Question 7 of 7: Transient Stability under a Sustained Fault

Problem 7: Transient Stability under a Sustained Fault (5 + 10 + 5 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 initial power angle; whether the machine stays in synchronism when a three-phase fault at the middle of Line 3 is never cleared; and the maximum angle the rotor reaches during the resulting swing.

Approach. Build the pre-fault and during-fault transfer reactances, draw the two power-angle curves, and apply the equal-area criterion. Because the fault is sustained there is no third (post-fault) curve: the faulted characteristic serves as both the accelerating and the decelerating curve, and the critical angle is read off it.

1. Part (a) — Pre-fault transfer reactance and initial angle. The two identical 0.08 lines are in parallel and the rest is in series: $$X_{pre}=0.03+0.05+\frac{(0.08)(0.08)}{0.08+0.08}=0.03+0.05+0.04=0.12$$ $$P_{\max,1}=\frac{EV}{X_{pre}}=\frac{(1.42)(1.00)}{0.12}=11.8333\ \text{pu}$$ Setting the electrical power equal to the mechanical input at steady state, $$\sin\delta_0=\frac{P_m}{P_{\max,1}}=\frac{2.241}{11.8333}=0.189380 \;\Rightarrow\;\boxed{\delta_0=10.917^{\circ}}$$ The machine is operating well down the stable side of its characteristic, at only 18.9 per cent of its pre-fault pull-out power.
2. Part (b) — Transfer reactance during the fault. A bolted three-phase fault at the mid-point of Line 3 grounds a new node F reached through half of that line, $0.08/2=0.04$, from the parallel bus. The remaining network is a star centred on that bus, with arms $$x_a=0.03+0.05=0.08\ \text{(toward }E\text{)},\quad x_b=0.08=0.08\ \text{(Line 2, toward }V\text{)},\quad x_c=0.04\ \text{(toward F)}$$ Converting this star to a delta between $E$, $V$ and F, the arm that matters is the one directly joining $E$ to $V$: $$X_f=\frac{x_ax_b+x_bx_c+x_cx_a}{x_c} =\frac{(0.08)(0.08)+(0.08)(0.04)+(0.04)(0.08)}{0.04}=0.32$$ The other two delta arms run from $E$ and from $V$ to the faulted node, which is at zero potential; being shunt branches at fixed-voltage nodes they carry no part of the transferred power and are discarded. Hence $$P_{\max,2}=\frac{EV}{X_f}=\frac{1.42}{0.32}=4.4375\ \text{pu}$$ The transfer capability has fallen to 37.5 per cent of its pre-fault value — severe, but note that it has not fallen to zero, which it would have done had the fault been on the bus itself.
3. Test for stability before integrating anything. Because $P_{\max,2}=4.4375$ still exceeds $P_m=2.241$, the faulted characteristic can supply the mechanical input at some angle, so the machine is not automatically lost; if the reverse had been true the answer would be "unstable" with no further work. The two angles that bound the swing are $$\begin{aligned}\delta_1&=\sin^{-1}\frac{P_m}{P_{\max,2}}=\sin^{-1}(0.505014)=30.332^{\circ} \\ \delta_{lim}&=180^{\circ}-\delta_1=149.668^{\circ}\end{aligned}$$ Both are read off the faulted curve, because with a sustained fault that is the only characteristic the machine ever sees again.
4. Evaluate the accelerating area. Between $\delta_0$ and $\delta_1$ the mechanical input exceeds the electrical output and the rotor accelerates: $$A_1=\int_{\delta_0}^{\delta_1}\left(P_m-P_{\max,2}\sin\delta\right)d\delta =P_m(\delta_1-\delta_0)+P_{\max,2}\left(\cos\delta_1-\cos\delta_0\right)$$ With the angles in radians ($\delta_0=0.190531$, $\delta_1=0.529398$), $$A_1=(2.241)(0.338867)+(4.4375)(0.863111-0.981904)=0.23226$$
5. Evaluate the largest available decelerating area. Beyond $\delta_1$ the electrical output exceeds the input and the rotor decelerates, and the area remains available until the angle reaches $\delta_{lim}$, past which the machine would accelerate irrecoverably: $$A_{2,\max}=P_{\max,2}\left(\cos\delta_1-\cos\delta_{lim}\right)-P_m(\delta_{lim}-\delta_1) =2.99257$$
6. Apply the equal-area criterion. $$A_{2,\max}=2.99257\;\gg\;A_1=0.23226$$ $$\boxed{\text{The system remains stable: the available decelerating area exceeds the accelerating area by a factor of }12.9}$$ Physically, the fault is electrically remote enough — it sits at the midpoint of one of two parallel lines, so the healthy line and the far half of the faulted line still deliver power — and the machine was lightly loaded to begin with, at only 18.9 per cent of its pre-fault capability. Both facts work in the same direction.
7. Part (c) — Find the maximum angle of oscillation. The rotor swings until the accelerating and decelerating areas are exactly equal, so the net area from $\delta_0$ to the turning point $\delta_m$ vanishes: $$P_m(\delta_m-\delta_0)+P_{\max,2}\left(\cos\delta_m-\cos\delta_0\right)=0$$ $$(2.241)(\delta_m-0.190531)+(4.4375)(\cos\delta_m-0.981904)=0$$ This is transcendental and is solved numerically (bisection on the interval $[\delta_1,\delta_{lim}]$, in which the left-hand side changes sign exactly once): $$\boxed{\delta_m=0.892748\ \text{rad}=51.151^{\circ}}$$ The swing stops 98.5° short of the limiting angle 149.668°, confirming part (b) from a second direction. In the absence of damping the rotor would then oscillate about 30.332° indefinitely; real damping torque, and in practice the protection that would clear the fault long before the first swing completed, bring it to rest there.

Final results.