Set 13 — Medium and Long Lines Compared
Twenty worked problems on choosing between the models of Sets 9, 11 and 12. They are not rivals but successive truncations of one series in the electrical length \(\gamma l\), and the choice between them is therefore a question about a single dimensionless number rather than about kilometres. This set quantifies the errors, locates the boundaries, and shows where the conventional length-based rules break down — for cables, at harmonic frequencies, and near resonance.
One series, three truncations. Short line: \(A = 1\), \(B = Z\), \(C = 0\). Nominal-\(\pi\): \(A = 1 + \frac{ZY}{2}\), \(B = Z\), \(C = Y(1+\frac{ZY}{4})\). Exact: \(\cosh\gamma l\), \(Z_c\sinh\gamma l\), \(\sinh(\gamma l)/Z_c\).
The criterion is \(\gamma l\), not the length. Errors scale with powers of the electrical length, so a rule in kilometres is only a proxy — valid for overhead lines at 50 Hz because \(\beta\) is nearly the same for all of them, and invalid for cables and harmonics.
Error growth. The nominal-\(\pi\)'s error in \(B\) grows as \((\gamma l)^2\) and its error in \(A\) as \((\gamma l)^4\). A doubling of length therefore quadruples one and multiplies the other by sixteen.
The conventional boundaries. Short below about 80 km (\(\beta l < 5^\circ\)), medium to about 250 km (\(\beta l < 16^\circ\)), long beyond. They correspond to errors of roughly 0.4%, 1.4% and more.
Cables are electrically longer than their length suggests. A cable's velocity is about \(0.63c\) against \(0.94c\) overhead, so its wavelength is only two thirds as long and a given kilometre counts for half as much again.
Harmonics multiply the electrical length. \(\beta\) scales with frequency, so a 400 km line is 26° long at 50 Hz and 75° long at the third harmonic — close to quarter-wave resonance, with a voltage magnification of nearly four.
Cascaded sections converge quadratically. Splitting a line into \(n\) nominal-\(\pi\) sections reduces the error by \(n^2\). It is the route when hyperbolic functions are unavailable, and it is how transient programs represent a line.
A line has \(z = 0.16 + j0.40\ \Omega\)/km and \(y = j3.0\times10^{-6}\) S/km. For a length of 300 km, compute the ABCD constants by all three models and set them side by side.
The two governing quantities, from Set 12:
Total lumped parameters:
Short line: \(A = 1\), \(B = 129.24\angle68.20^\circ\), \(C = 0\).
Nominal-\(\pi\):
Exact:
Collecting:
Tabulate the nominal-\(\pi\) error in \(|A|\) against length for the line of Problem 1, and deduce the scaling law.
Computing \(|\cosh\gamma l|\) and \(|1 + \mathbf{Z}\mathbf{Y}/2|\) at each length:
Each doubling of length multiplies the error by roughly 16 to 22.
The scaling law. From Set 12 Problem 16, the neglected term is the next in the cosh series:
A fourth-power law, so a doubling should multiply the error by 16 — which the table confirms, the excess coming from the next term again.
Checking at 400 km, where \(|\gamma l| = 0.4547\):
against the observed 0.195% — agreeing to the accuracy of a one-term estimate.
Repeat Problem 2 for \(|B|\), and explain why the error is so much larger than in \(A\).
The nominal-\(\pi\) takes \(B = \mathbf{Z}\) flat, while the exact value is \(Z_c\sinh\gamma l\):
Each doubling multiplies the error by about four — a second-power law rather than a fourth.
Why. The exact \(B\) expands as
and the nominal-\(\pi\) keeps only the leading \(\mathbf{Z}\). It therefore discards a term of order \((\gamma l)^2\), where in \(A\) it discarded one of order \((\gamma l)^4\).
Checking at 400 km:
against the observed 3.53%.
Tabulate the nominal-\(\pi\) error in \(|C|\), and explain why it is smaller than the error in \(B\) although both are second-order.
The values:
Also a second-power law — each doubling multiplies the error by four — but the coefficient is smaller.
Why smaller. The exact \(C\) expands as
while the nominal-\(\pi\) takes \(\mathbf{Y}(1 + \mathbf{Z}\mathbf{Y}/4)\) — it does include a correction, but with the wrong coefficient.
The discrepancy is therefore the difference between the two coefficients:
Exactly half the error in \(B\), as the coefficients \(1/12\) and \(1/6\) predict.
Express the conventional length-based model boundaries in terms of electrical length, and state the conditions under which the length-based rule is valid.
For an overhead line at 50 Hz, \(\beta \approx 1.1\times10^{-3}\) rad/km, so:
Why a length rule works at all. The phase constant is
and Set 6 Problem 16 showed that \(v \approx c\) for every overhead line regardless of its construction, because \(LC \approx \mu_0\varepsilon_0\). So \(\beta\) is nearly a universal constant at a given frequency, and kilometres translate directly into degrees.
The three conditions for validity:
At 60 Hz, for instance, the medium-line boundary falls from 250 to about 208 km for the same electrical length.
The 300 km line of Problem 1 delivers 100 MW at 220 kV, 0.9 power factor lagging. Find the voltage regulation predicted by each model.
The load, per phase:
Computing \(V_S = |A\mathbf{V}_R + B\mathbf{I}_R|\) and then \(\%\text{reg} = (V_S/|A| - V_R)/V_R\) for each:
Two errors that partly cancel. The short-line model overstates \(V_S\) by 4.6% — it has no shunt admittance to divert current from \(B\). But it also takes \(|A| = 1\), missing the 5.6% Ferranti rise. The two mistakes work in opposite directions and its regulation comes out only 1.2 points low.
The nominal-\(\pi\) gets \(|A|\) almost exactly right but overstates \(|B|\) by 1.8%, so it overstates the regulation by 0.6 points.
Find the Ferranti rise predicted by each model for the line of Problem 1 at 150, 300 and 600 km.
The rise is \(1/|A| - 1\) in every case:
The short-line model predicts zero at every length, because \(A = 1\) exactly. This is not an error that shrinks as the line shortens — it is the wrong answer at any length whatever.
The nominal-\(\pi\) tracks the exact model closely, and — worth noting — it overstates the rise slightly at every length:
This is the safe direction for design, though at much greater lengths it will reverse, since the lumped model has no pole.
Find the maximum power transfer predicted by each model for the line of Problem 1 at 150, 300 and 600 km, with both terminals at 220 kV.
From Set 11 Problem 19, the maximum receiving-end power is
where \(\alpha\) and \(\beta\) are here the angles of \(A\) and \(B\).
Evaluating:
in MW.
The short line and nominal-\(\pi\) agree exactly, because both take \(B = \mathbf{Z}\) and the power limit depends chiefly on \(|B|\). Their agreement is not evidence of correctness — both are wrong together.
Both understate the limit, by 1.4% at 300 km and 10.5% at 600 km:
The understatement follows directly from their overstatement of \(|B|\), since \(P_{\max} \propto 1/|B|\).
Assess the three models for a 150 km line and recommend one.
Electrical length:
Errors of the nominal-\(\pi\) against exact:
Under load at 100 MW, 0.9 lagging:
Recommendation: nominal-\(\pi\). Its errors are under half a per cent in every constant, its regulation is within 0.07 points of the exact value, and its Ferranti prediction is within 0.01 points.
The short-line model would give \(V_S\) 1.2% high and no Ferranti effect at all — acceptable only if the Ferranti rise is irrelevant to the study, which for a line that will be energised before being loaded it is not.
Repeat Problem 9 for a 300 km line — the borderline case.
Electrical length:
Beyond the conventional 16° boundary, so formally a long line.
Errors of the nominal-\(\pi\):
Under the same 100 MW load:
The verdict depends on the study. For regulation and Ferranti the nominal-\(\pi\) is within 0.6 points and 0.04 points — entirely adequate. For the power limit it is 2.8% low, which for a stability margin is marginal.
Recommendation: nominal-\(\pi\) for load-flow and voltage studies; exact constants for stability and power-limit work. The extra cost is one evaluation of two hyperbolic functions.
Repeat Problems 9 and 10 for a 600 km line, and state what the lumped models get badly wrong.
Electrical length:
The constants:
Under the same 100 MW, 0.9 lagging load at 220 kV:
What goes badly wrong. Three failures, in order of severity:
Recommendation: exact constants only. At \(\beta l = 38^\circ\) neither lumped model is defensible for any quantity.
Represent the 600 km line as two, then four, nominal-\(\pi\) sections in cascade, and find the error in \(|B|\) each time.
Each of \(n\) sections has \(\mathbf{Z}/n\) and \(\mathbf{Y}/n\), and the overall constants are the \(n\)-fold matrix product of the section matrices.
For \(n = 2\), each section covers 300 km with \(\mathbf{Z}_1 = 48 + j120\) and \(\mathbf{Y}_1 = j9.0\times10^{-4}\):
Against the exact \(|B| = 240.32\ \Omega\):
Down from 8.16% with a single section — a reduction of 4.6 times.
Continuing:
The error falls almost exactly as \(1/n^2\): from \(n = 2\) to \(n = 4\) it drops by a factor of 4.07, and from 4 to 8 by 4.01.
How many nominal-\(\pi\) sections are needed to represent the 600 km line with an error below 1% in \(|B|\)? Derive a general rule.
From the table of Problem 12, the error passes below 1% between \(n = 2\) (1.91%) and \(n = 3\) (0.84%):
The general rule. Each section has electrical length \(\gamma l/n\), and from Problem 3 its error in \(B\) is
Setting this equal to the target \(\varepsilon_{\text{target}}\) and solving:
For the 600 km line, \(|\gamma l| = 0.6821\) and \(\varepsilon_{\text{target}} = 0.01\):
Equivalently, and more memorably: each section should be electrically shorter than
A cable has \(R = 0.03\ \Omega\)/km, \(L = 0.15\) mH/km and \(C = 0.17\ \mu\text{F}\)/km. Find its \(\gamma\), \(Z_c\), wavelength and velocity, and compare with the overhead line.
Per-km parameters at 50 Hz:
Note the striking contrast with the overhead line: the reactance is eight times smaller and the susceptance eighteen times larger.
Propagation constant and characteristic impedance:
Wavelength and velocity:
Comparing directly:
Find the model boundaries for the cable of Problem 14, and compute its Ferranti rise at 50, 100 and 200 km.
The boundaries in electrical length are the same as for any line — \(5^\circ\) and \(16^\circ\). Converting to kilometres with \(\beta = 1.6584\times10^{-3}\) rad/km:
Against the overhead values of 80 and 250 km:
Both ratios equal the wavelength ratio \(3789/5628 = 0.673\), as they must.
The Ferranti rise at each length:
Comparing with the overhead line, where 200 km gave 2.45%: the cable's rise at 200 km is that of a 300 km overhead line.
The 400 km overhead line carries harmonic currents. Find its electrical length and \(|A|\) at the fundamental and at the 3rd, 5th, 7th and 11th harmonics.
At harmonic order \(h\) both the reactance and the susceptance scale with frequency:
The resistance is taken as unchanged, neglecting skin effect — which understates the damping at high orders.
Evaluating \(\gamma_h l\) and \(A = \cosh\gamma_h l\) at each order:
The pattern is not monotonic. \(|A| = |\cosh\gamma l|\) behaves as \(|\cos\beta l|\) damped by the losses, so it passes through minima near \(\beta l = 90^\circ, 270^\circ\) and maxima near \(0^\circ, 180^\circ\).
The 7th harmonic, at \(\beta l = 176^\circ\), is almost exactly a half wavelength — and there \(|A| \approx 1\), so the line is transparent to it.
Find the harmonic order at which the 400 km line reaches quarter-wave resonance, the voltage magnification at the third harmonic, and the practical consequences.
Quarter-wave resonance occurs at \(\beta_h l = 90^\circ\). Since \(\beta_1 l = 25.58^\circ\):
Between the third and fifth harmonics — and since only odd harmonics are present in a balanced three-phase system, the third is the nearest.
Magnification at the third harmonic. From Problem 16, \(|A| = 0.2655\):
A third-harmonic voltage injected at the sending end appears at the open receiving end nearly four times larger.
Why it does not diverge. Two things limit it: the line is not exactly at resonance (75.5° rather than 90°), and the resistance damps it. Even so, a 2% third-harmonic distortion at the source becomes 7.5% at the far end.
The practical consequences:
The problem is worst on a lightly loaded long line, since load damps the resonance — exactly the condition that also produces the worst Ferranti effect.
Compute the equivalent-\(\pi\) parameters of the 600 km line and compare with its nominal-\(\pi\), showing that the equivalent form removes the need to choose a model at all.
From Set 12 Problem 10, the exactly equivalent lumped parameters are
With \(\gamma l = 0.1290 + j0.6699\) for 600 km:
Against the nominal-\(\pi\), which uses \(\mathbf{Z} = 258.49\angle68.20^\circ\) and \(\mathbf{Y}/2 = 9.0\times10^{-4}\angle90^\circ\):
The series impedance differs by 7.6% and the shunt admittance by only 0.2% — consistent with the errors found in Problems 3 and 4.
The point. The equivalent-\(\pi\) is a lumped circuit of three elements, exactly like the nominal one, and slots into a \(Y\)-bus identically. But it reproduces the hyperbolic constants exactly at any length.
Different studies depend on different constants. Determine which constant dominates each of the common studies, and hence the model each requires.
Mapping studies to constants:
Combining with the error laws of Problems 2 to 4, where the nominal-\(\pi\) errs as \((\gamma l)^4/24\) in \(A\), \((\gamma l)^2/6\) in \(B\) and \((\gamma l)^2/12\) in \(C\):
So the boundary depends on the study by nearly a factor of three — from about 250 km for power-limit work to 700 km for Ferranti estimates.
The conventional single boundary at 250 km is therefore the most conservative of these, chosen so that one rule serves every purpose.
Set out a procedure for choosing a line model, and apply it to four cases: a 40 km 33 kV feeder; a 180 km 220 kV line; a 500 km 400 kV line; and a 120 km 132 kV cable.
The procedure:
Case 1 — 40 km, 33 kV feeder. \(\beta l = 2.6^\circ\), \(|\gamma l| = 0.045\):
Short line. The charging current is negligible and the feeder is never operated unloaded for long.
Case 2 — 180 km, 220 kV line. \(\beta l = 11.5^\circ\), \(|\gamma l| = 0.205\):
Nominal-\(\pi\). Comfortably inside the medium band for every study.
Case 3 — 500 km, 400 kV line. \(\beta l = 32.0^\circ\), \(|\gamma l| = 0.568\):
Exact or equivalent-\(\pi\). And the line needs compensation regardless, which the model must represent.
Case 4 — 120 km, 132 kV cable. Using the cable's \(\beta = 1.658\times10^{-3}\): \(\beta l = 11.4^\circ\), \(|\gamma l| = 0.207\):
Nominal-\(\pi\), though the length alone would have suggested a short line by the overhead rule. The charging current is the dominant concern here in any case — at 120 km this cable is beyond its critical length.
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 \(|\gamma l| = 0.25\). Estimate the nominal-\(\pi\) error in \(B\).
Show answer
\((0.25)^2/6 = 0.0104 = \mathbf{1.04\%}\).P2. For the same line, estimate the error in \(A\).
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\((0.25)^4/24 = 1.63\times10^{-4} = \mathbf{0.016\%}\) — sixty times smaller.P3. A 250 km line has \(\beta = 1.1\times10^{-3}\) rad/km. Which model?
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\(\beta l = 0.275\) rad \(= 15.8^\circ\) — just inside the medium band. Nominal-\(\pi\), with about 1.3% error in \(B\).P4. Why do the short-line and nominal-\(\pi\) models give identical power limits?
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Both take \(B = \mathbf{Z}\), and \(P_{\max} \propto 1/|B|\). Their agreement is no evidence of correctness — Problem 8.P5. How many nominal-\(\pi\) sections give under 0.5% error in \(B\) for a line with \(|\gamma l| = 0.68\)?
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\(n \ge 0.68/\sqrt{6(0.005)} = 0.68/0.173 = 3.9\), so \(n = \mathbf{4}\).P6. A cable has \(v = 0.63c\). What is its wavelength at 50 Hz?
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\(\lambda = v/f = 0.63(3\times10^5)/50 = \mathbf{3780}\) km, against about 5600 km overhead.P7. A 200 km cable and a 200 km overhead line — which is electrically longer, and by how much?
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The cable, by the inverse wavelength ratio \(5628/3789 = \mathbf{1.49}\) — it behaves like a 300 km overhead line.P8. A 300 km line has \(\beta_1 l = 19^\circ\). At what harmonic does it reach quarter-wave resonance?
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\(h = 90/19 = \mathbf{4.7}\) — dangerously close to the 5th harmonic, which is usually the dominant one.P9. Which constant governs the Ferranti effect, and which governs the power limit?
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Ferranti: \(\mathbf{A}\), through \(V_R = V_S/|A|\). Power limit: \(\mathbf{B}\), through \(P_{\max} \propto 1/|B|\) — Problem 19.P10. Why is the short-line model never acceptable for a no-load study, at any length?
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It sets \(A = 1\), predicting zero Ferranti rise. That is a category error, not a percentage error, and it does not shrink with length — Problem 7.P11. Write the equivalent-\(\pi\) parameters in terms of \(\gamma l\) and \(Z_c\).
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\(Z' = \mathbf{Z_c\sinh\gamma l}\) and \(Y'/2 = \mathbf{\tanh(\gamma l/2)/Z_c}\) — exact at any length.P12. A line at 60 Hz — how do the model boundaries in kilometres change?
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\(\beta \propto f\), so every boundary shrinks by \(50/60 = \mathbf{5/6}\): the medium limit falls from about 250 to 208 km.
Challenge Problems
Each of these needs an idea rather than a formula. Decide what the governing principle is before opening the answer.
C1. Problem 6 found the short-line model giving a nearly correct regulation through the cancellation of two large errors. Analyse when such cancellation occurs, why it is dangerous, and how to detect it.
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The two errors. The short-line model makes exactly two mistakes:
— It sets \(A = 1\) instead of \(|A| < 1\), which understates the no-load voltage and hence the regulation.
— It omits the shunt branch that would divert current from \(B\), which overstates \(V_S\) under load and hence overstates the regulation.
When they cancel. Writing the regulation as \((V_S/|A| - V_R)/V_R\), the short-line model has \(V_S\) too large by roughly \(|\gamma l|^2/2\) relative and \(|A|\) too large by the same order. The two effects are of similar magnitude and opposite sign, so cancellation is not accidental — it is structural, at moderate lagging loads.
It is exact when the two happen to balance, which for the 300 km line of Problem 6 occurred near 0.9 lagging. It fails when either term is suppressed:
— At no load, the second error vanishes (no current) and only the first remains: the model predicts 0% against 5.6%.
— At leading power factor, the impedance drop reverses sign and the errors add rather than cancel.
— At long lengths, the two errors grow at different rates — \(A\)'s as \((\gamma l)^4\), \(B\)'s effect as \((\gamma l)^2\) — so the balance is destroyed, as Problem 11 showed at 600 km.
Why it is dangerous. A model validated at one operating point appears trustworthy and is not. The agreement carries no information about the model's fidelity, because it arises from a coincidence rather than from the model capturing the physics.
How to detect it. Test the approximation at the extremes of the operating range, not at the nominal point — no load, full load, leading and lagging. If the error changes sign anywhere in that range, cancellation is occurring and the agreement at the nominal point is worthless. This is a general principle of approximation validation, not a fact about transmission lines.C2. The kilometre-based model boundaries work for overhead lines at power frequency. Establish exactly why, and enumerate every circumstance in which they fail.
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Why they work. The true criterion is \(\beta l\), andSet 6 Problem 16 showed that \(LC \approx \mu_0\varepsilon_0\) for every overhead line, because the logarithms cancel in the product. So \(v \approx c\) universally and \(\beta\) is nearly a constant of nature at a given frequency — about \(1.05\times10^{-3}\) rad/km at 50 Hz. Kilometres therefore translate into degrees at a fixed rate, independent of voltage class, conductor size, bundling or spacing.\[ \beta = \omega\sqrt{LC} = \frac{\omega}{v}, \qquad v = \frac{1}{\sqrt{LC}} \]
Where they fail:
— Cables. The dielectric has \(\varepsilon_r \approx 2.3\), so \(v \approx 0.63c\) and \(\beta\) is 50% larger. Boundaries shrink to two thirds — Problem 15.
— Other frequencies. \(\beta \propto f\). At 60 Hz boundaries shrink by 5/6; at the fifth harmonic by a factor of five — Problem 16.
— Transients. At surge frequencies of tens of kilohertz, \(\beta l\) for even a single span exceeds the boundaries, which is why travelling-wave methods are mandatory there.
— Series-compensated lines. A capacitor cancels part of \(x\), reducing the effective \(\beta\) and making the line electrically shorter than its length — the deliberate purpose of the compensation.
— Shunt-compensated lines. Reactors reduce the effective \(b\), again shortening the line electrically. A 600 km line with 60% shunt compensation behaves like a 380 km one.
— Gas-insulated lines. Different \(L\) and \(C\) again.
— Zero-sequence networks. Different \(z\) and \(y\) entirely, and hence a different \(\beta\) — the earth-return path changes both.
The general statement: the length rule is valid whenever the line's \(LC\) product is that of air at power frequency. Every failure above is a case where it is not. Computing \(\gamma\) directly costs one square root and is never wrong.C3. Modern practice uses the equivalent-\(\pi\) universally, which appears to make this entire set obsolete. Assess that claim.
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The claim has real force. A load-flow program computes \(Z' = Z_c\sinh\gamma l\) and \(Y'/2 = \tanh(\gamma l/2)/Z_c\) once at data preparation, then treats every line identically. No model selection occurs, no approximation is made, and the computational cost is two hyperbolic functions per line — negligible. In that sense the short and nominal-\(\pi\) models are indeed obsolete as computational tools.
What survives, and why it matters:
— Judgement about results. An engineer who knows that \(|A| = 0.79\) means a 26% Ferranti rise can recognise an implausible load-flow output. One who has only ever seen \(Z'\) and \(Y'\) as numbers in a data file cannot.
— Hand estimation. Feasibility studies, examination work, and sanity checks on a program's output all need a calculation that can be done without software. \(V_S \approx V_R + IR\cos\phi + IX\sin\phi\) is still the fastest route to a number.
— Understanding the exact solution. The hyperbolic constants are opaque; the series expansion of Problem 16 makes them intelligible by connecting them to the lumped models. Nobody develops intuition about \(\cosh\gamma l\) directly.
— Knowing what the equivalent-\(\pi\) still cannot do. It is exact for terminal behaviour at one frequency in the steady state — and says nothing about the interior voltage profile (Set 12 Challenge C2), about transients, or about other frequencies. An engineer who believes the equivalent-\(\pi\) is simply "exact" will use it where none of those conditions hold.
— Transient programs. These cannot use the frequency-domain equivalent at all and fall back on cascaded lumped sections — Problems 12 and 13 — so the cascade analysis remains live.
Verdict: obsolete as arithmetic, essential as understanding. The situation is exactly that of long division after the calculator: nobody performs it, and everybody must know what it does. The genuine risk of the equivalent-\(\pi\)'s universality is that it conceals the electrical length entirely, so a user never learns that a 120 km cable and a 40 km feeder are in different regimes — which is the specific error Problem 20 was constructed to expose.
Multiple-Choice Questions
MCQ 1. The correct criterion for choosing a line model is:
(a) the length in km (b) the voltage class (c) the electrical length \(\gamma l\) (d) the conductor sizeShow answer
(c). The length rule is a proxy valid only for overhead lines at power frequency — Problem 5.MCQ 2. The nominal-\(\pi\) error in \(B\) grows as:
(a) \(\gamma l\) (b) \((\gamma l)^2\) (c) \((\gamma l)^3\) (d) \((\gamma l)^4\)Show answer
(b), as \((\gamma l)^2/6\). The error in \(A\) is fourth-power — Problems 2 and 3.MCQ 3. Which constant does the short-line model get qualitatively wrong?
(a) \(A\) (b) \(B\) (c) \(C\) (d) noneShow answer
(c). Setting \(C = 0\) denies the charging current entirely — a category error rather than a percentage one.MCQ 4. The short-line and nominal-\(\pi\) models give the same power limit because:
(a) both are exact (b) both take \(B = Z\) (c) both take \(A = 1\) (d) coincidenceShow answer
(b). \(P_{\max} \propto 1/|B|\), so they inherit the same error — Problem 8.MCQ 5. Splitting a line into \(n\) nominal-\(\pi\) sections reduces the error by:
(a) \(n\) (b) \(n^2\) (c) \(\sqrt n\) (d) \(2^n\)Show answer
(b) \(n^2\) — each section's error is quadratic in its own electrical length — Problem 12.MCQ 6. A cable's wavelength compared with an overhead line's is about:
(a) the same (b) two thirds (c) half (d) twiceShow answer
(b) two thirds, since \(v \approx 0.63c\) against \(0.94c\) — Problem 14.MCQ 7. At the \(h\)-th harmonic, a line's electrical length is:
(a) unchanged (b) \(h\) times larger (c) \(h^2\) times larger (d) \(\sqrt h\) times largerShow answer
(b). \(\beta \propto f\), so a 400 km line is 26° at 50 Hz and 75° at the third harmonic — Problem 16.MCQ 8. A 400 km line reaches quarter-wave resonance at about which harmonic?
(a) 1.5 (b) 3.5 (c) 7 (d) 11Show answer
(b) 3.52, giving a magnification of 3.77 at the third harmonic — Problem 17.MCQ 9. The equivalent-\(\pi\) circuit is:
(a) an approximation better than nominal-\(\pi\) (b) exact at any length (c) valid only for short lines (d) the same as nominal-\(\pi\)Show answer
(b) exact, since it is constructed to reproduce the hyperbolic constants — Problem 18.MCQ 10. The nominal-\(\pi\) serves Ferranti studies to a greater length than power-limit studies because:
(a) \(A\) converges faster than \(B\) (b) \(B\) converges faster (c) they are equal (d) it does notShow answer
(a). \(A\)'s error is fourth-order and \(B\)'s second-order, so the usable length differs by nearly threefold — Problem 19.MCQ 11. At 60 Hz the model boundaries in kilometres:
(a) are unchanged (b) increase by 6/5 (c) decrease by 5/6 (d) decrease by 25/36Show answer
(c). \(\beta \propto f\), so a given electrical length is reached at a shorter physical length.MCQ 12. Two models agreeing on a result is evidence of correctness when:
(a) always (b) never (c) they differ in the assumption being tested (d) they use the same dataShow answer
(c). The short line and nominal-\(\pi\) agree on the power limit only because both share \(B = Z\) — Problem 8.
Key Formulas
| Quantity | Relation | Notes |
|---|---|---|
| Short line | \(A = 1\), \(B = Z\), \(C = 0\) | Correct to \((\gamma l)^0\) |
| Nominal-π | \(A = 1 + \frac{ZY}{2}\), \(B = Z\), \(C = Y(1+\frac{ZY}{4})\) | \(A\) correct to \((\gamma l)^2\) |
| Exact | \(\cosh\gamma l\), \(Z_c\sinh\gamma l\), \(\sinh\gamma l/Z_c\) | All orders |
| Equivalent-π | \(Z' = Z_c\sinh\gamma l\), \(Y'/2 = \tanh(\gamma l/2)/Z_c\) | Exact, lumped form |
| Error in \(A\) | \(\approx |\gamma l|^4/24\) | Fourth power |
| Error in \(B\) | \(\approx |\gamma l|^2/6\) | Second power — sets the boundary |
| Error in \(C\) | \(\approx |\gamma l|^2/12\) | Half that in \(B\) |
| Boundaries | short \(\beta l < 5^\circ\), medium \(< 16^\circ\) | 80 and 250 km overhead at 50 Hz |
| Cascade convergence | error \(\propto 1/n^2\) | \(n \ge |\gamma l|/\sqrt{6\varepsilon}\) |
| Section length rule | \(|\gamma l/n| \le \sqrt{6\varepsilon}\) | 14° per section for 1% |
| Cable velocity | \(v = c/\sqrt{\varepsilon_r} \approx 0.63c\) | Boundaries shrink to two thirds |
| Harmonic scaling | \(\beta_h = h\beta_1\) | Resonance at \(h = 90^\circ/\beta_1l\) |
| Frequency scaling | boundaries \(\propto 1/f\) | 60 Hz gives 5/6 of the 50 Hz lengths |
| Study dependence | Ferranti \(\to A\); power limit \(\to B\); charging \(\to C\) | Decides the usable length |
Common Mistakes
Choosing a model by length alone. Valid only for overhead lines at power frequency — Problem 5 and Challenge C2.
Applying overhead boundaries to a cable. A cable is electrically 50% longer per kilometre — Problem 15.
Using power-frequency constants at a harmonic. \(\beta\) scales with order, and a 400 km line is near resonance at the third — Problems 16 and 17.
Judging a model by its error in \(A\). \(A\) converges fourth-order and flatters the model; \(B\) sets the real boundary — Problem 3.
Treating agreement between two models as confirmation. Not when they share the assumption in question — Problem 8.
Trusting an approximation validated at one operating point. Cancellation of errors is common and does not survive a change of load — Problem 6 and Challenge C1.
Using the short-line model for any no-load study. It predicts zero Ferranti rise at every length — Problem 7.
Confusing the equivalent-\(\pi\) with the nominal-\(\pi\). One is exact, the other a two-term truncation — Problem 18.
Cascading sections that are individually too long. Each section must satisfy the medium-line criterion in its own right — Problem 13.
Ignoring compensation when assessing electrical length. Series and shunt compensation both shorten a line electrically.
Assuming the equivalent-\(\pi\) is exact for everything. It is exact for terminal behaviour at one frequency in the steady state, and nothing else — Challenge C3.
Applying positive-sequence \(\gamma\) to a zero-sequence study. The earth return changes both \(z\) and \(y\), and hence \(\beta\).
The three line models have been placed on one axis — the electrical length — and the choice between them reduced to an error estimate against the constant a given study depends on. The equivalent-\(\pi\) removes the choice entirely for computation, leaving the ladder as the route to understanding rather than to arithmetic.
Set 14 takes the phenomena this analysis exposed and treats them as design problems: the Ferranti effect and its compensation, surge impedance loading as an operating target, and the tuned line that Set 12's quarter-wave resonance suggested. Set 15 closes Part 3 with the complex power relations and the receiving-end circle diagram. From Set 16 the line becomes one entry in an admittance matrix and is not examined again.