Conductor Sizing 3

[fzlo]

I'm quite new to the engineering field and I was told to check if the sizing for the motor was done correct. The motor was a 214KW 480V with FLA of 330A. I thought that since the motor only had an RPM of 1000, I would need to size the conductor for 125% of the FLA and not the FLC. I was later given the datasheet for the motor and saw multiple power ratings and now I am currently unsure how the power rating was selected for the circuit. I was also then told that it was controlled by a motor drive which has listed the current at 438A. Does the conductor have to be sized for the input current of the VFD for this or am I supposed to use the FLA of the motor?

 

[waross]

125% of the nameplate Amps of the motor.

Bill
--------------------
Ohm's law
Not just a good idea;
It's the LAW!

 

[TugboatEng]

Yes, the power is lower at the low RPM but the peak amps is still the same as the nameplate.

Your datasheet adds some confusion because it's showing power ratings at different duty cycles. The nameplate should show the S1 rating which is 325 amps.

The VFD needs to meet or exceed the motor rating. A VFD rated for 438 amps exceeds the rating. During commissioning you will input the motor nameplate data into the VFD.

 

[fzlo]

So for the conductor sizing, I would need to size it for 125% of 325A which is 406.25A. Is that correct?

Just a follow-up question, someone also mentioned about sizing the conductor based on the percent overload of the motor from the motor torque/current of the duty cycle. I thought that the overload sizing was a completely different thing and you would only apply it if the 125% was not enough to run the motor?

 

[jraef]

IF you were in North America, the cables FEEDING the VFD are required to be 125% of the VFD Input Amps (430.122), which may be different from the OUTPUT amp rating (which is how you buy them) REGARDLESS of what motor is connected to the output. So in this case you would need to read the data sheet for that specific VFD to find the Input Amps. As an example of something close to what you have, I am looking at a VFD rated for 430A @ 480V, which is the OUTPUT amps. But the INPUT is only 406A. So conductors feeding that VFD would be sized at 125% of 406A, or 508A minimum conductor ampacity.

The cables FROM the VFD to the motor follow the same normal rules, so 125% of the motor FLC from the NEC charts, adjusted IF the actual motor FLA on the nameplate is higher tan the NEC chart. Over here, we would call that a 300HP motor, so our charts call for the FLC to be 361A, but you say the actual FLA on the motor is 330A, so we would have to go with 361A as the conductor size basis, x 1.25 = 451A minimum conductor ampacity.


" We are all here on earth to help others; what on earth the others are here for I don't know." -- W. H. Auden

 

[fzlo]

jraef, thank you for the more detailed explanation. I do just want to ask about the FLC. Wouldn't the FLC be at 302A since 170kW is only at 227HP, and would use the 250HP value for it?

 

[waross]

My calculations agree with jraef's numbers.
Where did 170 kW come from?
The mechanical power is given. divide by the PF and eff to get electrical kW.
You may have multiplied instead of dividing.

Bill
--------------------
Ohm's law
Not just a good idea;
It's the LAW!

 

[fzlo]

I apologize. You are indeed correct that I had mistakenly multiplied the pf instead of dividing it. Thank you for correcting that.