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domeng

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How do you calculate the available fault current on the secondary side of an oped delta 2 transformer bank (120/240 volts -3 phase -4 wire) system.
 
LarryFine

LarryFine

Master Electrician Electric Contractor Richmond VA
Location
Henrico County, VA
Occupation
Electrical Contractor
First, you need some info on the transformers from the POCO, the secondary impedance mainly, and usually presume 'infinite bus' on the primary.

Usually, the POCO gives you everything up to the point of attachment, or wherever your conductors meet theirs. You should be in contact with a POCO field engineer already.

You then take the service drop size and length, conductors to the meter and the main, and everything else in the current pathway to the point you need to size.

It's been a few years since I did one, but I do remember using a calculator available around here somewhere. I'm sure you'll get other responses. Stand by.
 
mcclary's electrical

mcclary's electrical

Senior Member
Location
VA
LarryFine said:
First, you need some info on the transformers from the POCO, the secondary impedance mainly, and usually presume 'infinite bus' on the primary.

Usually, the POCO gives you everything up to the point of attachment, or wherever your conductors meet theirs. You should be in contact with a POCO field engineer already.

You then take the service drop size and length, conductors to the meter and the main, and everything else in the current pathway to the point you need to size.

It's been a few years since I did one, but I do remember using a calculator available around here somewhere. I'm sure you'll get other responses. Stand by.
Click to expand...

I alaways get that number from them,, and I've asked a few times how it's figured and I seem to get different answers. I've been told it depends how far you ar from a substation, how many step downs it goes through, of course, the size of the xfmr, and many other things all contribute to that final fault current number. I was told even if the change a xfmr with a xfmr of the same size, it still changes the fault current slightly, since no 2 xfmrs are identical
 
D

domeng

Member
mcclary's electrical said:
I alaways get that number from them,, and I've asked a few times how it's figured and I seem to get different answers. I've been told it depends how far you ar from a substation, how many step downs it goes through, of course, the size of the xfmr, and many other things all contribute to that final fault current number. I was told even if the change a xfmr with a xfmr of the same size, it still changes the fault current slightly, since no 2 xfmrs are identical
Click to expand...
Utility co will only give me sizes of transformers.(I) 100 KVA AND (1) 50 KVA connected in an open delta config.
 
S

skeshesh

Senior Member
Location
Los Angeles, Ca
Stopmoving said:
The POCO is covering their butts. I've had this happen before, you need to get the information from them...in writing, for your own protection.
Click to expand...

I'd take his advice. Utility here charges something like 500-1000 bux to provide available fault current and the clients sometimes don't like it but it's just necessary.
 
bob

bob

Senior Member
Location
Alabama
You will need to get the %Z for the transformers as has been stated.
One of the most popular uses of open delta is to supply a small three phase load and a large single phase load. Typically, a small transformer is installed alongside an existing 120/240 volt lighting transformer.
Calculate the available fault current for the large transformer. This will be the available fault current on "A" phase.
Calculate the available fault current on the small transformer. This will be the available fault current on "C" phase.
Add the currents vectorially. This will be the available fault current on "B" phase.
 
D

domeng

Member
Is this additive since you have two transformers connected on secondary to give us 120/240-3phase-4wire or do we use the largest of the two?
 
kwired

kwired

Electron manager
Location
NE Nebraska
For those that say you need to know information on the primary side of the transformer I think one should assume an infinite supply for the purpose of selecting equipment. Knowing all the details is fine if you are just wanting to know the real numbers.

Why should you have to redesign something down the road because the utility made some changes because someone opend a new plant just across the road and now your available fault current has increased to the point that you need to change it. Or worse yet the available current has increased and nobody has even noticed there may be a problem because of it.
 
bob

bob

Senior Member
Location
Alabama
domeng said:
Thanks! This fits in with a response I got from Square D engineering
Click to expand...

What was their response? I assumed the infinite bus method for the calculation. Please provide your transformer sizes and the results.
 
Last edited:
mivey

mivey

Senior Member
kwired said:
For those that say you need to know information on the primary side of the transformer I think one should assume an infinite supply for the purpose of selecting equipment. Knowing all the details is fine if you are just wanting to know the real numbers.

Why should you have to redesign something down the road because the utility made some changes because someone opend a new plant just across the road and now your available fault current has increased to the point that you need to change it. Or worse yet the available current has increased and nobody has even noticed there may be a problem because of it.
Click to expand...
Not only that, but reduce the actual %Z by 10% to allow for transformer change-out and impedance tolerance.

While the primary configuration might change, the utility should not put you on a much larger tranformer. The infinite bus will take care of the primary changes.
 
bob

bob

Senior Member
Location
Alabama
Looking at my post #11 assume a 50 kva %Z 2.5 and a 15 kva %Z 2.5 both 240 volt.
50 kva fault 208 amps/0.025 = 8333 amps Ia
15 kva fault 62.5 amps/0.025 = 2500 amps Ic
Ia + Ib + Ic = 0
Ib@240 = -Ia @0 - Ic@120
Ib = -8333 - (-2500x0.5 + J2500x0.866)
Ib = -8333 - (-1750 + J2165)
Ib = -8333 + 1750 - J2165.
Ib = sqrt(-6583? - J2165?)
Ib = 6930 amps.
Does this look correct?
 
mivey

mivey

Senior Member
bob said:
Looking at my post #11 assume a 50 kva %Z 2.5 and a 15 kva %Z 2.5 both 240 volt.
50 kva fault 208 amps/0.025 = 8333 amps Ia
15 kva fault 62.5 amps/0.025 = 2500 amps Ic
Ia + Ib + Ic = 0
Ib@240 = -Ia @0 - Ic@120
Ib = -8333 - (-2500x0.5 + J2500x0.866)
Ib = -8333 - (-1750 + J2165)
Ib = -8333 + 1750 - J2165.
Ib = sqrt(-6583? - J2165?)
Ib = 6930 amps.
Does this look correct?
Click to expand...
Use 1250 instead of 1750
 
bob

bob

Senior Member
Location
Alabama
mivey said:
Use 1250 instead of 1750
Click to expand...
Thanks


Looking at my post #11 assume a 50 kva %Z 2.5 and a 15 kva %Z 2.5 both 240 volt.
50 kva fault 208 amps/0.025 = 8333 amps Ia
15 kva fault 62.5 amps/0.025 = 2500 amps Ic
Ia + Ib + Ic = 0
Ib@240 = -Ia @0 - Ic@120
Ib = -8333 - (-2500x0.5 + J2500x0.866)
Ib = -8333 - (-1250 + J2165)
Ib = -8333 + 1250 - J2165.
Ib = sqrt(-7083? - J2165?)
Ib = 6744 amps.
 
mivey

mivey

Senior Member
bob said:
Thanks


Looking at my post #11 assume a 50 kva %Z 2.5 and a 15 kva %Z 2.5 both 240 volt.
50 kva fault 208 amps/0.025 = 8333 amps Ia
15 kva fault 62.5 amps/0.025 = 2500 amps Ic
Ia + Ib + Ic = 0
Ib@240 = -Ia @0 - Ic@120
Ib = -8333 - (-2500x0.5 + J2500x0.866)
Ib = -8333 - (-1250 + J2165)
Ib = -8333 + 1250 - J2165.
Ib = sqrt(-7083? - J2165?)
Ib = 6744 amps.
Click to expand...
Ib = sqrt(-7083? - J2165?) = 7407
 
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