Fault Current Calculation 2f2r25

Job Name Company Name

Personalize by entering the 'Job Name' and 'Your Company Name'

Step 1b: Insert the kVA Available Fault Current Calculation rating (PF) power factor and

Utility Fault Current

0 amperes

kVA = E = trans. FLA =

I = kVA x 1000 = trans. FLA 0 Step 1a: Insert available

Isca = trans. fault FLAcurrent x 100 ampere rating = transformer Z local utility from

or

Isca = ampere short-circuit current RMS symmetrical.

Point to Point Method Length Step 2: Select type of (distance) conductor and conduit 2xLxI (ASC) 'f ' factor =

N x C x E L-N

Multiplier

# conductors per phase

Phase conductor constant Volt Line to Line Step 3: Insert the Length of one conductor Neutral conductor constant Volt Line to Neutral

M =Step14: Insert the number of conductors per phase. 1 +f Line to Line Line to Neutral Fault Current at Service Equipment

PF = Z = Isca =

L = Isca = N = C = E L-L = f = C = EL-N = f =

300 0 0

by:John Sokolik (Z) impedence of transformer.

(Note: When you insert the kVA rating, power factor and impedence you must select a voltage from the select the Voltage and Phase from dropdown box before the value will be calculated.)

90% 3% % 3,000 amperes

150 3,000 1 0 Phase Conductor 0 Volt 0.000 0 Neutral Conductor 0 Volt 0.000

Step 5a: phase M Select = 1.000 conductor size size. M = 1.000

Finished product Isca x M = fault current at terminals of main disconnect L- L = CALCULATED FAULT Isca x M = fault current at terminals of main disconnect L- N = CURRENT

Step 5b: Select neutral conductor size.

3,000 amperes 0 amperes

Fault Current from Service disc. to "_____" Single Phase Feeder 2xLxI 'f ' factor =

N x C x E L-N

Length (distance) (ASC) # conductors per phase

Continue by entering next Phase conductor constant point to point Volt Line to Line Neutral conductor constant Volt Line to Neutral Multiplier

L = Isca = N = C = E L-L = f = C = EL-N = f =

3,000 Phase

0

0 Phase Conductor

0 Volt 0.000 0 Neutral Conductor 0 Volt 0.000

M=

1 1+f

Line to Line Line to Neutral

M M

= =

1.000 1.000

3,000 amperes 0 amperes

Isca x M = fault current at terminals of the L- L = Isca x M = fault current at terminals of the L- N = Calculation does not include motor contribution

Branch Circuit Fault fro Single Phase Branch 2xLxI 'f ' factor =

N x C x E L-N

Point to Point Length (distance) (ASC) # conductors per phase

Phase conductor constant Volt Line to Line Neutral conductor constant Volt Line to Neutral Multiplier M=

1 1+f

Line to Line Line to Neutral

L = Isca = N = C = E L-L = f = C = EL-N = f = M M

= =

3,000 Phase

0

0 Phase Conductor

0 Volt 0.000 0 Neutral Conductor 0 Volt 0.000 1.000 1.000

Isca x M = fault current at terminals of the L- L = Isca x M = fault current at terminals of the L- N = Calculation does not include motor contribution

3,000 amperes 0 amperes

ion

1b: Insert the kVA g (PF) power factor and Ver. 4.00 pedence of transformer.

When you insert the kVA rating, actor and impedence you must voltage from the select the and Phase from dropdown box he value will be calculated.)

12

1

tep 5b: Select neutral conductor size.

Neutral

1

1

Neutral

1

1

Project Name Your Company Name

Available Fault Current Calculation by:John Sokolik

Utility Fault Current

32,000 amperes

kVA = E = trans. FLA =

I = kVA x 1000 = trans. FLA E x 1.732 Isca =

trans. FLA x 100 x PF transformer Z

480 0

PF = Z =

=

Isca = ampere short-circuit current RMS symmetrical.

Ver. 7.2

[email protected]

Isca

=

0 amperes

Point to Point Method 'f ' factor =

1.732 x L x I

Length (distance) (ASC)

N x C x E L-N

# conductors per phase

Phase conductor constant Volt Line to Line Neutral conductor constant Volt Line to Neutral

Multiplier M=

1 1+f

Line to Line Line to Neutral Fault Current at Service Equipment

L = Isca = N = C = E L-L = f = C = EL-N = f = M M

= =

Isca x M = fault current at terminals of main disconnect L- L = Isca x M = fault current at terminals of main disconnect L- N =

100 32,000 1 12,122 Phase Conductor 480 Volt 0.953 12,122 Neutral Conductor 277 Volt 2.476 0.512 0.288

Three Single conductors in PVC raceway AWG

### ### ### ### ### ### ### ### ### ### ### ### ### ### ### ### 400 kcmil 500 kcmil 600 kcmil 750 kcmil 1000 kcmil

16,389 amperes 9,206 amperes

Neutral ### ### 220/127V

Fault Current from Service Equipment to __ Single Phase Feeder 2xLxI 'f ' factor =

Length (distance) (ASC)

L = Isca =

### ### ### 16,389 Phase

9,206 Neutral

380/220V

'f ' factor =

N x C x E L-N

# conductors per phase

Phase conductor constant Volt Line to Line Neutral conductor constant Volt Line to Neutral Multiplier M=

1 1+f

Line to Line Line to Neutral

N C E L-L f C EL-N f

= = = = = = =

0 Phase Conductor 480 Volt 0.000 0 Neutral Conductor 277 Volt 0.000

M M

= =

1.000 1.000

Isca x M = fault current at terminal of the L- L = Isca x M = fault current at terminal of the L- N = Calculation does not include motor contribution

1

1

### ### ### ### ### ### ### ### ### ###

16,389 amperes 9,206 amperes

AL/PVC CU/MET AL/MET ### ### ### ###