Load calc HELP!!

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rexowner

Senior Member
Location
San Jose, CA
Occupation
Electrician
I was reviewing my calcs again and looking at the Example D3(a) Industrial Feeders ina common Raceway.

in that latter part Ungrounded Feeder Condutors they used the 99,000VA instead of the 113,200VA that they had gotten above. Why did they not have use the 113,200VA???

Because 99,000VA is the actual load (i.e. 100% of the
continuous load, not 125%), which is what is used
to apply the 310.15/16 correction factors.
 

Smart $

Esteemed Member
Location
Ohio
well thats what i dont understand, how can this only apply to the 100 % of the continuous load and 125%?? where is the exception that allows this??

Because of this...
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.

Note the determined wire size is 90?C 2/0 with an ampacity of 195A. It's 75?C equivalent is 175A, which exceeds the minimum required ampacity of the noncontinuous load plus 125% of the continuous load (136A).
 
Because of this...


Note the determined wire size is 90?C 2/0 with an ampacity of 195A. It's 75?C equivalent is 175A, which exceeds the minimum required ampacity of the noncontinuous load plus 125% of the continuous load (136A).

Ok i made a typo, that should have said,

well thats what i dont understand, how can this only apply to the 100 % of the continuous load and NOT 125%?? where is the exception that allows this??
 

Smart $

Esteemed Member
Location
Ohio
Ok i made a typo, that should have said,

Same answer.

First off, it is applied to the noncontinuous load plus 125% of the continuous load... only they are working it backwards (from the way you are perceiving the requirement) in the example.
 
Same answer.

First off, it is applied to the noncontinuous load plus 125% of the continuous load... only they are working it backwards (from the way you are perceiving the requirement) in the example.

I suppose thats what im not getting, if you have a non continuous load (42,400va) plus 125 of a continous load (1.25 * 56,600=70750) that would equal 70750+42400va= 113150 va

then its 113150/0.7/0.96=168377.97619va

168377.98va/(480 *1.73)=202.767 amps

So what am im missing here that they can use the 99,000va and not the 113150?? the language says otherwise.
 

Smart $

Esteemed Member
Location
Ohio
I suppose thats what im not getting, if you have a non continuous load (42,400va) plus 125 of a continous load (1.25 * 56,600=70750) that would equal 70750+42400va= 113150 va

then its 113150/0.7/0.96=168377.97619va

168377.98va/(480 *1.73)=202.767 amps

So what am im missing here that they can use the 99,000va and not the 113150?? the language says otherwise.

You're applying adjustment and correction to the load calculation. The requirement says the ampacity must be equal to or greater before the application of adjustments and corrections.

70,750VA + 42,400VA = 113,150VA
113150VA ? (480V ? √3) = 136A

The ampacity of the conductors must be equal or greater than 136.

The reason they use 99,000 is because that is the actual load, and "reverse engineer" the application of adjustment and correction factors for 90?C rated conductors. The result is 177A. Choosing the closest greater than standard ampacity of a 90?C-rated conductor indicates a minimum 2/0 conductor is necessary. Now that we have a size, it must independently meet both the "100% N'cont' + 125% Cont'" requirement and the 75?C termination requirement. Ampacity of a 75?C-rated conductor is 175A before the application of any adjustment or correction factors. 175A is greater than 136A. Requirement is therefore met using 2/0 90?C-rated conductors.

Also note that if you apply the adjustment and correction factors to the "100% N'cont' + 125% Cont'" value, you may end up using one common size larger conductor than you really have to...!!!
 

rexowner

Senior Member
Location
San Jose, CA
Occupation
Electrician
I suppose thats what im not getting, if you have a non continuous load (42,400va) plus 125 of a continous load (1.25 * 56,600=70750) that would equal 70750+42400va= 113150 va

then its 113150/0.7/0.96=168377.97619va

168377.98va/(480 *1.73)=202.767 amps

So what am im missing here that they can use the 99,000va and not the 113150?? the language says otherwise.

Brother,

The language in 215.2(A)(1) calls for 125% of the
continuous load to be considered only "before the
application of any adjustment factors."

215(A)(1) says *nothing* about the ampacity calculation
*after adjustment factors* -- it simply doesn't address
it and doesn't apply to this separate calculation.

As the example points out, "the conductors must
independently meet requirements for termination...
and [raceway] conditions"
(my emphasis, because this is a separate, independent
calculation) Again, by definition, since this is an
"independent" calculation, it has nothing to do with
the 125% requirement, otherwise it wouldn't be
an "independent" calculation.

HTH.
 
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