Note: I am leery of any internet voltage drop calculator where one does not know the algorithm

*(edit to add)* Assumption: System is 208/120Y, 3ph, 5W, three phases, Neutral, EGC.

Note: I didn't check any of the fill calcs.

I'm a slow poster - Acknowledging and adding to ww and infinity posts

Here's what I have:

post 1 said:

So I went to adjust the R2 formula at the bottom of table 8 note 2, to adjust the K factor so I can present my calculation in a more accurate and professional way, and I could use some guidance.

Since you are using Table 8, you are saying the load is straight resistive. There is no inductive component, the pf is 1. That is fine. It will influence how the Vd is calculated. Table 9 does not apply.

Infinity is dead-on. The feeder OCP is 1200A, the conductors must be 1200A or more. With 6 parallel, each conductor must be 200A (or more). That equates to 250 AL or larger.

post 6 said:

The NEC handbook pointed out that we are permitted to use the ampacity of the 90°C column of 310.15(B)(16) for derating rating conductors for things like # of CCC, Ambient Temp, and voltage drop; so a basic example is 4 CCC get derated by 80%; therefor #10THHN can be derated from 35 amps to 28 amps.

I am trying to generate a calculation based on the 90°C rating so we reduce to anything less than 800kcmil.

When you look at the ampacity of an 800kcmil AL in the 75°C column it is rated at 395 amps, and when I look at the 90°C column I see that a 700kcmil can carry 425 amps, which is more than the 75°C column and saves the day

This makes no sense at all. The conductors are way oversize the require for the OCP. 75C, 90C - it doesn't matter. The conductors are already big enough for 200 amps. I don't have a handbook, however, I can not think of any reason why the handbook would reference the "90C column" as affecting "Voltage Drop". If you didn't miss-read that, I have no idea of what they are referencing.

Now let's look at the Voltage Drop:

post 6 said:

Since the K rating of 21.2 is based on 75°C, I wanted to calculate the actual K value at 90°C.

To do this, first I need to find the adjusted resistance of 700kcmil, using the formula of note 2 table 8 by subtracting the 15°C - 75°C.

This allowed me to calculate a K of 18.06; and when I use the CM formula for 2%, I end up being able to run the 700kcmil feeders.

I think I understand why you are doing this:

The DC resistance (table 8) for 75C, 750kcmil AL is right on the ragged edge for the required resistance to get the Vd down to 2%. Doing the calc single line to neutral, no current on the neutral, .

02 x 120 = 2.4 V drop.

2.4V/1082.7 = .0022 ohms for the 6 parallel conductors

.0022 x 6 = 0133 ohms/470 feet (for each conductor)

Which gives (.0133/470) x (1000/470) = .0283 ohms/1000 feet.

750kcmil AL = .0282 ohms/1000ft. (at 75C)

750kcmil AL at .0282 ohms/1000 meets spec.

This I don't get:

DC resistance for 75C, 700kcmil is just a little bit too much resistance. Table 8 does not have an entry for 700kcmil. So you use the k-factor and area to come up with 75C, 700kcmil resistance = (750/700) x .0282 = .0302 ohms/1000 at 75C. whoops, too much. Your calcs show that if you drop the conductor temp 15C, the resistance drops enough to get the 700kcmil resistance under .0283 ohms/1000ft. Could be true. I did not check your calcs.

Questions:

This does not have anything to do with the 90C column.

How did you decide the conductor temp would be 15C lower than 75C? The conductor temp will certainly be less than 75C. If you use 750kcmil (ampacity 435A), with a load of 1082.7A/6 = 180A load.

Of course the conductor temp will be less than 75C. How much less, I don't know. I see a pretty complicated mechanical engineering type heat loss calc to figure it out. Soil temps, thermal resistances, and lots of stuff I don't know.

Unless I am clear screwed up and you are trying to calculate the 700kcmil resistance at 90C. And that makes no sense at all - why would you care?

You have the answer and supporting calcs: 750kcmil AL will work and meet the 2% Vd spec.

This you didn't ask, and I would never give advice concerning revenue matters, so please consider the following as gentle suggestions from someone that is having a terrible time seeing your work from my side of your monitor. Unless you have a point that completely eludes me (which is entirely possible):

Leave out anything to do with the 90C column

Leave out anything to do with any guesses at the conductor temp being less than 75C.

Leave out any references to any internet calculators. Show your calcs referencing Table 8. Unless the engineer of record likes a particular one - then use that one.

Random underground rumblings from

the worm

*(edit to add)* I just read this over. It appears a bit jumbled. Hopefully I understood and there is something useful.