Continuous load calculation

[yired29]

What minimum size OCPD and copper THHN conductors are required to supply a continuous load of 100 amps when installed in a raceway with 6 current carrying conductors? All terminations are 75?.


A) #3 with 100 amp OCPD
B) #3 with 125 amp OCPD
C) #2 with 100 amp OCPD
D) #2 with 125 amp OCPD
E) #1 with 100 amp OCPD
F) #1 with 125 amp OCPD
G) 1/0 with 100 amp OCPD
H) 1/0 with 125 amp OCPD

This is not a test question I am just interested in how many different answers we will get.

 

[iwire]

1/0 - 125

 

[yired29]

1/0 - 125

Will you please show the math for wire size.

 

[hmspe]

Oops. I hit G when I meant to hit H. Sorry for messing up the survey.

 

[iwire]

Will you please show the math for wire size.

You have a 100 amp continuous load, that requires a conductor rated 125%.

215.2 Minimum Rating and Size.
(A) Feeders Not More Than 600 Volts.
(1) General. Feeder conductors shall have an ampacity not
less than required to supply the load as calculated in Parts
III, IV, and V of Article 220. The minimum feeder-circuit
conductor size, before the application of any adjustment or
correction factors, shall have an allowable ampacity not
less than the noncontinuous load plus 125 percent of the
continuous load.

100 amps x 1.25% = 125 amps.

So we need a conductor with a rating of at least 125 amps.

1/0 @ 90C = 170 amps but with a 80% derate for 6 condutors

170*.80=136 amps

A 1 AWG would come out below 125 amps.

 

[iwire]

Oops. I hit G when I meant to hit H. Sorry for messing up the survey.

I fixed it, your vote is now H.

 

[Rick Mack]

Will you please show the math for wire size.

210.20(A) OCP selected from 240.6(A) to be not less than 125% of Continuous
Load. 100A * 1.25 = 125A
210.19(A)(1) Wire selected from T310.16, 75C column because of termination limits,
based on 125% of Continuous load before any adjustments = #1 CU
T310.15(B)(2)(a) Derate #1 by 80%. 130A * .80 = 104A = adjusted ampacity
This is adequate for the Load To Be Served [210.19(A)(1)]
240.4(B) In this case, the next higher size OCP above the adjusted ampacity
is allowed to be used = 110 A. However, we have already determined
that 125 OCP is required, so go up one size in wire = 1/0 CU.
150A * .80 = 120A which is adequate for Load To Be Served (100A)
and next higher OCP is 125A.
Answer: 1/0 with 125A OCP

 

[yired29]

210.20(A) OCP selected from 240.6(A) to be not less than 125% of Continuous
Load. 100A * 1.25 = 125A
210.19(A)(1) Wire selected from T310.16, 75C column because of termination limits,
based on 125% of Continuous load before any adjustments = #1 CU
T310.15(B)(2)(a) Derate #1 by 80%. 130A * .80 = 104A = adjusted ampacity
This is adequate for the Load To Be Served [210.19(A)(1)]
240.4(B) In this case, the next higher size OCP above the adjusted ampacity
is allowed to be used = 110 A. However, we have already determined
that 125 OCP is required, so go up one size in wire = 1/0 CU.
150A * .80 = 120A which is adequate for Load To Be Served (100A)
and next higher OCP is 125A.
Answer: 1/0 with 125A OCP

THHN are 90 degree conductors look at 110.14 (C) we can adjust from the 90 degree ampacity.

 

[iwire]

210.20(A) OCP selected from 240.6(A) to be not less than 125% of Continuous
Load. 100A * 1.25 = 125A
210.19(A)(1) Wire selected from T310.16, 75C column because of termination limits,

Assuming he used a 90C conductor we can use the 90C column for derating.

T310.15(B)(2)(a) Derate #1 by 80%. 130A * .80 = 104A = adjusted ampacity
This is adequate for the Load To Be Served [210.19(A)(1)]
240.4(B)

No, it is not adequate for the Load To Be Served, the conductor is a required by 215.2(A) to be rated at least 125% of the load.

Answer: 1/0 with 125A OCP

Right answer but you arrived at it in the wrong way.

 

[LarryFine]

Right answer but you arrived at it in the wrong way.

Nobody said we had to show our work. grin

 

[Rick Mack]

THHN are 90 degree conductors look at 110.14 (C) we can adjust from the 90 degree ampacity.

110.14(C) is used when we do not know the termination rating. We are told in this case
that the termination rating is 75C. This determines the column we make our initial
selection from for the conductor.

You are right though, I should have the 90C column for the derate.

 

[yired29]

110.14(C) is used when we do not know the termination rating. We are told in this case
that the termination rating is 75C. This determines the column we make our initial
selection from for the conductor.

You are right though, I should have the 90C column for the derate.

Do you still think it should be a 1/0 with 125 amp OCPD?

 

[Rick Mack]

Assuming he used a 90C conductor we can use the 90C column for derating.



No, it is not adequate for the Load To Be Served, the conductor is a required by 215.2(A) to be rated at least 125% of the load.




Right answer but you arrived at it in the wrong way.

210.19(A)(1) and 215.2 have similar wording. I see this as a Branch Circuit, i.e. from the last OCP to the utilization device.
210.19(A)(1) First sentence says "Branch circuit conductors shall have an ampacity not less than the maximum load to be served." which in this case is 100A. It then goes on
to say to select the conductor before any adjustments are made based on 125% of
continuous load. The conductor would be selected from the 75C column = #1 @130A.

You are right. I should have used the 90C column for the derate because of THHN.
150A * .80 = 120A Using 240.4(B), the next higher OCP would be 125A which agrees
with the 100A * 1.25 = 125A we used to size the OCP [210.20(A)].

So the Answer should be: #1 with 125A OCP

 

[yired29]

This seems like a simple question but I have found some info that has changed my mind. Please take a look at the ROC from the 96 code cycle page 46 and 47 comment 2-103 at this link.

http://www.nfpa.org/Assets/files/AboutTheCodes/70/NEC-ROCA-1995-Articles 90 to 400.pdf

Also look at this link from square d

http://www.squared.com/us/products/gendoc.nsf/07a0210021262d45862564b5006e4f84/ee5709d4e2de1823852565de0058e80b/$FILE/0110HO9503.pdf

 

[yired29]

This link was written by Jim Pauley from Square D who happens to be on CMP 2 which deals with article 210 and 215.


http://www.squared.com/us/products/gendoc.nsf/07a0210021262d45862564b5006e4f84/ee5709d4e2de1823852565de0058e80b/$FILE/0110HO9503.pdf


Based on the information in the 96 ROC and this article along with this link
http://www.nfpa.org/assets/files/AboutTheCodes/70/70-A2010-ROP.pdf
to the 2011 ROP page 153, proposal 2-195

I believe this is enough proof that the minimum size wire for this installation would be a #1 THHN conductor protected by a 125 amp OCPD. The ROP and ROC allow us to understand the intent of the code.

In the past I would have thought the minimum size conductor would have been a 1/0 but with this info I have changed my mind.

 

[Rick Mack]

This link was written by Jim Pauley from Square D who happens to be on CMP 2 which deals with article 210 and 215.


http://www.squared.com/us/products/gendoc.nsf/07a0210021262d45862564b5006e4f84/ee5709d4e2de1823852565de0058e80b/$FILE/0110HO9503.pdf


Based on the information in the 96 ROC and this article along with this link
http://www.nfpa.org/assets/files/AboutTheCodes/70/70-A2010-ROP.pdf
to the 2011 ROP page 153, proposal 2-195

I believe this is enough proof that the minimum size wire for this installation would be a #1 THHN conductor protected by a 125 amp OCPD. The ROP and ROC allow us to understand the intent of the code.

In the past I would have thought the minimum size conductor would have been a 1/0 but with this info I have changed my mind.

Thanks for the additional references. I teach a NEC class to Industrial Maintenance students at York Technical College in Rock Hill, SC. Sizing conductors can be one of the most challenging topics to explain clearly since adjusted ampacity and OCP selection both factor in.

Looking at the 2011 ROP, page 153, 2-195, I find that I will have to revise my answer
to 1/0 THHN. According to "2-195" wording, the adjustied ampacity for #1 THHN =
150A * .80 = 120A, which is less than 125% of continuous load, and could not be used.
Using 1/0 THHN = 170A * .80 = 136A, which is "not less than" 125% of continuous load.

It seems that the "adjusted ampacity not less than 125% of continuous load" will always
be larger than the "maximum load to be served", which in our example would be the
continuous load of 100A. The "and" between these two conditions means we have to meet both conditions and will require us to take the larger value. The "load to be served"
seems to be a non-issue.

Thanks again for the good discussion.

 

[yired29]

Thanks for the additional references. I teach a NEC class to Industrial Maintenance students at York Technical College in Rock Hill, SC. Sizing conductors can be one of the most challenging topics to explain clearly since adjusted ampacity and OCP selection both factor in.

Looking at the 2011 ROP, page 153, 2-195, I find that I will have to revise my answer
to 1/0 THHN. According to "2-195" wording, the adjustied ampacity for #1 THHN =
150A * .80 = 120A, which is less than 125% of continuous load, and could not be used.
Using 1/0 THHN = 170A * .80 = 136A, which is "not less than" 125% of continuous load.

It seems that the "adjusted ampacity not less than 125% of continuous load" will always
be larger than the "maximum load to be served", which in our example would be the
continuous load of 100A. The "and" between these two conditions means we have to meet both conditions and will require us to take the larger value. The "load to be served"
seems to be a non-issue.

Thanks again for the good discussion.

I do not believe you read the ROP page 153 2-195 completely. There was only one thing that changed with that proposal and it had nothing to do with this discussion. The minimum size should be a #1 THHN. Read the panel meeting action. The proposal was only accepted in part.

 

[yired29]

Thanks for the additional references. I teach a NEC class to Industrial Maintenance students at York Technical College in Rock Hill, SC. Sizing conductors can be one of the most challenging topics to explain clearly since adjusted ampacity and OCP selection both factor in.

Looking at the 2011 ROP, page 153, 2-195, I find that I will have to revise my answer
to 1/0 THHN. According to "2-195" wording, the adjustied ampacity for #1 THHN =
150A * .80 = 120A, which is less than 125% of continuous load, and could not be used.
Using 1/0 THHN = 170A * .80 = 136A, which is "not less than" 125% of continuous load.

It seems that the "adjusted ampacity not less than 125% of continuous load" will always
be larger than the "maximum load to be served", which in our example would be the
continuous load of 100A. The "and" between these two conditions means we have to meet both conditions and will require us to take the larger value. The "load to be served"
seems to be a non-issue.

Thanks again for the good discussion.

The best explantion to this was the 96 ROC and the article by Jim Pauley from Square D. With that information #1 with 125 amp OCPD

 

[Rick Mack]

I do not believe you read the ROP page 153 2-195 completely. There was only one thing that changed with that proposal and it had nothing to do with this discussion. The minimum size should be a #1 THHN. Read the panel meeting action. The proposal was only accepted in part.

You are right (again). I didn't finish reading the panel statement.
I agree the part rejected should not be there.

I printed off Jim Pauley's paper but have not gotten through it yet.
Thanks for the feedback.

 

[HEYDOG]

Most breakers are only designed to operate at 80% of their rating when used on a continious load. 100 amp continous load = 100 x 1.25 = 125 amp breaker.
(.8 x 125 = 100)
Then you must increase wire size to match breaker size. If there wasn't more than three current carrying conductors and all termination were rated for 75 degree "C" you could use # 1 "THHN" but there are six current carrying conductors so there has to be a further adjustment of .8. Thus 125/.8 = 156 amps. A 1/0 "THHN" is good for 170 amps while a # 1 THHN is only rated for a 150 amps. I'm going with the 1/0.