Shop lighting

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Old 09-12-13, 11:20 AM
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Shop lighting

Can I wire safely on two circuits the following: 12 fixtures, 6 lamp t-5s, 54 watt bulbs, 360 watt input, 3 amps line current
And if so, how?
 
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Old 09-12-13, 12:33 PM
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What size circuit? Individual switches or multiple lights per switch? If multiple lights per switch how many per switch?
 

Last edited by ray2047; 09-12-13 at 02:13 PM.
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Old 09-12-13, 02:44 PM
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Curious as to why you would need 72 lights in a shop, if I read your Post correctly. 12 fixtures at 6 lamps each???
 
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Old 09-12-13, 05:18 PM
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Welcome to the forums!

6 fixtures with 6 54W lamps in each is 1,944W of total load. That's more than the 1,920W that a 20A circuit should supply as a continuous load.

That doesn't square with
360 watt input, 3 amps line current
Can you explain your plan a bit more?
 
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Old 09-12-13, 07:38 PM
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Is this for a barn? (just an assumption based on your name)
6-bulb T5's are most common for very high ceilings. At 8' they would be blindingly bright, IMO.
 
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Old 09-12-13, 08:07 PM
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I installed 100 of those fixtures in a warehouse last year. They were at 18' and were blinding.
Shiny silver reflector and very impressive light output.

Each fixture requires 360 watts so you would need 3 circuits to power them.

If you could supply them with 208v or 240v then two circuits would work.


There is 324 watts of fluorescent and the ballast uses approx. 35 watts for a total of 360 watts.
 
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Old 09-14-13, 02:44 PM
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12 fixtures @ 3 amps-per-fixture = 36 amps , total load.

A "continual load" is a load that operates for 3 or 3+ hours.

For a "continual load" , the rating of the circuit-breaker = the load on the breaker X 125% ( Art 210.20 A ) = the ampacity of the circuit conductors.

15 X 1.25 = 18.75 amps which will allow a 20 amp breaker
 
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Old 09-14-13, 04:36 PM
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15 X 1.25 = 18.75 amps which will allow a 20 amp breaker
And 15A / 3A/fixture = 5 fixtures. So that's two fully-loaded 20A circuits with two fixtures left over.

Originally Posted by Nashkat1
6 fixtures with 6 54W lamps in each is 1,944W of total load. That's more than the 1,920W that a 20A circuit should supply as a continuous load.
And that's without the ballast load.

Originally Posted by PJmax
Each fixture requires 360 watts so you would need 3 circuits to power them... There is 324 watts of fluorescent and the ballast uses approx. 35 watts for a total of 360 watts.
 
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Old 09-15-13, 11:18 AM
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4 fixtures-per-circuit , each fixture = 3 amps, = a 12 amp circuit-load.

12 X 1.25 = 15 amps = three 15 amp circuits for 12 fixtures.

This is a "continual load" calculation , but even if this is not a "continual load" condition, a limit of 12 amps on each 15 circuit is a Sound & Safe circuit-design.

If all circuit-conductors from the panel to the circuits division-point are in a raceway , that's a five #14 conductors raceway-fill which introduces the de-rating possibilty .
 
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Old 09-15-13, 04:30 PM
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If all circuit-conductors from the panel to the circuits division-point are in a raceway, that's a five #14 conductors raceway-fill which introduces the de-rating possibilty.
That's either 6 or 4 14AWG current-carrying conductors, depending on whether a multiwire branch circuit is installed for two of the ungrounded conductors. Regardless, that puts more than three current-carrying conductors in one raceway, which requires derating the conductors.

The good news is that Table 310.15(B)(3)(a) only requires a derating factor of 80% for 4-6 current-carrying conductors in the same raceway and Table 310.15(B)(16) (formerly Table 310.16) lists the ampacity of 14 AWG copper THWN rated for use in an ambient temperature of 75[SUP]o[/SUP] C. as 20, and the ampacity of 14 AWG copper THHN rated for use in an ambient temperature of 90[SUP]o[/SUP] C. as 25.

Starting with 20A, 20A * .8 = 16A * .8 = 12.8A for the THWN. It's even better for the THHN, of course, if none of the run will be in a wet location. This looks doable.
 
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