Question about calculating voltage drop on 3 phase inductive circuits

[mivey]

Another typo...

Forgot closing parenthesis for sqrt function.

The story of my life.

 

[mivey]

Not sure what I missed but the point is that you can't compare the vector difference with the magnitude difference. This is close:

Z = 0.4 + j0.0108 = 0.4<1.55?
Vs = 120<0?
I = 15<-30.345?
IZ = 6.0<-28.8?
Vr = Vs - IZ = 114.8<1.443?

VD = |Vs|-|Vr| = 120-114.8 = 5.2 volts /= 6.0<-28.8?


also,

Θ = Θ_I - Θ_Vr = -30.345? - 1.443? = -31.788?

Cos (Θ) = 0.85 = p.f.

I fat-fingered this calc because I got the wrong answer for VD_exact. Since I couldn't resist, the corrected numbers for this example are:

Z = 0.4 + j0.0108 = 0.4001<1.5466?
Vs = 120<0?
I = 15<-30.3448?
IZ = 6.002<-28.7982?
Vr = Vs - IZ = 114.777<1.4435?

VD (exact) = |Vs|-|Vr| = 120-114.777 = 5.2234 volts /= 6.002<-28.7982?

also,

Θ = Θ_I - Θ_Vr = -30.3448? - 1.4435? = -31.7883?

Cos (Θ) = 0.85 = p.f.


and using effective Z:

VD (approx) = I * (RcosΘ + XsinΘ) = 15 * (0.4*0.85 + 0.0108*0.52678) = 15 * Zeff = 15 * 0.34569 = 5.1853

 

[Smart $]

I fat-fingered this calc because I got the wrong answer for VD_exact. Since I couldn't resist, the corrected numbers for this example are:

...

Example of what?

Your first posted calculation was in response to Bes'...

OK.
Voltage drop is IZ. Current times impedance. If R ≤ Z always, why, how, can IR ever exceed IZ?
Maybe an example of a practical application with actual numbers would be helpful.

So what scenario do the numbers represent? I see nothing which compares IR to IZ, especially showing IZ is less than IR... which is the gist of the originating question. And how does it relate to wire size?

 

[Haji]

I see nothing which compares IR to IZ, especially showing IZ is less than IR... which is the gist of the originating question.

Exact 'Z' can not be less than 'R'. But an approximate 'Z' (=RcosΘ + XsinΘ) can be. To see that, consider this

VD (approx) = I * (RcosΘ + XsinΘ) = 15 * (0.4*0.85 + 0.0108*0.52678) = 15 * Zeff = 15 * 0.34569 = 5.1853

Here R=0.4>0.34569(=Zeff)

But it, lower Z, may not lead to choosing a wrong size wire.

And how does it relate to wire size?

 

[mivey]

Example of what?

That voltage drop is not IZ. Voltage drop is a difference in magnitudes given by |Vs|-|Vr| or by |I| * Zeff. As I stated:

Voltage drop is IZ.

Not quite. One is a voltage vector the other is a comparison of magnitudes.

I also pointed out that you don't know the angle of I. Usually you have |I| |Vsource| and the power factor at the load. As I stated:

You know the magnitude of I but you don't know the angle. Iterating should find the value.

So what scenario do the numbers represent?

It was really about the math. Another example would be:
Vs - IZ = Vs - Vfeeder = Vr: 120<0? - 10<30? = 111.45<2.57?

The vector difference gives |Vs - Vr| = |IZ| = 10

The voltage drop is 120 - 111.45 = 8.55

FWIW, the numbers were for 100 feet of #12-1 at 167?F in PVC, 85% lagging p.f. load, single-phase 120 volt source.

I see nothing which compares IR to IZ, especially showing IZ is less than IR... which is the gist of the originating question.

The gist is that Z and Zeff are two different things. One might think that vdrop = IZ but it does not since vector differences and magnitude difference are two different things. The confusion is the two different uses of the term voltage drop.

And how does it relate to wire size?

I covered that in my other posts with the tables of percentages.

Let them get this straight first:

|Vs - Vr| /= |Vs|-|Vr|

and the rest should take care of itself.

 

[mivey]

Exact 'Z' can not be less than 'R'. But an approximate 'Z' (=RcosΘ + XsinΘ) can be.

'Z' and 'Zeff' are two different things. Exact and approximate 'Zeff' can both be less than 'R'. 'Z' can never be less than 'R'.

 

[Haji]

That voltage drop is not IZ. Voltage drop is a difference in magnitudes given by |Vs|-|Vr| or by |I| * Zeff. As I stated:



I also pointed out that you don't know the angle of I. Usually you have |I| |Vsource| and the power factor at the load. As I stated:


It was really about the math. Another example would be:
Vs - IZ = Vs - Vfeeder = Vr: 120<0? - 10<30? = 111.45<2.57?

The vector difference gives |Vs - Vr| = |IZ| = 10

The voltage drop is 120 - 111.45 = 8.55

FWIW, the numbers were for 100 feet of #12-1 at 167?F in PVC, 85% lagging p.f. load, single-phase 120 volt source.

The gist is that Z and Zeff are two different things. One might think that vdrop = IZ but it does not since vector differences and magnitude difference are two different things. The confusion is the two different uses of the term voltage drop.

I covered that in my other posts with the tables of percentages.

Let them get this straight first:

|Vs - Vr| /= |Vs|-|Vr|

and the rest should take care of itself.

Both vector difference and magnitude difference of two voltages point to one thing: VOLTAGE DROP, though the former is accurate and the latter is approximate.

 

[mivey]

Both vector difference and magnitude difference of two voltages point to one thing: VOLTAGE DROP, though the former is accurate and the latter is approximate.

They are pointing to two different things. Work a few examples with various angles. It should become clear to you.

Alternatively, draw two vectors. The vector difference is a line from tail-to-tail. The voltage drop is the difference in the vector lengths. Take a 120 volt source and an inductive feeder with resistive load:

Vs = 120<0?
Zfeeder = 1<90?
Zload = 1<0?
Ztotal = Zfeeder + Zload = 1<90? + 1<0? = 1.4142<45?
I = Vs / Ztotal = 120<0? / 1.4142<45? = 84.8528<-45?
Vfeeder = I * Zfeeder = 84.8528<-45? * 1<90? = 84.8528<45?
Vload = I * Zload = 84.8528<-45? * 1<0? = 84.8528<-45?

Vector difference = Vs - Vload = 120<0? - 84.8528<-45? = 84.8528<45?

Vdrop = |Vs| - |Vload| = 120 - 84.8528 = 35.1472 volts

this is not the same as |Vs - Vload| = 84.8528 volts

 

[Haji]

(1)You have taken the angle of Zfeeder to be 90. Do values change, if we take lower and more practical values for the angle of Zfeeder?

(2)

They are pointing to two different things.

No. They are pointing to two different values of the same thing i.e the voltage drop.
(3)

Vector difference = Vs - Vload = 120<0? - 84.8528<-45? = 84.8528<45?

Vdrop = |Vs| - |Vload| = 120 - 84.8528 = 35.1472 volts

this is not the same as |Vs - Vload| = 84.8528 volts

Yes. You are correct.
Incidentally,
'|Vs| - |Vload|' is not same as 'I*(RcosΘ + XsinΘ)' which is 'I*Zeff'.

 

[mivey]

(1)You have taken the angle of Zfeeder to be 90. Do values change, if we take lower and more practical values for the angle of Zfeeder?

Of course the values change but the concept does not. {add: see example in post #45}

(2)No. They are pointing to two different values of the same thing i.e the voltage drop.

No. You are confusing the two different uses of the term 'voltage drop'.

The vector difference tells you the voltage impressed across the feeder impedance. The magnitude difference (what we are calling 'voltage drop' per the NEC discussion) tells you the difference in the magnitude of the voltage impressed across the load vs the magnitude of the source EMF. One is a voltage vector from tail-to-tail of the source and load voltage vectors. The other is the difference in the source and load voltage vector lengths. Two different things.

(3)Yes. You are correct.

Then you should see they represent two different things.

Incidentally, '|Vs| - |Vload|' is not same as 'I*(RcosΘ + XsinΘ)' which is 'I*Zeff'.

One is exact and the other is an approximation but they represent the same thing: the difference in the source and load voltage vector lengths.

 

[Haji]

You are confusing the two different uses of the term 'voltage drop'.

The vector difference tells you the voltage impressed across the feeder impedance. The magnitude difference (what we are calling 'voltage drop' per the NEC discussion) tells you the difference in the magnitude of the voltage impressed across the load vs the magnitude of the source EMF. One is a voltage vector from tail-to-tail of the source and load voltage vectors. The other is the difference in the source and load voltage vector lengths. Two different things.

You are correct. You mean there is no relationship between the two?

One is exact and the other is an approximation but they represent the same thing: the difference in the source and load voltage vector lengths.

NO. Take your own example at post#48.
Since feeder impedance angle is 90 degrees, the feeder has R=0, X=1, cosΘ=0 and sinΘ=1. So feeder voltage drop= I*Zeff=84.8528*(0+1)= 84.8528 V, whereas |Vs| - |Vload| = 120 - 84.8528 = 35.1472 volts only.

 

[mivey]

You are correct. You mean there is no relationship between the two?

The relationship is that they are based on the same two voltage vectors (V_source and V_load). However, they are two different calculations using those two vectors.

NO. Take your own example at post#48.

Yes. Nothing changes other than demonstrating that the approximation is not so good for this extreme case so the exact solution is the one that should be used. The approximate is ok when there is a small angle between the voltage vectors. This can be seen in the illustration in the IEEE red book (figure 3-11).

Since feeder impedance angle is 90 degrees, the feeder has R=0, X=1, cosΘ=0 and sinΘ=1. So feeder voltage drop= I*Zeff=84.8528*(0+1)= 84.8528 V, whereas |Vs| - |Vload| = 120 - 84.8528 = 35.1472 volts only.

Θ = 0? (the load is resistive).

Vdrop approximate = I*(RcosΘ+XsinΘ) = 84.8528*(0*1 + 1*0) = 0 volts. The approximation is not so good so we should use the exact solution which gives us 35.1472 volts. The voltage impressed on the feeder is 84.8528 volts and is not the voltage drop we seek.

 

[Smart $]

That voltage drop is not IZ. Voltage drop is a difference in magnitudes given by |Vs|-|Vr| or by |I| * Zeff. As I stated:



I also pointed out that you don't know the angle of I. Usually you have |I| |Vsource| and the power factor at the load. As I stated:


It was really about the math. Another example would be:
Vs - IZ = Vs - Vfeeder = Vr: 120<0? - 10<30? = 111.45<2.57?

The vector difference gives |Vs - Vr| = |IZ| = 10

The voltage drop is 120 - 111.45 = 8.55

The voltage drop (magnitude) is all we are concerned with, not the vector difference. The load only realizes V_R without any vectorial respect to V_S.

I seldom use vector math, so I appreciate the opportunity to revisit. However, it takes me a while to hone in on the gist... i.e. interpretation is not readily forthcoming. Nowadays I grasp textually-well-explained versions much easier :happyyes:

FWIW, the numbers were for 100 feet of #12-1 at 167?F in PVC, 85% lagging p.f. load, single-phase 120 volt source.

It's a generally-accepted practice to state parameters before the exercise. When I look at the math I say to myself, what the heck is he talking about :huh:

The gist is that Z and Zeff are two different things. One might think that vdrop = IZ but it does not since vector differences and magnitude difference are two different things. The confusion is the two different uses of the term voltage drop.

I never believed or thought Z and Z_EFF were the same, nor was I aware of two different uses of the term voltage drop (for practical applications regarding wire size and circuit length). The only thing that has confused me thus far in this discussion is what I consider your divergence from the issue.

I covered that in my other posts with the tables of percentages.

I see no correlation. I understand that you understand your posts and how they correlate. However, readers including myself (and I consider myself fairly comprehensive) cannot always perceive another's unwritten words.

Let them get this straight first:

|Vs - Vr| /= |Vs|-|Vr|

and the rest should take care of itself.

Okay... I'll wait and see

 

[mivey]

The voltage drop (magnitude) is all we are concerned with, not the vector difference. The load only realizes V_R without any vectorial respect to V_S.

I agree. Besoeker brought up the IZ vector.

It's a generally-accepted practice to state parameters before the exercise. When I look at the math I say to myself, what the heck is he talking about :huh:

Plenty of room in the Q&A session. Sometimes I just type as I think. Oh well.

I never believed or thought Z and Z_EFF were the same

I know you didn't but others have.

nor was I aware of two different uses of the term voltage drop (for practical applications regarding wire size and circuit length).

The vector difference is also called voltage drop, but in a different context. That is the IZ reference.

The only thing that has confused me thus far in this discussion is what I consider your divergence from the issue.

Then I guess you did not follow the other poster's questions about Z can't be less than R and that voltage drop = IZ

I see no correlation.

The list of percentages I gave were directly related to what you said about R & effective Z. The other had nothing to do with what you said as you were not discussing Z instead of Zeff.

 

[Smart $]

... Besoeker brought up the IZ vector.

...

The vector difference is also called voltage drop, but in a different context. That is the IZ reference.

I don't believe he did. Yes he brought up "IZ", but in the context of discussion I do not think he meant vectorial IZ... but rather E=IR (magnitude only) transformed to V_D=IZ_EFF for the purpose at hand. Once you brought up vector Z, the discussion drifted.

 

[mivey]

I don't believe he did. Yes he brought up "IZ", but in the context of discussion I do not think he meant vectorial IZ... but rather E=IR (magnitude only) transformed to V_D=IZ_EFF for the purpose at hand. Once you brought up vector Z, the discussion drifted.

I will simply refer you to post #15:

Z=(R²+X²)^1/2

That my friend is vector Z, not Zeff. That is also the significance of saying that Z can't be less than R and that voltage drop = I * Z (not Z of the Zeff kind).

 

[GoldDigger]

I will simply refer you to post #15:
That my friend is vector Z, not Zeff.

Actually it is the magnitude of vector Z. While Zeff is a difference between two magnitudes.

I finally realized what was going on about ten posts ago, and I accept that the difference between the magnitude of the source voltage and the magnitude of the load voltage is not the same as the magnitude of the vector difference unless the three voltage vectors are parallel.
This is easy to see by looking at any triangle. The length of the third side is not equal to either the sum or difference of the lengths of the other two sides.

One gives you the voltage dropped across the wires while the other gives you the difference in the voltage seen by the load.

And to finally satisfy my concern about why the effect shows up so greatly with large wires, I have only to recognize that the ratio between the DC resistance and the Inductance of the conductor (and therefore X_I of the series impedance) varies considerably with size, changing the direction of that vector.

 

[Besoeker]

One gives you the voltage dropped across the wires while the other gives you the difference in the voltage seen by the load.

That point presents the point well and deserves merit.
:thumbsup:

 

[Smart $]

I will simply refer you to post #15:

Z=(R²+X²)^1/2

That my friend is vector Z, not Zeff. That is also the significance of saying that Z can't be less than R and that voltage drop = I * Z (not Z of the Zeff kind).

See it as you will... but I've yet to see him indicate any value in true vector form in this discussion. However, I didn't see your response to his post as a divergent.

That didn't come until after the originating question in post #30:

OK.
Voltage drop is IZ. Current times impedance. If R ≤ Z always, why, how, can IR ever exceed IZ?
Maybe an example of a practical application with actual numbers would be helpful.

Again no indication of vector values.

IMO, anyone that understands we are only concerned with |V_S| - |V_R| would know that V_D = "IZ" ≠ |IZ|

I think it's more that Haji, prior to post #30, stimulated a vector anlysis and it carried over into your replies to Bes'.

Now can we get back on track? The objective was and remains to show an instance of IR > IZ_EFF, where IZ_EFF = V_D = |V_S| - |V_R|.

 

[GoldDigger]

Now can we get back on track? The objective was and remains to show an instance of IR > IZ_EFF, where IZ_EFF = V_D = |V_S| - |V_R|.

Will you accept a limiting case and agree that the real world will lie somewhere between the extremes?

Try this on for size: Load has a PF of 0 with X_I=6 ohms. W(ith a 120 volt V_S, the current is 20 Amps.
Assume wiring with R=2 ohms and X_L=0.
(Unperturbed voltage drop across the wires would be 40 volts. Actual voltage drop will be less, since the current will go down, but not by much, so let's do the numbers.)
Overall impedance is 6 + 2j. The current will now be 120/6.32, which is 19 amps. The voltage across the load will be 19 x 6 = 114 volts.
The drop in voltage seen by the load is 6 volts, so Zeff would be .32 ohms while R=2 ohms. (The drop across the wiring is 38 volts.)

If the load had a PF of 1, the vectors would all line up and the series resistance would be 8 ohms, current would be 15 ohms, voltage across the load would be 90 volts and voltage across wiring would be 30 volts, so Zeff would be 30/15 = 2 ohms = R.

If you put in more realistic numbers, and use a reasonable PF, the vector directions get messier and the math gets harder to do, but the result will still be, for some wire impedances, that Zeff < R.