Underground Cables
Burying a circuit replaces metres of air with millimetres of solid dielectric, and every consequence of that substitution — a radial field that is fiercest at the conductor surface, a capacitance thirty times larger than an overhead line's, and heat that can only escape by conduction — follows from the same cylindrical geometry.
- Why a cable's insulation resistance falls as it is made longer while its conductor resistance rises, and why both come from the same integral.
- How the single formula \(g(x)=V/\big(x\ln(R/r)\big)\) governs stress, capacitance and insulation thickness together, and why the stress is worst where the metal is.
- Why the cheapest cable for a given voltage has \(R/r=e\) — the most economical conductor size — and what it costs to depart from it in either direction.
- How capacitance grading and intersheath grading flatten the stress profile, and the factor \((1+\alpha)/2\) by which they raise the working voltage.
- How the two bridge measurements on a three-core belted cable give \(C_s\), \(C_c\) and the working value \(C_N=C_s+3C_c\).
- Why cables are thermally limited where overhead lines are not, and why an a.c. cable has a critical length beyond which it carries only its own charging current.
- How Chapter 5's system comparison changes when the limiting voltage is measured between conductors instead of to earth — and why three-phase still wins.
Why the Ground Is a Different Place
Chapter 2 divided transmission and distribution into two families by the manner of running the conductors: overhead or underground. Chapters 5 to 8 then developed the overhead family in full — the conductor hanging in air, its insulation supplied free by the atmosphere and by distance, its heat carried away by wind, its parameters fixed by the geometry of the tower. Every one of those advantages disappears when the circuit goes into a trench.
An overhead line at 132 kV keeps its conductors some three metres from the nearest earthed steelwork. A 132 kV cable has to do the same job across perhaps eighteen millimetres of polymer. The dielectric strength required is therefore not of the order of a few kilovolts per centimetre, as in air, but of the order of a hundred — and the material that provides it must also be flexible enough to bend round a drum, stable for forty years at 90 °C, and free of the microscopic voids in which partial discharge starts. That single requirement dictates almost everything else about a cable.
Three consequences follow immediately, and they are the reasons this chapter exists as a separate piece of theory rather than as a footnote to Chapter 7.
The field is intense and radial. In an overhead line the electric field at the conductor surface is a few kilovolts per centimetre and the surrounding medium is self-healing: if it breaks down, corona forms and the air recovers (Chapter 15). In a cable the field at the conductor surface is a hundred times greater and the medium does not recover. Insulation design is therefore a matter of computing the stress at every radius and keeping it below the permissible value everywhere — the subject of Sections 9-5 and 9-6.
The capacitance is enormous. Chapter 7 obtained \(C_n = 2\pi\varepsilon/\ln(D_{eq}/r)\) for an overhead line, where \(D_{eq}\) is metres and \(r\) is centimetres, so the logarithm is around 7 and the capacitance around \(0.009\ \mu\text{F/km}\). In a cable \(R/r\) is between 2 and 4, the logarithm is below 1.4, and the permittivity is three or four times that of air. The capacitance comes out between \(0.15\) and \(0.4\ \mu\text{F/km}\) — twenty to forty times larger. Everything charging current touches is magnified by that factor.
The heat cannot get out. An overhead conductor loses heat by convection and radiation into moving air, and can be run at 75 °C without difficulty. A buried cable must pass its losses outward through the insulation, the sheath, the serving and then several metres of soil, each of which is a thermal resistance in series. The permissible current is set by that thermal path, not by the metal — Section 9-8.
Against these penalties stand the reasons cables are used at all: no right-of-way corridor, no visual intrusion, immunity to wind, ice, lightning and falling trees, no risk of contact, and a fault rate per kilometre roughly a tenth that of an overhead line. What is paid for them is capital cost — between five and twenty times the cost of an equivalent overhead circuit at transmission voltages — and repair time, since a cable fault must first be located along a buried route and then excavated, a matter of days rather than the hours an overhead patrol needs.
| Property | Overhead line | Underground cable |
|---|---|---|
| Insulating medium | Air, plus insulator strings at the supports | Solid or impregnated dielectric, continuous |
| Capital cost (132 kV, per km) | Reference | Five to twenty times higher |
| Capacitance | \(\approx 0.009\ \mu\text{F/km}\) | \(0.15\) to \(0.4\ \mu\text{F/km}\) |
| Inductive reactance | \(\approx 0.4\ \Omega/\text{km}\) | \(\approx 0.1\ \Omega/\text{km}\) (conductors close together) |
| Limiting factor on current | Conductor temperature in moving air | Heat conduction through insulation and soil |
| Fault frequency | Higher, mostly transient and self-clearing | Lower, but always permanent |
| Repair after a fault | Hours; the fault is visible | Days; the fault must be located and dug out |
| Practical a.c. length | Unlimited (subject to Chapters 12 and 28) | Limited by charging current — see below |
The last row deserves an argument rather than an assertion. A cable's charging current per unit length is \(I_c=\omega C V_{ph}\), and it is drawn whether the cable is loaded or not. Because \(C\) is proportional to length, so is \(I_c\); at some length the charging current alone equals the conductor's thermal rating and the cable can deliver no useful power at all. That length — the critical length — is around 40 to 60 km at 400 kV for a land cable, and it is the reason every long submarine link in the world is built as HVDC (Chapter 38), where \(\omega=0\) and the charging current vanishes.
How a Cable Is Built
A power cable is a set of concentric layers, each of which exists to solve one problem. Working outward from the centre:
The conductor. Stranded copper or aluminium, exactly as in Chapter 5 and for the same reason — flexibility. In multi-core cables the strands are compacted into a sector shape rather than a circle, because three 120° sectors fill a circular cable far better than three circles do, and the space factor decides the overall diameter and hence the cost of everything outside it.
The conductor screen. A thin semiconducting tape or extruded layer over the stranding. A stranded surface is not a smooth cylinder; at every strand crevice the field would be locally intensified and a small air pocket would sit exactly where the stress is highest. The semiconducting screen sits at conductor potential, presents a smooth equipotential cylinder to the dielectric, and eliminates the pockets. Without it, the clean radial-field theory of the next three sections would not describe a real cable.
The insulation. Impregnated paper for the traditional cable, PVC up to about 3.3 kV, and cross-linked polyethylene (XLPE) for practically everything built today up to 500 kV. Its thickness is set by the permissible stress, not by the mechanical requirement.
The insulation screen. A second semiconducting layer, this time in contact with an earthed metallic tape. It forces the outer boundary of the dielectric to be an equipotential surface, which is what makes the field purely radial.
The metallic sheath. Lead or aluminium, extruded seamlessly over the screen. It excludes moisture — impregnated paper absorbs water vapour greedily and loses its strength when it does — holds the outside of the dielectric at earth potential, and provides a return path for fault current.
The bedding, armouring and serving. Bedding is a cushion of compounded fibrous material protecting the soft sheath from the armour. The armour is galvanised steel wire or tape and takes the mechanical abuse of laying and of anything that is later dug into the route. The serving is a final fibrous or PVC jacket protecting the armour from corrosion.
Cables are classified by the voltage they are built for, and the classification matters because the construction changes at each step: low-tension up to 1.1 kV, high-tension to 11 kV, super-tension to 33 kV, extra-high-tension to 66 kV, and above that the pressure cables — oil-filled or gas-pressure — in which the dielectric is held under pressure so that no void can form as the cable heats and cools.
The other classification is by how the three phases are arranged. In a belted cable the three cores are laid up together, each with its own insulation, and a common belt of insulation is applied over all three before the sheath. It is cheap, and it is used to 11 kV. Above that it fails, for a reason worth understanding: with only one earthed surface — the sheath, some distance away — the field between the cores is not radial. It runs partly along the layers of the impregnated paper, and paper is several times weaker along its laminations than across them. The tangential component of stress, plus the voids in the fillers between the cores, sets the ceiling.
The remedy is to give every core its own earthed screen. In the H-type (Hochstädter) cable each core is wrapped in metallised perforated paper and all three screens touch, so each core sees its own concentric earthed cylinder and its field becomes radial again; the perforations let the impregnating oil move. In the S.L. (separate lead) cable each core gets its own lead sheath. Either way the three-core cable becomes three single-core cables sharing an armour, and the theory of Sections 9-3 to 9-6 applies to each core exactly.
Everything that follows in this chapter — stress, grading, insulation resistance, capacitance from a single \(\ln(R/r)\) — assumes a radial field. That assumption is exact for a single-core or screened cable and only approximate for a belted one, which is why the belted cable's capacitance has to be measured rather than computed (Section 9-7).
Insulation Resistance
No dielectric is perfect. A small conduction current leaks radially outward from the conductor, through the insulation, to the earthed sheath, and the resistance it meets is one of the standard acceptance tests on a new cable and the standard diagnostic on an old one. Computing it takes one integral.
Take a cable of length \(l\), conductor radius \(r\) and inner sheath radius \(R\), the insulation having resistivity \(\rho\). Consider a cylindrical shell of radius \(x\) and thickness \(dx\). The leakage current crosses it through an area \(2\pi x l\) over a path length \(dx\), so its resistance is
The shells are traversed one after another by the same leakage current, so they are in series and their resistances add:
Note the length. The conductor resistance of Chapter 5 was \(\rho l/a\), proportional to length; the insulation resistance is inversely proportional to it. The reason is not subtle: a longer cable offers more leakage paths in parallel, exactly as a wider conductor offers more conduction paths. A 1 km cable with 500 MΩ of insulation resistance becomes 250 MΩ when a second identical kilometre is jointed on.
Two practical points follow. Insulation resistivity is a strong function of temperature — it can fall by an order of magnitude between 20 °C and 60 °C — so a measured value means nothing unless the temperature is quoted with it. And because \(R_{\text{ins}}\) depends on \(\ln(R/r)\) while the capacitance of the next section depends on \(1/\ln(R/r)\), their product
is independent of the geometry and of the length altogether. It is the dielectric's own time constant, and it is why a disconnected high-voltage cable stays charged for a long time after the supply is removed, and must always be discharged to earth before anyone touches it.
Capacitance, Charging Current and Dielectric Loss
Chapter 7 began with Gauss's law applied to a long charged cylinder and obtained the field at radius \(x\) from a charge \(q\) per unit length. That result transfers to the cable unchanged, with the permittivity of free space replaced by that of the dielectric:
Here the sheath carries the charge \(-q\) and confines the whole of the flux to the annulus, so unlike the overhead line of Chapter 7 there is no need to worry about the return path or about images in the earth — the cable is its own return. Integrating the field along a radius gives the potential difference between conductor and sheath, and the capacitance follows by definition:
The numerical form uses \(2\pi\varepsilon_0 = 0.05563\ \mu\text{F/km}\) and \(1/18 = 0.05556\). For impregnated paper \(\varepsilon_r\approx3.5\), for XLPE \(\varepsilon_r\approx2.3\), and \(R/r\) is rarely outside \(1.8\) to \(4\).
Put a typical set of figures in. With \(\varepsilon_r=3.5\) and \(R/r=2\), \(C = 3.5/(18\times0.693) = 0.281\ \mu\text{F/km}\). The three-phase overhead line of Chapter 7, with \(D_{eq}=3.5\) m and \(r=1\) cm, gives \(\ln(D_{eq}/r)=5.86\) and \(C_n=0.0095\ \mu\text{F/km}\). The cable's capacitance is thirty times the line's, and there is nothing to be done about it: the whole point of a cable is to put the earthed surface close to the conductor.
Chapter 7 also derived what that capacitance does. The cable draws a leading current even with its far end open, and generates reactive power:
The second expression is the same one Chapter 7 obtained for the overhead line, and it is worth remembering in that form because \(V_{LL}\) is the number written on the drawing. Its consequence is severe. A 33 kV cable with \(C=0.28\ \mu\text{F/km}\) generates \(314\times0.28\times10^{-6}\times(33\times10^3)^2 = 95.8\) kVAr per kilometre — enough that a moderately long urban cable network is a substantial reactive source in its own right, one that raises the voltage at light load and must be accounted for in the reactive planning of Chapter 34.
Real dielectrics are not lossless capacitors. Polarisation lags the applied field, and the current the cable draws leads the voltage not by \(90^\circ\) but by \(90^\circ-\delta\), the small angle \(\delta\) being the dielectric loss angle. Resolving that current into its quadrature and in-phase parts, the in-phase part is \(I_c\tan\delta\), and the power it carries is dissipated inside the insulation:
Dielectric Stress and the Most Economical Size
The field strength inside the insulation is called the dielectric stress, written \(g\) and quoted in kV/cm. It is the quantity the insulation must survive, and the whole design of a cable is the business of keeping it below the permissible value at every radius. Eliminating \(q\) between the two results of the last section gives it directly in terms of the applied voltage:
The permittivity has cancelled. Stress in a homogeneous cable depends only on the applied voltage and the geometry, never on the material — a fact that surprises students and that Section 9-6 exploits, because the only way to change the profile is to make the dielectric non-homogeneous.
Since \(g\) varies as \(1/x\), it is greatest where \(x\) is smallest, that is, at the conductor surface, and least at the sheath:
Insulation must be chosen for \(g_{max}\), which occurs at the conductor surface. Everything outside that radius is stressed less than it can stand, so the further the sheath is from the conductor, the more of the dielectric is doing nothing. \(V\) here is the voltage between conductor and sheath; for a three-phase system that is the phase voltage, and since breakdown responds to the instantaneous field, it is the peak value that must be used with a peak-valued stress limit.
This immediately poses a design question. The conductor radius \(r\) is fixed by the current the cable must carry — but only from below; nothing prevents making it larger. Suppose the working voltage \(V\) and the permissible stress \(g_{max}\) are given, and ask what conductor radius leads to the smallest cable. Writing the requirement \(g_{max}=V/(r\ln(R/r))\) as an expression for the overall radius,
Two competing effects are visible in that formula. Make \(r\) very small and the exponent blows up, so the insulation has to be enormously thick. Make \(r\) very large and the exponent tends to unity, but the leading factor \(r\) itself grows without limit. Somewhere between the two the overall size is least, and differentiating locates it:
The same condition arrives from the other direction: with \(R\) fixed, \(g_{max}=V/(r\ln(R/r))\) is least when \(r\ln(R/r)\) is greatest, and \(\frac{d}{dr}\big[r\ln(R/r)\big]=\ln(R/r)-1=0\) gives \(R/r=e\) again. Minimising the size for a given stress and minimising the stress for a given size are the same problem.
The optimum is a flat one, which is fortunate, because \(r\) also has to satisfy the current rating and the standard conductor sizes are discrete. Example 2 shows that departing from \(R/r=e\) by a factor of \(1.5\) in either direction raises \(g_{max}\) by only about fifteen per cent. What the rule really provides is a check: a cable design in which \(R/r\) is far from \(e\) is either wasting dielectric or courting breakdown, and the designer should know which.
When the calculated economical radius exceeds what the current demands, the conductor is not made solid at that size — it is built hollow, or over a helical spacer, or with a central oil duct. The metal that would fill the middle carries almost no current anyway once skin effect is taken into account (Chapter 5), so nothing is lost by leaving it out.
Grading: Capacitance and Intersheath
Section 9-5 leaves an uncomfortable result. The stress falls as \(1/x\), so the material at the conductor surface is worked to its limit while the material at the sheath is barely worked at all — and in a cable with \(R/r=e\) the outer layer is stressed at only \(37\%\) of the inner. Since the insulation must be paid for, transported and cooled through, that is waste, and the thicker the dielectric the worse it gets. Grading is the collective name for the tricks that flatten the profile.
The stress formula \(g(x)=q/(2\pi\varepsilon x)\) shows the two available handles. The stress at a given radius can be reduced either by increasing the permittivity \(\varepsilon\) there, or by reducing the charge \(q\) that the layer has to contain — that is, by breaking the single capacitor into several capacitors in series and holding their junctions at fixed potentials. The first is capacitance grading; the second is intersheath grading.
Capacitance grading. Replace the homogeneous dielectric by concentric layers of different materials, the innermost having the highest permittivity. With three layers of relative permittivity \(\varepsilon_{r1},\varepsilon_{r2},\varepsilon_{r3}\) filling the annuli \(r\to r_1\), \(r_1\to r_2\) and \(r_2\to R\), the charge \(q\) is common to all of them because they are in series, and the stress in layer \(i\) at radius \(x\) is \(q/(2\pi\varepsilon_0\varepsilon_{ri}x)\). The stress within each layer is still worst at that layer's inner boundary, so there are three local maxima:
Setting all three equal — which is the best that can be done, since the largest of them is what limits the cable — gives the grading condition, and the working voltage follows by integrating the stress across the three layers in turn:
Since \(r < r_1 < r_2\), the condition forces \(\varepsilon_{r1} > \varepsilon_{r2} > \varepsilon_{r3}\): the strongest, densest dielectric goes next to the conductor where the field is fiercest. Compare with the ungraded voltage \(V_0 = g_{max}\,r\ln(R/r)\) for the same \(r\), \(R\) and \(g_{max}\); the ratio is the gain grading has bought.
Two practical limits keep capacitance grading modest. There are not many dielectrics whose permittivities span a useful range while all having adequate strength, so the number of steps is small. And permittivity varies 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 must be checked at both.
Intersheath grading. The second method keeps one dielectric throughout and inserts thin metallic cylinders — intersheaths — at radii \(r_1\) and \(r_2\), held at fixed potentials \(V_1\) and \(V_2\) by tappings on the supply transformer. Each annulus is now a separate cable in miniature, with its own conductor (the intersheath inside it) and its own earthed sheath (the intersheath outside it), carrying only its own share of the total voltage. The stress in each annulus is worst at its inner boundary, exactly as before:
Two conditions are wanted: the three maxima equal, and the design as compact as possible. Take the radii in geometric progression, \(r_1/r = r_2/r_1 = R/r_2 = \alpha\), so that every logarithm equals \(\ln\alpha\). The three expressions then reduce to \((V-V_1)/r=(V_1-V_2)/r_1=V_2/r_2\), and adding the three numerators recovers the total voltage:
With \(r=1\) cm and \(R=4\) cm, one intersheath at \(r_1=2\) cm gives \(\alpha=2\) and a factor of \(1.5\): the same cable, the same maximum stress, half as much again of working voltage. Two intersheaths in a cable with \(R/r=8\) give \(\alpha=2\) and \((1+2+4)/3=2.33\).
Intersheath grading is elegant on paper and awkward in service. The intersheaths must be held at their design potentials at every instant, which means extra transformer tappings and extra terminations; damage to one intersheath throws the whole stress distribution onto the remaining layers; and because the intersheaths are metallic cylinders separated by dielectric, they draw charging currents among themselves that circulate along the cable and add loss. Modern practice therefore prefers to avoid grading altogether by using a single dielectric strong enough to be worked hard — which is another way of saying that XLPE replaced the problem rather than solving it.
Capacitance of Three-Core Belted Cables
Section 9-2 explained why the field in a belted cable is not radial. The consequence for capacitance is that no single \(\ln(R/r)\) will describe it. What can be said is that the electrostatic behaviour of three conductors inside an earthed sheath is completely captured by six capacitances: three between pairs of cores, all equal by symmetry and written \(C_c\), and three from each core to the sheath, again equal and written \(C_s\).
The three \(C_c\) form a delta between the cores; the three \(C_s\) form a star from the cores to the sheath, whose star point is the sheath itself, held at earth potential. To combine them, convert the delta of \(C_c\) into its equivalent star. For capacitances the delta–star transformation runs the opposite way to the one used for impedances in Chapter 3: a delta of \(C_c\) is equivalent to a star of \(3C_c\), because a star capacitance of \(3C_c\) presents \(3C_c/2\) between two terminals, which is exactly what a delta of \(C_c\) presents (\(C_c\) direct, in parallel with two \(C_c\) in series).
Both stars now have their common point at the neutral potential — the sheath is at earth, and for a balanced three-phase supply the electrical neutral of the cores is also at earth — so they are simply in parallel:
\(C_N\) is the capacitance actually seen by each phase, and the only one that appears in a circuit calculation: the charging current per phase is \(I_c=\omega C_N V_{ph}\) and the charging reactive power is \(Q_c=\omega C_N V_{LL}^{2}\), exactly the expressions of Chapter 7 and of Section 9-4.
Since \(C_s\) and \(C_c\) cannot be computed from the geometry with any confidence, they are measured. Two bridge tests suffice, and each is designed so that one of the two unknowns disappears.
Test 1: all three cores bunched, measured against the sheath. Joining the cores together short-circuits every \(C_c\) — both plates of each are at the same potential, so none of them can hold charge. What remains is the three \(C_s\) in parallel:
Test 2: two cores joined to the sheath, measured against the third. Now the test terminal is core 1, and everything else — cores 2 and 3 and the sheath — is at the other terminal. Three capacitances bridge the gap: \(C_s\) from core 1 to the sheath, and \(C_c\) from core 1 to each of the other two cores. They are all in parallel:
Heating and the Current Rating
An overhead conductor is rated by asking how hot it becomes in still air at 40 °C. A cable is rated by asking how hot it becomes at the end of a chain of thermal resistances: insulation, bedding, serving, and then the ground itself out to the surface. The chain is long, and it is the reason a cable of a given cross-section carries perhaps a third to a half of what the same metal carries in the open air.
Three losses feed heat into the cable. The conductor loss \(I^2R\), with \(R\) the a.c. resistance — larger than the d.c. value because of the skin effect of Chapter 5 and, in a multi-core cable, because of the proximity effect of the neighbouring cores. The dielectric loss \(W_d=\omega C V_{ph}^2\tan\delta\) of Section 9-4, which is generated in the insulation itself, at the very start of the thermal path. And the sheath loss, produced by the currents the alternating flux induces in a metallic sheath — eddy currents in every case, and, if the sheaths of a three-phase single-core installation are bonded at both ends, large circulating currents as well. The remedy for the last is cross-bonding: the sheaths are transposed in sections so that the three induced e.m.f.s cancel around the loop, exactly as the conductors themselves are transposed in Chapter 6 for the analogous reason.
Heat flow obeys the same mathematics as current flow, so the thermal resistance of the insulation is obtained by the integral of Section 9-3 with the electrical resistivity replaced by the thermal resistivity \(\rho_t\) in kelvin-metres per watt:
The buried cable's thermal resistance to the surface adds to this, and it is by far the larger term: dry soil has \(\rho_t\) between \(1\) and \(3\) K·m/W, against about \(5\) for the paper or polymer, but the soil path is metres long where the insulation path is millimetres. Summing the chain and requiring that the conductor temperature stay below its permitted value \(\theta_{max}\) — 70 °C for PVC, 90 °C for XLPE — gives the rating:
The dielectric loss subtracts from the temperature budget before the current gets any of it. At 11 kV that subtraction is negligible; at 400 kV it can take a fifth of the available rise, which is the practical reason EHV cables are built with low-loss dielectrics and, often, with active cooling.
Everything about a cable installation therefore turns on the thermal environment rather than on the cable. Laying several circuits in one trench makes each one heat the others, and every circuit is derated by a grouping factor. Laying in ducts inserts a layer of stagnant air, worth a further derating. Dry soil around a heavily loaded cable can lose its moisture, and its thermal resistivity can then double — a slow failure mechanism that has destroyed cables which were correctly rated for the soil as originally surveyed. Against this, a cable that is loaded only during the day may be given a higher cyclic rating than its continuous one, because the surrounding earth stores heat overnight; and the largest EHV circuits are cooled by pumping oil or water along the route, which converts the thermal problem into a hydraulic one.
Comparing Systems on the Underground Basis
Section 5-2 compared ten ways of arranging conductors — d.c., single-phase, two-phase and three-phase, with and without earthed neutrals — under four conditions held identical: the same power \(P\), the same length \(l\), the same total loss \(W\), and the same maximum voltage. The first three conditions are the same here. The fourth is not, and the whole of this section is the consequence.
An overhead conductor's nearest earthed neighbour is the tower, several metres away through air, and its nearest live neighbour is another phase, several metres away through the same air. The insulation that is closest to failing is the conductor-to-earth path, so the honest comparison holds the maximum voltage to earth constant. A cable core's nearest earthed neighbour is its own sheath, millimetres away, and in a belted cable its nearest live neighbour is an adjacent core, also millimetres away. The dielectric between cores is what is worked hardest, so the honest comparison holds the maximum voltage between conductors constant.
The change looks small and is not. Earthing the mid-point of a two-wire d.c. system was worth a factor of four in Chapter 5, because it doubled the working voltage at no cost in insulation to earth. On the underground basis it is worth nothing at all: the voltage between the two conductors is \(V_m\) whether the mid-point is earthed or not, and \(V_m\) is what is fixed.
The recipe is Chapter 5's, unchanged: find the current from the power and the voltage the arrangement actually offers, write the loss over the \(n\) conductors, solve for the area the loss budget permits, and multiply by the total length of conductor.
Two-wire d.c., the reference. The voltage between the conductors is \(V_m\), so \(I_1=P/V_m\) and
Earthing the mid-point changes nothing: the conductors sit at \(+V_m/2\) and \(-V_m/2\), their difference is still \(V_m\), and the volume is still \(K\). The three-wire d.c. system takes the same two outers and adds a neutral of half section which carries nothing when the load is balanced, so it pays \(2.5a_1l = 1.25K\) for the privilege of offering two voltage levels.
Single-phase, two-wire. The peak voltage between the conductors is \(V_m\), so the r.m.s. voltage available is \(V_m/\sqrt2\), and the current is larger by \(1/\cos\phi\) as well:
Mid-point earthing again buys nothing, so the mid-point-earthed single-phase system also comes to \(2K/\cos^2\phi\); adding an idle half-neutral makes it \(2.5K/\cos^2\phi\).
Two-phase, four-wire. Two independent single-phase circuits in quadrature, each carrying \(P/2\). The largest voltage between any two conductors is that across a pair, since the two phases are \(90^\circ\) apart and the conductors of different phases differ by only \(V_m/\sqrt2\) at the peak. Each pair therefore works at the full \(V_m\) while carrying half the power, so each current is half of \(I_4\) and each area is half of \(a_4\) — but there are four conductors instead of two, and the volume comes to \(2K/\cos^2\phi\) again.
Two-phase, three-wire. Here the two outers are \(90^\circ\) apart with a common return, so the voltage between the outers is \(\sqrt2\) times the outer-to-neutral voltage. Fixing the largest conductor-to-conductor voltage at \(V_m\) leaves each outer at a peak of \(V_m/\sqrt2\) to the neutral, that is \(V_m/2\) r.m.s. The outer current is \(I_8=(P/2)/\big((V_m/2)\cos\phi\big) = P/(V_m\cos\phi)\), and the neutral carries the phasor sum of two equal currents at \(90^\circ\), namely \(\sqrt2 I_8\). Holding the current density constant makes the neutral \(\sqrt2\) times the section of an outer and its resistance \(R_8/\sqrt2\):
since \((2+\sqrt2)^2 = 6+4\sqrt2 = 11.657\) and \(11.657/4 = 2.914\).
Three-phase, three-wire. The maximum voltage between conductors is now the line voltage, so \(V_m\) is the peak line value and the peak phase value is \(V_m/\sqrt3\), giving an r.m.s. phase voltage of \(V_m/\sqrt6\). Each conductor carries a third of the power:
Adding a neutral of half section, idle for a balanced load because \(1+a+a^2=0\) (Chapter 3), gives \(3.5a_9l = 1.75K/\cos^2\phi\). The complete comparison, with Chapter 5's overhead results alongside for contrast:
| System | Conductors | Underground basis | At \(\cos\phi=0.8\) | Overhead basis (Ch. 5) |
|---|---|---|---|---|
| D.C. two-wire, one earthed | 2 | \(K\) | 1.000 \(K\) | \(K\) |
| D.C. two-wire, mid-point earthed | 2 | \(K\) | 1.000 \(K\) | \(K/4\) |
| D.C. three-wire | 2.5 | \(1.25K\) | 1.250 \(K\) | \(0.3125K\) |
| 1-phase two-wire | 2 | \(2K/\cos^2\phi\) | 3.125 \(K\) | \(2K/\cos^2\phi\) |
| 1-phase two-wire, mid-point earthed | 2 | \(2K/\cos^2\phi\) | 3.125 \(K\) | \(0.5K/\cos^2\phi\) |
| 1-phase three-wire | 2.5 | \(2.5K/\cos^2\phi\) | 3.906 \(K\) | \(0.625K/\cos^2\phi\) |
| 2-phase four-wire | 4 | \(2K/\cos^2\phi\) | 3.125 \(K\) | \(0.5K/\cos^2\phi\) |
| 2-phase three-wire | 3.41 | \(2.914K/\cos^2\phi\) | 4.553 \(K\) | \(1.457K/\cos^2\phi\) |
| 3-phase three-wire | 3 | \(\mathbf{1.5K/\cos^2\phi}\) | 2.344 \(K\) | \(0.5K/\cos^2\phi\) |
| 3-phase four-wire | 3.5 | \(1.75K/\cos^2\phi\) | 2.734 \(K\) | \(0.583K/\cos^2\phi\) |
The three-way tie of the overhead comparison was created by mid-point earthing, which raised the working voltage of the single-phase and two-phase systems at no insulation cost. Underground that trick is unavailable, both rivals fall back to \(2K/\cos^2\phi\), and three-phase is left alone at the top.
Two further readings of the table are worth having. The ratio of the underground figure to the overhead figure is \(1\) for the plain single-phase two-wire system, \(3\) for both three-phase systems, \(2\) for the two-phase three-wire, and \(4\) for every mid-point-earthed arrangement — and in each case it is exactly the square of the factor by which the available voltage has been cut. And the d.c. systems, which won the overhead comparison outright, still lead underground, for the same reason as before: no \(\sqrt2\), no power factor. That they were not built is a matter of the transformer, not of the conductor, and Chapter 38 revisits the question now that power electronics has supplied the missing component.
Worked Examples
Problem. A single-core cable 1 km long has a conductor radius of 1.0 cm and an inner sheath radius of 2.0 cm. The impregnated-paper dielectric has resistivity \(4.5\times10^{14}\ \Omega\cdot\)cm, \(\varepsilon_r=3.5\) and \(\tan\delta=0.005\). The cable forms one phase of a 33 kV, 50 Hz three-phase system. Find the insulation resistance, the capacitance, the leakage current, the charging current and the dielectric loss.
Solution. With \(R/r=2\), \(\ln(R/r)=0.6931\) throughout. Work the insulation resistance in centimetres, so \(l=10^5\) cm:
That is 496 MΩ. The capacitance follows from the numerical form of Section 9-4:
The leakage current, by contrast, is \(19\,053/(4.96\times10^{8}) = 3.84\times10^{-5}\) A, or 38 µA — forty-four thousand times smaller. A cable's insulation conducts almost nothing and displaces a great deal, which is why the capacitance and not the leakage governs its behaviour. The dielectric loss is
For comparison, if this cable carried 250 A through a conductor of \(0.08\ \Omega\)/km the conductor loss would be \(250^2\times0.08 = 5000\) W/km — thirty times the dielectric loss. Repeat the calculation at 220 kV, where \(V_{ph}\) is 38 times larger, and \(W_d\) grows by a factor of \(38^2\) while \(I^2R\) does not grow at all.
Problem. A single-core cable is to work at 19.05 kV r.m.s. between conductor and sheath — one phase of a 33 kV system — in a dielectric whose permissible stress is 30 kV/cm peak. Find the most economical conductor radius, the inner sheath radius, the insulation thickness and the minimum stress. Then show what \(g_{max}\) becomes if the conductor radius is made 0.6 cm or 1.4 cm with the same overall radius.
Solution. Breakdown responds to the instantaneous field, so the peak voltage is the one to use: \(V = \sqrt2\times19.05 = 26.95\) kV. The economical design has \(R/r=e\), so \(\ln(R/r)=1\) and
So the conductor diameter is about 18 mm and the dielectric is 15.4 mm thick; the outer layer is stressed at \(11.04/30 = 37\%\) of the inner, which is \(1/e\), as it must be.
Now hold \(R=2.442\) cm and move \(r\):
Both are over the limit, and both by a modest amount — the optimum is genuinely flat, and a design that misses it by a third in either direction pays only six to fifteen per cent in stress. What the rule guarantees is that no other radius does better, so a cable with \(R/r\) far from \(2.718\) is worth a second look.
Problem. A single-core cable has a conductor radius of 0.8 cm and an inner sheath radius of 2.5 cm. Two dielectrics are available, of relative permittivity 4 and 2.5, each safe to 40 kV/cm peak. Find the radius at which they should meet and the peak working voltage, and compare with the same cable filled with one dielectric.
Solution. The grading condition \(\varepsilon_{r1}r=\varepsilon_{r2}r_1\) places the boundary at
The stronger dielectric — the one with the higher permittivity — goes inside, where the field is worst. Integrating the stress across the two annuli:
With a single dielectric of either kind, the same \(r\) and \(R\) give
Grading has raised the safe voltage from 36.5 kV to 49.3 kV peak — from 25.8 kV to 34.9 kV r.m.s. — a gain of 35% for no extra material at all, only a different arrangement of it.
Problem. A cable has \(r=1\) cm and \(R=4\) cm in a dielectric permitting 60 kV/cm peak. Find the peak working voltage without grading; then insert one intersheath at the best radius, find the new working voltage and the potential at which the intersheath must be held.
Solution. Ungraded,
With one intersheath the two annuli should have equal ratios, so \(\alpha=\sqrt{R/r}=\sqrt4=2\) and the intersheath goes at \(r_1=2\) cm. Equal maximum stress in the two annuli requires
Check both annuli: inner, \((124.8-83.2)/(1\times0.6931) = 60.0\) kV/cm; outer, \(83.2/(2\times0.6931) = 60.0\) kV/cm. Equal, as designed. The improvement factor is \(124.8/83.2 = 1.50 = (1+\alpha)/2\), and the figure in Section 9-6 is precisely this calculation drawn to scale.
Note what the intersheath potential turns out to be: 83.2 kV, the same number as the ungraded cable's entire working voltage. That is a coincidence of these figures, not a general result — but it does make the point that the intersheath sits at a high potential and must be terminated, insulated and maintained as a live part.
Problem. On a 5 km, 11 kV, 50 Hz three-core belted cable, the capacitance measured between the three cores bunched together and the sheath is 0.90 µF; the capacitance measured between one core and the other two joined to the sheath is 0.60 µF. Find \(C_s\), \(C_c\) and \(C_N\), and hence the charging current per phase and the total charging kVAr.
Solution. The first test short-circuits the three \(C_c\) and leaves the three \(C_s\) in parallel; the second puts \(C_s\) in parallel with two \(C_c\):
The short-cut formula agrees: \(C_N = 1.5C_b - C_a/6 = 0.90 - 0.15 = 0.75\ \mu\)F. Then
That is 0.15 µF/km and 5.7 kVAr/km. An 11 kV overhead line of the same length has about 0.0094 µF/km and generates 0.36 kVAr/km — sixteen times less. On a large urban network of such cables the aggregate is not a curiosity but a planning quantity: a hundred kilometres of this cable is a 570 kVAr capacitor permanently connected to the 11 kV busbar.
Problem. A three-phase, three-wire underground system is to deliver 10 MW at 0.8 power factor over 10 km with a total line loss of 5% of the power delivered, the maximum voltage between conductors being 33 kV peak. Copper has \(\rho = 1.72\times10^{-8}\ \Omega\)m. Find the conductor cross-section and the total volume of copper. Compare with a two-wire d.c. system on the same basis, and with the same three-phase line judged on Chapter 5's overhead basis.
Solution. \(W = 0.05\times10\ \text{MW} = 500\) kW. Using the three-phase result of Section 9-9 with \(V_m = 33\,000\) V peak:
That is 98.7 mm². Confirm it the long way: the r.m.s. phase voltage is \(33\,000/(\sqrt3\sqrt2) = 13\,472\) V, so \(I = 10^{7}/(3\times13\,472\times0.8) = 309.3\) A; the permitted resistance per conductor is \(R = W/(3I^2) = 5\times10^{5}/(3\times309.3^{2}) = 1.742\ \Omega\); and \(a = \rho l/R = 1.72\times10^{-4}/1.742 = 9.87\times10^{-5}\) m². The volume is
The yardstick for the same duty is
Three results worth carrying away. The three-phase system needs 2.34 times the copper of a two-wire d.c. system doing the same job. It needs exactly three times as much copper when the 33 kV is measured between conductors as when it is measured to earth, because the available phase voltage falls by \(\sqrt3\) and volume goes as the inverse square of voltage. And in absolute terms, 26 tonnes of copper before a single metre of insulation, sheath or armour has been bought is why Section 9-1's cost ratio of five to twenty is not an exaggeration.
Chapter Summary
\(R_{\text{ins}} = \dfrac{\rho}{2\pi l}\ln\dfrac{R}{r}\) — inversely proportional to length, unlike conductor resistance.
\(C = \dfrac{2\pi\varepsilon_0\varepsilon_r}{\ln(R/r)} = \dfrac{\varepsilon_r}{18\ln(R/r)}\ \mu\text{F/km}\) — thirty times an overhead line's.
\(g(x) = \dfrac{V}{x\ln(R/r)}\), worst at the conductor, and independent of the permittivity.
\(R/r = e\), \(r = V/g_{max}\) — the flattest optimum in the subject, and a design check.
\(\varepsilon_{r1}r=\varepsilon_{r2}r_1\) for capacitance grading; \((1+\alpha)/2\) gained per intersheath.
\(C_N=C_s+3C_c=1.5C_b-C_a/6\) from two bridge readings; belted construction stops at 11 kV.
Thermal, not electrical: \(\theta_{max}-\theta_{amb} = (I^2R+W_d)S_{\text{total}}\), with the soil dominating \(S\).
Fixing voltage between conductors, three-phase three-wire wins outright at \(1.5K/\cos^2\phi\).
Practice Problems
Take \(2\pi\varepsilon_0 = 0.0556\ \mu\text{F/km}\) and \(f=50\) Hz unless stated otherwise. Where a stress limit is quoted as a peak value, remember to convert the r.m.s. working voltage before using it.
- A single-core cable 2 km long has a conductor diameter of 1.5 cm and an insulation thickness of 1.2 cm. The dielectric has \(\rho = 5\times10^{14}\ \Omega\cdot\)cm and \(\varepsilon_r = 3.2\). Find the insulation resistance and the capacitance, and verify that their product equals \(\rho\varepsilon\).
- A 33 kV, three-phase, 50 Hz cable system 12 km long has a capacitance of 0.25 µF/km per phase. Find the charging current per phase and the total charging MVAr, and state what fraction of a 400 A thermal rating the charging current consumes.
- A single-core cable working at 38 kV peak between conductor and sheath has a conductor radius of 1.2 cm and a sheath radius of 3.0 cm. Find \(g_{max}\) and \(g_{min}\), and determine the conductor radius that would minimise \(g_{max}\) for the same overall size.
- A cable is to be designed for 66 kV r.m.s. between conductor and sheath in a dielectric permitting 50 kV/cm peak. Determine the most economical conductor diameter, the overall diameter of the insulation, and the minimum stress in the finished design.
- A cable of conductor radius 0.9 cm and sheath radius 2.7 cm is graded with two dielectrics of relative permittivity 5 and 3, both safe at 38 kV/cm peak. Find the boundary radius, the peak working voltage, and the percentage gain over the ungraded cable.
- A cable with \(r=1.5\) cm and \(R=6\) cm uses a dielectric permitting 55 kV/cm peak. Find the working voltage with no intersheath, with one intersheath, and with two, in each case placing the intersheaths so that the maximum stresses are equal, and quote the intersheath potentials.
- Measurements on a three-core belted cable give 1.2 µF between the three cores bunched and the sheath, and 0.8 µF between one core and the remaining two joined to the sheath. Find \(C_s\), \(C_c\) and \(C_N\), and the charging kVAr at 22 kV, 50 Hz.
- Repeat Section 9-9's derivation for the single-phase three-wire and the three-phase four-wire systems, and confirm the entries \(2.5K/\cos^2\phi\) and \(1.75K/\cos^2\phi\). Then state, for each of the ten systems, the ratio of its underground volume to its overhead volume, and explain each ratio in one sentence.