Fault Current

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kingpb said:
As long as the cable carries the proper listing; and the AHJ approves of its use, it can be utilized.
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Agree.
EPDM wire applications: "Lead Wire, UL 3340, UL 3374, UL 3399; Appliance Wiring; Power Cables; Welding Cables ;Direct Burial Cables."
 
Again with the derail, but I suggest that in the same way you are reviewing the engineering of the available short circuit current, you review the engineering of the selection of the secondary conductors.

A very common mistake is to take current ratings supplied by manufacturers of cables for specialized uses, and try to apply them for NEC compliant installations.

For example: http://www.farnell.com/datasheets/1804767.pdf is a type of EPDM welding cable, presumably different from what you have. Depending on where you look on that datasheet, a 4/0 cable could be used at 405A or 600A...but if you use it in a conduit in an NEC installation, it is type RHH with an ampacity of 260A (at 90C; if you have 75C terminations then your ampacity drops to 230A); on the other hand if you have 'single conductors in free air' then your ampacity might be 405A.

This just strikes me as another part of your installation to double check. It is quite possible that with the wire ratings, the type of install, and 'engineering supervision' what you have is perfectly correct.

I promise that I'll help with the available short circuit calcs when the single line comes available :)

-Jon
 
winnie said:
Again with the derail, but I suggest that in the same way you are reviewing the engineering of the available short circuit current, you review the engineering of the selection of the secondary conductors.

A very common mistake is to take current ratings supplied by manufacturers of cables for specialized uses, and try to apply them for NEC compliant installations.

For example: http://www.farnell.com/datasheets/1804767.pdf is a type of EPDM welding cable, presumably different from what you have. Depending on where you look on that datasheet, a 4/0 cable could be used at 405A or 600A...but if you use it in a conduit in an NEC installation, it is type RHH with an ampacity of 260A (at 90C; if you have 75C terminations then your ampacity drops to 230A); on the other hand if you have 'single conductors in free air' then your ampacity might be 405A.

This just strikes me as another part of your installation to double check. It is quite possible that with the wire ratings, the type of install, and 'engineering supervision' what you have is perfectly correct.

I promise that I'll help with the available short circuit calcs when the single line comes available :)

-Jon
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Bottom line - I don't want to re-engineer the system. My interpretation of the Code leads me to believe that what was done is compliant. I'm going to review my connected load because that really is the driving factor of what I do. The secondary conductors are enclosed in a conduit, but the conduit is a bout 4 ft long, conductors are about 8 ft. I'm not concerned with overheating the conductors; I put a data logger on the system at start-up to get a worse-case scenario of the connected load and the highest leg was in the vicinity of 100 amps.
 
buffalonymann said:
the table you referenced does not list EPDM wire, not sure how you arrived at 230A
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It would not be. EPDM is the chemical description of a type of rubber. When you look at the datasheet for the cables, it may list the 'UL' insulation type that the wire meets the requirements for. So for example this EPDM wire meets the requirements for UL RHH and RHW https://www.awcwire.com/portals/0/pdf/specs/awc_I67_dlo-welding.pdf but this one doesn't meet any UL specs https://www.southwire.com/ProductCatalog/XTEInterfaceServlet?contentKey=prodcatsheet204

When you look at the NEC tables, you will see the UL spec types. So to evaluate NEC compliance you look for the UL insulation type in the tables.

buffalonymann said:
Bottom line - I don't want to re-engineer the system. My interpretation of the Code leads me to believe that what was done is compliant. I'm going to review my connected load because that really is the driving factor of what I do. The secondary conductors are enclosed in a conduit, but the conduit is a bout 4 ft long, conductors are about 8 ft. I'm not concerned with overheating the conductors; I put a data logger on the system at start-up to get a worse-case scenario of the connected load and the highest leg was in the vicinity of 100 amps.
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If you have a 100A connected load, on wire that has a 230A NEC rating (which is known to be very conservative), protected by 400A OCPD...I would be comfortable that the installation would work reliably for a very long time. I do not believe it is NEC compliant though.

-Jon
 
The code makes a distinction of length of conduit where conduit fill (and therefore temperature rise considerations) may be ignored. That is 18". Take from that what you will.

winnie said:
If you have a 100A connected load, on wire that has a 230A NEC rating (which is known to be very conservative), protected by 400A OCPD...I would be comfortable that the installation would work reliably for a very long time. I do not believe it is NEC compliant though.
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Second this.
 
buffalonymann said:
My interpretation of the Code leads me to believe that what was done is compliant.
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Two sets of 4/0 transformer secondary conductors are not properly protected by a 300A primary c/b on a 480-240V delta/delta transformer, per NEC requirements.

Nor would one set of 4/0 transformer secondary conductors be properly protected by a 200A primary c/b.
 
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kingpb said:
No offense intended to the OP; but I find it somewhat odd that a SLD sketch isn't being provided since it would be quite simple to do.
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I believe that buffalonymann is working on getting the necessary information for the SLD; drawing the sketch is the easy part; sometimes getting things like the transformer information can be quite difficult. When he gets the info I will do my best to help him with his originally requested calculations, despite my own attempts to derail the thread :)

-Jon
 
winnie said:
I believe that buffalonymann is working on getting the necessary information for the SLD; drawing the sketch is the easy part; sometimes getting things like the transformer information can be quite difficult. When he gets the info I will do my best to help him with his originally requested calculations, despite my own attempts to derail the thread :)

-Jon
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OK, have the one line - xfmr secondary is 2500kva - 480/277 with 6.33% Z. I came up with about 52,000 amps. the feeder CB from the MDP is 100kA - travels about 150 ft to another CB rated at 30 kA.
 
Okay. I get for the transformer:
The normal full load current of the transformer is 2500000/480/sqrt(3) = 3007A
3007/0.0633=47500A short circuit current

Then I followed https://iaeimagazine.org/magazine/2015/07/07/calculating-short-circuit-current/
which calculates everything in terms of %impedance; you have a formula for %impedance of the conductors, and another formula that gives you short circuit current in terms of adding up the various impedance values

Assuming 2 #500 conductors in parallel per phase, 140 feet of conductor, using an impedance of 0.029 ohm/kft per conductor gives a circuit impedance:
0.029 * 2500 / (480^2) * 140/2 * 100 = 2.2

Assuming utility infinite bus I get:
2500 * 100 / (sqrt(3) * 0.48 * (6.33 + 2.2) = 35200A

Missing from this calculation is the impedance feeding the 2500 KVA transformer, which could bring down the available short circuit current at the second 800A breaker. However given current information it appears to be slightly under-rated.

-Jon
 
winnie said:
Okay. I get for the transformer:
The normal full load current of the transformer is 2500000/480/sqrt(3) = 3007A
3007/0.0633=47500A short circuit current

Then I followed https://iaeimagazine.org/magazine/2015/07/07/calculating-short-circuit-current/
which calculates everything in terms of %impedance; you have a formula for %impedance of the conductors, and another formula that gives you short circuit current in terms of adding up the various impedance values

Assuming 2 #500 conductors in parallel per phase, 140 feet of conductor, using an impedance of 0.029 ohm/kft per conductor gives a circuit impedance:
0.029 * 2500 / (480^2) * 140/2 * 100 = 2.2

Assuming utility infinite bus I get:
2500 * 100 / (sqrt(3) * 0.48 * (6.33 + 2.2) = 35200A

Missing from this calculation is the impedance feeding the 2500 KVA transformer, which could bring down the available short circuit current at the second 800A breaker. However given current information it appears to be slightly under-rated.

-Jon
Click to expand...

It is utility power feeding the primary at 34.5kV
 
winnie said:
Okay. I get for the transformer:
The normal full load current of the transformer is 2500000/480/sqrt(3) = 3007A
3007/0.0633=47500A short circuit current

Then I followed https://iaeimagazine.org/magazine/2015/07/07/calculating-short-circuit-current/
which calculates everything in terms of %impedance; you have a formula for %impedance of the conductors, and another formula that gives you short circuit current in terms of adding up the various impedance values

Assuming 2 #500 conductors in parallel per phase, 140 feet of conductor, using an impedance of 0.029 ohm/kft per conductor gives a circuit impedance:
0.029 * 2500 / (480^2) * 140/2 * 100 = 2.2

Assuming utility infinite bus I get:
2500 * 100 / (sqrt(3) * 0.48 * (6.33 + 2.2) = 35200A

Missing from this calculation is the impedance feeding the 2500 KVA transformer, which could bring down the available short circuit current at the second 800A breaker. However given current information it appears to be slightly under-rated.

-Jon
Click to expand...

Question - would you use 140 feet or 280 feet because the circuit loop is 280 feet?
 
buffalonymann said:
It is utility power feeding the primary at 34.5kV
Click to expand...

The additional information that you would need is the available fault current at 34.5kV. Often for the sake of simplifying the calculations this is assumed to be infinite, but really it is some finite value which reduces the available short circuit current.

-Jon
 
winnie said:
Assuming 2 #500 conductors in parallel per phase, 140 feet of conductor, using an impedance of 0.029 ohm/kft per conductor gives a circuit impedance:
0.029 * 2500 / (480^2) * 140/2 * 100 = 2.2
Click to expand...

I'm wondering where you got the impedance from. It doesn't seem right.

Also, the conductors from the transformer to the MDP will have an effect on the available short circuit current.
 
david luchini said:
I'm wondering where you got the impedance from. It doesn't seem right.
Click to expand...

Good catch. I was lazy and didn't pull out my copy of the NEC. I used https://www.mikeholt.com/img/product/pdf/1038225525sample.pdf.

When I look at Chapter 9 table 9, the reactance value for 500 kcmil conductors is 0.048 ohms to neutral per 1000 feet.

The calculation I gave is simplified and doesn't include resistance of the wires, but this is actually a small effect, since for short circuit the transformer is mostly inductive and adding the resistance is a 'vector' addition.

But if I re-run the calc using 0.048 I get:
0.048 * 2500 / (480^2) * 140/2 * 100 = 3.65

Note that this assumes copper conductors in steel conduit.

Assuming utility infinite bus I get:
2500 * 100 / (sqrt(3) * 0.48 * (6.33 + 3.65) = 30100A

Remember that this ignores the utility side impedance, the impedance of the conductors 'upstream' of the first breaker, and the resistance of the 140 foot feeder (which again will be only a small factor).

Tracing this all the way back to the original question, I think you can safely say that the desired new 200A breaker needs to be rated for a fault current of 30kA or needs to be 'series rated' with the 800A breaker to have a fault current rating of 30kA.

-Jon
 
winnie said:
Good catch. I was lazy and didn't pull out my copy of the NEC. I used https://www.mikeholt.com/img/product/pdf/1038225525sample.pdf.

When I look at Chapter 9 table 9, the reactance value for 500 kcmil conductors is 0.048 ohms to neutral per 1000 feet.

The calculation I gave is simplified and doesn't include resistance of the wires, but this is actually a small effect, since for short circuit the transformer is mostly inductive and adding the resistance is a 'vector' addition.

But if I re-run the calc using 0.048 I get:
0.048 * 2500 / (480^2) * 140/2 * 100 = 3.65

Note that this assumes copper conductors in steel conduit.

Assuming utility infinite bus I get:
2500 * 100 / (sqrt(3) * 0.48 * (6.33 + 3.65) = 30100A

Remember that this ignores the utility side impedance, the impedance of the conductors 'upstream' of the first breaker, and the resistance of the 140 foot feeder (which again will be only a small factor).

Tracing this all the way back to the original question, I think you can safely say that the desired new 200A breaker needs to be rated for a fault current of 30kA or needs to be 'series rated' with the 800A breaker to have a fault current rating of 30kA.

-Jon
Click to expand...

Jon,

I'm not familiar with the term 'series rated' Could you shed some light?
 
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