### Faults - Scenario 3

This post demonstrates how the Infinite Bus Method, shown in Appendix D Example D13 (Available Short Circuit Current) of the 2017 Philippine Electrical Code (PEC), can also be used to estimate fault currents for three-phase circuits.

SITUATION

A three-phase fault occurs down the line of a three-phase transformer rated 10MVA, 4.16 kiloVolts, and 5.5% impedance. What is the maximum symmetrical fault current produced?

ANALYSIS

1.) TRANSFORMER LOAD-SIDE SUB-CIRCUIT

Base power, 3ph: S,base = 10MVA

Base voltage, LL: V,base = 4.16kV

Impedance, given: Z = 5.5%

1.1.) Establishing mathematical relationships to avoid confusion:

S,base = S,3ph

V,base = V,LL

V,LN = V,LL / sqrt(3)

i,1ph = i,1L (assuming Y-config)

i,base = i,1L

S,3ph = 3 * V,LN * i,1ph

S,3ph = 3 * ( V,LL/sqrt(3) ) * i,1L

S,3ph = 3/sqrt(3) * V,LL * i,1L

S,3ph = [ 3/sqrt(3) * ( sqrt(3) / sqrt(3) ) ] * V,LL * i,1L

S,3ph = [ 3 * sqrt(3) / sqrt(3)^2 ] * V,LL * i,1L

S,3ph = [ 3 * sqrt(3) / 3 ] * V,LL * i,1L

S,3ph = sqrt(3) * V,LL * i,1L

S,base = sqrt(3) * V,base * i,base

1.2.) Base current, 1L:

i,base = S,base / (sqrt(3) * V,base)

i,base = 10MVA / (sqrt(3) * 4.16kV)

i,base = 1,387.861224 A

2.) FAULT CALCULATION

2.1.) Use rated voltage as reference.

V,pu = V,rated / V,base

V,pu = 4.16kV / 4.16kV

V,pu = 1 pu

2.2.) Short-circuit current, per-unit:

i,pu = V,pu / Z,pu

Z,pu = %Z / 100

i,pu = V,pu / (%Z / 100)

i,pu = (V,pu / %Z) * 100

i,pu = (1 / 5.5) * 100

i,pu = 1 / 0.055

i,pu = 18.18 pu

2.3.) Short-circuit current, 1L:

i,sc = i,pu * i,base

i,sc = 18.18 * 1,387.861224 A

i,sc = 25,231.317 A

CONCLUSION

The maximum symmetrical fault current produced is 25.23 kiloAmps at the secondary side of this transformer.

The Infinite Bus Method is straightforward and its equations for three-phase circuits can be summarized as follows:

S,base = three-phase power

V,base = line-line voltage

i,base (1 line) = S,base / (sqrt(3) * V,base)

V,pu = 1 pu

i,pu = (V,pu / %Z) * 100

i,pu = (1 / %Z) * 100

i,sc = i,pu * i,base

Note that for single-phase circuits, it looks like this:

S,base = single-phase power

V,base = single-phase voltage

i,base (1-phase) = S,base / V,base

V,pu = 1 pu

i,pu = (V,pu / %Z) * 100

i,pu = (1 / %Z) * 100

i,sc = i,pu * i,base

The steps in applying the Infinite Bus Method on three-phase circuits is very similar to those on single-phase circuits. They only differ in establishing the base values needed to estimate the fault current from per-unit quantities.

SITUATION

A three-phase fault occurs down the line of a three-phase transformer rated 10MVA, 4.16 kiloVolts, and 5.5% impedance. What is the maximum symmetrical fault current produced?

ANALYSIS

1.) TRANSFORMER LOAD-SIDE SUB-CIRCUIT

Base power, 3ph: S,base = 10MVA

Base voltage, LL: V,base = 4.16kV

Impedance, given: Z = 5.5%

1.1.) Establishing mathematical relationships to avoid confusion:

S,base = S,3ph

V,base = V,LL

V,LN = V,LL / sqrt(3)

i,1ph = i,1L (assuming Y-config)

i,base = i,1L

S,3ph = 3 * V,LN * i,1ph

S,3ph = 3 * ( V,LL/sqrt(3) ) * i,1L

S,3ph = 3/sqrt(3) * V,LL * i,1L

S,3ph = [ 3/sqrt(3) * ( sqrt(3) / sqrt(3) ) ] * V,LL * i,1L

S,3ph = [ 3 * sqrt(3) / sqrt(3)^2 ] * V,LL * i,1L

S,3ph = [ 3 * sqrt(3) / 3 ] * V,LL * i,1L

S,3ph = sqrt(3) * V,LL * i,1L

S,base = sqrt(3) * V,base * i,base

1.2.) Base current, 1L:

i,base = S,base / (sqrt(3) * V,base)

i,base = 10MVA / (sqrt(3) * 4.16kV)

i,base = 1,387.861224 A

2.) FAULT CALCULATION

2.1.) Use rated voltage as reference.

V,pu = V,rated / V,base

V,pu = 4.16kV / 4.16kV

V,pu = 1 pu

2.2.) Short-circuit current, per-unit:

i,pu = V,pu / Z,pu

Z,pu = %Z / 100

i,pu = V,pu / (%Z / 100)

i,pu = (V,pu / %Z) * 100

i,pu = (1 / 5.5) * 100

i,pu = 1 / 0.055

i,pu = 18.18 pu

2.3.) Short-circuit current, 1L:

i,sc = i,pu * i,base

i,sc = 18.18 * 1,387.861224 A

i,sc = 25,231.317 A

CONCLUSION

The maximum symmetrical fault current produced is 25.23 kiloAmps at the secondary side of this transformer.

The Infinite Bus Method is straightforward and its equations for three-phase circuits can be summarized as follows:

S,base = three-phase power

V,base = line-line voltage

i,base (1 line) = S,base / (sqrt(3) * V,base)

V,pu = 1 pu

i,pu = (V,pu / %Z) * 100

i,pu = (1 / %Z) * 100

i,sc = i,pu * i,base

Note that for single-phase circuits, it looks like this:

S,base = single-phase power

V,base = single-phase voltage

i,base (1-phase) = S,base / V,base

V,pu = 1 pu

i,pu = (V,pu / %Z) * 100

i,pu = (1 / %Z) * 100

i,sc = i,pu * i,base

The steps in applying the Infinite Bus Method on three-phase circuits is very similar to those on single-phase circuits. They only differ in establishing the base values needed to estimate the fault current from per-unit quantities.

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