Long Transmission Lines — Solved Problems

Set 12 — Long Transmission Lines

Twenty worked problems on the treatment that abandons lumping altogether. Applying Kirchhoff's laws to a differential element of line gives a second-order differential equation whose solution is a pair of travelling waves, and whose terminal behaviour is described by hyperbolic functions rather than polynomials. Everything from Sets 9 and 11 reappears as a limiting case, and two things appear that no lumped model contained at all: a resonance, and a wave that reflects.

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

i Method Recap

The two governing quantities. The propagation constant \(\gamma = \sqrt{zy} = \alpha + j\beta\) per unit length, and the characteristic impedance \(Z_c = \sqrt{z/y}\). Everything else is built from these two and the length.
The exact constants. \(A = D = \cosh\gamma l\), \(B = Z_c\sinh\gamma l\), \(C = \dfrac{\sinh\gamma l}{Z_c}\). Reciprocity follows from \(\cosh^2 - \sinh^2 = 1\) identically.
\(\alpha\) attenuates and \(\beta\) rotates. A wave travelling a distance \(x\) is multiplied by \(e^{-\alpha x}\) in magnitude and retarded by \(\beta x\) in phase. \(\alpha\) is in nepers per km, \(\beta\) in radians per km.
Wavelength and velocity. \(\lambda = 2\pi/\beta\) and \(v = \omega/\beta = f\lambda\). At 50 Hz an overhead line has \(\lambda \approx 5600\)–\(6000\) km, so even a 400 km line is only a fourteenth of a wavelength.
Two waves, not one. \(V(x) = V^{+}e^{\gamma x} + V^{-}e^{-\gamma x}\) — an incident wave and a reflected one. The reflection coefficient at the load is \(\Gamma = \dfrac{Z_R - Z_c}{Z_R + Z_c}\), zero when the line is terminated in its own characteristic impedance.
Terminated in \(Z_c\), nothing reflects. The voltage and current then vary as \(e^{\alpha x}\) alone, so the profile is a pure exponential and — for a lossless line — perfectly flat. That is surge impedance loading, seen exactly.
The equivalent-\(\pi\) is exact. \(Z' = Z_c\sinh\gamma l\) and \(\dfrac{Y'}{2} = \dfrac{1}{Z_c}\tanh\dfrac{\gamma l}{2}\) reproduce the hyperbolic constants exactly, which is how a long line enters a load flow.

VideoWalkthrough

Problem 1Exam levelPropagation Constant

A 400 kV line has \(z = 0.16 + j0.40\ \Omega\)/km and \(y = j3.0\times10^{-6}\) S/km. Find its propagation constant and identify the attenuation and phase constants. This line recurs throughout the set.

Solution

The propagation constant is the square root of the product of the per-unit-length parameters:

\[ \gamma = \sqrt{zy} \]

Forming the product, working in polar form to make the square root easy:

\[ z = 0.4308\angle68.20^\circ\ \Omega/\text{km}, \qquad y = 3.0\times10^{-6}\angle90^\circ\ \text{S/km} \]

\[ zy = 1.2924\times10^{-6}\angle158.20^\circ \]

Taking the square root — halve the angle, root the magnitude:

\[ \gamma = 1.1369\times10^{-3}\angle79.10^\circ\ \text{per km} \]

Resolving into rectangular components:

\[ \alpha = 1.1369\times10^{-3}\cos(79.10^\circ) = 2.150\times10^{-4}\ \text{Np/km} \]

\[ \beta = 1.1369\times10^{-3}\sin(79.10^\circ) = 1.1163\times10^{-3}\ \text{rad/km} \]

For the 400 km line:

\[ \gamma l = 0.4547\angle79.10^\circ = 0.0860 + j0.4465 \]

The angle of \(\gamma\) is close to 79°, not 90°, and the shortfall is entirely due to the line's resistance. For a lossless line \(zy\) would be purely negative real, giving \(\gamma\) at exactly 90° — pure phase shift with no attenuation. The 11° deficit here produces \(\alpha = 2.15\times10^{-4}\) Np/km, which over 400 km is 0.086 Np, or a magnitude change of \(e^{0.086} = 1.090\). That 9% is the whole of the line's attenuation, and Problem 13 shows exactly where it appears.

Answer\(\gamma = 1.1369\times10^{-3}\angle79.10^\circ\)/km; \(\alpha = 2.150\times10^{-4}\) Np/km, \(\beta = 1.1163\times10^{-3}\) rad/km

Problem 2Warm-upCharacteristic Impedance

Find the characteristic impedance of the line of Problem 1, and compare it with the lossless surge impedance \(\sqrt{L/C}\) of Set 7.

Solution

The characteristic impedance is the square root of the ratio rather than the product:

\[ Z_c = \sqrt{\frac{z}{y}} = \sqrt{\frac{0.4308\angle68.20^\circ}{3.0\times10^{-6}\angle90^\circ}} = \sqrt{143\,604\angle-21.80^\circ} \]

\[ Z_c = 378.95\angle-10.90^\circ\ \Omega \]

In rectangular form:

\[ Z_c = 372.1 - j71.7\ \Omega \]

Predominantly resistive with a small capacitive component — the negative angle is the signature of a line whose resistance is small but not zero.

The lossless comparison. Setting \(r = 0\), \(z\) becomes \(j\omega L\) and

\[ Z_c = \sqrt{\frac{j\omega L}{j\omega C}} = \sqrt{\frac{L}{C}} = \sqrt{\frac{0.40/\omega}{3.0\times10^{-6}/\omega}} = \sqrt{\frac{0.40}{3.0\times10^{-6}}} = 365.1\ \Omega \]

a pure resistance, and 3.7% below the lossy value.

So resistance both raises the magnitude of \(Z_c\) and gives it a negative angle, but neither effect is large.

Note the striking contrast with \(\gamma\). The propagation constant is a product and the characteristic impedance a ratio, so resistance affects them quite differently: it changed the angle of \(\gamma\) by 11° and produced the entire attenuation, while it moved \(|Z_c|\) by less than 4%. This is why surge impedance can be quoted as a real number in Set 7 without apology, while attenuation cannot be discussed at all without \(r\).

Answer\(Z_c = 378.95\angle-10.90^\circ\ \Omega = 372.1 - j71.7\ \Omega\), against 365.1 \(\Omega\) lossless

Problem 3Exam levelExact Constants

Find the exact ABCD constants of the 400 km line of Problem 1.

Solution

The exact constants, from the solution of the wave equation:

\[ A = D = \cosh\gamma l, \qquad B = Z_c\sinh\gamma l, \qquad C = \frac{\sinh\gamma l}{Z_c} \]

With \(\gamma l = 0.0860 + j0.4465\), the hyperbolic functions of a complex argument expand as

\[ \cosh(a + jb) = \cosh a\cos b + j\sinh a\sin b \]

\[ \sinh(a + jb) = \sinh a\cos b + j\cosh a\sin b \]

Evaluating the real hyperbolic and circular functions:

\[ \cosh(0.0860) = 1.0037, \quad \sinh(0.0860) = 0.0861, \quad \cos(0.4465) = 0.9019, \quad \sin(0.4465) = 0.4318 \]

Hence:

\[ \cosh\gamma l = (1.0037)(0.9019) + j(0.0861)(0.4318) = 0.9052 + j0.0372 = 0.9060\angle2.35^\circ \]

\[ \sinh\gamma l = (0.0861)(0.9019) + j(1.0037)(0.4318) = 0.0777 + j0.4334 = 0.4403\angle79.84^\circ \]

The constants:

\[ A = D = 0.9060\angle2.35^\circ \]

\[ B = (378.95\angle-10.90^\circ)(0.4403\angle79.84^\circ) = 166.87\angle68.94^\circ\ \Omega \]

\[ C = \frac{0.4403\angle79.84^\circ}{378.95\angle-10.90^\circ} = 1.1620\times10^{-3}\angle90.74^\circ\ \text{S} \]

These are the true constants; everything in Set 11 was an approximation to them. Note that \(|A| = 0.906\), so this line has a Ferranti rise of 10.4% — and that \(\angle C = 90.74^\circ\) rather than exactly 90°, the excess being the resistance. The hyperbolic functions do all the work: they are the exact solution of the differential equation, and no series truncation has occurred anywhere.

Answer\(A = D = 0.9060\angle2.35^\circ\), \(B = 166.87\angle68.94^\circ\ \Omega\), \(C = 1.1620\times10^{-3}\angle90.74^\circ\) S

Problem 4Warm-upReciprocity

Show that the exact constants satisfy \(AD - BC = 1\) identically, without reference to any particular line.

Solution

Substituting the hyperbolic forms:

\[ AD - BC = \cosh^{2}\gamma l - \left(Z_c\sinh\gamma l\right)\left(\frac{\sinh\gamma l}{Z_c}\right) \]

The characteristic impedance cancels exactly:

\[ = \cosh^{2}\gamma l - \sinh^{2}\gamma l \]

And this is the fundamental hyperbolic identity:

\[ \cosh^{2}\theta - \sinh^{2}\theta = 1 \quad\text{for any complex } \theta \]

\[ \Rightarrow\quad AD - BC = 1 \]

Checking numerically with the Problem 3 values:

\[ A^{2} = 0.8208\angle4.70^\circ = 0.8180 + j0.0673 \]

\[ BC = (166.87)(1.1620\times10^{-3})\angle(68.94 + 90.74) = 0.1939\angle159.68^\circ = -0.1819 + j0.0673 \]

\[ A^{2} - BC = 0.9999 + j0.0000 \approx 1\ \checkmark \]

Reciprocity is not a property this line happens to have — it is built into the structure of the solution. The \(Z_c\) cancels because \(B\) and \(C\) are defined as its product and quotient with the same \(\sinh\), and the remainder is a trigonometric identity. The same cancellation is why Set 11's nominal-\(\pi\) also satisfied it exactly despite being an approximation: the approximation was constructed to preserve reciprocity, which is a stronger constraint than accuracy.

Answer\(AD - BC = \cosh^2\gamma l - \sinh^2\gamma l = 1\), identically for any line

Problem 5Exam levelExact vs Nominal-π

Compute the nominal-\(\pi\) constants for the 400 km line and find the error in each against the exact values of Problem 3.

Solution

Total parameters and their product:

\[ \mathbf{Z} = 64 + j160\ \Omega, \qquad \mathbf{Y} = j1.2\times10^{-3}\ \text{S}, \qquad \mathbf{Z}\mathbf{Y} = -0.192 + j0.0768 \]

The nominal-\(\pi\) constants:

\[ A = 1 + \frac{\mathbf{Z}\mathbf{Y}}{2} = 0.904 + j0.0384 = 0.9048\angle2.43^\circ \]

\[ B = \mathbf{Z} = 172.33\angle68.20^\circ\ \Omega \]

\[ C = \mathbf{Y}\left(1 + \frac{\mathbf{Z}\mathbf{Y}}{4}\right) = 1.1426\times10^{-3}\angle91.16^\circ\ \text{S} \]

Against the exact values:

\[ \begin{array}{lccc} & \text{Exact} & \text{Nominal-}\pi & \text{Error} \\ \hline |A| & 0.9060 & 0.9048 & 0.20\% \\ |B|\ (\Omega) & 166.87 & 172.33 & 3.53\% \\ |C|\ (\times10^{-3}\ \text{S}) & 1.1620 & 1.1426 & 1.81\% \end{array} \]

Comparing with the 200 km line of Set 11, where the errors were 0.01%, 0.80% and 0.40%:

\[ \begin{array}{ccc} \text{Length} & \text{Error in } B & \text{Ratio} \\ \hline 200\ \text{km} & 0.80\% & — \\ 400\ \text{km} & 3.53\% & 4.4\times \end{array} \]

A doubling of length quadrupled the error, confirming the quadratic growth predicted by the series expansion of Set 11 Challenge C1.

Three and a half per cent in \(B\) is the point at which the lumped model stops being acceptable. \(B\) determines the voltage drop under load and the power transfer limit, so a 3.5% error propagates directly into both. The conventional boundary of 250 km corresponds to an error of about 1.4%, which is tolerable; beyond it the exact solution — or the equivalent-\(\pi\) of Problem 10, which is the exact solution wearing a lumped disguise — becomes necessary.

AnswerErrors of 0.20% in \(A\), 3.53% in \(B\) and 1.81% in \(C\) — quadrupled from the 200 km case

Problem 6Warm-upWavelength

Find the wavelength and velocity of propagation on the line of Problem 1 at 50 Hz, and express the 400 km length as a fraction of a wavelength.

Solution

The wavelength follows from the phase constant alone:

\[ \lambda = \frac{2\pi}{\beta} = \frac{6.2832}{1.1163\times10^{-3}} = 5628\ \text{km} \]

Velocity of propagation:

\[ v = f\lambda = 50(5628) = 2.814\times10^{5}\ \text{km/s} = 2.814\times10^{8}\ \text{m/s} \]

\[ \frac{v}{c} = \frac{2.814\times10^{8}}{2.998\times10^{8}} = 0.939 \]

Somewhat below the 0.99\(c\) of Set 7, because that calculation neglected resistance while \(\beta\) here includes it.

The line as a fraction of a wavelength:

\[ \frac{l}{\lambda} = \frac{400}{5628} = 0.0711 = \frac{1}{14.1} \]

\[ \beta l = 0.4465\ \text{rad} = 25.58^\circ \]

A four-hundred-kilometre line is a fourteenth of a wavelength, and that single number explains everything about which model to use. A lumped circuit element is one whose dimensions are negligible against a wavelength; at \(1/14\) the approximation is straining, which is exactly what the 3.5% error of Problem 5 records. A 1400 km line would be a quarter wavelength and the lumped model would not merely be inaccurate but qualitatively wrong — Problem 19 examines that case.

Answer\(\lambda = 5628\) km, \(v = 2.814\times10^8\) m/s \(= 0.94c\); the line is \(1/14\) of a wavelength

Problem 7Exam levelFerranti Exactly

Find the exact open-circuit voltage rise of the 400 km line, compare it with the nominal-\(\pi\) prediction, and give the lossless closed form.

Solution

On open circuit, from Set 11 Problem 6:

\[ \frac{V_R}{V_S} = \frac{1}{|A|} = \frac{1}{|\cosh\gamma l|} \]

With the exact \(|A| = 0.9060\) from Problem 3:

\[ \frac{V_R}{V_S} = \frac{1}{0.9060} = 1.1037 \quad\Rightarrow\quad \text{rise} = 10.37\% \]

The nominal-\(\pi\) gave \(|A| = 0.9048\):

\[ \frac{1}{0.9048} = 1.1052 \quad\Rightarrow\quad 10.52\% \]

Overstating the rise by 0.15 points — an error small in absolute terms but in the wrong direction, since the lumped model has no resonance and must eventually understate.

The lossless closed form. With \(\alpha = 0\), \(\gamma l = j\beta l\) and

\[ \cosh(j\beta l) = \cos\beta l \quad\Rightarrow\quad \frac{V_R}{V_S} = \frac{1}{\cos\beta l} = \sec\beta l \]

Evaluating with \(\beta l = 25.58^\circ\):

\[ \sec(25.58^\circ) = \frac{1}{0.9019} = 1.1088 \quad\Rightarrow\quad 10.88\% \]

Slightly larger than the lossy value, because resistance damps the rise.

Three answers within half a point of each other — and the agreement is misleading. At 400 km all three models are still in their region of agreement. Push to 1000 km and the lossless secant gives 45%, the exact hyperbolic 38%, and the nominal-\(\pi\) only 30% — the lumped model's error growing without bound while the other two remain close. The secant form is the one to carry, because it makes the quarter-wave singularity visible in a way the numerical answers never do.

AnswerExact 10.37%, nominal-\(\pi\) 10.52%, lossless \(\sec\beta l - 1 = 10.88\%\)

Problem 8Exam levelUnder Load

The 400 km line delivers 300 MW at 400 kV, 0.95 power factor lagging. Find the sending-end voltage and current.

Solution

Receiving-end quantities per phase:

\[ \mathbf{V}_R = \frac{400\times10^{3}}{\sqrt3} = 230\,940\angle0^\circ\ \text{V} \]

\[ I_R = \frac{300\times10^{6}}{\sqrt3(400\times10^{3})(0.95)} = 455.8\ \text{A}, \qquad \mathbf{I}_R = 455.8\angle-18.19^\circ \]

Sending-end voltage:

\[ A\mathbf{V}_R = 0.9060\angle2.35^\circ(230\,940) = 209\,232\angle2.35^\circ \]

\[ B\mathbf{I}_R = 166.87\angle68.94^\circ \times 455.8\angle-18.19^\circ = 76\,059\angle50.75^\circ \]

\[ \mathbf{V}_S = (209\,057 + j8580) + (48\,146 + j58\,899) = 257\,203 + j67\,479 \]

\[ = 265\,899\angle14.70^\circ\ \text{V per phase} \]

\[ V_{S,\text{line}} = \sqrt3(265\,899) = 460.6\ \text{kV} \]

Sending-end current:

\[ C\mathbf{V}_R = 1.1620\times10^{-3}\angle90.74^\circ(230\,940) = 268.4\angle90.74^\circ = -3.47 + j268.4 \]

\[ D\mathbf{I}_R = 0.9060\angle2.35^\circ \times 455.8\angle-18.19^\circ = 412.9\angle-15.84^\circ = 397.3 - j112.7 \]

\[ \mathbf{I}_S = 393.8 + j155.7 = 423.4\angle21.56^\circ\ \text{A} \]

Sending-end power factor — note the current leads the voltage:

\[ \phi_S = 14.70^\circ - 21.56^\circ = -6.86^\circ \quad\Rightarrow\quad \cos\phi_S = 0.993\ \text{leading} \]

A sending-end voltage of 460 kV to deliver 400 kV, and a leading sending-end current — this line is in trouble on both counts. 460 kV is 15% above nominal and beyond the insulation class; the leading current means the generator must absorb reactive power. Both are symptoms of an uncompensated 400 km line, and both are fixed by the same measure: shunt reactors, which reduce the effective \(Y\) and bring \(|A|\) back towards unity.

Answer\(V_S = 460.6\) kV at \(14.70^\circ\), \(\mathbf{I}_S = 423.4\angle21.56^\circ\) A, \(\cos\phi_S = 0.993\) leading

Problem 9Warm-upRegulation and Efficiency

Find the voltage regulation and transmission efficiency of the line of Problem 8.

Solution

Regulation. On no load the receiving voltage rises to \(V_S/|A|\):

\[ V_{R,\text{NL}} = \frac{265\,899}{0.9060} = 293\,487\ \text{V per phase} \]

\[ \%\text{reg} = \frac{293\,487 - 230\,940}{230\,940}\times100 = 27.08\% \]

Efficiency. Sending-end power from the sending-end quantities:

\[ P_S = 3V_SI_S\cos\phi_S = 3(265\,899)(423.4)(0.993) = 335.4\ \text{MW} \]

\[ \eta = \frac{300}{335.4}\times100 = 89.46\% \]

The loss is 35.4 MW — nearly 12% of the delivered power. At \(r = 0.16\ \Omega\)/km this line carries a single unbundled conductor; a bundled EHV design of the same voltage would have perhaps a quarter of that resistance and a correspondingly better efficiency. The regulation, however, is set by \(X\) and by the Ferranti term, and bundling would barely touch it.

Both figures are unacceptable by any operating standard: regulation should be under 10% and efficiency above 95%.

27% regulation is not a line that needs adjustment; it is a line that cannot be operated as built. The remedy is not more copper — the loss is high but the regulation is dominated by the Ferranti term and by \(|B| = 167\ \Omega\). What such a line receives in practice is shunt reactors at both ends and often at an intermediate switching station, together with series compensation, which together bring \(|A|\) towards 1 and \(|B|\) down by half. That combination is standard on every 400 km-plus EHV line in service.

AnswerRegulation \(= 27.08\%\), \(P_S = 335.4\) MW, \(\eta = 89.46\%\) — both unacceptable

Problem 10Exam levelEquivalent-π

Find the equivalent-\(\pi\) circuit of the 400 km line, and show that it reproduces the exact constants.

Solution

The equivalent-\(\pi\) is a lumped network chosen so that its ABCD constants are exactly the hyperbolic ones. Matching \(B\) first, since a \(\pi\) has \(B = Z'\):

\[ Z' = Z_c\sinh\gamma l = B = 166.87\angle68.94^\circ\ \Omega \]

Matching \(A\), for which a \(\pi\) gives \(A = 1 + Z'Y'/2\):

\[ 1 + \frac{Z'Y'}{2} = \cosh\gamma l \quad\Rightarrow\quad \frac{Y'}{2} = \frac{\cosh\gamma l - 1}{Z_c\sinh\gamma l} \]

The half-angle identity \(\dfrac{\cosh\theta - 1}{\sinh\theta} = \tanh\dfrac{\theta}{2}\) simplifies this to

\[ \frac{Y'}{2} = \frac{1}{Z_c}\tanh\frac{\gamma l}{2} \]

Evaluating with \(\gamma l/2 = 0.0430 + j0.2233\):

\[ \tanh\frac{\gamma l}{2} = 0.2311\angle78.72^\circ \]

\[ \frac{Y'}{2} = \frac{0.2311\angle78.72^\circ}{378.95\angle-10.90^\circ} = 6.098\times10^{-4}\angle89.62^\circ\ \text{S} \]

Comparing with the nominal-\(\pi\), which used \(Y/2 = 6.0\times10^{-4}\angle90^\circ\) and \(Z = 172.3\angle68.20^\circ\):

\[ \begin{array}{lcc} & \text{Nominal-}\pi & \text{Equivalent-}\pi \\ \hline Z\ (\Omega) & 172.33\angle68.20^\circ & 166.87\angle68.94^\circ \\ Y/2\ (\text{S}) & 6.000\times10^{-4}\angle90^\circ & 6.098\times10^{-4}\angle89.62^\circ \end{array} \]

The equivalent-\(\pi\) is exact, and that is how every long line enters a load flow. A load-flow program cannot integrate a differential equation for each branch; it needs lumped impedances to assemble into a \(Y\)-bus. The equivalent-\(\pi\) gives it exactly that, at no cost in accuracy — the hyperbolic functions are evaluated once when the data is prepared, and thereafter the line is a three-element lumped network like any other. This is the reconciliation promised in Set 11 Problem 11: the \(\pi\) is not the best model of a line, it is the shape every model is converted into.

Answer\(Z' = 166.87\angle68.94^\circ\ \Omega\), \(Y'/2 = 6.098\times10^{-4}\angle89.62^\circ\) S — exactly equivalent

Problem 11Exam levelTerminated in Z_c

The 400 km line is terminated in its own characteristic impedance. Find the receiving-end current, the power delivered, and the ratio of sending- to receiving-end voltage.

Solution

With \(\mathbf{Z}_R = Z_c\) the current follows directly:

\[ \mathbf{I}_R = \frac{\mathbf{V}_R}{Z_c} = \frac{230\,940\angle0^\circ}{378.95\angle-10.90^\circ} = 609.4\angle10.90^\circ\ \text{A} \]

Leading by 10.90° — the load looks slightly capacitive because \(Z_c\) does.

Power delivered:

\[ P_R = 3V_RI_R\cos(10.90^\circ) = 3(230\,940)(609.4)(0.982) = 414.6\ \text{MW} \]

Close to the nominal SIL of \(V_L^2/|Z_c| = (400)^2/378.95 = 422.2\) MW, the small difference being the power factor of \(Z_c\).

The voltage ratio. Substituting \(\mathbf{I}_R = \mathbf{V}_R/Z_c\) into the two-port relation:

\[ \mathbf{V}_S = \mathbf{V}_R\cosh\gamma l + Z_c\sinh\gamma l\left(\frac{\mathbf{V}_R}{Z_c}\right) = \mathbf{V}_R(\cosh\gamma l + \sinh\gamma l) = \mathbf{V}_Re^{\gamma l} \]

So the ratio of magnitudes is the attenuation alone:

\[ \frac{|\mathbf{V}_S|}{|\mathbf{V}_R|} = e^{\alpha l} = e^{0.0860} = 1.0898 \]

\[ V_{S,\text{line}} = 400(1.0898) = 435.9\ \text{kV} \]

By the same argument \(\mathbf{I}_S = \mathbf{I}_Re^{\gamma l}\), so the current ratio is identical and the ratio \(V/I\) is \(Z_c\) at every point along the line.

Terminating in \(Z_c\) collapses the two-wave solution to one — nothing reflects. The line looks, from any point, exactly like an infinite line, and the voltage decays smoothly by \(e^{-\alpha x}\) towards the load. For a lossless line \(\alpha = 0\) and the profile is perfectly flat, which is the idealisation behind the surge impedance loading of Set 7. Here the 8.98% rise from receiving to sending end is the line's total attenuation, and it is the cleanest possible measurement of it.

Answer\(\mathbf{I}_R = 609.4\angle10.90^\circ\) A, \(P_R = 414.6\) MW, \(V_S/V_R = e^{\alpha l} = 1.0898\)

Problem 12Warm-upVoltage Profile

Tabulate the voltage magnitude at 100 km intervals along the line of Problem 11, measuring \(x\) from the receiving end.

Solution

The general solution at a distance \(x\) from the receiving end:

\[ \mathbf{V}(x) = \mathbf{V}_R\cosh\gamma x + \mathbf{I}_RZ_c\sinh\gamma x \]

With the surge-impedance termination this collapses, as Problem 11 showed, to

\[ \mathbf{V}(x) = \mathbf{V}_Re^{\gamma x} \quad\Rightarrow\quad |\mathbf{V}(x)| = V_Re^{\alpha x} \]

Evaluating with \(\alpha = 2.150\times10^{-4}\) Np/km and \(V_R = 400\) kV line:

\[ \begin{array}{ccc} x\ (\text{km}) & e^{\alpha x} & |V|\ (\text{kV}) \\ \hline 0 & 1.0000 & 400.00 \\ 100 & 1.0217 & 408.69 \\ 200 & 1.0439 & 417.57 \\ 300 & 1.0666 & 426.65 \\ 400 & 1.0898 & 435.92 \end{array} \]

A smooth exponential rise from load to source, of 2.17% per hundred kilometres — no oscillation, and no maximum in the interior.

The profile at surge impedance loading is monotonic and gentle, and that is exactly why SIL is the loading a line "wants". At any other loading the two waves of Problem 14 interfere and the profile develops curvature — sagging below the endpoints under heavy load, bulging above them under light load. It is that bulge, not the endpoint values, that decides insulation coordination on a long compensated line, and finding it requires the full expression rather than the ABCD constants, which describe only the ends.

Answer400.0, 408.7, 417.6, 426.7, 435.9 kV — a pure exponential at \(e^{\alpha x}\)

Problem 13Warm-upAttenuation

Interpret the attenuation constant physically. Convert \(\alpha\) for this line into decibels per 100 km, and find the distance over which a wave falls to half its amplitude.

Solution

Physical meaning. A wave travelling a distance \(x\) has its amplitude multiplied by \(e^{-\alpha x}\). One neper is the distance over which the amplitude falls by a factor \(e\).

In decibels. Since 1 Np \(= 20\log_{10}e = 8.686\) dB:

\[ \alpha = 2.150\times10^{-4}\ \text{Np/km} = 1.868\times10^{-3}\ \text{dB/km} = 0.187\ \text{dB per 100 km} \]

Half-amplitude distance:

\[ e^{-\alpha x} = 0.5 \quad\Rightarrow\quad x = \frac{\ln 2}{\alpha} = \frac{0.6931}{2.150\times10^{-4}} = 3224\ \text{km} \]

Over the actual 400 km the attenuation is

\[ e^{-\alpha l} = e^{-0.0860} = 0.9176 \quad\Rightarrow\quad 8.2\%\ \text{loss of amplitude} \]

Three thousand kilometres to halve a wave — power lines are extraordinarily low-loss waveguides. A telephone cable of the same era attenuated by half in a few kilometres; a coaxial cable at radio frequency does so in metres. The reason is that \(\alpha \approx r/2Z_c\) for a low-loss line, and a power conductor is chosen to have very small \(r\) against a \(Z_c\) of several hundred ohms. It is also why travelling-wave fault locators work: a surge injected at one end is still measurable after hundreds of kilometres and several reflections.

Answer\(\alpha = 0.187\) dB per 100 km; half amplitude after 3224 km; 8.2% loss over 400 km

Problem 14Challenge-liteTwo Waves

Resolve the voltage on the line of Problem 8 into incident and reflected waves at the receiving end, and find their magnitudes at the sending end.

Solution

The general solution written as two exponentials:

\[ \mathbf{V}(x) = \mathbf{V}^{+}e^{\gamma x} + \mathbf{V}^{-}e^{-\gamma x} \]

With \(x\) measured from the receiving end, \(\mathbf{V}^{+}\) is the wave travelling towards the load and \(\mathbf{V}^{-}\) the one reflected back.

At \(x = 0\) the two must reproduce the receiving-end conditions. Solving the pair of equations for voltage and current:

\[ \mathbf{V}^{+} = \frac{\mathbf{V}_R + Z_c\mathbf{I}_R}{2}, \qquad \mathbf{V}^{-} = \frac{\mathbf{V}_R - Z_c\mathbf{I}_R}{2} \]

Evaluating with \(\mathbf{V}_R = 230\,940\angle0^\circ\) and \(\mathbf{I}_R = 455.8\angle-18.19^\circ\):

\[ Z_c\mathbf{I}_R = (378.95\angle-10.90^\circ)(455.8\angle-18.19^\circ) = 172\,725\angle-29.09^\circ \]

\[ \mathbf{V}^{+} = 195\,501\angle-12.40^\circ\ \text{V}, \qquad \mathbf{V}^{-} = 57\,993\angle46.39^\circ\ \text{V} \]

At the sending end the incident wave has grown by \(e^{\alpha l}\) and the reflected one has decayed by the same factor:

\[ |\mathbf{V}^{+}(l)| = 195\,501(1.0898) = 213\,058\ \text{V} \]

\[ |\mathbf{V}^{-}(l)| = 57\,993(0.9176) = 53\,214\ \text{V} \]

Their phasor sum must reproduce the sending-end voltage of Problem 8:

\[ 213\,058\angle13.18^\circ + 53\,214\angle20.80^\circ = 265\,896\angle14.70^\circ\ \checkmark \]

The reflected wave is 30% of the incident one, and it is what makes the profile of Problem 12 curve. Under surge-impedance termination it would be zero and the two-wave picture would collapse to one. Here the load is inductive rather than matched, so a substantial wave returns towards the source — and the interference between the two is the entire content of the Ferranti effect, the voltage profile and the reactive imbalance. The travelling-wave picture is not an alternative to the ABCD description; it is the same solution written so that the mechanism is visible.

Answer\(\mathbf{V}^{+} = 195.5\angle-12.4^\circ\) kV, \(\mathbf{V}^{-} = 58.0\angle46.4^\circ\) kV; at the sending end 213.1 and 53.2 kV

Problem 15Exam levelReflection Coefficient

Find the reflection coefficient of the load in Problem 8, and evaluate it for an open circuit, a short circuit and a matched termination.

Solution

The reflection coefficient is the ratio of the two waves at the load:

\[ \Gamma = \frac{\mathbf{V}^{-}}{\mathbf{V}^{+}} = \frac{\mathbf{V}_R - Z_c\mathbf{I}_R}{\mathbf{V}_R + Z_c\mathbf{I}_R} = \frac{Z_R - Z_c}{Z_R + Z_c} \]

writing \(Z_R = \mathbf{V}_R/\mathbf{I}_R\) for the load impedance.

The load impedance in Problem 8:

\[ Z_R = \frac{230\,940\angle0^\circ}{455.8\angle-18.19^\circ} = 506.7\angle18.19^\circ\ \Omega \]

Hence:

\[ \Gamma = \frac{506.7\angle18.19^\circ - 378.95\angle-10.90^\circ}{506.7\angle18.19^\circ + 378.95\angle-10.90^\circ} = \frac{254.5\angle64.6^\circ}{857.9\angle5.8^\circ} = 0.297\angle58.8^\circ \]

Agreeing with the ratio \(57\,993/195\,501 = 0.297\) of Problem 14.

The three limiting cases:

\[ \begin{array}{lcl} \text{Open circuit } (Z_R \to \infty) & \Gamma = +1 & \text{full reflection, in phase} \\ \text{Short circuit } (Z_R = 0) & \Gamma = -1 & \text{full reflection, inverted} \\ \text{Matched } (Z_R = Z_c) & \Gamma = 0 & \text{no reflection} \end{array} \]

The open-circuit case is the Ferranti effect in wave language: the reflected wave adds to the incident one at the load, doubling it there, and the two interfere along the line to give the \(\sec\beta l\) profile.

The same reflection coefficient governs power-frequency behaviour and lightning surges — only the frequency differs. At 50 Hz it explains the Ferranti effect and the voltage profile; at surge frequencies it explains why an incoming lightning wave doubles at an open circuit-breaker, why it inverts at an earthed structure, and why a cable joined to an overhead line reflects at the junction because their surge impedances differ by a factor of ten. That last case is the reason cable–line junctions are a preferred location for surge arresters.

Answer\(\Gamma = (Z_R - Z_c)/(Z_R + Z_c) = 0.297\angle58.8^\circ\); \(+1\) open, \(-1\) short, 0 matched

Problem 16Challenge-liteThe Short-Line Limit

Show that the exact constants reduce to the nominal-\(\pi\) and then to the short-line constants as \(\gamma l \to 0\), and identify the order at which each model is correct.

Solution

Expanding the hyperbolic functions, writing \(\theta = \gamma l\) so that \(\theta^2 = \mathbf{Z}\mathbf{Y}\):

\[ \cosh\theta = 1 + \frac{\theta^{2}}{2} + \frac{\theta^{4}}{24} + \cdots = 1 + \frac{\mathbf{Z}\mathbf{Y}}{2} + \frac{(\mathbf{Z}\mathbf{Y})^{2}}{24} + \cdots \]

For \(B\), using \(Z_c = \sqrt{\mathbf{Z}/\mathbf{Y}}\) and \(\sinh\theta = \theta + \theta^3/6 + \cdots\):

\[ B = Z_c\sinh\theta = \sqrt{\frac{\mathbf{Z}}{\mathbf{Y}}}\left(\sqrt{\mathbf{Z}\mathbf{Y}} + \frac{(\mathbf{Z}\mathbf{Y})^{3/2}}{6}\right) = \mathbf{Z}\left(1 + \frac{\mathbf{Z}\mathbf{Y}}{6} + \cdots\right) \]

And similarly:

\[ C = \frac{\sinh\theta}{Z_c} = \mathbf{Y}\left(1 + \frac{\mathbf{Z}\mathbf{Y}}{6} + \cdots\right) \]

The hierarchy. Comparing term by term:

\[ \begin{array}{lccc} & A & B & C \\ \hline \text{Exact} & 1 + \frac{ZY}{2} + \frac{(ZY)^2}{24} & Z\left(1 + \frac{ZY}{6}\right) & Y\left(1 + \frac{ZY}{6}\right) \\ \text{Nominal-}\pi & 1 + \frac{ZY}{2} & Z & Y\left(1 + \frac{ZY}{4}\right) \\ \text{Short line} & 1 & Z & 0 \end{array} \]

So the short line is correct to order \((\gamma l)^0\), the nominal-\(\pi\) to order \((\gamma l)^2\) in \(A\) but only \((\gamma l)^0\) in \(B\), and the exact solution to all orders.

Note that the nominal-\(\pi\)'s \(C\) uses \(1/4\) where the correct coefficient is \(1/6\) — it is not merely truncated but wrong at the first correction, which is why its error in \(C\) (1.81%) is comparable with its error in \(B\) (3.53%) despite \(B\) having no correction at all.

Three models, one series, and the choice is simply how many terms to keep. This is the cleanest possible statement of the relationship between Sets 9, 11 and 12: they are not rival descriptions but successive truncations of the same expansion in the electrical length. It also explains why the boundaries between them fall where they do — at \(\gamma l \approx 0.09\) the neglected \((\gamma l)^2\) term is 0.4%, and at \(\gamma l \approx 0.28\) it is 4%.

AnswerShort line correct to \((\gamma l)^0\), nominal-\(\pi\) to \((\gamma l)^2\) in \(A\); exact to all orders

Problem 17Exam levelLength Dependence

For the per-km parameters of Problem 1, tabulate the exact constants and the Ferranti rise at 50, 100, 200, 400, 800 and 1400 km, and comment on the trends.

Solution

Evaluating \(A = \cosh\gamma l\), \(B = Z_c\sinh\gamma l\) and \(C = \sinh(\gamma l)/Z_c\) at each length:

\[ \begin{array}{cccccc} l\ (\text{km}) & \beta l\ (^\circ) & |A| & |B|\ (\Omega) & |C|\ (\times10^{-3}) & \text{Ferranti} \\ \hline 50 & 3.2 & 0.9985 & 21.53 & 0.150 & 0.15\% \\ 100 & 6.4 & 0.9940 & 43.00 & 0.299 & 0.60\% \\ 200 & 12.8 & 0.9761 & 85.48 & 0.595 & 2.45\% \\ 400 & 25.6 & 0.9060 & 166.87 & 1.162 & 10.37\% \\ 800 & 51.2 & 0.6504 & 302.38 & 2.106 & 53.75\% \\ 1400 & 89.5 & 0.3057 & 396.24 & 2.759 & 227\% \end{array} \]

Three distinct trends. \(|A|\) falls monotonically and accelerates, driving the Ferranti rise; \(|B|\) grows but sub-linearly, flattening towards \(|Z_c|\); \(|C|\) does the same towards \(1/|Z_c|\).

The saturation of \(B\) is worth noting. Doubling from 200 to 400 km nearly doubles it, but from 800 to 1400 km it rises only 31%:

\[ B \to Z_c \quad\text{as}\quad \beta l \to 90^\circ, \quad\text{since}\quad \sinh(j\pi/2) = j \]

Meanwhile the Ferranti rise diverges. At 1400 km, \(\beta l = 89.5^\circ\) and the line is within half a degree of quarter-wave resonance.

The two quantities that matter most move in opposite directions, and that is the defining difficulty of very long a.c. lines. The series impedance \(B\) saturates, so the power transfer limit stops worsening; but \(|A|\) collapses, so the no-load voltage rise becomes unmanageable. A 1400 km a.c. line is not limited by its ability to carry power — it is limited by the impossibility of energising it. This asymmetry is the single strongest technical argument for HVDC beyond about 600 km, since a d.c. line has no \(\beta\) and therefore no resonance at any length.

Answer\(|A|\) falls 0.9985 to 0.306 and the Ferranti rise grows 0.15% to 227%, while \(|B|\) saturates towards \(|Z_c|\)

Problem 18Warm-upThe Lossless Line

Write the ABCD constants of a lossless line and evaluate them for the 400 km line, taking \(Z_c = 365.1\ \Omega\) and \(\beta l = 25.58^\circ\). Compare with the exact values.

Solution

With \(r = 0\) we have \(\alpha = 0\) and \(\gamma l = j\beta l\). The hyperbolic functions of a pure imaginary argument become circular:

\[ \cosh(j\beta l) = \cos\beta l, \qquad \sinh(j\beta l) = j\sin\beta l \]

So the constants become:

\[ A = D = \cos\beta l, \qquad B = jZ_c\sin\beta l, \qquad C = j\frac{\sin\beta l}{Z_c} \]

All real or purely imaginary — \(A\) real, \(B\) and \(C\) imaginary. A lossless line has no in-phase component anywhere.

Evaluating at \(\beta l = 25.58^\circ\):

\[ A = \cos(25.58^\circ) = 0.9019 \]

\[ B = j(365.1)\sin(25.58^\circ) = j157.6 = 157.6\angle90^\circ\ \Omega \]

\[ C = j\frac{0.4317}{365.1} = j1.182\times10^{-3} = 1.182\times10^{-3}\angle90^\circ\ \text{S} \]

Against the exact values:

\[ \begin{array}{lccc} & \text{Exact} & \text{Lossless} & \text{Error} \\ \hline |A| & 0.9060 & 0.9019 & 0.45\% \\ |B|\ (\Omega) & 166.87 & 157.6 & 5.6\% \\ |C|\ (\times10^{-3}) & 1.1620 & 1.1820 & 1.7\% \end{array} \]

Reciprocity still holds exactly:

\[ AD - BC = \cos^{2}\beta l - (jZ_c\sin\beta l)\left(\frac{j\sin\beta l}{Z_c}\right) = \cos^{2} + \sin^{2} = 1\ \checkmark \]

The lossless model gets \(A\) to half a per cent and \(B\) to only six, and that division of accuracy is exactly what makes it useful. Stability studies (Sets 24 to 29) depend on \(A\) and on the angle relationship, where the lossless form is excellent and gives the clean \(P = V_SV_R\sin\delta/X\). Voltage-drop and loss calculations depend on \(B\) and on the resistance, where it is not usable at all. Knowing which quantity a given study depends on decides whether the simplification is legitimate.

Answer\(A = \cos\beta l = 0.9019\), \(B = jZ_c\sin\beta l = j157.6\ \Omega\), \(C = j\sin\beta l/Z_c\); errors 0.45%, 5.6%, 1.7%

Problem 19Challenge-liteQuarter-Wave Resonance

Find the length at which this line becomes a quarter wavelength, and examine its behaviour there on open circuit, on short circuit and with a load.

Solution

Quarter-wave length:

\[ \beta l = \frac{\pi}{2} \quad\Rightarrow\quad l = \frac{\pi/2}{1.1163\times10^{-3}} = 1407\ \text{km} \]

Equivalently \(\lambda/4 = 5628/4 = 1407\) km.

On open circuit. The lossless constants give \(A = \cos(90^\circ) = 0\), so

\[ \frac{V_R}{V_S} = \frac{1}{|A|} \to \infty \]

The real line, with losses, gives \(|A| = 0.306\) and a rise of 227% — enormous but finite. Resistance is the only thing preventing an unbounded voltage.

On short circuit. The input impedance of a lossless line terminated in \(Z_R\) is

\[ Z_{in} = Z_c\frac{Z_R + jZ_c\tan\beta l}{Z_c + jZ_R\tan\beta l} \]

\[ \beta l = 90^\circ \Rightarrow \tan\beta l \to \infty \Rightarrow Z_{in} = \frac{Z_c^{2}}{Z_R} \]

So the quarter-wave line is an impedance inverter: a short circuit at one end appears as an open circuit at the other, and vice versa.

With a load. The same relation gives

\[ Z_{in} = \frac{Z_c^{2}}{Z_R} \quad\Rightarrow\quad P = \frac{V_S V_R}{Z_c} \quad\text{independent of } \delta \]

A quarter-wave line transfers power at constant magnitude regardless of the angle — a curiosity exploited in "tuned power lines", which is the subject of Set 14.

1407 km is not a hypothetical length — it is comparable with the distances Indian, Chinese, Brazilian and Russian grids actually span. No a.c. line is ever built anywhere near it uncompensated: the practical limit is around 600 km with heavy shunt and series compensation, which is equivalent to shortening the line electrically. Beyond that HVDC takes over, precisely because it has no \(\beta\) and therefore no resonance at any length whatsoever.

Answer\(\lambda/4 = 1407\) km; open-circuit rise 227% (infinite if lossless); \(Z_{in} = Z_c^2/Z_R\) — an impedance inverter

Problem 20Exam levelComplete Study

Assemble a complete assessment of the 400 km, 400 kV line: its constants, its behaviour at 300 MW and on no load, its SIL, and a recommendation.

Solution

Constants and derived quantities, from Problems 1 to 6:

\[ \begin{array}{ll} \gamma & 1.1369\times10^{-3}\angle79.10^\circ\ \text{/km} \\ \alpha,\ \beta & 2.150\times10^{-4}\ \text{Np/km},\ 1.1163\times10^{-3}\ \text{rad/km} \\ Z_c & 378.95\angle-10.90^\circ\ \Omega \\ \lambda,\ \beta l & 5628\ \text{km},\ 25.58^\circ \\ A = D & 0.9060\angle2.35^\circ \\ B & 166.87\angle68.94^\circ\ \Omega \\ C & 1.1620\times10^{-3}\angle90.74^\circ\ \text{S} \end{array} \]

At 300 MW, 0.95 lagging (Problems 8 and 9):

\[ V_S = 460.6\ \text{kV}, \quad \cos\phi_S = 0.993\ \text{leading}, \quad \%\text{reg} = 27.1\%, \quad \eta = 89.5\% \]

On no load (Problem 7):

\[ V_R = 400(1.1037) = 441.5\ \text{kV}\ \text{for a 400 kV sending voltage} \]

Surge impedance loading (Problem 11):

\[ \text{SIL} = \frac{(400)^{2}}{378.95} = 422\ \text{MW} \]

The line runs at 300 MW, or 71% of SIL, so it is a net generator of reactive power — confirmed by the leading sending-end current.

The verdict. Three separate failures:

\[ \begin{array}{lll} \text{Regulation } 27\% & \text{against a 10\% limit} & \text{Fails} \\ \text{Efficiency } 89.5\% & \text{against a 95\% target} & \text{Fails} \\ \text{No-load } 441.5\ \text{kV} & \text{against a 420 kV insulation class} & \text{Fails} \end{array} \]

Recommendation. Shunt reactors of roughly 50% compensation at both ends, reducing the effective \(Y\) and bringing \(|A|\) from 0.906 towards 0.95 — which alone cuts the no-load voltage to about 421 kV. Series compensation of 40–50% to bring \(|B|\) from 167 to about 100 \(\Omega\), addressing the regulation. An intermediate switching station at 200 km, halving the electrical length of each section and permitting sequential energisation.

An uncompensated 400 km EHV line fails every criterion, and that is not a defect of this particular design. It is why no such line exists: compensation is not an improvement added to a working line but a precondition for the line working at all. Set 14 takes up the compensation strategies quantitatively, and Set 38 shows the alternative — beyond about 600 km, HVDC dispenses with \(\beta\), with the Ferranti effect, with charging current and with the stability limit, all at once.

AnswerReg 27.1%, \(\eta\) 89.5%, no-load 441.5 kV, SIL 422 MW — fails on all three counts; needs shunt and series compensation

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 \(z = 0.20 + j0.45\ \Omega\)/km and \(y = j3.2\times10^{-6}\) S/km. Find \(\gamma\).

Show answer
\(z = 0.4924\angle66.04^\circ\); \(zy = 1.5757\times10^{-6}\angle156.04^\circ\); \(\gamma = \mathbf{1.2553\times10^{-3}\angle78.02^\circ}\)/km.
P2. Find \(Z_c\) for that line.

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\(z/y = 153\,875\angle-23.96^\circ\); \(Z_c = \mathbf{392.3\angle-11.98^\circ}\ \Omega\).
P3. Find its wavelength and velocity at 50 Hz.

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\(\beta = 1.2553\times10^{-3}\sin78.02^\circ = 1.2278\times10^{-3}\); \(\lambda = 2\pi/\beta = \mathbf{5117}\) km; \(v = \mathbf{2.56\times10^8}\) m/s.
P4. A 300 km line has \(\gamma l = 0.06 + j0.33\). Find \(\cosh\gamma l\).

Show answer
\(\cosh(0.06)\cos(0.33) + j\sinh(0.06)\sin(0.33) = (1.0018)(0.9461) + j(0.0601)(0.3239) = \mathbf{0.9478 + j0.0195}\).
P5. Find the Ferranti rise for that line.

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\(|A| = 0.9480\), so rise \(= 1/0.9480 - 1 = \mathbf{5.49\%}\).
P6. A line has \(\alpha = 3\times10^{-4}\) Np/km. Over what distance does a wave halve?

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\(\ln2/\alpha = 0.6931/3\times10^{-4} = \mathbf{2310}\) km.
P7. A line with \(Z_c = 400\ \Omega\) feeds a load of \(600\ \Omega\) resistive. Find \(\Gamma\).

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\((600-400)/(600+400) = \mathbf{0.20}\) — a fifth of the incident wave returns.
P8. What is \(\Gamma\) at an open-circuited end, and what does it imply for the voltage there?

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\(\Gamma = \mathbf{+1}\): the reflected wave equals the incident one and they add, doubling the voltage at the open end.
P9. Write the equivalent-\(\pi\) shunt admittance in terms of \(\gamma l\) and \(Z_c\).

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\(Y'/2 = \mathbf{\tanh(\gamma l/2)/Z_c}\), from the half-angle identity — Problem 10.
P10. A lossless line has \(\beta l = 30^\circ\) and \(Z_c = 380\ \Omega\). Write its ABCD constants.

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\(A = D = \cos30^\circ = \mathbf{0.866}\); \(B = j380\sin30^\circ = \mathbf{j190}\ \Omega\); \(C = j\sin30^\circ/380 = \mathbf{j1.316\times10^{-3}}\) S.
P11. At what electrical length does a line become a quarter wavelength, and what is \(A\) there for a lossless line?

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\(\beta l = 90^\circ\); \(A = \cos90^\circ = \mathbf{0}\), so the open-circuit voltage is unbounded — Problem 19.
P12. Why does a line terminated in \(Z_c\) have a flat voltage profile if lossless?

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Nothing reflects, so \(V(x) = V_Re^{\gamma x}\). With \(\alpha = 0\) that is a pure phase rotation and \(|V|\) is constant — Problems 11 and 12.

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 wave equation for a transmission line from Kirchhoff's laws applied to a differential element, solve it, and show how the ABCD constants follow.

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The element. Take a length \(dx\) at distance \(x\) from the receiving end, with series impedance \(z\,dx\) and shunt admittance \(y\,dx\). Kirchhoff's laws give
\[ \frac{dV}{dx} = zI, \qquad \frac{dI}{dx} = yV \]
Decoupling. Differentiating the first and substituting the second:
\[ \frac{d^2V}{dx^2} = z\frac{dI}{dx} = zyV = \gamma^2V \]
— the wave equation, with \(\gamma^2 = zy\). Similarly \(d^2I/dx^2 = \gamma^2I\).

General solution.
\[ V(x) = V^{+}e^{\gamma x} + V^{-}e^{-\gamma x} \]
and from \(I = (1/z)\,dV/dx\), using \(\gamma/z = \sqrt{y/z} = 1/Z_c\):
\[ I(x) = \frac{1}{Z_c}\left(V^{+}e^{\gamma x} - V^{-}e^{-\gamma x}\right) \]
Note the minus sign: the reflected wave carries current in the opposite direction, which is the whole content of the reflection.

Applying boundary conditions at \(x = 0\):
\[ V^{+} = \frac{V_R + Z_cI_R}{2}, \qquad V^{-} = \frac{V_R - Z_cI_R}{2} \]
Substituting back:
\[ V(x) = V_R\frac{e^{\gamma x} + e^{-\gamma x}}{2} + Z_cI_R\frac{e^{\gamma x} - e^{-\gamma x}}{2} = V_R\cosh\gamma x + Z_cI_R\sinh\gamma x \]

\[ I(x) = \frac{V_R}{Z_c}\sinh\gamma x + I_R\cosh\gamma x \]
At \(x = l\) these are exactly the two-port relations, and reading off the coefficients gives \(A = D = \cosh\gamma l\), \(B = Z_c\sinh\gamma l\), \(C = \sinh(\gamma l)/Z_c\).

What was assumed: that \(z\) and \(y\) are uniform along the line, that the line is in sinusoidal steady state at one frequency, and that it is a two-conductor system with no coupling to anything else. The first fails at a cable–line junction, the second for transients, the third for coupled circuits — and each failure requires a different treatment rather than a correction to this one.
C2. The ABCD constants describe only the terminals, yet the maximum voltage on a compensated long line often occurs in the interior. Establish where it occurs and why, and explain the consequence for insulation coordination.

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The profile. From the general solution, the magnitude at distance \(x\) from the receiving end is
\[ |V(x)| = |V_R\cosh\gamma x + Z_cI_R\sinh\gamma x| \]
This is a sum of two terms whose relative phase changes with \(x\) — hence interference, hence the possibility of an interior maximum.

When an interior maximum exists. Writing the solution in wave form, \(|V(x)|^2 = |V^{+}|^2e^{2\alpha x} + |V^{-}|^2e^{-2\alpha x} + 2|V^{+}||V^{-}|\cos(2\beta x + \psi)\). The cosine term oscillates with period \(\lambda/2\), so:
— For \(l < \lambda/4\) (under 1400 km here) the cosine completes less than half a cycle and the profile is monotonic or has at most one turning point.
— For a lightly loaded line, \(|V^{-}|\) is large and the interference is strong; the maximum can sit well inside.
— With shunt reactors at both ends only, the compensation acts at the terminals while the line's own capacitance acts throughout — so the middle is under-compensated and bulges.

The classic case. A 400 km line with 100% terminal shunt compensation has both end voltages held at 1.0 p.u., yet the midpoint can reach 1.05 p.u. or more, because the distributed capacitance between the reactors is uncompensated. The ABCD constants report both ends at 1.0 and say nothing about it.

Consequences for insulation coordination:
— Line insulation must be rated for the interior maximum, not the terminal voltage. Insulator strings are sometimes graded along a line for this reason.
— Surge arrester placement at the terminals does not protect the middle.
— Intermediate compensation — a reactor at a mid-line switching station — is the direct remedy, and is why long lines are sectionalised.
— Corona (Set 8) is worst where the voltage is highest, so an interior maximum puts the corona hot spot where nobody is measuring.

The general lesson: a two-port description is complete for terminal behaviour and silent about everything else. When the interior matters — insulation, corona, intermediate compensation, fault location — the full distributed solution must be evaluated, and the ABCD constants are not merely insufficient but actively misleading, because they report a perfectly acceptable pair of endpoint voltages.
C3. Beyond about 600 km, HVDC displaces a.c. transmission. Assemble the argument from the results of this set, and identify what a.c. retains.

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What this set establishes against long a.c. lines:
— Ferranti effect. Problem 17: the no-load rise grows as \(\sec\beta l\), reaching 54% at 800 km and diverging at 1407. A d.c. line has \(\beta = 0\) and no rise whatever.
— Charging current. Set 6 Problem 20 found 164 MVAr from 300 km at 400 kV, all of it to be absorbed by reactors. A d.c. line's capacitance draws current only when the voltage changes — that is, never in steady state.
— Stability limit. \(P = V_SV_R\sin\delta/X\) falls as the line lengthens, and the angle margin must be preserved. A d.c. link has no angle and no synchronising torque to lose.
— Compensation cost. A 600 km a.c. line needs shunt reactors, series capacitors and intermediate switching stations. All of that capital is avoided.
— Conductor utilisation. A.c. suffers skin effect and must be rated for peak rather than r.m.s. insulation; d.c. uses the full cross-section and the insulation is rated to the d.c. level, so a d.c. line carries more power on the same towers.
— Cables. Set 6 Problem 14 found a critical length of 80 km for a.c. cables. For submarine crossings there is no a.c. option at all beyond that.

Against HVDC:
— Converter cost, a large fixed expense at each end. This is what sets the break-even distance — roughly 600 km overhead, 50 km submarine — below which a.c. is cheaper simply because it needs no converters.
— Tapping is difficult. An a.c. line can be tapped anywhere with a transformer; a multi-terminal d.c. scheme requires expensive d.c. breakers and coordinated control.
— Converter losses, roughly 0.7% per station, so 1.4% is lost before any line loss.
— Harmonics and reactive demand at line-commutated converters, requiring large filter and capacitor banks.
— No natural voltage transformation. A transformer is the cheapest device in power engineering and has no d.c. equivalent.

What a.c. retains: everything below the break-even distance, which is the overwhelming majority of the network; all distribution; all generation and utilisation, since machines are a.c. by nature; and the meshed, tappable, transformable structure that makes a grid a grid rather than a set of point-to-point links. HVDC is a specialised long-haul and asynchronous-interconnection technology layered on top of an a.c. system, not a replacement for it — and every HVDC terminal in the world connects to an a.c. network at both ends.

Self-Test

Multiple-Choice Questions

MCQ 1. The propagation constant of a line is:
(a) \(\sqrt{z/y}\) (b) \(\sqrt{zy}\) (c) \(zy\) (d) \(z/y\)

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(b). Option (a) is the characteristic impedance — a ratio rather than a product.
MCQ 2. The exact \(A\) constant of a line is:
(a) \(1 + ZY/2\) (b) \(\cosh\gamma l\) (c) \(\sinh\gamma l\) (d) \(Z_c\sinh\gamma l\)

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(b). Option (a) is its two-term expansion — Problem 16.
MCQ 3. The attenuation constant \(\alpha\) is measured in:
(a) radians per km (b) nepers per km (c) ohms per km (d) it is dimensionless

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(b). \(\beta\) takes radians per km; 1 Np \(= 8.686\) dB.
MCQ 4. At 50 Hz an overhead line's wavelength is about:
(a) 60 km (b) 600 km (c) 6000 km (d) 60 000 km

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(c). About 5600–6000 km, since \(v \approx c\) — Problem 6.
MCQ 5. A line terminated in its characteristic impedance has:
(a) no reflected wave (b) no incident wave (c) zero current (d) infinite voltage

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(a). \(\Gamma = 0\), and the profile becomes a pure exponential \(e^{\alpha x}\) — Problem 11.
MCQ 6. The reflection coefficient at an open-circuited end is:
(a) 0 (b) \(+1\) (c) \(-1\) (d) infinite

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(b) \(+1\). The waves add and the voltage doubles there. A short circuit gives \(-1\) — Problem 15.
MCQ 7. The equivalent-\(\pi\) shunt admittance is:
(a) \(Y/2\) (b) \(\tanh(\gamma l/2)/Z_c\) (c) \(\sinh(\gamma l)/Z_c\) (d) \(Z_c\tanh(\gamma l/2)\)

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(b). It reproduces the hyperbolic constants exactly, which is how a long line enters a load flow — Problem 10.
MCQ 8. For a lossless line, \(B\) equals:
(a) \(Z_c\cos\beta l\) (b) \(jZ_c\sin\beta l\) (c) \(Z_c\) (d) \(j\sin\beta l/Z_c\)

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(b). \(A = \cos\beta l\) is real and \(B\), \(C\) are purely imaginary — Problem 18.
MCQ 9. As a line approaches a quarter wavelength, its open-circuit receiving voltage:
(a) falls to zero (b) equals the sending voltage (c) grows without bound (d) is unaffected

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(c). \(A = \cos\beta l \to 0\) and \(V_R = V_S/|A| \to \infty\); losses keep it finite but enormous — Problem 19.
MCQ 10. A quarter-wave line acts as:
(a) an impedance inverter (b) a short circuit (c) an open circuit (d) a transformer of ratio 1:1

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(a). \(Z_{in} = Z_c^2/Z_R\), so a short at one end appears as an open at the other.
MCQ 11. The nominal-\(\pi\) model is correct in \(A\) to order:
(a) \((\gamma l)^0\) (b) \((\gamma l)^2\) (c) \((\gamma l)^4\) (d) all orders

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(b). It omits the \((ZY)^2/24\) term; the short line is correct only to \((\gamma l)^0\) — Problem 16.
MCQ 12. As a line grows very long, \(|B|\):
(a) grows without bound (b) saturates towards \(|Z_c|\) (c) falls to zero (d) oscillates

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(b). \(\sinh(j\pi/2) = j\), so \(B \to Z_c\) at a quarter wave — while \(|A|\) collapses — Problem 17.

Reference

Key Formulas

Quantity	Relation	Notes
Wave equation	\(d^2V/dx^2 = \gamma^2V\)	From KVL/KCL on \(dx\)
Propagation constant	\(\gamma = \sqrt{zy} = \alpha + j\beta\)	Per unit length
Characteristic impedance	\(Z_c = \sqrt{z/y}\)	Lossless: \(\sqrt{L/C}\), real
Exact constants	\(A = D = \cosh\gamma l\), \(B = Z_c\sinh\gamma l\), \(C = \sinh\gamma l/Z_c\)	\(AD-BC = \cosh^2-\sinh^2 = 1\)
Complex hyperbolics	\(\cosh(a+jb) = \cosh a\cos b + j\sinh a\sin b\)	\(\sinh(a+jb) = \sinh a\cos b + j\cosh a\sin b\)
Line profile	\(V(x) = V_R\cosh\gamma x + Z_cI_R\sinh\gamma x\)	\(x\) from the receiving end
Wave form	\(V(x) = V^{+}e^{\gamma x} + V^{-}e^{-\gamma x}\)	\(V^{\pm} = (V_R \pm Z_cI_R)/2\)
Reflection coefficient	\(\Gamma = (Z_R - Z_c)/(Z_R + Z_c)\)	\(+1\) open, \(-1\) short, 0 matched
Wavelength	\(\lambda = 2\pi/\beta\)	About 5600–6000 km at 50 Hz
Velocity	\(v = \omega/\beta = f\lambda\)	Near \(c\) overhead, \(0.66c\) in cable
Attenuation	1 Np \(= 8.686\) dB	Half amplitude at \(\ln2/\alpha\)
Matched termination	\(V_S/V_R = e^{\gamma l}\)	Magnitude ratio \(e^{\alpha l}\)
Equivalent-π	\(Z' = Z_c\sinh\gamma l\), \(Y'/2 = \tanh(\gamma l/2)/Z_c\)	Exactly equivalent
Lossless constants	\(A = \cos\beta l\), \(B = jZ_c\sin\beta l\)	\(C = j\sin\beta l/Z_c\)
Ferranti, lossless	\(V_R/V_S = \sec\beta l\)	Diverges at \(\beta l = 90^\circ\)
Quarter-wave line	\(Z_{in} = Z_c^2/Z_R\)	Impedance inverter
Input impedance	\(Z_{in} = Z_c\dfrac{Z_R + jZ_c\tan\beta l}{Z_c + jZ_R\tan\beta l}\)	Lossless

Diagnostics

Common Mistakes

Confusing \(\gamma = \sqrt{zy}\) with \(Z_c = \sqrt{z/y}\). A product and a ratio; resistance affects them quite differently — Problems 1 and 2.
Using \(z\) and \(y\) per km with \(l\) in metres. \(\gamma l\) must be dimensionless; a factor of 1000 gives an absurd answer.
Treating \(\cosh(a + jb)\) as \(\cosh a + j\cosh b\). The correct expansion mixes hyperbolic and circular functions — Problem 3.
Forgetting that \(\gamma l\) is complex. Both the attenuation and the phase must be carried; using \(|\gamma|l\) alone loses the distinction.
Taking \(\beta l\) in degrees where the formula wants radians. \(\lambda = 2\pi/\beta\) requires radians throughout.
Using the nominal-\(\pi\) beyond 250 km. Its error in \(B\) reaches 3.5% at 400 km and grows quadratically — Problem 5.
Confusing the equivalent-\(\pi\) with the nominal-\(\pi\). The first is exact, the second an approximation — Problem 10.
Expecting a flat profile at SIL on a real line. It is exponential at \(e^{\alpha x}\); flat only if lossless — Problems 11 and 12.
Applying the ABCD constants to interior points. They describe the terminals only, and the interior maximum can exceed both — Challenge C2.
Assuming the lossless model is adequate for voltage drop. It errs 5.6% in \(B\) while erring only 0.45% in \(A\) — Problem 18.
Overlooking the sign in the current wave. \(I(x) = (V^{+}e^{\gamma x} - V^{-}e^{-\gamma x})/Z_c\) — the minus is the reflection — Challenge C1.
Ignoring the quarter-wave singularity. No lumped model contains it, and it is real — Problem 19.

Looking Ahead

The line is now solved exactly. Two constants — \(\gamma\) and \(Z_c\) — and a length give everything: the terminal behaviour through the hyperbolic ABCD constants, the interior through the travelling-wave solution, and the resonance that no lumped model contained.

Set 13 compares the medium and long treatments directly on the same line, quantifying where each may be used. Set 14 takes up what this set exposed but did not resolve — the Ferranti effect, surge impedance loading and the tuned line — as design problems rather than analytical ones. Set 15 closes Part 3 with the complex power relations that the wave solution implies. From Set 16 the line ceases to be an object of study at all and becomes a single entry in an admittance matrix.