Question 7 of 7: Equal-Area Criterion and the Critical Clearing Angle

Problem 7: Equal-Area Criterion and the Critical Clearing Angle (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.

Find. The initial rotor angle, the power-angle curves during and after the fault, and the critical clearing angle beyond which the machine cannot be saved by opening the faulted line.

Approach. Compute the transfer reactance for each of the three network states (pre-fault, during fault, post-fault); the during-fault case needs a delta–star transform because the bolted fault ties an interior point of the network to the reference. Then apply the equal-area criterion, which gives the critical clearing angle in closed form.

1. Part (a) — pre-fault transfer reactance and initial angle. With both lines healthy the path from the machine internal voltage to the infinite bus is $$X_{\text{pre}}=0.20+0.05+\frac{(0.40)(0.40)}{0.40+0.40}+0.05=0.20+0.05+0.20+0.05=0.50\ \text{pu}$$ so the pre-fault power-angle curve is $$P_{\text{pre}}=\frac{EV}{X_{\text{pre}}}\sin\delta=\frac{(1.2)(1.0)}{0.50}\sin\delta =2.400\sin\delta$$ Setting this equal to the 1.0 pu load gives the initial operating point: $$\sin\delta_0=\frac{1.0}{2.400}=0.41667$$ $$\boxed{\delta_0=24.62^{\circ}=0.4297\ \text{rad}}$$
2. Part (b) — transform the faulted network. A bolted three-phase fault at F holds that point at reference potential, so F, the machine neutral and the infinite-bus neutral are all the same node. Call the sending bus $P$ and the receiving bus $Q$. Three branches now form a delta between $P$, $Q$ and the reference $N$: $$X_{PQ}=0.40\ (\text{the healthy line}),\quad X_{PN}=\alpha X=0.08,\quad X_{QN}=(1-\alpha)X=0.32$$ whose perimeter is $0.80$. Converting to a star: $$Z_P=\frac{(0.40)(0.08)}{0.80}=0.040,\quad Z_Q=\frac{(0.40)(0.32)}{0.80}=0.160,\quad Z_N=\frac{(0.08)(0.32)}{0.80}=0.032$$
3. Reduce the resulting T network to a transfer reactance. Adding the generator and transformer reactances to the appropriate star arms, $$X_a=0.25+Z_P=0.290\ \text{pu},\qquad X_b=0.05+Z_Q=0.210\ \text{pu}$$ with $Z_N=0.032$ pu remaining as the shunt arm to the reference. For a T network the transfer reactance between the two source terminals is $$X_{\text{transfer}}=X_a+X_b+\frac{X_aX_b}{Z_N} =0.290+0.210+\frac{(0.290)(0.210)}{0.032}=0.500+1.9031=2.4031\ \text{pu}$$ The shunt arm appears in the denominator, which is why a fault close to the sending bus (a small $Z_N$) is so much more severe than one further out. The during-fault curve is therefore $$\boxed{P_{\text{during}}=\frac{(1.2)(1.0)}{2.4031}\sin\delta=0.4994\sin\delta}$$ Because the peak of this curve, 0.499 pu, is well below the 1.0 pu mechanical power, the machine accelerates throughout the fault and no equilibrium exists while the fault is on — clearing is not optional.
4. Part (c) — post-fault curve. The fault is cleared by opening the faulted line, so line 1 disappears altogether and only line 2 carries the power: $$X_{\text{post}}=0.20+0.05+0.40+0.05=0.70\ \text{pu}$$ $$\boxed{P_{\text{post}}=\frac{(1.2)(1.0)}{0.70}\sin\delta=1.7143\sin\delta}$$ The system survives on this curve because its peak, 1.714 pu, still exceeds the 1.0 pu mechanical input; had it not, the machine would be lost regardless of how fast the breakers operated.
Figure 7.2 — the three power-angle curves with the accelerating area A1 (during the fault) and the decelerating area A2 (after clearing). The critical clearing angle is the abscissa at which the two areas are equal.
5. Part (d) — set up the equal-area balance. The largest angle from which the machine can still be recovered on the post-fault curve is the unstable equilibrium point of that curve: $$\delta_{\max}=180^{\circ}-\sin^{-1}\!\left(\frac{P_m}{P_{\text{post,max}}}\right) =180^{\circ}-\sin^{-1}\!\left(\frac{1.0}{1.7143}\right)=180^{\circ}-35.69^{\circ}$$ $$\delta_{\max}=144.31^{\circ}=2.5187\ \text{rad}$$ The accelerating area between $\delta_0$ and the clearing angle $\delta_{cr}$ must equal the decelerating area between $\delta_{cr}$ and $\delta_{\max}$: $$\int_{\delta_0}^{\delta_{cr}}\left(P_m-P_{2}\sin\delta\right)d\delta =\int_{\delta_{cr}}^{\delta_{\max}}\left(P_{3}\sin\delta-P_m\right)d\delta$$ where $P_2=0.4994$ and $P_3=1.7143$.
6. Solve for the critical clearing angle. Performing both integrations and collecting the terms in $\cos\delta_{cr}$ gives the standard closed form $$\cos\delta_{cr}=\frac{P_m(\delta_{\max}-\delta_0)-P_{2}\cos\delta_0 +P_{3}\cos\delta_{\max}}{P_{3}-P_{2}}$$ with the angles in radians in the first term. Substituting: $$\begin{aligned} P_m(\delta_{\max}-\delta_0) &= 1.0(2.5187-0.4297)=2.0890\\ P_{2}\cos\delta_0 &= (0.4994)(0.90906)=0.45394\\ P_{3}\cos\delta_{\max} &= (1.7143)(-0.81223)=-1.39240 \end{aligned}$$ $$\cos\delta_{cr}=\frac{2.0890-0.45394-1.39240}{1.7143-0.4994} =\frac{0.24265}{1.21494}=0.19972$$ $$\boxed{\delta_{cr}=78.48^{\circ}}$$ Numerically integrating the two areas at this angle returns $A_1=A_2=0.5857$ pu-radians, which closes the equal-area balance and confirms the result. The machine therefore has $78.48^{\circ}-24.62^{\circ}=53.86^{\circ}$ of swing available to it; converting that to a critical clearing time would need the inertia constant, which the question does not supply.

Final Results.