Question 7 of 7: Excitation Voltage and Transient Stability under a Sustained Fault

Problem 7: Excitation Voltage and Transient Stability under a Sustained Fault (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.

Find. The generator excitation voltage behind transient reactance in the pre-fault condition; and whether the machine retains synchronism when a bolted three-phase fault at the quarter point of one circuit is left on indefinitely.

Approach. Work backwards from the infinite bus. The two circuits in parallel fix the angle of bus 1 from the stated power transfer; the line current then follows, and pushing it through the machine and transformer reactance gives $E^{\prime}$. For part (b) the during-fault transfer reactance is obtained by a star–delta transform, and the equal-area criterion is applied — with the observation that a sustained fault admits only two power-angle curves, not three.

1. Part (a) — combine the two circuits and find the angle of bus 1. The two 0.60 pu circuits in parallel present $X_L=0.60/2=0.30$ pu between bus 1 and the infinite bus. Since the network is lossless, the 0.9 pu delivered to bus 1 also arrives at the infinite bus, so $$P=\frac{\left|V_1\right|V_{\infty}}{X_L}\sin\delta_1\;\Longrightarrow\;\sin\delta_1=\frac{0.9(0.30)}{1.05(1.0)}=0.257143\;\Longrightarrow\;\delta_1=14.901^{\circ}$$ so $\mathbf{V}_1=1.05\angle 14.901^{\circ}=1.01469+j0.27000$ pu.
2. Compute the line current. With the infinite bus as reference, $$\mathbf{I}=\frac{\mathbf{V}_1-\mathbf{V}_{\infty}}{jX_L}=\frac{\left(1.01469+j0.27000\right)-1.0}{j0.30}=\frac{0.01469+j0.27000}{j0.30}=0.9000-j0.04897\ \text{pu}$$ The real part is exactly 0.9000 pu, which is the arithmetic check that the angle was found correctly: with $V_{\infty}=1.0\angle 0^{\circ}$ the real part of the current is the power delivered.
3. Push the current through the machine and transformer to reach the internal emf. The two reactances in series between the internal node and bus 1 total $X_d^{\prime}+X_T=0.15+0.10=0.25$ pu, so $$\mathbf{E}^{\prime}=\mathbf{V}_1+j\left(0.25\right)\mathbf{I}=\left(1.01469+j0.27000\right)+j0.25\left(0.9000-j0.04897\right)$$ $$=\left(1.01469+0.012243\right)+j\left(0.27000+0.225000\right)=1.02694+j0.49500$$ $$\boxed{\ \mathbf{E}^{\prime}=1.1400\angle 25.735^{\circ}\ \text{pu}\ }$$ which is the 1.14 pu the question hands back in part (b) — a deliberate consistency check built into the paper.
4. Part (b) — establish the pre-fault power-angle curve and the initial angle. Referred to the internal emf, the total pre-fault reactance to the infinite bus is $0.25+0.30=0.55$ pu, so $$\begin{aligned}P_{\max,1}&=\frac{E^{\prime}V_{\infty}}{0.55}=\frac{1.14(1.0)}{0.55}=2.0727\ \text{pu}\\\delta_0&=\arcsin\!\frac{0.9}{2.0727}=25.735^{\circ}=0.44916\ \text{rad}\end{aligned}$$ reproducing the angle of $\mathbf{E}^{\prime}$ found in part (a), as it must.
5. Reduce the faulted network with a star–delta transform. With a bolted three-phase fault at $\alpha=0.25$ along circuit 2, that circuit splits into $0.6(0.25)=0.15$ pu from bus 1 to the fault point F and $0.6(0.75)=0.45$ pu from F onward. Bus 1 is now the centre of a star whose three arms run to the internal emf ($X_a=0.25$), to the infinite bus through the healthy circuit ($X_b=0.60$) and to the fault ($X_c=0.15$). Converting to a delta, the branch that directly joins the emf to the infinite bus is $$X_{\text{transfer}}=X_a+X_b+\frac{X_aX_b}{X_c}=0.25+0.60+\frac{(0.25)(0.60)}{0.15}=0.85+1.00\;\Longrightarrow\;\boxed{\ X_{\text{transfer}}=1.85\ \text{pu}\ }$$ The other two delta branches terminate on the fault point, which is at zero potential, so they carry no transfer power and are discarded. Note how severe the location is: the shunt arm $X_c$ sits in the denominator, so the closer the fault is to bus 1 the smaller $X_c$ becomes and the larger the transfer reactance grows.
6. Form the during-fault power-angle curve and compare it with the mechanical input. $$P_{\max,2}=\frac{E^{\prime}V_{\infty}}{X_{\text{transfer}}}=\frac{1.14(1.0)}{1.85}=0.6162\ \text{pu}$$ This is the decisive number. The mechanical input is $P_m=0.90$ pu, and the faulted network cannot transmit more than 0.6162 pu at any rotor angle: $$P_{\max,2}=0.6162<P_m=0.90$$
7. Apply the equal-area criterion — and observe that it cannot be satisfied. A sustained fault means the fault is never cleared, so there are only two curves: the pre-fault curve, which fixes $\delta_0$, and the during-fault curve, on which the machine must find a new equilibrium. An equilibrium requires $P_{\max,2}\sin\delta=P_m$, that is $\sin\delta=0.90/0.6162=1.4605$, which has no solution. The accelerating power $P_a=P_m-P_{\max,2}\sin\delta$ is therefore strictly positive for every angle, the decelerating area $A_2$ available beyond $\delta_0$ is exactly zero, and no accelerating area $A_1$ however small can be balanced: $$A_1=\int_{\delta_0}^{\delta}\left(P_m-P_{\max,2}\sin\delta\right)\mathrm{d}\delta>0\quad\text{for all }\delta>\delta_0,\qquad A_{2,\max}=0$$ $$\boxed{\ \text{The system is UNSTABLE under sustained fault conditions.}\ }$$ The rotor accelerates monotonically, passes $180^{\circ}$ and pole-slips; the machine loses synchronism and must be tripped.
8. State what this means for protection, and what would change the verdict. The conclusion is reached without integrating anything, and the test is worth remembering: for a sustained fault, $P_{\max,\text{fault}}\le P_m$ settles the question immediately. The engineering consequence is that stability here depends entirely on clearing the fault. Were the faulted circuit tripped out, the post-fault reactance would be $0.25+0.60=0.85$ pu, giving $P_{\max,3}=1.14/0.85=1.3412$ pu — comfortably above $P_m$ — so a critical clearing angle exists and the machine survives if the breakers are fast enough. That is precisely why transmission-line protection on generator outlet circuits is specified in cycles, not seconds.