22-Elec-B7 Power Systems Engineering · May 2017

Question 7 of 7: Transient Stability by the Equal-Area Criterion

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

Notes on this paper

Paper format. National Exams, May 2017 — 16-Elec-B7 Power Systems Engineering. Open-book, 3 hours, seven problems of equal value (25 points each); any five constitute a complete paper, and only the first five in the answer book are marked. All seven are solved here, because the set is a study resource rather than a sitting.

Reference texts. This exam code follows M. E. El-Hawary, Electrical Power Systems: Design and Analysis (IEEE Press) — the notation is his throughout: the bundle-conductor mean radii and long-line ABCD constants of Problem 1, the round-rotor power-angle relations of Problem 2, the three-winding equivalent circuit of Problem 3, the sequence-network fault reduction of Problem 6 and the equal-area criterion of Problem 7. Corroborating references: J. D. Glover, M. S. Sarma and T. J. Overbye, Power System Analysis and Design, 6th ed. (ch. 4–5 transmission-line parameters, ch. 6 power flow, ch. 7–9 symmetrical components and faults, 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 synchronous-machine and transformer chapters. Canadian practice references where protection and grounding are discussed: CSA C22.1 Canadian Electrical Code, Part I, CSA C22.3 for overhead systems, and IEEE C37 (protective relaying) and IEEE 142 Green Book (grounding).

Check: four judgement calls carried through the solutions, recorded here rather than hedged inside the answers.

(i) Problem 1 states neither a frequency nor a line length. Parts (a)–(c) need neither. Part (d) needs both, so 60 Hz is used (Canadian national exam) and the ABCD constants are given as closed-form functions of length, evaluated at a representative 400 km with a sensitivity table at 200 km and 600 km. The characteristic impedance and the surge-impedance loading do not depend on either quantity.

(ii) Problem 1 gives “bundle separation S” for a bundle whose figure shows eight sub-conductors on a circle. S is read as the distance between adjacent sub-conductors, so the bundle radius is $A = S/[2\sin(\pi/N)] = 0.6533\ \text{m}$ — the standard El-Hawary reading. If a marker instead reads S as the bundle diameter, every mean radius scales by the same factor and the method is unchanged.

(iii) Problem 5 says “the voltage at both sources is 1 p.u.” above a figure that draws three generators. All three are taken at 1.0 pu, which is the only reading that closes the network; the three internal e.m.f.s then merge into one node.

(iv) Problem 6: the zero-sequence source reactance is printed as $X_{0S}=j0.04$, possibly followed by a fifth digit. The value $j0.04$ is used. Reading it as $j0.045$ moves the zero-sequence Thévenin reactance from 0.19978 to 0.20051 pu and the fault current by 0.1 % — below the resolution of any answer here.

Problem 7: Transient Stability by the Equal-Area Criterion (25 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.

Quantity	Symbol	Value
Machine internal e.m.f.	$E$	1.25 pu
Infinite-bus voltage	$V$	1.00 pu
Machine plus step-up reactance	$X_g$	$j0.03$ pu
Line 1	$X_1$	$j0.05$ pu
Lines 2 and 3 (parallel)	$X_2=X_3$	$j0.08$ pu each
Mechanical (load) power, parts (a) / (b)–(d)	$P_m$	3.0 pu / 3.2 pu
Fault	—	bolted three-phase, mid-point of line 3
Clearing angle, parts (c)–(d)	$\delta_c$	$85^\circ$ (line 3 opened)

Find. The maximum swing angle under a sustained fault at 3.0 pu load; a demonstration that 3.2 pu is unstable under the same sustained fault; a demonstration that clearing at $85^\circ$ restores stability; and the resulting maximum swing angle.

Figure 8 — The single-machine infinite-bus circuit of Figure (6). The three-phase fault sits at the mid-point of line 3; clearing means opening the breakers at both of its ends, which leaves line 2 alone in service.

Approach. Build the three power-angle curves — prefault, during-fault and post-fault — from the transfer reactance in each state, obtaining the during-fault value by a delta–star transform that puts the fault shunt in the denominator. Then apply the equal-area criterion: the accelerating area between $\delta_0$ and the clearing angle must be recoverable as decelerating area before the rotor reaches the limit angle.

Prefault transfer reactance and the initial angle. Both lines are in service, so $X_{\text{pre}}=0.03+0.05+(0.08\parallel0.08)=0.03+0.05+0.04=0.12$ pu and$$P_{\max,\text{pre}}=\frac{EV}{X_{\text{pre}}}=\frac{1.25}{0.12}=10.4167\ \text{pu}.$$For part (a) with $P_m=3.0$ pu, $\sin\delta_0=3.0/10.4167=0.288$, so $\delta_0=16.738^\circ=0.292138\ \text{rad}$.
During-fault transfer reactance by delta–star transform. The fault grounds the mid-point of line 3, so from the machine-side node A there are three arms: $0.08$ pu to the infinite bus through line 2, $0.04$ pu to the grounded fault point, and $0.08$ pu back to the machine. Converting that star at A into a delta between the machine node, the bus and ground, the transfer arm is$$X_f=\frac{(0.08)(0.08)+(0.08)(0.04)+(0.04)(0.08)}{0.04}=\frac{0.0128}{0.04}=0.32\ \text{pu},$$the shunt arm to ground appearing in the denominator — which is why a fault nearer the sending end is far more severe. Hence $P_{\max,f}=1.25/0.32=3.90625$ pu.
Post-fault transfer reactance. Clearing removes line 3 entirely, leaving $X_{\text{post}}=0.03+0.05+0.08=0.16$ pu and$$P_{\max,\text{post}}=\frac{1.25}{0.16}=7.8125\ \text{pu}.$$The three curves — 10.4167, 3.90625 and 7.8125 pu peaks — are the whole of the stability analysis that follows.
Part (a) — equate the areas for the sustained fault. With the fault never cleared there are only two curves in play: the machine accelerates along $P_{\max,f}\sin\delta$ until that curve rises above $P_m$, then decelerates on the same curve. The swing stops where the net area vanishes:$$\begin{aligned}&\int_{\delta_0}^{\delta_{\max}}\bigl(P_m-P_{\max,f}\sin\delta\bigr)\,d\delta=0\\\Longrightarrow\ &P_m(\delta_{\max}-\delta_0)+P_{\max,f}\bigl(\cos\delta_{\max}-\cos\delta_0\bigr)=0,\end{aligned}$$with $\delta$ in radians.
Solve the transcendental equation. Substituting $P_m=3.0$, $P_{\max,f}=3.90625$, $\delta_0=0.292138$ rad and $\cos\delta_0=0.957630$,$$3.0\,(\delta_{\max}-0.292138)+3.90625\,(\cos\delta_{\max}-0.957630)=0,$$whose root between $\delta_0$ and the limit angle is$$\boxed{\delta_{\max}=1.676734\ \text{rad}=96.07^\circ.}$$It is below the limit angle $\delta_{\text{lim}}=180^\circ-\arcsin(3.0/3.90625)=180^\circ-50.17^\circ=129.83^\circ$, which is what confirms the stability the question asserts.
Part (b) — test 3.2 pu against the same sustained fault. Now $\sin\delta_0=3.2/10.4167$ gives $\delta_0=17.891^\circ=0.312249$ rad, and the faulted curve crosses $P_m$ at $\delta_1=\arcsin(3.2/3.90625)=55.005^\circ$, so $\delta_{\text{lim}}=124.995^\circ=2.181578$ rad. The maximum decelerating area available is the whole of the region from $\delta_1$ to $\delta_{\text{lim}}$, so the machine survives only if $A_1\le A_{2,\max}$:$$A_1=P_m(\delta_1-\delta_0)+P_{\max,f}(\cos\delta_1-\cos\delta_0)=0.59575\ \text{pu-rad},$$$$A_{2,\max}=P_{\max,f}(\cos\delta_1-\cos\delta_{\text{lim}})-P_m(\delta_{\text{lim}}-\delta_1)=0.57153\ \text{pu-rad}.$$
Declare the verdict for part (b). The accelerating area exceeds the largest decelerating area the faulted curve can offer:$$\boxed{A_1-A_{2,\max}=0.59575-0.57153=+0.02422\ \text{pu-rad}>0\ \Longrightarrow\ \text{unstable}.}$$The margin is only 4 per cent, so the machine is marginally unstable — it would swing past $\delta_{\text{lim}}$, at which point the faulted curve falls below $P_m$ again, the rotor re-accelerates and synchronism is irrecoverably lost. This is why the 3.0 pu case of part (a) is stable and the 3.2 pu case is not, despite the 7 per cent difference in load.

Figure 9 — Equal-area construction for parts (c) and (d). The rotor accelerates along the faulted curve (peak 3.906 pu) from δ₀ = 17.89° to the clearing angle 85°, shading area A₁; it then decelerates along the post-fault curve (peak 7.813 pu) until the equal area A₂ is accumulated at δmax = 89.62°, far short of the limit angle 155.82°.

Part (c) — accelerating area up to the clearing angle. Clearing at $\delta_c=85^\circ=1.483530$ rad, the rotor has accelerated along the faulted curve throughout:$$\begin{aligned}A_1&=P_m(\delta_c-\delta_0)+P_{\max,f}(\cos\delta_c-\cos\delta_0)\\&=3.2(1.171281)+3.90625(0.087156-0.951645),\end{aligned}$$i.e. $A_1=3.748098-3.376911=0.371187$ pu-rad. This is less than the 0.59575 pu-rad of part (b) because the swing is stopped at $85^\circ$ instead of continuing to $125^\circ$.
Compare it with the decelerating area now available. After clearing, the machine runs on the post-fault curve whose peak is 7.8125 pu, so the limit angle becomes$$\delta_{\text{lim}}=180^\circ-\arcsin\!\left(\frac{3.2}{7.8125}\right)=180^\circ-24.187^\circ=155.820^\circ=2.719577\ \text{rad},$$and the maximum decelerating area is$$\begin{aligned}A_{2,\max}&=P_{\max,\text{post}}(\cos\delta_c-\cos\delta_{\text{lim}})-P_m(\delta_{\text{lim}}-\delta_c)\\&=7.807976-3.955351=3.852625\ \text{pu-rad}.\end{aligned}$$Since$$\boxed{A_{2,\max}=3.8526\ \gg\ A_1=0.3712\ \text{pu-rad},}$$the system is comfortably stable. Equivalently, the critical clearing angle for these conditions is $143.5^\circ$, so clearing at $85^\circ$ leaves an enormous margin.
Part (d) — find where the swing actually stops. The rotor decelerates on the post-fault curve until the decelerating area equals $A_1$:$$P_{\max,\text{post}}\bigl(\cos\delta_c-\cos\delta_{\max}\bigr)-P_m\bigl(\delta_{\max}-\delta_c\bigr)=A_1,$$that is, $7.8125\,(0.087156-\cos\delta_{\max})-3.2\,(\delta_{\max}-1.483530)=0.371187$. Solving between $\delta_c$ and $\delta_{\text{lim}}$,$$\boxed{\delta_{\max}=1.564191\ \text{rad}=89.62^\circ.}$$The rotor overshoots the clearing angle by less than five degrees, which is the practical meaning of the very large stability margin found in part (c).
Close with the physical reading. Note the post-fault equilibrium angle is $\arcsin(3.2/7.8125)=24.19^\circ$, so the machine will oscillate about that value between roughly $24^\circ$ and $89.6^\circ$, damping onto it. Note also which quantity actually decides solvability: it is the post-fault peak, not the during-fault peak. Here the during-fault peak of 3.906 pu exceeds $P_m$, but even if it had not, the machine would still be recoverable so long as the post-fault peak stays above $P_m$ — whereas a post-fault peak below $P_m$ would make the machine unrecoverable however fast the breakers operated.

Final Results

Quantity	Symbol	Result
Prefault transfer reactance / peak power	$X_{\text{pre}}$, $P_{\max,\text{pre}}$	0.12 pu; 10.4167 pu
During-fault transfer reactance / peak power	$X_f$, $P_{\max,f}$	0.32 pu; 3.90625 pu
Post-fault transfer reactance / peak power	$X_{\text{post}}$, $P_{\max,\text{post}}$	0.16 pu; 7.8125 pu
(a) Maximum swing, sustained fault, $P_m=3.0$	$\delta_{\max}$	$96.07^\circ$ (limit $129.83^\circ$)
(b) Areas at $P_m=3.2$, sustained fault	$A_1$ vs $A_{2,\max}$	0.59575 vs 0.57153 pu-rad → unstable
(c) Areas with clearing at $85^\circ$	$A_1$ vs $A_{2,\max}$	0.37119 vs 3.85263 pu-rad → stable
Critical clearing angle for part (c) conditions	$\delta_{cr}$	$143.52^\circ$
(d) Maximum swing after clearing	$\delta_{\max}$	$89.62^\circ$

Concept & Theory

Topic. Transient stability of a single machine on an infinite bus: the equal-area criterion.

Key relations. $P_e=\dfrac{EV}{X}\sin\delta$ in each network state; swing equation $\dfrac{2H}{\omega_s}\dfrac{d^{2}\delta}{dt^{2}}=P_m-P_e$; stability requires $A_1=\int_{\delta_0}^{\delta_c}(P_m-P_e^{f})d\delta \le A_{2,\max}=\int_{\delta_c}^{\delta_{\text{lim}}}(P_e^{\text{post}}-P_m)d\delta$; $\delta_{\text{lim}}=\pi-\arcsin(P_m/P_{\max,\text{post}})$; delta-star transfer reactance $X=(z_az_b+z_bz_c+z_cz_a)/z_c$ with the shunt arm $z_c$ in the denominator.

Why this works. Multiplying the swing equation by $d\delta/dt$ and integrating once converts it into an energy statement: the kinetic energy the rotor gains while $P_m$ exceeds $P_e$ must be given back before the rotor angle passes the point where $P_e$ falls below $P_m$ again. Because the inertia constant cancels in that integration, the criterion needs no value of $H$ and no numerical integration — only the three power-angle curves. The angle at which the areas balance is the first swing peak; beyond $\delta_{\text{lim}}$ no restoring torque exists and the machine runs away. The delta-star transform is the load-bearing step, because a fault part-way along a line creates a shunt to ground that cannot be handled by series-parallel combination.

Common pitfall. Mixing radians and degrees in the $P_m(\delta_{\max}-\delta_0)$ term is the classic arithmetic failure — the trigonometric terms are dimensionless but that one is not. Conceptually, the commonest error is reading $\delta_{\text{lim}}$ off the wrong curve: for a sustained fault it belongs to the faulted curve, for a cleared fault to the post-fault curve, and using the prefault curve for either overstates the available area badly.

Source. El-Hawary, Electrical Power Systems: Design and Analysis, ch. 12 (power-system stability: the swing equation, the equal-area criterion and critical clearing); Glover, Sarma & Overbye, Power System Analysis and Design, 6th ed., §11.1–11.4; Grainger & Stevenson, Power System Analysis, ch. 16.

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