22-Elec-B7 Power Systems Engineering · December 2016

Question 4 of 7: Two-Bus Power Flow in Closed Form

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format. National Exams, December 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. This exam code follows M. E. El-Hawary, Electrical Power Systems: Design and Analysis (IEEE Press) — the notation is his throughout: the “cantilever” transformer equivalent of Problem 3, the two-reaction salient-pole construction of Problem 2, the ABCD two-port of Problem 1 and the sequence-network fault reduction of Problem 6. Corroborating references: J. D. Glover, M. S. Sarma and T. J. Overbye, Power System Analysis and Design, 6th ed. (ch. 3 transformers, ch. 4–5 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 C22.2 and the IEEE C62 / IEEE 142 (Green Book) series on system grounding.

Check: four judgement calls carried through the solutions, all recorded here rather than hedged in the answers.

(i) Problem 1 does not state a frequency. Parts (c) and (d) need one, and the paper is a Canadian national exam, so 60 Hz is used throughout — consistent with the 60 Hz stated in Problems 3 and 7. At 50 Hz the capacitance and surge impedance are unchanged (they do not involve frequency) and only the propagation constant of part (d) rescales by 5/6.

(ii) Problem 2 prints “neglecting armature reaction”. Armature reaction is precisely what the direct- and quadrature-axis reactances represent, so the phrase is read as its standard counterpart, neglecting armature resistance — the only reading under which the two given reactances can be used at all.

(iii) Problem 6 prints a part label “c-” with no question text. The five parts that do carry text (a, b, d, e, f) are marked 4 points each and total the 20 points the paper allots to every problem, so nothing is missing from the assessment; the sequence networks that the empty label most plausibly asked for are drawn anyway.

(iv) Problem 5 gives no reactance between buses 2 and 3. Figure (3) confirms the topology: bus 3 hangs radially off bus 1 through the j0.12 pu line, and the only other tie is line 1–2.

Problem 4: Two-Bus Power Flow in Closed Form (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.

Quantity	Symbol	Value
Slack bus	$V_1$	$1.00\angle 0^{\circ}$ pu
Line impedance	$z_{12}$	$j0.125$ pu (purely reactive)
Real power load at bus 2	$P_{D2}$	1.0 pu
Reactive power load at bus 2	$Q_{D2}$	0.125 pu
Real power injection at bus 2	$P_2$	$-1.0$ pu
Reactive power injection at bus 2	$Q_2$	$-0.125$ pu

Find. A derivation of the two given injection expressions from the general power-flow equations and again from first principles; the numerical value of the parameter $a$; the coefficients $b$, $c$, $d$ of the quartic in $|V_2|$; and the bus-2 voltage magnitude and angle.

Figure 4 — The two-bus system of Figure (2). With one slack bus and one load bus joined by a single purely reactive branch, the power-flow problem has a closed-form solution: no Gauss–Seidel or Newton–Raphson iteration is needed, and the quartic of part (c) is exact.

Approach. Build the two-bus admittance matrix, substitute it into the polar power-flow equations to obtain the stated forms and identify $a$; confirm the same result by writing $\mathbf{S}_2=\mathbf{V}_2\mathbf{I}_2^{*}$ directly; then eliminate $\theta_2$ between the two injection equations to leave a quadratic in $|V_2|^{2}$.

Part (a) — form the bus admittance matrix of the two-bus network. The single branch has admittance $y_{12}=1/(j0.125)=-j8$ pu, and there is no shunt element, so $$Y_{22}=y_{12}=-j8=8\angle{-90^{\circ}},\qquad Y_{21}=-y_{12}=+j8=8\angle{+90^{\circ}}$$ In the polar notation of the question, $\left|Y_{22}\right|=\left|Y_{21}\right|=8$ with $\psi_{22}=-90^{\circ}$ and $\psi_{21}=+90^{\circ}$.
Substitute into the general injection equations at bus 2. Writing the sum over $j=1,2$ with $\left|V_1\right|=1$ and $\theta_1=0$, $$P_2=\left|V_2\right|\left[\left|V_2\right|(8)\cos(90^{\circ})+(1)(8)\cos\left(\theta_2-90^{\circ}\right)\right]=8\left|V_2\right|\sin\theta_2$$ $$Q_2=\left|V_2\right|\left[\left|V_2\right|(8)\sin(90^{\circ})+(1)(8)\sin\left(\theta_2-90^{\circ}\right)\right]=8\left|V_2\right|\left[\left|V_2\right|-\cos\theta_2\right]$$ because $\cos(\theta_2-90^{\circ})=\sin\theta_2$ and $\sin(\theta_2-90^{\circ})=-\cos\theta_2$. These are exactly the printed forms, with the injections equal to minus the loads: $P_2=-1.0$ and $Q_2=-0.125$.
Part (b) — confirm the same result from first principles. The current leaving bus 2 towards bus 1 is $\mathbf{I}_2=\left(\mathbf{V}_2-\mathbf{V}_1\right)/(jx)$, so the complex power injected at bus 2 is $$\mathbf{S}_2=\mathbf{V}_2\mathbf{I}_2^{*}=\mathbf{V}_2\frac{\left(\mathbf{V}_2^{*}-\mathbf{V}_1^{*}\right)}{-jx}=\frac{j}{x}\left(\left|V_2\right|^{2}-\left|V_2\right|\angle\theta_2\right)$$ Expanding $\left|V_2\right|\angle\theta_2=\left|V_2\right|\cos\theta_2+j\left|V_2\right|\sin\theta_2$ and separating real and imaginary parts gives $P_2=\left|V_2\right|\sin\theta_2/x$ and $Q_2=\left|V_2\right|\left(\left|V_2\right|-\cos\theta_2\right)/x$, identical in form to part (a). Matching the two, $$\boxed{\ a=\frac{1}{x}=\frac{1}{0.125}=8\ \text{pu}\ }$$ so the parameter $a$ is nothing more mysterious than the branch susceptance magnitude.
Part (c) — eliminate the angle between the two injection equations. Rearrange each equation to isolate a trigonometric function: $$\begin{aligned}\left|V_2\right|\sin\theta_2&=\frac{P_2}{a}\\\left|V_2\right|\cos\theta_2&=\left|V_2\right|^{2}-\frac{Q_2}{a}\end{aligned}$$ Squaring and adding removes $\theta_2$ entirely, since $\sin^{2}+\cos^{2}=1$: $$\left|V_2\right|^{2}=\frac{P_2^{2}}{a^{2}}+\left(\left|V_2\right|^{2}-\frac{Q_2}{a}\right)^{2}$$
Expand and read off the three coefficients. Writing $u=\left|V_2\right|^{2}$ and collecting terms, $$u^{2}-\left(1+\frac{2Q_2}{a}\right)u+\frac{P_2^{2}+Q_2^{2}}{a^{2}}=0$$ which is the required form $b\left|V_2\right|^{4}-c\left|V_2\right|^{2}+d=0$ with $$\begin{aligned}b&=1\\c&=1+\frac{2Q_2}{a}=1+\frac{2(-0.125)}{8}=0.96875\\d&=\frac{P_2^{2}+Q_2^{2}}{a^{2}}=\frac{(-1.0)^{2}+(-0.125)^{2}}{64}=\frac{1.015625}{64}=0.01586914\end{aligned}$$ $$\boxed{\ b=1,\qquad c=0.96875,\qquad d=0.0158691\ }$$
Solve the quadratic in $u$ and choose the physical root. The discriminant is $c^{2}-4bd=0.93848-0.06348=0.87500$, so $$u=\frac{0.96875\pm\sqrt{0.875}}{2}=\frac{0.96875\pm0.935414}{2}=\{0.952082,\ 0.016668\}$$ The upper root gives $\left|V_2\right|=0.97575$ pu and the lower gives $0.12910$ pu. Both satisfy the algebra, but the lower root is the unstable low-voltage solution on the far side of the nose of the P–V curve, reached only at absurd current levels; the operating solution is always the upper root. That the discriminant is positive is itself the statement that this loading is feasible — it is a free voltage-stability index.
Part (d) — recover the angle and state both answers. From the real-power equation, $$\sin\theta_2=\frac{P_2}{a\left|V_2\right|}=\frac{-1.0}{8(0.975747)}=-0.128108\;\Longrightarrow\;\theta_2=-7.360^{\circ}$$ $$\boxed{\ \left|V_2\right|=0.9757\ \text{pu},\qquad \theta_2=-7.360^{\circ}\ }$$ The angle is negative because bus 2 absorbs real power, and the magnitude is depressed by 2.4 per cent because it also absorbs reactive power across a purely reactive branch.
Close the loop by back-substitution. Recomputing the injections from the answer, $8(0.975747)\sin(-7.360^{\circ})=-1.0000$ and $8(0.975747)\left[0.975747-\cos(-7.360^{\circ})\right]=-0.1250$, reproducing both loads exactly. Since the branch is lossless the slack bus must supply $P_1=+1.0$ pu of real power, and the reactive generation is $Q_1=\left|I\right|^{2}x+Q_{D2}$ with $\left|I\right|=\left|\mathbf{V}_1-\mathbf{V}_2\right|/x=1.0328$ pu, giving $Q_1=0.1333+0.125=0.2583$ pu — the reactive loss in the line is itself larger than the reactive load, which is the usual state of affairs on a heavily loaded reactive branch.

Quantity	Symbol	Result
Parameter a (branch susceptance)	$a$	8 pu
Quartic coefficients	$b,\ c,\ d$	1, 0.96875, 0.0158691
High-voltage (operating) root	$\left\|V_2\right\|$	0.9757 pu
Low-voltage (unstable) root	$\left\|V_2\right\|_{\text{low}}$	0.1291 pu
Bus-2 angle	$\theta_2$	$-7.360^{\circ}$
Line current magnitude	$\|I_{12}\|$	1.0328 pu
Slack-bus generation	$P_1,\ Q_1$	1.0000 pu, 0.2583 pu

Topic. Polar power-flow equations, and the closed-form two-bus solution.

Key relations. $P_i=\left|V_i\right|\sum_j\left|V_j\right|\left|Y_{ij}\right|\cos(\theta_i-\theta_j-\psi_{ij})$; $Q_i=\left|V_i\right|\sum_j\left|V_j\right|\left|Y_{ij}\right|\sin(\theta_i-\theta_j-\psi_{ij})$; $\mathbf{S}_i=\mathbf{V}_i\mathbf{I}_i^{*}$; elimination identity $u^{2}-\left(1+2Q_2/a\right)u+\left(P_2^{2}+Q_2^{2}\right)/a^{2}=0$ with $u=\left|V_2\right|^{2}$.

Why this works. The general power-flow problem is nonlinear and is normally solved by Newton–Raphson, but a two-bus system with one slack bus has only two unknowns and two equations, and because the unknowns enter only as $|V_2|\sin\theta_2$ and $|V_2|\cos\theta_2$ they can be eliminated algebraically. The resulting quadratic in $|V_2|^{2}$ is the analytic backbone of voltage-stability analysis: its two roots are the upper and lower halves of the P–V (nose) curve, they merge when the discriminant vanishes, and that merge point is the maximum loadability of the branch. Everything a large-system continuation power flow computes numerically is visible here in closed form.

Common pitfall. Sign conventions cause most of the lost marks. The injection at a load bus is the negative of the load, so $P_2=-1.0$ and $Q_2=-0.125$; using $+1.0$ flips the sign of $\theta_2$ and, worse, changes $c$ to 1.03125 and moves the answer. The second trap is reporting the low-voltage root, which is mathematically valid and physically unreachable.

Source. El-Hawary, Electrical Power Systems: Design and Analysis, ch. 8 (power-flow formulation, bus classification); Glover, Sarma & Overbye, §6.2–6.4 and §13 (voltage stability, P–V curves); Grainger & Stevenson, ch. 9.

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