22-Elec-B7 Power Systems Engineering · December 2016

Question 5 of 7: Balanced Three-Phase Fault and the Line 1–2 Current

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

Notes on this paper

Paper format. National Exams, December 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. This exam code follows M. E. El-Hawary, Electrical Power Systems: Design and Analysis (IEEE Press) — the notation is his throughout: the “cantilever” transformer equivalent of Problem 3, the two-reaction salient-pole construction of Problem 2, the ABCD two-port of Problem 1 and the sequence-network fault reduction of Problem 6. Corroborating references: J. D. Glover, M. S. Sarma and T. J. Overbye, Power System Analysis and Design, 6th ed. (ch. 3 transformers, ch. 4–5 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 C22.2 and the IEEE C62 / IEEE 142 (Green Book) series on system grounding.

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

(i) Problem 1 does not state a frequency. Parts (c) and (d) need one, and the paper is a Canadian national exam, so 60 Hz is used throughout — consistent with the 60 Hz stated in Problems 3 and 7. At 50 Hz the capacitance and surge impedance are unchanged (they do not involve frequency) and only the propagation constant of part (d) rescales by 5/6.

(ii) Problem 2 prints “neglecting armature reaction”. Armature reaction is precisely what the direct- and quadrature-axis reactances represent, so the phrase is read as its standard counterpart, neglecting armature resistance — the only reading under which the two given reactances can be used at all.

(iii) Problem 6 prints a part label “c-” with no question text. The five parts that do carry text (a, b, d, e, f) are marked 4 points each and total the 20 points the paper allots to every problem, so nothing is missing from the assessment; the sequence networks that the empty label most plausibly asked for are drawn anyway.

(iv) Problem 5 gives no reactance between buses 2 and 3. Figure (3) confirms the topology: bus 3 hangs radially off bus 1 through the j0.12 pu line, and the only other tie is line 1–2.

Problem 5: Balanced Three-Phase Fault and the Line 1–2 Current (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.

Quantity	Symbol	Value
Generator 1 reactance	$X_{G1}$	0.25 pu
Generator 2 reactance	$X_{G2}$	0.20 pu
Transformer T1	$X_{T1}$	0.04 pu
Transformer T2	$X_{T2}$	0.05 pu
Line 1–2	$X_{12}$	0.10 pu
Line 1–3	$X_{13}$	0.12 pu
Pre-fault voltage (flat start)	$E$	$1.0\angle 0^{\circ}$ pu at both machines
Fault	—	bolted balanced three-phase at bus 3

Find. The per-unit current flowing in line 1–2 while a bolted three-phase fault is held on bus 3.

Figure 5 — Positive-sequence reactance diagram. Each machine merges with its transformer into one branch from the reference to its bus; bus 3 hangs radially off bus 1 through line 1–3, so the entire fault current must pass through that line and the network reduces by inspection — no delta–star transform is needed.

Approach. Because a balanced three-phase fault excites only the positive-sequence network, and because both machines are represented as ideal 1.0 pu sources behind their reactances, the network reduces to a Thévenin equivalent at bus 3 by simple series–parallel combination. The fault current then divides between the two supply paths in inverse proportion to their reactances, and the share taken by the path through bus 2 is the current in line 1–2.

Merge each machine with its transformer. Generator 1 and T1 are in series between the reference and bus 1; generator 2 and T2 likewise between the reference and bus 2: $$X_{a}=X_{G1}+X_{T1}=0.25+0.04=0.29\ \text{pu},\qquad X_{b}=X_{G2}+X_{T2}=0.20+0.05=0.25\ \text{pu}$$ Both machine internal emfs are at $1.0\angle 0^{\circ}$ pu, so they can be tied to a single reference node — that is what licenses the series–parallel reduction that follows.
Collapse the two supply paths seen from bus 1. Looking back into the network from bus 1 there are exactly two routes to the reference: directly through $X_a=0.29$, and through line 1–2 and then generator 2, $X_{12}+X_b=0.10+0.25=0.35$ pu. In parallel, $$X_{\text{th},1}=\frac{(0.29)(0.35)}{0.29+0.35}=\frac{0.1015}{0.64}=0.158594\ \text{pu}$$ There is no third path, because bus 3 is a radial spur — nothing else ties buses 2 and 3 together.
Add the faulted line to reach the Thévenin reactance at bus 3. The whole of line 1–3 lies between bus 1 and the fault, in series with the equivalent just formed: $$X_{\text{th},3}=X_{13}+X_{\text{th},1}=0.12+0.158594\;\Longrightarrow\;\boxed{\ X_{\text{th},3}=0.278594\ \text{pu}\ }$$
Compute the total fault current at bus 3. With a pre-fault (flat) voltage of $1.0\angle 0^{\circ}$ and a purely reactive network, $$\mathbf{I}_F=\frac{1.0\angle 0^{\circ}}{j0.278594}=3.5895\angle{-90^{\circ}}\ \text{pu}$$ The current lags the pre-fault voltage by exactly $90^{\circ}$, as it must when every element is a pure reactance.
Divide that current between the two paths feeding bus 1. All of $\mathbf{I}_F$ flows through line 1–3, then splits at bus 1 between the local machine branch and the branch through line 1–2. The current divider assigns each branch the ratio of the other branch to the sum: $$\mathbf{I}_{12}=\mathbf{I}_F\frac{X_a}{X_a+X_{12}+X_b}=3.5895\angle{-90^{\circ}}\times\frac{0.29}{0.64}=3.5895\angle{-90^{\circ}}\times0.453125$$ $$\boxed{\ \mathbf{I}_{12}=1.6265\angle{-90^{\circ}}\ \text{pu}\ }$$ flowing from bus 2 towards bus 1, that is towards the fault.
Confirm the split and report the companion current. The branch through generator 1 carries $\mathbf{I}_{G1}=3.5895\times(0.35/0.64)=1.9630$ pu, and the two shares add back to $1.6265+1.9630=3.5895$ pu, which is the free check on the divider. Generator 1 supplies 55 per cent of the fault and generator 2 the remaining 45 per cent, despite generator 2 being the stronger machine — the 0.10 pu line in its path is what tips the balance.
Take the bus voltages during the fault as an independent cross-check. Bus 3 is held at zero by the bolted fault. Bus 1 sits one line-drop above it, and bus 2 one further drop above that: $$\left|V_1\right|=\left|I_F\right|X_{13}=3.5895(0.12)=0.4307\ \text{pu},\qquad \left|V_2\right|=\left|V_1\right|+\left|I_{12}\right|X_{12}=0.4307+0.1626=0.5934\ \text{pu}$$ and the same $\left|V_2\right|$ follows from the other side as $1-\left|I_{12}\right|X_b=1-1.6265(0.25)=0.5934$ pu. The two independent routes agreeing to four figures confirms both the reduction and the divider.

Quantity	Symbol	Result
Machine 1 branch (G1 + T1)	$X_a$	0.29 pu
Machine 2 branch (G2 + T2)	$X_b$	0.25 pu
Thévenin reactance at bus 3	$X_{\text{th},3}$	0.278594 pu
Total fault current at bus 3	$I_F$	$3.5895\angle{-90^{\circ}}$ pu
Current in line 1–2	$I_{12}$	$1.6265\angle{-90^{\circ}}$ pu
Current from generator 1	$I_{G1}$	$1.9630\angle{-90^{\circ}}$ pu
Bus 1 voltage during the fault	$\|V_1\|$	0.4307 pu
Bus 2 voltage during the fault	$\|V_2\|$	0.5934 pu

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

Key relations. $X_{\text{th}}$ by series–parallel reduction with all emfs tied to one node; $I_F=E/(jX_{\text{th}})$; current divider $I_k=I_F X_{\text{other}}/\sum X$; bus voltage during fault $V_k=E-I_k X_k$; $Z_{\text{bus}}$ formulation $I_F=V_k^{0}/Z_{kk}$.

Why this works. A balanced three-phase fault is the only fault that leaves the system symmetrical, so it excites the positive-sequence network alone and no symmetrical-component machinery is needed. Treating both machine emfs as $1.0\angle 0^{\circ}$ — the flat-start or classical assumption — puts them at the same potential, which means they can be merged into one reference node; that is what turns a two-source network into a single-source one and makes series–parallel reduction legitimate. The assumption neglects pre-fault load current, which is standard practice for breaker duty and relay setting because the fault current dominates it by an order of magnitude. The radial position of bus 3 is the second simplification: with only one path into the faulted bus, every ampere of fault current is line 1–3 current, and the topology never needs a delta–star transform.

Common pitfall. The classic slip is to invert the current divider, assigning each branch the ratio of its own reactance rather than the other branch's, which here would give 1.963 pu in line 1–2 instead of 1.626 pu. The check that catches it every time is that the smaller reactance must carry the larger current. A second error is to add $X_{13}$ into the parallel combination rather than in series after it.

Source. El-Hawary, Electrical Power Systems: Design and Analysis, ch. 10 (fault calculations, network reduction); Glover, Sarma & Overbye, §7.1–7.4 (three-phase faults, $Z_{\text{bus}}$ method); Grainger & Stevenson, ch. 10.

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