Master's Class Questions

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waywardson29570

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I am taking a Master's class at a vocational school and the following questions have been presented... rather vague... but want to make sure I am learning correctly.

Calculate the minimum VA capacity of a 155,000 sq ft school...

I understand that it's 155,000 x 3VA/sq' which equals 581,250 VA @ 1.25% for cont. load.

now... do I use Table 220.86 from here and get:
581,250/155,000 = 3.75/sq'
-3VA @ 100% = 3VA
-.75 x .75 = .5625
Total 3.5625/sq' x 155,000 = 522,187.5 VA (I want to make sure I'm not reading into and taking things too far based on this question)

Attached is the next question:

Part of this school would consist of a small building 50' x 60'. It is requested to install fluorescent light fixtures to illuminate the interior. If each fixture is to illuminate 60 sq', how many light fixtures will have to be installed?

This sounds straight forward to me...

50 x 60 sq' = 3000 sq'
3000 sq' / 60 sq'/fixture = 50 fixtures

My instructor; however, believes the 1.25% needs to come in to play here and I cannot grasp his logic.

Can anyone direct me further? Thanks in advance.
 
waywardson29570 said:
I am taking a Master's class at a vocational school and the following questions have been presented... rather vague... but want to make sure I am learning correctly.

Calculate the minimum VA capacity of a 155,000 sq ft school...

I understand that it's 155,000 x 3VA/sq' which equals 581,250 VA @ 1.25% for cont. load.

now... do I use Table 220.86 from here and get:
581,250/155,000 = 3.75/sq'
-3VA @ 100% = 3VA
-.75 x .75 = .5625
Total 3.5625/sq' x 155,000 = 522,187.5 VA (I want to make sure I'm not reading into and taking things too far based on this question)

Attached is the next question:

Part of this school would consist of a small building 50' x 60'. It is requested to install fluorescent light fixtures to illuminate the interior. If each fixture is to illuminate 60 sq', how many light fixtures will have to be installed?

This sounds straight forward to me...

50 x 60 sq' = 3000 sq'
3000 sq' / 60 sq'/fixture = 50 fixtures

My instructor; however, believes the 1.25% needs to come in to play here and I cannot grasp his logic.

Can anyone direct me further? Thanks in advance.
Click to expand...


Can't help with the first question as I don't do large commercial projects and therefore haven't studied that in a long time.

The second question, if the lights are going to be on 3 hours or longer then the load is considered a continuous load and would need the 125%. That's probably what your instructor is getting at. Most commercial/non dwelling lighting is considered continuous.
 
I appreciate your reply.... I understand the whole concept of the 1.25 for cont. load.... but the lighting question isn't referring to load... just square feet. My thinking is sq ft is sq ft no matter whether it's continuous or not in regards to how many luminaries are needed to cover the total sq footage...
 
waywardson29570 said:
I appreciate your reply.... I understand the whole concept of the 1.25 for cont. load.... but the lighting question isn't referring to load... just square feet. My thinking is sq ft is sq ft no matter whether it's continuous or not in regards to how many luminaries are needed to cover the total sq footage...
Click to expand...

But when asking for the VA capacity, the instructor could either be asking for the total VA of the lighting appliances themselves, or he could be asking for the calculated wiring capacity, where the VA is the service voltage times the supply/feeder ampacity. (Well, with appropriate factors for three phase, etc., but the key is that the VA sought may be the nominal supply VA, not the load VA.)
 
Thanks golddigger... maybe I am really looking at this with blinders on at the moment... I say that because I cannot see how the 1.25% comes in to play with the question of how many fixtures are required to cover a certain square footage when the square footage of the fixture is given. The question presented is not any type of load calculation just a sq footage with a sq footage divider...

very confused.... sorry....
 
waywardson29570 said:
... I cannot see how the 1.25% comes in to play with the question of how many fixtures are required to cover a certain square footage when the square footage of the fixture is given. The question presented is not any type of load calculation just a sq footage with a sq footage divider...
Click to expand...
Me neither. My question to the instructor would be, "125% of what?"

waywardson29570 said:
... very confused.... sorry....
Click to expand...
I don't think you are. What we have here is a failure to communicate - on the part of your instructor. If the instructor wants an answer in VA, the question should say that, not number of fixtures.

You could be a nice guy and gently lead him back to the reality of direct communication.

ice
 
waywardson29570 said:
Thanks golddigger... maybe I am really looking at this with blinders on at the moment... I say that because I cannot see how the 1.25% comes in to play with the question of how many fixtures are required to cover a certain square footage when the square footage of the fixture is given. The question presented is not any type of load calculation just a sq footage with a sq footage divider...

very confused.... sorry....
Click to expand...
As you stated the question in your first post, it asked for the required VA, not for the number of fixtures. Hence the room for either interpretation.

Now if there is a minimum light level required by building codes or contract, then it might make sense to put in a 125% allowance for the degradation of the lamps with time before you eventually replace them. But that would be a different 125% from the one associated with a continuous load.
 
yessir... I can understand from the first question giving the sq footage of the building... multiplying it by the 3VA/sq ft... then multiplying it by the 1.25% for the cont. load. That is clear to me. Now... once again... to get the VA capacity of the school... as the question stated, do I apply table 220.56 and break it down like I showed in my original post? Or, did I take it too far. My instructor couldn't advise me either way. His answer just stopped at the 3VA/sq' x 1.25. My additional came from a google search and got me in contact with electrical-knowhow.com... If this would be a state license test question... I don't want to take things further than necessary and possibly have the answer counted wrong. thanks again....
 
GoldDigger said:
As you stated the question in your first post, it asked for the required VA, not for the number of fixtures. Hence the room for either interpretation. ....
Click to expand...

waywardson29570 said:
... the following questions have been presented...

... Calculate the minimum VA capacity of a 155,000 sq ft school...

Attached is the next question:

... how many light fixtures will have to be installed? ....
Click to expand...

Hummmm .... Room? Where?

ice
 
waywardson29570 said:
I am taking a Master's class at a vocational school and the following questions have been presented... rather vague... but want to make sure I am learning correctly.

Calculate the minimum VA capacity of a 155,000 sq ft school...

I understand that it's 155,000 x 3VA/sq' which equals 581,250 VA @ 1.25% for cont. load.

now... do I use Table 220.86 from here and get:
581,250/155,000 = 3.75/sq'
-3VA @ 100% = 3VA
-.75 x .75 = .5625
Total 3.5625/sq' x 155,000 = 522,187.5 VA (I want to make sure I'm not reading into and taking things too far based on this question)

Attached is the next question:

Part of this school would consist of a small building 50' x 60'. It is requested to install fluorescent light fixtures to illuminate the interior. If each fixture is to illuminate 60 sq', how many light fixtures will have to be installed?

This sounds straight forward to me...

50 x 60 sq' = 3000 sq'
3000 sq' / 60 sq'/fixture = 50 fixtures

My instructor; however, believes the 1.25% needs to come in to play here and I cannot grasp his logic.

Can anyone direct me further? Thanks in advance.
Click to expand...

3VA/square foot comes from 220.12, which is the general lighting load. There is nothing in there that mentions adding 25% for continuous load, the load is 3VA times number of square feet. I would even go further and say that the Schools portion of the list only applies to general portions of your building and other specific portions like hallways/corridors or an auditorium if you have one need applied with different values. (look at the last listing in table 220.12).

Your second question is not a code question at all - it is just a simple math question of how many fixtures are needed if one lights up a certain defined area. No idea why any 125% factor would even be a part of this solution either. Your instructor is effectively saying the 60 square feet per fixture only gets 80% of illumination effectively done - I guess - makes no sense to me why??

Or he is jumping ahead with the 125% to load calculations - which are not a part of the question at this point.
 
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