Question 6 of 6: Transient stability by the equal-area criterion

Question 6: Transient stability by the equal-area criterion (20 marks)

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. Generator transient reactance $X_d'=j0.5$ pu behind an internal voltage $E'=1.5$ pu; step-up transformer $j0.1$ pu; two parallel lines of $j0.2$ pu each; infinite bus at $V=1.0$ pu. Mechanical input (equal to the pre-fault electrical output) $P_m=0.6$ pu. A solid three-phase fault occurs at the sending end of one line and is cleared by opening that line at both ends, leaving one line in service.

Find. The initial power angle; the three power-angle characteristics drawn on one set of axes with $P_m$, $\delta_0$, $\delta_{max}$ and the final angle marked; the equal-area argument on that diagram; and short written answers on critical clearing time, the link between power flow and stability studies, and circuit-breaker arcing.

Approach. Reduce the network to a single transfer reactance in each of the three conditions — pre-fault, during-fault and post-fault — take the power-angle curve in each, then apply the equal-area criterion to find the critical clearing angle.

1. Reduce the pre-fault network. The two lines are in parallel, so $$X_1=X_d'+X_T+\frac{X_L}{2}=0.5+0.1+\frac{0.2}{2}=0.70\ \text{pu}$$ $$P_{e1}=\frac{E'V}{X_1}\sin\delta=\frac{1.5\times 1.0}{0.70}\sin\delta=2.1429\sin\delta$$
2. Take the initial power angle. Before the fault the electrical output equals the mechanical input, so $$\sin\delta_0=\frac{P_m}{P_{max,1}}=\frac{0.6}{2.1429}=0.28 \quad\Longrightarrow\quad \boxed{\delta_0=16.26^\circ=0.2838\ \text{rad}}$$
3. Reduce the during-fault network. The fault is drawn at the sending end of the lower line, electrically on the bus that both lines leave. A solid three-phase fault holds that bus at zero volts, so there is no path from $E'$ to the infinite bus that avoids the short circuit: $$\boxed{P_{e2}=0\ \text{for all }\delta\ \text{during the fault}}$$ The whole of the mechanical input therefore accelerates the rotor for as long as the fault persists.
4. Reduce the post-fault network. Clearing removes the faulted line entirely, leaving one circuit: $$X_3=X_d'+X_T+X_L=0.5+0.1+0.2=0.80\ \text{pu},\qquad P_{e3}=\frac{1.5}{0.80}\sin\delta=1.875\sin\delta$$ The post-fault peak, 1.875 pu, comfortably exceeds $P_m=0.6$ pu, so a stable post-fault equilibrium exists and a finite critical clearing angle must therefore exist too. The new steady-state operating point — the final angle the machine settles to once the swing has damped out — is $$\delta_{final}=\sin^{-1}\!\frac{0.6}{1.875}=18.66^\circ$$
5. Locate the limiting angle on the post-fault curve. The rotor can be decelerated only while $P_{e3} > P_m$, which fails beyond the unstable equilibrium on the far side of the post-fault characteristic: $$\delta_{max}=180^\circ-\sin^{-1}\!\frac{0.6}{1.875}=180^\circ-18.66^\circ=161.34^\circ =2.8159\ \text{rad}$$

Using the diagram — the equal-area criterion. The swing equation is $$\frac{2H}{\omega_s}\frac{d^{2}\delta}{dt^{2}}=P_m-P_e(\delta)$$ Multiplying by $2\,d\delta/dt$ and integrating once turns it into a statement about areas: the rotor's kinetic energy relative to synchronous speed at any angle is proportional to $\int(P_m-P_e)\,d\delta$. The machine returns to synchronous speed — and therefore stops swinging — only where that integral returns to zero. On the figure this means the shaded accelerating area $A_1$, gained while the fault is on and $P_e=0$, must be given back as an equal decelerating area $A_2$ on the post-fault curve before the angle reaches $\delta_{max}$. If it can be, the machine is transiently stable; if the angle passes $\delta_{max}$ with energy still to give back, $P_e$ falls below $P_m$ again, the rotor re-accelerates and the machine goes out of step.

Writing the two areas for a clearing angle $\delta_c$, and using $P_e=0$ during the fault, $$A_1=\int_{\delta_0}^{\delta_c}\left(P_m-0\right)d\delta=P_m(\delta_c-\delta_0)$$ $$A_2=\int_{\delta_c}^{\delta_{max}}\left(P_{max,3}\sin\delta-P_m\right)d\delta =P_{max,3}\left(\cos\delta_c-\cos\delta_{max}\right)-P_m(\delta_{max}-\delta_c)$$ Setting $A_1=A_2$, the $P_m\delta_c$ terms cancel and the critical clearing angle follows in closed form: $$\cos\delta_{cr}=\frac{P_m(\delta_{max}-\delta_0)+P_{max,3}\cos\delta_{max}}{P_{max,3}}$$ Substituting $P_m=0.6$, $\delta_{max}=2.8159$ rad, $\delta_0=0.2838$ rad and $P_{max,3}=1.875$, $$\cos\delta_{cr}=\frac{0.6(2.8159-0.2838)+1.875\cos(161.34^\circ)}{1.875} =\frac{1.5193-1.7764}{1.875}=-0.13715$$ $$\boxed{\delta_{cr}=97.88^\circ}$$ At that clearing angle both areas evaluate to $0.8548$ pu-rad, which is the arithmetic confirmation that the criterion has been applied correctly.

Critical clearing time. The critical clearing angle just found is a geometric quantity; the protection engineer needs the corresponding time. The critical clearing time is the longest interval, measured from fault inception, for which the fault may remain on the system with the machine still able to return to synchronism after the breakers operate. For the special case here — where the during-fault electrical output is zero, so the accelerating power is the constant $P_m$ — the swing equation integrates exactly: $$\delta(t)=\delta_0+\frac{\pi f_0 P_m}{2H}t^{2} \quad\Longrightarrow\quad t_{cr}=\sqrt{\frac{2H(\delta_{cr}-\delta_0)}{\pi f_0 P_m}}$$ with the angles in radians. The paper does not state the inertia constant $H$, so a number cannot be given; for an illustrative $H=5$ MJ/MVA the expression gives $t_{cr}=\sqrt{2(5)(1.4246)/(\pi\cdot 60\cdot 0.6)}=0.355$ s, about 21 cycles (the swing to clearing is $\delta_{cr}-\delta_0=81.62^\circ=1.4246$ rad). It matters because it sets the total permitted fault-clearing time, and that budget must cover relay pickup and operating time, communication-aided tripping delay, breaker opening time and arc extinction. If the protection cannot clear inside it, the answer is not a faster relay setting but different equipment — a faster breaker, a communications-assisted scheme, single-pole tripping, or in the extreme a series capacitor or a braking resistor to change the power-angle curves themselves. Critical clearing time is therefore the quantity that connects a stability study to a protection specification, and it is the reason transmission breakers are specified at two or three cycles.

How the load-flow solution connects to a stability study. The two studies are sequential, not independent: the power flow supplies the initial conditions without which the swing equations cannot be started. Three specific things pass from one to the other. First, the converged power flow gives every bus voltage magnitude and angle, and every machine's real and reactive output, at $t=0^{-}$; from those the internal voltage magnitude $E'$ and initial rotor angle $\delta_0$ of each generator are computed by pushing the terminal quantities back through the machine's transient reactance — exactly the $1.5$ pu and $16.26^\circ$ used above. Second, the power flow fixes the mechanical input of every machine, because at $t=0$ the system is in equilibrium and $P_m=P_e$; that value is then held constant through the swing (the governor being too slow to act in the first second). Third, the power-flow case defines the network itself — the admittance matrix, the loads converted to constant-impedance or composite models, and the dispatch — which the stability program then modifies for the faulted and post-fault configurations. The corollary is that a stability study is only as good as the power-flow case it starts from: the same disturbance on a heavily loaded, low-voltage case and on a lightly loaded case gives entirely different critical clearing times, which is why stability is assessed against a set of stressed but credible dispatch cases rather than one nominal case.

Why an arc forms, and how it is extinguished. An arc forms because the current in an inductive circuit cannot be interrupted instantaneously. As the breaker contacts begin to part, the contact area shrinks to a few small points, the current density and the local temperature rise steeply, and the last metallic bridge vaporises. That metal vapour, together with the intense electric field across a gap that is still microscopic, produces thermionic and field emission of electrons from the cathode; the electrons ionise the vapour and the surrounding gas by collision, and a self-sustaining, highly conducting plasma column — the arc — carries the current across the widening gap. The system inductance sustains it: the energy $\tfrac12 LI^{2}$ stored in the circuit has to go somewhere, and any attempt to force the current to zero abruptly would generate a prohibitive $L\,di/dt$ voltage. On a.c. the arc is helped to die by the natural current zero that occurs twice per cycle, but it will restrike at the next half-cycle unless the gap recovers its dielectric strength faster than the transient recovery voltage rises across it.

Two methods used to extinguish the arc are as follows. High-pressure gas blast and quenching in SF₆. In a modern SF₆ puffer or self-blast breaker the opening stroke compresses sulphur hexafluoride and drives it axially along the arc through an insulating nozzle. The gas cools the plasma by convection, sweeps the ionised products out of the gap, and — because SF₆ is strongly electronegative — captures free electrons to form heavy, immobile negative ions. At the natural current zero the residual conductance collapses and the gap withstands the recovery voltage. Vacuum interruption. In a vacuum interrupter the contacts part inside a sealed envelope at about $10^{-6}$ torr, so there is no gas to ionise: the arc can only be sustained by metal vapour boiled off the contacts themselves. Contacts are shaped (spiral or axial-magnetic-field designs) so that the arc is kept diffuse and moving rather than allowed to root in one spot. At the current zero the vapour condenses on the shields in microseconds and dielectric strength is recovered extremely quickly — which is why vacuum dominates medium-voltage switchgear while SF₆ dominates transmission voltages. Older techniques achieving the same end are worth naming: oil breakers use the hydrogen liberated by decomposing the oil to cool and deionise the arc, air-blast breakers use compressed air, and air-break contactors use arc chutes with splitter plates that lengthen the arc and divide it into many series arcs, each needing its own cathode drop, until the supply can no longer sustain it.