Structure of Generation, Transmission and Distribution
A power system is a ladder of voltages, and the rung heights are not arbitrary: the volume of conductor needed to carry a given power falls as the square of the voltage, so the network climbs as high as insulation and switchgear costs allow and comes back down only where people have to touch it.
- The full chain from prime mover to service main, and the standard voltage at every rung of it.
- Why the volume of conductor material obeys \(\mathrm{Vol}\propto 1/(V^2\cos^2\phi)\) — derived, not asserted — and what that single result decides about system layout.
- How transmission efficiency and percentage line drop improve as \(1/V\) at constant current density.
- Why the climb stops: insulation, switchgear and corona costs rise with \(V\), giving an economic transmission voltage where terminal cost is twice conductor cost.
- Why generation is at 11 kV, transmission at 132–765 kV, distribution at 11 kV and utilisation at 400/230 V, and why transmission is 3-wire while secondary distribution is 4-wire.
- The genuine merits of d.c. transmission, and the precise reasons a.c. still won the network.
- What interconnection buys — and what it costs in the stability and protection problems of Parts 6 and 8.
From the Coalfield to the Lamp
Chapter 1 established the two facts that force a power system into existence: primary energy is where geology put it, and electrical energy cannot be stored on the way. This chapter turns those facts into a physical arrangement — what is built, at what voltage, and why the answer is the same in every country that has electrified.
In the first decades of the industry the answer was different. Demand was small, confined to lighting and a little heating, and a generating plant of a few hundred kilowatts sat inside the district it served. As demand grew that arrangement became untenable for reasons that are all economic. A large boiler and turbine are far more efficient than a small one; a station built at the pithead or the dam site avoids hauling fuel; and a single large machine serving many districts runs closer to full load, because the peaks of separate districts do not coincide. Every one of these arguments pushes the generating plant away from the consumer, and once it has moved, the energy has to be brought back.
What brings it back is a network of conductors, and it is conventional and useful to divide that network in two. Transmission carries bulk power over long distances between a small number of large nodes. Distribution spreads it from those nodes over a service area to a very large number of small consumers. The two have different economics, different topologies and different design rules, and they are separated in this book as they are separated in practice. The whole assembly — the conveyance of electric power from the generating station to the consumer's premises — is the electric supply system.
Generation and transmission are carried out with three-phase, three-wire a.c.; distribution below 11 kV with three-phase, four-wire a.c. The overhead system is used almost everywhere because it costs a fraction of the underground alternative; underground cable is confined to dense cities, where the pole line would be intolerable and the risk to life unacceptable. Chapter 9 takes up cables in detail.
Four characteristics are shared by every power system regardless of size. It is three-phase and operates at essentially constant voltage. Generation and transmission plant is three-phase throughout, and industrial load is invariably three-phase; single-phase residential and commercial loads are distributed as evenly as possible among the three phases so that the system remains balanced, which is what permits the per-phase analysis of Chapter 3. Electricity is generated by synchronous machines, driven by prime movers that convert fossil, nuclear or hydraulic energy into mechanical energy. And the transmission network is not one system but several subsystems at different voltages, joined by transformers.
The Voltage Ladder
Trace a kilowatt-hour from the turbine shaft to a lamp filament and it passes through five distinct stages, each at its own voltage and each separated from the next by a transformer.
Generating station. Three-phase alternators operating in parallel produce power at 11 kV, occasionally 6.6 kV on older or smaller machines and up to 21 kV on very large ones. A step-up transformer raises this to the transmission voltage, which is chosen according to the length of the line and the power to be sent — 66, 132, 220, 400 or 765 kV.
Primary transmission. The high-voltage line runs by three-phase, three-wire overhead construction to the outskirts of the load centre. This is the stage the whole of Parts 2 and 3 is about.
Secondary transmission. At a receiving station on the edge of the city the voltage is stepped down to 33 kV, and power is distributed at that level to substations placed at strategic points within the city. The line is still three-phase, three-wire.
Primary distribution. At each substation the voltage falls to 11 kV, again three-phase and three-wire, and these lines run along the main roads. Consumers with a demand above roughly 50 kW are supplied at 11 kV and provide their own substation; it is cheaper for the utility, and it puts the transformer losses on the consumer's side of the meter.
Secondary distribution. Distribution substations — pole-mounted or ground-mounted transformers, sited every few hundred metres — step 11 kV down to a three-phase, four-wire 400 V system. Between any two lines the voltage is 400 V; between any line and the neutral it is \(400/\sqrt3 = 230\) V. Single-phase lighting and domestic loads connect between a line and the neutral, and three-phase loads across the three lines directly.
A fourth conductor across 300 km would be a third more overhead line for no return. Across the last 200 m to a house it is unavoidable — and it is the reason a domestic supply is 230 V while the same transformer's line voltage is 400 V.
Why High Voltage: The Volume of Conductor Material
All the transformers in the ladder exist to serve one result, and it is worth deriving carefully because everything about the geography of a power system follows from it.
Suppose a power \(P\) is to be delivered over a three-phase, three-wire line of length \(l\) at line voltage \(V\) and power factor \(\cos\phi\). Let the conductor have resistivity \(\rho\) and cross-sectional area \(a\), and let the total permissible line loss be \(W\), which we hold fixed — that is the honest comparison, since a line that loses more is not delivering the same service. The line current follows from Chapter 1 and the three-phase relation of Chapter 3:
The factor of 3 for the three conductors has cancelled against the \(3\) hidden in \(I^2\). Solve for the area the loss budget permits, then multiply by the three conductors and by the length to obtain the quantity that is actually bought:
Everything else in the expression — \(P\), \(l\), \(\rho\), \(W\) — is fixed by the job or by the material. Only \(V\) and \(\cos\phi\) are at the engineer's disposal, and both enter squared. Doubling the transmission voltage divides the copper or aluminium bill by four.
Two consequences deserve to be drawn out. The first is quantitative: raising a line from 132 kV to 400 kV reduces the conductor requirement to \((132/400)^2 = 10.9\%\) of its former value, a saving so large that it pays for the transformers, the taller towers, the longer insulator strings and the bigger switchgear several times over. Example 1 works the numbers.
The second is that \(\cos\phi\) appears in exactly the same way as \(V\). A line operated at 0.7 power factor needs \((1/0.7)^2 = 2.04\) times the conductor of the same line at unity — which is the capital-cost half of the argument that Chapter 1 made on running cost alone. Reactive compensation and high transmission voltage are two forms of the same economy.
Efficiency, Voltage Drop, and the Limit to Voltage
The conductor-volume argument fixed the loss and asked for the material. Turn the question round: fix the material — more precisely, fix the current density \(J = I/a\), which is what a conductor's thermal rating really constrains — and ask what happens to efficiency and voltage drop as \(V\) rises.
The current \(I\) has cancelled, leaving the loss fraction proportional to \(1/V\). The same cancellation governs the voltage drop. The drop along one conductor is
This is the striking one. At a given current density the absolute drop in volts is the same whatever the line voltage — it depends only on the material, the length and how hard the copper is worked. Expressed as a percentage of the line voltage it therefore falls as \(1/V\). A 5660 V drop is catastrophic on an 11 kV line and negligible on a 400 kV one.
The first is a capital saving, the second a running saving, the third a quality-of-supply improvement that feeds directly into the regulation calculations of Chapter 13.
If that were the whole account, transmission would be done at a million volts. It is not, because a second family of costs rises with \(V\) rather than falling. Insulation must be thicker and insulator strings longer; clearances to ground, to steelwork and between phases grow, so the towers grow with them; transformers, circuit breakers, isolators, instrument transformers and surge arresters all become larger and more expensive; and corona sets in, demanding bundled conductors of larger diameter than the loss calculation would justify.
Set the two families against each other. Write the conductor cost, which falls as \(1/V^2\), and the terminal and insulation cost, which rises roughly in proportion to \(V\):
The optimum is where the saving in conductor material from one more kilovolt is exactly offset by the extra cost of insulating it — and, as the last line shows, that balance falls where the terminal equipment costs twice what the conductor does. The curve is flat near its minimum, which is fortunate: it means the standard voltages can be a short list rather than a continuum, and a line need only be built at the nearest standard rung.
For a first estimate before any costing is done, the empirical rule known as Still's formula is used:
Generation and the 11 kV Bus
At the top of the ladder stands the synchronous alternator, driven by a prime mover that has converted heat, falling water or wind into shaft power. Steam turbines dominate thermal and nuclear stations and turn at 3000 rpm on a 50 Hz system with a two-pole machine; hydro sets turn far more slowly and carry many poles, since the frequency is fixed by \(f = pN/120\) with \(p\) poles and \(N\) in rpm. The machine is the subject of Chapter 26, and its behaviour during a fault is the subject of Chapter 21; here only its terminal voltage concerns us.
If high voltage is so valuable, why generate at a mere 11 kV and pay for a step-up transformer? Because the alternator's insulation problem is far harder than a line's. Its conductors are not suspended in air with metres of clearance but packed into slots in a laminated iron core, separated from earthed steel by a few millimetres of solid insulation, and subject to vibration, thermal cycling and the mechanical forces of a short circuit. The winding voltage a machine can carry is limited by the slot, not by economics, and 11 kV to about 21 kV is what the slot allows. A transformer, by contrast, can be immersed in oil and built to any voltage required — so it is far cheaper to generate low and transform up than to build a machine that generates high.
| Station type | Prime mover | Typical unit size | Role in the system |
|---|---|---|---|
| Thermal (coal, gas) | Steam or gas turbine | 210–800 MW | Base and intermediate load; slow to start, cheap to run at full output |
| Nuclear | Steam turbine | 500–1200 MW | Base load only; economics demand continuous full output |
| Hydro | Water turbine | 50–500 MW | Peaking and frequency regulation; seconds to load |
| Gas turbine / diesel | Combustion turbine | 10–250 MW | Peaking and reserve; expensive fuel, very fast start |
| Wind and solar | None synchronous | 1–5 MW per unit | Energy when available; connected through converters — Chapter 39 |
The differences in this table are what make Part 7 a subject at all. If every machine cost the same to run and started equally quickly, there would be no dispatch problem. Because they do not, Chapter 31 must decide how to share load among the units that are running and Chapter 32 which units to run at all.
Transmission: Primary, Secondary, Overhead, Underground
Primary transmission is three-phase, three-wire, overhead, and at the highest voltage the system uses. Its conductors are bare aluminium stranded around a steel core (ACSR), suspended from steel lattice towers on strings of porcelain or glass discs, with one or two earth wires strung above them to intercept lightning. Chapter 5 describes the construction; Chapters 6 to 8 compute the inductance and capacitance that the geometry produces.
Secondary transmission at 33 or 66 kV performs the same function on a smaller scale, moving bulk power from the receiving station to substations inside the city. The distinction between primary and secondary is one of role rather than physics: both are three-wire three-phase lines and both are analysed by exactly the same methods in Part 3.
The choice between overhead line and underground cable is decided almost entirely on cost. An underground circuit costs several times its overhead equivalent at distribution voltage and an order of magnitude more at transmission voltage, because the insulation that air provides free must be manufactured, and because a cable fault takes days to locate and repair rather than hours. Against that, cable eliminates the danger of contact with live conductors, removes the visual intrusion of pole lines from crowded streets, and is immune to wind, ice and lightning.
| Overhead line | Underground cable | |
|---|---|---|
| Insulation | Air, free and self-healing after a flashover | Manufactured dielectric; a failure is permanent |
| Cost | Low | Several times higher; rises steeply with voltage |
| Capacitance | Small — spacing is metres | Large — spacing is millimetres; charging current limits length (Chapter 9) |
| Fault behaviour | Most faults transient; auto-reclosing restores supply | Every fault permanent; location and repair take days |
| Where used | All transmission, most distribution | Dense urban distribution, water crossings, station entries |
The capacitance row is the one that has technical rather than commercial consequences. A cable's conductors are separated by a millimetre or two of dielectric instead of several metres of air, so its shunt capacitance per kilometre is one or two orders of magnitude larger. The resulting charging current is drawn whether or not any load is connected, and beyond a few tens of kilometres at transmission voltage it consumes the whole current rating of the cable. That single fact is why long submarine links are built as HVDC (Chapter 38) — a d.c. cable draws no steady charging current at all.
Distribution: Feeders, Distributors and Service Mains
Below 11 kV the design problem changes character. Transmission moves a large power between two points and is judged on efficiency; distribution delivers small powers to very many points and is judged on voltage regulation and on the cost of the copper spread over a service area. Its three kinds of conductor are distinguished by whether load is tapped from them along their length.
No consumer is ever connected directly to a feeder. The distinction is not bureaucratic: the design criterion is genuinely different, which is why the two are sized by different calculations.
The arrangement of the distributors decides the reliability. A radial system, in which each distributor is fed from one end only, is the cheapest and the simplest to protect, but a fault anywhere along it darkens everything beyond the fault, and the voltage drop is worst at the far end. A ring main, fed at two or more points as in the figure, halves the worst-case drop and allows any one section to be isolated without loss of supply — at the cost of more conductor and considerably more complicated protection, since fault current can now arrive from either direction. Chapter 36 returns to that difficulty when directional relays are introduced.
Voltage drop, not loss, is the binding constraint on a distributor, and statutory limits are typically \(\pm6\%\) at the consumer's terminals. Example 5 works out the standard result for a distributor loaded uniformly along its length, which is the model for a street of similar houses.
A.C. against D.C.
The ladder of Section 2-2 is an a.c. structure, and it is worth asking why, since on several technical counts direct current is plainly superior.
A d.c. link needs two conductors instead of three. It has no inductance, no capacitance and no phase angle, so there is no reactive voltage drop and no charging current — a d.c. line's regulation is better and a d.c. cable can be any length. There is no skin effect, so the whole cross-section of the conductor carries current instead of a surface layer. For a given insulation level the working voltage is higher, because insulation is stressed by the peak of an a.c. wave and \(V_{\text{peak}}=\sqrt2\,V_{\text{rms}}\); a d.c. line can therefore run at \(\sqrt2\) times the r.m.s. voltage of an a.c. line with the same insulators. Corona loss and radio interference are lower. Dielectric losses in cables vanish. And, decisively for long links, two a.c. systems joined by d.c. need not stay in synchronism with one another — there are no stability limits and no synchronising problems of the kind Part 6 is devoted to.
Against that stands one fact of overwhelming practical weight. Direct voltage cannot be transformed. Power cannot be generated at high d.c. voltage, because commutation in a d.c. machine limits it; it cannot be stepped up for transmission or down for distribution by any device as simple, as cheap and as efficient as a transformer; and d.c. circuit breakers are hard to build, because there is no natural current zero at which to extinguish the arc. A.c., meanwhile, can be generated at high voltage, transformed up and down at will with better than 99% efficiency, and interrupted by a breaker that need only hold off the arc for the few milliseconds until the current passes through zero of its own accord.
| Issue | D.C. transmission | A.C. transmission |
|---|---|---|
| Conductors | Two | Three |
| \(L\), \(C\), phase angle | Absent — no reactive drop, no charging current | Present; limits cable length and line loading |
| Skin effect | None; full cross-section used | Raises the effective resistance |
| Insulation for a given working voltage | Less — stressed at \(V\), not \(\sqrt2 V\) | More |
| Corona and interference | Lower | Higher |
| Stability and synchronising | No limit; asynchronous tie possible | Power transfer limited by angle — Part 6 |
| Changing voltage level | Converter stations, costly and lossy | Transformer: cheap, passive, >99% efficient |
| Circuit breaking | No natural current zero | Current zero twice per cycle |
| Generation at high voltage | Limited by commutation | Straightforward |
Interconnection: The System as One Machine
Nothing so far requires more than one generating station feeding one set of loads. In practice every station of consequence is tied to every other through the transmission network, and the reasons are the same economic ones that moved the station away from the city in the first place.
Interconnection lets the cheapest plant on the whole system serve the load, rather than the cheapest plant in one district — which is precisely the problem Chapter 31 formulates and solves. It reduces the reserve capacity that must be held idle, since a spare unit shared among many systems covers the loss of any one of them. It smooths the aggregate load, because the peaks of different regions and different industries occur at different times, so the ratio of peak to average demand improves and every unit runs closer to its best efficiency (Chapter 30). And it lets seasonal resources be traded — hydro in the monsoon, thermal in the dry months.
The price is that the network becomes a single dynamical object. Every synchronous machine on it turns in step with every other, locked by the electrical angle across the lines that join them; the frequency is a system-wide variable, and a disturbance at one bus is felt, attenuated but not localised, at every other. A fault must therefore be cleared in a fraction of a second or the machines will lose synchronism and the system will separate — the subject of Chapters 27 to 29. Protection must be selective enough to remove only the faulted element, since tripping more than necessary can propagate a disturbance into a cascade (Chapters 35 to 37). And the frequency must be regulated continuously against a load that never stops changing, which is the automatic generation control of Chapter 33.
Parts 6, 7 and 8 of this book exist because that bargain was accepted everywhere. The single-line diagram of Chapter 3 and the per-unit system of Chapter 4 are the tools that make an interconnected system of thousands of buses possible to write down at all.
Worked Examples
Problem. 50 MW is to be transmitted 100 km at 0.9 power factor with a total line loss of 5%. Using hard-drawn aluminium of resistivity \(\rho = 2.83\times10^{-8}\ \Omega\,\mathrm{m}\) and density 2700 kg/m³, find the conductor cross-section, the volume and the mass of conductor required at 132 kV, and compare with 400 kV.
Solution. The permitted loss is \(W = 0.05\times50 = 2.5\) MW. From Section 2-3, with \(P = 50\times10^6\) W, \(l = 10^5\) m and \(\cos\phi = 0.9\):
At 400 kV nothing changes except \(V^2\), so both area and volume scale by \((132/400)^2 = 0.1089\):
The 400 kV line needs under 11% of the metal — 145 tonnes less on a route of only 100 km. Against that must be set two 400 kV transformers, taller towers and larger switchgear, which is why the comparison must be made over the whole scheme and not on conductor alone. In practice a 400 kV line is never built with a 22 mm² conductor: corona would be intolerable at that diameter, so the conductor is made much larger and bundled (Chapters 8 and 15), and the realised loss is far below 5%.
Problem. A 200 km three-phase line works its aluminium conductors at a current density \(J = 1.0\ \mathrm{A/mm^2}\), with \(\rho = 2.83\times10^{-8}\ \Omega\,\mathrm{m}\) and a load power factor of 0.9. Find the transmission efficiency and the percentage line drop at 220 kV, and again at 400 kV.
Solution. Convert the current density to SI: \(J = 1.0\ \mathrm{A/mm^2} = 10^{6}\ \mathrm{A/m^2}\). The absolute drop per conductor does not depend on the voltage at all:
At 220 kV the phase voltage is \(220/\sqrt3 = 127.0\) kV, and at 400 kV it is 230.9 kV, so
For the efficiency, use the result of Section 2-4:
The approximation \(\eta \approx 1 - \sqrt3\rho lJ/(V\cos\phi)\) gives 95.05% and 97.28%, close enough for a first estimate. Note that both the drop and the loss fraction scale exactly as \(1/V\), and that the ratio \(400/220 = 1.818\) reproduces itself in \(4.46/2.45 = 1.82\) — a useful check on any arithmetic of this kind.
Problem. A generating station exports 120 MVA. Find the line current at each rung of the ladder — 11 kV, 220 kV, 33 kV and 11 kV again. Then show what would happen if the 100 km transmission were attempted at generator voltage, taking a conductor of 200 mm² and \(\rho = 2.83\times10^{-8}\ \Omega\,\mathrm{m}\).
Solution. For a balanced three-phase system \(|S| = \sqrt3\,V I\), so \(I = |S|/(\sqrt3\,V)\):
| Stage | Voltage | Line current for 120 MVA |
|---|---|---|
| Generator bus | 11 kV | \(120\times10^6/(1.732\times11\times10^3) = 6298\ \mathrm{A}\) |
| Primary transmission | 220 kV | \(120\times10^6/(1.732\times220\times10^3) = 314.9\ \mathrm{A}\) |
| Secondary transmission | 33 kV | \(120\times10^6/(1.732\times33\times10^3) = 2100\ \mathrm{A}\) |
| Primary distribution | 11 kV | \(6298\ \mathrm{A}\) |
The conductor resistance over 100 km is
The first figure is not merely bad, it is impossible: the line would have to dissipate fourteen times the power it was asked to deliver. The second is 3.5% of the transfer, entirely normal. The two differ by the factor \((220/11)^2 = 400\), and \(1684/400 = 4.21\) confirms it. This single comparison is the whole justification for the step-up transformer at the station.
Problem. Compare the volume of conductor material required by a two-wire d.c. system and a three-phase, three-wire a.c. system carrying the same power over the same distance with the same total loss and the same maximum voltage to earth. Evaluate the ratio at 0.8 power factor.
Solution. Let the maximum voltage to earth be \(V_m\) in both cases; this is what the insulation is designed for.
Two-wire d.c. The conductors sit at \(+V_m\) and \(-V_m\), so the load sees \(2V_m\):
Three-phase a.c. The insulation is stressed by the peak of the phase voltage, so \(V_{\text{ph}} = V_m/\sqrt2\) in r.m.s. terms:
The d.c. line needs 32% of the a.c. conductor at 0.8 power factor and half of it even at unity. Two effects combine: two conductors instead of three, and a \(\sqrt2\) advantage in working voltage for the same insulation. This is a genuine and large saving, and it is exactly why HVDC becomes economic once the line is long enough for the conductor saving to outweigh the cost of the two converter stations — typically beyond 600–800 km overhead, or a few tens of kilometres by submarine cable.
Problem. A distributor 300 m long is fed at one end and carries a load of 0.5 A per metre distributed uniformly along it. The resistance of the go-and-return conductors is \(0.0002\ \Omega\) per metre. Find the voltage drop at the far end, and compare it with the drop that would occur if the whole load were concentrated at the far end.
Solution. Let \(i\) be the current per unit length and \(r\) the resistance per unit length, with the feeding point at \(x=0\). At a distance \(x\), the conductor still has to carry the load of everything beyond it, so the current there is \(i(l-x)\). The drop in an element \(dx\) is \(i(l-x)\,r\,dx\), and integrating along the whole length,
The total current is \(I = il = 150\) A and the total resistance \(R = rl = 0.06\ \Omega\), so if that whole current had to be carried the whole way — the load concentrated at the far end — the drop would be \(IR = 150\times0.06 = 9\) V, exactly twice as much. Writing the result as \(\Delta V = \tfrac12 (il)(rl)\) shows why: a uniformly loaded distributor behaves as though its entire load were concentrated at its mid-point. On a 230 V service the 4.5 V drop is 1.96%, comfortably inside a \(\pm6\%\) limit, whereas an end-concentrated 9 V would be 3.9% and would leave little margin for the drop already incurred upstream.
Problem. A scheme is to transmit 200 MW over 250 km. Costing gives the conductor as \(A/V^2\) with \(A = 2.662\times10^{6}\) (₹ crore·kV²) and the terminal equipment as \(BV\) with \(B = 0.5\) ₹ crore per kV. Find the economic voltage, tabulate the total cost at the standard levels, and check the answer against Still's formula.
Solution. From Section 2-4,
| \(V\) (kV) | Conductor \(A/V^2\) | Terminals \(BV\) | Total (₹ crore) |
|---|---|---|---|
| 132 | 152.8 | 66.0 | 218.8 |
| 200 | 66.6 | 100.0 | 166.6 |
| 220 | 55.0 | 110.0 | 165.0 |
| 250 | 42.6 | 125.0 | 167.6 |
| 400 | 16.6 | 200.0 | 216.6 |
At the optimum the terminal cost of 110 is exactly twice the conductor cost of 55, as the general condition of Section 2-4 requires. The minimum is also very flat: anywhere between 200 and 250 kV the total is within 1.6% of the best, which is why utilities can standardise on a handful of voltages instead of costing each line individually.
Still's formula, with \(L = 250\) km and \(P = 200\,000/3 = 66\,667\) kW per phase, gives
Both routes point at the same rung, and the standard 220 kV level is adopted.
Chapter Summary
11 kV generation → 132/220/400 kV primary transmission → 33 kV secondary → 11 kV primary distribution → 400/230 V service.
Everything above 400 V is three-phase three-wire; secondary distribution adds a neutral for single-phase consumers.
\(\text{Vol}=3P^2\rho l^2/(WV^2\cos^2\phi)\), so it falls as \(1/V^2\) and as \(1/\cos^2\phi\).
At fixed current density the drop is \(\rho lJ\) volts regardless of \(V\); as a percentage it and the loss both fall as \(1/V\).
Insulation, switchgear and corona costs rise with \(V\); the optimum \(V^*=(2A/B)^{1/3}\) puts terminal cost at twice conductor cost.
Stator slot insulation, not economics, limits generation voltage — so generate low and transform up.
Feeders carry constant current and are sized for capacity; distributors are tapped along their length and are sized for voltage drop.
D.C. wins on conductors, insulation, cables and stability; a.c. wins because voltage can be transformed and current interrupted.
Problems
Take \(\rho = 2.83\times10^{-8}\ \Omega\,\mathrm{m}\) for aluminium and \(1.72\times10^{-8}\ \Omega\,\mathrm{m}\) for copper unless told otherwise, and assume balanced three-phase, three-wire transmission. State clearly whether a voltage quoted is line-to-line or line-to-neutral before substituting it anywhere.
- A three-phase line delivers 25 MW at 0.85 lagging over 60 km with 6% loss. Find the conductor cross-section and the total volume of aluminium at 66 kV, and state by what factor both change if the line is built at 132 kV instead.
- Show from the result of Section 2-3 that improving the power factor of a fixed load from 0.75 to 0.95 permits the same conductor to carry \((0.95/0.75)^2\) times the power at the same loss. Evaluate the percentage increase.
- A 150 km line works its conductors at \(1.2\ \mathrm{A/mm^2}\) with a load power factor of 0.85. Compute the drop per conductor in volts, the percentage drop, and the transmission efficiency at 132 kV and at 220 kV.
- Repeat the derivation of Section 2-3 for a single-phase two-wire line and show that the volume of conductor required is twice that of the three-phase three-wire line carrying the same power at the same line voltage, loss and power factor.
- A distributor 400 m long is fed at both ends at the same voltage and carries a uniformly distributed load of 0.6 A per metre, with \(r = 0.00025\ \Omega\) per metre. Find the point of minimum voltage and the drop there, and compare it with the same distributor fed from one end only.
- Costing for a 300 km scheme gives a conductor cost of \(A/V^2\) with \(A = 5.4\times10^{6}\) (₹ crore·kV²) and terminal cost \(BV\) with \(B = 0.6\) ₹ crore per kV. Determine the economic voltage and the total cost there, and verify that the terminal cost is twice the conductor cost. Which standard voltage would you adopt?
- A 500 kVA, 11 kV/400 V distribution transformer supplies a four-wire secondary. Find the full-load line current on each side, and the current in the neutral if the three phase loads are 180 A, 150 A and 150 A, all at unity power factor and displaced by 120°.
- List the technical advantages of d.c. transmission and, for each one, identify the a.c. counterpart that outweighs it in a meshed network. Then explain in your own words why a 60 km submarine crossing is built as HVDC while a 60 km overhead line is not.