×
INTELLIGENT WORK FORUMS
FOR ENGINEERING PROFESSIONALS

Log In

Come Join Us!

Are you an
Engineering professional?
Join Eng-Tips Forums!
  • Talk With Other Members
  • Be Notified Of Responses
    To Your Posts
  • Keyword Search
  • One-Click Access To Your
    Favorite Forums
  • Automated Signatures
    On Your Posts
  • Best Of All, It's Free!
  • Students Click Here

*Eng-Tips's functionality depends on members receiving e-mail. By joining you are opting in to receive e-mail.

Posting Guidelines

Promoting, selling, recruiting, coursework and thesis posting is forbidden.

Students Click Here

Jobs

"Neutral" Transformer sizing

"Neutral" Transformer sizing

"Neutral" Transformer sizing

(OP)
Dear all,

We have come across with below situation.

The power system is available with 3 wire/22kV. Now, we want to use the neutral wire for the using area at 22kV. Due to so long distance and mountaneous area. We force to use one another transformer 22/0.4kV_Wye/Delta connected to the system. The purpose of this transformer is nothing more than taking advantage of its primary side neutral point to run along with the original 22kV to the using point.  

Question for this is: How to correctly size this transformer capacity? Take note that we will just only use its neutral point.

Any suggestion is welcome.

Regards,

Hien   

RE: "Neutral" Transformer sizing

Let me try. Here it is:

CODE

  Phase to Neutral Voltage X expected phase to ground fault current X "K" = 3-phase kVA
K is a constant that depends on the allowed fault duration:

CODE

time          K
10 sec      0.064
1 min       0.104
2 min       0.139
3 min       0.170
4 min       0.196
5 min       0.220
This information is taken form the Westinghouse "Electrical Transmission and Distribution Reference Book" published in 1964.
 

RE: "Neutral" Transformer sizing

If you are sizing the transformer for steady state single phase loads on the 22 kV system, then the transformer must be three times the maximum single phase load.
Example: With a 10A load on "A" phase the 10A must return through the "A" phase transformer winding. The voltage will be 22kV / 1.72 = 12.7kV
Single phase load = 10A x 12.7kV = 120 KVA
Transformer bank =  x 120 KVA
A corresponding current will circulate in the delta secondary. 10A x (22kV/.4kV) = 550A
Phase angle errors may cause a neutral displacement.
Phase angle errors may cause circulating currents in the delta winding.

Bill
--------------------
"Why not the best?"
Jimmy Carter

Red Flag This Post

Please let us know here why this post is inappropriate. Reasons such as off-topic, duplicates, flames, illegal, vulgar, or students posting their homework.

Red Flag Submitted

Thank you for helping keep Eng-Tips Forums free from inappropriate posts.
The Eng-Tips staff will check this out and take appropriate action.

Reply To This Thread

Posting in the Eng-Tips forums is a member-only feature.

Click Here to join Eng-Tips and talk with other members!


Resources