HVAC Unit loads and demand factors

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david luchini said:
I'd rather use the #3 feeder conductors for an actual current of 94.5A.
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I just want to specify the amount that is going to work per the code. As an engineer, I am not paying for the #2 or #3, unless I specify the smaller amount and it is found out to not meet code, then I will have to pay for it, the client will be unhappy, someone can get hurt, and property damage can happen. So that is why it is important to me to get the code required size.

David, just curious if you are an engineer, electrician, code inspector, or something else?
 
jjs said:
I just want to specify the amount that is going to work per the code. As an engineer, I am not paying for the #2 or #3, unless I specify the smaller amount and it is found out to not meet code, then I will have to pay for it, the client will be unhappy, someone can get hurt, and property damage can happen. So that is why it is important to me to get the code required size.

David, just curious if you are an engineer, electrician, code inspector, or something else?
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I'm an engineer. Nothing prevents you from calculating the load at higher than actual. Of course you can't go smaller than the required calculated load.

In my last example, you could run a #2, or #1 or #1/0 feeder conductor. The Code required size would be #3, based on the load calc from 430.24.
 
david luchini said:
I'm an engineer.
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As long as you stuck with the back and forth, I should have figured that. :)
My kids got me a t-shirt about arguing with an engineer and wrestling with a pig. What most people don't realize is that engineers can come to different conclusions and not get mad at the other person. We might think they are completely wrong in their conclusion, but we don't need to take it personally and start calling people names that disagree with us. So far I have not been convinced enough to change how I currently do it. At worst my service and feeder MAY be oversized slightly. I am happier with that conclusion than it being undersized and there being a problem. With the way there are incremental sizes of equipment, on most jobs it probably will not make the equipment a different size either way. I appreciate your time. I am disappointed that only one other person besides you jumped in to give their perspective on it. I would still like to hear from some more members, maybe they can add something new to the discussion that neither of us has brought up so far. I like to have things completely settled in my mind, I don't care if my original way was the right way or the wrong way, I just want to move forward with the right way. I hope that is clear as mud. Again, thanks for your time, having the discussion is easiest way for me to flesh out the answer.
 
david luchini said:
220.50 directs you to 430.24 for the calculation of motor loads for feeders and services.

430.24 tells you to take 125% of the largest motor (as determined by 430.6(A)) plus the sum of the full load currents of the other motors (as determined by 430.6(A)) as the load for several motors.

430.6(A) tells you that the current rating of the motor shall be determined by the Tables 430.247 thru 430.250.

440.6 MODIFIES the provisions of 430.6 when dealing with Air-Conditioning and Refrigerating Equipment. (If you recall we talked about how Art 430 applies to Art 440 equipment, except as modified by Art 440.)

440.6(A) says that the Rated Load Current of the compressor shall be used as the current rating (instead of the current rating from Tables 430.247 thru 430.250.)

440.6(B) says that the marked full load current of fans or blowers in Art 440 Equipment shall be used as the current rating (instead of the current rating from Tables 430.247 thru 430.250.)


So, imagine you had a 480V feeder supplying a 5HP pump, a 1HP fan, and a piece of Air Conditioning Equipment that had a 5HP compressor with an RLA of 6.5A and a 1HP blower with an FLA of 1.8A.

Per 220.50 and 430.24 (and 440.6) the load on the feeder would be:

5HP pump = 7.6A*1.25 = 9.5A
1HP fan ......................= 2.1A
5HP Comp ..................= 6.5A
1HP blower .................= 1.8A
Total.......................... = 19.9A

The MCA of the Air Conditioning Equipment (which would be 9.93A) does not figure into the load calculation for the feeder at all.
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I'm a newbie but seems both of you JJS & David are correct. Where ever you loop it will come up with almost the same answer.

440.35 will direct you to 440.4(B) which will direct you to 440.34 which will then direct you to 440.33 that says (Highest RLA x 0.25) + all RLA & FLA

while 430.24 says (Highest FLA x 1.25) + all FLA in the group.

I.E
Comp 1 = 12.20
Comp 2 = 16.70
OFM 1 FLA = 3.90
OFM 2 FLA = 3.90
IFM FLA = 6.40

Solution 1, using 440.33
= (16.7 x 0.25) + 12.20 +16.70 + 3.90 +3.90 + 6.40
= 46.15

Solution 2, using 430.24
=(16.7 x 1.25) + 12.20 + 3.90 + 3.90 + 6.40
=47.15

This is a typical PACU which compressors are designed either with the same rating (RLA) to balance the loading conditions or with just a minimal difference such in given example.
 
david luchini said:
Your math is wrong. Both of these solutions equal 47.275.
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The equation using 440.33 with the resultant figure (46.15) is correct in terms of the given numbers presented in the equation.

However, OP (either a typo) or misreading the article (430.24.) had written the multiplier 0.25 instead of 1.25.

The math is correct, the way it was presented in terms of how the article should be presented is different. (0.25 vs 1.25) .
I would be glad to hear from OP regarding this.

This borders on pedantry. . . and you know better than that.
 
myspark said:
The equation using 440.33 with the resultant figure (46.15) is correct in terms of the given numbers presented in the equation.
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No, it's not.

myspark said:
However, OP (either a typo) or misreading the article (430.24.) had written the multiplier 0.25 instead of 1.25.
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Replacing the 1.25 with 0.25 in the 430.24 solution would give a drastically wrong answer.

myspark said:
B]The math is correct, the way it was presented in terms of how the article should be presented is different. (0.25 vs 1.25) .[/B]
I would be glad to hear from OP regarding this.
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The math is wrong...the way it is presented in terms of how the article is presented IS different.

I'd be glad to hear from the OP as well. His point seemed to be that using two different methods would produce "almost the same answer."
In fact, he presented the same equation in two different manners and somehow came up with different answer each time. Neither of which answer is correct.
 
david luchini said:
No, it's not.



Replacing the 1.25 with 0.25 in the 430.24 solution would give a drastically wrong answer.



The math is wrong...the way it is presented in terms of how the article is presented IS different.

I'd be glad to hear from the OP as well. His point seemed to be that using two different methods would produce "almost the same answer."
In fact, he presented the same equation in two different manners and somehow came up with different answer each time. Neither of which answer is correct.
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It was a misplaced cells in excel in placing the formula. Thank you for the correction, both of them are 47.275.
 
david luchini said:
No, it's not.



Replacing the 1.25 with 0.25 in the 430.24 solution would give a drastically wrong answer.



The math is wrong...the way it is presented in terms of how the article is presented IS different.

I'd be glad to hear from the OP as well. His point seemed to be that using two different methods would produce "almost the same answer."
In fact, he presented the same equation in two different manners and somehow came up with different answer each time. Neither of which answer is correct.
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You are just parroting what I already said.
The math is correct. . . the "ipso facto" details that OP used does not match with the NEC guidelines.

I am eager to hear from OP. . . and there is no need for resorting to punitive insinuations.
 
myspark said:
You are just parroting what I already said.
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I haven't parroted what you have said...I've disagreed with you.

myspark said:
The math is correct. . . the "ipso facto" details that OP used does not match with the NEC guidelines.
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I'm still disagreeing with you...as the math was not correct. Fortunately, the OP has discovered his error. Also, statements of fact are not "punitive insinuations."
 
The way I interpret 440.33 is below

1. Sum of all RLA (or branch circuit selection whichever is greater) of all motor compressors
2. Sum of FLC of all other motors (indoor & out door)
3. and 25 % of highest motor compressor or FLC in the group

Item 1 = need to add all RLA of all motor compressors involved
Item 2 = need to add all FLC fan motors involved
Item 3 = says 25% of highest RLA or FLC. (It didn't say 125% of the highest)

If my interpretation is correct then the collective discussion from JJ & David seems to be on the same page. (Based on a typical PACU nameplate data I got)
 
myspark said:
However, OP (either a typo) or misreading the article (430.24.) had written the multiplier 0.25 instead of 1.25.

The math is correct, the way it was presented in terms of how the article should be presented is different. (0.25 vs 1.25) .
I would be glad to hear from OP regarding this.
Click to expand...


Comp 1 = 12.20
Comp 2 = 16.70
OFM 1 FLA = 3.90
OFM 2 FLA = 3.90
IFM FLA = 6.40

Solution 1, using 440.33
= (16.7 x 0.25) + 12.20 +16.70 + 3.90 +3.90 + 6.40
= 47.275

Solution 2, using 430.24
=(16.7 x 1.25) + 12.20 + 3.90 + 3.90 + 6.40
=47.275

I didn't use 0.25 multiplier in 430.24.
 
david luchini said:
I haven't parroted what you have said...I've disagreed with you.


I'm still disagreeing with you...as the math was not correct. Fortunately, the OP has discovered his error. Also, statements of fact are not "punitive insinuations."
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OP hasn't responded and for you to make that harsh assessment (that he is wrong) is just another one of your hypocritical remarks . . . as always.
You can disagree with me come hell and high water. . . but it doesn't mean your statement is valid.
 
Thomas&Kulas said:
Comp 1 = 12.20
Comp 2 = 16.70
OFM 1 FLA = 3.90
OFM 2 FLA = 3.90
IFM FLA = 6.40

Solution 1, using 440.33
= (16.7 x 0.25) + 12.20 +16.70 + 3.90 +3.90 + 6.40
= 47.275

Solution 2, using 430.24
=(16.7 x 1.25) + 12.20 + 3.90 + 3.90 + 6.40
=47.275

I didn't use 0.25 multiplier in 430.24.
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No, you didn't use 0.25 multiplier in 430,24. You used it in 440.33.
Read your first equation.

Mabuhay!
 
myspark said:
You can disagree with me come hell and high water. . . but it doesn't mean your statement is valid.
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I'm sorry that you have problems with facts. Surely you are in jest when you suggest that statements of fact are "hypocritical remarks."


I'm afraid I will have to disagree with again. You said this in post #53...

myspark said:
OP hasn't responded and for you to make that harsh assessment (that he is wrong)
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The OP said this in post #48...

Thomas&Kulas said:
It was a misplaced cells in excel in placing the formula. Thank you for the correction, both of them are 47.275.
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The fact is, the OP responded and admitted that he was wrong. I'm not really sure why you feel the need to argue this.

(16.7 x 0.25) + 12.20 +16.70 + 3.90 +3.90 + 6.40 is fairly simple arithmetic...I would say it is indisputable that that expression does not reduce to 46.15.
 
Thomas&Kulas said:
If my interpretation is correct then the collective discussion from JJ & David seems to be on the same page. (Based on a typical PACU nameplate data I got)
Click to expand...

I think you've misunderstood what JJS was arguing. He was concerned with FEEDER sizing, not with Branch Circuit sizing.
 
Thomas&Kulas said:
I'm a newbie but seems both of you JJS & David are correct. Where ever you loop it will come up with almost the same answer.

440.35 will direct you to 440.4(B) which will direct you to 440.34 which will then direct you to 440.33 that says (Highest RLA x 0.25) + all RLA & FLA

while 430.24 says (Highest FLA x 1.25) + all FLA in the group.

I.E
Comp 1 = 12.20
Comp 2 = 16.70
OFM 1 FLA = 3.90
OFM 2 FLA = 3.90
IFM FLA = 6.40

Solution 1, using 440.33
= (16.7 x 0.25) + 12.20 +16.70 + 3.90 +3.90 + 6.40
= 46.15

Solution 2, using 430.24
=(16.7 x 1.25) + 12.20 + 3.90 + 3.90 + 6.40
=47.15

This is a typical PACU which compressors are designed either with the same rating (RLA) to balance the loading conditions or with just a minimal difference such in given example.
Click to expand...

Same thing was calculated in both those examples.

First one has 16.7 x 0.25, then later adds 16.7
Second one has 16.7 x 1.25
The rest of the each example are identical

(16.7 x .25) + 16.7 gives you same result as 16.7 x 1.25.

For some reason he had a math error in both examples as the answer to both should have been 47.275.
 
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