Voltage Drop: Single Phase (L-L) Loads on wye

[xguard]

I have single phase 208 Volt lights staggered phase to phase at varying distances.
For example:
Light 1 phase A-C
Light 2 phase B-C
Light 3 phase A-B

The name plate of the light gives me 2.3 Amps @ 208 volts.

Is this as simple as 2.3 Amps * 208 Volts = 478.4 Volt-Amps

Then apply half of that to each line: 478.4 / 2= 239.2 Volt-Amps

Phase A B C
Light 1 239 239
Light 2 239 239
Light 3 239 239

I guess what’s giving me trouble is:
1. These loads are varying distances. If this was a feeder to a panel I think it would make more sense to me.
a. If I calculate the voltage drop between light 1 and light 2 then line B will have more current on it than line C. If it was a feeder I could multiply 478 volt-amps times 3 and then figure out the line current is 4 amps (The B phase contribution from Light 3 doesn’t add the full 2.3 amps). I would use 4 amps to find the voltage drop in this segment.
b. Now if I move to the segment between Light 2 and Light 3 the line current is 2.3 amps. I would use 2.3 amps to find voltage drop in this segment.

I feel like I’m missing or confusing something but can’t put my finger on it.

Please let me know if this is a valid approach to finding the line current for the purposes of calculating voltage drop.

Thank you

 

[xguard]

I have single phase 208 Volt lights staggered phase to phase at varying distances.
For example:
Light 1 phase A-C
Light 2 phase B-C
Light 3 phase A-B

The name plate of the light gives me 2.3 Amps @ 208 volts.

Is this as simple as 2.3 Amps * 208 Volts = 478.4 Volt-Amps

Then apply half of that to each line: 478.4 / 2= 239.2 Volt-Amps

Phase A B C
Light 1 239 239
Light 2 239 239
Light 3 239 239

I guess what’s giving me trouble is:
1. These loads are varying distances. If this was a feeder to a panel I think it would make more sense to me.
a. If I calculate the voltage drop between light 1 and light 2 then line B will have more current on it than line C. If it was a feeder I could multiply 478 volt-amps times 3 and then figure out the line current is 4 amps (The B phase contribution from Light 3 doesn’t add the full 2.3 amps). I would use 4 amps to find the voltage drop in this segment.
b. Now if I move to the segment between Light 2 and Light 3 the line current is 2.3 amps. I would use 2.3 amps to find voltage drop in this segment.

I feel like I’m missing or confusing something but can’t put my finger on it.

Please let me know if this is a valid approach to finding the line current for the purposes of calculating voltage drop.

Thank you

Fewer bites than I had fishin last Sunday ?

 

[GoldDigger]

Splitting the KVA in half and adding up on the phases, then applying the three phase load formula will be a good approximation as long as the load is balanced.
Taking the current that you get and using it in a single wire calculation will underestimate the VD because of the unbalanced sections, but I do not think that will be a big enough difference to worry about.
The alternative would be to use a program or spreadsheet to do the precise vector calculation on each segment in turn.

Sent from my XT1585 using Tapatalk

 

[ramsy]

Light 1 phase A-C
Light 2 phase B-C
Light 3 phase A-B

The name plate of the light gives me 2.3 Amps @ 208 volts.

Is this as simple as 2.3 Amps * 208 Volts = 478.4 Volt-Amps

Since each leg powers 1/2 * 2 lights, Amps = 2.3 per leg.

3Ø Power = 208 * 2.3 * Sqrt(3) * Pf

Balanced load formula not shown = 2.3A per leg @ Pf=1 (Check nameplate for Power factor, not Ballast factor)

Solving for distance and 3% VD, #12 Awg can be <= 900ft 1-way, regardless of how lights are staggered.

Now adjust x 1.25 for continuous load = 2.875A, used for Derating & Temp. Rise: #12 is good at 30.4°C.

NEC 220.18(B) inductive lighting nameplates, with "total-ampere ratings" do not require x 1.25 inductive load adjustments like motors in 430 Sec II