EE  457  FINAL  PROJECT  Fault  Analysis  II  

Yanyi  He,  Eurydice  B  Ulysses  &  Mengqian  Ding  

 

Task1  

The  map  of  our  system  is  shown  below:  

 

 

The  one-­‐‑line  diagram  of  our  system  is  shown  below:  

   

 

Table  I:  Bus  Voltages  and  Angles  

Bus number Bus name Bus voltage (pu) Angle (degrees) 1 OWL 1.0 0 2 SWIFT 1.0 -1.58 3 PARROT 1.0 -3.08 4 LARK 0.9620 -6.57 5 JAY 0.9630 -6.79 6 RAVEN 0.9670 -6.76 7 WREN 0.9612 -7.15 8 ROBIN 0.9687 -6.32 9 SISKIN 0.9704 -4.22 10 JUNCO 0.9036 -10.02 11 QUAIL 0.9761 -4.60 12 HERON 0.9840 -4.19 13 EGRET 0.8955 -11.59 14 GULL 0.9926 -2.68 15 CROW 0.9691 -6.26 16 YANYI 0.9532 -7.92 17 EURY 0.9528 -6.39  

 

Table  II:  Real  and  Reactive  Power  Flow  and  Currents  at  all  Lines    

From Bus Name

To Bus Name

Real Power Flow (MW)

Reactive Power Flow (MVAR)

Current (Amps)

1 Owl 9 Siskin 111.0 26.4 409 11 Quail 78.2 9.7 282 14 Gull 5.9 -1.2 21 2 Swift 11 Quail 89.6 24.3 333 12 Heron 81.2 13.9 295 14 Gull 29.2 4.2 106 3 Parrot 6 Raven 68.1 28.0 322 12 Heron 17.1 13.9 110 15 Crow 64.9 31.8 333 4 Lark 5 Jay 35.8 -7.6 55 9 Siskin -65.8 -2.4 271

5 Jay 4 Lark -12.7 6.5 53 6 Raven -4.3 -12.1 48 7 Wren 34.2 1.4 127 8 Robin -27.2 -11.6 110 11 Quail -90.0 -14.3 340 6 Raven 3 Parrot -84.3 -25.7 327 5 Jay 4.3 10.7 43 7 Wren 5 Jay -34.2 -2.0 128 15 Crow -55.8 -18.0 219 8 Robin 5 Jay 27.2 10.4 108 12 Heron -67.2 -15.4 255 9 Siskin 1 Owl -109.4 -21.1 412 4 Lark 73.2 2.8 271 17 Siskin 26.2 13.3 109 10 Junco 13 Egret 10.6 .3 98 17 Siskin -25.6 -10.3 255 11 Quial 1 Owl -77.0 -8.7 285 2 Swift -88.7 -22.3 336 5 Jay 90.7 15.9 338 12 Heron 2 Swift -80.5 -13.0 297 3 Parrot -27.3 -17.3 118 8 Robin 67.7 15.3 253 13 Egret 10 Junco -10.5 -.3 98 16 Crow -19.5 -9.7 204 14 Gull 1 Owl -5.9 -2.8 23 2 Swift -29.1 -7.2 108 15 Crow 3 Parrot -86.1 -29.5 337 7 Wren 56.0 17.6 217 16 Crow 20.1 11.9 86 16 Crow 13 Egret 20.1 11.1 202 15 Crow -20.1 -11.1 202 17 Siskin 9 Siskin -26.2 -12.1 254 10 Junco 26.2 12.1 254

 

 

 

 

 

 

Table  III:  Postive  &  zero  sequence  line  impedance    

From Bus Name

To Bus Name

Positive sequence impedance (Magnitude)

Zero sequence impedance ((Magnitude))

Positive sequence impedance  

Zero sequence impedance  

1 Owl 9 Siskin 0.0684 0.2053 0.0119+j0.0674   0.035700+j0.2022  

11 Quail 0.1046 0.3138 0.0182+j0.103   0.054600+j0.309  

14 Gull 0.0804 0.2413 0.014+j0.0792   0.042000+j0.2376  

2 Swift 11 Quail 0.0631 0.1840 0.0107+j0.0604   0.032100+j0.1812  

12 Heron 0.0579 0.1736

0.01+j0.057   0.030000+j0.171  

14 Gull 0.0684 0.2053 0.0119+j0.0674   0.035700+j0.2022  

3 Parrot 6 Raven 0.0787 0.2361 0.013699+0.0775   0.041097+j0.2325  

12 Heron 0.0787 0.2361

0.0137+0.0775   0.041100+j0.2325  

15 Crow 0.0672 0.2017 0.0117+j0.0662   0.035100+j0.1986  

4 Lark 5 Jay 0.0250 0.075 0.005898+j0.0243   0.017694+j0.0729  

9 Siskin 0.0536 0.16.9 0.0093+j0.05281   0.027900+j0.15843  

5 Jay 6 Raven 0.0321 0.0964 0.0076+j0.03121   0.022800+j0.09363  

7 Wren 0.0178 0.0533 0.0042+j0.1728   0.012600+j0.05184  

8 Robin 0.0324 0.0973 0.0077+j0.03152   0.023100+j0.09456  

11 Quail 0.0556 0.1667 0.009696+j0.0547   0.029097+j0.1641  

7 Wren 15 Crow 0.0416 0.1249 0.007198+j0.041   0.021594+j0.123  

8 Robin 12 Heron 0.0279 0.08.8 0.004899+j0.0275   0.014697+j0.0825  

10 Junco 13 Egret 0.2225 0.6675 0.0637+j0.21319   0.191100+j0.63957  

17 Siskin 0.2519 0.7556 0.0721+j0.24133   0.216300+0.72399  

13 Egret 16 Crow 0.3407 1.0220 0.0975+j0.3264   0.292500+0.9792  

 

Table  IV:  Transformer    Impedance  

Transformer (9-17) 0.1333282 Transformer(15-16) 0.1333282

 

Table  V:  Phase  A  Current  

Bus Number 3 Phase Fault SLG Fault 1 52.3227 16.8305 2 56.016 17.6774 3 51.1001 17.2831 4 22.1130 10.0570 5 31.9231 12.9230 6 22.1770 10.1870 7 24.0667 10.7653 8 24.7105 10.8939 9 19.3930 9.0002

10 3.4744 2.3026 11 27.7341 11.6282 12 28.9724 12.0504 13 3.2859 2.1259 14 20.6743 9.3405 15 20.2304 9.4847 16 6.1588 5.6915 17 6.1066 5.6298

 

The bus with the highest fault current is Bus 2 (3-Phase fault). The bus with lowest fault current is Bus 13 (SLG fault).

Bus 1 has larger three-phase fault current than single line ground fault current. However there are no buses that have larger SLG fault currents than three-phase fault.

Bus number Three-Phase Line-Line DLG SLG 1 32.3227 10.1440 101.1608 16.8305

 

For the sequence voltages at the fault bus, refer to the attached report.

Ranking the fault types from highest to lowest fault currents: DLG > 3-Phase > SLG > LL

The sequence voltages (p.u) under SLG fault

Bus number Positive Negative Zero 1 (rectangular) 0.9126-j0.0621 -0.8699+j0.0957 :-0.0427-j0.0336 ()

1 (polar) 0.9147<-3.89 0.875<173.7 0.001435<90.0

Thevenin equivalent positive and zero sequence impedances (rectangular form) (p.u)

Bus number Positive Negative Zero 1 .001128+j0.019079 0.110417+j0.110197 0.000095+j0.009680

 

Under SLG fault, Phase A line current in each line connected to the faulted bus . Bus 1 is connected with Bus 9, 11 and 14.

Bus 9 Bus 11 Bus 14 Total Phase A current 1.65058<-59.71 1.905<-55.5 1.871<-51.3 3.47<-27.6

 

The total line current is not equal to Bus 1’s phase A current because Bus 1 also has generators, the fault current at Bus 1 should be equal to the total Phase A fault current of all connected lines and generator.

For the 3-phase fault, find the effect on the fault current of line outages of lines connected to the fault bus . Bus 1 is connected to Bus 9, 11 and 14. From the table, we can see that with the line outages, the fault currents at Bus 1 are reduced. Because the fault current from the outage line is not added to the total fault current at Bus 1.

Outage line Fault current 9 48.9463

11 48.3921 14 47.9810

 

We select a rural 161-KV line connecting Bus 1 and Bus 9.

Phase a with 3 phase fault 0.00958 0.04907i Phase a with SLG fault 0.1406 + 0.2372i Phase b &c with LL 0.1179 + 0.1100i Phase b & c with DLG 0.0337 + 0.0677i  

 

 

 

 

 

 

Task 2

For the selected rural 161-kv line connecting Bus 1 with Bus 9

(1) Apparent impedance : The same as before

Phase a with 3 phase fault 0.00958 0.04907i Phase a with SLG fault 0.1406 + 0.2372i Phase b &c with LL 0.1179 + 0.1100i Phase b & c with DLG 0.0337 + 0.0677i

   

(2) Using the positive sequence line impedance from Table IV since we studied the line 1-9, and Bus 9 connects to Bus 4. The Zone 1 apparent impedance is 0.8*0.0684 = 0.0547 The Zone 2 apparent impedance is 1.2*0.0684 = 0.0821 The Zone 3apparent impedance is 1.2*0.0684+ 1.5* 0.0536 = 0.1625 The Zone 1 impedance is 0.8*(0.0119 + j0.0674) = 0.0095 + 0.0539i The Zone 2 impedance is 1.2*(0.0119 + j0.0674) = 0.0143 + 0.0809i The Zone 3 impedance is (0.0119 + j0.0674) + 1.5* (0.0093 + j0.05281) = 0.0258 + 0.1466i  Z_base = 161*161/100 = 259.21. Line current is 271 amps, and voltage at Bus 9 is 0.9704 p.u. The load impedance is 0.9704*161*1000/271/259.21 = 2.2241 >> protection impedance  

(3) Impedance Relay of Bus 1:                        

Task 3 In EE 456 Project, We considered all the N-1 line contingencies. We have 21 lines and have 3 PV diagrams. The PV diagram is with respect to Bus 7.  From Line 1 to Line 8

   From Line 9 to Line 15

   From Line 16

         

 

Task 4  With regular generators, Transformer fault at bus 15. Series 1 is relative rotor angle of rotor 3. Series 2 is relative rotor angle of rotor 2. We trip the line 15-3 to stabilize the post fault power systems.  

   

With regular generators, Transformer fault at bus 9. Series 1 is relative rotor angle of rotor 3. Series 2 is relative rotor angle of rotor 2. We trip the line 9-1 to stabilize the post fault power systems.

   

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 With wind generator at Bus 20. The fault is at bus 9. Red line is the Wind rotor’s angle.  

     

Conclusion During the course of this project we were exposed to different type of analysis used in the power industry. New observations were made in addition to a comprehensive overview of the knowledge learn during our power system class.  

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