Question 1 of 7: Bundle-Conductor Line Parameters and the Long-Line ABCD Constants

Problem 1: Bundle-Conductor Line Parameters and the Long-Line ABCD Constants (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.

Given.

Find. The series inductance and the shunt capacitance per metre per phase, first with the earth ignored and then with the earth included as an image plane at depth $h_1$; then the two-port constants $A$, $B$, $C$, $D$ of the distributed-parameter (long-line) model.

Approach. Replace each bundle by an equivalent single conductor — the geometric mean radius $\mathrm{GMR}_b$ for inductance and the capacitive mean radius $\mathrm{CGMR}_b$ for capacitance — combine the three phase positions into one equivalent spacing $\mathrm{GMD}$, then apply the standard logarithmic formulas; for part (c) add the image-charge correction, and for part (d) build $\gamma$ and $Z_c$ from $L$ and $C$ and evaluate the hyperbolic two-port constants.

1. Part (a) — place the eight sub-conductors on their circle. For $N$ sub-conductors spaced equally on a circle of radius $A$, the distance between adjacent sub-conductors is $S = 2A\sin(\pi/N)$, so$$A=\frac{S}{2\sin(\pi/N)}=\frac{0.500}{2\sin(22.5^\circ)}=\frac{0.500}{0.765367}=0.65328\ \text{m}.$$The sub-conductor radius is $r = 3.6/2 = 1.8\ \text{cm} = 0.018\ \text{m}$, and its own inductive mean radius is $r^{\prime}=r\,e^{-1/4}=0.018\times0.778801=0.0140184\ \text{m}$.
2. Form the bundle mean radii. Every sub-conductor sees the other $N-1$ at distances whose product is $N A^{N-1}$ (a standard property of points equally spaced on a circle), so$$\mathrm{GMR}_b=\bigl[N\,r^{\prime}A^{N-1}\bigr]^{1/N},\qquad \mathrm{CGMR}_b=\bigl[N\,r\,A^{N-1}\bigr]^{1/N}.$$With $A^{7}=0.050798\ \text{m}^{7}$,$$\begin{aligned}\mathrm{GMR}_b&=\bigl[8(0.0140184)(0.050798)\bigr]^{1/8}=\bigl[5.6967\times10^{-3}\bigr]^{1/8}=0.52413\ \text{m},\\\mathrm{CGMR}_b&=\bigl[8(0.018)(0.050798)\bigr]^{1/8}=\bigl[7.3149\times10^{-3}\bigr]^{1/8}=0.54076\ \text{m}.\end{aligned}$$The two differ by the fixed factor $e^{0.25/N}=e^{0.03125}=1.03174$, exactly as the bundle theory requires.
3. Combine the three phase positions. The configuration is flat and horizontal with equal spacings $D_1$, so the two outer phases are $2D_1$ apart and the equivalent (transposed) spacing is the geometric mean of the three pair distances:$$\mathrm{GMD}=\sqrt[3]{D_1\cdot D_1\cdot 2D_1}=D_1\sqrt[3]{2}=16(1.259921)=20.1587\ \text{m}.$$Transposition is assumed, which is what makes a single per-phase value meaningful.
4. Inductance per metre per phase. With both radii in the same units,$$\begin{aligned}L&=2\times10^{-7}\ln\!\frac{\mathrm{GMD}}{\mathrm{GMR}_b}=2\times10^{-7}\ln\!\frac{20.1587}{0.52413}\\&=2\times10^{-7}\ln(38.4653)=2\times10^{-7}(3.649830),\end{aligned}$$so that$$\boxed{L=7.2993\times10^{-7}\ \text{H}\,\text{m}^{-1}\ \text{per phase}\;=\;0.72993\ \text{mH}\,\text{km}^{-1}.}$$That is roughly 25 per cent below the 0.97 mH km$^{-1}$ typical of a single-conductor line, which is the whole point of bundling.
5. Part (b) — capacitance with the earth neglected. The capacitive formula is the same logarithm with $\mathrm{CGMR}_b$ in place of $\mathrm{GMR}_b$ and the permittivity in front:$$C=\frac{2\pi\varepsilon_0}{\ln\bigl(\mathrm{GMD}/\mathrm{CGMR}_b\bigr)}=\frac{2\pi(8.854\times10^{-12})}{\ln(20.1587/0.54076)}=\frac{5.56325\times10^{-11}}{3.618444},$$giving$$\boxed{C=1.5375\times10^{-11}\ \text{F}\,\text{m}^{-1}\ \text{per phase}\;=\;1.5375\times10^{-2}\ \mu\text{F}\,\text{km}^{-1}.}$$
6. Part (c) — add the earth as a plane of image charges. A perfectly conducting earth is replaced by mirror charges of opposite sign at depth $h_1$. Each conductor is $H_{ii}=2h_1=42.00\ \text{m}$ from its own image, and conductor $i$ is a distance $H_{ij}=\sqrt{(h_i+h_j)^2+d_{ij}^{\,2}}$ from the image of conductor $j$:$$\begin{aligned}H_{12}=H_{23}&=\sqrt{42^2+16^2}=44.944\ \text{m},\\ H_{13}&=\sqrt{42^2+32^2}=52.802\ \text{m}.\end{aligned}$$The correction subtracts the logarithm of the ratio of the two geometric means,$$C_{\text{earth}}=\frac{2\pi\varepsilon_0}{\ln\dfrac{\mathrm{GMD}}{\mathrm{CGMR}_b}-\ln\dfrac{\sqrt[3]{H_{12}H_{23}H_{13}}}{\sqrt[3]{H_{11}H_{22}H_{33}}}}.$$
7. Evaluate the image correction. The two means are $\sqrt[3]{H_{12}H_{23}H_{13}}=\sqrt[3]{(44.944)^2(52.802)}=47.424\ \text{m}$ and $\sqrt[3]{H_{11}H_{22}H_{33}}=42.000\ \text{m}$, so the correction term is $\ln(47.424/42.000)=\ln(1.12915)=0.121553$ and the denominator falls from 3.618444 to 3.496891:$$\boxed{C_{\text{earth}}=\frac{5.56325\times10^{-11}}{3.496891}=1.5909\times10^{-11}\ \text{F}\,\text{m}^{-1}.}$$The earth raises the capacitance by 3.47 per cent — it always raises it, because the image charges pull extra charge onto the conductor at the same voltage.
8. Part (d) — build the distributed constants. Neglecting the series resistance (none is given), $z=j\omega L$ and $y=j\omega C$ per metre with $\omega=2\pi(60)=376.99\ \text{rad}\,\text{s}^{-1}$. The characteristic impedance and the propagation constant are then$$Z_c=\sqrt{\frac{z}{y}}=\sqrt{\frac{L}{C}}=\sqrt{\frac{7.2993\times10^{-7}}{1.5375\times10^{-11}}}=217.89\ \Omega\ (\text{purely resistive}),$$$$\gamma=\sqrt{zy}=j\omega\sqrt{LC}=j(376.99)(3.35003\times10^{-9})=j1.26293\times10^{-6}\ \text{m}^{-1},$$i.e. $\alpha=0$ and $\beta=1.26293\times10^{-3}\ \text{rad}\,\text{km}^{-1}$ ($0.072361^\circ\ \text{km}^{-1}$). The wave velocity is $v=1/\sqrt{LC}=2.985\times10^{8}\ \text{m}\,\text{s}^{-1}$, 99.6 per cent of the speed of light, which is the arithmetic check that $L$ and $C$ belong to the same geometry.
9. Assemble the long-line two-port. The exact distributed-parameter model gives$$\begin{aligned}A=D&=\cosh\gamma\ell=\cos\beta\ell,\\ B&=Z_c\sinh\gamma\ell=jZ_c\sin\beta\ell,\\ C&=\frac{\sinh\gamma\ell}{Z_c}=j\frac{\sin\beta\ell}{Z_c}.\end{aligned}$$the right-hand forms following because $\gamma$ is purely imaginary for a lossless line. At the representative length $\ell=400\ \text{km}$, $\beta\ell=0.505173\ \text{rad}=28.944^\circ$ and$$\boxed{A=D=0.87509,\qquad B=j105.45\ \Omega,\qquad C=j2.2211\times10^{-3}\ \text{S}.}$$The reciprocity check $AD-BC=A^2-BC=0.765783+0.234217=1.0000$ closes exactly, as it must for any passive symmetric two-port.
10. Report the length sensitivity and the natural loading. Because no length is stated, the table below evaluates the same three formulas at 200, 400 and 600 km. Finally the surge-impedance loading, which is length-independent, is$$\mathrm{SIL}=\frac{V_{LL}^{2}}{Z_c}=\frac{(1100\times10^{3})^{2}}{217.89}=5.55\times10^{9}\ \text{W}=5553\ \text{MW},$$a figure entirely consistent with a real 1100 kV UHV circuit.

Final Results