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.
You looking for this?
http://www.bussmann.com/7/Arc-FlashCaluclatorNotes.html
Okay, how about this one?You have to konw the fault current to use this calc. It's for arc flash to determine what PPE should be worn. IMO
Utility co will only give me sizes of transformers.(I) 100 KVA AND (1) 50 KVA connected in an open delta config.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
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.
Thanks! This fits in with a response I got from Square D engineering
Not only that, but reduce the actual %Z by 10% to allow for transformer change-out and impedance tolerance.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.
Use 1250 instead of 1750Looking 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?
ThanksUse 1250 instead of 1750
Ib = sqrt(-7083? - J2165?) = 7407Thanks
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.