Scott (or anyone else): I glanced through you very informative technical resource area, but I could not find any examples of short circuit/fault current calculations. Would you mind showing me an example of one when you find the time? Thank you very much.

One simplified calculator is listed at http://www.bussmann.com/apen/software/ Available short-circuit currents are primarily controlled by various transformer impedances in power systems, which also cause voltage drop in AC-electrical circuits.

#128731 - 11/10/0308:17 AMRe: Short circuit calc's

Scott: Thanks. Again, at your convienance. I hope I didn't come across as complaining about your very informative section, that was certainly not the intent.

Bjarney: Thanks. I have reviewed the site and found it quite helpful

Ryan Jackson, Salt Lake City

#128733 - 01/05/0405:10 AMRe: Short circuit calc's

Easiest way is ignore primary and secondary conductor impedances since the transformer impedance is usually quite large compared to the wire impedance. Then it's just Ohm's law, I = E/Z, or I = (1/%Z)(FLA).

In other words, just divide FLA by Transformer PU impedance(%Z/100)

Example: For a typical 25 kva, 12kv to 120/240v transformer, Z% = 2.4%. I (SC) = 104.2/.024 = 4340A

BTW, some new pole mounted and some padmount transformers have impedances less than 2 percent.

#128735 - 02/07/0412:03 AMRe: Short circuit calc's

Here's some "Quick 'N Dirty" SCA calc formulas, using the basic "Point-to-Point" method.

Available short-circuit symmetrical RMS current is found using the 6 steps described below:

Step 1; Determine the Transformer's Full-Load Amperes (I·fl):

3Ø Transformer

I·fl = KVA × 1000 / E·l-l × 1.732

1Ø Transformer

I·fl = KVA × 1000 / E·l-l

Step 2; Find the Transformer multiplier (M·tr):

M·tr = 100 / (%Z × 0.9) where: %Z is the Transformer's Impedance.

Step 3; Determine the Transformer let-thru short circuit current (I·sc):

I·sc = I·fl × M·tr

FYI: Induction Motor contribution may be added to the above figure at this point. An estimate for Motor contribution to SCA is: 4 to 6 × FLA of Motor(s). LRA may also qualify.

Step 4; Calculate the "F" factor:

3Ø Faults:

F = 1.732 × L × I·3ø / C × E·l-l

1Ø L-L Faults on 1ph center tapped transformer:

F = 2 × L × I·l-l / C × E·l-l

1Ø L-N Faults on 1ph center tapped transformer:

F = 2 × L × I·l-n / C × E·l-n

where: "L" = Length (feet) of conductor to the fault. "C" = Constant from "Table C" of the Bussman manual (will add this later - big database!). "I" = Available short circuit current at beginning of circuit. "L-L" = Line-to-Line. "L-N" = Line-to-Center Tapped "Neutral" conductor. *Note: L-N fault current is higher than the L-L fault current at the secondary terminals of a 1Ø center tapped transformer. The short circuit current available for this case in step 4 should be adjusted at the transformer terminals as follows: I·l-n = 1.5 × I·l-l at transformer terminals.

Step 5; Calculate "M" (multiplier):

M = 1 / 1 + F

Step 6; Calculate the available short circuit symmetrical RMS current at the point of fault (I·afc):

I·afc = I·sc × M

I'll post a few example SCA scenarios later when posting the "C" database.

Scott35

Scott " 35 " Thompson Just Say NO To Green Eggs And Ham!

#128737 - 05/04/0406:21 PMRe: Short circuit calc's