Question 3 of 6: Two-generator supply and load-flow bus types

Question 3: Two-generator supply and load-flow bus types (20 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 69 kV load bus carries a three-phase load of 500 MVA at 0.8 power factor lagging, i.e. $S_L=400+j300$ MVA. Two generators feed that bus, each through a purely reactive line of $j2\ \Omega$ per phase (from the figure). Generator 1 delivers $S_1=240+j180$ MVA to the load bus. The load-bus voltage is taken as the reference, $V=69\angle 0^\circ$ kV line-to-line, i.e. $39\,837\angle 0^\circ$ V per phase.

Find. (I) the phasor current in line 1; (II) the terminal voltage of G1; (III) the reactive power delivered by G2 at its own terminals; and, in part B, the variable assignment that distinguishes a slack bus from a generator bus, plus what to inspect in a converged power-flow solution.

[Figure not reproduced: Figure Q3 — the two-machine system redrawn from the paper. Both lines are purely reactive, so all the loss is reactive and the real power splits without loss. See the official exam paper.]

Approach. Work per phase with the load bus as the angular reference. Each line current follows from the complex power it carries at that bus; the terminal voltage follows from a single voltage-drop equation; and G2's terminal reactive output is the reactive power it delivers to the bus plus the $I^{2}X$ absorbed by its own line.

1. Part A(I) — convert G1's delivered power into a current. Complex power per phase is $S_{1\phi}=V I^{*}$, so with the three-phase quantity and the phase voltage $$I_1=\left(\frac{S_1}{3V_{\phi}}\right)^{\!*} =\left(\frac{(240+j180)\times 10^{6}}{3\times 39\,837\angle 0^\circ}\right)^{\!*}$$ The magnitude of $S_1$ is $\sqrt{240^{2}+180^{2}}=300$ MVA at $\tan^{-1}(180/240)=36.87^\circ$ lagging, so $$|I_1|=\frac{300\times 10^{6}}{\sqrt{3}\times 69\,000}=2\,510.2\ \text{A}$$ $$\boxed{I_1=2\,510.2\angle-36.87^\circ\ \text{A}=2\,008.2-j1\,506.1\ \text{A}}$$ The current lags the load-bus voltage, as it must for a lagging delivery.
2. Part A(II) — add the line drop to the bus voltage. Travelling from the load bus back towards the machine, the terminal voltage is the bus voltage plus the drop across the line impedance: $$V_{t1}=V_{\phi}+jX_{L}I_1=39\,837+j2\,(2\,008.2-j1\,506.1)$$ $$=39\,837+3\,012.3+j4\,016.3=42\,849+j4\,016\ \text{V}$$ $$\boxed{V_{t1}=43.04\angle 5.35^\circ\ \text{kV per phase}=74.54\ \text{kV line-to-line}}$$ The machine therefore runs $8.0\ \%$ above the 69 kV bus, which is what pushing 180 MVAr through $j2\ \Omega$ costs.
3. Part A(III) — find what line 2 must carry. The load bus obeys a complex power balance, so line 2 supplies whatever line 1 does not: $$S_2=S_L-S_1=(400+j300)-(240+j180)=160+j120\ \text{MVA}\;(200\ \text{MVA at }0.8\ \text{pf})$$ The corresponding current has the same power-factor angle as line 1 because the two deliveries happen to share a power factor: $$|I_2|=\frac{200\times 10^{6}}{\sqrt{3}\times 69\,000}=1\,673.5\ \text{A},\qquad I_2=1\,673.5\angle-36.87^\circ\ \text{A}$$
4. Add the reactive loss of line 2 to get the terminal quantity. The question asks for the reactive power at G2's terminals, which is larger than the 120 MVAr arriving at the load bus by the reactive power consumed in the line itself: $$Q_{loss,2}=3\,|I_2|^{2}X_L=3\times(1\,673.5)^{2}\times 2=16.80\ \text{MVAr}$$ $$\boxed{Q_{G2}=120+16.80=136.80\ \text{MVAr}\quad\text{(with }P_{G2}=160\ \text{MW)}}$$ Because the line has no resistance, the real power is unchanged at 160 MW. As a check, the terminal voltage of G2 is $V_{t2}=39\,837+j2\,I_2=41\,931\angle 3.66^\circ$ V per phase (72.63 kV line-to-line), and $3V_{t2}I_2^{*}=160.0+j136.80$ MVA, which reproduces both components exactly.

Part B — bus types in the power-flow formulation. Every bus in a power-flow study carries four quantities: the voltage magnitude $|V|$, the voltage angle $\delta$, the net real injection $P$ and the net reactive injection $Q$. Two of the four are specified as data and the remaining two are computed; which pair is specified is exactly what defines the bus type.

At a swing (slack) bus the voltage magnitude and the voltage angle are the specified quantities — conventionally $1.0$ pu at $0^\circ$ — and the real and reactive injections are the unknowns. Only one such bus exists per island. It is needed for two reasons. It provides the angular reference against which every other bus angle is measured, since the power-flow equations depend only on angle differences. And it absorbs the system's mismatch: the network losses are not known until the solution exists, so the total generation cannot be scheduled in advance, and the slack machine takes up whatever real power (and reactive power) is left over when the iteration converges.

At a generator bus, also called a voltage-controlled or PV bus, the specified quantities are the scheduled real injection $P$ and the regulated voltage magnitude $|V|$; the unknowns are the reactive injection $Q$ and the voltage angle $\delta$. The machine's excitation system is what holds $|V|$, so the reactive output is whatever that regulation demands. Because a real exciter has limits, a PV bus is tested at every iteration against its $Q_{min}$ and $Q_{max}$; if the computed reactive output violates a limit, the bus is converted to a load (PQ) bus with $Q$ fixed at the violated limit and $|V|$ released as a new unknown. For completeness, an ordinary load (PQ) bus specifies $P$ and $Q$ and solves for $|V|$ and $\delta$.

Part B — what to look for in the returned solution. A converged power flow is not the same thing as an acceptable one. At least the following should be checked before the case is believed:

Bus voltage magnitudes. Every bus should sit inside its operating band — typically 0.95 to 1.05 pu on the transmission system, and tighter still at delivery points governed by CSA CAN3-C235 service-voltage limits. A bus that is low signals a reactive deficiency; one that is high signals excess charging on a lightly loaded line.

Branch loadings. Every line and transformer flow should be compared with its continuous thermal rating and, on long lines, with the stability-limited transfer. An overloaded branch is the first thing a contingency will turn into a cascade, so both the base case and the credible single-contingency (N−1) cases must be clean.

Generator reactive output. Each machine's $Q$ should lie inside its capability curve. If several units are pinned at their reactive limits the voltage schedule is only nominally being held, and the system has no reserve left to ride through a disturbance.

Slack-bus real power and losses. The slack machine's output should be a plausible dispatch, not an absorbing sink or an implausibly large source; and the total $I^{2}R$ loss should be a small percentage of the load (typically 2–5 %). A wild slack output or an implausible loss almost always means a data error rather than a system problem.

Angle differences and convergence quality. Angular separations across individual branches should stay modest (a common screening figure is below about $30^\circ$ to $40^\circ$), because a large separation is a warning of weak coupling and poor transient margin. Finally the mismatch tolerance actually achieved, and the number of iterations, should be inspected — a case that only just converged, or one that needed an unusual number of iterations, deserves suspicion.