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G1 1 G1 ratings: 100 MVA 138 kV R = 01 pu XS = 09 pu X' = 020 pu X" = 010 pu X2 = 010 pu Xg0 = 005 pu
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M2 2 M2 ratings: 50 MVA 138 kV R = 01 pu XS = 11 pu X' = 030 pu X" = 018 pu X2 = 015 pu Xg0 = 010 pu
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T1 ratings: 100 MVA 138/110 kV R = 001 pu X1 = 005 pu X2 = 005 pu X0 = 005 pu
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L1 impedance: R = 15 X1 = 75 X2 = 75 X0 = 125
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T2 ratings: 50 MVA 120/144 kV R = 001 pu X1 = 005 pu X2 = 005 pu X0 = 005 pu
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Figure P13-2 One-line diagram of the power system in Problem 13-8 SOLUTION To simplify this problem, we will pick the base quantities for this power system to be 100 MVA and 138 kV at generator G1 , which is in Region 1 Therefore, the base quantities are:
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Sbase,1 = 100 MVA Vbase,1 = 138 kV
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I base,1 = S3 ,base 3VLL,base 1
LL ,base 1
100, 000, 000 VA = 4184 A 3 (13,800 V )
Z base,1
(V =
S3 ,base
(13,800 V ) 2
100,000,000 VA
= 1904
Sbase,2 = 100 MVA 110 kV Vbase,2 = 138 kV = 110 kV 138 kV
I base,2 = S3 ,base 3VLL ,base 2
LL ,base 2
100,000, 000 VA = 5249 A 3 (110,000 V )
Z base,2
(V =
S3 ,base
(110,000 V ) 2
100,000,000 VA
= 121
Sbase,3 = 100 MVA 144 kV Vbase,3 = 110 kV = 132 kV 120 kV
I base,3 = S3 ,base 3VLL,base 3 = 100,000,000 VA = 4374 A 3 (13, 200 V )
Z base,3
(V =
LL ,base 3
S3 ,base
(13, 200 V ) 2
100,000,000 VA
= 1742
The per-unit impedances expressed on the system base are: 2 Vgiven Snew per-unit Z new = per-unit Z given Vnew Sgiven
(10-8)
RG1 = 010 pu X G1 = 010 pu X G1 = 020 pu X S ,G1 = 090 pu X 2,G1 = 010 pu X g 0,G1 = 005 pu RT 1 = 001 pu X 1,T 1 = 005 pu X 2,T 1 = 005 pu X 0,T 1 = 005 pu RT 2 120 kV 100 MVA = ( 001 pu ) = 00238 pu 110 kV 50 MVA
120 kV 100 MVA X 1,T 2 = ( 005 pu ) = 01190 pu 110 kV 50 MVA X 2,T 2 X 0,T 2 RM 2 120 kV 100 MVA = ( 005 pu ) = 01190 pu 110 kV 50 MVA 120 kV 100 MVA = ( 005 pu ) = 01190 pu 110 kV 50 MVA
2 2 2
138 kV 100 MVA = ( 010 pu ) = 02186 pu 132 kV 50 MVA
138 kV 100 MVA X M 2 = ( 018 pu ) = 03935 pu 132 kV 50 MVA 138 kV 100 MVA X M 2 = ( 030 pu ) = 06558 pu 132 kV 50 MVA X S ,M 2 X 2, M 2 138 kV 100 MVA = (110 pu ) = 24045 pu 132 kV 50 MVA 138 kV 100 MVA = ( 015 pu ) = 03279 pu 132 kV 50 MVA
2 2 2 2
138 kV 100 MVA X g 0, M 2 = ( 010 pu ) = 02186 pu 132 kV 50 MVA
Rline,pu = X 1,line,pu X 0,line,pu
Rline, 15 = = 0124 pu Z base 121 X 75 = X 2,line,pu = 1,line, = = 06198 pu Z base 121 X 125 = 0,line, = = 10331 pu Z base 121
The assumption that the system is initially unloaded means that the internal generated voltages of the generator and motor are = 100 0 before the fault The resulting per-unit sequence diagrams of the power system is shown below Note that the effect of transformer T2 on the zero sequence diagram
T1 L1 T2
j005
j06198
j01190 00238
010 jX" = j010 jX' = j020 jXS = j090
Positive Sequence
jX" = j03935 jX' = j06558 jXS = j24045 100 0
100 0
j005
j06198
j01190 00238
jX2 = j010
Negative Sequence
jXS = j03279 M2
j005
j10331
j01190 00238
jX0 = j005
Zero Sequence
jXS = j02186 M2
(a) Note that the fault occurs at Bus 3 , which is the high-voltage side of transformer T2 For the symmetrical three phase fault, we only need to short the positive sequence network at Bus 3 and calculate the current flows The input file required for program faults to calculate the fault current is shown below
% File describing the power system of Problem 13-8a, including % resistances % % System data has the form: %SYSTEM name baseMVA SYSTEM Prob13_8a 100 % % Bus data has the form: %BUS name volts BUS One 100 BUS Two 100 BUS Three 100 BUS Four 100 % % Note that transformers T1 and T2 are treated as "transmission lines" % here Transmission line data has the form: %LINE from to Rse Xse Gsh Bsh X0 Vis LINE One Two 00100 00500 0000 0000 00500 3 LINE Two Three 01240 06198 0000 0000 10331 3 LINE Three Four 00238 01190 0000 0000 01190 0 % % Generator data has the form: %GENERATOR bus R Xs Xp Xpp X2 X0 GENERATOR One 010 090 020 010 010 005 % % Motor data has the form: %MOTOR bus R Xs Xp Xpp X2 X0 MOTOR Four 02186 24045 06558 03935 03279 02186 % % type data has the form: %FAULT bus Calc Type Calc_time (0=all;1=sub;2=trans;3=ss) FAULT Three 3P 1
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