22-Elec-B7 Power Systems Engineering · Undated paper

Question 7 of 7: Equal-Area Stability under a Sustained Mid-Line Fault

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

Notes on this paper

Paper format. National Exams, May 2019 — 16-Elec-B7 Power Systems Engineering, 3 hours, open book, any non-communicating calculator permitted. The paper prints seven questions, headed “Problem 1” to “Problem 7”, each carrying printed sub-part splits that total 20 points; the rubric states that any five questions constitute a complete paper, that only the first five questions as they appear in your answer book will be marked, and that all questions are of equal value. All seven are solved in full here, because the set is a study resource rather than a three-hour sitting. page 1 reads “National Exams May 2019” and every page footer reads “16-Elec-B7/May 2019”.

Reference texts. M. E. El-Hawary, Electrical Power Systems: Design and Analysis, rev. ed., IEEE Press / Wiley — the reference text named in the Engineers Canada syllabus for this examination code, and the source of the “cantilever” transformer model used in Problem 3. Supporting texts: J. D. Glover, M. S. Sarma and T. J. Overbye, Power System Analysis and Design, 6th ed.; J. J. Grainger and W. D. Stevenson, Power System Analysis; P. M. Anderson, Analysis of Faulted Power Systems; P. Kundur, Power System Stability and Control. Canadian practice context (grounding philosophy, protection coordination) follows CSA C22.1 and IEEE Std 142 as applied by Canadian utilities and industrial plants.

Note on the printed values. Every number used below is the one the paper prints, including B = j70, X_eq = 3.0 Ω, a utility X₀ of 0.1 and a grounding resistor of 8 Ω; the printed paper governs throughout.

Check — three engineering readings that a marker may take differently.

(1) Problem 5. The prose says “the voltage at both sources is 1 p.u.” above a Figure (3) that plainly draws four generators, G₁ to G₄. Every drawn source is taken at 1.0 pu, which is the only reading that closes the network.

(2) Problem 6. Figure (4-c) draws the 5 000 kVA generator alongside the resistance-grounded utility transformer, while the sentence introducing it mentions only the transformer. The generator is retained for part (e), consistent with the figure. It is ungrounded either way, so it changes only the positive- and negative-sequence networks; the zero-sequence answer is identical under both readings.

(3) Problem 2. “Neglecting armature reaction” cannot mean discarding x_d and x_q, which are the armature-reaction reactances — the question would then have no data. It is read here in its usual examination sense: neglect saturation and armature resistance, and use the ideal two-reaction model.

Question 7: Equal-Area Stability under a Sustained Mid-Line Fault (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. A single machine on an infinite bus through three identical lines.

Problem 7 data
Quantity	Value (pu)
Generator transient reactance $X_d'$	0.50
Transformer reactance $X_{tr}$	0.30
Each of the three lines	0.75
Internal EMF $E$ / infinite bus $V$	1.25 / 1.00
Mechanical input $P_m$	0.6 (parts a, b) and 0.3 (parts c, d)
Fault	bolted three-phase at the mid-point of line 1, sustained

Find. The initial power angle at each loading, and whether the machine remains in synchronism when the fault is never cleared.

Figure 7.1 — Figure (5) of the paper. Generator behind its transient reactance and transformer, feeding an infinite bus over three identical parallel lines.

Approach. Build the pre-fault transfer reactance by inspection, and the during-fault transfer reactance by a star–delta transform of the network seen with the fault point earthed. A sustained fault means only two power-angle curves ever exist, so the limiting angle must be read off the during-fault curve and stability is decided by comparing the accelerating area with the largest decelerating area that curve can offer.

Pre-fault transfer reactance. The three lines are in parallel between bus 1 and the infinite bus: $$X_{pre} = X_d' + X_{tr} + \frac{0.75}{3} = 0.50 + 0.30 + 0.25 = 1.05\ \text{pu}$$ $$P_{max,pre} = \frac{EV}{X_{pre}} = \frac{1.25 \times 1.00}{1.05} = 1.19048\ \text{pu}$$
Part (a) — the initial power angle at 0.6 pu. Steady state means electrical output equals mechanical input: $$\sin\delta_0 = \frac{P_m}{P_{max,pre}} = \frac{0.6}{1.19048} = 0.50400$$ $$\boxed{\delta_0 = 30.27^\circ}$$
During-fault transfer reactance — the star–delta step. With a bolted three-phase fault at the mid-point of line 1, bus 1 sees three branches: $0.80$ pu back to the internal EMF, $0.375$ pu (half of line 1) down to the earthed fault point, and $0.375$ pu (lines 2 and 3 in parallel) forward to the infinite bus. That is a star centred on bus 1 whose three terminals are the EMF node, the reference and the infinite bus. Converting it to a delta, the transfer arm between EMF and infinite bus is $$X_f = \frac{X_AX_B + X_BX_C + X_CX_A}{X_C} = \frac{(0.80)(0.375) + (0.375)(0.375) + (0.375)(0.80)}{0.375} = \frac{0.740625}{0.375}$$ $$\boxed{X_f = 1.975\ \text{pu}} \qquad\Longrightarrow\qquad P_{max,f} = \frac{1.25}{1.975} = 0.63291\ \text{pu}$$ The shunt arm to the fault sits in the denominator, which is why a fault nearer the machine (a smaller shunt arm) is far more severe than one out at the far end.
Part (b) — set up the equal-area comparison at 0.6 pu. Because the fault is never cleared there are only two curves: the machine leaves the pre-fault curve at $\delta_0$ and lives on the during-fault curve thereafter. The during-fault peak, 0.63291 pu, still exceeds $P_m = 0.6$ pu, so an equilibrium does exist on the faulted curve at $$\delta_1 = \arcsin\!\left(\frac{0.6}{0.63291}\right) = 71.44^\circ$$ and the limiting angle, read off that same faulted curve, is $$\delta_{lim} = 180^\circ - \delta_1 = 108.56^\circ$$ Beyond $\delta_{lim}$ the faulted curve has fallen back below $P_m$ and the machine accelerates again without bound.
Evaluate the two areas. The accelerating area runs from $\delta_0$ to $\delta_1$, where the faulted curve lies below $P_m$: $$A_1 = \int_{\delta_0}^{\delta_1}\!\left[P_m - P_{max,f}\sin\delta\right]d\delta = P_m(\delta_1 - \delta_0) + P_{max,f}(\cos\delta_1 - \cos\delta_0)$$ $$= 0.6(1.24657 - 0.52824) + 0.63291(0.31836 - 0.86348) = 0.43100 - 0.34501 = 0.08599$$ The largest decelerating area the faulted curve can supply runs from $\delta_1$ to $\delta_{lim}$: $$A_{2,max} = P_{max,f}(\cos\delta_1 - \cos\delta_{lim}) - P_m(\delta_{lim} - \delta_1) = 0.40299 - 0.38902 = 0.01419$$ Both angle differences must of course be in radians in the $P_m$ terms, which is the classic slip in this calculation.

Figure 7.2 — equal-area construction at P_m = 0.6 pu. The accelerating area A_1 (red) is six times the largest available decelerating area A_2 (green): the machine cannot be retarded before it passes the limiting angle.

Part (b) — the verdict at 0.6 pu. Since $$A_{2,max} = 0.01419 \;\ll\; A_1 = 0.08599 \qquad (\text{a ratio of only } 0.165)$$ $$\boxed{\text{The system is UNSTABLE under a sustained fault at } P_m = 0.6\ \text{pu}}$$ The rotor still has six sevenths of its excess kinetic energy left when it reaches $108.56^\circ$; past that angle the faulted curve delivers less than $P_m$ again, the acceleration resumes, and the machine pulls out of step. Note carefully that the existence of a during-fault equilibrium at $71.44^\circ$ did not guarantee stability — it only guaranteed that the question needed integrating rather than answering by inspection.
Part (c) — the initial power angle at 0.3 pu. Nothing about the network has changed, only the dispatch: $$\sin\delta_0 = \frac{0.3}{1.19048} = 0.25200$$ $$\boxed{\delta_0 = 14.60^\circ}$$
Part (d) — the verdict at 0.3 pu. The two curves are unchanged, so $$\delta_1 = \arcsin\!\left(\frac{0.3}{0.63291}\right) = 28.29^\circ, \qquad \delta_{lim} = 151.71^\circ$$ $$A_1 = 0.3(0.49382 - 0.25473) + 0.63291(0.88056 - 0.96773) = 0.07173 - 0.05517 = 0.01653$$ $$A_{2,max} = 0.63291(0.88056 + 0.88056) - 0.3(2.64777 - 0.49382) = 1.11457 - 0.64619 = 0.46841$$ Now $$A_{2,max} = 0.46841 \;\gg\; A_1 = 0.01653 \qquad (\text{a ratio of } 28.3)$$ $$\boxed{\text{The system is STABLE under a sustained fault at } P_m = 0.3\ \text{pu}}$$ Solving $A_1(\delta_{max}) = 0$ for the turning point gives a maximum swing of $\delta_{max} = 42.61^\circ$, comfortably short of the $151.71^\circ$ limit: the rotor advances about 28 electrical degrees, is retarded, and settles into a damped oscillation about the new faulted-curve equilibrium at $28.29^\circ$.

Figure 7.3 — the same construction at P_m = 0.3 pu. Now A_2 exceeds A_1 by a factor of 28 and the rotor turns back at 42.6°.

The two halves of the question are the same network at two dispatches, and the boundary between them is worth quoting. Solving $A_{2,max}(P_m) = A_1(P_m)$ numerically gives a critical sustained-fault loading of 0.5428 pu: below it the machine survives a permanent mid-line fault on one of three circuits, above it the machine is lost however long one waits. The 0.6 pu case sits 11 % above that boundary and the 0.3 pu case 45 % below it. In practice, of course, no such fault is ever left standing — the operational reading of part (b) is that this machine at 0.6 pu depends on the protection actually clearing the fault, and its critical clearing angle would be the next question to ask.

Question 7 — final results
Quantity	$P_m = 0.6$ pu	$P_m = 0.3$ pu
Pre-fault transfer reactance / peak power	1.05 pu / 1.19048 pu
During-fault transfer reactance / peak power	1.975 pu / 0.63291 pu
Initial power angle $\delta_0$	30.27°	14.60°
During-fault equilibrium $\delta_1$	71.44°	28.29°
Limiting angle $\delta_{lim} = 180^\circ - \delta_1$	108.56°	151.71°
Accelerating area $A_1$	0.08599 pu·rad	0.01653 pu·rad
Maximum decelerating area $A_{2,max}$	0.01419 pu·rad	0.46841 pu·rad
Area ratio $A_{2,max}/A_1$	0.165	28.3
Verdict	UNSTABLE	STABLE ($\delta_{max} = 42.61^\circ$)
Critical sustained-fault loading	0.5428 pu

Topic. The equal-area criterion for first-swing transient stability.

Key relations. Swing equation $\dfrac{2H}{\omega_s}\dfrac{d^2\delta}{dt^2} = P_m - P_e$; equal-area criterion $\displaystyle\int_{\delta_0}^{\delta_{max}}(P_m - P_e)\,d\delta = 0$. Closed forms: $A_1 = P_m(\delta_1-\delta_0) + P_{max,f}(\cos\delta_1 - \cos\delta_0)$, $A_{2,max} = P_{max,f}(\cos\delta_1 - \cos\delta_{lim}) - P_m(\delta_{lim}-\delta_1)$, $\delta_{lim} = \pi - \delta_1$. Star–delta transfer arm $X = (X_AX_B + X_BX_C + X_CX_A)/X_C$.

Why this works. Multiplying the swing equation by $d\delta/dt$ and integrating shows that the rotor's kinetic energy relative to synchronous speed is the integral of $(P_m - P_e)$ over angle. The machine returns to synchronous speed exactly when that integral vanishes — hence “equal areas”. The construction needs no inertia constant and no time-stepping, which is why it survives as the first test applied to any single-machine problem. The sustained-fault variant is the simplest case of all: there is no post-fault curve to return to, so the entire decelerating budget must come from the faulted curve between $\delta_1$ and $\pi - \delta_1$, a fixed and usually small amount. That is why sustained faults are so much more onerous than cleared ones, and why the answer at 0.6 pu flips despite an equilibrium existing.

Common pitfall. Two. Reading $\delta_{lim}$ off the pre-fault or post-fault curve instead of the sustained-fault curve — here that would give $180^\circ - 30.27^\circ = 149.7^\circ$ and a spurious “stable” verdict. And mixing degrees into the $P_m(\delta_{lim}-\delta_1)$ term while the cosines are evaluated in degrees; the angle differences must be in radians, and that single term is where the error hides because everything else still looks reasonable.

Source. El-Hawary, Electrical Power Systems: Design and Analysis, ch. 12 (power system stability); Kundur, Power System Stability and Control, §13.1–13.3; Anderson & Fouad, Power System Control and Stability, ch. 2.

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