Question 7 of 7: Equal-area criterion and the critical clearing angle

Question 7: Equal-area criterion and the critical clearing angle

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 voltage E behind a transient reactance, feeding an infinite bus through line 1 and then two parallel lines, with a three-phase fault at F on line 3 immediately adjacent to the receiving busbar.

Find. The initial power angle at 1.0 pu load; the power-angle curve while the fault is on; the power-angle curve after line 3 is opened; and the critical clearing angle.

Approach. Each of the three network conditions — pre-fault, during-fault and post-fault — reduces to a single transfer reactance between E and V, and each therefore has its own sinusoidal power-angle curve $P = (EV/X)\sin\delta$. The initial angle comes from the pre-fault curve at $P = P_m$, and the critical clearing angle follows from equating the accelerating area under the during-fault curve to the maximum decelerating area available under the post-fault curve.

1. Part (a), reduce the pre-fault network. With both parallel lines in service they combine in parallel and add to the two series reactances: $$X_{\text{pre}}=0.20+0.05+\frac{(0.50)(0.40)}{0.50+0.40}+0.05=0.20+0.05+0.22222+0.05=0.52222\ \text{pu}$$ $$P_{\max,1}=\frac{EV}{X_{\text{pre}}}=\frac{(1.2)(1.0)}{0.52222}=2.29787\ \text{pu}$$
2. Solve for the initial angle. In the steady state the electrical output equals the mechanical input, $P_e = P_m = 1.0$ pu, so $$\sin\delta_0=\frac{P_m}{P_{\max,1}}=\frac{1.0}{2.29787}=0.43519$$ $$\boxed{\delta_0 = 25.80^\circ = 0.45025\ \text{rad}}$$
3. Part (b), examine the network during the fault. The fault at F sits at the receiving end of line 3, that is at the busbar where lines 2 and 3 come together. A bolted three-phase fault there holds that busbar at zero voltage. Both parallel lines then terminate on a short circuit, and the infinite bus behind the receiving transformer is also connected to that same zero-voltage point, so there is no path from E to V that does not pass through the fault. The transfer reactance is infinite and
   $$\boxed{P_{e,\text{fault}} = 0\ \text{for all }\delta}$$ Physically the machine still delivers current, but into a purely reactive short circuit, so the real power it exports is zero. This is the worst case for stability: the full 1.0 pu of mechanical input goes into accelerating the rotor from the instant the fault appears.
4. Part (c), reduce the post-fault network. Clearing opens line 3 at both ends, removing both the fault and that line. Only line 2 remains in the parallel section: $$X_{\text{post}}=0.20+0.05+0.50+0.05=0.80\ \text{pu}$$ $$P_{\max,3}=\frac{EV}{X_{\text{post}}}=\frac{1.2}{0.80}=1.500\ \text{pu}$$ so the post-fault characteristic is $$\boxed{P_{e,\text{post}}=1.500\sin\delta\ \text{pu}}$$ It still peaks above the 1.0 pu mechanical input, so a stable post-fault equilibrium exists at $\delta = \sin^{-1}(1/1.5) = 41.81^\circ$; had $P_{\max,3}$ fallen below 1.0 pu the machine could not have been stabilised at any clearing time.
5. Part (d), fix the limit of the decelerating swing. The largest angle from which the rotor can still be pulled back is the unstable equilibrium on the post-fault curve, $$\delta_{\max}=180^\circ-\sin^{-1}\frac{P_m}{P_{\max,3}}=180^\circ-41.81^\circ=138.19^\circ=2.41188\ \text{rad}$$ Beyond this the post-fault electrical power again falls below $P_m$ and the rotor accelerates without limit.
6. Write the equal-area condition. The accelerating area between $\delta_0$ and the clearing angle, and the decelerating area between clearing and $\delta_{\max}$, must be equal: $$\int_{\delta_0}^{\delta_{cr}}\left(P_m-0\right)d\delta=\int_{\delta_{cr}}^{\delta_{\max}}\left(P_{\max,3}\sin\delta-P_m\right)d\delta$$ Evaluating both integrals and collecting the terms in $\delta_{cr}$ gives the closed-form result $$\cos\delta_{cr}=\frac{P_m\left(\delta_{\max}-\delta_0\right)+P_{\max,3}\cos\delta_{\max}}{P_{\max,3}}$$ where $\delta_{\max}$ and $\delta_0$ are in radians.
7. Evaluate the critical clearing angle. Substituting the numbers, $$\cos\delta_{cr}=\frac{(1.0)(2.41188-0.45025)+(1.500)(-0.74536)}{1.500}=\frac{1.96163-1.11803}{1.500}=0.56239$$ $$\boxed{\delta_{cr} = 55.78^\circ}$$ Checking the answer by direct integration, the accelerating area is $A_1 = P_m(\delta_{cr}-\delta_0) = 0.52328$ pu-rad and the decelerating area available between $55.78^\circ$ and $138.19^\circ$ is also $0.52328$ pu-rad, so the two balance exactly.
8. Interpret the result. The rotor may swing a further $55.78^\circ - 25.80^\circ = 29.98^\circ$ beyond its initial position before the breakers must have opened. Converting that to a clearing time requires the swing equation, $M\,d^{2}\delta/dt^{2}=P_m$ during a zero-output fault, which integrates to $t_{cr}=\sqrt{4H(\delta_{cr}-\delta_0)/(\pi f_0P_m)}$; for a typical inertia constant $H = 4$ MJ/MVA at 60 Hz this gives 0.211 s, or roughly 13 cycles — comfortably within the reach of modern protection, which clears a transmission fault in three to five cycles.