I = VA / (E x 1.732 x Z)?

[kwired]

To add to what GD said and on a little simpler level (don't know how much you actually know about, you apparently are a student or apprentice of this trade) the typical loadcenter circuit breaker you will find in homes and small business is designed and tested to be able to withstand a 10,000 amp fault. If the supply to the loadcenter is capable of delivering more then 10,000 amps during a fault condition, you probably don't want to be around when that fault happens. Even the 10kA fault that it is supposed to withstand will likely leave you needing clean underwear if you are nearby when such event happens unexpectedly.


When I was in college our class happened to go to Square D's QO breaker plant in Lincoln, NE. This was maybe 1988 or 1989. My favorite point of that visit was the test lab. They let us observe them test a three pole breaker subject to what we were told was a 10,000 amp bolted three phase fault. Don't recall what amp setting the test breaker was, but it had at least 8 or 6 awg conductors on it and all three leads bolted together, snapped into a small loadcenter and closed the heavy doors on a bigger vault. Then turned on the power - "BOOM". opened the vault door and smoke was still rolling out of the loadcenter. Reset the breaker, and it did physically operate though it looked like something you wouldn't want to ever use again.

 

[Sahib]

The Z for 2 phase shorting case is to be twice the value for the 3 phase shorting case.

 

[Phil Corso]

Sahib...

Since the 'Z' comes from the %impedance, then, using your approach I(2) = I(3)/2, where the subscripts represent, respectively, 2-phase, and 3-phase fault currents!

The solution comes from the fact that Ia=0, and Ib = -Ic! Using a very useful mathematical tool called, Symmetrical-Components, results in the answer I(2) = 0.866 x I(3)!

Additional info available on request!

Regards, Phil

 

[Ingenieur]

The Z for 2 phase shorting case is to be twice the value for the 3 phase shorting case.

You don't need fortesque's symmetrical components (although fun lol) phasors will do
assume
100 kva
240/3
1.8% or 0.018 pu

z = 240^2 / 100000 x 0.018 = 0.0104 ohm
the v^2/S term is the base Z
Derived from S = v^2 / Z

For single phase from kvl and kcl with terminals shorted
basically single ph with 2 coils of Z
fault i = 240/(2 x 0.0104) = 11574 A

when deriving for 3 phase fault the sum at the fault node must be 0
so balanced each ph contributes 1/3
so we need consider only 1 coil
fault i = v/(sqrt 3 Z) = 240/(1.732 0.0104)= 13365

same as S/(sqrt3 v Z pu) = 100000/(1.732 240 0.018) = 13365

You can do the same kvl/kcl phasor analysis including the lines
Zl line
Zx xfmr
3 ph ... v/(sqrt3(Zl + Zx))
1 line to gnd ... v/(sqrt3(2 Zl + Zx))
L-L ... v/(2(Zl + Zx))

sc's basically convert/transforms from a phase reference frame to a sequence reference
positive abc at 120 deg
negative acb at 120 deg
zero all in phase
if no faults only pos
gnd fault you get some zero
line-line some neg

once transformed becomes algebra
then convert back to phase
do e with a matrix using an operator
j is 1/90 deg is common to phasor/complex
a is used for sc = 1/120 deg