Question 4 of 7: Bus types, fast-decoupled power flow, and a three-bus solution

Question 4: Bus types, fast-decoupled power flow, and a three-bus solution (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) — the three bus types. Every bus in a power-flow study carries four variables — voltage magnitude $|V|$, voltage angle $\delta$, net injected real power $P$ and net injected reactive power $Q$ — of which exactly two are specified and two are computed. Which pair is specified defines the bus type.

Part (b) — fast-decoupled power flow. The fast-decoupled method is a simplification of Newton–Raphson that exploits two properties of transmission networks. First, transmission branches are strongly inductive ($X\gg R$), so real power is governed almost entirely by voltage angles and reactive power almost entirely by voltage magnitudes: the off-diagonal blocks $\partial P/\partial|V|$ and $\partial Q/\partial\delta$ of the Jacobian are small and are set to zero. Second, in normal operation the angle differences across branches are small and the magnitudes are close to 1 pu, so $\cos\delta_{ik}\approx1$, $\sin\delta_{ik}\approx0$ and $|V_i|\approx1$.

Applying both approximations reduces the coupled Newton update to two independent, real, constant, symmetric systems $$\Delta P/|V| = B'\,\Delta\delta,\qquad \Delta Q/|V| = B''\,\Delta|V|$$ in which $B'$ and $B''$ are built from the network susceptances alone. Because they never change, they are factorised once at the start and re-used at every iteration, so each iteration costs only a forward-and-back substitution instead of a fresh Jacobian and a fresh factorisation. The method converges more slowly per iteration than full Newton (geometrically rather than quadratically) but each iteration is roughly an order of magnitude cheaper, which is why it remains the workhorse for contingency screening and real-time security analysis, where thousands of cases must be solved per minute. Its accuracy degrades on distribution-level networks where $R/X$ is large, and on heavily loaded systems where angle differences are no longer small.

Given. The three-bus system of Figure (2), with lossless series reactances $x_{12}=0.10$ pu and $x_{23}=0.08$ pu and no shunt elements, and the partially completed table.

Find. The missing voltage magnitude at bus 2 and all five missing power injections.

Approach. Because the paper supplies every angle and two of the three magnitudes, no iteration is required at all: the real-power injection at bus 2 is linear in $|V_2|$ on a lossless network, so one division gives $|V_2|$, after which the remaining injections follow by direct substitution into the power-flow equations.

1. Write the injection equations for a lossless network. With every branch a pure reactance, $G_{ik}=0$ and $B_{ik}=-1/x_{ik}$ for $i\ne k$, so the standard polar power-flow equations collapse to $$P_i=|V_i|\sum_{k\ne i}\frac{|V_k|}{x_{ik}}\sin(\delta_i-\delta_k)$$ $$Q_i=|V_i|\left[\sum_{k\ne i}\frac{|V_i|}{x_{ik}}-\sum_{k\ne i}\frac{|V_k|}{x_{ik}}\cos(\delta_i-\delta_k)\right]$$ The self-susceptance contributes nothing to $P$, which is the structural fact that makes the next step a one-liner.
2. Solve for the bus-2 voltage magnitude. Applying the $P$ equation at bus 2, where $\delta_2-\delta_1=-3^\circ$ and $\delta_2-\delta_3=-8^\circ$, $$\begin{aligned} P_2&=|V_2|\left[\frac{|V_1|}{x_{12}}\sin(-3^\circ)+\frac{|V_3|}{x_{23}}\sin(-8^\circ)\right]\\ &=|V_2|\left[-0.523360+-1.774457\right]=-2.297817\,|V_2| \end{aligned}$$ Because $|V_2|$ factors out completely, setting $P_2=-2.4$ gives it directly: $$\boxed{|V_2|=\frac{-2.4}{-2.297817}=1.044470\ \text{pu}}$$ No Gauss–Seidel or Newton iteration is needed; the paper has removed the nonlinearity by handing over $\delta_2$.
3. Compute the slack-bus injection. With every voltage now known, bus 1 follows by substitution: $$P_1=|V_1|\frac{|V_2|}{x_{12}}\sin(\delta_1-\delta_2)=0.5466\ \text{pu}$$ $$Q_1=|V_1|\left[\frac{|V_1|}{x_{12}}-\frac{|V_2|}{x_{12}}\cos(\delta_1-\delta_2)\right]=-0.4304\ \text{pu}$$ so $\boxed{P_1=0.5466\ \text{pu},\ Q_1=-0.4304\ \text{pu}}$. The slack bus exports real power and absorbs reactive power, the latter because bus 2 is running above 1.00 pu and pushes reactive power back along the line.
4. Compute the bus-2 reactive injection and the bus-3 injections. The same two formulas applied at buses 2 and 3 give $$Q_2=0.9279\ \text{pu}$$ $$P_3=1.8534\ \text{pu},\qquad Q_3=-0.1824\ \text{pu}$$ so $\boxed{Q_2=0.9279,\quad P_3=1.8534,\quad Q_3=-0.1824\ \text{(pu)}}$. Bus 3 has the largest angle and so is the strongest exporter of real power; it, like bus 1, absorbs reactive power.
5. Close the table with the two conservation identities. A lossless network must satisfy $\sum_i P_i=0$ exactly, and here $0.5466-2.4+1.8534=0.0000$. The reactive sum is not zero but must equal the reactive absorbed by the series reactances, $$\sum_i Q_i=\sum_{\text{branches}}|I|^{2}x$$ The branch currents are $|I_{12}|=0.6957$ and $|I_{23}|=1.8258$ pu, giving $0.6957^{2}(0.10)+1.8258^{2}(0.08)=0.3151$ pu, which matches $-0.4304+0.9279+-0.1824=0.3151$ pu. Both identities validate all five computed injections at once, and neither uses any information beyond the completed table.