Available fault current

[stevenal]

So it happened again today when I gave an engineer an estimate of fault current at the terminals of a 120/240 V single phase transformer. He was agast. Dealt with lots of other utilities, and never heard of a 75kVA that could deliver so much current. First he had a hard time believing that half winding line to ground faults deliver about 1.5 X the full winding line to line value. He's going to check with some manufacturers on this point.

I assume infinite bus on the primary. I use a low limit impedance from our specs, not the nameplate, since transformer may be replaced at some point. I lower that value by a factor of 7.5% based on ANSI/IEEE impedance tolerances. Voltage is increased 5% also based on standards.

What's going on? Am I overly conservative? Why aren't others giving similar figures? On what basis are they ignoring the line to ground (most common type of fault)multiplier. What do you do?

 

[rbulsara]

There are tolerances built in all other equipment rating too and its not that someone going to select an equipment very close to a calcualted value or if existing someone will repalce unit rated 10kA just because a calculation show 10,050 kA.

What about not having a perfect fault any way? There is always some fault impedance, usually ignored.

 

[BJC]

stevenal

Do you have any numbers and examples of what your calculating? I did a job recently with a 75 kVA transformer (Z=1.5 %) and an AFC of 18 kA. That I didn't consider excessive since panels rated at 22kAIC are pretty common.

 

[dpc]

General Electric Publication GET-3550 has some estimated maximum fault currents for single-phase transformers - you might compare your numbers.

For a 75 kVA 120/240V single-phase transformer connected to an infinite bus on the primary, they estimate 22,331 A for a line-to-neutral fault at tranformer terminals.

Is that close to what you come up with?

 

[stevenal]

BJC
I also used 1.5%. Full winding fault is 23.7KA. Half winding is 37.1kA.

http://www.bussmann.com/apen/software/s_circuit.asp

says 20.8kA for the full winding fault. They evidently don't use the tolerances that I do and ignore the half winding fault. 18kA sounds pretty light. Did you include source or cable impedance?

 

[stevenal]

DPC,

I get a similar result if I use 2.3% instead 1.5% and back off the tolerances.

I like these old scanned documents. How relevant do you suppose this data is to the newer high efficiency low impedance designs.

 

[dpc]

The GE data for 75 kVA was based on 2.24 percent. Don't forget the resistance of the transformer - it becomes increasingly signficant in the smaller sizes.

For a 120/240V system, the impedance between the transformer and the service will limit fault current a bit.

The GE Pub referenced has been updated quite a few times over the years. I don't know if the data for distribution xfmrs has changed that much or not.

 

[rbulsara]

What makes L-N fault 1.5 times??

Isc= Full Amp/%Z

for 75 kvA, 240V L-L, 1.5% Z yeilds

(75*1000/240)/.015 =20,833A

For L-N fault only half the winding is involved and the kVA rating is also cut in half, %z remains the same. The L-N fault would come out half the L-L SCC. Am I missing something?

 

[jbartos]

Suggestion: To assume an infinite bus upstream of 75kVA transformer is possible, however, it is rare in many instances since the 4.16kV or 13.8kV conductors are relatively small in their cross-sectional area and the run to some upstream network transformer is fairly long, e.g. miles. So, all those high short circuit current values would be valid for the transformer being located relatively close to the next upstream transformer, which is possible, however, it is not so frequent.

 

[stevenal]

rbulsara,

Yes you are. kVA doesn't matter directly, only voltage and actual impedance. We base the % impedance on the kVA, though, just to muddy things up.

My first thoughts were: half winding means half the voltage and half the impedance > it's the same. By this way of thinking all the impedance is in the secondary windings. The other extreme is to consider all the impedance in the primary. This impedance is reflected to the secondary by the square of the turns ratio. A line to ground fault links half the number of turns that the full winding fault does, so the reflected impedance is 1/4 of that for a line to line fault. Half voltage and quarter impedance means double the fault current. The real solution is to accuratly divy up the impedance amongst the windings. Test data doesn't provide this split, though. I used the Westinghouse method of estimating a multiplier by separating the r and x parts. In most cases, though, it comes out to about 1.5 times. One spreadsheet I downloaded someplace just uses 1.5.

 

[stevenal]

jbartos,

I prefer to approximate using an infinite bus. Source impedances are small in relation to the transformer impedances in question. I'd also hate to have to explain to customers that the improvements we're continually making in the system might require them to rewire.

 

[rbulsara]

stevenal:

Yes, I agree.
I missed cutting the %z in half. Rated current remains the same, voltage is cut in half.

Other way of seeing this (which I should have done), half the voltage is required to flow the rated current through the half the shorted secondary winding, compared to the same requried to force the rated current through the full shorted secondary winding. This amounts to half the %Z.

 

[jbartos]

Suggestion to stevenal (Electrical) Oct 9, 2003 marked ///\\jbartos,
I prefer to approximate using an infinite bus. Source impedances are small
///Agree.\\ in relation to the transformer impedances in question.
///There are also transmission line conductor impedances, beside the actual source impedance, meant to be large upstream transformer or generator.\\ I'd also hate to have to explain to customers that the improvements we're continually making in the system might require them to rewire.
///Actually, the customers would like the smaller short circuit current levels since the infinite bus as the input for short circuit calculation tends to be very conservative. Obviously, if the power distribution is substantially improved, the short circuit currents would be increased. However, the Utilities are well prepared for those cases by using current limiting fuses.\\

 

[rbulsara]

stevnal:

On the other hand does what i said (in my second post) make sense...I have to think this again.

Say 50V was required to circulate rated amp through a shorted 240V winding, same voltage will be required to circulate the rated current through half the winding as only half the voltage is induced in the secondary!
Volts per turn does not change!

Someone need to test this, as not much written literature is available. I still doubt more current at half the voltage.

 

[alehman]

I tend to agree with rbulsara.

Side note - low impedance / high efficiency 3-phase pad-mount transformers are also becomming common is smaller sizes (<225 kVA). Fault currents in these small systems are sometimes much higher than past experience.

 

[jghrist]

It is frustrating that I can't find any references that cover this calculation. How about this for an approach that differs from Stevenal's? Assume 1/2 of the impedance is in the primary and 1/2 in the secondary. Assume 1/2 of the secondary impedance is in each of the two windings that are in series for the 240 volt impedance test. Transfer all of the impedances to the secondary side (0.01152 ohm for 1.5% Z). For a 120 volt phase fault on a single winding, the driving voltage is 1/2 that of a 240 volt fault, and the impedance is 3/4 Z. Therefore the fault is 2/3 that of a 240 volt fault = 120/(0.75·0.01152) = 13889A.

If the two secondary windings were paralleled, then the impedance would be Z/2 + Z/8 = Z·5/8. Current would be 120/(Z·5/8) = 16667A, 80% of the 240 volt fault.

 

[stevenal]

jgrist,

The impedance changes for the half winding fault. Different turns ratio means you've cut that impedance assigned to the primary and transfered to secondary to 1/4 (turns ratio squared) the value if would be for a full winding fault. In your example, half winding fault impedance is (1/2)*0.5^2+1/4 =0.375X that of the full winding fault. With half the voltage, current is 1.333X. An easier way might be to do the calculation on the primary side for each case, then transfer the result across the turns ratio of the transformer for each case.

I agree, there is not much written. The GE pdf that DPC suggested does touch on it. They claimed it could be up to 2X. They also touch on the r and x factors that Westinghouse uses. The Westinghouse reference is the old green &quot;Distribution Systems&quot; book. See pages 222 and 223. They don't relate it directly to fault current. The ZH1-2 value they are calculating is theoretically what one would measure on the primary with X1 and X2 shorted, so it is applicable to fault calcs.

Guess I'm right, the half winding multiplier is being ignored. Even jghrist is unaware.

 

[jghrist]

I think you are right. I really wasn't too happy with the way I treated the primary winding impedance. I tried briefly to transfer everything to the primary side but did not get sensible answers; probably did something wrong.

Alas, I haven't had access to the old green Westinghouse distribution systems book for many years. When I did, it was very dated but contained a lot of things not available elsewhere.

It would be interesting to do an impedance voltage test on a distribution transformer with X1 and X2 shorted. Also with the two windings in parallel. Seems that you would need this for a bank of single phase transformers serving a 120/208 load.

 

[stevenal]

http://www.edreference.com/UpdateEDR/

contains another fault current calculator that simply uses a 1.5X multiplier. Still doubtful rbulsara and alehman? I was too at one time.

jghrist,

You had to point that out about the parallel windings didn't you? Back to my spreadsheet.

 

[rbulsara]

Stevnal:

I am still not convinced. While I am not saying you are right or wrong. I am also trying to reason. I am surprised if this was such a big deal, it would have been well documented somewhere in so called standard tables that are available for all kinds of transformers and SCC they can provide. Utiltiy companies would have made big deal about it and all residential panelboards would have to be 42kAIC rated minimum and which is not the case. 10kA rated would almost become a no-no.

I would like some tests. Increase in current may be due to considering the parallel windings. I know one of the persons who is involved in developement of EDR and will check out.

For one thing, if you look at the equivalent circuit of a transformer, (See Std. Elect. Engineer's Hand Book) the reactance of the transformer in ohms is a fixed value X, a result of mutual and leakage flux. The reactance of a tranformer is NOT Xs+Xp , adjusted to turn ratios. The nameplate %Z is a test result value and not a straight inductance measurement. Only conversion to turn ratio apply to Rs and Rp and load impedance Z which is zero for our discussion.

When half the winding is linked, I would think the mutual leakage flux would be half? Some one need to verify this.
If it is so, %X remains the same. You can play around with values of R which is not a overriding factor at all.

I am not in my office, if i get sometime I will have some manufacturer test this for us.

Also a per unit value is the same for either voltage. For a minute neglect the R.

For 75kvA,240V xfmr, the base Z = 240V/312.5A. And for 1.5% per unit reactance, hte ohmic value will 0.01152ohms .

For the half winding, the base Z = 120V/312.5A. If the half the linkage were to cut the mutual reactance in half, it will be half of .01152 ohm and the per unit value will be (.01152*0.5)/(120/312.5)=.015 or 1.5%, the same per unit value.