Long Lines and the Wave Equation
Stop pretending the capacitance sits anywhere in particular, write Kirchhoff's laws for a slice of line of length \(dx\), and the transmission line turns into a differential equation whose solution is two waves travelling in opposite directions — from which the exact ABCD constants, the surge impedance and the Ferranti effect all follow.
- How the two first-order equations \(dV/dx = zI\) and \(dI/dx = yV\) come from one slice of line, and why differentiating either gives a wave equation.
- Why its solution introduces exactly two new constants — the characteristic impedance \(Z_c=\sqrt{z/y}\) and the propagation constant \(\gamma=\sqrt{zy}\) — and what each one physically is.
- Why the exact ABCD constants are \(A=D=\cosh\gamma l\), \(B=Z_c\sinh\gamma l\), \(C=\sinh(\gamma l)/Z_c\), and how their series expansions contain Chapter 11's nominal circuits as truncations.
- Three ways to evaluate a hyperbolic function of a complex argument, and when each is worth using.
- How the solution splits into an incident and a reflected travelling wave, what \(\alpha\), \(\beta\), \(\lambda\) and the wave velocity mean, and when the reflected wave vanishes.
- How to convert the exact constants into an equivalent-π circuit that a load-flow program can store, and how big the corrections are.
- Why the lossless line has a flat voltage profile at surge impedance loading, and what \(\mathrm{SIL}=V_{LL}^{2}/Z_s\) is worth.
The Limit of Lumping
Chapter 11 ended with a sequence and a suspicion. The sequence was one nominal-\(\pi\), then two half-length ones in cascade, then four, then sixteen — each subdivision cutting the error by four. The suspicion was that the sequence converges, and that whatever it converges on is the line itself.
There is nothing mysterious about why it converges. The error of a nominal circuit comes entirely from putting the capacitance in the wrong place: the real charging current enters gradually along the line, and the lumped model injects it in one or two abrupt gulps. The shorter the section, the less distance there is over which the model can be wrong, and the smaller the discrepancy. Taking the section length to zero removes the error altogether — at the price of an infinite number of sections, which is to say at the price of a differential equation.
That price is worth paying twice over. The first return is accuracy, and past about 250 km at 50 Hz there is no alternative: at 400 km a single nominal-\(\pi\) misstates \(B\) by nearly 8%, which is no longer a rounding. The second return is more interesting. A lumped circuit has terminals and nothing in between; it can say what the voltage is at the two ends and is silent about the middle. The distributed solution produces \(V(x)\) and \(I(x)\) as functions of position, and with them come questions a lumped model cannot even be asked: where along a lightly loaded line is the voltage highest, how fast does a disturbance travel, what happens where two lines of different construction meet.
As always the criterion is electrical, not geographical. What matters is \(\beta l\), the phase shift the line imposes, measured against the radian. Everything in Part 3 has been an argument about how large \(\beta l\) is: negligible in Chapter 10, small in Chapter 11, no longer small here.
The reference line of Chapter 11 — 200 km, 220 kV, \(z=0.16+j0.40\ \Omega\)/km, \(y=j3.0\times10^{-6}\) S/km — is carried through this chapter as well, so that the exact answers quoted there can be seen to be earned. Its nominal answers are on record: \(A=0.97605\angle0.563^\circ\), \(B_\pi = 86.16\angle68.20^\circ\ \Omega\), \(V_{s(LL)} = 234.84\) kV under the standard load. By the end of Section 12-5 the same quantities will have been computed without approximation.
The Equations of a Uniform Line
Consider one phase of a uniform line, with the neutral as the return, and measure distance \(x\) from the receiving end. Measuring from the far end rather than the near end is a choice of convenience: the receiving-end conditions are what a load problem gives, so making them the boundary conditions at \(x=0\) keeps the algebra short.
Over a length \(dx\) the line has a series impedance \(z\,dx\) and a shunt admittance \(y\,dx\), where \(z = r+j\omega L\) and \(y = g+j\omega C\) are the per-unit-length values Chapters 5 to 7 computed. The leakage conductance \(g\) is set to zero for an overhead line, as Chapter 5 argued: insulator leakage is small, variable with weather, and impossible to characterise usefully.
Two laws applied to the element give two equations. The voltage rises across the series impedance by the current times that impedance, and the current rises by whatever leaks off through the shunt admittance:
Strictly the shunt current should be \(y\,dx\,(V_x + dV_x)\), but the extra term is \(y\,dx\,dV_x\), a product of two infinitesimals, and vanishes in the limit. This is the only approximation made in the entire chapter, and it is not an approximation at all — it is the definition of the derivative.
The two equations are coupled: each derivative involves the other variable. Uncouple them by differentiating one and substituting the other.
Voltage and current satisfy the same second-order equation, so the same pair of exponentials describes both; only the constants differ. This is the steady-state, single-frequency form of the transmission-line wave equation, and every property of a long line is a property of its two solutions.
Solving Them: the Characteristic Impedance and the Propagation Constant
A linear second-order equation with constant coefficients has exponential solutions. Trying \(V_x = Ke^{sx}\) gives \(s^{2}=\gamma^{2}\), so \(s=\pm\gamma\) and the general solution is a combination of the two:
The current is not free to be chosen; it is fixed by the first of the two original equations. Differentiating and dividing by \(z\):
\(\gamma\) is the propagation constant: it says how a wave changes as it advances one kilometre — its real part \(\alpha\) how much amplitude is lost, its imaginary part \(\beta\) how much phase is gained. \(Z_c\) is the characteristic impedance, the ratio of voltage to current in a wave travelling in one direction only. Note carefully that \(\gamma\) and \(Z_c\) depend on \(z\) and \(y\) but not on the length; they are properties of the line's cross-section and the frequency alone.
The two solution constants can now be written compactly:
Look at what \(Z_c\) means before assigning \(C_1\) and \(C_2\) any values. Suppose \(C_2=0\), so that only the first exponential survives. Then \(V_x/I_x = Z_c\) everywhere along the line, at every \(x\). A wave of that kind sees the same impedance no matter how far it has travelled, because the line ahead of it always looks the same. It is in exactly this sense that \(Z_c\) is the impedance "of the line" — not of a length of it, but of the infinite line, which is the only length whose input impedance is independent of where you measure it.
For an overhead line at 50 Hz, \(z\) is dominated by \(j\omega L\) and \(y\) is purely \(j\omega C\), so \(z/y \approx L/C\) is nearly real and \(Z_c\) is nearly a real resistance of a few hundred ohms with a small negative angle contributed by \(r\). Numbers appear in Example 1.
The Terminal Conditions and the Hyperbolic Solution
The constants \(C_1\) and \(C_2\) are fixed by what happens at one end. Because \(x\) was measured from the receiving end, the conditions there are simply \(V_x=V_R\) and \(I_x=I_R\) at \(x=0\). Setting \(x=0\) in both solutions makes every exponential unity:
Substitute back and collect the terms in \(V_R\) and in \(I_R\):
These two equations are the complete solution: they give the voltage and current at every point of the line, not merely at its ends. Setting \(x=l\) picks out the sending end and produces the ABCD matrix.
\(A=D\), because a uniform line is symmetrical. And \(AD-BC = \cosh^{2}\gamma l - Z_c\sinh\gamma l\cdot\dfrac{\sinh\gamma l}{Z_c} = \cosh^{2}\gamma l-\sinh^{2}\gamma l = 1\) — reciprocity, delivered by a hyperbolic identity rather than by algebra on \(YZ\). Chapter 10's two checks survive intact into the exact theory.
Two dimensional remarks are worth making because they catch errors. The argument \(\gamma l\) is dimensionless — \(\gamma\) is per kilometre and \(l\) is kilometres — as it must be, since one cannot take the cosh of an impedance. And \(B=Z_c\sinh\gamma l\) has the dimension of \(Z_c\), an impedance, while \(C=\sinh(\gamma l)/Z_c\) has the dimension of an admittance: the same pattern as every ABCD matrix since Chapter 10.
Nothing in this derivation used the length. The equations hold for a line of any length whatever, and the short and medium models must therefore be recoverable from them. Section 12-5 shows how.
Evaluating cosh and sinh of a Complex Angle
The formulae of Section 12-4 are exact but not yet usable: \(\gamma l = \alpha l + j\beta l\) is complex, and \(\cosh\) of a complex number is not something a calculator's hyperbolic key will produce. Three methods are in use, and each is the right one in different circumstances.
Method 1 — the addition formulae. Expand \(\cosh(\alpha l+j\beta l)\) using the standard identity, then convert the hyperbolic functions of the imaginary part into circular ones with \(\cosh jθ = \cos θ\) and \(\sinh jθ = j\sin θ\):
Everything on the right is a real function of a real argument. The one trap is the angle: \(\beta l\) is in radians, and feeding it to a calculator set to degrees is the commonest single error in long-line problems. This method is exact, needs no assumption about length, and is the one to use for a genuinely long line.
Method 2 — the series. Both functions have everywhere-convergent power series, and because \((\gamma l)^{2} = zy\,l^{2} = \mathbf{Y}\mathbf{Z}\), only even powers of \(\gamma l\) ever appear once the \(Z_c\) or \(1/Z_c\) is folded in. The result is a series in the same product \(\mathbf{Y}\mathbf{Z}\) that Chapter 11 used:
The whole of Part 3 is therefore one formula read to successively higher order. It also explains why the nominal circuits carry \(1/4\) where the exact series carries \(1/6\): a lumped element cannot reproduce a \(1/6\), and the T and \(\pi\) circuits bracket it from either side, which is precisely the mirror-error result of Section 11-5.
The series converges very fast — for the reference line \(|YZ| = 0.0517\), so the third term is \(10^{-4}\) and the fourth is \(4\times10^{-6}\). Two terms beyond the leading one give six-figure accuracy up to about 400 km, which makes Method 2 the practical choice for hand calculation and the natural one for a computer, since it needs no trigonometry at all.
Method 3 — polar form. Write each exponential in magnitude-and-angle form directly:
This is Method 1 rearranged, and it is the form to reach for when the travelling-wave interpretation of Section 12-6 is wanted, because the two exponentials are the two waves. It is also the form in which the calculation is least error-prone on a calculator with a polar-to-rectangular key.
Incident and Reflected Waves
Return to the exponential form of the solution before it was folded into hyperbolic functions, and put the constants back in:
So far this is algebra. It becomes physics on restoring the time dependence. A phasor \(\mathbf{V}\) stands for the instantaneous quantity \(\sqrt2\,|\mathbf{V}|\cos(\omega t + \angle\mathbf{V})\), so writing \(\phi_1 = \angle(V_R+I_RZ_c)\) and \(\phi_2 = \angle(V_R-I_RZ_c)\), the instantaneous voltage at position \(x\) is
Each term is a function of two variables, time and distance, and each describes a wave. Follow a point of constant phase on the first term: \(\omega t+\beta x = \text{constant}\) requires \(dx/dt = -\omega/\beta\). The point moves in the direction of decreasing \(x\) — and since \(x\) was measured from the receiving end, decreasing \(x\) means moving toward the receiving end. This is the incident wave, launched at the source and travelling to the load. Its amplitude carries \(e^{\alpha x}\), which is largest at large \(x\) (the sending end) and smallest at \(x=0\): the wave loses amplitude as it travels, exactly as it should.
The second term has \(\omega t-\beta x=\) constant, so \(dx/dt = +\omega/\beta\): it travels back toward the source. This is the reflected wave, and its amplitude \(e^{-\alpha x}\) is largest at \(x=0\), where it is born, and decays as it retreats. The voltage measured anywhere on the line is the sum of the two.
The reflected wave exists only because the load does not match the line. Its amplitude is proportional to \(V_R-I_RZ_c\), which vanishes when \(V_R/I_R = Z_c\) — that is, when the load impedance equals the characteristic impedance. The ratio of the two amplitudes at the receiving end has a name:
\(\Gamma_R=0\) for a matched load \(Z_R=Z_c\) — the line delivers its surge impedance loading and nothing comes back. \(\Gamma_R=+1\) on open circuit (\(Z_R\to\infty\)): the wave returns with its voltage unchanged in sign, and incident and reflected add, which is the Ferranti effect seen as an interference. \(\Gamma_R=-1\) on short circuit: the returning voltage wave is inverted so that the two cancel at the fault, as they must.
Attenuation, Phase, Wavelength and Velocity
The two parts of \(\gamma=\alpha+j\beta\) are worth separating and naming, because between them they contain all the physical content of the propagation constant.
| Symbol | Name | Unit | Meaning |
|---|---|---|---|
| \(\alpha\) | Attenuation constant | neper/km | Fractional loss of amplitude per kilometre: amplitude falls by \(e^{-\alpha}\) each km. Zero if \(r=g=0\). |
| \(\beta\) | Phase constant | rad/km | Phase lag gained per kilometre. \(\beta l\) is the electrical length of the line. |
| \(\lambda = 2\pi/\beta\) | Wavelength | km | Distance over which the phase advances by one full cycle. |
| \(v = \omega/\beta = f\lambda\) | Wave velocity | km/s | Speed of a point of constant phase — approaching \(3\times10^{5}\) km/s for an overhead line. |
One neper is a loss factor of \(e\); the conversion to decibels is \(1\ \text{Np} = 20\log_{10}e = 8.686\) dB. For the reference line \(\alpha = 2.15\times10^{-4}\) Np/km, so over its 200 km the incident wave loses \(\alpha l = 0.043\) Np, which is \(0.37\) dB or 4.2% of its amplitude. Transmission lines are, by the standards of any other kind of cable, extraordinarily low-loss waveguides.
The phase constant is the more consequential of the two, because it sets both the wavelength and the electrical length that has governed every classification in Part 3. For the reference line \(\beta = 1.116\times10^{-3}\) rad/km, and therefore
This is the deep reason the lumped models of Chapters 10 and 11 work as well as they do, and it is the reason no ordinary line ever runs at \(\beta l = \pi/2\), where \(\cos\beta l\) would vanish. A quarter-wave line at 50 Hz would be 1400 km long; the tuned lines of Chapter 14, for which \(\beta l = \pi\), belong to the same arithmetic and explain why they are a theoretical curiosity rather than an engineering practice at power frequency.
The velocity deserves one further remark. For a lossless line, \(\gamma = j\omega\sqrt{LC}\) exactly, so \(\beta=\omega\sqrt{LC}\) and \(v = 1/\sqrt{LC}\), independent of frequency. Chapters 6 and 7 gave \(L = 2\times10^{-7}\ln(D_{eq}/D_s)\) and \(C = 2\pi\varepsilon_0/\ln(D_{eq}/r)\) per metre; if \(D_s\) were equal to \(r\), the two logarithms would cancel in the product and \(\sqrt{LC}=\sqrt{\mu_0\varepsilon_0}\), giving exactly the speed of light. They do not quite cancel, because \(D_s = 0.7788r\) accounts for internal flux linkage, so a real overhead line propagates at 94% to 98% of \(c\). The internal inductance of the conductor is the entire discrepancy — a satisfying closure of an argument begun in Chapter 6.
The Equivalent-π Circuit
The exact constants are a matrix, and a matrix is not a circuit. For load flow (Chapter 16 onwards), for fault studies, and for anything that must be drawn on an impedance diagram, a two-port has to be represented by elements. Chapter 11's nominal-\(\pi\) was a guess at those elements; here they can be computed, and the result is the equivalent-\(\pi\), which reproduces the exact line at the design frequency with no error at all.
Let the equivalent circuit have a series impedance \(Z'\) and a shunt admittance \(Y'/2\) at each end. Section 11-4 derived its constants, so the requirement is three equations:
The first is immediate. The second gives the shunt element once \(Z'\) is known:
using the half-angle identity \((\cosh\theta-1)/\sinh\theta = \tanh(\theta/2)\). Only two of the three constants were used, so the third is a genuine test: substituting into \(C=Y'(1+Y'Z'/4)\) and simplifying with \(\sinh\gamma l = 2\sinh(\gamma l/2)\cosh(\gamma l/2)\) gives \(1+Y'Z'/4 = \cosh^{2}(\gamma l/2)\) and hence \(C = (2/Z_c)\sinh(\gamma l/2)\cosh(\gamma l/2) = \sinh(\gamma l)/Z_c\), which is exactly what it must be. The equivalent-\(\pi\) is not an approximation.
The second form of each is the useful one: it says the equivalent circuit is the nominal circuit multiplied by a correction factor that tends to unity as \(\gamma l\to0\). For small \(\gamma l\), \(\sinh\theta/\theta \approx 1+\theta^{2}/6\) and \(\tanh(\theta)/\theta\approx1-\theta^{2}/3\), so the series impedance shrinks by about \(|YZ|/6\) and the shunt admittance grows by about \(|YZ|/12\).
Two cautions attach to the equivalent circuit. It is exact only at the frequency for which \(\gamma\) was evaluated: a network built to be an equivalent-\(\pi\) at 50 Hz is not one at 250 Hz, so harmonic studies must return to the distributed equations. And its shunt branch is not a pure capacitance — the correction factor is complex, so \(Y'/2\) carries a small real part. That real part is not a physical conductance; it is the bookkeeping by which a two-element circuit accounts for a loss that is actually spread along the conductor.
The corresponding equivalent-T follows by the same argument applied to Chapter 11's T constants, and is quoted for completeness: \(Y'' = \sinh(\gamma l)/Z_c\) and \(Z''/2 = Z_c\tanh(\gamma l/2)\), so the series halves are multiplied by \(\tanh(\gamma l/2)/(\gamma l/2)\) and the shunt by \(\sinh(\gamma l)/(\gamma l)\) — the two correction factors exchanged, which is the same mirror symmetry that governed the nominal circuits.
The Lossless Line and the Surge Impedance
On an EHV line the resistance is deliberately made small — that is what bundling in Chapter 8 achieves — and a great deal of insight is bought cheaply by setting \(r=0\) and \(g=0\) altogether. Then \(z=j\omega L\), \(y=j\omega C\), and both new constants simplify dramatically:
\(Z_s\) is real: a pure resistance, called the surge impedance or natural impedance. It is 350–400 \(\Omega\) for a single-conductor overhead line, about 250–300 \(\Omega\) for a bundled EHV line, and 30–50 \(\Omega\) for a cable, whose capacitance is enormous and inductance small (Chapter 9). With \(\alpha=0\) the hyperbolic functions become circular:
Every earlier result can now be read off in one line. The no-load rise is \(1/\cos\beta l\), which for small \(\beta l\) is \(1+(\beta l)^{2}/2\) — Chapter 11's Ferranti formula, now derived rather than estimated. Maximum power transfer occurs at \(\delta=\angle B = 90^\circ\) and equals \(V_sV_r/(Z_s\sin\beta l)\), so a line one wavelength-fraction long can transfer less power the longer it is, exactly as Section 11-8 found.
The most illuminating special case is a load exactly equal to \(Z_s\). Put \(I_R = V_R/Z_s\) into the general solution:
At SIL the magnitudes \(|V_x|\) and \(|I_x|\) are constant along the whole line: only the phase changes with distance. The voltage profile is flat, the power factor is unity everywhere, the regulation is zero, and no reflected wave exists. The line neither generates nor absorbs reactive power, because the reactive power produced by the shunt capacitance, \(V^{2}\omega C\) per unit length, exactly equals that consumed by the series inductance, \(I^{2}\omega L = V^{2}\omega L/Z_s^{2} = V^{2}\omega C\).
SIL is therefore the natural unit in which to quote a line's loading, and it is used that way throughout the industry: a line at 0.5 SIL is lightly loaded and will exhibit a voltage rise; at 1.5 SIL it is heavily loaded and will need reactive support at the receiving end. For the reference line, \(L = x/\omega = 1.273\) mH/km and \(C = b/\omega = 9.55\) nF/km, so \(Z_s = \sqrt{L/C} = 365\ \Omega\) and \(\mathrm{SIL} = 220^{2}/365 = 132.6\) MW — against the 60 MW it was asked to carry in Chapter 11, or 0.45 SIL. A lightly loaded line, exactly as its \(9.4\%\) regulation and \(2.5\%\) Ferranti rise imply.
Worked Examples
Problem. The 200 km, 220 kV, 50 Hz line has \(z = 0.16+j0.40\ \Omega\)/km and \(y = j3.0\times10^{-6}\) S/km. Find \(\gamma\), \(Z_c\) and the exact ABCD constants, and verify \(AD-BC=1\).
Solution. Both new constants come from products and quotients of \(z\) and \(y\), so put both in polar form first: \(z = 0.43081\angle68.199^\circ\ \Omega\)/km and \(y = 3.0\times10^{-6}\angle90^\circ\) S/km.
Now the electrical length. Multiplying \(\gamma\) by 200 km:
Check: \(A^{2} = 0.952643+j0.018591\) and \(BC = (85.48\angle68.383^\circ)(5.9521\times10^{-4}\angle90.184^\circ) = 0.050876\angle158.566^\circ = -0.047357+j0.018591\), so \(A^{2}-BC = 1.000000\;\checkmark\)
Chapter 11 predicted these to within \(0.008\%\) in \(A\) and \(0.4\%\) to \(0.8\%\) in \(B\) and \(C\), which the table in Section 11-5 now stands confirmed against.
Problem. The line of Example 1 delivers 60 MW at 220 kV, 0.9 power factor lagging. Find the sending-end voltage and current, the regulation and the efficiency, and compare with the nominal-circuit results of Chapter 11.
Solution. The receiving-end quantities are unchanged from Chapter 11: \(V_r = 127\,017\) V per phase and \(\mathbf{I}_r = 174.96\angle-25.84^\circ = 157.46-j76.26\) A.
Against Chapter 11: the nominal-T gave 234.52 kV and the nominal-\(\pi\) 234.84 kV, bracketing this 234.64 kV as promised; regulation 9.22% and 9.37% against 9.26%; efficiency 95.83% and 95.97% against 95.95%. The whole exercise of distributing the parameters has moved the sending-end voltage by 0.14% and the regulation by a tenth of a percentage point. At 200 km that is the honest size of the prize, and it is worth knowing so that the exact method is applied where it pays — at 400 km, in Example 4, it changes the answer by ten times as much.
Problem. For the same line, recompute \(A\), \(B\) and \(C\) by the series of Method 2, and then find the elements of the exact equivalent-\(\pi\) circuit.
Solution. With \(\mathbf{Z}=32+j80\), \(\mathbf{Y}=j6\times10^{-4}\) and \(\mathbf{Y}\mathbf{Z} = -0.048+j0.0192\), the square is \((YZ)^{2} = 0.0019354-j0.0018432\).
Three terms reproduce Example 1 to five figures, and the fourth term of the \(A\) series is \((YZ)^{3}/720 \approx 1.4\times10^{-7}\) — beyond any conceivable need.
For the equivalent-\(\pi\), form the two correction factors. With \(\gamma l = 0.22737\angle79.099^\circ\), \(\sinh\gamma l = 0.22556\angle79.283^\circ\), \(\gamma l/2 = 0.11369\angle79.099^\circ\) and \(\tanh(\gamma l/2) = 0.11414\angle79.007^\circ\):
So the exact circuit differs from the nominal one by a series impedance \(0.80\%\) smaller and a shunt admittance \(0.40\%\) larger — the ratio of two to one that Section 11-5 predicted from the \(1/6\) and \(1/12\) coefficients, appearing here as a statement about circuit elements rather than about constants. The \(Z'\) obtained is identical to the \(B\) of Example 1, as it must be, since \(B=Z'\) for any \(\pi\) network.
Problem. A 400 km, 400 kV, 50 Hz line has \(z = 0.03+j0.32\ \Omega\)/km and \(y = j3.6\times10^{-6}\) S/km. Find \(\gamma\), \(Z_c\), the ABCD constants, the open-circuit receiving voltage and charging current at 400 kV, the SIL, and the maximum power transfer with both ends at 400 kV.
Solution. \(z = 0.32140\angle84.644^\circ\), \(y = 3.6\times10^{-6}\angle90^\circ\).
Open-circuit the far end and hold 400 kV at the sending end. With \(V_s = 400\,000/\sqrt3 = 230\,940\) V per phase:
Chapter 11's estimate \((\beta l)^{2}/2 = 0.429795^{2}/2 = 9.24\%\) is within three-quarters of a point of the exact \(9.98\%\); the difference is the fourth-order term the estimate discarded, which at this length is no longer entirely negligible. A 40 kV overvoltage on an unloaded 400 kV line is far outside any insulation margin, which is why lines of this class are never energised without shunt reactors connected.
The maximum is at \(\delta = 84.8^\circ\), which no operator would go near; at a practical \(\delta=30^\circ\) the same formula gives 623 MW, or \(1.16\) SIL. That is the real loadability of a 400 km line, and it is set by angle, not by conductor temperature — the crossover Chapter 10 anticipated has now happened.
Problem. For the reference line at the load of Example 2, find the amplitudes and phases of the incident and reflected voltage waves at the receiving end, the reflection coefficient, and the wavelength and wave velocity. What load would remove the reflected wave?
Solution. Everything follows from \(I_RZ_c\):
The two add back to \(127\,017\angle0^\circ\), which is the check that they have been split correctly. The reflected wave is a substantial fraction of the incident one:
To make \(\Gamma_R\) vanish the load impedance must equal \(Z_c = 378.95\angle-10.90^\circ\ \Omega\), which per phase at 127 kV means \(V_r^{2}/|Z_c| = 42.6\) MW per phase, or \(127.7\) MW three-phase, drawn at \(\cos(10.90^\circ) = 0.982\) leading. Using the lossless surge impedance \(Z_s = 365\ \Omega\) instead gives the conventional \(\mathrm{SIL} = 220^{2}/365 = 132.6\) MW at unity power factor. The line is being asked to carry 60 MW at 0.9 lagging — less than half its natural loading, and of the wrong character — so a large reflected wave is exactly what should be expected. Add capacitors to bring the load toward \(Z_c\) and the reflection, the regulation and the voltage rise all fall together.
Problem. A 200 km line is tested at 50 Hz from its sending end. With the far end open the input impedance is \(1640\angle-89.62^\circ\ \Omega\); with the far end short-circuited it is \(87.57\angle67.82^\circ\ \Omega\). Determine \(Z_c\) and \(\gamma l\), and hence \(\gamma\) per kilometre.
Solution. Express each measurement in terms of the ABCD constants. On open circuit \(\mathbf{I}_R=0\), so \(\mathbf{V}_S=A\mathbf{V}_R\) and \(\mathbf{I}_S=C\mathbf{V}_R\); on short circuit \(\mathbf{V}_R=0\), so \(\mathbf{V}_S=B\mathbf{I}_R\) and \(\mathbf{I}_S=D\mathbf{I}_R\):
The product eliminates the hyperbolic function and the quotient eliminates \(Z_c\):
Invert the hyperbolic tangent with \(\tanh^{-1}w = \tfrac12\ln\dfrac{1+w}{1-w}\):
These are the constants of the reference line, recovered to four figures from two impedance measurements without any knowledge of \(r\), \(x\) or \(b\). Note that the angle inside the logarithm was taken in radians, and note the ambiguity the method carries: \(\tanh\) repeats every \(j\pi\) in its argument, so \(\gamma l\) is determined only to within \(j\pi\), and the correct branch is chosen by knowing the line's length to be well under half a wavelength. For a power line that is never in doubt. Finally, from \(\gamma\) and \(Z_c\) the per-kilometre parameters follow as \(z = \gamma Z_c\) and \(y = \gamma/Z_c\), so the test determines everything about the line.
Chapter Summary
\(dV/dx = zI\) and \(dI/dx=yV\) from one slice; differentiating gives \(d^{2}V/dx^{2}=\gamma^{2}V\).
\(\gamma=\sqrt{zy}=\alpha+j\beta\) per km and \(Z_c=\sqrt{z/y}\) — both independent of length.
\(A=D=\cosh\gamma l\), \(B=Z_c\sinh\gamma l\), \(C=\sinh(\gamma l)/Z_c\); \(AD-BC=1\) by identity.
\(A=1+\tfrac12 YZ+\tfrac{1}{24}(YZ)^{2}\), \(B=Z(1+\tfrac16 YZ)\) — Chapters 10 and 11 are truncations.
Incident \(\propto e^{\alpha x}e^{j\beta x}\) toward the load, reflected \(\propto e^{-\alpha x}e^{-j\beta x}\) back; \(\Gamma_R = (Z_R-Z_c)/(Z_R+Z_c)\).
\(\lambda=2\pi/\beta\approx6000\) km at 50 Hz; \(v=\omega/\beta\) is 94–98% of \(c\), the shortfall being internal flux.
\(Z'=Z\dfrac{\sinh\gamma l}{\gamma l}\), \(\dfrac{Y'}{2}=\dfrac{Y}{2}\dfrac{\tanh(\gamma l/2)}{\gamma l/2}\) — exact, but only at one frequency.
Loaded with \(Z_s=\sqrt{L/C}\), a lossless line has a flat voltage profile, unity power factor and no reflection.
Practice Problems
Keep \(\beta l\) in radians and check \(AD-BC=1\) before using any set of constants. Difficulty rises down the list.
- A 300 km, 50 Hz line has \(z=0.20+j0.45\ \Omega\)/km and \(y=j3.2\times10^{-6}\) S/km. Compute \(\gamma\), \(Z_c\), \(\alpha l\) and \(\beta l\), and hence the exact ABCD constants. Verify \(AD-BC=1\).
- For the line of Problem 1, delivering 80 MW at 220 kV and 0.9 power factor lagging, find the sending-end voltage and current, the regulation and the efficiency.
- Repeat Problem 1 using the three-term series of Method 2, and state the percentage difference from your exact answers for \(A\) and for \(B\).
- Compute the wavelength and the wave velocity of the line of Problem 1, express its length as a fraction of a wavelength, and state the length that would make it a quarter-wave line at 50 Hz.
- The line of Problem 1 is energised at 220 kV with the far end open. Find the receiving-end voltage, the charging current, and the reactive power drawn in MVAr. Compare the rise with the estimate \((\beta l)^{2}/2\).
- Find the elements \(Z'\) and \(Y'/2\) of the exact equivalent-\(\pi\) of the line of Problem 1, and state each as a percentage of the corresponding nominal value.
- A 500 km, 400 kV line has \(L = 0.97\) mH/km and \(C=11.8\) nF/km, with the resistance neglected. Find \(Z_s\), \(\beta\), the SIL, the no-load voltage rise at 400 kV, and the power transferred at a load angle of \(30^\circ\) with both ends at 400 kV.
- Open- and short-circuit tests on a 250 km line at 50 Hz give \(Z_{oc}=1120\angle-88.1^\circ\ \Omega\) and \(Z_{sc}=112\angle70.4^\circ\ \Omega\). Determine \(Z_c\), \(\gamma l\), \(\gamma\), and hence the series impedance \(z\) and shunt admittance \(y\) per kilometre.