# NCEES #136



## Rei (Feb 22, 2010)

I don't think #136 is in the index. Anyhow, I don't agree with the answer so hopefully someone can help me out.

Voltage drop = I|Z|

where Z=Rcos(theta)+Xsine(theta) and |Z|=sqr[(Rcos(theta))^2 + (Xsine(theta))^2]

unity power factor: Z = .05cos(0)+0.05sine(0) = 0.05 + j0 and |Z| = 0.05

zero power factor: Z= .05cos(90)+0.05sine(90) = 0 + j0.05 and |Z| = 0.05

0.707 power factor: .05cos(45)+0.05sine(45) = 0.0354 + j0.0354 and |Z| = 0.05

therefore, the voltage drop would be the same regardless of the power factor and the answer should be (a) voltage drop is independent of the load power factor.

What did I do wrong? The answer is actually (d).


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## Flyer_PE (Feb 22, 2010)

The power factor of the load has no affect on the impedance of the cable. It will always be 0.05 + j0.05.

However, changing the power factor of the load will alter the voltage drop across that impedance.

Maybe take a slightly different swing at it than they did.

Set VSource= 100 Volts

Find VLoad for each of the power factors:

1. Unity Power Factor

VLoad=VSource - I*Z = (100+j0)V - (50 + j0)(.05 + .05) = (97.5 - j2.5) V = 97.53 V at -1.47 Deg.

2. Zero Power Factor

VLoad=VSource - I*Z = (100+j0)V - (0 - j50)(.05 + .05) = (97.5 + j2.5) V = 97.53 V at +1.47 Deg.

3. Power Factor 0.707 (Theta = 45 degrees)

VLoad=VSource - I*Z = (100+j0)V - (50 angle -45 deg.)(.05 + .05) = (96046 + j0) V = 96.46V


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## z06dustin (Feb 22, 2010)

man Flyer is on a roll today.


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## Rei (Feb 22, 2010)

Flyer_PE said:


> The power factor of the load has no affect on the impedance of the cable. It will always be 0.05 + j0.05.However, changing the power factor of the load will alter the voltage drop across that impedance.
> 
> Maybe take a slightly different swing at it than they did.
> 
> ...


Thanks. I guess I have always misused Table 9. Since they have colums for 0.85 power factor, I thought we would adjust the power factor other than 0.85 on the impedances.


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## Rei (Feb 22, 2010)

Flyer: I just saw a post (can't remember the name) from #512 and that person also used Z = R cos theta + j X sin theta to find the cable impedance which is something I've always thought to use. I will double check when I get home, but I think I found it in the EC&amp;M handbook.


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## nuclear bus (Feb 24, 2010)

Rei said:


> Flyer: I just saw a post (can't remember the name) from #512 and that person also used Z = R cos theta + j X sin theta to find the cable impedance which is something I've always thought to use. I will double check when I get home, but I think I found it in the EC&amp;M handbook.


That was probably me. Sorry to add to the confusion. That equation is used to find the adjusted Z to calculate the voltage drop in a conductor, it's not technically to find the cable impedance. I know that sounds confusing. Flyer PE said it best:

"The power factor of the load has no affect on the impedance of the cable. It will always be 0.05 + j0.05.

However, changing the power factor of the load will alter the voltage drop across that impedance."

You are NOT finding the cable Z by using that equation I posted that I got out of the NEC. The cable has an impedance value on it's own, even if current is not flowing, not connected to anything. This is not the value you are finding. Rather, it is an intermediate equation used in finding the voltage drop once sources and loads are connected to either end. The other discussion involved looking up the values in Table 9, so I just assumed that you were going to plug and chug those values into the equation they give there as an example to calculate voltage drop. I apologize if that caused any confusion. I don't really know what more I can add.


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## Nik (Mar 8, 2010)

I think you might want to rethink for the 2nd calculation. It says "Zero Load Power Factor" . Zero load power factor is nothing but, no load condition. At which voltage drop will be minimal. Or, ideal case, Power Factor = 1 , as all the components in the transmission line (low freq.) are assumed resistive.

Please correct me if I am wrong. I do not understand why -j50 would count as "Zero power factor"

Thanks,

Nikhil



Flyer_PE said:


> The power factor of the load has no affect on the impedance of the cable. It will always be 0.05 + j0.05.However, changing the power factor of the load will alter the voltage drop across that impedance.
> 
> Maybe take a slightly different swing at it than they did.
> 
> ...


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## Flyer_PE (Mar 9, 2010)

^Zero power factor doesn't mean zero current, it just means no real power and thus no work is being done. The angle between the voltage and current waveforms is 90o (Cos 90o=0). In this case, it's a purely inductive load.


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## libbym78 (Oct 11, 2010)

Ok, I am have a question on this problem. Sometimes, to keep my numbers straight, I do different parts of the problem separately. In this case, I wanted to find the voltage drop value first and then subtract that from the starting voltage (assume 100)

So I punch this into my calc

Unity

(50&lt;0)*(.05+.05j)=3.53&lt;45

0 PF

(50&lt;-90)*(.05+.05j)=3.53&lt;-45

.707 pf

(50&lt;-45)*(.05+.05j)=3.53&lt;0

This seems to indicate that the Vdrop is not dependant on the PF. I would then take these values and subtract them from 100V--for all of them getting 96.46V

But when I include the 100V in the equation:

I get the following:

100-((50&lt;0)*(.05+.05j))=97.53&lt;-1.47

0 PF

100-9(50&lt;90)*(.05+.05j)0=97.53&lt;1.46

.707 pf

100-((50&lt;-45)*(.05+.05j))=96.46&lt;0

Why does having the 100- included in the formula change anything? Is my calculator making an error, or is there a legitimate reason that I have to calculate the ending voltage (as opposed to the amount that it drops). What do others get, when the omit the 100-

Any help would be much appreciated!

Thanks

Mike


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## Flyer_PE (Oct 11, 2010)

^You're manipulating vectors. The calculator is taking the 100 as 100 at an angle of 0o.


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## BamaBino (Aug 11, 2011)

I see a mechanical analogy to this problem in that when you want to lob an object (such as a mortar) the maximum distance then you lob it at 45 degrees.


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## evlar (Oct 9, 2011)

A-ha. For all cases, the voltage magnitude across the cable does not change, but when the power factor is 0.707, the voltage magnitude transferred to the load is least. Thanks for clarifying this up!


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## madblur (Feb 17, 2014)

I realize this post was from a few years ago, but I have a question about this same NCEES question. I am able to follow all of the equations above. However, I believe what I am stuck on is how the question is worded. The question asks "which of the following statements concerning the phase-to-neutral voltage drop at the receiving end of the feeder is most correct".

The answer, as previously noted, is (D) - "The voltage drop will be larger for a lagging load power factor of .707 than for either a unity load power factor or a zero load power factor."

I interpreted "at the receiving end of the feeder" to mean the point beyond where the line voltage drop would occur.

In the above example, VLoad=96.46, seems to me like this would be the smallest voltage drop "at the receiving end" (which I would consider to be the balanced load excluding the phase conductors) and the largest voltage drop in the phase conductors. I know I am missing something, or maybe just misinterpreting the question. If somebody could shed some light on this, I would greatly appreciate it.


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## DK PE (Feb 19, 2014)

madblur said:


> _I interpreted "at the receiving end of the feeder" to mean the point beyond where the line voltage drop would occur._
> 
> I'm not sure what you mean by this but think you may be making more complex than needed. The line impedance is of course distributed uniformly but is modeled with lumped. The load end of line connects to the actual load. The drop is Vsource - VLoad
> 
> ...


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## DK PE (Feb 19, 2014)

I'm not sure what you mean by this but think you may be making more complex than needed. The line impedance is of course distributed uniformly but is modeled with lumped. The load end of line connects to the actual load. The drop is Vsource - VLoad

Some of the math above assumes the source V/_ 0 degrees and current in phase (50 + j0) for a unity power factor load. This is not correct as the only case the source voltage and current are in phase would be for a load with a negative reactance component to cancel out the line positive reactance.

Go back to the solution given in the sample exam book. I think it will be come clear that using the standard IEEE red book IRcos (phi) +... equation you can work it out.

This above almost makes me think you are reading smallest voltage at the load not smallest drop? Sorry if I misinterpreted your question.

Sorry messed up formatting the "quote"


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## madblur (Feb 22, 2014)

Sorry for wording my questions so confusingly. I do believe I was overthinking this. After looking at your response, think I understand where I was confused. I knew the line impedance is uniform across the line and is simply modeled as a lump impedance. I was questioning whether the Voltage drop they were referring to was (1) from the source end of the line to the load end of the line, or (2) the voltage drop across the "actual load" connected.

You're statement that you thought I was "reading smallest voltage at the load not smallest drop" was basically a correct assumption. Since the question is "concerning the phase-to-neutral voltage drop at the receiving end of the feeder," I understood this to mean the voltage potential at the "load end of the line" that will drop across the "actual load" connected.

I now realize that the question is referring to the voltage drop between the source end of the line and the load end of the line. The solution in the book refers to the equations for effective impedance for 3 phase conductors and for a good approximation for line-to-neutral voltage drop {across the conductors} (in Chapter 9, Table 9 of the NEC). In other words, the question is basically asking: "What load power factor will yield the largest phase-to-neutral voltage drop between the source end of the line and the load end of the line?"


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## DK PE (Feb 22, 2014)

madblur said:


> Sorry for wording my questions so confusingly. I do believe I was overthinking this. After looking at your response, think I understand where I was confused. I knew the line impedance is uniform across the line and is simply modeled as a lump impedance. I was questioning whether the Voltage drop they were referring to was (1) from the source end of the line to the load end of the line, CORRECT or (2) the voltage drop across the "actual load" connected. Nope..
> 
> You're statement that you thought I was "reading smallest voltage at the load not smallest drop" was basically a correct assumption. Since the question is "concerning the phase-to-neutral voltage drop at the receiving end of the feeder," I understood this to mean the voltage potential at the "load end of the line" that will drop across the "actual load" connected.
> 
> I now realize that the question is referring to the voltage drop between the source end of the line and the load end of the line. (YES YES YES!!) The solution in the book refers to the equations for effective impedance for 3 phase conductors and for a good approximation for line-to-neutral voltage drop {across the conductors} (in Chapter 9, Table 9 of the NEC). In other words, the question is basically asking: "What load power factor will yield the largest phase-to-neutral voltage drop between the source end of the line and the load end of the line?"




Now you've got it. To really understand this, I recommend you beg/borrow a copy of IEEE Red book and find and study the phasor diagram that creates the basis for the equation used in Table 9. It will take a few minutes to review the trigonometry and comprehend but then this will all make sense.


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## yaoyaodes (Jul 19, 2021)

For zero load power factor, How do I know if it's leading or lagging?


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## Zach Stone P.E. (Jul 20, 2021)

yaoyaodes said:


> For zero load power factor, How do I know if it's leading or lagging?


The power factor comes out to zero only when the load is either 100% capacitive (-jX = -90º) or 100% inductive (+jX = +90º).

Ex: 

PF = cos(ø)
PF = cos(90º) or cos(-90º)
PF = 0

You can tell if it is lagging or leading by the impedance. Capacitive loads have a negative j in front of the reactance (-jX) and inductive loads have a positive j in front of the reactance (+jX).

Capacitive loads are a leading power factor. 
Inductive loads are a lagging power factor.

Most loads are inductive and lagging. The only leading power factor loads are capacitor banks or synchronous motors that are operating in a state of over excitation to supply reactive power.


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