Question 4 of 7: The Slack Bus, and a Three-Bus Power-Flow Solution

Question 4: The Slack Bus, and a Three-Bus Power-Flow Solution (20 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) — why the mathematical reference bus is called the slack bus.

The mathematical reason for fixing one voltage angle is trivial: the power-flow equations depend only on angle differences, so the solution is undetermined up to an arbitrary common rotation and one angle must be pinned down. Fixing the magnitude at the same bus removes a second degeneracy and gives the per-unit system a voltage datum. Neither of those explains the name.

The name comes from a physical constraint that cannot be satisfied in advance. Conservation of power requires that total generation equal total load plus total transmission losses. The load is known from the forecast, but the losses are quadratic in the line currents and therefore are not known until after the flow has been solved — they typically run to two or three per cent of the load and vary with the dispatch itself. If the scheduler tried to specify the active power output of every generator, the specified quantities would over-determine the problem and, except by accident, would be inconsistent with the loss that the resulting flow actually produces. One machine must therefore be left unscheduled so that it can take up whatever difference remains — it takes up the slack. Its active power output is an output of the study, not an input.

The same bus must also absorb the reactive mismatch, because the reactive losses in the series reactances and the reactive generation of the line charging are equally unknown before the solution. Practically, therefore, the slack bus is chosen to be a bus that can genuinely behave this way: a large machine or plant with substantial spinning reserve in both directions, a stiff bus with a high short-circuit level so that holding $|V|$ constant is realistic, and, in an interconnected study, usually the tie to the neighbouring system, which really can supply or absorb the residue. Choosing a small or remote machine as slack produces a numerically valid but physically meaningless answer, since the computed slack output may exceed the machine’s rating. In a Canadian interconnected study the slack is typically placed at a major hydro plant or at the interconnection with the adjacent control area.

Given. A three-bus network in which all three lines have the same series admittance, with the bus data of Figure (2) as tabulated below.

Find. The voltage magnitude at bus 2, and then the active power injected into the network at bus 2.

Approach. Build the bus admittance matrix, which is purely imaginary because the network is lossless, then write the reactive-power injection at bus 2. Because every angle in the network is already given, that equation contains $|V_2|$ as its only unknown and is a plain quadratic; the active power then follows by substitution.

1. Part (b) — assemble the bus admittance matrix. Each line contributes $y_L=-j8.0$ pu, so the off-diagonal entries are $Y_{ik}=-y_L=+j8.0$ and each diagonal is the negative sum of the two lines meeting there, $Y_{ii}=-j16.0$. Writing $Y_{ik}=G_{ik}+jB_{ik}$, $$B=\begin{bmatrix}-16&8&8\\ 8&-16&8\\ 8&8&-16\end{bmatrix}\ \text{pu}\qquad G=0$$ The absence of any resistance is what makes the problem tractable by hand, and it also means the network consumes no real power — a fact used as a closure check at the end.
2. Write the reactive injection at bus 2. For a network with $G=0$ the polar power-flow equations reduce to $$P_i=\sum_k |V_i||V_k|B_{ik}\sin(\theta_i-\theta_k) \qquad Q_i=-\sum_k |V_i||V_k|B_{ik}\cos(\theta_i-\theta_k)$$ Applying the second to bus 2 and separating the self term, $$Q_2=-\left[B_{22}|V_2|^{2}+B_{21}|V_2||V_1|\cos\theta_{21}+B_{23}|V_2||V_3|\cos\theta_{23}\right]$$
3. Substitute the given angles and reduce to a quadratic in the one unknown. The angle differences are $\theta_{21}=-8.0^{\circ}-0.0^{\circ}=-8.0^{\circ}$ and $\theta_{23}=-8.0^{\circ}-7.5^{\circ}=-15.5^{\circ}$, whose cosines are $0.990268$ and $0.963630$. Hence $$-2.1=16|V_2|^{2}-\left[8(1.000)(0.990268)+8(1.100)(0.963630)\right]|V_2|$$ $$16|V_2|^{2}-16.402093|V_2|+2.1=0$$ Note that the unknown appears linearly in the two mutual terms and quadratically only in the self term — this is why giving the angle turns a normally iterative problem into a one-line algebraic one.
4. Solve, and choose the physical root. The discriminant is $16.402093^{2}-4(16)(2.1)=269.0287-134.4=134.6286$, whose square root is $11.60295$, so $$|V_2|=\frac{16.402093\pm11.60295}{32}=0.875158\ \ \text{or}\ \ 0.149973$$ $$\boxed{|V_2|=0.87516\ \text{pu}}$$ The upper root is the operating solution; the lower one is the unstable low-voltage solution on the far side of the nose of the P-V curve, at which the same reactive demand would be met by an enormous current. That the discriminant is comfortably positive is itself a free voltage-stability indication — had it been near zero the bus would have been at its loadability limit.
5. Interpret the depressed voltage. A magnitude of 0.875 pu is well below the neighbouring buses at 1.000 and 1.100 pu, and the reason is the reactive injection of $-2.1$ pu: bus 2 is absorbing 2.1 pu of reactive power through two lines of only $0.125$ pu reactance each. Such a bus would in practice carry shunt compensation, and this calculation is exactly how the size of that compensation would be settled.
6. Part (c) — active power injected at bus 2. Substituting the solved magnitude into the real-power equation, and noting that the self term vanishes because $\sin 0=0$, $$P_2=|V_2|\left[|V_1|B_{21}\sin\theta_{21}+|V_3|B_{23}\sin\theta_{23}\right]$$ $$\begin{aligned} P_2&=0.875158\left[(1.000)(8)(-0.139173)+(1.100)(8)(-0.267238)\right] \\ &=0.875158(-3.465083) \end{aligned}$$ $$\boxed{P_2=-3.0325\ \text{pu}}$$ The negative sign is correct and expected: bus 2 lags both of its neighbours, so real power flows into it, and as an injection that is a negative number. Bus 2 is a load bus drawing 3.03 pu of active power and 2.10 pu of reactive power.
7. Close the whole table with two conservation identities. Completing the injections at the other two buses gives $S_1=-0.1742+j0.3422$ pu and $S_3=3.2067+j3.2140$ pu. The first check is that $\sum P_i=0$ exactly, because the network is lossless; the computed sum is $4\times10^{-16}$ pu. The second is that $\sum Q_i$ must equal the reactive absorption of the three series reactances, $\sum |I_{ik}|^{2}x_{ik}$; the injections sum to $1.45615$ pu and the line currents of $1.44487$, $1.35815$ and $2.77795$ pu give $0.26096+0.23057+0.96462=1.45615$ pu. Both identities validate all six injections at once, not merely the two the question asked for.