Question 1 of 7: Ferranti effect and the exact long-line model

Question 1: Ferranti effect and the exact long-line model

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

The Ferranti effect is the rise of the open-circuit receiving-end voltage of a transmission line above its sending-end voltage. It is a property of the distributed shunt capacitance of the line. With the far end open there is no load current, but the line's own capacitance still draws a charging current from the source, and that current is leading — it flows into the line, not out of it. A leading current flowing through the series inductive reactance of the line produces a voltage rise rather than the familiar drop, so the voltage grows steadily from the sending end towards the open receiving end.

The exact statement follows from the two-port model. With the receiving end open, $I_R = 0$, so $V_S = A V_R$ and

Because $\cosh(\gamma\ell)$ has magnitude less than unity for a lossless line of practical length (its real argument is small and its imaginary argument makes the cosine term fall below one), $|V_R|$ exceeds $|V_S|$. The rise grows roughly with the square of the line length and directly with frequency, so it is negligible on a 50 km distribution feeder and severe on a 400 km EHV line or a long cable, where the shunt capacitance per kilometre is an order of magnitude larger than on an overhead line.

Operationally the effect matters at four points. First, on energisation: closing a breaker onto a long unloaded line can drive the far-end busbar and the open-end equipment well above rated voltage, stressing insulation and saturating instrument and station-service transformers. Second, at light load, typically overnight, when the load falls below the line's surge-impedance loading the same mechanism raises system voltage everywhere and forces operators to absorb reactive power. Third, it drives equipment selection: shunt reactors, switched at the line ends or connected permanently through the tertiary of a transformer, are sized specifically to cancel the line charging; generators may be run under-excited, and surge arresters and breakers must be rated for the elevated open-end voltage. Fourth, it is the reason line-energisation studies precede commissioning, and the reason a long line is normally closed at the sending end onto a reactor-compensated far end rather than left open. On this line the open-circuit rise is $1/|A| = 1/0.8928 = 1.120$, that is 12.0 per cent above the sending-end voltage.

Parts (b) and (c) — the exact line model

Given. A 250 km, 138 kV three-phase line whose distributed constants and receiving-end operating point are listed below.

Find. The characteristic impedance, propagation constant and its real and imaginary parts; then the sending-end voltage, current, power factor and the transmission efficiency, using the exact (hyperbolic) long-line model rather than a nominal-π approximation.

Approach. Form $Z_c=\sqrt{z/y}$ and $\gamma=\sqrt{zy}$ from the per-kilometre constants, build the ABCD constants from $\cosh\gamma\ell$ and $\sinh\gamma\ell$, then push the known receiving-end phasors through the two-port to obtain $V_S$ and $I_S$.

1. Put the distributed constants in polar form. $z = 0.15+j0.70 = 0.71589\angle 77.905^\circ\ \Omega/\text{km}$ and $y = 5.0\times10^{-6}\angle 90^\circ\ \text{S/km}$. Working in polar form makes both square roots one-line operations.
2. Characteristic impedance. $$Z_c=\sqrt{\frac{z}{y}}=\sqrt{\frac{0.71589\angle 77.905^\circ}{5.0\times10^{-6}\angle 90^\circ}} =\sqrt{143\,178\angle(-12.095^\circ)}$$ which gives $\boxed{Z_c = 378.39\angle-6.047^\circ\ \Omega = 376.28 - j39.86\ \Omega}$. The small negative angle is typical: a real line is slightly resistive, so $Z_c$ sits just below the real axis instead of on it.
3. Propagation constant. By the same route, $$\gamma=\sqrt{zy}=\sqrt{(0.71589\angle 77.905^\circ)(5.0\times10^{-6}\angle 90^\circ)} =1.89195\times10^{-3}\angle 83.953^\circ\ \text{km}^{-1}$$ Resolving into rectangular components gives the attenuation and phase constants, $$\boxed{\alpha = 1.9932\times10^{-4}\ \text{Np/km},\;\; \beta = 1.8814\times10^{-3}\ \text{rad/km}}$$ Over the whole line, $\gamma\ell = 0.049829 + j0.470354$, so $\alpha\ell = 0.04983$ Np (0.433 dB) and $\beta\ell = 0.47035$ rad $= 26.949^\circ$. The line is therefore about one-thirteenth of a wavelength long — short enough that a nominal-π model would not be badly wrong, but long enough that the exact model is worth the extra two hyperbolic functions.
4. Assemble the ABCD constants. With $\cosh\gamma\ell = \cosh\alpha\ell\cos\beta\ell + j\sinh\alpha\ell\sin\beta\ell$ and $\sinh\gamma\ell = \sinh\alpha\ell\cos\beta\ell + j\cosh\alpha\ell\sin\beta\ell$, $$A=D=\cosh\gamma\ell = 0.89280\angle 1.450^\circ$$ $$B=Z_c\sinh\gamma\ell = 172.52\angle 78.360^\circ\ \Omega$$ $$C=\frac{\sinh\gamma\ell}{Z_c}=1.20494\times10^{-3}\angle 90.454^\circ\ \text{S}$$ The reciprocity check $AD-BC = 1.0000$ confirms the arithmetic before any of it is used.
5. Fix the receiving-end phasors. Per phase, $V_R = 132\,000/\sqrt{3} = 76\,210\ \text{V}\angle 0^\circ$, and the current follows from the delivered power, $$I_R=\frac{P_R}{\sqrt3\,V_{R,LL}\cos\varphi_R}\angle-\varphi_R =\frac{15\times10^{6}}{\sqrt3(132\,000)(0.85)}\angle-31.788^\circ = 77.186\angle-31.788^\circ\ \text{A}$$
6. Sending-end voltage. Substituting into $V_S = AV_R + BI_R$, $$V_S=(0.89280\angle 1.450^\circ)(76\,210\angle 0^\circ)+(172.52\angle 78.360^\circ)(77.186\angle-31.788^\circ)$$ $$V_S=68\,036\angle 1.450^\circ + 13\,317\angle 46.572^\circ = 77\,172+j11\,390\ \text{V}$$ so $|V_S| = 78\,009$ V per phase and $\boxed{V_S = 135.12\ \text{kV line-to-line at }\angle 8.397^\circ}$, a rise of 2.4 per cent over the 132 kV receiving-end voltage.
7. Sending-end current. From $I_S = CV_R + DI_R$, $$I_S=(1.20494\times10^{-3}\angle 90.454^\circ)(76\,210)+(0.89280\angle 1.450^\circ)(77.186\angle-31.788^\circ)$$ $$I_S = (-0.73+j91.83)+(59.47-j34.82) = 58.73+j57.01\ \text{A}$$ giving $\boxed{I_S = 81.87\angle 44.145^\circ\ \text{A}}$. The charging current is what turns a 77 A load current into an 82 A sending-end current at a very different angle.
8. Sending-end power factor. The angle between $V_S$ and $I_S$ is $8.397^\circ - 44.145^\circ = -35.747^\circ$; the current leads the voltage, so $$\boxed{\cos\varphi_S = 0.8116\ \text{leading}}$$ This is not an error. The line's total charging is $B_{\text{tot}} = 5.0\times10^{-6}(250) = 1.25\times10^{-3}$ S, which at 132 kV supplies about 21.8 MVAr, far more than the 9.30 MVAr the load absorbs, so the line exports reactive power back to the source. The surge-impedance loading is $V^2/|Z_c| = 138^2/378.39 = 50.3$ MW, and the line is carrying only 15 MW — well under a third of SIL, which is exactly the regime in which a line behaves capacitively.
9. Sending-end power and efficiency. From $S_S = 3V_S I_S^{*}$, $P_S = 15.550$ MW and $Q_S = -11.193$ MVAr, so the real loss is $15.550 - 15.000 = 0.550$ MW and $$\eta=\frac{P_R}{P_S}=\frac{15.000}{15.550}\times 100 = \boxed{96.47\ \text{per cent}}$$