Question 4 of 7: Bus Types and a Three-Bus Power-Flow Table

Question 4: Bus Types and a Three-Bus Power-Flow Table (20 marks)

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) — the three bus types. A power-flow bus carries four quantities — $|V|$, $\delta$, $P$ and $Q$ — of which exactly two are specified and two are solved for. Which two are specified defines the type.

The slack (swing, reference) bus specifies $|V|$ and $\delta$, and solves for $P$ and $Q$. There is exactly one in a conventional formulation, and it serves two distinct purposes. It provides the angular reference, because the power-flow equations involve only angle differences and would otherwise be under-determined. And it absorbs the mismatch: network losses are not known until the solution exists, so one generator must be left free to make up whatever the rest of the system does not supply. It is normally assigned to the largest generator or the strongest interconnection.

The generator or voltage-controlled bus (PV bus) specifies $P$ and $|V|$, and solves for $Q$ and $\delta$. Real power is set by the turbine governor and the dispatch, while voltage magnitude is held by the automatic voltage regulator acting on excitation — so those are the two the operator actually commands. The reactive output is whatever the machine must produce to hold that voltage, and it is checked after each iteration against the excitation and stator limits of the capability curve; if a limit binds, the bus is converted to a PQ bus with $Q$ fixed at the limit and $|V|$ released.

The load or PQ bus specifies $P$ and $Q$, and solves for $|V|$ and $\delta$. Nearly every bus in a real network is of this type: a load draws a known real and reactive demand from a forecast, and what its voltage settles to is precisely what the study is meant to discover. A bus with no injection at all is a PQ bus with $P = Q = 0$.

Applying that classification here: bus 1 is the slack bus, since $|V|$ and $\delta$ are both given and both injections are asked for; bus 2 is a PQ load bus, drawing 2.5 pu; and bus 3 is a PV generator bus, holding 1.05 pu. The table hands us the converged angles, which is what turns a normally iterative problem into one line of algebra.

Given. Two purely reactive branches, $x_{12} = 0.1$ pu and $x_{23} = 0.08$ pu, with $V_1 = 1.00\angle 0^\circ$, $\delta_2 = -3.5^\circ$, $P_2 = -2.5$ pu, $|V_3| = 1.05$ pu and $\delta_3 = +5.5^\circ$.

Find. $|V_2|$, and then all six remaining injections $P_1$, $Q_1$, $Q_2$, $P_3$, $Q_3$.

Approach. On a lossless network the real-power injection at a bus has no self-term, so $P_2$ is exactly linear in $|V_2|$ once every angle is known. One division gives the missing magnitude; every other entry then follows by direct substitution.

1. Part (b) — write the injection equations for a purely reactive network. With $G_{ij} = 0$ throughout, the standard polar power-flow equations collapse to $$P_i = |V_i|\sum_{k \ne i}\frac{|V_k|}{x_{ik}}\sin(\delta_i - \delta_k), \qquad Q_i = \sum_{k \ne i}\left[\frac{|V_i|^2}{x_{ik}} - \frac{|V_i||V_k|}{x_{ik}}\cos(\delta_i - \delta_k)\right]$$ The critical structural point is that a purely susceptive $Y_{ii}$ contributes nothing to real power, so no $|V_2|^2$ term survives in $P_2$.
2. Solve for the missing magnitude by one division. Bus 2 connects to buses 1 and 3 only: $$P_2 = |V_2|\left[\frac{|V_1|}{x_{12}}\sin(\delta_2 - \delta_1) + \frac{|V_3|}{x_{23}}\sin(\delta_2 - \delta_3)\right]$$ $$= |V_2|\left[\frac{1.00}{0.1}\sin(-3.5^\circ) + \frac{1.05}{0.08}\sin(-9.0^\circ)\right] = |V_2|\left[-0.61049 - 2.05320\right] = -2.66369\,|V_2|$$ Setting this equal to the given $-2.5$ pu, $$\boxed{|V_2| = \frac{-2.5}{-2.66369} = 0.93855\ \text{pu}}$$ No iteration and no quadratic: the angles were the hard part and the paper supplied them.
3. Bus 1 — the slack injections. Bus 1 sees only bus 2: $$P_1 = \frac{|V_1||V_2|}{x_{12}}\sin(\delta_1 - \delta_2) = \frac{(1.00)(0.93855)}{0.1}\sin(3.5^\circ)$$ $$\boxed{P_1 = 0.57297\ \text{pu}}$$ $$Q_1 = \frac{|V_1|^2}{x_{12}} - \frac{|V_1||V_2|}{x_{12}}\cos(\delta_1 - \delta_2) = 10.0000 - 9.36797$$ $$\boxed{Q_1 = 0.63202\ \text{pu}}$$
4. Bus 3 — the generator injections. Bus 3 sees only bus 2: $$P_3 = \frac{|V_3||V_2|}{x_{23}}\sin(\delta_3 - \delta_2) = \frac{(1.05)(0.93855)}{0.08}\sin(9.0^\circ)$$ $$\boxed{P_3 = 1.92703\ \text{pu}}$$ $$Q_3 = \frac{|V_3|^2}{x_{23}} - \frac{|V_3||V_2|}{x_{23}}\cos(\delta_3 - \delta_2) = 13.78125 - 12.16680$$ $$\boxed{Q_3 = 1.61446\ \text{pu}}$$
5. Bus 2 — the remaining reactive entry. Bus 2 has two branches, so both self-terms appear: $$Q_2 = |V_2|^2\!\left(\frac{1}{x_{12}} + \frac{1}{x_{23}}\right) - \frac{|V_2||V_1|}{x_{12}}\cos(\delta_2-\delta_1) - \frac{|V_2||V_3|}{x_{23}}\cos(\delta_2-\delta_3)$$ $$= 0.88087(22.5) - 9.36797 - 12.16680 = 19.81963 - 21.53477$$ $$\boxed{Q_2 = -1.71512\ \text{pu}}$$ The load bus draws 2.5 pu of real power and 1.715 pu of reactive power, i.e. a load power factor of $2.5/\sqrt{2.5^2 + 1.715^2} = 0.825$ lagging — an ordinary industrial figure, which is a useful sanity check that no sign has been dropped.
6. Close the table with the two conservation identities. These validate all six injections at once and cost one line each. Real power must balance exactly on a lossless network: $$\sum_i P_i = 0.57297 - 2.5 + 1.92703 = 0.00000$$ Reactive power must equal the total absorbed by the series reactances: $$\sum_i Q_i = 0.63202 - 1.71512 + 1.61446 = 0.53136$$ $$\sum_{\text{lines}} |I|^2x = (0.85308)^2(0.1) + (2.39424)^2(0.08) = 0.07277 + 0.45859 = 0.53136$$ Both close to five decimals, which is stronger evidence than re-deriving any single entry.