Fixture Circuit Calc

A

Alwayslearningelec

Senior Member
Location
NJ
Occupation
Estimator
Hello I have a calculation question.

Voltage is 120v.

Panel is located 500’ away

Total length 2,800’

Engineer wants them fed with (4) 20a circuits.

Engineer has come back and said (4) 20A circuits is not enough to feed these and we should add additional circuits as needed.

Not considering voltage drop I believe I can get 1900 watts on a circuit. I would need 5 circuits to feed these correct?

Fixture wattage below. Thanks

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If you can distribute the load evenly figuring it as a continuations load you would need 6 circuits.
 
Alwayslearningelec said:
That's what I thought and why I asked. So why is it watts?
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You need to take some sort of a basic electricity course to really get it, but I will answer you anyway. Volts × amps is watts in a DC circuit. In an AC circuit, volts and amps are virtually never in perfect unison with each other, meaning that when voltage is at it peak, it either leads or lags the current. This causes "losses" in the amount of work that can be done. It is directly proportional. The "phase angle" from peak to zero is 90º. If the phase angle was 9º out that would be 10% out, that would mean 10% less power. It doesn't matter if the phase angle is positive or negative, same result. So, phase angle is expressed as power factor as a percentage from 1-100%. Watts in an AC circuit = volts × amps × power factor.

Sorry, you asked. It makes my head hurt just typing it.
 
Strathead said:
You need to take some sort of a basic electricity course to really get it, but I will answer you anyway. Volts × amps is watts in a DC circuit. In an AC circuit, volts and amps are virtually never in perfect unison with each other, meaning that when voltage is at it peak, it either leads or lags the current. This causes "losses" in the amount of work that can be done. It is directly proportional. The "phase angle" from peak to zero is 90º. If the phase angle was 9º out that would be 10% out, that would mean 10% less power. It doesn't matter if the phase angle is positive or negative, same result. So, phase angle is expressed as power factor as a percentage from 1-100%. Watts in an AC circuit = volts × amps × power factor.

Sorry, you asked. It makes my head hurt just typing it.
Click to expand...
lol, thank you. Makes my head hurt reading.
 
Alwayslearningelec said:
lol, thank you. Makes my head hurt reading.
Click to expand...
Well, the upshot is that while the spec is 3.7W / LF for the lights, you need to check the VA / LF to do the calculation in post #8 (or equivalently, the power factor, as W = VA * PF). So you don't quite have enough info. If the power factor is 0.9, then the effect on post #8's calculation is to make the answer 5.4/0.9 = 6 circuits exactly. If the power factor is below 0.9, you'll need 7 circuits.

Cheers, Wayne
 
If your power source is 3 phase, it would be real nice to subdivide the loads into groups of 3, so 6 circuits. If you have to go to 7, I would split it up into 9 to balance the load and even out the voltage drop.
 
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