Question 1 of 7: Bundle Conductors and ABCD Performance of an 1800 kV Line

Question 1: Bundle Conductors and ABCD Performance of an 1800 kV Line (20 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) — what a bundle conductor line is. A bundle conductor line replaces each single phase conductor with two, three, four or more sub-conductors held a fixed distance apart — typically 400 to 500 mm — by spacer dampers every few tens of metres along the span. Electrically the bundle behaves as one conductor whose effective radius is far larger than that of any of its members, because the current shares among sub-conductors that are separated by many conductor diameters. The quantity that captures this is the bundle geometric mean radius, which for an N-conductor bundle on a circle of radius R is $\mathrm{GMR}_b = \left[N\,r'\,R^{N-1}\right]^{1/N}$, with $r' = r e^{-1/4}$ for inductance and $r$ itself for capacitance.

The consequences follow from that one geometric fact. Series inductance falls, because $L = 2\times10^{-7}\ln(\mathrm{GMD}/\mathrm{GMR}_b)$ and the denominator has grown; a four-conductor bundle typically cuts reactance by about 30 % against a single conductor of the same total aluminium. Shunt capacitance rises for the same reason, since $C = 2\pi\varepsilon_0/\ln(\mathrm{GMD}/\mathrm{CGMR})$. Taken together these move the surge impedance $Z_c = \sqrt{L/C}$ down and the surge impedance loading $\mathrm{SIL} = V^2/Z_c$ up, so a bundled line carries substantially more power at the same voltage and the same angular separation. That is the reason no line above about 300 kV is built unbundled.

The second motivation is surface gradient. Corona onset is governed by the electric field at the conductor surface, and spreading the charge over several sub-conductors reduces the peak gradient by roughly the same factor by which the effective radius grows. Bundling therefore suppresses corona loss, radio interference and audible noise — the constraints that actually set conductor size at extra-high voltage, well before thermal rating does. The costs are mechanical: heavier towers, larger wind and ice loading, spacer maintenance, and a greater short-circuit force between sub-conductors during a fault, which is what sets the spacer interval.

Given. A symmetric, reciprocal three-phase line with $A = D = 0.95\angle 0^\circ$ and $B = j70\ \Omega$. Nominal system voltage 1800 kV; the line delivers $S_R = 2000$ MVA at $V_R = 1750$ kV (line-to-line) at unity power factor.

Find. The shunt constant $C$, then the sending-end voltage and the sending-end current at that receiving-end load.

Approach. Recover $C$ from the reciprocity identity $AD - BC = 1$, then run the two-port forward once with the receiving-end phasors as the input.

1. Part (b) — recover C from reciprocity. Every passive, bilateral two-port satisfies $AD - BC = 1$, and a line that is geometrically symmetric end-to-end has $D = A$. Hence $$C = \frac{A^2 - 1}{B} = \frac{(0.95)^2 - 1}{j70} = \frac{-0.0975}{j70}$$ Dividing by $j$ multiplies by $-j$, so the negative sign cancels and $$\boxed{C = j1.3929\times10^{-3}\ \text{S} = 1.3929\times10^{-3}\angle 90.00^\circ\ \text{S}}$$ The free check on this class of problem is the angle: on a real line $C$ is very nearly a pure shunt susceptance and must land within about a degree of $+90^\circ$. It lands on exactly $90^\circ$ here, because the printed $A$ is purely real and $B$ purely imaginary — the paper has posited an ideal lossless line.
2. Reduce the receiving-end data to phasors. Take $V_R$ as the reference. Per phase, $$V_R = \frac{1750}{\sqrt{3}} = 1010.363\ \text{kV}\angle 0^\circ$$ and at unity power factor the current is in phase with it: $$I_R = \frac{S_R}{\sqrt{3}\,V_{LL}} = \frac{2000\times10^6}{\sqrt{3}\times1750\times10^3} = 659.83\ \text{A}\angle 0^\circ$$
3. Part (c) — the sending-end voltage. The first two-port row is $V_S = A V_R + B I_R$. Both terms are easy here because $I_R$ is real: $$V_S = 0.95(1010.363\times10^3) + (j70)(659.83) = 959.845\times10^3 + j46.188\times10^3\ \text{V}$$ $$\boxed{V_S = 960.955\angle 2.755^\circ\ \text{kV per phase} = 1664.42\ \text{kV line-to-line}}$$
4. Read the result before believing it. The sending-end voltage has come out below the receiving-end voltage, 1664 kV against 1750 kV, and that is physically correct rather than a sign slip. The surge impedance is $Z_c = \sqrt{B/C} = \sqrt{j70/(j1.3929\times10^{-3})} = 224.18\ \Omega$, so the surge impedance loading at 1800 kV is $\mathrm{SIL} = (1800)^2/224.18 = 14\,453$ MW (13 661 MW referred to the 1750 kV operating voltage). The line is delivering 2000 MVA, about 14 % of natural loading. Far below SIL the line's own charging current outruns the load current, the far end floats up, and the sending end must sit lower — the Ferranti effect in its partial-load form.
5. Part (d) — the sending-end current. The second two-port row is $I_S = C V_R + D I_R$ with $D = A$: $$I_S = (j1.3929\times10^{-3})(1010.363\times10^3) + 0.95(659.83) = 626.84 + j1407.32\ \text{A}$$ $$\boxed{I_S = 1540.58\angle 65.99^\circ\ \text{A}}$$ The shunt term contributes 1407 A of pure charging current against a load-side term of only 627 A, which is the same 14 %-of-SIL story read off the current instead of the voltage.
6. Close with the sending-end power balance. The angle between $V_S$ and $I_S$ is $65.99^\circ - 2.755^\circ = 63.24^\circ$ with the current leading, so $$\mathrm{pf}_S = \cos 63.24^\circ = 0.450\ \text{leading}, \qquad S_S = 3V_S I_S^{*} = 2000.0 - j3965.5\ \text{MVA}$$ The real part reproduces the 2000 MW delivered at the far end to the last digit, which is the correct closure for a line whose $A$ is real and whose $B$ and $C$ are purely reactive: it is lossless, so the transmission efficiency is 100 % by construction. The sending end absorbs 3965 MVAr of reactive power — the charging the lightly loaded line is generating and pushing back into the source.