Lighting Calculations

[kody916]

A recent project was given to me where had to do lighting design for a three floor building. It was a small size building and I figured out that adding 20 fixtures in each level will be sufficient. This means total of 60 fixtures for the whole building. Each fixture was rated 120v, 100w. There was an existing 120/240v lighting panel in the building. My electrical calculations are below
Total watt for first floor 100 x 20 = 2000W
Total current = 2000/120
= 16.6 A
Wire size based on NEC rule =16.6*1.25 = 20.75 which means 20A breaker and size #12 wire.
The issue that I have is about the voltage drop. How should I calculate it?
Should I take the distance of the farthest fixture when it comes to the length of circuit or should I take the distance of the middle fixture for circuit length? The fixtures are evenly spaced. I am being allowed to use only 3 circuits from the existing panel (one for each floor).
Comments / modifications awaited.

Thanks

 

[david luchini]

A recent project was given to me where had to do lighting design for a three floor building. It was a small size building and I figured out that adding 20 fixtures in each level will be sufficient. This means total of 60 fixtures for the whole building. Each fixture was rated 120v, 100w. There was an existing 120/240v lighting panel in the building. My electrical calculations are below
Total watt for first floor 100 x 20 = 2000W
Total current = 2000/120
= 16.6 A
Wire size based on NEC rule =16.6*1.25 = 20.75 which means 20A breaker and size #12 wire.
The issue that I have is about the voltage drop. How should I calculate it?
Should I take the distance of the farthest fixture when it comes to the length of circuit or should I take the distance of the middle fixture for circuit length? The fixtures are evenly spaced. I am being allowed to use only 3 circuits from the existing panel (one for each floor).
Comments / modifications awaited.

Thanks

You can't put 16.6A of continuous load on a 20A breaker.

 

[kody916]

You can't put 16.6A of continuous load on a 20A breaker.

I know that i cant use 20A breaker when i have continuous load of 16.6A. but breakers come in standard sizes. 5A, 10A, 15A, 20A,
and so on....
there is no breaker rated for 16.6A or 17A.
thats y i had to go with 20A breaker

 

[david luchini]

I know that i cant use 20A breaker when i have continuous load of 16.6A. but breakers come in standard sizes. 5A, 10A, 15A, 20A,
and so on....
there is no breaker rated for 16.6A or 17A.
thats y i had to go with 20A breaker

Breakers are also rated 25, 30, 35, 40, 45, 50, etc.

210.20(A) says where a branch circuit supplies continuous loads...the rating of the overcurrent device shall not be less than 125% of the continuous load.

20 is less than 20.75. You cannot put 16.6A continuous load on a 20A breaker.

 

[sparky999]

Correct me if im wrong, but breakers are only rated for 80%......You could order a 100% breaker , but it may not be cost effective.

20 *80% = 16A.....

you're over by .6A

 

[kody916]

Breakers are also rated 25, 30, 35, 40, 45, 50, etc.

210.20(A) says where a branch circuit supplies continuous loads...the rating of the overcurrent device shall not be less than 125% of the continuous load.

20 is less than 20.75. You cannot put 16.6A continuous load on a 20A breaker.

here the total current is 16.6A. Per NEC wire size is determined by multiplying the total ampacity by 1.25.

which means 16.6 * 1.25 = 20.75A.

Now the breaker has to protect you cable. Which means the the breaker should have ampacity of 20.75A based on the above said calculation.
but since there is no such 20.75A breaker, the next available breaker to be selected has to be 25A.

so if i am getting it right, you want the breaker to be 25A and not 20A because the 20A breaker violates 125% rule.

please reply and thanks for ur response

 

[kody916]

Correct me if im wrong, but breakers are only rated for 80%......You could order a 100% breaker , but it may not be cost effective.

20 *80% = 16A.....

you're over by .6A

Regarding the ?80% rule?.

The thought is and per the UL listing a breaker is to be loaded to no more then 80% of its rating.

Say we have a continuous load of 20A. To size the cable we use 1.25X20A = 25A, A #10 wire is needed.
The breaker must be sized to protect the cable in this instance. 30A breaker is required.

20A/30A = 66.6% loaded.

If we used a 25A breaker
20A/25A = 80% loaded breaker.

If the breaker is sized per the 125% wire the breaker will always be loaded at max 80%. If a smaller breaker then the 125% wire size the cable will be protected but the load may not work. The code is not to guarantee something will work, it will guarantee the cables will not overheat at start a fire.

This only applies to continuous loads, non-continuous loads can be run at 100% the rated breaker/cable capacity.

Per the UL listing for small breakers:

1. CIRCUIT BREAKERS, MOLDED-CASE AND CIRCUIT BREAKER ENCLOSURES (DIVQ)
   
   MAXIMUM LOAD
   Unless otherwise marked, circuit breakers should not be loaded to exceed 80 percent of their current rating, where in normal operation the load will continue for 3 hours or more.

 

[david luchini]

here the total current is 16.6A. Per NEC wire size is determined by multiplying the total ampacity by 1.25.

which means 16.6 * 1.25 = 20.75A.

Now the breaker has to protect you cable. Which means the the breaker should have ampacity of 20.75A based on the above said calculation.
but since there is no such 20.75A breaker, the next available breaker to be selected has to be 25A.

so if i am getting it right, you want the breaker to be 25A and not 20A because the 20A breaker violates 125% rule.

please reply and thanks for ur response

You are on the right track. However, I don't think 210.3 will let you use a 25A branch circuit.

You must also see if your lighting is permitted to be supplied by 30A branch circuit per 210.23(B).

It might be easier to find a light fixture with lower wattage.

 

[david luchini]

Regarding the ?80% rule?.

The thought is and per the UL listing a breaker is to be loaded to no more then 80% of its rating.

Say we have a continuous load of 20A. To size the cable we use 1.25X20A = 25A, A #10 wire is needed.
The breaker must be sized to protect the cable in this instance. 30A breaker is required.

A 30A breaker is NOT required for a 20A continuous load (although it would be permitted.) A 25A breaker is the required minimum breaker size. 20*1.25 = 25.

 

[Dennis Alwon]

If I were restricted to one circuit then I would tell them they need to delete one fixture. Also depending on the distance from panel to the end run you may want to use 10 wire for voltage drop (VD)