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Old 09-25-2017, 08:59 AM   #1
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Default Available Fault Current

Need help ASAP. With the situation in south Florida, FLP NOT able to compute a AFC letter and I need to build a commercial service quick. My question: I'm building a small 3ph service for irrigation. Feed from open Delta two transformer system. I'm going to compute the primary at infinite since FPL is on hurricane duty. How do I input the transformer data since the 3ph two transformer delta system nameplates are 1ph 120/240. Oh yes, if you are not familure with AFC calcs please don't answer. And don't need any advise on building the service. Thanks in advance
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Old 09-25-2017, 10:45 AM   #2
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Most utilities use the infinite bus as the basis for the primary FC. Since you say you have an open delta setup - many utilities use this rule - Open Delta services have maximum fault currents equal to or less than single-phase services with the same transformer size and voltage. Use the single-phase table and the larger of the two transformers to identify the maximum secondary fault current.

This table works well. Answers are close enough.
This is the one I have bookmarked, I'm sure there are others:

http://www.ed3online.org/files/Const...012-06-05).pdf
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Old 09-25-2017, 11:12 AM   #3
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With an open ∆, each transformers contribution is independent of the other because they are connected to each other at only one end, not both. Therefore, the maximum available fault current will be the larger of the 2 transformers.

Divide the impedance by the full-load current and you'll have fault current.

Example, suppose the largest transformer is 50KVA, 240 volts and the impedance is 5.2%

50,000 240 = 208. 208 .052 = 4000 amps.
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Old 09-25-2017, 03:57 PM   #4
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This stuff is also to be found in Ugly's.

Pages 58-61.

All of the formulas are right there.
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