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

Problem 4: Bus Types and a Three-Bus Power Flow (5 + 5 + 5 + 5 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) — Bus types in the conventional power-flow formulation

Every bus in a power-flow study carries four variables — 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, and the classification exists because the power-flow equations are nonlinear and need exactly $2n$ independent specifications to be determinate.

There is exactly one slack bus in a study, and it is needed for two reasons. The angles are only meaningful relative to something, so one bus must supply the angular reference; and the network losses are unknown until the solution is found, so one generator must be left free to absorb the mismatch between total generation and total load plus losses. In practice the largest and most stable machine in the system is chosen. A voltage-controlled bus reverts to a load bus during the iteration if its computed reactive output would exceed the generator's limit — the machine can no longer hold the scheduled voltage, so $Q$ is fixed at the limit and $|V|$ is released. In this problem bus 1 is the slack bus, bus 3 has both its magnitude and its angle stated (so it behaves as a second reference), and bus 2 is a load bus whose angle has been given to the candidate, which is what makes the whole system solvable in closed form instead of by Newton–Raphson iteration.

Parts (b), (c) and (d) — Solving the three-bus network

Given.

Find. The voltage magnitude and the reactive injection at bus 2, and the real and reactive power generated at buses 1 and 3.

Approach. Build the bus admittance matrix, write the injection equations in polar form, and exploit the fact that with $\delta_2$ given the real-power equation at bus 2 is linear in $|V_2|$ — so no iteration is needed anywhere in this problem.

1. Assemble the bus admittance matrix. With pure series reactances and no shunt elements the branch admittances are $$\begin{aligned}y_{12}&=\frac{1}{j0.20}=-j5.0 \\ y_{23}&=\frac{1}{j0.08}=-j12.5\end{aligned}$$ so, using $Y_{ii}=\sum y_{ik}$ and $Y_{ik}=-y_{ik}$, the susceptance entries needed below are $$B_{11}=-5.0,\quad B_{12}=B_{21}=+5.0,\quad B_{22}=-17.5,\quad B_{23}=B_{32}=+12.5,\quad B_{33}=-12.5$$
2. Part (b) — Write the real-power injection at bus 2. For a purely reactive network the general polar injection equation $P_i=\sum_k |V_i||V_k|\left(G_{ik}\cos\theta_{ik}+B_{ik}\sin\theta_{ik}\right)$ collapses to the sine terms only. The self term vanishes because $\sin 0 = 0$, leaving $$P_2=|V_2|\left[|V_1|B_{21}\sin(\delta_2-\delta_1) +|V_3|B_{23}\sin(\delta_2-\delta_3)\right]$$ Every quantity in the bracket is known, so the equation is linear in $|V_2|$.
3. Solve for the voltage magnitude at bus 2. Substituting $\delta_2=-5^{\circ}$ and $\delta_3=32^{\circ}$, $$\begin{aligned} |V_1|B_{21}\sin(-5^{\circ})&=(1.0)(5.0)(-0.087156)=-0.435779\\ |V_3|B_{23}\sin(-37^{\circ})&=(1.05)(12.5)(-0.601815)=-7.898822 \end{aligned}$$ $$-4.0=|V_2|(-8.334601)\;\Rightarrow\;\boxed{|V_2|=0.4799\ \text{pu}}$$ The result is low, and it should be: 4.0 pu is a heavy load to draw through reactances of $j0.20$ and $j0.08$, and the deep angular spread already implied by $\delta_3=32^{\circ}$ is the signature of a stressed network. It is the correct answer to the question as posed, not an arithmetic slip.
4. Compute the reactive injection at bus 2. The companion equation for a lossless network is $Q_i=-\sum_k |V_i||V_k|B_{ik}\cos\theta_{ik}$, and here the self term does not vanish: $$\begin{aligned} Q_2&=-\left[B_{21}|V_2||V_1|\cos(-5^{\circ})+B_{22}|V_2|^{2} +B_{23}|V_2||V_3|\cos(-37^{\circ})\right]\\ &=-\left[2.390504+(-4.030774)+5.030639\right] \end{aligned}$$ $$\boxed{Q_2=-3.3904\ \text{pu}}$$ The negative sign means bus 2 absorbs 3.390 pu of reactive power, so the load there is 0.763 power factor lagging — unsurprising at this voltage, since the reactive demand of the two lines is charged against the load bus.
5. Part (c) — Evaluate the generation at bus 1. Bus 1 connects only to bus 2, so its injections involve a single off-diagonal term: $$\begin{aligned} P_1&=|V_1||V_2|B_{12}\sin(\delta_1-\delta_2)=(1.0)(0.4799)(5.0)\sin(5^{\circ})=0.2091\\ Q_1&=-\left[B_{11}|V_1|^{2}+B_{12}|V_1||V_2|\cos(5^{\circ})\right] =-\left[-5.0+2.390504\right]=2.6095 \end{aligned}$$ $$\boxed{\begin{aligned}P_1&=0.2091\ \text{pu} \\ Q_1&=2.6095\ \text{pu}\end{aligned}}$$ Bus 1 delivers almost no real power — its angle is only 5° ahead of bus 2 across a comparatively large reactance — but it supplies substantial reactive power, which is what holds its own terminal voltage at 1.0 pu against the depressed voltage next door.
6. Part (d) — Evaluate the generation at bus 3. By the same pair of equations across the $j0.08$ line: $$\begin{aligned} P_3&=|V_3||V_2|B_{32}\sin(\delta_3-\delta_2)=(1.05)(0.4799)(12.5)\sin(37^{\circ})=3.7909\\ Q_3&=-\left[B_{33}|V_3|^{2}+B_{32}|V_3||V_2|\cos(37^{\circ})\right]=8.7506 \end{aligned}$$ $$\boxed{\begin{aligned}P_3&=3.7909\ \text{pu} \\ Q_3&=8.7506\ \text{pu}\end{aligned}}$$ Bus 3 carries essentially the whole load, as its 37° lead over bus 2 across the stiffer line demands.
7. Check the two balances. The network is lossless in real power, so the three real injections must sum to zero, and the reactive injections must sum to the reactive absorbed by the two line reactances: $$P_1+P_2+P_3=0.2091-4.0+3.7909=-0.000000\ \checkmark$$ $$Q_1+Q_2+Q_3=7.9697\quad\text{versus}\quad I_{12}^{2}x_{12}+I_{23}^{2}x_{23}=1.3706+6.5991=7.9697\ \checkmark$$ Both close to the last figure carried, which confirms the admittance signs and the angle conventions simultaneously — the single most valuable check available in a power-flow calculation.

Final results.