Three Phase Wattage question - AGAIN!

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DLTravis

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OK ? a little beating a dead horse here. :roll:

I?ve been asked by my employer to figure out how much it is costing us to light the exterior of our building. I am trying to get an accurate count of the fixtures and wattages of these fixtures. I think this will net the most accurate number. However, that task is not as easy as it sounds.

I?ve seen similar questions asked countless times here and elsewhere and I find the answers confusing. Usually the poster is trying to figure out Amp load to calculate wire size etc.

Isn?t there an easy formula to figure out wattage using these measured numbers?

Measured Current and Voltages
Va-b ≈ 211
Va-c ≈ 210
Vb-c ≈ 210

Ia ≈ 65
Ib ≈ 42
Ic ≈ 55

This was probably covered in my schooling 13 years ago. The mind tends to forget things seldom used?:grin:
 
There actually isn't enough information there to give a accurate answer, but there is enough information to give an approximate answer.

To get an accurate answer, you need to know the power factor of the load.

To get an approximate answer, you ignore power factor, and assume that the supply voltages are balanced.

You calculate the line-neutral voltage, and then multiply the line-neutral voltage by the sum of all the line currents.

Thus:
210/1.732= 121V

121 * (65 + 42 + 55 ) = 19.6 KVA

So you can say 'Something less than 20 KW.'

To get more accurate you will need a measurement of power factor, and to get more accurate than that you will need a proper watt meter.

-Jon
 
The KVA could be calculated from the above figures, but not the wattage since we dont know the power factor.

I would calculate the VA , by multiplying each line current by the phase to neutral voltage, which in this case is 120 volts nominal, or about 121.5 volts actual.
phase A 65 amps at 121.5 volts= 7,898 VA
phase B 42 amps at 121.5 volts= 5,103 VA
phase C 55 amps at 121.5 volts= 6,683 VA

This gives a total of 19,684 VA or 19.7 KVA.

If the lamps are incandescent, then watts=VA if, as is more likely these are discharge lamps, then we need to know the power factor to calculate the load in KW and therefore the running costs.
 
Thanks for all the help :grin:

I thought it was more simple than I was making it out to be. I should have payed more attention is skool :cool:
 
In case you were wondering :rolleyes:
State-Capitol-MN-3.jpg
 
alternate method of guestimating

alternate method of guestimating

Hey Guys!! Why couldn't you add up all the wattages of the bulbs then multiply by one hour.

for instance

(400watt MH) X ( 1/1000 kilowatt/watt) X (1 HOUR) X (4380 hrs/peryear) X ( $0.10) = $ 175.2


Does this sounds right???

seems a little high to me..
 
rt66electric said:
Why couldn't you add up all the wattages of the bulbs then multiply by one hour.

Does this sounds right???
Click to expand...
Because the bulb wattage does not reflect the actual consumption of the entire fixture; i.e., the bulb and the ballast.

Some fluorescent ballasts actually show a line current less than P/E would suggest, but most discharge lighting uses more than the bulb rating.
 
rt66electric said:
Hey Guys!! Why couldn't you add up all the wattages of the bulbs then multiply by one hour.

for instance

(400watt MH) X ( 1/1000 kilowatt/watt) X (1 HOUR) X (4380 hrs/peryear) X ( $0.10) = $ 175.2


Does this sounds right???

seems a little high to me..
Click to expand...

See OP :grin:


Here’s an update. I was able to come up with a fairly accurate number. You all might find it interesting that it costs about $3.02/hr. to light the place up (exterior accent lighting only).
 
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gar said:
090212-1554 EST

What is your cost per KWH, and how did you arrive at the cost per hour?

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Click to expand...


We pay $.07/KWH

As I stated in my original post I was also working on getting an accurate count on the devices. After much looking I was able to gather all of that info. All of the fixtures are made by the same Mfg. they claim an 85-90% efficiency.

Added all the bulb wattages * 1.1 (to add 10%) divide by 1000 multiply by .07 "viola"

This turned out to be easier than trying to locate all of the feeders and take measurements. I'm sure this is not spot on accurate but it should be close enough for government work :grin:
 
gar said:
090212-1625 EST

How did the original voltage and current measurements from your first post relate to this project?

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Click to expand...


If you read the OP you will see. Those numbers were from one of the main contactors in the system. Originally it looked to be an easier task to take readings to get a more accurate number. However, more digging revealed it was going to be difficult to separate out only the accent lighting and find all of the feeds. Being a good lazy man I took the easy, albeit less accurate, path.:wink:
 
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