Why higher kVA/MVA transformer have higher fault level on LV side? 1

[Msingh15]

Hi Guys,

I came across this situation while looking at some fault level data.

Why are fault levels higher on LV side of higher rating transformer despite having higher Z%?

If we have a 100kVA 11/0.415kV Z=4% & 1000kVA 11/0.415kV Z=6% transformer. according to theory and formulas we have the fault current at 415V side is higher for 1000kVA transformer.

shouldn't this be other way around since 100kVA has only 4% of impedance?

My question here is, what physical component in two different rating transformers dictate the fault levels?

Is it the number of windings? Since, a transformer is a type of inductor and during fault scenario the voltage tends to collapse and inductor interacts and try to maintain that voltage and we see more fault current in higher rating transformer (1000kVA)?

Any thoughts?

Regards,

Manny

 

[waross]

The full name is percent impedance voltage.
Percent impedance voltage is the percent of rated voltage that will drive rated current through shorted secondary terminals.
Rated current divided by the %imp gives the available fault current.
If the 1000 KVA transformer was rerated at 100 KVA, the %imp would be 0.6%
The rerate would not change the available fault current but it would change the percentages.


Bill
--------------------
Ohm's law
Not just a good idea;
It's the LAW!

 

[cranky108]

With the same % impedance, a higher MVA transformer will have higher fault current.

It does not have to be that way, as you can specify a maximum fault current level, but will result in higher losses.

That said, the management of fault current levels is an art, and is largly the specified impedance of a transformer.

 

[AC-Power]

I am only an electrician, but I've always been skeptical of nameplate xfmr's impedance (Z%) data. The core would have to be saturated (xfmr under FLA) to determine this. I'm not sure if xfmr manufacturers are doing this, or simply deriving it by simply reading winding resistance and adding estimated impedance by comparison to historical trend data. In other words, if you've a difference of 2% impedance, base your calc's on 6% as an added safety factor.

 

[waross]

The full name of the parameter is the "Percent impedance voltage".
It is designed and then verified by testing.
Simple test:
The secondary terminals are shorted and the applied voltage is increased until the current equals rated full load current.
That voltage, expressed as a percentage of rated voltage is the "Percent impedance voltage" or %imp.

A fully offset fault current may approach 2*√2 (2.83) times the "Available Sort Circuit Current" as calculated from the %imp.
The amount of offset depends on the point on wave of the fault and the X/R ratio of the transformer.
This factor is considered when switchgear is rated and tested for a given fault level.
One of the reasons that it is done this way is to make it easy for electricians to select switchgear that is adequately rated for the fault current of a transformer under worst case conditions.

Bill
--------------------
Ohm's law
Not just a good idea;
It's the LAW!

 

[LionelHutz]

In your example the transformer rating went up 10X and the impedance went up 1.5X. Fault current is proportional to kVA and inversely proportional to impedance, or fault current is proportional to kVA/%Z. Since the kVA went up more than impedance went up, the fault current also goes up.