Short Transmission Lines — Modelling

Set 9 — Short Transmission Lines — Modelling

Twenty worked problems on the model that discards the shunt admittance of Sets 6 and 7 entirely. What remains is a series \(R + jX\) between two buses — an ordinary a.c. circuit, solvable with the phasor algebra of any first course. The interest lies less in the arithmetic than in two things it exposes: how small the difference between the exact and approximate solutions really is, and how much the answer depends on the power factor rather than on the line.

Textbook Chapter 10 · 20 solved · 12 practice · 3 challenge · 12 MCQs

i Method Recap

The model. \(\mathbf{V}_S = \mathbf{V}_R + \mathbf{I}\mathbf{Z}\), with \(\mathbf{Z} = R + jX_L\) per phase and the shunt branch discarded. The current is the same at both ends — the defining simplification, and the thing that fails first as the line lengthens.
Always take \(\mathbf{V}_R\) as the reference phasor. Then \(\mathbf{I} = I(\cos\phi_R - j\sin\phi_R)\) for a lagging load, \(+j\) for leading. Every sign error in this set comes from getting that one line wrong.
The approximate formula. \(V_S \approx V_R + IR\cos\phi_R + IX_L\sin\phi_R\). It resolves the impedance drop along \(\mathbf{V}_R\) and ignores the quadrature component, which enters only through its square. The error is typically under 0.05% and it is safe for anything but a deliberate accuracy exercise.
Sending-end power factor. \(\phi_S = \phi_R + \alpha\) where \(\alpha\) is the angle of \(\mathbf{V}_S\) ahead of \(\mathbf{V}_R\), or directly \(\cos\phi_S \approx (V_R\cos\phi_R + IR)/V_S\).
Efficiency is reckoned on power sent, regulation on voltage received. \(\eta = P_R/(P_R + 3I^2R)\) and \(\%\text{reg} = (V_{R,\text{NL}} - V_{R,\text{FL}})/V_{R,\text{FL}}\). For a short line the no-load receiving voltage is simply \(V_S\), since no current flows.
Work per phase, in phase quantities. Divide a three-phase line voltage by \(\sqrt3\) at the start and multiply back at the end. The impedance quoted is always per phase; the power quoted is almost always three-phase.
ABCD for a short line: \(A = D = 1\), \(B = Z\), \(C = 0\). The zero in \(C\) is the discarded shunt branch, and the whole of Sets 11 and 12 consists of putting it back.

VideoWalkthrough

Problem 1Warm-upLoad Current

A single-phase overhead line delivers 1100 kW at 33 kV at 0.8 power factor lagging. Find the line current as a phasor, taking the receiving-end voltage as reference.

Solution

The magnitude of the current follows from the power delivered:

\[ I = \frac{P_R}{V_R\cos\phi_R} = \frac{1100\times10^{3}}{33\,000 \times 0.8} = \frac{1.1\times10^{6}}{26\,400} = 41.67\ \text{A} \]

Taking the receiving-end voltage as the reference phasor:

\[ \mathbf{V}_R = 33\,000 + j0\ \text{V} \]

The load is lagging, so the current lags the voltage and its imaginary part is negative:

\[ \cos\phi_R = 0.8 \quad\Rightarrow\quad \sin\phi_R = 0.6 \]

\[ \mathbf{I} = I(\cos\phi_R - j\sin\phi_R) = 41.67(0.8 - j0.6) = 33.33 - j25.00\ \text{A} \]

Checking the magnitude:

\[ |\mathbf{I}| = \sqrt{(33.33)^{2} + (25.00)^{2}} = \sqrt{1111 + 625} = 41.67\ \text{A}\ \checkmark \]

The sign of the imaginary part is the single most consequential decision in this set. Lagging means the current lags the voltage, so with \(\mathbf{V}_R\) along the real axis the current dips below it: \(-j\). A leading load takes \(+j\), and Problem 10 shows how completely that reverses the behaviour of the line. Writing this line down carefully once, at the start, prevents almost every error that follows.

Answer\(\mathbf{I} = 33.33 - j25.00 = 41.67\angle-36.87^\circ\) A

Problem 2Exam levelExact Phasor Solution

The line of Problem 1 has a total resistance of 10 \(\Omega\) and an inductive reactance of 15 \(\Omega\). Find the sending-end voltage exactly, and the angle by which it leads the receiving-end voltage.

Solution

The short-line model is simply Kirchhoff's voltage law around the series impedance:

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

Forming the impedance drop:

\[ \mathbf{I}\mathbf{Z} = (33.33 - j25)(10 + j15) \]

\[ = 333.3 + j500 - j250 - j^{2}375 = (333.3 + 375) + j(500 - 250) = 708.3 + j250 \]

The real part has two contributions — \(IR\cos\phi\) from the resistance and \(IX\sin\phi\) from the reactance acting on the reactive current — and both are positive for a lagging load. This is why lagging loads produce the largest voltage drops.

Adding the receiving-end voltage:

\[ \mathbf{V}_S = 33\,000 + 708.3 + j250 = 33\,708.3 + j250\ \text{V} \]

Magnitude:

\[ V_S = \sqrt{(33\,708.3)^{2} + (250)^{2}} = 33\,709\ \text{V} \]

Angle ahead of \(\mathbf{V}_R\):

\[ \alpha = \tan^{-1}\frac{250}{33\,708.3} = \tan^{-1}(0.00742) = 0.42^\circ \]

The quadrature component of 250 V changed the magnitude by less than one volt. Adding 250 in quadrature to 33 708 gives 33 709 — because the contribution enters as \(\sqrt{a^2+b^2} \approx a(1 + b^2/2a^2)\), and \(b/a\) here is 0.0074, whose square is 55 parts per million. That observation is the entire justification for the approximate formula of Problem 3, and it explains why the approximation survives even at power factors and impedances far worse than these.

Answer\(\mathbf{V}_S = 33\,708 + j250\) V, \(V_S = 33\,709\) V at \(\alpha = 0.42^\circ\)

Problem 3Warm-upApproximation

Recompute the sending-end voltage of Problem 2 using the approximate formula, and find the error.

Solution

The approximate expression takes only the component of the impedance drop that lies along \(\mathbf{V}_R\):

\[ V_S \approx V_R + IR\cos\phi_R + IX_L\sin\phi_R \]

Substituting:

\[ V_S \approx 33\,000 + 41.67(10)(0.8) + 41.67(15)(0.6) \]

\[ = 33\,000 + 333.4 + 375.0 = 33\,708.4\ \text{V} \]

Against the exact 33 709 V from Problem 2:

\[ \text{Error} = \frac{33\,709 - 33\,708.4}{33\,709}\times100 = 0.002\% \]

The two terms are worth reading separately. The resistance contributes 333 V and the reactance 375 V — and the reactance term is the larger despite \(\sin\phi_R\) being smaller than \(\cos\phi_R\), because \(X_L > R\).

Two parts per hundred thousand, and the approximation is doing real work. It converts a complex multiplication and a square root into two products and a sum, which matters when the unknown is \(V_R\) rather than \(V_S\) — as in Problem 17, where the exact form gives a messy implicit equation and the approximate form gives a simple quadratic. The error only becomes visible when \(IZ\) approaches a substantial fraction of \(V_R\), which on a short line it never does.

Answer\(V_S \approx 33\,708.4\) V — an error of 0.002%

Problem 4Warm-upSending-End Power Factor

Find the sending-end power factor of the line of Problem 2, by two methods.

Solution

Method 1 — from the angles. The current lags \(\mathbf{V}_R\) by \(\phi_R\), and \(\mathbf{V}_S\) leads \(\mathbf{V}_R\) by \(\alpha\), so the current lags \(\mathbf{V}_S\) by the sum:

\[ \phi_S = \phi_R + \alpha = 36.87^\circ + 0.42^\circ = 37.29^\circ \]

\[ \cos\phi_S = \cos(37.29^\circ) = 0.7956\ \text{lagging} \]

Method 2 — directly. The in-phase component of \(\mathbf{V}_S\) relative to the current, divided by its magnitude:

\[ \cos\phi_S = \frac{V_R\cos\phi_R + IR}{V_S} = \frac{33\,000(0.8) + 41.67(10)}{33\,708.4} \]

\[ = \frac{26\,400 + 416.7}{33\,708.4} = \frac{26\,816.7}{33\,708.4} = 0.7958\ \text{lagging} \]

The two agree to four figures, the small difference being the approximation of Problem 3.

The power factor got worse along the line — from 0.800 to 0.796 — and it always does for a lagging load. The line's own reactance absorbs vars, \(I^2X_L\) of them, which must be supplied from the sending end on top of the load's own demand. So the sending end sees more reactive power for the same real power, and a lower power factor. On a long or heavily loaded line this degradation is severe, and it is the reason shunt capacitors are installed at load centres rather than at generating stations.

Answer\(\cos\phi_S = 0.796\) lagging, against 0.800 at the load

Problem 5Warm-upEfficiency

Find the line losses, the power sent and the transmission efficiency of the line of Problem 2.

Solution

The line is single-phase, and the 10 \(\Omega\) is the total resistance of both conductors, so:

\[ P_{\text{loss}} = I^{2}R = (41.67)^{2}(10) = 1736(10) = 17\,364\ \text{W} = 17.36\ \text{kW} \]

Power sent:

\[ P_S = P_R + P_{\text{loss}} = 1100 + 17.36 = 1117.36\ \text{kW} \]

Transmission efficiency:

\[ \eta = \frac{P_R}{P_S}\times100 = \frac{1100}{1117.36}\times100 = 98.45\% \]

Checking against the sending-end quantities directly, which must agree:

\[ P_S = V_SI\cos\phi_S = 33\,708.4(41.67)(0.7958) = 1.1175\times10^{6}\ \text{W}\ \checkmark \]

The reactance produced a 375 V drop and consumed no real power at all. Only the resistance dissipates; the reactance merely shifts the phase and exchanges reactive power. This is why a line can have poor regulation and excellent efficiency at the same time — the two are governed by different components of the same impedance, and improving one by adding series capacitance leaves the other untouched.

AnswerLosses 17.36 kW, \(P_S = 1117.4\) kW, \(\eta = 98.45\%\)

Problem 6Exam levelThree-Phase Line

A three-phase short line delivers 5 MW at 33 kV, 0.85 power factor lagging. Each conductor has \(R = 4\ \Omega\) and \(X_L = 8\ \Omega\). Find the sending-end voltage, the sending-end power factor and the efficiency.

Solution

Work per phase. Convert the line voltage at once:

\[ V_R = \frac{33\,000}{\sqrt3} = 19\,052\ \text{V per phase} \]

Line current from the three-phase power:

\[ I = \frac{P_R}{\sqrt3\,V_L\cos\phi_R} = \frac{5\times10^{6}}{\sqrt3(33\,000)(0.85)} = 102.9\ \text{A} \]

\[ \cos\phi_R = 0.85 \quad\Rightarrow\quad \sin\phi_R = 0.5268 \]

Sending-end voltage per phase, by the approximate formula:

\[ V_S \approx 19\,052 + 102.9(4)(0.85) + 102.9(8)(0.5268) \]

\[ = 19\,052 + 349.9 + 433.7 = 19\,835.6\ \text{V per phase} \]

As a line voltage:

\[ V_{S,\text{line}} = \sqrt3(19\,835.6) = 34\,357\ \text{V} = 34.36\ \text{kV} \]

Sending-end power factor:

\[ \cos\phi_S = \frac{V_R\cos\phi_R + IR}{V_S} = \frac{19\,052(0.85) + 102.9(4)}{19\,835.6} = \frac{16\,194 + 411.6}{19\,835.6} = 0.8372 \]

Losses in three conductors, and efficiency:

\[ P_{\text{loss}} = 3I^{2}R = 3(102.9)^{2}(4) = 127.1\ \text{kW} \]

\[ \eta = \frac{5000}{5000 + 127.1}\times100 = 97.52\% \]

The factor of three appears exactly once, in the loss, and the \(\sqrt3\) exactly twice — once in and once out. Everything between is single-phase arithmetic on a per-phase equivalent circuit. Converting at the start and converting back at the end is the discipline that makes three-phase problems no harder than single-phase ones; carrying line quantities through the middle is what makes them error-prone. Note that the 4 \(\Omega\) is per conductor here, whereas in Problem 5 the 10 \(\Omega\) was the total loop resistance of a single-phase line.

Answer\(V_S = 34.36\) kV, \(\cos\phi_S = 0.837\) lagging, \(\eta = 97.52\%\)

Problem 7Exam levelMaximum Length

What is the maximum length of a single-phase line with copper conductors of 0.775 cm² cross-section over which 200 kW at unity power factor and 3300 V may be delivered at 90% transmission efficiency? Take the resistivity of copper as \(1.725\ \mu\Omega\)cm.

Solution

Power sent and lost:

\[ P_S = \frac{200\,000}{0.9} = 222\,222\ \text{W}, \qquad P_{\text{loss}} = 222\,222 - 200\,000 = 22\,222\ \text{W} \]

Line current at unity power factor:

\[ I = \frac{200\,000}{3300} = 60.61\ \text{A} \]

The loss occurs in two conductors, so with \(R\) the resistance of one:

\[ P_{\text{loss}} = 2I^{2}R \quad\Rightarrow\quad R = \frac{22\,222}{2(60.61)^{2}} = \frac{22\,222}{7347} = 3.025\ \Omega \]

Now the geometry. Working consistently in centimetres:

\[ R = \frac{\rho l}{a} \quad\Rightarrow\quad l = \frac{Ra}{\rho} = \frac{3.025(0.775)}{1.725\times10^{-6}} \]

\[ = \frac{2.344}{1.725\times10^{-6}} = 1.359\times10^{6}\ \text{cm} \]

Converting:

\[ l = 1.359\times10^{4}\ \text{m} = 13.6\ \text{km} \]

Thirteen kilometres at 3.3 kV, and the voltage is what limits it. The same conductor at 33 kV would carry the same power at a tenth the current, a hundredth the loss, and therefore a hundred times the distance — 1360 km, at which point the short-line model would have failed long before the conductor did. This is Set 2's \(1/V^2\) argument reappearing as a distance rather than a volume, and it is why 3.3 kV is a distribution voltage and not a transmission one.

Answer\(R = 3.025\ \Omega\) per conductor, giving a maximum length of \(13.6\) km

Problem 8Warm-upABCD Parameters

Write the ABCD parameters of a short transmission line of series impedance \(Z\), verify the reciprocity condition, and state what the value of \(C\) represents physically.

Solution

The two-port relations, in the standard convention with current flowing into the load:

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

For a short line the model gives \(\mathbf{V}_S = \mathbf{V}_R + \mathbf{I}_RZ\) and \(\mathbf{I}_S = \mathbf{I}_R\). Matching coefficients:

\[ A = 1, \qquad B = Z, \qquad C = 0, \qquad D = 1 \]

\[ \begin{bmatrix} \mathbf{V}_S \\ \mathbf{I}_S \end{bmatrix} = \begin{bmatrix} 1 & Z \\ 0 & 1 \end{bmatrix}\begin{bmatrix} \mathbf{V}_R \\ \mathbf{I}_R \end{bmatrix} \]

Reciprocity:

\[ AD - BC = (1)(1) - (Z)(0) = 1\ \checkmark \]

This holds for every passive line model and is the single most useful check on a set of computed ABCD constants.

What \(C = 0\) means. \(C = \mathbf{I}_S/\mathbf{V}_R\) with the receiving end open-circuited — that is, the current the line draws when it is energised but unloaded. For a short line this is taken as zero, which is precisely the statement that the shunt capacitance of Sets 6 and 7 has been discarded.

Its consequence is that \(A = 1\), so the no-load receiving voltage equals the sending voltage exactly, and there is no Ferranti effect in this model.

Every difference between the three line models lives in \(C\), and everything else follows from it. The short line sets \(C = 0\); the nominal-\(\pi\) of Set 11 gives it the value \(Y(1 + ZY/4)\); the exact model of Set 12 gives \(\sinh(\gamma l)/Z_c\). Since \(A = D = 1 + ZY/2\) in the \(\pi\) model and \(\cosh(\gamma l)\) exactly, the departure of \(A\) from unity is the direct measure of how much shunt admittance matters — and hence of which model is required.

Answer\(A = D = 1\), \(B = Z\), \(C = 0\); \(AD - BC = 1\); \(C = 0\) is the discarded shunt branch

Problem 9Exam levelVoltage Regulation

Find the voltage regulation of the three-phase line of Problem 6, and explain why the no-load receiving-end voltage equals the sending-end voltage for a short line.

Solution

Voltage regulation is defined as the rise in receiving-end voltage when the load is removed, as a fraction of the full-load value:

\[ \%\text{reg} = \frac{V_{R,\text{NL}} - V_{R,\text{FL}}}{V_{R,\text{FL}}}\times100 \]

Why \(V_{R,\text{NL}} = V_S\) here. On no load the current is zero, so there is no drop across \(Z\) and the receiving end sits at the sending-end voltage. In ABCD terms, \(V_{R,\text{NL}} = V_S/A\) and \(A = 1\) for a short line.

Substituting the per-phase values from Problem 6:

\[ \%\text{reg} = \frac{19\,835.6 - 19\,052}{19\,052}\times100 = \frac{783.6}{19\,052}\times100 = 4.11\% \]

Line voltages give the same answer, since the \(\sqrt3\) cancels:

\[ \frac{34.36 - 33.00}{33.00}\times100 = 4.12\%\ \checkmark \]

4.1% regulation with 97.5% efficiency — the two say quite different things about the same line. Efficiency asks how much power is lost and depends on \(R\) alone; regulation asks how much the voltage sags and depends chiefly on \(X_L\), which here contributed 434 of the 784 V. A line can be efficient and badly regulated, or the reverse. Utility practice usually limits regulation to about 5%, and on a line where that binds it is series capacitance — cancelling \(X_L\) — rather than more copper that provides the remedy.

AnswerRegulation \(= 4.11\%\); \(V_{R,\text{NL}} = V_S\) because \(A = 1\) when the shunt branch is discarded

Problem 10Exam levelLeading Power Factor

Repeat Problem 6 with the load at 0.85 power factor leading, and compare the sending-end voltage and regulation with the lagging case.

Solution

The current magnitude is unchanged at 102.9 A, but it now leads, so its imaginary part is positive:

\[ \mathbf{I} = 102.9(0.85 + j0.5268) = 87.47 + j54.21\ \text{A} \]

In the approximate formula the reactive term changes sign, because \(\sin\phi_R\) is now negative in the convention used:

\[ V_S \approx V_R + IR\cos\phi_R - IX_L\sin\phi_R \]

\[ = 19\,052 + 349.9 - 433.7 = 18\,968.2\ \text{V per phase} \]

As a line voltage:

\[ V_{S,\text{line}} = \sqrt3(18\,968.2) = 32.85\ \text{kV} \]

Regulation:

\[ \%\text{reg} = \frac{18\,968.2 - 19\,052}{19\,052}\times100 = -0.44\% \]

The sending-end voltage is lower than the receiving-end voltage, and the regulation is negative. Comparing the two cases:

\[ \begin{array}{lcc} \text{Load p.f.} & V_S\ (\text{kV}) & \%\text{reg} \\ \hline 0.85\ \text{lagging} & 34.36 & +4.11 \\ 0.85\ \text{leading} & 32.85 & -0.44 \end{array} \]

The same power, the same current, the same line — and a swing of 1.5 kV in the sending-end voltage. A leading current draws reactive power out of the line's inductance rather than pushing it in, so the \(IX_L\sin\phi\) term reverses. This is the entire principle of power-factor correction and of the synchronous condenser: reactive current, correctly signed, is a voltage-control mechanism far more effective than anything done to the conductor. It also foreshadows the Ferranti effect, where the line's own capacitance supplies the leading current and produces the same voltage rise with no load at all.

Answer\(V_S = 32.85\) kV, regulation \(-0.44\%\) — the receiving voltage exceeds the sending voltage

Problem 11Warm-upUnity Power Factor

Repeat Problem 6 with the load at unity power factor, and compare the current, regulation and efficiency with the 0.85 lagging case.

Solution

At unity power factor the same 5 MW needs less current:

\[ I = \frac{5\times10^{6}}{\sqrt3(33\,000)(1.0)} = 87.48\ \text{A} \]

The reactive term vanishes entirely, since \(\sin\phi_R = 0\):

\[ V_S \approx 19\,052 + 87.48(4)(1) + 0 = 19\,052 + 349.9 = 19\,401.9\ \text{V per phase} \]

\[ V_{S,\text{line}} = \sqrt3(19\,401.9) = 33.61\ \text{kV} \]

Regulation and losses:

\[ \%\text{reg} = \frac{349.9}{19\,052}\times100 = 1.84\% \]

\[ P_{\text{loss}} = 3(87.48)^{2}(4) = 91.8\ \text{kW}, \qquad \eta = \frac{5000}{5091.8}\times100 = 98.20\% \]

Setting the two cases side by side:

\[ \begin{array}{lcccc} \text{p.f.} & I\ (\text{A}) & V_S\ (\text{kV}) & \%\text{reg} & \eta\ (\%) \\ \hline 0.85\ \text{lag} & 102.9 & 34.36 & 4.11 & 97.52 \\ 1.00 & 87.5 & 33.61 & 1.84 & 98.20 \end{array} \]

Correcting the power factor from 0.85 to unity cut the regulation by more than half and the losses by 28%, without touching the line. The current fell by 15% — which alone accounts for the loss reduction through \(I^2\) — but the regulation improved far more than that, because the entire \(IX_L\sin\phi\) term disappeared. This is why power-factor correction is almost always the cheapest available improvement to a distribution feeder, and why tariffs penalise poor power factor: the utility's regulation and losses both suffer from a problem the consumer creates.

Answer\(I = 87.5\) A, \(V_S = 33.61\) kV, regulation 1.84%, \(\eta = 98.20\%\)

Problem 12Exam levelRegulation vs Power Factor

For the line of Problem 6 carrying a fixed current of 102.9 A, tabulate the voltage regulation at power factors of 0.6, 0.8, 0.85 and 1.0 lagging, and 0.85, 0.8 and 0.6 leading.

Solution

Express the drop as percentages of the receiving voltage. At \(I = 102.9\) A and \(V_R = 19\,052\) V:

\[ \%R = \frac{IR}{V_R}\times100 = \frac{102.9(4)}{19\,052}\times100 = 2.16\% \]

\[ \%X = \frac{IX_L}{V_R}\times100 = \frac{102.9(8)}{19\,052}\times100 = 4.32\% \]

The regulation is then a simple combination, with the sign of the reactive term set by whether the load lags or leads:

\[ \%\text{reg} \approx \%R\cos\phi \pm \%X\sin\phi \qquad (+ \text{ lagging},\ - \text{ leading}) \]

Evaluating:

\[ \begin{array}{lccc} \text{p.f.} & \%R\cos\phi & \pm\%X\sin\phi & \%\text{reg} \\ \hline 0.60\ \text{lag} & 1.30 & +3.46 & +4.75 \\ 0.80\ \text{lag} & 1.73 & +2.59 & +4.32 \\ 0.85\ \text{lag} & 1.84 & +2.28 & +4.11 \\ 1.00 & 2.16 & 0 & +2.16 \\ 0.85\ \text{lead} & 1.84 & -2.28 & -0.44 \\ 0.80\ \text{lead} & 1.73 & -2.59 & -0.86 \\ 0.60\ \text{lead} & 1.30 & -3.46 & -2.16 \end{array} \]

The regulation falls monotonically from +4.75% at 0.6 lagging to \(-2.16\%\) at 0.6 leading, passing through zero somewhere between unity and 0.85 leading — the point located exactly in Problem 13.

The power factor swings the regulation over a range of seven percentage points on a line whose impedance never changed. Regulation is not a property of a line but of a line and its load, which is why manufacturers quote a transformer's regulation at a stated power factor and why a feeder that behaves perfectly on a resistive load can misbehave badly on a motor load of the same kVA. The \(\%R\) and \(\%X\) pair is the compact way to carry this: two numbers describe the line, and the load supplies the angle.

Answer\(\%R = 2.16\), \(\%X = 4.32\); regulation ranges from \(+4.75\%\) at 0.6 lagging to \(-2.16\%\) at 0.6 leading

Problem 13Challenge-liteZero Regulation

Find the power factor at which the line of Problem 6 has zero voltage regulation, and show that the condition depends only on the line's \(R/X\) ratio.

Solution

Zero regulation requires \(V_S = V_R\), so the approximate drop must vanish. With a leading load:

\[ IR\cos\phi - IX_L\sin\phi = 0 \]

The current cancels immediately, which is the first indication that the answer will not depend on the loading:

\[ R\cos\phi = X_L\sin\phi \quad\Rightarrow\quad \tan\phi = \frac{R}{X_L} \]

For this line:

\[ \tan\phi = \frac{4}{8} = 0.5 \quad\Rightarrow\quad \phi = 26.57^\circ\ \text{leading} \]

\[ \cos\phi = 0.894\ \text{leading} \]

Verifying with the percentages of Problem 12:

\[ 2.16(0.894) - 4.32(0.447) = 1.931 - 1.931 = 0\ \checkmark \]

Note that the condition \(\tan\phi = R/X_L\) says the load's impedance angle is the complement of the line's: the line has \(\tan\theta = X_L/R = 2\), the load \(\tan\phi = 0.5\), and \(\theta + \phi = 90^\circ\).

The zero-regulation power factor is a property of the line alone, and it is always leading. Every real line has \(X_L > 0\), so \(\tan\phi = R/X_L\) is finite and positive and the load must lead. For a transmission line with \(X/R = 20\) the required power factor is 0.9988 leading — almost purely capacitive reactive support, which is precisely what a static var compensator provides. For a distribution feeder with \(X/R = 1\) it is 0.707 leading, an entirely practical condition, which is why capacitor banks on distribution feeders can genuinely hold the far-end voltage flat.

Answer\(\tan\phi = R/X_L = 0.5\), giving \(\cos\phi = 0.894\) leading — independent of the load current

Problem 14Warm-upPercentage Impedance

Show that the percentage resistance and reactance of Problem 12 are identical to the per-unit impedance of the line on a base equal to the load rating, and interpret them.

Solution

The load in Problem 6 is 5 MW at 0.85 power factor, so its apparent power is

\[ S = \frac{5}{0.85} = 5.882\ \text{MVA} \]

Taking this as the base MVA with 33 kV as the base voltage:

\[ Z_B = \frac{(33)^{2}}{5.882} = \frac{1089}{5.882} = 185.1\ \Omega \]

The line impedance in per-unit on that base:

\[ R_{pu} = \frac{4}{185.1} = 0.0216, \qquad X_{pu} = \frac{8}{185.1} = 0.0432 \]

These are exactly the 2.16% and 4.32% of Problem 12, as they must be:

\[ \frac{IR}{V_R} = \frac{R}{V_R/I} = \frac{R}{Z_B} = R_{pu}\ \checkmark \]

since \(V_R/I\) at rated conditions is the base impedance.

Interpretation. \(\%R\) is the fraction of the receiving voltage dropped by resistance at rated current, and \(\%X\) the same for reactance. Together they say everything about the line's regulating behaviour, with the load supplying only the angle at which they combine.

Percentage impedance and per-unit impedance are the same quantity, and recognising it collapses two topics into one. A transformer's "6% impedance" is its per-unit impedance on its own rating, which is why Set 3's base-change formula applies to it directly and why Set 4 Problem 4 could convert it without ever thinking about ohms. The regulation formula \(\%R\cos\phi + \%X\sin\phi\) is therefore not a line formula at all — it applies unchanged to a transformer, a cable or a generator, whenever a series impedance sits between a source and a load.

Answer\(R_{pu} = 0.0216\), \(X_{pu} = 0.0432\) — identical to \(\%R\) and \(\%X\) on the load's own base

Problem 15Exam levelSending-End P and Q

For the line of Problem 6, find the real and reactive power at the sending end, and verify the sending-end power factor found there.

Solution

Real power. The load's power plus the resistive loss:

\[ P_S = P_R + 3I^{2}R = 5000 + 127.1 = 5127.1\ \text{kW} \]

Reactive power at the load:

\[ Q_R = P_R\tan\phi_R = 5000\tan(31.79^\circ) = 5000(0.6197) = 3098.7\ \text{kVAr} \]

Reactive power absorbed by the line:

\[ Q_{\text{line}} = 3I^{2}X_L = 3(102.9)^{2}(8) = 254.2\ \text{kVAr} \]

Total at the sending end:

\[ Q_S = Q_R + Q_{\text{line}} = 3098.7 + 254.2 = 3352.9\ \text{kVAr} \]

Verifying the sending-end power factor:

\[ \cos\phi_S = \frac{P_S}{\sqrt{P_S^{2} + Q_S^{2}}} = \frac{5127.1}{\sqrt{5127.1^{2} + 3352.9^{2}}} = \frac{5127.1}{6126.1} = 0.837\ \checkmark \]

Agreeing with Problem 6.

The line absorbed 254 kVAr and dissipated 127 kW — exactly twice as much reactive as real, because \(X_L = 2R\). Both scale as \(I^2\), so their ratio is fixed by the line's \(X/R\) and never changes with loading. On a transmission line with \(X/R = 20\) the reactive absorption is twenty times the real loss, which is why heavily loaded transmission lines are enormous consumers of reactive power and why voltage collapse — the subject of Set 34 — is a reactive phenomenon rather than a real one.

Answer\(P_S = 5127\) kW, \(Q_S = 3353\) kVAr, giving \(\cos\phi_S = 0.837\) lagging

Problem 16Exam levelPer-Unit

Rework Problem 6 entirely in per-unit on a base of 100 MVA, 33 kV, and confirm the sending-end voltage.

Solution

Base impedance:

\[ Z_B = \frac{(33)^{2}}{100} = 10.89\ \Omega \]

\[ \mathbf{Z}_{pu} = \frac{4 + j8}{10.89} = 0.3673 + j0.7346\ \text{p.u.} \]

The load in per-unit — its apparent power on the 100 MVA base:

\[ S_{pu} = \frac{5/0.85}{100} = \frac{5.882}{100} = 0.0588\ \text{p.u.} \]

Taking \(\mathbf{V}_R = 1.0\angle0^\circ\) p.u., the current is numerically the apparent power:

\[ \mathbf{I}_{pu} = 0.0588\angle-31.79^\circ = 0.0500 - j0.0310\ \text{p.u.} \]

The impedance drop:

\[ \mathbf{I}\mathbf{Z} = (0.0500 - j0.0310)(0.3673 + j0.7346) \]

\[ = 0.01837 + j0.03673 - j0.01139 + 0.02277 = 0.04114 + j0.02534 \]

Sending-end voltage:

\[ \mathbf{V}_S = 1.0 + 0.04114 + j0.02534 = 1.04114 + j0.02534\ \text{p.u.} \]

\[ |\mathbf{V}_S| = \sqrt{1.08397 + 0.00064} = 1.0414\ \text{p.u.} \]

Converting back:

\[ V_{S,\text{line}} = 1.0414(33) = 34.37\ \text{kV}\ \checkmark \]

Agreeing with Problem 6, and the regulation is read straight off as 4.14% with no further arithmetic.

In per-unit the regulation is \(|V_S| - 1\), and no \(\sqrt3\) appeared anywhere. That is the practical payoff of Set 3's construction: the base was built so that a three-phase line voltage of 33 kV is 1.0 p.u. and the per-phase arithmetic follows automatically. Note also that the per-unit impedance came out at 0.37 + j0.73 — very large, because a 100 MVA base is inappropriate for a line carrying 5.9 MVA. The answer is right, but the numbers would sit in the familiar bands only on a base closer to the line's own rating, as Problem 14 showed.

Answer\(\mathbf{V}_S = 1.0414\angle1.39^\circ\) p.u. \(= 34.37\) kV — regulation 4.14%

Problem 17Exam levelInverse Problem

A three-phase line delivers 3600 kW at 0.8 power factor lagging. The sending-end voltage is 33 kV and each conductor has \(R = 5.31\ \Omega\) and \(X_L = 5.54\ \Omega\). Determine the receiving-end voltage, the line current and the transmission efficiency.

Solution

Here the receiving voltage is unknown, so the current cannot be found first. Express it in terms of \(V_R\), working per phase with 1200 kW:

\[ I = \frac{1200\times10^{3}}{V_R(0.8)} = \frac{1.5\times10^{6}}{V_R} \]

Sending-end voltage per phase:

\[ V_S = \frac{33\,000}{\sqrt3} = 19\,052\ \text{V} \]

Substituting into the approximate formula:

\[ 19\,052 = V_R + \frac{1.5\times10^{6}}{V_R}(5.31)(0.8) + \frac{1.5\times10^{6}}{V_R}(5.54)(0.6) \]

\[ = V_R + \frac{1.5\times10^{6}}{V_R}(4.248 + 3.324) = V_R + \frac{1.1358\times10^{7}}{V_R} \]

Multiplying through by \(V_R\) gives a quadratic:

\[ V_R^{2} - 19\,052V_R + 1.1358\times10^{7} = 0 \]

\[ V_R = \frac{19\,052 \pm \sqrt{(19\,052)^{2} - 4(1.1358\times10^{7})}}{2} = \frac{19\,052 \pm 17\,820}{2} \]

Two roots, 18 436 V and 616 V. The larger is the operating solution:

\[ V_R = 18\,436\ \text{V per phase}, \qquad V_{R,\text{line}} = \sqrt3(18\,436) = 31.93\ \text{kV} \]

Current, losses and efficiency:

\[ I = \frac{1.5\times10^{6}}{18\,436} = 81.36\ \text{A} \]

\[ P_{\text{loss}} = 3(81.36)^{2}(5.31) = 105.4\ \text{kW}, \qquad \eta = \frac{3600}{3705.4}\times100 = 97.16\% \]

The discarded root is not merely spurious — it is a real operating point, and a catastrophic one. At \(V_R = 616\) V the line would carry 2436 A to deliver the same 3600 kW, dissipating far more than it delivers. That the quadratic has two roots is the algebraic signature of voltage collapse: a constant-power load can be supplied at a high voltage with low current or a low voltage with high current, and as the load grows the two roots approach each other until they merge and no solution exists at all. That merging point is the nose of the P–V curve of Set 34, and it is visible here in the simplest possible line model.

Answer\(V_R = 31.93\) kV, \(I = 81.4\) A, \(\eta = 97.16\%\)

Problem 18Warm-upEfficiency vs Power Factor

For the line of Problem 6 delivering a fixed 5 MW at a fixed 33 kV, find the efficiency at power factors of 0.7, 0.85 and 1.0, and derive the general dependence.

Solution

With \(P_R\) and \(V_R\) fixed, the current varies inversely with the power factor:

\[ I = \frac{P_R}{\sqrt3\,V_L\cos\phi} \propto \frac{1}{\cos\phi} \]

\[ P_{\text{loss}} = 3I^{2}R \propto \frac{1}{\cos^{2}\phi} \]

Scaling from the 127.1 kW at 0.85 found in Problem 6:

\[ P_{\text{loss}}(\cos\phi) = 127.1\left(\frac{0.85}{\cos\phi}\right)^{2} \]

Evaluating:

\[ \begin{array}{cccc} \cos\phi & I\ (\text{A}) & P_{\text{loss}}\ (\text{kW}) & \eta\ (\%) \\ \hline 0.70 & 125.0 & 187.4 & 96.39 \\ 0.85 & 102.9 & 127.1 & 97.52 \\ 1.00 & 87.5 & 91.8 & 98.20 \end{array} \]

Improving the power factor from 0.7 to 1.0 halves the losses and raises the efficiency by 1.8 percentage points.

This is Set 2 Problem 9 seen from the other end. There the conclusion was that conductor volume varies as \(1/\cos^2\phi\) for a fixed loss; here the loss varies as \(1/\cos^2\phi\) for a fixed conductor. Both statements are the same physics — the reactive current occupies the conductor without delivering power — and a utility can spend the penalty either as capital in copper or as revenue in losses. Correcting the power factor avoids paying it at all, which is why it is almost always the first move.

Answer\(\eta = 96.39\%\), \(97.52\%\) and \(98.20\%\) at 0.7, 0.85 and 1.0; loss \(\propto 1/\cos^2\phi\)

Problem 19Challenge-liteValidity

Establish quantitatively when the short-line model may be used, using the line parameters of Set 7 — \(X = 0.32\ \Omega\)/km and \(B = 3.5\times10^{-6}\) S/km — and state the error incurred at the boundary.

Solution

The model discards the shunt admittance, so it is valid when the charging current is negligible against the load current. Take a 400 kV line at its surge impedance loading of 525 MW:

\[ I_{\text{load}} = \frac{525\times10^{6}}{\sqrt3(400\times10^{3})} = 758\ \text{A} \]

Charging current per km:

\[ I_c = V_{ph}B = \frac{400\,000}{\sqrt3}(3.5\times10^{-6}) = 231\,000(3.5\times10^{-6}) = 0.808\ \text{A/km} \]

Setting the charging current at 5% of the load current:

\[ l = \frac{0.05(758)}{0.808} = \frac{37.9}{0.808} = 47\ \text{km} \]

The equivalent statement in electrical length. From Set 7, the ratio of charging to natural loading is \(\beta l\), so a 5% criterion is

\[ \beta l \le 0.05\ \text{rad} \quad\Rightarrow\quad l \le \frac{0.05}{1.05\times10^{-3}} = 48\ \text{km} \]

The two routes agree, as Set 7 Challenge C1 established they must.

In practice the short-line model is used up to about 80 km, accepting \(\beta l \approx 0.085\) rad and an error in \(A\) of

\[ |A| \approx 1 - \frac{(\beta l)^{2}}{2} = 1 - 0.0036 \quad\Rightarrow\quad 0.36\% \]

The error grows as the square of the electrical length, which is why the boundary is so forgiving. At 50 km the error in \(A\) is 0.14%; at 80 km, 0.36%; at 160 km it is still only 1.4%. The conventional 80 km limit is therefore conservative, and the model degrades gracefully rather than failing suddenly. What does not degrade gracefully is the no-load behaviour: with \(C = 0\) the model predicts no Ferranti rise at all, which is qualitatively wrong at any length, and that — not the loaded accuracy — is the real reason to abandon it.

AnswerValid to about 50 km on a 5% criterion, used to 80 km with a 0.36% error in \(A\); error \(\propto (\beta l)^2\)

Problem 20Exam levelComplete Study

A 66 kV, 60 km three-phase line has \(R = 0.16\ \Omega\)/km and \(X_L = 0.40\ \Omega\)/km per phase, and delivers 20 MW at 0.9 power factor lagging. Carry out a complete short-line study: sending-end voltage, power factor, regulation, losses, efficiency and per-unit impedance on 100 MVA.

Solution

Step 1 — line impedance and per-phase voltage:

\[ R = 0.16(60) = 9.6\ \Omega, \qquad X_L = 0.40(60) = 24.0\ \Omega, \qquad \frac{X}{R} = 2.5 \]

\[ V_R = \frac{66\,000}{\sqrt3} = 38\,105\ \text{V per phase} \]

Step 2 — line current:

\[ I = \frac{20\times10^{6}}{\sqrt3(66\,000)(0.9)} = \frac{20\times10^{6}}{102\,884} = 194.4\ \text{A} \]

\[ \cos\phi_R = 0.9 \quad\Rightarrow\quad \sin\phi_R = 0.4359 \]

Step 3 — sending-end voltage:

\[ V_S \approx 38\,105 + 194.4(9.6)(0.9) + 194.4(24.0)(0.4359) \]

\[ = 38\,105 + 1679.6 + 2033.7 = 41\,818\ \text{V per phase} \]

\[ V_{S,\text{line}} = \sqrt3(41\,818) = 72.43\ \text{kV} \]

Step 4 — regulation and sending-end power factor:

\[ \%\text{reg} = \frac{41\,818 - 38\,105}{38\,105}\times100 = 9.74\% \]

\[ \cos\phi_S = \frac{38\,105(0.9) + 194.4(9.6)}{41\,818} = \frac{34\,295 + 1866}{41\,818} = 0.865 \]

Step 5 — losses and efficiency:

\[ P_{\text{loss}} = 3(194.4)^{2}(9.6) = 1088\ \text{kW}, \qquad \eta = \frac{20\,000}{21\,088}\times100 = 94.84\% \]

Step 6 — per-unit on 100 MVA, 66 kV, where \(Z_B = 66^2/100 = 43.56\ \Omega\):

\[ \mathbf{Z}_{pu} = \frac{9.6 + j24.0}{43.56} = 0.220 + j0.551\ \text{p.u.} \]

Verdict. At 60 km the short-line model is comfortably valid, but a regulation of 9.7% is well beyond the usual 5% limit. The line needs either a higher sending-end voltage, power-factor correction at the load, or series compensation.

A perfectly ordinary line, correctly analysed, that fails its regulation limit — and the model that revealed it is the simplest in the book. Nothing here required the shunt admittance, the wave equation or a load flow. That is worth remembering as Sets 11 and 12 add machinery: the added complexity buys accuracy on long lines and the no-load case, but the essential behaviour of a loaded line — voltage falls, power factor worsens, losses go as \(1/\cos^2\phi\) — is already fully visible in \(\mathbf{V}_S = \mathbf{V}_R + \mathbf{I}\mathbf{Z}\).

Answer\(V_S = 72.43\) kV, \(\cos\phi_S = 0.865\), reg 9.74%, loss 1088 kW, \(\eta = 94.84\%\), \(Z_{pu} = 0.220 + j0.551\)

Practice

Practice Problems

Work each on paper before opening the answer. The answer is given so you can check yourself; the method is deliberately not.

P1. A three-phase line delivers 2 MW at 11 kV, 0.8 p.f. lagging. Find the line current.

Show answer
\(I = 2\times10^6/(\sqrt3 \cdot 11\,000 \cdot 0.8) = \mathbf{131.2}\) A.
P2. That line has \(R = 3\ \Omega\), \(X_L = 5\ \Omega\) per phase. Find \(V_S\) per phase by the approximate formula.

Show answer
\(V_R = 6351\) V; \(V_S = 6351 + 131.2(3)(0.8) + 131.2(5)(0.6) = 6351 + 314.9 + 393.6 = \mathbf{7059.5}\) V.
P3. Find the regulation of that line.

Show answer
\((7059.5 - 6351)/6351 = \mathbf{11.15\%}\) — well above the usual 5% limit.
P4. Find its losses and efficiency.

Show answer
\(3(131.2)^2(3) = 154.9\) kW; \(\eta = 2000/2154.9 = \mathbf{92.81\%}\).
P5. A short line has \(R = 5\ \Omega\), \(X = 12\ \Omega\). At what power factor is its regulation zero?

Show answer
\(\tan\phi = R/X = 5/12 = 0.4167\), so \(\phi = 22.6^\circ\) and \(\cos\phi = \mathbf{0.923}\) leading.
P6. Write the ABCD parameters of a line of impedance \((8 + j20)\ \Omega\) and verify \(AD - BC = 1\).

Show answer
\(A = D = 1\), \(B = 8 + j20\), \(C = 0\); \(AD - BC = 1 - 0 = \mathbf{1}\ \checkmark\).
P7. A line has \(\%R = 3\) and \(\%X = 6\). Find its regulation at 0.8 p.f. lagging and at 0.8 leading.

Show answer
Lagging: \(3(0.8) + 6(0.6) = \mathbf{+6.0\%}\). Leading: \(3(0.8) - 6(0.6) = \mathbf{-1.2\%}\).
P8. Why does the sending-end power factor differ from the receiving-end one, and in which direction for a lagging load?

Show answer
The line's reactance absorbs \(3I^2X_L\) of vars, which must be supplied from the sending end. The sending-end power factor is therefore worse (lower) — Problem 4.
P9. A line delivers 1 MW at 0.9 p.f. with 40 kW of loss. Find its efficiency and the loss at 0.75 p.f. for the same power and voltage.

Show answer
\(\eta = 1000/1040 = \mathbf{96.15\%}\). At 0.75: \(40(0.9/0.75)^2 = \mathbf{57.6}\) kW.
P10. Why does the inverse problem (finding \(V_R\) from \(V_S\)) give two roots?

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A constant-power load can be supplied at high voltage with low current or low voltage with high current. The roots merge at the voltage-collapse point — Problem 17.
P11. Up to roughly what length is the short-line model used, and what is the error in \(A\) there?

Show answer
About 80 km at 50 Hz, where \(\beta l \approx 0.085\) and the error in \(|A|\) is \((\beta l)^2/2 = \mathbf{0.36\%}\).
P12. A short line shows negative regulation. What does that tell you about the load?

Show answer
It is leading, and more strongly than the zero-regulation power factor \(\tan\phi = R/X\). The receiving voltage exceeds the sending voltage — Problem 10.

Challenge

Challenge Problems

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

C1. Derive the approximate voltage-drop formula from the exact phasor expression, obtain the leading correction term, and establish precisely when the approximation fails.

Show answer
Exact. With \(\mathbf{V}_R = V_R\angle0\) and \(\mathbf{I} = I\angle-\phi\):
\[ \mathbf{V}_S = V_R + I(\cos\phi - j\sin\phi)(R + jX) = \underbrace{V_R + IR\cos\phi + IX\sin\phi}_{a} + j\underbrace{(IX\cos\phi - IR\sin\phi)}_{b} \]

\[ V_S = \sqrt{a^2 + b^2} \]
The expansion. Since \(b \ll a\):
\[ V_S = a\sqrt{1 + (b/a)^2} \approx a + \frac{b^2}{2a} \]
So the approximate formula is \(V_S \approx a\) and the leading correction is \(b^2/2a\) — second order in the quadrature drop, which is why it is so small.

Checking against Problem 2: \(a = 33\,708.3\), \(b = 250\), so the correction is \(250^2/(2 \times 33\,708) = 0.93\) V — exactly the 33 709 minus 33 708.4 observed.

When it fails. The relative error is \((b/a)^2/2\). Writing \(b = IZ\sin(\theta - \phi)\) where \(\theta\) is the impedance angle, and \(a \approx V_R\):
\[ \text{error} \approx \frac{1}{2}\left(\frac{IZ}{V_R}\right)^2\sin^2(\theta - \phi) \]
It therefore fails when:
— \(IZ/V_R\) is large, i.e. heavy loading or a high-impedance line. At \(IZ/V_R = 0.3\) the error reaches 4.5%.
— \(\theta - \phi \to 90^\circ\), i.e. the load angle is far from the line angle. Worst at unity power factor on a highly inductive line, where \(b\) is maximal.

Practical rule: the approximation is good to better than 1% whenever the regulation itself is under about 15%, which covers essentially every acceptable line. It is not reliable for computing the angle \(\alpha\), which depends on \(b\) directly rather than on its square — and \(\alpha\) is exactly what power transfer depends on.
C2. Voltage regulation and transmission efficiency are both measures of line quality, yet a line can be excellent at one and poor at the other. Explain the mechanism, and set out what a designer does about each.

Show answer
Different components of the same impedance.
— Efficiency depends on \(R\) alone: \(\eta = P_R/(P_R + 3I^2R)\). Reactance dissipates nothing.
— Regulation depends on both, as \(\%R\cos\phi + \%X\sin\phi\), and for a transmission line with \(X/R = 10\) to 20 the reactance term dominates at any realistic power factor.

The four cases:
— High \(R\), low \(X\) (a cable, or a small distribution conductor): poor efficiency, tolerable regulation.
— Low \(R\), high \(X\) (a long EHV line): excellent efficiency, poor regulation. Problem 20's line, at \(X/R = 2.5\), already showed 94.8% efficiency with 9.7% regulation.
— Both low: a short, heavy line — ideal and expensive.
— Both high: an undersized long feeder, failing on both counts.

The remedies differ completely:
— To improve efficiency: more conductor (lower \(R\)), higher voltage (lower \(I\)), or power-factor correction (lower \(I\)). Kelvin's law of Set 2 governs the first.
— To improve regulation: series capacitors cancelling \(X_L\); shunt capacitors supplying the load's vars locally so \(\sin\phi \to 0\); tap-changing transformers raising \(V_S\); or a leading power factor as in Problem 13. Adding copper barely helps, since \(R\) is not the dominant term.

The key asymmetry: series compensation improves regulation dramatically and efficiency not at all, while extra copper improves efficiency and regulation only slightly. A designer who responds to a regulation problem by ordering a larger conductor is spending money on the wrong parameter — which is exactly the diagnosis Problem 20 invites.
C3. The short-line model sets \(C = 0\). Examine what is lost by that, distinguishing errors of magnitude from errors of kind, and identify the situation in which the model is not merely inaccurate but qualitatively wrong.

Show answer
Errors of magnitude — small and well-behaved. Under load, the shunt admittance diverts a current \(V B l\) that does not reach the load. Problem 19 found this to be 5% of the load current at about 50 km, and the resulting error in \(A\) grows as \((\beta l)^2/2\) — 0.14% at 50 km, 1.4% at 160 km. These are ordinary approximation errors: they shrink smoothly as the line shortens, and they can be bounded in advance.

Errors of kind — qualitative and unbounded. With \(C = 0\) the model asserts \(A = 1\) exactly, so:
— No Ferranti effect. The model predicts \(V_R = V_S\) on open circuit at any length. The truth is \(V_R = V_S/\cos(\beta l)\), which rises and diverges as \(\beta l \to 90^\circ\). This is not a small error; it is the wrong sign of behaviour.
— No charging current. The model says an unloaded line draws nothing. A 300 km, 400 kV line draws 164 MVAr (Set 6 Problem 20) — a quantity that determines whether the line can be energised at all without a reactor.
— No resonance. The exact model has poles at \(\beta l = 90^\circ\); the short-line model has none, so it cannot represent the tuned-line phenomena of Set 14.
— Reactive balance reversed. The model has the line absorbing vars at every loading. A real line generates them below SIL — the entire basis of Set 7's surge impedance loading.

The situation where it is qualitatively wrong: the lightly loaded or unloaded long line. There the load current, which the model handles well, is small or absent, while the charging current, which the model denies entirely, is the whole story. A 100 km line at full load is well served by the short-line model; the same line at no load is not described by it at all.

The lesson: the validity of a model depends on the operating condition as well as the geometry. "Short line" is a statement about a line and its duty together, not about a length.

Self-Test

Multiple-Choice Questions

MCQ 1. In the short-line model the shunt admittance is:
(a) lumped at the receiving end (b) split between both ends (c) neglected entirely (d) distributed

Show answer
(c). Options (a) and (b) describe the nominal-\(\pi\) and nominal-T models of Set 11.
MCQ 2. The ABCD parameters of a short line are:
(a) \(A=D=1, B=Z, C=0\) (b) \(A=D=1, B=0, C=Y\) (c) \(A=D=Z, B=1, C=0\) (d) all unity

Show answer
(a). \(C = 0\) is the discarded shunt branch — Problem 8.
MCQ 3. For a lagging load, the sending-end power factor compared with the receiving-end one is:
(a) higher (b) lower (c) equal (d) leading

Show answer
(b) lower. The line's reactance absorbs vars that must be supplied from the sending end — Problem 4.
MCQ 4. The approximate voltage-drop formula neglects:
(a) the resistance (b) the reactance (c) the quadrature component of the drop (d) the load current

Show answer
(c). It enters only through its square, so the error is second order — Challenge C1.
MCQ 5. Voltage regulation is zero when the load power factor satisfies:
(a) \(\tan\phi = X/R\) lagging (b) \(\tan\phi = R/X\) leading (c) \(\cos\phi = 1\) (d) never

Show answer
(b). Always leading, and independent of the load current — Problem 13.
MCQ 6. For a fixed power and voltage, the line loss varies with power factor as:
(a) \(\cos\phi\) (b) \(\cos^2\phi\) (c) \(1/\cos\phi\) (d) \(1/\cos^2\phi\)

Show answer
(d). \(I \propto 1/\cos\phi\) and loss \(\propto I^2\) — Problem 18.
MCQ 7. Transmission efficiency depends on:
(a) \(R\) only (b) \(X\) only (c) both equally (d) neither

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(a). Reactance dissipates no real power. Regulation, by contrast, is usually dominated by \(X\) — Challenge C2.
MCQ 8. A negative voltage regulation indicates a load that is:
(a) lagging (b) leading (c) resistive (d) impossible

Show answer
(b) leading, beyond the zero-regulation power factor — Problem 10.
MCQ 9. The percentage resistance of a line equals its per-unit resistance on a base of:
(a) 100 MVA (b) the line's rated voltage and the load's rating (c) 1 MVA (d) the generator's rating

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(b). \(V_R/I\) at rated conditions is the base impedance — Problem 14.
MCQ 10. Solving for \(V_R\) given \(V_S\) and a constant-power load yields:
(a) one root (b) two roots, one physical (c) no roots (d) infinitely many

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(b). The lower root is a genuine high-current operating point; the two merge at voltage collapse — Problem 17.
MCQ 11. The short-line model is normally used up to about:
(a) 8 km (b) 80 km (c) 250 km (d) 800 km

Show answer
(b) 80 km, where \(\beta l \approx 0.085\) rad and the error in \(A\) is 0.36% — Problem 19.
MCQ 12. The short-line model is qualitatively wrong for:
(a) a heavily loaded short line (b) an unloaded long line (c) a leading load (d) unity power factor

Show answer
(b). It predicts no charging current and no Ferranti rise, which is the wrong kind of behaviour rather than a small error — Challenge C3.

Reference

Key Formulas

Quantity	Relation	Notes
The model	\(\mathbf{V}_S = \mathbf{V}_R + \mathbf{I}\mathbf{Z}\)	\(\mathbf{I}_S = \mathbf{I}_R\); shunt discarded
Load current, 3-φ	\(I = P_R/(\sqrt3 V_L\cos\phi_R)\)	Line quantities
Current phasor	\(\mathbf{I} = I(\cos\phi_R \mp j\sin\phi_R)\)	\(-\) lagging, \(+\) leading
Approximate \(V_S\)	\(V_R + IR\cos\phi_R \pm IX_L\sin\phi_R\)	Error \(\approx (b/a)^2/2\)
Correction term	\(b^2/2a\), \(b = IX\cos\phi - IR\sin\phi\)	Second order — Challenge C1
Sending-end p.f.	\(\cos\phi_S \approx (V_R\cos\phi_R + IR)/V_S\)	Or \(\phi_S = \phi_R + \alpha\)
Line loss, 3-φ	\(3I^2R\)	Per-conductor \(R\)
Efficiency	\(\eta = P_R/(P_R + 3I^2R)\)	Depends on \(R\) only
Voltage regulation	\((V_{R,NL} - V_{R,FL})/V_{R,FL}\)	\(V_{R,NL} = V_S\) since \(A = 1\)
Regulation, compact	\(\%R\cos\phi \pm \%X\sin\phi\)	\(\%R = IR/V_R\), \(\%X = IX/V_R\)
Zero regulation	\(\tan\phi = R/X_L\), leading	Independent of current
ABCD	\(A = D = 1,\ B = Z,\ C = 0\)	\(AD - BC = 1\)
Sending-end power	\(P_S = P_R + 3I^2R\), \(Q_S = Q_R + 3I^2X_L\)	Ratio fixed by \(X/R\)
Loss vs power factor	loss \(\propto 1/\cos^2\phi\)	Fixed \(P_R\) and \(V_R\)
Validity	\(\beta l \lesssim 0.085\) rad, \(l \lesssim 80\) km	Error in \(A\) \(\approx (\beta l)^2/2\)

Diagnostics

Common Mistakes

Getting the sign of \(j\sin\phi\) wrong. Lagging takes \(-j\) with \(\mathbf{V}_R\) as reference. This single line decides the sign of the whole answer — Problems 1 and 10.
Using line voltage where phase voltage is meant. Divide by \(\sqrt3\) at the start and multiply back at the end — Problem 6.
Confusing per-conductor with total resistance. A single-phase line quotes the loop; a three-phase line quotes one conductor. Compare Problems 5 and 6.
Multiplying the three-phase loss by three twice. \(3I^2R\) already covers all three phases.
Taking efficiency as \(1 - \text{loss}/P_R\). It is \(P_R/(P_R + \text{loss})\) — the two differ increasingly as the loss grows.
Assuming \(V_{R,NL} = V_S\) for any line. True only when \(A = 1\), which is exactly the short-line assumption — Problem 9.
Taking the smaller root of the inverse quadratic. It is a real but catastrophic operating point — Problem 17.
Applying the approximate formula to find the angle \(\alpha\). The approximation discards precisely the quadrature term that \(\alpha\) depends on.
Expecting more copper to fix a regulation problem. Regulation is usually dominated by \(X\), which copper does not change — Challenge C2.
Reporting a positive regulation for a strongly leading load. It is negative beyond \(\tan\phi = R/X\), and the arithmetic will say so if the sign is handled correctly.
Using the short-line model on an unloaded long line. It denies the charging current entirely, which is the whole of the answer there — Challenge C3.
Quoting regulation without stating the power factor. Problem 12 found it ranging over seven percentage points on one unchanged line.

Looking Ahead

Everything in this set followed from one equation, \(\mathbf{V}_S = \mathbf{V}_R + \mathbf{I}\mathbf{Z}\), and the essential behaviour of a loaded line is already fully contained in it: the voltage falls, the power factor worsens, and the losses vary as the inverse square of the power factor.

Set 10 stays with the short line and pursues regulation and efficiency in their own right — including the compensation methods that Challenge C2 introduced. Set 11 then restores the shunt admittance as a lumped element at one or both ends, which gives \(A \ne 1\) and with it the Ferranti effect that this model could not represent. Set 12 abandons lumping for the wave equation and finds that the nominal-\(\pi\) was itself an approximation, exact only in the limit of vanishing electrical length.