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#13773 09/11/02 03:27 PM
Joined: Jul 2001
Posts: 311
F
Member
Please check my math:

(22) 1500 watt metal halide light standard fixtures @208 volts, 8.2 amps each. They will be balanced on a 120/208 volt, three phase, 4 wire service.

37,523 watt load divided by 208 x 1.73=104 amps.
The load would be considered continuous thus multiply by 1.25 = 130 amps total load.

Am I correct?

Joined: Jan 2001
Posts: 1,044
Tom Offline
Member
I dunno.

Try this, 21 fixtures installed so they balance out on all 3 phases equals 114.8 amps, add one more fixture on 2 phases brings the total to 123 amps multiply by 1.25 = 153.75 amps

you'll either need a 150 amp panel (your calculation) or a 200 amp panel (my calculation). The price difference is probably non-existant. The only cost that might matter would be the feeder size.


Few things are harder to put up with than the annoyance of a good example.
Joined: Sep 2002
Posts: 131
T
Member
Frank, I did the calculation and came up with the same amperage as you did. I beleive you are correct

Joined: Jul 2002
Posts: 175
E
Member
Each fixture - 208v x 8.2a = 1705.6 va
22 fixtures + 2 (to balance load) = 24 fixtures
1705.6 va x 24 = 40 934.4 va

40 934.4 va divide by 208 = 196.8 amps
divide by 1.73 = 113.7 amps
113.7 x 1.25 (continuous load) = 142.5 amps

Ed

Joined: Mar 2001
Posts: 2,056
R
Member
Frank,

Don't forget to add (10%?) for ballasts.

Joined: Jul 2001
Posts: 311
F
Member
Is that for power factor?

Joined: Jan 2001
Posts: 1,044
Tom Offline
Member
Am I missing something? Why is everyone trying to change this calculation to volt-amps when Frank has stated that each fixture draws 8.2 amps?

Doing the Volt-Amp thing might be advantageuos if this load could be equally shared by all three phases, but since it can't, I feel the resulting answers are too low. The math is correct, so I'll assume that everyone slipped off their boots to use their toes [Linked Image] and they all did their gazintas correctly.

There will be 15 fixtures on A phase x 8.2 = 123 amps, 15 fixtures on B phase x 8.2 amps = 123 amps

C phase only has 14 fixtures x 8.2 amp=114.8 amp.

125 amps x 125% = 153.75 amps, too much for 150 amp feeder or 150 amp panel.


Few things are harder to put up with than the annoyance of a good example.
Joined: May 2001
Posts: 160
C
Member
A 1500 watt fixture at 208 volts will draw 7.2amps continuous.Why are we talking 8.2 amps and adding 125% for a continuous load?
What am I missing?
Chris

Joined: Apr 2002
Posts: 7,283
Likes: 3
Member
Chris:
A 1500 watt metal halide sport light draws 8.2 amps at 208 voltsat full illumination, and 4.70 amps starting. Source of info is Lumark/Cooper Lighting.

That is a high power factor, CWA.

HotLine1
John

PS to Frank: Put in a 200 amp feeder and panel please!


John
Joined: May 2001
Posts: 160
C
Member
John,
Thanks for the info.Does Lumark/Cooper have a WEB site with tech info for their products?
Chris

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