Automated Fixtures 208v Power Load Balancing

[Jon8-)]

I am sure you have answered this question before. I searched for sometime here on CB but couldn't find the full answer.

This is what I found here on CB posted by Derekleffew back in 2008


"At 208V, moving all motors inside a VL3000Spot (never tested a 3500Spot) will only increase the current draw approximately 1A, an almost insignificant amount.

Calculating the total current required when operating at 208V 3Ø ∆ is not as simple as the above posts would make it appear.
Assume a VL3000 draws 11A, across TWO of the three hot legs. For calculation purposes, consider twelve(12) units, distributed equally across all three legs: 4 on ØX, 4 on ØY, 4 on ØZ. The total on any one leg is (4*11=44A), right?
WRONG! The total(s) on each leg must be multiplied by the constant: 1.732 [=SQRT(3)], to determine the actual load, in order to account for the "across both legs" portion, (as It's actually: 4 on X-Y, 4 on Y-Z, 4 on Z-X).
Thus, assuming the legs are perfectly balanced, ~76.2 Amps/Leg, 120/208VAC 3Ø Wye-connected service, is required for 12x VL3000s.

Two quick ways to calculate, but ways that WILL get you into trouble--as MLs often don't come in multiples of three, nor are always circuited perfectly balanced; are:A) Multiply the total number of fixtures by the current draw of each, then divide the result by 1.732, [or SQRT(3) if you enjoy lots of insignificant digits].
OR,
B) Multiply the total number of fixtures by the current draw of each, divide by 3, and multiply that result by 1.732, [or SQRT(3) if you enjoy lots of insignificant digits]."

This all makes sense, but when I try to use the same formula using only 2 x VL3000 fixtures the total Amps per leg does not make any sense. I used Pual Pelletier's LD calculator to calculate the following. One VL3000 is pulling 11A on X and 11A on Y, The other VL3000 is pulling 11A on Y and 11A on Z. Why is the total load 11A on X, 19A on Y and 11A on Z? . I don't understand how it is calculating the 2 x 11A on the Y leg. I would assume you would add the two 11A from each fixture to get 22A total for the Y leg and then as mentioned above multiply this by 1.732. Also why do you not Multiply the other 2 legs by 1.732? Was the LD Calc wrong? Can you please explain how you would calculate total load per legs if you only had 2 x VL3000? Also when you look up the specs on a VL3000 Spot fixture it states the following:

Standard AC power distribution from 200– 264 VAC, 50/60 HZ. The unit requires 7 to 12 A depending on the AC supply voltage.

So I am trying to figure out how you get 11A total for the fixture. I came close to 11A using the following equation. 1200w lamp divided by 208v = 5.7A then multiply that by 1.732 = 9.8 then add 1A for the internal motor power and that gives you 10.8A round to 11A. Is this correct?


Thank you for your expertise.


Jon

 

[JD]

Simply put, the power flow is occurring at different times. There is some overlap and cancellation, so it is not double or 0 as it would be with single phase.
For example, HNH with equal loads bout 180 degrees out of phase (single phase) would result in almost 0 current flow on N. HNH with the hots in phase would result in X2 the current flow on N.
BUT... 3 phase power has a phase lag of 120 degrees, so there is some cancellation on N.

 

[Jon8-)]

Thanks JD for explaining this in technical terms but I really need to know how to calculate the example I gave with the two VL3000 fixtures. What is the formula or equation I should used to get me the total amps per leg. I also wanted to know if I was calculating how much amps a single fixture is given that it has a 1200w lamp. Can you answer any of these questions?

 

[JD]

Ok, lets break your question down into two separate issues:

1) Why does a 1200 watt fixture draw 11 amps at 208 volts?
The fixture draw is based on the worst case scenario, factoring the lamp, ballast, and supporting electronics at peak draw, which would be right after the lamp is struck. It has nothing to do with the 1.732 calculation.

2) With two fixtures; One connected from Phase A to Phase B, and the second between Phase B and Phase C, what is the current draw on each phase and why?
Since Phase A and Phase C are at the tail end, the current on both of those legs would be 11 amps. The common connect point is Phase B, which would be 11 X 1.732 = 19.05 amps (or darn close to 19 amps) so the formula works. The reason it is not 22 as mentioned above, is that there is some phase cancellation where current is actually flowing from phase A to phase C right through both fixtures.

Note: there is actually a third issue going on as the lamps are ballasted that drives it to the higher number. The issue is called "Power Factor" and has to do with the fact that peak current draw does not line up with peak voltage, but that gets a little lengthily to explain. Using the 11 amp number covers the PF equation too.

I hope this helps.

 

[David Ashton]

the wattage of a discharge lamp unit is not very useful, the VA rating is what really matters and there are two VA ratings that really matter,there is the switch on rating which is the power drawn while firing up and the running rating which is the power drawn during use.
If the fitting has an incandescent lamp in it then the VA is close to the wattage but these are not common.

 

[derekleffew]

... Also when you look up the specs on a VL3000 Spot fixture it states the following:

Standard AC power distribution from 200– 264 VAC, 50/60 HZ. The unit requires 7 to 12 A depending on the AC supply voltage.

So I am trying to figure out how you get 11A total for the fixture. ...

All we know for sure, according to Vari*Lite, is that
at 200V, the unit uses 12A,
and at
at 264V, the unit uses 7A.
One can do the maths, and assuming the relationship of voltage to amperage is linear (which, in this case, it isn't BTW), derive that
at 208V, the unit uses 11A.
But this is really only an approximation; "close enough for summer stock," as they say.
One cannot determine the amperage draw of a discharge lamp solely from the wattage and voltage, without also knowing the power factor, efficiency of the ballast, and other details.

For comparison purposes, the Martin MAC2000 Profile II (Electronic Ballast) uses the same exact lamp as the VL3K. Martin's manual states:

...[100-120V data deleted]
208 V / 50 Hz . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1470 W, 10.5 A
208 V / 60 Hz . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1470 W, 10.3 A
230 V / 50 Hz . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1480 W, 9.6 A
230 V / 60 Hz . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1470 W, 9.4 A
250 V / 50 Hz . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1480 W, 8.6 A
250 V / 60 Hz . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1480 W, 8.5 A

So since the MAC2000 draws 10.3A vs. the VL3K at 11A, is the MAC2K better? Brighter/Faster/Louder?

 

[Chris15]

So since the MAC2000 draws 10.3A vs. the VL3K at 11A, is the MAC2K better? Brighter/Faster/Louder?

... or simply less efficient?

 

[Jon8-)]

All we know for sure, according to Vari*Lite, is that
at 200V, the unit uses 12A,
and at
at 264V, the unit uses 7A.
One can do the maths, and assuming the relationship of voltage to amperage is linear (which, in this case, it isn't BTW), derive that
at 208V, the unit uses 11A.
But this is really only an approximation; "close enough for summer stock," as they say.
One cannot determine the amperage draw of a discharge lamp solely from the wattage and voltage, without also knowing the power factor, efficiency of the ballast, and other details.

For comparison purposes, the Martin MAC2000 Profile II (Electronic Ballast) uses the same exact lamp as the VL3K. Martin's manual states:

So since the MAC2000 draws 10.3A vs. the VL3K at 11A, is the MAC2K better? Brighter/Faster/Louder?

Derek,

Thanks for Explaining this. I was already using this method to guestimate the total amps per fixture. It def. is an easy way to quickly find the total amps.
Not all fixtures specs spell everything thing out in this way though. I was wondering if there was a mathematical way to get the correct answer every time even if you don't know how many amps the fixture pulls at 200v & 264v. Chris ended up giving me the formula needed to calculate the fixtures total amperage as long as the manufacture gives you the PF.

Thanks,

Jon