|
0 members (),
161
guests, and
10
robots. |
Key:
Admin,
Global Mod,
Mod
|
|
|
Joined: Jan 2003
Posts: 11
OP
Member
|
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 ?
|
|
|
|
Joined: Oct 2000
Posts: 2,723 Likes: 1
Broom Pusher and Member
|
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!
|
|
|
|
Joined: Aug 2001
Posts: 599
Member
|
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?
|
|
|
|
Joined: Jan 2003
Posts: 11
OP
Member
|
Ok, I wasn't working the problem right. Looks like it is back to the drawing board. thanks
|
|
|
|
Joined: Oct 2000
Posts: 2,723 Likes: 1
Broom Pusher and Member
|
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!
|
|
|
|
Joined: Apr 2002
Posts: 197
Member
|
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.
|
|
|
|
Joined: Apr 2002
Posts: 2,527
Moderator
|
1.6%Z? That is one stiff transformer.
Padmount? Delta-Wye?
|
|
|
|
Joined: Jan 2003
Posts: 11
OP
Member
|
After the 1.732 you multiply by 100
Where and how do you get that figure?
|
|
|
|
Joined: Apr 2002
Posts: 2,527
Moderator
|
It's likely that is needed to convert 1.6% into 0.016 per-unit impedance.
|
|
|
|
Joined: Apr 2002
Posts: 197
Member
|
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).]
|
|
|
Tom
Shinnston, WV USA
Posts: 1,044
Joined: January 2001
|
|
|
|
|