# NEW NCEES problem # 512



## WantsPE (Mar 29, 2009)

Can someone please throw some light on why the answer shows Z 500ft as (0.035+j 0.049) /4??

I understand the Z1000ft value and to me, the answer for Z 500ft should be Z1000ft /2 and since there are 2 sets of conductors, shouldnt we multiply Z by 2 to get the value for Z 500ft for 2 sets of conductors??

I know I am missing something here, I just cannot figure out what it is...

Thank you..


----------



## jcreit (Mar 29, 2009)

WantsPE said:


> Can someone please throw some light on why the answer shows Z 500ft as (0.035+j 0.049) /4??I understand the Z1000ft value and to me, the answer for Z 500ft should be Z1000ft /2 and since there are 2 sets of conductors, shouldnt we multiply Z by 2 to get the value for Z 500ft for 2 sets of conductors??
> 
> I know I am missing something here, I just cannot figure out what it is...
> 
> Thank you..


Correct if I am wrong. I believe the reason why we have to divide it by 2 is first; it has 2 conductors per line and secondly, parallel resistances divides by 2.


----------



## niurou (Mar 29, 2009)

WantsPE said:


> Can someone please throw some light on why the answer shows Z 500ft as (0.035+j 0.049) /4??I understand the Z1000ft value and to me, the answer for Z 500ft should be Z1000ft /2 and since there are 2 sets of conductors, shouldnt we multiply Z by 2 to get the value for Z 500ft for 2 sets of conductors??
> 
> I know I am missing something here, I just cannot figure out what it is...
> 
> Thank you..


The thicker the wire the lower the impedance, two cable is twice "thicker" than one cable. so impedance is half.

Man, this is the most basic parallel and series impedance calculation...


----------



## smelltheglove (Feb 14, 2010)

I know this thread is almost a year old but.....

Can anyone shed any light on why the R and X values were not multiplied by Cosine theta and Sine theta? I realize this is only an approximation of the Z for that length of wire, but the power factor in the question is 0.9 and the table lists effective Z for PF's of .85. Besides, it pulls the R and X values from the other part of the table anyway.

Thanks in advance.


----------



## Flyer_PE (Feb 14, 2010)

The question is asking for the impedance of the cable only. The power factor (or any other characteristic for that matter) of the load doesn't affect the cable impedance.


----------



## nuclear bus (Feb 21, 2010)

smelltheglove said:


> I know this thread is almost a year old but.....
> Can anyone shed any light on why the R and X values were not multiplied by Cosine theta and Sine theta? I realize this is only an approximation of the Z for that length of wire, but the power factor in the question is 0.9 and the table lists effective Z for PF's of .85. Besides, it pulls the R and X values from the other part of the table anyway.
> 
> Thanks in advance.


I know where you're coming from, but I think you're just getting one step ahead. Before you can calculate the voltage drop you need to first determine the cable impedance, then correct it for the power factor of the load. So first you figure out the cable impedance as 0.035 + j0.049 / 4 using the first two columns for X and R in Table 9 of the NEC handbook. That's all this problem asks for. But, if you had to go on and calculate the voltage drop, then you'd need to correct the impedance using the power factor Z = R cos theta + j X sin theta. Hope this helps. I know it helps me just trying to explain it


----------



## Rei (Feb 22, 2010)

nuclear bus said:


> smelltheglove said:
> 
> 
> > I know this thread is almost a year old but.....
> ...


Now you get me confused. Will you go to my post on question #136 and tell me the different between my calculation and Flyer's calculation? I use Z = R cos theta + j X sin theta same as you to account for the power factor value which would proof that the answer is A and not D. Flyer, however, said the power factor is account for the current and not the impedance values which from his calculation, answer D is correct.


----------



## Rei (Feb 23, 2010)

Flyer, how come you don't answer my question. Are you saying using Z = R cos theta + j X sin theta to find the cable impedances other than 0.85 pf is wrong?


----------



## Flyer_PE (Feb 23, 2010)

Rei said:


> Flyer, how come you don't answer my question. Are you saying using Z = R cos theta + j X sin theta to find the cable impedances other than 0.85 pf is wrong?


I didn't realize there was a question directed at me. The formula you have there is wrong.

Z = R+jX

R = Z cos theta

X = Z sin theta

Regardless of what load is supplied by the cable, the impedance of that cable does not change. All that column in Chapter 9 Table 9 does is give a quick and dirty method for estimating the voltage drop to a typical motor without actually doing the vector addition. One of the things to keep in mind is that engineers are NOT the target audience of the NEC. It is intended for electricians.


----------

