Network Type Of Fault: By: Erwin J. Salvador, M.eng.ee, Pee 3442

NETWORK TYPE OF FAULT

By: Erwin J. Salvador, M.Eng.EE, PEE 3442

Purpose of Short-Circuit Studies • A Short-Circuit Study can be used to determine any or all of the following: – Verify protective device close and latch capability – Verify protective device Interrupting capability – Protect equipment from large mechanical forces (maximum fault kA) – I2t protection for equipment (thermal stress)

– Selecting ratings or settings for relay coordination

Elements That Contribute Current to a Short-Circuit • Generator

• Power Grid • Synchronous Motors • Induction Machines • Lumped Loads (with some % motor load) • Inverters

Elements Do Not Contribute Current in PowerStation • Static Loads • Motor Operated Valves • All Shunt Y Connected Branches

Sources and Models of Fault Currents Sources •Synchronous Generators •Synchronous Motors & Condensers •Induction Machines •Electric Utility Systems (Power Grids)

Models All sources are modeled by an internal voltage behind its impedance. E = Prefault Voltage R = Machine Armature Resistance X = Machine Reactance (X”d, X’d, Xd)

Synchronous Generators Synchronous Generators are modeled in three stages. Synchronous Reactance Transient Reactance Subtransient Reactance

Synchronous Motors & Condensers

Electric Utility Systems

Induction Machines

The fault current contribution tends to Treated the same as synchronous motors except they do not contribute to Act as a generator to supply fault remain constant. the fault after 2 sec. current. This current diminishes as the magnetic field in the machine decays.

½ Cycle Network This is the network used to calculate momentary short-circuit current and protective device duties at the ½ cycle after the fault.

1 ½ to 4 Cycle Network This network is used to calculate the interrupting short-circuit current and protective device duties 1.5-4 cycles after the fault.

30-Cycle Network This is the network used to calculate the steady-state short-circuit current and settings for over current relays after 30 cycles.

Calculation Methods 1) The ANSI standards handle the AC Decay by varying machine impedance during a fault.

2)The ANSI standards handle the the dc offset by applying multiplying factors. The ANSI Terms for this current are: •Momentary Current •Close and Latch Current •First Cycle Asymmetrical Current

Fault Current Spectrum

Reactance Representation for Utility and Synchronous Machine ½ Cycle

1 ½ to 4 Cycle

30 Cycle

X”d

X”d

X”d

X”d

X”d

X’d

Hydro-Gen with Amortisseur winding

X”d

X”d

X’d

Hydro-Gen without Amortisseur winding

0.75*X”d

0.75*X”d

X’d

X”d

X”d

X”d

1.5*X”d

Utility

Turbo Generator

Condenser Synchronous Motor

Reactance Representation for Induction Machine ½ Cycle

1 ½ to 4 Cycle

>1000 hp , <= 1800 rpm

X”d

1.5*X”d

>250, at 3600 rpm

X”d

1.5*X”d

All others, >= 50 hp

1.2*X”d

3.0*X”d

< 50 hp

1.67*X”d

Device Duty and Usage of Fault Currents from Different Networks ½ Cycle Currents (Subtransient Network)

1 ½ to 4 Cycle Currents (Transient Network)

HV Circuit Breaker

Closing and Latching Capability

Interrupting Capability

LV Circuit Breaker

Interrupting Capability

---

--Fuse

Interrupting Capability

SWGR / MCC

Bus Bracing

---

Relay

Instantaneous Settings

---

30 Cycle currents are used for determining overcurrent settings.

• • •

The one line diagram is a graphical representation of the power system The diagram should show all sources of short circuit current and all other significant elements Essential data such as equipment ID, rated MVA, kV, impedances, and other data found in every equipment should also be reflected on the diagram

•

Positive Sequence Impedances - The positive sequence impedance of synchronous machines varies depending on the time period after the machine experiences fault

•

Negative Sequence Reactance - Approximately a few percent higher or lower than the positive impedance value

•

Zero Sequence Reactance - Usually much less than the others (about 40 to 60% of any of the above)

EQUIVALENT ZERO-SEQUENCE MODELS FOR SYNCHRONOUS MACHINES ZERO - SEQUENCE NETWORKS OF POWER SYSTEM COMPONENTS

❖ Negative Sequence Impedances • Generally assumed to be equal to the positive sequence reactance ❖ Zero Sequence Impedances • Usually less than the others • For this reason, the machines are never solidly grounded.

SYNCHRONOUS

Machines/Generators Connection

Equivalent Zero Sequence Circuit

GROUNDING System Configuration

Z0

n SOLIDLY GROUNDED

a

NEUTRAL

I0 GROUND

b

I0 3Z

( Resistor or Reactor )

Z

Z0

n

IMPEDANCE GROUNDED NEUTRAL

GROUND

WYE

c

UNGROUNDED -WYE OR DELTA CONNECTION ( Machine /Generators )

DELTA

n

Z0 I0 GROUND

EQUIVALENT ZERO-SEQUENCE MODELS FOR TWO WINDING TRANSFORMERS Two Winding Transformers

❖ The positive and negative sequence impedances of all transformers are identical.

Three Phase Connection

a

Zero Sequence Circuit

a

a’

b

b’ c’

a’

c a b

❖ The zero sequence impedance is either equal to the other two sequence reactances or infinite except for three-phase, coretype transformers.

Z0 a

Z

a’ n

b

Z0

a

b’ c’

a’

3Zn

c a c

d

a’

b

b’

c

c’

a

a’

b

b’

c

c’

b

Z0 a

a’

a e

Z0

ZnL

ZnH

b’ c’

a’

a

a

3ZnH

Z0

a’

3ZnL

a’

c

f

a

a’

b

b’ c’

Z0 a

a’

c

g

a

a’

b

b’ c’

Z0 a

a’

c

h

a

a’

b

b’ c’

c

Z0 a

a’

The Per-Unit value of any quantity (volts, amps, ohms, watts, volt-ampere, etc. ) is defined as the ratio of the quantity to its base value expressed as decimal.

Per Unit Quantity

Actual Quantity Base Quantity

Advantages of Per-Unit Values: • Manufacturers usually specify the impedance of an apparatus in per-unit values.

• The Per-Unit impedance expressed in proper base is same on the either side of the transformer.

• Per-Unit impedances of machines of the same type and widely different rating usually lie within a narrow range.

• The way which transformers are connected in 3phase does not affect the per-unit impedance.

𝑉𝑎 = (𝑉𝑎1 + 𝑉𝑎2 + 𝑉𝑎0 ) 𝑉𝑏 = (𝑉𝑏1 + 𝑉𝑏2 + 𝑉𝑏0 ) 𝑉𝑐 = (𝑉𝑐1 + 𝑉𝑐2 + 𝑉𝑐0 )

Note: The formulas can also be applied for current components

Formulas:

𝑉𝑎1 =

1 (𝑉 3 𝑎

+ a𝑉𝑏 + 𝑎2 𝑉𝑐 )

𝑉𝑏1 = 𝑎2 𝑉𝑎1

𝑉𝑐1 = a𝑉𝑎1 Where: a = 1∠120⁰ 𝑎2 = 1∠𝟐𝟒0⁰

•consist of three phasors equal in magnitude, displaced 120 electrical degrees from each other, and with the same phase sequence as the original one

Note: The formulas can also be applied for current components

Formulas:

𝑉𝑎2 =

1 (𝑉 3 𝑎

+ 𝑎2 𝑉𝑏 + 𝑎𝑉𝑐 )

𝑉𝑏2 = a𝑉𝑎1 𝑉𝑐2 = 𝑎2 𝑉𝑎1 Where: a = 1∠120⁰ 𝑎2 = 1∠𝟐𝟒0⁰

•consist of three phasors equal in magnitude, displaced by 120 electrical degrees from each other, and having opposite phase sequence as the original

Note: The formulas can also be applied for current components

Formulas:

𝑉𝑎0 =

1 (𝑉 3 𝑎

+ 𝑉𝑏 + 𝑉𝑐 )

𝑉𝑏0 = 𝑉𝑎0 𝑉𝑐0 = 𝑉𝑎0 •consists of three phasors equal in magnitude and with zero displacement with each other

In a three phase system, the sum of the line currents is equal to the current In in the return path through the neutral.

I N = I a + Ib + Ic then ,

But:

I a0

1 = (I a + I b + I c ) 3

I N = 3I a 0

Note: If there is no path through the neutral of a three-phase system, then IN = 0 and the line currents contain no zerosequence components

A very important tool in fault analysis is the network reduction by Wye to Delta and Delta to Wye Transformation The Formulas are as follows:

I. Delta to Wye Transformation

𝑍1 =

𝑍12 (𝑍31 ) 𝑍12 +𝑍23 +𝑍31

𝑍2 =

𝑍12 (𝑍23 ) 𝑍12 +𝑍23 +𝑍31

𝑍3 =

𝑍23 (𝑍31 ) 𝑍12 +𝑍23 +𝑍31

II. Wye to Delta Transformation

𝑍12 =

𝑍1 𝑍2 +𝑍2 𝑍3 +𝑍3 𝑍1 𝑍3

𝑍23 =

𝑍1 𝑍2 +𝑍2 𝑍3 +𝑍3 𝑍1 𝑍1

𝑍31 =

𝑍1 𝑍2 +𝑍2 𝑍3 +𝑍3 𝑍1 𝑍2

Sequence Networks

SLG Fault Sequence Network Connections Formulas:

𝐼𝑆𝐿𝐺 = 3𝐼𝑎𝑜 𝑥 𝐼𝑏𝑎𝑠𝑒 , Amps. Where: 𝐼𝑎𝑜

𝐸𝑎 = 𝑍1 + 𝑍2 + 𝑍0 + 3𝑍𝑓

Z1 = Positive Sequence Impedance Z2 = Negative Sequence Impedance Zg0 = Zero Sequence Impedance ZF = Fault Impedance; equal to zero if bolted fault Ea = Driving voltage equal to 1.0 at pre-fault Note: ZF = 0 if bolted fault : ZN = 0 if solidly grounded

L-L Fault Sequence Network Connections Formulas:

𝐼𝐿𝐿 = 𝐼𝑏 =

3𝐼𝑎1 𝑥 𝐼𝑏𝑎𝑠𝑒 , Amps.

Where: 𝐼𝑎1

𝐸𝑎 = 𝑍1 + 𝑍2 + 𝑍𝑓

𝐼𝑐 = −𝐼𝑏 Z1 = Positive Sequence Impedance Z2 = Negative Sequence Impedance ZF = Fault Impedance; equal to zero if bolted fault Ea = Driving voltage equal to 1.0 at pre-fault

DLG Fault Sequence Network Connections Formulas: •

For solidly grounded neutral and no impedance in the fault:

𝐼𝐷𝐿𝐺 •

3𝐸𝑎 𝑥 𝑍2 = 𝑍1 𝑍0 + 𝑍1 𝑍2 + 𝑍2 𝑍0

For solidly grounded neutral and with fault through an impedance:

𝐼𝐷𝐿𝐺

3𝐸𝑎 𝑥 𝑍2 = 𝑍1 𝑍2 + (𝑍1 𝑍2 )(3𝑍𝑥𝑍0 )

Where:

I DLG = 3I a 0 xI BASE

 Ea − I a1Z1  I a0 = −   Z 0 + 3Z F   Ea − I a1Z1  I a 2 = −   Z 2 + 3Z F  Note: ZF = 0 if bolted fault : ZN = 0 if solidly grounded

Ea I a1 = Z 2 (Z 0 + 3Z F ) Z1 + Z 2 + Z 0 + 3Z F

Z 0 = Z g 0 + 3Z N Z1 = Positive Sequence Impedance Z2 = Negative Sequence Impedance Zg0 = Zero Sequence Impedance ZF = Fault Impedance; equal to zero if bolted fault Ea = Driving voltage equal to 1.0 at pre-fault

1) The one line diagram of a simple power system is shown in the figure below. The neutral of each generator is grounded through a current limiting reactor of j1/6 per unit on a 100MVA base. Determine the following fault currents for 3-phase, SLG, LL and DLG at bus 3 at bolted fault. Bus 3

L13

T1

L23 T2

G1

G2

Bus 1

L12

Bus 2

Positive Sequence Diagram

Convert Delta to Wye

Conversion from Delta to Wye:

𝑋=

(𝑗0.10)(𝑗0.2) 2 =j 𝑗0.1 + 𝑗0.2 + 𝑗0.25 55

(𝑗0.10)(𝑗0.25) 1 𝑌= =j 𝑗0.1 + 𝑗0.2 + 𝑗0.25 22

𝑍=

(𝑗0.20)(𝑗0.25) 1 =j 𝑗0.1 + 𝑗0.2 + 𝑗0.25 11

𝐴 = 𝑍𝐴 + 𝑋 = 𝑗

41 220

𝐵 = 𝑍𝐵 + 𝑍 = 𝑗

53 220

𝐶=

1 1 1 + 41 53 𝑗 220 𝑗 220

273 =𝑗 20680

𝑍𝑇𝐻1 = 𝐶 + 𝑍 = 𝑗

283 1880

Negative Sequence Diagram

𝑍𝑇𝐻2 = 𝑗 Same computation as Positive Sequence

283 1880

Zero Sequence Diagram

Equivalent Zero Sequence Circuit

Equivalent Zero Sequence Circuit

G1 & G2 ZERO SEQUENCE IMPEDANCE

𝑍0 = 𝑍𝑔0 + 3𝑍𝑁 𝑍0 = 𝑗0.05 + 3

1 = 𝑗0.55 6

Convert Delta to Wye

(𝑗0.25)(𝑗0.15) 3 𝑋= =j 𝑗0.25 + 𝑗0.15 + 𝑗0.45 68 (𝑗0.15)(𝑗0.45) 27 𝑌= =j 𝑗0.25 + 𝑗0.15 + 𝑗0.45 340

𝑍=

(𝑗0.45)(𝑗0.25) 9 =j 𝑗0.25 + 𝑗0.15 + 𝑗0.45 68

𝐴 = 𝑍𝐴 + 𝑋 = 𝑗

219 340

31 𝐵 = 𝑍𝐵 + 𝑍 = 𝑗 170

𝐶=

1 1 1 + 219 31 𝑗 340 𝑗 170

=𝑗

6789 47770

𝑍𝑇𝐻0

249 =𝐶+𝑍=𝑗 1124

Three Phase Fault Calculation 𝐼3∅ 𝑝. 𝑢. =

1 𝑍𝑇𝐻

𝐼3∅ 𝑝. 𝑢. =

1 = −j6.643 283 𝑗 1880

𝐼3∅ 𝑎𝑐𝑡𝑢𝑎𝑙 = 𝐼𝐹 𝑝. 𝑢. ×

𝐼3∅ 𝑎𝑐𝑡𝑢𝑎𝑙 = 6.643 ×

𝑆𝐵𝐴𝑆𝐸

3 𝑉𝐵𝐴𝑆𝐸 100𝑀𝑉𝐴

3 69𝑘𝑉

= 5,558. 461A

Three Phase Network

SLG Fault Equivalent Network

Single Line to Ground Fault Calculation 𝐼𝑆𝐿𝐺 𝑝. 𝑢. = 3 × 𝐼𝐴0 = 3

𝐼𝑆𝐿𝐺 𝑝. 𝑢. = 3

𝐸 𝑋1 + 𝑋2 + 𝑋0 + 3𝑍𝐹

1 < 0° 283 283 249 𝑗 +𝑗 +𝑗 + 3(𝑗0) 1880 1880 1124

𝐼𝑆𝐿𝐺 𝑝. 𝑢. = −5.741 𝑝𝑢

𝐼𝑆𝐿𝐺 𝑎𝑐𝑡𝑢𝑎𝑙 = 𝐼𝑆𝐿𝐺 𝑝. 𝑢. ×

100𝑀𝑉𝐴 3 69𝑘𝑉

= 4,803.722A

L-L Fault Equivalent Network

Line to Line Fault Calculation 𝐼𝐿−𝐿 𝑝. 𝑢. = −𝑗 3 𝐼𝑎1

𝐼𝑎1 =

𝐼𝑎1

𝐸 = 27,792.761 A 𝑍1 + 𝑍2 + 𝑍𝐹

1 < 0° 940 = = −𝑗 283 283 𝑗 1880 (2)

𝐼𝐿−𝐿 𝑝. 𝑢. = −5.753 𝐼𝐿−𝐿 𝑎𝑐𝑡𝑢𝑎𝑙 = 𝐼𝐹 𝑝. 𝑢. ×

100𝑀𝑉𝐴

3 69𝑘𝑉

= 4,813.847 A

DLG Fault Equivalent Network

Double Line to Ground Fault Calculation 𝐼𝐷𝐿𝐺 𝑝. 𝑢. = 3 × 𝐼𝐴0

𝐸 − 𝐼𝐴1 𝑍1 =3 − 𝑍0 + 3𝑍𝐹

𝐼𝐴1 = 3

𝐼𝐴1 =

𝐸 𝑍 (𝑍 + 3𝑍𝐹 ) 𝑍1 + 𝑍 2 + 0𝑍 + 3𝑍 2 0 𝐹

1 283 249 𝑗 (𝑗 + 𝑗0) 283 1880 1124 𝑗 1880 + 283 249 𝑗 1880 + 𝑗 1124 + 𝑗0

𝐼𝐴1 = −𝑗4.164

𝐼𝐴0 = −

283 1 − (−𝑗4.164)(𝑗 1880) 249 𝑗 1124

𝐼𝐴0 = −𝑗1.685 𝐼𝐷𝐿𝐺 𝑝. 𝑢. = 3 × 𝐼𝐴0 = −𝑗5.054 𝐼𝐷𝐿𝐺 𝑎𝑐𝑡𝑢𝑎𝑙 = 𝐼𝐷𝐿𝐺 𝑝. 𝑢. ×

100𝑀𝑉𝐴

3 69𝑘𝑉

= 4,228.660 A

2) The power system shown in the figure develops a symmetrical fault at point F on the 132 kV overhead – line transmission system. What is the three – phase fault MVA level at F?

Positive Sequence Impedance Diagram

Positive Sequence Impedance Diagram

𝑍𝑝.𝑢.

𝑀𝑉𝐴𝑛𝑒𝑤 𝑘𝑉𝑜𝑙𝑑 = 𝑝. 𝑢. × × 𝑀𝑉𝐴𝑜𝑙𝑑 𝑘𝑉𝑏𝑎𝑠𝑒

2

𝑍𝐺1 = 𝑍𝐺2

100𝑀𝑉𝐴 132𝑘𝑉 = 𝑗0.17 × × 75𝑀𝑉𝐴 132𝑘𝑉

𝑍𝐺3 = 𝑍𝐺4

100𝑀𝑉𝐴 132𝑘𝑉 = 𝑗0.15 × × 37.5𝑀𝑉𝐴 132𝑘𝑉

𝑍𝑇1 = 𝑍𝑇2

100𝑀𝑉𝐴 132𝑘𝑉 = 𝑗0.12 × × 75𝑀𝑉𝐴 132𝑘𝑉

𝑍𝑇3 = 𝑍𝑇4

100𝑀𝑉𝐴 132𝑘𝑉 = 𝑗0.10 × × 37.5𝑀𝑉𝐴 132𝑘𝑉

𝑍𝑇5 = 𝑍𝑇6

100𝑀𝑉𝐴 132𝑘𝑉 = 𝑗0.10 × × 30𝑀𝑉𝐴 132𝑘𝑉

𝑍𝑇7 = 𝑍𝑇8 = 𝑍𝑇9

𝑍𝐴𝐶 = 𝑍𝐵𝐹 = 𝑗 0.015ൗ𝑘𝑚 × 32.2𝑘𝑚 = 𝑗0.483

2

=𝑗

17 75

𝑍𝐴𝐵 = 𝑗 0.015ൗ𝑘𝑚 × 24.2𝑘𝑚 = 𝑗0.363

2

= 𝑗0.4

𝑍𝐶𝐷 = 𝑗 0.015ൗ𝑘𝑚 × 16.1𝑘𝑚 = 𝑗0.2415

= 𝑗0.16

𝑍𝐷𝐹 = 𝑗 0.015ൗ𝑘𝑚 × 8.05𝑘𝑚 = 𝑗0.12075

2

2

2

4 =𝑗 15

1 =𝑗 3

100𝑀𝑉𝐴 132𝑘𝑉 = 𝑗0.12 × × 60𝑀𝑉𝐴 132𝑘𝑉

2

= 𝑗0.2

Simplifying the network:

−0.25849 𝐴= = 𝑗0.7121 𝑗0.363 𝐵=

−0.25849 = 𝑗0.53518 𝑗0.483

𝐶=

−0.25849 = 𝑗2.6295 29 𝑗 295

−0.99823 = 𝑗0.37963 𝑗2.6295 −0.99823 𝑌= = 𝑗7.89 𝑗0.12651 𝑋=

𝑍=

−0.99823 = 𝑗4.1335 𝑗0.2415

−4.6821 = 𝑗9.6935 𝑗0.483 −4.6821 𝑏= = 𝑗44.9983 𝑗0.10405 −4.6821 𝑐= = 𝑗0.59341 𝑗7.89

𝑎=

𝑋𝑇𝐻1 = 𝑗0.1757

Negative Sequence Impedance Diagram

Negative Sequence Impedance Diagram

𝑍𝑝.𝑢.

𝑀𝑉𝐴𝑛𝑒𝑤 𝑘𝑉𝑜𝑙𝑑 = 𝑝. 𝑢. × × 𝑀𝑉𝐴𝑜𝑙𝑑 𝑘𝑉𝑏𝑎𝑠𝑒

2

𝑍𝐺1 = 𝑍𝐺2

100𝑀𝑉𝐴 132𝑘𝑉 = 𝑗0.12 × × 75𝑀𝑉𝐴 132𝑘𝑉

𝑍𝐺3 = 𝑍𝐺4

100𝑀𝑉𝐴 132𝑘𝑉 = 𝑗0.10 × × 37.5𝑀𝑉𝐴 132𝑘𝑉

𝑍𝑇1 = 𝑍𝑇2

100𝑀𝑉𝐴 132𝑘𝑉 = 𝑗0.12 × × 75𝑀𝑉𝐴 132𝑘𝑉

𝑍𝑇3 = 𝑍𝑇4

100𝑀𝑉𝐴 132𝑘𝑉 = 𝑗0.10 × × 37.5𝑀𝑉𝐴 132𝑘𝑉

𝑍𝑇5 = 𝑍𝑇6

100𝑀𝑉𝐴 132𝑘𝑉 = 𝑗0.10 × × 30𝑀𝑉𝐴 132𝑘𝑉

𝑍𝑇7 = 𝑍𝑇8 = 𝑍𝑇9

𝑍𝐴𝐶 = 𝑍𝐵𝐹 = 𝑗 0.015ൗ𝑘𝑚 × 32.2𝑘𝑚 = 𝑗0.483

2

=𝑗

8 25

𝑍𝐴𝐵 = 𝑗 0.015ൗ𝑘𝑚 × 24.2𝑘𝑚 = 𝑗0.363

=𝑗

4 15

𝑍𝐶𝐷 = 𝑗 0.015ൗ𝑘𝑚 × 16.1𝑘𝑚 = 𝑗0.2415

2

2

= 𝑗0.16 2

2

𝑍𝐷𝐹 = 𝑗 0.015ൗ𝑘𝑚 × 8.05𝑘𝑚 = 𝑗0.12075

4 =𝑗 15

1 =𝑗 3

100𝑀𝑉𝐴 132𝑘𝑉 = 𝑗0.12 × × 60𝑀𝑉𝐴 132𝑘𝑉

2

= 𝑗0.2

Simplifying the network:

−0.2676 𝐴= = 𝑗0.73719 𝑗0.363 𝐵=

−0.2676 = 𝑗0.55404 𝑗0.483

𝐶=

−0.2676 = 𝑗2.4532 6 𝑗 55

−0.9199 = 𝑗0.37496 𝑗2.4532 −0.9199 𝑌= = 𝑗7.5708 𝑗0.1215

𝑋=

𝑍=

−0.9199 = 𝑗3.8089 𝑗0.2415

−4.46667 = 𝑗9.24776 𝑗0.483 −4.46667 𝑏= = 𝑗44.41351 𝑗0.10405 −4.46667 𝑐= = 𝑗0. 58999 𝑗7.89

𝑎=

𝑋𝑇𝐻2 = 𝑗0.1751

Zero Sequence Impedance Diagram

Equivalent Zero Sequence Circuit

Simplifying the circuit:

𝑋1 =

1 1 1 1 + + 𝑋𝑇1 𝑋𝑇2 𝑋𝑇7

1 𝑋2 = 1 1 1 + + 𝑋𝑇3 𝑋𝑇5 𝑋𝑇6

𝑋3 =

1

1 1 + 𝑋𝑇4 𝑋𝑇8

Convert Delta to Wye

𝐴=

(𝑋𝑎𝑏 )(𝑋1 ) 𝑋𝑎𝑏 + 𝑋1 + 𝑋2

(𝑋2 )(𝑋1 ) 𝐶= 𝑋𝑎𝑏 + 𝑋1 + 𝑋2 𝐵=

(𝑋2 )(𝑋𝑎𝑏 ) 𝑋𝑎𝑏 + 𝑋1 + 𝑋2

Add up impedances

Add up impedances

Convert Delta to Wye

𝑍𝑇𝐻0 = 𝑗0.1884508366

Three Phase Fault Calculation 𝐼3∅

1 𝑝. 𝑢. = 𝑍𝑇𝐻

𝐼3∅ 𝑝. 𝑢. =

1 = −j5.692 𝑗0.1757

𝐼3∅ 𝑎𝑐𝑡𝑢𝑎𝑙 = 𝐼𝐹 𝑝. 𝑢. ×

𝐼3∅ 𝑎𝑐𝑡𝑢𝑎𝑙 = 5.692 ×

𝑆𝐵𝐴𝑆𝐸 3 𝑉𝐵𝐴𝑆𝐸

100𝑀𝑉𝐴 3 132𝑘𝑉

= 2,489.394A

SLG Fault Equivalent Network

Single Line to Ground Fault Calculation

𝐼𝑆𝐿𝐺 𝑝. 𝑢. = 3 × 𝐼𝐴0

𝐼𝑆𝐿𝐺

𝐸 =3 𝑋1 + 𝑋2 + 𝑋0 + 3𝑍𝐹

1 < 0° 𝑝. 𝑢. = 3 𝑗0.1757 + 𝑗0.1751 + 𝑗0.1885 + 3(𝑗0)

𝐼𝑆𝐿𝐺 𝑝. 𝑢. = −5.56277 𝑝𝑢

𝐼𝑆𝐿𝐺 𝑎𝑐𝑡𝑢𝑎𝑙 = 𝐼𝑆𝐿𝐺 𝑝. 𝑢. ×

100𝑀𝑉𝐴

3 132𝑘𝑉

= 2,433.0794 A

L-L Fault Equivalent Network

Line to Line Fault Calculation 𝐼𝐿−𝐿 𝑝. 𝑢. = −𝑗 3 𝐼𝑎1 𝐸 = 27,792.761 A 𝑍1 + 𝑍2 + 𝑍𝐹 1 < 0° = = −𝑗2.8506 𝑗0.1757 + 𝑗0.1751

𝐼𝑎1 = 𝐼𝑎1

𝐼𝐿−𝐿 𝑝. 𝑢. = −𝑗4.9374

𝐼𝐿−𝐿 𝑎𝑐𝑡𝑢𝑎𝑙 = 𝐼𝐹 𝑝. 𝑢. ×

100𝑀𝑉𝐴 3 132𝑘𝑉

= 2,159.566 A

DLG Fault Equivalent Network

Double Line to Ground Fault Calculation 𝐼𝐷𝐿𝐺 𝑝. 𝑢. = 3 × 𝐼𝐴0 𝐼𝐴1

𝐸 − 𝐼𝐴1 𝑍1 =3 − 𝑍0 + 3𝑍𝐹

𝐸 = 𝑍 (𝑍 + 3𝑍𝐹 ) 𝑍1 + 𝑍 2 + 0𝑍 + 3𝑍 2 0 𝐹

𝐼𝐴1 =

1 𝑗0.1751(𝑗0.1885 + 𝑗0) 𝑗0.1757 + 𝑗0.1751 + 𝑗0.1885 + 𝑗0

𝐼𝐴1 = −𝑗3.752676

𝐼𝐴0 = −

1 − (−𝑗3.752676)(𝑗0.1757) 𝑗0.1885

𝐼𝐴0 = 𝑗1.80719 𝐼𝐷𝐿𝐺 𝑝. 𝑢. = 3 × 𝐼𝐴0 = 𝑗5.4216

𝐼𝐷𝐿𝐺 𝑎𝑐𝑡𝑢𝑎𝑙 = 𝐼𝐷𝐿𝐺 𝑝. 𝑢. ×

100𝑀𝑉𝐴 3 132𝑘𝑉

= 2,371.3185 A

REFERENCES: 1. IEEE Std 551 2006 Recommending Practice for CALCULATIING SHORT-CIRCUIT CURRENTS IN INDUSTRIAL AND COMMERCIAL POWER SYSTEMS; Violet Book Institute of Electrical and Electronics Engineers, Inc. 2. IEEE Std 399 1997 Recommending Practice for INDUSTRIAL AND COMMERCIAL POWER SYSTEMS ANALYSIS ; Brown Book Institute of Electrical and Electronics Engineers, Inc. 3. IEEE Std 242 2001 Recommending Practice for PROTECTION AND COORDINATION OF INDUSTRIAL AND COMMERCIAL POWER SYSTEMS; Buff Book Institute of Electrical and Electronics Engineers, Inc.

4. IEEE Std 1015 1997 Recommending Practice for APPLYING LOW-VOLTAGE CIRCUIT BREAKERSUSED IN IDUSTRIAL AND COMMERCIAL POWER SYSTEMS; Blue Book 5. ANALYSIS OF FAULTED POWER SYSTEMS by: Paul Anderson Copyright 1973 6. SYMMETRICAL COMPONENTS by: C.F. Wagner and R.D. Evans Copyright 1933 by McGRAWHILL BOOK COMPANY

REFERENCES: 7. SYMMETRICAL COMPONENTS FOR POWER SYSTEMS ENGINEERING By: J. Lewis Blackburn Copyright 1993 Marcel Dekker Inc.

8. ELEMENTS OF POWER SYSTEM ANALYSIS 4th Edition By: William Stevenson, Jr. Copyright 1982 by McGraw-Hill, Inc. 9. ELECTRICAL POWER SYSTEMS 4th Edition By: C.L. Wadhwa Copyright 2005 by New Age InternationalV 10. POWER SYSTEM ANALYSIS By: Hadi Saadat Copyright 1999 by McGraw-Hill Co.

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