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 Post subject: Short circuit calculation for split winding transformerPosted: Tue Jun 25, 2013 1:59 pm

Joined: Tue Jul 10, 2012 11:18 am
Posts: 3
Hi All,
a first step in arc flash is to calculate short circuit.
i have in my system a transformer with a split secondary winding winding
the rating of the transformer
Primary 3000KVA
Secondary 1: 1500kVA
Secondary 2: 1500kVA
ZI% = 6%
ZY%= 5.75
ZX%= 6.01
primary voltage 25000V
secondary 480V

My software (using ANSI / IEEE method) calculated the fault as
3000/(1.73*0.48*0.0575)

i think it should be calculated differently, and i should use 1500kV in my circulations for each side.

Please let me know if you share my opinion or maybe im completely wrong.
Should you have a link to a reliable source it would supper helpful.

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 Post subject: Posted: Fri Jun 28, 2013 6:43 am
 Plasma Level

Joined: Wed May 07, 2008 5:00 pm
Posts: 816
Location: Rutland, VT
I think 1500 kVA should have been used. The reason is that the bus attached to the secondary winding will only see the fault current that can be passed by the 1500 kVA winding.
What software are you using?

_________________
Barry Donovan, P.E.
www.workplacesafetysolutions.com

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 Post subject: Posted: Sun Jun 30, 2013 9:58 am
 Plasma Level

Joined: Mon Sep 17, 2007 5:00 pm
Posts: 1464
Location: Scottsdale, Arizona
I assume this is a delta-delta / delta-wye transformer? Is it for a rectifier? Just curious. I used to perform these types of calculations for large DC rectifier traction power systems and this was common for cancellation of 5th and 7th harmonics. wbd nailed it with his answer.

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Jim Phillips, P.E.
Brainfiller.com

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 Post subject: Posted: Mon Jul 01, 2013 8:59 am
 Sparks Level

Joined: Tue Oct 26, 2010 5:00 pm
Posts: 184
Location: Maple Valley, WA.
Many of the larger short circuit programs have the capability of modeling a three winding transformer. I know that SKM PowerTools for windows will handle this.

_________________
Robert Fuhr, P.E.; P.Eng.
PowerStudies

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 Post subject: Posted: Mon Jul 01, 2013 10:16 am
 Sparks Level

Joined: Tue Apr 17, 2012 8:19 am
Posts: 241
Location: Charlotte, NC
Jim Phillips (brainfiller) wrote:
I assume this is a delta-delta / delta-wye transformer? Is it for a rectifier? Just curious. I used to perform these types of calculations for large DC rectifier traction power systems and this was common for cancellation of 5th and 7th harmonics. wbd nailed it with his answer.

I have seen grid-tie applications that use transformers like this. You have an inverter on each secondary and the primary ties to the grid.

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 Post subject: Posted: Mon Jul 01, 2013 10:34 am
 Sparks Level

Joined: Tue Oct 09, 2012 5:00 pm
Posts: 53

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 Post subject: Posted: Tue Jul 02, 2013 4:46 am
 Plasma Level

Joined: Wed May 07, 2008 5:00 pm
Posts: 816
Location: Rutland, VT
EasyPower is another program that I know can handle 3 winding transformers.

_________________
Barry Donovan, P.E.
www.workplacesafetysolutions.com

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 Post subject: Posted: Tue Jul 02, 2013 8:32 am
 Sparks Level

Joined: Mon Jun 07, 2010 9:40 am
Posts: 119
wbd wrote:
I think 1500 kVA should have been used. The reason is that the bus attached to the secondary winding will only see the fault current that can be passed by the 1500 kVA winding.

????

The %Z is based on the KVA Base, and typically the KVA base is the full rating of the transformer. [ If the %z is based on the per winding KVA then 1500 Would be correct. Contact the XF mfg to verify which value is the base kva if the name plate isn't clear. ]

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 Post subject: Posted: Mon Jul 15, 2013 8:36 am

Joined: Fri Apr 22, 2011 1:51 pm
Posts: 6
The impedance of a transformer is determined by short circuiting a winding (LV in this case Winding 1) and passing winding rated current and determining the input voltage. Thus each winding (1.5MVA) would have its impedance based on the above test. This should be confirmed by the certified test report for the transformer (as noted by JK above). With a three winding transformer there should be three impedance components H-X, H-Y and X-Y. Sequence networks are then needed to determine the fault currents.

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