Question 4 of 7: Shunt Compensation and a Two-Bus Power Flow

Problem 4: Shunt Compensation and a Two-Bus Power Flow (25 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 circuits

Advantages. A shunt capacitor bank supplies reactive power locally, at the bus where the load consumes it, so that reactive current no longer has to be imported over the line. The immediate consequences are a rise in the local voltage (the whole basis of part b of this question), a reduction in the current the line carries and therefore a reduction in the $I^{2}R$ losses, and a release of thermal capacity in lines and transformers that can then be used to carry real power. Because vars are being generated where they are needed, the sending-end machines are relieved of the duty and can run nearer unity power factor, improving their real-power capability. Capacitors are also, per var, by far the cheapest reactive-power source available: they have no rotating parts, no losses to speak of (a fraction of a per cent), a small footprint, and can be installed in modest steps and switched to follow the daily load curve. In Canadian utility practice, shunt capacitors are the standard first remedy for a distribution feeder or a sub-transmission bus that sags under peak load.

Disadvantages. The output of a capacitor falls with the square of the voltage, $Q_C = V^{2}/X_C$, so it delivers least support exactly when it is needed most — during a voltage collapse or a depressed-voltage fault recovery it withdraws rather than helps, and heavy reliance on shunt capacitors is itself a recognised precursor of voltage instability. The support is also discrete: banks come in fixed steps, so the voltage is corrected in jumps and switching causes transients, inrush currents between paralleled banks, and possible restrike at the breaker. Capacitors resonate with the system inductance, and a resonant frequency landing near the fifth or seventh harmonic will amplify existing harmonic currents, so a harmonic study and often a detuning reactor are required (the relevant guidance is IEEE Std 519 for harmonic limits and IEEE Std 18 for the capacitors themselves). Left in service at light load they cause over-voltage and can drive self-excitation of nearby induction machines or ferroresonance with transformer magnetising branches. Finally, capacitors do nothing for transient stability — their contribution is a voltage-dependent shunt, not a change in the transfer reactance — which is why series compensation or a static var compensator is chosen instead where the dynamic performance is what matters.

Parts (b), (c) and (d) — the two-bus flow

Given. Bus 1 is the slack bus at $V_1 = 1.00\angle 0^{\circ}$ pu. The line admittance is $y = -j10$ pu, so the line is a pure series reactance $z = j0.10$ pu. The net load at bus 2 is $S_D = 3 + j0.75$ pu. The permitted voltage band at bus 2 is 0.95 to 1.05 pu, and a switchable capacitor bank is available there.

Find. The bus-2 voltage magnitude with the bank off and the resulting switching decision; then, with the voltage held at the appropriate limit, the reactive power the bank must supply and the corresponding bus-2 angle.

Approach. Write the complex power injected at bus 2 in terms of $|V_2|$ and $\delta_2$, separate real and imaginary parts, then eliminate $\delta_2$ using $\sin^{2}+\cos^{2}=1$ to leave a quadratic in $|V_2|^{2}$.

1. Part (b) — write the injected power at bus 2. The bus admittance matrix of a two-bus system with a single series branch of admittance $y$ has $Y_{22}=y$ and $Y_{21}=-y$, so the current injected at bus 2 is $I_2=-yV_1+yV_2$ and $$S_2=V_2I_2^{*}=V_2\left(-y^{*}V_1^{*}+y^{*}V_2^{*}\right)$$ With $y=-j10$ (hence $y^{*}=+j10$), $V_1=1.0\angle 0^{\circ}$ and $V_2=V\angle\delta$, the real and imaginary parts separate cleanly: $$P_2=10V\sin\delta,\qquad Q_2=10V^{2}-10V\cos\delta$$ The injected quantities are the negatives of the load, so $P_2=-3.0$ pu and $Q_2=-0.75$ pu with the bank off.
2. Reduce to one quadratic. Rearranging the two equations to isolate the trigonometric terms, $$V\sin\delta=-P_Dx=-0.30,\qquad V\cos\delta=V^{2}+Q_Dx=V^{2}+0.075$$ where $x = 0.10$ pu is the line reactance. Squaring and adding removes $\delta$ entirely. Writing $u=V^{2}$, $$u=(0.30)^{2}+(u+0.075)^{2} \ \Longrightarrow\ u^{2}-0.85u+0.095625=0$$ The discriminant is $0.85^{2}-4(0.095625)=0.340$, so $$u=\frac{0.85\pm\sqrt{0.340}}{2}=0.71655\ \text{or}\ 0.13345$$
3. Take the physical root and decide the switching. The two roots are the two intersections of the load characteristic with the system P–V curve: the upper one is the normal high-voltage operating point, the lower one lies on the far side of the nose of the curve and is not a stable operating condition. Taking the upper root, $$V_2=\sqrt{0.71655}$$ $$\boxed{|V_2|=0.8465\ \text{pu}}$$ This is well below the 0.95 pu floor, so $$\boxed{\text{the capacitor bank must be switched ON}}$$ As a check on the algebra, substituting $|V_2| = 0.8465$ and $\delta_2 = -20.76^{\circ}$ back into the two injection equations reproduces $-3.000$ and $-0.750$ pu.
4. Part (c) — hold the voltage at the appropriate limit. The violated limit is the lower one, so the bank is sized to bring bus 2 exactly to $|V_2| = 0.95$ pu. The real-power equation fixes the angle first, because the capacitor changes no real power: $$\sin\delta_2=\frac{-P_Dx}{V_2}=\frac{-0.30}{0.95}=-0.31579 \ \Longrightarrow\ \cos\delta_2=0.94883$$ The reactive power now injected at bus 2 is $$Q_2=\frac{V_2^{2}-V_2\cos\delta_2}{x} =\frac{(0.95)^{2}-(0.95)(0.94883)}{0.10} =\frac{0.9025-0.90139}{0.10}=0.01112\ \text{pu}$$
5. Size the capacitor. That injection is the difference between what the bank supplies and what the load absorbs, $Q_2 = Q_C - Q_D$, so $$Q_C=Q_2+Q_D=0.01112+0.750$$ $$\boxed{Q_C=0.7611\ \text{pu}}$$ On a 100 MVA base that is 76.1 Mvar. Almost all of it goes to cancelling the load’s own 0.75 pu demand; the remaining 0.011 pu is what the line itself now needs, and the fact that this residue is so small is the signature of a short, purely reactive line.
6. Part (d) — the bus-2 angle. The angle was already determined by the real-power equation in step 4, since the capacitor does not alter the real-power transfer: $$\delta_2=\sin^{-1}(-0.31579)$$ $$\boxed{\delta_2=-18.41^{\circ}}$$ Compensation has moved the angle from $-20.76^{\circ}$ to $-18.41^{\circ}$: the same 3.0 pu of real power now flows across the line at a smaller angular separation, which is a direct improvement in the steady-state stability margin as well as in the voltage.

Final Results.