Question 4 of 7: Shunt capacitor switching on a two-bus system

Question 4: Shunt capacitor switching on a two-bus system

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 capacitors are the cheapest source of reactive power on a network, and their advantages follow from that. They raise voltage at the point of connection by injecting vars locally instead of importing them over the line, which in turn unloads the line of reactive current, reduces the $I^{2}R$ loss and releases thermal capacity for real power. They correct the power factor seen by the upstream network, which defers reinforcement and, on a distribution utility's tariff, reduces demand charges. They are modular, so banks can be added in steps as load grows, and they can be switched — automatically on voltage, time or var flow — to follow the daily load cycle. Losses are low, roughly 0.1 to 0.5 W per kvar, there are no rotating parts, maintenance is minimal, and installation on an existing substation is straightforward.

The disadvantages are equally characteristic. Capacitor output falls with the square of voltage, $Q_C = BV^{2}$, so the support weakens exactly when a voltage collapse makes it most needed — the opposite of the behaviour wanted from a voltage-control device, and a recognised contributor to voltage instability. The compensation is fixed in steps rather than continuous, so voltage is regulated coarsely unless many small banks are used. Capacitors form a resonant circuit with the system inductance and can amplify harmonics; a bank tuned near a characteristic harmonic of nearby converters will overheat and fail, so detuning reactors and a harmonic study are usually needed. Switching a bank produces a large inrush current and a voltage transient that can trip sensitive loads or reignite in the breaker, and back-to-back switching between adjacent banks is particularly severe. Over-compensation at light load raises voltage too far, adding to the Ferranti rise of Question 1. Finally, capacitors can interact with induction motors on load rejection to give self-excitation, and with series inductance to give ferroresonance.

In practice, then, shunt capacitors are the workhorse of steady-state var supply while dynamic support — static var compensators, STATCOMs, or the synchronous condensers of Question 2 — is reserved for the fast and voltage-independent duty.

Parts (b) to (e) — the two-bus load-flow calculation

Given. The two-bus system of Figure (1), with bus 1 as the slack bus and a switchable capacitor at bus 2.

Find. Whether the capacitor bank must be in service; the resulting bus-2 voltage magnitude and angle; and the real power, reactive power and power factor supplied by the slack bus.

Approach. Write the injected complex power at bus 2 in terms of the two unknowns $x = |V_2|\cos\delta_2$ and $w = |V_2|\sin\delta_2$. Those two equations are linear in $x$ and $w$ once $|V_2|^{2}$ is treated as a third variable, and eliminating $x$ and $w$ leaves one quadratic in $|V_2|^{2}$ — so the flow is solved in closed form, with no iteration, for both the capacitor-out and capacitor-in cases.

1. Build the bus admittance matrix. With a single line and a shunt at bus 2, $$Y_{11}=y_{12}=1-j10,\quad Y_{12}=Y_{21}=-y_{12}=-1+j10,\quad Y_{22}=y_{12}+jB_c$$ The load is a demand, so the injection at bus 2 is $S_2 = -P_D - jQ_D = -4.3 + j0.6336$ pu.
2. Write the injected power in Cartesian form. Putting $V_2 = x + jw$ and $V_1 = 1$ into $S_2 = V_2\!\left(Y_{21}V_1+Y_{22}V_2\right)^{*}$ and separating real and imaginary parts gives the pair $$\begin{aligned} -x+10w+u &= -4.3 \\ -10x-w+(10-B_c)u &= 0.6336 \end{aligned}$$ where $u = |V_2|^{2} = x^{2}+w^{2}$. These are the standard polar power-flow equations rewritten in rectangular coordinates.
3. Eliminate to a single quadratic. The first equation gives $x = 10w+u+4.3$. Substituting into the second removes $x$ and, remarkably, also removes the $10u$ terms, leaving a linear relation between $w$ and $u$: $$-101w - B_cu - 43 = 0.6336\quad\Longrightarrow\quad w = \frac{-B_cu-43.6336}{101}$$ Back-substituting both into $u = x^{2}+w^{2}$ produces one quadratic in $u$, which is solved exactly.
4. Part (b), test the system with the capacitor out. Setting $B_c = 0$ gives $w = -0.432016$ and $x = u - 0.020158$, and the quadratic $u^{2}-1.040316u+0.187044 = 0$ has roots $u = 0.809158$ and $u = 0.231159$. The lower root is the unstable low-voltage solution and is discarded, so $$|V_2| = \sqrt{0.809158} = 0.8995\ \text{pu}$$ This is below the 0.95 pu floor, so $\boxed{\text{the capacitor bank must be switched ON}}$. The margin is not marginal — the bus is 5.3 per cent below the permitted minimum, and heavy: the 4.3 pu load is being drawn through a line of only $z = 1/(1-j10) = 0.00990+j0.09901$ pu.
5. Part (c), solve again with the capacitor in service. With $B_c = 0.83$ the linear relation becomes $w = -0.0082178u-0.4320158$ and $x = 0.9178218u-0.0201584$, and the quadratic $$0.8424644u^{2}-1.0299073u+0.1870440=0$$ has roots $u = 1.000602$ and $u = 0.221887$. Taking the operable high-voltage root, $$\boxed{|V_2| = 1.0003\ \text{pu}}$$ which sits comfortably inside the 0.95 to 1.05 band, so the single 0.83 pu bank is sufficient and no second step is needed.
6. Part (d), recover the angle. From the converged root, $w = -0.440239$ and $x = 0.898216$, so $$\delta_2=\tan^{-1}\frac{w}{x}=\tan^{-1}\frac{-0.440239}{0.898216}$$ $$\boxed{\delta_2 = -26.11^\circ}$$ The negative sign is the expected direction: bus 2 lags the slack bus because real power flows from bus 1 to bus 2. The magnitude is large for a single line, which again reflects how heavily loaded it is.
7. Part (e), compute the slack-bus injection. The current leaving bus 1 is $I_1 = Y_{11}V_1+Y_{12}V_2$, which evaluates to $$I_1=(1-j10)(1)+(-1+j10)(0.898216-j0.440239)=4.50417-j0.57760\ \text{pu}$$ so, with $V_1 = 1\angle0^\circ$, $$S_1=V_1I_1^{*}=\boxed{P_1 = 4.5042\ \text{pu},\; Q_1 = 0.5776\ \text{pu}}$$ and the slack-bus power factor is $$\cos\varphi_1=\cos\left(\tan^{-1}\frac{0.5776}{4.5042}\right)=\boxed{0.9919\ \text{lagging}}$$
8. Close the balance as a check. The real power delivered exceeds the load by $4.5042-4.3000 = 0.2042$ pu, and the line loss computed independently is $|I|^{2}R = (20.621)(0.009901) = 0.20417$ pu, which agrees. On the reactive side the line absorbs $|I|^{2}X = 2.0417$ pu, the capacitor supplies $B_c|V_2|^{2} = 0.8305$ pu and the load returns 0.6336 pu, and $2.0417 - 0.8305 - 0.6336 = 0.5776$ pu is exactly the slack-bus reactive output. Both balances closing is strong evidence the quadratic root was the right one.