Maximum fault current of center-tapped single-phase transformer

[Ahmed Abdelfattah]

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

 

[pipe_bender]

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

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.

 

[jim dungar]

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.

 

[electrofelon]

AFC for single phase transformers is typically higher L-N. A good rule of thumb is to use 1.5X the L-L value. Look over the first page of this document:

https://www.eaton.com/content/dam/eaton/products/electrical-circuit-protection/fuses/solution-center/bus-ele-tech-lib-electrical-formulas.pdf

 

[don_resqcapt19]

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)

 

[wwhitney]

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)

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

 

[D. Castor]

The impedance value on the namepate is across the full 240 V winding.

 

[baserunner4]

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.

 

[MyCleveland]

Another member…”Mivey”, did a detailed discussion on the 1/2 winding impedance. I could not find it.
I believe early 2018. It might have been a discussion in a different thread topic.

 

[jim dungar]

...the secondary behaved as a constant impedance

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.

 

[electrofelon]

Will a two winding transformer behave differently than a single winding with a center tap in regards to L-N fault current? I believe the typical utility transformer is the former.

 

[tortuga]

The Bussmann FC² app shows a line to line available fault current of 7,894 and line to neutral one of 11,841.

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

 

[jim dungar]

Is this where you subtract the 10% factor?

The tolerance for an %Z is usually applied before it is used in a formula.

Such as %Z = 2.2 with 10% tolerance yields a range of 2.2x1.1. to 2.2X.9: so the value used would be %Z = 1.98 for the worst case.

 

[tortuga]

The tolerance for an %Z is usually applied before it is used in a formula.

Such as %Z = 2.2 with 10% tolerance yields a range of 2.2x1.1. to 2.2X.9: so the value used would be %Z = 1.98 for the worst case.

Okay thanks for clarifying
my numbers still get the same result, using 37.5kVA or the 38 it rounded to they don't match the results from the app:

240 * .0198 = 4.752
37500/4.752 = 7891.4

120 * .0198 = 2.376
37500/2.736 = 15782.8

7,891.4 vs 7,894 off by 2.6 not a big deal but
15782.8 vs 11841 is off by 3941.8
This time I did it in a spreadsheet
if I change the voltage from 120V to 160V it gets close but not sure why the app would change the voltage.
In the app I am just selecting 'transformer' single phase so there should be no wire or conduit at that point.

 

[tortuga]

Looking a little closer at the document Ethan posted
They use a different multiplier for L-N faults 0.6 and only half the kVA:



So
.022 * .06 = .0132
37500/2 = 18750
120 * .0132 = 1.584
18750 / 1.584 = 11837.1
now I am only off the app by a more reasonable 3.9 VA

 

[don_resqcapt19]

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

It is my understanding that the app lowers the impedance by 10% if you check that box. The transformer impedance is not exact and can be off by 10%, so this gives you a worst case.

Not sure if the app actually rounds the 37.5 to 38 in the calculations as I get 7,102 by hand and 7,105 when telling the app to use the nameplate impedance without change.

Not sure about your numbers. If I divide 37500 by 5.28 I get 7102.

 

[wwhitney]

Looking a little closer at the document Ethan posted
They use a different multiplier for L-N faults 0.6 and only half the kVA:

So that image you posted has columns for L-N multipliers for both %X (reactance) and %R (resistance). The %R multiplier is always 0.75, which makes sense if the primary and secondary windings are sized so that their resistances match. As a single phase L-N fault will use all of the primary winding and half of the secondary winding, or 75% of the total windings.

But the %X multiplier column is 0.6 for up to 100 kVA, and 1.0 for over 100 kVA. Why isn't this always 1.0?

So

The way the calculation would go for a 2.2%Z 37.5 kVA transformer is:

X/R ratio = 1.4 (from table)
L-L %X = sin(arctan(1.4))*%Z = 1.79%
L-L %R = cos(arctan(1.4))*%Z = 1.28%
(Note that 1.79²+1.28²=2.2², as expected. Also, sin(arctan(x)) and cos(arctan(x)) can be simplified to algebraic expressions, but I didn't bother.)

Using the multipliers from the table,
L-N %X = 0.6 * L-L = 1.07%
L-N %R = 0.75 * L-L = 0.96%
Z = sqrt(R²+X²)
L-N %Z = 1.44%

L-N AFC = 37500/2/120/1.44% = 10851A

Cheers, Wayne

 

[tortuga]

It is my understanding that the app lowers the impedance by 10% if you check that box. The transformer impedance is not exact and can be off by 10%, so this gives you a worst case.

Not sure if the app actually rounds the 37.5 to 38 in the calculations as I get 7,102 by hand and 7,105 when telling the app to use the nameplate impedance without change.

Not sure about your numbers. If I divide 37500 by 5.28 I get 7102.

I also get 7102 when doing it that way

Don the numbers I was referring to not matching up was the ones from the app with the 10% box checked:

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)

240 * .0198 = 4.752
37500/4.752 = 7891.4