430-22 (C) Wye-Start Delta-Run Motor

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430.22 (C) States "Wye-Start, Delta Run Motor: For a wye-start delta-run connected motor, the selection of the the branch circuit conductors on the line side of the controller shall be based on the highest of the full-load current rating shown on the nameplate. The selection of the branch-cirucuit conductors between the controller and the motor shall be based on 58% of the full load current.

I am assuming starter and controller in this context are the same definition. Thereby if the Nameplate Amperage on the motor is 540 amperes times 1.25 equals 675 Amperes so that 3 sets of 4/0 THHN conductors at 75 degree C limititation from terminal connections would provide 690 amperes on line side of controller and the load side connections from the controller would then be 392 amperes (58% of 675) and three sets of 1/0 THHN conductors from 75 degree C column would suffice providing 450 amperes of connection.

My assumuptions are the overload protection at the controller would be based on the 392 amperes via 430-32 at 1.15% for all other motors as well as the Overcurrent Device at the controller being rated using dual element fuses would be 175% over 392 at 686 amperes rounded to next common size of 700 or for a inverse time circuit breaker 1200 amperes. With the motor feeder to the contoller being rated at 675 amperes with fused protection on the feeder @ 175% with time delay fuses would be 1176 rounded up to 1200 Amperes.

I believe I'm on the right path with this one however if someone see's holes in my logic, please advise.

Thank you in advance for your help.
Steve
 

Jraef

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You are forgetting that on a Wye-Start Delta-Run motor, you will have 6 leads going from the starter to the motor. Each set of those 6 leads will need to be sized for 58% of the motor FLA ( x 1.25).

The OL relay is only in the 3 leads coming off of the main (1M) contactor, also sized for 58% of the motor nameplate FLA.

[img http://ecatalog.squared.com/catalog/174/html/sections/16/images/17416045.1.jpg]
 
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