Twofold question here:
First, can i splice THHN copper to an existing run of aluminum SE cable in order to extend a feeder to a subpanel? I find nothing in the NEC disallowing this, as long as I use an approved copper-to-aluminum device and some No-Alox. Still, it seems weird.
Then, if i CAN do this, how would I calculate the voltage drop? would i first calculate the aluminum for just the length of the run that is a SE, and then calculate the copper for just the length that is THHN? Or would i calculate both the copper and the aluminum for the entire length of the run?
I appreciate any and all feedback. thanks.

You would need to calculate the resistance for the footage of Aluminum and the footage of copper separately and then add the tow "resistors" together and apply the voltage drop formula that you've chosen to use. Use Table 8 in the back of the NEC for resistance of conductors.

Depending on the distance of copper run, you might find that the added computation complexity of combining both conductor material types is insignificant. As I remember, Copper usually carries 40% more current per size than aluminum. So, this would make the aluminum run "the point of largest impedance". Although the copper run will have a VD value, by percentage it will be very small compared to the aluminum. I would extend the VD computation an additional "X" feet with aluminum factors to build a nice conservative circuit.

However, with that said, if the copper run distance is larger than the aluminum, then disregard the above. If you post the distances I'll compute the total VD using IEEE exact formula for you as this formula is a bit complex.

Also, don't forget to allow correction factors for ambient temperature in your VD computation. The NEC tables are set at 30ºC(86ºF) which is not practical anywhere in the USA. I personnally use 40ºC(104 - 105ºF) as a standard ambient with additional temperature rise for attic and roof mount circuits. The Chapter 8 and 9 tables are "borrowed" from IEEE Std 141 and are derived at 75ºC.

Not to sound like a commerical (this is really true), but using software that takes these and other factors into consideration is a lot easier and more accurate for VD computations.

Diver IMO in most cases electricians do not need to get that deep into it.

For one thing as far as the NEC is concerned you can have as much voltage drop as you want to live with. (Other than fire pumps)

If I was working this problem I would figure each section separately and simply look at the total result.

Connecting copper to aluminum is no problem with the correct connector and No-Ox is optional.

Bob


[This message has been edited by iwire (edited 09-22-2005).]

Hi iwire,

You are correct in regards to the NEC and VD. NEC only mentions VD parameters in fine print which are non-enforceable. However, the equipment manufacturer does specify a VD range and that is why, in my opinion, this type of accuracy is important.

Thanks for your responses. Diver: I'm curious what Vd formula you use that is so confusing. I use this:
cmils = (2 x K x amperage x distance)
divided by voltage drop.
Where:
Cmils is the wire size in circular mils.
For copper, "approximate K" = 12.6.
For aluminum, "approximate K" = 21.2.
And my voltage drop in this situation would be 3% x 240v = 7.2v.

If, say,the aluminum portion of my run is 45 feet and the copper portion is 82 feet, my question is whether i calculate the aluminum circ mils by using a distance of 45' or by using the total distance of 127'. And likewise for the copper: would I plug 82' into the equation or 127'? Remember that there is no OCPD between the aluminum and the copper; they are simply spliced so that they are essentially acting as one feeder.

Does any of this change anyone's response?

Hi tripp,

I use the IEEE Std 141 exact formulae:

Vd = V + IRcos(theta) + IXsin(theta) - sqrt(V^2 - (IXcos(theta) - IRsin(theta))^2)
where:
Vd = Voltage drop (Line to Neutral)
V = Voltage (source)
I = Current in amperes (A)
R = AC Resistance from NEC® Chapter 9 Table 9
X = AC Reactance from NEC® Chapter 9 Table 9
distance (L) is considered from the Resistance & Reactance Tables
where Ohms per unit / 1000 * L in same unit = R or X
theta = Arccos(device or circuit Power Factor) = angle of phase offset

Line to Line is computed by Line to Neutral VD / Sqrt(3) for 3 phase circuits.

You'll also have to get into some sin wave destruction formulae for power factor averaging when computing the Vd across a panel's bus bar. Total(kW) / Total(kVA) should be close enough for the bus bar PF average.

I forgot to mention that since resistance(impedance) is very sensative to temperature you must multiply the resistance value by the following factor:

R2 = R1[1 + a(T2 - 75)]
where:
R2 = Adjusted Conductor Resistance
R1 = Table Conductor Resistance @ 75ºC
a = Conductor Material Resistivity
a(cu) = 0.00323 and a(AL) = 0.00330
T2 = Ambient Temperature in Celsius (TA)

convert temperature in ºF to ºC with:
TºC = (TºF - 32) / 1.8

Also:
You would take the VD across the aluminum first. Then use the lessened voltage from the aluminum run to start the copper VD computation. Most engineers would compute the run as all aluminum to keep conservative.

And:
Regardless of the voltage drop formula you use, The fact that you take VD into consideration puts you in the top 1% of electrican's in my opinion!!!

The NEC only references VD in fine print notes. HOWEVER, all electrical devices used in the USA must have a UL label. United Labratories tests these device under "normal" conditions within a VD limit from its nameplate rated voltage. If you, the electrican, install a circuit that is not capable of supplying the correct voltage within the devices voltage range, the following will happen.
1. The UL approval is VOIDED.
2. The conductors will produce added heat. Heat, like raditation, is accumulative (just ask a fireman for verification). This means that a fire will eventually result in a given amount of time from not considering voltage drop.
3. When this happens who does the insurance company go after.......

Personally, as an engineer, I need to be as accurate as possible to not only ensure a problem free installation, but to also cover my butt in case there is a future problem. The IEEE Std 141 Exact Formula is the only VD formula that is recognized as correct both nationally and internationally.

Volts computes VD with all the necessary environmental and device considerations and sizes the conductors to the correct NEC ampacity table, the correct voltage drop and the correct terminal temperature of the device....in less time than you can write a single sentence.

Diver Dan- In all due respect to your obvious education, I don't know vary many electricians out there that have enough background to use your formula. It certianly is proper for the engineers when an exact voltage drop is calculated, but for the average job out there, most electricians get by very well just by using the simple formula found in any electricians handbook and using Table 8 in Chapter 9 of the NEC. They don't even use Table 9 and consider impedance. Most electricians (myself included) use only the minimum amount of math they have to use to get by. They are in trouble when they have to do voltage drop calculations on Sensitive Electronic Equipment Article 647 Section 647.4(D) where voltage drop limits are in place by code, not optional.

Diver I have to go with George's post.

I fully understand that as an engineer you have to be precise, that is your profession.

Most electricians do not need to be that accurate for VD, and regardless of what the books tell us the real world has shown us that equipment is much more tolerant to voltage variations then you would think.

Voltage drop is not a safety issue, it is more of an efficiency issue.

Just out of curiosity have you ever checked your VD predictions against the real thing after the work is done?

George, who the heck uses Article 647?

I have never seen that article applied, it does not apply to much.


Bob

Hi guys,

I understand what you are saying and partically agree. Engineers do need to be as percise as possible as they are one of the prime targets for insurance companies in regards to law suits. And, whenever an electrican or contractor performs a "Design / Build" installation, he is now infact the engineer and a potentical target.

Knowing that we all want to do the best job possible and at one point we are only as good as our tools permit us, Volts becomes another tool option. Except, if doing Design / Build projects, it usually pays for itself in the first job.

Also and very important, voltage drop can very much be a safety issue!!!

Please explain how voltage drop can be a safety issue if the rest of the installation follows the NEC.

Bob

Okay, I have to ask, what POCO provides a stable enough system that a "precise" VD can be calculated?

Roger

Your right Roger, and that's why we can use the simple approach we do for voltage drop and get away with it. NOW Article 647 on the other hand may be somewhat more stable. As Bob pointed out it's not a popular Article. I only know of one contractor who gets involved with "Technical Power" as it is called and it gets kinda accurate. They use Technical Power also in some audio systems.

Holy Toledo, Diver! I'm almost sorry i asked what formula you use! Except that it's good to know it exists and who to go to if i really want to know! Still, i think i'll stick with my Uglies book for now.

Nevertheless, since i sent you my distances in my last post, did you calculate my voltage drop? I'm curious to see if we match up.. Also, still not sure if you think i should calculate my cmils for each conductor (aluminum and copper) using the total distance involved or whether i should calculate the size of each based only on the distance that conductor runs.

Hi All,

Roger, you're right!!! The best VD computaton is only an approximation. POCO, conductor age, connections, circuit breaker age, etc. are all non-calculated contributors that cause computation vs actual VD variations.

iwire, A simple case is when you run conductors over a roof or attic where the ambient temperature can reach high levels. Unless the conductors are sized with regards to VD, this will cause a large VD that will start the insulation degradation process that will lead to a fire.

Tripp, I forgot about your VD, sorry.
I can see that the runs are 45' for aluminum and 87' for copper. Please post the conductor size, ambient temperature, and the raceway type and I'll knock it up right now.

I just check back to get Tripps info and though I add a few more notes.

As I stated prior, regardless of the VD formulae you're using, the fact that you all are using VD in conductor sizing gains my respect and attention. I would like to note that I forgot to mention probably the largest contributor to VD variation is ambient temperature. The only problem I have with 2KID as a bases of VD is that it doesn't take temperature into account. Yesterday I had a conversation with a contractor who went stricky by 310.16 and wondered why Volts upsized a conductor. I point out to him the roll that temperature played in VD and that the NEC does account for it with the temperature Correction Factors located just below the 310.16 table, something that he had never noticed or done prior.

I have to copy Divers formula and in about 7 weeks, when I give the entry level guys (In Basic Electricity) the KIL way, and that spins there heads; wait for Diver's, that may be incentive to learn KIL.

Interesting..

John

Hi Diver - here are my numbers:

aluminum:45'
copper: 82' (not 87')

wire size: inverting KIL to find cmils, and using a desired Vd of 3%, i got #4 for the SE aluminum cable and #6 for the copper.

Ambient temp: good question. we're in Montana; the wire will be run through a basement crawlspace (alum.) in no conduit; and then outside along a cement pad in PVC sch.80 (copper).

Take it from there.

Don't forget the following rule:

250.122(B) Increased in Size Where ungrounded conductors are increased in size, equipment grounding conductors, where installed, shall be increased in size proportionately according to circular mil area of the ungrounded conductors.

Thanks Joe - i have every intention of increasing EGC accordingly. Thanks for being on top of it.

Hi Diver


Diver I mean no disrespect, however I personally do not believe your above statement is true. If the circuit has the NEC required overcurrent protection that conductor in the attic will never become hot enough to start a fire or damage the insulation assuming you considered the proper ambient when selecting the conductor as required by the NEC.

99% of the time reduced voltage at the load will result in reduced current used by the load.

In the few cases that a load may draw more current with less voltage (say a motor) the motor overload protection will open before damaging the conductors.

Now don't get me wrong, I run a lot of 10 AWG for 20 amp circuits, I don't like the idea of my money being wasted heating undersized conductors.

That said I do not believe VD is a safety issue as long as the rest of the NEC is applied correctly.

Now you have me confused, those temperature correction factors have nothing to do with voltage drop.

Those temperature correction factors have to do with lowering (or raising) the allowable conductor ampacity depending on the ambient temp. This is regardless of circuit length and is required by the NEC.

Bob

DiverDan,

I hope that it doesn't look like we are forming a bandwagon to call you 'wrong'; your notes are overall quite informative and useful. I think that several of us are picking up on points that we disagree with to a certain extent; I just wanted to note the overall positive nature of your writing...and then dive into one of those little points *grin*

I am going to call that an obfuscatory equation, math which hides the truth behind algebraic manipulations. It might be the right equation for a computer program to use, but it doesn't make things clear to humans.

Also, I take issue with calling that equation 'exact', even if the IEEE does so. At best it is a much better approximation to reality. But it presumes sinusoidal applied voltage and linearity of the various interacting materials, and can only be close, not exact. But that is me picking very small nits

The most basic voltage drop equation is simply Ohm's law, E=I*R where E is the voltage dropped across the resistance, and I is the current flowing. You simply substitute in the resistance of the wire in question, and this gives you the voltage drop. The resistance of the wire is determined by the resistance per unit length of the wire and total length of the run, and is the same resistance parameter used in the IEEE equation. This basic equation is accurate enough for most voltage drop calculations that an electrician will encounter, but it misses on significant points.

In particular, in AC circuits, the voltage drop will include both the DC resistance of the wire and the AC inductance of the wire, and the current flow is not necessarily in phase with the voltage.

But with a simple little trick, we can rescue Ohm's law. We simply replace all of the 'real' quantities (single numbers representing values) with 'complex' quantities (numbers of the form x + jy where j is the square root of -1) The combination of resistance and inductive reactance (R and X) becomes the 'complex impedance' R + jX. The combination of in phase current (D) and out of phase current (Q) becomes the complex current D + jQ. Q is positive for _leading_ reactive current, eg in capacitive circuits.

Now we simply apply Ohm's law using these 'complex' quantities. To deal with 'complex' multiplication, remember that j is the square root of -1, so whenever you get j^2 just replace it with -1

E = I * R becomes E = (R + jX) * (D + jQ) = RD - XQ + j(RQ + DX)

Getting more complex, but we see that the core is still E = I * R, ordinary Ohm's law, but when we expand out the calculation we see the reactive component of impedance and the power factor of the current flow. By writing the equation in complex form, we get the intuitive: Voltage drop is caused by the current flowing through the impedance.

If I have gotten my signs correct, then the above equation should be reducible to the IEEE equation (meaning that they should be mathematically the same). All those trig functions (cos(theta), sin(theta), arccos(power factor) )?? they come out of the fact that rather than reporting complex current in the x + jy form, current is described in terms of total current and power factor. The trig functions are essentially converting amps and power factor into the D + jQ representation of current flow. I have not checked in detail, but the IEEE equation might collapse if the load has a _leading_ power factor; eg current flowing into a capacitor. The arccos(PF) equation would treat a leading PF the same as a trailing PF, but the real circuit would respond differently.

Sorry to run-on. Here is a nice tutorial that I found on using complex math to describe impedance: http://www.ibiblio.org/obp/electricCircuits/AC/AC_11.html
The entire textbook is worth a look.

-Jon

P.S. On the voltage drop heating the wire issue: since both voltage drop and wire heat are caused by the same resistance, consideration of one is closely related to consideration of the other. Most of the time I consider them separate issues, since voltage drop depends upon the _total_ resistance of the wire, but doesn't care about resistance per unit length, whereas wire heating depends upon resistance per unit length, but doesn't care about the distance or _total_ resistance.

A 1 foot length of 12ga wire has a resistance of 1.589 milli-ohms. Run 500A through this wire and you have a voltage drop of 0.8V, generally considered acceptable in a 120V circuit. Of course, not only would the insulation degrade on this wire, but the copper itself would quickly fail

Am missing something here? Has the current or voltage been posted for this problem? Maybe I could learn something here. Someone tell me the voltage and current and let's see if I can do my own calculation for voltage drop.

[This message has been edited by George Little (edited 09-24-2005).]

Diver Dan, I would also like to say I have no interest in giving you a hard time.

I think it is natural that an EE and electrician would look at this issue very differently.

Most of the time the jobs I work are designed by EEs and I can not help but appreciate the large amount of copper that is required by the EEs.

Bob

Hah, George, at least someone is staying on target.

Tripp,

Voltage drop depends upon current. Since we don't know the actual current flow in the conductors that you are proposing, we can't calculate the voltage drop.

The equation that you posted does not calculate voltage drop; it takes as its _input_ the desired voltage drop, and the distance, and gives as its output the size of conductor that you need. Of course, this equation is simply the voltage drop equation re-arranged.

In other words, if you use the simple voltage drop equation E = I R, you can re-arrange by dividing by I, and get R = E/I, which tells you what resistance is required to get a particular desired voltage drop.

Your equation goes one step further, and takes the inherent resistivity of the conductor into account, to give you the conductor size. You will need to double check; some of these equations use as their constants values that take into account 'round trip distance', others use 'one way' distances. Same equation, different constants, and you need to match the constants that you use to the question that you are answering.

The key point to keep in mind is that voltage drops that are electrically in series simply add up. If you have a 100 foot length of aluminium wire, with a voltage drop of 1V, and a 200 foot length of copper with a voltage drop at 1.5V, then the total voltage drop is 2.5V.

So with the normal E = I R equation, you would first find the resistance of your aluminium conductor, and calculate the voltage drop, and then you would find the resistance of the copper conductor, calculate it's voltage drop, and add the two up.

With your reverse direction equation, you will need to first 'distribute' the total allowed voltage drop to each portion of the circuit. Say you want a total of 3% voltage drop, with 45 feet of Al followed by 87 feet of Cu. You might arbitrarily assign 1% voltage drop to the Al portion, and 2% to the Cu portion. Or could be 0.5% to the Al portion and 2.5% to the Cu portion. Whatever; you just pick two reasonable looking numbers that add up to your desired total.

Then you simply work your equation with the individual section voltage drops, for the length of the individual sections only. So you would run the equation with 45 feet of Al and a 1% voltage drop, and separately run the equation for 87 feet of Cu and a 2% voltage drop, and the two conductors electrically in series will give a 3% voltage drop at the design current.

-Jon

As none of the needed info has been posted to figure the VD I think we have stayed right on target.

This is what I see.

1)The AL SE part of this circuit is existing. The voltage drop in that part of the circuit we will have to live with.

2)The CU part of the circuit is to be added and we can make changes to help combat voltage drop.

If Tripp provides us with just the circuit current and voltage along with the wire sizes and lengths we could get a pretty decent answer.

It's all just a guesstimation.

Bob

Okay guys - just when i thought the brainiacs had forgotten i even had a question! The truth is, yes, i really am trying to figure out what size wire to run. I know what voltage drop i want, because i follow NEC guidelines - 3% for feeder or branch circuit, 5% for total. I also know my Vd formulas (or thought i did till i became acquainted with Diver ), and know that i can transpose any algebraic equation to solve for whatever unknown i want. Hence,

Vd = (2 x K x I x D) divided by cmils

becomes

Cmils = (2 x K x I x D) divided by Vd.

And when solving for cmils, you use what's called "approximate K" values; for aluminum this value is 21.2; for copper it is 12.6.

Distances are 45' alum; 82' copper. In the above formula, D id for one-way distance. (Note: for three-phase calcualtions, the "2" in the above equation would be replaced with "1.73").

In my case I am pulling a feeder to a subpanel, for a 2-pole 50 amp. My intent so far is to pull SE #4 alum, spliced to THHN copper stranded #6. In this situation, either method i chose (whether calculating, say, the alum [or cop] using just the distance of that material's run [45' or 82'] or using the total distance [127'] in each calculation) gave me the same answer because regardless or what answer the Vd formula gives you, you still have to abide by ampacities in T.310.16 for minimum. In some other scenario my answers may have not matched up, depending on how i computed distance, so MY QUESTION REMAINS: which way do i compute distance for this Vd formula: when calculating the alum do i use the distance of just the alum portion of the feeder; or do i use the distance of the entire feeder? same question for the copper.
Thanks.

I thought that I answered that question. You calculate each part entirely separately. The voltage drops add up.

To size the Al section, you use the length of the Al section and the voltage drop apportioned to the Al section.
To size the Cu section, you use the length of the Cu section and the voltage drop apportioned to the Cu section.

If you want to keep the voltage drop below 3%, then you must apportion this 3% between the two sections. If you have 10 sections, then you apportion your voltage drop between all of the components, and then calculate for each component separately.

-Jon

OH MY GOD!!!! I just woke up, read the news on MSN about Rita, checked out this post and WOW!!! You east coasters have been very busy!!!

With half closed eyes and only one downed soda I'll try to answer some of these posts.

Tripp, I ran your info through Volts with nice results. I assumed a load of 75A as this is the max for #4 Al and #6 Cu conductors. I used the Earth(buried) raceway selection at 60ºF for the Al and PVC Schedule 80 at 105ºF for the Cu. I also assumed 208V, 3 phase.

Impedances computed at .02571... for Al and .05566... for Cu. Totaling at .08137... for the 127' total run. Using Vd = I * R and %Vd = Vd / V, I computed a 2.93 % Vd for the total run.

Gad, it's still to early for me and these numbers!

Joe, Volts automatically adjusts the ground conductors per 250.122 when increasing or decreasing conductors.

I'm sorry but I forgot the name of the gentleman using real and imagionary(j) numbers. I'm not going to double check your figures but I am very impressed!!! Volts does use similar formula for computing sin wave distruction in busbars.

And for the gentleman who stated that these types of formulae are okay to use in software...That's the point of using software in place of quick rule of thumb formula. I used to compute VD by hand using "IEEE Std 141 Exact Formule" and it drove me nuts, especially when a change was made and I had to redo everything again, and again.

And NEC only suggests and does not enforce VD. However, since ambient temperature is a major contributor to VD, the NEC included the temperature Correction Factors to adjust the conductor size accordingly.

I hope that I didn't leave much out, but I need another soda and some more wakeup time.

[This message has been edited by DiverDan (edited 09-24-2005).]

Okay- I'll ask it one more time- What is "I" and what is "E". Unless I have these values I can not calculate voltage drop. Maybe I'm green I've only been an electrician for 40 years.

Tripp, George has asked for the I & E values, please let us know what they are so that those who may want to can perform the calculations for comparisons.

With out knowing these numbers even Dan had to assume, and that pretty much reduces his high dollar precise software to a guessing program.

Roger

Ok, after another soda I have found a large error in my assumptions.

Since this is SE cable, I am assuming that the conductor connects to a CT then a circuit breaker. Both these devices have a 75ºC terminal temperature rating. With that the Al and Cu conductors carry a max of 65A(I) for #4 & #6 respectively.

I am also assuming V(E) at 208, ambient for the ground at or less than 36" depth at 60ºF and above ground at 105ºF, 3 phase power. Soil thermal conditivity(RHO) is also assumed at 90.

The conductor impedance(Z) values are .02571 for Al and .05566 for Cu at 45' and 82' respectivly.

Ztotal = 0.08137 ohms
Vd = I * Z = 65 * 0.08137 = 5.28905
%Vd = Vd / V(E) * 100 = 5.28905 / 208 * 100 = 2.54% Vd with the above assumptions. If V(E) = 480 or the SE cable is connected differently or any of my other assumptions are different, then these values are wrong.

And I agree with George Little, there are too many assumptions.
Opps, I guess I was writting this when Roger posted, I also agree with Roger.

[This message has been edited by DiverDan (edited 09-24-2005).]

Well, i did give you guys my "I" in the last post, but i'm sorry i forgot to give "E". As i mentioned earlier, amperage is 50amps. Voltage is 240. This is a residental addition - power to a detached shop.

Well I used the NEC Table 8 and given the length of 45' #4 Aluminum and 82' of #6 copper, 240v. and 50a. and came up with voltage drop of 6.3v.

If I'm right this is less than the recommended 3% for feeders.

Someone either agree or disagree with me

Quote:
___________________________________________
"However, since ambient temperature is a major contributor to VD, the NEC included the temperature Correction Factors to adjust the conductor size accordingly."
___________________________________________

Ambient Temperature Has Nothing to do with Voltage Drop of the wire. Voltage Drop Increases Wire temp and then you have to take that into account.
% Volts Dropped is % Power Wasted. The Customer IS Paying for the power wasted in the wires. As far as I'm concerned all we as electricians have to worry about is Volts Dropped After the Meter. Who Cares about what the POCO supplies to us at the meter. We need to keep Voltage drop to a minimum after the meter. During my apprentiship days we did alot of calculations along these lines. Upsizing wires WILL pay for itself within a couple of years, Especially with continuous loads.
Pretty much Any wire run over 100' needs to be upsized! I can't stress this any more The Customer IS paying for the power wasted in the wire due to Voltage drop. They Pay for it but don't get to use it! Think about that when you use 5% Volts Dropped. That is basically 5% tacked onto their electricity bill. Code is a minimum and in this case just reccomended, It's hard to bid jobs this way. Ain't it great being an Electrician Competing to do cheaper work!

Justin

That is a not really the case, a 5% VD will not directly mean a 5% increase in the electrical usage. It will not even be close, the wasted power due to VD will be an very small percentage of the total bill.

That aside I do agree the customer ends up paying for electricity not directly use by them and if they are willing to pay up front for larger conductors it is a better job overall.

Bob

DiverDan


Here you are mistaken, the temperature correction factors are not in the NEC for voltage drop reasons.

The correction factors modify the rated ampacity of a conductor in order to keep the conductors insulation operating within it's design when the ambient temp is above or below 78–86 F.



Bob

I'm explaining how the wheel works here and that is not what I want to do.

I don't know who to use the quote tags so please bear with me.

From Justin:
Ambient Temperature Has Nothing to do with Voltage Drop of the wire. Voltage Drop Increases Wire temp and then you have to take that into account.

This is just plain wrong. Do some research.


From Bob,
Here you are mistaken, the temperature correction factors are not in the NEC for voltage drop reasons.

You have mistaken what I said. Voltage drop increases as ambient temperature increases. If you had ever used the above temperature formulae you would see how this happens. Conductors must be upsized to compensate for high VD. So, knowing the the NEC does not stress VD, the temperature correction factors signaled a need to develop the square of the temperature differences formula which is how the temperature correction factors are derived. All three, the VD temperature formulae, the square of the temperature differences, and the NEC temperature correction factors accomplish the same thing!

As a commentary:
Your "beliefs" do not change physics. It would only benefit your expertise to do a little more research into these subjects.

And while you're researching, lookup Neher-McGrath as their formulae was used to establish the NEC ampacity tables in the same way that the above temperature adjustment VD formulae was initially used to establish the NEC temperature correction factors.

[This message has been edited by DiverDan (edited 09-25-2005).]

Well, i am very satisfied with all the feedback i received and i thank everyone for their attention and furthermore i am continuing to enjoy the ongoing discussion.

the more i know, the more i see there is to know; or...so much to learn, so little time to spend on the computer.

DiverDan- There is no argument that temperature plays a part in voltage drop. Here's how - Since resistance increases with temperature, voltage drop increases. the amount of increase is not significant (IMHO) Would you perform your magic and see what voltage drop you come up with for the problem as presented in this thread? thank You in advance.

Sorry Dan I missed one of your post where you had worked the problem using "Volts" Is the Volts program Mac compatible?

My conclusion:

DiverDan's "Volts" program came up with 2.93% VD or 7.032v.

Using NEC Table 8 and simple VD formula I came up with 2.62% VD or 6.3v.

So the difference is 0.732v.

I assume we both used 240v. as the voltage.

I think we can live with these differences.

Thanks Dan for your insight.

[This message has been edited by George Little (edited 09-25-2005).]

George Little, my previous post's computed values DO NOT APPLY to this situation as they were based on I at 65A, 208VAC, 3-phase. As a degreed engineer of 30 years, I am liable for my computations and therefore protective that I am not miss quoted nor my computations miss used.

And, Quote:
There is no argument that temperature plays a part in voltage drop. Here's how - Since resistance increases with temperature, voltage drop increases. the amount of increase is not significant (IMHO)

"the amount of increase is not significant (IMHO)"
This is also just plain wrong!
Why not use the NEC the way that it is intended to be used with the temperature adjustment factors or try the IEEE temperature adjustment formulae with your VD computations. You'll be suprised how significant the role of ambient temperature can be in VD and conductor sizing.

As I stated earlier:
Your "beliefs" do not change physics.

For those who wish further discussion and information I can be reached at info@dolphins-software.com.

[This message has been edited by DiverDan (edited 09-26-2005).]

OK Dan you want to step up to the plate and surprise us or do you just want to keep on saying the same thing over again?

Put some figures down to prove your point.

Please demonstrate the 'major effect' ambient temp has on VD using an example.

A branch circuit, straight 240 volt, single phase, 100 amp load current (a unit heater), 200' one way length run in PVC.

What size copper conductor do I need to stay under 3% drop at 100 F ambient.

Now the same circuit in an ambient temp of 0 F.

Is there a significant difference?

Will we be surprised?

Bob


[This message has been edited by iwire (edited 09-26-2005).]

Here's a 1 hr. video by George Newton that discusses Voltage Drop, etc.
http://easylink.playstream.com/gnewton100/voltdrop/voltage_drop141.wmv
www.electrician.com

And when your done with Bob's (iwire) problem let me know what your voltage drop answer is for Tripp's problem. And I'll ask again is the "Volt's" software Mac compatible?

IMHO we are going to have to be very careful with wording and reading to make this a useful conversation; there are several subtle issues here. I've been working on this post for about an hour, running around in circles.

1) Voltage drop change with ambient temperature.

2) Wire size change with ambient temperature.

As I read DiverDan's posts, I believe that he is saying that the ampacity derating mandated for elevated temperature in table 310.16 is caused primarily by 1).

I disagree with this. When a conductor is operated at maximum ampacity, the internal self heating of the conductor raises the temperature to the maximum allowed by the insulation system. Once the wire is heated up to its operating temperature, it is at its operating temperature, _not_ at ambient temperature. The resistance of the copper wire is _not_ dependant upon the ambient temperature; it is dependant upon the _copper wire_ temperature. This means that for any given conductor insulation system, the resistance of the copper _at maximum ampacity_ can be taken as a constant.

Conductor ampacity is rated at 30C ambient with a given maximum allowed conductor temperature.

If we presume 90C conductors, then operation at 60C requires an ampacity correction of 0.71.

In both cases the conductor itself is presumed to be at 90C.

This is a reduction in allowed current capacity of 1/root(2) and a reduction in allowed internal self heating of 1/2.

Now let us look at the temperature change of resistance.
Let us presume a conductor that is not subject to significant self heating. This, for example, would be a long conductor sized for voltage drop rather than for maximum ampacity. The change in resistance of this conductor between 30C and 60C is not insignificant:
The temperature correction for resistance is
R2 = R1 [ 1 + K(T2 - 75) ] which is an approximate linear correction for resistances tabulated at 75C
At 30C the resistance of the conductor is 85% of its 75C value
At 60C the resistance of the conductor is 95% of its 75C value

In other words, comparing the same conductor at 30C versus 60C, the voltage drop at 60C will be about 12% greater. (meaning that if you had a 3% VD at 30C, you would expect a 3.36% VD at 60C.) The power dissipated per unit length at the same current would be 12% greater.

But the thermal ampacity correction shows that the heat dissipation capability of the conductor is far more significant.

Now jumping over to Bob's example. The answer to Bob's question would incorporate a number of factors which would totally hide the point of the discussion, with wire size being set by OCPD requirements (125% requirements), and factors of derating for wire temperature rating versus terminal temperature rating, maximum ampacity of heater circuits, etc. I'm not going to answer the question as stated. Instead I will focus entirely on the NEC 310.16 ampacities and use Bob's question as the basis.

Say we have 75C wire and 75C terminations, and the conductor must safely carry 100A. Without considering ambient thermal issues, table 310.16 says that a #3 conductor can be used.

At 0F, the temperature correction factor for 75C wire is 1.05, meaning that the ampacity of this #3 wire is 105A; we are just fine. In fact, the table doesn't go down to 0F; at 0F the temperature correction factor is probably on the order of 1.4!

At 100F, the temperature correction factor for 75C wire is 0.88. We now must use a #2 conductor with a derated ampacity of 102A.

Now let us consider voltage drop.

At 100F, using the #2 conductors, we are quite close to the thermal ampacity of the wire, so we can assume that the wire is nearly at 75C. The resistive component of the voltage drop will be:
(100A * 0.194 Ohm/kFT * 0.4kFT) / 240V = 3.23%

At 0F, using the #3 conductors, we are not even close to the thermal ampacity of the wire. Since we would be at the thermal ampacity at 30C, we can estimate that the conductor temperature will be 75C - (30C - 0F) = 75C - (30C - -17C) = 28C
The resistive component of the voltage drop will be:
(100A * 0.245 Ohm/kFT * 0.4kFT * [1 + 0.00323 * (28 - 75)] ) / 240V = 3.46%

So using Bob's example numbers, a #2 conductor would be sufficient for ampacity at 100F, but not sufficient for voltage drop. A #3 conductor would be sufficient for ampacity at 0F, but not sufficient for voltage drop.

But now let us figure the voltage drop of the #2 conductor in the 0F environment:
A #2 conductor (75C rating) carrying 100A is at its thermal ampacity at an ambient of 40C. This means that the conductor, in a 40C ambient, will be at 75C when carrying 100A. So in a -17C environment, we can expect a conductor temperature of 18C. The voltage drop is
(100A * 0.194 Ohm/kFt * 0.4kFt * [1 + 0.00323 * (18 - 75)]) / 240V = 2.64%

In this example, the temperature coefficient of resistance means a _significant_ difference in voltage drop, enough to mean the difference between needing to use a #1 conductor and a #2 conductor. However it seems pretty clear to me that the ampacity limits set by 310.16 are not changed by the temperature coefficient resistance of copper.

-Jon

Jon, first I thank you for explaining what I believed but did not know how to express.


Yes, that came to me as I was driving to work this morning.

I think my question would have been better if I simply said what is the voltage drop of a certain conductor at 0 F and 86 F.

Thanks again for spending your time on this, during my ride in I was realizing how complicated this little problem was.

Bob

I'll jump back into this after work this evening.

Winnie- I follow your approach and that means you did an excellent job of explaining. Tough to do, I've taught code classes for a number of years and so I appreciate your detail. There is however a "Fly in the ointment" as they say. Based on Bob's problem our conductor sizing also needs to comply with NEC Article 424.3 Section 423.3(B). This IMHO would require us to upsize the conductors by 25%;

I came up with a 1/0 conductor and a voltage drop of 4.88v. and would be within the recommended 3% max.

It's 2:00 AM and I'm not going to have the time for this thread.

George Little, This site's book store has all the information you need regarding Volts and computer compatability.

For all you others, Volts is offered with a 10-day free trial. Its Help file contains voltage drop and many other formulae and examples of use. Download links and program requirements and information are also available from ECN's book store.

Sorry, but with my new schedule, this is the best I can do.