Fuse Clearing Time for SLG fault on TRF Secondary

[ENGEIT]

I am a new member to the site, but has always refer to the site for answer to many of my electrical related issues. Recently, I ran into a problem with regarding to a high side transformer fuse clearing time for SLG fault on the secondary of a transformer. Below is the information to my issue.

Transformer: 10000kVA, 67kV Delta-12.47kV Wye-Grounded, 7.13%
High Side Fuse: S&C 100E-SMD-1A Slow
SLG Fault on Secondary: 5519A

To my understanding, I though the high side fuse will only see ~58% of the SLG fault on the secondary of transformer for A-Phase and C-Phase due to shifting of transformer damage curve ([5519*(12.47/67)*0.58]=595.7A @ 67kV) . Based on 595.7A current, the S&C 100E-SMD-1A Slow will clear the fault at ~1.8112 seconds. However, I was told that the 1.8112 seconds clearing time is incorrect. And I was told that the fuse should clear based on the following current 1027A ([5519*(7.2/67*1.732)]) not 595.7A. Either I am not fully understand the concept or whatever I was being taught in school was incorrect. Please advise.

Thanks,

 

[jghrist]

I believe that what you were being taught in school is correct. What is the basis for the formula that gives 1027A? Look at it this way. The current in the secondary winding is 5519A. The turns ratio is 67/7.2. Therefore the current in the primary winding is 5519*(7.2/67) = 593.1. This is also the line current in the two phases connected to the primary winding because there is no current in the other primary windings. The difference between 593.1 and 595.7 is that 0.58 is an approximation of 1/sqrt(3).

 

[Parchie]

Lemme help.
Let's make it clearer by stating the voltage ratings of your primary and secondary windings:
[ul]
[li]Primary winding = 67 kV per phase[/li]
[li]Secondary winding = 12.47/1.732 = 7.2 kV per phase.[/li]
[/ul]Your secondary line current of 5519 A will be seen as equal to a primary phase current of = 5519 x (12.47/1.732)/67 = 593 A.
Remember, that value is what runs thru the primary windings of you 100 MVA transformer.
Since your primary fuse is installed at the lines supplying the primary windings, it follows that the current it sees will be the line current of your transformer primary, which is equal to Iphase x 1.732 = 593*1.732 = 1027 A!

Still, refering to my copy of S&C clearing times, 100E Slow clears in 0.57523 seconds @ 1027A, not 1.812 seconds! (100E Slow clears in 1.80556 seconds @ 593 A)

 

[ScottyUK]

Parchie,

Your last line of calcs doesn't look quite right: the [√]3 factor should not be present. The only current flowing at the 67kV side is the current in the primary winding feeding the faulted secondary. This must be the case to maintain ampere-turn balance in the core. Sometimes physically drawing out the current flows in each pair of windings can help if it's hard to visualise.

 

[jghrist]

the current it sees will be the line current of your transformer primary, which is equal to Iphase x 1.732 = 593*1.732 = 1027 A!

Please explain how the line current could equal 1.732 times Iphase when there is only current in one phase winding.

 

[Parchie]

Thanks for the correction. I got lost there! With a 1 pu SLG fault at wye secondary, the primary delta winding which is magnetically coupled to the grounded secondary phase will have a 0.58pu phase current and two lines will see this current.
@ENGEIT, are you sure 5519 Amps is the SLG fault current?

 

[dpc]

ENGEIT,

Your primary current value look correct to me. I've learned to always agree with jghrist). 5519 A is reasonable - must include some source impedance. It will also depend on what is used for the transformer zero sequence reactance.