The code states:
"AC Adjustable Voltage Motors. For motors used in alternating-current, adjustable voltage, variable torque drive system, the ampacity of conductors, or ampere rating of switches, branch-circuit short-circuit and ground- fault protection, etc., shall be based on maximum operating current marked on the motor or control nameplate, or both. If the maximum operating current does not appear on the nameplate, the ampacity determination shall be based on 150 percent of values given in Tables 430-149 and 430-150."
According to these two last tables the motor may be of Induction Type [squirrel cage or wound-rotor] or Synchronous Type Unity Power Factor [pf=1or pf=0.8-0.9 ].
The supply voltage may be variable using a variable voltage transformer or all kind of variable voltage driver as VFD or ASD or vector controller or else
Regards
It seems to imply a motor with an AC drive attached although the reference to adjustable voltage does raise question in my mind, I understand the voltage is selected either 380 or 460 for example and frequency and current are adjustable with the drive.
I'll watch for answer on this from the experts also. nec makes my head ache!
Thank you for your responses, but 430.6(C) makes no mention of adjustable speed, or adjustable frequency. In doiing so, is the Code not allowing this section to be used to size motor conductors to VFD driven motors?
I believe that the adjusable voltage motors of 430.6(C) are motors that have the speed controlled by voltage reduction only. This is known as "primary voltage control" and is only effective with high slip motors. It is pretty much obsolete, but has been used with variable torque (fan and centrifugal pump) loads and hoists.
My search and understanding of 430.6(C) is becoming less elusive. Thank you CJCPE.
I am trying to understand the Code in selecting the minimum size of motor circuit conductors from a VFD to a motor. 430.120 does not modify provisions of Part 1 to Part IX, so no help there. The manufacturer is referring to 430.6(C) to size.
430.6(C) makes no mention of adjustable speed, or adjustable frequency. In doing so, is the Code not allowing this section to be used to size motor conductors to VFD driven motors?
A manuacturer's package documentation is sizing these motor circuit conductors and I suspect they are xxlarge #3/0. And have documented the motor as 107A FLA.
The VFD rated output is 205A,
This Reliance motor has been described as "special", the motor is a 75HP, 380V, 3 Ph, variable Torque, 15 Min Duty, Code K and Design Y, suited for 75 to 135Hz. No FLA or LRA information is on the nameplate.
Let me correct my earlier response "430.6 (C) has nothing to do with AFDs, AFAIK." Sorry for any confusion.
You need to size the motor conductors from VFD to motor per standard requirements of 430.6 - 125% of motor FLA, per the table in the NEC - unless you know the actual FLA are greater. Conductors to the VFD are sized at 125% of the rated input to the VFD.
But you have a very special motor, so I'd check with Reliance to get an actual FLA. NEC requires use of the FLA in the NEC tables as a minimum size. So even if your nameplate FLA are less, you need to use the NEC value.
From the NEC Table 430-150 one can see the current is inverse proportional with the voltage. So, for 380 V and 75 hp for an Induction Motor the full-load current will be 116.2 A
This could be the rated current for a 75 hp Induction Motor with pf=.85 and efficiency 86% for 380 V.
According to 430.6(c) the ampacity determination shall be based on 150 percent of the value given in Table 430-150.
That means 1.5*116.2 =174.3 A
This is a conservative value as it is only 15 minutes duty, then according to 430-22(a) one could use only 85% of the Nameplate Current Rating.
That means 0.85*174.3 = 148.1 A.
According to Table 430-7(b) if code= K the kilovolt-Ampere per HP for locked-rotor will be 8-8.99 for Y connection.
Then let say 9*75=675 kVA
Since ratedV=380 V then Ilockedrot=675/sqrt (3)/.380=1025.6A
But the VFD maintains a lower start current by keeping V/frq. constant so the starting current will be about 1.5*Irated =116.2*1.5=174.3 A
This is my opinion.
Regards
Thank you, 7anoter4. I see you are up quite early also, we posted within minutes of one another. You answered while I typed. Also thanks for last post it makes things very clear.
