General Lighting

[cbuck025]

Once the total VA is calculated using the volt/sqft table how is it finished to determine your general lighting load and the number of CB's you will need? Does this include your receptacles?

Example: 2100 sqft home @ 3VA/sqft equals 6300VA

voltage: 120/240 single phase
General lighting load equals 26.25 amps or 52.5 amps, what exact voltage should i use? 120 or 240. This does not include my receptacles does it? Or is this just telling me that I can use 2 15Amp breakers to satisfy my general lighting load? Is there something generic that says how many 15 Amp receptacles are permitted on a 20Amp circuit? In the industry, is it standard practice to wire the receptacles in series or parallel. Thanks for your help, just trying to brush up on things. The military doesn't give me the chance to use my NEC knowledge too often.
Thanks, Chris

 

[wyatt]

Re: General Lighting

divied the total 6300va by 1800va for 15 amp and 2400va for 20 amp don't forget to add 125% to the total 6300va if the lights are contiuous load.

[ February 14, 2006, 09:18 PM: Message edited by: wyatt ]

 

[wyatt]

Re: General Lighting

in a dweling genral use recpticals are included in this calculation. you could in therorey have 1000 recpt on one 15 amp braker. as long as you have the min circuits for the SQ foot and they are balanced.

 

[infinity]

Re: General Lighting

as long as you have the min circuits for the SQ foot and they are balanced.

What do you mean by balanced? Is this a code requirement or a design option?

 

[wyatt]

Re: General Lighting

Take a look at 210.11 b see if you get the same thing I get from it?

 

[ramdiesel3500]

Re: General Lighting

Wyatt
I never really thought about 210.11(B) that way, but now that you mention it, I believe that is what it means. If you evenly distribute the load, it should be balanced when you are done. It would have been nice if they had explicitly stated that the load should be evenly divided between phases.

 

[charlie b]

Re: General Lighting

It's a bit convoluted, but here's the path to your answer:
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• <font size="2" face="Verdana, Helvetica, sans-serif">210.11 tells you to provide branch circuits for the loads that are calculated per 220.10.</font>

<font size="2" face="Verdana, Helvetica, sans-serif"></font>

• <font size="2" face="Verdana, Helvetica, sans-serif">220.10 says to calculate loads per 220.12, 220.14, and 220.16.</font>

<font size="2" face="Verdana, Helvetica, sans-serif"></font>

• <font size="2" face="Verdana, Helvetica, sans-serif">220.12 is where the "3 watts per square foot" comes into play.</font>

<font size="2" face="Verdana, Helvetica, sans-serif"></font>

• <font size="2" face="Verdana, Helvetica, sans-serif">220.14(J) says that the general purpose receptacle, the bathroom receptacles, and a few others (but not the kitchen and not the laundry) are included in the "3 watts per square foot."</font>

<font size="2" face="Verdana, Helvetica, sans-serif"></font>

• <font size="2" face="Verdana, Helvetica, sans-serif">220.16 is for adding loads to existing buildings, so lets ignore it for now.</font>

<font size="2" face="Verdana, Helvetica, sans-serif">That tells me that in your hypothetical 2100 square foot home, the 6300 VA can be used to power all the lights and general purpose receptacles. Each circuit that supplies these loads will be a 120 volt circuit, so you divide the 6300 VA by 120 volts, to get a total of 52.5 amps. That 120 volts is the basis of the 1800 VA (for a 15 amp circuit) and the 2400 (for a 20 amp circuit) that wyatt mentioned.

Now I am going to deviate from code requirements, and add a touch of design choice. You can avoid any question of whether a load is or is not continuous, and whether you do or do not have to account for an extra 25% of its load, by choosing to limit the circuit loading to 80% of its rating. That would be 12 amps for a 15 amp circuit and 16 amps for a 20 amp circuit. So if you are using 15 amp circuits, you will need at least 5 (i.e., 52.5 / 12 = 4.375, round up to 5), and if you are using 20 amp circuits, you will need at least 4 (i.e., 52.5 / 16 = 3.28, round up to 4).

 

[wyatt]

Re: General Lighting

First I would like to say that from a design point our the way I would lay out a house is not as I have said in the post I was trying to point out that the code dose a bad job on res. branch circuit loads. No house is equaly loaded through out bed room recpt are low. Lighting load very, a 12X12 room can have 120va to 1000va. but 12x12=144x3=432va?