Need help solving Example D3(a)

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antenna2001

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I am preparing for the Colorado State Licensing Examination and I am working on learning to solve the examples on the back of the 2005 NEC.

In Example D3, for a commercial building, the minimum size of the feeder is based on the sum of the noncontinuous load plus 125% of the continuous load (See the bottom left of page 70-719).

In Example D3(a), for an industrial installation, the minimum size of the feeder is based on the actual load, without the 125% factor applied to the continuous load (See the 6th line below the heading "Ungrounded Feeder Conductors" on page 70-720).

I can't find any section in the code that states that these two installations should be treated differently. Every section I read says that the procedure in Example D3 is the correct one.

Does anyone know why these two examples are solved differently?

Thanks
 
Examples

Examples

I Believe if you look a little further down on page 70-720 on the left (sub totals, actual load) near top on left you will see that 25% was added to the 56,600 continuous load( 14,200).It should actual be 14,150. 125% is the same as 100% plus 25%. 125% of the 56,600 is 70,750. 56,600 + 14150=70,750. Hope this helps and I hope I am correct
 
Example D

Example D

I believe both examples do use 125% of the continuous load. In example D3 it is pretty clear on the bottom of page 70-719 near the bottom on the left. However, in example D3(a) the load is taken at 100% but then 25% of the continuous load was added in under subtotals on page 70-720 (left top). I believe there is a miscalculation because 25% of the 56,600 is really 14,150 not 14,200 as stated. So, 125% of 56,600= 70,750 and the total of 56,600 and 14,150= 70,750. Hope I answered your question.
 
I still don't see it.

I still don't see it.

Dennis,
In the section that you mention, 99,000 is the total before the addition of the 25%. Then, 25% is added to obtain 113,200. The problem is that when they actually do the size calculation (about 1/2 way down the page), they use the 99,000 VA number:
99,000 VA / 0.7 / 0.96 = 147,000 VA
Also, 3 lines before this equation, you see the phrase "based on actual load".
I don't see them adding the 25% anywhere after that.

Thanks for your reply and please follow up so we can decide if I am overlooking something.

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Dennis Alwon said:
I believe both examples do use 125% of the continuous load. In example D3 it is pretty clear on the bottom of page 70-719 near the bottom on the left. However, in example D3(a) the load is taken at 100% but then 25% of the continuous load was added in under subtotals on page 70-720 (left top). I believe there is a miscalculation because 25% of the 56,600 is really 14,150 not 14,200 as stated. So, 125% of 56,600= 70,750 and the total of 56,600 and 14,150= 70,750. Hope I answered your question.
Click to expand...
 
Eample D

Eample D

The ungrounded feeder conductors are based on the actual load for some reason. I am also confused, perhaps it is because the correction factor of temperature and condition of use in a raceway exceeds the load of the 125% of the continuous load. If the load were less perhaps then one would use the 125% load. I am guessing at this point-- I am not, by any means, efficient with my calculations. Hopefully one of the experts can help us both on this one.
 
Dennis Alwon said:
The ungrounded feeder conductors are based on the actual load for some reason. I am also confused, perhaps it is because the correction factor of temperature and condition of use in a raceway exceeds the load of the 125% of the continuous load. If the load were less perhaps then one would use the 125% load. I am guessing at this point-- I am not, by any means, efficient with my calculations. Hopefully one of the experts can help us both on this one.
Click to expand...
Perhaps 215.2(A)(1), second sentence will clear it up:
The minimum feeder-circuit conductor size, before the application of any adjustments or correction factors, shall have an allowable ampacity not less than the noncontinuous load plus 125 percent of the continuous load.
Click to expand...
The example chose 2/0 @ 90?, rated ampacity of 195A... and a correlated 75? rated ampactiy of 175A. The calculated load @ 100% noncontinuous, 125% continuous was 136A.

If you expect the code to be completely integrated and use absolute logic, you will be disappointed.
 
Unintegrated code

Unintegrated code

Smart $,
I am in the process of analizing your logic. In the midtime, let me say that I think that because of the liability that might result from following (or not following) the code, it is necessary that its logic be as absolute and as integrated as possible. Everything is ok until a building goes on fire due to a miscalculated feeder. The stuff we deal with in electricity can not be left to chance. We must know what we are doing.



"If you expect the code to be completely integrated and use absolute logic, you will be disappointed."[/QUOTE]
 
Smart $,
I tend to agree with you. 215.2 requires application of the 125% factor. This brings us back to the original question. Why are they not applying the factor in the answer?

Now, if we were to follow the prescription of 215.2, the calculations should be as follows:

215.2(A)(1) says that the minimum conductor size must be based on 125% continuous + noncontinuous before ANY correction factors are applied. The total noncontinuous load that they calculated include correction factors for the motors. These correction factors are unacceptable here. Therefore, a new noncontinuous figure without correction factors is necessarry:

Receptacles: ..................................................3960 VA
Compressor: .................................................11431 VA
Grinder: .................................................2494 VA
3 Welders at 17.94 A each: 3 * 17.94 A * 480 V = 25834 VA

Total Noncontinuous load: ...........................43629 VA

Total continuous (does not change): 56600 VA

Noncontinuous + 125 % continuous = 43629 + 1.25 * 56600 = 114379 VA

Converting to Amps: 114379/480 = 238 A

310.16(Table) at 90 degC produces a minimum conductor size of 300 kcmil rated at 255 A. Then, 310.15(A)(2) requires a 70% adjustment factor of table 310.15(B)(2)(a) applied. This gives

255 A * 0.70 = 179 A

Then, the final minimum acceptable conductor size is 300 kcmil rated at 179 A.


Please everyone check this logic and tell me what am I missing. 300 kcmil seems excessive but that is what the doctor seems to have ordered. I'd never make it as a pharmacist!

Thanks
 
antenna2001 said:
Now, if we were to follow the prescription of 215.2, the calculations should be as follows:

215.2(A)(1) says that the minimum conductor size must be based on 125% continuous + noncontinuous before ANY correction factors are applied. The total noncontinuous load that they calculated include correction factors for the motors. These correction factors are unacceptable here.
Click to expand...
The statement in 215.2(A)(1), "...before the application of any adjustments or correction factors..." is regarding ONLY the feeder-circuit conductors.

I also noticed you are converting to amps incorrectly:
antenna2001 said:
Converting to Amps: 114379/480 = 238 A
Click to expand...
The example is a 480V, 3? system. To convert to amps you must divide by (480 x √3).
 
Smart $ said:
The statement in 215.2(A)(1), "...before the application of any adjustments or correction factors..." is regarding ONLY the feeder-circuit conductors.

I also noticed you are converting to amps incorrectly:

The example is a 480V, 3? system. To convert to amps you must divide by (480 x √3).
Click to expand...

Smart $,
Thanks for your reply. I am now using the sqrt(3) in the calculation and I think all the given loads affect the feeders. Therefore, I still included them. My original question is still unanswered and I found another question regarding the use of factors (see the last sentences of this posting). I am begining to think that the code solution is wrong but I am reluctant to come to this conclussion unless I have other opinions. Please check me again.

Following the prescription of 215.2, the calculations should be as follows:

215.2(A)(1) says that the minimum conductor size must be based on 125% continuous + noncontinuous before ANY correction factors are applied. The total noncontinuous load that they calculated include correction factors for the motors. These correction factors are unacceptable here. Therefore, a new noncontinuous figure without correction factors is necessary:

Receptacles: .................................................. 3960 VA
Compressor: .................................................1 1431 VA
Grinder: .................................................2 494 VA
3 Welders at 17.94 A each: 3 * 17.94 A * 480 V = 25834 VA

Total Noncontinuous load: ...........................43629 VA

Total continuous (does not change): 56600 VA

Noncontinuous + 125 % continuous = 43629 + 1.25 * 56600 = 114379 VA

Converting to Amps: 114379/(480 * sqrt(3) ) = 138 A

310.16(Table) at 90 degC produces a minimum conductor size of 2/0 rated at 150 A. Then, 310.15(A)(2) requires a 70% adjustment factor of table 310.15(B)(2)(a) applied. This gives

150 A * 0.70 = 105 A

Then, the minimum acceptable conductor size is 2/0 rated 105 A


But if this is correct, then the 177 A that the code solution gives is wrong. Is it?

Also, I just noticed that when they do the calculation
99,000 VA / 0.7 / 0.96 = 147,000 VA
They are using the correction dividers instead of correction factors. I think it should be
99,000 VA * 0.7 * 0.96 = 66,528 VA

Am I seen this wrong? In the other examples they use the factors as factors and here they use them as dividers. This can't be right!

Please everyone. I need all the criticisms you can think of.
Thanks
 
I'll run through this one time...

I'll start from combined loads, as I do not see ANY problem with the determination of individual loads and the application of demand factors.
56,600 VA, Continuous load
42,400 VA, Noncontinuous load
99,000 VA, Actual load [actually, Calculated load]
119 A = 99,000 / (480 x √3), Converted to amperes​
14,200 VA, 25% of continuous load
113,200 VA, 100% noncontinuous plus 125% continuous loads
136 A = 113,200 / (480 x √3), Converted to amperes​
Ungrounded Feeder Conductors Size Determination
We know the feeder-circuit conductors ampacity will be corrected for ambient temperature. Also, they will be adjusted for eight (8) current carrying conductors in one raceway.
[derated ampacity] = [Table ampacity] x [ambient correction factor] x [CCC in raceway adjustment]​
Where [derated ampacity] result must be equal or exceed the load amperes.

Conversely, we can work from the load amperes:
[minimum Table ampacity] = [calculated load] / [ambient correction factor] / [CCC in raceway adjustment]​
We work using this method. However, we must keep in mind we have not included yet a 125% adjustment for continuous loads.
[147,000 VA or 177 A] = [99,000 VA or 119 A] / 0.96 / 0.7​
Based on the result we choose 2/0 AWG Cu rated at 195 A @ 90?C as the conductors' size as 1/0 is rated at 170 A @ 90?C.

What is the load amperage adjusted for 125% continuous? 136 A.

Is 195 A > 136 A? Yes!

We also know we cannot exceed the table's 75?C rating for the same size conductor because of terminal ratings.

What is the Table 75?C rating for 2/0 AWG Cu? 175 A

Is 175A > 136 A? Yes!

We have met the requirement of the feeder conductors size being rated for the noncontinuous load plus 125% of the continuous load.
 
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Smart $:
Brilliant!; Awsome! I keep a file that I call "Master Moves" where I place developments like yours that I normally can't arrive at except after unusual amounts of thinking. Your answer is certainly going to make it into that file. Although I think that I eventually would have arrived at your solution, chances are that I would have never seen it. That is why I was reluctant to immediately assume that the code solution was wrong.

I appreciate the time that you took to go through the solution step-by-step. I want you to know that your time was not wasted. It will prevent me from making potentially dangerous mistakes in the future, when I will be applying this theory to real-life problems.

Thanks again

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Smart $ said:
I'll run through this one time...
Click to expand...
 
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