Calculating load for lighting

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dsbrown

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I've been discussing panel schedules in our office. Currently when filling out the load for lighting, the panel schedule that is being used in the offices takes the lighting at a demand factor 125% (you enter the VA load of the light fixtures on a circuit and the schedule multiplies that value by 1.25 to calculate the total VA load on the panel). The two sections that are in question (2005 NEC) are 220.42 and 220.20(A).

Some of the group contended that because of 220.20(A) stating that continuous load is multiplied by 125% to calculate the size of overcurrent protection serving the circuit (and lighting being a continuous load) that the 125% also applied to the VA load entered into the panel schedule.

The other faction contended that, refering to paragraph and table 220.42, that the demand factor used should be 100% (office building application) and that the VA value entered on the on the panel schedule should only be multiplied by one.

Are there any other sections that apply to this question (we didn't note any others that applied directly to this question)? And what is the value that lighting load VA needs to be multiplied by to calculate the total VA load on a panel?

Thanks in advance,
 
I will probably be corrected, but I beleive that you take the individual branch circuits X 1.25 when continuous and add them up. In the case of an office, there is no diversity allowed so you then multiply that answer by 100%.

If you had a hospital or a hotel/motel, you couldn't diversify the continuous load, because of the * that indicates continuous loads can't be diversified. You could diversify noncontinuous loads, but not continuous ones.

Don't forget that in spite of the load you installed you must use the unit values in 220.12 if they are higher than the actual load installed. If you have any type of energy conservation code, you probably can't install as much light as 220.12 requires you to plan for anyhow.

Finally my 2005 doesn't have a 220.20, did you mean 210.20?

Jim T
 
When calculating lighting loads in other than dwellings you generally multipy the VA demand of 220.12 based on the sq. ft. area, by 125%, article 220 deals with branch circuit feeder and service calculations. commercial and industrial lighting loads are considered as continuous . When you size the BC conductors and OCPD you technically reference 210.19 and 20,when you size feeders you technically apply 215.2 and 3 .and 230 .42 for sevices. that being said they pretty much all say the same thing. the minimum circuit capacity is set at 125% of the continuios load plus 100% of the non- continuios load. And the OCPD unless listed with the assembly it is installed in generally restricts the applied continuios load to 80% of the device continuios current rating(ignore spelling)

Charlie
 
cpal said:
When calculating lighting loads in other than dwellings you generally multipy the VA demand of 220.12 based on the sq. ft. area, by 125%, article 220 deals with branch circuit feeder and service calculations. commercial and industrial lighting loads are considered as continuous .

Charlie
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Charlie

I've never multiplied 220.12 numbers by 125%, although I regularly multiply actual wattages by 125% when continuous. Can you explain why the 220.12 numbers don't already have the 125% built into the table?

Jim T
 
Hi Jim,

The table does not include the 125% because not everything in the table is always a continuous load. The 125% is an option you have to add because you are familiar with the building and it's use. Same as a continuous fixed appliance verse one that is used only on occasion.
 
dsbrown said:
I've been discussing panel schedules in our office...
Click to expand...
Here’s how I’d do it (I think… first real attempt here, so please limit criticism to a constructive nature). First off, when doing panel schedules restrict oneself to Sections I and II of Article 220. The only reason for mentioning Section I is if you have to go outside Section II for specialized application loads as referenced in Table 220.3. Since we are talking about general illumination, we can restrict the discussion to Table 220.12 and 210.20(A)

First step for general illumination is Table 220.12 for the type of occupancy. In the case of office buildings the minimum load calculation is 3.5VA per sq. ft. For the sake of discussion, let’s use an actual area measure, say 24,000 sq. ft. Multiply that area by 3.5 and I get 84 kVA.

Now let’s say I have four subpanels handling lighting and we’ll make it easy in that each covers an area of approximately 6,000 sq. ft. Therefore each subpanel has to have available at least 21 kVA for general illumination. The lighting will be 277V, so 21 kVA divided by 277V is 75.8A.

Referring to 210.20(A), unless a 100% rated assembly is used (not), I multiply this value by 125% and get 94.7A. Typically, office lighting is run on 20A branch circuits, so I need a minimum of 5 branch circuits for the general lighting handled by this subpanel. If the actual load is greater than 75.8A, I would multiply the actual load by 125% and divide by 20A to determine the minimum number of 20A branch circuits. Now I divvy up the lighting to any reasonable number of circuits so long as that number is equal to or more than the determined minimum and does not exceed the maximum connected load of 16A on any one circuit.

As to how to specify these loads for the panel schedule, I'd use the actual connected load. Panel schedules are used to depict branch circuit loading, the total connected load on each phase, and an indication of how balanced the loads are amongst the phases (in addition to other items not directly related to this discussion). These values also serve as direct indicators for potential load additions in the future. If I want to depict demand factors and the resulting adjusted load, I'd do so in a table or table extension to one side. Feeder calculations, if shown, would be entirely separate as the load values used may not directly reflect the actual connected loads, subtotals, or totals in the panel schedule. For instance, in the above example, what if the general lighting load for one panel was only 65A. I would still have to calculate the feeder using the 75.8A minimum. If anyone can or has devised a panel schedule that indicates all of the information without being a total jumble of numbers, I’d like to see it :D
 
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jtester said:
Charlie

I've never multiplied 220.12 numbers by 125%, although I regularly multiply actual wattages by 125% when continuous. Can you explain why the 220.12 numbers don't already have the 125% built into the table?

Jim T
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An argument could be made if you read

220.42 General Lighting. The demand factors specified in
Table 220.42 shall apply to that portion of the total branchcircuit
load calculated for general illumination. They shall
not be applied in determining the number of branch circuits
for general illumination.
220.42 requires that continuios loads be increased by 25%i, for such calculations.


Don't forget the receptacle loads are addressed here

220.44 Receptacle Loads — Other Than Dwelling
Units. Receptacle loads calculated in accordance with
220.14(H) and (I) shall be permitted to be made subject to
the demand factors given in Table 220.42 or Table 220.44.


Charlie
 
websparky said:
Hi Jim,

The table does not include the 125% because not everything in the table is always a continuous load. The 125% is an option you have to add because you are familiar with the building and it's use. Same as a continuous fixed appliance verse one that is used only on occasion.
Click to expand...

Dave

210.20 the continuous load rule, which has been cited here, applies to branch circuits, not feeders, panels, services, etc. If you look at Feeder and Service Load Calculations, 220 Part III, it does not instruct you to use the loads determined from 210, it refers you to loads determined from 220 Parts I and II. I can't find a reference requiring Table 220.12 to be multiplied by 1.25.

Jim T
 
haskindm said:
Jim,
Take a look at 215.2A which states that continuous loads must be considered for feeders.
Click to expand...

I'm going to have to think about this for a while.

The current energy code only allows me to install 1 va/sq.ft in an office, and 220.12 requires me to calculate it at 3.5 va/sq.ft. I can buy multiplying my load, 1 va/sq ft if that is what I installed, by 1.25, but I am still struggling to believe that I need to multiply the 3.5 va/sq.ft. by 1.25.

The actual load that I installed is certainly continuous, but the 3.5 va/sq.ft. that is in the table isn't.

It's too early for a beer, I'm going to heat up my coffee and think about this a bit longer.

Jim T
 
Jim,
Just to keep you scratching your head...
What about show-window lighting and sign circuits? They are certainly continuous loads. It is generally accepted that the value for these in article 220 must be multiplied by 125% (1.25). Check out the examples in Chapter 9 (I know that the examples are not enforceable, but they give an idea of the thinking of the code making panel).
 
haskindm said:
Jim,
Just to keep you scratching your head...
What about show-window lighting and sign circuits? They are certainly continuous loads. It is generally accepted that the value for these in article 220 must be multiplied by 125% (1.25). Check out the examples in Chapter 9 (I know that the examples are not enforceable, but they give an idea of the thinking of the code making panel).
Click to expand...

The examples in Chapter 9 took the wind out of my sails. I think I'll take the rest of the day off, go build a new fly rod, and maybe feel better.

Thanks

Jim Tester
 
Smart $ said:
Now let?s say I have four subpanels handling lighting and we?ll make it easy in that each covers an area of approximately 6,000 sq. ft. Therefore each subpanel has to have available at least 21 kVA for general illumination. The lighting will be 277V, so 21 kVA divided by 277V is 75.8A.
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Well, after reading this above, I can tell you are not an electrical engineer! As a matter of fact, this is covered in 3rd. year apprenticeship classes. Which explains the rambling that follows the above.

On a constructive note: Respond to posts that you actually comprehend.
 
websparky said:
Well, after reading this above, I can tell you are not an electrical engineer! As a matter of fact, this is covered in 3rd. year apprenticeship classes. Which explains the rambling that follows the above.

On a constructive note: Respond to posts that you actually comprehend.
Click to expand...
03fzing.jpg

Nice burn, Zinger. :D
 
websparky said:
Well, after reading this above, I can tell you are not an electrical engineer! As a matter of fact, this is covered in 3rd. year apprenticeship classes. Which explains the rambling that follows the above.

On a constructive note: Respond to posts that you actually comprehend.
Click to expand...
I don't claim to be an electrical engineer though I see you do. If an attitude comparable to yours comes with the title, I know I don't ever want to be one! Nevertheless, I'll hazard a guess that you muttle through some issues all the same. Particularly, though just my humble opinion, you seem to be a bit short on civility. You certainly have quite a twist going on the meaning of constructive criticism. Are these requirements of the title, too?

To all, please provide a concise explanation of my error(s)? And please, restrict the discussion to the topic at hand.
 
To be honest, I'm not quite sure of the proper route on this. Smarter guys than I am are not clear either, so I won't feel bad for admitting it.

Smart, I believe the "4 panels" decision so early in the reasoning was incorrect. I would determine the lighting load required, and then from that determine the branch circuits required, and then see where that put me.

Also, the use of multiwire circuits in a 480/277 system would result in the numbers looking more like this:

3.5 kVA x 24,000 sq.ft. = 84000
84000 / 480V = 175A
175A / 20A = 8.75

Since this is representing groups of three 20A breakers, we'll multiply it back out:

8.75 x 3 = 26.25 ..... which would mean that we're looking at 27 breakers to fill 175A worth of "calculated lighting load". That wouldn't justify more than a panel. ;)

So goes the magic of three phase.

Edit to add:

This would be the NEC way. If the actual connected load was less, then we'd behave a little differently at the feeder and branch circuit level, but I believe the service would still have to have sufficient ampacity for 175A @ 480V for lighting loads.
 
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Dumb $,

I will extend the exact same courtesies that you have extended to me in previous post about "phases". Remember that thread? You know, the one where you would not answer other poster's questions about your still unfounded theories.

In this world of forums respect comes from honest replies. Like anywhere else in this life, respect is earned not merely granted. The way you have burst on the scene without identifying yourself leaves most of us to take you as you portay yourself, a jacka**.

If you want to be respected you first have to be trusted.

If you want to be educated, by a book and read-up on the subject.
 
jtester said:
The current energy code only allows me to install 1 va/sq.ft in an office, and 220.12 requires me to calculate it at 3.5 va/sq.ft. I can buy multiplying my load, 1 va/sq ft if that is what I installed, by 1.25, but I am still struggling to believe that I need to multiply the 3.5 va/sq.ft. by 1.25.

The actual load that I installed is certainly continuous, but the 3.5 va/sq.ft. that is in the table isn't.
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I see this going several different ways:
The 3.5 VA number is greater than the actual connected load.
In this case,
  • The actual connected load is multiplied by 125%
  • The actual connected load is compared to the 3.5 VA number
  • The greater of the two is applied to the service.
  • The actual connected load (x 125%) is used at the branch circuit / feeder level
  • The panel schedule uses this number for balancing, circuit selection, etc

The 3.5 VA number is less than the actual connected load.
In this case,
  • The actual connected load is multiplied by 125%
  • This corrected actual load is used in the general lighting load portion of the service
  • The rest is the same as the prior scenario.
You would never multiply the 3.5 VA number by 125% in either scenario, as it isn't required by any section.

Thoughts?
 
georgestolz said:
To be honest, I'm not quite sure of the proper route on this. Smarter guys than I am are not clear either, so I won't feel bad for admitting it.

Smart, I believe the "4 panels" decision so early in the reasoning was incorrect. I would determine the lighting load required, and then from that determine the branch circuits required, and then see where that put me.

Also, the use of multiwire circuits in a 480/277 system would result in the numbers looking more like this:

3.5 kVA x 24,000 sq.ft. = 84000
84000 / 480V = 175A
175A / 20A = 8.75

Since this is representing groups of three 20A breakers, we'll multiply it back out:

8.75 x 3 = 26.25 ..... which would mean that we're looking at 27 breakers to fill 175A worth of "calculated lighting load". That wouldn't justify more than a panel. ;)

So goes the magic of three phase.

Edit to add:

This would be the NEC way. If the actual connected load was less, then we'd behave a little differently at the feeder and branch circuit level, but I believe the service would still have to have sufficient ampacity for 175A @ 480V for lighting loads.
Click to expand...
I'm not getting your method. Your method actually ends up requiring more single pole branch circuits (27) than mine (20 total for 4 panels x 5-20A BC's/panel), and that is even after rounding mine up 4 times.

As for "That wouldn't justify more than a panel." it was not my intention to imply general illumination was the only circuits handled by these four panels... there would most likely be other 480/277 utilization equipment connected to these panels.

As an aside, thank you for the civil response :)
 
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