Transformer short circuit fault calculation clarification

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Discussion Overview

The discussion revolves around the calculation of short circuit fault currents in a transformer system, specifically addressing the use of phase voltage in these calculations. Participants explore the implications of transformer and cable impedances in the context of a three-phase bolted short circuit scenario.

Discussion Character

  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant presents a formula for calculating fault current at the load, suggesting that the phase voltage is used to represent the voltage drop from the neutral point of the WYE transformer to one phase.
  • Another participant provides a formula for calculating transformer impedance in ohms, indicating that the per unit impedance is derived from the transformer's rated voltage and power.
  • A later reply emphasizes that the use of phase voltage aligns with the winding impedance and the star connection of the load, suggesting that this approach yields the current in the secondary winding.
  • It is noted that while using line voltages and equivalent delta connections could also work, it may be less direct and usable compared to the phase voltage method.

Areas of Agreement / Disagreement

Participants express differing views on the appropriateness of using phase voltage versus line voltage in the calculations, indicating that there is no consensus on the best approach for this scenario.

Contextual Notes

Participants do not resolve the underlying assumptions regarding the choice of voltage representation in the fault current calculations, leaving these aspects open for further discussion.

lavalin
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Hi,

I'm trying to perform a short circuit fault calculation on this transformer and load I have when a 3 phase bolted short (worse case) occurs.

The transformer specs are below:
Z(transformer)=5%, Z(cable)= .00306 + j.0027, Delta/Star 6.6kV/400V

I can understand calculating the short circuit available current at the load:
Ifault(load) = Vphase/Z(transformer)
where Z(transformer) is the impedance of one winding, and the voltage across it would be 400/1.73 ~=230V
Ifault(load) = 230/.05
= 4.6kA

Now calculating the fault down at the load is where I get a little confused:
The example I'm following states that the load current fault for a 3 phase bolted short is:
Ifault(load) = Vphase/(Z(transformer) + Z(cable))

My only guess/understanding of why they use Vphase is that Vphase represents the voltage drop from the neutral (mid point) of the WYE transformer to the end of 1 phase? I've drawn a diagram to help clarify what I mean, its hosted at:
http://img83.imageshack.us/my.php?image=61884112bb8.jpg

Can someone please explain to me why they use Vphase in the load fault calculation?
 
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If = Vph / Ztr(ohm)

Ztr(ohm) = %Z / Zb

Zb = V2/S

%Z : Transformer per unit impedance
V : Transformer Rated voltage (V)
S : Transformer Rated power (VA) :smile:


--------------------------------------
Creative thinking is enjoyable, Then think about your surrounding things and other thought products. http://electrical-riddles.com
 
Thanks for your reply. My question was however clarification on why they use phase voltage when they're calculating at the load end, total fault current that the transformer can deliver.
 
Hi! I would say:

They use the phase voltage because this fits with the winding impedance, and the load is in star connection as well. This gives you the current in a secondary winding.

Within this logic, you could equally well compute everything with line voltages and equivalent delta connections. It would be consistent, but less direct and less usable.
 

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