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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 Full-Load Amperes of the Transformer: I(FLA) = KVA × 1000 ÷ E(L-L) × 1.73 [ "I(FLA)" = Full-Load Amperes, "E(L-L)" = Line-To-Line Voltage ] 2: Find Transformer's Multiplier: Mpl. = 100 ÷ 0.9 × %Z [ "Mpl." = Multiplier, "0.9" figures worst-case scenario tolerance of Impedance percentage ] 3: Determine Transformer's Let-Through 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 " ThompsonJust 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: Bussmann-83,531.135 EDR-83,532 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 " ThompsonJust 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? Delta-Wye?

#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 per-unit 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 with-out 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 01-27-2003).]

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