Maximum fault current of center-tapped single-phase transformer

A

Ahmed Abdelfattah

Member
Location
Edmonton, Alberta, Canada
Greetings,

I have a project that is fed from a 37.5kVA transformer that has a primary voltage of 7.2kV and secondary single-phase output of 240/120V. The transformer appears to be center-tapped. The impedance of the transformer is specified as 2.2%.

When calculating the maximum fault current that will be let-thru the transformer, if found that I have different scenarios leading to different fault currents:

Scenario #1: Assuming the the impedance is calculated across the 240V terminals:
Scenario 1A: The maximum fault current based on 240V is 37.5x10^3/(240*0.022)=7,102.3A
Scenario 1B: The maximum fault current based on 120V is 37.5x10^3/(120*0.011)=28,409.1A

Scenario #2:
Assuming the the impedance is calculated across the 120V terminals:
Scenario 2A: The maximum fault current based on 240V is 37.5x10^3/(240*0.044)=3,551.1A
Scenario 1B: The maximum fault current based on 120V is 37.5x10^3/(120*0.022)=14,204.6A

I appreciate if someone can guide me into the correct way for calculating the maximum transformer fault current, which voltage level and impedance to consider, and why.

Thank you
 
Ahmed Abdelfattah said:
Greetings,

I have a project that is fed from a 37.5kVA transformer that has a primary voltage of 7.2kV and secondary single-phase output of 240/120V. The transformer appears to be center-tapped. The impedance of the transformer is specified as 2.2%.

When calculating the maximum fault current that will be let-thru the transformer, if found that I have different scenarios leading to different fault currents:

Scenario #1: Assuming the the impedance is calculated across the 240V terminals:
Scenario 1A: The maximum fault current based on 240V is 37.5x10^3/(240*0.022)=7,102.3A
Scenario 1B: The maximum fault current based on 120V is 37.5x10^3/(120*0.011)=28,409.1A

Scenario #2:
Assuming the the impedance is calculated across the 120V terminals:
Scenario 2A: The maximum fault current based on 240V is 37.5x10^3/(240*0.044)=3,551.1A
Scenario 1B: The maximum fault current based on 120V is 37.5x10^3/(120*0.022)=14,204.6A

I appreciate if someone can guide me into the correct way for calculating the maximum transformer fault current, which voltage level and impedance to consider, and why.

Thank you
Click to expand...
I'd wager Scenario #1 its A: The maximum fault current based on 240V as the fault see's 240V.
and Scenario #2 its B The maximum fault current based on 120V as its line to ground fault.
 
Impedance is across the transformer line terminals. It does not involve any type of taps. Yes, it is possible for manufacturers to perform extra impedance tests, but no one is going to pay for it on standard production units.

The basic formula, of 3-phase %Z/I, does not often apply to center tapped configurations because these transformers do not use 3 same sized units.
 
The Bussmann FC² app shows a line to line available fault current of 7,894 and line to neutral one of 11,841. (note based on a -10% factor for the impedance, the worst case)
 
don_resqcapt19 said:
The Bussmann FC² app shows a line to line available fault current of 7,894 and line to neutral one of 11,841. (note based on a -10% factor for the impedance, the worst case)
Click to expand...
That's a ratio of 1.5. If the secondary behaved as a constant impedance (in ohms, not %) source independent of voltage, the L-N AFC would be 1/2 the L-L AFC. So this implies that the secondary source impedance as a 240V source is 3 times the source impedance as a 120V source.

Is there a simple explanation of where this factor of 3 comes from?

Cheers, Wayne
 
The correct way is to use the transformer’s full secondary voltage (240V) and the given 2.2% impedance. Transformer impedance is specified based on the full winding, not half (120V side). So, Scenario 1A is correct. No adjustment to impedance is needed for calculating L-N or L-L faults.
 
wwhitney said:
...the secondary behaved as a constant impedance
Click to expand...
That is not how transformer impedance works.

The proper term is %IZ, and it is defined as the % of primary voltage needed to drive 100% of secondary current through a bolted short circuit fault. The magnetic circuit of the transformer is part of the impedance.
The test maximum fault current through the secondary winding does not change if the short is L-N or L-L.
 
don_resqcapt19 said:
The Bussmann FC² app shows a line to line available fault current of 7,894 and line to neutral one of 11,841.
Click to expand...
Hmmm I confirmed your numbers in the app but my numbers on paper are not matching up
for a 37.5 kVA 2.2% I get 7891.4 and 15782.8
240 * .022 = 5.28
Is this where you subtract the 10% factor?
5.28 - .528 = 4.752
37500/4.752 = 7891.4

The app seems to round 37.5 kVA to 38 kVA but then if I change to a 38kVA
I get
7996.6 and 15993.3
 
You must log in or register to reply here.
Top