NCEES Practice Exam Zeff Use

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Aniket Shah

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Hi Everyone, quick question I had on Zeff equation shown in the NEC under table 9.

I understand that we use that equation only when pf is not equal to 0.85. However, on many of NCEES practice exam questions, they use pf's that are not 0.85, yet they use the values directly off table 9 without using the Zeff equation.

Would someone mind explaining why NCEES does not use the Zeff equation even though the pf is not 0.85?

Thanks in advance!
 
Off the top of my head (since I don't have the NEC book in front of me right now):

Zeff = R cos(theta) + X sin(theta), where theta = cos^-1 (p.f.)

The Zeff values given in Table 9 are based on p.f. = 0.85, and for other power factor values you have to first calculate theta before plugging in the Zeff equation.

The effective impedance Zeff value is only an approximation. I don't remember the exact details, but I do remember Zach Stone's Electrical PE Review once covered the fact that Zeff is only an approximation.

I usually use the individual R and X values from NEC Chapter 9 Table 9, since they are more accurate and many more practice problems that I have seen involving voltage drop and NEC Chapter 9 Table 9 use the individual R and X values instead of the Zeff values.

Hope this helps...
 
Ah yes this helps a lot. Thanks! Ill just use the R and X values directly off table 9 then. Thanks for the in-depth explanation. Much appreciated!
 
Hi Everyone, quick question I had on Zeff equation shown in the NEC under table 9.

I understand that we use that equation only when pf is not equal to 0.85. However, on many of NCEES practice exam questions, they use pf's that are not 0.85, yet they use the values directly off table 9 without using the Zeff equation.

Would someone mind explaining why NCEES does not use the Zeff equation even though the pf is not 0.85?

Thanks in advance!
Just as you've noticed, NCEES is a bit inconsistent with their approach to solving voltage drop problems using the NEC Chapter 9 Table 9 Values.

Sometimes they use the effective impedance, and other times they cherry pick the R and XL values from the table to form a complex impedance.

According to the NEC, which is where these values come from and the second-largest subject on the power PE exam, the R and XL values are to be used to calculate a new effective impedance to be used for approximate voltage drop calculations any time the power factor of the circuit is not 0.85. If the power factor of the circuit is 0.85, then the effective impedance values are to be used directly from the table.

There is no mention in the NEC of using the R and XL values to form a complex impedance. The NEC does not deal with complex numbers.

In the newer version of the NCEES practice exam, NCEES has doubled down on their approach stating in one of the solutions that either approach is acceptable.

I would err on the side of using the effective impedance values at all times, either directly from the table for 0.85 PF or calculating the new effective impedance with R and XL when the PF is not 0.85 since that is exactly how NEC instructions you to use the table values.

The only time I would use the R and XL table values directly as a complex impedance for a problem on the PE exam is if the answer choices are complex numbers that would require using a complex number for line impedance to arrive at.
 
Great thank you everyone! Ill practice with the reference manual formula and see what I get.

Zach, I will err on using the Zeff equation for calculating VD.

Thanks again!
 

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