22-Elec-B7 Power Systems Engineering · December 2016
Question 1 of 7: Bundle Conductors, Line Capacitance and the ABCD Constants
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Exams, December 2016 — 07-Elec-B7 Power Systems Engineering. Open-book, 3 hours duration. Seven problems are printed; any five constitute a complete paper, only the first five appearing in the answer book are marked, and all questions are of equal value (20 points each). All seven are solved in full below, because this set is a study resource rather than a timed sitting.
Reference texts. This exam code follows M. E. El-Hawary, Electrical Power Systems: Design and Analysis (IEEE Press) — the notation is his throughout: the “cantilever” transformer equivalent of Problem 3, the two-reaction salient-pole construction of Problem 2, the ABCD two-port of Problem 1 and the sequence-network fault reduction of Problem 6. Corroborating references: J. D. Glover, M. S. Sarma and T. J. Overbye, Power System Analysis and Design, 6th ed. (ch. 3 transformers, ch. 4–5 transmission lines, ch. 6 power flow, ch. 7–9 faults and symmetrical components, ch. 11 transient stability); J. J. Grainger and W. D. Stevenson, Power System Analysis; and S. J. Chapman, Electric Machinery Fundamentals, 5th ed., for the machine and transformer chapters. Canadian practice references where grounding and protection are discussed: CSA C22.1 Canadian Electrical Code, Part I, CSA C22.2 and the IEEE C62 / IEEE 142 (Green Book) series on system grounding.
Check: four judgement calls carried through the solutions, all recorded here rather than hedged in the answers.
(i) Problem 1 does not state a frequency. Parts (c) and (d) need one, and the paper is a Canadian national exam, so 60 Hz is used throughout — consistent with the 60 Hz stated in Problems 3 and 7. At 50 Hz the capacitance and surge impedance are unchanged (they do not involve frequency) and only the propagation constant of part (d) rescales by 5/6.
(ii) Problem 2 prints “neglecting armature reaction”. Armature reaction is precisely what the direct- and quadrature-axis reactances represent, so the phrase is read as its standard counterpart, neglecting armature resistance — the only reading under which the two given reactances can be used at all.
(iii) Problem 6 prints a part label “c-” with no question text. The five parts that do carry text (a, b, d, e, f) are marked 4 points each and total the 20 points the paper allots to every problem, so nothing is missing from the assessment; the sequence networks that the empty label most plausibly asked for are drawn anyway.
(iv) Problem 5 gives no reactance between buses 2 and 3. Figure (3) confirms the topology: bus 3 hangs radially off bus 1 through the j0.12 pu line, and the only other tie is line 1–2.
Problem 1: Bundle Conductors, Line Capacitance and the ABCD Constants (20 points)
Find. A description of bundling and its electrical consequences; a proof that the capacitive and inductive mean radii of an $N$-conductor bundle differ by the fixed factor $e^{0.25/N}$; the shunt capacitance per metre and the characteristic impedance of the 16-bundle line; and the two-port constants $A$ and $B$ of the 250 km line.
Figure 1 — Cross-section of one phase of the 16-sub-conductor bundle. Every sub-conductor sits on a circle of radius R, so the geometric mean of the N – 1 distances from one sub-conductor to the others reduces to R (N)1/N − 1, and the only difference between the inductive and capacitive mean radii is the self term.
Part (a) — what a bundle is, and what it does to the line. A bundle conductor line carries each phase not on one large conductor but on two, three, four or (as here) sixteen smaller sub-conductors, held a fixed distance apart by spacer-dampers every 50–80 m along the span and connected in parallel at every tower. Electrically the bundle behaves as one conductor whose equivalent radius is set by the bundle geometry, not by the metal, and that equivalent radius is very much larger than any physically buildable single conductor.
Three consequences follow, and they are the reason every transmission line above roughly 230 kV in Canada is bundled. First, the series inductance falls: $L=2\times10^{-7}\ln(\text{GMD}/\text{GMR})$, and enlarging the GMR shrinks the logarithm, typically taking 20–30 per cent off the reactance of a twin-bundle line and more for larger bundles. Second, the shunt capacitance rises for the same reason, since $C=2\pi\varepsilon_0/\ln(\text{GMD}/\text{CGMR})$. Together these lower the surge impedance $Z_c=\sqrt{L/C}$ and raise the surge-impedance loading $\text{SIL}=V_{LL}^{2}/Z_c$, which is the single most useful measure of how much power a line of given voltage can carry — bundling therefore buys transmission capacity directly. Third, and the reason bundling was invented, the electric field at the conductor surface is shared among the sub-conductors, so the surface gradient drops well below the Peek disruptive value; corona loss, radio interference and audible noise collapse with it, and at 2000 kV no unbundled conductor could avoid continuous corona at all.
The price is mechanical and operational rather than electrical. A bundle presents a much larger wind and ice profile, so towers and foundations grow; spacer-dampers are needed to stop sub-conductor clashing under short-circuit pinch forces and aeolian vibration; and the larger charging current makes light-load Ferranti voltage rise worse, which is why long UHV lines are fitted with shunt reactors. Bundling also raises the thermal rating by putting more aluminium in the air with a better surface-to-volume ratio for cooling.
Part (b) — write the two mean radii of the bundle, which differ only in the self term. For a bundle of $N$ identical sub-conductors of radius $r$ equally spaced on a circle of radius $R$, both mean radii are the geometric mean of the $N^{2}$ distances from each sub-conductor to every sub-conductor (itself included). The product of the $N-1$ distances from one sub-conductor to the others is $N R^{\,N-1}$, a standard identity for points equally spaced on a circle. The self distance is the only thing that changes: for inductance it is the self-GMR of a solid round wire, $r^{\prime}=r\,e^{-1/4}$, whereas for capacitance it is the actual radius $r$, because charge resides on the surface and there is no internal flux linkage to account for. Hence $$\text{GMR}=\left[\,N\,r^{\prime}R^{\,N-1}\right]^{1/N},\qquad \text{CGMR}=\left[\,N\,r\,R^{\,N-1}\right]^{1/N}$$
Divide the two — everything geometric cancels. The factor $N R^{\,N-1}$ is common to both, so the ratio depends only on the ratio of the self terms, raised to the $1/N$ power: $$\frac{\text{CGMR}}{\text{GMR}}=\left(\frac{r}{r^{\prime}}\right)^{1/N}=\left(\frac{r}{r\,e^{-1/4}}\right)^{1/N}=\left(e^{1/4}\right)^{1/N}\;\Longrightarrow\;\boxed{\ \frac{\text{CGMR}}{\text{GMR}}=e^{0.25/N}\ }$$ The result is independent of $R$, of $r$ and of the phase spacing, which is why it can be applied to any bundle. Note that it is greater than unity: the capacitive radius always exceeds the inductive one, and the gap closes as $N$ grows — for $N=16$ the two differ by only 1.57 per cent.
Part (c) — recover the inductive logarithm from the given inductance. The per-phase inductance of a transposed line is $L=2\times10^{-7}\ln\!\left(\text{GMD}/\text{GMR}\right)\ \text{H}\,\text{m}^{-1}$, so the logarithm is read straight off the datum: $$\ln\!\frac{\text{GMD}}{\text{GMR}}=\frac{L}{2\times10^{-7}}=\frac{0.744\times10^{-6}}{2\times10^{-7}}=3.72$$ Neither GMD nor GMR is needed separately, which is the whole point of framing part (c) after part (b).
Convert the logarithm to its capacitive counterpart using part (b). Since $\text{CGMR}=\text{GMR}\,e^{0.25/N}$, taking logarithms gives $\ln(\text{GMD}/\text{CGMR})=\ln(\text{GMD}/\text{GMR})-0.25/N$. With $N=16$, $$\ln\!\frac{\text{GMD}}{\text{CGMR}}=3.72-\frac{0.25}{16}=3.72-0.015625=3.704375$$
Evaluate the shunt capacitance per metre. For a transposed three-phase line the line-to-neutral capacitance is $C=2\pi\varepsilon_0/\ln(\text{GMD}/\text{CGMR})$, so $$C=\frac{2\pi\left(8.854\times10^{-12}\right)}{3.704375}=\frac{5.5633\times10^{-11}}{3.704375}\;\Longrightarrow\;\boxed{\ C=1.5018\times10^{-11}\ \text{F}\,\text{m}^{-1}\ }$$ that is $0.01502\ \mu\text{F}\,\text{km}^{-1}$, roughly 60 per cent above the $0.0090$–$0.0095\ \mu\text{F}\,\text{km}^{-1}$ of an ordinary single-conductor line, exactly the enlargement the 16-bundle is there to produce.
Take the characteristic (surge) impedance of the lossless line. With the series resistance and the shunt conductance neglected, $Z_c=\sqrt{z/y}=\sqrt{j\omega L/(j\omega C)}=\sqrt{L/C}$, which is purely real and independent of frequency: $$Z_c=\sqrt{\frac{0.744\times10^{-6}}{1.5018\times10^{-11}}}=\sqrt{4.9542\times10^{4}}\;\Longrightarrow\;\boxed{\ Z_c=222.58\ \Omega\ }$$ The corresponding surge-impedance loading is $\text{SIL}=V_{LL}^{2}/Z_c=(2000\times10^{3})^{2}/222.58=1.797\times10^{10}\ \text{W}$, about $17\,970$ MW. The figure is enormous because the question posits an experimental 2000 kV line; it is the natural loading such a line would carry, not an error.
Part (d) — form the propagation constant of the lossless 250 km line. For $R=G=0$, $\gamma=\sqrt{zy}=j\omega\sqrt{LC}=j\beta$, so the line is a pure phase shifter. With $\omega=2\pi(60)=376.99\ \text{rad}\,\text{s}^{-1}$, $$\beta=\omega\sqrt{LC}=376.99\sqrt{\left(0.744\times10^{-6}\right)\left(1.5018\times10^{-11}\right)}=376.99\left(3.3426\times10^{-9}\right)=1.2601\times10^{-6}\ \text{rad}\,\text{m}^{-1}$$ Over $\ell=250\ \text{km}=2.5\times10^{5}\ \text{m}$ the electrical length is $\beta\ell=0.31504\ \text{rad}=18.050^{\circ}$, comfortably inside the range where the exact model still matters but is not yet extreme.
Read the two-port constants off the exact long-line solution. For a lossless line $A=D=\cosh(\gamma\ell)=\cos(\beta\ell)$ and $B=Z_c\sinh(\gamma\ell)=jZ_c\sin(\beta\ell)$, so $$A=\cos(18.050^{\circ})=0.9508\angle 0^{\circ}\ \text{(pu)},\qquad B=j\,(222.58)\sin(18.050^{\circ})=j\,68.97\ \Omega$$ and therefore $$\boxed{\ A=D=0.9508\angle 0^{\circ},\qquad B=68.97\angle 90^{\circ}\ \Omega\ }$$ The remaining constant follows for free, $C=j\sin(\beta\ell)/Z_c=j1.3921\times10^{-3}\ \text{S}$, and the reciprocity identity closes the check: $AD-BC=\cos^{2}(\beta\ell)+\sin^{2}(\beta\ell)=1$ exactly, as it must for any passive bilateral two-port.
