Shop lighting
#1
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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?
And if so, how?
#4
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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
Can you explain your plan a bit more?
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
#6
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.
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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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
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
#8
15 X 1.25 = 18.75 amps which will allow a 20 amp breaker
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.
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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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 .
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 .
#10
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.
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.