### Faults - Scenario 2

This post is inspired by Example 10.2 of "Elements of Power System Analysis, 4th Edition" by William D. Stevenson, Jr.

SITUATION

A synchronous motor rated 30MVA, 13.2kV, with a SUBTRANSIENT reactance of 20%, draws 20MW with leading power factor of 80% from a feeder line with 10% reactance calculated from the machine base ratings.

The line's terminal voltage is 12.8kV, and at the sending end is a synchronous generator also rated 30MVA, 13.2kV, with a SUBTRANSIENT reactance of 20%.

When a 3-phase fault occurs at the motor terminals, what are the symmetrical fault currents at the generator, the motor and the fault point?

ANALYSIS

1.) SOURCE SUB-CIRCUIT

Base power, 3ph: S,base = 30MVA

Base voltage, LL: V,base = 13.2kV

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 = 30MVA / (sqrt(3) * 13.2kV)

i,base = 1.312 kA

2.) NORMAL OPERATION

2.1.) Line-to-line voltage as reference, per-unit:

V,pu-> = (12.8kV < 0 deg) / 13.2kV = 0.97 pu < 0 deg

2.2.) Motor apparent power, 3ph:

S,mtr,3ph-> = (P / pf) < [ -arccos(pf) ]

-- Power negative angle means "leading".

S,mtr,3ph-> = (20MW / 0.8) < [ -arccos(0.8) ]

S,mtr,3ph-> = 25 MVA < (-36.87 deg)

2.3.) Motor apparent power, per-unit:

S,mtr,pu-> = S,mtr,3ph-> / S,base

S,mtr,pu-> = [ 25MVA < (-36.87 deg) ] / 30MVA

S,mtr,pu-> = 5/6 pu < (-36.87 deg)

2.4.) Full-load current, per-unit:

S,mtr,pu-> = (V,pu->) (i,pu->)*

(i,pu->)* = S,mtr,pu-> / V,pu->

(i,pu->)* = [ 5/6 pu < (-36.87 deg) ] / (0.97 pu < 0 deg)

(i,pu->)* = 0.86 pu < (-36.87 deg)

i,pu-> = [ 0.86 pu < (-36.87 deg) ]*

-- Conjugate (i->)* means "to reverse angle sign".

i,pu-> = 0.86 pu < (36.87 deg)

-- Current positive angle means "current leads voltage".

2.5.) Generator internal EMF, per-unit:

0 = -E,gen,pu-> + i,pu-> (Z,gen,pu-> + Z,line,pu->) + V,pu->

E,gen,pu-> = i,pu-> (Z,gen,pu-> + Z,line,pu->) + V,pu->

E,gen,pu-> = (0.86 < 36.87) (j0.2 + j0.1) + (0.97 < 0)

E,gen,pu-> = (0.8152 + j0.2064) pu

E,gen,pu-> = 0.8409 pu < 14.21 deg

2.6.) Motor internal EMF, per-unit:

0 = -V,pu-> + i,pu-> (Z,mtr,pu->) + E,mtr,pu->

E,mtr,pu-> = V,pu-> - i,pu-> (Z,mtr,pu->)

E,mtr,pu-> = 0.97<0 - (0.86 < 36.87) (j0.2)

E,mtr,pu-> = (1.0732 - j0.1376) pu

E,mtr,pu-> = 1.082 pu < (-7.31 deg)

3.) FAULT INCIDENT

3.1.) Generator fault current, per-unit:

0 = -E,gen,pu-> + i,f,gen,pu-> (Z,gen,pu-> + Z,line,pu->)

i,f,gen,pu-> = E,gen,pu-> / (Z,gen,pu-> + Z,line,pu->)

i,f,gen,pu-> = (0.8409 < 14.21) / (j0.2 + j0.1)

i,f,gen,pu-> = (0.6881 - j2.7172) pu

i,f,gen,pu-> = 2.803 pu < (-75.79 deg)

3.2.) Generator fault current, 1L:

i,f,gen-> = (i,f,gen,pu->) (i,base)

i,f,gen-> = [ 2.803 < (-75.79) ] (1.312kA)

i,f,gen-> = 3.678 kA < (-75.79 deg)

3.3.) Motor fault current, per-unit:

0 = -E,mtr,pu-> + i,f,mtr,pu-> Z,mtr,pu->

i,f,mtr,pu-> = E,mtr,pu-> / Z,mtr,pu->

i,f,mtr,pu-> = (1.082 < (-7.31)) / (j0.2)

i,f,mtr,pu-> = (-0.6884 - j5.366) pu

i,f,mtr,pu-> = 5.41 pu < (-97.31 deg)

3.4.) Motor fault current, 1L:

i,f,mtr-> = (i,f,mtr,pu->) (i,base)

i,f,mtr-> = (5.41 < (-97.31)) (1.312kA)

i,f,mtr-> = 7.098 kA < (-97.31 deg)

3.5.) Total fault current, per-unit:

i,f,sum,pu-> = i,f,gen,pu-> + i,f,mtr,pu->

i,f,sum,pu-> = 2.803 < (-75.79) + 5.41 < (-97.31)

i,f,sum,pu-> = (-0.0003 - j8.0833) pu

i,f,sum,pu-> = 8.0833 pu < (-90 deg)

3.6.) Total fault current, 1L:

i,f,sum-> = (i,f,sum,pu->) (i,base)

i,f,sum-> = (8.0833 < (-90)) (1.312kA)

i,f,sum-> = 10.605 kA < (-90 deg)

CONCLUSIONS

The total symmetrical fault current is 10.605 kiloAmps.

Generator contribution to the fault is only 3.678 kiloAmps, but motor contribution to the fault is 7.098 kiloAmps.

Contribution by the generator is lesser due to the presence of feeder line impedance.

Motors become generators during fault incidents, and may cause bigger damage if the fault occurs near the motor because only this machine's internal impedance is present.

SITUATION

A synchronous motor rated 30MVA, 13.2kV, with a SUBTRANSIENT reactance of 20%, draws 20MW with leading power factor of 80% from a feeder line with 10% reactance calculated from the machine base ratings.

The line's terminal voltage is 12.8kV, and at the sending end is a synchronous generator also rated 30MVA, 13.2kV, with a SUBTRANSIENT reactance of 20%.

When a 3-phase fault occurs at the motor terminals, what are the symmetrical fault currents at the generator, the motor and the fault point?

ANALYSIS

1.) SOURCE SUB-CIRCUIT

Base power, 3ph: S,base = 30MVA

Base voltage, LL: V,base = 13.2kV

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

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

i,base = 30MVA / (sqrt(3) * 13.2kV)

i,base = 1.312 kA

2.) NORMAL OPERATION

2.1.) Line-to-line voltage as reference, per-unit:

V,pu-> = (12.8kV < 0 deg) / 13.2kV = 0.97 pu < 0 deg

2.2.) Motor apparent power, 3ph:

S,mtr,3ph-> = (P / pf) < [ -arccos(pf) ]

-- Power negative angle means "leading".

S,mtr,3ph-> = (20MW / 0.8) < [ -arccos(0.8) ]

S,mtr,3ph-> = 25 MVA < (-36.87 deg)

2.3.) Motor apparent power, per-unit:

S,mtr,pu-> = S,mtr,3ph-> / S,base

S,mtr,pu-> = [ 25MVA < (-36.87 deg) ] / 30MVA

S,mtr,pu-> = 5/6 pu < (-36.87 deg)

2.4.) Full-load current, per-unit:

S,mtr,pu-> = (V,pu->) (i,pu->)*

(i,pu->)* = S,mtr,pu-> / V,pu->

(i,pu->)* = [ 5/6 pu < (-36.87 deg) ] / (0.97 pu < 0 deg)

(i,pu->)* = 0.86 pu < (-36.87 deg)

i,pu-> = [ 0.86 pu < (-36.87 deg) ]*

-- Conjugate (i->)* means "to reverse angle sign".

i,pu-> = 0.86 pu < (36.87 deg)

-- Current positive angle means "current leads voltage".

2.5.) Generator internal EMF, per-unit:

0 = -E,gen,pu-> + i,pu-> (Z,gen,pu-> + Z,line,pu->) + V,pu->

E,gen,pu-> = i,pu-> (Z,gen,pu-> + Z,line,pu->) + V,pu->

E,gen,pu-> = (0.86 < 36.87) (j0.2 + j0.1) + (0.97 < 0)

E,gen,pu-> = (0.8152 + j0.2064) pu

E,gen,pu-> = 0.8409 pu < 14.21 deg

2.6.) Motor internal EMF, per-unit:

0 = -V,pu-> + i,pu-> (Z,mtr,pu->) + E,mtr,pu->

E,mtr,pu-> = V,pu-> - i,pu-> (Z,mtr,pu->)

E,mtr,pu-> = 0.97<0 - (0.86 < 36.87) (j0.2)

E,mtr,pu-> = (1.0732 - j0.1376) pu

E,mtr,pu-> = 1.082 pu < (-7.31 deg)

3.) FAULT INCIDENT

3.1.) Generator fault current, per-unit:

0 = -E,gen,pu-> + i,f,gen,pu-> (Z,gen,pu-> + Z,line,pu->)

i,f,gen,pu-> = E,gen,pu-> / (Z,gen,pu-> + Z,line,pu->)

i,f,gen,pu-> = (0.8409 < 14.21) / (j0.2 + j0.1)

i,f,gen,pu-> = (0.6881 - j2.7172) pu

i,f,gen,pu-> = 2.803 pu < (-75.79 deg)

3.2.) Generator fault current, 1L:

i,f,gen-> = (i,f,gen,pu->) (i,base)

i,f,gen-> = [ 2.803 < (-75.79) ] (1.312kA)

i,f,gen-> = 3.678 kA < (-75.79 deg)

3.3.) Motor fault current, per-unit:

0 = -E,mtr,pu-> + i,f,mtr,pu-> Z,mtr,pu->

i,f,mtr,pu-> = E,mtr,pu-> / Z,mtr,pu->

i,f,mtr,pu-> = (1.082 < (-7.31)) / (j0.2)

i,f,mtr,pu-> = (-0.6884 - j5.366) pu

i,f,mtr,pu-> = 5.41 pu < (-97.31 deg)

3.4.) Motor fault current, 1L:

i,f,mtr-> = (i,f,mtr,pu->) (i,base)

i,f,mtr-> = (5.41 < (-97.31)) (1.312kA)

i,f,mtr-> = 7.098 kA < (-97.31 deg)

3.5.) Total fault current, per-unit:

i,f,sum,pu-> = i,f,gen,pu-> + i,f,mtr,pu->

i,f,sum,pu-> = 2.803 < (-75.79) + 5.41 < (-97.31)

i,f,sum,pu-> = (-0.0003 - j8.0833) pu

i,f,sum,pu-> = 8.0833 pu < (-90 deg)

3.6.) Total fault current, 1L:

i,f,sum-> = (i,f,sum,pu->) (i,base)

i,f,sum-> = (8.0833 < (-90)) (1.312kA)

i,f,sum-> = 10.605 kA < (-90 deg)

CONCLUSIONS

The total symmetrical fault current is 10.605 kiloAmps.

Generator contribution to the fault is only 3.678 kiloAmps, but motor contribution to the fault is 7.098 kiloAmps.

Contribution by the generator is lesser due to the presence of feeder line impedance.

Motors become generators during fault incidents, and may cause bigger damage if the fault occurs near the motor because only this machine's internal impedance is present.

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