komponen simetri
DESCRIPTION
Teknik elektroTRANSCRIPT
Nama:Febry Ardianto
NRP:2208100178
Tugas ANALISA SISTEM TENAGA
SIMULASI MENGGUNAKAN MATLAB
HUBUNG SINGKAT 3 FASA PADA BUS 4
>>
%Problem : SHORT CIRCUIT
%
The Fault Current, bus voltage and line current is use
%
the lgfault, llfault, and dlgfault functions
% Impedansi Urutan
zdata1=[040.00000.20000
0 5 0.00000.20000
120.00000.10000
130.00000.10000
140.00000.05000
230.00000.10000
250.00000.05000];
zdata0=[020.00000.05000
040.00000.14000
050.00000.14000
120.00000.30000
130.00000.30000
140.00000.05000
230.00000.30000
259999999999999];
% Jenis Gangguan
zdata2 = zdata1
Zbus0 = zbuild(zdata0)
Zbus1 = zbuild(zdata1)
Zbus2 = Zbus1;
disp('(a) Symmetrical three-phase fault')
symfault(zdata1, Zbus1)
zdata2 =
04.000000.2000
05.000000.2000
1.00002.000000.1000
1.00003.000000.1000
1.00004.000000.0500
2.00003.000000.1000
2.00005.000000.0500
Zbus0 =
0.0000 + 0.1080i0.0000 + 0.0216i0.0000 + 0.0648i0.0000 + 0.0795i-0.0000 + 0.0000i
0.0000 + 0.0216i0.0000 + 0.0443i0.0000 + 0.0330i0.0000 + 0.0159i-0.0000 + 0.0000i
0.0000 + 0.0648i0.0000 + 0.0330i0.0000 + 0.1989i0.0000 + 0.0477i-0.0000 + 0.0000i
0.0000 + 0.0795i0.0000 + 0.0159i0.0000 + 0.0477i0.0000 + 0.0955i-0.0000 + 0.0000i
-0.0000 + 0.0000i-0.0000 + 0.0000i-0.0000 + 0.0000i-0.0000 + 0.0000i0.0000 + 0.1400i
Zbus1 =
0 + 0.1397i 0 + 0.1103i 0 + 0.1250i 0 + 0.1118i 0 + 0.0882i
0 + 0.1103i 0 + 0.1397i 0 + 0.1250i 0 + 0.0882i 0 + 0.1118i
0 + 0.1250i 0 + 0.1250i 0 + 0.1750i 0 + 0.1000i 0 + 0.1000i
0 + 0.1118i 0 + 0.0882i 0 + 0.1000i 0 + 0.1294i 0 + 0.0706i
0 + 0.0882i 0 + 0.1118i 0 + 0.1000i 0 + 0.0706i 0 + 0.1294i
(a) Symmetrical three-phase fault
Warning: See help sprintf for valid escape sequences.
> In symfault at 38
Enter Faulted Bus No. -> 4
Enter Fault Impedance Zf = R + j*X in complex form (for bolted fault enter 0). Zf = 0
Balanced three-phase fault at bus No. 4
Total fault current = 7.7273 per unit
Bus Voltages during fault in per unit
BusVoltageAngle
No.Magnitudedegrees
10.13640.0000
20.31820.0000
30.22730.0000
40.00000.0000
50.45450.0000
Line currents for fault at bus No. 4
FromToCurrentAngle
BusBusMagnitudedegrees
142.7273-90.0000
211.8182-90.0000
230.9091-90.0000
310.9091-90.0000
G45.0000-90.0000
4F7.7273-90.0000
G52.7273-90.0000
522.7273-90.0000
Another fault location? Enter 'y' or 'n' within single quote ->HUBUNG SINGKAT 1 FASA PADA BUS 4
>>
%Problem : SHORT CIRCUIT
%
The Fault Current, bus voltage and line current is use
%
the lgfault, llfault, and dlgfault functions
% Impedansi Urutan
zdata1=[040.00000.20000
0 5 0.00000.20000
120.00000.10000
130.00000.10000
140.00000.05000
230.00000.10000
250.00000.05000];
zdata0=[020.00000.05000
040.00000.14000
050.00000.14000
120.00000.30000
130.00000.30000
140.00000.05000
230.00000.30000
259999999999999];
% Jenis Gangguan
zdata2 = zdata1
Zbus0 = zbuild(zdata0)
Zbus1 = zbuild(zdata1)
Zbus2 = Zbus1;
disp('(b) Line-to-ground fault' )
lgfault(zdata0, Zbus0, zdata1, Zbus1, zdata2, Zbus2)
zdata2 =
04.000000.2000
05.000000.2000
1.00002.000000.1000
1.00003.000000.1000
1.00004.000000.0500
2.00003.000000.1000
2.00005.000000.0500
Zbus0 =
0.0000 + 0.1080i0.0000 + 0.0216i0.0000 + 0.0648i0.0000 + 0.0795i-0.0000 + 0.0000i
0.0000 + 0.0216i0.0000 + 0.0443i0.0000 + 0.0330i0.0000 + 0.0159i-0.0000 + 0.0000i
0.0000 + 0.0648i0.0000 + 0.0330i0.0000 + 0.1989i0.0000 + 0.0477i-0.0000 + 0.0000i
0.0000 + 0.0795i0.0000 + 0.0159i0.0000 + 0.0477i0.0000 + 0.0955i-0.0000 + 0.0000i
-0.0000 + 0.0000i-0.0000 + 0.0000i-0.0000 + 0.0000i-0.0000 + 0.0000i0.0000 + 0.1400i
Zbus1 =
0 + 0.1397i 0 + 0.1103i 0 + 0.1250i 0 + 0.1118i 0 + 0.0882i
0 + 0.1103i 0 + 0.1397i 0 + 0.1250i 0 + 0.0882i 0 + 0.1118i
0 + 0.1250i 0 + 0.1250i 0 + 0.1750i 0 + 0.1000i 0 + 0.1000i
0 + 0.1118i 0 + 0.0882i 0 + 0.1000i 0 + 0.1294i 0 + 0.0706i
0 + 0.0882i 0 + 0.1118i 0 + 0.1000i 0 + 0.0706i 0 + 0.1294i(b) Line-to-ground fault
Line-to-ground fault analysis
Enter Faulted Bus No. -> 4
Enter Fault Impedance Zf = R + j*X in complex form (for bolted fault enter 0). Zf = 0
Single line to-ground fault at bus No. 4
Total fault current = 8.4679 per unit
Bus Voltages during the fault in per unit
Bus -------Voltage Magnitude-------
No. Phase aPhase bPhase c
10.14450.95780.9578
20.45700.91520.9152
30.30080.93500.9350
40.00000.95570.9557
50.60150.91680.9168
Line currents for fault at bus No. 4
FromTo-----Line Current Magnitude----
BusBusPhase aPhase bPhase c
142.89060.09810.0981
211.92700.06540.0654
230.96350.03270.0327
310.96350.03270.0327
4F8.46790.00000.0000
521.99250.99620.9962
Another fault location? Enter 'y' or 'n' within single quote ->
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