22-Elec-B7 Power Systems Engineering · May 2016

Question 7 of 7: Transient stability under a sequence of faults and switchings

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format. National Exams, May 2016 — 07-Elec-B7 Power Systems Engineering. Open-book, 3 hours duration. Seven problems are printed; any five constitute a complete paper, only the first five appearing in the answer book are marked, and all questions are of equal value (20 points each). All seven are solved in full below, because this set is a study resource rather than a timed sitting.

Reference texts. The syllabus for this code follows M. E. El-Hawary, Electrical Power Systems: Design and Analysis (IEEE Press) — the notation used throughout (the ABCD two-port and its equivalent pi, the two-reaction salient-pole construction, the sequence-network fault reduction) is his, and the phrase “all power formulae in the text” in Problem 1 refers to that book. Corroborating references: J. D. Glover, M. S. Sarma and T. J. Overbye, Power System Analysis and Design, 6th ed. (ch. 3 transformers, ch. 5 transmission lines, ch. 6 power flow, ch. 7–9 faults and symmetrical components, ch. 11 transient stability); J. J. Grainger and W. D. Stevenson, Power System Analysis; and S. J. Chapman, Electric Machinery Fundamentals, 5th ed., for the machine and transformer chapters. Canadian practice references where grounding and protection are discussed: CSA C22.1 Canadian Electrical Code, Part I, CSA C22.3 No. 1 for overhead systems, and IEEE C37 series for protective relaying.

Check: four data points read off the printed figures. (i) In Figure (2) of Problem 4 the three branch reactances are j0.25, j0.20 and j0.10 pu, with bus 4 hanging off bus 2 through the j0.10 branch — the topology is taken from the drawing. (ii) Figure (3) of Problem 5 shows three sources, not two: generators at bus 1 and bus 4 and a third machine below bus 5. The sentence “the voltage at both sources is 1 p.u.” is taken to mean all sources are at 1.0 pu, which is the only reading that lets the network be solved. (iii) In Figure (4-a) of Problem 6, transformer T2 carries its delta on the generator side and T1 carries its delta on the motor side; motor 1 is an ungrounded wye and motor 2 is grounded through a reactor. Those four winding connections, not the reactance values, decide the answer. (iv) Problem 7 gives the switching angles in radians (1.2, 1.25 and 1.9 rad).

Check: Problem 3(a) as printed is over-determined and internally inconsistent. Rated kVA, the load power factor, the efficiency and the primary voltage are four data for two unknowns, and at 0.85 power factor lagging no positive series reactance satisfies them all. The inconsistency is quantified and resolved inside the answer rather than hidden.

Problem 7 — Transient stability under a sequence of faults and switchings (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 transient reactance $X_d^{\prime}=0.6$ pu behind a 0.20 pu transformer, feeding an infinite bus through three identical 0.6 pu lines; $E=2.4$ pu, $V=1.0$ pu, mechanical input $P_m=1.0$ pu held constant. The disturbance is a sequence of four network states.

Stage	Network state	Angle range
0	pre-fault: all three lines in service	steady state at $\delta_0$
1	three-phase fault at the mid-point of line 1	$\delta_0 \to \delta_1 = 1.2$ rad
2	line 1 tripped at both ends; lines 2 and 3 in service	$\delta_1 \to \delta_2 = 1.25$ rad
3	three-phase fault at the mid-point of line 2	$\delta_2 \to \delta_3 = 1.9$ rad
4	line 2 tripped; line 3 alone remains	$\delta_3 \to$ maximum swing

Find. (a) the initial rotor angle; (b) whether the machine remains in synchronism through the whole sequence, by the equal-area criterion.

Approach. Build the transfer reactance for each of the five network states — two of them need a delta-to-star transform because the fault sits part-way along a line — then accumulate accelerating and decelerating areas across the four stages and test whether the running total can be brought back to zero before the post-fault curve reaches its unstable equilibrium.

Part (a) — the pre-fault transfer reactance and initial angle. With three 0.6 pu lines in parallel the corridor is 0.2 pu, so $X_{pre}=0.6+0.2+0.6/3=1.0000$ pu and $P_{max}=EV/X_{pre}=2.4000$ pu. Setting the electrical output equal to the mechanical input, $$\delta_0=\sin^{-1}\!\left(\frac{P_m}{P_{max}}\right)=\sin^{-1}\!\left(\frac{1.0}{2.4}\right)=\boxed{0.4298\ \text{rad}=24.624^{\circ}}$$
Part (b) — transfer reactance while the line-1 fault persists. The fault splits line 1 into two 0.3 pu halves meeting at F; lines 2 and 3 in parallel give 0.3 pu directly from bus 1 to bus 2. The three 0.3 pu arms form a delta on (bus 1, bus 2, F) whose star equivalent has arms of $(0.3)(0.3)/0.9=0.1000$ pu each. Adding the 0.8 pu source reactance to the bus-1 arm gives a T network with a shunt arm to the fault, so $$X_{f1}=X_a+X_b+\frac{X_aX_b}{X_N}=0.9000+0.1000+\frac{(0.9000)(0.1000)}{0.1000}=1.9000\ \text{pu}$$ so the during-fault characteristic is $P_{e}=1.2632\sin\delta$.
The three remaining network states. After line 1 is tripped the corridor is two lines in parallel, $X=0.8+0.3=1.10$ pu, giving $P_{max}=2.1818$ pu. When the second fault appears at the middle of line 2, the delta is now (0.3, 0.3, 0.6) with star arms $0.1500$, $0.1500$ and shunt $0.0750$ pu, whence $X_{f2}=3.0000$ pu and $P_{max}=0.8000$ pu. Finally, with line 3 alone, $X=0.8+0.6=1.40$ pu and $P_{max}=1.7143$ pu.
Accumulate the area over the first fault. Each area is $\int(P_m-P_{max}\sin\delta)\,d\delta=P_m\Delta\delta+P_{max}\left[\cos\delta\right]$, evaluated with $\delta$ in radians. From $\delta_0=0.4298$ to $\delta_1=1.2$ rad on the $1.2632\sin\delta$ curve, $$A_1=(1.0)(1.2-0.4298)+1.2632(\cos 1.2-\cos 0.4298)=+0.079654\ \text{pu-rad}$$ a small accelerating area, because the during-fault peak of $1.263$ pu still exceeds $P_m$ over most of the interval.
The short decelerating window, and the second fault. Between $\delta_1=1.2$ and $\delta_2=1.25$ rad the machine runs on the two-line curve and decelerates: $A_2=-0.052623$ pu-rad. The second fault then drops the peak to $0.8000$ pu, below $P_m$, so the machine accelerates continuously from $\delta_2=1.25$ to $\delta_3=1.9$ rad: $A_3=+0.139110$ pu-rad. The running total at the instant line 2 clears is $$A_{net}=0.079654-0.052623+0.139110=+0.166142\ \text{pu-rad}$$
Test the decelerating area available on the surviving line. The post-clearing curve is $1.7143\sin\delta$, whose unstable equilibrium lies at $\delta_{max}=\pi-\sin^{-1}(1.0/1.7143)=2.5188$ rad $=144.31^{\circ}$. The area available between $\delta_3=1.9$ rad and that limit is $$A_{2,max}=\int_{1.9}^{2.5188}\left(1.7143\sin\delta-P_m\right)d\delta=0.219422\ \text{pu-rad}$$
Apply the equal-area criterion. The available decelerating area $0.2194$ pu-rad exceeds the accumulated accelerating area $0.1661$ pu-rad, so the swing is arrested before the rotor angle reaches the unstable equilibrium: $$\boxed{A_{2,max}=0.2194 > A_{net}=0.1661\ \Longrightarrow\ \text{the system is STABLE}}$$
Locate the maximum swing. Solving $A_{net}+\int_{1.9}^{\delta}\left(P_m-1.7143\sin\delta\right)d\delta=0$ for the first return to zero net area gives $$\delta_{swing}=2.2311\ \text{rad}=127.83^{\circ}$$ against the limit of 144.31°, so the machine turns back with about 16.5° of margin — positive but not generous. It then oscillates about the new equilibrium $\sin^{-1}(1/1.7143)=35.69^{\circ}$ and, with damping, settles there.

Figure 10 — the equal-area construction. Shaded red areas accelerate the rotor and shaded green areas decelerate it; the swing stops where the green area to the right of 1.9 rad has cancelled the net red area to its left, at 2.231 rad, comfortably short of the 2.519 rad unstable equilibrium.

Check: what decides solvability here is the post-fault peak, not the during-fault peak. The second fault leaves a during-fault peak of 0.800 pu, below the 1.0 pu mechanical input, so the machine accelerates without interruption for the whole 0.65 rad of that stage. That is normal for a cleared fault and says nothing about stability. The test that matters is that the final curve peaks at 1.714 pu, above $P_m$, so a decelerating area exists at all. Had the last line also been lost, no clearing time however short would have saved the machine.

Problem 7 — final results
Quantity	Symbol	Result
Pre-fault transfer reactance	$X_{pre}$	1.0000 pu
Pre-fault peak power	$P_{max,0}$	2.4000 pu
Initial power angle	$\delta_0$	0.4298 rad (24.62°)
During fault on line 1	$X_{f1}$, $P_{max}$	1.9000 pu, 1.2632 pu
Line 1 out, lines 2 and 3 in	$X$, $P_{max}$	1.10 pu, 2.1818 pu
During fault on line 2	$X_{f2}$, $P_{max}$	3.0000 pu, 0.8000 pu
Line 3 alone (post-clearing)	$X$, $P_{max}$	1.40 pu, 1.7143 pu
Net accelerating area to $\delta_3$	$A_{net}$	0.1661 pu-rad
Decelerating area available	$A_{2,max}$	0.2194 pu-rad
Maximum rotor swing	$\delta_{swing}$	2.2311 rad (127.83°)
Unstable equilibrium on the final curve	$\delta_{max}$	2.5188 rad (144.31°)
Verdict	—	Stable

Topic. The equal-area criterion applied to a multi-stage disturbance.

Key relations. Swing equation $\frac{2H}{\omega_s}\frac{d^{2}\delta}{dt^{2}}=P_m-P_{max}\sin\delta$; first integral $\int(P_m-P_e)\,d\delta = 0$ at the instant of maximum swing; areas $A=P_m\Delta\delta+P_{max}\left[\cos\delta\right]_{\delta_a}^{\delta_b}$; delta-to-star for a mid-line fault, transfer reactance $X=X_a+X_b+X_aX_b/X_N$.

Why this works. The equal-area criterion is the first integral of the swing equation for a single machine against an infinite bus, and it works because the kinetic energy the rotor gains while its electrical output is below the mechanical input must all be given back before the rotor stops advancing. Nothing in that argument depends on there being only two network states, so a sequence of faults and switchings is handled by simply summing the signed areas stage by stage in the order they occur. The mid-line fault is what forces the delta-to-star step: with the shunt arm to the fault appearing in the denominator of the transfer reactance, a fault closer to the sending bus is far more severe than one near the middle.

Common pitfall. Evaluating $P_m\,\Delta\delta$ with the angle in degrees while the cosine terms are in radians — the single most common arithmetic error in this calculation, and the one a numerical re-integration of the areas catches instantly. The second is applying the shortcut “during-fault peak below $P_m$ means unstable”, which is true only for a sustained fault; here it would give the wrong verdict.

Source. El-Hawary, Electrical Power Systems: Design and Analysis, power-system stability chapter; Glover, Sarma and Overbye, 6th ed., ch. 11; Kundur, Power System Stability and Control, ch. 13 for the multi-swing generalisation.

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