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Topic Title: Transformer percentage impedence
Topic Summary: Calculation of fault current
Created On: 02 June 2013 11:32 AM
Status: Read Only
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 02 June 2013 11:32 AM
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umair84

Posts: 6
Joined: 30 May 2013

If percentage impedence of transformer is 8% then how much is its fault current?
 02 June 2013 08:46 PM
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Avatar for timothyboler                                      .
timothyboler

Posts: 229
Joined: 25 July 2008

Assuming an infinate bus, zero fault impedance and zero other series impedance of any kind:

Ifault(per unit) = V(pu)/Z(pu)

Ifault(pu) = 1.0/0.08 = 12.5(pu)


So would be 12.5 times the base (nominal) current of the transformer secondary at base (nominal) voltage.

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Everyone loves a fireman - but hates the fire inspector.
 03 June 2013 08:37 PM
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umair84

Posts: 6
Joined: 30 May 2013

thanks for reply.will your calculation be true if percentage impedence of transformer is 8% and we have to determine short circuit current for three phase short circuit occur on secondary side of transformer?
 03 June 2013 10:08 PM
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Avatar for timothyboler                                      .
timothyboler

Posts: 229
Joined: 25 July 2008

Sounds like a homework question :-/

The 8% or (0.08) is a per unit (or relative) value; relative to either the base voltage and base current of either the secondary or primary values.

Therefore the absolute value of fault current will depend on the base voltage and base current that you want to refer to. Remember the term "secondary" just means at the load-side of the transformer (could be the HV or LV side depending on whether it's a step-up or step-down transformer) so yes it would only make sense for the percentage impedance to be used to calculate the "secondary" fault current.

If the fault was on the primary side we wouldn't care about the impedance of the transformer as it wouldn't be in the path of fault current. I would suggest you do some background reading on the per unit system and look at some simple fault calcs.

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Everyone loves a fireman - but hates the fire inspector.
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