Question 4 of 7: Bus Classification and a Three-Bus Power-Flow Solution

Problem 4: Bus Classification and a Three-Bus Power-Flow Solution (25 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.

(a) Bus types in the conventional power-flow formulation. Each bus carries four quantities — voltage magnitude $|V|$, voltage angle $\delta$, net real injection $P$ and net reactive injection $Q$ — and exactly two are specified in advance, the remaining two being computed. Three types are recognised. The slack (swing, reference) bus, of which there is exactly one, specifies $|V|$ and $\delta$ (the latter almost always at zero, to fix the angular reference) and leaves $P$ and $Q$ to be found; it must exist because network losses are unknown until the solution is complete, so one generator has to absorb the mismatch. A load or PQ bus specifies $P$ and $Q$ from the forecast demand and leaves $|V|$ and $\delta$ unknown; the great majority of buses in a real network are of this type, and a bus with no load or generation at all is a PQ bus with both injections zero. A generator, voltage-controlled or PV bus specifies $P$ (from the dispatch) and $|V|$ (from the automatic voltage regulator) and leaves $Q$ and $\delta$ unknown; its reactive output is checked against the machine capability limits at each iteration and, if a limit binds, the bus is converted to PQ at that limit for the remaining iterations. In this problem bus 1 is the slack ($V_1=1.0\angle0^\circ$), bus 2 is a load bus, and bus 3 is voltage-controlled with both its magnitude and angle given, which is what makes the network solvable in closed form.

Given.

Find. $|V_2|$ and $Q_2$ with $\delta_2$ taken as $-12^\circ$, then the complex generation at buses 1 and 3.

Approach. Build the bus admittance matrix, write the injection at bus 2 in polar form and exploit the fact that a purely susceptive $Y_{22}$ contributes nothing to real power, so $P_2$ is linear in $|V_2|$ and one division answers part (b). With all three voltages then known, parts (c) and (d) are direct evaluations of $S_i=V_iI_i^{*}$.

1. Part (b) — assemble the bus admittance matrix. With $y_{12}=1/(j0.10)=-j10$ and $y_{23}=1/(j0.08)=-j12.5$,$$\mathbf{Y}=\begin{bmatrix}-j10 & j10 & 0\\ j10 & -j22.5 & j12.5\\ 0 & j12.5 & -j12.5\end{bmatrix}\ \text{pu}.$$Every element is purely imaginary because no resistance or shunt is given.
2. Write the real-power injection at bus 2. In polar form $P_i=\sum_k |V_i||V_k|\bigl(G_{ik}\cos\theta_{ik}+B_{ik}\sin\theta_{ik}\bigr)$ with $\theta_{ik}=\delta_i-\delta_k$. The self term carries $\sin 0 = 0$ and every $G$ is zero, so$$P_2=|V_2|\left[\frac{|V_1|}{x_{12}}\sin(\delta_2-\delta_1)+\frac{|V_3|}{x_{23}}\sin(\delta_2-\delta_3)\right],$$which is linear in $|V_2|$ — the reason the paper hands you $\delta_2$.
3. Solve for the bus-2 voltage magnitude. Substituting the numbers, with $\sin(-12^\circ)=-0.207912$ and $\sin(-44^\circ)=-0.694658$,$$\begin{aligned}P_2&=|V_2|\bigl[10(-0.207912)+13.125(-0.694658)\bigr]\\&=|V_2|(-2.07912-9.11739)=-11.19651\,|V_2|,\end{aligned}$$and setting this equal to the given $-3.75$ pu,$$\boxed{|V_2|=\frac{-3.75}{-11.19651}=0.33493\ \text{pu}.}$$
4. Reactive injection at bus 2. The companion polar expression is $Q_i=\sum_k |V_i||V_k|\bigl(G_{ik}\sin\theta_{ik}-B_{ik}\cos\theta_{ik}\bigr)$, so$$Q_2=-B_{22}|V_2|^{2}-|V_2|\left[\frac{|V_1|}{x_{12}}\cos(\delta_2-\delta_1)+\frac{|V_3|}{x_{23}}\cos(\delta_2-\delta_3)\right],$$which evaluates to$$\begin{aligned}Q_2&=22.5(0.33493)^{2}-0.33493\bigl[10(0.978148)+13.125(0.719340)\bigr]\\&=2.52395-6.43822,\end{aligned}$$so that$$\boxed{Q_2=-3.9143\ \text{pu}\quad(\text{i.e. the load absorbs }3.9143\ \text{pu of vars}).}$$
5. Flag what this answer means physically. A voltage of 0.335 pu is far below any admissible operating range; it is nevertheless the unique value consistent with the stated $\delta_2=-12^\circ$, because with the angle fixed the real-power equation is linear and has exactly one root. The reason is visible in the data: bus 3 leads bus 2 by $44^\circ$ across only $j0.08$ pu, an enormously stiff, heavily-stressed transfer. The answer is reported as computed, with the observation that this is a study point on the lower branch of the network’s P–V characteristic rather than a realistic dispatch.
6. Part (c) — generation at bus 1. All three complex voltages are now known, so the injections follow from $S_i=V_i\bigl(\sum_k Y_{ik}V_k\bigr)^{*}$. At bus 1 only the line to bus 2 exists, so$$S_1=V_1\left(\frac{V_1-V_2}{j0.10}\right)^{*},\qquad V_2=0.33493\angle-12^\circ=0.32761-j0.06964\ \text{pu}.$$Evaluating, $I_{12}=(V_1-V_2)/(j0.10)=(0.67239+j0.06964)/(j0.10)=0.69635-j6.72393$, whence$$\boxed{P_1=0.6964\ \text{pu},\qquad Q_1=6.7239\ \text{pu}.}$$
7. Part (d) — generation at bus 3. The same construction on the 2–3 line, with $V_3=1.05\angle32^\circ=0.89043+j0.55642$,$$S_3=V_3\left(\frac{V_3-V_2}{j0.08}\right)^{*},$$gives $I_{32}=(0.56282+j0.62606)/(j0.08)=7.82575-j7.03525$ and therefore$$\boxed{P_3=3.0536\ \text{pu},\qquad Q_3=10.6191\ \text{pu}.}$$
8. Close the two conservation checks. The network is lossless, so the real injections must sum to zero: $0.6964+(-3.75)+3.0536=0.0000$ — exact, which validates $|V_2|$ and both generations at once. The reactive injections do not sum to zero but must equal the reactive absorption of the two reactances: $Q_1+Q_2+Q_3=6.7239-3.9143+10.6191=13.4287\ \text{pu}$, and $|I_{12}|^2x_{12}+|I_{23}|^2x_{23}=(6.7599)^2(0.10)+(10.5233)^2(0.08)=4.5696+8.8591=13.4287$, equal to the injection sum within the rounding of the last digit. The very large reactive requirement is the direct consequence of the depressed bus-2 voltage noted above.

Final Results