Voltage drop and the 90 degree column

KP2

Senior Member
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.:D

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

Moderator
Staff member
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

Curmudgeon still using printed IEEE Color Books
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

Senior Member
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

Moderator
Staff member
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

Senior Member
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

Moderator
Staff member
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

Senior Member
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

Moderator
Staff member
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

Curmudgeon still using printed IEEE Color Books
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.
 
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infinity

Moderator
Staff member
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

Curmudgeon still using printed IEEE Color Books
Hope not.

However:
post 6 said:
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

Senior Member
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

Senior Member
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

PE (Electrical), NABCEP certified
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

Electron manager
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

Senior Member
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.:D
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

Senior Member
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

PE (Electrical), NABCEP certified
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. :D

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.
 
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winnie

Senior Member
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
 

kwired

Electron manager
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
But you are trying to find a way to reduce conductor size so it will fit in your raceway.


Yes higher temp insulated conductor will have more ampacity per the tables, but at same time will increase VD if you use that additional capacity. Same conductor will still have same VD characteristics regardless of what insulation is on it, all the higher temp rating does is allow higher insulation temp before compromising insulation integrity becomes a concern.

You are adding one benefit but taking another away. May not be 1:1 difference so details are important.
 

iceworm

Curmudgeon still using printed IEEE Color Books
Winnie, kw -
Thank you. I don't have a handbook. I had no idea what to make of the quote.

Never really trusted the handbook anyway. All that extra is just an opinion
 

wwhitney

Senior Member
A few comments:

1) If voltage drop depends strictly on copper/aluminum area, then my previous suggestion of using 12 sets instead of 6 sets was a step in the wrong direction, as with a fixed insulation thickness, smaller conductors are less efficient in terms of copper/aluminum area vs insulated conductor area.

2) If 1082.7A is the NEC calculated load, then I would think there is a reasonable argument for reducing that figure for voltage drop calculations. "Everyone" knows the NEC calculated load is a significant overestimate. You still have to meet it for ampacity, but as voltage drop is not an NEC issue, you can use other procedures for determining the real load for voltage drop.

3) 2014 Table C.9A says that (5) 750 kcmil compact conductors with THWN or XHHW insulation will fit in a 4" PVC conduit, Schedule 80 (worst case, as you haven't specified the conduit material). Is that not adequate for the original parameters of your question? With 6 sets, I get that 250 kcmil is the smallest aluminum conductor in each set, and for 1200A, 250 kcmil aluminum is the smallest EGC. So your EGC will have to match your ungrounded conductor size in each set.

4) How does neutral size figure into 3 phase voltage drop calculations? If it doesn't, then you could shrink the neutral and fit in (4) 800 kcmils (ungrounded and EGC) and a smaller neutral. At least 400 kcmil should fit per the tables, but you might fit 500 kcmil. I haven't done the calculations.

Cheers,
Wayne
 

synchro

Senior Member
3) 2014 Table C.9A says that (5) 750 kcmil compact conductors with THWN or XHHW insulation will fit in a 4" PVC conduit, Schedule 80 (worst case, as you haven't specified the conduit material). Is that not adequate for the original parameters of your question? With 6 sets, I get that 250 kcmil is the smallest aluminum conductor in each set, and for 1200A, 250 kcmil aluminum is the smallest EGC. So your EGC will have to match your ungrounded conductor size in each set.

Cheers,
Wayne
Agree with Wayne that these should fit based on the table. If you are not comfortable being this close to the edge on fitting the conductors, I wonder if there would be enough benefit by going copper just for the EGC. Of course going all copper would help but be big $$.
 

KP2

Senior Member
Thanks again for keeping this moving folks; the PVC runs are SCH40.

I realize that VD is not required; but in multifamily most EOR seem to require something, and this project is strict.

So far, I am up to 750kcmil feeders with a 600kcmil EGC and they won't fit in the 4" PVC

I see 3 potential possibilities remaining;

1. If there is a way to reduce the EGC 1 size it will fit, or perhaps I made a mistake and did the increase wrong.

2. I was not provided with the full calculation, so I do not know what the neutral load is, but reducing the neutral down a size will seal the deal as well.

3. Compact conductors.

Thanks Again
Kevin

MC-S2(6) sets of 4" C with 4 - 750kcmil & 1 - 600kcmil EGC
Supply Voltage 3Ø (V)208
VD=(1.732xLxRxI)/1000
1Ø=2 3Ø=1.7321.732
Resistance0.0282750kcmil AL
Length (ft)480
Demand (A)1082.7
Number of sets6
Voltage Drop (V)4.23
Percent of Voltage Dropped2.0
Increase for EGC
DetailsFeeder Ampacity (A)# of setsWire size (CM)Total CM
Feeder before adjustment120045000002000000
Adjusted Feeder 67500004500000
Size Ratio for increased EGC2.25
EGC before adjustment (CM)250000 Wire size (CM)
Adjusted EGC562500 600000
Conduit Fill# of conductorsArea (in^2)Total Area (in^2)
750kcmil41.05324.2128
600kcmil10.87090.8709
5.0837
4" PVC equals5.022
 

KP2

Senior Member
Looking again at my steps for the increase in the EGC, perhaps I should have just used 1 set of conductors; so 750/500 = 1.5 for a size ratio.

That would equal 250,000 CM X 1.5 = 375,000cm --> a 400kcmil would be the increase this way. that helps too.

I read 250.122(B) to equal the circular mill of the ungrounded conductors as the sum of the CM of feeders, because I was going from 4 to 6 sets with the increase.

Is this the correct increase method for this situation?

Thanks
Kevin
 

winnie

Senior Member
1) I read 250.122(B) the same way you do. IMHO you go by the _total_ change in cross sectional area. So if your baseline is 4 parallel 500s to get the required ampacity and you change to 6 parallel 750s then the increase in cross sectional area is 2.25x.

However_ code doesn't actually specify the baseline that you need to use. For your installation, 12 parallel 1/0s, 6 parallel 250s, 4 parallel 500s, 3 parallel 900s or 2 parallel 1750s are all 'reasonable' in the sense that they get you to more than 1200A using NEC tabulated ampacity. Doing the calculation from a different baseline might help, though I wouldn't use conductors that you can't actually purchase or install in your conduit as a reasonable baseline.

2) Check the area of the conductors; I thought that most aluminium conductors were supplied as compact stranded. http://www.cmewire.com/catalog/sec12-lvi-al/lvi-al-05-rw90-xlpe.pdf gives a diameter of 1.08in for 750s, which is an area of 0.916 square inches Sizes listed at 'wire and cable your way' are also smaller; using their conductor sizes you could just barely fit 4 900s and a 500 EGC in each of your conduits.

3) You could use copper for your EGC to make it smaller.

-Jon
 
To the people who are saying that VD is not an NEC requirement, that may be true, but if the area he is in has an energy code, them he has to follow that. In FLA where I am from the energy code dictates the VD requirement. It says in our energy code that a maximum of 5% VD is allowed and does not say it has to be 2/3, 3/2, 4/1, 1/4 etc. Just 5% max. Is the 2% a job specific requirement?
 

KP2

Senior Member
Hello folks, I submitted my voltage drop calculations (using 600kcmil at about 2.5% voltage drop) to the EOR; and this is what he cam back with;

"
Kevin,



Need to carry calculations to next level . meaning, calculate voltage drop from meter center(s) to unit load centers; and from unit load center to furthest device on a branch circuit.

Total drop for entire run (service to MC to load center to outlet) should not exceed 5% total.

Thanks,"


I just don't see how this is supposed to produce a realistic voltage drop result.

I understand if this was for Site Poles, or a large feeder to multiple motors; but these are apartments with loads that will never reach 100% of the calculated demand load of 100% of the units on the feeder.

Even with a motor, we use the name plate amperage not the FLC at 125% like the conductors; it's like I need to prepare for Armageddon.

Am I to calculate this voltage drop request the following way;
1. To get started I would use 100% demand of 100% of all the units to calculate the voltage drop on the feeders to the meter center.
Approximately 2.5%

2. Calculate the voltage drop of each Units sub-feed at 100% of the demand for each unit; the largest voltage drop of this set of feeders will have it's percentage added to the percent of drop from step 1. I should have this below 3% already since the unit sub feed is typically my only voltage drop concern.

3. Calculate the Voltage drop of each branch circuit of each units loadcenter; then take the largest percentage of voltage drop within that loadcenter and add it to the voltage drop of that unit's sub feed (should be less than 3%) and add it the 2.5% drop on the meter center feeder and pray they are all less than 5%.


This sounds as silly as how I started this tread, but I think it is what is expected of me.

Any thoughts would be greatly appreciated, this just seems so ambiguous to me.

Thanks
Kevin
 

MyCleveland

Senior Member
KP2
.....it's like I need to prepare for Armageddon. I like that.
How did you come up with the 2.5%VD using 600 AL ? Did you use 60C for the wire R values ?
I get the following results (% VD) using your description of the service and load amps.
60C 75C
PF 1.0 2.47 2.60
PF .99 2.84 2.97
PF .98 2.98 3.11
You see where this is heading....
Which brings up the question, can you use the 60C R values if you are using the 75C ampacity values, I think not.
In your case though, your service wire is so oversized (230%), there must be another way to calculate because conductors will never come close to rated ampacity.

Is the Service point where all the meters are located ? or are there feeders to various meter stacks ?
Not sure you will be able to meet the 5% total if you are already at 2.5 - 3.0 at the service.

When I calculate the apartment feeder VD I use 66% of the unit load. If I used 100% I would be up sizing quite a bit.
 

KP2

Senior Member
Thanks MyCleveland,

KP2
How did you come up with the 2.5%VD using 600 AL ?
I used the resistance values Table 8.
Meter Center Feeder Voltage Drop
MC-S2 - (6) sets of 600kcmil Aluminum Conductors
VD=(1.732xLxRxI)/1000
1Ø=2 3Ø=1.7321.732
Resistance of 600kcmil AL0.0353
Length (Ft)480
Demand (A)1082.7
Number of sets6
Voltage Drop (V)5.30
Supply Voltage 3Ø (V)208
Percent of Voltage Drop (%)2.5

Did you use 60C for the wire R values ?
No, if I can, how?

I submitted using 600kcmil which gave me 2.5%, my previous example posted was with 750kcmil; I figured I would go in with the 600's since I feel they are adequate.

I get the following results (% VD) using your description of the service and load amps.
60C 75C
PF 1.0 2.47 2.60
PF .99 2.84 2.97
PF .98 2.98 3.11
You see where this is heading....
Which brings up the question, can you use the 60C R values if you are using the 75C ampacity values, I think not.
Not exactly, other than I am ignoring PF.
Can I use 60C R values, if how?

In your case though, your service wire is so oversized (230%), there must be another way to calculate because conductors will never come close to rated ampacity.
I sure hope so.

Is the Service point where all the meters are located ? or are there feeders to various meter stacks ?
This is for a feeder to a meter stack; there are three in total.

When I calculate the apartment feeder VD I use 66% of the unit load. If I used 100% I would be up sizing quite a bit.
Any chance you have documentaion to lower the ampacity for the voltage drop calculation; 66% of the demand would be a huge help.

Thanks
Kevin
 

KP2

Senior Member
Yeah, at 66% of the load, the Voltage drop is at 1.6%

That would save the day for sure, if I could validate it.
Thanks
Kevin
 

MyCleveland

Senior Member
The resistance value of the conductor is unrelated to the ampacity selection.
David
Tables 8 & 9 in Chapter 9, values are based on 75C wire. Table 9 gives a correction formula for adjustment to other temps.
Ampacity tables give us selections of 60-75 or 90C values.
My point was if you are using ampacity from say the 60C column, you then adjust values in table 9 to 60C, or corrected R value.
Agree or Disagree ?
 

MyCleveland

Senior Member
Yeah, at 66% of the load, the Voltage drop is at 1.6%

That would save the day for sure, if I could validate it.
Thanks
Kevin
KP2
I did not mean taking 66% of the total bldg load.
I do this for the apartment feeder ONLY....meter stack to apartment.
If your adjusted load for the meter stack is 1083 amps you must use this for your calc.
 

david luchini

Moderator
Staff member
David
Tables 8 & 9 in Chapter 9, values are based on 75C wire. Table 9 gives a correction formula for adjustment to other temps.
Ampacity tables give us selections of 60-75 or 90C values.
My point was if you are using ampacity from say the 60C column, you then adjust values in table 9 to 60C, or corrected R value.
Agree or Disagree ?
I do not agree. The Table values are based on a 75C conductor operating temperature, not a 75C conductor insulation rating.

If you had identical 200A loads with 3/0 feeders of identical length in identical ambient temperatures, one feeder with THWN (75C insulation) and one feeder with THHN (90C insulation), the voltage drop would be the same.
 

powerpete69

Member
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.
In addition to the nicely written description above, don't forget to put a 4/0 Ground conductor in each of the conduits per NEC 250.122.
Also, if the conduits are large enough, it would be more cost effective to run 75C, 350 MCM in 4 conduits which satisfies your 1200 amp breaker per NEC 310.13(A), Leave the two conduits spare if possible. A 3" conduit would satisfy for 350 MCM for 3 wire or 4 wire system including the 4/0 ground.
Assuming you are willing to use copper.
 

MyCleveland

Senior Member
I do not agree. The Table values are based on a 75C conductor operating temperature, not a 75C conductor insulation rating.

If you had identical 200A loads with 3/0 feeders of identical length in identical ambient temperatures, one feeder with THWN (75C insulation) and one feeder with THHN (90C insulation), the voltage drop would be the same.
I agree with your statement.
My point was if you are using the 60C amapcity you can then use an adjusted R to a 60C operating temp, or if using the 90C ampacity you then need to adjust to a 90C R value.
 

ggunn

PE (Electrical), NABCEP certified
I agree with your statement.
My point was if you are using the 60C amapcity you can then use an adjusted R to a 60C operating temp, or if using the 90C ampacity you then need to adjust to a 90C R value.
I am a little confused as to what you are driving at, but the ampacity tables have to do with the stability of different types of insulation, nothing more. They all refer to the same internal metal wire and they have nothing to do with voltage drop. They do not imply that conductors from different columns will reach different temperatures while carrying the same amount of current. They only tell you how much current a conductor can carry before the insulation starts to melt; and the different columns refer to different types of insulation.
 

KP2

Senior Member
David
Tables 8 & 9 in Chapter 9, values are based on 75C wire. Table 9 gives a correction formula for adjustment to other temps.
Ampacity tables give us selections of 60-75 or 90C values.
My point was if you are using ampacity from say the 60C column, you then adjust values in table 9 to 60C, or corrected R value.
Agree or Disagree ?
I did start off this thread thinking I could adjust the R value since we will use XHHW-2 conductors and adjust the temperature by 15°C, but folks did not agree. I did not present this very clearly in the beginning, I know that did not help.
 

KP2

Senior Member
KP2
I did not mean taking 66% of the total bldg load.
I do this for the apartment feeder ONLY....meter stack to apartment.
If your adjusted load for the meter stack is 1083 amps you must use this for your calc.
So the term "Adjusted load" what does that mean; am I being to literal in hoping that the demand load and the adjusted load are not the same?

The demand load is 1083 amps, which is the result of the feeder calculation based on 220.84.
 

ggunn

PE (Electrical), NABCEP certified
I did start off this thread thinking I could adjust the R value since we will use XHHW-2 conductors and adjust the temperature by 15°C, but folks did not agree. I did not present this very clearly in the beginning, I know that did not help.
I hope you got your answer, which is "no". The resistance of a conductor does not depend on what type of insulation it has.
 

KP2

Senior Member
In addition to the nicely written description above, don't forget to put a 4/0 Ground conductor in each of the conduits per NEC 250.122.
Also, if the conduits are large enough, it would be more cost effective to run 75C, 350 MCM in 4 conduits which satisfies your 1200 amp breaker per NEC 310.13(A), Leave the two conduits spare if possible. A 3" conduit would satisfy for 350 MCM for 3 wire or 4 wire system including the 4/0 ground.
Assuming you are willing to use copper.
Thank you for this; the EOR has specified 6 sets of 600kcmil CU with a 300kcmil ground.

We reminded the GC that we qualified Aluminum, and here we are.

Compared to your 350's I see just how over engineered these feeders are in the first place.

Now to prove it, and not step on toes in the process.

Thanks Again everyone for the help.
Kevin
 

KP2

Senior Member
I hope you got your answer, which is "no". The resistance of a conductor does not depend on what type of insulation it has.
Correct, however the temperature of the conductors have everything to do with the resistance of the conductor.
 

powerpete69

Member
Incidentally, demand load is how much amps you run on average over a 8 hour day for example.
Connected load is the load with everything on at the same time which is just an mathematical exercise and not reality.
Are you sure your average amps will be 1083 amps? I am doubting it. Also, your 1200 amp breaker is likely a 80% breaker which means that it will trip over 960 amps within an hour or two....or less. It will not trip right away. It will just trip when it gets too hot similar to your wires getting too hot.
Also, why are you using 2% voltage drop? Why can't it be 5% voltage drop? A 480 Volt system feeds 460 Volt loads, that's a more than acceptable 5%. 240V panel feeds 230V loads, again 5% voltage drop. Take a 110V receptacle fed by a 120V panel. That's 9% voltage drop and everything is just fine. Not sure what 208V loads are, but the point is that everything is built for voltage drop since most loads are some distance away.

If you really have a demand load of 1082 (again, I doubt), use a voltage drop calculator to figure your voltage drop. You may want to upsize a hair over 350 MCM using my 5% figure, but again, that breaker is going to trip a few times a day if you really have the 1082 A demand load. Which essentially means that (4 conduits) 350 MCM should work fine and meets code.
 

ggunn

PE (Electrical), NABCEP certified
Correct, however the temperature of the conductors have everything to do with the resistance of the conductor.
Of course, although it is a relatively minor effect and has nothing to do with which ampacity column the conductor type is in. The ampacity tables have nothing to do with voltage drop.
 

MyCleveland

Senior Member
I am a little confused as to what you are driving at, but the ampacity tables have to do with the stability of different types of insulation, nothing more. They all refer to the same internal metal wire and they have nothing to do with voltage drop. They do not imply that conductors from different columns will reach different temperatures while carrying the same amount of current. They only tell you how much current a conductor can carry before the insulation starts to melt; and the different columns refer to different types of insulation.
The different columns also represent the different max operating temps...60-75-90C.
Each temp selected will result in a different R value and therefore a different VD result.
 

wwhitney

Senior Member
Of course, although it is a relatively minor effect and has nothing to do with which ampacity column the conductor type is in.
Maybe not so minor. Suppose your initial design uses a conductor at its full 75C ampacity (based on 30C ambient), and you decide to use two sets to reduce voltage drop. If the heat dissipation is linear in temperature over ambient (may be an erroneous model), then to first order the resistance has halved, so if the operating temperature with one set is 75C, the operating temperature with two sets will be only 52C.

At 20C, the temperature coefficient of resistivity of Al is 0.0038. If the coefficient is close to that for the 52C to 75C range, then the resistivity at 52C will only be about 92% of that at 75C. So using 2 sets would reduce the voltage drop to 46% of original, rather than 50% of original. [To be more accurate, you'd have to iterate, as the lower resistivity would mean an even lower operating temperature, etc. That's why I rounded 52.5C down to 52C.]

OK, so maybe it is minor. But for extreme cases like in the OP, it may start to be significant.

Cheers, Wayne
 

MyCleveland

Senior Member
Thanks Wayne
I am terrible at this...trying to type out when I would rather talk it through.
My thought was because he has so much wire capacity, there must be some way to account for that in a temp adjustment of R giving a better VD result but I don't know how to guide KP2 in this calc.

I only chimed in when i could not duplicate KP2's VD calc without using a 60C R value...and rambled from that point.
 

powerpete69

Member
Also, why are you using 2% voltage drop? Why can't it be 5% voltage drop? A 480 Volt system feeds 460 Volt loads, that's a more than acceptable 5%. 240V panel feeds 230V loads, again 5% voltage drop. Take a 110V receptacle fed by a 120V panel. That's 9% voltage drop and everything is just fine. Not sure what 208V loads are, but the point is that everything is built for voltage drop since most loads are some distance away.
Please disregard this portion of the post, my apologies. I was thinking that we were at the last load, but we are not. You will need additional distances from whatever gear we are feeding here. So the max 2% probably is realistic.
 

ggunn

PE (Electrical), NABCEP certified
The different columns also represent the different max operating temps...60-75-90C.
Each temp selected will result in a different R value and therefore a different VD result.
Not for the same amount of current. You can push more current through a 90 degree wire than you can through a 75 degree wire, true enough, but if you push more current through any wire, of course Vd is greater, even if R were to stay the same.
 
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