How many receptacles to install in an area on a circuit?

[ritelec]

Laying out circuits residentially I would figure a receptacle at 180 va..

well found out a couple years back that was not correct.

The amount of receptacles installed are based on the area as per load calculations for general lighting @ 3va.

That being said. How do you do that?

Guessing here....

Lets say I want to run a 15 amp circuit.

15a x 120v = 1800va

1800va / 3va = 600sq feet

25' x 24' = 600 sq feet

If I where just installing receptacle outlets, would that mean that if I where placing receptacles in a 25' x 24' (600 sq ') room that circuit could not leave that room and feed other rooms?

Or that one circuit could feed receptacles in four 12.5' x 12' ( 150 sq ' x 4 = 600 sq ' ) rooms? (someone please help with the math here, my math is showing four rooms but it looks like 25 and 24 halfed at 12.5 and 12 should be two rooms not four ???)

Is this correct?

Thank you

 

[macmikeman]

3 watts per square foot residential is for calculating the service size. 180 volt amps per receptacle outlets is for general purpose non dwelling installations.

Here is your mactip of the day! Nobody says you can't use the 180 volt amps for your outlet numbering , and it isn't a bad number usually, but remember any fool can overload anything if they are a big enough fool.

 

[ActionDave]

The 3VA/ft' tells you how many circuits you need. You can put as many recpts as you want on a circuit.

 

[jeremy.zinkofsky]

You don't need to do it that way. There are no restrictions on where to place receptacles in commercial facilities or how to circuit them. The 3VA per sq ft. Is only a guideline. You, of course, are limited on the number of receptacles that can be put on a single circuit and that's why the NEC assumes a minimum 180VA per receptacle. Just figure out how many receptacles you are allowed to put on a given breaker and place them wherever you want. So, using your example, a 15A breaker loaded to 80% (by code) allows you 12A. Using S = VI (or S = VI*sqrt3 for three phase) you will have 120V * 12A = 1440VA at your disposal. So you can only put 1440VA / 180VA = 8 outlets on 15A breaker.

 

[ritelec]

The 3VA/ft' tells you how many circuits you need. You can put as many recpts as you want on a circuit.

ok......

So in that 25 by 24 foot room, 1 circuit is required for just that room right? I can put 100 receptacles on that circuit but just in that one room. If I add another receptacle in another area it shouldn't be on that circuit.
Is that correct?

 

[macmikeman]

You don't need to do it that way. There are no restrictions on where to place receptacles in commercial facilities or how to circuit them. The 3VA per sq ft. Is only a guideline. You, of course, are limited on the number of receptacles that can be put on a single circuit and that's why the NEC assumes a minimum 180VA per receptacle. Just figure out how many receptacles you are allowed to put on a given breaker and place them wherever you want. So, using your example, a 15A breaker loaded to 80% (by code) allows you 12A. Using S = VI (or S = VI*sqrt3 for three phase) you will have 120V * 12A = 1440VA at your disposal. So you can only put 1440VA / 180VA = 8 outlets on 15A breaker.

OP states it is residential in his first post. The Nec does not require 180 va minimum in dwellings and your post is correct if it was a commercial building.

 

[ritelec]

You don't need to do it that way. There are no restrictions on where to place receptacles in commercial facilities or how to circuit them. The 3VA per sq ft. Is only a guideline. You, of course, are limited on the number of receptacles that can be put on a single circuit and that's why the NEC assumes a minimum 180VA per receptacle. Just figure out how many receptacles you are allowed to put on a given breaker and place them wherever you want. So, using your example, a 15A breaker loaded to 80% (by code) allows you 12A. Using S = VI (or S = VI*sqrt3 for three phase) you will have 120V * 12A = 1440VA at your disposal. So you can only put 1440VA / 180VA = 8 outlets on 15A breaker.

I'm sorry.... I should have clarified residential , I believe the 180va is commercial, correct?

 

[ritelec]

I'm thinking how to word this........

Lets say the 25 x 24 room with 100 receptacles in it is correct. That no other receptacles can be installed on that circuit outside of that room.

Now, add six 100watt lighting outlets.

Could one 15amp circuit still supply those 100 receptacles and those six 100watt lighting outlets in that one room only?

 

[ActionDave]

ok......

So in that 25 by 24 foot room, 1 circuit is required for just that room right? I can put 100 receptacles on that circuit but just in that one room. If I add another receptacle in another area it shouldn't be on that circuit.
Is that correct?

You divide your square footage by 3VA and get the number of circuits you need for the house and then you divide them up as you see fit. Going room by room is one way, and a good way, but it's a design issue that you control so you can adapt as you see fit.

 

[ActionDave]

I'm thinking how to word this........

Lets say the 25 x 24 room with 100 receptacles in it is correct. That no other receptacles can be installed on that circuit outside of that room.

Now, add six 100watt lighting outlets.

Could one 15amp circuit still supply those 100 receptacles and those six 100watt lighting outlets in that one room only?

Yes.

 

[ritelec]

okay..

Thank you

 

[ActionDave]

You divide your square footage by 3VA and get the number of circuits you need for the house and then you divide them up as you see fit. Going room by room is one way, and a good way, but it's a design issue that you control so you can adapt as you see fit.

BTW this leaves out any local codes. New Mexico for instance limits the number of recpts that you can put on a 15 or 20A circuit. It's stupid. You end up with an insane number of branch circuits, most with less than a 1/4A load on them.

 

[jeremy.zinkofsky]

OP states it is residential in his first post. The Nec does not require 180 va minimum in dwellings and your post is correct if it was a commercial building.

Sorry, didn't see the residential bit. It is much easier to figure out circuiting in a dwelling because the NEC basically tells you where to put outlets I.e. every six feet etc. Just make sure that there aren't too many outlets on a single breaker by doing a quick & conservative calculation and wire each branch of outlets in a way that makes sense. If a single room is too large then you have to split up the outlets onto more than one circuit or run the risk of the client tripping breakers all the time.

 

[JFletcher]

Or that one circuit could feed receptacles in four 12.5' x 12' ( 150 sq ' x 4 = 600 sq ' ) rooms? (someone please help with the math here, my math is showing four rooms but it looks like 25 and 24 halfed at 12.5 and 12 should be two rooms not four ???)

Is this correct?

Thank you

If you halved the measurement on both walls of one room, you wind up with 1/2 L x 1/2 W = 1/4 A (area), thus a 25' x 24' room split into 12.5' x 12' dimensions gives 4 rooms (the reciprocal of 1/4th). 1/2 squared = 1/4th. If you made that 25' x 24' room into 8 1/3' x 8' rooms (1/3rd size), you would have 9 equal areas 1/3 x 1/3 = 1/9. 1/9th x 9 = 1 (the room you started with).

If if halved all three dimensions in a room (L x W x H), you would wind up with 8 'rooms' (1/2 x 1/2 x 1/2 = 1/8). and if reduced to 1/3rd size, you'd have 27 rooms (think of a Rubic's Cube: the cube is the main room, each face is 1/3rd the size of the room, so there are 27 blocks in a Rubic's Cube).

 

[kwired]

3VA/ft2 includes general lighting loads, not just general receptacles.

For a dwelling you still need at least 2 SABC's, a laundry circuit (unless laundry is provided elsewhere on primarily multifamily applications), probably a 20 amp bathroom circuit, and branch circuits for any fixed appliances that require separate circuits. The rest can be divided over that 3 VA/ft2, but has no minimum or maximum number of outlets on any one circuit.

 

[iwire]

If you halved the measurement on both walls of one room, you wind up with 1/2 L x 1/2 W = 1/4 A (area), thus a 25' x 24' room split into 12.5' x 12' dimensions gives 4 rooms (the reciprocal of 1/4th). 1/2 squared = 1/4th. If you made that 25' x 24' room into 8 1/3' x 8' rooms (1/3rd size), you would have 9 equal areas 1/3 x 1/3 = 1/9. 1/9th x 9 = 1 (the room you started with).

If if halved all three dimensions in a room (L x W x H), you would wind up with 8 'rooms' (1/2 x 1/2 x 1/2 = 1/8). and if reduced to 1/3rd size, you'd have 27 rooms (think of a Rubic's Cube: the cube is the main room, each face is 1/3rd the size of the room, so there are 27 blocks in a Rubic's Cube).

:huh::huh:

No wonder I don't do residential, it is far to complicated.

 

[roger]

:huh::huh:

No wonder I don't do residential, it is far to complicated.

At least it is if we had to use that formula.

Roger

 

[1964element]

One for each bedroom
One for each bathroom
Multiple for kitchen
One laundry
One garage
Its not that hard.


Sent from my E6782 using Tapatalk

 

[ritelec]

:huh::huh:

No wonder I don't do residential, it is far to complicated.

At least it is if we had to use that formula.

Roger

drum roll cymbal crash

TOO FUNNY !

 

[K8MHZ]

If you halved the measurement on both walls of one room, you wind up with 1/2 L x 1/2 W = 1/4 A (area), thus a 25' x 24' room split into 12.5' x 12' dimensions gives 4 rooms (the reciprocal of 1/4th). 1/2 squared = 1/4th. If you made that 25' x 24' room into 8 1/3' x 8' rooms (1/3rd size), you would have 9 equal areas 1/3 x 1/3 = 1/9. 1/9th x 9 = 1 (the room you started with).

If if halved all three dimensions in a room (L x W x H), you would wind up with 8 'rooms' (1/2 x 1/2 x 1/2 = 1/8). and if reduced to 1/3rd size, you'd have 27 rooms (think of a Rubic's Cube: the cube is the main room, each face is 1/3rd the size of the room, so there are 27 blocks in a Rubic's Cube).

There are only 26 blocks in a Rubik's Cube. Two planes of 9, and one, the center, plane of 8. Funny, I just tore mine apart and put it back together a few days ago, so I know how many blocks there are for certain, and can list the color combinations of each block.