Short Circuit Calculation

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mepengineer

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I can figure out the short circuit calculation , see attach. trying to find oiut how you get the 4A 12,367A from the calculation
dont i need the impedance of transformer? little confuse

thanks
 

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GoldDigger

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You do not need to know the impedance of the transformer directly since result 4 gives you the open circuit voltage and short circuit current at that point. From that you can calculate the Thevinin equivalent series resistance and use it in the next calculation.
What is not clear to me from the one line drawing is whether the current is line to line or line to neutral (if they are different).
 
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jrohe

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You do not need to know the impedance of the transformer directly since result 4 gives you the open circuit voltage and short circuit current at that point. From that you can calculate the Thevinin equivalent series resistance and use it in the next calculation.
What is not clear to me from the one line drawing is whether the current is line to line or line to neutral (if they are different).

The image also does not indicate the conductor material (copper or aluminum), whether it is three single conductors or a 3-conductor cable, or the conduit type (steel or nonmagnetic). All of these factors play into calculating the available fault current.
 

Pharon

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It's a safe bet that they're looking for the symmetrical three phase bolted fault, which is generally viewed as the worst case scenario.

In some cases it isn't, but given no other information, that's the way I'd calculate it (and assuming copper conductors).
 

GoldDigger

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Just looking at the drawing, I would not have identified that as there phase. A three phase symmetric fault would have half the line resistance in the additional segment of wire compared to single phase single phase single sided.
 

topgone

Senior Member
I can figure out the short circuit calculation , see attach. trying to find oiut how you get the 4A 12,367A from the calculation
dont i need the impedance of transformer? little confuse

thanks
Yep.
That's a three-phase fault on a 300 kVA transformer with a 2.08 % impedance (Fault current =833/0.0208 = 40kA)!
For a 50 feet line sized #1 AWG, you'll get around 12kA for a 3-phase fault at the line's end. Here's why:
C value of #1 conductor = 7293; f = 1.732 x 50' x 40,000/(7293 x 1 x 208) = 2.28; M = 1/(1+2.28) = 0.304; Therefore, fault current at end of the 50-foot line = 40,000 x 0.304 = 12,181A.
Crystal!
 

Julius Right

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Electrical Engineer Power Station Physical Design Retired
I agree with topgone, of course. I only intend to explain the use of parameter C.
Since [approx.] Ztrf=EL-L/SQRT(3)/Isc4 then Isc4A=EL-L/SQRT(3)/(Ztrf+Zcb) ?also approximate-and by dividing with Ztrf Isc4A=EL-L/SQRT(3)/Ztrf/(1+Zcb/Ztrf)
or Isc4A=Isc4/(1+Zcb/Ztrf). Now if Zcb/Ztrf=f
f=Zcb/ (EL-L/SQRT(3)/Isc4)=sqrt(3)*Isc4*Zcb/(EL-L)
Zcb=zcb1000ft/1000*L/n. If C=1000/zcb1000ft then f=sqrt(3)*L*Isc4/(EL-L*C*n).
This way is very approximate - I agree with mepengineer. However it is a very expedite way.:D
 

Phil Corso

Senior Member
MepEngineer ?

Your problem is the classic ?Fault-Reduction? question that?s occasionally presented in the PE Exam! Using my GI.FI.E.S (pronounced Jiffys) method, here is my solution:

Part I:


GI
ven: Vs in Volts, I4, and I4A in Amperes

FI
nd: Reactance, Xr, to reduce I4 to I4A (At this point cable parameters are unnecessary!)

E
quation: Xr = ( Vs/Sqrt3 )*[ (1/I4A ? (1/I4) ] Ohm/ph

S
olution: Xr = (208/Sqrt3)*[ (1/12,367)-(1/40,000) ] = 0.00670 Ohm/ph => 6.70 mOhm/ph

MepEngineer? the above the answer is sufficient, IFF, it was the answer required on the PE Exam! However your sketch presents an incongruous situation! The cable shown, i.e., #1 AWG, having a length of 50-ft, will not yield the same reactance! Instead, the value will be as shown below:

Part II:


GI
ven: #1 AWG Cable, Length, L in ft, and XL, per NEC, in ?/k-ft

FI
nd: Cable Total Reactance, XT, in ? (Ignoring fact cable is 3 times more resistive!)

E
quation: XT = (L/1,000)*XL

S
olution: XT = (50/1,000)*0.046 = 0.0023 Ohm => 2.30 mOhm/ph

MepEngineer? are you sure about the parameters shown in your illustration?

Regards, Phil Corso
 
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Phil Corso

Senior Member
GoldDigger... No! Not in the secondary, but two primary lines will. Let me explain:

Consider a two-winding Xfmr; one Delta-connected, the other Wye-connected! Now consider the D-winding connected to a source and the Y-winding connected to a static load. The following chart illustrates primary and secondary current distribution in percent of a 3-ph Y-side terminal fault:

Phase Line-Ground Fault Line-Line Fault
| Prim'y | Secnd'y | | Prim'y | Secnd'y |
A 58% 100% 100% 87%
B 58% 0% 50% 87%
C 0% 0% 50% 0%

Ex 1) If secnd'y Ph-A shorts to ground, it will (*generally*) carry 100% of the current associated with a 3-ph fault, while B & C phases carry none! Two primary lines each carry 58% and the third none!

Ex 2) If two secondary lines short, then each will carry 58% of the current associated with a 3-phase fault, but the third will carry none! One primary line will carry 100%, while each of the other two will carry 50%!

Please note the asterisked (*) term "generally"! It means that for certain Xfmrs the line-to-ground fault, will produce a higher fault-current than the 3-ph fault! But that's another topic!

Caveat:
The above only cover D-Y xfmrs! Other winding arrangements, i.e, Y-D, Y-Y, etc, have different distributions!!

Regards, Phil
 
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GoldDigger

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Phil, how do you get from the OP's one line that this is a delta-wye transformer? Or even delta delta for that manner? I am not up to speed on the symbols used.
 

Phil Corso

Senior Member
GoldDigger... I didn't!

Your question prompted me to explain the conditions, under which, doubling the cable-impedance was warranted for a line-to-neutral (ground) fault! There are none for the problem presented by the OP! And I gave an example of a frequently used configuration!

Phil

Ps: Took the opportunity to re-align the Table!

Phase........Line-Ground Fault............Line-Line Fault
................| Prim'y | Secnd'y |.........| Prim'y | Secnd'y |
...A...............58%....100%..............100%........87%
...B...............58%........0%................50%........87%
...C................0%.........0%................50%..........0%

 
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mepengineer

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Location
New York, NY
thanks everyone for all you help . i have a better understanding now. i have found another problem see attach.
but this is on a 480v . lets keep the discussion and ideas open.
 

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topgone

Senior Member
thanks everyone for all you help . i have a better understanding now. i have found another problem see attach.
but this is on a 480v . lets keep the discussion and ideas open.

Let's do it in another way this time.
From tables:
R/1000 ft of 250MCM = 0.0457 ohms/1000ft.
X/1000 ft = 0.0399 ohms/ 1000 ft.
Z = 0.060667 ohms/1000 ft or a conductor impedance = 0.060667 x50 /1000 = 3.0333 x 10^-3 ohms

The system impedance can be computed from the Isc = 40kA:
MVAsc = 1.732 x 40,000 x 480 = 32.255 MVA and
the system impedance Zs = 480^2/32.255 = 6.928 x 10^-3 ohms
Then, we compute for the short circuit amps available at the end of the 250MCM:
= 40,000 x Zs/(Zs+Zcond) = 40,000 x 6.928/(6.928 + 3.0333) =28,819.66 Amps.

Similar results compared to using C value of 16483 in the method used earlier!
 

mepengineer

Member
Location
New York, NY
Yep.
That's a three-phase fault on a 300 kVA transformer with a 2.08 % impedance (Fault current =833/0.0208 = 40kA)!
For a 50 feet line sized #1 AWG, you'll get around 12kA for a 3-phase fault at the line's end. Here's why:
C value of #1 conductor = 7293; f = 1.732 x 50' x 40,000/(7293 x 1 x 208) = 2.28; M = 1/(1+2.28) = 0.304; Therefore, fault current at end of the 50-foot line = 40,000 x 0.304 = 12,181A.
Crystal!


is this from where i start from the secondary of the transformer i go 50' away on #1 awg and this is what i end up with (12,181A)?
 

topgone

Senior Member
is this from where i start from the secondary of the transformer i go 50' away on #1 awg and this is what i end up with (12,181A)?
Correct. Take the cue from my latest calc above your post. As mentioned by @Phil Corso, the short circuit current reduces corresponding with the additional impedance inserted by the additional line (in your case, #1 AWG). The reduction ratio being Zsystem/(Zsystem + Zconductor).

Also, readback @Julius Right's explanation on how the C value, the f and M factors come into play with these SC calculations. Hope to be of help.
 
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