Question 1 of 7: Ferranti Effect and Long-Line ABCD Performance

Question 1: Ferranti Effect and Long-Line ABCD Performance (25 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Part (a) — the Ferranti effect. On a long transmission line the distributed shunt capacitance draws a charging current that flows through the series inductance of the line even when nothing is connected at the far end. That charging current leads the voltage by very nearly 90°, so the voltage it develops across the series reactance adds to the receiving-end voltage instead of subtracting from it. The result is the Ferranti effect: on an open-circuited or very lightly loaded line the receiving-end voltage rises above the sending-end voltage, and the rise grows with the square of the line length and with the square of the frequency.

The two-port statement is compact. With the far end open, $I_R = 0$, so $V_S = A V_R$ and therefore $|V_R| = |V_S| / |A|$. For this line $|A| = 0.95$, so an open-ended 765 kV line sits about $1/0.95 = 1.0526$, i.e. 5.3 % above its sending-end voltage. On a longer or a series-compensated circuit the figure reaches 10–15 %.

It matters operationally for three reasons. First, sustained overvoltage stresses insulation, bushings and surge arresters, and it saturates transformers connected at the open end, which then draw heavily distorted magnetising current. Second, the effect is at its worst exactly when the system is least able to absorb it — at light load overnight and immediately after energising a line, when the generators nearest the line may already be running at their under-excitation limits. Third, it dictates equipment: Canadian 500 kV and 735 kV circuits carry shunt reactors, frequently with a neutral reactor, precisely to absorb the line charging and hold the open-end voltage down, and the energisation sequence in the switching order exists to manage the same phenomenon. The practical test a system operator applies is the comparison of line loading with the surge impedance loading: below SIL the line is a net source of reactive power and the voltage profile rises, above SIL it is a net sink and the profile falls.

Given.

Find. The shunt constant $C$, then the sending-end voltage and current, the sending-end power factor and the transmission efficiency at the stated receiving-end load.

Approach. Recover $C$ from the reciprocity condition $AD - BC = 1$ with $D = A$, then evaluate the two ABCD equations at the stated receiving-end operating point and take the ratio of real powers.

1. Part (b) — recover $C$ from reciprocity. A transmission line is a passive, reciprocal, symmetric two-port, so $AD - BC = 1$ and $D = A$. Solving for $C$, $$C=\frac{A^{2}-1}{B}$$ Squaring the given $A$ gives $A^{2} = 0.9025\angle 4^\circ = 0.900301 + j0.062955$, so $$\begin{aligned}A^{2}-1 &= -0.099699 + j0.062955 \\ &= 0.117912\angle 147.729^\circ\end{aligned}$$ Dividing by $B$, $$\boxed{\begin{aligned}C &= \frac{0.117912\angle 147.729^\circ}{195\angle 87.5^\circ} \\ &= 6.0468\times10^{-4}\angle 60.229^\circ\ \text{S}\end{aligned}}$$ In rectangular form $C = 3.0024\times10^{-4} + j5.2487\times10^{-4}$ S.
2. Sanity-check the recovered constant. On a physical line the shunt constant is very nearly a pure susceptance, so $C$ should come out within about a degree of $+90^\circ$; here it is $60.23^\circ$, which means the printed $2^\circ$ angle on $A$ has loaded the model with a large equivalent shunt conductance. The characteristic impedance that follows, $$\begin{aligned}Z_c &= \sqrt{B/C} = \sqrt{\frac{195\angle 87.5^\circ}{6.0468\times 10^{-4}\angle 60.229^\circ}} \\ &= 567.88\angle 13.64^\circ\ \Omega\end{aligned}$$ gives a surge impedance loading of $765^2/567.9 = 1030.5$ MW. Keep that number in view: the load about to be applied is 1176 MVA, well above SIL, so the sending-end voltage must come out higher than the receiving-end voltage.
3. Part (c) — reduce the receiving-end load to per-phase quantities. Working line-to-neutral on the receiving end, $$\begin{aligned}|V_R| &=\frac{720\ \text{kV}}{\sqrt3}=415.692\ \text{kV} \\ \angle V_R &= 0^\circ\end{aligned}$$ The apparent power is $S_R = P_R/\cos\phi_R = 1000/0.85 = 1176.471$ MVA, so the line current is $$\begin{aligned}|I_R| &= \frac{S_R}{\sqrt3\,|V_{R,LL}|} \\ &= \frac{1176.471\times10^{6}}{\sqrt3\,(720\times10^{3})} = 943.383\ \text{A}\end{aligned}$$ With $\cos\phi_R = 0.85$ lagging, $\phi_R = 31.788^\circ$ and $$I_R = 943.383\angle -31.788^\circ\ \text{A}$$
4. Evaluate the first ABCD equation for the sending-end voltage. Substituting into $V_S = A V_R + B I_R$, the two terms are $$\begin{aligned}A V_R &= (0.95\angle 2^\circ)(415.692\angle 0^\circ) = 394.908\angle 2^\circ \\ &= 394.667 + j13.782\ \text{kV} \\ B I_R &= (195\angle 87.5^\circ)(943.383\angle -31.788^\circ) \\ &= 183.960\angle 55.712^\circ = 103.635 + j151.990\ \text{kV}\end{aligned}$$ Adding, $$\boxed{\begin{aligned}V_S &= 498.302 + j165.772 \\ &= 525.153\angle 18.401^\circ\ \text{kV per phase}\end{aligned}}$$ which is $\sqrt3 \times 525.153 = \mathbf{909.59}$ kV line-to-line.
5. Evaluate the second ABCD equation for the sending-end current. With $D = A$, $I_S = C V_R + A I_R$, and the two terms are $$\begin{aligned}C V_R &= (6.0468\times10^{-4}\angle 60.229^\circ)(415\,692) \\ &= 251.359\angle 60.229^\circ = 124.81 + j218.18\ \text{A} \\ A I_R &= (0.95\angle 2^\circ)(943.383\angle -31.788^\circ) \\ &= 896.214\angle -29.788^\circ = 777.79 - j445.24\ \text{A}\end{aligned}$$ so that $$\boxed{I_S = 902.60 - j227.05 = 930.72\angle -14.120^\circ\ \text{A}}$$
6. Sending-end power factor. The power-factor angle is the angle by which the current lags the voltage at the same terminals: $$\begin{aligned}\phi_S &= \angle V_S - \angle I_S = 18.401^\circ - (-14.120^\circ) \\ &= 32.521^\circ\end{aligned}$$ $$\boxed{\cos\phi_S = 0.8432\ \text{lagging}}$$
7. Transmission efficiency. The three-phase sending-end real power is $$\begin{aligned}P_S &= 3\,|V_S||I_S|\cos\phi_S \\ &= 3(525.153\times10^{3})(930.72)(0.8432) = 1236.39\ \text{MW}\end{aligned}$$ and the efficiency is the ratio of the powers at the two ends, $$\boxed{\eta=\frac{P_R}{P_S}=\frac{1000}{1236.39}=80.88\ \%}$$

The 236 MW of loss deserves a word, because it is far more than the series resistance alone could account for. The series arm carries $\text{Re}(B) = 195\cos 87.5^\circ = 8.51\ \Omega$, worth about 22.7 MW at this current. The remaining 155.6 MW is charged to $\text{Re}(C) = 3.00\times 10^{-4}$ S acting across the receiving-end voltage — the artefact of the printed $2^\circ$ angle on $A$ flagged at the top of this paper. The voltage rise itself is genuine and expected: at 1176 MVA the line is loaded to 1.14 times its surge impedance loading, so it is a net absorber of reactive power and the sending end must sit above the receiving end.