Question 1 of 7: Power Factor of System Equipment; Bundled-Conductor Line Parameters and the Long-Line Model

Question 1: Power Factor of System Equipment; Bundled-Conductor Line Parameters and the Long-Line 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) — how the power factor of transformers and induction motors varies with loading.

Both machines are shunt-excited inductive devices, and that single fact governs the whole shape of the answer. The current drawn by either one separates into a magnetising component, which is fixed by the applied voltage and frequency and is almost independent of the load, and a load component, which is very nearly in phase with the voltage and grows in proportion to the useful output. The power factor is simply the ratio of the in-phase part to the total, so it must start low and rise as the load component overtakes the fixed magnetising component.

At no load a distribution transformer draws only its exciting current, typically one to three per cent of rated current and lagging the voltage by close to eighty degrees, so the no-load power factor is of the order of 0.1 to 0.2. An induction motor is worse because its magnetising current must cross an air gap: the no-load current is typically 25 to 40 per cent of rated current at a power factor of 0.05 to 0.15. As load is applied the in-phase component climbs steeply while the magnetising component barely moves, and the power factor rises rapidly through the lower half of the loading range. It reaches a maximum somewhere between about three quarters of rated load and rated load — roughly 0.80 to 0.90 for a typical squirrel-cage induction motor and, for a transformer, essentially the power factor of the load it is feeding, since the exciting current has by then become a small perturbation.

Beyond that maximum the power factor falls again, and for a different reason. The series leakage reactance now carries a large current, and the reactive power absorbed in the leakage path grows with the square of the current. In an induction motor the slip also increases, which raises the rotor leakage reactance seen at the terminals relative to the rotor resistance, so the machine becomes more inductive as it approaches breakdown torque. The practical consequence for a power system is that lightly loaded plant is the dominant source of poor system power factor: an oversized motor running at a quarter of its rating is not merely inefficient, it is a large source of reactive demand. This is why utilities in Canada levy power-factor or kVA-demand charges, why correct sizing matters, and why shunt capacitor banks are switched at the load rather than at the substation whenever the reactive current would otherwise be carried through the distribution feeder.

Given. A 250 km, completely transposed, 60 Hz three-phase line with its phases in a flat horizontal arrangement and a four-conductor bundle per phase, as tabulated below.

Find. The total series inductance and the total shunt capacitance of the line, and then the generalised constants $A$ and $B$ of the exact distributed-parameter (long-line) two-port model.

Approach. Compute the geometric mean distance between phases, replace each bundle by an equivalent single conductor through its own geometric mean radius (the GMR-based radius for inductance and the physical-radius-based one for capacitance), obtain $L$ and $C$ per metre, scale to the whole line, then form $Z$ and $Y$ and evaluate $A=\cosh\gamma\ell$ and $B=Z_c\sinh\gamma\ell$.

1. Part (b) — geometric mean distance of the flat configuration. For a completely transposed line the three inter-phase distances enter only through their geometric mean. With adjacent phases 10 m apart the outer pair is 20 m apart, so $$D_{eq}=\sqrt[3]{D_{ab}D_{bc}D_{ca}}=\sqrt[3]{(10)(10)(20)}=\sqrt[3]{2000}=12.5992\ \text{m}$$ Transposition is what licenses this single equivalent spacing; without it the three phases would have unequal inductances.
2. Equivalent radius of the four-conductor bundle for inductance. Treat the bundle as one composite conductor and take the geometric mean of all distances from one sub-conductor to the group, including its own GMR. For four sub-conductors at the corners of a square of side $S$ the distances are $D_s$, $S$, $S$ and $S\sqrt{2}$, so $$D_{SL}=\sqrt[4]{D_s\,S\,S\,S\sqrt{2}}=\sqrt[4]{\sqrt{2}}\;\sqrt[4]{D_s S^{3}}=1.09051\sqrt[4]{(0.0115)(0.4)^{3}}$$ The fourth root of $7.36\times10^{-4}$ is $0.164710$ m, and multiplying by the bundle factor gives $$\boxed{D_{SL}=0.179617\ \text{m}}$$ The bundle behaves as a conductor roughly eighteen centimetres in radius, against a sub-conductor GMR of barely one centimetre — this is the whole point of bundling.
3. Series inductance per phase. The inductance of one phase of a transposed three-phase line is $$L=2\times10^{-7}\ln\!\left(\frac{D_{eq}}{D_{SL}}\right)\ \text{H/m}=2\times10^{-7}\ln\!\left(\frac{12.5992}{0.179617}\right)=2\times10^{-7}\ln(70.1447)$$ The logarithm is $4.250561$, so $L=8.501122\times10^{-7}$ H/m, that is $$\boxed{L=0.850112\ \text{mH/km}\qquad L_{total}=0.212528\ \text{H}}$$ The corresponding series reactance is $X_L=2\pi(60)(0.212528)=80.1212\ \Omega$ for the whole 250 km.
4. Part (c) — equivalent radius of the bundle for capacitance. Capacitance is governed by the charge on the conductor surface, so the physical outside radius $r$ replaces the GMR in the same geometric mean: $$D_{SC}=\sqrt[4]{\sqrt{2}}\;\sqrt[4]{r S^{3}}=1.09051\sqrt[4]{(0.01)(0.4)^{3}}=\boxed{0.173450\ \text{m}}$$ It comes out slightly smaller than $D_{SL}$ here, and the reason is worth noting: for an ordinary solid round conductor $D_s\approx0.7788r$, so the capacitive radius would normally be the larger of the two. In this problem the data give $D_s=1.15r$, which is characteristic of a hollow or expanded-core conductor, and the usual order is reversed.
5. Shunt capacitance per phase, earth effects neglected. $$C=\frac{2\pi\varepsilon_0}{\ln(D_{eq}/D_{SC})}=\frac{2\pi(8.854\times10^{-12})}{\ln(12.5992/0.173450)}=\frac{5.56325\times10^{-11}}{4.285585}$$ so $C=1.298129\times10^{-11}$ F/m and $$\boxed{C=0.0129813\ \mu\text{F/km}\qquad C_{total}=3.24532\ \mu\text{F}}$$ The corresponding shunt admittance of the whole line is $Y=j\omega C_{total}=j1.223457\times10^{-3}\ \text{S}$, a charging susceptance the line will impose on the system whether or not it carries load.
6. Part (d) — total series impedance from the stated X/R ratio. With $X/R=8$ the resistance follows directly from the reactance already computed, $$\begin{aligned} R&=\frac{X_L}{8}=\frac{80.1212}{8}=10.0151\ \Omega \\ Z&=10.0151+j80.1212=80.7447\,/\,82.876^{\circ}\ \Omega \end{aligned}$$ and, with $G$ negligible, $Y=j1.223457\times10^{-3}$ S as found above.
7. Propagation constant and characteristic impedance. The exact model works with the two derived quantities $$\gamma\ell=\sqrt{ZY}\qquad Z_c=\sqrt{Z/Y}$$ Substituting, $ZY=-9.80248\times10^{-2}+j1.22530\times10^{-2}$, whose square root is $$\begin{aligned} \gamma\ell&=0.019530+j0.313698 \\ \alpha\ell&=0.019530\ \text{Np},\qquad \beta\ell=0.313698\ \text{rad}=17.9736^{\circ} \end{aligned}$$ and $$\boxed{Z_c=256.403-j15.963=256.899\,/\,{-3.5625^{\circ}}\ \Omega}$$ The small negative angle on $Z_c$ and the small real part of $\gamma\ell$ are both signatures of a low-loss line, as they should be at $X/R=8$.
8. Generalised constants A and B. For the distributed-parameter model $$A=D=\cosh(\gamma\ell)\qquad B=Z_c\sinh(\gamma\ell)$$ Evaluating the hyperbolic functions of the complex argument, $$\boxed{A=0.951380+j0.006027=0.951400\,/\,0.3630^{\circ}}$$ $$\boxed{B=9.6903+j78.8388=79.432\,/\,82.9928^{\circ}\ \Omega}$$ The magnitude of $A$ is below unity by about five per cent, which is the usual order for a 250 km line at 60 Hz, and $B$ is about two per cent smaller in magnitude than the lumped series impedance $Z$ — the distributed model always predicts a slightly smaller effective series impedance than the nominal-pi model.
9. Free reciprocity check. The line is a passive reciprocal two-port, so $AD-BC=1$ must hold exactly. Recovering the remaining constant from $C=\sinh(\gamma\ell)/Z_c$ gives $C=1.203568\times10^{-3}\,/\,90.118^{\circ}\ \text{S}$, and $A^{2}-BC=1.000000$ to six decimal places. That $C$ sits within a fifth of a degree of $+90^{\circ}$ confirms it is essentially a pure shunt susceptance, which is what a lossless-dominated line must produce.