Finding total Watts in flourecent lighting

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jocool1313

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I have an occupancy sensor that has a max load of 800 watts. With flourecent lighting to find the total watts do I ...

1.) Add all the watts from the lamps in each fixture? Ie 32 watts per lamp 3 lamps per fixture = 96 watts

2.) use P=I*E and get P=.5 amps * 120 volts or p=.5*120 or P=60 watts?

I believe i need to use option 2 since a ballast alters the incoming voltage and that is how you can get 96 watts of power with only .5 amps.

Am I correct?

Thanks,
Joe
 
chris kennedy

chris kennedy

Senior Member
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Bob is correct as per 220.18(B). But when figuring what an OS can handle, I take the total lamp wattage ?V?.85 then compare that to the ballast rating of the OS.

Welcome to the Forum.
 
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Volta

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Location
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iwire said:
Where is that outlined in 220?
Click to expand...

I don't think it is.

The best I can do is 220.5, as a list of nominal voltages. I don't know the sq ft being illuminated, so 220.12 is out. I assume this load is for general illumination, so 220.14 is out.

Of course we are not yet considering the feeder or service, as the added load of the OS device is negligible, so Parts III and IV are out.

Otherwise, I dont see where 220 applies. We need to know that the load does not exceed the device's rating per 110.3(B), probably. This guessing method is just that, but sounded reasonable to me, until someone actually removes the ballast cover.
 
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gar

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100606-2025 EST

Using 800 W as a criteria for allowed load on a device is probably an invalid use of that specification. Is there more information provided? For example is the device rated for 800 W of tungsten load at 120 V, or an 800 W resistive load for any voltage from 0 to 120 V, or a constant resistance of 120^2/800 = 14,400/800 = 18 ohms at any voltage from 0 to 120 V, or an 800 W motor at 120 V?

More correctly it probably should be rated for current with some additional qualifications. For example a tungsten load of 800 W at 120 V is about 6.67 A after the filament is up to temperature. Probably about 1 cycle. The short time inrush might be close to 100 A for a set of lamps starting at room temperature.

Fluorescents have a different type of inrush and also are not a pure resistive load. Steady state current will be higher than that calculated from the wattage of the lights.

The type of output switch in the device determines the failure mode. A mechanical contact has different characteristics than that of a solid state device.

Then also consider that the manufacture may use some specmanship in defining their ratings.

Follow what others have told you about looking at the code, but you need to understand how devices work and what the ratings mean.

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V

Volta

Senior Member
Location
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jbelectric777 said:
But he "knows" his actual connected load. So each ballast current X 120 then the sum X 1.25
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I don't think that we know the load yet.

When we do figure that out, I think we can apply 100% of it to the switch.

If the inductive rating is 800 watts, then we can continuously supply 800 watts of inductive load through it, in my opinion.
 
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gar

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If I connect an inductor, for example an unloaded transformer, to an AC source you will find a large disparity of power (watts) to VA. As an example I measured 0.260 A @ 120 = 31 VA, while the power is 4 W, or about a 7 to 1 ratio.

Two different Slimline fixtures that I have are quite different. Both have identical new bulbs in them. One is 137 VA and 134 W, very good PF, the other is 110 VA and 97 W. These are both magnetic ballast units. Bought at different times and at least one may have a new ballast relative to the original.

.
 
W

WastefulMiser

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Location
ANSI World
gar said:
100606-2246 EST

Two different Slimline fixtures that I have are quite different. Both have identical new bulbs in them. One is 137 VA and 134 W, very good PF, the other is 110 VA and 97 W. These are both magnetic ballast units. Bought at different times and at least one may have a new ballast relative to the original.
Click to expand...

gar, what did you use to measure real power?
 
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gar

Senior Member
Location
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EE
100607-0545 EST

WastefulMiser:

I used a "Kill-A-Watt".

I have run tests on these to see if they can handle wide variations in power factor.

For example I can put a 2.25 mfd oil filled capacitor on it and read 0.11 A and 0 W. The "Kill-A-Watt" only quantizes to 1 W. Calculated current for 2.25 mfd is about 0.101 A, but I have a distorted sine wave. A different, but average reading type, reads 0.127 A. My RMS meter is not at home so I can't use it for comparison.

An electrolytic AC motor starter capacitor, 21-25 mfd, reads 1.28 A and 17 W. This I expect to have higher losses. I do not have a larger capacitance oil filled unit for comparison.

An unloaded drill press motor, except for spindle, reads 6 A, 380 W, and 720 VA.

On a 100 W incandescent bulb the readings are 104 W and 104 VA.
On one standard CFL 26 W and 43 VA.

The "Kill-A-Watt" units are not high quality and do have problems, but are a useful inexpensive instrument.

.
 
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