22-Elec-B7 Power Systems Engineering · May 2016

Question 4 of 7: Completing the bus table of a four-bus network

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 4 — Completing the bus table of a four-bus network (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.

Given. Four buses joined by three lossless branches, with one unknown voltage magnitude and one specified injection.

Bus	$\|V_i\|$ (pu)	$\delta_i$	$P_i$ (pu)	$Q_i$ (pu)
1	1.00	0°	?	?
2	?	−10°	−0.95	?
3	1.02	−12°	?	?
4	0.98	−8°	?	?

Branch reactances (from Figure 2): $x_{12}=0.25$, $x_{23}=0.20$, $x_{24}=0.10$ pu, all purely inductive, so the network is lossless in real power.

Find. The seven missing entries: $|V_2|$ and the eight injections implied by the specified voltage profile.

Figure 5 — the four-bus network of Problem 4. Bus 2 is the junction of all three branches, which is why one scalar equation at that bus is enough to fix its voltage magnitude.

Approach. Every branch is a pure reactance, so the real-power injection at bus 2 is exactly linear in $|V_2|$; solve that one equation for $|V_2|$, then evaluate the four complex injections from $S_i = V_i\left(\sum_k Y_{ik}V_k\right)^{*}$.

Build the bus admittance matrix. Each branch contributes $y_{ik}=1/(jx_{ik})$: $y_{12}=-j4$, $y_{23}=-j5$, $y_{24}=-j10$ pu. The off-diagonals are $Y_{ik}=-y_{ik}$ and the diagonals are the sums of the branch admittances at the bus, so $$Y_{22}=-j(4+5+10)=-j19,\qquad Y_{21}=j4,\quad Y_{23}=j5,\quad Y_{24}=j10$$
Write the real injection at bus 2 in terms of the branch flows. For a lossless branch the real power leaving bus $i$ towards bus $k$ is $\dfrac{|V_i||V_k|}{x_{ik}}\sin(\delta_i-\delta_k)$. Collecting the three branches at bus 2 and writing $v=|V_2|$, $$P_2=-\frac{(1)v}{0.25}\sin 10^{\circ}+\frac{(1.02)v}{0.20}\sin 2^{\circ}-\frac{(0.98)v}{0.10}\sin 2^{\circ}$$ the three terms being the flows to buses 1, 3 and 4 respectively. Every term carries the factor $v$ once and only once, because $Y_{22}$ is purely susceptive and so contributes nothing to real power.
Solve the linear equation. Evaluating the coefficients gives $P_2 = -0.858620\,v$, and setting this equal to the specified $P_2=-0.95$ pu, $$v=\frac{-0.95}{-0.858620}\qquad\Longrightarrow\qquad \boxed{|V_2|=1.1064\ \text{pu}}$$ There is no second root and no iteration: the linearity is a property of the lossless network, not an approximation.
Evaluate the bus-1 injection. With $V_2$ now known, $S_1=V_1\left(Y_{11}V_1+Y_{12}V_2\right)^{*}$, giving $$\boxed{P_1=0.7685\ \text{pu},\qquad Q_1=-0.3585\ \text{pu}}$$ Bus 1 exports real power and absorbs reactive power, which is what a machine at the light end of a heavily capacitive-free reactive corridor must do.
Evaluate the remaining injections the same way. Repeating $S_i=V_i\left(\sum_kY_{ik}V_k\right)^{*}$ at buses 2, 3 and 4, $$\begin{aligned} S_2 &= -0.95 + j2.4252 \\ S_3 &= -0.1969 - j0.4373 \\ S_4 &= 0.3784 - j1.2324\end{aligned}$$ all in per unit, with the sign convention that a positive value is an injection into the network and a negative value is a load.
Check the answer against the two conservation laws. The real injections sum to $0.000000$ pu, zero to machine precision — correct, because a network of pure reactances dissipates nothing. The reactive injections sum to $0.3970$ pu, which must equal the total reactive absorption $\sum |I_{ik}|^{2}x_{ik}$ of the three branches: 0.179779 + 0.044222 + 0.173046 = 0.397047 pu. Both identities close, which validates the whole table.

Check: the answer is dictated by the printed angle set, and it is an unusual operating point. Holding bus 2 at −10° while its neighbours sit at −12° and −8° forces a large reactive injection there (+2.425 pu) and a bus voltage of 1.106 pu, above the 1.00 pu swing bus. Physically that means bus 2 must carry substantial local reactive support — a capacitor bank or a compensator — alongside its 0.95 pu real load. Nothing in the arithmetic is wrong; the question specifies a voltage profile and asks what injections sustain it, which is the reverse of the usual power-flow question, and the profile it specifies is a demanding one.

Problem 4 — the completed bus table
Bus	$\|V_i\|$ (pu)	$\delta_i$	$P_i$ (pu)	$Q_i$ (pu)
1	1.0000	0°	0.7685	-0.3585
2	1.1064	−10°	−0.9500	2.4252
3	1.0200	−12°	-0.1969	-0.4373
4	0.9800	−8°	0.3784	-1.2324

Branch flows implied by the completed table
Branch	$P$ leaving the first bus (pu)	$Q$ leaving the first bus (pu)	Reactive absorption $\|I\|^{2}x$ (pu)
1 → 2	0.7685	-0.3585	0.1798
2 → 3	0.1969	0.4816	0.0442
2 → 4	-0.3784	1.4054	0.1730

Topic. The power-flow equations read backwards: from a specified voltage profile to the injections that sustain it.

Key relations. $S_i=V_i\left(\sum_kY_{ik}V_k\right)^{*}$; for a lossless branch $P_{ik}=\frac{|V_i||V_k|}{x_{ik}}\sin(\delta_i-\delta_k)$ and $Q_{ik}=\frac{|V_i|^{2}-|V_i||V_k|\cos(\delta_i-\delta_k)}{x_{ik}}$; conservation checks $\sum_i P_i = 0$ and $\sum_i Q_i = \sum_{branches}|I|^{2}x$.

Why this works. A conventional power flow specifies P and Q at load buses and solves iteratively for the voltages, because the equations are nonlinear in the unknown magnitudes and angles. This question inverts the exercise: all four angles and three of the four magnitudes are given, so almost every equation becomes an evaluation rather than an unknown. The one genuine unknown, the magnitude at bus 2, appears linearly in its own real-power equation because the self-admittance term contributes only reactive power on a lossless network — and that is precisely why no Gauss-Seidel or Newton iteration is called for.

Common pitfall. Mixing the injection convention with the load convention half-way through the table, so that some entries are negative loads and others are positive loads. State the convention once (here: positive means injected into the network) and hold to it. The other frequent slip is taking the self-admittance as the sum of the branch impedances rather than admittances.

Source. El-Hawary, Electrical Power Systems: Design and Analysis, power-flow chapter; Glover, Sarma and Overbye, 6th ed., ch. 6; Grainger and Stevenson, Power System Analysis, ch. 9.

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