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IG1 = ( 0630 815 )( 4184 A ) = 2636 815 A
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The subtransient, transient, and steady-state motor currents will be the same as the current flowing from Bus 4 to Bus 3 These currents are given in per-unit above The actual motor currents are found by multiplying by the base current in Region 3:
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I 2 = (1764 647 )( 4374 A ) = 7716 647 A M I 2 = (1232 726 )( 4374 A ) = 5389 726 A M I M 2 = ( 0394 845 )( 4374 A ) = 1723 845 A
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(c) The inclusion of the resistances changed the subtransient fault current by (3250-2999)/2999 100% = 84% It looks like we could neglect the resistances and still be within 10% of the proper fault current value 12-8 Assume that a symmetrical three-phase fault occurs at the terminals of motor M 1 on the low-voltage side of transformer T1 in the power system shown in Figure P12-3 Make the assumption that the power system is operating at rated voltage, and that it is initially unloaded Calculate the subtransient, transient, and steady-state fault current on the high-voltage side of the transformer, the low-voltage side of the transformer, and in the motor
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Region 1 Power System: V = 138 kV short-circuit MVA = 500 MVA T1
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Region 2 M1 1
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T1 ratings: 50 MVA 138/138 kV R = 001 pu X = 005 pu
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M1 ratings: 50 MVA 138 kV R = 01 pu XS = 11 pu X' = 030 pu X" = 018 pu
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Figure P12-3
One-line diagram of the power system in Problem 12-8
SOLUTION To simplify this problem, we will pick the base quantities for this power system to be 50 MVA and 138 kV at the high-voltage side of transformer T1 , which is in Region 1 Therefore, the base quantities are:
Sbase,1 = 50 MVA Vbase,1 = 138 kV
I base,1 = S3 ,base
3VLL,base 1
LL ,base 1
50,000,000 VA = 2092 A 3 (138,000 V )
Z base,1
(V =
S3 ,base
(138,000 V ) 2
50,000,000 VA
= 3809
Sbase,2 = 50 MVA
138 kV Vbase,2 = 138 kV = 138 kV 138 kV
I base,2 = S3 ,base
3VLL,base 2
LL ,base 2
50, 000, 000 VA = 2092 A 3 (13,800 V )
Z base,2
(V =
S3 ,base
(13,800 V ) 2
50,000,000 VA
= 3809
The per-unit impedances given in the problem are all correctly expressed on the system base without conversion The per-unit impedance of the power system can be calculated from Equation (12-26):
Short-circuit MVA pu = Vnominal,pu I SC ,pu
(12-26)
Since the voltage is assumed to be equal to the rated value, Vnominal,pu = 100 Therefore, the short-circuit current is
I SC ,pu = Short-circuit MVA pu = 500 MVA = 100 50 MVA
ZTH =
V 100 = = 010 pu I SC 100
(12-28) The resulting per-unit per-phase
If the impedance is treated as a pure reactance, Z TH = j 010 pu equivalent circuit is shown below:
T1 j005 1 001 2
010 j110 j030 j018
j010
100 0 Power System 100 0
The input file for this power system is shown below
% File describing the power system of Problem 12-8 % % System data has the form: %SYSTEM name baseMVA SYSTEM Prob12_8 100 % % Bus data has the form: %BUS name volts BUS One 100 BUS Two 100 % % Note that transformer T1 is treated as a "transmission line" % here Transmission line data has the form: %LINE from to Rse Xse Gsh Bsh X0 Vis LINE One Two 00100 00500 0000 0000 0000 0 % % Generator data has the form: %GENERATOR bus R Xs Xp Xpp X2 X0 GENERATOR One 010 090 020 010 000 000 % % Motor data has the form: %MOTOR bus R Xs Xp Xpp X2 X0 MOTOR Two 010 110 030 018 000 000 % % type data has the form: %FAULT bus Calc Type Calc_time (0=all;1=sub;2=trans;3=ss) FAULT Two 3P 0
The resulting outputs are shown below
>> faults prob_12_8_fault Input summary statistics: 27 lines in system file 1 SYSTEM lines 2 BUS lines 1 LINE lines 1 GENERATOR lines 1 MOTOR lines 1 TYPE lines Results for Case Prob12_8 Symmetrical Three-Phase Fault at Bus Two Calculating Subtransient Currents |====================Bus Information============|=====Line Information=====| Bus Volts / angle Amps / angle | To | Amps / angle | no Name (pu) (deg) (pu) (deg) | Bus | (pu) (deg) | |==========================================================================| 1 One 0274/ 2494 0000/ 000 Two 5376/ -5375 2 Two 0000/ 000 10212/ -5716 One 5376/ 12625 |==========================================================================|
Results for Case Prob12_8 Symmetrical Three-Phase Fault at Bus Two Calculating Transient Currents |====================Bus Information============|=====Line Information=====| Bus Volts / angle Amps / angle | To | Amps / angle | no Name (pu) (deg) (pu) (deg) | Bus | (pu) (deg) | |==========================================================================| 1 One 0187/ 1244 0000/ 000 Two 3661/ -6625 2 Two 0000/ 9000 6816/ -6871 One 3661/ 11375 |==========================================================================|
Results for Case Prob12_8 Symmetrical Three-Phase Fault at Bus Two Calculating Steady-State Currents |====================Bus Information============|=====Line Information=====| Bus Volts / angle Amps / angle | To | Amps / angle | no Name (pu) (deg) (pu) (deg) | Bus | (pu) (deg) | |==========================================================================| 1 One 0053/ -471 0000/ 000 Two 1046/ -8340 2 Two 0000/ 000 1046/ -8340 One 1046/ 9660 |==========================================================================|
The subtransient, transient, and steady-state fault currents on the high-voltage side of the transformer are the currents flowing from Bus 1 to Bus 2, referred to the high-voltage side They are calculated by the program above The actual fault currents are found by multiplying by the base current in Region 1:
IT 1 = ( 5376 538 )( 2092 A ) = 1125 538 A IT 1 = ( 3661 663 )( 2092 A ) = 766 663 A
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