Where did these values come from?

[wwhitney]

How can Effective Z at 0.85 PF for Uncoated Copper Wires be of a lower impedance value than Alternating-Current Resistance for Uncoated Copper Wires?

Good question. The answer lies in Note 2 to the table, which tells you that Effective Z is just R cos(theta) + X sin(theta), where theta is the power factor angle, i.e. tan(theta) = 0.85. It further notes that this value is for calculating voltage drop.

The short answer is that for calculating the magnitude of fault current, where there is no load impedance in the circuit, you can ignore the Effective Z column. You just want the total impedance |Z| = sqrt(R^2 + X^2) for Ohm's Law.

The long answer is a bit more complicated and I'm not sure I have all the details squared away. But the basic idea is that if you are just interested in calculating voltage drop on the supply wiring, then for a load with a non-unity power factor, the reactance of the supply wiring may "cancel" some of the reactance of the load and give you less voltage drop on the supply wiring than you would have for a purely resistive load. I think.

Cheers, Wayne

 

[mbrooke]

Good question. The answer lies in Note 2 to the table, which tells you that Effective Z is just R cos(theta) + X sin(theta), where theta is the power factor angle, i.e. tan(theta) = 0.85. It further notes that this value is for calculating voltage drop.

The short answer is that for calculating the magnitude of fault current, where there is no load impedance in the circuit, you can ignore the Effective Z column. You just want the total impedance |Z| = sqrt(R^2 + X^2) for Ohm's Law.

But doesn't a short circuit also produce lagging power factor due to an X/R ratio? Does the transformer have a short circuit power factors being that its windings are inductive around an iron core? For that matter what is the power factor for a transformer in its short circuited state?

Faults further away from the transformer would be more resistive (closer to unity) I imagine, while faults nearer to the transofmer would be more reactive (further from unity into the lagging plain) .

If cosine phi has an effect on voltage drop, I would imagine it has the same effect on current magnitude during a short circuit?

Of course it gets even more complicated when a breaker also deals with an X/R ratio limitation on interrupting- though I don't think this holds true for LV breakers as they already assume a huge X/R disparity during their AIC listing while in engineering?

I bit off more than I can chew, but its starting to make some sense. How do you eat a whole candy store? One bite at a time.

The long answer is a bit more complicated and I'm not sure I have all the details squared away. But the basic idea is that if you are just interested in calculating voltage drop on the supply wiring, then for a load with a non-unity power factor, the reactance of the supply wiring may "cancel" some of the reactance of the load and give you less voltage drop on the supply wiring than you would have for a purely resistive load. I think.

Cheers, Wayne

Could reactance off set some of the magnetism which pushes electrons to the surface of a wire? Either the table is wrong (I've always been skeptical of its accuracy), there is more to physics than I thought (good possibility) or both (most likely scenario).

Cheers back!

 

[mbrooke]

No.

You can do tables and simple calculations on max fault current easily. Back before calculators this is how we did it. That is the essence of the ANSI X only method...recognizing that for most real world cases we don’t care about R or we can just do fudge factor approximation.

Faults are not that variable.

Starting out we can talk about various types of what are just sparks. The current is severely limited. This is where we get things like neon and discharge lighting. No damage is done.

Once we exceed a threshold current a curious thing happens. Below the threshold voltage matters and we get all kinds of strange things like negative resistance. But above the threshold, the impedance is almost independent of voltage.. We are in the realm of arcs or bolted faults.

In a power arc or bolted fault we see both a sudden exponentially decaying DC current called the assymmetrical current due to stored energy, primarily in motors and transformers, and a symmetrical (AC) current. The asymmetrical fault current depends on phase angle which is random but within a couple cycles it dies down. The symmetrical current us easy to calculate based on the system impedance. Impedance at the fault is small or soon will be with the temperatures involved.

Touch voltage as you said is easy too. The biggest variable is skin resistance. We usually use the minimum. It is voltage dependent. Wet skin is about 1 Kohm but gradually decreases at higher voltages down to 600 ohms. So North American modelers use 1,000 ohms. European modelers frequently incorrectly use 600 ohms but to be correct they need to then iteratively adjust based on voltage. This is upper bounding but it’s shock which is what we normally do...no reason to worry about accuracy.

Can you elaborate more on what you mean by some faults being severally limited, others arc, and of course I'm familiar with bolted faults.

We can convert the whole system to sequence voltages and currents because it’s easier to do fault calls in that form. Then a ground fault is a short to ground. A 3 phase fault affects all three phases. There is also line-ground-line rounding out ALL fault types. Then we have arcing and bolted faults. Arcing faults are going to be single phase or three phase. If they are single phase and enclosed it WILL be three phase within 1 cycle so no need to figure out any other combination. Sequence math is symmetrical so we don’t have to grind out the math for each phase or combination of phases. The math was invented for this purpose. This isn’t upper bounding, it’s exact results, within about +/-10% of experimental results.

Which is what makes this so difficult for electricians, even engineers. If we can calculate all these values in advance for every scenario we can come to a simple table listed in ohms. Using addition math with basic ohms law anyone with a basic understanding of mathematics can calculate the short circuit current at the service and anywhere in a power system there after.

A lot of threads here pertain to short circuit value inquires and there respective AIC with no easy answer or consensus being reached for the electrician out in the field.

Given that short circuit values are essential to properly safe guarding property and life, they must absolutely be listed in the code. Expensive Software, apps, long IEEE equations, calling the POCO, confusing answers without consensus, ect should not be an electrician's first line in obtaining values essential to the job at hand.

 

[wwhitney]

But doesn't a short circuit also produce lagging power factor due to an X/R ratio? Does the transformer have a short circuit power factors being that its windings are inductive around an iron core? For that matter what is the power factor for a transformer in its short circuited state?

OK, I have no idea about these questions, I just know about the mathematics for an idealized constant voltage source. I haven't even studied simple transformer models like an idealized voltage source with one or more internal impedances.

But it may be helpful to work out a few examples. Say you have an idealized 120V AC source, infinite available current. Then you attach a 1 ohm resistor to the voltage source. V=IR tell you the current will be 120A. If you instead attach an inductor (or capacitor) with a reactance of 1 ohm, then |Z| = 1 ohm still, so the current will still be 120A. If you attach two identical 1 ohm devices in series, then |Z| = 2, and the current will be 60A. But if you attach a 1 ohm resistor in series with a 1 ohm inductor (or capacitor), then |Z| = sqrt(2), and the current will be 120/sqrt(2) = 84.9A.

Now I need to work out an example with wire impedance in series with load impedance, and show that for purposes of voltage drop, the wire's "effective" scalar impedance can be less than its resistance for loads with non-zero reactance. But it's taking a bit longer than I have time for now, perhaps I'll try again later.

Cheers, Wayne

 

[mbrooke]

OK, I have no idea about these questions, I just know about the mathematics for an idealized constant voltage source. I haven't even studied simple transformer models like an idealized voltage source with one or more internal impedances.

Not to worry, I have no idea either. My theory is that a shorted transformer is not purely resistive on the secondary, but a combination of the two mostly toward reactance. That lagging power factor flowing through the fault loop will (IMHO) raise or perhaps even lower the impedance of the wire based on its size.

But it may be helpful to work out a few examples. Say you have an idealized 120V AC source, infinite available current. Then you attach a 1 ohm resistor to the voltage source. V=IR tell you the current will be 120A. If you instead attach an inductor (or capacitor) with a reactance of 1 ohm, then |Z| = 1 ohm still, so the current will still be 120A. If you attach two identical 1 ohm devices in series, then |Z| = 2, and the current will be 60A. But if you attach a 1 ohm resistor in series with a 1 ohm inductor (or capacitor), then |Z| = sqrt(2), and the current will be 120/sqrt(2) = 84.9A.

I agree here. Resistance and inductance plays a huge role and is basically unpredictable unless calculated.

Now I need to work out an example with wire impedance in series with load impedance, and show that for purposes of voltage drop, the wire's "effective" scalar impedance can be less than its resistance for loads with non-zero reactance. But it's taking a bit longer than I have time for now, perhaps I'll try again later.

Cheers, Wayne

I'm egar to know.

 

[mbrooke]

The resistance of the 3 awg and 8 awg wire is just from a typical DC resistance table, like in Ugly's book. or NEC chapter 9 table 8.

But what about the transformer? I want to make a table.