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#128365 - 01/26/03 12:22 AM Short Circuit 1000kva 3 ph 1.6% Impedance 277/480  
HouseOfPower  Offline
Member
Joined: Jan 2003
Posts: 11
Houston Texas
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 ?


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#128366 - 01/26/03 02:50 AM Re: Short Circuit 1000kva 3 ph 1.6% Impedance 277/480  
Scott35  Offline

Broom Pusher and
Member
Joined: Oct 2000
Posts: 2,707
Anaheim, CA. USA
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 " 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  
Nick  Offline
Member
Joined: Aug 2001
Posts: 599
Riverside, CA
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  
HouseOfPower  Offline
Member
Joined: Jan 2003
Posts: 11
Houston Texas
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  
Scott35  Offline

Broom Pusher and
Member
Joined: Oct 2000
Posts: 2,707
Anaheim, CA. USA
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  
Gwz  Offline
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  
Bjarney  Offline
Moderator
Joined: Apr 2002
Posts: 2,527
West-Southern Inner-Northeast ...
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  
HouseOfPower  Offline
Member
Joined: Jan 2003
Posts: 11
Houston Texas
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  
Bjarney  Offline
Moderator
Joined: Apr 2002
Posts: 2,527
West-Southern Inner-Northeast ...
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  
Gwz  Offline
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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