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#128729 11/07/03 08:46 PM
Joined: Aug 2003
Posts: 1,374
R
Ryan_J Offline OP
Moderator
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.


Ryan Jackson,
Salt Lake City
#128730 11/07/03 09:50 PM
Joined: Apr 2002
Posts: 2,527
B
Moderator
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 11/10/03 08:17 AM
Joined: Oct 2000
Posts: 2,721
Broom Pusher and
Member
SCA calcs (formulas) were something I should have compiled and posted in the Reference section, but strangely enough have yet to do so!
[Linked Image]

I'll see what can be compiled from the Bussman SPD documents for the three commonly used methods, then post ASAP.

Scott35


Scott " 35 " Thompson
Just Say NO To Green Eggs And Ham!
#128732 11/10/03 10:26 AM
Joined: Aug 2003
Posts: 1,374
R
Ryan_J Offline OP
Moderator
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 [Linked Image]


Ryan Jackson,
Salt Lake City
#128733 01/05/04 05:10 AM
Joined: Jul 2002
Posts: 8,407
Member
Ryan,
As a rule, PSCC(Prospective Short Circuit Current) is governed by this formula, as a very simplistic way of looking at it:
I(pscc)= V(oc) / Z.


[This message has been edited by Trumpy (edited 01-05-2004).]

#128734 02/06/04 07:02 PM
Joined: May 2003
Posts: 26
I
Member
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 02/07/04 12:03 AM
Joined: May 2003
Posts: 13
M
Member
There are different method of calculating
short circuits.
It depends what component are installed
on the system.

One book that help me a lot, whem I am
doing short circuit calculation is

"Electrical System analysis and design for
Industrial Plant by Irwin Lazar".

The examples are very easy to understand
and very applicable to modern electrical
system that exist today.

#128736 02/14/04 12:37 AM
Joined: Oct 2000
Posts: 2,721
Broom Pusher and
Member
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 05/04/04 06:21 PM
Joined: May 2004
Posts: 7
F
New Member
"Short Circuit Capacity: Basic Calculations and Transformer Sizing"
http://www.powerqualityanddrives.com/short_circuit_transformer/


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