22-Elec-B7 Power Systems Engineering · Undated paper

Question 5 of 7: Three-Phase Bolted Fault on a Meshed Four-Generator System

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 5: Three-Phase Bolted Fault on a Meshed Four-Generator System (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. All reactances per unit on a common base, read from Figure (3).

Problem 5 network data
Element	Reactance (pu)	Connects
$G_1$ + its transformer	0.20 + 0.04 = 0.24	source to bus 1
$G_2$ + its transformer	0.25 + 0.06 = 0.31	source to bus 1
$G_3$ + its transformer	0.25 + 0.06 = 0.31	source to bus 2
$G_4$ + its transformer	0.25 + 0.06 = 0.31	source to bus 2
Tie line	0.20	bus 1 to bus 2
Line 1–3	0.20	bus 1 to bus 3
Line 2–3	0.20	bus 2 to bus 3
All internal source voltages	1.00∠0° pu	—

Check — the prose says “both sources”, the figure draws four. The question sentence reads “assume that the voltage at both sources is 1 p.u.” while Figure (3) plainly draws $G_1$ to $G_4$. Every drawn machine is taken at 1.0 pu here; that is the only reading under which the network closes, and it makes all four internal EMFs a single node, which is what permits the delta–star reduction below. Under any other reading the problem would need generator loadings that the paper never supplies.

Find. (a) the total fault current at bus 3; (b) each generator's contribution and the bus 1 and bus 2 voltages during the fault.

Figure 5.1 — Figure (3) of the paper. Four generators, each behind its own transformer; buses 1 and 2 are tied directly and both reach the faulted bus 3 through a 0.2 pu line.

Approach. With all four EMFs equal, the machines collapse into a single 1.0 pu node behind two parallel-combined source arms. The source node, bus 1 and bus 2 then form a delta; one delta–star transform turns the whole network into a series-parallel ladder, and back-substitution gives every bus voltage and every branch current by hand.

Part (a) — combine the machines behind each bus. All internal EMFs are equal, so their internal nodes may be merged into one reference source node $N$ at 1.0 pu. At bus 1 the two paths to $N$ are in parallel: $$x_a = \frac{(0.24)(0.31)}{0.24 + 0.31} = 0.135273\ \text{pu}$$ At bus 2, $G_3$ and $G_4$ are identical: $$x_b = \frac{(0.31)(0.31)}{0.62} = 0.155000\ \text{pu}$$
Transform the source delta into a star. The three elements $x_a$ ($N$ to bus 1), $x_b$ ($N$ to bus 2) and the tie $x_c = 0.20$ (bus 1 to bus 2) form a delta. With $\Sigma = x_a + x_b + x_c = 0.490273$, $$Z_N = \frac{x_ax_b}{\Sigma} = 0.042767, \qquad Z_1 = \frac{x_ax_c}{\Sigma} = 0.055183, \qquad Z_2 = \frac{x_bx_c}{\Sigma} = 0.063230$$ all in per unit. The tie line is now absorbed into the star and never has to be handled separately.

Figure 5.2 — the delta formed by the two source arms and the bus 1–2 tie, converted to its equivalent star. After the transform the network is a simple series-parallel ladder.

Collapse the ladder to the Thevenin reactance at the fault. From the star centre, two arms reach bus 3: $$Z_1 + x_{13} = 0.255183, \qquad Z_2 + x_{23} = 0.263230$$ In parallel these give $0.129571$ pu, and adding the star arm to the source node, $$\boxed{X_{th} = Z_N + 0.129571 = 0.172339\ \text{pu}}$$ The fault current follows at once, since the prefault voltage at bus 3 is 1.0 pu: $$\boxed{I_f = \frac{1.0}{j0.172339} = 5.8025\angle -90^\circ\ \text{pu}}$$ This is a bolted fault, so bus 3 sits at exactly zero volts and the whole prefault voltage appears across $X_{th}$.
Part (b) — divide the current between the two feeding lines. The two arms of the parallel pair share $I_f$ in inverse proportion to their reactances: $$I_{1\to3} = I_f\frac{0.263230}{0.518413} = 2.9463\ \text{pu}, \qquad I_{2\to3} = I_f\frac{0.255183}{0.518413} = 2.8562\ \text{pu}$$ The two sum back to 5.8025 pu, which is a free check on the division.
Read the bus voltages straight off those line currents. Bus 3 is held at zero by the bolted fault, and the only element between bus 1 and bus 3 is the 0.2 pu line, so $$\boxed{V_1 = I_{1\to3}\,x_{13} = 2.9463(0.20) = 0.58926\ \text{pu}}$$ $$\boxed{V_2 = I_{2\to3}\,x_{23} = 2.8562(0.20) = 0.57125\ \text{pu}}$$ Bus 1 holds slightly higher because it is the electrically stiffer of the two — its source arm is 0.1353 pu against 0.1550 pu at bus 2.
Distribute the contributions back to the individual machines. Each machine drives its own bus through its own series reactance, so $$I_{G1} = \frac{1 - V_1}{0.24} = \frac{0.41074}{0.24} = 1.7114\ \text{pu}, \qquad I_{G2} = \frac{1 - V_1}{0.31} = 1.3250\ \text{pu}$$ $$I_{G3} = I_{G4} = \frac{1 - V_2}{0.31} = \frac{0.42875}{0.31} = 1.3831\ \text{pu}$$ All four are at $-90^\circ$ on a purely reactive network.
Close with two independent checks. The four machine contributions must total the fault current: $$1.7114 + 1.3250 + 1.3831 + 1.3831 = 5.8025\ \text{pu}\ \checkmark$$ And the tie line must carry the difference between what bus 1 receives and what it sends into the fault: $$I_{1\to2} = \frac{V_1 - V_2}{x_c} = \frac{0.58926 - 0.57125}{0.20} = 0.09007\ \text{pu}$$ Bus 1 receives $1.7114 + 1.3250 = 3.0364$ pu from its machines and sends $2.9463$ pu down line 1–3, leaving $0.0901$ pu — which is exactly the tie-line current. A nodal-admittance solve of the untransformed five-branch network reproduces $V_1$, $V_2$ and $I_f$ to nine decimals, confirming the reduction.

Question 5 — final results (all currents at −90°, on the common base)
Quantity	Value (pu)
(a) Thevenin reactance at bus 3	0.172339
(a) Total fault current $I_f$	5.8025
(b) Contribution of $G_1$	1.7114
(b) Contribution of $G_2$	1.3250
(b) Contribution of $G_3$	1.3831
(b) Contribution of $G_4$	1.3831
(b) Voltage at bus 1	0.58926
(b) Voltage at bus 2	0.57125
Current in line 1–3 / line 2–3	2.9463 / 2.8562
Current in the bus 1–2 tie	0.09007 (bus 1 → bus 2)

Topic. Symmetrical three-phase fault analysis by network reduction.

Key relations. Thevenin equivalent at the fault: $I_f = V_{pf}/jX_{th}$ with $V_{pf} = 1.0$ pu. Delta–star: $Z_N = x_ax_b/\Sigma$, $Z_1 = x_ax_c/\Sigma$, $Z_2 = x_bx_c/\Sigma$ with $\Sigma = x_a+x_b+x_c$. Current division $I_k = I_f Z_{other}/(Z_1+Z_2)$; bus voltage during a bolted fault $V_k = I_{k\to f}x_{kf}$; machine contribution $(1 - V_k)/x_k$.

Why this works. A balanced three-phase fault excites only the positive-sequence network, so no symmetrical-component machinery is needed — the single-phase positive-sequence diagram is the whole problem. Because every source is at the same 1.0 pu, all internal nodes are one node, and superposition is unnecessary: the prefault load current is ignored (the classical assumption) and the entire prefault voltage drives the fault. The delta–star transform is what makes a meshed network hand-solvable; once it is done the reduction is a ladder, and the back-substitution recovers every bus voltage without ever forming or inverting a bus impedance matrix. Reserve the matrix method for the independent check.

Common pitfall. Trying to series-parallel a meshed network. Buses 1 and 2 are joined both directly and through bus 3, so no pair of elements is in simple series or parallel until the delta is transformed; students who force a series-parallel combination typically drop the tie line and overstate $I_f$. The second pitfall is computing generator contributions from the star arms rather than from the machines' own reactances — the star is a mathematical equivalent and its arms carry no physical machine.

Source. El-Hawary, Electrical Power Systems: Design and Analysis, ch. 10 (faults on power systems); Anderson, Analysis of Faulted Power Systems, ch. 2; Glover, Sarma & Overbye, ch. 7.

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