Commercial Calculation 2

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jca0108

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I'm having trouble calculating the general lighting and receptacles loads for banks and office buildings. In the codebook, under 220.14 (k) it tells me to either use 180v A for each receptacle or use 1vA per square ft. That I understand,in Mike's exam prep book on page 466, there is a example of 220.14 (k). I also understand that. What I dont understand is when I took the calculation practice test questions, 25 and 26 on page 486. I know there is a typo in answer choice b. I followed 220.14 (k) and dont get the same thing. The answer key added 220.14 (k)(1) and (2) together. Why?

Thanks,

Juan
 

Dennis Alwon

Moderator
Staff member
Location
Chapel Hill, NC
Occupation
Retired Electrical Contractor
jca0108 said:
The answer key added 220.14 (k)(1) and (2) together. Why?

Thanks,

Juan

I don't have this book to follow your example but I would agree with you. It is either (K)(1) or (K)(2) not both.
 

jca0108

Member
Question # 25 : What is the service calculated load for the general lighting and receptacles for a 30,000 sq ft bank?
(A) 111kVA
(B) 162kVA ---This is the answer
(C) 123kVA
(D) 175kVA

My answer: 30,000-sq ft
* 1VA- 220.14 (K)(2)
_________
30,000 VA

Since the question doesnt give a number of receptacles, I cant use 220.14 (K)(1). I feel the question is missing something . Because how can 30,000 sq ft be used for lighting and receptacle loads. None of his examples in the book , or DVD, cover that angle of only having the square footage, and solving the problem.
 

Smart $

Esteemed Member
Location
Ohio
Hmmm.... the question asks for general lighting, too

30,000 sqft ? 3.5 VA/sqft = 105,000 VA for general lighting—Table 220.12
30,000 sqft ? 1.0 VA/sqft = 30, 000 VA for receptacles—220.14(K)(2)

Total =135,000 VA
 

Dennis Alwon

Moderator
Staff member
Location
Chapel Hill, NC
Occupation
Retired Electrical Contractor
Smart $ said:
Hmmm.... the question asks for general lighting, too

30,000 sqft ? 3.5 VA/sqft = 105,000 VA for general lighting?Table 220.12
30,000 sqft ? 1.0 VA/sqft = 30, 000 VA for receptacles?220.14(K)(2)

Total =135,000 VA

That would explain it. :grin:
 

ray cyr

Senior Member
Location
Yakima, Wash.
Remember continuous duty on the general lighting:
[30,000(sq.ft.)x3.5(from 220.12)x1.25(cd)=131250va]+30,000(220.14k)= 161250
the book answer is rounded up to 162kva
 

cadpoint

Senior Member
Location
Durham, NC
Smart $ said:
Hmmm.... the question asks for general lighting, too

30,000 sqft ? 3.5 VA/sqft = 105,000 VA for general lighting—Table 220.12
30,000 sqft ? 1.0 VA/sqft = 30, 000 VA for receptacles—220.14(K)(2)

Total =135,000 VA

No,
Lights VA +125% = (continious)
105000 + 125% = 131250 +30,000 = (B) 162kVA ---This is the answer

Seems Mike got it right ... 210.20(A) Continious and Noncontinuous Loads

Via
220.3 Application of Other articles
 
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Smart $

Esteemed Member
Location
Ohio
cadpoint said:
No,
Lights VA +125% = (continious)
105000 + 125% = 131250 +30,000 = (B) 162kVA ---This is the answer

Seems Mike got it right ... 210.20(A) Continious and Noncontinuous Loads

Via
220.3 Application of Other articles
I don't know how many times I have to say it, "LOAD" is not "AMPACITY".

You only multiply by 125% to determine "AMPACITY".

The question asks for the "service calculated load".

Article 220 tells you exactly how to determine the general lighting and receptacle loads. Nowhere does it require any number to be multiplied by 125% in determining the load.
 

cadpoint

Senior Member
Location
Durham, NC
OK, You can write Mike.

If you want to hold to that and lights are not being continous, fine.

Its not how I was taught.

If I used the wrong Technical Term and voiding anything by its usage of whats said in the NEC 08' Code, excuse me.
Oh and I added 125% not multiplied...
220.3
 
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ray cyr

Senior Member
Location
Yakima, Wash.
Smart $ said:
I don't know how many times I have to say it, "LOAD" is not "AMPACITY".

You only multiply by 125% to determine "AMPACITY".

The question asks for the "service calculated load".

Article 220 tells you exactly how to determine the general lighting and receptacle loads. Nowhere does it require any number to be multiplied by 125% in determining the load.

So calculating the load and calculating the size of the service entrance conductors are two different calculations? OK, so to get this answer the question should have been "What is the service calculated load for the service entrance conductors for the general lighting and receptacles for a 30,000 sq ft bank?
 

ray cyr

Senior Member
Location
Yakima, Wash.
Isn't that kind of implied in the title of the article: Article 220 Branch-Circuit, Feeder, and Service calculations? Don't we calculate the size of these things? When we calculate the size of these don't we multiply continuous duty by 1.25?
I'm actually not trying to be a smart aleck, I am trying to learn from people who are much more knowledgable than I, to do that I have to ask questions.:smile:
 

Smart $

Esteemed Member
Location
Ohio
ray cyr said:
Isn't that kind of implied in the title of the article: Article 220 Branch-Circuit, Feeder, and Service calculations? Don't we calculate the size of these things? When we calculate the size of these don't we multiply continuous duty by 1.25?
I'm actually not trying to be a smart aleck, I am trying to learn from people who are much more knowledgable than I, to do that I have to ask questions.:smile:
You are correct. It is implied. But you have to consider the wording of the question. What if it were a two part question that asked first for the calculated load and then secondly asked for the service conductor ampacity, would you give the same answer?
 

Smart $

Esteemed Member
Location
Ohio
ray cyr said:
So calculating the load and calculating the size of the service entrance conductors are two different calculations?
Not different calculations but rather one is part of the other... or a prerequisite. You have to calculate the service load to determine the required minimum size of the service conductors. For the latter, you multiply continuous loads by 125% and add the non-continuous loads to arrive at a minimum ampacity rating for service conductors for which the service equipment is not rated for 100% continuous operation (which is essentially always).

OK, so to get this answer the question should have been "What is the service calculated load for the service entrance conductors for the general lighting and receptacles for a 30,000 sq ft bank?
I would have worded it something like, "What is the required minimum ampacity of service conductors for general lighting and receptacles of a 30,000 sqft bank?" But that is only if I was after the answer provided. I would never ask such a question because we all know that these two amassed loads will likely never be the only two loads on a service at a bank (unless it just happens to be one of those basic service ones that are inside another store).
 

Dennis Alwon

Moderator
Staff member
Location
Chapel Hill, NC
Occupation
Retired Electrical Contractor
Smart $ said:
I would have worded it something like, "What is the required minimum ampacity of service conductors for general lighting and receptacles of a 30,000 sqft bank?" But that is only if I was after the answer provided. I would never ask such a question because we all know that these two amassed loads will likely never be the only two loads on a service at a bank (unless it just happens to be one of those basic service ones that are inside another store).
What about asking to provide the req. minimum ampacity of a feeder conductor for general lighting and recep. of a 30,000 sq. ft. bank? What would you do then? Would you put in the 125%
 

Smart $

Esteemed Member
Location
Ohio
cadpoint said:
OK, You can write Mike.

If you want to hold to that and lights are not being continous, fine.

Its not how I was taught.

If I used the wrong Technical Term and voiding anything by its usage of whats said in the NEC 08' Code, excuse me.
Oh and I added 125% not multiplied...
220.3
If you add, you add 25%... not 125%?for which you will end up with 225% :D

Lighting is considered a continuous load... I'm not saying anything to the contrary... really!!! A load calculation determines the load in volt-amperes. It does not determine minimum ampacity of conductors. That requires additional calculation.
 

Smart $

Esteemed Member
Location
Ohio
Dennis Alwon said:
What about asking to provide the req. minimum ampacity of a feeder conductor for general lighting and recep. of a 30,000 sq. ft. bank? What would you do then? Would you put in the 125%
Yes... :wink:

PS: ...but I would convert to amperes :D
 
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