Fire Pump Conductors

[jpujols01]

I'm installing a fire pump 50 hp 208 v, according with the table 430.250 the FLA is 143 amps.
Then 143 amps x 1.25= 178.75 amps, I suppose to use one 3/0 per phase, but the fire pump control had a wye /delta configuration contact relays ( M1, M2 ) that feed the fire pump.
Because of this configuration I know that I have to run two set of cables in one conduit from the fire control panel to the pump but I'm little bit confused if have to use two set of 3/0 or divide the FLA in two and run 2 set of let's say #1 which will give about 200amps.
Please help, will appreciate....thanks.



J Pujols.

(Title Changed to Match Subject)

 

[infinity]

For WYE start-Delta run you do not need to use the same size #3/0 conductors because you will have six conductors instead of three from the controller to the pump motor. In that configuration the conductors will only carry 58% of the rated current but you still need to add the 125% factor so you end up with 72% of the FLA from the table in the NEC. (58*125%=72%)

 

[jpujols01]

Ok, then I have to do this 143 amps x 0.72= 102.96 amps, so 2 set of # 2cu plus 1 # 6 for ground will be the conductors required in this case.


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[an_elec_eng]

For circuit breaker can we use 58% to reduce lrc or fla current????

 

[gadfly56]

For circuit breaker can we use 58% to reduce lrc or fla current????

Sure, if your disconnect has six fuses in it.