Question 7 of 7: Transient Stability under a Sustained Fault

Question 7 (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. A machine of internal voltage E = 1.44 p.u. reaches a 1.00 p.u. bus through j0.03, then line 1 (j0.05), then two parallel circuits (lines 2 and 3) of j0.08 each; the active load is 3 p.u.

Find. The initial power angle; whether the machine stays in step for a sustained three-phase fault at the mid-point of line 3; and the maximum angle of oscillation.

Approach. Get the prefault and during-fault transfer reactances (the latter needs a star-to-delta transform), then settle the swing with the equal-area criterion.

1. Part (a) — prefault transfer reactance and initial angle. With both parallel circuits healthy the reactance from the internal voltage to the infinite bus is $$\begin{aligned}X_{pre}&=x_{g}+x_{1}+\dfrac{x_{2}x_{3}}{x_{2}+x_{3}} \\ &=0.03+0.05+0.04=0.1200\end{aligned}$$ so the prefault power-angle curve has amplitude $P_{max,pre}=EV/X_{pre}=12.0000$ p.u. Setting the electrical power equal to the 3 p.u. active load, $$\begin{aligned}\sin\delta_{0}&=\dfrac{P_{m}}{P_{max,pre}}=\dfrac{3}{12.0000} \; \\ &\Rightarrow\;\boxed{\delta_{0}=14.4775°=0.252680\ \text{rad}}\end{aligned}$$
2. Part (b) — during-fault transfer reactance. A three-phase fault at the mid-point of line 3 grounds a node inside the parallel section, so the network is no longer reducible by series and parallel combination. Calling the junction of the two parallel circuits node A, the three branches meeting there are 0.08 towards the source, 0.08 towards the bus along line 2, and 0.040 towards the fault along the first half of line 3. Transforming that star into a delta puts the fault branch in the denominator: $$\begin{aligned}X_{f}&=\dfrac{Z_{AE}Z_{AV}+Z_{AV}Z_{AF}+Z_{AF}Z_{AE}}{Z_{AF}} \\ &=\dfrac{0.012800}{0.040}=\boxed{0.3200\ \text{p.u.}}\end{aligned}$$ The other two delta branches connect each source to the fault and so shunt current to earth without transferring any power; only $X_{f}$ enters the power-angle curve.
3. Test whether the machine can transmit anything at all. The faulted curve has amplitude $$\begin{aligned}P_{max,f}&=\dfrac{EV}{X_{f}}=\dfrac{(1.44)(1.00)}{0.3200} \\ &=4.5000\ \text{p.u.}\end{aligned}$$ Since 4.5000 exceeds the mechanical input of 3 p.u., the machine is not automatically unstable; had it been smaller, the rotor would accelerate without limit and no integration would be needed. The faulted curve crosses $P_{m}$ at $$\begin{gathered}\delta_{1}=\sin^{-1}\!\left(\dfrac{3}{4.5000}\right)=41.8103°, \\ \delta_{lim}=180°-\delta_{1}=138.1897°\end{gathered}$$ Because the fault is sustained, there are only two curves and both $\delta_{1}$ and $\delta_{lim}$ are read off the faulted one.
4. Apply the equal-area criterion. The rotor accelerates from $\delta_{0}$ to $\delta_{1}$ and decelerates beyond it. Comparing the available areas before solving anything transcendental, $$\begin{aligned}&A_{1} \\ &=P_{m}(\delta_{1}-\delta_{0})-P_{max,f}\left(\cos\delta_{0}-\cos\delta_{1}\right) \\ &=0.428138\end{aligned}$$ $$\begin{aligned}&A_{2,max} \\ &=P_{max,f}\left(\cos\delta_{1}-\cos\delta_{lim}\right) -P_{m}(\delta_{lim}-\delta_{1}) \\ &=1.661792\end{aligned}$$ Since $A_{1}\lt A_{2,max}$, the decelerating area available before the limiting angle is reached is more than enough to absorb the accelerating area: $$\boxed{\text{the system remains stable under the sustained fault}}$$
5. Part (c) — maximum angle of oscillation. The swing stops where the net area vanishes, that is where $$\begin{aligned}&\int_{\delta_{0}}^{\delta_{max}}\left(P_{m}-P_{max,f}\sin\delta\right)d\delta \\ &=0 \; \\ &\Longrightarrow\; P_{m}\left(\delta_{max}-\delta_{0}\right) +P_{max,f}\left(\cos\delta_{max}-\cos\delta_{0}\right) \\ &=0\end{aligned}$$ Substituting the numbers gives the transcendental equation $$3\,\delta_{max}+4.5000\cos\delta_{max}-5.115147=0$$ whose root between $\delta_{0}$ and $\delta_{lim}$ is $$\boxed{\delta_{max}=1.291432\ \text{rad}=73.9936°}$$ That is comfortably short of the 138.190° limit, so the rotor turns back and the machine oscillates about $\delta_{1}$ — undamped in this idealised model, and damped in practice by the amortisseur and load-frequency effects.