22-Elec-B7 Power Systems Engineering · December 2018
Question 7 of 7: Equal-Area Stability under a Sustained Fault
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Exams, December 2018 — 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”; the rubric states that any five questions constitute a complete paper and that all questions are of equal value. Every one of the seven is solved in full here, because the set is a study resource rather than a three-hour sitting. Sub-part point values are quoted as printed; note that the printed splits total 25 points on Problems 1, 2, 4, 5 and 6, 20 points on Problem 3, and Problem 7 carries no printed split at all — under the “equal value” rubric each question is worth one fifth of the paper regardless.
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; S. J. Chapman, Electric Machinery Fundamentals, 5th ed. Canadian practice context (protection philosophy, grounding) follows CSA and IEEE C37 relaying practice as applied by Canadian transmission utilities.
Check — two places where the printed data are not internally consistent, and what was done about each.
(1) Problem 1. The pair A = 0.95∠2°, B = 195∠87.5° forces a shunt parameter with a large real part (Re C = 3.00 × 10−4 S). No real transmission line has a shunt conductance of that size — on a physical line C lands within about a degree of +90°. The consequence is a transmission efficiency of only 80.9 %, of which about 156 MW of the 236 MW loss is charged to that fictitious conductance. The arithmetic below is exactly what the printed constants give and is what the marker will be looking for; the physical caveat is stated so the low efficiency is not read as a slip.
(2) Problem 6. Table (2) lists the line negative-sequence reactance as 0.320 pu against a positive-sequence 0.30 pu. A transmission line is a static, symmetric element, so physically X− must equal X+. The printed 0.320 is used throughout (the paper governs). Taking 0.30 instead moves the answer to Problem 6(b) from 1.4169∠145.58° pu to 1.4212∠146.04° pu — a change of 0.3 %, so neither reading changes any conclusion.
Question 7: Equal-Area Stability under a Sustained Fault (25 marks)
Given. $E = 1.25$ pu behind $X'_d = 0.50$ pu, transformer $X_{Tr} = 0.30$ pu, three parallel lines each $x = 0.45$ pu joining bus 1 to bus 2, infinite bus $V = 1.00$ pu. Mechanical loading 0.5 pu in parts (a)–(b) and 0.3 pu in parts (c)–(d). A bolted three-phase fault occurs at the midpoint of line 1 and is sustained — the breakers do not clear it.
Find. The initial rotor angle at each loading, and whether the machine remains transiently stable under the sustained fault in each case.
The circuit of Figure (5), with the sustained three-phase fault at the midpoint of line 1. Because the fault is never cleared there are only TWO power-angle curves in this problem, not the usual three.
Approach. Build the prefault transfer reactance by simple series–parallel arithmetic and read the initial angle off the prefault power-angle curve. For the faulted network the midpoint fault creates a delta of three branches between bus 1, bus 2 and the reference, so one delta–star transformation gives the during-fault transfer reactance. Because the fault is sustained there is no third curve, and the equal-area criterion is applied between the prefault operating angle and the limiting angle read off the faulted curve.
Prefault transfer reactance and power-angle curve. The three lines are in parallel, so
$$X_{pre}=X'_d+X_{Tr}+\frac{x}{3}=0.50+0.30+0.15=0.95\ \text{pu}$$
$$P_{max,pre}=\frac{EV}{X_{pre}}=\frac{(1.25)(1.00)}{0.95}=1.315789\ \text{pu}$$
Part (a) — initial rotor angle at 0.5 pu loading. At steady state the electrical output equals the mechanical input, so
$$\sin\delta_0=\frac{P_m}{P_{max,pre}}=\frac{0.5}{1.315789}=0.380000$$
$$\boxed{\delta_0=22.33^\circ}$$
Part (b) — build the during-fault network. A bolted fault at the midpoint of line 1 splits that line into two halves of $j0.225$ each and holds the midpoint F at zero potential. Lines 2 and 3 remain in parallel between bus 1 and bus 2, contributing $0.45/2 = 0.225$ pu. Bus 1, bus 2 and the reference node F therefore form a delta whose three sides are all $j0.225$ — and no series–parallel move connects the source to the infinite bus without passing through the shorted node, so the transformation is unavoidable.
The during-fault delta and its star equivalent. Because all three delta branches are equal at j0.225, every star arm comes out at j0.075, and the shunt arm to the faulted node lands in the DENOMINATOR of the transfer reactance.
Transform and compute the during-fault transfer reactance. With the three delta sides summing to $0.675$, every star arm is
$$\frac{(0.225)(0.225)}{0.675}=0.075\ \text{pu}$$
Adding the generator and transformer to the bus 1 arm gives a T network with a sending arm $X_a$, a receiving arm $X_b$ and a shunt arm $X_c$ to the reference:
$$\begin{aligned}X_a &=0.50+0.30+0.075=0.875 \\ X_b &=0.075 \\ X_c &=0.075\end{aligned}$$
The transfer reactance of a T network with a shunt arm is
$$\begin{aligned}X_f &= X_a+X_b+\frac{X_aX_b}{X_c} \\ &= 0.875+0.075+\frac{(0.875)(0.075)}{0.075}\end{aligned}$$
$$\begin{aligned}\boxed{X_f=1.825\ \text{pu}} \\ P_{max,f} &=\frac{1.25}{1.825}=0.684932\ \text{pu}\end{aligned}$$
The shunt arm sitting in the denominator is the structural reason a fault closer to the sending bus is far more severe: a smaller $X_c$ inflates the transfer reactance without limit.
Locate the two angles that bound the swing. Under a sustained fault the machine has a new equilibrium on the faulted curve, reached where the faulted curve crosses the mechanical input:
$$\begin{aligned}\sin\delta_1 &=\frac{P_m}{P_{max,f}}=\frac{0.5}{0.684932}=0.730000 \\ \Longrightarrow\quad \delta_1 &=46.89^\circ\end{aligned}$$
Beyond that point the machine decelerates, and it can do so only until the faulted curve falls back through $P_m$ on its way down:
$$\delta_{lim}=180^\circ-\delta_1=133.11^\circ$$
Note that $\delta_{lim}$ is read off the faulted curve, not the prefault one — there is no post-fault curve to read, because the fault is never cleared.
Equal-area construction at 0.5 pu loading. The accelerating area A1 between 22.33 and 46.89 degrees is comfortably smaller than the decelerating area available out to 133.11 degrees.
Apply the equal-area criterion at 0.5 pu. The accelerating area from $\delta_0$ to $\delta_1$ is
$$A_1=P_m(\delta_1-\delta_0)-P_{max,f}\left(\cos\delta_0-\cos\delta_1\right)$$
with the angles in radians, $\delta_0 = 0.389796$ and $\delta_1 = 0.818322$:
$$\begin{aligned}A_1 &= 0.5(0.428526)-0.684932(0.924986-0.683447) \\ &= 0.214263-0.165438 = 0.048825\end{aligned}$$
The maximum decelerating area available is
$$A_{2,max}=P_{max,f}\left(\cos\delta_1-\cos\delta_{lim}\right)-P_m(\delta_{lim}-\delta_1)$$
$$\begin{aligned}A_{2,max} &= 0.684932(1.366894)-0.5(1.504949) \\ &= 0.936229-0.752474 = 0.183755\end{aligned}$$
Since $A_{2,max}=0.1838 > A_1=0.0488$,
$$\boxed{\text{the system is STABLE, with an area margin of }3.76:1}$$
Numerical integration of both areas reproduces 0.048825 and 0.183755, which is the check that catches the classic degrees-for-radians slip in the $P_m(\delta_{lim}-\delta_1)$ term.
Part (c) — initial rotor angle at 0.3 pu loading. The prefault curve is unchanged, so
$$\sin\delta_0=\frac{0.3}{1.315789}=0.228000$$
$$\boxed{\delta_0=13.18^\circ}$$
Part (d) — repeat the equal-area test at 0.3 pu. The faulted curve is also unchanged, so only the two bounding angles move:
$$\begin{aligned}\sin\delta_1 &=\frac{0.3}{0.684932}=0.438000 \;\Longrightarrow\; \delta_1=25.98^\circ \\ \delta_{lim} &=154.02^\circ\end{aligned}$$
In radians $\delta_0 = 0.230023$, $\delta_1 = 0.453373$ and $\delta_{lim} = 2.688220$, giving
$$\begin{aligned}A_1 &= 0.3(0.223350)-0.684932(0.973661-0.898975) \\ &= 0.067005-0.051155 = 0.015850\end{aligned}$$
$$\begin{aligned}A_{2,max} &= 0.684932(1.797950)-0.3(2.234847) \\ &= 1.231473-0.670454 = 0.561018\end{aligned}$$
Since $A_{2,max}=0.5610 \gg A_1=0.0159$,
$$\boxed{\text{the system is STABLE, with an area margin of }35.4:1}$$
Equal-area construction at 0.3 pu loading. Lowering the mechanical input both shrinks the accelerating area and enlarges the decelerating one, which is why the margin improves by an order of magnitude for a 40 percent reduction in load.
Both loadings survive, so the interesting answer is the margin rather than the verdict. The reason the machine survives at all is that the fault does not collapse the transfer completely: $P_{max,f} = 0.685$ pu still exceeds both mechanical loadings, so an equilibrium exists on the faulted curve and the rotor has somewhere to settle. Pushing the loading upward, the equal-area balance $A_1 = A_{2,max}$ is reached at a mechanical input of 0.5853 pu. That single number frames both answers: at 0.5 pu the machine sits 15 % below the sustained-fault stability limit, while at 0.3 pu it sits 49 % below it. Above 0.5853 pu no amount of waiting saves the machine, and above 0.685 pu the faulted curve cannot even carry the mechanical input, so the rotor accelerates monotonically from the instant of the fault.
