FLC of motor when HP is not listed

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I am sizing the conductors, overcurrent, and overload protection for a motor. The HP of the motor (3.5), is not listed in the table 430.250 to determine the FLC. I have contacted the manufacturer and was told to calculate everything based of the FLA. Not sure how to determine the FLC that is not listed or if I should calculate using the FLA. Attached is the nameplate of the motor.
GetFileAttachment
 
INS class : F
Phase : 3
Rated Voltage : 230/460V
Rated Freq : 60Hz
Rated Effy : 86.5%
Time : cont
Frame : FF165
Ip55 AMB 40c
HP(KW) : 3.5 / (2.55)
Rated current : 7 / 4 A
Rated speed : 3450
Service factor : 1.10
Rated power factor : 92%
Bearings : 6205z/6203z
 

drktmplr12

Senior Member
Location
South Florida
Occupation
Electrical Engineer
From what i've read in the nec, the FLA shouldn't be used to size the conductors, short circuit, ground fault protection.

overload protection MUST be based on the nameplate FLA.

conductors and short circuit protection wont change from 3.5 to 5 HP, at 460V 3 phase. Run #12's and install a listed motor controller, unless its a long distance where voltage drop must be considered.

OP didn't state the voltage.
 

Jraef

Moderator, OTD
Staff member
Location
San Francisco Bay Area, CA, USA
Occupation
Electrical Engineer
overload protection MUST be based on the nameplate FLA.

conductors and short circuit protection wont change from 3.5 to 5 HP, at 460V 3 phase. Run #12's and install a listed motor controller, unless its a long distance where voltage drop must be considered.

OP didn't state the voltage.
Bingo.

This is a somewhat common issue now as EU suppliers send equipment over here. The motors are rated in kW sizes that are common for them, but don't translate to the right HP values as motors over here would have. In this case I suspect that this is a standard 2.2kW 230/400V 50Hz design, so 2.2kW translates to about 3HP mechanically. But when used at 60Hz over here, the speed increases by 20%, so the HP increases by the same amount and since 2.2kW is actually 2.95 HP, a 20% increase actually becomes 3.54HP and they are just rounding down, or because of some tested value or decrease in efficiency.

But to satisfy the NEC requirements for using the charts in Article 430 based on the HP values, not the FLA, you must round up to the nearest HP, so 5HP in this case.
 

Besoeker

Senior Member
Location
UK
Bingo.

This is a somewhat common issue now as EU suppliers send equipment over here. The motors are rated in kW sizes that are common for them, but don't translate to the right HP values as motors over here would have. In this case I suspect that this is a standard 2.2kW 230/400V 50Hz design, so 2.2kW translates to about 3HP mechanically. But when used at 60Hz over here, the speed increases by 20%, so the HP increases by the same amount and since 2.2kW is actually 2.95 HP, a 20% increase actually becomes 3.54HP and they are just rounding down, or because of some tested value or decrease in efficiency.

But to satisfy the NEC requirements for using the charts in Article 430 based on the HP values, not the FLA, you must round up to the nearest HP, so 5HP in this case.
Yet more reasons to adopt SI.......:p
 

don_resqcapt19

Moderator
Staff member
Location
Illinois
Occupation
retired electrician
...
But to satisfy the NEC requirements for using the charts in Article 430 based on the HP values, not the FLA, you must round up to the nearest HP, so 5HP in this case.
What section says I have to do it that way? Why can't I interpolate between the values given in the table?
 
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