Question 7 of 7: Transient Stability under a Sustained Fault, by the Equal-Area Criterion

Question 7: Transient Stability under a Sustained Fault, 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 $E$ feeding an infinite bus of voltage $V$ through a transient reactance, one series line and two parallel lines, as drawn in Figure (5).

Find. The initial power angle in each case, and whether the machine retains synchronism if the fault is never cleared.

Approach. Compute the pre-fault transfer reactance by series and parallel combination, take the initial angle from the pre-fault power-angle curve, then obtain the during-fault transfer reactance by a star-to-delta transformation about the faulted node and apply the equal-area criterion between the two curves only, since a sustained fault means there is no third, post-fault curve.

1. Pre-fault transfer reactance and power-angle curve. With both parallel lines healthy the reactance from the internal EMF to the infinite bus is $$\begin{aligned} X_{pre}&=x_d'+x_1+\frac{x_2x_3}{x_2+x_3}=0.01+0.02+\frac{(0.03)(0.03)}{0.06} \\ &=0.01+0.02+0.015=0.045\ \text{pu} \end{aligned}$$ so the pre-fault curve is $$P_{e}=\frac{EV}{X_{pre}}\sin\delta=\frac{(1.2)(1.0)}{0.045}\sin\delta=\boxed{26.667\sin\delta\ \text{pu}}$$ The system is very stiff — a total reactance of under five per cent means a pull-out power of nearly twenty-seven times base — which is why loads of 6 and 8 pu still correspond to small angles.
2. During-fault transfer reactance requires a star-to-delta transformation. A bolted three-phase fault holds the mid-point of line 3 at zero potential, so that point becomes a connection to the reference. Around the junction node where the two parallel lines separate there are now three arms: $a=x_d'+x_1=0.03$ back to the EMF, $b=x_2=0.03$ forward to the infinite bus, and $c=0.015$ down to the faulted point. Converting that star to a delta, the branch that joins the EMF to the infinite bus is $$X_{f}=\frac{ab+bc+ca}{c}=\frac{(0.03)(0.03)+(0.03)(0.015)+(0.015)(0.03)}{0.015}=\frac{0.0018}{0.015}$$ $$\boxed{X_{f}=0.12\ \text{pu}}\qquad P_{e,fault}=\frac{1.2}{0.12}\sin\delta=10.000\sin\delta\ \text{pu}$$ The other two delta branches run from each source to the reference; they carry current into the fault but transfer no real power between the two machines, so they play no part in the power-angle curve. Note that a series-parallel combination cannot produce this result — the shunt arm must land in the denominator, and that is exactly what makes a fault close to the sending end so much more severe than one far from it.
3. Part (a) — initial power angle at 6 pu. Before the fault the machine sits on the pre-fault curve at the angle that matches the mechanical input, $$\sin\delta_0=\frac{P_m}{26.667}=\frac{6.0}{26.667}=0.22500 \quad\Longrightarrow\quad \boxed{\delta_0=13.0029^{\circ}=0.226943\ \text{rad}}$$
4. Part (a) — does an equilibrium exist during the fault? The during-fault peak is 10.0 pu, comfortably above the mechanical input of 6.0 pu, so the faulted curve does cross the $P_m$ line and there is a new equilibrium angle $$\sin\delta_1=\frac{6.0}{10.0}=0.600 \quad\Longrightarrow\quad \delta_1=36.8699^{\circ}=0.643501\ \text{rad}$$ Had $P_{max,f}$ been at or below $P_m$ the machine would accelerate without limit and the answer would be “unstable” with no integration needed at all. Because the fault is sustained there is no third curve, so the limiting angle is read off the faulted curve: $\delta_{max}=180^{\circ}-\delta_1=143.1301^{\circ}=2.498092$ rad.
5. Part (a) — accelerating and decelerating areas. $$A_1=\int_{\delta_0}^{\delta_1}\left(P_m-P_{max,f}\sin\delta\right)d\delta=P_m(\delta_1-\delta_0)+P_{max,f}(\cos\delta_1-\cos\delta_0)$$ $$\begin{aligned} A_1&=6.0(0.643501-0.226943)+10.0(0.800000-0.974359) \\ &=2.499348-1.743587=0.75576 \end{aligned}$$ $$A_{2,max}=P_{max,f}(\cos\delta_1-\cos\delta_{max})-P_m(\delta_{max}-\delta_1)$$ $$\begin{aligned} A_{2,max}&=10.0(0.800000+0.800000)-6.0(1.854591) \\ &=16.000000-11.127543=4.87246 \end{aligned}$$ Both angle differences must be in radians; this is the one place in the calculation where working in degrees silently destroys the answer.
6. Part (a) — verdict. Since $A_{2,max}=4.8725$ greatly exceeds $A_1=0.7558$, $$\boxed{\text{the machine remains stable at }P_m=6.0\ \text{pu}}$$ with a margin of about six and a half to one in area. The rotor swings out from $13.00^{\circ}$, passes through the new equilibrium at $36.87^{\circ}$ and reaches a maximum angle where the areas balance — well short of the $143.13^{\circ}$ limit — then oscillates about $36.87^{\circ}$, settling there if any damping is present.
7. Part (b) — initial power angle at 8 pu. On the same pre-fault curve, $$\sin\delta_0=\frac{8.0}{26.667}=0.30000 \quad\Longrightarrow\quad \boxed{\delta_0=17.4576^{\circ}=0.304693\ \text{rad}}$$ and during the fault $$\begin{aligned} \sin\delta_1&=\frac{8.0}{10.0}=0.800 \\ \delta_1&=53.1301^{\circ}=0.927295\ \text{rad} \\ \delta_{max}&=126.8699^{\circ}=2.214297\ \text{rad} \end{aligned}$$
8. Part (b) — areas and verdict. $$\begin{aligned} A_1&=8.0(0.927295-0.304693)+10.0(0.600000-0.953939) \\ &=4.980820-3.539392=1.44143 \end{aligned}$$ $$\begin{aligned} A_{2,max}&=10.0(0.600000+0.600000)-8.0(2.214297-0.927295) \\ &=12.000000-10.296018=1.70398 \end{aligned}$$ $A_{2,max}$ still exceeds $A_1$, so $$\boxed{\text{the machine remains stable at }P_m=8.0\ \text{pu, but only just}}$$ The ratio has collapsed from 6.45 to 1.18. Raising the load by a third has consumed more than eighty per cent of the stability margin, and the rotor now swings to $108.6^{\circ}$ before the areas balance, against $63.6^{\circ}$ in case (a). That is deep into the region where any modelling optimism — neglected damping, a slightly larger fault reactance, governor lag — would change the verdict.
9. Locate the true limit, since the two cases straddle it so closely. Setting $A_1=A_{2,max}$ and solving $$P_m\left(\pi-\delta_1-\delta_0\right)=P_{max,f}\left(\cos\delta_0+\cos\delta_1\right)$$ with $\delta_0=\arcsin(P_m/26.667)$ and $\delta_1=\arcsin(P_m/10.0)$ gives a critical mechanical input of $$\boxed{P_{m,crit}=8.155\ \text{pu}}$$ So case (a) is safe by a wide margin, case (b) sits within two per cent of the sustained-fault stability limit, and any load above about 8.16 pu would lose synchronism if this fault were never cleared. In practice, of course, protection would clear a mid-line three-phase fault in three to five cycles, and the post-fault curve with line 3 removed, $$P_{e,post}=\frac{1.2}{0.06}\sin\delta=20.0\sin\delta$$ would restore a large stability margin at both loadings.