Underground Cables

The two radii. The conductor diameter is 2.4 cm, so \(r = 1.2\) cm, and the insulation thickness carries the boundary out to the inner face of the sheath:

This one logarithm is used again in Problems 2, 15, 16 and 17. Every property of this cable hangs on it.

The integral. Shells of radius \(x\) and thickness \(dx\) are crossed in series by the same leakage current, so their resistances add:

Keep \(\rho\) and \(l\) in the same units. With \(\rho\) in Ω·cm, \(l\) must be in centimetres: 3 km is \(3\times10^5\) cm.

Substituting:

That is 175.2 MΩ, or 525.7 MΩ·km — the figure a manufacturer would quote, since the length-independent form is the useful one.

The leakage current is drawn at phase voltage, the sheath being earthed:

109 µA. Problem 2 finds the charging current of the same cable to be 4.76 A — some forty thousand times larger. A cable conducts almost nothing and displaces a great deal.

Doubling the length halves the resistance. The second 3 km is 175.2 MΩ of leakage path connected in parallel with the first, not in series with it:

The conductor resistance of the same joint doubles. The two resistances of a cable move in opposite directions with length, and writing \(\rho l\) where \(\rho/l\) belongs is the commonest arithmetic error in this chapter.

The capacitance, from the numerical form of the chapter, which folds \(2\pi\varepsilon_0 = 0.05563\ \mu\)F/km into the constant 18:

The charging current:

Leading, and drawn with the far end open. Beside it the 109 µA of leakage from Problem 1 is invisible.

The reactive power, most conveniently in the line-voltage form, because that is the number written on the drawing:

That is 90.6 kVAr per kilometre for the three phases together.

The comparison with the overhead line:

Twenty-eight times, and there is nothing to be done about it: the entire point of a cable is to put an earthed surface within millimetres of the conductor, and that is precisely what makes the capacitance large. Set 6 computed the overhead figure; the logarithm there is 5.9, here it is 0.73.

What that means on a network. Thirty kilometres of this cable is a permanently connected 2.7 MVAr capacitor. At light load it raises the voltage rather than depressing it, and the reactive planning of an urban 33 kV system has to count cable capacitance as a source. Problem 20 follows the same quantity to the point where it becomes a limit rather than a nuisance.

The capacitance gives the geometry. Per kilometre, \(C = 0.52/2 = 0.26\ \mu\)F/km, and the capacitance formula inverts directly:

Note what has not been found: \(r\) and \(R\) separately. Capacitance fixes only their ratio, because the field is scale-invariant — a cable twice the size with the same ratio has exactly the same capacitance per unit length.

The insulation resistance then gives the resistivity, the geometry now being known:

With \(l = 2\times10^5\) cm, so the answer comes out in Ω·cm. A healthy impregnated-paper cable sits between \(10^{14}\) and \(10^{15}\); a reading two decades below that is a wet or contaminated dielectric.

The time constant needs no geometry at all. Multiplying the two integrals cancels the logarithm and the length together:

Check it against the material constants: \(\rho\varepsilon = (6.39\times10^{12}\ \Omega\text{m})(3.5\times8.854\times10^{-12}) = 198\) s. The two agree, as they must.

And that is the safety point. Disconnect the cable at full voltage and its charge decays with a time constant of 3.3 minutes:

On a 33 kV cable, 5% of the peak phase voltage is still 1.3 kV — lethal. A disconnected cable is not a safe cable, and the discharge rod and earth are applied before anything else, every time.

Why the product is worth remembering. Two bench measurements that each depend on unknown dimensions combine into one that depends only on the material. It is a standard trick of the subject: form the combination in which the unknown geometry cancels, and what remains identifies the material.

Use the peak voltage. Breakdown responds to the instantaneous field, and the voltage across this dielectric is the phase voltage, the sheath being earthed:

Two conversions, both easy to forget, and forgetting either understates the stress by \(\sqrt3\) or \(\sqrt2\). The line voltage never appears in a single-core stress calculation.

The stress profile. Eliminating the charge between \(g = q/(2\pi\varepsilon x)\) and \(V = (q/2\pi\varepsilon)\ln(R/r)\):

The permittivity has cancelled. A cable filled with any homogeneous dielectric whatever has this same profile — which is exactly why grading, in Problems 8 to 11, has to make the dielectric non-homogeneous to change anything.

The two extremes:

The ratio is the geometry and nothing else. The insulation must be specified for 42.4 kV/cm everywhere, though the outer layer never sees more than 18.2.

Where the stress reaches 30 kV/cm:

So the inner 0.62 cm of the 2.0 cm wall is worked above 30 kV/cm — 31% of the thickness.

But the volume fraction is smaller, and it is the honest figure, because material is bought by volume and the inner shells are the small ones:

Only 22% of the dielectric does the demanding work; the remaining 78% is present because it has to be somewhere between the conductor and the sheath, not because it is needed at that stress.

That waste is the case for grading. The profile is fixed by \(1/x\) as long as the dielectric is uniform, so the only two ways to flatten it are to vary the permittivity with radius (Problems 8 and 9) or to break the annulus into several capacitors in series at held potentials (Problems 10 and 11).

The working voltage across the dielectric, as a peak value:

Why an optimum exists. Writing the stress requirement as an expression for the overall radius,

A small conductor blows up the exponent and demands an enormous wall; a large conductor tames the exponent but the leading factor \(r\) grows without limit. Between the two the cable is smallest.

Differentiating:

The exponential never vanishes, so the bracket must. The condition is \(R/r = e\) for every cable ever designed, irrespective of voltage, material or size.

The dimensions follow immediately:

A 36 mm conductor inside a 31 mm wall of dielectric — which is a recognisable 132 kV cable, and about 98 mm over the insulation screen before sheath, armour and serving are added.

The minimum stress, with \(\ln(R/r) = 1\):

Always 36.8%, in every economically designed cable, because the ratio of the stresses is the ratio of the radii and that has been fixed at \(e\). The outer dielectric is worked at just over a third of its capability — the irreducible waste that Problems 8 to 11 attack.

A check on the design. Confirm the stress at the conductor directly, without using the optimum:

And a practical caution. Nothing in the derivation mentions current. If the 132 kV circuit needs only 400 A, a 36 mm solid conductor is far more copper than the load requires — so it is not made solid. A hollow conductor, or one wound over a helical spacer or a central oil duct, provides the radius without the metal, and skin effect means the missing centre was carrying almost nothing anyway.

The problem is now the other way round. Problem 5 fixed \(g_{max}\) and minimised \(R\); here \(R\) is fixed and \(g_{max}\) is to be minimised:

Differentiating \(f\), remembering that \(R\) is now the constant:

The identical condition. Minimising the size for a given stress and minimising the stress for a given size are the same problem seen from two sides, because both are governed by the single product \(r\ln(R/r)\).

The dimensions and the stress:

No conductor radius whatever, inside a 4 cm duct at 66 kV, can bring the conductor-surface stress below 36.6 kV/cm. That is a hard floor set by the duct.

Departing from it in both directions:

A conductor 39% too small costs 9.6%; one 50% too large costs 11.9%. Both are modest, and both are one-sided — the optimum is a genuine minimum with zero slope, so small departures cost almost nothing and only large ones bite.

The two failure modes are not symmetric, though. The stress penalty is similar either way, but what goes wrong differs:

The second is the dangerous one. Problem 17 shows that the thick wall of the first case costs rating through \(S_{\text{ins}}\), which is an economic penalty; the thin wall of the second removes the margin that manufacturing imperfection eats into, which is a reliability one.

What the duct really decides. Given \(R = 4.0\) cm and a dielectric permitting, say, 40 kV/cm, the highest voltage the duct can ever carry is

So a 66 kV circuit fits and a 132 kV circuit cannot, no matter how the conductor is chosen. Reusing an existing duct route puts a ceiling on the voltage of any future reinforcement, and that ceiling is worth computing before the duct is laid.

The peak voltage across the dielectric is that of Problem 6:

Invert the stress condition for the outer radius, the conductor radius now being given:

Note that \(R\) appears inside a logarithm, so the thickness grows exponentially as the conductor is made thinner. There is no linear intuition to fall back on here.

The economical design at the same stress:

The comparison:

A conductor 26% too small needs 23% more dielectric thickness — and, since the cable's cross-sectional area goes as \(R^2\), 10% more of everything outside it: sheath, armour, serving, drum space and trench.

So the extra copper pays for itself, which is the whole content of the economical-size rule in practice:

2.56 cm² more metal buys 0.53 cm off the wall over the whole route, plus a shorter thermal path and a lower conductor resistance the load never asked for. Which side wins depends on the copper price and the route length, and that calculation is exactly why standard cable ranges exist at all.

A check that the design is sane. Its \(R/r = 3.85\), against the economical 2.72 — high enough that Problem 5's warning applies, and the outer dielectric works at \(1/3.85 = 26\%\) of the inner rather than 37%. The design is safe and slightly wasteful, which is the acceptable direction to err.

Why grading works at all. The two layers are capacitors in series, so they carry the same charge per unit length, and the stress in layer \(i\) is

Raising \(\varepsilon_r\) in the inner layer lowers the stress there for the same charge. The permittivity that cancelled in Problem 4 no longer cancels, because it is no longer the same everywhere.

Each layer is worst at its own inner face, so there are two local maxima, and the best design makes them equal:

The condition forces \(\varepsilon_{r1} > \varepsilon_{r2}\): the denser, higher-permittivity dielectric goes inside, where the field is worst. Putting them the other way round makes the cable worse than ungraded.

The working voltage is the integral of the stress across both annuli, each starting from \(g_{max}\) at its inner face:

The ungraded cable, same \(r\), same \(R\), same permissible stress:

A gain of 31.5% for no extra material at all — the same volume of dielectric, differently arranged.

What each version can serve, converting back through peak and \(\sqrt3\):

The ungraded cable falls just short of a 66 kV system; the graded one clears it with 20% in hand. The grading has moved the cable across a voltage class boundary, which is the commercially significant thing it does.

Check the second layer's stress at its outer face, to confirm nothing is over-stressed where it was not examined:

Half the limit, so the profile is still a sawtooth rather than flat. Two layers cannot do better than this; Problem 9 adds a third.

Why it is rarely done. Few dielectrics have usefully different permittivities and comparable strength; and permittivity drifts with temperature at different rates in different materials, so a cable graded exactly at 20 °C is no longer exactly graded at 70 °C. The design has to be checked at both ends of the temperature range, and the grading gain quoted for the worse of the two.

The grading condition chains across all three layers, since the charge is common to capacitors in series:

The boundaries are decided entirely by the permittivities available; the designer does not get to choose them once the materials are chosen. Notice how close together they are — both boundaries fall in the inner 22% of the wall.

The working voltage, integrating the stress across the three annuli in turn:

Against the ungraded cable:

A gain of 47.3%, against the 31.5% two layers bought in Problem 8. Grading is worth more when the annulus is thicker, because there is more under-worked dielectric to reclaim.

The share each layer holds — and this is the instructive part:

The outermost layer holds nearly three-quarters of the voltage — because it is 1.87 cm thick while the other two together are 0.53 cm. The grading condition sets the boundaries, and the boundaries land wherever the permittivity ratios put them, not where the designer would like.

Which shows where the scheme fails. The outer layer starts at \(g_{max}\) and falls to

— 42% of the limit at the sheath. Most of the original waste is still there, sitting in the outermost layer, and the two extra materials have done almost nothing about it.

A fourth layer would have to go outside, where the waste is, and it needs a permittivity below 3. Suppose \(\varepsilon_{r4} = 2.3\) (XLPE):

Another 15% — worthwhile, and obtainable only because a fourth material with a genuinely lower permittivity existed. Layers added inside would need permittivities above 5, which for a solid cable dielectric of adequate strength is where the catalogue ends.

The limit of the method, stated plainly. Perfect grading — constant stress at every radius — requires \(\varepsilon_r \propto 1/x\), a continuously varying permittivity from 5 at the conductor to 1.25 at the sheath. No such material exists, so grading is always a staircase approximation to a curve, and the number of steps is set by the number of usable dielectrics rather than by anything electrical.

The ungraded cable first, as the benchmark:

What an intersheath does. A thin metallic cylinder at radius \(r_1\), held at a fixed potential by a transformer tapping, splits the annulus into two cables in series. Each has its own conductor and its own earthed sheath and carries only its share of the voltage:

Each is worst at its own inner boundary, exactly as before — the profile becomes a sawtooth with two teeth instead of one long ramp.

Placing it. Two conditions are wanted: equal maxima, and the most compact design. Taking the radii in geometric progression makes both logarithms equal:

Equalising the two maxima then reduces to a statement about the radii alone, since \(\ln\alpha\) cancels:

The working voltage is the sum of the two shares:

The improvement is \(124.77/83.18 = 1.500 = (1+\alpha)/2\), exactly the general result for two equal geometric steps.

Check every boundary, which is the step that catches sign and placement errors:

The stress jumps discontinuously at the intersheath, from 25 back up to 50. That discontinuity is the grading: the metal cylinder terminates one field and starts another.

In system terms, the graded cable serves

A 132 kV circuit with margin, where the ungraded cable manages only 102 kV between lines. And the intersheath itself sits at 58.8 kV rms — a live part, needing its own termination, its own bushing and its own maintenance.

Three annuli in geometric progression:

Every logarithm is now \(\ln 2 = 0.6931\), which is what makes the arithmetic collapse.

Equal maxima in the three annuli require the three voltage shares to be proportional to the three inner radii:

Adding the three shares recovers the total:

And \((1+\alpha+\alpha^2)/3 = (1+2+4)/3 = 2.333\), confirming the general form.

The two potentials, working inward from the earthed sheath:

The general rule. With \(n\) equal geometric steps of ratio \(\alpha = (R/r)^{1/n}\), every share is \(g_{max}r_i\ln\alpha\) and the radii form a geometric series:

A ratio of a sum to a count — the mean of the geometric series. Since the terms increase, the mean always exceeds 1, so grading can never make a cable worse when it is done correctly.

The second question. For \(R/r = 4\), ask for \(V/V_0 = 2\) with \(\alpha = 4^{1/n}\):

The gain saturates. Even with infinitely many intersheaths — a perfectly uniform stress of \(g_{max}\) everywhere — the factor is only \((R/r-1)/\ln(R/r) = 3/1.386 = 2.164\). Doubling is therefore possible but expensive: \(n = 9\) gives 2.002, so eight intersheaths are needed, each holding about 11% of the voltage and each requiring its own tapping and termination.

Which is the practical verdict on the scheme. The first intersheath buys 50%, the second a further 14%, the third a further 6%. Diminishing returns arrive immediately, while the cost — a tapping, a bushing, a termination and a maintained live part — is the same for each. Two is the most that has ever been built, and even that is now historical.

Why six capacitances and not one. The field in a belted cable is not radial — there is one earthed boundary for three cores — so no single \(\ln(R/r)\) describes it. What can be said is that the electrostatics of three conductors inside an earthed sheath is completely captured by three core-to-core capacitances \(C_c\) in delta and three core-to-sheath capacitances \(C_s\) in star. They are measured, not computed.

Test 1 — all three cores bunched, against the sheath. Joining the cores puts both plates of every \(C_c\) at the same potential, so none of them can hold charge. Only the three \(C_s\) remain, in parallel:

Test 2 — two cores joined to the sheath, against the third. Everything except core \(a\) is now one terminal, so three capacitances bridge the gap: \(C_s\) from \(a\) to the sheath and \(C_c\) from \(a\) to each of the other two, all in parallel:

Combining them into the working value. A delta of \(C_c\) presents \(C_c\) directly plus two \(C_c\) in series between any two terminals, that is \(1.5C_c\); a star of \(3C_c\) presents two of \(3C_c\) in series, also \(1.5C_c\). The delta is therefore a star of \(3C_c\) — note that for capacitance the transformation runs the opposite way to the impedance case of Set 1:

Both stars have their common point at neutral potential — the sheath is at earth and, for a balanced supply, so is the electrical neutral of the cores — so they are simply in parallel.

The short cut, which skips \(C_s\) and \(C_c\) altogether:

Test 3 — two cores joined, measured against the third, sheath isolated. Now the sheath is a floating node. From core \(a\) to the joined pair \(bc\) there are two parallel routes: the two \(C_c\) directly, and \(C_s\) from \(a\) to the floating sheath in series with the two \(C_s\) of \(b\) and \(c\) in parallel:

The series combination is the point: an isolated sheath still conducts charge from core to core, it simply does so through two capacitances instead of one.

Per kilometre, for comparison with catalogue data:

Squarely in the 0.15–0.4 µF/km band the chapter quotes, and Problem 13 turns it into a charging current.

Only \(C_N\) enters a circuit calculation. The six capacitances of Problem 12 were a modelling device; what each phase actually draws current through is the single equivalent value:

The charging current:

Leading the phase voltage by very nearly 90°; Problem 16 measures how far short of 90° it actually falls.

The reactive power, in the line-voltage form:

Against the overhead line at the same voltage:

Twenty times, matching the ratio of the capacitances \(0.185/0.0094 = 19.7\) exactly, since everything else in the expression is common.

A hundred kilometres of it:

A 0.7 MVAr capacitor bank permanently connected to the 11 kV busbar, which cannot be switched out because it is the network itself. On a large urban distribution system this is not a curiosity but a planning quantity: it raises the voltage at light load, it must be counted in the reactive balance, and it changes the resonant frequency of the whole system.

And it inverts a familiar rule. An overhead line at light load absorbs reactive power below its surge impedance loading and produces it above; an 11 kV cable network produces reactive power at every load it will ever see, because its inductive reactance is about a quarter of an overhead line's while its capacitance is twenty times larger. Set 31 works the same comparison through for a long line, where the balance between the two is the whole subject.

Write the third test in terms of the two unknowns. With the sheath floating, core \(a\) reaches the joined pair by two routes in parallel — two \(C_c\) directly, and \(C_s\) in series with the parallel pair \(2C_s\) through the floating sheath:

One equation, two unknowns. It should not be enough — and yet the quantity actually wanted comes out of it.

Compare it with the quantity wanted:

The two expressions are proportional — the same linear combination, scaled by \(2/3\). The third test measures \(C_N\) directly, up to a known constant, and needs no other reading at all.

The number:

Exactly the value Problem 12 obtained from two good readings. The broken sheath has cost nothing.

Why it works, which is more interesting than the algebra. Both the third test and the working condition put the three cores in a state where the sheath carries no net charge to earth: in the test because it is isolated, in service because a balanced three-phase set of core voltages sums to zero and holds the sheath at neutral potential anyway. The test therefore reproduces the service condition almost exactly, which is why one number suffices.

Why \(C_s\) and \(C_c\) individually cannot be recovered. Any pair satisfying \(\tfrac{2}{3}C_s + 2C_c = 0.740\) reproduces the reading:

All three are indistinguishable to this test and all three give the same \(C_N\). A single measurement can determine a single combination of two unknowns, and the combination it determines here happens to be the useful one.

When the split does matter. Not for charging current or reactive power, both of which need only \(C_N\). It matters for the unbalanced case — a single-phase fault, or an open conductor — where the sheath is not at neutral potential and the two capacitances behave differently. For those studies the sheath must be repaired and the full set of tests performed.

Where the loss comes from. Polarisation lags the applied field, so the current the cable draws leads the voltage not by 90° but by \(90^\circ - \delta\). Resolving it, the in-phase component is \(I_c\tan\delta\) and the power it carries is dissipated inside the insulation:

Note what it does not contain: the load current. Dielectric loss is present the moment the cable is energised and is unchanged whether it carries 300 A or nothing.

At 33 kV:

A ratio of 60 to 1. At distribution voltage the dielectric loss is a rounding error in the loss account.

The same paper cable at 220 kV, which is the whole point:

The conductor loss has not changed at all — the current is the same. The dielectric loss is now 75% of it, and it is generated inside the insulation, at the very start of the thermal path where it does the most damage.

With XLPE at 220 kV. Two things improve together: the permittivity falls from 3.5 to 2.3, and the loss tangent falls from 0.005 to 0.0004:

A nineteenfold reduction, of which a factor of 12.5 is the loss tangent and 1.52 the permittivity. Back to 3.9% of the conductor loss — negligible again.

The loss angle itself, as it would be quoted on a test certificate:

Which is why the loss tangent is quoted rather than the angle: 0.005 against 0.0004 is a readable difference, 17 minutes of arc against 1.4 is not.

And the reason it is not merely an economic question. The loss tangent of impregnated paper rises with temperature, so:

A positive feedback loop. Below a critical voltage it converges to a slightly elevated temperature; above it, it does not converge at all and the dielectric fails thermally. Problem 18 computes where that ceiling lies.

The current is almost entirely capacitive, so the capacitance follows from its magnitude:

Matching Problem 2's computed 0.2649. The approximation \(I \approx I_c\) is safe because the in-phase component is smaller by a factor of \(\tan\delta\), and its effect on the magnitude is smaller still — of order \(\tan^2\delta\), or one part in forty thousand.

The power factor, which for a cable is a very small number rather than a nearly-unity one:

The loss angle is the complement:

For angles this small \(\cos\phi\), \(\sin\delta\) and \(\tan\delta\) are numerically identical to four figures, which is why the measured power factor may be quoted directly as the loss tangent without apology.

Cross-check against the loss formula:

Which is 151 W/km per phase, and 1.81 kW for the three phases over the 4 km — the figure Problem 15 obtained by a different route.

What 1500 W would have meant. The same current, so the same capacitance and the same geometry, but:

Two and a half times the expected value with the capacitance unchanged. A rise in \(\tan\delta\) at constant \(C\) is the classic signature of a dielectric that has absorbed moisture or is discharging internally in voids — the geometry is intact, the material is not. This is precisely the diagnostic separation Problem 3 made with insulation resistance, arrived at from the a.c. side.

How it is measured in practice. Not with a wattmeter — 604 W in an apparent 121 kVA is 0.5%, far beyond a wattmeter's accuracy at that power factor. A Schering bridge balances the cable against a loss-free standard capacitor and reads \(\tan\delta\) directly from a resistance ratio, resolving parts in \(10^5\). The wattmeter arithmetic above is the principle; the bridge is the instrument.

Heat flow obeys the same mathematics as current flow, so the thermal resistance of the annulus is Problem 1's integral with the thermal resistivity in place of the electrical one — and per unit length, so no \(l\) appears:

The same logarithm for the third time. This is the chapter's central economy: fix \(r\) and \(R\) and three of the cable's four defining properties are fixed with them.

The chain, which is a series circuit because all the heat crosses every layer:

The soil is 73% of it, though it is thermally the better material — because its path is metres long where the insulation's is 13 mm. A cable is rated by its surroundings far more than by itself.

The temperature budget, in watts per metre:

Every watt per metre generated anywhere inside the cable must cross the whole 1.509 K·m/W, and 43.08 W/m is all the cable can shed.

The current. With \(R = 0.10\ \Omega\)/km \(= 1.0\times10^{-4}\ \Omega\)/m and \(W_d = 0.151\) W/m:

What the dielectric loss costs here:

One ampere in 656 — 0.15%. At 33 kV the dielectric loss is genuinely negligible in the rating, and it is customary to omit it entirely below 66 kV. Problem 18 shows how quickly that stops being true.

Why \(W_d\) is subtracted rather than added to the losses. The current is the unknown; the dielectric loss is not. It takes its share of the 65 K budget first, whatever the loading, and only what is left is available to be turned into current. Writing \(I = \sqrt{(\theta_{max}-\theta_{amb})/RS}\) and then adding \(W_d\) to the total loss afterwards is the standard error, and it overstates the rating.

A sense of scale. The same 300 mm² of copper in an overhead ACSR conductor in moving air carries some 600 A at 75 °C with no thermal resistance worth the name; here the metal itself could carry far more, and 655 A is what the soil permits. Doubling the copper would not double the rating, because the heat still has to cross the same ground — which is why underground circuits are duplicated rather than enlarged.

The dried soil. Only \(S_{\text{total}}\) changes:

A fall from 655 A to 498 A — 24% of the rating gone, and nothing about the cable has changed.

The scaling is the useful part. Neglecting the tiny \(W_d\) term, the rating goes as \(1/\sqrt{S_{\text{total}}}\):

So doubling the soil resistance costs only 24%, not 50%, because the insulation's share of the chain is unchanged. The square root is what makes cable ratings relatively forgiving of thermal uncertainty — and also what makes them stubborn: recovering that 24% needs the total thermal resistance back down again, and there is no cheap way to do that once a cable is buried.

And it is a self-inflicted failure. The cable dries the soil by heating it; the drier soil raises the temperature further; the process runs away slowly over a summer. It is why heavily loaded circuits are laid in a bed of stabilised backfill — a sand-cement mixture whose thermal resistivity does not rise when it loses moisture — rather than in the spoil that came out of the trench.

The same cable at 220 kV. From Problem 15 the paper dielectric now dissipates 6714 W/km, that is 6.714 W/m:

655 A becomes 603 A — an 8% loss of rating bought by nothing but the voltage. At 33 kV the same term cost 0.15%.

The ceiling. Set the dielectric loss equal to the whole permitted heat flow and solve for voltage:

At 557 kV this paper cable reaches 90 °C with its far end open and no load current at all. It is not a cable at that voltage; it is a heater.

And the true ceiling is lower still, because \(\tan\delta\) rises with temperature. The feedback loop of Problem 15 diverges well before the static calculation says it should:

Repeat the last line's arithmetic with \(\omega C\tan\delta = 5.469\times10^{-5}\times0.0004 = 2.188\times10^{-8}\) and the ceiling moves to 2430 kV — which is to say it does not exist. The material, not the geometry, is what caps a paper cable's voltage.

Where the sheath voltage comes from. The alternating flux of the three conductors links the sheath loop, inducing a longitudinal e.m.f. along each sheath. With the sheaths bonded at both ends that loop is closed, and a circulating current flows. The mutual reactance between a conductor and its own sheath, for the flat formation, is

The circulating current is the induced e.m.f. divided by the sheath's own impedance:

Forty-seven per cent of the load current, flowing in a metal that carries no load and does no useful work.

The losses:

The sheath very nearly doubles the cable's losses. In the compact algebraic form, \(\lambda_1 = \dfrac{R_s}{R}\cdot\dfrac{X^{2}}{R_s^{2}+X^{2}} = 4\times0.2249 = 0.900\) — the same 0.90.

What that does to the rating. The thermal calculation of Problem 17 sees the total loss, so the effective resistance is \(R(1+\lambda_1)\):

Bonding both ends has thrown away 27% of the circuit's capacity.

Cross-bonding recovers most of it. The route is divided into groups of three minor sections and the sheaths transposed at each joint, so that each sheath spends one section in each position and the three induced e.m.f.s — 120° apart — sum to zero around the loop:

Exactly the identity that made a balanced three-phase neutral current vanish in Set 1, and exactly the reason conductors themselves are transposed on an overhead line. No circulating current can flow; only eddy losses remain, typically \(\lambda_1 \approx 0.04\).

The improvement:

35% more current from a change in bonding alone — no more copper, no different cable. Which is why every EHV single-core circuit of any length is cross-bonded, and why the alternative, single-point bonding, is used only on short routes where the standing voltage at the free end stays within the sheath insulation's rating.

Why not steel-wire armour. On a three-core cable the three conductor currents sum to zero, so the net flux linking the armour is small and steel armour is harmless. On a single-core a.c. cable the conductor current is unbalanced by definition, and steel around it sees the full alternating flux:

All three are severe, and the losses are dissipated in the outermost layer where the thermal path is shortest but the temperature limit of the serving is lowest. Single-core a.c. cables therefore use aluminium wire armour, or non-magnetic tape, or no armour at all — and a steel-armoured single-core cable is a classic specification error, not a design choice.

The charging current per kilometre:

Drawn whether the cable is loaded or not, and proportional to length because the capacitance is.

The critical length is where that alone fills the conductor:

At 83 km the cable is thermally full at the sending end with its far end open. It can transmit nothing.

Below that, the two currents are in quadrature — the charging component leads by 90°, the load component is nearly in phase — so they combine as the sides of a right triangle rather than adding:

Which is why the useful current in the figure holds up so well at first and then collapses: the loss is second-order until \(I_c\) becomes comparable with the rating.

At 40 km:

Against 831 MVA at zero length — so half the critical length costs only 12% of the capacity. The penalty is negligible until it is sudden.

At 132 kV, the same cable, the same capacitance:

Three times the length, for exactly the ratio of the voltages — the critical length varies as \(1/V\), so the higher the voltage the shorter the cable may be. That is the exact opposite of the overhead-line rule, where higher voltage buys distance.

What can be done about it, and what cannot. Shunt reactors at both ends halve the charging current the conductor has to carry, roughly doubling the critical length; mid-point compensation on a land cable does better still:

The last line is decisive. A land route can be broken every 50 km for a compensating station; a route across a strait cannot. The charging current of a long submarine a.c. cable is therefore irreducible.

Hence d.c. With \(\omega = 0\) the charging current vanishes identically, the capacitance charges once at energisation and then draws nothing, and the cable's whole thermal rating is available for useful current at any length whatever:

Which is why every submarine link beyond about 50 km in the world is HVDC — not because the converters are desirable, but because the alternative does not exist. Set 33 of the book's later parts takes up the converter side of that argument.