Fault current for larger transformers

[GMIEE]

Looking at a list of fault currents provided to me by the utility, a large hangar with a 500 kva 12.47kv/480v transformer has a fault current of 11kA.

The small building next door with a 150kva transformer 12.47kv/208v has a fault current of 33kA.

These are fed from the same substation. Why is it that the fault current at the transformer secondary is so much smaller on the BIGGER transformer?

 

[davidbeach]

The bigger transformer has a higher impedance, 5.5%; while the smaller transformer has a smaller impedance and a lower secondary voltage, 1.3%. Sounds about right.

 

[magoo2]

To add to what davidbeach points out, the transformer standards (ANSI C57) call for 5.75% impedance on ratings of 500 kVA and above. Below that, there is no specified %Z and you'll find a fairly wide range of impedance values from around 1.2% to 3% commonly available.

 

[rbulsara]

Also consider the different voltage levels. The fault MVA at 480V, 11kA is about 9 MVA and at 208V,33kA is about 12MVA.

So despite the SC current is three times larger the actual difference in fault MVA is not 3 times. Above explanation of the impedance diffrence is still valid.

 

[nightfox1925]

At 33kA through fault from a 150kVA, 208V transformer secondary voltage, assuming an infinite source and 150kVA base, the %Z = 1.26%

At 11kA through fault from a 500kVA, 480V transformer secondary voltage, assuming an infinite source and 150kVA base, the %Z = 5.46%

This sounds fair enough.

 

[nightfox1925]

Oooops I'm sorry, the Z=5.46% is based from a 500kVA base.

 

[sberbece]

GMIEE,
You can estimate the 3 phase, symmetrical short circuit current yourself using the following formulas:
Ik [A] = Un[V]/(sqrt(3)*ZT[ohm])
Where :
ZT = z[%]*Un[V]*Un[V]/(100*ST[VA], z – transformer impedance, S-transformer’s rating, Un – the voltage level where you want to calculate the SC.

Of course, the resulted estimated short circuit value will be higher than the real value, because is assuming an infinite power source connected on the HV side of the transformer.

Regards,
Stefan

 

[waross]

Rated current divided by the per unit impedance will give the available fault current or the "Available Short Circuit Current".
Note, the term "available" does not follow the normal definition, but refers to a calculated value that is used for selecting equipment.
The transformers are rated at operating temperature. The fault current produced by a cold transformer will be greater.
The asymetrical current component is also ignored in these calculations.
However, when equipment is rated for "Available Fault Current" or "Available Short Circuit Current" the rating is adjusted to compensate for these simplifications.
A switch rated for 10kA "Available Short Circuit Current" will safely withstand the actual current produced by a transformer with an "Available Short Circuit Current" of 10 kA or less. The actual fault current will generally be greater. Sometimes much greater.
"Available Short Circuit Current" is a value that is simple to calculate. Equipment rated in "Available Short Circuit Current" is easy to match safely to systems and transformers. The differences between the simple value derived by dividing the rated current by the per unit impedance and the actual fault current is calculated and considered by the equipment manufacturers when they rate their equipment.
Protection engineers and designers conserned with the mechanical forces produced by interacting magnetic fields must use the more difficult calculations to predict the actual current that will result from a fault.

Bill
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