Main Feeder overload protection device

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Engr.Mohamed

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hello ,

it's my first post here , with questionon how i calaculate the main feeder overload device protection for multi motor.

i returned to the nec 430 article.
whcih stated that

the sum of the largest
branch-circuit protective device plus the sum of the fullload
currents of the other motors, .

my question here if the result not according to the standard breakers. so i will go the the next large breaker or go down to the next low breaker.


i have 2 chiller system each with 1600 AMP Inverse Time Breaker and FLA for each is 842 AMP. also i have one motor with FLA 220 AMP.

According to the NEC

OCP for the feeder = 1600 "the largest OCP"+842 "FLA"+220"FLA = 2662 AMP

now go up the the next larg or do down to next small.

AS if you see the EXAMPLE D8 , the code goes down not up??



i hope to find answer

thanks dears
 
Short answer, 430.62 does not allow you to "round up", but there may be flaws in your calculations since you are using Instantaneous Breaker numbers.
I assume we are addressing the feeder short-circuit-Ground-Fault protection since "overload" is on a individual motor basis.
Note that 430.62 which has the wording you quote, also notes that for instantaneous trip breakers you are to use the values allowed in 430.52 for the proper device.
You will note in 430.52(C)(3) that and Instantaneous trip breaker is only allowed as part of a listed combination motor controller and not for feeder protection, so, if you were dealing with motor feeders, you would calculate the max inverse time or fuse OCP per 430.52 and add the FLA of the other loads.
You feeder protection would need to be a Inverse Time breaker or a fuse and the values based on 430.52.

Before continuing, since you stated "chillers", so I think the appropriate article would be Art 440. If so, the the equipment nameplate show a MOCP ? (See 440.4)
 
Last edited:
augie47 said:
Short answer, 430.62 does not allow you to "round up", but there may be flaws in your calculations since you are using Instantaneous Breaker numbers.
I assume we are addressing the feeder short-circuit-Ground-Fault protection since "overload" is on a individual motor basis.
Note that 430.62 which has the wording you quote, also notes that for instantaneous trip breakers you are to use the values allowed in 430.52 for the proper device.
You will note in 430.52(C)(3) that and Instantaneous trip breaker is only allowed as part of a listed combination motor controller and not for feeder protection, so, if you were dealing with motor feeders, you would calculate the max inverse time or fuse OCP per 430.52 and add the FLA of the other loads.
You feeder protection would need to be a Inverse Time breaker or a fuse and the values based on 430.52.

Before continuing, since you stated "chillers", so I think the appropriate article would be Art 440. If so, the the equipment nameplate show a MOCP ? (See 440.4)
Click to expand...
..

i check the AR 440.4 in it i found how to calculate the branch cuitct overload device for the compressors .

now i need to calculate the main feeder over-current protection for the 2 chillers each chiller have 2 compressor with another motor 60 HP and 5 HP on one Breaker

i will attach my calculation not regarding to this matter .
also i attach the chiller data sheet.
my calculation regarding to the main feeder over current protection
http://www.mediafire.com/?vvv44yyimgpqol7


the issue with me that the consultant need to increase the size of the breaker to 3000 AMP from 2500 AMP
and according to my calculation 2500 AMP is enough
 

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the fram of the breaker is 2500 amp only so if i changed the breker i will change the frame and the maon busbars.

but my concern is it's required t change the main breaker or not..

as the nec for the main feeder OCP we will go down not go up.. but for the branch circuitwe will go up
 
My calculations show a feeder OCP of 2382, which would not allow a 2500 amp breaker unless your conductor ampacity was 2500 amps.
 
Max OCP Unit 1 = 1600 amp Inverse Time Breaker + FLA Unit 2 (385 + 385 +6 + 6)
 
In that case 1600 + 385 + 385 + 36 = 2406.. still less than 2500.
 
H**L fire :D It helps when asking a question if you provide all pertinent data.

Fire pumps are a whole different animal (see Art 695) and in all probability would not be allowed on a circuit with any other equipment.
 
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