Question 1 of 7: ABCD to equivalent pi, parallel lines, and the machine behind them
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Exams, May 2016 — 07-Elec-B7 Power Systems Engineering. Open-book, 3 hours duration. Seven problems are printed; any five constitute a complete paper, only the first five appearing in the answer book are marked, and all questions are of equal value (20 points each). All seven are solved in full below, because this set is a study resource rather than a timed sitting.
Reference texts. The syllabus for this code follows M. E. El-Hawary, Electrical Power Systems: Design and Analysis (IEEE Press) — the notation used throughout (the ABCD two-port and its equivalent pi, the two-reaction salient-pole construction, the sequence-network fault reduction) is his, and the phrase “all power formulae in the text” in Problem 1 refers to that book. Corroborating references: J. D. Glover, M. S. Sarma and T. J. Overbye, Power System Analysis and Design, 6th ed. (ch. 3 transformers, ch. 5 transmission lines, ch. 6 power flow, ch. 7–9 faults and symmetrical components, ch. 11 transient stability); J. J. Grainger and W. D. Stevenson, Power System Analysis; and S. J. Chapman, Electric Machinery Fundamentals, 5th ed., for the machine and transformer chapters. Canadian practice references where grounding and protection are discussed: CSA C22.1 Canadian Electrical Code, Part I, CSA C22.3 No. 1 for overhead systems, and IEEE C37 series for protective relaying.
Check: four data points read off the printed figures. (i) In Figure (2) of Problem 4 the three branch reactances are j0.25, j0.20 and j0.10 pu, with bus 4 hanging off bus 2 through the j0.10 branch — the topology is taken from the drawing. (ii) Figure (3) of Problem 5 shows three sources, not two: generators at bus 1 and bus 4 and a third machine below bus 5. The sentence “the voltage at both sources is 1 p.u.” is taken to mean all sources are at 1.0 pu, which is the only reading that lets the network be solved. (iii) In Figure (4-a) of Problem 6, transformer T2 carries its delta on the generator side and T1 carries its delta on the motor side; motor 1 is an ungrounded wye and motor 2 is grounded through a reactor. Those four winding connections, not the reactance values, decide the answer. (iv) Problem 7 gives the switching angles in radians (1.2, 1.25 and 1.9 rad).
Check: Problem 3(a) as printed is over-determined and internally inconsistent. Rated kVA, the load power factor, the efficiency and the primary voltage are four data for two unknowns, and at 0.85 power factor lagging no positive series reactance satisfies them all. The inconsistency is quantified and resolved inside the answer rather than hidden.
Problem 1 — ABCD to equivalent pi, parallel lines, and the machine behind them (20 points)
Given. Two lines in parallel between buses 1 and 2, with the data below. All quantities are per unit on a common base and both lines are lossless (every parameter is purely imaginary).
Quantity
Symbol
Value
Line 1 — two-port constant
$A_1 = D_1$
0.996
Line 1 — series constant
$B_1$
$j0.02$ pu
Line 2 — series impedance
$Z_2$
$j0.05$ pu
Line 2 — half shunt admittance
$Y_2/2$
$j0.02$ pu
Bus 2 voltage magnitude
$|V_2|$
1.00 pu
Bus 2 active load
$P_2$
1.25 pu at 0.85 pf lagging
Machine synchronous reactance
$X_s$
0.03 pu
Find. (a) the equivalent-pi elements of line 1; (b) the single equivalent line replacing the pair; (c) the voltage magnitude, phase angle, active and reactive power at bus 1; and (d) the excitation voltage and torque angle of the round-rotor machine sitting behind $X_s$.
Figure 1 — the network of Problem 1. Line 1 is handed to you as a two-port; line 2 is handed to you as an equivalent pi. Part (a) converts the first into the second so that the two can be paralleled.
Approach. Invert the standard nominal-pi relations $A = 1 + ZY/2$ and $B = Z$ to recover line 1’s pi elements, add the two pi sections in parallel (series impedances in parallel, shunt admittances in sum), then run the resulting ABCD two-port backwards from the known receiving-end condition and finally push the result through the machine reactance.
Part (a) — recover $Z_1$ directly from $B_1$. For a pi section with series impedance $Z$ and shunt admittance $Y/2$ at each end, the two-port constants are $A = D = 1 + \frac{ZY}{2}$, $B = Z$ and $C = Y\left(1 + \frac{ZY}{4}\right)$. The middle relation is immediate: $$Z_1 = B_1 = j0.02\ \text{pu}$$
Solve the $A$ relation for the shunt branch. Rearranging $A_1 = 1 + Z_1 Y_1/2$ gives $\dfrac{Y_1}{2} = \dfrac{A_1 - 1}{Z_1}$, and substituting the numbers, $$\frac{Y_1}{2}=\frac{0.996-1}{j0.02}=\frac{-0.004}{j0.02}=j0.2 \ \text{pu}$$ so the equivalent-pi elements of line 1 are $\boxed{Z_1 = j0.02\ \text{pu},\ Y_1 = j0.40\ \text{pu}}$. The sign is the useful check: dividing a negative real number by $+j$ returns $+j$, so the shunt is capacitive, which is what line charging must be.
Confirm with the reciprocity identity. A passive two-port satisfies $AD - BC = 1$, so with $A = D$ the third constant follows as $C_1 = (A_1^2-1)/B_1 = 0.3992\,j$ pu. The direct evaluation $C_1 = Y_1\left(1 + Z_1Y_1/4\right) = j0.40(1-0.002) = j0.3992$ agrees, which proves the pi conversion is self-consistent rather than merely plausible.
Part (b) — parallel the two series arms. Two pi sections joining the same pair of buses share their series arms in parallel and their shunt arms in sum: $$Z_{eq}=\frac{Z_1Z_2}{Z_1+Z_2}=\frac{(j0.02)(j0.05)}{j0.07}=j0.014286\ \text{pu}$$ which is $j1/70$ pu exactly.
Sum the shunt arms end by end. Each end of the equivalent line carries both half-shunts, so $$\frac{Y_{eq}}{2}=\frac{Y_1}{2}+\frac{Y_2}{2}=j0.2+j0.02=\boxed{j0.22\ \text{pu}}$$ alongside the series arm $Z_{eq}=j0.014286$ pu of the previous step. The equivalent two-port constants that follow are $A_{eq}=D_{eq}=1+Z_{eq}Y_{eq}/2=0.996857$ and $C_{eq}=j0.439309$ pu.
Part (c) — convert the bus-2 load into a current phasor. Taking $V_2 = 1.0\angle 0^{\circ}$ pu for the moment, the load angle is $\varphi=\cos^{-1}0.85=31.788^{\circ}$, so $Q_2 = P_2\tan\varphi = 1.25(0.619745)=0.7747$ pu and $$I_2=\left(\frac{S_2}{V_2}\right)^{*}=(1.25+j0.7747)^{*}=1.2500-j0.7747\ \text{pu}$$
Run the two-port from the receiving end. The sending-end pair follows from $V_1 = A_{eq}V_2 + B_{eq}I_2$ and $I_1 = C_{eq}V_2 + D_{eq}I_2$: $$\begin{aligned} V_1 &= 0.996857(1.0)+j0.014286\,(1.2500-j0.7747) = 1.007924+j0.017857 \\[2pt] I_1 &= j0.439309(1.0)+0.996857(1.2500-j0.7747) = 1.246071-j0.332937 \end{aligned}$$ In polar form $V_1 = 1.0081\angle 1.0150^{\circ}$ pu and $I_1 = 1.2898\angle -14.9594^{\circ}$ pu.
Form the sending-end complex power. With $S_1 = V_1 I_1^{*}$, $$\boxed{S_1 = 1.2500 + j0.3578\ \text{pu}}$$ The real part reproduces the load exactly, as it must on a lossless network; the whole of the reactive difference, $0.3578-0.7747=-0.4169$ pu, is the net of the series reactive absorption against the charging supplied by the two shunt arms. Re-referencing to bus 1 as the phase reference, as the question directs, simply subtracts 1.0150° from every angle, putting bus 2 at $-1.0150^{\circ}$.
Part (d) — step behind the synchronous reactance. The round-rotor machine is represented by a constant emf $E$ behind $jX_s$, so $$E = V_1 + jX_sI_1 = (1.007924+j0.017857) + j0.03(1.246071-j0.332937) = 1.017912+j0.055239$$
Read the excitation magnitude and torque angle. Converting to polar form and measuring the angle from the terminal voltage (bus 1 is the machine terminal and the stated reference), $$\boxed{|E| = 1.0194\ \text{pu},\qquad \delta = 2.0912^{\circ}}$$ The free check is the round-rotor power expression: $P = \dfrac{|E||V_1|}{X_s}\sin\delta = \dfrac{(1.0194)(1.0081)}{0.03}\sin 2.0912^{\circ} = 1.2500$ pu, which is the load the machine is carrying. A torque angle of only two degrees is the expected consequence of a 0.03 pu synchronous reactance — this is a very stiff machine.
Figure 2 — the single equivalent pi of part (b). Both original shunt arms land on each end, so the equivalent line is far more capacitive than either line alone.
