There is 23 feet of AWG 1 conductor from the open delta transformers to the service panel where the bolted short circuit and arc flash occurred. How do you include the impedance of the conductor with the transformers short circuit current of 6971 amps to get a final lower bolted short circuit current at panel? How do you compute for it? The arc flashed breakers were in OFF position and no load downstream of it as detailed in the link below detailing how the arc flash occured.

The latest IEEE 1584-2018 concludes from additional experiments that 2kA is enough to cause arc flash if the panel has any breakers. They admit the original IEEE 1584-2002 assumed empty panel hence had flaws. My actual case proved it.

related to: Short circuit current of 3-phase open delta transformer

and the actual arc flashed breakers: Was this caused by arc flash or just short circuit without any arc flash?

enter image description here

  • \$\begingroup\$ How are the rhs conductors joined under the black tape? If just « shoved » together then go pay for a qualified electrician and keep everyone safe. Using cardboard as protection is not best practise. \$\endgroup\$ – Solar Mike Mar 16 at 1:07
  • \$\begingroup\$ No, the small black wire is just a ground wire and at back of the live wire and separately taped, so they are not in contact. The cardboard is just temporary to lower the live wire to the terminal. \$\endgroup\$ – Jtl Mar 16 at 1:10
  • \$\begingroup\$ Given the other pictures you show of that device then I suggest you don’t continue using it... \$\endgroup\$ – Solar Mike Mar 16 at 1:42
  • \$\begingroup\$ It was already replaced in 2015. It was just file photo to let me understand how to compute for bolted short circuit current especially when the new IEEE 1584-2018 stated 2k can cause arc flash. \$\endgroup\$ – Jtl Mar 16 at 2:01
  • \$\begingroup\$ You should show what you are actually working with so we have a relevant base on which to provide answers. \$\endgroup\$ – Solar Mike Mar 16 at 2:03

You know the short circuit current at the transformer, and you know the open circuit voltage at the transformer, so calculating the magnitude of the supply impedance at the transformer is trivial.

You know the cross sectional area of the cable, you know the material and you can look up the resistivity, so cable resistance is trivial.

You don't know if the supply impedance is mostly resistive, but it only makes at most a factor of root two, so don't worry about it unless there is some really unusual stuff hung of that transformer.

From there it is just sums.

  • \$\begingroup\$ What formulas do you use to get the impedances of wires? \$\endgroup\$ – Jtl Mar 16 at 8:39
  • \$\begingroup\$ Also isn't it supply impedance mostly inductive.. why did you say "mostly resistive"? \$\endgroup\$ – Jtl Mar 17 at 1:48

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