Set 10 — Short Transmission Lines — Regulation
Twenty worked problems on the quantity that most often decides whether a line is acceptable. Set 9 computed regulation; this set attacks it. The three available remedies — series capacitance, shunt capacitance and tap changing — act on entirely different terms of the same expression, and choosing between them requires knowing which term dominates. The set closes with the maximum power a short line can transfer, which turns out to be governed by the same impedance from the other direction.
Regulation in compact form. \(\%\text{reg} \approx \%R\cos\phi \pm \%X\sin\phi\). Two numbers describe the line and the load supplies the angle. Every remedy in this set attacks one of the three factors.
Series compensation attacks \(\%X\). A capacitor in series subtracts from the line reactance, so \(X_{\text{net}} = X_L - X_C\). It cannot reduce the regulation below the \(\%R\cos\phi\) floor, and it raises the fault current through the compensated section.
Shunt compensation attacks \(\sin\phi\). A capacitor across the load supplies its reactive demand locally, so the line carries less current at a better power factor. It reduces the losses as well, which series compensation does not.
Tap changing attacks neither — it raises \(V_S\). It corrects the receiving voltage without improving the line at all, so the losses and the reactive absorption are unchanged. It is a remedy for the symptom, and often the right one.
Capacitor sizing. To move a load of \(P\) from \(\cos\phi_1\) to \(\cos\phi_2\) requires \(Q_C = P(\tan\phi_1 - \tan\phi_2)\). This one line answers most compensation questions.
Power transfer. \(P_R = \dfrac{V_SV_R}{Z}\cos(\theta - \delta) - \dfrac{V_R^2}{Z}\cos\theta\), maximum at \(\delta = \theta\). For a line with \(R \to 0\) this collapses to the familiar \(V_SV_R\sin\delta/X\).
Know which term dominates before choosing a remedy. On a transmission line \(\%X \gg \%R\) and the reactive term rules; on a distribution feeder they are comparable and the resistive term can dominate at good power factors. The same remedy can be decisive on one and useless on the other.
A 66 kV, 60 km three-phase line has \(R = 9.6\ \Omega\) and \(X_L = 24.0\ \Omega\) per phase and delivers 20 MW at 0.9 power factor lagging. Find its percentage resistance and reactance, and hence its voltage regulation. This line recurs throughout the set.
Per-phase voltage and line current:
Percentage resistance and reactance at this current:
With \(\cos\phi = 0.9\) and \(\sin\phi = 0.4359\) lagging:
Reading the two contributions separately is the whole point of this decomposition: the reactance supplies 5.34 of the 9.75 points, and the resistance 4.41.
A short three-phase line of impedance \((6 + j8)\ \Omega\) per phase has sending- and receiving-end voltages of 120 kV and 110 kV for a load at 0.9 power factor lagging. Determine the power output and the sending-end power factor.
Both voltages are given and the current is unknown — the reverse of Problem 1. Convert to per-phase:
Using the approximate formula with \(I\) as the unknown:
Collecting the current:
Power output:
Sending-end power factor:
The regulation implied is \((120-110)/110 = 9.1\%\), which is what fixed the current in the first place.
For the line of Problem 1, find the sending-end power factor and the reactive power absorbed by the line.
Sending-end voltage per phase, from the percentages of Problem 1:
Sending-end power factor:
Reactive power absorbed by the line:
Against the load's own reactive demand:
The line adds 28% to the reactive burden the sending end must supply — which is the direct cause of the power factor falling from 0.900 to 0.865.
For the line of Problem 1 at fixed current, tabulate the regulation at power factors from 0.6 lagging to 0.6 leading, and identify the zero-regulation point.
With \(\%R = 4.90\) and \(\%X = 12.25\) from Problem 1:
The zero-regulation power factor, from Set 9 Problem 13:
Checking it against the table, which brackets it between unity (+4.90) and 0.90 leading (−0.93):
A series capacitor providing 50% compensation is installed in the line of Problem 1. Find the new reactance, the new regulation, and the capacitor's reactive rating.
Fifty per cent compensation means the capacitor cancels half the line reactance:
The current is unchanged at 194.4 A — the load still demands 20 MW at 0.9 power factor — so the new percentage reactance is simply halved:
New regulation:
Capacitor rating — three phases, each carrying the full line current:
The improvement is from 9.75% to 7.08% — a reduction of 2.67 points, exactly the reactive contribution that was cancelled.
Find the regulation of the line of Problem 1 with 100% series compensation, and hence the lowest regulation series compensation alone can achieve. Comment on what limits it.
With \(X_C = X_L = 24.0\ \Omega\) the net reactance vanishes:
The regulation reduces to the resistive term alone:
Tabulating the degrees of compensation:
The floor is 4.41%, set entirely by \(\%R\cos\phi\), and no series capacitance whatever can go below it.
What limits compensation in practice. Full compensation is never used. Beyond about 70% the line becomes a net capacitance, and three problems appear: subsynchronous resonance, in which the compensated line's electrical resonance couples to a turbine-generator's torsional modes; ferroresonance with transformer magnetising inductance; and the loss of a well-defined fault-current direction, which confuses distance protection.
A shunt capacitor bank is to raise the power factor of the 20 MW load of Problem 1 from 0.9 to 0.95 lagging. Find its rating, and the new line current.
The load's reactive demand at each power factor:
The capacitor supplies the difference:
The real power is unchanged, so the new line current follows from the improved power factor:
Down from 194.4 A, a reduction of 5.2%.
The loss reduction that comes free with it:
Find the voltage regulation of the line of Problem 1 after the shunt correction of Problem 7.
Both the current and the angle have changed, so the percentages must be recomputed at \(I = 184.2\) A:
With \(\cos\phi = 0.95\) and \(\sin\phi = 0.3122\):
Against the original 9.75%, an improvement of 1.72 points.
Note that the resistive contribution is unchanged at 4.41%. The current fell by 5.2% but \(\cos\phi\) rose by the same 5.6%, and the product \(I\cos\phi\) — which is fixed by the real power — cannot change at all:
Find the capacitor rating needed to correct the load of Problem 1 to unity power factor, and the resulting regulation and losses.
To unity, the capacitor must supply the whole of the load's reactive demand:
New current:
Regulation, with the reactive term now vanishing entirely:
Exactly the floor of Problem 6, reached by an entirely different route.
Losses:
Compare series and shunt compensation for the line of Problem 1 on every relevant criterion, and state when each is preferred.
Collecting the results of Problems 5 to 9 for equal regulation of 4.41%:
The mechanisms differ fundamentally. A series capacitor changes the line; a shunt capacitor changes the load. The first reduces the impedance between two points, the second reduces the current flowing between them.
Series compensation is preferred when the line is long and highly inductive, when the objective is to raise the power transfer limit (Problem 18), and when the reactive rating must be small. It self-regulates: its effect grows as \(I^2\), so it acts hardest at heavy load.
Shunt compensation is preferred when losses matter, when the reactive demand is at an identifiable load centre, when switching in blocks to follow the load is wanted, and on distribution feeders where \(X/R\) is low. Its weakness is that its output falls as the square of the voltage, so it fails just as the voltage collapses.
Instead of compensating the line of Problem 1, the sending-end transformer is tapped to hold the receiving-end voltage at 66 kV. Find the required tap setting, and state what the tap changer does not improve.
From Problem 3, holding \(V_R\) at 66 kV requires a sending-end voltage of
The tap ratio required, relative to nominal:
Standard on-load tap changers offer \(\pm10\%\) in steps of 1.25%, so the nearest available tap is \(+10\%\):
A slight overshoot of 0.23%, which is the granularity of the tap and is entirely acceptable.
What the tap changer does not improve. The current is unchanged at 194.4 A, so:
Moreover the tap changer must draw 72.6 kV from a system that may not have it to give — it moves the voltage problem upstream rather than solving it.
An 11 kV distribution feeder 5 km long has \(R = 0.54\ \Omega\)/km and \(X_L = 0.35\ \Omega\)/km and supplies 2 MW at 0.85 power factor lagging. Find its regulation, and the regulation achievable by full shunt correction and by full series compensation.
Feeder impedance — note that here \(R > X_L\):
Per-phase voltage and current:
Percentages and regulation, with \(\sin\phi = 0.5268\):
Full shunt correction to unity. The current falls to \(I = 2\times10^6/(\sqrt3 \cdot 11\,000) = 105.0\) A:
Full series compensation. The reactive term vanishes at the original current:
Both remedies reach the same 4.46%, an improvement of only 1.79 points from 6.25%.
A 400 kV transmission line has \(\%R = 0.5\) and \(\%X = 10\) at its rated load. Compare the effectiveness of series compensation on it with the distribution feeder of Problem 12, at 0.9 power factor lagging.
Transmission line, uncompensated, at \(\cos\phi = 0.9\):
Fully series compensated:
Distribution feeder, from Problem 12:
The general result. The fraction of the regulation that compensation can remove is
— a function of the line's \(X/R\) and the load angle only, independent of the loading.
Evaluating at \(\cos\phi = 0.9\), so \(\cot\phi = 2.065\):
The shunt capacitor of Problem 9 costs \(\text{₹}500\) per kVAr installed. If energy costs \(\text{₹}5\) per kWh and the loss load factor is 0.4, find the annual saving in losses and the simple payback period.
From Problem 9, the loss reduction at full load:
Annual energy saved, using the loss load factor to convert peak loss to average:
Annual saving:
Capital cost of the bank:
Simple payback:
And this counts only the loss saving. The released capacity — the line now carries 175 A instead of 194 A for the same power, freeing 10% of its thermal headroom — and the improved regulation are both obtained at no additional cost.
Derive the receiving-end power of a short line in terms of the terminal voltages, the impedance \(Z\angle\theta\) and the load angle \(\delta\), and evaluate its parameters for the line of Problem 1.
Take \(\mathbf{V}_R = V_R\angle0\) and \(\mathbf{V}_S = V_S\angle\delta\). The current from the line equation:
The complex power delivered to the load:
Taking the real part:
For the line of Problem 1:
Checking the lossless limit: with \(R \to 0\) we get \(Z \to X\), \(\theta \to 90^\circ\), \(\cos\theta \to 0\), and
the familiar form used throughout the stability analysis of Set 24.
Find the maximum power the line of Problem 1 can deliver with both terminal voltages held at 66 kV, and the load angle at which it occurs.
From Problem 15, \(P_R\) is maximised when \(\cos(\theta - \delta) = 1\), that is at
The maximum, per phase, with \(V_S = V_R = 38\,105\) V:
For three phases:
Against the actual 20 MW delivered, the line operates at
Comparing with the lossless estimate, which is what is usually quoted:
The resistance has cut the true maximum to 58% of the lossless figure — a very large correction, because \(X/R\) here is only 2.5.
Find the load angle at which the line of Problem 1 delivers its actual 20 MW with both terminal voltages at 66 kV.
Per phase, the delivered power is
Substituting into the power transfer equation of Problem 15:
Evaluating the constant term and rearranging:
Hence:
A useful cross-check against the phasor solution. From Set 9's method the sending-end voltage had a quadrature component; here the load angle of 7.55° is the angle between the two terminal voltages when both are held at 66 kV, which is a different and larger quantity than the 1.4° found when \(V_S\) was free to rise.
Find the effect of 50% series compensation on the maximum power of the line of Problem 1, and repeat for a line of the same reactance but \(R = 1.2\ \Omega\). Explain the difference.
The Problem 1 line, \(R = 9.6\ \Omega\). With \(X\) halved to 12 \(\Omega\):
Against 105.9 MW uncompensated — an improvement of just 0.5%.
The low-resistance line, \(R = 1.2\ \Omega\), \(X = 24\ \Omega\). Uncompensated:
With 50% compensation, \(X = 12\ \Omega\):
The explanation. Halving \(X\) halves \(Z\) only when \(X \gg R\). On the high-resistance line, \(Z\) fell from 25.85 only to 15.37 — a factor of 1.68, not 2 — and \(\cos\theta\) rose from 0.37 to 0.62, cancelling most of the gain.
For the low-resistance line of Problem 18 (\(R = 1.2\ \Omega\), \(X_L = 24\ \Omega\)), tabulate the maximum power against the degree of series compensation \(k = X_C/X_L\), and state what limits \(k\).
With \(X_{\text{net}} = (1 - k)X_L\), and using \(P_{\max} = 3V_R^2(1 - \cos\theta)/Z\) throughout:
The gain accelerates as \(k \to 1\) because \(P_{\max} \propto 1/Z\) and \(Z \to R\).
What limits \(k\). Practical compensation stops at 0.5 to 0.7, for reasons that are entirely non-electrical in origin:
Subsynchronous resonance is the binding constraint and the one that has caused real damage: the compensated line's electrical resonance at \(f\sqrt{k}\) can couple to a turbine-generator shaft's torsional natural frequency and destroy it, as happened at Mohave in 1970.
The line of Problem 1 must meet a 5% regulation limit while delivering 20 MW. Evaluate every remedy, recommend one, and justify the choice.
The starting point. Regulation 9.75%, of which 4.41 points are resistive and 5.34 reactive. The target is 5%.
Option 1 — series compensation. The reactive contribution must fall to \(5.00 - 4.41 = 0.59\) points:
Eighty-nine per cent compensation — beyond the practical limit of Problem 19, and inviting subsynchronous resonance. Rejected.
Option 2 — shunt correction. Even at unity power factor the regulation is 4.41%, which meets the target with a margin of 0.59 points. The required bank, from Problem 9:
Correcting only to 0.98 lagging would give \(\%\text{reg} = 4.50(0.98) + 11.25(0.199) = 4.41 + 2.24 = 6.65\%\) — insufficient. Full correction is needed. Feasible.
Option 3 — tap changing. A \(+10\%\) tap holds \(V_R\) at 66 kV, meeting the customer's requirement but leaving the losses at 1088 kW and the reactive absorption at 2.72 MVAr. Feasible but treats only the symptom.
Option 4 — larger conductor. To halve \(\%R\) requires roughly twice the metal, and it would leave the reactive term almost untouched:
Still failing, at very large cost. Rejected.
Recommendation: shunt capacitors, 9.69 MVAr at the load. It achieves 4.41%, comfortably inside the limit; it reduces losses by 206 kW, paying back in 1.34 years (Problem 14); it releases 10% of the line's thermal capacity; and it introduces none of the resonance or protection complications of series compensation.
The one caution is light load: with the full bank connected and little load, the leading current would raise the receiving voltage. The bank should therefore be switched in blocks, following the load.
Practice Problems
Work each on paper before opening the answer. The answer is given so you can check yourself; the method is deliberately not.
P1. A line has \(\%R = 3\), \(\%X = 9\). Find its regulation at 0.8 p.f. lagging.
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\(3(0.8) + 9(0.6) = 2.4 + 5.4 = \mathbf{7.8\%}\).P2. What is the lowest regulation series compensation alone can give that line at 0.8 p.f. lagging?
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The resistive floor: \(\%R\cos\phi = 3(0.8) = \mathbf{2.4\%}\).P3. A 5 MW load at 0.8 p.f. lagging is to be corrected to 0.95. Find the capacitor rating.
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\(Q_C = 5(\tan36.87^\circ - \tan18.19^\circ) = 5(0.750 - 0.329) = \mathbf{2.11}\) MVAr.P4. By what factor do the losses fall when a load is corrected from 0.8 to 0.95 p.f. at constant power?
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\((0.8/0.95)^2 = 0.709\), so losses fall by \(\mathbf{29\%}\).P5. A line has \(X/R = 8\) and a load at 0.85 p.f. What fraction of its regulation can series compensation remove?
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\(1/(1 + (1/8)\cot\phi)\); \(\cot\phi = 1.614\), so \(1/(1+0.202) = \mathbf{83\%}\).P6. A series capacitor of 15 \(\Omega\) carries 300 A in a three-phase line. Find its reactive rating.
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\(3I^2X_C = 3(300)^2(15) = \mathbf{4.05}\) MVAr.P7. A line has \(Z = 10 + j30\ \Omega\). Find \(Z\), \(\theta\) and \(\cos\theta\) for the power transfer equation.
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\(Z = 31.6\ \Omega\), \(\theta = \tan^{-1}3 = 71.57^\circ\), \(\cos\theta = R/Z = \mathbf{0.316}\).P8. For that line with both terminals at 132 kV, find the maximum three-phase power.
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\(V_{ph} = 76\,210\) V; \(P = 3(76\,210)^2(1-0.316)/31.6 = \mathbf{377}\) MW.P9. Why does a tap changer not reduce line losses?
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It raises \(V_S\) but leaves the current unchanged — the load still demands the same power at the same corrected voltage. Losses go as \(I^2R\) — Problem 11.P10. A shunt capacitor's output falls as \(V^2\). Why is this a serious drawback?
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Its support weakens exactly when the voltage sags and it is most needed — the opposite of a series capacitor, whose effect grows with current.P11. What limits series compensation to about 70%?
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Subsynchronous resonance chiefly — the series \(LC\) resonance can couple to turbine-generator torsional modes. Also ferroresonance, protection mis-measurement and raised fault levels — Problem 19.P12. A feeder has \(X/R = 0.5\) and poor regulation. What should be done?
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Not reactive compensation — the resistive term dominates. Use a larger conductor, a higher voltage, or move the transformer closer to the load — Problem 12.
Challenge Problems
Each of these needs an idea rather than a formula. Decide what the governing principle is before opening the answer.
C1. Series and shunt compensation both reduced the regulation of Problem 1 to the same 4.41% floor. Establish why that floor exists, prove it cannot be beaten by either method, and identify what can beat it.
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The floor. Regulation is \(\%R\cos\phi + \%X\sin\phi\). Expanding the resistive term:Every quantity on the right is fixed by the line's resistance, the delivered power and the operating voltage. Neither compensation method can change any of them.\[ \%R\cos\phi = \frac{IR\cos\phi}{V_R}\times100 = \frac{R}{V_R}\cdot\frac{P_R}{\sqrt3 V_L}\times100 \]
Why series compensation cannot beat it. A series capacitor changes only \(X\). At \(X_{\text{net}} = 0\) the reactive term vanishes and the resistive term stands alone. Going further — overcompensating so \(X_{\text{net}} < 0\) — would make the reactive term negative and could indeed reduce the total, but the line then behaves as a net capacitance with all the resonance hazards of Problem 19, and the improvement is bought at severe risk.
Why shunt compensation cannot beat it. A shunt capacitor changes only \(\phi\). At unity power factor \(\sin\phi = 0\) and again the resistive term stands alone. Overcorrecting into leading territory would make the reactive term negative, but the leading current then raises the receiving voltage at light load — a real operational hazard, and the reason capacitor banks are switched in blocks.
What can beat it. Only measures that change \(R\), \(P_R\) or \(V_L\):
— A larger conductor, reducing \(R\) directly. Expensive: halving \(R\) needs twice the metal for 2.2 points here.
— A higher operating voltage, which reduces the term as \(1/V^2\) — the most effective measure by far, and the reason the problem barely exists at 400 kV.
— A shorter electrical distance, by moving the supply point closer to the load.
— Deliberate overcompensation, accepting the light-load risk and managing it by switching.
The structural point: reactive compensation manipulates the reactive power flow, and the resistive voltage drop is caused by real power flow. They are orthogonal, and no amount of the first will address the second.C2. Series and shunt capacitors both inject reactive power, yet they behave oppositely under fault, under light load and as the voltage falls. Set out the differences and explain them from the circuit topology alone.
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The topological difference: a series capacitor carries the line current; a shunt capacitor sees the bus voltage. Every behavioural difference follows.
Output. Series: \(Q = 3I^2X_C\), rising as the square of the current. Shunt: \(Q = V^2/X_C\), falling as the square of the voltage.
Under heavy load. Series output rises — self-regulating, acting hardest when needed. Shunt output falls slightly as the bus voltage sags. Advantage: series.
Under light load. Series output falls to near zero — harmlessly. Shunt output stays at full value and can drive the voltage dangerously high, requiring switching. Advantage: series.
As the voltage collapses. Shunt output falls as \(V^2\), so at 0.8 p.u. voltage it delivers only 64% of its rating — it withdraws support precisely during a collapse. Series compensation is unaffected by voltage and its effect actually strengthens as the current rises. Decisive advantage: series, and the reason SVCs and STATCOMs were developed — a STATCOM's output falls only as \(V\), not \(V^2\).
Under fault. Series must carry the full fault current, requiring MOV or spark-gap bypass, and it lowers the impedance so it raises the fault level. Shunt sees a collapsed voltage and delivers almost nothing — harmless. Advantage: shunt.
On losses. Series leaves the current unchanged, so losses are unchanged. Shunt reduces the current and hence the losses — 19% in Problem 9. Advantage: shunt.
Resonance. Series introduces an \(LC\) resonance in the current path — subsynchronous resonance, ferroresonance. Shunt introduces a parallel resonance with the source inductance, which matters for harmonics but is far less hazardous. Advantage: shunt.
Insulation. Series sits at line potential and must be insulated to it, on a platform. Shunt sits at bus potential with one terminal earthed. Advantage: shunt.
Conclusion: series for power transfer and stability on long inductive lines; shunt for voltage support and loss reduction at load centres. Real systems use both, and the choice is made by which of the above columns the particular problem falls in.C3. Problem 16 found the true maximum power to be 58% of the lossless estimate. Derive the general correction factor, and establish when the lossless formula of Set 24 may be trusted.
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The two expressions. With equal terminal voltages \(V\):The ratio. Writing \(Z = X/\sin\theta\) and \(\cos\theta = R/Z\):\[ P_{\max} = \frac{3V^2}{Z}(1 - \cos\theta), \qquad P_{\max,\text{lossless}} = \frac{3V^2}{X} \]Expressing in terms of \(n = X/R\), where \(\sin\theta = n/\sqrt{1+n^2}\) and \(\cos\theta = 1/\sqrt{1+n^2}\):\[ \frac{P_{\max}}{P_{\max,\text{lossless}}} = \frac{X}{Z}(1 - \cos\theta) = \sin\theta(1 - \cos\theta) \]Evaluating:\[ \frac{P_{\max}}{P_{\max,\text{lossless}}} = \frac{n}{\sqrt{1+n^2}}\left(1 - \frac{1}{\sqrt{1+n^2}}\right) \]The 0.583 at \(X/R = 2.5\) confirms Problem 16's 58%.\[ \begin{array}{cc} X/R & \text{Ratio} \\ \hline 1 & 0.207 \\ 2.5 & 0.583 \\ 5 & 0.784 \\ 10 & 0.891 \\ 20 & 0.947 \\ 50 & 0.980 \end{array} \]
Asymptotics. For large \(n\), expanding gives ratio \(\approx 1 - 1/n - 1/2n^2\), so the error is approximately \(1/n = R/X\) — a memorable rule: the lossless formula overestimates the power limit by roughly the reciprocal of the X/R ratio.
When it may be trusted:
— \(X/R > 20\) (EHV transmission): error under 5%. Use freely.
— \(X/R = 10\) to 20 (HV transmission): error 5–11%. Acceptable for screening, not for a stability margin.
— \(X/R = 3\) to 10 (sub-transmission): error 11–50%. Do not use.
— \(X/R < 3\) (distribution): error above 50%. Meaningless.
The wider point: the entire equal-area and swing-equation apparatus of Sets 24 to 29 rests on \(P = EV\sin\delta/X\). That is legitimate because transient stability is a transmission phenomenon, where \(X/R\) is large. Applying the same machinery to a distribution network — as is sometimes attempted for distributed generation studies — is not a small approximation but a category error.
Multiple-Choice Questions
MCQ 1. A series capacitor improves regulation by reducing:
(a) \(\%R\) (b) \(\%X\) (c) \(\sin\phi\) (d) the load currentShow answer
(b). It subtracts from the line reactance and leaves the current unchanged — Problem 5.MCQ 2. A shunt capacitor improves regulation by reducing:
(a) \(\%R\) (b) \(\%X\) (c) \(\sin\phi\) and the current (d) the line lengthShow answer
(c). Supplying the load's vars locally lowers both the angle and the current — Problem 7.MCQ 3. The lowest regulation obtainable by compensation alone is:
(a) zero (b) \(\%R\cos\phi\) (c) \(\%X\sin\phi\) (d) half the originalShow answer
(b). Neither method can change \(I\cos\phi\), which is fixed by the real power — Problems 6, 8 and Challenge C1.MCQ 4. To correct a load of \(P\) from \(\phi_1\) to \(\phi_2\) requires a capacitor of:
(a) \(P(\cos\phi_1 - \cos\phi_2)\) (b) \(P(\tan\phi_1 - \tan\phi_2)\) (c) \(P(\sin\phi_1 - \sin\phi_2)\) (d) \(P\tan\phi_2\)Show answer
(b). The capacitor supplies the difference in reactive demand at constant real power.MCQ 5. A tap-changing transformer improves:
(a) the receiving voltage only (b) the losses (c) the power factor (d) all threeShow answer
(a). The current is unchanged, so losses, reactive absorption and power factor are all untouched — Problem 11.MCQ 6. On a distribution feeder with \(X/R < 1\), reactive compensation for voltage support is:
(a) highly effective (b) largely ineffective (c) dangerous (d) mandatoryShow answer
(b). The resistive term dominates and neither method touches it — Problems 12 and 13.MCQ 7. A series capacitor's reactive output varies as:
(a) \(V^2\) (b) \(I^2\) (c) \(V\) (d) it is constantShow answer
(b) \(3I^2X_C\) — self-regulating, acting hardest at heavy load. A shunt capacitor varies as \(V^2\).MCQ 8. Series compensation is normally limited to about 70% because of:
(a) cost (b) subsynchronous resonance (c) insulation (d) lossesShow answer
(b). The series resonance can couple to turbine-generator torsional modes — Problem 19.MCQ 9. The maximum power of a short line occurs at a load angle:
(a) \(\delta = 90^\circ\) (b) \(\delta = \theta\), the impedance angle (c) \(\delta = 0\) (d) \(\delta = 45^\circ\)Show answer
(b). Option (a) is the lossless special case, where \(\theta \to 90^\circ\) — Problem 16.MCQ 10. Neglecting resistance overestimates the maximum power by roughly:
(a) \(R/X\) (b) \(X/R\) (c) \((R/X)^2\) (d) it underestimates itShow answer
(a). About 5% at \(X/R = 20\) and 42% at \(X/R = 2.5\) — Challenge C3.MCQ 11. Series compensation raises the power limit substantially only when:
(a) the load is leading (b) \(X \gg R\) (c) the line is short (d) alwaysShow answer
(b). Problem 18 found +89% at \(X/R = 20\) but only +0.5% at \(X/R = 2.5\).MCQ 12. The chief weakness of shunt capacitors during a voltage collapse is that their output:
(a) rises uncontrollably (b) falls as \(V^2\) (c) becomes inductive (d) is unaffectedShow answer
(b). They withdraw support exactly when it is most needed — the reason SVCs and STATCOMs exist.
Key Formulas
| Quantity | Relation | Notes |
|---|---|---|
| Regulation, compact | \(\%R\cos\phi \pm \%X\sin\phi\) | \(+\) lagging, \(-\) leading |
| Percentage impedance | \(\%R = IR/V_R\), \(\%X = IX/V_R\) | Per-unit on the load's own base |
| Resistive floor | \(\%R\cos\phi\) | Unbeatable by any compensation |
| Series compensation | \(X_{\text{net}} = X_L(1-k)\), \(k = X_C/X_L\) | Practical \(k \le 0.7\) |
| Series capacitor rating | \(Q_C = 3I^2X_C\) | Carries load and fault current |
| Shunt capacitor sizing | \(Q_C = P(\tan\phi_1 - \tan\phi_2)\) | At constant real power |
| Shunt capacitor output | \(Q_C = V^2/X_C\) | Falls as \(V^2\) — its weakness |
| Loss ratio after correction | \((\cos\phi_1/\cos\phi_2)^2\) | Shunt only; series leaves \(I\) unchanged |
| Removable fraction | \(1/[1 + (R/X)\cot\phi]\) | Decides whether to compensate |
| Tap ratio | \(t = V_{S,\text{required}}/V_{S,\text{nominal}}\) | Corrects voltage only |
| Power transfer | \(P_R = \frac{V_SV_R}{Z}\cos(\theta-\delta) - \frac{V_R^2}{Z}\cos\theta\) | \(\cos\theta = R/Z\) |
| Maximum power | \(P_{\max} = \frac{V_R^2}{Z}(1-\cos\theta)\) at \(\delta=\theta\) | Per phase, equal terminal voltages |
| Lossless limit | \(P_{\max} = V_SV_R/X\) | Overestimates by about \(R/X\) |
| Correction factor | \(\sin\theta(1-\cos\theta)\) | 0.58 at \(X/R = 2.5\); 0.95 at 20 |
Common Mistakes
Forgetting to recompute \(\%R\) and \(\%X\) after shunt correction. The current changes, so both percentages change — Problem 8.
Expecting series compensation to reduce losses. The current is unchanged; only shunt correction reduces it — Problem 10.
Expecting compensation to reach zero regulation. The \(\%R\cos\phi\) floor is untouchable — Challenge C1.
Sizing a shunt capacitor from \(\cos\phi\) rather than \(\tan\phi\). The formula is \(P(\tan\phi_1 - \tan\phi_2)\).
Rating a series capacitor on the load rather than the current. It is \(3I^2X_C\), and it must also survive the fault current.
Applying reactive compensation to a low-\(X/R\) feeder. Problem 12 found it removing only 29% of the regulation.
Assuming a tap changer solves the underlying problem. It corrects the voltage and nothing else, and can worsen the upstream var balance — Problem 11.
Using \(P_{\max} = V_SV_R/X\) on a line with low \(X/R\). A 42% overestimate at \(X/R = 2.5\) — Problem 16.
Taking \(\delta = 90^\circ\) at maximum power. It is \(\delta = \theta\); the 90° is the lossless special case.
Confusing \(\theta\), the impedance angle, with \(\phi\), the load angle. \(\cos\theta = R/Z\) is a property of the line alone.
Leaving a full capacitor bank connected at light load. The leading current raises the receiving voltage; banks must be switched in blocks — Problem 20.
Compensating beyond 70% without considering resonance. Subsynchronous resonance has destroyed turbine shafts — Problem 19.
The short line has now been analysed and remedied. Its regulation is understood, its compensation options compared, and its power limit computed — all from a single series impedance and the assumption that the current is the same at both ends.
That assumption is what Set 11 abandons. Restoring the shunt admittance as a lumped element at the ends gives the nominal-\(\pi\) and nominal-T models, in which \(A \ne 1\) and the sending and receiving currents differ. The immediate consequence is that the no-load receiving voltage is no longer the sending voltage — the Ferranti effect appears, and with it the whole reactive behaviour of a long line that the short model could not represent. Set 12 then shows that even the nominal-\(\pi\) is an approximation, and derives the exact solution from the wave equation.