Voltage drop

[Geber]

I have a question about the Table 9 that was used by some in trying to answer this. The resistance for 1000 feet of 12 AWG copper wire is 2.0 Ω. The effective Z at 85% power factor is 1.7 Ω. I'm trying to recall what I remember about RL circuits and see how to fit that with the terminology in the NEC. But maybe someone has a quick explanation of why the effective Z is less than R.

Also, I notice the caption of Table 9 says "for 600-Volt Cables" but later in the caption says "Three Single Conductors in Conduit". I'm confused, is it a cable or a conduit?

 

[EnriqueB]

I look forward to any ansawer you get. I may learn something from it.
Thank you

I already sent an email to him; I'll keep you posted.

 

[Carultch]

I have a question about the Table 9 that was used by some in trying to answer this. The resistance for 1000 feet of 12 AWG copper wire is 2.0 Ω. The effective Z at 85% power factor is 1.7 Ω. I'm trying to recall what I remember about RL circuits and see how to fit that with the terminology in the NEC. But maybe someone has a quick explanation of why the effective Z is less than R.

Also, I notice the caption of Table 9 says "for 600-Volt Cables" but later in the caption says "Three Single Conductors in Conduit". I'm confused, is it a cable or a conduit?

Smart$ has a diagram that shows it.

Voltage drop - chapter 9 tables & effective Z

Hi, I'm struggling to understand why the effective Z of conductors, which is used to calculate voltage drop, changes with respect to the power factor. For a given current I, I don't understand why the wire resistance could possibly be affected by the PF. Furthermore, I tried calculating the...

forums.mikeholt.com

Essentially, we're taking a projection of the voltage drop, that is aligned with the voltage that the load receives, rather than the actual vector of voltage drop. What we're interested in, is the difference between the magnitude of the source voltage, and the magnitude of the voltage that is received by the load. When the voltage drop is not aligned with the source voltage, as it is when you account for complex impedance, the difference is slightly less than the maximum possible difference if they were aligned (as we'd calculate in a resistive case).

 

[Geber]

Calrultch, looking at the diagram you pointed to by Smart $, we can say that the voltage received at the load has a lower magnitude than at the source, and is at a slightly different point in the cycle (that is, has a phase shift). From the point of view of how well the load operates, we care about the change in magnitude, but not about the phase change. So we treat the voltage at the load as the x, or real, axis, and treat the real part of IZ as the voltage drop.

 

[wwhitney]

When the voltage drop is not aligned with the source voltage, as it is when you account for complex impedance, the difference is slightly less than the maximum possible difference if they were aligned (as we'd calculate in a resistive case).

To expand on this, there are two complex impedances at play, the wire and the load, which are in series. For voltage drop in the wire, all we really need to know is the load current phase angle; the analysis works for any load with a consistent current phase angle, even it's not necessarily a fixed impedance.

Basically the voltage drop phase angle relative to the source voltage will be the sum of the load current phase angle plus the wire's impedance angle (arctan(X/R)). But the load typically only cares about the magnitude of the voltage it will see, not its phase, and to compute that to first order we only care about the component of the voltage drop parallel to the source voltage. That means for the case of 0 load current phase angle, aka a resistive load or a power factor 1 load, we can compute the relevant component of the voltage drop using only the wire resistance. In other words for this case Z effective, the value you need to use in a voltage drop formula, is just the resistance R.

When the load current phase angle isn't zero, i.e. the power factor is less than 1, then both the wire resistance and the wire reactance contribute to the relevant voltage drop, and Z effective becomes a mix of X and R. As the phase angle increase from 0 to 90 degrees, i.e. the power factor drops from 1 to 0, the contribution of R to Z effective decreases, and the contribution of X increases. The formula is in the footnote to Chapter 9 Table 9.

The relevance of this to the OP is that while wire resistance per unit length is inversely proportional to the conductor's cross-sectional area, the wire reactance per unit length hardly changes with conductor size. And for large conductors (500 kcmil is large) they are approximately the same magnitude. Which implies that for small conductors like the #12 in the OP, the wire reactance is basically negligible compared to the resistance. The upshot is that for voltage drop calculations on small wires, the lower the power factor, the lower the effective Z, and the lower the voltage drop.

Cheers, Wayne

 

[EnriqueB]

I have the answer from them, they made a correction on the book.


Enrique, thank you very much for providing the information I requested. Answer choice (b) was changed to 3.80V and should have been reflected in your digital product. Here’s the rational used to answer this question:


E

 

[EnriqueB]

I look forward to any ansawer you get. I may learn something from it.
Thank you

 

[ggunn]

For a balanced system, the current on the neutral is zero, so all the Vd is on the phase conductors since zero amps means zero volts.

Vd = IR, and %Vd = (Vd/V)(100%) where V is the phase to neutral voltage.

I like this method because it comes from first principles (Ohm's Law), it is very simple to implement, and it works for both single and three phase.

 

[wwhitney]

I like this method because it comes from first principles (Ohm's Law), it is very simple to implement, and it works for both single and three phase.

It works for 3-phase, but you have to understand that the voltage drop is towards the neutral point (for the balanced case that the current is in phase with the voltage to the neutral point). So to convert that to an L-L voltage drop, you need to take the voltage drop in one wire from the V=IR formula and multiply by sqrt(3) (rather than by 2).

Cheers, Wayne

 

[Tulsa Electrician]

Thank you for taking the time to ask and provide update.

Now that that we have the correct answer I'll share how I came up with the incorrect ansawer.

I read the question as a trick question and use three wires or 1.732 which fits the ansawer.
By using 1.732 it changes the footage for resistance. I also used the three phase formula instead of single phase.

However I knew that did not seam correct. So I went back to ohms and Kirchoff first law aka current law (I = E/R) and turned load into a series circuit and it provided the correct ansawer.

What I did was change the math to support the answer then did math check and it was proven to be incorrect.

I always like to prove an ansawer two ways. When is does not match I know I have an issue with the formula used. Which I should have done before answering so sorry for that.

For the 16 amps balanced three wire load, I turned that into two 7.5 ohm resister in series which are additive.

To start 240/16= 15 ohms
Then divide that by 2 for line to netural load or two 7.5 ohm resisters in series. I did this due to the 120/240 nature of the question. This way the math works as a math check.
120/7.5= 16
240/15= 16
E/I =R
The netural current would be zero for the balanced load. This follows Kirchoff first law

Ohms and Kirchoff first law I= E/R 240 volt/15 ohms= 16 amps (I) of load now we have to factor in the conductors feeding this load and the voltage drop based on the conductors resistance.

In this case we used simple VD=I *R
We have I (amps) so now we find R total for conductor. Using table 8 for #12 solid. R/1000 or 1.93/1000 = .00193 per foot
.00193*125 = .24125 for conductor
16*.24125= 3.86 for VD

So what I learned was everyone makes mistakes and sometimes it is good to question was does not make sense.

So thank you for your effort and thank you Mike holt team to identify and make corrections.

Tulsa

 

[ggunn]

It works for 3-phase, but you have to understand that the voltage drop is towards the neutral point (for the balanced case that the current is in phase with the voltage to the neutral point). So to convert that to an L-L voltage drop, you need to take the voltage drop in one wire from the V=IR formula and multiply by sqrt(3) (rather than by 2).

Cheers, Wayne

Of course, but %Vd is usually what is of concern. As long as you compare the single conductor Vd with V to N voltage you are good whether it is single phase or three phase. No pesky sqrt(3).

 

[wwhitney]

Of course, but %Vd is usually what is of concern. As long as you compare the single conductor Vd with V to N voltage you are good whether it is single phase or three phase. No pesky sqrt(3).

Agreed. But in a 3-phase 3-wire circuit, it's not obvious that to get %VD you need to compare the single wire VD formula to the L-N voltage. Particularly when the voltage system may not have an N in the first place!

Cheers, Wayne

 

[ggunn]

Agreed. But in a 3-phase 3-wire circuit, it's not obvious that to get %VD you need to compare the single wire VD formula to the L-N voltage. Particularly when the voltage system may not have an N in the first place!

Cheers, Wayne

Was I not clear about that? Yes, using the single conductor approach requires that you calculate %Vd as a ratio of Vd to V-N voltage, but once you have it it's the same number as you would get calculating it any other way, be it single phase or three phase. In a delta configuration you ignore the fact that there is no neutral and calculate Vd as if there were one, i.e., 277V for a 480V three phase system and 120V for a 208V one.

It's simple and it works.