Question 1 of 7: Transposition and the Exact Long-Line ABCD Model

Problem 1: Transposition and the Exact Long-Line ABCD Model (20 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) — what a transposed line is, and why it is needed. The three phase conductors of an overhead line do not occupy electrically equivalent positions on the structure. On a flat or a triangular configuration one conductor is nearer to the other two, or nearer the earth, than its neighbours are to each other, so the self and mutual inductances and the self and mutual capacitances of the three phases are not equal. Left alone, that geometric asymmetry makes the series impedance matrix and the shunt admittance matrix non-symmetric, and a perfectly balanced set of applied voltages then produces unbalanced currents — a standing negative-sequence and zero-sequence content that exists purely because of the tower geometry.

Transposition is the practice of rotating the three phases through the three physical positions in equal thirds of the route length: each conductor occupies position 1 for a third of the line, position 2 for a third, and position 3 for the remaining third, the changeovers being made at transposition structures. Averaged over a full transposition cycle every phase then sees the same mean geometry, so the three self impedances become equal and the three mutual impedances become equal.

Two consequences make it worth the cost. First, the line becomes electrically symmetric, so the negative- and zero-sequence voltages it generates from balanced load current fall essentially to zero; on a long EHV circuit an untransposed line can produce a percent or more of negative-sequence voltage, which causes double-frequency rotor heating in every generator and large induction motor connected to it, and which can pick up negative-sequence relays. Second — and this is why the calculation in parts (b) and (c) is legitimate at all — symmetry is exactly the condition under which the symmetrical-component transformation diagonalises the impedance matrix. Only then can a three-phase line be represented by a single per-phase positive-sequence series impedance $z$ and shunt admittance $y$, which is the model the question hands us. In modern practice full transposition is often replaced by rotational transposition (the phase order is rotated between successive line sections, or between the two circuits of a double- circuit tower) because the transposition structure itself is expensive and is a recognised source of outages; the electrical objective is unchanged.

Given. A 500 km, 765 kV, 60 Hz three-phase line with distributed constants, delivering 2300 MVA at 0.85 power factor lagging to a receiving-end bus held at 97.5 % of rated voltage.

Find. The exact long-line constants $A$, $B$, $C$, $D$, and then the sending-end voltage, current, power factor and transmission efficiency at full load.

Approach. Form the propagation constant $\gamma=\sqrt{zy}$ and the characteristic impedance $Z_c=\sqrt{z/y}$, evaluate the hyperbolic functions of $\gamma\ell$ to get $A$, $B$ and $C$, then push the receiving-end phasors through $V_S = A V_R + B I_R$ and $I_S = C V_R + D I_R$ and take the powers at each end.

1. Part (b) — form the propagation constant and the characteristic impedance. Both follow directly from the per-kilometre constants, and both are computed in polar form because the square roots are then trivial: $$\gamma=\sqrt{zy},\ Z_c=\sqrt{z/y},\ z = 0.30037\angle87.138^\circ\ \Omega/\text{km},\ y = 5.000\times10^{-6}\angle90^\circ\ \text{S/km}.$$ Multiplying and dividing the polar forms, $$zy = 1.50187\times10^{-6}\angle177.138^\circ,\ z/y = 6.0074\times10^{4}\angle-2.862^\circ,$$ so that $$\gamma = 1.22551\times10^{-3}\angle88.569^\circ\ \text{km}^{-1},\ Z_c = 245.102\angle-1.4312^\circ\ \Omega.$$
2. Scale the propagation constant to the whole line. Multiplying by $\ell = 500$ km gives the electrical length of the circuit, $$\gamma\ell = 0.612755\angle88.5688^\circ = 0.0153045 + j0.6125637,$$ i.e. an attenuation of $\alpha\ell = 0.01530$ Np (about 1.5 %) and a phase shift of $\beta\ell = 0.61256$ rad $= 35.097^\circ$. A line this long is a genuinely distributed circuit: the 35° of electrical length is what makes $A$ fall well below unity.
3. Evaluate the hyperbolic functions. Splitting $\gamma\ell$ into its real and imaginary parts, $$\cosh\gamma\ell=\cosh\alpha\ell\cos\beta\ell+j\sinh\alpha\ell\sin\beta\ell = 0.818273+j0.008800,$$ $$\sinh\gamma\ell=\sinh\alpha\ell\cos\beta\ell+j\cosh\alpha\ell\sin\beta\ell = 0.012522+j0.575034 = 0.575171\angle88.7525^\circ.$$
4. Assemble the ABCD constants. The exact long-line definitions are $A=D=\cosh\gamma\ell$, $B=Z_c\sinh\gamma\ell$ and $C=\sinh(\gamma\ell)/Z_c$, so $$\boxed{\begin{aligned} A &= D = 0.81832\angle0.6162^\circ \\ B &= 140.975\angle87.3213^\circ\ \Omega \\ C &= 2.34666\times10^{-3}\angle90.1837^\circ\ \text{S} \end{aligned}}$$ The two useful sanity checks both pass: reciprocity gives $AD-BC = 1.0000\angle0^\circ$ to five figures, and $C$ sits within a fifth of a degree of $+90^\circ$, which is what a nearly pure shunt susceptance must do.
5. Part (c) — put the receiving-end load into phasor form. Working per phase with $V_R$ as reference, $$V_R=\frac{745.875}{\sqrt3}=430.631\angle0^\circ\ \text{kV},\ \varphi_R=\cos^{-1}0.85=31.788^\circ,$$ $$I_R=\frac{S_R}{\sqrt3\,V_{R,LL}}\angle-\varphi_R=\frac{2300}{\sqrt3(745.875)}\angle-31.788^\circ = 1.78033\angle-31.788^\circ\ \text{kA}.$$ The current lags because the load is inductive.
6. Apply the first ABCD equation for the sending-end voltage. Each term is evaluated in polar form and then added rectangularly: $$A V_R = 352.394\angle0.616^\circ\ \text{kV},\ B I_R = 250.983\angle55.533^\circ\ \text{kV},$$ $$V_S = (352.374+j3.790)+(142.039+j206.923)\ \text{kV},$$ $$\boxed{V_S = 537.442\angle23.083^\circ\ \text{kV per phase} = 930.88\ \text{kV line-to-line}}$$ That is 21.7 % above the 765 kV rating, and the reason is visible in the two terms: the $B I_R$ drop is over 70 % as large as $A V_R$ because the line is carrying its rated MVA at only 0.85 power factor.
7. Apply the second ABCD equation for the sending-end current. With $D=A$, $$C V_R = 1.01054\angle90.184^\circ\ \text{kA},\ A I_R = 1.45688\angle-31.172^\circ\ \text{kA},$$ $$I_S = (-0.00324+j1.01054)+(1.24653-j0.75410) = 1.24329+j0.25644,$$ $$\boxed{I_S = 1.26946\angle11.654^\circ\ \text{kA}}$$ The shunt charging current $C V_R$ is over 1 kA in its own right, and it is almost in phase quadrature with $V_R$; that is what pulls the sending-end current angle from $-31.2^\circ$ up to $+11.7^\circ$.
8. Take the sending-end power factor and the two end powers. The power-factor angle is the difference of the two phasor angles, $$\varphi_S=23.083^\circ-11.654^\circ=11.429^\circ,\ \cos\varphi_S=0.9802\ \text{lagging},$$ $$S_S=\sqrt3\,V_{S,LL}I_S=\sqrt3(930.88)(1.26946)=2046.8\ \text{MVA},$$ $$P_S=2046.8\cos\varphi_S=2006.2\ \text{MW},\ P_R=2300(0.85)=1955.0\ \text{MW}.$$
9. Divide to get the transmission efficiency. Efficiency is the ratio of the two real powers, nothing more: $$\boxed{\eta=\frac{P_R}{P_S}=\frac{1955.0}{2006.2}=0.9745 = 97.45\ \text{per cent}}$$ The 51.2 MW of loss is the $3I^2R$ dissipation in $0.015\ \Omega/\text{km}\times500\ \text{km}=7.5\ \Omega$ of resistance per phase, evaluated at the current profile along the line.
10. Interpret the result against the surge impedance loading. The line's SIL is $$P_{\text{SIL}}=\frac{V_{\text{rated}}^{2}}{|Z_c|}=\frac{(765)^2}{245.1}=2388\ \text{MW},$$ so 2300 MVA is close to SIL in magnitude — but at 0.85 power factor only 1955 MW of it is real power, and the missing 1212 Mvar has to be pushed through $150\ \Omega$ of series reactance. That is why the sending end must be held 21.7 % high. In Canadian EHV practice a circuit operated this way would carry shunt-capacitor or SVC support at the receiving end (and probably series compensation), so that the line runs nearer to unity power factor and the sending-end voltage stays within the ±5 % operating band.

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