G1 500 MVA 13.8 KV x"=0.2 X₂ = 0.2 Xo 0.1 T1 Bara 1 SC AY 500 MVA 13.8 kV/ 500 KV X=0.12 0 T2 j50 st Bara 2 150 G2 150 S Bara 4 750 MVA 18 KV 750 MVA 500 kV/ 18 KV x=0.1 x=0.18 X2 = 0.18 Xo =0.09 Bara 3 Y fault condition in busbar 1. wwwwwwwwww T3 1000 MVA 500 kV/20 kV X=0.1 G3 1000 MVA 20 KV x" =0.17 X₂=0.2 Xo =0.09 For the power system with a single line diagram; a) By choosing the base values of 1000 MVA and 20 kV in the Generator G3 region, the system is zero, Draw the reactance diagrams of the positive and negative sequence circuits. (for each line x0 = x1 = x2 = 50 ohm) - b) Find the Thevenin equivalents according to the wwwwwwwwwwwwww

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Chapter7: Alternating Current, Power Distribution, And Voltage Systems
Section: Chapter Questions
Problem 22RQ
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KANA
G1
500 MVA
13.8 KV
x₂ = 0.2
Xo -0.1
Bara 1
SC
AY
500 MVA
13.8 KV/ 500 KV
x=0.12
-0
150
Bara 2
00
www
150 St
G2
150 St
Bara 4
Bara 3
750 MVA
500 kV/ 18 KV
x=0.1
750 MVA
18 KV
x"=0.18
X2 -0.18
Xo =0.09
DL
1000 MVA
500 kV/20 kV
X=0.1
G3
1000 MVA
20 kV
x"=0.17
X₂ -0.2
Xo =0.09
For the power system with a single line diagram;
a) By choosing the base values of 1000 MVA and 20
kV in the Generator G3 region, the system is zero,
Draw the reactance diagrams of the positive and
negative sequence circuits. (for each line x0 =
x1 = x2 = 50 ohm)
b) Find the Thevenin equivalents according to the
fault condition in busbar 1.
Transcribed Image Text:KANA G1 500 MVA 13.8 KV x₂ = 0.2 Xo -0.1 Bara 1 SC AY 500 MVA 13.8 KV/ 500 KV x=0.12 -0 150 Bara 2 00 www 150 St G2 150 St Bara 4 Bara 3 750 MVA 500 kV/ 18 KV x=0.1 750 MVA 18 KV x"=0.18 X2 -0.18 Xo =0.09 DL 1000 MVA 500 kV/20 kV X=0.1 G3 1000 MVA 20 kV x"=0.17 X₂ -0.2 Xo =0.09 For the power system with a single line diagram; a) By choosing the base values of 1000 MVA and 20 kV in the Generator G3 region, the system is zero, Draw the reactance diagrams of the positive and negative sequence circuits. (for each line x0 = x1 = x2 = 50 ohm) b) Find the Thevenin equivalents according to the fault condition in busbar 1.
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