Question 4 of 7: Exact Two-Bus Power Flow and Reactive Compensation

Question 4: Exact Two-Bus Power Flow and Reactive Compensation (25 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.

Given. A two-bus system with $V_1 = 1.0\angle 0^\circ$ pu held at bus 1 (the slack bus), a purely reactive tie $Z_L = j0.10$ pu, and at bus 2 a load $S_{D2}$ plus a controllable shunt reactive source $Q_{G2}$. Case (a)–(b): $S_{D2} = 2.0 + j1.2$ pu. Case (c)–(d): $S_{D2} = 0.5 - j0.7$ pu, i.e. a capacitive load. Acceptable voltage band $0.95 \le |V_2| \le 1.05$ pu.

Find. The exact voltage magnitude and angle at bus 2 in each case, and the type and amount of reactive generation needed at bus 2 to bring the voltage inside the band.

Approach. The word “exactly” in the question is an instruction not to iterate. With a purely reactive tie the two injection equations can be rearranged into a single quadratic in $u = |V_2|^{2}$, whose upper root is the physical high-voltage solution and whose lower root lies on the far side of the P–V nose. Parts (b) and (d) then run the same equations backwards with $|V_2|$ fixed at the relevant limit.

Taking bus 2's net injection as $S_2 = -S_{D2} + jQ_{G2}$ and writing $\delta$ for the angle of $V_2$ relative to bus 1, the two injections across a reactance $x$ are

Rearranging gives $|V_2|\sin\delta = P_2 x$ and $|V_2|\cos\delta = u - Q_2 x$ with $u = |V_2|^{2}$, and squaring and adding eliminates $\delta$ altogether:

1. Part (a) — assemble the injection and form the quadratic. With $Q_{G2} = 0$ the net injection at bus 2 is $S_2 = -2.0 - j1.2$ pu, so $P_2 = -2.0$ and $Q_2 = -1.2$. Substituting with $x = 0.10$, $$\begin{aligned}u^{2} \\ \quad \\ \quad \\ \quad \\ \quad \\ &\quad -\left[2(-1.2)(0.1)+1\right]u+(0.1)^{2}\left[(-2.0)^{2}+(-1.2)^{2}\right]=0\end{aligned}$$ $$u^{2}-0.76\,u+0.0544=0$$
2. Solve and select the physical root. The discriminant is $0.76^{2}-4(0.0544)=0.5776-0.2176=0.36$, so $$u=\frac{0.76\pm 0.60}{2}=0.6800\ \ \text{or}\ \ 0.0800$$ The upper root is the normal operating solution; the lower one sits on the unstable lower branch of the P–V curve and is discarded. Hence $$\boxed{|V_2|=\sqrt{0.68}=0.8246\ \text{pu}}$$
3. Recover the angle. From the two rearranged relations, $$\begin{aligned}|V_2|\sin\delta &=P_2x=-0.2000 \\ |V_2|\cos\delta &=u-Q_2x=0.68+0.12=0.8000\end{aligned}$$ so $$\boxed{\delta_2=\arctan\!\left(\frac{-0.2000}{0.8000}\right)=-14.04^\circ}$$ The check is immediate: $\sqrt{0.8^{2}+0.2^{2}} = \sqrt{0.68}$, which reproduces the magnitude. A useful by-product is the discriminant itself, 0.36 here — it is a free voltage-stability index, and it would vanish at the nose of the P–V curve.
4. Part (b) — run the equations backwards at the lower limit. At 0.8246 pu the bus is far below the 0.95 pu floor, so reactive support is needed. The cheapest compliant operating point is the floor itself. The real-power equation is unaffected by $Q_{G2}$, so the angle follows first: $$\begin{aligned}\sin\delta &=\frac{P_2x}{|V_2|}=\frac{-0.2000}{0.95}=-0.210526 \\ \Longrightarrow\quad \delta &=-12.15^\circ\end{aligned}$$ Then the reactive injection required at the bus is $$\begin{aligned}Q_2 &= \frac{|V_2|^{2}-|V_2|\cos\delta}{x} = \frac{0.9025-0.928709}{0.10} \\ &= -0.2621\ \text{pu}\end{aligned}$$ Since $Q_2 = Q_{G2} - Q_{D2}$ with $Q_{D2} = 1.2$ pu, $$\boxed{Q_{G2}=Q_2+Q_{D2}=-0.2621+1.2=0.9379\ \text{pu (capacitive)}}$$ This is the minimum; holding 1.00 pu instead would need 1.4020 pu and holding 1.05 pu would need 1.9172 pu, so the admissible range of settings is 0.9379 to 1.9172 pu. A shunt capacitor bank of about 0.94 pu is the answer.
5. Part (c) — repeat with the capacitive load. Now $S_{D2} = 0.5 - j0.7$, so the net injection is $S_2 = -0.5 + j0.7$ pu: the capacitive load makes the reactive injection at bus 2 positive. With $P_2 = -0.5$ and $Q_2 = +0.7$, $$u^{2}-\left[2(0.7)(0.1)+1\right]u+(0.1)^{2}\left[0.25+0.49\right]=0$$ $$u^{2}-1.14\,u+0.0074=0$$ The discriminant is $1.2996-0.0296=1.2700$, so $u = 1.133471$ or $0.006529$, and taking the upper root, $$\boxed{|V_2|=\sqrt{1.133471}=1.0646\ \text{pu}}$$ The angle follows as before from $|V_2|\sin\delta = -0.0500$ and $|V_2|\cos\delta = 1.133471-0.07 = 1.063471$: $$\boxed{\delta_2=-2.69^\circ}$$
6. Part (d) — identify the type of compensation and size it. At 1.0646 pu the bus now sits above the 1.05 pu ceiling, so the requirement has reversed: the bus needs reactive power to be absorbed, not supplied. The device is a shunt reactor (or a synchronous condenser run under-excited, or a static var compensator operating in its inductive range). Fixing $|V_2| = 1.05$ pu, $$\begin{aligned}\sin\delta &=\frac{-0.0500}{1.05}=-0.047619 \\ \Longrightarrow\quad \delta &=-2.73^\circ\end{aligned}$$ $$Q_2=\frac{1.1025-1.048809}{0.10}=+0.5369\ \text{pu}$$ With $Q_{D2} = -0.7$ pu for the capacitive load, $Q_2 = Q_{G2} + 0.7$, so $$\boxed{Q_{G2}=0.5369-0.7=-0.1631\ \text{pu}}$$ The negative sign is the answer: a shunt reactor absorbing 0.1631 pu of reactive power at bus 2 brings the voltage down to exactly 1.05 pu.

The two halves of this question are the same algebra run in opposite directions, and together they make the operating point of the shunt device the whole story. An inductive load pulls the far bus down and calls for capacitors; a capacitive load pushes it up and calls for reactors. That is why a modern static var compensator is built to do both, and why a transmission operator switches capacitor banks in at the evening peak and switches reactors in overnight.