Question about NEC Table 9 to calculate voltage drop

[nukem2k5]

I've looked at several other threads and websites 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 disregard the phase angle? More fundamentally, 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 (that is, using " j " in the value of the line impedance X_L).

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_L = j 0.044 Ω, per Table 9. The voltage values below are drop from source to load, (not the full single-circuit roundtrip, which requires multiplying by 2).

A) Using the normal method with complex numbers, V=IZ = I(R+jX) (that is, 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 with the Current phase angle: (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 with 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)?

I'm preparing for the PE exam next week and this is one thing I've never fully understood, since various practice problems seem to solve it using any of the three ways stated above. I suppose I don't really understand the concept of "Effective" impedance and why using actual impedance - along with current and voltage phase angles - isn't correct.

 

[GoldDigger]

One detail that is probably relevant and is demonstrated by the example numbers that you show is this:


There are two perfectly legitimate measurements that you could refer to as voltage drop.
One is the actual drop in voltage across the circuit wiring being considered. That is, the magnitude of the current times the ohmic resistance of the wire (or the AC impedance of the wire if you are using large enough wire for that to be significantly different). This tells you how much power will be dissipated in the wiring.
The other is the change in magnitude of the voltage seen at the load terminals. If the load current is highly reactive this will be much smaller than the first value.

Which of the two is the more useful measurement depends on why you need the VD number.
For NEC purposes, even though voltage drop is not directly regulated, the first number seems to be the one that is being discussed.
If you are dealing with a voltage sensitive load, the second number may be more important.

 

[nukem2k5]

One detail that is probably relevant and is demonstrated by the example numbers that you show is this:


There are two perfectly legitimate measurements that you could refer to as voltage drop.
One is the actual drop in voltage across the circuit wiring being considered. That is, the magnitude of the current times the ohmic resistance of the wire (or the AC impedance of the wire if you are using large enough wire for that to be significantly different). This tells you how much power will be dissipated in the wiring.
The other is the change in magnitude of the voltage seen at the load terminals. If the load current is highly reactive this will be much smaller than the first value.

Which of the two is the more useful measurement depends on why you need the VD number.
For NEC purposes, even though voltage drop is not directly regulated, the first number seems to be the one that is being discussed.
If you are dealing with a voltage sensitive load, the second number may be more important.

I think this sparked something in my brain. V=IZ gives me a voltage vector which is the vectoral difference between the sending and receiving voltages. The second/third equations are intended to give me the magnitude difference of the sending and receiving voltages as if they were superimposed on each other. Is that right? If so, the diagram from IEEE Red Book would make a bit more sense, even though I'm not quite grasping the math. In using the NEC equation, which I believe is taken straight from IEEE 141, are we supposed to use the phase angle of the current or not? By that, I mean do we use 50 A or 50∟-49.5° A in the equation?

 

[nukem2k5]

I see now that I got the sign of the angle wrong in the equations, leading to the big disparity. The angle in the equation should be positive because of the lagging circuit. Doing so results in voltage magnitude for (B) and (C) of 5.57, which is close to 6.39 and within reasonable tolerance for an "approximation" type of equation, which I believe this NEC equation is supposed to be.

That being said, when using the equation from Table 9 Note 2, are we supposed to include the angle of the current, or just use its magnitude? V = IRcosΘ + IXsinΘ

 

[dkarst]

... snip...V = IRcosΘ + IXsinΘ

In this relationship from IEEE red book, I normally represents the RMS current with no angle. The phasor math is taken care of via the figure.

Just as an aside, I would recommend understanding the phasor diagram you posted... It uses the load voltage as the reference which is typically not common, but phasor diagrams are very helpful.

 

[nukem2k5]

In this relationship from IEEE red book, I normally represents the RMS current with no angle. The phasor math is taken care of via the figure.

Just as an aside, I would recommend understanding the phasor diagram you posted... It uses the load voltage as the reference which is typically not common, but phasor diagrams are very helpful.

Using what I recall of trigonometry, I interpret that diagram to mean that the length (magnitude) of the "IZ" vector (which represents the complex voltage drop along the cable) is slightly longer than the sum of (IRcosɸ + IXsinɸ) if they were both laid on the same axis. The latter is simply the X-axis component of the IZ vector. In other words, if "IZ" were rotated clockwise to be at 0degrees, it would reach slightly farther than where the "estimated voltage drop" line ends. Is that correct?

If so, then it makes sense to me that the result of my equation A (6.39 V) is slightly greater in magnitude than the results in B & C (5.57 V). Ultimately, the NEC/IEEE method is simplifying the math by comparing just the magnitudes of the voltages only on one axis. Maybe that's "good enough" but it's not as exact as using the actual (calculated or observed) angles and doing vector math. Or am I still missing something?

 

[nukem2k5]

I would recommend understanding the phasor diagram you posted... It uses the load voltage as the reference which is typically not common, but phasor diagrams are very helpful.

I think the reason they use load voltage as the reference is because the IEEE equation uses the load power factor angle (NEC says to use the power factor of the circuit). If we know the PF of the load, we can easily show the load-end voltage vector, the circuit's current vector, and the angle between them. At that point, the angle between the current and the sending-end voltage is still unknown, and is wholly dependent on the complex impedance of the cable, which is exactly what this equation is trying to avoid using in the first place, by giving us an "effective Z" to use in simple arithmetic.

The IEEE nomenclature seems more accurate since we can't know the circuit power factor angle without doing the very thing the equation is trying to avoid/simplify.

 

[Smart $]

...

A) Using the normal method with complex numbers, V=IZ = I(R+jX) (that is, 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 with the Current phase angle: (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 with 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)?
...

In A, you calculated the actual voltage across the conductor (IZ) as a vector. However, we are only interested in the voltage across the load at its terminals, hence the effective voltage drop is the difference in magnitude only between source voltage and load terminal voltage. It makes no difference to the load that the voltage angle at its terminals is shifted from the voltage angle at the source. Note in the diagram you posted that all the "dimensioned" terms are transposed (i.e. projected) to the load voltage vector, the magnitude of which is the sum of IRcos(theta) + IXsin(theta)... with some error from rectangular projection rather than polar. Look up IEEE exact voltage drop for comparison.

 

[Phil Corso]

Nukem...

R and X values are in Ohms/1,000 ft! Equivalent Impedance, Z = (Ckt Length/1,000) x (R + jX) shown in NEC, Table 9!

Regards, Phil Coro

 

[nukem2k5]

In A, you calculated the actual voltage across the conductor (IZ) as a vector. However, we are only interested in the voltage across the load at its terminals, hence the effective voltage drop is the difference in magnitude only between source voltage and load terminal voltage. It makes no difference to the load that the voltage angle at its terminals is shifted from the voltage angle at the source. Note in the diagram you posted that all the "dimensioned" terms are transposed (i.e. projected) to the load voltage vector, the magnitude of which is the sum of IRcos(theta) + IXsin(theta)... with some error from rectangular projection rather than polar. Look up IEEE exact voltage drop for comparison.

That diagram is from IEEE-141, as is the equation. If you were to rotate the sending-end voltage vector clockwise to be on the same axis as the receiving-end (in other words, disregard the phase angle difference between them), the difference in voltage magnitudes would still be greater than the sum of the x-axis components of the voltage drop, which are shown in the diagram as IRcosΘ and IXsinΘ. This equation appears to be a simplification that gets us within a close tolerance of a real value (which, I believe, is the "error" shown in the diagram), while avoiding having to do complex algebra, which many average electricians reading the NEC may not know how to do. That's the conclusion I've arrived at, anyway.

 

[nukem2k5]

Nukem...

R and X values are in Ohms/1,000 ft! Equivalent Impedance, Z = (Ckt Length/1,000) x (R + jX) shown in NEC, Table 9!

Regards, Phil Coro

Thanks, Phil. You're right. In my example, I assumed a 1000' circuit length to slightly simplify the math (it's actually 2000' total but I was only interested in the voltage drop at the load terminals, not total circuit Vdrop).

 

[junkhound]

The original post obviously had sign errors (as later acknowledged)

Additionally, it appears IEEE 141 has never been corrected, and that the 1986 simplification and error continues to be published? (I cannot access a current IEEE 141 to check, but the diagram posted earlier is similar to the 1986 version of IEEE 141)

Reference: [h=2]Exact voltage drop calculations ; R. J. Titus ; Industry Applications Society Annual Meeting, 1992., Conference Record
Abstract:
The derivation of the IEEE STD 141-1986 exact voltage drop formula is presented to correct an existing error in the standard. A updated and more accurate formula for calculating a voltage drop is then presented as a proposed replacement to the current standard's formula. The two methods are contrasted with an example from an industrial application.
Published in: Industry Applications Society Annual Meeting, 1992., Conference Record of the 1992 IEEE
[/h]BTW, 141 back then at least specifically stated that the calculation was 'one way' so that it could be used for single or 3 phase systems.

 

[Julius Right]

In my opinion, I*Z=(I*cosfi-jI*sinfi)*(R+jX)=I*R*cos(fi)+I*X*sin(fi)+j[I*X*cos(fi)-I*R*sin(fi)] does not represent the actual VD[voltage drop]. [VD=actual voltage drop=Es[voltage at source end of the cable]-ER [voltage at the output end of the cable].
See: IEEE 141/1993 Figure 3-11 Phasor diagram of voltage relations for voltage-drop calculations
I*R*cos(fi)+I*X*sin(fi)=AB
j[I*X*cos(fi)-I*R*sin(fi)]=BC
abs(I*Z)=AC
If OC=Es and OD=OC then OD-OA=VD
By-the-way, the angle in the above formula it is positive always.

 

[Smart $]

... the 1986 simplification and error continues to be published?...

I think the diagram is basically the same.

The formula has changed to what is known as the Exact Method.

E_VD = E_S + IRcosΦ + IXsinΦ - √[E_S² - (IXcosΦ - IRsinΦ)²]

Basically, the formula just includes the voltage shown as error on the diagram above, which is E_S - √[E_S² - (IXcosΦ - IRsinΦ)²].

 

[rian0201]

It is a lot easier if you complex notation and a complex calculator.. You just type in everything and viola!!! But remember to conjugate the necessary items.


Sent from Mars

 

[JoeStillman]

It is a lot easier if you complex notation and a complex calculator.. You just type in everything and viola!!! But remember to conjugate the necessary items.


Sent from Mars

That's how I passed the exam in '87, with an HP41C, and the calculators are even smarter now.

 

[junkhound]

Even pretty straightforward with a slide rule in '67.





Don't know what a slide rule is ??

Grandson left his ipod at home, needed to make a call - DW pointed to the 50 YO wall dial phone on the kitchen wall: "Grandma, how do you use this thing" ?