Transformer short circuit fault calculation clarificationby lavalin Tags: calculation, circuit, clarification, fault, transformer 

#1
Nov508, 06:51 PM

P: 6

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? 



#2
Nov608, 12:52 AM

P: 219

If = V_{ph} / Z_{tr}(ohm)
Z_{tr}(ohm) = %Z / Z_{b} Z_{b} = V^{2}/S %Z : Transformer per unit impedance V : Transformer Rated voltage (V) S : Transformer Rated power (VA)  Creative thinking is enjoyable, Then think about your surrounding things and other thought products. http://electricalriddles.com 



#3
Nov608, 04:09 PM

P: 6

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.




#4
Nov1208, 03:31 PM

P: 660

Transformer short circuit fault calculation clarification
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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