3 phase LED branch circuits- finding amps

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Isaiah

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Just wanna be sure on this…I have two, three phase, three pole 480V branch CBs from a panelboard feeding two strings of 35 floodlight LEDs each. The vendor data shows average system watts of 459 per fixture. Initially I assumed I should calculate the amps as if it were a typical three phase circuit meaning W divided by V x PF x 1.73. But the fixtures themselves are only single phase. So should the branch circuit also be calculated at single phase 480V, instead of three phase?


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Depends how the strings are wired. Are they two wire or three wire? If you round up to 1 amp per light and you balance 35 lights over 3 phases you'd something like 24,23 and 23 Amps approximately on the three phases. If its two wire you'd have 35,35, 0 per string.
So with two wire your highest loaded phase could have 70A and three wire 48A
 
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tortuga said:
Depends how the strings are wired. Are they two wire or three wire? If you round up to 1 amp per light and you balance 35 lights over 3 phases you'd somthing like 24,24 and 22 Amps approximately on the three phases. If its two wire you'd have 35,35, 0 per string.
So with two wire your highest loaded phase could have 70A and three wire 48A
Click to expand...

They are wired a with three wires plus ground


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480V Line to Line connected, so a delta connected load so to speak.
Each light will put about 1 Amp on two legs.
 
Isaiah said:
Just wanna be sure on this…I have two, three phase, three pole 480V branch CBs from a panelboard feeding two strings of 35 floodlight LEDs each. The vendor data shows average system watts of 459 per fixture. Initially I assumed I should calculate the amps as if it were a typical three phase circuit meaning W divided by V x PF x 1.73. But the fixtures themselves are only single phase. So should the branch circuit also be calculated at single phase 480V, instead of three phase?


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Click to expand...
Please clarify how are the loads connected. Are the LEDs connected in wye (277V) or in delta (480V)? 35/11 =11.67, take 11 LEDs on two of the phases and 13 LEDs on the third phase. Assuming a unity PF the line currents are:
  • On the first arrangement (wye), the line currents will be Ia = Ib = 18.22A, Ic = 21.53A, In = 3.31A.
  • In the second arrangement (delta), the currents will be Ia = 19.9 A, Ib = 18.22A, and Ic = 19.9A.
 
topgone said:
Please clarify how are the loads connected. Are the LEDs connected in wye (277V) or in delta (480V)? 35/11 =11.67, take 11 LEDs on two of the phases and 13 LEDs on the third phase. Assuming a unity PF the line currents are:
  • On the first arrangement (wye), the line currents will be Ia = Ib = 18.22A, Ic = 21.53A, In = 3.31A.
  • In the second arrangement (delta), the currents will be Ia = 19.9 A, Ib = 18.22A, and Ic = 19.9A.
Click to expand...

Line to line 480V across all three phases=A/B, A/C and B/C…no neutral to 277V


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Isaiah said:
But the branch calculation is still using sq root of 3. Correct?
Click to expand...
To size the branch circuit I use the highest loaded leg.
And then add the continuous factor.
For the service / feeder load calc I would just keep it in VA
and add the continuous factor to the VA.
 
First you need to confirm that the strings are wired as a three phase load. I know you say you have 3 wires, but confirm that the manufacturer of the string used all 3 wires and tried to balance things.

If you have N matched single phase loads distributed across 3 phases in as balanced a fashion as possible, my approach is to first find the nearest multiple of 3 greater than or equal to N.

Take this multiple of 3, times the watts per individual load, and then use the 3 phase current equation.

In your case I would use 36 * 459 / V / pf/ 1.732

Jon
 
winnie said:
First you need to confirm that the strings are wired as a three phase load. I know you say you have 3 wires, but confirm that the manufacturer of the string used all 3 wires and tried to balance things.

If you have N matched single phase loads distributed across 3 phases in as balanced a fashion as possible, my approach is to first find the nearest multiple of 3 greater than or equal to N.

Take this multiple of 3, times the watts per individual load, and then use the 3 phase current equation.

In your case I would use 36 * 459 / V / pf/ 1.732

Jon
Click to expand...

I agree with your calculation-
thanks
Isaiah


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