# Lighting Layout and Single Line Diagrams for Building Designs?

Discussion in 'Homework Help' started by melvinlacia, Apr 28, 2012.

1. ### melvinlacia Thread Starter New Member

Mar 7, 2011
3
0
Here is a single line diagram of an electrical design for buildings.
http://imageshack.us/photo/my-images/855/picture1om.jpg/

I started doing the illumination calculation (zonal cavity method) and layouted the necessary number of lighitng fixtures on the drawing.
Now I wanted to wire them with switches of course and connect them to a power source. So by wiring you group the fixtures and be connected to a specific circuit. There would only be limited number of fixtures to be connected to a certain circuit depending on the capacity of the circuit breakers.

My problem is that I am not sure that given that Single line diagram above, how to calculate the total load to be connected to a certain circuit and thus the number of fixtures to be connected. There are 5 circuits shown on the diagram.

For example, i have a fixture that says" 4x36 Watts fluorescent lamp". So how many lamps should be connected to circuit one(the on the leftmost side on the diagram).

Please teach me a general solution to solve this. Thank you very much...

Mar 7, 2011
3
0

3. ### MrChips Moderator

Oct 2, 2009
18,374
5,804
If someone understood the problem and had an answer they would respond.

4. ### melvinlacia Thread Starter New Member

Mar 7, 2011
3
0
feel free to ask if you want to clarify something on my post. please if you have some thoughts even you're not very sure, just post it...I might realize something or get something from it... thank you...

5. ### MrChips Moderator

Oct 2, 2009
18,374
5,804
I am not an electrician and I don't know what 50AT/100AF 3pole means.

If you have 4x36 watt lamps, that is equal to 144 watts.
If you have a 50A@120VAC service line, that is 50 x 120 = 6000 watts

Hence the maximum number of fixtures allowed = 6000/144 = 40.

(Check with a certified electrician.)

6. ### JoeJester AAC Fanatic!

Apr 26, 2005
4,031
1,774
If the 50AT are 50 amperes (11 kW @ 220V) both of them at maximum draw will exceed your supply of 100A. You have a 50 + 50 + 20 + 20 +20 or a total of 160A .... even if your loads were 80% of max, that would be 128 amperes .... still exceeding your supply of 100 amperes.

I am not an electrician, although I do know that typically the loading should be no more than 80% of the rating of the wires and breakers.

7. ### strantor AAC Fanatic!

Oct 3, 2010
4,993
3,079
FYI
AT = amp trip
AF = amp frame
Modern circuit breakers are manufactured in a limited number of mechanical "frame" sizes, and then outfitted with interchangeable trip units.

A 50 Amp Frame circuit breaker configured with a 40 Amp trip unit will provide protection for a nominally 40 A circuit. Depending on the breaker, it could be outfitted with up to a 50 Amp trip unit, since 50 A is the maximum continuous rating of the breaker current-carrying parts.

So you have 2 50A breakers and 3 20A breakers. Breakers are sized for the maximum current carrying capacity of the wires used. Breakers are for protecting the wires, not the loads. So your breaker ratings don't tell us anything about the lamps, other than the maximum amount you could put on any one breaker, as MrChips demonstrated in post #5. However, MrChips was doing calculations for 120V, and your drawing shows 208V.

4x36 Watts fluorescent lamp = 144W
144W / 208V = .692A
50A (breaker) / .692A = 72.25

So, theoretical maximum of 72 flourescent fixtures on one 50A breaker. however That's not really a good idea. It's good to leave some "head room". These are not pure resistive loads; I don't know much about them, but they may have some inrush current which (if you had 72 on one breaker) may try to draw much more than 50A when switched on, tripping the breaker.

8. ### Nichesound New Member

May 19, 2009
2
0
hmmm 72 fixtures on a 50 amp breaker...hmmmm sounds suspect...typically a 20amp single pole or 15 amp single pole loaded up to 80% of the circuit is the recommended application for designing a lighting circuit I would think. You have to protect the wire that is correct.