Question 1 of 7: Line Parameters and the ABCD Two-Port Model

Problem 1: Line Parameters and the ABCD Two-Port 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) — effect of ambient temperature on Z and Y. The two line constants respond to temperature in completely different ways, and the whole of the answer is that Z is temperature-sensitive while Y is essentially not.

The series impedance is $Z = R + j\omega L$ per unit length. The resistance term is the one that moves. Metallic conduction resistivity rises almost linearly with conductor temperature over the working range, and the standard engineering form is the inferred-zero-resistance relation

with $T = 234.5\,{}^{\circ}\text{C}$ for annealed copper, $241\,{}^{\circ}\text{C}$ for hard-drawn copper and $228\,{}^{\circ}\text{C}$ for hard-drawn aluminium and ACSR. For aluminium at a 20 °C reference this is about $0.40\ \%$ per degree, so a 30 °C rise in ambient temperature — which, at constant current, is carried straight through to the conductor because the conductor sits at ambient plus its own $I^2R$ rise — increases $R$ by roughly 12 %. Two second-order effects add to it: the a.c. resistance rises slightly faster than the d.c. value because skin depth $\delta = \sqrt{2\rho/(\omega\mu)}$ grows with $\rho$, and the spiralling of ACSR strands lengthens the current path as the conductor elongates.

The inductance $L$ is fixed by geometry through $L = 2\times 10^{-7}\ln(D_m/D_s)$: the geometric mean distance between phases and the self-GMD of the conductor. Neither changes appreciably with temperature. Thermal elongation increases sag, which lowers the conductor at mid-span and can change the effective spacing slightly, but the logarithm makes the sensitivity negligible. So the series impedance rises, almost entirely through $R$, the $X/R$ ratio falls, and the $I^2R$ loss and voltage drop both increase — which is exactly why a hot day reduces a line's ampacity rating.

The shunt admittance $Y = G + j\omega C$ behaves differently. The capacitance $C = 2\pi\varepsilon_0/\ln(D_m/r)$ is again a pure geometry term, and the permittivity of air is insensitive to temperature; the increased sag lowers the conductors toward ground and raises the capacitance to earth by a fraction of a percent, which is the only real mechanism. The conductance $G$ represents leakage across insulator surfaces and corona loss. It is dominated by humidity, surface contamination and air density rather than by temperature as such, and it is so small on a clean line that it is routinely neglected. There is one genuine temperature link worth stating: air density $\delta_{air} = \dfrac{3.92\,b}{273 + t}$ falls as temperature rises, which lowers the corona-onset gradient in Peek's law and so increases corona loss and $G$ on a heavily stressed line.

Given.

Find. The two remaining generalised constants $B$ and $C$ (part b), then the sending-end current, the sending-end power factor and the transmission efficiency (part c).

Approach. Recover $I_r$ from the stated load, invert the first ABCD equation for $B$, use reciprocity for $C$, then evaluate the second ABCD equation for $I_s$ and compare sending- and receiving-end real power.

1. Part (b) — convert the load into a receiving-end current phasor. The complex power convention $S = V I^{*}$ gives $I_r = \left(S_r/V_r\right)^{*}$. With the power factor angle $\varphi_r = \cos^{-1}0.85 = 31.79^\circ$ and a lagging load the current lags the voltage, so $$I_r=\left(\frac{1.25\angle 31.79^\circ}{1.0\angle 0^\circ}\right)^{*} = 1.2500\angle-31.79^\circ\ \text{pu}.$$
2. Solve the first ABCD equation for B. Rearranging $V_s = A V_r + B I_r$ isolates the only unknown: $$B=\frac{V_s-A V_r}{I_r} =\frac{1.1500\angle9.00^\circ-0.9800\angle0.20^\circ}{1.2500\angle-31.79^\circ}.$$ The numerator is $0.2354\angle48.55^\circ$ pu, so $$\boxed{B=0.1884\angle80.34^\circ\ \text{pu}\ \ (\Omega\ \text{on the chosen base})}$$ The phase angle of about $80^\circ$ is the signature of an overhead line: $B$ is the equivalent series impedance of the nominal-$\pi$ or long-line model, and it is strongly inductive.
3. Use reciprocity to get C. Every passive, bilateral two-port satisfies $AD-BC=1$, and a symmetrical line has $D=A$, which is exactly the constraint the question prints. Hence $$C=\frac{A^{2}-1}{B} =\frac{0.0402\angle170.40^\circ}{0.1884\angle80.34^\circ} \;\Longrightarrow\; \boxed{C=0.2134\angle90.06^\circ\ \text{S}\ \text{(pu)}}$$ Its angle is within a twentieth of a degree of $+90^\circ$, i.e. $C \approx j0.2134$, which is precisely what the model demands: $C$ is the total shunt susceptance of the line, and on a lossless-shunt line it is purely capacitive. That the arithmetic delivers it without being told to is a strong check that $A$, $V_s$ and the load data are mutually consistent.
4. Part (c) — sending-end current from the second ABCD equation. With both constants known, $$I_s=C V_r + A I_r =0.2134\angle90.06^\circ\times 1.0 +0.9800\angle0.20^\circ\times1.2500\angle-31.79^\circ$$ Adding the two contributions in rectangular form, $(-0.0002+0.2134j) + (1.0435-0.6417j)$, gives $$\boxed{I_s=1.1278\angle-22.32^\circ\ \text{pu}}$$ Note that the shunt term $C V_r$ is almost purely imaginary and therefore reduces the net lagging component of $I_s$: the line's own charging current partially compensates the inductive load, which is why $|I_s|$ is only 1.1278 pu although $A|I_r| = 1.2250$ pu.
5. Sending-end power factor. The power-factor angle is the angle by which $I_s$ lags $V_s$: $$\varphi_s=\angle V_s-\angle I_s =9^\circ-\left(-22.32^\circ\right) =31.32^\circ$$ $$\boxed{\cos\varphi_s=0.8543\ \text{lagging}}$$ The sending end is very slightly better than the receiving end (0.854 against 0.850) for the reason just given — the distributed capacitance supplies part of the reactive demand.
6. Real power at each end. Sending-end complex power is $S_s = V_s I_s^{*} = 1.2970\angle31.32^\circ$ pu, so $P_s = 1.1080$ pu and $Q_s = 0.6742$ pu. The receiving-end real power follows directly from the given load, $P_r = S_r\cos\varphi_r = 1.25\times 0.85 = 1.0625$ pu.
7. Transmission efficiency. Efficiency is the ratio of useful delivered power to power dispatched into the line: $$\eta=\frac{P_r}{P_s}\times 100\,\% =\frac{1.0625}{1.1080}\times 100\,\% \;\Longrightarrow\;\boxed{\eta=95.90\,\%}$$ The loss, $P_s-P_r = 0.0455$ pu, is about 4 % of the dispatched power — a realistic figure for a long line carrying near its surge-impedance loading.

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