SCA at XFMR

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jkim780

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This might be a silly question but here we go. When you do quick calcs for available fault current calcualtion at transformer instead of pick up the phone and asking the Utility company you use the following formula.

I(SCA)=KVA/V(L-L)*1.732/Z% or I(FLA)/Z%

Can you explain to me how this formula gives you the SCA at tranformer? I heard people said this is 'Ohms law' but I don't see the ohms law in here? Ohms law is V=I*Z and what we have here is I/Z. How does I/Z gives you the SCA at XFMR?

Did I make myself clear?
 
Re: SCA at XFMR

In this case Z is in [Per Unit Value]

Z [%] =Z[actual]/Z[base] x 100

The Ohm's law is still in charge.
 
Re: SCA at XFMR

In this case Z is in [Per Unit Value] - no dimension

Z[%] = Z[Actual] x 100 / Z [Base]

The Ohm's law is still in charge
 
Re: SCA at XFMR

The %Z is an approximate value for a size and type of transformer. Representative transformers have been tested by shorting out the secondary bushings and measuring the current while applying voltage to the HV side of the transformer. When the current is 100% on the secondary side, a voltage reading is taken on the primary side and the ratio between that voltage and the primary rated voltage is the impedance of the transformer.

Without impedance, you would just multiply the primary fault current by the turns ratio to get the available fault current on the secondary side. :D
 
Re: SCA at XFMR

Pardon my ignorance but I still don't understand.

Oliver, how is "per unit value Ohm's law?

Charlie, now I know what the impedance is. But how does I/Z give you SCA at transformer?
 
Re: SCA at XFMR

As I understand it the % of impedance is a value of primary voltage required to cause full load secondary current if the secondary had a bolted short at it's terminals.

A 2% Z transformer 480/120/208 75 kVA has a pri FLC of 90 A and a sec FLA of 208 A + or -.

2% of Pri Voltage is 9.6V

If 9.6 Volts are applied to the pri of a trans that has a bolted short on the sec the sec winding will only allow 208A to flow within the sec coils.
This will not harm the transformer.????

All Short Circuit Calculations in the classroom are performed with assuming an infinite buss pri.(no voltage drop and a no loss of energy)

Probably not true!!

If the original 480 volts are applied to that shorted transformer (and you can be sure that 480 will be applied), then the sec current will be 50 times higher or 100/2 = 50 the secondary current flow with the bolted short will be 10400A or 50 times greater than it was with the 9.6V applied 208A X 50 = 10,400A.

This is a crude method but it will alert an installer if the existing switch gear is near able to deal with a change out when the new energy efficient transformer has a lower percent of impedance.


Charlie Palmieri

[ March 18, 2005, 06:01 PM: Message edited by: cpal ]
 
Re: SCA at XFMR

The Per Unit Value is DIMENSIONLESS . It is like a numerical constant. That's why the Ohm's law is in charge.

For more information on Power System Representation see 'Power System Analysis' by Charles A. Gross.
 
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