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

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

Part (a) — the three bus types. Each bus of a power-flow problem carries four quantities: the voltage magnitude $|V|$, the voltage angle $\delta$, the net real power injection $P$ and the net reactive power injection $Q$. Two of the four are specified as data and two are computed; which two are specified is what defines the bus type.

Two practical riders complete the answer. A PV bus reverts to a PQ bus during the solution if its computed $Q$ hits a machine limit: $Q$ is then fixed at the limit and $|V|$ is released. And a bus with neither generation nor load (a pure junction or a switching station) is simply a PQ bus with $P = Q = 0$.

In the system of Figure 2 bus 1 is the slack bus ($V_1 = 1.0\angle 0^\circ$, $P_1$ and $Q_1$ unknown), bus 3 is a voltage-controlled generator bus whose solved state $V_3 = 1.05\angle 32^\circ$ is given here, and bus 2 is a load bus carrying $P_2 = -4$ pu with $|V_2|$ and $\delta_2$ to be found. The question hands us $\delta_2 = -8.5^\circ$, which converts the non-linear power-flow solution into a direct calculation.

Given.

Find. $|V_2|$ and $Q_2$, then the generation $P_1 + jQ_1$ and $P_3 + jQ_3$.

Approach. Write the standard polar power-flow injection equations for a purely reactive network; the $P_2$ equation is then linear in $|V_2|$ and solves in one line, after which every remaining quantity is an explicit evaluation.

1. Specialise the injection equations to a lossless network. For a line of pure series reactance $x_{ik}$ the general polar forms $P_i=\sum_k |V_i||V_k|\left(G_{ik}\cos\delta_{ik}+B_{ik}\sin\delta_{ik}\right)$ and $Q_i=\sum_k |V_i||V_k|\left(G_{ik}\sin\delta_{ik}-B_{ik}\cos\delta_{ik}\right)$ collapse to $$P_i=\sum_{k\neq i}\frac{|V_i||V_k|}{x_{ik}}\sin\delta_{ik}, \qquad Q_i=\sum_{k\neq i}\left(\frac{|V_i|^{2}}{x_{ik}} -\frac{|V_i||V_k|}{x_{ik}}\cos\delta_{ik}\right)$$ with $\delta_{ik}=\delta_i-\delta_k$. Real power is carried by angle difference, reactive power by magnitude difference — the two decouple, which is the whole reason a fast-decoupled solver works.
2. Solve the bus-2 real-power equation for the magnitude. Bus 2 connects only to buses 1 and 3, and $|V_2|$ factors out of both terms: $$P_2=|V_2|\left[\frac{|V_1|\sin(\delta_2-\delta_1)}{x_{12}} +\frac{|V_3|\sin(\delta_2-\delta_3)}{x_{23}}\right]$$ Evaluating the bracket with $\delta_2-\delta_1 = -8.5^\circ$ and $\delta_2-\delta_3 = -40.5^\circ$, $$\frac{1.00\sin(-8.5^\circ)}{0.08} +\frac{1.05\sin(-40.5^\circ)}{0.20} =-1.8459-3.4113=-5.2572$$ so $$|V_2|=\frac{-4.0}{-5.2572} \;\Longrightarrow\;\boxed{|V_2|=0.7609\ \text{pu}}$$
3. Reactive injection at bus 2. Substituting the same angles and the magnitude just found into the $Q$ equation, $$Q_2=\frac{|V_2|^{2}}{x_{12}}+\frac{|V_2|^{2}}{x_{23}} -\frac{|V_2||V_1|\cos(\delta_2-\delta_1)}{x_{12}} -\frac{|V_2||V_3|\cos(\delta_2-\delta_3)}{x_{23}}$$ $$\boxed{Q_2=-2.3129\ \text{pu}}$$ The sign convention is injection, so a negative value means bus 2 absorbs 2.3129 pu of reactive power: the load is drawing about 0.58 pu of lagging vars for every 1 pu of real power, i.e. a power factor of roughly 0.87 lagging at that bus.
4. Part (c) — generation at the slack bus. Bus 1 sees only bus 2, so $$P_1=\frac{|V_1||V_2|}{x_{12}}\sin(\delta_1-\delta_2) =\frac{1.00\times0.7609}{0.08}\sin(8.5^\circ) \;\Longrightarrow\;\boxed{P_1=1.4058\ \text{pu}}$$ $$Q_1=\frac{|V_1|^{2}}{x_{12}}-\frac{|V_1||V_2|}{x_{12}}\cos(\delta_1-\delta_2) =\frac{1}{0.08}-\frac{0.7609}{0.08}\cos(8.5^\circ) \;\Longrightarrow\;\boxed{Q_1=3.0937\ \text{pu}}$$ The slack machine is delivering nearly 3.1 pu of reactive power for 1.4 pu of real power — a consequence of the depressed voltage at bus 2, which forces the reactive support to come from both ends.
5. Part (d) — generation at bus 3. Identically, with $\delta_3-\delta_2 = 40.5^\circ$ and $x_{23} = 0.20$, $$P_3=\frac{1.05\times0.7609}{0.20}\sin(40.5^\circ) \;\Longrightarrow\;\boxed{P_3=2.5942\ \text{pu}}$$ $$Q_3=\frac{1.05^{2}}{0.20}-\frac{1.05\times0.7609}{0.20} \cos(40.5^\circ) \;\Longrightarrow\;\boxed{Q_3=2.4751\ \text{pu}}$$
6. Balance checks. The lines are lossless, so the real powers must sum to zero: $P_1+P_2+P_3 = 1.4058 - 4 + 2.5942 = 0.000000$ pu. The reactive powers must sum to the $I^{2}X$ absorbed by the two reactances: $$\sum Q=3.2559\ \text{pu} \quad\text{versus}\quad |I_{12}|^{2}x_{12}+|I_{32}|^{2}x_{23} =0.9238+2.3321 =3.2559\ \text{pu},$$ with $|I_{12}| = 3.3981$ pu and $|I_{32}| = 3.4148$ pu. Both balances close, which validates every number above.