AC Series Voltage Drop

[cestone]

Hello All,

I was reading an article on series voltage drop in EC&M and part of the example has me baffled and hope some of you can make sense of it for me.

The example is a string of five luminaires equally spaced at 200 feet. Voltage is 240volt using #10 AWG copper in PVC with each luminaire having a load of 1.354 amperes. The length of conductors from the panel to the first luminaire is also 200 feet.

The writer states the following: "the % Vd equals 100 x (Vd x 240). The % Vd for each segment is equal to (0.650V ? 240V) x 100=0.271%.

The respective currents for the five segments are 6.77amperes, 5.416amperes, 4.062amperes, 2.708amperes, and 1.354amperes.

The respective cumulative % Vd for the five segments are 1.354%, 2.437%, 3.250%, 3.792%, and 4.063% for the final luminaire position.

If using #10 AWG conductors for the entire 1,000' to the last luminaire, wouldn't the Vd decrease from the first luminaire at (1.354% Vd) to the last luminaire? How would the writer come up with 4.063% Vd for the last 200' of the string using #10 AWG?

Thank you all in advance for your help and comments.

Errol

 

[david luchini]

Voltage drop did decrease from the first luminaire to the last luminaire.

The voltage drop on the first 200' (from source to first luminaire) is 1.354%

The voltage drop for the next 200' is 1.083%
For the next 200' is 0.813%
For the next 200' is 0.542%
For the last 200' is 0.271%

The voltage drop from the source to the last fixture is 4.063%, which is the sum of the five segment drops. But the last drop is clearly the smallest.

 

[IamJonscranium]

Are you asking if the rate of voltage drop changes over the length of the feed?


ETA: Nevermind, answered above.


Just to clear it up, voltage drop decreases because, well, the voltage decreases? Sort of like how a leak slows down as the bucket empties.

 

[cestone]

AC Series Voltage Drop

Thank you David

I understand now how the writer came up with each of the five cumulative %Vd for each segment.

I understand needing to know the cumulative %Vd at the last device of 4.063% in accordance with the NEC fpn, but is there any need to know the %Vd of the second to the last segment or the third to the last segment?

 

[david luchini]

Just to clear it up, voltage drop decreases because, well, the voltage decreases? Sort of like how a leak slows down as the bucket empties.

I would say that this is incorrect. The voltage drop decreases because the current decreases. For instance, the example has a 1.354% drop from the (240V) source to the first fixture with a 6.77A load. This means the drop is 3.2496V.

Imagine if the circuit was 5 equally spaced luminaires that had a load of 1.354Amps each on a 208V circuit. The voltage drop from the (208V) source to the first fixture with a 6.77A load (on the same #10 conductors) will be 3.2496V, the same as in the original example. Of course, since the source voltage is now lower, the first segment percentage voltage drop will be 1.562%.

I understand needing to know the cumulative %Vd at the last device of 4.063% in accordance with the NEC fpn, but is there any need to know the %Vd of the second to the last segment or the third to the last segment?

I'd say that you cannot know the vd to the last fixture unless you first know the vd for each of the segments before it.