Voltage drop and the 90 degree column

[KP2]

Hello folks, I was wondering if someone can help me make some adjustment for voltage drop based on the 90°C column.

Some quick details; I am bound to 6 sets of conduits and need to supply 1082.7 amps at a distance of 480' from the 208 volt 3Ø supply without exceeding 2% voltage drop with aluminum.

I would like to select my wire size based on the 90°C column, and my first attempt was to size the run based on the 75°C information, and then use the conductor ampacity from the 75°C column to select a conductor from the 90°C.

The CM formula equals 6 sets of 764.5kcmil so that rounds up to 800kcmil, this equals 395 amps per conductor, so we slide over to the 90°C column and can see that a 700kcmil is good for 425A. I would like to use the 700's but I need to have a better way to validate my slide rule here.

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.

I believe I can subtract 15°C from the 75°C and recalculate the resistance to solve for the exact K.

I get a K of 18.06 which gives me the 700kcmil feeders I'm looking for.

Thanks in advance.
Kevin

 

[infinity]

If your calculation says 700 kcmil you're likely going to need to use 750 kcmil conductors. 700 kcmil conductors are pretty rare, usually special order, and would then be more expensive than 750's. You need to factor in that the terminations will be 75° C and use the 75° C conductor ampacity in determining conductor size. Also what size OCPD will you be using to protect these conductors?

 

[iceworm]

KP - I 'm feeling slow today. Generally, the conductor size is picked by ampacity, or picked by voltage drop. Usually the two don't run into each other. However, before that:

When you say "6 sets of conduits" I'm translating that to "there are three phase conductors in each conduit, giving six parallel conductors for each phase". If that is so, then each parallel conductor is carrying 1182.7/6 = 180A for each of the 6 conductors.

Given that - the rest of your post doesn't make any sense.

I tried "6 conductors" instead of 6 conduits. That gives 2 parallel per phase - which of course is 1082.7/2 = 541A per parallel conductor.

So, I'm missing something.

 

[electrofelon]

KP - I 'm feeling slow today. Generally, the conductor size is picked by ampacity, or picked by voltage drop. Usually the two don't run into each other. However, before that:

When you say "6 sets of conduits" I'm translating that to "there are three phase conductors in each conduit, giving six parallel conductors for each phase". If that is so, then each parallel conductor is carrying 1182.7/6 = 180A for each of the 6 conductors.

Given that - the rest of your post doesn't make any sense.

I tried "6 conductors" instead of 6 conduits. That gives 2 parallel per phase - which of course is 1082.7/2 = 541A per parallel conductor.

So, I'm missing something.

Ice, I am very confused also and do not understand the OP. Perhaps he can re-explain what he is trying to do.

 

[infinity]

This is how I read it:

6 conduits
1082.7 amps load
480'
208 volts, 3Ø
2% max VD

What is the minimum aluminum conductor size in each of the 6 parallel raceways? Since there is no mention of load type you would also need to factor in the OCPD size to enure that the conductors are large enough to meet the size of the OCPD. For 1082 amps the next standard size would be 1200 amps but with a 480' run that shouldn't be a problem.

 

[KP2]

Sorry to be confusing; but here is the situation;

We have a 208V 3Ø feeder to a meter bank for a multifamily project.

The feeder has a 1200 amp main disconnect and is 470' in length and will be fed with Aluminum conductors. The demand load is calculated to be 1082.7 amps.

At 3% Voltage Drop you can run 6 sets of 600kcmil to this load, so we installed 6 conduits under the slab.

Now, I am finding out that we needed to be at 2%. When I make that adjustment to the voltage drop calculation, the result is 6 sets of 800kcmil, based on the 75°C column. (Thanks to Bill Brahmford for the calculator) After resizing the EGC, this feeder will not fit in the 4" PVC installed.

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. I'm fine with 750kcmil as well since we are under a bit of a time crunch now, thanks for tipping me off to that. :thumbsup:

Now for the proof that 700kcmil is adequate so I can submit this to the EOR.

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 hope this helps, and please ask any other questions, I will check back.

Thank you all
Kevin

 

[infinity]

VD compensation is not required by the NEC for this feeder and is a design issue. VD aside you'll still need to have 1200 amps worth of conductors. 6 sets of 250 kcmil aluminum (205 amps*6 = 1230 @ 75° C) is the minimum size conductors required by the NEC for a 1200 amps OCPD. Since that is the minimum required to next step is to find 6 sets of a certain size that will satisfy your 2% VD. According to my Southwire VD calculator you cannot get down to 2% with only 6 sets @208 volts.

 

[wwhitney]

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

Insulation temperature rating (90C versus 75C) is not going to help you directly with voltage drop. If the voltage drop formula tells you that for 6 sets you need to use 800 kcmil to hit your voltage drop requirement, you need to use 800 kcmil. There's no getting around that with just 6 sets.

Where the 90C insulation rating could help you is to by using 2 sets per conduit. Check what size you need for voltage drop with 12 sets, check if 2 sets of those will fit in one conduit, and then you can use the 90C ampacity when applying the 0.80 factor to check that the 12 sets have at least 1200A ampacity.

Cheers, Wayne

 

[infinity]

I used 12 sets in my Southwire calculator and came up with 12 sets of 600 kcmil with a VD of 1.94%. That's 6-600's in each raceway with an EGC if you do not require a neutral which in not going to fit. At this distance and load two transformers one at each end might be a better design.

 

[iceworm]

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:

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.

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:

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.

 

[infinity]

Now let's look at the Voltage Drop:

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)
​

Not sure if it changes anything but he did state a length of 480'.​

 

[iceworm]

Hope not.

However:

The feeder has a 1200 amp main disconnect and is 470' in length and will be fed with Aluminum conductors. The demand load is calculated to be 1082.7 amps

 

[electrofelon]

As has been mentioned, VD is (usually) not a code requirement. I would consider the ACTUAL load for the VD calc (or make a guess). Using the NEC load calc amps would be wasting money as the true load will often be about half of of that.

 

[synchro]

As has been mentioned, VD is (usually) not a code requirement. I would consider the ACTUAL load for the VD calc (or make a guess). Using the NEC load calc amps would be wasting money as the true load will often be about half of of that.

Agree. And unless the multifamily project has a single centralized HVAC system with a large compressor motor, it's likely that the magnitude of the short term load variation as a percent of the total load will be reduced compared to a single residence. Such short term variation is usually the most objectionable due to visible light flicker, etc.

 

[ggunn]

I did not slog through all the posts in this thread, but when I do voltage drop calculations I do it from first principles, i.e., Vd=ImaxR = (2)(D)(Imax)(r/1000) where r is the resistance per 1000'. I calculate Imax from the supply voltage and the load, and the resistance per 1000' does not depend on from which column I pick the wire. Am I missing something? It's quite possible.

 

[kwired]

Sorry to be confusing; but here is the situation;

We have a 208V 3Ø feeder to a meter bank for a multifamily project.

The feeder has a 1200 amp main disconnect and is 470' in length and will be fed with Aluminum conductors. The demand load is calculated to be 1082.7 amps.

At 3% Voltage Drop you can run 6 sets of 600kcmil to this load, so we installed 6 conduits under the slab.

Now, I am finding out that we needed to be at 2%. When I make that adjustment to the voltage drop calculation, the result is 6 sets of 800kcmil, based on the 75°C column. (Thanks to Bill Brahmford for the calculator) After resizing the EGC, this feeder will not fit in the 4" PVC installed.

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. I'm fine with 750kcmil as well since we are under a bit of a time crunch now, thanks for tipping me off to that. :thumbsup:

Now for the proof that 700kcmil is adequate so I can submit this to the EOR.

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 hope this helps, and please ask any other questions, I will check back.

Thank you all
Kevin

Your voltage drop on a particular conductor size is going to stay the same regardless of what kind of insulation it has on it. All that increasing insulation temp rating does is allow the conductor to operate at a higher temperature without compromising the insulation.

If you have a specification for a certain voltage drop level then you are stuck with a particular minimum size conductor to meet that specification. One thing you have though it will cost more is you can use copper instead of aluminum and you likely can still meet your voltage drop requirement in your already installed raceways.

 

[winnie]

Some quick details; I am bound to 6 sets of conduits and need to supply 1082.7 amps at a distance of 480' from the 208 volt 3Ø supply without exceeding 2% voltage drop with aluminum.

I would like to select my wire size based on the 90°C column, and my first attempt was to size the run based on the 75°C information, and then use the conductor ampacity from the 75°C column to select a conductor from the 90°C.

The CM formula equals 6 sets of 764.5kcmil so that rounds up to 800kcmil, this equals 395 amps per conductor, so we slide over to the 90°C column and can see that a 700kcmil is good for 425A. I would like to use the 700's but I need to have a better way to validate my slide rule here.

I think that I've finally figured out what you are trying to do, but unfortunately it is not a correct approach.

It looks as though you've calculated your required conductor cross section on the basis of voltage drop at 75C, then looked up the 75C ampacity of such a conductor, then looked for that same (or higher) amp rating in the 90C column, and want to use that 90C conductor.

Unfortunately that is not how voltage drop works. All that matters is the cross section of the wire and its resistance. If you need to use 800kcmil conductors with 75C insulation, you would still need to use 800kcmil conductors with 90C insulation.

With that said, temperature does play a role in your case. The resistance of both Copper and Aluminium go up with temperature, so if you run the conductors cooler their resistance is lower.

As iceworm calculated, you are right on the edge assuming 750 kcmil conductors operating at 75C, looking only at resistance (not reactance). But at 60C your resistance (and thus resistive voltage drop) will be about 6% lower. Since you are using these conductors at well below their 60C amapacity, it is reasonable to assume that the conductor temperature is below 60C.

Although honestly, if you are so close to the edge of your requirements that you need to consider the temperature coefficient of resistance in your voltage drop calculations, then you'd probably look at reactance and drop in the supply transformer as well. Unless the VD requirement is set by some energy code that is trying to minimize loss in the feeders.

-Jon

 

[KP2]

Thank you all for the replies on this hope for a reduction in conductor size.

In the 2014 handbook on page 213 regarding the increase to the EGC 250.122(B) the blue type reads;
"Generally, the minimum-sized EGC is selected from Table 250.122 based on the rating or setting of the feeder or branch-circuit OCPD(s). Where the ungrounded circuit conductors are increased in size to compensate for voltage drop or for any other reason related to proper circuit operation, the EGC's must be increased proportionately. In some cases, use of a conductor with a higher insulation temperature rating allows for compliance with ampacity adjustment and correction requirements without having to increase the circular mil area of the conductor"

The blue commentary continues to demonstrate the Size Ratio calculation.

Based on that statement, I was given hope that I could do something about my voltage drop problem. The way I hoped this would apply is by reducing the Size Ratio by selecting the conductor from the next temp rating.

I understood that if I had to calculate the voltage drop at 95°C, I could do that using the R2 formula, and then the typical voltage drop formula. I just tried to decrease the temperature and recalculate K.

Anyone have feedback on how that comment applies, it may be all I need.

Thanks again
Kevin

 

[ggunn]

I guess I am confused. I don't see how the procedure for determining the size increase for an EGC affects voltage drop. Once you increase CCC size for your target voltage drop you apply the ratio of the increase to the minimum size EGC.
I guess you could decide the size EGC you want and use it to back calculate CCC sizing, but that would be the tail wagging the dog.

Also, voltage drop calculations are about the metal in the wire, which is the same no matter which column you are looking at. The columns are only about the heat tolerance of different types of insulation.

 

[winnie]

In the 2014 handbook on page 213 regarding the increase to the EGC 250.122(B) the blue type reads;
"Generally, the minimum-sized EGC is selected from Table 250.122 based on the rating or setting of the feeder or branch-circuit OCPD(s). Where the ungrounded circuit conductors are increased in size to compensate for voltage drop or for any other reason related to proper circuit operation, the EGC's must be increased proportionately. In some cases, use of a conductor with a higher insulation temperature rating allows for compliance with ampacity adjustment and correction requirements without having to increase the circular mil area of the conductor"

Anyone have feedback on how that comment applies, it may be all I need.

This quote applies as follows: Imagine you have a bunch of 20A circuits, which typically are run using 12ga wire, protected by a 12ga EGC. Say you had a bunch of 75C rated wire and had to run 9 'current carrying conductors' in a single conduit. Derating rules would push you up to 10ga wire and you would also be required to use a 10ga EGC. But if instead you used 90C wire you would be derating from a 30A ampacity and you could stick with 12ga wire.

Note however that the voltage drop in these hotter 12ga conductors would be greater than the same conductors operating at a lower temperature.

The quote is not saying that voltage drop is the same in conductors of the same ampacity but different temperature ratings, rather it is noting that you can use the higher temperature rating to reduce the conductor size for purpose of derating, and thus avoid having to increase the size of the EGC.

Reducing conductor size will always increase voltage drop.

-Jon