Question 4 of 7: Shunt Capacitors, and a Linearised Three-Bus Load Flow

Question 4 (Problem 4): Shunt Capacitors, and a Linearised Three-Bus Load Flow (5 + 10 + 10 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) — shunt capacitors on transmission lines. Shunt capacitor banks are the cheapest source of reactive power on a transmission system, and their attractions follow from that one fact. They raise the voltage at the bus where they are installed and along the feeders behind it; they unload the upstream circuits of reactive current, which releases real-power capacity in lines and transformers that were previously carrying magnetising current; they cut the $I^{2}R$ loss in those circuits, because the current magnitude falls even though the real power does not; they improve the power factor seen by the supplier and so avoid reactive-demand charges; and they are modular, cheap per kvar, essentially maintenance free, and can be switched in blocks to follow the daily load curve.

The drawbacks all stem from the fact that a capacitor is a constant-impedance device, not a constant-kvar source. Its output falls with the square of the voltage, so it delivers least reactive support exactly when a depressed system needs it most — the opposite of the behaviour required for voltage stability. It forms a parallel resonance with the inductive source impedance, and if that resonance lands near a load harmonic (typically the 5th or 7th on an industrial bus) the harmonic current is magnified, overheating the capacitors and distorting the bus voltage. Energising a bank draws a large inrush transient, and back-to-back switching of adjacent banks is worse; the transient can also be magnified at a downstream low-voltage capacitor. Under light load an over-compensated line suffers a voltage rise. Finally, capacitors need their own unbalance protection and discharge resistors, and switched banks demand a control scheme that will not hunt.

Given. A three-bus network in which every branch has z_L = j0.5 p.u.; bus 1 is the slack bus at 1.0∠0°, bus 2 holds |V₂| = 1.0 while generating P_G2 = 0.2 p.u., and bus 3 carries a load of 0.3 + j0.1 p.u.

Find. The magnitude |V₃| from the imaginary part of the bus-3 equation, then the two angles θ₂ and θ₃ from the real parts at buses 2 and 3.

Approach. Form the Y-bus, multiply each equation by the conjugate voltage, apply the first-order expansion of the exponential and separate real from imaginary parts.

1. Build the bus admittance matrix. Every branch has the same series admittance $y=1/(j0.5)=-j2.0$, and the three buses are fully interconnected, so $$\begin{gathered}Y_{ik}=-y=+j2.0\ (i\neq k), \\ Y_{ii}=2y=-j4.0\end{gathered}$$ with all shunt elements absent. Writing $b=2.0$ keeps the algebra readable.
2. Part (b) — write the bus-3 equation and multiply out. Starting from the form the question supplies and multiplying both sides by $V_{3}^{*}=V_{3}e^{-j\theta_{3}}$, $$\begin{aligned}&P_{3}-jQ_{3} \\ &=V_{3}e^{-j\theta_{3}}\left[\,jbV_{1}+jbV_{2}e^{j\theta_{2}} -j2bV_{3}e^{j\theta_{3}}\right]\end{aligned}$$ With $V_{1}=|V_{2}|=1.0$ and the small-angle substitution $e^{j\theta}\simeq 1+j\theta$ applied to each exponential, $$\begin{aligned}&P_{3}-jQ_{3} \\ &=jbV_{3}(1-j\theta_{3})+jbV_{3}\!\left[1+j(\theta_{2}-\theta_{3})\right] -j2bV_{3}^{2}\end{aligned}$$
3. Separate the real and imaginary parts. Collecting terms gives one equation of each kind: $$\operatorname{Re}:\;P_{3}=bV_{3}\left(2\theta_{3}-\theta_{2}\right)$$ $$\operatorname{Im}:\;-Q_{3}=2bV_{3}\left(1-V_{3}\right)$$ The imaginary part contains only $V_{3}$, which is precisely why the question directs the magnitude to be taken from it. The load draws 0.3 + j0.1 p.u., so $P_{3}=-0.30$ and $Q_{3}=-0.10$.
4. Solve the quadratic for the bus-3 voltage magnitude. Rearranging the imaginary part, $$\begin{aligned}2bV_{3}^{2}-2bV_{3}-Q_{3}&=0\; \\ &\Longrightarrow\; 4.0V_{3}^{2}-4.0V_{3}+0.10=0\end{aligned}$$ $$\begin{aligned}V_{3}&=\dfrac{4.0\pm\sqrt{4.0^{2}-4(4.0)(0.10)}}{2(4.0)} \\ &=0.974342\ \text{or}\ 0.025658\end{aligned}$$ The question asks for the larger root, and it is also the only physically sensible one — the small root corresponds to the low-voltage branch of the power-flow solution, which no operating system sits on: $$\boxed{|V_{3}|=0.974342\ \text{p.u.}}$$
5. Part (c) — write the bus-2 equation the same way. Repeating the multiplication for bus 2, with $|V_{2}|=1.0$, $$\begin{aligned}&P_{2}-jQ_{2} \\ &=e^{-j\theta_{2}}\left[\,jb-j2be^{j\theta_{2}} +jbV_{3}e^{j\theta_{3}}\right]\end{aligned}$$ and after the same linearisation, $$\begin{gathered}\operatorname{Re}:\;P_{2}=b\left[\theta_{2}(1+V_{3})-V_{3}\theta_{3}\right], \\ \operatorname{Im}:\;Q_{2}=b\left(1-V_{3}\right)\end{gathered}$$ Nothing is connected to bus 2 but the generator, so $P_{2}=P_{G2}=0.20$.
6. Solve the two real equations simultaneously. Substituting $V_{3}=0.974342$ into the two real parts gives a linear pair in the two unknown angles: $$3.948683\,\theta_{2}-1.948683\,\theta_{3}=0.2$$ $$-1.948683\,\theta_{2}+3.897367\,\theta_{3}=-0.3$$ Solving, $$\boxed{\theta_{2}=0.016810\ \text{rad}=0.9632°,\qquad \theta_{3}=-0.068570\ \text{rad}=-3.9288°}$$ The signs are the physical check: bus 2 exports power so its angle leads the slack bus, while bus 3 is the only load and its angle lags both sources. As a by-product the bus-2 reactive output is $Q_{2}=b(1-V_{3})=0.051317$ p.u.