Questions Regarding the minimum size of conductor for the inverter output circuit

F

Frank6172

Member
Location
Maryland
Occupation
Engineer
Hi All,

Here is the scenario: 2017 NEC, 208V, 3-phase, 4W, WYE-WYE, 35 degree C 2% high ambient temperature in ASHRAE, 3-current carrying conductors(L1,2 and 3) in the conduit. All lugs and breakers the conductors connected to are rated at 75 degree C. Continuous output current is 279A and we are using CU THWN-2 wire. We know the OCPD size should be 350A since it's the next size up of (279x1.25=348.75A). What is the minimum size of conductors? The AC run is very short so please don't consider voltage drop here. I got 500 kcmil but my colleague is saying 400 kcmil is fine? The 690.8(B) is telling me the conductor ampacity should be greater than (279x1.25=348.75A) and I looked up the 75 degree C column of 310.15(B)(16) due to the article 110.14 (C) (1) (b).

Thanks very much in advance!
 
400 kcmil only good for 335A but 240.4(B) allows you to use the next standard breaker size as long as does not correspond a standard amp rating, which it does not as the next standard amp rating 380A. Your friend is right, 400 is good because of this. Sorry for the bad news
 
Frank6172 said:
Continuous output current is 279A and we are using CU THWN-2 wire. We know the OCPD size should be 350A since it's the next size up of (279x1.25=348.75A)
Click to expand...
If the bold is the minimum conductor size how can a 335 amp, 400 kcmil conductor be used?
 
You don’t use the output current to size the conductors. You use the motor HP to get the FLC from the appropriate table in the NEC (usually 430.152), x 1.25 as the MINIMUM conductor ampacity. We don’t know the motor HP value here.

If it is something like a chiller that doesn’t provide a HP value, just a RLA (Running Load Amps), the chiller data should give you the MCA for the motor, which is what you use. If not, only THEN can you use the motor RLA value to determine conductor size.

As to the breaker, if there is a VFD involved, the VFD tech data should provide you with breaker size data, based on the UL listing of the VFD. You must (per 110.3.B) follow that. Also note that the VFD input conductor size has separate rules (430.122).
 
jwoody said:
'Infinity" is right, my input should not apply. The conductors should at least be rated for the current. 500 is correct. Sorry for the wrong info
Click to expand...
Thanks very much for your response! My thought is that 240.4(B) allows us to choose the next higher standard OCPD size based on the conductor ampacity. We should not do the reverse the engineering to have the 350A OCPD to determine the conductor sizing.
 
Jraef said:
You don’t use the output current to size the conductors. You use the motor HP to get the FLC from the appropriate table in the NEC (usually 430.152), x 1.25 as the MINIMUM conductor ampacity. We don’t know the motor HP value here.

If it is something like a chiller that doesn’t provide a HP value, just a RLA (Running Load Amps), the chiller data should give you the MCA for the motor, which is what you use. If not, only THEN can you use the motor RLA value to determine conductor size.

As to the breaker, if there is a VFD involved, the VFD tech data should provide you with breaker size data, based on the UL listing of the VFD. You must (per 110.3.B) follow that. Also note that the VFD input conductor size has separate rules (430.122).
Click to expand...
I don't think the motor situation applies in our case. We are sizing the interactive inverter output circuit using the article 690.8(A) and 690.8(B).

Thanks!
 
2017 NEC (per the OP) 690.8(A)(3) says that for an inverter output circuit, the maximum current is the inverter output current rating, or 279A for the OP.

2017 NEC 690.8(B)(1) tells us that for termination considerations (which implies 75C ratings given the info in the OP), the minimum conductor size has an ampacity at least 125% * 279A = 349A, without adjustment or correction. 400 kcmil Cu has a 75C ampacity of 335A, too little, so we need to go to 500 kcmil.

2017 NEC 690.8(B)(2) tells us that for non-termination considerations (which means we can use the 90C ratings), the minimum conductor size has an ampacity of at least 279A after adjustment or correction. With 35C ambient and 90C rated conductors, the correction factor is sqrt(55/60) = 0.957, so we need a 90C table ampacity of at least 291A. So for this requirement 300 kcmil Cu would comply.

The larger requirement of 500 kcmil Cu per 690.8(B)(1) controls.

Cheers, Wayne
 
wwhitney said:
2017 NEC (per the OP) 690.8(A)(3) says that for an inverter output circuit, the maximum current is the inverter output current rating, or 279A for the OP.

2017 NEC 690.8(B)(1) tells us that for termination considerations (which implies 75C ratings given the info in the OP), the minimum conductor size has an ampacity at least 125% * 279A = 349A, without adjustment or correction. 400 kcmil Cu has a 75C ampacity of 335A, too little, so we need to go to 500 kcmil.

2017 NEC 690.8(B)(2) tells us that for non-termination considerations (which means we can use the 90C ratings), the minimum conductor size has an ampacity of at least 279A after adjustment or correction. With 35C ambient and 90C rated conductors, the correction factor is sqrt(55/60) = 0.957, so we need a 90C table ampacity of at least 291A. So for this requirement 300 kcmil Cu would comply.

The larger requirement of 500 kcmil Cu per 690.8(B)(1) controls.

Cheers, Wayne
Click to expand...
Thanks very much for your detailed and professional response! I have reached out an agreement with my colleague that 500 kcmil needs to be used.
 
You must log in or register to reply here.
Top