Question 1 of 7: Transmission Capacity and the ABCD Two-Port

Problem 1: Transmission Capacity and the ABCD Two-Port (5 + 5 + 10 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) — Transmission capacity of a line

The transmission capacity of a line is the largest steady-state power it can carry continuously without violating any of three independent limits, and the capacity of a given line is whichever of the three binds first. The thermal limit is the current at which conductor annealing, sag and statutory ground clearance become the constraint; it governs short lines and is set by the conductor material and the ambient conditions. The voltage-drop limit is the loading at which the receiving-end voltage falls (or, on a lightly loaded line, rises) outside the operating band, typically ±5 per cent; it governs lines of intermediate length. The steady-state stability limit is the loading at which the machines at the two ends can no longer hold synchronism, and because the transferred power is $P=\dfrac{|V_S||V_R|}{X}\sin\delta$, this limit is inversely proportional to the series reactance and so governs long lines such as the 240-mile circuit in this problem. A convenient benchmark for a long line is its surge-impedance loading, $\text{SIL}=V_{LL}^{2}/Z_c$, at which the line neither absorbs nor produces reactive power; a 760-kV line is typically operated at one to two times SIL.

At the system design stage the capacity is raised by reducing the series reactance or raising the voltage. Bundling the conductors — the four-conductor bundle specified here — increases the geometric mean radius, which lowers the series reactance and raises the surge impedance loading by roughly 20 to 30 per cent relative to a single conductor of the same total aluminium, while simultaneously reducing the surface gradient and therefore corona loss and radio interference. Raising the nominal voltage is the most powerful single measure, since capacity scales with the square of voltage; building a second circuit on the same right-of-way, installing series capacitors to cancel part of the inductive reactance, or choosing an HVDC link (which has no angular stability limit at all) are the other standard design choices.

During operation the capacity is raised without rebuilding the line. Reactive support at the receiving end — switched shunt capacitors, a static var compensator or a STATCOM — holds the receiving-end voltage up and so pushes back the voltage-drop limit; on this class of line, shunt reactors are switched in at light load for the opposite reason, to absorb the line charging. Dynamic (weather-adjusted) line rating exploits the fact that the static thermal rating is computed for a still, hot summer day: with wind speed and ambient temperature telemetered, a Canadian utility can legitimately claim substantially more current on a cold, windy winter evening, which is exactly when the peak occurs. Generation re-dispatch, phase-shifting transformers to steer flow away from the constrained circuit, and special protection schemes that arm a fast generator runback all raise the permissible loading against the stability limit.

Parts (b) and (c) — Recovering the two-port and using it

Given.

Find. The constants $A$ and $C$ of the two-port from the reference operating point, and then the sending-end voltage, current and power factor when the load changes to 1400 MVA at 0.85 power factor lagging.

Approach. Convert the three-phase data to the per-phase (line-to-neutral) model in which $B$ is defined, solve the first ABCD equation algebraically for $A$, recover $C$ from the reciprocity constraint, and then run both equations forward at the new load.

1. Part (b) — Put the data on a per-phase basis. Because $B$ is an impedance per phase, the two-port equations relate line-to-neutral voltages and line currents: $$V_R=\frac{700}{\sqrt{3}}=404.145\ \text{kV},\quad V_S=\frac{760.5}{\sqrt{3}}\angle 19.66^{\circ}=439.075\angle 19.66^{\circ}\ \text{kV}$$ The receiving-end current follows from the three-phase apparent power, and its angle is the negative of the load angle because the power factor is lagging: $$I_R=\frac{S_R}{\sqrt{3}\,V_{R,LL}}\angle-\cos^{-1}(0.95) =\frac{1500\times 10^{6}}{\sqrt{3}\,(700\times 10^{3})}\angle-18.195^{\circ} =1{,}237.18\angle -18.195^{\circ}\ \text{A}$$
2. Solve the sending-end voltage equation for the voltage-ratio constant. The product of the known series arm with the known current is $$BI_R=(125.9\angle 87.5^{\circ})(1{,}237.18\angle -18.195^{\circ})=155{,}761\angle 69.305^{\circ}\ \text{V}=55.045+j145.710\ \text{kV}$$ Rearranging $V_S = AV_R + BI_R$ and substituting the rectangular forms $V_S = 413.479+j147.721$ kV gives $$A=\frac{V_S-BI_R}{V_R}=\frac{358.435+j2.011\ \text{kV}}{404.145\ \text{kV}} =\boxed{0.88691\angle 0.3214^{\circ}}$$ A magnitude near 0.89 with an angle of a few tenths of a degree is exactly what a 240-mile line at this voltage should show, which is the first sanity check on the arithmetic.
3. Recover the shunt constant from reciprocity. The line is a symmetric passive two-port, so $AD - BC = 1$ with $D = A$, giving $C = (A^{2}-1)/B$. Squaring the result of the previous step, $$\begin{aligned}A^{2}&=0.786560+j0.008826 \\ A^{2}-1&=0.213622\angle 177.632^{\circ}\end{aligned}$$ $$C=\frac{A^{2}-1}{B}=\frac{0.213622\angle 177.632^{\circ}}{125.9\angle 87.5^{\circ}}=\boxed{1.69676\times 10^{-3}\angle 90.132^{\circ}\ \text{S}}$$ The angle is the free check that the given data are self-consistent: for a line whose shunt path is essentially pure capacitance, $C$ must come out within about a degree of $+90^{\circ}$, and 90.13° does. The magnitude corresponds to a total shunt susceptance of 1.6968×10⁻³ S over 240 miles, or about 7.07 µS per mile, the expected order for a bundled 760-kV circuit.
4. Part (c) — Compute the new receiving-end current. The receiving-end voltage is unchanged, but the load falls to 1400 MVA at a poorer power factor: $$I_R'=\frac{1400\times 10^{6}}{\sqrt{3}\,(700\times 10^{3})} \angle-\cos^{-1}(0.85)=1{,}154.70\angle -31.788^{\circ}\ \text{A}$$
5. Evaluate the sending-end voltage. The two terms of $V_S = AV_R + BI_R'$ are $$\begin{aligned} AV_R&=(0.88691\angle 0.3214^{\circ})(404.145\times 10^{3})=358{,}435+j2{,}011\ \text{V}\\ BI_R'&=(125.9\angle 87.5^{\circ})(1{,}154.70\angle -31.788^{\circ})=145{,}377\angle 55.712^{\circ}\ \text{V}=81{,}899+j120{,}112\ \text{V} \end{aligned}$$ Adding them and converting back to a line-to-line magnitude, $$V_S=440{,}334+j122{,}123\ \text{V}=456{,}955\angle 15.501^{\circ}\ \text{V (line-to-neutral)}$$ $$\boxed{V_{S,LL}=\sqrt{3}\,(456{,}955)=791.47\ \text{kV at }15.501^{\circ}}$$
6. Evaluate the sending-end current. The second equation uses the constant just recovered, and the shunt term is what makes the sending-end current smaller than the receiving-end current here: $$\begin{aligned} CV_R&=(1.69676\times 10^{-3}\angle 90.132^{\circ})(404.145\times 10^{3})=-1.581+j685.737\ \text{A}\\ AI_R'&=(0.88691\angle 0.3214^{\circ})(1{,}154.70\angle -31.788^{\circ})=1{,}024.12\angle -31.467^{\circ}\ \text{A} \end{aligned}$$ $$\boxed{I_S=871.930+j151.143=884.93\angle 9.834^{\circ}\ \text{A}}$$
7. Extract the sending-end power factor and check the energy balance. The power-factor angle is the angle by which the sending-end current lags the sending-end voltage: $$\phi_S=15.501^{\circ}-(9.834^{\circ})=5.667^{\circ} \;\Rightarrow\;\boxed{\text{pf}_S=\cos5.667^{\circ}=0.9951\ \text{lagging}}$$ The line-charging current supplied by the shunt term has cancelled most of the load's reactive demand as seen from the sending end, which is why the power factor there is so much better than the 0.85 at the load. Confirming with a power balance, $$\begin{aligned}P_S&=\sqrt{3}\,V_{S,LL}I_S\cos\phi_S=1{,}207.2\ \text{MW} \\ P_R&=1400(0.85)=1{,}190\ \text{MW}\end{aligned}$$ so the line loss is 17.2 MW, which agrees with $3I^{2}R$ using the resistive part of $B$, $R=125.9\cos 87.5^{\circ}=5.492\ \Omega$, at a mean current near 1,020 A. The loss is 1.44 per cent of the delivered power, a credible figure for 240 miles at this voltage.

Final results.