Sizing Motor Conductors for VFD application

[JUSTEINEE]

Hello,

I am a young engineer with limited design experience. Any insight you can give me would be appreciated but please include specific code references so I can try to follow along and fix my mistakes.

I am trying to size motor circuit conductors for a 300HP 480V 3Phase motor with a VFD controller. The motor nameplate current is 343.4A and the drive is rated at 420A.

I have started reading article 430 but it did not take me long to get confused. I believe that I can use 430.6 (C) to size the ampacity of my motor conductors (My branch/feeder conductors to the VFD must be sized per 430.120). If I understand what 430.6 (C) says, the motor circuit conductors shall be based on the maximum operating current marked on the motor or control nameplate or both without any adjustments. If not marked use 150% of corresponding value found in 430.250.

Therefore, in my case I would use the motor nameplate current of 343.4 and size to 500 kcmil at 75 degree C. However, it also says I could use the controller?s maximum current, which would be 420A or 600 kcmil at 75 degree C. If I did not know either value I would have to size to 150% of 430.250 which would be 361*1.5=541.5 or 1000 kcmil at 75 degree C or parallel 500 kcmil at 75 degree C. Does the "or" imply I can choose either option? If so, I am going to choose the smallest conductor and save myself time and money. Am I properly protecting the motor conductors either way? Why would it say motor, or controller, or both?

Turning further in article 430, I came across 430.22, which says that my motor circuit conductors shall be 125% of the motor's full-load current. Going to 430.250 this would be 361*1.25=451.25A or 700 kcmil at 75 degree C. Does this Negate 430.6 (C)?

Then I went to 430.120. I believe I understand the intention of this section but I am confused on how it correlates with 430.6 (C). The conductors feeding the VFD shall be not less than 125% of the rated input or 420*1.25 = 525 or 1000 kcmil at 75 degree C or parallel 400 kcmil at 75 degree C.

Someone please help me make sense of this. I do not know if my thoughts have been clearly conveyed but I am grasping at straws.

Thanks in advance for your time and patience.

 

[petersonra]

Hello,

I am a young engineer with limited design experience. Any insight you can give me would be appreciated but please include specific code references so I can try to follow along and fix my mistakes.

I am trying to size motor circuit conductors for a 300HP 480V 3Phase motor with a VFD controller. The motor nameplate current is 343.4A and the drive is rated at 420A.

I have started reading article 430 but it did not take me long to get confused. I believe that I can use 430.6 (C) to size the ampacity of my motor conductors (My branch/feeder conductors to the VFD must be sized per 430.120). If I understand what 430.6 (C) says, the motor circuit conductors shall be based on the maximum operating current marked on the motor or control nameplate or both without any adjustments. If not marked use 150% of corresponding value found in 430.250.

Therefore, in my case I would use the motor nameplate current of 343.4 and size to 500 kcmil at 75 degree C. However, it also says I could use the controller?s maximum current, which would be 420A or 600 kcmil at 75 degree C. If I did not know either value I would have to size to 150% of 430.250 which would be 361*1.5=541.5 or 1000 kcmil at 75 degree C or parallel 500 kcmil at 75 degree C. Does the "or" imply I can choose either option? If so, I am going to choose the smallest conductor and save myself time and money. Am I properly protecting the motor conductors either way? Why would it say motor, or controller, or both?

Turning further in article 430, I came across 430.22, which says that my motor circuit conductors shall be 125% of the motor's full-load current. Going to 430.250 this would be 361*1.25=451.25A or 700 kcmil at 75 degree C. Does this Negate 430.6 (C)?

Then I went to 430.120. I believe I understand the intention of this section but I am confused on how it correlates with 430.6 (C). The conductors feeding the VFD shall be not less than 125% of the rated input or 420*1.25 = 525 or 1000 kcmil at 75 degree C or parallel 400 kcmil at 75 degree C.

Someone please help me make sense of this. I do not know if my thoughts have been clearly conveyed but I am grasping at straws.

Thanks in advance for your time and patience.

The motor conductors are just normal motor conductors. The FLC value used is from the tables in the NEC, not the nameplate.

The VFD input is per 430, section X (430.120).

 

[LISHAJI]

The motor conductors are just normal motor conductors. The FLC value used is from the tables in the NEC, not the nameplate.

The VFD input is per 430, section X (430.120).

Makes sense, the VFD input as per 430.122 and motor input from VFD as per 430.6 or use the larger rating of either through out!!

 

[JUSTEINEE]

So If I am understanding the responses thus far. You are telling me to ignore 430.6 (C) and act like it does not exist. I should use the codes that reference general motor applications not VFD applications and size my motor circuit conductors to 125% of the corresponding value of 430.250.

 

[jojo]

I agree with you, LISHAJI. You are sizing motor conductors "for a 300HP 480V 3Phase motor with a VFD controller. The motor nameplate current is 343.4A and the drive is rated at 420A."

With NEC 430.6 (C) " .. If maximum operating current does not appear on the nameplate, the ampacity determination shall be based on 150% of the values given in Tables 430.249 and 430.250." Using Table 430.250, a 300Hp 480V 3-phase motor has a full-load current of 361 amperes. To determine the minimum conductor size = 1.5 x 361amperes = 541.4 amperes. that is, 2 sets of 3-350kcmil + 1/0GND.

With NEC 430.122 (A) "Branch /Feeder Circuit Conductors: Circuit conductors supplying power conversion equipment included as part of an adjustable-speed drive system shall have an ampacity NOT less than 125% of the rated input to the power conversion equipment." Given the rating of the drive at 420A; determine the conductor ampacity = 1.25 x 420 = 525 amperes. Again, 2 sets of 3-350kcmil + 1/0GND.

Makes sense..