Medium Transmission Lines
Once the charging current is too large to throw away, the line stops being a single impedance and becomes a genuine two-port — and the whole of its behaviour follows from deciding where along the line to pretend the capacitance sits.
- Why the shunt admittance \(Y\) cannot be dropped beyond about 80 km, tested with a number rather than a length in kilometres.
- How lumping a distributed \(Y\) at one point (nominal T) or at two points (nominal \(\pi\)) produces two different circuits, and the ABCD constants of each, derived from Kirchhoff's laws.
- Why both circuits give \(A=D=1+\tfrac12 YZ\), why \(AD-BC=1\) in both cases, and what each gets right and wrong when measured against the exact line of Chapter 12.
- Why regulation must now be written \(\big(|V_s|/|A| - |V_r|\big)/|V_r|\), and why the short line's simpler form was a special case.
- Where the Ferranti rise comes from, why it is approximately \((\beta l)^2/2\), and why a lightly loaded line is the dangerous one.
- How \(P_r\) and \(Q_r\) are written for a general two-port, and how Chapter 10's \(V_sV_r\sin\delta/X\) reappears as the special case \(A=1\).
- Why cascading two half-length nominal circuits cuts the error by four, and where that argument leads in Chapter 12.
When the Charging Current Stops Being Small
Chapter 10 rested on a single approximation and stated it plainly: the shunt admittance \(Y=j\omega C\) is set to zero. Everything that followed — the one-line phasor equation, the equality \(\mathbf{I}_s=\mathbf{I}_r\), the ABCD matrix with \(A=D=1\) and \(C=0\) — is a consequence of that one deletion. The chapter also gave the test that decides whether the deletion is permissible: compare the charging current with the load current, or equivalently compare the electrical length \(\beta l\) with one radian.
Apply that test to a line long enough to matter. Take a 200 km, 220 kV, 50 Hz overhead line with the parameters that Chapters 6 and 7 would produce for a single conductor per phase at that voltage: \(r = 0.16\ \Omega\)/km, \(x = 0.40\ \Omega\)/km and a shunt susceptance \(b = 3.0\ \mu\)S/km. This line will be used throughout the chapter, and Chapter 12 will solve it exactly, so it is worth writing its totals down once.
The phase voltage is \(220\,000/\sqrt3 = 127\,017\) V. If the whole of \(Y\) were connected at the receiving end it would draw
A 60 MW load at 0.9 power factor draws \(60\times10^{6}/(\sqrt3\times220\,000\times0.9) = 175.0\) A. The charging current is 44% of the load current — not the 1.6% of Chapter 10's short line. Deleting it would be an error of the same order as the answer, and the reactive power it represents, \(V_{LL}^{2}|Y| = (220\times10^{3})^{2}\times6.0\times10^{-4} = 29.0\) MVAr, happens to equal almost exactly the reactive power the load absorbs. A model that cannot see \(Y\) cannot see any of that.
The second, load-independent form of the test agrees. Chapter 12 will show that the exact \(A\) constant is \(\cosh(\gamma l)\), which is close to \(\cos(\beta l)\) when the losses are small. For this line \(\beta = 1.116\times10^{-3}\) rad/km, so \(\beta l = 0.2233\) rad \(=12.79^\circ\) and \(\cos(\beta l) = 0.9752\). The short-line model asserts \(A=1\); it is wrong by 2.5%, where at 80 km it was wrong by 0.4%. The error grows as \(l^{2}\), so doubling the length quadruples it, and by 200 km it has passed out of the range in which anyone would accept it.
The two boundaries have different characters. The lower one is the point at which neglecting \(Y\) altogether becomes unacceptable; the upper one is the point at which lumping \(Y\) becomes unacceptable and Chapter 12's distributed solution is required. Between them the line is a two-port with four non-trivial constants and nothing worse.
Lumping a Distributed Admittance
The line's capacitance is not at any one place. Every metre of conductor has capacitance to earth and to the other phases, so the charging current enters the line gradually: none of it at the far end, all of it by the time the near end is reached. Representing that continuously is exactly what Chapter 12 does, and it costs a differential equation. The medium-line model buys simplicity by pretending the capacitance is concentrated at a small number of points.
Two lumpings are in universal use, and they differ only in where the concentration is placed.
- Put the whole of \(Y\) at the middle of the line and split the series impedance into two halves, one on each side of it. The result looks like the letter T, and is the nominal-T circuit.
- Put half of \(Y\) at each end and leave the series impedance whole between them. The result looks like the letter \(\pi\), and is the nominal-\(\pi\) circuit.
The word nominal matters. These circuits are not equivalent to the line; they are approximations to it that happen to use the nominal (total) values \(Z\) and \(Y\). Chapter 12 constructs the equivalent-\(\pi\), whose elements are corrected values that reproduce the line exactly at the working frequency. The difference between nominal and equivalent is the difference between a convenient guess at where to put the capacitance and a calculation of what to put there.
Both circuits are built from the same two totals, \(Z\) and \(Y\), and the same rule for combining them, and both will be reduced by the same procedure: write Kirchhoff's laws with \(\mathbf{V}_r\) and \(\mathbf{I}_r\) as the known quantities, and eliminate everything else until \(\mathbf{V}_s\) and \(\mathbf{I}_s\) stand alone. What emerges is the ABCD matrix that Chapter 10 introduced.
Nothing about that statement depended on the line being short. It follows from linearity, from reciprocity, and from the line looking the same from both ends. Whatever the two nominal circuits produce must satisfy both identities, and checking that it does is the first test of any derivation below.
The Nominal-T Circuit
Draw the receiving end on the right. The load current \(\mathbf{I}_r\) leaves at the receiving terminals, having flowed through the right-hand half-impedance \(\mathbf{Z}/2\) from the mid-point. Call the mid-point voltage \(\mathbf{V}_C\), since it is the voltage across the lumped capacitance.
Three equations describe the circuit completely. The first is Kirchhoff's voltage law along the right-hand half, the second is Kirchhoff's current law at the mid-point node, and the third is the voltage law along the left-hand half:
Substitute the first into the second to remove \(\mathbf{V}_C\) from the current equation:
That is already the second row of the matrix: \(C=\mathbf{Y}\) and \(D = 1+\tfrac12\mathbf{Y}\mathbf{Z}\). Now substitute both results into the third equation:
\(A=D\), as the symmetry of the circuit demands — it is unchanged if the two ends are swapped. The determinant is \(\left(1+\tfrac12 YZ\right)^{2} - YZ\left(1+\tfrac14 YZ\right) = 1 + YZ + \tfrac14 Y^{2}Z^{2} - YZ - \tfrac14 Y^{2}Z^{2} = 1\), so reciprocity holds exactly, not approximately.
Every term has a reading. \(C=\mathbf{Y}\) says that the current that leaks off the line is the whole admittance times the mid-point voltage — and on open circuit, when \(\mathbf{I}_r=0\), that is the entire sending-end current. The factor \(\left(1+\tfrac14 YZ\right)\) multiplying \(\mathbf{Z}\) in \(B\) says that a load current does not see the full series impedance: part of the drop it would have caused is offset by the charging current flowing the other way through the second half of \(\mathbf{Z}\). Since \(YZ\) is negative and real to first order (\(Y\) is nearly \(+j|Y|\) and \(Z\) is nearly \(+j|Z|\), so their product is nearly \(-|Y||Z|\)), that factor is slightly less than one and \(B\) is smaller than \(\mathbf{Z}\).
Setting \(\mathbf{Y}=0\) collapses the matrix to \(\begin{bmatrix}1&\mathbf{Z}\\0&1\end{bmatrix}\), which is Chapter 10 exactly. That is the sense in which the short line is a special case rather than a separate theory.
The Nominal-π Circuit
The alternative lumping splits \(Y\) rather than \(Z\). Half the admittance, \(\mathbf{Y}/2\), hangs at the receiving-end busbar and half at the sending-end busbar, with the whole series impedance \(\mathbf{Z}\) between them. There is now no mid-point node to name, and the derivation is one step shorter.
The current in the series branch is the load current plus whatever the receiving-end capacitance draws, and the sending-end voltage follows from one application of the voltage law along that branch:
So \(A = 1+\tfrac12\mathbf{Y}\mathbf{Z}\) again, and \(B=\mathbf{Z}\) exactly — the load current sees the whole series impedance, because in this circuit nothing stands between the two ends but \(\mathbf{Z}\). The sending-end current is the series current plus the charging current of the sending-end half:
Compare with the T matrix: \(A\) and \(D\) are identical, while the correction factor \(\left(1+\tfrac14 YZ\right)\) has moved from \(B\) to \(C\). The two circuits are, in a precise sense, each other's mirror image — a fact that Section 11-5 turns into a quantitative statement about their errors.
The nominal-\(\pi\) is the one used in practice, and there are three reasons for it. Its shunt branches attach at the busbars, where the real capacitor banks, reactors and other lines also attach, so a network assembled from \(\pi\) sections has all its shunt elements at nodes that already exist — which is exactly what the \(Y\)-bus of Chapter 16 needs, and why the \(\pi\) model is what a load-flow program stores for every line. Its \(B\) constant is the plain series impedance, which keeps hand calculation simple. And it has no fictitious internal node, whereas the T circuit's mid-point is a node the real line does not possess.
Comparing the Two, and Checking Both
Two different circuits claim to represent the same line and give different answers. Which is right, and by how much is the other wrong? The question can only be settled against the exact solution, which Chapter 12 obtains by solving the line's differential equations. That solution gives
Set the three results side by side. Both nominal circuits reproduce \(A\) correctly as far as the term in \(YZ\), missing only \((YZ)^{2}/24\); since \(|YZ|\approx(\beta l)^{2}\), that missing term is of order \((\beta l)^{4}/24\), which for the reference line is \(1\times10^{-4}\). The constant \(A\) is therefore essentially exact in both circuits, and any difference between them must live in \(B\) and \(C\).
It does, and symmetrically. The exact \(B\) carries the factor \(\left(1+\tfrac16 YZ\right)\); the T circuit supplies \(\left(1+\tfrac14 YZ\right)\) and overshoots by \(YZ/12\), while the \(\pi\) circuit supplies \(1\) and undershoots by \(YZ/6\). For \(C\) the roles are exchanged: the \(\pi\) circuit errs by \(YZ/12\) and the T circuit by \(YZ/6\).
Neither circuit is uniformly better: each is twice as accurate as the other in one constant and twice as bad in the other, and the errors have opposite signs. Because \(YZ\approx-(\beta l)^{2}\), every error grows as the square of the length — the reason the medium-line model has an upper length limit at all.
Numbers make the point better than the algebra. For the reference line, \(\mathbf{Y}\mathbf{Z} = j6\times10^{-4}(32+j80) = -0.0480+j0.0192\), so
Chapter 12 will find the exact constants of this line to be \(A=0.97613\angle0.559^\circ\), \(B=85.48\angle68.38^\circ\ \Omega\) and \(C=5.952\times10^{-4}\angle90.18^\circ\) S. The table sets all three models against each other.
| Constant | Nominal T | Nominal π | Exact (Ch. 12) | T error | π error |
|---|---|---|---|---|---|
| \(A=D\) | \(0.97605\angle0.563^\circ\) | \(0.97605\angle0.563^\circ\) | \(0.97613\angle0.559^\circ\) | \(-0.008\%\) | \(-0.008\%\) |
| \(B\ (\Omega)\) | \(85.13\angle68.48^\circ\) | \(86.16\angle68.20^\circ\) | \(85.48\angle68.38^\circ\) | \(-0.41\%\) | \(+0.80\%\) |
| \(C\ (\mu\text{S})\) | \(600.0\angle90.00^\circ\) | \(592.8\angle90.28^\circ\) | \(595.2\angle90.18^\circ\) | \(+0.80\%\) | \(-0.41\%\) |
| \(AD-BC\) | \(1\) exactly | \(1\) exactly | \(1\) exactly | — | — |
The predicted factor of two, and the predicted change of sign, both appear. The practical conclusion is that at 200 km either circuit is good to under one per cent in every constant, which is finer than the parameters \(r\), \(x\) and \(b\) are themselves known — Chapter 5's resistance depends on a temperature that is only estimated, and Chapter 7's capacitance on a conductor height that sags with the load. Arguing about the third significant figure of \(B\) is arguing below the noise floor of the data.
Regulation and Efficiency for a Two-Port
The definitions of Chapter 10 stand unchanged; only the way they are computed alters, and in each case the alteration traces back to a single fact — the line no longer carries the same current at both ends.
Voltage regulation is the rise in receiving-end voltage when the load is thrown off with \(\mathbf{V}_s\) held constant. Throwing off the load sets \(\mathbf{I}_r=0\), and the first ABCD relation then reads \(\mathbf{V}_s = A\mathbf{V}_r\). Unlike the short line, the receiving end does not rise to \(V_s\): it rises to \(V_s/A\), and since \(|A|\lt1\) that is higher than \(V_s\).
Setting \(|A|=1\) recovers Chapter 10's \(\big(|V_s|-|V_r|\big)/|V_r|\). Because \(|A|\lt1\), dividing by it raises the no-load voltage and so raises the computed regulation above what the short-line formula would have given for the same \(V_s\) and \(V_r\) — the line's own capacitance works against it at light load even as it helps at heavy load.
Transmission efficiency is still \(P_r/P_s\), but \(P_s\) can no longer be written as \(P_r\) plus \(3I^{2}R\) with a single \(I\), because the current varies along the line. Two routes are open. The direct one computes the sending-end power from the sending-end quantities:
The indirect one adds up the losses branch by branch, and the answer depends on the circuit chosen because the two circuits distribute the current differently. In the nominal-\(\pi\), all the resistance is in one branch carrying \(\mathbf{I}_Z\); in the nominal-T, half the resistance carries \(\mathbf{I}_r\) and half carries \(\mathbf{I}_s\):
Agreement between the direct and the indirect route is the arithmetic check on the whole calculation, exactly as it was in Chapter 10, and it is worth performing every time: it catches an error in \(\cos\phi_s\), which is the quantity students most often get wrong, because \(\phi_s\) is now the angle between two phasors that have both moved.
One consequence of the shunt branch deserves stating on its own. On a lightly loaded medium line \(|\mathbf{I}_s|\) can be smaller than \(|\mathbf{I}_r|\), because the charging current at the receiving end partly cancels the lagging load current before the series impedance is reached. Example 1 shows exactly that: 175 A enters the load and only 154 A leaves the sending-end busbar. The line is not creating power; it is supplying reactive power locally, so less current has to be pushed through the impedance to deliver the same watts. That is the same effect a capacitor bank produces in Chapter 10's Example 6, except that here the line provides it free.
The Line on No Load: the Ferranti Rise
Set \(\mathbf{I}_r=0\) and ask what the open-circuited line does. The two ABCD relations reduce to \(\mathbf{V}_s = A\mathbf{V}_r\) and \(\mathbf{I}_s = C\mathbf{V}_r\), so
Take the reference line's numbers. \(|A| = 0.97605\), so \(|\mathbf{V}_{r,\text{NL}}| = |\mathbf{V}_s|/0.97605 = 1.0245\,|\mathbf{V}_s|\). Energising the line at 220 kV and leaving the far end open leaves 225.4 kV standing at that open end: the receiving voltage exceeds the sending voltage by 2.45%. This is the Ferranti effect, and it is the first phenomenon in this book that the short-line model was structurally incapable of showing.
The mechanism is visible in the circuit. With no load, the only current flowing is the charging current drawn by the shunt capacitance, and that current leads the voltage by 90°. It flows through the series inductive reactance, where a leading current produces a voltage rise in the direction of flow rather than a drop — the same sign reversal that made regulation negative at a leading power factor in Chapter 10, section 10-6, now produced by the line itself rather than by the load.
The size of the rise follows from a one-line estimate. Neglect the resistance, so that \(\mathbf{Z}\approx j\omega Ll\) and \(\mathbf{Y}\approx j\omega Cl\) and \(\mathbf{Y}\mathbf{Z} \approx -\omega^{2}LCl^{2} = -(\beta l)^{2}\), with \(\beta=\omega\sqrt{LC}\) as in Section 11-1. Then
For the reference line \(\beta l = 0.2233\) rad and the estimate gives \(2.49\%\) against the computed \(2.45\%\). At 300 km it becomes \(5.6\%\), at 400 km \(10.0\%\). Since it does not involve the load at all, it cannot be cured by anything done at the load end: the remedy is a shunt reactor, which absorbs the charging current before it reaches the reactance. Chapter 14 develops the effect in full and Chapter 34 sizes the reactors.
The open-circuit sending current is worth a moment too. With \(C = 5.928\times10^{-4}\angle90.28^\circ\) S and \(V_{r,\text{NL}}=130\,134\) V per phase, \(|\mathbf{I}_s| = 77.1\) A at almost exactly \(90^\circ\) leading — a line drawing 77 A while delivering nothing at all. That current is a real load on the generator's stator and on every transformer between, and it is why an unloaded transmission line is not a harmless thing to leave energised.
Power Transfer in ABCD Form
Chapter 10 asked what flows when both ends are held at fixed voltage, and answered \(P=V_sV_r\sin\delta/X\). That answer used \(\mathbf{I}=(\mathbf{V}_s-\mathbf{V}_r)/\mathbf{Z}\), which is only true when \(A=1\) and \(C=0\). The general version replaces it.
Write \(A = |A|\angle\alpha\), \(B=|B|\angle\beta\), \(\mathbf{V}_s = |V_s|\angle\delta\) and \(\mathbf{V}_r=|V_r|\angle0^\circ\). Solving the first ABCD relation for the receiving current and forming the complex power gives
With \(A=1\), \(\alpha=0\) and \(B=\mathbf{Z}=|Z|\angle\theta\) these become Chapter 10's expressions term for term. The maximum is reached, as before, when the first cosine is unity — that is, at \(\delta=\beta\), the angle of \(B\) rather than the angle of \(Z\).
Evaluate it for the reference line with both ends held at 220 kV. With \(|A|=0.97605\), \(\alpha=0.563^\circ\), \(|B|=86.16\ \Omega\), \(\beta=68.20^\circ\) and \(|V_s|=|V_r|=127\,017\) V, the first term is \(127\,017^{2}/86.16 = 1.872\times10^{8}\) W per phase and the second is \(0.97605\times1.872\times10^{8}\times\cos67.64^\circ = 0.695\times10^{8}\) W, so
Set that beside the thermal rating. A single 220 kV circuit rated at 400 A carries \(\sqrt3\times220\times0.4 = 152\) MVA. The stability limit is still above the thermal limit, but only by a factor of 2.3, where Chapter 10's 20 km line had a margin of 6.5. The trend is unmistakable: \(|B|\) grows roughly in proportion to length, \(P_{r,\max}\) falls in inverse proportion, and somewhere past 300 km the two limits cross. Chapter 14 names the crossing point in terms of the surge impedance loading, and Chapter 28 turns it into a stability criterion.
Cascading, and Where the Nominal Circuits Stop
Chapter 10 noted that two-ports in cascade multiply. That rule is worth restating here because it is what makes the ABCD form indispensable rather than merely tidy: a line, a transformer at each end, a series capacitor, and a second line in tandem all reduce to one matrix product.
The order matters — matrix multiplication does not commute — and the convention is that the leftmost matrix is the one nearest the sending end, because the column vector it eventually multiplies is the receiving-end pair. Each factor has unit determinant, so the product does too, and the check \(AD-BC=1\) survives any amount of cascading.
That rule immediately suggests a way to push the medium-line model past its natural limit. If a 300 km line is too long for one nominal-\(\pi\) circuit, represent it as two 150 km nominal-\(\pi\) circuits in cascade. Each section then has a quarter of the \(YZ\) product of the whole, so each section's error is a quarter of the whole line's, and there are two of them — a net improvement of a factor of four. Example 6 works the arithmetic and confirms it exactly.
Repeating the subdivision is the obvious next step: four sections of 75 km, sixteen of 18.75 km, and so on, each halving of the section length quartering the error. In the limit of infinitely many infinitesimal sections the model stops being an approximation at all. That limit is a differential equation, and taking it is precisely what Chapter 12 does.
The criterion is on \(\beta l\), never on kilometres alone. A 60 Hz line reaches the same \(\beta l\) in \(5/6\) of the distance. An underground cable, whose capacitance per kilometre is thirty times an overhead line's (Chapter 9), reaches it in perhaps a fifth of the distance, which is why a 50 km cable already demands the treatment that a 250 km overhead line does.
One honest limitation remains. The nominal circuits give the terminal behaviour and nothing else. They say what happens at the two ends; they say nothing about the voltage at the middle of the line, because their internal nodes are fictitious. If the question is where along a 400 km line the voltage is highest at light load, or how a surge propagates, no lumped circuit of any refinement will answer it — the answer is a function of distance, and only the distributed model produces functions of distance. That is the second reason for Chapter 12, and the more interesting one.
Worked Examples
Problem. The 200 km, 220 kV, 50 Hz line of Section 11-1 has \(\mathbf{Z}=32+j80\ \Omega\) and \(\mathbf{Y}=j6.0\times10^{-4}\) S. It delivers 60 MW at 220 kV and 0.9 power factor lagging. Using the nominal-\(\pi\) circuit, find the ABCD constants, the sending-end voltage and current, the sending-end power factor, the voltage regulation and the efficiency.
Solution. Work per phase with \(\mathbf{V}_r\) as reference. First the constants, from \(\mathbf{Y}\mathbf{Z} = j6\times10^{-4}(32+j80) = -0.0480+j0.0192\):
Check: \(AD-BC = (0.97605\angle0.563^\circ)^{2} - (86.16\angle68.20^\circ)(5.928\times10^{-4}\angle90.28^\circ) = 0.95248+j0.01874 - (-0.04752+j0.01874) = 1.000\ \checkmark\)
Now the receiving-end quantities. With \(\cos\phi_r=0.9\), \(\sin\phi_r=0.43589\):
The sending current is less than the receiving current, 154 A against 175 A, for the reason Section 11-6 gave. The sending power factor is set by the angle between \(\mathbf{V}_s\) and \(\mathbf{I}_s\):
Check the loss independently. The series-branch current is \(\mathbf{I}_Z = \mathbf{I}_r+\mathbf{V}_r\mathbf{Y}/2 = 157.46-j76.26+j38.11 = 157.46-j38.16 = 162.02\) A, so \(P_{\text{loss}} = 3\times162.02^{2}\times32 = 2.520\) MW and \(P_s = 60+2.52 = 62.52\) MW — agreeing with the terminal computation to four figures.
Problem. Repeat Example 1 using the nominal-T circuit and compare. The exact answer, from Chapter 12, is \(V_{s(LL)}=234.64\) kV with \(\eta=95.95\%\) and \(9.26\%\) regulation.
Solution. \(A\) and \(D\) are unchanged; only \(B\) and \(C\) move:
so \(V_{s(LL)} = \sqrt3\times135\,400 = 234.52\) kV, and
Set the three results side by side:
| Quantity | Nominal T | Exact | Nominal π |
|---|---|---|---|
| \(V_{s(LL)}\) (kV) | 234.52 | 234.64 | 234.84 |
| \(|\mathbf{I}_s|\) (A) | 154.45 | 154.20 | 154.07 |
| Regulation | 9.22% | 9.26% | 9.37% |
| Efficiency | 95.83% | 95.95% | 95.97% |
The exact value lies between the two nominal results in every row, as Section 11-5 predicted, and the whole spread is 0.14% on the voltage. Either circuit answers the engineering question; neither is worth defending against the other.
Problem. The reference line is energised at 220 kV with its far end open. Find the receiving-end voltage, the sending-end current, and check the rise against the \((\beta l)^{2}/2\) estimate given \(\beta = 1.116\times10^{-3}\) rad/km.
Solution. With \(\mathbf{I}_r=0\), the ABCD relations reduce to two divisions:
The rise is \((130\,134-127\,017)/127\,017 = 2.45\%\). The estimate gives
The two agree to two parts in a hundred of each other, the small difference being the resistance, which the estimate ignored. Note what the 77.1 A costs: at 220 kV it is \(\sqrt3\times220\times0.0771 = 29.4\) MVA of purely reactive current, circulating to deliver nothing. On a 400 kV line of the same length the same calculation gives about three times as much, which is why long EHV lines carry permanently connected shunt reactors.
Problem. For the loaded line of Example 1, account for every MVAr: what the load absorbs, what the series reactance consumes, what each shunt branch generates, and what the sending end must therefore supply. Verify against \(Q_s\) computed from the terminal quantities.
Solution. Take each branch in turn, all three-phase.
Against the terminal computation, \(Q_s = 3|\mathbf{V}_s||\mathbf{I}_s|\sin\phi_s = 3\times135\,586\times154.07\times\sin3.93^\circ = 4.29\) MVAr. The books balance.
Read the numbers rather than merely checking them. The line generates 31.1 MVAr and consumes 6.3, a net production of 24.8 MVAr against a load demand of 29.1 — so it supplies 85% of its own load's reactive requirement, and the generator at the far end is asked for only 4.3 MVAr. At half this load the series consumption falls as the square of the current while the shunt generation is unchanged, and the sending end would have to absorb reactive power. That reversal, and the voltage rise that accompanies it, is the operational face of the Ferranti effect, and it is why reactive dispatch on a lightly loaded network is a harder problem than on a heavily loaded one (Chapter 34).
Problem. A 160 km, 132 kV, 50 Hz three-phase line has \(r=0.20\ \Omega\)/km, \(x=0.45\ \Omega\)/km and \(b=2.9\ \mu\)S/km. It delivers 30 MW at 132 kV, 0.9 power factor lagging. Using the nominal-\(\pi\) circuit find the ABCD constants, the sending-end conditions, the regulation and the efficiency.
Solution. The totals are \(\mathbf{Z} = 160(0.20+j0.45) = 32+j72\ \Omega\) and \(\mathbf{Y} = j160\times2.9\times10^{-6} = j4.64\times10^{-4}\) S, so \(\mathbf{Y}\mathbf{Z} = -0.033408+j0.014848\).
so \(V_{s(LL)} = \sqrt3\times84\,091 = 145.65\) kV and \(\phi_s = 5.45^\circ+11.56^\circ = 17.00^\circ\), \(\cos\phi_s = 0.9563\) lagging.
Cross-check from the terminals: \(P_s = 3\times84\,091\times132.04\times0.9563 = 31.86\) MW \(\checkmark\). Twelve per cent regulation is unacceptable for supply to consumers, and the culprit is identifiable: \(I_ZX = 139\times72 = 10\,009\) V against \(I_ZR = 4448\) V, so more than two-thirds of the drop is reactive. Correcting the load to unity power factor would remove the \(X\sin\phi_r\) term entirely and cut the regulation to about 6% — the cheapest remedy, as always, and the one Chapter 10's Example 6 costed.
Problem. A 300 km line has the same per-kilometre parameters as the reference line: \(z = 0.16+j0.40\ \Omega\)/km and \(y = j3.0\times10^{-6}\) S/km. Compute \(A\) and \(B\) treating it as (a) one nominal-\(\pi\) of 300 km and (b) two cascaded nominal-\(\pi\) circuits of 150 km each. The exact values from Chapter 12 are \(A = 0.94641+j0.02121\) and \(B = 46.29+j118.19\ \Omega\).
Solution (a). \(\mathbf{Z}=48+j120\ \Omega\), \(\mathbf{Y}=j9\times10^{-4}\) S, so \(\mathbf{Y}\mathbf{Z} = -0.1080+j0.0432\):
Solution (b). Each 150 km section has \(\mathbf{Z}_1 = 24+j60\), \(\mathbf{Y}_1 = j4.5\times10^{-4}\), \(\mathbf{Y}_1\mathbf{Z}_1 = -0.0270+j0.0108\), and therefore
The sections are identical, so the cascade is the square of one matrix:
Compare with the exact \(|A| = 0.94665\) and \(|B| = 126.93\ \Omega\):
| Model | \(|A|\) | Error in \(|A|\) | \(|B|\) (Ω) | Error in \(|B|\) |
|---|---|---|---|---|
| One 300 km π | 0.94625 | \(-0.042\%\) | 129.24 | \(+1.82\%\) |
| Two 150 km π | 0.94655 | \(-0.011\%\) | 127.50 | \(+0.45\%\) |
| Exact | 0.94665 | — | 126.93 | — |
Both errors fall by a factor of four when the section length is halved, exactly as Section 11-9 argued they must. Four sections of 75 km would bring the error in \(B\) to about \(0.11\%\), sixteen sections to \(0.007\%\), and the sequence is plainly converging on something. Chapter 12 finds what it converges on without taking a single step of the sequence.
Chapter Summary
\(Y\) must be kept once the charging current is a serious fraction of the load current, i.e. once \(\beta l\) exceeds about \(0.09\) rad.
\(A=D=1+\tfrac12 YZ\), \(B=Z(1+\tfrac14 YZ)\), \(C=Y\) — all of \(Y\) at the mid-point.
\(A=D=1+\tfrac12 YZ\), \(B=Z\), \(C=Y(1+\tfrac14 YZ)\) — half of \(Y\) at each busbar.
T errs by \(YZ/12\) in \(B\) and \(YZ/6\) in \(C\); π the reverse, with opposite sign. The exact value lies between them.
\(\big(|V_s|/|A|-|V_r|\big)/|V_r|\) — the no-load receiving voltage is \(V_s/A\), not \(V_s\).
An open line rises by about \((\beta l)^{2}/2\); \(2.5\%\) at 200 km, \(10\%\) at 400 km. Cured by shunt reactors.
\(P_r = \dfrac{|V_s||V_r|}{|B|}\cos(\beta-\delta)-\dfrac{|A||V_r|^{2}}{|B|}\cos(\beta-\alpha)\), maximum at \(\delta=\beta\).
ABCD matrices multiply; halving the section length quarters the error, and the limit is Chapter 12.
Practice Problems
Work per phase with the receiving-end voltage as reference, state which circuit was used, and check \(AD-BC=1\) before going any further. Difficulty rises down the list.
- A 150 km, 110 kV, 50 Hz line has \(z=0.18+j0.42\ \Omega\)/km and \(b=2.8\ \mu\)S/km. Compute \(\mathbf{Z}\), \(\mathbf{Y}\) and the ABCD constants for both the nominal-T and the nominal-\(\pi\) circuit, and verify that \(AD-BC=1\) in each case.
- For the line of Problem 1, delivering 25 MW at 110 kV and 0.85 power factor lagging, find the sending-end voltage and current by the nominal-\(\pi\) circuit, and the regulation and efficiency.
- Repeat Problem 2 with the nominal-T circuit and state the percentage difference in \(V_s\) between the two answers. Which of your two results is the larger, and does that agree with Section 11-5?
- The line of Problem 1 is energised at 110 kV with the far end open. Find the receiving-end voltage, the sending-end current, and the reactive power the line then draws from the source in MVAr.
- A 220 kV, 200 km line has \(\mathbf{Z}=40+j100\ \Omega\) and \(\mathbf{Y}=j1.0\times10^{-3}\) S. Find \(P_{r,\max}\) and the angle at which it occurs when both ends are held at 220 kV, and compare it with a thermal rating of 450 A.
- For the line of Example 5 (160 km, 132 kV), determine the capacitor rating at the receiving-end busbar that would raise the load power factor to 0.98 lagging, and recompute the regulation and efficiency. Compare the improvement with that obtained in Chapter 10, Example 6.
- A 250 km line with \(z = 0.14+j0.38\ \Omega\)/km and \(y=j3.2\times10^{-6}\) S/km is to be modelled as two cascaded 125 km nominal-\(\pi\) sections. Compute \(A\) and \(B\) for the cascade and for a single 250 km nominal-\(\pi\), and state the difference between them as a percentage.
- A generating station feeds a 200 km medium line through a transformer of series impedance \(j15\ \Omega\) referred to the line side. Write the ABCD matrix of the transformer, form the cascade with the reference line of Example 1, and find the overall \(A\), \(B\), \(C\), \(D\). Verify \(AD-BC=1\) for the product and explain in one sentence why it must hold.