Derate 400amp parallel feed

[27hillcrest]

I have a project that involves installing a 400amp sub panel. To feed the panel the code book is telling me to install a 600 MCM. (I am not equipped to pull 600) Instead of pulling the 600 I would like to install parallel conductors. My plan is to install 4/0 parallel conductors in 1-3" emt. Can any of you see if I am missing anything? Thanks in advance.

 

[bob]

If the caculated load is not greater that 380 amps you can use 500 kcm.
If you are talking about 4/0 in the same conduit that will not have 400 amp capacity. You must derate the cable to 80%.
2 - 4/0 = 460 amps x 0.80 = 384 amps. That just 4 amps over the 500 kcm. If the conductors are installed in separate conduits then it would be ok. In fact you could use 3/0 at 200 amps.

 

[27hillcrest]

panel loads

panel loads

The panel will be used to power up 2 Liebert CRAC units. (computer room air conditioners (110amp breaker) and 2 outside condenser packs. (20 amp breakers. There will be nothing else out of this panel. So You are telling me 500 will be ok? The feeds will be 480v 3phase

 

[infinity]

If the caculated load is not greater that 380 amps you can use 500 kcm.
If you are talking about 4/0 in the same conduit that will not have 400 amp capacity. You must derate the cable to 80%.
2 - 4/0 = 460 amps x 0.80 = 384 amps. That just 4 amps over the 500 kcm. If the conductors are installed in separate conduits then it would be ok. In fact you could use 3/0 at 200 amps.

If he's using a 90 degree conductor than you're a little off on your derating. #4/0 is rated at 260 amps in the 90 degree column which is used for derating.

260(2)= 520 amp(80%)=416 amps

 

[pierre]

It would seem that you would be able to install 500's and only have to pull one set of conductors.

 

[infinity]

It would seem that you would be able to install 500's and only have to pull one set of conductors.

I would agree, but it sounded like he wanted to avoid pulling large conductors. Personally since he's using only one conduit I would pull the 500's as Pierre has stated. Seems just as easy as pulling 2 sets of #4/0's in one conduit.

 

[27hillcrest]

500mcm

500mcm

I will be using THHN. Lets take out the connected load for a minute. If I were to install a 400 panel I will need to size the conductors according to the 75 degree column. Mainly because most breaker terminations are rated 75 degree. Where in the NEC does it allow me to size the panel to the connected load? I thought you should always size the panel to the full load rate of the panel. Am I over building things?

 

[pierre]

If your load is smaller than 400 amps, refer to 240.6 and NEMA rated panels to find the next smaller panel that will work with your computed load.

 

[27hillcrest]

Re: 500mcm

Re: 500mcm

I will be using THHN. Lets take out the connected load for a minute. If I were to install a 400 panel I will need to size the conductors according to the 75 degree column. Mainly because most breaker terminations are rated 75 degree. Where in the NEC does it allow me to size the panel to the connected load? I thought you should always size the panel to the full load rate of the panel. Am I over building things?

 

[infinity]

240.4(B) allows you to go up to the next standard size OCPD when the conductors' ampacity doesn't correspond to a standard size listed in 240.6. So since 500kcmil THHN is rated at 380 amps at 75 degrees you could go up to the next standard size OCPD which would be 400 amps. This is unless your computed load were to exceed 380 amps, if it doesn't than you're OK with the 400 amp OCPD protecting your 380 amp conductors.

 

[bob]

If he's using a 90 degree conductor than you're a little off on your derating. #4/0 is rated at 260 amps in the 90 degree column which is used for derating.

260(2)= 520 amp(80%)=416 amps

Infinity
You are correct. I appreciate your pointing out my mistake. I knew better but posted it incorrectly.

 

[bob]

Hillcrest

The panel will be used to power up 2 Liebert CRAC units. (computer room air conditioners (110amp breaker) and 2 outside condenser packs. (20 amp breakers. There will be nothing else out of this panel.

You need to get the Full Load Amps off the name plate on the 2 Liebert Units and the 2 outside units. The size of the breaker does not provide the needed information. You can size the breaker at 175% of the FLA for A/C units. If that is the case the actual load may only be about 150 amps.

 

[sceepe]

I will be using THHN. Lets take out the connected load for a minute. If I were to install a 400 panel I will need to size the conductors according to the 75 degree column. Mainly because most breaker terminations are rated 75 degree. Where in the NEC does it allow me to size the panel to the connected load? I thought you should always size the panel to the full load rate of the panel. Am I over building things?

You should size your feeders based on the overcurrent protection not the panel rating. If you put in a 400 amp panel with a 200 amp main breaker you must use at least 200 amp wire. You don't have to use 400 amp wire.

However, you are limiting the future capacity of the panel. If someone wants to add load to the panel they will have to change out the main breaker and repull the feeder. I guess what I am trying to say is that it is a design issue not a code issue.

 

[27hillcrest]

Thanks for all the help. The name plate rating for each unit is a 110amp breaker. The ouside units name plate rating is 20 amp. Looks like the 500 will work!!

 

[bob]

You need to get the Full Load Amps off the name plate on the 2 Liebert Units and the 2 outside units. The size of the breaker does not provide the needed information.

 

[spsnyder]

You need to find the FLA ratings on the units. Unless you are designing for future expansion, I think you'll find that you are over-designing and pulling conductors larger than you need.

 

[A/A Fuel GTX]

If he's using a 90 degree conductor than you're a little off on your derating. #4/0 is rated at 260 amps in the 90 degree column which is used for derating.

260(2)= 520 amp(80%)=416 amps

Since this is a 3 phase feeder, wouldn't we use 70% as a multiplier since we would have 8 current carrying conductors if they were run in parallel?

 

[27hillcrest]

The full load amps off the Liebert is 101.3 amps. Liebert recommends 110amp breaker for each unit. The cooling fan packs have a full load rating of 9.8amp. Liebert recommends a 20 amp feed for those. So total that will be feed out of this panel will be 2X101.3 + 2X9.8 = 222.2. The breaker I have available to feed this is a 400amp. Going back to the original question of parallel feeds in one conduit. What sizes would you all feed this with? parallel 3/0? parallel 4/0? I don't have any pulling equipment for the 500 although there has been some great deals on ebay lately :lol: , I hope I'm not being a pest here??

 

[spsnyder]

This would be considered a continuos load, therefore 1.25 *222.2 = 278 Amps. The next standard size breaker would be a 300 amp breaker. The minimum conductors (1 set) would be 300 MCM (285amps in the 75-degree column). If you wanted to run parallel in one conduit, not considering derating would be 8 No. 1/0 (150 amps in the 75-degree column). Now checking for the derating assuming a 90-degree conductor is used... (80% of 170 (from 90-degree column)) *2 = 272 amps. Therefore have to go up to 2/0 because for the derating. Double checking....80% of 195amps * 2 = 312 amps. "So in brevity I shall be brief..." At minimum, you need a 300 amp breaker with (2) sets of 2/0 CU. Anything larger would be accounting for future expansion.

Hope that helps.

 

[27hillcrest]

spsnyder,
Thanks for the detailed overview. I greatly appreciate it!