Voltage drop with Universal Voltage Fixtures

I

I^2R

Member
Location
NH
Hi,
Calculating a 20A 277V pole lighting circuit along the driveway of a large industrial site.
16 fixtures, total of 1,975 foot run. Calculation (see below) gives 6.91% voltage drop at the farthest fixture. Typically, I'd consider this too much, however the fixtures are modern "Universal Voltage" and accept 120-277V. The far fixture would be getting 257V and would be perfectly happy. Aside from violating the "recommendation" in the NEC regarding voltage drop, I don't see a problem. At these distances, upsizing the wire more does not give significant improvement, unless I go to silly sizes ($$$). I'm already at #8 to the first fixture and #10 from there.
The only other real issue I can see would be the effect on clearing time of the breaker, but I don't really know how to quantify that. The total current is only 6.75A, so I could drop the breaker to 15A for faster tripping.

A similar scenario is playing out with the indoor circuits for the 180 high output highbay fixtures.

Thoughts?

P.S., the spreadsheet is an iterative calculation, so it takes into account the increased amperage due to decreasing voltage along the circuit.

1749051151307.png
 
I don't see an issue with the voltage situation. Even with a single-voltage fixture, the worst thing would be that the lights may be dimmer at the far end. I am assuming the drivers can handle lower voltages.
 
Those modern light fixtures probably have Switch Mode Power Supplies.
The nice thing about most SMPS's is they don't much care about line voltage.
If the line voltage drops, they just draw more current for the same output power.
and the opposite happens if the line voltage increases.
 
If you have a single distant constant power load, and the voltage drop works out to 5% while operating, then the ratio of power lost as I2*R to power delivered is 5%/95% = 5.3%.

For your application, you could easily calculate the I2*R losses in each segment, total those, and compare that to the power delivered to the fixtures. If it's less than 5.3%, that complies with the "spirit" of the 5% voltage drop limit, as least as far as energy conservation goes. [Edit: I get 187W / 19*93W = 10.6% if I didn't make any calculator typos.]

Cheers, Wayne
 
Last edited:
zbang said:
You might run into energy conservation codes which limit the drop to minimize losses, but that's not an electrical matter as such.
Click to expand...
Yes, the 2018 IECC limits voltage drop to 5% on the feeder and branch circuit combined.

So not really allowed in NH, or a few other states, and probably a few areas that have locally adopted an energy code:

State Portal | Building Energy Codes Program

www.energycodes.gov www.energycodes.gov

But it amazes me how many contractors act like they have never heard of an energy code. :unsure:
 
steve66 said:
Yes, the 2018 IECC limits voltage drop to 5% on the feeder and branch circuit combined.
Click to expand...
2021 IECC C405.10 Voltage Drop:

"The total voltage drop across the combination of customer-owned service conductors, feeder conductors and branch circuit conductors shall not exceed 5 percent."

Nothing corresponding in the residential IECC, so it applies only to the commercial IECC.

Cheers, Wayne
 
You might consider running this as 4 wires and 480/277V, with the fixtures distributed across all three phases. This means running 4 wires but they are thinner; you get ~2% drop with 4 #12 wires.

Also since your system is being used well below the ampacity of the conductors, you can reasonably use the 25C resistance rather than the 75C resistance. That helps your numbers a bit.

-Jonathan
 
I'm not able to reproduce the numbers in the OP. Treating the fixtures as constant current (based on the power rating and nominal 277V) gives me a total current of 6.37906A. If I then use the currents in each segment to calculate the voltage drop in each segment to updated the voltages at each fixture to recompute the currents, I get a total current of 6.41882A. Repeating the process gives a total current of 6.41900A. Further iterations do not change any of those significant figures. [So convergence seems rapid, and the correction of 40 mA from the constant current approximation seems negligible.]

Edit: the above is using the resistance values per kft in the OP. Those values are slightly higher than the values in the NEC, both Chapter 9 Table 8 and Table 9.

Cheers, Wayne
 
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