Short Circuit Calculation (What Is Correct Method?)

mbrooke

Batteries Not Included
Location
United States
Occupation
Electricity
It wasn’t advice, it was the truth

I know its the truth in the sense the AHJ doesn't care, but I think its wrong (unethical) to play off someones ignorance knowing the equipment could explode with unfused conductors inside it.
 

mbrooke

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Location
United States
Occupation
Electricity
See this post. I just don’t understand why some POCOs use charts that assume a particular Z that is generally a lot lower than actual impedance. I made a comment here about the ease of some modeling software that is super easy to give fault numbers that are real.

Thanks :)

So why don't all POCOs do this or give a detailed PDF showing the assumed Z of the trafo and assumed temps involved?

There really needs to be some type of table in the NEC so electricians can do the same.
 

kingpb

Senior Member
Power companies are going to give you a number that is conservative; you just need to know which side of the transformer they are giving it to you on. I've been given number on high side and also low side. Typically, the value is going to assume infinite bus any way.

The easiest way to do quick SC calcs is the MVA Method. Search online, you can download the IEEE document.
 

Hv&Lv

Senior Member
Location
-
Occupation
Engineer/Technician
Thanks :)

So why don't all POCOs do this or give a detailed PDF showing the assumed Z of the trafo and assumed temps involved?

There really needs to be some type of table in the NEC so electricians can do the same.
One word.

Liability...
 

Dell3c

Member
Location
WA
Occupation
Electrician
I have seen two different ways of running a short circuit (fault) calculation to determine ratings for gear. Normally designing commercial buildings, selecting AIC ratings.

The first method Ive seen used by many places around town is Isc=V/Z. In this case you calculate the total impedance starting from NEC 'table 9' and then divide the voltage by total calculated impedance. However I cant seem to find where this full calculation is written out or described as the acceptable method.

The second method is the point-to-point method used by Eaton/Bussman calc:

Is one of these a more accurate/correct method? Have you seen AHJ's every require one over the other?
(example) the interrupting capacity or fault current rating of a 25KVA transformer w/a 1.5% impedance, supplied by 120/240V single phase secondary.
FLA= (KVA x 1000)/V
FLA= (2500 /240) =104A

Finding Interrupting capacity.
Fault Current = (FLA / impedance)
Fault Current = (104A / .015)=6933A
Fault Current is 6933A
 

mbrooke

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Location
United States
Occupation
Electricity
(example) the interrupting capacity or fault current rating of a 25KVA transformer w/a 1.5% impedance, supplied by 120/240V single phase secondary.
FLA= (KVA x 1000)/V
FLA= (2500 /240) =104A

Finding Interrupting capacity.
Fault Current = (FLA / impedance)
Fault Current = (104A / .015)=6933A
Fault Current is 6933A

Incorrect, its double that value for a close in 120 volt fault.
 

david luchini

Moderator
Staff member
Location
Connecticut
Occupation
Engineer
That shows the fault current for a 120V secondary or a 240V secondary, not for a L-N fault on a 120/240V secondary.

Just as for the 3 phase, they show the fault current for a 208V, 240V, and 480V secondary.
 

mbrooke

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Location
United States
Occupation
Electricity
That shows the fault current for a 120V secondary or a 240V secondary, not for a L-N fault on a 120/240V secondary.

Just as for the 3 phase, they show the fault current for a 208V, 240V, and 480V secondary.

Right, but I'm thinking that the secondary 120 volt winding will have the full leakage reactance from the core.
 

mbrooke

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Location
United States
Occupation
Electricity
The chart you referenced tells you that the maximum short circuit current for a 50kVA transformer with a 120V secondary and a 2% impedance will be 20,828A. 240V is irrelevant to that value.

240 volts is because while the turns ratio increases the core can only transfer so much magnetism.
 

david luchini

Moderator
Staff member
Location
Connecticut
Occupation
Engineer
240 volts is because while the turns ratio increases the core can only transfer so much magnetism.
I must not be following your point. If you have a (pick a primary) 12000V-120V single phase, 50kVA transformer with a %Z of 2.0, the max available short circuit current will be 20,828A, as the list you posted shows. 240V is irrelevant to a 120V secondary transformer.

I think you're trying to introduce something into that Alabama Power chart that isn't there?
 

mbrooke

Batteries Not Included
Location
United States
Occupation
Electricity
I must not be following your point. If you have a (pick a primary) 12000V-120V single phase, 50kVA transformer with a %Z of 2.0, the max available short circuit current will be 20,828A, as the list you posted shows. 240V is irrelevant to a 120V secondary transformer.
Correct, both windings in parellel.

But when both are in series the fault current from X1 to X0 will be double vs X1 to X2.
 

david luchini

Moderator
Staff member
Location
Connecticut
Occupation
Engineer
Maybe this will be more clear:

50kVA, 2.0%Z, 120V secondary: max short circuit current = 20,828A

50kVA, 2.0%Z, 240V secondary: max short circuit current = 10,414A.

50kVA, 2.0%Z, 120/240V secondary: max short circuit current L-L = 10,414A.
..........................................................................max short circuit current L-N = 15,621A.
 

tortuga

Code Historian
Location
Oregon
Occupation
Electrical Design
For single phase I use
Transformer FLA = KVA X 1000 /Volts

Available fault current = 100% / impedance X Transformer FLA


F= 2 X L X I / C X E
M = 1 /1 + F

Available Fault current = I X M


Where
L is length in feet
I = Fault current
F = Factor
M = multiplier line to line
C = Constant values from a table of conductor sizes (like bussman ) X number of conductors
E = voltage

Then compare with other software tools as needed.

Edit: plugged in Davids numbers above and got 10,417.
 
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