electricalperson
Senior Member
- Location
- massachusetts
according to 695.6C2 , when sizing conductors for a fire pump motor we follow 430.22 and also follow the voltage drop rule of 695.7
if i have a fire pump rated at 25hp 208v i would size that according to table 430.250
74.8 x 1.25 = 93.5a so i would need a #3 awg conductor
the confusion sets in with article 695.7 voltage drop. in the example in the book it has a 25 hp 208v 3 phase fire pump motor located 175 feet away from the service. the controller is 150 feet away from the service.
so the VD to the controller cant be more than 15 percent and the motor cant be more than 5% when operating at 115% full load current.
so after all the math is done, i dont have any problem with the formula at all
( cmil = (1.732 x k x LRC x D)/VD )
after all the math is done i come up with a #3 to feed the controller
what i dont understand is the 115% part. in the next example we size conductors to the motor from the controller. he takes the FLC from table 430.250 and multiplies it by 1.15 instead of 1.25 so we come up with 86amps instead of the 93 amps using the 1.25 multiplier.
the formula is cmil = (1.732 x 12.9 x 86 x 175)/10.4 (the 10.4v is 5% of 208v per 695.7)
the answer is a #4 AWG conductor.
the question is how come in the first example when sizing conductors to the motor we use 125% according to article 430.22 and when we sized the conductors from the controller to the motor we use 115% instead? i am very confused and i hope i did not confuse anyone else with this long thread.
if i have a fire pump rated at 25hp 208v i would size that according to table 430.250
74.8 x 1.25 = 93.5a so i would need a #3 awg conductor
the confusion sets in with article 695.7 voltage drop. in the example in the book it has a 25 hp 208v 3 phase fire pump motor located 175 feet away from the service. the controller is 150 feet away from the service.
so the VD to the controller cant be more than 15 percent and the motor cant be more than 5% when operating at 115% full load current.
so after all the math is done, i dont have any problem with the formula at all
( cmil = (1.732 x k x LRC x D)/VD )
after all the math is done i come up with a #3 to feed the controller
what i dont understand is the 115% part. in the next example we size conductors to the motor from the controller. he takes the FLC from table 430.250 and multiplies it by 1.15 instead of 1.25 so we come up with 86amps instead of the 93 amps using the 1.25 multiplier.
the formula is cmil = (1.732 x 12.9 x 86 x 175)/10.4 (the 10.4v is 5% of 208v per 695.7)
the answer is a #4 AWG conductor.
the question is how come in the first example when sizing conductors to the motor we use 125% according to article 430.22 and when we sized the conductors from the controller to the motor we use 115% instead? i am very confused and i hope i did not confuse anyone else with this long thread.
