faultv6.2.xls

Job Name

Company Name

Available Fault Current Calculation by:John Sokolik

Utility Fault Current amperes kVA = 300 E = 240

trans. FLA = 1250E

= PF = 90%

transformer Z Z = 3% %46,296 amperes

Point to Point MethodLength (distance) L = 150

'f ' factor = 2 x L x I (ASC) 46,296

# conductors per phase N = 1 Phase conductor constant C = 13,923 Phase Conductor

240 Volt f = 4.156

Neutral conductor constant C = 13,923 Neutral ConductorMultiplier 120 Volt

f = 8.313

1 + f Line to Line M = 0.194Line to Neutral M = 0.107

Fault Current at Service Equipment

8,978 amperes7,457 amperes

Fault Current from Service disc. to panel "_____"

Single Phase Feeder Length (distance) L = 40

'f ' factor = 2 x L x I (ASC) 8,978 Phase 7,457

# conductors per phase N = 1 Phase conductor constant C = 2,430 Phase Conductor

240 Volt f = 1.231

Neutral conductor constant C = 1,995 Neutral Conductor120 Volt

Multiplier f = 2.492

Line to Line M = 0.448

I = kVA x 1000 = trans. FLA

Isca = trans. FLA x 100

Isca = ampere short-circuit current RMS symmetrical. Isca =

Isca = N x C x E L-N

Volt Line to Line E L - L =

Volt Line to Neutral E L - N =

M = 1

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




M = 1

Step 1b: Insert the kVA rating (PF) power factor and (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.)

or

Step 1a: Insert available fault current ampere rating

from local utility

Step 3: Insert the Length of one conductor

Step 4: Insert the number of conductors per phase.

Step 2: Select type of conductor and conduit

Step 5b: Select neutral conductor size.

Step 5a: Select phase conductor size size.

Finished product CALCULATED FAULT

CURRENT

Personalize by enteringthe 'Job Name' and

'Your Company Name'

Continue by entering next point to point

F8

Clear Contents from this cell if using Step 1b.

I8

If you have inserted the Utility Fault Current amps, inserting the kVA in this cell is not required. If available fault current is not known, insert transformer kVA rating here.

I9

This value will be filled in when you 'Select Voltage & Phase' from the first dropdown box

I14

If you have inserted the Utility Fault Current amps inserting trans. impedence in this cell is not required. If transformer fault current is not known, insert transformer impedence here.

I18

Enter the length of one phase conductor in the group

I20

Enter the number of electrical conductors per phase between the transformer and the equipment

I21

Select a conductor and raceway from the top drop down box Then select phase conductor size from AWG drop down box.

I24

Select neutral conductor size from AWG drop down box.

I40

Data from above calculation

K40


I42

To select a conductor check the appropriate box to the right of the largest phase conductor

1 + f Line to Neutral M = 0.286

4,024 amperes2,135 amperes

Calculation does not include motor contribution

Branch Circuit Fault fr Point to Point

Single Phase Branch Length (distance) L = 30

'f ' factor = 2 x L x I (ASC) 4,024 Phase 2,135

# conductors per phase N = 1

Phase conductor constant C = 617 Phase Conductor

240 Volt

f = 1.630

Neutral conductor constant C = 617 Neutral Conductor

120 VoltMultiplier f = 1.730

Line to Line M = 0.380

1 + f Line to Neutral M = 0.366

1,530 amperes

782 amperes

Calculation does not include motor contribution

Isca x M = fault current at terminals of the panel L- L = Isca x M = fault current at terminals of the panel L- N =

Isca =

N x C x E L-N



M = 1

Isca x M = fault current at terminals of the panel L- L =

Isca x M = fault current at terminals of the panel L- N =

I63


K63


I65

To select a conductor check the appropriate box to the right of the largest phase conductor

Available Fault Current CalculationVer. 4.00

Neutral

Step 1b: Insert the kVA rating (PF) power factor and (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.)

Step 5b: Select neutral conductor size.

V34

=choose(U34,0,208,208,480,480) Link to cell I9 for select voltage

P35

Neutral , Choose formula ref point

V35

=choose(U34,0,240,208,208,480,480) Link to cell I9 for select voltage

V36

=choose(U34,0,208,208,480,480) Link to cell I9 for select voltage

V37

=CHOOSE(U34,0,I8*1000/I9,I8*1000/I9,I8*1000/I9*1.732,I8*1000/I9,I8*1000/I9*1.732) Link to cell I9 for select voltage

V38

=choose(U34,0,120,120,120,277,277) Link to nuetral voltage

V39

=choose(U34,0,240,208,208,480,480) Link to line to line voltage

V41


V44


P49

Link cell to conductor size

V49


Neutral

V67


P72


V72


W79

when cell = -1 then feeder is 3ph

X79

if cell = 1 then branch is 1ph

W80

when cell = -1 then feeder is 3ph

X80

if cell = 1 then branch is 1ph

Job Name

Your Company Name

Available Fault Current Calculation by: John Sokolik Ver. 6.6 Three Single conductors

Utility Symmetrical Fault Current 26,456 amperes kVA = in PVC raceway E = 240 AWG

trans. FLA = 0 ###E ###

###

= PF = ###

transformer Z Z = ###0 amperes ###

###Point to Point Method ###

Length (distance) L = 150 ###

'f ' factor = 2 x L x I (ASC) 26,456 ###

# conductors per phase N = 1 ### Phase conductor constant C = 13,923 Phase Conductor ###

240 Volt ### f = 2.375 ###

Neutral conductor constant C = 13,923 Neutral Conductor ###Multiplier 120 Volt ###

f = 4.750 400 kcmil500 kcmil

1 + f Line to Line M = 0.296 600 kcmilLine to Neutral M = 0.174 750 kcmil

Fault Current at Service Equipment 1000 kcmil

7,838 amperes6,901 amperes Neutral

###

###

[email protected]

I = kVA x 1000 = trans. FLA

Isca = trans. FLA x 100 x PF

Isca = ampere short-circuit current RMS symmetrical. Isca =




M = 1

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

mailto:[email protected]

F8

When the fault current is provided by the governing utility insert the value here. The KVA, power factor (PF) and impedence (Z) of the transformer is not required to continue. Go to L= ,and insert footage of conductor length. Use 'Edit', 'Clear', Comments' to clear any input cell.

I8

If you have inserted the Utility Fault Current amps, inserting the kVA in this cell is not required. If available fault current is not known, insert transformer kVA rating here.

I13

Inserting a (PF) power factor is only required when using method 1b, see example sheet tab in lower left of the sheet. The (PF) default for this cell should be 100% unless the power factor is known.

I14

If you have inserted the Utility Fault Current amps inserting trans. impedence in this cell is not required. If transformer fault current is not known, insert transformer impedence here.

I18

Enter the length of one phase conductor in the group

I20

Enter the number of electrical conductors per phase between the transformer and the service equipment

I21

Select a conductor and raceway from the top drop down box Then select phase conductor size from AWG drop down box.

I24

Select neutral conductor size from AWG drop down box.

faultv6.2.xls

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