Determining the number of lighting fixtures
If I am lighting up a 50' x 50' room to achieve 40 foot candles, ... in an industrial environment
Everyone will complain that it is way too dark to work or for safety.
Recommended Lighting Levels in Buildings
Room type Light level Light Level Power Density
(Foot Candles) (lux) watts/sq.ft.
-------------------------+-------------------+-----------------+-----------------
Corridor 5-10 FC 50-100 lux 0.66
Laboratory (Classroom) 50-75 FC 500-750 lux 1.43
Laboratory (Professional) 75-120 FC 750-1200 lux 1.81
Workshop 30-75 FC 300-750 lux 1.59
Unfortunately there's no way to 'MathJax a table' or adjust that so it looks better. See the article link above for more information about acceptable lighting levels for various circumstances. Here's a quote from that article:
"Lighting in our living and workplaces is critically important for our ability to accomplish tasks efficiently and safely. In addition, proper light levels prevent eye strain, which allows us to work comfortably for longer periods of time.".
I need to install a 100K lumen fixture, or 10 10K lumen fixtures.
A single light is suitable for a small closet, in a large room you would end up with a bright central portion and dark corners, many shadows from any objects in the room too.
But the equation doesn't take into account the height at which the fixtures are installed and how far they need to be positioned from each other. Also, the equation only applies to ideal conditions, say an empty room. But in an industrial environment where you have machines and equipment in the way, the equation will yield an estimate that is lower than what's actually required to achieve the desired ft-candles.
Position a fixture above and behind each person's 'operating position' while they are at the machine. Additional lighting would go in aisles where materials are transported to and from each machine, these levels can be lower than what is needed at the machines. If emergency exits are 'off in the corner', far from any lights, an additional fixture would be needed in that area.
Any remaining spots that don't have a fixture above the area should have one.
Now you have your 50'x50' area with unevenly distributed fixtures that provide task specific lighting efficiently (not evenly) distributed.
In a setting like an office or school classroom you would go for 'even distribution' over individualized lighting since the makeup of any random 5'x5' area is highly similar.
To save a bit you could have slightly greater spacing nearer the windows and use slightly lower wattage but lamps are only available at specific wattages and ceiling joists to hang the lights are spaced at particular intervals - thus (in schools and offices) grid ceilings are usually populated with a light every other row, though modern lighting does have dimming to provide a savings that is less often used in industrial settings due to the high ceilings and greater cost of dimming high powered lights.
That gives you an idea what you want. You can take that to a lighting designer whom probably will be willing to assist you if you want to purchase and have them install over two dozen lights.
If you want to do it all yourself you will need a CAD for lighting design.
I recommend DIALux because it's easy to use and free. Simply draw your room, choose the fixture you are going to use (or one that is similar, for a guesstimate), and adjust the spacing to achieve a specific level of light at a specific 'working height' (sitting or standing) off of the floor.
Best Answer
The lamps aren't necessarily more efficient. It's just that there are less losses in power transmission.
Correct.
You have to remember that they can be somewhat out of phase giving a power-factor < 1 which needs to be taken into consideration for power (watts) calculations but you are on the right track.
Figure 1. The voltage difference between two phases is a sinewave.
Since there is no neutral connection you can look at it either way. Current will alternate between A and B.
Current will also alternate between B and C.
Figure 2. Two currents flowing in opposite directions cancel out as indicated at point (1) when the B phase is at zero volts (assuming resistive load). At other times B has to carry the sum of the return currents with a peak value when B is at maximum.
A load connected between two phases has only two wires and sees a sinusoidal voltage applied to it. As far as the load is concerned it has a single-phase supply. (It knows nothing about neutral.)
Single-phase breakers would leave the other phase connected and the potentially faulty circuit still live. Good practice would be to use a 2-pole breaker on each load or a 3-phase breaker to isolate everything at once.