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 Post subject: Dual Rated Xfmr's & Fault Current
PostPosted: Fri Sep 07, 2012 7:33 am 
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I recently came across a project that is utilizing an oil filled xfmr w/fans and is dual rated so the ratings read: 3000/3360; 3750/4200. When you have a transfomer like this (5.75%Z) at what value do you calculate the fault current at? If I understand correctly the xfmr.'s dual rating is directly related to the temperature so the impedance of the xfmr does not change as the windings are the windings correct? Does ANSI have a method that the %z is based on a paticular rating or would this be defined by the manufactuer?


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PostPosted: Fri Sep 07, 2012 10:40 am 
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PostPosted: Mon Sep 10, 2012 8:32 am 
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mike01 wrote:
I recently came across a project that is utilizing an oil filled xfmr w/fans and is dual rated so the ratings read: 3000/3360; 3750/4200. When you have a transfomer like this (5.75%Z) at what value do you calculate the fault current at? If I understand correctly the xfmr.'s dual rating is directly related to the temperature so the impedance of the xfmr does not change as the windings are the windings correct? Does ANSI have a method that the %z is based on a paticular rating or would this be defined by the manufactuer?

I agree, use base 3000 kVA and base %Z


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PostPosted: Mon Sep 10, 2012 11:24 am 
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The nameplate %Z is based on the nominal (base) rating (i.e. no fans and no temperature rise) of the transformer. In your case, the 3,000 KVA rating is what should be used with the nameplate 5.75 %Z in order to calculate the impedance of the transformer. The impedance is calculated by the manufacturer based on the ANSI C57 standards. As long as it is an ANSI rated transformer, the manufacturer will be in step with the ANSI standard. This will not be true for non-ANSI rated transformers.


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PostPosted: Wed Sep 12, 2012 12:34 pm 
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The percent impedance is obtain via a short circuit test. The secondary terminals are shorted and the %Z is actually the %IZ (percent voltage) that is necessary on the primary to create rated full load current to flow on the secondary. The rated full load current is based on the base / air cooled rating so that means the % impedance is also based on the base rating.

This is used for the infinite bus short circuit calculation.

If %Z (primary voltage) results in the rated full load current (FLA)
Then 100% voltage results in the maximum short circuit amperes (SCA) on the secondary.
  • %Z results in FLAsecondary
  • 100% results in SCAsecondary
Rearranging the two relationships produces the well known (?) equation:
  • Maximum SCA = (FLAsecondary x 100) / %Z
Many refer to this as the infinite bus short circuit calculation.

The FLAsecondary is the based on the lower/base kVA rating as others have noted.


Here are two examples to illustrate what would happen if the higher kVA rating is used:

Example 1: Short circuit current using the 3000 / 3360 kVA AA/FA and 5.75%Z - I assumed 480 volts

AA/FA where AA is the air cooled (base) rating and FA is the fan cooled rating

Using the air cooled rating (AA):
  • FLA = 3000kVA / [0.48kV x sqrt(3)] = 3608.4 Amps
  • SCA = [3608.4 x 100] / 5.75% = 62,760 amps (rounded up)
Example 2: If the 3360 kVA FA fan cooled rating was used:

The %Z would be higher. Remember it is the % Primary Voltage that results in the FLA. If the larger kVA is used, the larger FLA is also used which results in the %Z = 6.44% (scaled by the kVA ratings)

In this case:
  • FLA = 3360 kVA/[0.48kV x sqrt(3)] = 4041.5 Amps
  • SCA = [4041.5 x 100] / 6.44% = 62,760 (rounded up)
If the transformer was tested with the larger kVA rating, the %Z would be greater but the end result of the short circuit calculations would be the same.

Hope this helps shed a little more light on this topic (or confusion - it's a fine line o_O )

_________________
Jim Phillips, P.E.
Brainfiller.com


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