Available Fault Current Calculation Question

[industrial1003]

I have a utility transformer that appears to be a delta primary, wye secondary (3 hots in, 3 hots and neutral out). Utility was nice enouth to provide me with transformer impedance and X/R ratios to perform the required calculations. I was able to do the calculation for the bolted L-L-L fault at the secondary terminals but do not know how to determine the L-N fault current at the same location considering it is a 3 phase transformer not 1 phase (not sure the single phase transformer formula is applicable as it will result in a much higher AFC and suposidly a 3 phase bolted fault is the worst case condition)? I utilized one of the more simplistic formulas from Cooper Bussmann to perform the L-L-L, but none of their white papers address the L-N fault on a three phase system with the exception of a footnote in one document that said it could be 50 - 125% of L-L-L. Unfortunately I do not have access to any of the IEEE color books that might address this situation and none of the apps I have tried for this show the L-N fault value on a 3 phase system. Any help would be appreciated.

 

[steve66]

I think the L-N fault is only higher than the 3 phase fault in single phase systems.

Here is a foot note from one of the busman books:

Note 5
.
The L-N fault current is higher than the L-L fault current at the secondary

terminals of a single-phase center-tapped transformer. The short-circuit

current available (I) for this case in Step 4 should be adjusted at the

transformer terminals as follows: At L-N center tapped transformer terminals,

I

L-N = 1.5 x IL-L at Transformer Terminals.

At some distance from the terminals, depending upon wire size, the L-N fault

current is lower than the L-L fault current. The 1.5 multiplier is an approximation

and will theoretically vary from 1.33 to 1.67. These figures are based on

change in turns ratio between primary and secondary, infinite source available,

zero feet from terminals of transformer, and 1.2 x %X and 1.5 x %R for

L-N vs. L-L resistance and reactance values. Begin L-N calculations at

transformer secondary terminals, then proceed point-to-point.

 

[industrial1003]

Steve66,

Thanks for the reply. I had seen that as well and had grouped it under the single phase equations that may not pertain (not really sure). It seemed to be more directed to the specific winding configuation of a single phase transformer (a portion of a series winding), and in the tech sheet I had looked at was only indicated as a note when calculating the "F" factor for the multiplier used in adjusting due to the conductor run length. After much research from various articles I saw that the single phase fault currents will always be higher when considering the same transformer KVA as mathmatically you end up increasing dividend (transformer FLA), and additionally when talking about line to neutral voltages, it always reduces the divisor so the result is always greater. That was the reason I was not sure the information would apply given an actual 3 phase system not a single phase transformer tapped off the three phase system (something those equations do work for). I guess that is what is causing my concern. If I used the single phase values, it would almost tripple my AFC from around 25,000 3ph to around 70,000 single phase (both values including SF of 4 X FLA for Motor Contribution) and even my switch gear does not have that high of a withstand rating (65,000KVA at service disconnects). The utility had given me numbers that were both about 20,000 for L-L-L and L-N (doesn't mean they are correct but I am assuming they are since utility transformers are typically tested for their values), and when making the label I wanted to account for the motor contribution so calculating the L-N AFC seemed like a logical step although proved to be impossible to find an example of thus far.

 

[Bugman1400]

Is the neutral grounded? If so, solid ground or resistive ground?
What are the xfmr impedances that you were given? And, what size is the xfmr?

 

[industrial1003]

Bugman1400,

The transformer info I was given is as follows:

1000 KVA
34500 V primary
480Y/277V secondary
%Z 5.75
X1/R1 9.74
X0/R0 9.13
LLL Fault Test Data 19921
LN Fault Test Data 19790

 

[industrial1003]

Bugman 1400

Sorry, I forgot the first part of your question. Neutral is grounded in the gear with a bus bar jumper which runs to main ground on the I beam next to the gear. I do not know if/how the transformer outside the building is grounded (existing installation), but was assuming it was only my secondary side in the gear to avoid parallel paths to ground.

 

[Bugman1400]

It appears that your fault numbers are based on an infinite bus so, they must be from the mfr. If you are interested in knowing how to calc the 3PH and SLG (single-line to ground) fault values, you should read up on sequence networks and how to draw the sequence network for a 3PH and SLG fault. Basically, for a 3PH fault the sequence network is reduced since the fault is balanced. This mean that the negative and zero sequence networks can be removed and makes for a much easier calc and visual. For the SLG fault, all three sequence networks are included (positive, negative, and zero). Since this is such a small xfmr, you can use the Z%=5.75 for all three sequence networks.
If you are looking for a rule of thumb to approximate SLG faults to 3PH fault values, you have to be careful that you don't make wrong assumptions. It is key to know if the system is solid or resistively grounded. A resistive ground system will drop the SLG fault value dramatically....sometimes down to just 10 Amps. For solidly grounded systems, sometimes the SLG fault value is larger than the 3PH fault value.
You should be able to Google or YouTube hundreds of example of how a 3PH and SLG fault is calc'd.

 

[steve66]

I did a few trail runs with SKM software, and this seems to be true:


For a utility supply with a limited capacity, the single Line to ground fault current is higher than the 3 phase current at the transformer secondary. As the utility capacity increases, the 3 Phase and L-G fault currents get closer and closer together. Also, as we increase cable length on the secondary, the L-G decreases faster than the 3P, and at some point it becomes less than the 3P.

Since you are assuming an infinite primary source, I think your worst case is going to be the 3 phase fault. At the transformer secondary, both 3P and L-G will be about equal, and at the main switchboard the L-G will be less than the 3P.


I don't generally bother to calculate the single phase fault, and I normally just assume it is going to be less. I think circuit breaker AIC ratings are based on a 3 phase fault anyway, so I'm not sure the single phase would matter even if it were a little higher.

 

[industrial1003]

Thanks guys. You are both correct in the fact that for the calculations I am using an infinite primary. I appreciate the assistance. If I can find a verified example I will attach it to the post or the link.

 

[Phil Corso]

Gentlepeople...

In referring to solidly-grounded systems, then I rigorously disagree. Whether or not the magnitude of single-phase to ground (L-G) fault is greater than a 3-phase (L-L-L) is dependent on the magnitude of the Zero-sequence impedance of the network compared to Positive and Negative impedances!

L-G faults are seldom greater if the source is a 2-winding transformer! For alternators it is common to find that the L-G fault magnitude exceeds 3-phase fault magnitude! And, if the network under consideration involves 3-winding transformers, special scrutiny is required because resultant impedance could be less than zero, thus increasing on L-G fault magnitude!

Regards, Phil Corso

 

[Phil Corso]

Gentlrpeople...

Anyone interested in a more techno-cratic explanation?

Phil Corso

 

[GoldDigger]

Always.

 

[Bugman1400]

Gentlepeople...

In referring to solidly-grounded systems, then I rigorously disagree. Whether or not the magnitude of single-phase to ground (L-G) fault is greater than a 3-phase (L-L-L) is dependent on the magnitude of the Zero-sequence impedance of the network compared to Positive and Negative impedances!

L-G faults are seldom greater if the source is a 2-winding transformer! For alternators it is common to find that the L-G fault magnitude exceeds 3-phase fault magnitude! And, if the network under consideration involves 3-winding transformers, special scrutiny is required because resultant impedance could be less than zero, thus increasing on L-G fault magnitude!

Regards, Phil Corso

Oh but Phil, you didn't say what kind of 3-wdg xfmrs. I don't think anybody disagrees with your second sentence above but, I'm sure you know that. For alternators (I assume you are referring to generators), I find that most are impedance grounded and the L-G faults would never come close to the magnitude of the phase faults.

 

[Phil Corso]

Gentlepeople...

We're getting well beyond "Electricity for the Uninitiated", but I'll try. There are many types of electrical failures in a 3-phase system. Some of the well-known types are:

o LLL (3-phase short).

o LL (line-to-line short)

o LLG (double-line to ground short, and,

o SLG (single-line to ground short).

Of course many members of the list are familiar with the first and last. The current in a 3-phase short is essentially determined by the source phase-to-ground voltage, En, divided by the sum of the circuit element impedances between the source and the point of failure. This is called a 'symmetrical' fault.

All of the others are considered 'asymmetrical' faults, and their determination is not as simple as the LLL. Thus, electrical engineers resort to a mathematical-tool called Symmetrical Components (developed by C.L. Fortescue of Westinghouse (?) in 1918). It assigns to each cable, generator, transformer, etc, three impedances. They are: Z1 (the positive-sequence); Z2 (the negative-sequence); and Zo, of course, the zero-sequence. Then by judiciously combining these three impedances, any of the faults described above can be solved. Let the fault-current equal If. Then,

o For LLL, the current, If(3) = En / Z1.

o For LG, it’s a little more involved, but the current, If(1) = 3 x En / (Z1+Z2+Zo).

In most industrial plant networks Z1 and Z2 are equal. Then… if Zo equals Z1… If(1) = If(3). Typically, however, if alternator’s are involved, their Zo is less than Z1, thus LG fault-current magnitude is greater than LLL fault-current magnitude!

Please note the above are valid only when the network has a solid-connection between neutral and ground! The effects of inserting resistor or inductance elements from neutral-to-ground, are not covered. Also excluded are the effects of distributed ground capacitance!

Comments, Good, Bad, Ugly are invited!

Phil Corso

 

[Phil Corso]

Bugman... oh well, I did invite comments:

1) An alternator is the correct term for a syncronous-generator!

2) Alternators, including combustion engine 'generators', less than 600V are seldom "resistance" or "impedance" grounded!

3) I'm referring to transformer having a primary winding, a secondary winding, and a tertiary winding! What type did you have in mind?

Phil

 

[mivey]

I think the L-N fault is only higher than the 3 phase fault in single phase systems.

Or when the source zero sequence impedance is blocked. The OP has a delta-wye transformer so the source zero sequence impedance is blocked.

 

[mivey]

For a utility supply with a limited capacity, the single Line to ground fault current is higher than the 3 phase current at the transformer secondary.

Not necessarily for a wye-wye bank.

As the utility capacity increases, the 3 Phase and L-G fault currents get closer and closer together. Also, as we increase cable length on the secondary, the L-G decreases faster than the 3P, and at some point it becomes less than the 3P.

Because the zero sequence impedance grows faster than the positive sequence impedance.

I don't generally bother to calculate the single phase fault, and I normally just assume it is going to be less. I think circuit breaker AIC ratings are based on a 3 phase fault anyway, so I'm not sure the single phase would matter even if it were a little higher.

One should always look unless you know you have plenty of source impedance. The L-G and L-L-G can be higher if close to the source.

 

[mivey]

Thanks guys. You are both correct in the fact that for the calculations I am using an infinite primary.

A good conservative approach for sizing. Not so good for arc flash.

 

[mivey]

Gentlepeople...

In referring to solidly-grounded systems, then I rigorously disagree. Whether or not the magnitude of single-phase to ground (L-G) fault is greater than a 3-phase (L-L-L) is dependent on the magnitude of the Zero-sequence impedance of the network compared to Positive and Negative impedances!

L-G faults are seldom greater if the source is a 2-winding transformer! For alternators it is common to find that the L-G fault magnitude exceeds 3-phase fault magnitude! And, if the network under consideration involves 3-winding transformers, special scrutiny is required because resultant impedance could be less than zero, thus increasing on L-G fault magnitude!

Regards, Phil Corso

Well I should have read ahead. Looks like you have addressed it already.

 

[mivey]

In most industrial plant networks Z1 and Z2 are equal. Then… if Zo equals Z1… If(1) = If(3). Typically, however, if alternator’s are involved, their Zo is less than Z1, thus LG fault-current magnitude is greater than LLL fault-current magnitude!

The L-L-G can be the worst.