Your voltage drop is being figured at 120V, i.e. Line to Neutral. The drop multiplier is 1... not 2, if you're disregarding neutral voltage drop.
As it appears, your total drop is about half of what the picture shows... at 120V.
Series Voltage Drop All Inclusive
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As it appears, your total drop is about half of what the picture shows... at 120V.
Phil Corso
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SprayMax...
Look up the thread "Pole Lights"
Phil Corso
Look up the thread "Pole Lights"
Phil Corso
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- Placerville, CA, USA
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- Retired PV System Designer
If the calculation has all 14 poles on the same phase, it is the wrong calculation!
Not necessarily. The circuit has 14 other poles on the second phase, and 14 other poles on the third. Figuring up the voltage drop of one phase of line-to-neutral loads may not divulge the big picture of the whole circuit, but there is nothing wrong with doing it as such.
Phil,
I noticed you tend to refer to other threads quite a bit in reply. If you would be so kind in the future, please provide a link rather than just a title. I have to assume you've already reviewed the thread you reference such that it correlates with the current thread... so grabbing the link should be essentially effortless... whereas citing a thread title forces all patrons to use a search engine with a much greater chance of not getting to the thread you reference.
I like the path this conversation is progressing. This disagreement is what I am looking for. Sure all of the examples for vd always show a single phase load line to neutral or a three phase load with all three phases connected but they seldom show what happens when you send a three phase feed and start tapping off of it in single phase arrangements. This is the discussion I'm looking for.
Smart $ yes, you have the system down right.
So for the arrangement shown, I can just use the "1" factor since the current cancel in the neutral. Ok, so the VD is not as bad as shown. That's good to know. So with a three phase feed and all the lights balanced on the three phases the last light in each phase will be seeing about 111V.
Now what if we do account for the voltage drop in between poles?
Phil, could you please link the thread you mentioned.
thank you
Assuming we're still talking L-N loads that are coincidental, calculate as I*Z*L where I is the current of current pole only, Z is the impedance per 1000ft, and L is the distance from previous pole or source (x1000ft). 'tis a good approximation.
Great at response Smart $, I'll run the numbers once I get to a computer.
Thanks!
now shifting gears a little bit, not looking at the arrangement I previously posted but altering it a bit. Suppose we look at all 84 fixtures and we now connected the Single phase - Line to Line. Again looking at 84 fixtures / 3 combinations (A-B, B-C, C-A). Again 2 fixtures per pole. This brings us to 14 poles for the A-B arrangement. You need to do a series voltage drop calc, how would you approach it? VD = ??????? Per segment?
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I agree with that as long as the calculation explicitly ignores the neutral drop. But I think the OP was using the two wire single phase VD. Which overstates the drop by almost a factor of two.
(Single wire drop times 2.)
I agree. His calculation method is not clear, and the reason I have to state which method I assume he wants to be using in my replies.
Per segment. Example:
http://forums.mikeholt.com/showthread.php?t=82885&p=644296#post644296
If you use the same size wire through the entire circuit and have a balanced load, I imagine you could use something like a load center calculation to get close... but I have never really considered how to go about determining the load center on the three phase circuit.
Here's how it is done on a single phase two-wire circuit:
gar
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spraymax6:
Having read some of the posts I feel you have not been clear on the statement of the problem, nor of the assumptions you have made.
Using the assumption that the three phase loading is balanced. This is not quite correct, but probably sufficiently close for your real world purposes. Then there is no difference in the voltage drop along a hot wire between wye and delta loading. This means that we assume the voltage drop on the neutral is not significant.
But there is a big differencre if you use three separate neutrals for what otherwise would be a wye connection.
In your spread sheet you use a line to neutral voltage of 120 V, but in post 1 you stated 277 V. This is a huge error. Also in your spread sheet your cummulative current column is upside down. The largest line current occurs at the input of the distribution system, not the end. Anoher problem is that it is likely that current per fixture drops as you progress toward the end of the line. But again for practical reasons your constant current value per light is sufficiently good. You don't need high accuracy for your study. Also displayed values in your spread sheet have too many significant figures. Cut the displayed decimal places to possibly 2 or 1 places for current.
Fundamentally you need to know the voltage drop along one hot wire. Only if you use a separate neutral for each phase do you multiply by 2, and this is becuse there is voltage drop on two wires. In this case you probably should not call it a neutral. Back to assumption of a delta load, or a wye with zero neutral current, then voltage drop on that one hot wire is resistance of that wire times the current in that wire. There is no factor of 2 or 1.732 . the easy way to visualize that current is to assume a wye load and zero neutral current.
Don't just plug numbers into some equation. You need to know how the equation relates to your circuit.
.
spraymax6:
Having read some of the posts I feel you have not been clear on the statement of the problem, nor of the assumptions you have made.
Using the assumption that the three phase loading is balanced. This is not quite correct, but probably sufficiently close for your real world purposes. Then there is no difference in the voltage drop along a hot wire between wye and delta loading. This means that we assume the voltage drop on the neutral is not significant.
But there is a big differencre if you use three separate neutrals for what otherwise would be a wye connection.
In your spread sheet you use a line to neutral voltage of 120 V, but in post 1 you stated 277 V. This is a huge error. Also in your spread sheet your cummulative current column is upside down. The largest line current occurs at the input of the distribution system, not the end. Anoher problem is that it is likely that current per fixture drops as you progress toward the end of the line. But again for practical reasons your constant current value per light is sufficiently good. You don't need high accuracy for your study. Also displayed values in your spread sheet have too many significant figures. Cut the displayed decimal places to possibly 2 or 1 places for current.
Fundamentally you need to know the voltage drop along one hot wire. Only if you use a separate neutral for each phase do you multiply by 2, and this is becuse there is voltage drop on two wires. In this case you probably should not call it a neutral. Back to assumption of a delta load, or a wye with zero neutral current, then voltage drop on that one hot wire is resistance of that wire times the current in that wire. There is no factor of 2 or 1.732 . the easy way to visualize that current is to assume a wye load and zero neutral current.
Don't just plug numbers into some equation. You need to know how the equation relates to your circuit.
.
Smart $, I think I got the most accurate calculation by doing what I show in the attached PDF. I looked at the starting group of poles, first pole with the first Phase A connection, then pole 2 with the first Phase B connection, and then pole 3 with the first phase C connection. I saw that the current heading back to the source leaving the Phase C pole would be 2.94A @ -120 deg, then I looked at the current after the pole 2 with the first phase B connection. The current there now will be 2.94@120deg (Phase B current) and 2.94@-120 (Phase C current). Phasor addition yields the current leaving there to be 2.94@180deg. Now at the first pole (first phase A connection) the current is zero because Phase As current of 2.94@0deg cancels the last bit of current. The same holds true for the rest of the 13 Phase groups. Finally I add all the neutral voltage drop together and then add that to the phase voltage drop that I got previously (the correct one, using a factor of 1 instead of 2) and I get the most accurate representation.
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Per segment. Example:
http://forums.mikeholt.com/showthread.php?t=82885&p=644296#post644296
If you use the same size wire through the entire circuit and have a balanced load, I imagine you could use something like a load center calculation to get close... but I have never really considered how to go about determining the load center on the three phase circuit.
Here's how it is done on a single phase two-wire circuit:
Smart $, thanks for this. I will take some time to digest this information. This looks great though.
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Gar,
Yes you are right, that was sloppy on my end. I first started out this post by trying to show all possible ways to wire roadway lights to a three phase feed. In that case I chose 480/277 arbitrarily. My real application was 208/120V and started using that because it was easier to explain the answer I was looking for. The information got mixed up, that is my fault. Yes another point of confusion was the spread sheet. The way the formula was written made it more convenient to list it that way, but you are right it should be reverse.
"But there is a big differencre if you use three separate neutrals for what otherwise would be a wye connection."
Yes this is exactly what I was trying to show in one of my later arrangements in my original PDFs. I wanted to prove that when you share the neutral on a three phase feed, even though the connection is a single phase line to neutral, it will be less than the "2" factor. Where I got lost is how much less than the "2" factor, did the "1.732" rear its head again. The answer turned up to be its a factor of "1".
Phil Corso
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To All...
In the Thread I referenced, I offered an EXCEL spread-sheet covering evenly-spaced Loads (lamp posts)!
Contact me for a copy!
In the Thread I referenced, I offered an EXCEL spread-sheet covering evenly-spaced Loads (lamp posts)!
Contact me for a copy!
You don't sum the neutral voltage drop though the entire length. The neutral voltage drop is only applied from pole to pole.
As I mentioned earlier, it is not an accurate calculation assuming neutral currents between pole 1 and pole 3 cancel entirely and there is zero current between pole 3 and pole 4. But if you do consider that as an accurate model for the sake of calculation brevity, the zero current segments are indicative of the neutral voltage drop not summing through the entire circuit... i.e. each group starts over, so none is carried.
The best approximation is just I*Z*L of the preceding or following segment... but because they do not sum through the circuit, the only segment we would be concerned with would be the second to last and last pole... and you would add that to the voltage drop on the line at the last pole.
Smart $: so the addition would be extremely negligible in that case. Something like 0.64V for a total VD of 8.86V
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I disagree a bit. There will be a voltage drop on the neutral from pole 1 through pole three. For each pole set it will be based only it the single pole currents even though the rest of the pole currents (balanced) are "passing through".
But taking that as your new starting voltage you will get the same drop again for each set of three poles. Those three pole drops WILL add up along the string.
But taking that as your new starting voltage you will get the same drop again for each set of three poles. Those three pole drops WILL add up along the string.
I'm assuming we are using the model with zero current on every third segment (1, 4, 7, 10, ..., 40).
Zero current, zero voltage drop... on that neutral segment. So the 3Ø sets to the load side only experience the voltage drop of the line conductors, and the neutral voltage drop within that set.
If you want to be more accurate and use network analysis, yes there is some voltage drop accumulated along the neutral... but the difference realized isn't worth the computation necessary... that is, if doing by hand, so to speak. Simulation software can do it rather easily.
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