Where did these values come from?

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All right- correct me if I'm wrong... The graphic has 583 amps... This means a 100 amp breaker will take 3 seconds (average) (6 seconds max) to clear the fault according to these time current curves:

https://library.industrialsolutions.abb.com/publibrary/checkout/GES-6203B?TNR=Time Current Curves|GES-6203B|PDF&filename=GES-6203B.pdf


Assuming a 40/60 voltage drop to the concrete someone is standing on this would provide about 72 volts potential assuming an infinite source.

Now, 72 volts on a 100 ohm hand to feet value:


Means 720 milliamps of current will pass through the body during those 3-6 seconds of clearing.

This would put us in the AC-4 region of the body graph:


OK, so... It looks like I'm missing something 🤔🤔 I'm lost guys. 😲

Source impedance to high? Wire impedance not right? Hmmm...
 
Frank DuVal said:
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.
Click to expand...

Thanks!

So in reality, the current would be lower than 583 amps...

So can we work back and guess what size transformer is supplying this setup? Its Z?

583 seems kind of low... but then again Mike might be on to something.
 
Frank DuVal said:
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.
Click to expand...

Ok, so 0.245 ohms at 1000 feet DC for #3, 0.25 ohms 1000 feet 3 AWG AC... what do I do with 0.047 ohms ohms XL? This is the part I'm hung up on.
 
The two segments are in series, so their impedances add.

If you know the impedance of the #3 leg as resistance and reactance, and the impedance of the #8 leg likewise, you just add the resistances to get the total resistance, and add the reactances to get total reactance. Then the magnitude of the total impedance is sqrt(resistance^2 + reactance^2). And |V| = |I| * |Z| gives you the magnitude of the current.

Cheers, Wayne
 
wwhitney said:
The two segments are in series, so their impedances add.

If you know the impedance of the #3 leg as resistance and reactance, and the impedance of the #8 leg likewise, you just add the resistances to get the total resistance, and add the reactances to get total reactance. Then the magnitude of the total impedance is sqrt(resistance^2 + reactance^2). And |V| = |I| * |Z| gives you the magnitude of the current.

Cheers, Wayne
Click to expand...


Easy, I can do this!

Same Still applies for 3 phase?

How about metal conduit?

And the trafo?

I keep feeling like I need positive, negative, zero sequence components along with a geometric mean radius. Or am I over thinking this?
 
mbrooke said:
I might be thinking of this: https://en.wikipedia.org/wiki/Cartesian_coordinate_system

Its common for transmission lines. Not sure if calling it Z= R+jX would be appropraite here.
The thing about AC or DC resistance only is that it produces high fault current values (vs actual) especially with larger conductors.
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The mathematicians wrote complex numbers in the form a + i b where i represents the square root of -1. Electricians simply could not tolerate seeing i represent anything other than current, so they chose to write a + j b. The distance of a point from the origin in the complex plane can be represented as d**2 = (a + i b) (a - i b) = a**2 + b**2 (the product of the complex number and its complex conjugate.)
 
GoldDigger said:
The mathematicians wrote complex numbers in the form a + i b where i represents the square root of -1. Electricians simply could not tolerate seeing i represent anything other than current, so they chose to write a + j b. The distance of a point from the origin in the complex plane can be represented as d**2 = (a + i b) (a - i b) = a**2 + b**2 (the product of the complex number and its complex conjugate.)
Click to expand...

Thanks.

Any idea how to add transformer impedance to the mix?

I'd like it so I can assume infinite at the primary, then add trafo Z, Sevice Z, feeder Z and then branch Z together to get a final L-G value.

Something that any electrician can do with a table.
 
mbrooke said:
Thanks.

Any idea how to add transformer impedance to the mix?

I'd like it so I can assume infinite at the primary, then add trafo Z, Sevice Z, feeder Z and then branch Z together to get a final L-G value.

Something that any electrician can do with a table.
Click to expand...
The same formulas apply to series and parallel impedances as apply for series and parallel resistances, you just use the complex numbers instead of real numbers in the calculation.
 
GoldDigger said:
The same formulas apply to series and parallel impedances as apply for series and parallel resistances, you just use the complex numbers instead of real numbers in the calculation.
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Can I have a break down of that process? 😇

I want to convert sequence component results into a single table that anyone can use for calculating fault current and breaker disconnection times.
 
mbrooke said:
Can I have a break down of that process? 😇

I want to convert sequence component results into a single table that anyone can use for calculating fault current and breaker disconnection times.
Click to expand...
I don't believe that is possible.

Too many variables, so that you would need a huge book of tables (you would need a couple of dimensions for transformer size and Z, then circuit size, then feeder and circuit conductors, then length, then wiring method, temperature....).

Instead of tables you would need to condense all of the information into a set of calculations, perhaps run in a spreadsheet or app.

But even then you would need to deal with the basic inaccuracy of your data. Simply the one point you raised: the impedance of the fault current path once you include all of the possible parallel paths through 'grounded metal' would be a huge and impossible to calculate factor. And then there is the impedance of the fault itself; it could be a 'bolted fault' or a re-striking arc, or a high impedance fault of some sort until the heat burns it into one of the others.

At best you can estimate the maximum fault current and breaker clearing time, but at best this is a bounding value, not a hard and accurate value.

-Jon
 
winnie said:
I don't believe that is possible.

Too many variables, so that you would need a huge book of tables (you would need a couple of dimensions for transformer size and Z, then circuit size, then feeder and circuit conductors, then length, then wiring method, temperature....).

Instead of tables you would need to condense all of the information into a set of calculations, perhaps run in a spreadsheet or app.

But even then you would need to deal with the basic inaccuracy of your data. Simply the one point you raised: the impedance of the fault current path once you include all of the possible parallel paths through 'grounded metal' would be a huge and impossible to calculate factor. And then there is the impedance of the fault itself; it could be a 'bolted fault' or a re-striking arc, or a high impedance fault of some sort until the heat burns it into one of the others.

At best you can estimate the maximum fault current and breaker clearing time, but at best this is a bounding value, not a hard and accurate value.

-Jon
Click to expand...

I disagree if I may- you can set on a fixed temperature ie 75*C. Z for Remoex, Z for MC/AC and Z for EMT/IMC/RMC like its set up now.

A list of common KVA sizes and there typical Zs in a column.


On can, for example: a single L-G fault choose 0.016 ohms for the POCO transformer, 0.002 ohms for the service, 0.035 ohms for the feeder (phase + EGC) and 0.25 ohms for the branch circuit (phase + EGC) to its furthest point.

Simple ohms law would take you to the required destination.


This would allow an Electrician to calculate short circuit values in the field with simple math instead of complex engieering software even if it results in slight error, but still under the AIC equipment rating in most cases.

I notice there is a lack of short circuit and disconnection times missing from most prints and no way to get those values in the field easily.
 
Fair enough.

And if your goal is to calculate _maximum_ bounding values to make sure you don't exceed the fault handling capacity of components then I think tables would work.

But if you want to calculate fault clearance times, I think you would have a bigger problem. Because of the variability of the faults themselves, I would expect huge variability in fault clearance times.

Calculating touch voltages during a fault I think could work, again as a maximum bounding value. Anything that reduces the circuit impedance (say parallel ground paths) would also tend to reduce the touch voltage. Similarly if the fault itself shows high impedance and reduced current flow (thus extending clearance time) then touch voltage will be reduced.

So I guess if you make it clear that you are calculating reasonable bounding values rather then exact values, you could do it with tables.

-Jon
 
winnie said:
Fair enough.

And if your goal is to calculate _maximum_ bounding values to make sure you don't exceed the fault handling capacity of components then I think tables would work.
Click to expand...

Agree.

But if you want to calculate fault clearance times, I think you would have a bigger problem. Because of the variability of the faults themselves, I would expect huge variability in fault clearance times.
Click to expand...

Hmmm, you bring up a good point... Is puttering what you have in mind?

Calculating touch voltages during a fault I think could work, again as a maximum bounding value. Anything that reduces the circuit impedance (say parallel ground paths) would also tend to reduce the touch voltage. Similarly if the fault itself shows high impedance and reduced current flow (thus extending clearance time) then touch voltage will be reduced.

So I guess if you make it clear that you are calculating reasonable bounding values rather then exact values, you could do it with tables.

-Jon
Click to expand...

Agree to the concept, I like it! :)
 
wwhitney said:
The two segments are in series, so their impedances add.

If you know the impedance of the #3 leg as resistance and reactance, and the impedance of the #8 leg likewise, you just add the resistances to get the total resistance, and add the reactances to get total reactance. Then the magnitude of the total impedance is sqrt(resistance^2 + reactance^2). And |V| = |I| * |Z| gives you the magnitude of the current.

Cheers, Wayne
Click to expand...

Alright, got this for #10:


#10 copper PVC--------- (0.05x0.05) + (1.2x1.2)= 1.2010

#10 copper Steel Conduit ------ (0.063x0.063) + (1.2x1.2) = 1.2016

Effective Z at 0.85 PF for Uncoated Copper Wires ------ 1.1

Why would these values be higher than those listed under: Effective Z at 0.85 PF for Uncoated Copper Wires?
 
wwhitney said:
The two segments are in series, so their impedances add.

If you know the impedance of the #3 leg as resistance and reactance, and the impedance of the #8 leg likewise, you just add the resistances to get the total resistance, and add the reactances to get total reactance. Then the magnitude of the total impedance is sqrt(resistance^2 + reactance^2). And |V| = |I| * |Z| gives you the magnitude of the current.

Cheers, Wayne
Click to expand...

Alright- now I'm really going Huh?!? o_O 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? Would not AC resistance remain the same despite power factor? How does physics make this work when? I always thought resistance was the function of a conductor's temperature and not its magnetic field? Would that mean an incandescent light bulb in a powerful magnetic field glow dimmer?

https://imgur.com/a/8hlwmZJ


I think this might be for another thread... I'll open a new one if you'd like and you can post your reply there. Just let me know if that needs to be the case. I'm clueless.
 
Forgive me if I'm missing the obvious in that table- I'm dyslexic- so sometimes I'll literally swap numbers and figures.
 
winnie said:
Fair enough.

And if your goal is to calculate _maximum_ bounding values to make sure you don't exceed the fault handling capacity of components then I think tables would work.

But if you want to calculate fault clearance times, I think you would have a bigger problem. Because of the variability of the faults themselves, I would expect huge variability in fault clearance times.

Calculating touch voltages during a fault I think could work, again as a maximum bounding value. Anything that reduces the circuit impedance (say parallel ground paths) would also tend to reduce the touch voltage. Similarly if the fault itself shows high impedance and reduced current flow (thus extending clearance time) then touch voltage will be reduced.

So I guess if you make it clear that you are calculating reasonable bounding values rather then exact values, you could do it with tables.

-Jon
Click to expand...

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.

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.

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.
 
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