240.86(C) Motor contribution and series rating

[msshel0]

We are doing a controls/power distribution retrofit on a piece of equipment. I have attached a .pdf of a simplified one line for the equipment. I have a 800A fused main disconnect with a class L Fuse. I have calculated the available fault current @ 48,114A. I'm feeding 2 Panels: a main control panel (MCP) and a power distribution panel (PDP). The MCP has a 600A main with 460.3A of Motor load. The PDP has a 400A main with 104A motor load. The let-through current of the 800A fuse is 17,000A. If I take my motor contributions into account at 6X the full load current, the motors add roughly 3,400A to my fault current. Bringing my available fault current to 21,400A on the load side of the fused disconnect. So, at that point I would think that bringing my MCP and PDP to 25kA would be sufficient. However, since I have 460.3A of motor load in my MCP does that mean my PDP has to be rated > 46030A due to 240.86(C) of the NEC? If that is the case, can somebody explain to me why the motor contribution can only be 1% of the lower rated breaker's rating? Am I mis-applying the code?

Thanks for any help,
Matt Shelton
Electrical Engineer

 

[Dennis Alwon]

Bump-- any help here-- Out of my experience

 

[jim dungar]

As you noted, when using a 'tested' series-rating the NEC limits the motor rated FLA contribution, between the devices, to not more than 1% of the downstream device's AIC rating (i.e. 100A of motors for a 10kAIC device). Backing into this number requires understanding how circuit breakers are rated and how series testing is done.

This is not the same as determining motor fault contribution, either before or after, the 'series devices'. This calculation involves the nominal 6x value you are calculating.

Simply adding two SCA values together can give a reasonable result, but in an actual power study we consider the X/R ratios as well as the current.

 

[msshel0]

Thank you for the response. So to clarify, the answer is that I'm talking about 2 different subjects:

PDP-rate at > 46.3kA
1) fault current calcs let me rate it at 22 kA
2) 240.86(C) requires me to rate it for, at least 46.3 kA

MCP-rate at > 22kA
1) fault current calcs require me to rate it at, at least 22 kA
2) 240.86(C) would require it to be at least 10.4 kA

It looks as though you're saying my fault current calculations are too simplistic. Can you recommend a good reference material? Finally, do you know where 1% (in 240.86(c)) comes from. It feels like a number somebody just pulled out of the air.

Again, thanks for any help

 

[jim dungar]

Thank you for the response. So to clarify, the answer is that I'm talking about 2 different subjects:

Yes these are two different subjects that may be interrelated.