportant, but it does not need to be measured to better than a factor of two for most studies.

Problem 12.3. Explain why GMR is used for inductance calculations but the physical radius is used for capacitance.

Solution 12.3

The GMR is a bookkeeping device for the conductor's internal inductance — the flux that links current flowing inside the conductor itself. Substituting $r{e}^{-1/4}$ for $r$ in the external-inductance formula produces exactly the extra ${\mu }_0/8\pi$ per unit length that the internal flux contributes, so one logarithm covers both effects.

Capacitance has no internal counterpart. Charge resides entirely on the conductor's surface and there is no field inside a conductor in electrostatic equilibrium, so there is nothing extra to fold in. The physical surface is where the charge is, so the physical radius is the right one.

A useful mnemonic: GMR is smaller than $r$ because the extra internal inductance is equivalent to a conductor that appears slightly thinner. Using GMR for capacitance would overstate it, since a thinner conductor holds less charge at a given potential.

Problem 12.4. A three-phase line has two earth wires. Starting from a $5\times 5$ impedance matrix, what steps produce the data Chapter 10's sequence entry wants?

Solution 12.4

1. Kron-reduce out the two earth wires. Partition the $5\times 5$ into a $3\times 3$ phase block ${\mathbf{Z}}_{pp}$, a $3\times 2$ coupling block ${\mathbf{Z}}_{pg}$, and a $2\times 2$ earth-wire block ${\mathbf{Z}}_{gg}$, then form ${\mathbf{Z}}_{pp}-{\mathbf{Z}}_{pg}{\mathbf{Z}}_{gg}^{-1}{\mathbf{Z}}_{gp}$. This is exact, and it is valid because the earth wires are held at essentially zero potential.
2. Balance the result. The reduced $3\times 3$ will not have identical diagonals or identical off-diagonals, because the earth wires sit asymmetrically over the phases. Average the diagonals to a single $Z_s$ and the off-diagonals to a single $Z_m$ — which is what transposition does physically.
3. Take the eigenvalues from Chapter 10: $Z_0=Z_s+2Z_m$ and $Z_1=Z_s-Z_m$, and the same for the capacitance matrix with its sign pattern.
4. Divide out $\omega$ to get $L$ and $C$ per unit length, then enter with the length through Bergeron — Sequence RLC.

Step 2 is the only approximation in the chain, and it is the one that matches the assumption Chapter 10's model is built on.

Problem 12.5. A 3 km cable has $Z_C=30\mathrm{\Omega }$ and a propagation velocity of $1.7\times {10}^8\text{m/s}$. Find $\tau$, decide the model, and comment on the junction where it meets a $350\mathrm{\Omega }$ overhead line.

Solution 12.5

$$\tau =\frac{3000}{1.7\times {10}^8}=1.76\times {10}^{-5}\text{s}=17.6\mu \text{s}.$$

Whether it can be a travelling-wave line depends on the step. At $\mathrm{\Delta }t=20\mu \text{s}$ it cannot — $\tau <\mathrm{\Delta }t$, so Chapter 9's fallback demotes it to a $\pi$. At $\mathrm{\Delta }t=2\mu \text{s}$ it gets $N=9$ steps of delay and is modelled properly. Note that a 3 km overhead line would have $\tau =10\mu \text{s}$; the cable's slower velocity makes even a short run worth resolving.

At the junction, a surge arriving from the line sees a large impedance drop:

$$\mathrm{\Gamma }=\frac{30-350}{30+350}=\frac{-320}{380}=-0.84.$$

Most of the arriving wave reflects, inverted, and the voltage transmitted into the cable is only $1+\mathrm{\Gamma }=0.16$ of the incident — a sixfold reduction. Cables are strong reflectors and, incidentally, good surge attenuators for whatever lies beyond them. The corollary is that the overhead line just before the junction sees a nearly-cancelling reflection and the current there roughly doubles.

Problem 12.6. A colleague enters $Z_1=0.40\mathrm{\Omega }$ into the sequence surge-impedance form, taking the value from the reactance column of a datasheet. What will they observe?

Solution 12.6

A line that behaves nothing like a transmission line, in a way that is easy to misdiagnose.

With $Z_1=0.4\mathrm{\Omega }$ and, say, ${\tau }_1=500\mu \text{s}$, the implied parameters are $L_1=Z_1{\tau }_1=200\mu \text{H}$ and $C_1={\tau }_1/Z_1=1250\mu \text{F}$. The capacitance is roughly a thousand times what any line has, so the model will draw an enormous charging current, and the near-zero surge impedance means it will look like a short circuit to any arriving surge — $\mathrm{\Gamma }\approx -1$ at the sending end, so voltages collapse rather than propagate.

The diagnosis is the velocity check: $\nu =d/\tau$ should come out near $3\times {10}^5\text{km/s}$ for an aerial mode. Any value wildly away from that means the surge impedance and the travel time are inconsistent, which is exactly what happens when a series impedance is entered where a surge impedance belongs.

12.8 References

• J. R. Carson, "Wave propagation in overhead wires with ground return", Bell System Technical Journal, 1926 — the ground-return correction integral.
• A. Deri, G. Tevan, A. Semlyen and A. Castanheira, "The complex ground return plane: a simplified model for homogeneous and multi-layer earth return", IEEE Trans. PAS, 1981 — the complex depth of penetration.
• L. M. Wedepohl and D. J. Wilcox, "Transient analysis of underground power transmission systems", Proc. IEE, 1973 — the coaxial loop formulation for cables.
• H. W. Dommel, Electromagnetic Transients Program (EMTP) Theory Book, Bonneville Power Administration — line-constants and cable-constants calculations.
• J. Arrillaga and N. R. Watson, Power Systems Electromagnetic Transients Simulation, IET Power and Energy Series 39 — overhead line and cable parameter derivation.

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