Question 7 of 7: Transient stability under a sustained fault — equal-area criterion

Question 7: Transient stability under a sustained fault — equal-area criterion (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.

Given. A single machine on an infinite bus, with the pre-fault, during-fault and (because the fault is sustained) post-fault power-angle characteristics as follows.

Find. (a) the pre-fault rotor angle; (b) whether synchronism survives the sustained fault, with the equal-area argument that settles it.

Approach. Set the pre-fault electrical output equal to the mechanical input to obtain $\delta_0$, locate the new equilibrium $\delta_1$ on the faulted curve, and compare the accelerating area between $\delta_0$ and $\delta_1$ with the largest decelerating area the faulted curve can supply before the rotor reaches $\delta_{\max}=180^\circ-\delta_1$.

1. Find the initial operating angle. Before the fault the machine is in equilibrium, so the electrical output equals the mechanical input: $$P_m=P_{\max,1}\sin\delta_0\quad\Longrightarrow\quad \sin\delta_0=\frac{4.5}{12.50}=0.36000$$ $$\boxed{\delta_0=21.100^\circ=0.90975\ \text{rad above the origin, i.e. }0.36827\ \text{rad}}$$ The machine is running at less than a third of its pull-out power, which is a comfortable steady-state margin and says nothing yet about transient behaviour.
2. Establish which curves apply. The fault is sustained: it is never cleared, so no third characteristic ever appears and the machine lives on the faulted curve $4.7\sin\delta$ for all time after inception. This is the fact that decides the whole question, and it is worth checking in the wording before anything else, because on a cleared fault the post-fault curve returns towards $12.50\sin\delta$ and the answer would be entirely different.
3. Locate the faulted-curve equilibrium and the limiting angle. On the faulted curve the rotor would be in equilibrium where $$4.7\sin\delta_1=4.5\quad\Longrightarrow\quad \delta_1=73.225^\circ$$ Because $P_{\max,2}=4.7>P_m=4.5$, such an equilibrium exists at all — had the faulted peak been below 4.5 pu the machine would accelerate monotonically and be lost with no integration needed. The stable equilibrium is $\delta_1$; its unstable mirror image, past which the rotor can never be recovered, is $$\delta_{\max}=180^\circ-\delta_1=106.775^\circ$$
4. Compute the accelerating area. Between $\delta_0$ and $\delta_1$ the mechanical input exceeds the electrical output, so the rotor gains kinetic energy: $$A_1=\int_{\delta_0}^{\delta_1}\!\!\left(P_m-P_{\max,2}\sin\delta\right)d\delta =P_m(\delta_1-\delta_0)+P_{\max,2}\left(\cos\delta_1-\cos\delta_0\right)$$ With $\delta_1-\delta_0=0.90975$ rad, $\cos\delta_0=0.93295$ and $\cos\delta_1=0.28861$, $$A_1=4.5(0.90975)+4.7\,(0.28861-0.93295)=\boxed{A_1=1.06548\ \text{pu}\cdot\text{rad}}$$ The angle difference here must be in radians; using degrees in the $P_m\,\Delta\delta$ term is the single most common arithmetic error in this calculation.
5. Compute the largest available decelerating area. Beyond $\delta_1$ the electrical output exceeds the input and the rotor decelerates, but only until $\delta_{\max}$, after which the output falls below the input again and recovery becomes impossible: $$A_{2,\max}=\int_{\delta_1}^{\delta_{\max}}\!\!\left(P_{\max,2}\sin\delta-P_m\right)d\delta =2P_{\max,2}\cos\delta_1-P_m\left(\delta_{\max}-\delta_1\right)$$ using $\cos\delta_{\max}=-\cos\delta_1$. Substituting, $$A_{2,\max}=2(4.7)(0.28861)-4.5(106.775-73.225)\frac{\pi}{180} =\boxed{A_{2,\max}=0.07796\ \text{pu}\cdot\text{rad}}$$
6. Apply the equal-area criterion. Stability requires that the decelerating area available be at least as large as the accelerating area already accumulated. Here $$A_1=1.06548\ \gg\ A_{2,\max}=0.07796\ \text{pu}\cdot\text{rad}$$ — the accelerating area is more than thirteen times the largest area available to absorb it. Therefore $$\boxed{\text{The system will NOT remain stable under a sustained fault.}}$$ The rotor sweeps past $\delta_{\max}=106.775^\circ$ still travelling forward, the electrical output then falls below the mechanical input again, acceleration resumes without bound, and synchronism is lost on the first swing. Numerical integration of the two areas at the answer returns 1.06548 and 0.07796 pu·rad, confirming the closed forms.
7. Quantify what would have saved it. Two contrasts make the result concrete rather than merely negative. If the fault could not be cleared, the faulted-curve peak would have to be raised to at least 5.416 pu — more than the machine's own steady load of 4.5 pu by a wide margin — before $A_{2,\max}$ would match $A_1$. If instead the fault were cleared by tripping the faulted circuit and the full $12.50\sin\delta$ characteristic were restored, the critical clearing angle would be $$\cos\delta_{cr}=\frac{P_m(\delta_{\max,3}-\delta_0)+P_{\max,3}\cos\delta_{\max,3} -P_{\max,2}\cos\delta_0}{P_{\max,3}-P_{\max,2}} \quad\Longrightarrow\quad \delta_{cr}=132.05^\circ$$ which is a generous margin. The engineering conclusion is the standard one: this machine is not saved by any amount of inertia, only by speed of clearing, which is precisely why transmission protection is specified in cycles.