Line Performance: Regulation and Efficiency

Part 2 of this book computed the parameters of a line: resistance from the conductor, inductance from the geometry in Chapter 6, capacitance in Chapter 7, both modified by bundling in Chapter 8. Part 3 has been turning those parameters into a circuit. Chapter 10 kept only the series impedance and got a model good to 80 km. Chapter 11 lumped the shunt admittance at one or two points and reached 250 km. Chapter 12 distributed it properly, solved the wave equation, and produced constants that are exact at any length.

All three chapters ended with the same two-port relation, differing only in what was put into the four constants:

Everything a planner or an operator wants to know about a transmission line is a question about that matrix and the load hung on its right-hand terminals. There are three such questions, and they are not independent.

How much does the voltage move? Consumers' equipment is designed for a nominal voltage and tolerates perhaps \(\pm 5\%\) around it. If the receiving-end voltage sags when the load comes on and soars when it goes off, either the load suffers or a tap-changer and a capacitor bank must be bought to hide the problem. The measure of that movement is voltage regulation.

How much power is thrown away? Every ampere in the conductor heats it. That heat is bought fuel that never reaches a customer, and on a 400 kV interconnector carrying 600 MW for twenty years its cumulative value is comparable to the cost of the line itself. The measure is transmission efficiency.

How much can be sent at all? Chapter 10 showed that a short line is limited by conductor heating, Chapter 12 that a long one is limited by the angle \(\delta\) long before the conductor is warm. That is the loadability question, and Chapter 14 takes it up with the surge impedance as its yardstick.

This chapter answers the first two exactly, for any line, from the ABCD constants alone. The reference line of Chapters 11 and 12 — 200 km, 220 kV, 50 Hz, \(z = 0.16+j0.40\ \Omega\)/km, \(y = j3.0\times10^{-6}\) S/km, delivering 60 MW at 0.9 power factor lagging — runs through the whole discussion so that every new formula can be checked against a number already known.

Regulation is defined by a thought experiment. Hold the sending-end voltage at whatever value delivers rated voltage to the load at full load. Now disconnect the load, leaving the sending-end voltage untouched. The receiving-end voltage will move. Regulation is that movement, expressed as a fraction of the full-load value.

To evaluate it we need the no-load receiving voltage. Set \(\mathbf{I}_r = 0\) in the two-port relation. The first row collapses immediately:

The constant \(A\) alone decides the no-load voltage. That is worth pausing over, because \(A\) is the one constant of the four that the short-line model of Chapter 10 discarded — there \(A=1\) exactly, and the no-load voltage is simply the sending-end voltage. Every departure of \(|A|\) from unity is a statement about the shunt capacitance, and Chapter 12 showed that \(A=\cosh\gamma l\), which for a real line has a magnitude less than one. A line therefore always ends up with a no-load voltage higher than its sending-end voltage. That is the Ferranti effect, and Chapter 14 is devoted to it.

Two conventions must be kept straight, because textbooks and examination papers use both. The definition above holds \(|V_s|\) fixed and asks what \(|V_r|\) does; this is the physical statement, since the sending end is normally a generator bus held at a set voltage. The alternative holds \(|V_r|\) fixed at rated value and computes the \(|V_s|\) required — which is what the worked examples of Chapters 10 to 12 did, because it needs no iteration. The two give the same regulation figure provided the same load is used, because the ratio \(|V_s|/(|A||V_r|)\) is what the formula actually contains. Working from a fixed \(|V_r|\) is therefore the practical route, and the "hold \(V_s\) fixed" language is simply how the answer is interpreted.

Regulation is not a property of the line. It is a property of the line and the load it is carrying, and the load's power factor matters as much as its magnitude. Chapter 10 derived the approximate drop for a short line by projecting the impedance drop onto the receiving voltage; with \(\mathbf{V}_r\) as reference and \(\mathbf{I}_r = I\angle-\phi\) for a lagging load,

The in-phase part of the drop dominates the magnitude; the quadrature part contributes only at second order and is why the approximation is good to a fraction of a percent for any realistic line. Read the bracket. The resistive term \(R\cos\phi\) is always positive: real power always costs voltage. The reactive term \(X\sin\phi\) changes sign with the power factor, and since \(X\) is two to ten times \(R\) on an overhead line, it is the term that decides the outcome.

At a lagging power factor the load draws reactive power through the line's reactance and the drop is large. At unity power factor only the resistive term survives. At a leading power factor the reactive term subtracts, and if the load leads strongly enough the two terms cancel exactly.

The cancellation can be made exact. Write \(Z = R+jX = |Z|\angle\theta\), take \(\mathbf{V}_r\) as reference and let the current lead by \(\phi\), so \(\mathbf{I}_r = I\angle\phi\). Then \(\mathbf{V}_s = V_r + I|Z|\angle(\theta+\phi)\) and

Regulation can therefore be negative, and on a line serving a capacitor-compensated industrial load it routinely is. Negative regulation is not automatically a virtue: it means that throwing the load off makes the voltage fall, which is just as much a disturbance as a rise. The design target is a small regulation of either sign, not a large negative one.

Efficiency has the definition one would expect,

and the whole difficulty is in getting \(P_{\text{loss}}\) right. On a short line it is easy, because the same current flows in every element of the conductor: \(P_{\text{loss}} = 3I^{2}R\), and there is nothing else. That single formula, pushed a little, produces the most useful scaling law in transmission engineering. Substitute \(I = P_r/(\sqrt3\,V_{LL}\cos\phi)\):

Three other loss mechanisms exist and must at least be named. Leakage across insulator strings is represented by the conductance \(g\) of Chapter 12's line model; on a clean, dry overhead line it is negligible, on a salt-polluted coastal line it is not. Dielectric loss in the insulation is negligible overhead but significant in the cables of Chapter 9, where it grows as \(V^{2}\omega C\tan\delta\) and sets the practical length limit of an a.c. cable. Corona loss, the subject of Chapter 15, is zero below a threshold voltage and then rises steeply; on an EHV line in foul weather it can exceed the \(I^{2}R\) loss at light load. All three share a property that matters for Section 13-6: they depend on the voltage, not on the current, and so are roughly constant while the load varies.

On a medium or long line the sending-end and receiving-end currents are not equal, because the shunt capacitance injects current all along the way. Chapter 12 wrote the current at a distance \(x\) from the receiving end as

and the true loss is the integral of \(3r|I(x)|^{2}\) over the length, plus \(3g|V(x)|^{2}\) if leakage is modelled:

This integral is never evaluated in practice, and it does not have to be. Conservation of energy already supplies the answer: whatever is dissipated inside the line is exactly the difference between what enters and what leaves.

The tempting shortcuts are all wrong, and it is worth knowing by how much. Using \(3I_r^{2}R\) on the reference line at its 60 MW load gives 2.94 MW against a true 2.53 MW — sixteen per cent high, because the receiving-end current is the largest current anywhere on the line. Using \(3I_s^{2}R\) gives 2.28 MW, ten per cent low. Averaging the squares, \(\tfrac32 R\big(|I_s|^{2}+|I_r|^{2}\big)\), gives 2.61 MW, three per cent high, and this is the best of the shortcuts because the current profile is close to linear over 200 km. None of them is a substitute for \(P_s - P_r\).

The current profile itself repays a glance. At 0.9 lagging the reference line carries 175.0 A at the receiving end and only 154.2 A at the sending end, falling monotonically along the way — the charging current, which leads the voltage by roughly \(90^\circ\), partially cancels the lagging load current, and the cancellation is most complete near the sending end where the accumulated charging current is largest. At unity power factor the opposite happens: the charging current adds in quadrature to a load current that has no reactive component, and the profile rises from 157.5 A at the load to 171.7 A at the source.

A line does not run at one load. It follows the daily and seasonal demand curve of Chapter 30, and its efficiency changes as it does. Split the loss into the part that varies with current and the part that does not:

Efficiency is then \(\eta = P_r/(P_r + P_0 + kP_r^{2})\). Differentiate with respect to \(P_r\) and set the derivative to zero. The numerator of \(d\eta/dP_r\) is proportional to \(P_0 - kP_r^{2}\), so:

The rule holds cleanly whenever the fixed loss is genuinely independent of the load current, as corona and leakage are. It fails in an instructive way when the "fixed" loss is itself an \(I^{2}R\) loss in the same conductor, which is what the charging current produces. On the reference line the no-load loss is 184 kW of \(I^{2}R\) caused by 77 A of charging current. Adding a lagging load does not simply add to that current, it partly cancels it, so the loss initially falls: at 5 MW the total loss is 159 kW, below the no-load figure. The efficiency peak, at 15.1 MW and 98.50%, therefore does not satisfy "variable equals fixed" at all. The lesson is not that the rule is wrong but that it applies to independent loss terms, and the charging loss is not independent.

For planning, a single instantaneous efficiency is less useful than the energy lost over a year. If the line's peak loss is \(P_{\text{loss,max}}\), the annual energy loss is

where LF is the load factor of Chapter 30 and LSF the loss-load factor. The quadratic dominance in that empirical relation is the \(I^{2}\) again: a line with a load factor of 0.6 has a loss factor of only 0.43, because it spends most of its time well below peak, where the loss falls quadratically. Example 5 puts a rupee figure on it.

Chapters 10, 11 and 12 offered four models of the same 200 km line. Now that regulation and efficiency have precise definitions, the four can be put side by side on the same load and the cost of each simplification read directly. The load throughout is 60 MW at 220 kV, 0.9 lagging, so \(\mathbf{I}_r = 174.95\angle-25.842^\circ\) A and \(V_r = 127\,017\) V per phase.

Three observations survive the arithmetic. First, the two nominal circuits bracket the exact answer in every column, the T underestimating and the π overestimating — exactly as Section 11-5 predicted from the signs of the neglected series terms. Averaging them would give 234.68 kV against a true 234.64 kV.

Second, the short-line model happens to get the regulation nearly right, at 9.06% against 9.26%, and this is luck rather than merit. It errs twice in opposite directions: it overstates \(|V_s|\) by 2.3% because it lets the full load current flow through the whole reactance, and it takes \(|A|=1\) instead of 0.976, which understates the no-load voltage by the same order. The two mistakes very nearly cancel in the ratio.

Third, no such cancellation protects the efficiency. The short-line model gives 95.33% against 95.95%, and its 2.94 MW loss is 16% too large, because it puts the receiving-end current — the largest current on the line — into every metre of conductor. Regulation forgives a bad model; efficiency does not. A quantity that depends on the square of a current is far less tolerant of a wrong current profile than one that depends on a ratio of voltages.

Everything so far has taken the load as given and computed the sending end. The operator's question is usually the reverse: both bus voltages are held at rated value by generators and tap-changers, and what is wanted is the range of real and reactive power the line can then deliver, and what reactive support the receiving bus needs. The circle diagram answers all of that in one figure.

Solve the two-port relation for the receiving-end current and form the complex power. Write \(A = |A|\angle\alpha\), \(B = |B|\angle\beta\), take \(\mathbf{V}_r = |V_r|\angle0\) as reference and let the sending-end voltage lead by the load angle, \(\mathbf{V}_s = |V_s|\angle\delta\):

Read this as geometry in the \(P\)–\(Q\) plane. The second term is a fixed vector: it does not contain \(\delta\), so with both voltages held it never moves. The first term has fixed length and an angle that sweeps as \(\delta\) changes. A fixed point plus a vector of constant length turning about it traces a circle.

Separating real and imaginary parts gives the two working formulas, which are just Chapter 11's power-transfer expressions written in a new order:

Four readings come off the diagram at once. The horizontal coordinate of any point on the circle is the real power delivered; the vertical coordinate is the reactive power delivered to the line's receiving terminals. The topmost point of the circle is the maximum real power that can be pushed through with those two voltages, reached at \(\delta = \beta\). The vertical gap between the circle and the load's actual reactive demand is the reactive power that a capacitor bank or synchronous condenser must supply if the receiving voltage is to be held at the assumed value. And the load angle at any operating point is \(\beta\) minus the angle subtended at the centre — Chapter 28 will use exactly that angle as the state variable of the stability problem.

The formulas of this chapter identify the levers precisely. Regulation is driven by \(I\big(R\cos\phi + X\sin\phi\big)\); efficiency by \(P_rR/(V^{2}\cos^{2}\phi)\). Only three quantities appear in both: the current, the reactance, and the power factor. Each can be attacked.

Shunt capacitors at the receiving end. A capacitor bank supplies the load's reactive demand locally, so the line no longer carries it. The current falls from \(P/(\sqrt3 V\cos\phi)\) toward \(P/(\sqrt3 V)\), which cuts the \(I^{2}R\) loss by \(\cos^{2}\phi\) and removes the \(IX\sin\phi\) term from the drop simultaneously. This is the cheapest and commonest remedy, and Chapter 34 sizes it properly. Its weakness is that the reactive output falls as \(V^{2}\), so it delivers least support exactly when the voltage has collapsed and support is most needed.

Series capacitors. Inserting \(-jX_C\) in series cancels part of \(X\). With a compensation degree \(k = X_C/X\), the effective reactance becomes \(X(1-k)\), and both the reactive term of the drop and the denominator of the power-transfer formula shrink together. Chapter 14 shows that this also shortens the line electrically and raises its natural loading. The risks — subsynchronous resonance, and the fault current that must be bypassed around the capacitor — belong to Chapter 38.

Shunt reactors. These do the opposite: they absorb the charging reactive power of a lightly loaded line and suppress the Ferranti rise. They improve the no-load voltage at the cost of a slightly worse full-load one, and on an EHV line they are switched in and out as the load follows the daily curve.

Higher voltage, or a bundle. Both attack the problem at its root. Doubling \(V\) at constant power quarters the loss fraction and halves the current, and hence halves the drop. Bundling reduces \(X\) by about 30% and \(R\) in proportion to the number of subconductors — the whole argument of Chapter 8, now with a performance number attached.

Take 50 Hz throughout, use the model appropriate to the stated length, and quote regulation and efficiency to two decimal places. Problems 6 to 8 are of GATE standard.