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Hi Diver 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. 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.
Bob Badger Construction & Maintenance Electrician Massachusetts




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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. 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
Bob Badger Construction & Maintenance Electrician Massachusetts




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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 runon. 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 milliohms. 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




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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 09242005).]
George Little




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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
Bob Badger Construction & Maintenance Electrician Massachusetts




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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 rearranged.
In other words, if you use the simple voltage drop equation E = I R, you can rearrange 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




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Hah, George, at least someone is staying on target. 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
Bob Badger Construction & Maintenance Electrician Massachusetts




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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 oneway distance. (Note: for threephase 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 2pole 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.




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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




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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 09242005).]
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