Segmental voltage drop on a three-phase circuit feeding road lighting fixtures

[scr]

Please take a look at the diagram below:



I want to calculation the three-phase voltage drop on each 130 ft segment. I just want to confirm from you people that would I get a reasonably accurate voltage drop if I just add up the fixture currents like this:

390 ft	0.5 A
260 ft	1.0 A
130 ft	1.5 A

...then use these values in the usual three-phase voltage drop formula to find the voltage drop on each segment, and lastly add them up to get the total three phase voltage drop on the circuit?

Also, would the above method also applicable to the scenario where each fixture was connected phase to neutral, instead of phase to phase like in my example (assuming no unbalanced currents)?

Any input will be appreciated.

 

[Dennis Alwon]

Why not just use 1 amp for the entire 390 feet? or is this an exercise on How To rather than a real situation???

 

[GoldDigger]

Please take a look at the diagram below:

View attachment 9871

I want to calculation the three-phase voltage drop on each 130 ft segment. I just want to confirm from you people that would I get a reasonably accurate voltage drop if I just add up the fixture currents like this:

390 ft	0.5 A
260 ft	1.0 A
130 ft	1.5 A

...then use these values in the usual three-phase voltage drop formula to find the voltage drop on each segment, and lastly add them up to get the total three phase voltage drop on the circuit?

Also, would the above method also applicable to the scenario where each fixture was connected phase to neutral, instead of phase to phase like in my example (assuming no unbalanced currents)?

Any input will be appreciated.

You asked for input, so here it goes:

1. The current in the individual phase lines from different loads will not completely cancel for a balanced load as it would for wye loads to a neutral, but the load currents will still add as vectors rather than adding magnitudes.
Two 1A loads from A to C and from B to C will only cause a current of sqrt(3) A in the common portion of the C conductor.
2. Because of the shared conductors, you really have to do the math for each segment of the combined circuit rather than treating them as three separate systems.
In particular, the 390 foot length to the most distant load will see a voltage drop in both phase wires, not just in one. But the voltage drop in the shared phase wire will be different in each of the 130 foot segments.
So you really have to do the calculation for each 130 foot section and then add the voltage drops as vectors to get the magnitude of the final VD at the most distant load.

If you calculated the VD to the farthest load as the current times the resistance of two 390 foot wires you would at least have a conservative estimate.

 

[scr]

I've edited the diagram above and added the neutral:



As can be seen the loads are now connected between phase and neutral. So will it be okay now to add all the currents starting from the last fixture up to the first, and then find the three-phase voltage drop on each segment, like this:

390 ft	0.5 A
260 ft	1.0 A
130 ft	1.5 A

 

[GoldDigger]

I've edited the diagram above and added the neutral:

View attachment 9872

As can be seen the loads are now connected between phase and neutral. So will it be okay now to add all the currents starting from the last fixture up to the first, and then find the three-phase voltage drop on each segment, like this:

390 ft	0.5 A
260 ft	1.0 A
130 ft	1.5 A

I would do it this way:
1. The voltage drop for the end load just caused by the last 130 feet is
260 feet at .5 amps. Half of that VD will be on C and half of it will be on N.
2. The voltage drops just caused by the middle segment are
130 feet at .5 A on C, causing that voltage drop just on C. plus 130 feet at .5 x sqrt(3) A causing a voltage drop on N from both loads. (The square root of three account for the vector sum of the two loads using it at over that segment.
Plus 130 feet at .5 A on B only.
3. The first segment will have a voltage drop of 130 at .5A on the A wire, B wire and C wire independently.
The voltage drop on the neutral will be zero since the load will be balanced through that segment.

Now do not bother with any three-phase voltage drop formulas and just add the voltage drops on each of the three segments for each phase separately. The voltage drop at each load will be the sum of the voltage drops in its individual phase plus the drop in the shared neutral up to that point.

However!!! the phase line voltage drop and the neutral line voltage drop will not add up neatly since they are vectors too.
Nor will the two segment voltage drops in the neutral add up simply, since the voltage drops on the second and third segment are not in phase either.

Short answer: Use a good circuit simulator and plug all of the values in. <sigh>

The actual math is left as an exercise for you.

Knowing individual drops will tell you exactly how much power is wasted on each wire in each segment, but will not tell you what voltage drop a particular load will see.

 

[Smart $]

I would do it this way:
1. The voltage drop for the end load just caused by the last 130 feet is
260 feet at .5 amps. Half of that VD will be on C and half of it will be on N.
2. The voltage drops just caused by the middle segment are
130 feet at .5 A on C, causing that voltage drop just on C. plus 130 feet at .5 x sqrt(3) A causing a voltage drop on N from both loads. (The square root of three account for the vector sum of the two loads using it at over that segment.
Plus 130 feet at .5 A on B only.
3. The first segment will have a voltage drop of 130 at .5A on the A wire, B wire and C wire independently.
The voltage drop on the neutral will be zero since the load will be balanced through that segment.

Now do not bother with any three-phase voltage drop formulas and just add the voltage drops on each of the three segments for each phase separately. The voltage drop at each load will be the sum of the voltage drops in its individual phase plus the drop in the shared neutral up to that point.

However!!! the phase line voltage drop and the neutral line voltage drop will not add up neatly since they are vectors too.
Nor will the two segment voltage drops in the neutral add up simply, since the voltage drops on the second and third segment are not in phase either.

Short answer: Use a good circuit simulator and plug all of the values in. <sigh>

The actual math is left as an exercise for you.

Knowing individual drops will tell you exactly how much power is wasted on each wire in each segment, but will not tell you what voltage drop a particular load will see.

You're making it more difficult than it has to be... on the basic level*

The Vd contribution for each line conductor is straight forward. Line current times resistance for the distance.

Vd on neutral in the basic sense is fairly straight forward also. Balanced loads means 0A on first 130', so no additional Vd for closest load. Current on neutral from closest to middle load will be that of the closest load, so line current times resistance for the distance to middle load. This adds to Vd for middle load and farthest load. The current on the last segment of the neutral is that of the farthest load, so line current of farthest load times resistance for distance of last segment and gets added to farthest load's Vd.

*If you want to get truly technical. starting with identical loads, the Vd's actually unbalance the load currents so there will be current on the first neutral segment... and each load will not draw exactly 0.5A. However superfluous, an exercise for the advanced student...

 

[GoldDigger]

I would not worry about the changes in load caused by the voltage drop differences either.
But I would note that the neutral voltage drop across segment 2 will not be in phase with the neutral voltage drop across segment 3, so you cannot just add the two numbers.
The effect is small, but it is there.
Similarly if the voltage drop on the phase wire is out of phase with the neutral voltage drop, the voltage drop seen by the load will be lower than the sum of the two.

Tapatalk!

 

[Smart $]

I would not worry about the changes in load caused by the voltage drop differences either.
But I would note that the neutral voltage drop across segment 2 will not be in phase with the neutral voltage drop across segment 3, so you cannot just add the two numbers.
The effect is small, but it is there.
Similarly if the voltage drop on the phase wire is out of phase with the neutral voltage drop, the voltage drop seen by the load will be lower than the sum of the two.

That's correct... but it would be closer to actual than just Line to load Vd.

 

[scr]

Thank you all for your answers.

However, I must still find the voltage drop on the whole circuit (there are around 30 lamps on the circuit, each lamp being alternatingly connected between a phase and neutral, like this: A-N ... 130 ft ... B-N ... 130 ft ... C-N ... 130 ft ... A-N ... 130 ft ... B-N ... and so on).

As much I would love to find out a way to calculate the voltage drop with as much accuracy as possible, still I must ask you people: For the time being, what could be the quick and easy way to find VD of the said circuit that could be the closest to the true value?

 

[GoldDigger]

IMHO the simplest good approximation is that for the phase lines you simply figure the voltage drop for each phase depending on the light's position in the 390' unit and then add the voltage drop for each additional complete 390' unit going back to the source combining the current from the end luminaire with the current for each additional luminaire on that phase with each additional block.
There will be a simple algebraic formula for this arithmetic series.
For the neutral wire, the voltage drop for each three segment block will be a constant regardless of the position of that block in the string, but it will be different for each phase.

Tapatalk!

 

[scr]

I've just realized that I shouldn't have written the 0.5 A that is shown besides each light. I've redrawn the diagram with the actual wattage and power factor for each lamp:



This should clear up any confusing caused by previous diagrams. Please consider this latest diagram and disregard the previous ones.