# NEC Table 9 Note 2



## EEVA PE (Oct 1, 2011)

1. Are the two equations mentioned in NEC Table 9 note 2 equivalent for effective Z.

Z = R cos () + X cos () and Z = R x PF + X Sin[arccos(PF)]

They appear the same to me, but if they different when do you use a specific equation instead of the other.

2. When using the above equations as the impedance input to find voltage drop, can I assume one does not include the lead or lag angle in the current since it has already been included in effective Z.


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## Flyer_PE (Oct 1, 2011)

EEVA said:


> 1. Are the two equations mentioned in NEC Table 9 note 2 equivalent for effective Z.
> Z = R cos () + X cos () and Z = R x PF + X Sin[arccos(PF)]
> 
> They appear the same to me, but if they different when do you use a specific equation instead of the other.
> ...


1. The two equations are the same. Power factor (PF) is the cosine of the impedance angle.

2. Not always. For voltage drop, V=IZ where both I and Z are vectors. If the impedance is primarily resistive (angle near 0), you can likely get away with ignoring it. But on the exam, you have to be aware of how much accuracy is required to get the right answer.


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## nukem2k5 (Oct 17, 2016)

Flyer_PE said:


> 2. Not always. For voltage drop, V=IZ where both I and Z are vectors. If the impedance is primarily resistive (angle near 0), you can likely get away with ignoring it. But on the exam, you have to be aware of how much accuracy is required to get the right answer.


Necro'ing this thread.  I've looked at several other threads (on EB and Mike Holt forums) where this NEC Table 9 Note 2 formula is discussed, and I'm still slightly confused.  

When using the Effective Z from the table (for 0.85 power factor) -- or when calculating a new Effective Z (for some other power factor) -- in order to calculate voltage drop on a line, do we use only the magnitude of the current, and not the phase angle?  I don't understand why we use the concept of "effective Z"  rather than using the actual impedance (R + j X -- notice the j) given in Table 9 along with proper vector math (wherein the j is used in the value of the line impedance).  

To clarify: let's assume #1/0 uncoated copper in PVC, 1000' from source to load, with I = 50 ∟-49.5 A (that is, PF = 0.65 lag, and assuming source voltage angle = 0).  For this cable, R = 0.12 Ω and X = j 0.044 Ω.

A)   Using the normal method (with j on the reactance):        (50 ∟-49.5 A) (0.12 + j 0.044 Ω)  =   *6.39 ∟-29.4  V*

B)   Using the NEC (IEEE red book) method:       (50 ∟-49.5 A) [ 0.12 cos(-49.5) + 0.044 sin(-49.5) Ω ]  =   *2.22 ∟-49.5  V*

C)   Using the NEC (IEEE red book) method and only the current magnitude:      (50 A) [ 0.12 cos(-49.5) + 0.044 sin(-49.5) ]  =   *2.22 ∟0  V*

Am I missing something?  Why is the answer in (A) much different from those in (B) and (C)?


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## nukem2k5 (Oct 17, 2016)

Perhaps the difference in the equations above is that (A) is the vectoral difference between the sending/receiving voltage phasors, whereas (B) and (C) are effectively the difference in the magnitude, as if the two phasors were superimposed on each other?


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## Flyer_PE (Oct 18, 2016)

nukem2k5 said:


> A)   Using the normal method (with j on the reactance):        (50 ∟-49.5 A) (0.12 + j 0.044 Ω)  =   *6.39 ∟-29.4  V*
> 
> B)   Using the NEC (IEEE red book) method:       (50 ∟-49.5 A) [ 0.12 cos(-49.5) + 0.044 sin(-49.5) Ω ]  =   *2.22 ∟-49.5  V*
> 
> ...


A the the most accurate answer.  B and C are quick approximations but you did the math wrong on both.  They are not  vector equations (no "j" in front of the inductance value).

So C becomes (50)*[0.12 cos(49.5) + 0.044 sin(49.5)] = 50*(0.078 + .033) = 50 * 0.111 = 5.55 V.


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## nukem2k5 (Oct 18, 2016)

Flyer_PE said:


> A the the most accurate answer.  B and C are quick approximations but you did the math wrong on both.  They are not  vector equations (no "j" in front of the inductance value).
> 
> So C becomes (50)*[0.12 cos(49.5) + 0.044 sin(49.5)] = 50*(0.078 + .033) = 50 * 0.111 = 5.55 V.


I didn't use j in equations for B &amp; C.  Also, the current angle in your calculation should be negative because it's a lagging circuit and I stated that the voltage phasor reference angle is assumed to be 0.  For ABC rotation, that'd make the current angle -49.5, right, or do I have something backwards?  

So when using the equation from the NEC, are we supposed to include the current phase angle (as in *), or not (as in [C]) ?
*


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## nukem2k5 (Oct 18, 2016)

Oh!  It should be positive, since the value used there should be (θv - θi) = (0 - (-49.5)) = (0 + 49.5) = 49.5

In that case, the answers for * (**5.57 **∟* *49.5 V)** and [C] (**5.57 **V)** are closer to [A] and within reasonable tolerance for an "approximation" type of equation.  Still, are supposed to use the current angle or just the magnitude in the NEC equation?
*


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## Flyer_PE (Oct 18, 2016)

In NEC space, you really don't care much about the angle.

Take a close look at Note 2 for Table 9.  They define theta as the power factor angle for the circuit.  That angle is actually the conjugate of the current angle so it would be a positive number for any inductive circuit.


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## nukem2k5 (Oct 19, 2016)

Flyer_PE said:


> In NEC space, you really don't care much about the angle.


This is the conclusion I came to.  I figure the intent of the equation in NEC (and IEEE 141) is to approximate and simplify the calculation by working with "real" numbers and not worrying about phase angle (except that it's inherent to the power factor).  



Flyer_PE said:


> That angle is actually the conjugate of the current angle so it would be a positive number for any inductive circuit.


I was having a premature senior moment earlier when I used a negative value in the equations.  I think of the power factor angle as being Θv - Θi.  It's only equal to the conjugate of the current angle if you're using 0 as the voltage angle, right?


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## nukem2k5 (Oct 19, 2016)

Flyer_PE said:


> Take a close look at Note 2 for Table 9.  They define theta as the power factor angle for the circuit.


When we use the equation, we usually know (assume) the PF of the load at the receiving end, but the power factor of the total circuit (load and cable) -- which would be "seen" at the sending end -- can only be truly be known if we know the complex impedance of the circuit (that is, of the load and of the cable), and then find the complex current using I = Vsending / (Zcable +Zload), with V as the reference vector (i.e. angle of 0).    Using the load power factor in this equation, as instructed in IEEE-141, allows us to assign a phase angle to the receiving-end Voltage vector (in their case, 0 degrees) as well as to the circuit's Current vector; but still, the phase angle of the sending-end Voltage relative to the Current in the circuit (which ultimately gives us the power factor of the circuit, as requested by NEC) is very much dependent on the resistive and reactive characteristics of the cable.  It seems like a like a chicken-and-egg scenario.

IEEE uses the receiving-end voltage as the reference, since (I presume) we only "know" the receiving-end (load) power factor (e.g. motor nameplate value), and must do this vector math using the cable's Resistance and Reactance to actually know the power factor of the total circuit.  NEC seems flawed since it calls for the "power factor angle of the circuit" even though, I think, in practice, we usually use the load power factor, since that's all we "know."

Perhaps this is understood by the authors and they choose to disregard it, since the load power factor presumably has a much larger influence than the cable in the total circuit power factor (and thus the angles of voltage and current), and it keeps things simpler for the average reader.

Thoughts?


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## knight1fox3 (Oct 19, 2016)

nukem2k5 said:


> IEEE uses the receiving-end voltage as the reference, since (I presume) we only "know" the receiving-end (load) power factor (e.g. motor nameplate value), and must do this vector math using the cable's Resistance and Reactance to actually know the power factor of the total circuit.  NEC seems flawed since it calls for the "power factor angle of the circuit" even though, I think, in practice, we usually use the load power factor, since that's all we "know."
> 
> Perhaps this is understood by the authors and they choose to disregard it, since the load power factor presumably has a much larger influence than the cable in the total circuit power factor (and thus the angles of voltage and current), and it keeps things simpler for the average reader.
> 
> Thoughts?


As per usual, we engineers tend to over-think things. This being a prime example. One thing to keep in mind about the NEC when applying it to an actual industrial application, is that it makes assumptions so as to provide conservative direction. Indeed the power factor could be a more accurate calculation based on a particular system, but they try to keep it more generalized to keep from getting into specifics that may not apply to every application. This would also be dictated by a point of common coupling and where does that end? The application power supply? The utility? And so on. NEC exists to maintain uniform standardization across the board, and as you pointed out, to keep it fairly simple and easy to apply in that regard. Anything beyond that is on a higher, more in-depth level which typically entails detailed engineering analysis around a specific application.


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## nukem2k5 (Oct 19, 2016)

knight1fox3 said:


> As per usual, we engineers tend to over-think things. This being a prime example.


Indeed.


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## roy167 (Mar 20, 2019)

If the PF is leading . if you use Rcos(o) + X Sin(o)

Do you use negative angle in sin? for cos angle it won't make any difference as Cosine of +/- angle is the same.


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