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#128365  01/26/03 12:22 AM
Short Circuit 1000kva 3 ph 1.6% Impedance 277/480


I have another question.
On a short circuit current of a 1000 kva, 3 phase transformerbank with a 1.6% impedance operating at 277/480
First I did: 1000 x 1000= 1,000,000 / 277
=3610 3610/ 1.732= 2084 2084/ .016= 130,272.38?
Did I goof up with the impedance? Shoul it be divided by the .016 or 1.6 ?



#128366  01/26/03 02:50 AM
Re: Short Circuit 1000kva 3 ph 1.6% Impedance 277/480


I came up with an SCA of 83,619.64 Amps at the Transformer (3Ø Bolted Fault, Infinite Primary, 0.9 of the %Z for tolerances).
Here's the formula I used to acheive the resulting 83619.64 amps:
1: Determine the FullLoad Amperes of the Transformer:
I(FLA) = KVA × 1000 ÷ E(LL) × 1.73 [ "I(FLA)" = FullLoad Amperes, "E(LL)" = LineToLine Voltage ]
2: Find Transformer's Multiplier:
Mpl. = 100 ÷ 0.9 × %Z [ "Mpl." = Multiplier, "0.9" figures worstcase scenario tolerance of Impedance percentage ]
3: Determine Transformer's LetThrough Current:
I(SCA) = I(FLA) × Mpl.
So using your Transformer's ratings, here's what it looks like:
1: I(FLA) = 1204.2 Amps ( 1000 × 1000 ÷ 480 × 1.73 )
2: Mpl. = 69.44 ( 100 ÷ 0.9 × 1.6 )
3: I(SCA) = 83619.644444 Amperes ( 1204.2 × 69.44 )
Check my math and let me know if there are any errors.
Scott35 S.E.T.
Scott " 35 " Thompson Just Say NO To Green Eggs And Ham!



#128367  01/26/03 01:06 PM
Re: Short Circuit 1000kva 3 ph 1.6% Impedance 277/480


Scott, Your math checks out. Just out of curiousity I ran this through some software programs. Here are the results: Bussmann83,531.135 EDR83,532 This is using your .9 of %Z value. Pretty close huh?



#128368  01/26/03 01:15 PM
Re: Short Circuit 1000kva 3 ph 1.6% Impedance 277/480


Ok, I wasn't working the problem right. Looks like it is back to the drawing board. thanks



#128369  01/26/03 03:22 PM
Re: Short Circuit 1000kva 3 ph 1.6% Impedance 277/480


Nick,
Closer than I even expected!!!
I ran the numbers by hand! (feel free to make jokes or laugh!).
Scott35 S.E.T.
Scott " 35 " Thompson Just Say NO To Green Eggs And Ham!



#128370  01/26/03 10:07 PM
Re: Short Circuit 1000kva 3 ph 1.6% Impedance 277/480

Member

Joined: Apr 2002
Posts: 197


Given: KVA = 1000 E = 480V 3Ø = 1.732 % Z = 1.6 .9 = factor if KVA great than 25
using continuous calculator input the formula would be:
KVA * 1000 / 480V / 1.732 = * 100 / 1.6%Z / .9 = SCA
1000*1000/480/1.732=*100/1.6/.9= 83531.135061158155846377555384484 short circuit amperes.
That is by using the calculator in Windows Acessories on the computer programs file.
Just think    Could have used 1.732050807568877293527446341115059 for the square root of 3.
Then it would be 83528.684778591690467180089771695 Short Circuit Amperes.



#128371  01/26/03 10:58 PM
Re: Short Circuit 1000kva 3 ph 1.6% Impedance 277/480


1.6%Z? That is one stiff transformer.
Padmount? DeltaWye?



#128372  01/26/03 11:09 PM
Re: Short Circuit 1000kva 3 ph 1.6% Impedance 277/480


After the 1.732 you multiply by 100
Where and how do you get that figure?



#128373  01/26/03 11:14 PM
Re: Short Circuit 1000kva 3 ph 1.6% Impedance 277/480


It's likely that is needed to convert 1.6% into 0.016 perunit impedance.



#128374  01/27/03 08:33 AM
Re: Short Circuit 1000kva 3 ph 1.6% Impedance 277/480

Member

Joined: Apr 2002
Posts: 197


That is not the formula.
It is a way to keep using the ( hand ) calculator without stopping to enter a previous calculation into the final answer.
Of course the 1.6%Z sounds low for that KVA.
Maybe it's one of the super efficient transformer (super cooled coils) or that ( amorrous steel) [ I know that is not the correct spelling, but it is a special conductor/or/core used for some experimental transformers. It was mentioned in the EC&M magazine about 10 years ago.].
The local PoCo has 50KVA pad mounts with 0.9 %Z.
These 0.9 %Z units are 7.2KV/120/240V 1Ø 3W pad mounts.
[This message has been edited by Gwz (edited 01272003).]




