UNSYMMETRICAL FAULTS

Unsymmetrical Faults

Unsymmetrical Faults Most faults that occur in a power system are unsymmetrical. These include: Single line – ground Line – line Double line - ground

Unsymmetrical Faults Unsymmetrical faults cause unbalance in the system. Symmetrical components can therefore be used in the analysis of the faults. The standard 2 methods can be used: Thevenin’s method 𝑍𝑏𝑢𝑠 method

Unsymmetrical Faults The following notation is used in this analysis: 𝐼𝑓𝑎 , 𝐼𝑓𝑏 , 𝐼𝑓𝑐 - Fault current from each line into the fault 𝑉𝑗𝑎 , 𝑉𝑗𝑏 , 𝑉𝑗𝑐 - Line to ground voltages at bus j during the fault 𝑉𝑓 - Prefault voltage which is the line to neutral voltage before the fault occurs

Unsymmetrical Faults

Unsymmetrical Faults

Unsymmetrical Faults Positive sequence

Unsymmetrical Faults Negative sequence

Unsymmetrical Faults Zero sequence

Unsymmetrical Faults Bus voltage changes during the fault

Unsymmetrical Faults Bus voltage changes during the fault

Similar vectors will be obtained for the negative and zero sequence.

Unsymmetrical Faults Bus voltages during the fault

Unsymmetrical Faults Bus voltages during the fault

Unsymmetrical Faults Considering only the faulted bus k

Unsymmetrical Faults Faults may involve fault impedance 𝑍𝑓 between the faulted lines or between the faulted line and ground

Unsymmetrical Faults

𝑍𝑓 = 0 is referred to as a bolted fault This is the worst case scenario (highest fault current)

Unsymmetrical fault analysis process 1. Draw the network showing the fault Single line-ground faults 2. Analysis the network and set up current and voltage equations Single line-ground faults 3. Use the standard symmetrical equations of voltage to calculate symmetrical components Single line-ground faults Single line-ground faults

Unsymmetrical fault analysis process 4.

Using symmetrical components calculate the current and voltage at the fault (characteristic equations) Single line-ground faults 5. Form an equivalent circuit using the positive, negative and zero circuits to reflect the characteristic equations in 4 Single line-ground faults 6. Analyse the equivalent circuit and solve for the fault current Single line-ground fault - example

Single line-ground faults

At faulted bus k 𝐼𝑓𝑏 = 0 𝐼𝑓𝑐 = 0 𝑉𝑘𝑎 = 𝑍𝑓 𝐼𝑓𝑎

Single line-ground faults The symmetrical components of the current 𝐼𝑓𝑎 can the be calculated 𝐼𝑓𝑎(0) 1 1 (1) = 1 1 𝑎 𝐼𝑓𝑎 3 2 (2) 1 𝑎 𝐼𝑓𝑎 𝐼𝑓𝑎

(0)

= 𝐼𝑓𝑎

(1)

= 𝐼𝑓𝑎

(2)

1 𝑎2 𝑎

=

Or 𝐼𝑓𝑎 = 3 𝐼𝑓𝑎(0) = 3 𝐼𝑓𝑎

𝐼𝑓𝑎 0 0

𝐼𝑓𝑎 3 (1)

= 3 𝐼𝑓𝑎(2)

Single line-ground faults The symmetrical components of the voltage at the faulted bus k become: 𝑉𝑘𝑎(0) = −𝑍𝑘𝑘 (0) 𝐼𝑓𝑎(0) 𝑉𝑘𝑎(1) = 𝑉𝑓 −𝑍𝑘𝑘 (1) 𝐼𝑓𝑎(1) 𝑉𝑘𝑎(2) = −𝑍𝑘𝑘 (2) 𝐼𝑓𝑎(2)

Single line-ground faults The voltage at the faulted bus k becomes:

𝑉𝑘𝑎 = 𝑉𝑘𝑎(0) + 𝑉𝑘𝑎(1) + 𝑉𝑘𝑎(2) = −𝑍𝑘𝑘 (0) 𝐼𝑓𝑎(0) + 𝑉𝑓 −𝑍𝑘𝑘 (1) 𝐼𝑓𝑎(1) −𝑍𝑘𝑘 (2) 𝐼𝑓𝑎(2) = 𝑉𝑓 - (𝑍𝑘𝑘 (0) + 𝑍𝑘𝑘 (1) + 𝑍𝑘𝑘 (2) ) 𝐼𝑓𝑎(0) = 3 𝐼𝑓𝑎(0) 𝑍𝑓 Therefore 𝐼𝑓𝑎

(0)

= 𝐼𝑓𝑎

(1)

= 𝐼𝑓𝑎

(2)

=

𝑉𝑓 𝑍𝑘𝑘 (0) + 𝑍𝑘𝑘 (1)

+ 𝑍𝑘𝑘 (2)+3𝑍𝑓

Single line-ground faults 𝐼𝑓𝑎

(0)

= 𝐼𝑓𝑎

(1)

= 𝐼𝑓𝑎

(2)

=

𝑉𝑓 𝑍𝑘𝑘 (0) + 𝑍𝑘𝑘 (1)

Represents the following circuit

+ 𝑍𝑘𝑘 (2)+3𝑍𝑓

Single line-ground faults The Thevenin equivalent circuits of the three sequence networks are connected in series with a fault impedance of 3𝑍𝑛 and the prefault voltage source 𝑉𝑓

Single line-ground fault - example

Single line-ground fault - example

Single line-ground fault - example Positive sequence network

Negative sequence

Single line-ground fault - example Zero sequence network

Single line-ground fault - example The sequence networks are used to determine the 𝑍𝑏𝑢𝑠 matrix for each sequence 𝑍𝑏𝑢𝑠 (1) = 𝑍𝑏𝑢𝑠 (2) = 𝑗0.1437 𝑗0.1211 𝑗0.1211 𝑗0.1696 𝑗0.0789 𝑗0.1104 𝑗0.0563 𝑗0.0789

𝑗0.0789 𝑗0.1104 𝑗0.1696 𝑗0.1211

𝑗0.0563 𝑗0.0789 𝑗0.1211 𝑗0.1437

Single line-ground fault - example The sequence networks are used to determine the 𝑍𝑏𝑢𝑠 matrix for each sequence 𝑍𝑏𝑢𝑠 (0) = 𝑗0.1553 𝑗0.1407 𝑗0.0493 𝑗0.0347

𝑗0.1407 𝑗0.1999 𝑗0.0701 𝑗0.0493

𝑗0.0493 𝑗0.0701 𝑗0.1999 𝑗0.1407

𝑗0.0347 𝑗0.0493 𝑗0.1407 𝑗0.1553

Single line-ground fault - example Since the fault occurs at bus 3, 𝑍33 of the 𝑍𝑏𝑢𝑠 matrices are used in the analysis. The series connected equivalent circuit is

Single line-ground fault - example 𝐼𝑓𝐴 = =

(0)

= 𝐼𝑓𝐴

𝑍33(0) +

(1)

= 𝐼𝑓𝐴

(2)

=

𝑉𝑓 𝑍𝑘𝑘 (0) + 𝑍𝑘𝑘 (1)

1∠00 𝑍33(1) + 𝑍33(2) +3(0)

+ 𝑍𝑘𝑘 (2)+3𝑍𝑓

1∠00 𝑗0.1999+𝑗0.1696+𝑗0.1696

= -j1.8549 Therefore total fault current is 𝐼𝑓𝐴 = 3 𝐼𝑓𝐴(0) = -j5.5648 pu

Single line-ground fault - example 𝐼𝑏𝑎𝑠𝑒 = =

100 x 106 3 x 345 x 103

= 167.35 A

Therefore 𝐼𝑓𝐴 = -j5.5648 (167.35 A) = -j 931.37 A

Single line-ground fault - example The line to ground voltages at the terminals of machine 2: Machine 2 is connected to bus 4 on the low voltage side of transformer 𝑇2 Sequence voltages of phase a at bus 4 are 𝑉4𝑎(0) = −𝑍43(0) 𝐼𝑓𝐴(0) 𝑉4𝑎(1) = 𝑉𝑓 −𝑍43(1) 𝐼𝑓𝐴(1) 𝑉4𝑎(2) = −𝑍43(2) 𝐼𝑓𝐴(2)

Single line-ground fault - example 𝑉4𝑎(0) = - (j0.1407)(-j1.8549) = j0.2610 𝑉4𝑎(1) = 1∠00 - (j0.1211)(-j1.8549) = 0.7754 𝑉4𝑎(2) = - (j0.1211)(-j1.8549) = -0.2246

Single line-ground fault - example 𝑉4𝑎(0) = - (j0.1407)(-j1.8549) = j0.2610 𝑉4𝑎(1) = 1∠00 - (j0.1211)(-j1.8549) = 0.7754 𝑉4𝑎(2) = - (j0.1211)(-j1.8549) = -0.2246

𝑉𝑎 𝑉𝑏 𝑉𝑐

=

1 1 1

1 𝑎2 𝑎

1 𝑎 𝑎2

𝑉4𝑎(0) 𝑉4𝑎(1) 𝑉4𝑎(2)

Single line-ground fault - example 𝑉4𝑎(0) = - (j0.1407)(-j1.8549) = 0.2610 𝑉4𝑎(1) = 1∠00 - (j0.1211)(-j1.8549) = 0.7754 𝑉4𝑎(2) = - (j0.1211)(-j1.8549) = -0.2246 1 1 1 0.2610 𝑉𝑎 2 = 0.7754 1 𝑎 𝑎 𝑉𝑏 2 −0.2246 1 𝑎 𝑎 𝑉𝑐 0.2898∠00 = 1.0187∠ −121.80 1.0187∠121.80

Single line-ground fault - example Line to ground base voltage at machine 2 =

𝑉𝑎 𝑉𝑏 𝑉𝑐

3.346∠00 = 11.763∠ −121.80 kV 11.763121.80

20 3

kV

Single line-ground fault - example Subtransient current out of phase c of machine 2: First we determine the subtransient symmetrical currents in phase a

𝐼𝑎(0) = =

𝑉4𝑎(0)

𝑗𝑋0 𝑗0.2610 𝑗0.04

= j6.525

Single line-ground fault - example 𝐼𝑎

𝑉𝑓−𝑉4𝑎(1) = 𝑗𝑋" 1 − 0.7754 𝑗0.2

(1)

=

= -j1.123

𝐼𝑎

(2)

=

=

𝑉4𝑎(2) 𝑗𝑋2

0.2246 𝑗0.2

= -j1.123

Single line-ground fault - example 1 1 1

1 𝑎2 𝑎

1 𝑎 𝑎2

𝐼𝑎(0) 𝐼𝑎(1) 𝐼𝑎(2)

𝐼𝑎 𝐼𝑏 = 𝐼𝑐 𝐼𝑐 = 𝐼𝑎(0) + 𝑎 𝐼𝑎(1) + 𝑎2 𝐼𝑎(2) = -j6.525 + (1∠1200 )(-j10123) + (1∠2400 )(-j1.123) = -j5.402 pu 𝐼𝑏𝑎𝑠𝑒 = =

100 x 106 3 x 20 x 103

= 2886.8A

𝐼𝑐 = (-j5.402 )(2886.8A) = -j15594 A

Line - line faults

At the point at bus k 𝐼𝑓𝑎 = 0 𝐼𝑓𝑏 = -𝐼𝑓𝑐 𝑉𝑘𝑏 - 𝑉𝑘𝑐 = 𝑍𝑓 𝐼𝑓𝑏

Line - line faults The symmetrical components of the current 𝐼𝑓𝑎 can the be calculated 𝐼𝑓𝑎(0) 1 1 (1) = 1 1 𝑎 𝐼𝑓𝑎 3 2 (2) 1 𝑎 𝐼𝑓𝑎 𝐼𝑓𝑎(0) = 0 𝐼𝑓𝑎(1) = - 𝐼𝑓𝑎(2)

1 𝑎2 𝑎

0 𝐼𝑓𝑏 −𝐼𝑓𝑏

Line - line faults 𝐼𝑓𝑎(0) = 0 Implies that no zero sequence current is being injected into the network because of the fault. Therefore line-line fault analysis do not involve the zero sequence network

Line - line faults 𝐼𝑓𝑎(1) = - 𝐼𝑓𝑎(2) Implies that the positive and negative sequence networks are connected in parallel as shown below:

Line - line faults From the combined circuit

𝐼𝑓𝑎

(1)

= −𝐼𝑓𝑎

(2)

=

𝑉𝑓 𝑍𝑘𝑘 (1)

+ 𝑍𝑘𝑘 (2) +𝑍𝑓

Line - line faults -example

line

1. The currents in the fault 2. The line-line voltages at the fault bus 3. The line-line voltages at the terminals of machine 2

Line - line faults -example

Line - line faults -example Positive sequence network

Negative sequence

Line - line faults -example Zero sequence network

Line - line faults -example The sequence networks are used to determine the 𝑍𝑏𝑢𝑠 matrix for each sequence 𝑍𝑏𝑢𝑠 (1) = 𝑍𝑏𝑢𝑠 (2) = 𝑗0.1437 𝑗0.1211 𝑗0.1211 𝑗0.1696 𝑗0.0789 𝑗0.1104 𝑗0.0563 𝑗0.0789

𝑗0.0789 𝑗0.1104 𝑗0.1696 𝑗0.1211

𝑗0.0563 𝑗0.0789 𝑗0.1211 𝑗0.1437

Line - line faults -example Since the fault occurs at bus 3, 𝑍33 of the 𝑍𝑏𝑢𝑠 matrices are used in the analysis. The combined equivalent circuit is

Line - line faults -example 𝐼𝑓𝐴

(1)

= −𝐼𝑓𝐴

(2)

=

=

1∠00 𝑍33(1) + 𝑍33(2)

=

1∠00 𝑗0.1696+𝑗0.1696

= -j2.9481

𝑉𝑓 𝑍𝑘𝑘 (1)

+ 𝑍𝑘𝑘 (2)+𝑍𝑓

Line - line faults -example 𝐼𝑓𝐴 = 𝐼𝑓𝐴 0 + 𝐼𝑓𝐴(1) + 𝐼𝑓𝐴(2) = 0 - j2.9481 + j2.9481 = 0 (as expected) 𝐼𝑓𝐵 =

𝐼𝑓𝐵 (0) + 𝐼𝑓𝐵 (1) + 𝐼𝑓𝐵 (2)

= 𝑎2 𝐼𝑓𝐴(1) + a 𝐼𝑓𝐴(2) = (1∠2400 )(-j2.9481) + (1∠1200 )(j2.9481) = -5.106 𝐼𝑓𝐶 =

𝐼𝑓𝐶 (0) + 𝐼𝑓𝐶 (1) + 𝐼𝑓𝐶 (2)

= 𝑎 𝐼𝑓𝐴(1) + 𝑎2 𝐼𝑓𝐴(2) = (1∠1200 )(-j2.9481) + (1∠2400 )(j2.9481) = 5.106 𝐼𝑓𝐵 = - 𝐼𝑓𝐶 (as expected)

Line - line faults -example 𝐼𝑓𝐴 = 0 𝐴 𝐼𝑓𝐵 = 855 A 𝐼𝑓𝐶 = 855 A

Line - line faults -example Sequence voltages of phase A at bus 3 are

𝑉3𝑎(0) = 0 𝑉3𝑎(1) = 𝑉3𝑎(2) = 𝑉𝑓 −𝑍33(1) 𝐼𝑓𝐴(1) (see circuit) = 1∠00 - (j0.1696)(-j2.9481) = 0.5 Line - line faults -example

Line - line faults -example The line to ground voltages at bus 3 are 𝑉3𝐴 = 𝑉3𝐴(0) +𝑉3𝐴(1) + 𝑉3𝐴(2) = 1∠00 𝑉3𝐵 = 𝑉3𝐵(0) +𝑉3𝐵(1) + 𝑉3𝐵(2) =𝑎2 𝑉3𝐴(1) + 𝑎 𝑉3𝐴(2) = (1∠2400 )(0.5) + (1∠1200 )(0.5) =0.5∠1800

Line - line faults -example The line to ground voltages at bus 3 are 𝑉3𝐶 = 𝑉3𝐶 (0) +𝑉3𝐶 (1) + 𝑉3𝐶 (2) =𝑎𝑉3𝐴(1) +𝑎2 𝑉3𝐴(2) = (1∠1200 )(0.5) + (1∠2400 )(0.5) =0.5∠1800

𝑉3𝐵 = 𝑉3𝐶 (as expected)

Line - line faults -example The line to line voltages at bus 3 are 𝑉3𝐴𝐵 = 𝑉3𝐴 − 𝑉3𝐵 = 1.5∠00 𝑉3𝐵𝐶 = 𝑉3𝐵 − 𝑉3𝐶 = 0 𝑉3𝐶𝐴 = 𝑉3𝐶 − 𝑉3𝐴 = 1.5∠1800

Line - line faults -example The line to line voltages at bus 3 are: Since the phase voltage = 1 per unit, the base 345 voltage will be the phase voltage 𝑉𝑏 = = 199

kV 𝑉3𝐴𝐵 = 299∠00 kV 𝑉3𝐵𝐶 = 0 (as expected) 𝑉3𝐶𝐴 = 299∠1800 kV

3

Line - line faults -example The line to ground voltages at the terminals of machine 2: Machine 2 is connected to bus 4 on the low voltage side of transformer 𝑇2 Sequence voltages of phase a at bus 4 are 𝑉4𝑎(0) = −𝑍43(0) 𝐼𝑓𝐴(0) 𝑉4𝑎(1) = 𝑉𝑓 −𝑍43(1) 𝐼𝑓𝐴(1) 𝑉4𝑎(2) = −𝑍43(2) 𝐼𝑓𝐴(2)

Line - line faults -example 𝑉4𝑎(0) = 0 𝑉4𝑎(1) = 1∠00 - (j0.1211)(-j2.9481) = 0.643 𝑉4𝑎(2) = - (j0.1211)(-j2.9481) = 0.357

Line - line faults -example If phase shift is considered, and remembering that these voltages are on the low voltage side of the transformer 𝑉4𝑎(0) = 0 𝑉4𝑎(1) = 0.643 ∠−300 𝑉4𝑎(2) = 0.357 ∠300

Line - line faults -example 𝑉𝑎 𝑉𝑏 𝑉𝑐

=

1 1 1

1 𝑎2 𝑎

1 𝑎 𝑎2

𝑉4𝑎(0) 𝑉4𝑎(1) 𝑉4𝑎(2)

Line - line faults -example 𝑉4𝑎 𝑉4𝑏 𝑉4𝑐

=

1 1 1

1 𝑎2 𝑎

1 𝑎 𝑎2

0.8778∠ −9.40 = 0.8778∠ −170.60 0.286∠900

0 0.643 ∠ −300 0.357 ∠300

Line - line faults -example The line to line voltages at bus 4 are 𝑉4𝑎𝑏 = 𝑉4𝑎 − 𝑉4𝑏= 1.7322 𝑉4𝑏𝑐 = 𝑉3𝑏 − 𝑉3𝑐 = 0.9665∠-153.650 𝑉4𝑐𝑎 = 𝑉3𝑐 − 𝑉3𝑎 = 0.9665∠-153.650

Line - line faults -example base voltage at machine 2 =

𝑉4𝑎𝑏 𝑉4𝑏𝑐 𝑉4𝑐𝑎

20 3

kV

20∠00 = 11.2∠ −153.650 kV 11.2∠153.650

Double line – ground faults

At the point at bus k 𝐼𝑓𝑎 = 0 𝑉𝑘𝑏 = 𝑉𝑘𝑐 = (𝐼𝑓𝑏 + 𝐼𝑓𝑐 )𝑍𝑓

Double line – ground faults From the analysis equation 3𝐼𝑓𝑎(0) = 𝐼𝑓𝑎 + 𝐼𝑓𝑏 + 𝐼𝑓𝑐 Therefore

𝑉𝑘𝑏 = 𝑉𝑘𝑐 = 3𝐼𝑓𝑎(0) 𝑍𝑓

Double line – ground faults The symmetrical components of the voltage 𝑉𝑘𝑎 can the be calculated 𝑉𝑘𝑎(0) 1 1 1 𝑉𝑘𝑎 2 𝑉 (1) = 1 1 𝑎 𝑎 𝑉𝑘𝑎 𝑘𝑏 3 2 (2) 1 𝑎 𝑎 𝑉𝑘𝑏 𝑉𝑘𝑎 𝑉𝑘𝑎 −2𝑉𝑘𝑏 𝑉𝑘𝑎 = 3 2𝑉 𝑉 +𝑎𝑉 +𝑎 𝑘𝑎 𝑘𝑏 𝑘𝑏 𝑉𝑘𝑎(1) = 3 2 𝑉 +𝑎𝑉 𝑉 +𝑎 𝑘𝑎 𝑘𝑏 𝑘𝑏 𝑉𝑘𝑎(2) = 3 (0)

(1)

(2) (3)

Double line – ground faults From equation (2) and (3)

𝑉𝑘𝑎(1) = 𝑉𝑘𝑎(2) From equation (1) 3 𝑉𝑘𝑎(0) = 𝑉𝑘𝑎 − 2𝑉𝑘𝑏 = (𝑉𝑘𝑎(0) + 𝑉𝑘𝑎(1) + 𝑉𝑘𝑎(2) ) – 2(3𝐼𝑓𝑎(0) 𝑍𝑓 ) 2 𝑉𝑘𝑎(0) = 𝑉𝑘𝑎(1) + 𝑉𝑘𝑎(2) - 2(3𝐼𝑓𝑎(0) 𝑍𝑓 ) = 2 𝑉𝑘𝑎(1) - 2(3𝐼𝑓𝑎(0) 𝑍𝑓 ) 𝑉𝑘𝑎(1) = 𝑉𝑘𝑎(0) - 3𝐼𝑓𝑎(0) 𝑍𝑓

Double line – ground faults But 𝑉𝑘𝑎(1) = 𝑉𝑘𝑎(2) therefore 𝑉𝑘𝑎(2) = 𝑉𝑘𝑎(0) - 3𝐼𝑓𝑎(0) 𝑍𝑓

Also 𝐼𝑓𝑎 = 0 Implies 𝐼𝑓𝑎(0) + 𝐼𝑓𝑎(1) + 𝐼𝑓𝑎(2) = 0

Double line – ground faults These characteristic equations are obtained by connecting all three sequence networks in parallel

Double line – ground faults From the combined circuit

𝐼𝑓𝑎

(1)

=

𝑉𝑓 𝑍𝑘𝑘

(1)

𝑍𝑘𝑘 2 (𝑍𝑘𝑘 0 +3𝑍𝑓) + 𝑍𝑘𝑘 2 +𝑍𝑘𝑘 0 +3𝑍𝑓

Using current division of 𝐼𝑓𝑎(1) 𝐼𝑓𝑎(2) =-{

𝑍𝑘𝑘 0 +3𝑍𝑓 (1) } 𝐼 𝑍𝑘𝑘 (2) +𝑍𝑘𝑘 0 +3𝑍𝑓 𝑓𝑎

𝐼𝑓𝑎(0) =-{

𝑍𝑘𝑘 2 (1) } 𝐼 𝑍𝑘𝑘 (2) +𝑍𝑘𝑘 0 +3𝑍𝑓 𝑓𝑎

Double line – ground faults Note that if 𝑍𝑓 = ∞ (open circuit), the above equations reduce to those of a line-line fault

Double line – ground fault - example Find the subtransient current and line-line voltages at the fault when a double line-toground fault with 𝑍𝑓 = 0 occurs at the terminals of machine 2. Assume that the system is unloaded and operating at rated voltage when the fault occurs. Use the bus impedance matrices and neglect resistances.

Double line – ground fault - example

Single line-ground fault - example

Double line – ground fault - example Positive sequence network

Negative sequence

Double line – ground fault - example Zero sequence network

Line - line faults -example

Double line – ground fault - example The sequence networks are used to determine the 𝑍𝑏𝑢𝑠 matrix for each sequence 𝑍𝑏𝑢𝑠 (1) = 𝑍𝑏𝑢𝑠 (2) = 𝑗0.1437 𝑗0.1211 𝑗0.1211 𝑗0.1696 𝑗0.0789 𝑗0.1104 𝑗0.0563 𝑗0.0789

𝑗0.0789 𝑗0.1104 𝑗0.1696 𝑗0.1211

𝑗0.0563 𝑗0.0789 𝑗0.1211 𝑗0.1437

Single line-ground fault - example The sequence networks are used to determine the 𝑍𝑏𝑢𝑠 matrix for each sequence 𝑗0.19 0 0 𝑗0.08 (0) 𝑍𝑏𝑢𝑠 = 0 𝑗0.08 0 0

0 0 𝑗0.08 0 𝑗0.58 0 0 𝑗0.19

Double line – ground fault - example Since the fault occurs at bus 4, 𝑍44 of the 𝑍𝑏𝑢𝑠 matrices are used in the analysis. The combined equivalent circuit is

Double line – ground fault - example 𝐼𝑓𝑎 =

(1)

=

𝑉𝑓 𝑍𝑘𝑘

(1)

𝑍𝑘𝑘 2 (𝑍𝑘𝑘 0 +3𝑍𝑓) + 𝑍𝑘𝑘 2 +𝑍𝑘𝑘 0 +3𝑍𝑓

1∠00

j0.1437+ [

= -j4.4342

𝑗0.1437 𝑗0.19 𝑗0.1437+𝑗0.19

]

Double line – ground fault - example 𝐼𝑓𝑎(2) =-{ =-{

𝑍𝑘𝑘 0 +3𝑍𝑓 (1) } 𝐼 𝑍𝑘𝑘 (2) +𝑍𝑘𝑘 0 +3𝑍𝑓 𝑓𝑎

𝑗0.19 j0.19+𝑗0.1437]

= j2.5247

}(-j4.4342)

Double line – ground fault - example 𝐼𝑓𝑎(0) =-{ =-{

𝑍𝑘𝑘 2 (1) } 𝐼 𝑍𝑘𝑘 (2) +𝑍𝑘𝑘 0 +3𝑍𝑓 𝑓𝑎

𝑗0.1437 j0.1437+j0 .19]

= j1.9095

}(-j4.4342)

Double line – ground fault - example Current into the fault 𝐼𝑓𝑎 = 𝐼𝑓𝑎(0) + 𝐼𝑓𝑎(1) + 𝐼𝑓𝑎(2) = 0 (as axpected) 𝐼𝑓𝑏 =

𝐼𝑓𝑏 (0) + 𝐼𝑓𝑏 (1) + 𝐼𝑓𝑏 (2)

= 𝐼𝑓𝑎(0) + 𝑎2 𝐼𝑓𝑎(1) + a 𝐼𝑓𝑎(2) = j1.9095 + (1∠2400 )(-j4.4342) + (1∠1200 )(j2.5247) = 6.6726∠154.60 𝐼𝑓𝑐 =

𝐼𝑓𝐶 (0) + 𝐼𝑓𝐶 (1) + 𝐼𝑓𝐶 (2)

= 𝐼𝑓𝑎(0) + 𝑎 𝐼𝑓𝑎(1) + 𝑎2 𝐼𝑓𝑎(2) = 1.9095 + (1∠1200 )(-j2.9481) + (1∠2400 )(j2.9481) = 6.6726∠25.40

Double line – ground fault - example Total fault current 𝐼𝑓 = 𝐼𝑓𝑏 + 𝐼𝑓𝑐 = 5.7242∠900 OR 𝐼𝑓 = 𝐼𝑓𝑏 + 𝐼𝑓𝑐 = 3 𝐼𝑓𝑎 (0) = j5.7285 100000 = 3(20) 19264∠154.60 A

Base current =

𝐼𝑓𝑏 = 𝐼𝑓𝑐 = 19264∠125.40 A 𝐼𝑓 = 16536 ∠900 A

2887 A

Double line – ground fault - example Sequence components of voltage at the fault

𝑉4𝑎(1) = 𝑉4𝑎(2) = 𝑉4𝑎(0) = 𝑉𝑓 −𝑍44(1) 𝐼𝑓𝑎(1) (see circuit) = 1∠00 - (-j4.4342)(j0.1437) = 0.363 Double line – ground fault - example

Double line – ground fault - example The line to ground voltages at bus 4 are 𝑉4𝑎= 𝑉4𝑎(0) +𝑉4𝑎(1) + 𝑉4𝑎(2) = 0.363 + 0.363 + 0.363 = 1.089 𝑉4𝑏= 𝑉4𝑏 (0) + 𝑉4𝑏 (1) + 𝑉4𝑏 (2) = 𝑉4𝑎(0) + 𝑎2 𝑉4𝑎(1) + 𝑎 𝑉4𝑎(2) = 0.363 + (1∠2400 )(0.363) + (1∠1200 )(0.363) =0 (as expected)

Double line – ground fault - example The line to ground voltages at bus 4 are 𝑉4𝑐 = 𝑉4𝑐 (0) + 𝑉4𝑐 (1) + 𝑉4𝑐 (2) = 𝑉4𝑎(0) + 𝑎𝑉4𝑎(1) +𝑎2 𝑉4𝑎(2) = 0.363 + (1∠1200 )(0.363) + (1∠2400 )(0.363) =0 (as expected)

Double line – ground fault - example The line to line voltages at bus 4 are 𝑉4𝑎𝑏 = 𝑉4𝑎 − 𝑉4𝑏= 1.089∠00 𝑉4𝑏𝑐 = 𝑉4𝑏 − 𝑉4𝑐 = 0 𝑉4𝑐𝑎= 𝑉4𝑐 − 𝑉4𝑎= 1.089∠1800

Double line – ground fault - example The line to line voltages at bus 4 are: Since the phase voltage = 1 per unit, the base 20 voltage will be the phase voltage 𝑉𝑏 = = 11.5

kV 𝑉4𝑎𝑏= 12.5∠00 kV 𝑉4𝑏𝑐 = 0 (as expected) 𝑉4𝑐𝑎= 12.5∠1800 kV

3