Medium Transmission Lines — Solved Problems

Set 11 — Medium Transmission Lines

Twenty worked problems on the model that puts back what Set 9 discarded. Lumping the line's total shunt admittance at one or both ends turns the line from a series impedance into a genuine two-port, described by four constants rather than one. Two consequences follow immediately: the sending and receiving currents are no longer equal, and \(A \ne 1\) — which is precisely the Ferranti effect, invisible to the short-line model and unavoidable here.

Textbook Chapter 11 · 20 solved · 12 practice · 3 challenge · 12 MCQs

i Method Recap

The two-port description. \(\mathbf{V}_S = A\mathbf{V}_R + B\mathbf{I}_R\) and \(\mathbf{I}_S = C\mathbf{V}_R + D\mathbf{I}_R\). All four constants are complex; \(A\) and \(D\) are dimensionless, \(B\) is an impedance and \(C\) an admittance.
Nominal \(\pi\): the whole series impedance in the middle, half the shunt admittance at each end. \(A = D = 1 + \frac{ZY}{2}\), \(B = Z\), \(C = Y\left(1 + \frac{ZY}{4}\right)\).
Nominal T: the whole shunt admittance in the middle, half the series impedance each side. \(A = D = 1 + \frac{ZY}{2}\), \(B = Z\left(1 + \frac{ZY}{4}\right)\), \(C = Y\). The two models share \(A\) and differ in \(B\) and \(C\).
Reciprocity. \(AD - BC = 1\) for every passive line, and \(A = D\) for every symmetrical one. Checking both is the fastest way to catch an arithmetic error in a set of constants.
Regulation comes from \(A\), not from \(B\). On no load \(\mathbf{I}_R = 0\), so \(V_{R,\text{NL}} = V_S/|A|\) and \(\%\text{reg} = \dfrac{V_S/|A| - V_{R,\text{FL}}}{V_{R,\text{FL}}}\times100\).
The Ferranti effect is \(|A| < 1\). Since \(ZY\) is negative-real-dominated, \(1 + ZY/2\) has magnitude below unity, so an unloaded line's receiving voltage exceeds its sending voltage by \(1/|A|\).
Cascade networks multiply. Two two-ports in series have the matrix product of their ABCD matrices, in the order the power flows. Lines in parallel add their \(1/B\) terms — but the general result is easier through \(Y\)-parameters.

VideoWalkthrough

Problem 1Exam levelNominal-π

A 220 kV, 200 km three-phase line has \(R = 0.16\ \Omega\)/km, \(X_L = 0.40\ \Omega\)/km and \(B = 3.0\times10^{-6}\) S/km. Find its ABCD constants on the nominal-\(\pi\) model. This line recurs throughout the set.

Solution

Total series impedance and shunt admittance:

\[ \mathbf{Z} = (0.16 + j0.40)(200) = 32 + j80\ \Omega = 86.16\angle68.20^\circ\ \Omega \]

\[ \mathbf{Y} = j(3.0\times10^{-6})(200) = j6.0\times10^{-4}\ \text{S} \]

The product \(\mathbf{Z}\mathbf{Y}\) appears in every constant, so compute it once:

\[ \mathbf{Z}\mathbf{Y} = (32 + j80)(j6.0\times10^{-4}) = j0.0192 + j^{2}0.048 = -0.048 + j0.0192 \]

Note that it is predominantly negative real — which is what will make \(|A| < 1\) and produce the Ferranti effect of Problem 8.

The constants. For the nominal-\(\pi\), with half the shunt admittance at each end:

\[ A = D = 1 + \frac{\mathbf{Z}\mathbf{Y}}{2} = 1 + \frac{-0.048 + j0.0192}{2} = 0.976 + j0.0096 \]

\[ = 0.9760\angle0.564^\circ \]

\(B\) is simply the whole series impedance, since it sits between the two shunt branches:

\[ B = \mathbf{Z} = 32 + j80 = 86.16\angle68.20^\circ\ \Omega \]

And \(C\):

\[ C = \mathbf{Y}\left(1 + \frac{\mathbf{Z}\mathbf{Y}}{4}\right) = j6.0\times10^{-4}\left(1 + \frac{-0.048 + j0.0192}{4}\right) \]

\[ = j6.0\times10^{-4}(0.988 + j0.0048) = -2.88\times10^{-6} + j5.928\times10^{-4} \]

\[ = 5.928\times10^{-4}\angle90.28^\circ\ \text{S} \]

Four complex numbers now stand for a two-hundred-kilometre line, and nothing else about it will ever be needed. Every later calculation — regulation, efficiency, power transfer, the cascade with a transformer — uses only these. That is the point of a two-port description: it separates what the line is from what is done with it. Note that \(A\) is very nearly 1 and \(C\) very nearly \(\mathbf{Y}\), which is the sense in which a 200 km line is still "nearly short".

Answer\(A = D = 0.9760\angle0.56^\circ\), \(B = 86.16\angle68.20^\circ\ \Omega\), \(C = 5.928\times10^{-4}\angle90.28^\circ\) S

Problem 2Warm-upNominal-T

Find the ABCD constants of the same line on the nominal-T model, and compare them with the nominal-\(\pi\) values.

Solution

The nominal-T places the whole shunt admittance in the middle with half the series impedance on each side. \(A\) is unchanged:

\[ A = D = 1 + \frac{\mathbf{Z}\mathbf{Y}}{2} = 0.976 + j0.0096 = 0.9760\angle0.564^\circ \]

\(C\) is now simply the whole shunt admittance:

\[ C = \mathbf{Y} = j6.0\times10^{-4} = 6.0\times10^{-4}\angle90^\circ\ \text{S} \]

\(B\) carries the correction that \(C\) carried in the \(\pi\) model:

\[ B = \mathbf{Z}\left(1 + \frac{\mathbf{Z}\mathbf{Y}}{4}\right) = (32 + j80)(0.988 + j0.0048) \]

\[ = 31.616 + j0.1536 + j79.04 - 0.384 = 31.23 + j79.19 = 85.13\angle68.48^\circ\ \Omega \]

Comparing the two models:

\[ \begin{array}{lcc} & \text{Nominal-}\pi & \text{Nominal-T} \\ \hline A = D & 0.9760\angle0.56^\circ & 0.9760\angle0.56^\circ \\ B\ (\Omega) & 86.16\angle68.20^\circ & 85.13\angle68.48^\circ \\ C\ (\text{S}) & 5.928\times10^{-4}\angle90.28^\circ & 6.000\times10^{-4}\angle90^\circ \end{array} \]

They differ by 1.2% in \(B\) and 1.2% in \(C\), and not at all in \(A\).

Both models are approximations to the same line and neither is more correct. They lump a distributed admittance in different places, so they must disagree; that they agree exactly in \(A\) is a consequence of both being symmetrical networks with the same total \(\mathbf{Z}\mathbf{Y}\). Problem 17 compares both against the exact solution of Set 12 and finds the \(\pi\) slightly better in \(C\) and the T slightly better in \(B\) — a difference too small to matter, which is why the \(\pi\) is used almost universally, its shunt branches slotting naturally into a \(Y\)-bus.

Answer\(A = D = 0.9760\angle0.56^\circ\), \(B = 85.13\angle68.48^\circ\ \Omega\), \(C = 6.0\times10^{-4}\angle90^\circ\) S

Problem 3Warm-upReciprocity

Verify that \(AD - BC = 1\) for the nominal-\(\pi\) constants of Problem 1, and prove it holds for any nominal-\(\pi\).

Solution

Since \(A = D\), the condition is \(A^2 - BC = 1\). Squaring \(A\):

\[ A^{2} = (0.976 + j0.0096)^{2} = 0.952576 + j0.0187392 - 0.0000922 = 0.952484 + j0.018739 \]

Forming \(BC\):

\[ BC = (32 + j80)(-2.88\times10^{-6} + j5.928\times10^{-4}) \]

\[ = -9.216\times10^{-5} + j0.0189696 - j2.304\times10^{-4} - 0.047424 = -0.047516 + j0.018739 \]

Subtracting:

\[ A^{2} - BC = (0.952484 + 0.047516) + j(0.018739 - 0.018739) = 1.000 + j0\ \checkmark \]

The general proof. Substituting the nominal-\(\pi\) expressions:

\[ A^{2} - BC = \left(1 + \frac{ZY}{2}\right)^{2} - Z\,Y\left(1 + \frac{ZY}{4}\right) \]

\[ = 1 + ZY + \frac{(ZY)^{2}}{4} - ZY - \frac{(ZY)^{2}}{4} = 1 \]

Exactly, for any \(Z\) and \(Y\) — the terms cancel identically rather than approximately.

Reciprocity is a theorem about passive bilateral networks, not a property of this particular line. It holds for the nominal-T, for the exact hyperbolic solution, for a transformer, and for any cascade of them — which makes it the single most valuable check available. If a set of published ABCD constants fails \(AD - BC = 1\), either they contain a misprint or the network contains something non-reciprocal, and both are worth knowing before the constants are used. Note that the imaginary parts cancelled exactly here, which is a strong indication the arithmetic is right.

Answer\(A^2 - BC = 1.000 + j0\); the identity holds exactly for any \(Z\) and \(Y\)

Problem 4Exam levelSending-End Voltage

The line of Problem 1 delivers 100 MW at 220 kV, 0.9 power factor lagging. Find the sending-end voltage.

Solution

Receiving-end voltage per phase, taken as reference, and the load current:

\[ \mathbf{V}_R = \frac{220\times10^{3}}{\sqrt3} = 127\,017\angle0^\circ\ \text{V} \]

\[ I_R = \frac{100\times10^{6}}{\sqrt3(220\times10^{3})(0.9)} = 291.6\ \text{A} \]

\[ \mathbf{I}_R = 291.6\angle-25.84^\circ = 262.4 - j127.1\ \text{A} \]

The two terms of \(\mathbf{V}_S = A\mathbf{V}_R + B\mathbf{I}_R\):

\[ A\mathbf{V}_R = (0.976 + j0.0096)(127\,017) = 123\,969 + j1219 \]

\[ B\mathbf{I}_R = (32 + j80)(262.4 - j127.1) = 18\,565 + j16\,925 \]

Adding:

\[ \mathbf{V}_S = 142\,534 + j18\,144\ \text{V} \]

\[ |\mathbf{V}_S| = \sqrt{(142\,534)^{2} + (18\,144)^{2}} = 143\,685\ \text{V per phase} \]

As a line voltage, and its angle:

\[ V_{S,\text{line}} = \sqrt3(143\,685) = 248.87\ \text{kV}, \qquad \delta = \tan^{-1}\frac{18\,144}{142\,534} = 7.26^\circ \]

The \(A\mathbf{V}_R\) term is not \(\mathbf{V}_R\), and that is the whole difference from Set 9. Because \(|A| = 0.976\), the receiving voltage is scaled down before the impedance drop is added — the shunt admittance at the receiving end has diverted some current, so less flows through \(B\) than the load draws. The short-line model would have given \(\mathbf{V}_S = \mathbf{V}_R + \mathbf{I}\mathbf{Z} = 145\,582 + j18\,144\), or 254.1 kV — an overestimate of 2.1%, which is precisely the error the medium model exists to remove.

Answer\(V_S = 248.87\) kV at \(\delta = 7.26^\circ\)

Problem 5Exam levelSending-End Current

Find the sending-end current and power factor of the line of Problem 4.

Solution

The second two-port relation:

\[ \mathbf{I}_S = C\mathbf{V}_R + D\mathbf{I}_R \]

The charging term:

\[ C\mathbf{V}_R = (-2.88\times10^{-6} + j5.928\times10^{-4})(127\,017) = -0.366 + j75.30\ \text{A} \]

Almost purely imaginary and positive — a leading current of 75 A, which is the line's own charging current.

The load term:

\[ D\mathbf{I}_R = (0.976 + j0.0096)(262.4 - j127.1) = 257.3 - j121.5\ \text{A} \]

Adding:

\[ \mathbf{I}_S = 256.9 - j46.2 = 261.1\angle-10.20^\circ\ \text{A} \]

The sending-end power factor is the cosine of the angle between \(\mathbf{V}_S\) and \(\mathbf{I}_S\):

\[ \phi_S = 7.26^\circ - (-10.20^\circ) = 17.46^\circ \]

\[ \cos\phi_S = 0.954\ \text{lagging} \]

The sending-end power factor is 0.954 where the load's is 0.900 — it has improved, and the short-line model could never have predicted that. In Set 9 the power factor always worsened along the line, because the reactance absorbed vars. Here the shunt capacitance generates 75 A of leading current, which more than offsets the reactive absorption. The line is supplying part of its own load's reactive demand — the same phenomenon that Set 7 called operating below surge impedance loading, seen here in the currents rather than the powers.

Answer\(\mathbf{I}_S = 261.1\angle-10.20^\circ\) A, \(\cos\phi_S = 0.954\) lagging

Problem 6Exam levelRegulation

Find the voltage regulation of the line of Problem 4, and explain why \(A\) rather than \(B\) determines the no-load voltage.

Solution

On no load \(\mathbf{I}_R = 0\), so the two-port relation collapses to

\[ \mathbf{V}_S = A\mathbf{V}_R \quad\Rightarrow\quad V_{R,\text{NL}} = \frac{V_S}{|A|} \]

\(B\) multiplies \(\mathbf{I}_R\) and therefore disappears entirely — the series impedance carries no current when the far end is open.

With \(V_S = 143\,685\) V per phase held constant:

\[ V_{R,\text{NL}} = \frac{143\,685}{0.9760} = 147\,211\ \text{V per phase} \]

Regulation:

\[ \%\text{reg} = \frac{147\,211 - 127\,017}{127\,017}\times100 = \frac{20\,194}{127\,017}\times100 = 15.90\% \]

The two contributions are worth separating. Removing the load raises the receiving voltage from 127.0 kV to 147.2 kV, of which:

\[ \begin{array}{ll} \text{Loss of the } IZ \text{ drop} & 143.7 - 127.0 = 16.7\ \text{kV} \\ \text{Ferranti rise } (1/|A| - 1) & 147.2 - 143.7 = 3.5\ \text{kV} \end{array} \]

The short-line model would have given 13.1% here, understating the regulation by nearly three points. It sets \(A = 1\) and therefore predicts \(V_{R,\text{NL}} = V_S\), missing the Ferranti rise entirely. On a 200 km line that omission is 2.5 percentage points of regulation; on a 400 km line it would be over ten. Since regulation is usually the binding design constraint, this is the practical reason the medium model must be used beyond about 80 km.

Answer\(V_{R,\text{NL}} = 147.2\) kV/phase, regulation \(= 15.90\%\)

Problem 7Warm-upEfficiency

Find the sending-end power and the transmission efficiency of the line of Problems 4 and 5.

Solution

The sending-end power must be computed from the sending-end quantities, since \(\mathbf{I}_S \ne \mathbf{I}_R\) in this model:

\[ P_S = 3V_SI_S\cos\phi_S = 3(143\,685)(261.1)(0.954) \]

\[ = 1.0737\times10^{8}\ \text{W} = 107.37\ \text{MW} \]

Transmission efficiency:

\[ \eta = \frac{P_R}{P_S}\times100 = \frac{100}{107.37}\times100 = 93.13\% \]

The loss is 7.37 MW. Checking it against \(3I^2R\) with a current somewhere between the two ends — the mean of 261 and 292 is 276 A:

\[ 3(276)^{2}(32) = 7.31\ \text{MW}\ \checkmark \]

Agreeing to within 1%, which is as close as such an estimate can be expected to come.

The current varies along the line, so \(3I^2R\) has no single correct value of \(I\) to use. This is the first place where the medium model's abandonment of a single line current has a practical consequence: efficiency must be found as the difference of two end powers, not from a loss formula. The exact treatment of Set 12 integrates \(I^2R\) along the line properly; the nominal-\(\pi\)'s answer of 93.13% is a lumped approximation to that integral, and a good one at this length.

Answer\(P_S = 107.37\) MW, loss 7.37 MW, \(\eta = 93.13\%\)

Problem 8Exam levelFerranti Effect

The line of Problem 1 is energised from 220 kV with its far end open. Find the receiving-end voltage and explain the mechanism.

Solution

With \(\mathbf{I}_R = 0\):

\[ V_R = \frac{V_S}{|A|} = \frac{220}{0.9760} = 225.4\ \text{kV} \]

\[ \text{Rise} = \frac{1}{0.9760} - 1 = 2.45\% \]

The mechanism. The open-circuited line still draws charging current through its shunt capacitance. That current is leading, and it flows through the line's series inductance from the sending end.

A leading current through an inductance produces a voltage drop that opposes the applied voltage's direction of decrease — equivalently, \(\mathbf{I}X_L\) with \(\mathbf{I}\) leading by 90° adds to \(\mathbf{V}\) rather than subtracting:

\[ \mathbf{V}_R = \mathbf{V}_S - \mathbf{I}_c(R + jX_L), \qquad \mathbf{I}_c = jB\mathbf{V}_R \]

\[ \Rightarrow\quad \mathbf{V}_R(1 - BX_L + jBR) = \mathbf{V}_S \]

The term \(-BX_L\) is real and negative, so the bracket has magnitude below one and \(V_R > V_S\).

Recognising the bracket: with \(Y = jB\) and \(Z = R + jX_L\), it is \(1 + ZY\) — the same negative-real-dominated product found in Problem 1, and the reason \(|A| < 1\).

Named for Sebastian Ferranti, who met it in 1890 on the Deptford cables and found the far end at a higher voltage than the generator. The rise grows as the square of the line length — Problem 17 tabulates it — and reaches 10% at 400 km and far more beyond. It is why a long EHV line must never be energised from one end without shunt reactors, why line-charging breakers have a special duty rating, and why the reactive compensation of Set 6 Problem 19 is fitted before commissioning rather than after.

Answer\(V_R = 225.4\) kV against a sending 220 kV — a rise of 2.45%

Problem 9Warm-upNo-Load Current

Find the current the line of Problem 1 draws from the supply when energised on open circuit, and the reactive power it generates.

Solution

With \(\mathbf{I}_R = 0\) the second relation gives

\[ \mathbf{I}_S = C\mathbf{V}_R \]

and \(\mathbf{V}_R\) is now the raised voltage of Problem 8, not the nominal one.

Per phase, \(V_R = 225.4/\sqrt3 = 130\,135\) V:

\[ I_S = |C|V_R = (5.928\times10^{-4})(130\,135) = 77.1\ \text{A} \]

Leading by almost exactly 90°, since \(\angle C = 90.28^\circ\).

Reactive power drawn from the supply — negative, meaning generated:

\[ Q_S = \sqrt3\,V_{S,\text{line}}I_S = \sqrt3(220\times10^{3})(77.1) = 29.4\ \text{MVAr generated} \]

Cross-checking against the shunt admittance directly:

\[ Q = V_L^{2}B = (225.4\times10^{3})^{2}(6.0\times10^{-4}) = 30.5\ \text{MVAr} \]

The small difference is the reactive power absorbed by the series reactance carrying the charging current — \(3I^2X_L = 3(77)^2(80) = 1.4\) MVAr, which accounts for it exactly.

Nearly 30 MVAr flowing into a line that is delivering nothing. The generator at the sending end must absorb it, which for a synchronous machine means running underexcited — a condition limited by its own stability, and one that a single machine can sustain only up to a point. This is the operational reason long lines are energised with reactors already connected: not to protect the line, but to protect the machine that has to absorb what the line generates.

Answer\(I_S = 77.1\) A leading, generating about 29.4 MVAr

Problem 10Warm-upRegulation from A

A 240 kV line has \(A = 0.94\angle1^\circ\) and its sending-end voltage is held at 240 kV while it delivers full load at 220 kV. Find the voltage regulation.

Solution

The no-load receiving voltage:

\[ V_{R,\text{NL}} = \frac{V_S}{|A|} = \frac{240}{0.94} = 255.3\ \text{kV} \]

Regulation against the full-load value of 220 kV:

\[ \%\text{reg} = \frac{255.3 - 220}{220}\times100 = \frac{35.3}{220}\times100 = 16.0\% \]

Notice that neither \(B\) nor \(C\) nor the load current was needed. The full-load receiving voltage was given, and the no-load one follows from \(A\) alone.

Decomposing the 16.0%:

\[ \begin{array}{ll} \text{From the load drop } (240 \to 220) & \frac{240-220}{220} = 9.1\% \\ \text{From the Ferranti rise } (1/0.94) & 6.9\% \end{array} \]

Nearly half the regulation of this line has nothing to do with the load at all. An \(|A|\) of 0.94 corresponds to roughly a 300 km line, and its 6.4% Ferranti rise is a fixed property of the geometry that no amount of load management will alter. Only shunt reactors change \(|A|\) — by adding inductive admittance that partly cancels the line's capacitive \(Y\) and drives \(|A|\) back towards unity. That is the deeper reason compensation is mandatory on long lines rather than merely economic.

Answer\(V_{R,\text{NL}} = 255.3\) kV, regulation \(= 16.0\%\)

Problem 11Challenge-liteπ against T

Compare the nominal-\(\pi\) and nominal-T models of Problems 1 and 2 against the exact constants \(A = 0.9761\angle0.559^\circ\), \(B = 85.48\angle68.38^\circ\ \Omega\), \(C = 5.952\times10^{-4}\angle90.18^\circ\) S, and state which model is preferable.

Solution

Collecting all three sets:

\[ \begin{array}{lccc} & \text{Exact} & \text{Nominal-}\pi & \text{Nominal-T} \\ \hline |A| & 0.97613 & 0.97605 & 0.97605 \\ |B|\ (\Omega) & 85.48 & 86.16 & 85.13 \\ |C|\ (\times10^{-4}\ \text{S}) & 5.952 & 5.928 & 6.000 \end{array} \]

The errors:

\[ \begin{array}{lcc} & \text{Nominal-}\pi & \text{Nominal-T} \\ \hline A & 0.011\% & 0.011\% \\ B & +0.80\% & -0.41\% \\ C & -0.40\% & +0.81\% \end{array} \]

Both models err by well under 1% at this length, and they err in opposite directions: the \(\pi\) overstates \(B\) and understates \(C\), the T does the reverse. Neither is systematically more accurate.

Which is preferable, and why it is the \(\pi\). The reason has nothing to do with accuracy:

\[ \begin{array}{ll} \text{Bus-admittance assembly} & \text{The } \pi\text{'s shunt branches attach directly to buses} \\ \text{No internal node} & \text{The T introduces a fictitious mid-point node} \\ \text{Universal convention} & \text{Every load-flow program uses } \pi \end{array} \]

The nominal-T's mid-point node would have to be carried as an extra bus in a load flow, enlarging the admittance matrix for no benefit. Since Set 16 builds the \(Y\)-bus by inspection from \(\pi\) equivalents, the choice is settled there.

Two models of comparable accuracy, and the winner is decided by data structure rather than physics. This is a recurring pattern in power system analysis: the exact hyperbolic solution of Set 12 is more accurate than either, yet it too is converted to an equivalent-\(\pi\) before being used, precisely so that it slots into the same \(Y\)-bus machinery. The \(\pi\) is not the best model of a line; it is the shape that every other model gets converted into.

AnswerBoth err under 1% and in opposite directions; the \(\pi\) is preferred because it introduces no internal node

Problem 12Exam levelCascade

The 200 km line of Problem 1 is modelled as two 100 km nominal-\(\pi\) sections in cascade. Find the overall constants and compare with the single-section model.

Solution

Each half-section has half the impedance and half the admittance:

\[ \mathbf{Z}_1 = 16 + j40\ \Omega, \qquad \mathbf{Y}_1 = j3.0\times10^{-4}\ \text{S}, \qquad \mathbf{Z}_1\mathbf{Y}_1 = -0.012 + j0.0048 \]

\[ A_1 = D_1 = 0.994 + j0.0024, \quad B_1 = 16 + j40, \quad C_1 = 2.988\times10^{-4}\angle90.14^\circ \]

Cascade rule. Two-ports in series multiply as matrices, in the order the power flows:

\[ \begin{bmatrix} A & B \\ C & D \end{bmatrix} = \begin{bmatrix} A_1 & B_1 \\ C_1 & D_1 \end{bmatrix}\begin{bmatrix} A_2 & B_2 \\ C_2 & D_2 \end{bmatrix} \]

With both sections identical:

\[ A = A_1^{2} + B_1C_1, \qquad B = 2A_1B_1, \qquad C = 2A_1C_1, \qquad D = A \]

Evaluating:

\[ A = 0.97611\angle0.560^\circ, \qquad B = 85.65\angle68.34^\circ\ \Omega, \qquad C = 5.946\times10^{-4}\angle90.21^\circ\ \text{S} \]

Against the single-section values, and the exact ones from Problem 11:

\[ \begin{array}{lccc} & \text{Exact} & \text{One } \pi & \text{Two } \pi \\ \hline |A| & 0.97613 & 0.97605 & 0.97611 \\ |B|\ (\Omega) & 85.48 & 86.16 & 85.65 \\ |C|\ (\times10^{-4}) & 5.952 & 5.928 & 5.946 \end{array} \]

Splitting the line in two has cut the error in \(B\) from 0.80% to 0.20% — a factor of four for a doubling of sections.

The error falls as the square of the section length, so cascading converges on the exact solution quadratically. Four sections would give 0.05%, eight 0.012%. This is the practical route when the exact hyperbolic functions are unavailable, and it is exactly how a distributed line is represented in an electromagnetic transient program — as a chain of many short lumped sections, each short enough that its own lumping error is negligible.

Answer\(A = 0.97611\angle0.56^\circ\), \(B = 85.65\angle68.34^\circ\ \Omega\), \(C = 5.946\times10^{-4}\angle90.21^\circ\) S — error in \(B\) cut fourfold

Problem 13Exam levelTransformer in Cascade

A transformer of series impedance \(j15\ \Omega\) referred to the line side is connected at the receiving end of the line of Problem 1. Find the ABCD constants of the combination.

Solution

A series impedance is itself a two-port with the short-line constants of Set 9:

\[ \begin{bmatrix} A_T & B_T \\ C_T & D_T \end{bmatrix} = \begin{bmatrix} 1 & j15 \\ 0 & 1 \end{bmatrix} \]

The line comes first in the direction of power flow, so it is the left factor:

\[ \begin{bmatrix} A & B \\ C & D \end{bmatrix} = \begin{bmatrix} A_L & B_L \\ C_L & D_L \end{bmatrix}\begin{bmatrix} 1 & j15 \\ 0 & 1 \end{bmatrix} \]

Carrying out the multiplication:

\[ A = A_L, \qquad B = A_L(j15) + B_L, \qquad C = C_L, \qquad D = C_L(j15) + D_L \]

Evaluating with the Problem 1 constants:

\[ A = 0.976 + j0.0096 \]

\[ B = (0.976 + j0.0096)(j15) + 32 + j80 = -0.144 + j14.64 + 32 + j80 = 31.86 + j94.64 \]

\[ = 99.86\angle71.39^\circ\ \Omega \]

\[ C = 5.928\times10^{-4}\angle90.28^\circ\ \text{S} \]

\[ D = (-2.88\times10^{-6} + j5.928\times10^{-4})(j15) + 0.976 + j0.0096 = 0.9671 + j0.0096 \]

Note that \(A \ne D\) now — the combination is no longer symmetrical, because the transformer sits at one end only. Reciprocity still holds:

\[ AD - BC = 1.000\ \checkmark \]

Asymmetry is the signature of an unbalanced cascade, and it is diagnostic. \(A = D\) means the network looks the same from both ends; adding a transformer to one end breaks that, and \(D\) falls from 0.976 to 0.967 while \(A\) is untouched. Reciprocity survives because every element is passive and bilateral. If a computed cascade gives \(AD - BC \ne 1\) the multiplication was done in the wrong order or a matrix was transposed — the two commonest errors in this calculation.

Answer\(A = 0.9760\angle0.56^\circ\), \(B = 99.86\angle71.39^\circ\ \Omega\), \(C\) unchanged, \(D = 0.9671\angle0.57^\circ\)

Problem 14Warm-upParallel Lines

Two identical lines with the constants of Problem 1 are connected in parallel between the same two buses. Find the constants of the combination, and verify reciprocity.

Solution

Two identical two-ports in parallel share the same terminal voltages and split the current equally. For each half:

\[ \mathbf{V}_S = A\mathbf{V}_R + B\left(\frac{\mathbf{I}_R}{2}\right), \qquad \frac{\mathbf{I}_S}{2} = C\mathbf{V}_R + D\left(\frac{\mathbf{I}_R}{2}\right) \]

Rewriting in terms of the total currents:

\[ \mathbf{V}_S = A\mathbf{V}_R + \frac{B}{2}\mathbf{I}_R, \qquad \mathbf{I}_S = 2C\mathbf{V}_R + D\mathbf{I}_R \]

So the combination has:

\[ A' = A, \qquad B' = \frac{B}{2}, \qquad C' = 2C, \qquad D' = D \]

\[ A' = 0.9760\angle0.56^\circ, \quad B' = 43.08\angle68.20^\circ\ \Omega, \quad C' = 1.186\times10^{-3}\angle90.28^\circ\ \text{S} \]

Reciprocity:

\[ A'D' - B'C' = AD - \frac{B}{2}(2C) = AD - BC = 1\ \checkmark \]

Note what did not change: \(A\) is identical, so the Ferranti rise of the parallel pair is exactly that of one line. The impedance halved and the charging doubled, and their effects on \(A\) cancel.

Paralleling halves the impedance and doubles the charging, leaving the no-load voltage rise unchanged. That is worth knowing: a second circuit does nothing whatever for the Ferranti problem, though it halves the loaded voltage drop and doubles the power transfer limit. It also doubles the reactive generation, so the shunt reactor requirement doubles — a double-circuit line needs twice the compensation of a single one, for the same 2.45% rise.

Answer\(A' = A\), \(B' = B/2 = 43.08\angle68.20^\circ\ \Omega\), \(C' = 2C\); \(A'D' - B'C' = 1\)

Problem 15Challenge-liteMeasurement

Show how the ABCD constants of a line may be found from open- and short-circuit measurements at the sending end, and evaluate the tests for the line of Problem 1.

Solution

Open-circuit test. With the receiving end open, \(\mathbf{I}_R = 0\):

\[ \mathbf{V}_S = A\mathbf{V}_R, \qquad \mathbf{I}_S = C\mathbf{V}_R \quad\Rightarrow\quad Z_{OC} = \frac{\mathbf{V}_S}{\mathbf{I}_S} = \frac{A}{C} \]

Short-circuit test. With the receiving end shorted, \(\mathbf{V}_R = 0\):

\[ \mathbf{V}_S = B\mathbf{I}_R, \qquad \mathbf{I}_S = D\mathbf{I}_R \quad\Rightarrow\quad Z_{SC} = \frac{B}{D} \]

Evaluating for the Problem 1 line:

\[ Z_{OC} = \frac{0.9760\angle0.56^\circ}{5.928\times10^{-4}\angle90.28^\circ} = 1646\angle-89.72^\circ\ \Omega \]

\[ Z_{SC} = \frac{86.16\angle68.20^\circ}{0.9760\angle0.56^\circ} = 88.28\angle67.64^\circ\ \Omega \]

Recovering the constants. The two tests give two equations; with \(A = D\) and \(AD - BC = 1\) there are enough relations to solve completely:

\[ A = D = \frac{1}{\sqrt{1 - Z_{SC}/Z_{OC}}}, \qquad B = AZ_{SC}, \qquad C = \frac{A}{Z_{OC}} \]

The product also yields the characteristic impedance directly:

\[ Z_c = \sqrt{Z_{OC}Z_{SC}} = \sqrt{(1646)(88.28)}\angle\frac{-89.72 + 67.64}{2} = 381\angle-11.04^\circ\ \Omega \]

Compare the theoretical \(\sqrt{Z/Y} = 379\angle-10.9^\circ\ \Omega\) from Set 7 — agreeing to within the model's own accuracy.

The whole two-port can be characterised without knowing anything about the conductors. Two measurements at one end, and every constant follows — which is how commissioning tests on a real line are actually done, and how a cable of uncertain construction is characterised. The relation \(Z_c = \sqrt{Z_{OC}Z_{SC}}\) is the same result familiar from transmission-line theory at radio frequency, and it holds for any reciprocal symmetrical two-port whatever its internal structure.

Answer\(Z_{OC} = A/C = 1646\angle-89.7^\circ\ \Omega\), \(Z_{SC} = B/D = 88.3\angle67.6^\circ\ \Omega\), \(Z_c = \sqrt{Z_{OC}Z_{SC}} = 381\angle-11.0^\circ\ \Omega\)

Problem 16Warm-upInverse Relations

Derive the receiving-end quantities in terms of the sending-end ones, and use them to find the receiving-end voltage of the line of Problem 1 when 248.87 kV is applied and the sending-end current is \(261.1\angle-10.20^\circ\) A relative to \(\mathbf{V}_S\).

Solution

The two-port relations in matrix form:

\[ \begin{bmatrix} \mathbf{V}_S \\ \mathbf{I}_S \end{bmatrix} = \begin{bmatrix} A & B \\ C & D \end{bmatrix}\begin{bmatrix} \mathbf{V}_R \\ \mathbf{I}_R \end{bmatrix} \]

Inverting the matrix. Its determinant is \(AD - BC = 1\), so the inverse is simply the adjugate:

\[ \begin{bmatrix} \mathbf{V}_R \\ \mathbf{I}_R \end{bmatrix} = \begin{bmatrix} D & -B \\ -C & A \end{bmatrix}\begin{bmatrix} \mathbf{V}_S \\ \mathbf{I}_S \end{bmatrix} \]

\[ \mathbf{V}_R = D\mathbf{V}_S - B\mathbf{I}_S, \qquad \mathbf{I}_R = -C\mathbf{V}_S + A\mathbf{I}_S \]

That the determinant is unity is what makes this so simple. No division is needed anywhere, and the inverse network has the same constants with \(A\) and \(D\) interchanged and the off-diagonal terms negated.

Applying it, working in the frame where \(\mathbf{V}_S = 143\,685\angle0^\circ\) and \(\mathbf{I}_S = 261.1\angle-17.46^\circ\) — the angle between them being the sending-end power factor angle of Problem 5:

\[ D\mathbf{V}_S = (0.976 + j0.0096)(143\,685) = 140\,237 + j1379 \]

\[ B\mathbf{I}_S = (32 + j80)(249.1 - j78.4) = 14\,243 + j17\,419 \]

\[ \mathbf{V}_R = 125\,994 - j16\,040 = 127\,011\angle-7.26^\circ\ \text{V} \]

Recovering the 127 017 V of Problem 4 to within rounding, with the expected phase lag of 7.26°.

The reciprocity condition is not merely a check — it is what makes the network invertible without arithmetic. For a general two-port the inverse carries a factor \(1/(AD-BC)\); because that is exactly one for any passive line, the backward relations are as simple as the forward ones. This matters in load flow, where the direction of solution is not known in advance, and in protection, where a relay at one end must infer conditions at the other.

Answer\(\mathbf{V}_R = D\mathbf{V}_S - B\mathbf{I}_S\), \(\mathbf{I}_R = -C\mathbf{V}_S + A\mathbf{I}_S\); recovering \(V_R = 127.0\) kV/phase

Problem 17Exam levelA against Length

For a line with the per-km parameters of Problem 1, tabulate \(|A|\) and the Ferranti rise for lengths of 50, 100, 200, 300 and 400 km, and derive the scaling law.

Solution

Both \(\mathbf{Z}\) and \(\mathbf{Y}\) scale with length, so \(\mathbf{Z}\mathbf{Y}\) scales with its square:

\[ \mathbf{Z}\mathbf{Y} = zy\,l^{2}, \qquad A = 1 + \frac{zy\,l^{2}}{2} \]

where \(z\) and \(y\) are the per-km values.

Evaluating:

\[ \begin{array}{ccc} l\ (\text{km}) & |A| & \text{Ferranti rise (\%)} \\ \hline 50 & 0.99850 & 0.15 \\ 100 & 0.99400 & 0.60 \\ 200 & 0.97605 & 2.45 \\ 300 & 0.94625 & 5.68 \\ 400 & 0.90482 & 10.52 \end{array} \]

The scaling law. Since \(zy\) is dominated by its negative real part \(-xb\):

\[ |A| \approx 1 - \frac{xb\,l^{2}}{2} = 1 - \frac{(\beta l)^{2}}{2} \]

\[ \text{rise} = \frac{1}{|A|} - 1 \approx \frac{(\beta l)^{2}}{2} \]

recognising \(\beta = \sqrt{xb}\) as the phase constant of Set 7 — so the Ferranti rise is half the square of the electrical length in radians.

Checking at 400 km, where \(\beta l = 400(1.095\times10^{-3}) = 0.438\) rad:

\[ \frac{(0.438)^{2}}{2} = 0.0959 \quad\text{against the tabulated } 10.52\%\ \checkmark \]

The small excess is the next term in the expansion, which the nominal-\(\pi\) only partly captures.

Quadratic growth is what makes long lines qualitatively different rather than merely worse. A doubling of length quadruples the Ferranti rise, so the problem is negligible at 100 km, noticeable at 200, serious at 300 and unmanageable at 400 without compensation. The exact result is \(|A| = \cos(\beta l)\) for a lossless line, which reaches zero at \(\beta l = 90^\circ\) — a quarter wavelength, or 1500 km — where the open-circuit voltage becomes infinite. That singularity is the subject of Set 14's tuned lines, and the quadratic law here is simply its leading term.

AnswerRise grows as \((\beta l)^2/2\): 0.15% at 50 km rising to 10.5% at 400 km

Problem 18Exam levelReactive Balance

Find the reactive power at both ends of the line of Problems 4 and 5, and account for the difference.

Solution

At the receiving end, the load's demand:

\[ Q_R = P_R\tan\phi_R = 100\tan(25.84^\circ) = 48.43\ \text{MVAr} \]

At the sending end, from the computed \(V_S\), \(I_S\) and \(\phi_S\):

\[ Q_S = 3V_SI_S\sin\phi_S = 3(143\,685)(261.1)\sin(17.46^\circ) \]

\[ = 1.1255\times10^{8}(0.3000) = 33.77\ \text{MVAr} \]

The line supplies the difference:

\[ Q_R - Q_S = 48.43 - 33.77 = 14.66\ \text{MVAr generated by the line} \]

Accounting for it. The shunt capacitance generates, and the series reactance absorbs:

\[ Q_{\text{gen}} \approx V_L^{2}B = (230\times10^{3})^{2}(6.0\times10^{-4}) \approx 31.7\ \text{MVAr} \]

\[ Q_{\text{abs}} = 3I^{2}X_L \approx 3(276)^{2}(80) = 18.3\ \text{MVAr} \]

\[ \text{Net} \approx 31.7 - 18.3 = 13.4\ \text{MVAr generated} \]

Agreeing with the 14.66 MVAr to within the crudeness of the average-current and average-voltage estimates.

The line is therefore operating below its surge impedance loading, and it is a net reactive source.

Confirming the SIL prediction of Set 7 from an entirely independent calculation. This line's surge impedance is roughly \(\sqrt{80/(6\times10^{-4}\cdot 200/200)}\)… more directly, \(Z_c = 379\ \Omega\) from Problem 15, giving \(\text{SIL} = 220^2/379 = 128\) MW. At 100 MW the line is at 78% of SIL and must therefore generate vars — which it does, 14.7 MVAr of them. Above 128 MW it would absorb instead. The two frameworks, ABCD and surge impedance, are describing the same physics and agree.

Answer\(Q_R = 48.4\) MVAr, \(Q_S = 33.8\) MVAr — the line generates 14.7 MVAr, being below SIL

Problem 19Challenge-litePower Transfer

Derive the receiving-end power in terms of the ABCD constants, and find the maximum power and the load angle at 100 MW for the line of Problem 1 with both terminals at 220 kV.

Solution

Writing \(A = |A|\angle\alpha\), \(B = |B|\angle\beta\) and taking \(\mathbf{V}_S\) to lead \(\mathbf{V}_R\) by \(\delta\), solve the first relation for the current:

\[ \mathbf{I}_R = \frac{\mathbf{V}_S - A\mathbf{V}_R}{B} \]

The complex power at the receiving end:

\[ \mathbf{S}_R = \mathbf{V}_R\mathbf{I}_R^{*} = \frac{V_SV_R}{|B|}\angle(\beta - \delta) - \frac{|A|V_R^{2}}{|B|}\angle(\beta - \alpha) \]

\[ \boxed{P_R = \frac{V_SV_R}{|B|}\cos(\beta - \delta) - \frac{|A|V_R^{2}}{|B|}\cos(\beta - \alpha)} \]

The same shape as Set 10's short-line result, with \(|B|\) replacing \(Z\) and the second term carrying \(|A|\).

Maximum at \(\delta = \beta\). Per phase with \(V_S = V_R = 127\,017\) V, \(|B| = 86.16\), \(\beta - \alpha = 67.64^\circ\):

\[ \frac{V_R^{2}}{|B|} = \frac{1.6133\times10^{10}}{86.16} = 1.8724\times10^{8} \]

\[ P_{R,\max} = 3(1.8724\times10^{8})\big[1 - 0.9760\cos(67.64^\circ)\big] = 3(1.8724\times10^{8})(0.6288) \]

\[ = 353.1\ \text{MW} \]

Load angle at 100 MW. Per phase \(P_R = 33.33\) MW:

\[ \cos(\beta - \delta) = \frac{33.33\times10^{6} + 1.8724\times10^{8}(0.3712)}{1.8724\times10^{8}} = 0.5493 \]

\[ \beta - \delta = 56.69^\circ \quad\Rightarrow\quad \delta = 68.20 - 56.69 = 11.51^\circ \]

The line delivers 100 MW at 11.5° and could reach 353 MW at 68°, so it operates at 28% of its theoretical limit. That is typical and deliberate: operating practice keeps the angle across a line below about 30°, which here corresponds to roughly 200 MW. The margin exists so that a sudden disturbance — a fault cleared after a few cycles — advances the angle without pushing it past the maximum, which is the equal-area criterion of Set 29. Note also that this 353 MW exceeds the 128 MW surge impedance loading by nearly threefold; SIL is a reactive-balance figure, not a limit.

Answer\(P_{R,\max} = 353.1\) MW at \(\delta = 68.2^\circ\); \(\delta = 11.51^\circ\) at 100 MW

Problem 20Exam levelComplete Study

A 220 kV medium line has \(A = D = 0.936\angle0.98^\circ\), \(B = 142\angle76.4^\circ\ \Omega\) and \(C = (-5.18 + j914)\times10^{-6}\) S. It delivers 50 MW at 0.9 power factor lagging at 220 kV. Find the sending-end voltage and current, the regulation, and check reciprocity.

Solution

Step 1 — the load.

\[ \mathbf{V}_R = \frac{220\times10^{3}}{\sqrt3} = 127\,017\angle0^\circ\ \text{V} \]

\[ I_R = \frac{50\times10^{6}}{\sqrt3(220\times10^{3})(0.9)} = 145.8\ \text{A}, \qquad \mathbf{I}_R = 145.8\angle-25.84^\circ \]

Step 2 — sending-end voltage.

\[ A\mathbf{V}_R = 0.936\angle0.98^\circ(127\,017) = 118\,888\angle0.98^\circ = 118\,871 + j2033 \]

\[ B\mathbf{I}_R = 142\angle76.4^\circ \times 145.8\angle-25.84^\circ = 20\,704\angle50.56^\circ = 13\,155 + j15\,988 \]

\[ \mathbf{V}_S = 132\,026 + j18\,021 = 133\,251\angle7.77^\circ\ \text{V} \]

\[ V_{S,\text{line}} = \sqrt3(133\,251) = 230.8\ \text{kV} \]

Step 3 — sending-end current.

\[ C\mathbf{V}_R = (-5.18 + j914)\times10^{-6}(127\,017) = -0.658 + j116.09 \]

\[ D\mathbf{I}_R = 0.936\angle0.98^\circ \times 145.8\angle-25.84^\circ = 136.5\angle-24.86^\circ = 123.8 - j57.4 \]

\[ \mathbf{I}_S = 123.1 + j58.7 = 136.4\angle25.49^\circ\ \text{A} \]

Step 4 — sending-end power factor. The current now leads the sending voltage:

\[ \phi_S = 7.77^\circ - 25.49^\circ = -17.72^\circ \quad\Rightarrow\quad \cos\phi_S = 0.953\ \text{leading} \]

Step 5 — regulation.

\[ V_{R,\text{NL}} = \frac{133\,251}{0.936} = 142\,362\ \text{V}, \qquad \%\text{reg} = \frac{142\,362 - 127\,017}{127\,017}\times100 = 12.08\% \]

Step 6 — reciprocity check.

\[ A^{2} = 0.876\angle1.96^\circ, \qquad BC = 142\angle76.4^\circ \times 9.14\times10^{-4}\angle90.32^\circ = 0.1298\angle166.7^\circ \]

\[ A^{2} - BC = (0.8755 + j0.0300) - (-0.1263 + j0.0300) = 1.0018 + j0 \approx 1\ \checkmark \]

The sending-end power factor came out leading, and that is the headline of this line. Its charging current of 116 A exceeds the reactive component of the load current, so the generator at the sending end is being asked to absorb vars. With \(|A| = 0.936\) this is a line of roughly 300 km, well below its surge impedance loading at 50 MW, and it exhibits every long-line symptom: 12% regulation dominated by the Ferranti term, leading sending-end current, and a generator running underexcited. Every one of them is invisible to the short-line model, and every one is the reason Set 12 exists.

Answer\(V_S = 230.8\) kV at \(7.77^\circ\), \(\mathbf{I}_S = 136.4\angle25.49^\circ\) A (leading), regulation 12.08%, \(AD-BC = 1\)

Practice

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 \(Z = 40 + j100\ \Omega\) and \(Y = j8\times10^{-4}\) S. Find \(ZY\) and \(A\) for a nominal-\(\pi\).

Show answer
\(ZY = -0.08 + j0.032\); \(A = 1 + ZY/2 = \mathbf{0.960 + j0.016} = 0.9601\angle0.955^\circ\).
P2. Find \(C\) for that nominal-\(\pi\).

Show answer
\(C = Y(1 + ZY/4) = j8\times10^{-4}(0.98 + j0.008) = \mathbf{-6.4\times10^{-6} + j7.84\times10^{-4}}\).
P3. Find the Ferranti rise of that line.

Show answer
\(1/|A| - 1 = 1/0.9601 - 1 = \mathbf{4.16\%}\).
P4. A line has \(A = 0.90\angle1.5^\circ\). Its sending voltage is 400 kV. Find the no-load receiving voltage.

Show answer
\(400/0.90 = \mathbf{444.4}\) kV — a 44 kV rise, far beyond any insulation margin.
P5. Verify \(AD - BC = 1\) for \(A = D = 0.98\angle1^\circ\), \(B = 60\angle70^\circ\), \(C = 6.6\times10^{-4}\angle90^\circ\).

Show answer
\(A^2 = 0.9604\angle2^\circ\); \(BC = 0.0396\angle160^\circ\). \(A^2 - BC = 0.9598 + j0.0335 + 0.0372 - j0.0135 \approx \mathbf{0.997 + j0.020}\) — close enough to 1 given the rounded data.
P6. Two identical lines are in cascade, each with \(A_1, B_1, C_1, D_1\). Write the overall \(B\).

Show answer
\(B = A_1B_1 + B_1D_1 = \mathbf{2A_1B_1}\) when \(A_1 = D_1\).
P7. Two identical lines are in parallel. Write all four overall constants.

Show answer
\(A' = A\), \(B' = B/2\), \(C' = 2C\), \(D' = D\). Note \(A\) is unchanged, so the Ferranti rise is unchanged.
P8. A line's open-circuit impedance is \(2000\angle-88^\circ\ \Omega\) and its short-circuit impedance \(100\angle70^\circ\ \Omega\). Find \(Z_c\).

Show answer
\(Z_c = \sqrt{Z_{OC}Z_{SC}} = \sqrt{200\,000}\angle(-9^\circ) = \mathbf{447\angle-9^\circ}\ \Omega\).
P9. Why does regulation depend on \(A\) and not on \(B\)?

Show answer
On no load \(I_R = 0\), so the \(BI_R\) term vanishes and \(V_{R,NL} = V_S/|A|\) — Problem 6.
P10. A 300 km line has \(\beta = 1.1\times10^{-3}\) rad/km. Estimate its Ferranti rise.

Show answer
\(\beta l = 0.33\) rad; rise \(\approx (\beta l)^2/2 = \mathbf{5.4\%}\) — Problem 17.
P11. Why is the nominal-\(\pi\) preferred to the nominal-T?

Show answer
Not accuracy — both err under 1%. The \(\pi\) introduces no internal node, so its shunt branches attach directly to buses in a \(Y\)-bus — Problem 11.
P12. A line's sending-end current leads its sending-end voltage. What does that indicate?

Show answer
The line's charging current exceeds the load's reactive current — it is operating well below SIL and the generator must absorb vars — Problem 20.

Challenge

Challenge Problems

Each of these needs an idea rather than a formula. Decide what the governing principle is before opening the answer.

C1. Derive the nominal-\(\pi\) constants from the circuit, then show that they are the first two terms of the exact hyperbolic solution, and identify exactly what is neglected.

Show answer
From the circuit. The \(\pi\) has \(Y/2\) at each end and \(Z\) between. Working from the receiving end:
\[ \mathbf{I}_1 = \mathbf{I}_R + \frac{Y}{2}\mathbf{V}_R, \qquad \mathbf{V}_S = \mathbf{V}_R + Z\mathbf{I}_1 = \left(1 + \frac{ZY}{2}\right)\mathbf{V}_R + Z\mathbf{I}_R \]
giving \(A = 1 + ZY/2\) and \(B = Z\). Then
\[ \mathbf{I}_S = \mathbf{I}_1 + \frac{Y}{2}\mathbf{V}_S = Y\left(1 + \frac{ZY}{4}\right)\mathbf{V}_R + \left(1 + \frac{ZY}{2}\right)\mathbf{I}_R \]
giving \(C\) and \(D\).

The exact solution (Set 12) has \(A = \cosh\gamma l\), \(B = Z_c\sinh\gamma l\), \(C = \sinh(\gamma l)/Z_c\), with \(\gamma l = \sqrt{ZY}\) and \(Z_c = \sqrt{Z/Y}\).

Expanding. With \(\theta = \gamma l\):
\[ \cosh\theta = 1 + \frac{\theta^2}{2} + \frac{\theta^4}{24} + \cdots = 1 + \frac{ZY}{2} + \frac{(ZY)^2}{24} + \cdots \]
So \(A_{\pi}\) is exact through \((ZY)^1\) and omits \((ZY)^2/24\).
\[ Z_c\sinh\theta = \sqrt{\frac{Z}{Y}}\left(\theta + \frac{\theta^3}{6} + \cdots\right) = Z\left(1 + \frac{ZY}{6} + \cdots\right) \]
So \(B_{\pi} = Z\) omits \(Z\cdot ZY/6\). The nominal-T's \(B = Z(1 + ZY/4)\) overshoots, having \(1/4\) where \(1/6\) is correct.
\[ \frac{\sinh\theta}{Z_c} = Y\left(1 + \frac{ZY}{6} + \cdots\right) \]
So \(C_{\pi} = Y(1 + ZY/4)\) also has \(1/4\) against the correct \(1/6\).

What is neglected: terms of order \((ZY)^2 = (\gamma l)^4\) in \(A\), and the difference between \(1/4\) and \(1/6\) at order \((\gamma l)^2\) in \(B\) and \(C\). The latter is the larger error and explains the observed pattern of Problem 11: \(A\) is accurate to 0.011% while \(B\) and \(C\) are accurate only to about 0.8%. It also explains why the \(\pi\) and T err in opposite directions — one omits the correction entirely, the other overstates it by \(3/2\).
C2. The Ferranti effect grows as \((\beta l)^2/2\) for a lumped model but as \(\sec(\beta l)\) exactly. Explore the consequences, including what happens as \(\beta l \to 90^\circ\), and set out the practical countermeasures.

Show answer
The exact result. For a lossless line, \(A = \cos(\beta l)\), so
\[ \frac{V_R}{V_S} = \frac{1}{\cos\beta l} = \sec\beta l \]
Expanding: \(\sec x \approx 1 + x^2/2 + 5x^4/24\), whose leading term is exactly the lumped result.

Comparing at increasing length (\(\beta = 1.1\times10^{-3}\) rad/km):
\[ \begin{array}{cccc} l\ (\text{km}) & \beta l\ (\text{rad}) & \text{lumped }(\beta l)^2/2 & \text{exact } \sec\beta l - 1 \\ \hline 200 & 0.22 & 2.4\% & 2.5\% \\ 400 & 0.44 & 9.7\% & 10.9\% \\ 800 & 0.88 & 38.7\% & 59.2\% \\ 1200 & 1.32 & 87.1\% & 306\% \\ 1428 & 1.571 & 123\% & \infty \end{array} \]
The singularity. At \(\beta l = \pi/2\) — a quarter wavelength, about 1430 km at 50 Hz — \(\cos\beta l = 0\) and the open-circuit receiving voltage becomes unbounded. Physically the line is a quarter-wave resonator: the source sees an open circuit transformed to a short, drawing enormous current, and the far end sees a voltage antinode. Real losses keep it finite but the rise is still catastrophic.

Why the lumped model fails there. It has no poles. \(1 + ZY/2\) is a polynomial and cannot vanish for the physically realisable values of \(ZY\), so the nominal-\(\pi\) predicts a finite, and badly understated, rise at all lengths. Beyond about \(\beta l = 0.5\) rad (450 km) the discrepancy exceeds 10% and the lumped model must be abandoned.

Countermeasures:
— Shunt reactors, adding inductive admittance that partly cancels \(Y\), reducing the effective \(\beta\) and pushing the resonance further away. This is the primary remedy and it is designed in from the start.
— Intermediate switching stations, breaking the line into electrically shorter sections that are energised in sequence.
— Energising through a resistor or with pre-insertion resistors in the breaker, damping the transient.
— Sequential energisation with the far end earthed, then released.
— Never energise a long line from one end unloaded and uncompensated — the operational rule that all of the above serve.

The wider point: a lumped model can be wrong not merely by a percentage but by a whole qualitative feature. The resonance exists in the physics and in the exact solution and simply does not appear in the approximation, which is the strongest possible argument for knowing where a model's validity ends.
C3. The ABCD description reduces a line to four complex numbers. Examine what information is thereby lost, when that matters, and what the description cannot represent at all.

Show answer
What is retained. Everything about the terminal behaviour at one frequency, in the steady state, for the positive-sequence network. Any experiment performed only at the two ends is fully predicted.

What is lost — and when it matters:
— The voltage and current profile along the line. ABCD gives the ends only. This matters for insulation coordination, since the maximum voltage on a long compensated line may occur in the middle, and for locating the point of minimum voltage on a loaded feeder.
— Where the losses occur. Total loss is \(P_S - P_R\), but its distribution is invisible. Matters for thermal rating of individual sections.
— The internal structure. A nominal-\(\pi\), a nominal-T and the exact line with the same constants are indistinguishable from outside — the point made in Set 4 Challenge C3 about black boxes generally.

What cannot be represented at all:
— Other frequencies. The constants are computed at one \(\omega\). A harmonic study needs them recomputed, and \(\gamma l\) scales with frequency so a 5th harmonic sees five times the electrical length — a 300 km line is a 1500 km line to the 5th harmonic, and near quarter-wave resonance.
— Transients. Travelling waves, switching surges and lightning are time-domain phenomena; ABCD is a phasor description and says nothing about them.
— Unbalance. These are positive-sequence constants. Negative-sequence uses the same values, but zero-sequence needs an entirely separate set with different \(Z\) and \(Y\) — Set 22.
— Nonlinearity. Corona (Set 8) makes the effective capacitance voltage-dependent; ABCD assumes linearity.
— Mutual coupling to a parallel circuit on the same towers, which requires a coupled multi-port.

The judgement: for what it is designed to do — steady-state, single-frequency, balanced, terminal behaviour — the ABCD description is complete and exact, and its four numbers carry every bit of that information with nothing redundant. The \(AD - BC = 1\) constraint even shows that only three of the four are independent for a reciprocal line. The failures above are all failures of the assumptions, not of the representation, and each is addressed by building a different model rather than by adding constants to this one.

Self-Test

Multiple-Choice Questions

MCQ 1. For a nominal-\(\pi\) line, \(B\) equals:
(a) \(Z\) (b) \(Z(1 + ZY/4)\) (c) \(Y\) (d) \(1 + ZY/2\)

Show answer
(a). Option (b) is the nominal-T's \(B\) — Problems 1 and 2.
MCQ 2. The nominal-\(\pi\) and nominal-T models agree exactly in:
(a) \(A\) only (b) \(B\) only (c) \(C\) only (d) all four

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(a). Both give \(A = D = 1 + ZY/2\); they differ by about 1% in \(B\) and \(C\).
MCQ 3. The condition \(AD - BC = 1\) expresses:
(a) symmetry (b) reciprocity (c) losslessness (d) linearity

Show answer
(b). Symmetry is the separate condition \(A = D\), which the cascade of Problem 13 breaks while reciprocity survives.
MCQ 4. Voltage regulation of a medium line is determined by:
(a) \(B\) (b) \(C\) (c) \(A\) (d) \(D\) only

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(c). \(V_{R,NL} = V_S/|A|\), since \(I_R = 0\) removes the \(B\) term — Problem 6.
MCQ 5. The Ferranti effect arises because:
(a) \(|A| > 1\) (b) \(|A| < 1\) (c) \(B\) is large (d) \(C = 0\)

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(b). \(ZY\) is negative-real-dominated, so \(1 + ZY/2\) has magnitude below unity — Problem 8.
MCQ 6. The Ferranti rise grows with line length approximately as:
(a) \(l\) (b) \(l^2\) (c) \(\sqrt l\) (d) \(\ln l\)

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(b). Rise \(\approx (\beta l)^2/2\) — 0.15% at 50 km rising to 10.5% at 400 — Problem 17.
MCQ 7. Two identical lines in parallel have overall constants:
(a) \(A, 2B, C/2, D\) (b) \(A, B/2, 2C, D\) (c) \(2A, B, C, 2D\) (d) all halved

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(b). \(A\) is unchanged, so the Ferranti rise is unchanged — Problem 14.
MCQ 8. Two two-ports in cascade combine by:
(a) adding the constants (b) multiplying the ABCD matrices (c) averaging (d) adding admittances

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(b), in the order the power flows. Reversing the order gives a different and wrong answer for an asymmetric cascade — Problems 12 and 13.
MCQ 9. The open-circuit impedance measured at the sending end equals:
(a) \(A/C\) (b) \(B/D\) (c) \(B/A\) (d) \(D/C\)

Show answer
(a). With \(I_R = 0\), \(V_S/I_S = AV_R/CV_R = A/C\) — Problem 15.
MCQ 10. The characteristic impedance may be obtained from tests as:
(a) \(Z_{OC}/Z_{SC}\) (b) \(\sqrt{Z_{OC}Z_{SC}}\) (c) \(Z_{OC} - Z_{SC}\) (d) \(Z_{SC}\) alone

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(b). It holds for any reciprocal symmetrical two-port — Problem 15.
MCQ 11. A sending-end current that leads the sending-end voltage indicates the line is:
(a) above SIL (b) below SIL (c) at SIL (d) faulted

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(b) below SIL — the charging current exceeds the load's reactive current and the line is a net var source — Problems 5 and 20.
MCQ 12. The nominal-\(\pi\) is preferred to the nominal-T chiefly because:
(a) it is more accurate (b) it has no internal node (c) it is simpler (d) it needs fewer constants

Show answer
(b). Its shunt branches attach directly to buses, so it slots into a \(Y\)-bus without adding a node — Problem 11.

Reference

Key Formulas

Quantity	Relation	Notes
Two-port relations	\(\mathbf{V}_S = A\mathbf{V}_R + B\mathbf{I}_R\), \(\mathbf{I}_S = C\mathbf{V}_R + D\mathbf{I}_R\)	All four complex
Nominal-π	\(A = D = 1 + \frac{ZY}{2}\), \(B = Z\), \(C = Y(1+\frac{ZY}{4})\)	Half \(Y\) at each end
Nominal-T	\(A = D = 1 + \frac{ZY}{2}\), \(B = Z(1+\frac{ZY}{4})\), \(C = Y\)	Half \(Z\) at each side
Reciprocity	\(AD - BC = 1\)	Holds exactly, always
Symmetry	\(A = D\)	Broken by an asymmetric cascade
No-load voltage	\(V_{R,NL} = V_S/\|A\|\)	\(B\) does not appear
Regulation	\(\dfrac{V_S/\|A\| - V_{R,FL}}{V_{R,FL}}\times100\)
Ferranti rise	\(1/\|A\| - 1 \approx (\beta l)^2/2\)	Exactly \(\sec\beta l - 1\) lossless
No-load current	\(I_S = \|C\|V_R\)	Leading, at the raised \(V_R\)
Cascade	matrix product, in flow order	\(A = A_1A_2 + B_1C_2\) etc.
Identical parallel	\(A, B/2, 2C, D\)	\(A\) unchanged
Series impedance as two-port	\(A = D = 1, B = Z, C = 0\)	Set 9's short line
Open-circuit test	\(Z_{OC} = A/C\)	Receiving end open
Short-circuit test	\(Z_{SC} = B/D\)	Receiving end shorted
Characteristic impedance	\(Z_c = \sqrt{Z_{OC}Z_{SC}}\)	From measurement alone
Inverse relations	\(\mathbf{V}_R = D\mathbf{V}_S - B\mathbf{I}_S\)	No division, since \(AD-BC=1\)
Power transfer	\(P_R = \frac{V_SV_R}{\|B\|}\cos(\beta-\delta) - \frac{\|A\|V_R^2}{\|B\|}\cos(\beta-\alpha)\)	Max at \(\delta = \beta\)

Diagnostics

Common Mistakes

Mixing the \(\pi\) and T formulas. The \((1 + ZY/4)\) factor belongs to \(C\) in the \(\pi\) and to \(B\) in the T — Problems 1 and 2.
Using \(V_{R,NL} = V_S\). That is the short-line result. Here it is \(V_S/|A|\), and the difference is the whole Ferranti effect — Problem 6.
Computing efficiency from \(3I^2R\) with one current. \(\mathbf{I}_S \ne \mathbf{I}_R\) in this model; use \(P_S - P_R\) — Problem 7.
Forgetting that \(C\) is nearly imaginary but not exactly. Its small negative real part is what keeps \(AD - BC\) exactly unity.
Cascading in the wrong order. Matrix multiplication is not commutative; the element nearest the source comes first — Problem 13.
Expecting \(A = D\) after an asymmetric cascade. Symmetry is lost though reciprocity survives — Problem 13.
Doubling \(A\) for parallel lines. \(A\) is unchanged; only \(B\) and \(C\) scale — Problem 14.
Using the nominal-\(\pi\) beyond about 250 km. It has no resonance and understates the Ferranti rise badly at long lengths — Challenge C2.
Taking the sending-end power factor as necessarily lagging. On a lightly loaded medium line it is often leading — Problems 5 and 20.
Using nominal-\(\pi\) constants at a harmonic frequency. They are computed at one \(\omega\), and \(\gamma l\) scales with frequency — Challenge C3.
Applying positive-sequence constants to zero-sequence. Zero-sequence needs an entirely separate \(Z\) and \(Y\) — Set 22.
Skipping the \(AD - BC = 1\) check. It costs one line and catches nearly every arithmetic error in a set of constants — Problem 3.

Looking Ahead

The line is now a two-port, and the four constants that describe it carry everything the rest of this book will ask of it. The Ferranti effect has appeared, the sending and receiving currents differ, and the reactive balance of Set 7 has been confirmed from an independent direction.

But the model is still a lump. Challenge C1 showed that the nominal-\(\pi\) is the first two terms of a series, and Challenge C2 showed that it misses a resonance entirely. Set 12 derives the exact solution from the wave equation, replacing \(1 + ZY/2\) by \(\cosh\gamma l\) and \(Z\) by \(Z_c\sinh\gamma l\) — the point at which the 6000 km wavelength of Set 6 stops being a curiosity and becomes the governing length scale of the whole subject.