9.1 The single-line diagram of a three-phase power system is shown in Figure 9.17. Equipment ratings are given as follows: Synchronous generators: GI 1000 MVA 15 kV G2 1000 MVA 15 kV G3 500 MVA 13.8 kV 13.8 kV G4 750 MVA Transformers: TI 1000 MVA T2 1000 MVA T3 500 MVA T4 750 MVA Transmission lines: 1-2 765 kV 1-3 765 kV 2-3 765 kV XX₂ = 0.18, X, -0.07 per unit X=X₂-0.20, X, 0.10 per unit XX₂ = 0.15, X, = 0.05 per unit X-0.30, X₂= 0.40, X, 0.10 per unit 15 kV A/765 kV Y 15 kV A/765 kV Y 15kV Y/765kVY 15 kV Y/765 kVY X = 0.10 per unit X = 0.10 per unit X = 0.12 per unit X= 0.11 per unit Χ = 50 Ω, X = 150 Ω X = 40 Ω, X = 100 Ω X = 100 Ω Χ, " 40 Ω, Line 1-3 Line 1-2 T₂ Line 2-3 The inductor connected to Generator 3 neutral has a reactance of 0.05 per unit using generator 3 ratings as a base. Find fault current occurs at bus 3 if: 1- L-G fault actual 2- L-L-G fault 3- L-L fault base. using a 1000-MVA, 765-kV base in the zone of line 1-2 25 ase والمسيرا FI Pr

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9.1
The single-line diagram of a three-phase power system is shown in
Figure 9.17. Equipment ratings are given as follows:
Synchronous generators:
GI 1000 MVA 15 kV
G2 1000 MVA 15 kV
G3 500 MVA
G4 750 MVA
13.8 kV
13.8 kV
Transformers:
TI 1000 MVA
T2 1000 MVA
T3 500 MVA
T4 750 MVA
Transmission lines:
1-2 765 kV
1-3 765 kV
2-3 765 kV
X = X₂ = 0.18, X, = 0.07 per unit
X=X₂-0.20, X, 0.10 per unit
X = X₂ = 0.15, X, = 0.05 per unit
X=0.30, X₂ = 0.40, Xo = 0.10 per unit
15 kV A/765 kVY
15 kV A/765 kVY
15kV Y/765kV Y
15 kV Y/765 kVY
Χ
= 50 Ω,
X
= 40 Ω,
Χ = 40 Ω,
X =
X-0.10 per unit
X = 0.10 per unit
X = 0.12 per unit
X = 0.11 per unit
150 Ω
Χο = 100 Ω
X = 100 Ω
Line 1-2
Line 1-3
=
2
Line 2-3
3
The inductor
connected to
Generator 3 neutral
has a reactance of
0.05 per unit using
generator 3 ratings
as a base. Find
fault current
occurs at bus 3 if:
1- L-G fault actual
2- L-L-G fault 3-
L-L fault base.
using a 1000-MVA,
765-kV base in the
zone of line 1-2
T₂
ase
FIGURE 9.17
Problem 9.1
Transcribed Image Text:9.1 The single-line diagram of a three-phase power system is shown in Figure 9.17. Equipment ratings are given as follows: Synchronous generators: GI 1000 MVA 15 kV G2 1000 MVA 15 kV G3 500 MVA G4 750 MVA 13.8 kV 13.8 kV Transformers: TI 1000 MVA T2 1000 MVA T3 500 MVA T4 750 MVA Transmission lines: 1-2 765 kV 1-3 765 kV 2-3 765 kV X = X₂ = 0.18, X, = 0.07 per unit X=X₂-0.20, X, 0.10 per unit X = X₂ = 0.15, X, = 0.05 per unit X=0.30, X₂ = 0.40, Xo = 0.10 per unit 15 kV A/765 kVY 15 kV A/765 kVY 15kV Y/765kV Y 15 kV Y/765 kVY Χ = 50 Ω, X = 40 Ω, Χ = 40 Ω, X = X-0.10 per unit X = 0.10 per unit X = 0.12 per unit X = 0.11 per unit 150 Ω Χο = 100 Ω X = 100 Ω Line 1-2 Line 1-3 = 2 Line 2-3 3 The inductor connected to Generator 3 neutral has a reactance of 0.05 per unit using generator 3 ratings as a base. Find fault current occurs at bus 3 if: 1- L-G fault actual 2- L-L-G fault 3- L-L fault base. using a 1000-MVA, 765-kV base in the zone of line 1-2 T₂ ase FIGURE 9.17 Problem 9.1
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