Hi,(adsbygoogle = window.adsbygoogle || []).push({});

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?

**Physics Forums - The Fusion of Science and Community**

# Transformer short circuit fault calculation clarification

Know someone interested in this topic? Share a link to this question via email,
Google+,
Twitter, or
Facebook

Have something to add?

- Similar discussions for: Transformer short circuit fault calculation clarification

Loading...

**Physics Forums - The Fusion of Science and Community**