# fault calculation



## Rei (Apr 9, 2010)

I understand the 13.8kV base in G2 which is

140k/14k = 138k/Vbase

solve for Vbase = 13.8kV

I don't understand where they get 9.959kV base for G1. We are dealing with base value which we don't need to consider the square root of 3...at least for G2 they didn't use it.


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## Flyer_PE (Apr 9, 2010)

The square root of three comes from the fact that T1 is composed of three individual single-phase transformers. They give the turns ratio of the individual units rather than that of the resulting three-phase transformer.

Since the low side (10 kV) is connected delta, VLine=VPhase = 10 kV

The high side (80kV) is connected wye, VLine=sqrt3*VPhase = sqrt(3)*80 kV = 138.56 kV

In terms of line voltage, the T1 turns ration is 10:138.56.

Setting the high side voltage to 138 kV:

The voltage at G1 is 138 kv (10/138.56) = 9.959 kV


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## Rei (Apr 9, 2010)

Flyer_PE said:


> The square root of three comes from the fact that T1 is composed of three individual single-phase transformers. They give the turns ratio of the individual units rather than that of the resulting three-phase transformer.
> Since the low side (10 kV) is connected delta, VLine=VPhase = 10 kV
> 
> The high side (80kV) is connected wye, VLine=sqrt3*VPhase = sqrt(3)*80 kV = 138.56 kV
> ...


Wow...tricky. Thanks!


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