Circuit loading

3 replies [Last post]
x2jearly's picture
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Joined: 2014-07-03

Breaker loads:
I am circuiting 100 watt Metal Halide fixtures, 277 v, stanchion mounted on exterior platforms and walkways. What should I use for circuit loading:

a) 100w= 130 input wattage (from ballast chart) or
b) 1.15 max. input amps (from same chart).

I, along with the people I have worked with, always used the input wattage, but have recently been told to use the input amps.

130 input wattage vs 1.15 amps (318 watts) which is about 2.5 times more. Thats a lot more breakers.

Thanks!

 

ecmjacomen's picture
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Joined: 2013-10-08

The connected load is = to the input wattage from the ballast chart divided by the power factor. Assuming a 90% power factor, the connected load = 100 / 0.9 = 111.1 volt-amperes.
Because this is probably a continuous load (more than three hours "ON"), the calculated load is = 1,25 x 111,1 = 139 volt-amperes (equivalent to 139 / 277 = 0.5 amperes per fixture). Using a 30A CB per single-phase circuit, the maximum number of fixtures per circuit = 30 / 0.5 = 60 fixtures per circuit.

Linda Allen's picture
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Joined: 2013-12-06

I agree with the calculations stated by ecmajacomen with the exception of the number of fixtures that can be on one circuit. From the standpoint of highway lighting design, when running long circuits down roadways, one must also consider the voltage drop on the wire to be able to determine the number of fixtures on one circuit.

W A Werning's picture
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Joined: 2014-02-12

Use 220.14(D), Luminaires, for calculating the maximum volt-amperes if it is other than incandescent luminaire.

Formula: Single phase VA formula = V X A.
1.15A X 277V = 318.55VA.
Continuous duty: 319 X 125% = 399VA (each luminaire). This is disregarding the power factor.

A 20A circuit breaker should be used:
20A CB X 277V = 5,540 volt amperes circuit capacity.
5,540VA / 399VA = 13.88, rounded down to 13.

Check point: 13 X 399VA = 5,187VA (14 X 399VA = 5,586VA), violation NEC. Why use a 30A CB?

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