Transmission-line parameters and the two-port model: 22-Elec-B7 Power Systems Engineering, December 2019, Q2

Question 2: Transmission-line parameters and the two-port model (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.

Given.

Quantity	Value
System	345 kV, 60 Hz, double circuit; circuit 2 to be ignored
Sub-conductor	ACSR "Pigeon", diameter 12.75 mm, so $r=6.375$ mm
Bundle	2 sub-conductors, spacing $d=0.5$ m
Phase positions (from the tower drawing)	$A_1$ at 5 m from the centreline, 28 m above ground; $B_1$ at 5 m, 21.6 m; $C_1$ at 12 m, 21.6 m
Resistance	0.15 Ω/km per sub-conductor
Capacitance	0.038 µF/km per phase
Length in part (b)	50 km
Two-port constants, part (c)	$A=D=0.95\angle 0^\circ$, $B=33.2\angle 75^\circ\ \Omega$, $C=0.00027\angle 89^\circ$ S
Receiving-end condition, part (c)	345 kV rated, 200 MVA at unity power factor

Find. (a) the per-phase series inductance of the line; (b) the numerical per-phase equivalent circuit of a 50 km section, drawn; and (c) the sending-end voltage and current for the stated receiving-end load.

Figure Q2a — conductor positions of circuit 1 taken from the tower drawing (horizontal distances measured from the tower centreline; heights above ground). Each phase is a two-conductor bundle spaced 0.5 m.

Approach. Reduce the bundle to a single equivalent conductor through its geometric mean radius, reduce the unequal phase spacings to a single geometric mean distance, and apply the standard transposed-line inductance formula. Part (b) then only needs the series and shunt values of a 50 km section; part (c) is a direct evaluation of the two-port equations.

Part (a) — find the three phase-to-phase spacings from the tower geometry. Working in the plane of the tower cross-section with the centreline as the origin, $A_1=(5,\,28)$, $B_1=(5,\,21.6)$ and $C_1=(12,\,21.6)$ metres. Then $$D_{AB}=|28-21.6|=6.4\ \text{m},\qquad D_{BC}=|12-5|=7\ \text{m}$$ $$D_{AC}=\sqrt{7^{2}+6.4^{2}}=\sqrt{89.96}=9.4847\ \text{m}$$ The statement that "$A_1$ is directly above $B_1$" is what fixes the shared 5 m abscissa, and it is consistent with the 5 m dimension printed under phase $B_1$.
Reduce the unequal spacings to one geometric mean distance. For a transposed three-phase line the mutual coupling enters only through the cube root of the product of the three separations: $$D_{eq}=\sqrt[3]{D_{AB}\,D_{BC}\,D_{AC}}=\sqrt[3]{6.4\times 7\times 9.4847} =\sqrt[3]{424.92}=7.518\ \text{m}$$
Reduce the two-conductor bundle to one equivalent conductor. A solid round conductor of radius $r$ has the self geometric mean radius $r'=r\,e^{-1/4}=0.7788\,r$, which accounts for the internal flux linkage. With $r=6.375$ mm, $$r'=0.006375\times e^{-0.25}=4.9649\times 10^{-3}\ \text{m}$$ and for two sub-conductors spaced $d$ apart the bundle geometric mean radius is $$D_{sb}=\sqrt{r'\,d}=\sqrt{4.9649\times 10^{-3}\times 0.5}=0.049824\ \text{m}$$ Bundling has multiplied the effective radius by a factor of ten, which is exactly why 345 kV lines are bundled.
Apply the transposed-line inductance formula. With $\mu_0/2\pi=2\times 10^{-7}$ H/m, $$L=\frac{\mu_0}{2\pi}\ln\!\frac{D_{eq}}{D_{sb}} =2\times 10^{-7}\times\ln\!\frac{7.518}{0.049824} =2\times 10^{-7}\times\ln(150.86)$$ Since $\ln(150.86)=5.0163$, $$\boxed{L=1.0033\times 10^{-6}\ \text{H/m}=1.0033\ \text{mH/km per phase}}$$ Over the 50 km length of part (b) this is $L=50.166$ mH per phase, and the corresponding series reactance at 60 Hz is $X=2\pi(60)(0.050166)=18.91\ \Omega$.
Part (b) — assemble the series and shunt values for 50 km. The two sub-conductors of a bundle are in parallel, so the phase resistance is half the sub-conductor value: $$R=\frac{0.15}{2}\times 50=3.75\ \Omega,\qquad X=18.91\ \Omega,\qquad Z=3.75+j18.91\ \Omega\;(19.28\angle 78.78^\circ\ \Omega)$$ $$C=0.038\times 50=1.90\ \mu\text{F},\qquad Y=j2\pi(60)(1.90\times 10^{-6})=j7.163\times 10^{-4}\ \text{S}$$ Half of the shunt admittance is placed at each end, giving $Y/2=j3.581\times 10^{-4}$ S.
Choose and justify the model. At 50 km the line is short by the usual classification (below about 80 km), and the nominal-π constant works out to $$A=1+\frac{ZY}{2}=1+(3.75+j18.91)(j3.581\times 10^{-4})=0.9932+j0.0013$$ i.e. a magnitude of $0.9932$. The charging admittance therefore alters the terminal relations by about $0.7\ \%$. The nominal-π circuit shown below is the appropriate model because the question supplies a capacitance; dropping the two shunt arms reduces it to the short-line series model, and the boxed error above is the price of doing so.

Figure Q2b — nominal-π per-phase equivalent circuit of a 50 km section. Neglecting the two shunt arms leaves the short-line series model, an approximation good to about 0.7 % at this length.

Part (c) — set up the receiving-end quantities per phase. "Rated receiving-end voltage" means 345 kV line-to-line, and the two-port constants are per-phase quantities, so $$V_R=\frac{345\,000}{\sqrt{3}}=199\,186\angle 0^\circ\ \text{V}$$ At unity power factor the current is in phase with the voltage: $$I_R=\frac{S_R}{\sqrt{3}\,V_{R,LL}}=\frac{200\times 10^{6}}{\sqrt{3}\times 345\,000} =334.70\angle 0^\circ\ \text{A}$$
Evaluate the sending-end voltage. The two-port relation $V_S=A\,V_R+B\,I_R$ gives $$V_S=0.95\angle 0^\circ\times 199\,186+33.2\angle 75^\circ\times 334.70$$ $$=189\,227+11\,112\angle 75^\circ=189\,227+(2\,876+j10\,734)=192\,103+j10\,733\ \text{V}$$ $$\boxed{V_S=192.40\angle 3.198^\circ\ \text{kV per phase}=333.25\ \text{kV line-to-line}}$$
Evaluate the sending-end current. Similarly $I_S=C\,V_R+D\,I_R$: $$I_S=0.00027\angle 89^\circ\times 199\,186+0.95\times 334.70 =53.78\angle 89^\circ+317.96$$ $$=(0.94+j53.77)+317.96=318.90+j53.77\ \text{A}$$ $$\boxed{I_S=323.40\angle 9.571^\circ\ \text{A}}$$ The sending-end power-factor angle is $3.198^\circ-9.571^\circ=-6.37^\circ$, i.e. a power factor of 0.9938 leading.
Interpret the result before believing it. The sending-end voltage is lower than the receiving-end voltage, which looks wrong until the loading is compared with the surge impedance. From the given constants, $Z_c\approx\sqrt{B/C}=\sqrt{33.2\angle 75^\circ/0.00027\angle 89^\circ}=350.7\angle-7^\circ\ \Omega$, so the surge impedance loading is $$\mathrm{SIL}=\frac{V_{LL}^{2}}{|Z_c|}=\frac{(345\times 10^{3})^{2}}{350.7}=339.4\ \text{MW}$$ The line is carrying 200 MVA, only $59\ \%$ of its natural load, so the line's own charging current exceeds what the load absorbs and the Ferranti effect raises the far end. A leading sending-end power factor is the companion symptom of the same thing.

Check — the printed two-port constants are not self-consistent

A passive, reciprocal two-port must satisfy $AD-BC=1$. With the values printed on the paper, $AD-BC=0.95^{2}-(33.2\angle 75^\circ)(0.00027\angle 89^\circ)=0.9111\angle-0.155^\circ$, which is $9\ \%$ short of unity: the given $C$ is roughly an order of magnitude too small for the stated $A$ and $B$ (reciprocity would require $C=(A^{2}-1)/B=2.94\times 10^{-3}\angle 105^\circ$ S). The visible consequence is a power-balance failure — the sending end computes to $P_S=3|V_S||I_S|\cos(6.37^\circ)=185.5$ MW against a receiving-end 200 MW, which no passive line can do. The answers above are the exact evaluation of the constants as printed, which is what the question asks for. Note also that $V_S$ depends only on $A$ and $B$ and is therefore unaffected by the suspect $C$; only $I_S$ inherits the inconsistency, and its shunt term contributes just 53.8 A of the 323.4 A total.

Check — two readings on the tower drawing

The tower is dimensioned twice: 6.4 m / 1.8 m / 6.4 m across the topmost arm (which locates the two overhead shield wires, not the phases) and 5 m / 12 m along the bottom (which locate $B_1$ and $C_1$ from the centreline). The 5 m and 12 m figures are the ones used here, because they are the dimensions attached to the conductor drop-lines and because they are consistent with "$A_1$ directly above $B_1$". If instead the phases were placed at 8.2 m ($A_1$, $B_1$) and 6.4 m ($C_1$) as the topmost chain of dimensions might suggest, the horizontal separation $D_{BC}$ would collapse to 1.8 m, $D_{eq}$ would fall to 4.247 m and $L$ to 0.8891 mH/km, about $11\ \%$ lower. Also note that the formula used assumes the line is transposed; the drawing shows a fixed, unsymmetrical arrangement, so the answer is the per-phase average that transposition would produce.

Quantity	Result
(a) Phase spacings	$D_{AB}$ = 6.4 m, $D_{BC}$ = 7 m, $D_{AC}$ = 9.485 m
(a) Geometric mean distance	$D_{eq}$ = 7.518 m
(a) Bundle GMR	$D_{sb}$ = 0.049824 m
(a) Per-phase inductance	1.0033 µH/m = 1.0033 mH/km
(b) Series impedance, 50 km	3.75 + j18.91 Ω (19.28 ∠ 78.78° Ω)
(b) Shunt admittance, 50 km	j7.163 × 10⁻⁴ S, i.e. j3.581 × 10⁻⁴ S at each end
(b) Model	nominal-π (short-line series model within 0.7 %)
(c) Receiving-end current	334.70 ∠ 0° A
(c) Sending-end voltage	192.40 ∠ 3.198° kV per phase = 333.25 kV line-to-line
(c) Sending-end current	323.40 ∠ 9.571° A, pf 0.9938 leading

Topic. Series inductance of a bundled, unequally spaced overhead line, the lumped-parameter classification of line length, and the ABCD two-port representation.

Key relations. $L=2\times 10^{-7}\ln(D_{eq}/D_{sb})$ H/m with $D_{eq}=\sqrt[3]{D_{AB}D_{BC}D_{AC}}$; $r'=re^{-1/4}$ for a solid round conductor; $D_{sb}=\sqrt{r'd}$ for a two-conductor bundle; nominal-π constants $A=D=1+ZY/2$, $B=Z$, $C=Y(1+ZY/4)$; $V_S=AV_R+BI_R$, $I_S=CV_R+DI_R$; $Z_c=\sqrt{B/C}$ and $\mathrm{SIL}=V_{LL}^{2}/|Z_c|$.

Why this works. Inductance depends on flux linkage, and flux linkage depends on distance only logarithmically, so every geometric effect collapses into a ratio of two mean distances: one describing how far a phase is from its neighbours and one describing how far the current within a phase is spread out. Transposition is what allows the three unequal spacings to be replaced by their geometric mean, because over a full transposition cycle each conductor occupies each position for a third of the length. The exponential factor $e^{-1/4}$ in $r'$ is nothing more than the internal inductance $\mu_0/8\pi$ of a solid conductor rewritten as an equivalent hollow conductor of slightly smaller radius.

Common pitfall. Using the conductor diameter instead of the radius in $r'$ doubles the answer's argument and is the single most common slip; so is forgetting the $e^{-1/4}$ correction entirely, which would be right only for a hollow conductor. In part (b), halving the resistance for the two-conductor bundle is easy to miss — the bundle halves $R$ but only reduces $X$, it does not halve it. In part (c), remember that ABCD constants are per-phase: feeding 345 kV rather than $345/\sqrt{3}$ kV into $V_S=AV_R+BI_R$ inflates the answer by $\sqrt{3}$.

Source. Glover, Sarma & Overbye, Power System Analysis and Design, 6th ed., §4.4–4.7 (inductance, bundling, GMR/GMD) and §5.1–5.4 (short, medium and long line models, ABCD constants, surge impedance loading); El-Hawary, Ch. 4.

Question 2 of 6: Transmission-line parameters and the two-port model

Question 2: Transmission-line parameters and the two-port model (20 marks)