Bolted Three-Phase Fault on a Two-Machine Network: 22-Elec-B7 Power Systems Engineering, May 2013, Q5

Problem 5: Bolted Three-Phase Fault on a Two-Machine Network (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.

Element	Reactance (pu)	Between
$G_1$ + its transformer	$0.20 + 0.04 = 0.24$	internal e.m.f. of $G_1$ and bus 1
$G_2$ + its transformer	$0.25 + 0.06 = 0.31$	internal e.m.f. of $G_2$ and bus 2
line $\ell_{12}$	$0.10$	bus 1 and bus 2
line $\ell_{13}$	$0.10$	bus 1 and bus 3
line $\ell_{23}$	$0.10$	bus 2 and bus 3
$E_1 = E_2$	$1.0\angle 0^\circ$	both sources, on the same base

Find. The total current into a bolted three-phase fault at bus 3, the share of that current contributed by each machine, and the retained voltages at buses 1 and 2 while the fault is on.

Approach. A bolted symmetrical fault is a positive-sequence-only problem. Combine each machine with its transformer, build the positive-sequence reactance network, reduce it to a Thévenin equivalent at bus 3, then use superposition (pre-fault voltages plus the fault-current injection) to get the bus voltages and the branch flows.

Figure 5.1 — positive-sequence reactance diagram. Because both sources are 1.0∠0° pu the pre-fault network carries no current, so the pre-fault voltage everywhere — and in particular at bus 3 — is 1.0 pu.

Assemble the reactance network and note the pre-fault state. Each generator is represented by its e.m.f. behind the series combination of its own reactance and its step-up transformer, giving $x_{g1} = 0.20+0.04 = 0.24$ pu and $x_{g2} = 0.25+0.06 = 0.31$ pu. Because both e.m.f.s are stated as $1.0\angle 0^\circ$ pu and there is no load, no current circulates before the fault and the pre-fault voltage at every bus is $V^{(0)} = 1.0\angle 0^\circ$ pu. That is the Thévenin source voltage seen at bus 3.
Thévenin reactance at bus 3 by network reduction. Kill both sources (short the e.m.f.s to the reference) and look into bus 3. The two 0.1 pu legs run to bus 1 and bus 2; behind them sits a delta-free ladder in which bus 1 reaches reference through 0.24 pu, bus 2 through 0.31 pu, and the two buses are tied by 0.1 pu. Forming the bus admittance matrix $\mathbf{Y}_{bus}$ for buses 1, 2 and 3 (including the source shunts) and inverting it gives the impedance matrix, whose diagonal entry at bus 3 is the driving-point impedance: $$Z_{33}=j0.185514\ \text{pu} \;\Longrightarrow\;X_{th}=0.1855\ \text{pu}$$ The same number follows by hand: the parallel combination of $\left(0.24+0.1\right)$ pu through bus 1 and $\left(0.31+0.1\right)$ pu through bus 2 would give $0.1859$ pu, and the 0.1 pu tie between buses 1 and 2 shaves it to 0.1855 pu.
Part (a) — total fault current. For a bolted fault the fault impedance is zero, so $$I_f=\frac{V^{(0)}_3}{Z_{33}}=\frac{1.0\angle 0^\circ} {j0.185514} \;\Longrightarrow\;\boxed{I_f=5.3904\angle-90.00^\circ\ \text{pu}}$$ i.e. 5.3904 pu in magnitude, lagging the pre-fault voltage by 90° because the network is purely reactive. On a 100 MVA, 230 kV base that would be $5.390\times 100/(\sqrt{3}\times 230) = 1353$ A, which is the number a breaker would actually have to interrupt.
Bus voltages during the fault, by superposition. The faulted network is the pre-fault network plus an injection of $-I_f$ at bus 3, so $$V_i^{\,\text{fault}}=V_i^{(0)}-Z_{i3}\,I_f .$$ With $Z_{13} = j0.133622$ and $Z_{23} = j0.137405$ pu, $$\boxed{V_1=0.2797\angle 0^\circ\ \text{pu},\qquad V_2=0.2593\angle 0^\circ\ \text{pu}}$$ and of course $V_3 = 0$, which is what “bolted” means. Bus 1 holds slightly more voltage than bus 2 because it is fed through the stronger source (0.24 pu against 0.31 pu).
Part (b) — contribution of each machine. The current out of each machine is the drop across its own reactance divided by that reactance: $$I_{G1}=\frac{E_1-V_1}{jx_{g1}} =\frac{1.0-0.2797}{j0.24} =3.0012\angle-90.00^\circ\ \text{pu}$$ $$I_{G2}=\frac{E_2-V_2}{jx_{g2}} =\frac{1.0-0.2593}{j0.31} =2.3893\angle-90.00^\circ\ \text{pu}$$ $$\boxed{|I_{G1}|=3.0012\ \text{pu},\qquad |I_{G2}|=2.3893\ \text{pu}}$$ Their sum is 5.3904 pu, which reproduces $I_f$ exactly — the required check, since with no load every ampere leaving a machine must arrive at the fault.
Where the current actually flows. It is worth separating the machine contributions from the line flows, because they are not the same thing. The currents arriving at bus 3 along the two faulted-bus legs are $$I_{13}=\frac{V_1-V_3}{j0.1}=2.7972\ \text{pu},\qquad I_{23}=\frac{V_2-V_3}{j0.1}=2.5932\ \text{pu},$$ summing again to 5.3904 pu. The difference between $I_{G1}$ and $I_{13}$ is the 0.2040 pu that machine 1 pushes across the bus-1 to bus-2 tie line and which then reaches the fault through the bus-2 leg. A protection engineer needs exactly this split to set the directional and differential elements on each of the three lines.

Part	Quantity	Result
(a)	Thévenin reactance at bus 3, $X_{th}$	0.1855 pu
(a)	Total fault current, $I_f$	$5.3904\angle-90.00^\circ$ pu
(b)	Contribution of $G_1$	$3.0012\angle-90.00^\circ$ pu
(b)	Contribution of $G_2$	$2.3893\angle-90.00^\circ$ pu
(b)	Voltage at bus 1 during fault	0.2797 pu
(b)	Voltage at bus 2 during fault	0.2593 pu
check	Line flows into the fault $I_{13} + I_{23}$	2.7972 + 2.5932 = 5.3904 pu
check	Tie-line flow $I_{12}$	0.2040 pu

Topic. Symmetrical (three-phase) fault analysis by Thévenin equivalent and superposition.

Key relations. $I_f = V^{(0)}_k/(Z_{kk}+Z_f)$; $V_i^{\text{fault}} = V_i^{(0)} - Z_{ik}I_f$; $\mathbf{Z}_{bus} = \mathbf{Y}_{bus}^{-1}$; machine contribution $I_{G} = (E - V_{\text{terminal}})/jx_g$.

Why this works. A balanced three-phase fault leaves the system balanced, so only the positive-sequence network is involved and the whole problem is one linear circuit. The Thévenin construction works because the faulted network is the healthy network plus a single current injection at the fault bus: linearity lets us solve the healthy network (trivial here, since both e.m.f.s are equal and there is no load) and the injection-only network separately and add them. The column $Z_{ik}$ of the bus impedance matrix is exactly the set of transfer impedances that distributes that injection to every other bus, which is why $\mathbf{Z}_{bus}$ is the natural tool even for a three-bus problem.

Common pitfall. Two errors are common. The first is to add the machine reactances in parallel and ignore the tie line between buses 1 and 2, which here understates $X_{th}$ and overstates the fault current. The second is to assume the machine contributions divide in inverse proportion to $x_{g1}$ and $x_{g2}$ alone; they do not, because each machine also sees a different path to the fault, and part of $G_1$'s output reaches bus 3 through bus 2.

Source. El-Hawary, Electrical Power Systems: Design and Analysis, fault-analysis chapter; Glover, Overbye & Sarma, Power System Analysis and Design, 6th ed., §7.3–7.5 (bus impedance matrix and symmetrical faults); Grainger & Stevenson, Power System Analysis, Ch. 10.

Question 5 of 7: Bolted Three-Phase Fault on a Two-Machine Network

Problem 5: Bolted Three-Phase Fault on a Two-Machine Network (20 points)