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

Problem 4: Bus Classification and a Three-Bus Power-Flow Calculation (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 of a conventional power flow. Each bus carries four quantities — $|V|$, $\delta$, $P$ and $Q$ — of which exactly two are specified and two are solved for.

• Slack (swing, reference) bus. Known: $|V|$ and $\delta$ (the angle is fixed at zero to define the reference). Unknown: $P$ and $Q$. There is one such bus per island. It exists because the network losses are not known until the solution is complete, so one generator must be left free to make up whatever real and reactive power the balance demands.
• Voltage-controlled (PV, generator) bus. Known: $P$ (set by the turbine governor) and $|V|$ (held by the automatic voltage regulator). Unknown: $Q$ and $\delta$. If the solved $Q$ violates the machine's var limit, the bus is reclassified as a PQ bus at the violated limit and the solution is repeated.
• Load (PQ) bus. Known: $P$ and $Q$, both taken from the load forecast (negative injections for load). Unknown: $|V|$ and $\delta$. Most buses in a real study are of this type, including every bus with no generation at all.

In this problem bus 1 is the slack bus ($V_1=1.0\angle0^\circ$), bus 3 is a voltage-controlled bus whose magnitude and angle are both stated ($1.05\angle32^\circ$), and bus 2 is the load bus with $P_2=-4$ pu. The question removes the need to iterate by stipulating the bus-2 angle, which turns a Newton–Raphson solution into a set of direct substitutions.

Given. A three-bus system with two purely reactive tie lines, a slack machine at bus 1, a scheduled generator at bus 3, and a 4 pu real load at bus 2 whose angle is stipulated.

Find. $|V_2|$ and $Q_2$; then the real and reactive power generated at bus 1 and at bus 3.

[Figure not reproduced: Figure (2) redrawn. Both ties are purely reactive, so the network is lossless and the two generated real powers must sum to the 4 pu load exactly — a free check on the answer. See the official exam paper.]

Approach. Because both branches are purely reactive the general power-flow equations reduce to their lossless form; with $\delta_2$ stipulated, the bus-2 real-power equation is linear in $|V_2|$, and every remaining quantity is then a direct substitution.

1. Write the lossless injection equations. For a network whose branches are pure reactances, the real and reactive power injected at bus $k$ towards bus $m$ are $$P_{km}=\frac{|V_k||V_m|}{X_{km}}\sin(\delta_k-\delta_m),\ Q_{km}=\frac{|V_k|^{2}}{X_{km}}-\frac{|V_k||V_m|}{X_{km}}\cos(\delta_k-\delta_m).$$ The bus injection is the sum over all lines leaving that bus.
2. Part (b) — solve the bus-2 real-power equation for the voltage magnitude. Bus 2 is connected to buses 1 and 3, and the injection there is $-4$ pu. Both terms are proportional to $|V_2|$, so it factors out: $$-4=|V_2|\left[\frac{1.00}{0.08}\sin(-7^\circ)+\frac{1.05}{0.20}\sin(-7^\circ-32^\circ)\right],$$ $$-4=|V_2|\left[12.5(-0.121869)+5.25(-0.629320)\right]=|V_2|(-1.52336-3.30393),$$ $$\boxed{|V_2|=\frac{-4}{-4.82729}=0.8286\ \text{pu}}$$
3. Substitute back for the reactive injection at bus 2. With $|V_2|$ now known, $$Q_2=|V_2|^{2}\left(\frac{1}{0.08}+\frac{1}{0.20}\right)-\frac{|V_2|(1.00)}{0.08}\cos(-7^\circ)-\frac{|V_2|(1.05)}{0.20}\cos(-39^\circ),$$ $$Q_2=0.686571(17.5)-10.28197-3.38087=12.01500-13.66284,$$ $$\boxed{Q_2=-1.6456\ \text{pu (the bus 2 load absorbs 1.646 pu)}}$$
4. Part (c) — take the injection at the slack bus. Bus 1 has only the one tie, to bus 2, so $$P_1=\frac{|V_1||V_2|}{X_{12}}\sin(\delta_1-\delta_2)=12.5(0.828621)\sin(7^\circ),$$ $$Q_1=\frac{|V_1|^{2}}{X_{12}}-\frac{|V_1||V_2|}{X_{12}}\cos(7^\circ)=12.50000-10.28197,$$ $$\boxed{P_1=1.2623\ \text{pu},\ Q_1=2.2194\ \text{pu}}$$
5. Part (d) — take the injection at bus 3. Bus 3 also has a single tie, to bus 2, and the angle across it is $32^\circ-(-7^\circ)=39^\circ$: $$P_3=\frac{(1.05)(0.828621)}{0.20}\sin39^\circ=4.35026(0.629320),$$ $$Q_3=\frac{(1.05)^{2}}{0.20}-4.35026\cos39^\circ=5.51250-3.38087,$$ $$\boxed{P_3=2.7377\ \text{pu},\ Q_3=2.1317\ \text{pu}}$$
6. Check the balance the lossless network guarantees. With no series resistance anywhere, the generated real powers must equal the load exactly: $$P_1+P_3=1.26229+2.73771=4.00000\ \text{pu}=|P_2|.$$ The reactive powers do not balance, and should not: the series reactances absorb $Q_1+Q_3-|Q_2|=2.21945+2.13171-1.64562=2.70554$ pu, which is $\sum I^{2}X$ in the two lines.
7. Read the result as an operating engineer would. The stipulated $-7^\circ$ angle at bus 2 forces the load-bus voltage down to 0.829 pu, roughly 12 % below the acceptable lower limit of about 0.95 pu. The diagnosis is visible in the numbers: bus 2 draws 4 pu of real power and 1.65 pu of reactive power through two reactances that consume 2.71 pu of vars between them, and there is no reactive source at the load bus at all. Shunt capacitors or a static var compensator at bus 2 — or scheduling bus 3 harder so more of the transfer arrives over the shorter tie — is what a planning study would recommend next.