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. A machine of internal emf 1.2 pu feeding an infinite bus of 1.0 pu through a step-up reactance, one series line, and two identical parallel lines.

Find. The pre-fault load angle; whether the machine stays in synchronism for a sustained mid-line fault on line 3; and, if it does, the maximum angle the rotor swings to.

Approach. Compute the pre-fault transfer reactance and hence the pre-fault power-angle curve, get $\delta_0$ from $P_m=P_{\max}\sin\delta_0$, reduce the faulted network by a delta–star–delta pair of transformations to find the during-fault transfer reactance, then apply the equal-area criterion between the two curves.

1. Part (a) — form the pre-fault transfer reactance. Lines 2 and 3 are identical and in parallel, so $$X_{\text{pre}}=x_t+x_1+\tfrac12 x_2=0.03+0.05+0.04=0.12\ \text{pu},$$ $$P_{\max}=\frac{EV}{X_{\text{pre}}}=\frac{(1.2)(1.0)}{0.12}=10.0\ \text{pu}.$$
2. Solve for the initial power angle. In the steady state the electrical output equals the mechanical input: $$3.0=10.0\sin\delta_0\ \Rightarrow\ \sin\delta_0=0.30,$$ $$\boxed{\delta_0=17.458^\circ=0.30469\ \text{rad}}$$ The machine is lightly loaded relative to its pull-out capability — only 30 % of $P_{\max}$ — which is what will give it the margin found below.
3. Part (b) — set up the faulted network. A three-phase short at the mid-point of line 3 splits that line into two halves of $j0.04$ pu and grounds the junction. Calling the sending bus B (after $x_t+x_1=0.08$ pu), the infinite bus C, and the fault point F, the elements between B, C and F form a delta: $$x_{BC}=0.08\ \text{(line 2)},\ x_{BF}=0.04,\ x_{CF}=0.04.$$
4. Convert that delta to a star. With $\sum x=0.08+0.04+0.04=0.16$, $$Z_B=\frac{(0.08)(0.04)}{0.16}=0.02,\ Z_C=\frac{(0.08)(0.04)}{0.16}=0.02,\ Z_F=\frac{(0.04)(0.04)}{0.16}=0.01.$$ Calling the star centre N, the network is now a simple star with $0.08+0.02=0.10$ from the machine to N, $0.02$ from N to the infinite bus, and $0.01$ from N to ground.
5. Convert that star back to a delta to extract the transfer reactance. The E-to-V branch of the equivalent delta is what carries power, and it is $$X_f=\frac{(0.10)(0.02)+(0.02)(0.01)+(0.01)(0.10)}{0.01}=\frac{0.0032}{0.01}=0.32\ \text{pu},$$ $$P_{\max,f}=\frac{(1.2)(1.0)}{0.32}=3.75\ \text{pu}.$$ The grounded star arm $Z_F$ appears in the denominator, which is precisely why a fault electrically closer to the machine (smaller $Z_F$) would throttle the transfer harder.
6. Establish that an equilibrium exists at all during the fault. This is the decisive observation and it is worth making before any integration: $P_{\max,f}=3.75\ \text{pu}\gt P_m=3.0\ \text{pu}$, so the faulted curve still crosses the mechanical input line. It does so at $$\delta_1=\sin^{-1}\frac{3.0}{3.75}=53.130^\circ,\ \delta_{\text{lim}}=180^\circ-53.130^\circ=126.870^\circ.$$ Had $P_{\max,f}$ fallen below 3.0 pu the rotor would have accelerated monotonically and the machine would be unstable with no further calculation needed.
7. Compute the accelerating area. From $\delta_0$ to $\delta_1$ the input exceeds the output, and the area between them is $$A_1=\int_{\delta_0}^{\delta_1}\!\left(P_m-P_{\max,f}\sin\delta\right)d\delta=P_m(\delta_1-\delta_0)+P_{\max,f}(\cos\delta_1-\cos\delta_0),$$ $$A_1=3.0(0.92730-0.30469)+3.75(0.60000-0.95394)=1.86781-1.32727=0.54054.$$
8. Compute the largest decelerating area the faulted curve can supply. Beyond $\delta_1$ the output exceeds the input, and that remains true until $\delta_{\text{lim}}$, past which the machine can never recover: $$A_{2,\max}=\int_{\delta_1}^{\delta_{\text{lim}}}\!\left(P_{\max,f}\sin\delta-P_m\right)d\delta=P_{\max,f}(\cos\delta_1-\cos\delta_{\text{lim}})-P_m(\delta_{\text{lim}}-\delta_1),$$ $$A_{2,\max}=3.75(0.60000+0.60000)-3.0(2.21430-0.92730)=4.50000-3.86100=0.63899.$$
9. Compare the two areas and give the verdict. The available decelerating area exceeds the accelerating area, $$\boxed{A_{2,\max}=0.6390 \gt A_1=0.5405\ \Rightarrow\ \text{the system REMAINS STABLE}}$$ with a stability margin of $A_{2,\max}/A_1=1.18$. Physically, the fault does not remove the transfer path entirely — line 2 and the healthy half of line 3 keep 37.5 % of the pre-fault synchronising power available — and the machine was only 30 % loaded, so the swing is arrested before the rotor reaches the point of no return.
10. Part (c) — find the maximum swing angle by equating the two areas. The rotor stops when the net area vanishes, so $\delta_m$ solves $$P_m(\delta_m-\delta_0)+P_{\max,f}(\cos\delta_m-\cos\delta_0)=0,$$ $$3.0\,\delta_m+3.75\cos\delta_m=3.0(0.30469)+3.75(0.95394)=4.49135.$$ Solving this transcendental equation numerically on the interval $(\delta_0,\delta_{\text{lim}})$ gives $$\boxed{\delta_m=1.89333\ \text{rad}=108.480^\circ}$$ which is comfortably inside the $126.870^\circ$ limit — the same conclusion as step 9, reached by the other route.