Calculating total system Impedance

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mbrooke

mbrooke

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What is the proper way to calculate total circuit impedance? Do I obtain the Z of each wire segment then add all the Zs together or do I add up all the Rs and Xs for the total system and then do the math to obtain Z? Which way is right and which way is wrong and why would it matter?



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Total Z= 0.07433+0.49265+2.00073= 2.56771 ohms

OR

R= 0.062+0.49+2.0=2.552

X= 0.041+0.051+0.054=0.146

Z= 2.55617
 
mbrooke

mbrooke

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2.56771 ohms vs 2.55617 ohms doesn't seem like a big difference, but I'm betting one is right the other is wrong?
 
petersonra

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They are vector quantities so you have to add them up as vectors.

The individual Z values for each segment are actually a vector quantity that only expresses the magnitude.
 
petersonra

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mbrooke said:
That would mean both ways are incorrect? I'm confused.
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the second way is correct.

the first way expresses vector quantities using only its magnitude. a vector quantity always has a magnitude and angle. if the angles are the same you can add the magnitudes together and get the correct answer but in your case the angles are all different.
 
mbrooke

mbrooke

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petersonra said:
the second way is correct.

the first way expresses vector quantities using only its magnitude. a vector quantity always has a magnitude and angle. if the angles are the same you can add the magnitudes together and get the correct answer but in your case the angles are all different.
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Alright. But what happens if I use the prior way instead? I know you're correct, but in my world, it seems both ways give near identical results.
 
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Carultch

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You get close results, because X and R are in nearly the same proportion in each pairs of resistances and reactance. The "direction" of each sub-impedance is nearly the same, so it is a coincidence that adding up sub-impedance magnitudes gets you close to the correct calculation, as you would get if you separately add the resistances and reactances, and then use Pythagorean theorem to get total impedance magnitude.

Suppose you walk 300 ft at bearing 40 degrees, then 250 ft at bearing 50 degrees, then 100 ft at bearing 45 degrees.
A: How much distance did you walk in total (650 ft)?
B: How far are you from your starting point (as the crow flies), 647.9 ft?

The reason A and B are so close to each other, is that the bearings are so close together. Mix in bearings of other quadrants, you'll get a significantly different answer between gross distance traveled and net change in position. Adding up impedance is a similar mathematical problem as adding up displacement vectors. Resistance and reactance are represented as "perpendicular"in the mathematics that describe how they relate to each other.
 
synchro

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The second way you did it is correct as was mentioned, although it doesn't make much difference with the particular numbers you have.

Stepping back a little, an impedance is a complex quantity that can be represented either in rectangular form as Z = R + jX, or in polar form as Z = |Z| x e ʲᶿ where |Z| = √( R² + X²) and θ = arctan (X/R).

Using rectangular form is easiest when you want to add complex numbers, and polar form is easiest when you want to multiply complex numbers. Because you want to add impedances together then you should keep things in rectangular form.
So as petersonra mentioned your second way is correct because in this case you were adding the impedances in rectangular form (i.e., adding the respective R and X components of the complex impedance).

In the first way you did it, as petersonra said you were adding the magnitudes (i.e., one component of the polar form) of the individual series pairs of inductors and resistors. And so this will not work unless θ = arctan (Xk/Rk) is the same for all the pairs R1 and X1, R2 and X2, R3 and X3. Or in other words, only if each of the three complex impedance vectors is aligned with eachother along the same angle.
 
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mbrooke said:
Thank you for both replies.

Do things change when phase and ground are not the same size?
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No, as long as you are not considering the magnetic interactions between the two or more conductors of the circuit (see your other thread.) You still have multiple impedances in series, so you can either do vector addition using magnitudes and angles or you can sum the resistive components and the reactive components separately to construct the result components..
 
C

Carultch

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mbrooke said:
Do ground wires not carry fault current?
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During a fault, yes.

During normal operation, the ground wire and EGC system is an open circuit that is supposed to be a "dead end" dedicated to bonding all metal that doesn't intentionally carry current. It's like Resistor R6 in the following diagram. It has an impedance of its own, but it carries no current, and that impedance doesn't affect what the rest of the circuit does.
 
mbrooke

mbrooke

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Carultch said:
During a fault, yes.

During normal operation, the ground wire and EGC system is an open circuit that is supposed to be a "dead end" dedicated to bonding all metal that doesn't intentionally carry current. It's like Resistor R6 in the following diagram. It has an impedance of its own, but it carries no current, and that impedance doesn't affect what the rest of the circuit does.
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Circuit impedance is by far the most important during fault conditions. Are you saying its not? You come across as not being concerned about that???
 
C

Carultch

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mbrooke said:
Circuit impedance is by far the most important during fault conditions. Are you saying its not? You come across as not being concerned about that???
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It is important. I just didn't think that was part of the purpose of your calculation.

If it is to determine the system impedance during a fault, the details of the fault have to be defined. Such as where it happens within the circuit. The EGC would become another parallel path when that happens. The load and neutral conductor would be in series in path #1. The EGC would be in parallel to that, as a near-short circuit from the fault point back to the source.
 
mbrooke

mbrooke

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Carultch said:
It is important. I just didn't think that was part of the purpose of your calculation.
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It actually was the reason for it. I used a full size ground just to make the example simpler. My apologies for not being more explicit ahead of time.

If it is to determine the system impedance during a fault, the details of the fault have to be defined. Such as where it happens within the circuit. The EGC would become another parallel path when that happens. The load and neutral conductor would be in series in path #1. The EGC would be in parallel to that, as a near-short circuit from the fault point back to the source.
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Bolted fault, furthest point.

I would not include the neutral. Just a hot to case fault. Always assume no other parallel paths as I think this would present the highest impedance.
 
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