Question 1 of 7: Ferranti Effect and the Long-Line Model of a 765 kV Circuit

Problem 1: Ferranti Effect and the Long-Line Model of a 765 kV Circuit (25 points)

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 receiving-end voltage of a long transmission line above its sending-end voltage when the line is open-circuited or very lightly loaded. It is named for Sebastian Ziani de Ferranti, who observed it on the 10 kV Deptford cables in 1890. Physically it is the shunt capacitance of the line acting through the line’s own series inductance: with no load current to speak of, the only current the line draws is the capacitive charging current distributed along its length, and that current leads the voltage by ninety degrees. A leading current flowing through a series inductive reactance produces a voltage rise rather than a drop, so each successive element of the line adds a little to the voltage, and the receiving end sits highest.

The compact statement of the effect comes straight from the lossless long-line equations. At no load the receiving-end current is zero, so $V_S = A V_R$ with $A = \cosh(\gamma \ell)$, and for a lossless line this reduces to $V_R / V_S = 1/\cos(\beta \ell)$. Because $\cos(\beta \ell) < 1$ for any line shorter than a quarter wavelength, the receiving-end voltage always exceeds the sending-end voltage, and the excess grows roughly as the square of the line length. For this particular circuit $\beta \ell = 47.04^{\circ}$, so an open end would sit at $1/\cos 47.04^{\circ} = 1.467$ times the sending voltage — a 47 per cent rise.

It matters operationally for three reasons. First, sustained overvoltage stresses insulation, bushings and surge arresters, and drives transformers into saturation, so it is a plant-damage mechanism, not merely a metering curiosity. Second, it appears at exactly the worst moment: after a load rejection or a remote-end breaker trip, a fully energised line is suddenly unloaded and the receiving end can jump within a cycle. Third, it constrains commissioning and restoration practice — energising a long EHV line from one end is a routine step in a black start, and the Ferranti rise sets how much of the line may be energised at once. The countermeasures are shunt reactors (fixed or switched) at the line ends and at intermediate stations, which absorb the charging vars before they can drive the voltage up; static var compensators or STATCOMs where the response must be continuous; and operating rules that require the reactors in service before the line is energised and only permit them to be removed as load builds.

Parts (b) and (c) — long-line constants and a forward run

Given.

Find. The characteristic impedance, propagation constant and its two components, then the sending-end voltage, current, power factor and the transmission efficiency using the exact (distributed-parameter) model.

Approach. Form $Z_c=\sqrt{z/y}$ and $\gamma=\sqrt{zy}$ from the per-kilometre constants, build the exact ABCD matrix from $\gamma \ell$, then push the stated receiving-end phasors through it.

1. Part (b) — characteristic (surge) impedance. The characteristic impedance is the square root of the ratio of the series impedance to the shunt admittance, both taken per unit length: $$Z_c=\sqrt{\frac{z}{y}}=\sqrt{\frac{0.02+j0.54}{j7.8\times10^{-6}}} =\sqrt{\frac{0.54037\angle 87.879^{\circ}}{7.8\times10^{-6}\angle 90^{\circ}}}$$ The ratio inside the root is $69\,278\angle -2.121^{\circ}\ \Omega^2$, and taking the square root halves the angle: $$\boxed{Z_c=263.21\angle -1.061^{\circ}\ \Omega}$$ The angle is small and negative because the line is far more inductive than resistive; a purely lossless line would give a real surge impedance.
2. Propagation constant, attenuation constant and phase constant. The propagation constant is the geometric mean of the same two quantities: $$\gamma=\sqrt{zy}=\sqrt{(0.54037\angle 87.879^{\circ})(7.8\times10^{-6}\angle 90^{\circ})} =2.0530\times10^{-3}\angle 88.939^{\circ}\ \text{km}^{-1}$$ Resolving into rectangular form separates the two constants, since $\gamma=\alpha+j\beta$: $$\begin{aligned} \alpha &= 2.0530\times10^{-3}\cos 88.939^{\circ} = 3.800\times10^{-5}\ \text{Np/km}\\ \beta &= 2.0530\times10^{-3}\sin 88.939^{\circ} = 2.0527\times10^{-3}\ \text{rad/km} \end{aligned}$$ $$\boxed{\alpha=3.800\times10^{-5}\ \text{Np/km},\ \ \beta=2.0527\times10^{-3}\ \text{rad/km}}$$ Over the whole 400 km the electrical length is $\gamma \ell = 0.01520 + j0.82107$, that is 0.0152 nepers (only 0.132 dB of attenuation) and 0.82107 radians, or 47.04 degrees, of phase shift. The 0.132 dB confirms that a 765 kV line is an almost lossless waveguide over this distance; the 47 degrees is what makes it unmistakably a long line, well past the point where a nominal-pi model is trustworthy.
3. Assemble the exact two-port constants. The distributed-parameter solution of the line equations gives $$A=D=\cosh(\gamma \ell),\quad B=Z_c\sinh(\gamma \ell),\quad C=\frac{\sinh(\gamma \ell)}{Z_c}$$ Evaluating the hyperbolic functions of the complex argument $0.01520+j0.82107$: $$\begin{aligned} A &= D = 0.68161\angle 0.935^{\circ}\\ B &= 192.68\angle 88.129^{\circ}\ \Omega\\ C &= 2.7812\times10^{-3}\angle 90.250^{\circ}\ \text{S} \end{aligned}$$ Two free checks confirm the arithmetic before it is used. Reciprocity requires $AD-BC=1$, which these values satisfy to seven figures. And the angle of $C$ must sit within a degree of $+90^{\circ}$ because $C$ is essentially the line charging susceptance; at $90.25^{\circ}$ it does.
4. Part (c) — receiving-end phasors. Working per phase with the receiving-end voltage as reference, $$V_R=\frac{750\,000}{\sqrt3}=433.01\ \text{kV per phase}$$ and the load current follows from the three-phase real power: $$I_R=\frac{P_R}{\sqrt3\,V_{R,LL}\cos\phi} =\frac{100\times10^{6}}{\sqrt3\,(750\times10^{3})(0.955)}=80.61\ \text{A}$$ Lagging at 0.955 means the current sits at $-\cos^{-1}(0.955)=-17.25^{\circ}$, so $I_R=80.61\angle -17.25^{\circ}\ \text{A}$. Note how small this is: the line is carrying 100 MW where its surge-impedance loading is $V^2/Z_c=(765\ \text{kV})^2/263.21=2223\ \text{MW}$, so it is loaded to about 4.5 per cent of SIL.
5. Push the phasors back to the sending end. The two-port relation is $$\begin{bmatrix}V_S\\ I_S\end{bmatrix} =\begin{bmatrix}A&B\\ C&D\end{bmatrix} \begin{bmatrix}V_R\\ I_R\end{bmatrix}$$ Substituting the constants and the receiving-end phasors: $$\begin{aligned} V_S &= (0.68161\angle 0.935^{\circ})(433\,013\angle 0^{\circ}) + (192.68\angle 88.129^{\circ})(80.61\angle -17.25^{\circ})\\ &= 300.83\angle 3.715^{\circ}\ \text{kV per phase} \end{aligned}$$ $$\begin{aligned} I_S &= (2.7812\times10^{-3}\angle 90.250^{\circ})(433\,013\angle 0^{\circ}) + (0.68161\angle 0.935^{\circ})(80.61\angle -17.25^{\circ})\\ &= 1189.8\angle 87.713^{\circ}\ \text{A} \end{aligned}$$ In line terms $$\boxed{V_{S,LL}=\sqrt3\,(300.83)=521.05\ \text{kV},\qquad I_S=1189.8\ \text{A}}$$
6. Sending-end power factor and transmission efficiency. The angle between the sending-end voltage and current is $$\theta_S=3.715^{\circ}-87.713^{\circ}=-83.998^{\circ} \ \Longrightarrow\ \cos\theta_S=0.1046\ \text{leading}$$ The three-phase sending-end power is $$P_S=3V_SI_S\cos\theta_S=3(300\,827)(1189.8)(0.1046)=112.27\ \text{MW}$$ so the efficiency of transmission is $$\eta=\frac{P_R}{P_S}\times 100=\frac{100.00}{112.27}\times 100$$ $$\boxed{\eta=89.07\ \text{per cent},\qquad \cos\theta_S=0.105\ \text{leading}}$$ The line loss is the difference, 12.27 MW.

Final Results.