Question 2 of 7: Line Transposition and the Salient-Pole Machine

Problem 2: Line Transposition and the Salient-Pole Machine (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) — transposition. An overhead three-phase circuit is almost never built with its three phase conductors symmetrically placed with respect to one another and to ground: a flat horizontal configuration, a vertical configuration on a single pole, or a triangular configuration on a delta tower all put the three conductors at different mutual spacings. Because the inductance and capacitance of a phase depend on the logarithm of its distance to the other phases, unequal spacing makes the three phase impedances and admittances unequal. The circuit is then unbalanced even when the loads are balanced: it produces negative- and zero-sequence voltages, which cause double-frequency rotor heating in connected machines, spurious operation of ground relays, and telephone interference from residual currents.

Transposition is the practice of rotating the physical positions of the three phase conductors at intervals along the route so that each phase occupies each of the three available positions for one third of the total length. Over a complete transposition cycle the average distance from any one phase to the other two is the same for all three phases, and the line becomes electrically balanced: the series impedance matrix becomes fully symmetric, the mutual couplings are equalised, and the sequence impedances decouple. That is what licenses the single-phase, per-phase-equivalent analysis used everywhere else in this paper, including the geometric-mean-distance formula $D_m = \sqrt[3]{D_{ab}D_{bc}D_{ca}}$.

Mechanically it is done at a transposition structure. The three conductors are dead-ended on a heavier tower, and jumpers carry each phase across to the next position in a fixed rotation (a → b → c → a) — either on a purpose-built transposition tower with staggered arms, or in a substation, or, on a very long line, at the two one-third points of the route so that the cycle completes over the whole line. On modern lines full transposition is often omitted on short circuits, where the residual unbalance is tolerable, and is instead achieved statistically over a network by transposing only long or critical lines; where a double-circuit line is involved the two circuits are also phased so as to cancel one another's unbalance.

Given (parts b and c).

Find. The apparent power the machine delivers at its own terminals, then the excitation (internal) voltage $E_f$ and the torque angle $\delta$ measured from the terminal voltage.

Approach. The load is at the receiving end of the line, so the machine sits at the sending end: push the load voltage and current back through the ABCD model to get the machine terminal quantities, form $S = V_t I_a^{*}$, then apply the two-reaction (Blondel) construction for a salient-pole machine.

1. Recover the fourth line constant. Only $A$ and $B$ are printed, but reciprocity supplies $C$ exactly as in Problem 1: $$C=\frac{A^{2}-1}{B}=\frac{0.0402\angle170.40^\circ}{0.2000\angle85.00^\circ} =0.2009\angle85.40^\circ$$ Again the result is essentially $+j0.2009$, a pure shunt susceptance.
2. Machine terminal voltage. The load is the receiving end, so $$V_t=A V_L+B I_L =0.9800\angle0.20^\circ\times 1.0+0.2000\angle85.00^\circ\times 1.0 =1.0178\angle11.49^\circ\ \text{pu}.$$ The 11.5° advance is the transmission angle across the line.
3. Machine armature current. Similarly $$I_a=C V_L+A I_L=0.2009\angle85.40^\circ+0.9800\angle0.20^\circ =1.0167\angle11.56^\circ\ \text{pu}.$$
4. Apparent power output of the machine. With both terminal phasors known, $$S=V_t I_a^{*} =\left(1.0178\angle11.49^\circ\right)\left(1.0167\angle11.56^\circ\right)^{*}$$ $$\boxed{S=1.0348\angle-0.07^\circ\ \text{pu} = 1.0348\;-\;j0.0013\ \text{pu}}$$ so $|S| = 1.0348$ pu, $P = 1.0348$ pu and $Q = -0.0013$ pu. The machine therefore delivers about 1.035 pu of real power and essentially no net reactive power: the line's charging susceptance $C$ happens to supply almost exactly the reactive absorption of its own series reactance $B$ at this loading, so the machine runs at unity power factor even though it is 11.5° ahead of the load.
5. Part (c) — set up the two-reaction construction. A salient-pole rotor presents different reluctance along the pole axis ($d$) and between poles ($q$), so a single synchronous reactance will not do. With $r_a$ neglected, the torque angle is obtained from the phasor $E_q = V_t + jX_q I_a$, whose direction defines the $q$-axis: $$\tan\delta=\frac{I_a X_q\cos\varphi}{V_t+I_a X_q\sin\varphi}$$ where $\varphi$ is the angle by which $I_a$ lags $V_t$. Here $\varphi = -0.0724^\circ$ (a whisker leading), and $I_a X_q = 0.9659$ pu, so $$\tan\delta=\frac{0.9659} {1.0178-0.0012} =0.9501 \;\Longrightarrow\;\boxed{\delta=43.54^\circ}$$
6. Resolve the armature current onto the two axes. The current lies at $(\delta+\varphi)$ from the $q$-axis, so $$I_d=I_a\sin(\delta+\varphi)=1.0167 \sin(43.46^\circ)=0.6994\ \text{pu}, \qquad I_q=I_a\cos(\delta+\varphi)=0.7380\ \text{pu}.$$ Only $I_d$ magnetises along the pole axis, and only it is multiplied by $X_d$.
7. Excitation voltage. Projecting the phasor equation onto the $q$-axis with $r_a=0$, $$E_f=V_t\cos\delta+I_d X_d =1.0178\cos(43.54^\circ) +0.6994\times 1.15$$ $$\boxed{E_f=1.5422\ \text{pu}}$$ An excitation of 1.54 pu against a terminal voltage of 1.02 pu is a normal over-excited condition for a machine holding a 43.5° torque angle.
8. Independent check on the whole construction. The salient-pole power expression must reproduce the real power already found from $V_t I_a^{*}$: $$P=\frac{E_f V_t}{X_d}\sin\delta +\frac{V_t^{2}}{2}\left(\frac{1}{X_q}-\frac{1}{X_d}\right)\sin 2\delta =0.9401 +0.0947 =1.0348\ \text{pu},$$ which matches step 4 to four figures. The second term is the reluctance power, about 8 % of the total here — the part a round-rotor model would have missed.