Question 7 of 7: Transient Stability by the Equal-Area Criterion

Problem 7: Transient Stability by the Equal-Area Criterion (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.

Given.

Find. The pre-fault rotor angle; whether the machine remains in synchronism for a sustained three-phase fault at the mid-point of line 3; and the largest angle the rotor reaches while swinging.

Approach. Reduce the network to a single transfer reactance before and during the fault, form the two power-angle curves, and apply the equal-area criterion — the energy integral of the swing equation — to compare the accelerating and available decelerating areas.

1. Part (a) — pre-fault transfer reactance and initial angle. Lines 2 and 3 are in parallel, so $$X_{I}=0.03+0.05+\frac{0.08\times0.08}{0.08+0.08}=0.03+0.05+0.04 =0.1200\ \text{pu}$$ and the pre-fault power-angle curve is $P_{I}=\dfrac{EV}{X_I}\sin\delta$ with $$P_{\max,I}=\frac{1.25\times1.00}{0.12} =10.4167\ \text{pu}.$$ Setting $P_I = P_m = 3.10$ pu, $$\sin\delta_0=\frac{3.10}{10.4167} =0.29760 \;\Longrightarrow\;\boxed{\delta_0=17.31^\circ =0.3022\ \text{rad}}$$ The machine is loaded to only 30 % of its steady-state pull-out power, so it is comfortably stable in the steady state.
2. Part (b) — transfer reactance during the fault, by a star–delta transformation. A bolted three-phase fault at the mid-point of line 3 holds that point at reference potential. Splitting line 3 into two halves of $j0.04$ pu, the network between the machine and the infinite bus becomes a star whose three arms are $x_a = 0.03+0.05 = 0.08$ pu (machine side to the junction), $x_b = 0.08$ pu (line 2, junction to the bus) and $x_c = 0.04$ pu (junction to the faulted mid-point, i.e. to reference). The delta branch that directly bridges the machine and the bus is the one opposite the arm that goes to reference: $$X_{II}=\frac{x_ax_b+x_bx_c+x_cx_a}{x_c} =\frac{0.0064+0.0032+0.0032}{0.04} =0.3200\ \text{pu}$$ so during the fault $$P_{\max,II}=\frac{1.25\times1.00}{0.32} =3.9062\ \text{pu}.$$ The transfer capability collapses to about 37 % of its healthy value, but — crucially — it does not go to zero, because the fault is at the middle of one line rather than on the bus, and line 2 still provides a path.
3. Locate the two equilibrium angles on the faulted curve. Since $P_m = 3.10\ \text{pu} < P_{\max,II} = 3.9062$ pu, the faulted curve still crosses the mechanical input line, at $$\delta_1=\sin^{-1}\!\frac{3.10}{3.9062} =52.52^\circ \qquad\text{and}\qquad \delta_{\lim}=180^\circ-\delta_1=127.48^\circ .$$ $\delta_1$ is the stable equilibrium of the faulted network and $\delta_{\lim}$ the unstable one; the rotor accelerates below $\delta_1$ and decelerates between $\delta_1$ and $\delta_{\lim}$.

4. Accelerating area. The rotor starts at $\delta_0$ but the electrical power instantly falls to the faulted curve, so from $\delta_0$ to $\delta_1$ the machine accelerates: $$A_1=\int_{\delta_0}^{\delta_1}\left(P_m-P_{\max,II}\sin\delta\right)d\delta =P_m(\delta_1-\delta_0)+P_{\max,II}\left(\cos\delta_1-\cos\delta_0\right)$$ $$A_1=3.10\left(0.9167-0.3022\right) +3.9062\left(0.6084 -0.9547\right) =0.5525\ \text{pu}\cdot\text{rad}$$
5. Maximum decelerating area available. Beyond $\delta_1$ the electrical power exceeds the mechanical input, and the rotor can decelerate at most until it reaches the unstable equilibrium $\delta_{\lim}$: $$A_{2,\max}=\int_{\delta_1}^{\delta_{\lim}} \left(P_{\max,II}\sin\delta-P_m\right)d\delta =P_{\max,II}\left(\cos\delta_1-\cos\delta_{\lim}\right)-P_m(\delta_{\lim}-\delta_1)$$ $$A_{2,\max}=0.6981\ \text{pu}\cdot\text{rad}$$
6. Verdict. Comparing, $$A_{2,\max}=0.6981 \;>\;A_1=0.5525\ \text{pu}\cdot\text{rad}$$ $$\boxed{\text{The system REMAINS STABLE under the sustained fault}}$$ with a margin of $\left(A_{2,\max}-A_1\right)/A_1 = 26.3\,\%$. This is an unusual and instructive result: the machine survives a fault that is never cleared, because line 2 keeps a healthy path to the infinite bus and the pre-fault loading is light enough that the shrunken faulted curve can still absorb the swing. Had the fault been at the sending bus, $X_{II}$ would be infinite, $P_{\max,II}=0$, and no decelerating area would exist at all — the machine would pole-slip on the first swing.
7. Part (c) — maximum angle of oscillation. With no damping the rotor swings until the accelerating and decelerating areas are exactly equal. Setting $A_1(\delta_0\!\to\!\delta_m)=0$ over the whole excursion on the one faulted curve gives the single transcendental equation $$P_m\left(\delta_m-\delta_0\right) =P_{\max,II}\left(\cos\delta_0-\cos\delta_m\right)$$ $$3.10\left(\delta_m-0.3022\right) =3.9062\left(0.9547 -\cos\delta_m\right)$$ Solving by bisection between $\delta_1$ and $\delta_{\lim}$, $$\boxed{\delta_{\max}=105.34^\circ =1.8386\ \text{rad}}$$ which is comfortably short of the 127.48° limit — the geometric statement of the same margin found in step 6. The rotor then swings back, and in a real machine damping would spiral it into the new equilibrium at $\delta_1 = 52.52^\circ$.