Feeder Breaker Question (NEC 440.110 (C))

[jrt5054]

Hello All! This is my first post so please go easy on me. Long time reader of the forums however.

I have a wireway with several taps into fused disconnects feeding 2 chillers and 3 pump motors. It is currently fed from a 208V 3 phase 600A circuit breaker. We are replacing the chillers with slightly larger ones and I need to see if the 600A breaker is large enough or if we need to go to a larger size. Here is my thought process.

I am using the 2017 NEC and going through what it says in 440.110 (C).

Full-Load Current Calculation
Chiller 1 Compressor A - RLA = 48A
Chiller 1 Compressor B - RLA = 48A
Chiller 1 Compressor C - RLA = 122A
Chiller 2 Compressor A - RLA = 48A
Chiller 2 Compressor B - RLA = 48A
Chiller 2 Compressor C - RLA = 122A
Pump 1 - 3 HP (10.6A from 430.250)
Pump 2 - 3 HP (10.6A from 430.250)
Pump 3 - 3 HP (10.6A from 430.250)

Total full load current = 467.8A
Total Full load current * 115% = 537.97A

Locked Rotor Current Calculation
Chiller 1 Compressor A - LRA = 351A
Chiller 1 Compressor B - LRA = 351A
Chiller 1 Compressor C - LRA = 1010A
Chiller 2 Compressor A - LRA = 351A
Chiller 2 Compressor B - LRA = 351A
Chiller 2 Compressor C - LRA = 1010A
Pump 1 - 3 HP (71A from 430.251(B))
Pump 2 - 3 HP (71A from 430.251(B))
Pump 3 - 3 HP (71A from 430.251(B))

Total Locked Rotor Current = 3637A

Now here is where I am having a problem. When I go to look at Table 430.251(B) for the equivalent HP based on the total LRA I see that the table stops at 3207A. which isn't very close to the 3637A that I have calculated. The 3207 LRA is the equivalent of a 200HP motor according to the table. a 200HP motor has 528 FLA according to 430.250.

Does this mean that for any LRA over 3207A we just use a 200HP motor for the rest of the calculation?

Lets say for arguments sake that my total LRA was MUCH higher than 3207A. What do we do then?

Thanks All!

 

[Eddie702]

600 amp breaker is good for 480 amp continuous load so I think your ok there 600 X 80%==480 amps Not sure about the locked rotor

 

[jrt5054]

600 amp breaker is good for 480 amp continuous load so I think your ok there 600 X 80%==480 amps Not sure about the locked rotor

Unfortunately 440.110 requires you satisfy both the FRLA and LRA requirements. The LRA portion is what is giving me trouble

 

[Jraef]

Your mistake here is that you are reading the rules that apply to the BRANCH circuit, meaning the LAST device ahead of the load. When you have multiple different machines, this is a FEEDER circuit. All those rules about sizing for the LRA do not apply.

(There is no 440.110 by the way...)

 

[jrt5054]

Your mistake here is that you are reading the rules that apply to the BRANCH circuit, meaning the LAST device ahead of the load. When you have multiple different machines, this is a FEEDER circuit. All those rules about sizing for the LRA do not apply.

(There is no 440.110 by the way...)

Thanks for the response Jraef. And you are correct, there is no 440.110. I was flipping back and forth between 440 and 430 and got mixed up. The actual code I was referring to was 430.110

You make a good point about it being a feeder. I'll have to dig into that section of the code.

Thanks!

 

[mikeames]

430.24(1) + (2) Size the feeder

Take 125% of your largest motor = 122A

Then add all the other FLC of all the other motors on the feeder.

You need a feeder capable of 498.3A

Find the conductor size

When you have size round down to next standard OCPD size. You cant round up for a feeder only a BC

 

[jrt5054]

430.24(1) + (2) Size the feeder

Take 125% of your largest motor = 122A

Then add all the other FLC of all the other motors on the feeder.

When you have feeder size round down to next standard OCPD size. You cant round up for a feeder only a BC

I was reading 215.2 and 215.3 for the feeder and the circuit breaker and it says to take 125% of the continuous load and 100% of the non-continuous load and add them together. According to the mechanical engineer, all of the equipment is capable of running for 3 hours or more. My understanding is that the 215.2 and .3 method would produce a higher number than 430.24 and would therefore be the number i need to use.

Do you agree?

 

[mikeames]

Your situation is just motors which is a motor feeder covered in article 430. Article 215 is feeders in general.

 

[mikeames]

I should clarify my round down statement. IF after you find your OCPD and its in-between standard values round down.

430.110 is for sizing the disconnect. Not the feeder.

 

[augie47]

430.110 addresses the condition where multiple motors can not be started simultaneously.
Would that allow you to use a lesser number in this situation ?

 

[augie47]

I should clarify my round down statement. IF after you find your OCPD and its in-between standard values round down.

430.110 is for sizing the disconnect. Not the feeder.

I believe that is incorrect..430.110 is Part of IX, Disconnecting means.

 

[mikeames]

I believe that is incorrect..430.110 is Part of IX, Disconnecting means.

Yes I agree? 430.110 is not relevant to the OP.


He needs to take the total calculated load which I think is 498.3A (based on OP info) and then 430.61 refers you to table 430.52 where he then needs to apply the appropriate OCPD based on the motor types. If its 150% then the size of the OCPD/short circuit would be 498.3 x 1.5 = 747.45 round down to 700A

 

[augie47]

Definite foot in mouth.
I misread your post when I disagreed,.
I am of the opinion that 430.110 is applicable in general since 440 doesn't address feeders but I think there is a "loophole" in regard to simultaneous starting.

 

[jrt5054]

Ok here is what I have so far.

Feeder Wire Size

From 430.24 I took the largest motor size (122A) and multiplied by 1.25 to get 152.5. I then added 100% of the FLA of all other motor loads which give me 152.5+122+(48*4)+(10.6*3) = 498.3A. Which is good because the existing wire is rated for 620A

Feeder Protective Device

From 430.62 (as pointed to by 430.63) the breaker shall be greater than the following. I take the largest branch CB size as allowed by 430.52 or 440.22 (350A in my case taken from the chiller cut sheet) and add the full load amps of all other loads. So in this case it would be 350A+122+(48*2)+(10.6*3) = 599.8A

So am I right to say that the existing 600A breaker is technically too large?

 

[mikeames]

440.22 does not apply to you because its for branch circuits not feeders. We are only looking at the feeder. So CB size to the individual motors is not what we are talking about. Remember the the feeder will feed disconnects and overloads for each motor. We are sizing the short circuit rating for the feeder and the feeder size. Your existing 600 amp breaker is protecting only the feeder in case of a short or GF. It does nothing for the motors. The calcs above are meant to find the feeder size and if your 600a existing breaker is good enough. Your 600a breaker is not too large. If it holds it may be just ok but I came up with 700A.

 

[Jraef]

440.22 does not apply to you because its for branch circuits not feeders. We are only looking at the feeder. So CB size to the individual motors is not what we are talking about. Remember the the feeder will feed disconnects and overloads for each motor. We are sizing the short circuit rating for the feeder and the feeder size. Your existing 600 amp breaker is protecting only the feeder in case of a short or GF. It does nothing for the motors. The calcs above are meant to find the feeder size and if your 600a existing breaker is good enough. Your 600a breaker is not too large. If it holds it may be just ok but I came up with 700A.

Correct.
In fact, you could use an 800A or 1,000A feeder, so long as the conductors are sized for that. Don't keep mixing BRANCH and FEEDER rules. You are JUST a feeder here.