NEC / Complex Imaginary Voltage Drop Confusion

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pete3589

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I'm running through some problems and I think I'm using note 2 on the AC resistance and reactance chart incorrectly (pg. 768 of NEC 2014)

My previous method which appears to be wrong, is when I receive a voltage drop question that has a pf other than .85, after finding my resistance and reactance values (and adjusting for distances and/or x2 for single phase) that I would use the equation in  note 2 to calculate the effective Z, then multiply by current to get the drop. 

In complex imaginary questions (question 12, test 2 is an example) they do not use the Zeq equation in note 2 of the NEC and multiply the current with its phase angle to the R and X values determined from the table to get the answer. 

However, in other questions (question 10, test 2) they use Vdrop = I { RL cos (theta) + XL sin (theta)}.  If I were to use I with phase angle here and multiply by RL + XL I get a different answer. 

I'm sure I'm missing something simple that someone can help me out with.  Is it as simple as if I am given a current with a phase angle I should skip the Note 2 equation altogether?

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pete3589 said:
I'm running through some problems and I think I'm using note 2 on the AC resistance and reactance chart incorrectly (pg. 768 of NEC 2014)

My previous method which appears to be wrong, is when I receive a voltage drop question that has a pf other than .85, after finding my resistance and reactance values (and adjusting for distances and/or x2 for single phase) that I would use the equation in  note 2 to calculate the effective Z, then multiply by current to get the drop. 

In complex imaginary questions (question 12, test 2 is an example) they do not use the Zeq equation in note 2 of the NEC and multiply the current with its phase angle to the R and X values determined from the table to get the answer. 

However, in other questions (question 10, test 2) they use Vdrop = I { RL cos (theta) + XL sin (theta)}.  If I were to use I with phase angle here and multiply by RL + XL I get a different answer. 

I'm sure I'm missing something simple that someone can help me out with.  Is it as simple as if I am given a current with a phase angle I should skip the Note 2 equation altogether?

View attachment 9357

View attachment 9358
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The NEC table Zeq is only for pf 0.85. The drop in voltage will always be I<Theta1 X Z<Theta2. It will be a good idea to always use this equation, whatever may be the source for R and X (i.e whether we get R and X from table or from given in the question itself).

 
They didn't use that equation on the second question. Just multiplied by the current which was 230<-41.4

This is pretty much my question. Why 2 different methods?

 
They are wrong in first question. They have not followed the Note 2 of the NEC table in its right spirit. However the difference in answers will be very less and may be intelligent choice of the options corrects the mistake. If something else is there with someone else, please let me know. 

 
I know this is a late response, but for future posterity...

The NEC, Table 9, Note 2 method is an "approximation" of the line impedance and was provided for one simple reason - many times you only know what the magnitude of the current is when doing a design and don't know (or care to know) what the phase angle of the current is. So since you're simply using the magnitude of the current, it's "good enough" to use the sum of the component magnitudes of the impedance. If executing Note 2 as instructed, the equivalent impedance is the sum of the magnitudes of the component parts of R cos (0) and X sin (0), so you actually get a magnitude for your impedance only (no imaginary part of the impedance). So now just bring it home - figure up the correct length of the line based on Single or Three phase system and multiply by your current magnitude and that's your voltage drop magnitude.

Now, say that you're taking a test and you're given the magnitude and phase angle of the current. To be "technically" correct, you should multiply the given current vector by the impedance vector. In this case, you would treat the formula from Table 9 in the NEC as a Vector formula where the Impedance contains both the real and imaginary components: Z = R cos (0) + jX sin (0). Again, bringing it home by figuring in the correct line length and multiplying by the current VECTOR.

What does all this mean? Absolutely nothing. Both answers will come out "close enough" that one method is not really any better than the other method in real life applications. Try it out!

Figure up the voltage drop with the following information using the "approximation" method in the NEC: V = 480V, I = 350A, 3PH system, PF = 0.7, 300 feet of 500 KCMIL copper in PVC Conduit.

PF angle = arccos (0.7) = 45.57*

Resistance is: 0.027 Ohms/1000' - Per NEC Note 2: 0.00567 Ohms at 300 feet.

Reactance is: 0.039 Ohms/1000' - Per NEC Note 2: 0.00835504 Ohms at 300 feet.

Total (Per NEC Note 2), add these two up: 0.01402504 Ohms Impedance at 300 feet.

Multiply this by 350A = 4.90877V dropped.

----------

Now, same system, except current is 350A < 23*

Impedance is now: (0.00567 + j0.00835504), or (0.010 < 55.84*) at 300 feet.

Voltage drop: (350 < 23*) x (0.010 < 55.84*) = 3.534 < 78.83*

----------

We got answers that are within 1.4 volts of each other. IN REAL LIFE, this is really insignificant when you consider that the whole point of doing this is to find out what the voltage drop percentage is in order to meet the voltage drop recommendation provided by the NEC. At 480V, 3PH systems, this equates to: 1.02% voltage drop using the "approximated" impedance and 0.74% voltage drop with the exact impedance - both of which are well below the recommendation of 3% for feeders (also note that the "approximated" impedance provides the conservative result). Therefore, it really only matters when taking a Power PE exam.

If given a current vector on the exam, you should be using the impedance vector for your calculations. The choices should be far enough apart that you won't see both the approximated answer and the exact answer. Likely the wrong answers to choose from will consist of: you figured your line length wrong, or you misread the table, or you don't understand the material and you're way off the correct answer.

 
BigWheel said:
I know this is a late response, but for future posterity...

The NEC, Table 9, Note 2 method is an "approximation" of the line impedance and was provided for one simple reason - many times you only know what the magnitude of the current is when doing a design and don't know (or care to know) what the phase angle of the current is. So since you're simply using the magnitude of the current, it's "good enough" to use the sum of the component magnitudes of the impedance. If executing Note 2 as instructed, the equivalent impedance is the sum of the magnitudes of the component parts of R cos (0) and X sin (0), so you actually get a magnitude for your impedance only (no imaginary part of the impedance). So now just bring it home - figure up the correct length of the line based on Single or Three phase system and multiply by your current magnitude and that's your voltage drop magnitude.

Now, say that you're taking a test and you're given the magnitude and phase angle of the current. To be "technically" correct, you should multiply the given current vector by the impedance vector. In this case, you would treat the formula from Table 9 in the NEC as a Vector formula where the Impedance contains both the real and imaginary components: Z = R cos (0) + jX sin (0). Again, bringing it home by figuring in the correct line length and multiplying by the current VECTOR.

What does all this mean? Absolutely nothing. Both answers will come out "close enough" that one method is not really any better than the other method in real life applications. Try it out!

Figure up the voltage drop with the following information using the "approximation" method in the NEC: V = 480V, I = 350A, 3PH system, PF = 0.7, 300 feet of 500 KCMIL copper in PVC Conduit.

PF angle = arccos (0.7) = 45.57*

Resistance is: 0.027 Ohms/1000' - Per NEC Note 2: 0.00567 Ohms at 300 feet.

Reactance is: 0.039 Ohms/1000' - Per NEC Note 2: 0.00835504 Ohms at 300 feet.

Total (Per NEC Note 2), add these two up: 0.01402504 Ohms Impedance at 300 feet.

Multiply this by 350A = 4.90877V dropped.

----------

Now, same system, except current is 350A < 23*

Impedance is now: (0.00567 + j0.00835504), or (0.010 < 55.84*) at 300 feet.

Voltage drop: (350 < 23*) x (0.010 < 55.84*) = 3.534 < 78.83*

----------

We got answers that are within 1.4 volts of each other. IN REAL LIFE, this is really insignificant when you consider that the whole point of doing this is to find out what the voltage drop percentage is in order to meet the voltage drop recommendation provided by the NEC. At 480V, 3PH systems, this equates to: 1.02% voltage drop using the "approximated" impedance and 0.74% voltage drop with the exact impedance - both of which are well below the recommendation of 3% for feeders (also note that the "approximated" impedance provides the conservative result). Therefore, it really only matters when taking a Power PE exam.

If given a current vector on the exam, you should be using the impedance vector for your calculations. The choices should be far enough apart that you won't see both the approximated answer and the exact answer. Likely the wrong answers to choose from will consist of: you figured your line length wrong, or you misread the table, or you don't understand the material and you're way off the correct answer.
Click to expand...
Not exactly sure what line of reasoning you used (best I can guess is a very general, inaccurate approximation), but you never add reactance and resistance in the manner you presented above.  It is a vector sum, not a scalar sum.  Hence, the Pythagorean theorem should be dinging like a doorbell.

 
I do not have a copy of the NEC in front of me, but if memory serves, one of the "approximations" is only valid at 0.85 PF.  Also, this test is not about approximating! Work the problem legitimately and let the answer be "approximated" off your work, not the other way around!!  Good way to screw the pooch and be left sitting for another round 6 months later.  For instance, the test often gives answers that are close, and it's up to you to determine what the best answer is.  A gross approximation could put you on the wrong end of the spectrum.

 
The NEC Table 9 method is an "approximation." It even says so if you read fpn 2.

Table 9 works for all power factor values from 0 to unity, in fact - you use the left side of the table if the power factor is anything but .85 and the right side directly if it is .85.

It is okay to approximate, I promise.

TNPE said:
...if memory serves, one of the "approximations" is only valid at 0.85 PF.  
Click to expand...
Not wholly correct. The right side of the Table is a valid approximation of the equivalent impedance at 0.85 PF. These values are used directly without the use of fpn 2.

The values in the left half of the table are used in conjunction with fpn 2 to calculate the approximate equivalent impedance at all other values of PF. Read fpn 2 for proof.

TNPE said:
Also, this test is not about approximating! Work the problem legitimately and let the answer be "approximated" off your work, not the other way around!!
Click to expand...
It looks like you rushed through your reading of what I wrote. Reread paragraph 3 where you will hopefully conclude that I concur with this statement.

TNPE said:
For instance, the test often gives answers that are close, and it's up to you to determine what the best answer is.  A gross approximation could put you on the wrong end of the spectrum.
Click to expand...
It's not a gross approximation, as I clearly demonstrated in the provided example. Is it exact? No, hence the term "approximate." Again, see paragraph 3.

-------

I'm not sure what your line of work is in, but mine is facility design (I work for an Architectural Engineering firm). Knowing the NEC and calculating voltage drops in my feeders is a must and I'm well practiced in it and can confirm all hand calculations with professional tools I use every day.

I may very well have to take the PE exam again in October, but it won't be because I don't know the NEC or how to calculate voltage drop calculations.

tl;dr:

Read paragraph 3 again, read the NEC before telling someone they are wrong, and I agree that calculations on the exam should be exact.

 
BigWheel said:
The NEC Table 9 method is an "approximation." It even says so if you read fpn 2.

Table 9 works for all power factor values from 0 to unity, in fact - you use the left side of the table if the power factor is anything but .85 and the right side directly if it is .85.

It is okay to approximate, I promise.

Not wholly correct. The right side of the Table is a valid approximation of the equivalent impedance at 0.85 PF. These values are used directly without the use of fpn 2.

The values in the left half of the table are used in conjunction with fpn 2 to calculate the approximate equivalent impedance at all other values of PF. Read fpn 2 for proof.

It looks like you rushed through your reading of what I wrote. Reread paragraph 3 where you will hopefully conclude that I concur with this statement.

It's not a gross approximation, as I clearly demonstrated in the provided example. Is it exact? No, hence the term "approximate." Again, see paragraph 3.

-------

I'm not sure what your line of work is in, but mine is facility design (I work for an Architectural Engineering firm). Knowing the NEC and calculating voltage drops in my feeders is a must and I'm well practiced in it and can confirm all hand calculations with professional tools I use every day.

I may very well have to take the PE exam again in October, but it won't be because I don't know the NEC or how to calculate voltage drop calculations.

tl;dr:

Read paragraph 3 again, read the NEC before telling someone they are wrong, and I agree that calculations on the exam should be exact.
Click to expand...
Nothing I stated was incorrect, other than my recollection from memory and the 0.85 PF approximation approach (read where I said I didn't have a copy in front of me).  It is never ok to scalar add reactance and resistance for any other reason than approximating, and even then, I'd only find it valid if you were showing your electrician buddies a cute trick.  But why not do it exact?  Flexing your electrician shortcut skills?  Let's hope you didn't use those on the test and it bite you.  Also, I am a PE and practice in electric utilities, so take that for what it's worth.  

Happy approximating!! :beerchug:

 
As per usual, this is a perfect example of an engineer over-thinking things. I'm guilty of it myself at times, so don't take that as a dig at you.

You really need to review Table 9, read Note 2, and just come to grips with the fact that the "cute trick" you talk about isn't my trick. It's the "cute trick" the NEC prescribes for the approximation. I know you probably don't have a copy of the NEC, so I provided a link for you to review in my last post.

If you don't want to review it and possibly learn something you don't fully understand, then fine. Just drop the "I'm a PE, so there" act. It doesn't impress anyone - it actually makes you come of as immature and could leave people wondering what kind of engineer you really are that you feel you need to use that sort of petty logic to feel better about posting things that only prove how little you understand the NEC. 

Seriously, say what you just wrote out loud and try to imagine what it must sound like to people who know you're wrong...

It's honestly not your fault. If you really do practice on the utility side of the demarcation point, then you have about as much use for the NEC as I have for the NESC on my side of the demarcation point, but it's no excuse for insisting you're right when you are actually very wrong. 

Cheers ? 

 
BigWheel said:
As per usual, this is a perfect example of an engineer over-thinking things. I'm guilty of it myself at times, so don't take that as a dig at you.

You really need to review Table 9, read Note 2, and just come to grips with the fact that the "cute trick" you talk about isn't my trick. It's the "cute trick" the NEC prescribes for the approximation. I know you probably don't have a copy of the NEC, so I provided a link for you to review in my last post.

If you don't want to review it and possibly learn something you don't fully understand, then fine. Just drop the "I'm a PE, so there" act. It doesn't impress anyone - it actually makes you come of as immature and could leave people wondering what kind of engineer you really are that you feel you need to use that sort of petty logic to feel better about posting things that only prove how little you understand the NEC. 

Seriously, say what you just wrote out loud and try to imagine what it must sound like to people who know you're wrong...

It's honestly not your fault. If you really do practice on the utility side of the demarcation point, then you have about as much use for the NEC as I have for the NESC on my side of the demarcation point, but it's no excuse for insisting you're right when you are actually very wrong. 

Cheers ? 
Click to expand...
I stand by what I said, and nothing I've intimated is incorrect.  Why would you bring up an approximation on this board that isn't valid for this exam or for practice in industry?  I'd advise others reading your posts to take them with a pinch of salt.  Be sure to include this approximation on the first thing you stamp, whenever that time may be.  Also, I didn't say I'm a PE "so there".  I told you that I am a PE and indicated what field I work in.  I then left it up to you and any other reader to take it how you want.  However, one truth remains, I have successfully taken and passed the exam while you wait it out.  Therefore, I think I have an understanding of the exam and its contents.  I clearly said I do not have a copy of the NEC in front of me, but yes, I own multiple copies and they are at my disposal at work.  

You seemed to be so caught up on vectors in your original post.  An impedance of j10 is a vector, and for that matter, any impedance is a "vector."  So, please tell everyone how to use that approach if they aren't given an impedance vector when all values of impedance are a vector?  This is a rhetorical exercise for you, just in case you're not in to reading tea leaves and sarcasm on a web-based forum....

Be sure and scalar add reactance and resistance on your first project... Maybe it will be one that, in using this approach, exceeds the VD% and you call for a larger wire than necessary.  I bet your clients and employer will be pleased.....  

 
First: Yes, everything you have stated regarding the correct way to treat impedance vectors is correct. I have never denied that. While studying for the PE, I read a lot of your posts, which were very helpful with respect to my better understanding of content that I don't deal with or use on a daily basis. I personally feel that my weakest subjects are everything on your side of the demarcation point.

Let me apologize to you if you took away from anything I've posted that you are less of an engineer than anyone else here who contributes their valuable insight based on their unique experiences.

Having said all that - you seem to be stuck on this idea that treating vectors as scalars is something I came up with on my own to impress my electrician friends, or that I came up with it because I'm lazy or too stupid to be a Professional Engineer (admittedly, I'm reading into that last part what I think you mean and fully admit that I could be completely wrong regarding your true feelings).

In any case - It's becoming obvious to me that you're not going to dig into Table 9 and its notes to understand that treating the impedance vector as a scalar is, in fact, prescribed by the NEC for the purposes of APPROXIMATING THE VOLTAGE DROP of the feeder.

**READ THIS NEXT PARAGRAPH VERY CAREFULLY**

It might help if you understood that the NEC does NOT state what the maximum voltage drop of the low-voltage feeders shall be - rather, they RECOMMEND (for system stability) that the maximum voltage drop of our feeders be no more than 3%. Since the approximation that the NEC prescribes will produce the most conservative impedance value, it's no wonder that they would include it to cover every possible use of the NEC from design engineers (like me) who fully understand that the more accurate way of calculating impedance and voltage drops is to treat the impedance as a vector, to the electrical contractors (who probably don't have the necessary education to know or understand this), Construction Management Inspectors (CMIs), the Authority Having Jurisdiction (AHJ), to the final customer...any of the ones previously named could LITERALLY calculate the most conservative voltage drop of feeders on the back of a napkin over a plate of pasta at their favorite Italian restaurant while discussing the particulars of a proposed design since the math is conveniently reduced to simple arithmetic.

Here is a direct quote from Note 2 of Table 9 (bold text used by me to highlight the NEC's admitted limitations):

Note 2: Effective Z is defined as R cos(θ) + X sin(θ), where θ is the power factor angle of the circuit. Multiplying current by effective impedance gives a good approximation for line-to-neutral voltage drop. Effective impedance values shown in this table are valid only at 0.85 power factor (NOTE: The EFFECTIVE Impedance values referenced here are the two RIGHT columns of the table). For another circuit power factor (PF), effective impedance (Ze) can be calculated from R and XL values given in this table as follows: Ze = R × PF + XL sin[arccos(PF)] (NOTE: the R and XL values referenced in this sentence are referencing the LEFT three columns of the table).

Note that nowhere in this note is the imaginary "j" component of the vector included; therefore, the impedance approximation of the NEC is a SCALAR.

Now, just because I understand what the intent of the NEC was when they provided this approximation in no way implies that I endorse this as a way of doing real engineering work - it simply means that I acknowledge that the NEC does prescribe this as a way to conveniently calculate a conservative voltage drop for a feeder for the express purpose of determining whether or not the feeder length will cause the installation to exceed the NEC's recommendation of 3% of voltage drop after the construction is complete.

 
Admittedly, when I originally replied, I didn't notice your proclamation about it being an approximation.  Mea culpa.  That said, I entertained the rest cause your tones through text seemed condescending in nature.  I'm not proclaiming to know it all, nor am I the best in the world of power, but I'm no fool.  Therefore, if my responses seemed blunt in nature, that's why.

 
I love bacon!....and sausage, porkchops, ribs, bbq...

 
pete3589 said:
I'm running through some problems and I think I'm using note 2 on the AC resistance and reactance chart incorrectly (pg. 768 of NEC 2014)

My previous method which appears to be wrong, is when I receive a voltage drop question that has a pf other than .85, after finding my resistance and reactance values (and adjusting for distances and/or x2 for single phase) that I would use the equation in  note 2 to calculate the effective Z, then multiply by current to get the drop. 

In complex imaginary questions (question 12, test 2 is an example) they do not use the Zeq equation in note 2 of the NEC and multiply the current with its phase angle to the R and X values determined from the table to get the answer. 

However, in other questions (question 10, test 2) they use Vdrop = I { RL cos (theta) + XL sin (theta)}.  If I were to use I with phase angle here and multiply by RL + XL I get a different answer. 

I'm sure I'm missing something simple that someone can help me out with.  Is it as simple as if I am given a current with a phase angle I should skip the Note 2 equation altogether?

View attachment 9357

View attachment 9358
Click to expand...
Now that we've got all that behind us...

-------

In my opinion, neither of the provided methods are correct. Notice I said "method," and not "solution." Here's why:

In Problem 1 (the first picture above), they provide the current vector and the impedance vector and do not appear to word the problem statement in a way that references the NEC in any way; therefore, I would solve that problem using the current and impedance vectors given and would not bother to crack open the NEC and see what the approximation procedure is:

Z = [2*(220/1000)]*(0.78 + j0.052) = (0.343 + j0.023)

Vd = I * Z

Vd = (40<-10) * (0.343 + j0.023), or

Vd = (40<-10) * (0.344<3.81) = 13.76<-6.19 V

Vd% = (13.76V/120V)*100 = 11.5% <-- Sidebar: This would be an extremely undersized conductor for a feeder application, per the NEC recommendation for feeder voltage drops, but more especially for branch circuit voltage drops, which the NEC recommends should be less than 2%.

NOTE: The exact solution yielded an answer of 11.5%, while their "NEC approximated solution" yielded an answer of 11.0%. I don't know what choices you had to choose from, but hopefully you didn't have both 11.0% and 11.5% to choose from. If both choices were available, I would choose 11.5%, not 11.0% and conclude that the author of the question used the wrong method (it does happen, you know).

-----------

In problem 2 (the second picture above), the solution provided implies to me that the problem was worded in such a way that they WANT you to use Table 9 of the NEC to solve the problem (I didn't see the problem statement, so I'm making an assumption here). If this is the case, I would use Table 9 in conjunction with Note 2 to approximate the effective impedance to find the voltage drop (Note: since I don't see the original problem statement, I'm assuming that this is a 3PH system based on the voltage magnitude provided):

Ze = (500/1000) * [(0.038*0.75) + (0.040*sin(arccos(0.75)))]

Ze = 1/2*(0.0285 + 0.0265)

Ze = 0.0275 Ohms

Vd = 230A * 0.0275 = 6.24V

Vd% = (6.24/600)*100 = 1.04% <-- Sidebar: this 350-kcmil feeder would be an acceptable size for this application.

Vload = 600V - 6.24 = 593.76V

A couple of things to note here: 1. The exact solution method that the author used is almost exactly what the NEC approximated it should be, and 2. Notice that the problem only provided the magnitude of the current, and not the current vector.

Conclusion:

For the purposes of taking the Power PE Exam test, I would read the problem and use the method they asked for (explicitly or implicitly) and choose my best answer based on that calculation method. When the problem is worded in such a way as to "force you to use Table 9" for a solution, they are likely testing your ability to use the table correctly (meaning, did you choose the correct Resistance or Reactance based on the size and type of the conductor and the conduit used?).

I don't know what NCEES has in mind for "Table 9" questions, but if I was the one writing the test question, I would set it up and I would calculate four solutions that I would give you to choose from. The four solutions would include a "near" correct answer, of course, and the other three would be based on accidentally swapping Resistance and Reactance values, choosing the wrong conduit type (maybe), choosing the wrong conductor size (maybe), using the 0.85 PF values when the problem is set up for something other than 0.85 PF...these kinds of things.

 
So serious!

I guess you've never had Memphis BBQ....?  Dodged that one quicker than Hillary being quizzed about her rogue emails....

 
Let us try to remove  and quantify the confusion.

In real sense as per textbook definition VD= IEI-IVI. Now the best way would be to find this out and be happy. To  get VD on V axis we draw an arc touching V axis at C; so the exact VD = OC in the figure. But if want to take approximation let us take take offset of vector VD or IZ on V axis i.e. OB in the figure attached. It comes I R Cos theta+I X sin theta meaning approximate Z is R cos theta+ X sin theta. The error in using the formula is BC. It depends on you whether the error in calculations is acceptable to you or not. The error is not constant but will vary on R and X. that is angle theta itself. That is the reason the figures for .85 pf should not be taken for other pf's calculations because then error will increase. This is a case of meeting point of technicians and Engineers so confusion has to happen.

View attachment 9444

 
Please take AB = I X sin theta in figure and also in OA+AB. I am not able to take it out and edit.

 

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