calculating amp load of fixture on various voltages

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glennme

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Trying to consolidate a few posts into one. I'm working through a couple lighting load issues and have seen a couple ways to calculate anticipated current (meaning some are incorrect). In all cases, let's use a hypothetical 400W fixture that can be powered off a number of voltages. I think this is how one would calculate theoretical amp draw.

120V single phase: 400W / 120V = 3.33A
240V single phase: 400W / 240V = 1.67A
208V two ungrounded: 400W / 208V = 1.92A
277V three phase, 1 ungrounded: 400W / 277V = 1.44A
480V three phase, 2 ungrounded: 400W / 480V = 0.83A

One question arises when does the calculation get divided by the sq rt of 3 (1.73). I was thinking the last three cases would need to be divided by 1.73 so this seems to be where I'm missing a detail or two of understanding.

Thank you in advance.
 
Single phase L-L or poly phase L-N calculations with (1 un/grounded) don't use the vector adjustment √(?) or 1.732, but all others do.

The formula below is one measure used to check poly phase L-L ampere calcs.

I = √(A?+B?+A*B) / Vline / pf
(A,B,C values are in VA)

Whenever checking L-L legs of a 3-phase circuit, that ampere calculation always depends on what's going on with the third leg. VA / Volts / 1.732

Since each poly phase L-L measurement shows the total of all phases, if the 3rd leg also uses a 400 VA load, a 208Y full boat would show 3.33A at pf = 1, across any two phases even though each fixture only draws 1.11A and (VA / Volts / 1.732) = 1.11A.

If phase C is off or 0 VA, then the remaining two L-L 400 VA legs on the 208Y do strange things to that formula above. Phase A-B shows 2.31A, B-C shows 3.33A, and C-A shows 2.31A, even though theres nothing on C, with only two fixtures acting the same as before.

As I understand it, L-N and single phase L-L loads are not joined together with any other phases, therefore, make for simpler Panel Ampere schedules. While some shortcuts strive to simplify the poly phase calcs, anything except that formula above makes assumptions based on balanced loads and is not as reliable.
 
For the hypothetical fixture loads in the OP why would the 1.73 factor apply? None of the loads are actually 3 phase.
 
I agree, there is no such thing as a 3-phase light bulb. Also are these all incandescent fixtures? If they are HID fixtures the ballast/transformer losses will also need to be calculated.
 
This is good info. In actuality, we are using 400W metal halide bulbs for a warehouse application. Each runs on 277V L-N power. Part of our problem is we are seeing more load than the math would indicate so we are trying to sort this out with the electrician that performed the installation (new building).

On a given circuit, namely 1 ea L-N load, we have 6 fixtures and are reading 11A of load which is 1.8A per bulb. The calculations should indicate we have 400W / 277V = 1.44A. This is a 25% increase from expected. Would it be expected that ballast losses could be this high? Maybe voltage drop losses as well (20A circuits).

Each of our 4 panelboards are powered through a 600A fused disconnect. At this disconnect, we are reading 530A of load on each phase. I think the correct calculation for the load of 900 bulbs would be: 900 bulbs * 400W / 480V / 1.73 = 433A. Please advise if the math is correct. Note that the difference we are seeing is 25% which seems to be the same factor as on the individual circuit.

If you could confirm the math, that would be great. Then, does it make sense we have a 25% loss factor? If so, what could I check to determine the root cause of our extra load? Note only lighting is powered through the main disconnect.

Thanks.
 
As others have stated, you don't use the 1.732 factor for single phase loads, but you do see it when you are looking at the three phase loads that are aggregates of single phase loads. In your particular example, the factor is sort of hidden in the way that you are doing the calculations.

You have a 3 phase wye connected supply with 480V line-line. What is the line-neutral voltage? 480/1.732.

When you do the calculations on the 277V fixtures, you are really using (480/1.732)V / 400W.

On your 'losses', you will have real power consumed by the ballasts, so that the 400W fixture is drawing more than 400W. Additionally, you probably have power factor issues, where the ballasts are drawing current out of phase with the supply voltage. If the power factor is 0.8, then you would draw 400W of real power but 500AV of apparent power.

If the designer didn't consider power factor or ballast losses, then perhaps they also didn't worry about harmonics. I would put a meter on the neutral and see how high the neutral current is, both with all the lights on and with only 2 phases loaded.

If the designed did consider power factor and ballast losses, then it is possible that lower quality ballasts were 'value engineered' into the installation.

-Jon
 
glennme said:
277V three phase, 1 ungrounded: 400W / 277V = 1.44A
Click to expand...
glennme said:
480V three phase, 2 ungrounded: 400W / 480V = 0.83A
Click to expand...

I don?t know what you mean by ?277V three phase,? nor by ?480V three phase, 2 ungrounded.?

277 volts is not a three phase value. It is the single phase, line to neutral voltage in a 480/277 volt wye system. But it is not in and of itself three phase. But if you have a single phase, 277 volt, 400 watt load, then indeed its current would be 1.44 amps, as you calculated. You don?t use the 1.732 unless you are dealing with all three phases.

Most of the time I deal with 480 volt, three phase loads, there are three ungrounded conductors. I think you can only get two ungrounded conductors with a corner-grounded delta configuration, and I never deal with those. If you have a total of 400 watts connected to a 480 volt, three phase system, then you need to include the factor of 1.732 in the calculation. 400 divided by (480 times 1.732) equals 0.48 amps.
 
if you consider an input wattage of 450-475 for the HID ballasts, that doesn't seem too unrealistic.
 
ballast loss

ballast loss

The lighting fixture maufacturer should be able to provide you with the actual load for the fixture in question. Generally you need to include some loss for the ballast so a 400 watt lamp/fixture would draw about 460-475 watts when you include the ballast.
 
The ballast should have an amperage rating on it. You would use this value for calculating load not the bulb wattage. The manufacturer may also give you a power factor. If you have a 400-watt fixture with a 90% power factor the fixture would actually use 400/.90 or 444.44 watts.
 
glennme said:
On a given circuit, namely 1 ea L-N load, we have 6 fixtures and are reading 11A of load which is 1.8A per bulb.
Click to expand...

The first manufacturer I looked up states that their 400W MH ballast has 452W input and draws 1.75A at 277V. That would make about 485VA.
 
infinity said:
For the hypothetical fixture loads in the OP why would the 1.73 factor apply? None of the loads are actually 3 phase.
Click to expand...

I see the OP was answered quite nicely, but for my own clarity can someone confirm if this 1.73 factor or formula, I = √(A?+B?+A*B) / Vline / pf, applies to all ungrounded loads powered by 3? transformers?
 
Seems strange to me that there are two voltages in the parking lot (120/208 and 277/480). Is this a renovation and/or new construction ???????
 
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