×
INTELLIGENT WORK FORUMS
FOR ENGINEERING PROFESSIONALS

Are you an
Engineering professional?
Join Eng-Tips Forums!
• Talk With Other Members
• Be Notified Of Responses
• Keyword Search
Favorite Forums
• Automated Signatures
• Best Of All, It's Free!

*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.

# Calculation of the X0/R0 ratio for a Zig Zag transformer

 Forum Search FAQs Links MVPs

## Calculation of the X0/R0 ratio for a Zig Zag transformer

(OP)
I'm trying to calculate the X0/R0 ratio of a Zig Zag transformer. I got the following information:
-Winding connection: Zig Zag
-Short time neutral current: 3500 A 5 sec.
-Continuous neutral current : 250 A
-Voltage : 26.4 kV
-Zero sequence impedance: 7.5 ohms/ph
-No load losses: 1951 W
-Load losses : 18214 W

How do we calculate the X0/R0 ratio using these informations?

Thank!

### RE: Calculation of the X0/R0 ratio for a Zig Zag transformer

(OP)
Anyone has any hint for me? I looked at the standard IEEE C57.12.91, but could not find a specific example for a Zig-Zag transformer.

### RE: Calculation of the X0/R0 ratio for a Zig Zag transformer

Since the "load" losses should relate directly to the winding resistance, and the this load should consist entirely of zero sequence current, maybe you can approximate using this value. Once you have the resistance and the impedance you should be able to compute the X, and then the X/R.

### RE: Calculation of the X0/R0 ratio for a Zig Zag transformer

You have got the Z0 in Ohms. Unless you get the X/R ratio from the manufacturer it is my understanding that you cannot calculate X0/R0.

### RE: Calculation of the X0/R0 ratio for a Zig Zag transformer

@Tony2802,
I have verified your data and found these:
1) The information on the short-time neutral current of 3500A for 5 seconds and the continuous neutral current rating does not correspond to the IEEE 32 numbers that I found. The continuous current if we use IEEE 32 will yield 105A. The continuous current for a 10-second time rating is 3% of the ground fault (3500 X 0.03 = 105A). Instead, your data fits with a grounding duty for 1 minute (7% of ground fault current, 3500 X 0.07 =245 ~ 250A).
2) You can use the no-load and full-load losses to calculate the resistance. no-load losses = core losses and full-load losses = core losses + copper losses. Therefore your copper losses = 18,241 - 1951 = 16,263 W. Equate those losses to 3I2R and you get R which is equal to 0.813 ohms.
3) Using the Z0 = 7.5 ohms, you'll get x0 = square root (7.52-0.8132) = 7.46 ohms
I got x0/r0 = 9.2

#### 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.

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

Click Here to join Eng-Tips and talk with other members! Already a Member? Login

Close Box

# Join Eng-Tips® Today!

Join your peers on the Internet's largest technical engineering professional community.
It's easy to join and it's free.

Here's Why Members Love Eng-Tips Forums:

• Talk To Other Members
• Notification Of Responses To Questions
• Favorite Forums One Click Access
• Keyword Search Of All Posts, And More...

Register now while it's still free!