Voltage Drop & NEC

[N123]

Normally for single phase circuit voltage drop we use the following equation: (I * L * R / 1000) where I = load current, L = circuit length and R = resistance of the cable

For single phase, L = circuit length is equal to 2 * one-way length to account for the return length, so the equation can be re-written as (I * (one-way length * 2) * R / 1000).

For three phase circuit voltage drop we use the following equation: (1.732 * I * L * R / 1000) where I = load current, L = circuit length and R = resistance of the cable.

In this case is the length L one-way or two-way length? Taking a look at NEC Chapter 9 Table 9 calculation example, they are using a feeder with 100A continuous load, 240V source, 3 phase with a circuit length of 150 feet. They performed the following: Voltage drop (line to neutral) = 0.16 ohm * 150/1000 * 100A = 2.4V. To find line-to-line voltage drop they took 2.4V * 1.732 = 4.157V. Is this an error in the code? Should circuit length be multiplied by 2 here resulting in ~8.3V drop line-to-line? Does three phase circuit voltage drop only consider one-way length in the equation?

 

[don_resqcapt19]

The square-root of three in voltage-drop calculations is not there for the return current on the neutral as the 2 is in single-phase calculations, the √3 is required to convert the line-to-neutral voltage drop to a line-to-line voltage drop

 

[ggunn]

The square-root of three in voltage-drop calculations is not there for the return current on the neutral as the 2 is in single-phase calculations, the √3 is required to convert the line-to-neutral voltage drop to a line-to-line voltage drop

In the balanced case one can simply calculate the voltage drop by Ohm's Law in a phase conductor and compare it to the line to neutral voltage to get %Vd. Whether it is single phase or three phase does not matter.

 

[wwhitney]

Taking a look at NEC Chapter 9 Table 9 calculation example, they are using a feeder with 100A continuous load, 240V source, 3 phase with a circuit length of 150 feet. They performed the following: Voltage drop (line to neutral) = 0.16 ohm * 150/1000 * 100A = 2.4V. To find line-to-line voltage drop they took 2.4V * 1.732 = 4.157V. Is this an error in the code? Should circuit length be multiplied by 2 here resulting in ~8.3V drop line-to-line? Does three phase circuit voltage drop only consider one-way length in the equation?

The NEC example is not in error. Here's an explanation for an electrical engineer:

In AC, voltage and current are phasors. Resistive voltage drop is in phase with the current on the conductor. The one-way voltage drop formula gives the resistive voltage drop on a single conductor.

For a 2-wire circuit, a resistive load has current in phase with the voltage between the two wires. Each conductor drops voltage towards the midpoint voltage, so you get two voltage drops that add, or a factor of 2. For example on a split phase system L1, N, L2, with L1-N = L2-N = 120V, if the one way voltage drop for a resistive load supplied L1-L2 is 5V, then at the load the system voltages will be L1-N = L2-N = 115V. The total voltage drop is 10V, so the load will see 230V, rather than 240V.

For a 3-wire delta connected 3-phase circuit, the situation is more complicated. Even if the supply system has no neutral conductor, it is useful to consider the neutral point N as an abstract voltage reference. The simple case occurs when the line currents are all equal and are all in phase with the voltage L-N; for example 3 identical 2-wire resistive loads each connected L-L. Each load's current is in phase with the L-L voltage, but when you add the two currents to get the total current on a line conductor, the components perpendicular to the L-N voltage cancel and the total current is in phase with the L-N voltage. So for this simple case the voltage drop on each conductor, given by the one way formula, is towards N.

For example, on a 208Y/120V system with exactly nominal voltages, if the one way voltage drop in this simple case is 5V, the voltages at the load will all measure 115V L-N. If you are interested in the L-L voltage drop instead of the L-N voltage drop, then as usual for a balanced system, the L-L voltage will be sqrt(3) times the L-N voltage. That also means the L-L voltage drop will be sqrt(3) times the one way voltage drop.

It is possible to have a balanced 3-phase load where all the line currents are equal but not in phase with the voltage L-N; they would have a constant lagging or leading angle compared to the L-N voltage. In that case the voltage drop given by the one way formula would not be in phase with the L-N voltage, and for the above analysis you'd multiply the voltage drop by the cosine of the phase difference.

Cheers, Wayne

 

[wwhitney]

In the balanced case one can simply calculate the voltage drop by Ohm's Law in a phase conductor and compare it to the line to neutral voltage to get %Vd. Whether it is single phase or three phase does not matter.

Where for single phase, the balanced case refers to a split phase system L-N-L, with a 2-wire load connected L-L. Then the above is true when the currents are in phase with the L-N voltage.

Cheers, Wayne

 

[ggunn]

Where for single phase, the balanced case refers to a split phase system L-N-L, with a 2-wire load connected L-L. Then the above is true when the currents are in phase with the L-N voltage.

Cheers, Wayne

It also works for three phase. In single phase the factor of 2 for send and return is canceled by the factor of 2 for the L-L to L-N ratio. In three phase the same thing happens to the sqrt(3).

 

[wwhitney]

It also works for three phase. In single phase the factor of 2 for send and return is canceled by the factor of 2 for the L-L to L-N ratio. In three phase the same thing happens to the sqrt(3).

Yes, I didn't say otherwise. I was just amending what the term "balanced" has to mean in the single phase case for your statement to be true.

Cheers, Wayne