Calc for available Fault Curent? Texas Masters test :?:?

Status
Not open for further replies.

mark henderson

Senior Member
Location
Leander Texas
Ok on my test it asks for the available Fault current at the main for a building. What are my steps? What variables do I need?

Thanks, trying to make up 6 points.
 

kingpb

Senior Member
Location
SE USA as far as you can go
Occupation
Engineer, Registered
The simplest calc to determine worst case fault current at the main is to find the transformer nameplate data; specifically you need 3 pieces of information.

Transformer KVA
Transformer impedance, Z
Voltage at service

Step 1:
KVAsc = Transformer KVA/Transformer Z

Step 2:
Calculate sc current from KVAsc, for 3 phase divide KVAsc by voltage and sqrt 3, for single phase, simply divide by voltage.

Short example -
3 phase transformer, 500KVA, Z = 2.7%, voltage on service side is 208Y/120V

500KVA/0.027 = 18,520 KVA

18,520KVA/(208* sqrt 3) = 51.4kAsc
 

Smart $

Esteemed Member
Location
Ohio
Ok on my test it asks for the available Fault current at the main for a building. What are my steps? What variables do I need?

Thanks, trying to make up 6 points.
If service, ask POCO... :D


Really! ...it's dependent on details of "at the main for a building".
 

texie

Senior Member
Location
Fort Collins, Colorado
Occupation
Electrician, Contractor, Inspector
Ok on my test it asks for the available Fault current at the main for a building. What are my steps? What variables do I need?

Thanks, trying to make up 6 points.

I think we are going to need more details on the question. Do they want it at the terminals of the service disconnect or just a simple calc of the transformer fault current? As mentioned here, the fault current is easy to determine at the transformer terminals, but that won't tell you what it is at the terminals of the service disconnect. Can you give more details on the test question?
 

kingpb

Senior Member
Location
SE USA as far as you can go
Occupation
Engineer, Registered
I think we are going to need more details on the question. Do they want it at the terminals of the service disconnect or just a simple calc of the transformer fault current? As mentioned here, the fault current is easy to determine at the transformer terminals, but that won't tell you what it is at the terminals of the service disconnect. Can you give more details on the test question?

If the main is within a few feet of the LV side of the transformer, and the cable is sized to handle the full load current of the transformer; the small amount of impedance in the short cable will not alter, within any significance, the short circuit current, at least between equipment ratings.
 

mark henderson

Senior Member
Location
Leander Texas
Its hard to remember the whole question but I believe they asked at the breaker. Also I don't think they gave me the Transformer KVA, or Transformer impedance, Z.

SO if I don't know Impendence? or KVA

I think I get the rest.
480/277
500 kva
Z=2.7%
Is this correct below?

500/.027 =18518.518 Round 18520
18520/480= 38.58
38.58/1.732=22.28 ?
 

GoldDigger

Moderator
Staff member
Location
Placerville, CA, USA
Occupation
Retired PV System Designer
No. You find the rated full load current first, then increase by dividing by %Z.
End result is the same, of course. Your units for your answer are in kA.
 

kingpb

Senior Member
Location
SE USA as far as you can go
Occupation
Engineer, Registered
Rated full load current? another equation first? Not the breaker rating?

If you work in KVA or MVA, it doesn't matter which side of the transformer you are on. Yes, you can calculate current first by choosing which side of transformer you are on.

Say 500KVA @ 208V 3-ph current is 1388A, and short circuit is 1388/0.027 = 51.4KAIC. You get to the same value either way. I always work in KVA, or MVA, but that's me.

A transformer, because it is almost all reactance, basically only will allow the amount of fault current through it based on its own short circuit capability. This is not absolute, but engineering wise so darn close it is a very accepted practice.

When you divide the transformer kVA or MVA, by the transformer impedance, you are calculating the maximum possible short circuit capability, regardless of the impedance ahead (HV side) of the transformer. Meaning; that for this example you have 500KVA/0.027 = 18.52MVAsc. It would not matter if you had a system ahead of the transformer with a 9999MVAsc capability, the most you would see on the LV side is 18.52MVAsc.

You simply calculate the sc current (KAIC) by dividing by the appropriate voltage (and sqrt 3 if 3 phase). I used 208V in my example.

Yes, there would be some reduction due to reactance in the cable connecting the transformer LV side to the Main breaker, but if this distance is short, the limitations would not likely reduce the fault current enough to allow a change in rating of the breaker.

i.e. without cable 51.4kA (at 208V) and with cable maybe 49kA? Still would require 65kA equipment. Keeping in mind these are the maximum.

Typically, using the actual utility value on the HV side of transformer will not be of much benefit because the system fault current level is going to be greater than the transformer capability, and therefore the maximum is what you may see. I say typically, but not every case, there are always exceptions.

Further, the KVA value used in the calc should be based on what size you need for your service, NOT what the utility provides. There have been a lot of jobs that go south on that issue. The utility will most likely provide a transformer that is up to half what you determine. See how it changes the numbers:

Determine 500KVA is needed, utility provided a 300KVA instead. 300KVA/.027 = 11.11MVA, and at 208V that is 30.84KA. 42KAIC equipment would work.

That is a huge reduction in short circuit current. Your cable and breaker are all sized for continuous current based on 500KVA, but if you size your KAIC rating on the 300KVA, you could fall short in the future. Say the transformer burns up so utility brings out a 500KVA, still meets your continuous current rating but your KAIC rating of equipment is now obsolete. 42KAIC now needs to be 65KAIC. The owner is screwed, and worse, he doesn't even know he has a potential disaster on his hands.
 

Iron_Ben

Senior Member
Location
Lancaster, PA
That's a pretty tough question for a master electrician's test, it seems to me. You've gotten a lot of very good info in the responses to it. Let me add just a little.

As had been said, transformer kva, transformer impedance, and the voltage and phase are what you need. The POCO's system impedance is negligible compared to that of all but a really large transformer (bank) and can be ignored in all but a few cases.

The answer "ask the power company" what the available fault current is, is a good one. But there's at least one problem with it. The power company doesn't know the impedance of the transformer(s) at the top of the pole. Trust me, I worked for one for many years. The impedance for every transformer is on its nameplate, and we could send a man up in a bucket to read it, but we never did. I always figured 1.8% for a pole top unit. I've seen them in the yard a little higher and a little lower, but 1.8 is a good gamble. Note that if the transformer is pad mounted, single or three phase, the nameplate is easier to access and I unlocked hundreds of those over the years to get nameplate impedance for a customer. But send a guy up in a bucket to get it? Not going to do that.

The issue of the service entrance conductors is significant in a lot of cases. 100 feet of service wire is, usually, a significant impedance and the available fault current is a lot less at the end of the hundred feet than the beginning where it's bolted to the spades other terminals of the transformer. Now is the service one run of 1/0 triplex or three runs of 350 MCM? Obviously a lot more impedance In the former.

As was noted, the power company will almost always go a size or two (maybe 3) smaller than what connected load might seem to warrant. I can show you dozens of 1600 amp 480 services fed by 300 and 500 kva padmounts rather than 1000 or 1500 kva units. But in all the years I worked there, I can't ever recall a situation where we did a changeout and upsized. It was usually same size or down a size. Sometimes we swapped out just to get a unit with taps installed. We must have upsized a time or two but in all honesty I can't recall one at he moment. I like to think we wouldn't do it without discussing the ramifications of increased fault current with the customer or his engineer, but then there's always that midnight emergency changeout to consider when the crew is going to grab whatever is in the yard that will work.

Ben
 
Status
Not open for further replies.
Top