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 Post subject: Re: Quantity of carbonized particles before Arc Flash InitiaPosted: Sun Mar 17, 2019 4:11 pm

Joined: Wed Feb 20, 2019 3:06 am
Posts: 45
PaulEngr, and others,

Something puzzles me regarding differences of bolted short circuit current between two calculation methods.

Given single phase 240/120 transformers with 75kVA rating and 2% impedance

computing using the infinite bus short cut method

75,000/240/0.02=15625 amp

or the long cut (with derivation) method

kVa = 75,000
voltage =240v
The impedance of a load that would draw 75 kVA is 240^2/75000 = 0.768 ohms.
Transformer impedance = 0.02x 0.768 = 0.01536
Short circuit current = 240v/0.01536 = 15625 amp

Now consider 20 feet of wire has resistance of about 0.002 (from an online wire impedance calc)

To add them together one, one uses this
Total impedance = sqrt (0.002^2 + 0.01536^2)= 0.01549 amp.

bolted short circuit current 20 feet away AWG 1 is 240/0.01549= 15494 amp

The difference with adding 20 feet of AWG 1 wire is decrease of only 1% of the short circuit current.

Now consider the Arc Flash fault calculation method below:

http://www.cooperindustries.com/content ... rmulas.pdf

You can try it in excel spreadsheet at https://electrical-engineering-portal.c ... alculation

Input the same data.

With terminal short circuit current of 15625 amp.

The short circuit current 20 feet away AWG 1 is much lower 11595 amp.

Why is it much lower?

Did the arc flash method somehow give the 20 feet conductor an inductive value, but it's not part of the transformer winding.

Let's go to the Arc Flash Calculation method formulas.

at 1Ø Line-to-Neutral (L-N) Faults

f = [2 x L x I (l-l)] /[ C x n x E (l-)]

Where:

L = length (feet) of conductor to the fault.
C = constant from Table 4 of “C” values for conductors and
Table 5 of “C” values for busway.
n = Number of conductors per phase (adjusts C value for
parallel runs)
I = Available short-circuit current in amperes at beginning
of circuit.
E = Voltage of circuit.

(How did they determine C where the bolted short circuit current is much lower using this method but high when directly computing for it. How did they derive f?).

Then you get M = 1 / (1+f)

And multiply the transformer terminal short circuit current by M.

The result is much lower bolted short circuit current 20 feet away (awg 1) about 25% less compare to the ballpark computations. Why? Where is the hidden impedance in the wire. Instead of 0.002, it becomes higher like 0.010, how?

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 Post subject: Re: Quantity of carbonized particles before Arc Flash InitiaPosted: Mon Mar 18, 2019 5:08 pm

Joined: Wed Feb 20, 2019 3:06 am
Posts: 45
Ommi wrote:
PaulEngr, and others,

Something puzzles me regarding differences of bolted short circuit current between two calculation methods.

Given single phase 240/120 transformers with 75kVA rating and 2% impedance

computing using the infinite bus short cut method

75,000/240/0.02=15625 amp

or the long cut (with derivation) method

kVa = 75,000
voltage =240v
The impedance of a load that would draw 75 kVA is 240^2/75000 = 0.768 ohms.
Transformer impedance = 0.02x 0.768 = 0.01536
Short circuit current = 240v/0.01536 = 15625 amp

Now consider 20 feet of wire has resistance of about 0.002 (from an online wire impedance calc)

To add them together one, one uses this
Total impedance = sqrt (0.002^2 + 0.01536^2)= 0.01549 amp.

bolted short circuit current 20 feet away AWG 1 is 240/0.01549= 15494 amp

The difference with adding 20 feet of AWG 1 wire is decrease of only 1% of the short circuit current.

Now consider the Arc Flash fault calculation method below:

http://www.cooperindustries.com/content ... rmulas.pdf

You can try it in excel spreadsheet at https://electrical-engineering-portal.c ... alculation

Input the same data.

(images snipped)

With terminal short circuit current of 15625 amp.

The short circuit current 20 feet away AWG 1 is much lower 11595 amp.

Why is it much lower?

Did the arc flash method somehow give the 20 feet conductor an inductive value, but it's not part of the transformer winding.

Let's go to the Arc Flash Calculation method formulas.

at 1Ø Line-to-Neutral (L-N) Faults

f = [2 x L x I (l-l)] /[ C x n x E (l-)]

Where:

L = length (feet) of conductor to the fault.
C = constant from Table 4 of “C” values for conductors and
Table 5 of “C” values for busway.
n = Number of conductors per phase (adjusts C value for
parallel runs)
I = Available short-circuit current in amperes at beginning
of circuit.
E = Voltage of circuit.

(How did they determine C where the bolted short circuit current is much lower using this method but high when directly computing for it. How did they derive f?).

Then you get M = 1 / (1+f)

And multiply the transformer terminal short circuit current by M.

The result is much lower bolted short circuit current 20 feet away (awg 1) about 25% less compare to the ballpark computations. Why? Where is the hidden impedance in the wire. Instead of 0.002, it becomes higher like 0.010, how?

Is it possible the official arc flash current fault calculation method based on getting the constant c and solving for f and Isc is just wrong?

Maybe Jim Phillips has developed his own formulas different from the faulty official ones?

The short circuit current in the official method at 20 feet away (AWG 1) wire is 25% less while in direct logic based computations, it is only 1%. In fact an electrical engineer not familiar with arc flash calculations told me it was only 1% (he was the one who taught me how to compute manually) and he couldn't understand why the arc flash method becomes 25%.

I'm not asking for detail computations of loads, etc. I just want to know the fault current 20 feet away of service conductor (AWG 1) given any transformer source Isc. For my particular case, I have already computed the Isc of the transformers with help from an EE.

My building is only office load, so no need for motors, and all those miscellaneous equipments available in large commercial settings. Also I live in the Philippines so can't attend Jim classes in the USA as much as I want to.

So my questions are only limited to the previous message why the offical formulas based on getting constant C and solving for f and M and Isc gives too low short circuit current after 20 feet of service conductors.

PhilEngr, any ideas?

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 Post subject: Re: Quantity of carbonized particles before Arc Flash InitiaPosted: Tue Mar 19, 2019 10:57 am
 Plasma Level

Joined: Tue Oct 26, 2010 9:08 am
Posts: 2173
Location: North Carolina
Transformer impedance is almost purely inductive and I'm assuming you used the "resistance" column instead of the 60 Hz AC impedance column for the wire impedance? This is where we get into some confusion amongst short circuit models. The simplified methods usually assume that X/R >> 10 so that we can conveniently essentially ignore R and do all the calculations based on X.

A further issue is that arcing current is much lower than short circuit current. It is almost an adjustment unto itself relying on short circuit current, system voltage, and arc gap. The IEEE 1584-2002 model is given here:

https://www.ecmag.com/section/systems/a ... it-current

In reality there are a couple things going on. The first one arises from our simplification in assuming that arcing current is RMS when in fact it's not. It extinguishes and restrikes at a finite (nonzero) voltage. At low voltages it looks almost like a square wave whereas at higher voltages this effect almost disappears entirely. Thus the system voltage shows up in the empirical equation. Second for obvious reasons the arc gap has a pronounced (linear) effect on the arcing voltage in that it produces a voltage drop. Third the available fault current wll increase air temperature as it goes up which lowers the threshold arcing voltage which is in turn driving the arcing current (RMS perception).

Aside from this considering the Lee model which is that maximum power transfer occurs at the point where the arcing resistance (assumes arcs are purely resistive) equals the system resistance, which is a purely theoretical argument. So we get Iarc = 0.5*I-short circuit. This value cannot be less than this since maximum power transfer cannot exceed half the short circuit power transfer. With DC arcs it becomes more obvious that there are factors at play that decrease the arcing voltage (equivalently the arc resistance increases) to one of two stable points that are below the maximum power transfer point. This is an underlying effect on the AC arcs as well. You can see it very clearly in the "bimodal distribution" that shows up in IEEE 1584-2002 in the charts and tables. Sorry that this did not show up in the 2018 edition. I haven't seen the chart anywhere outside of the standard so I can't give you a link to what I mean.

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 Post subject: Re: Quantity of carbonized particles before Arc Flash InitiaPosted: Tue Mar 19, 2019 4:01 pm

Joined: Wed Feb 20, 2019 3:06 am
Posts: 45
PaulEngr wrote:
Transformer impedance is almost purely inductive and I'm assuming you used the "resistance" column instead of the 60 Hz AC impedance column for the wire impedance? This is where we get into some confusion amongst short circuit models. The simplified methods usually assume that X/R >> 10 so that we can conveniently essentially ignore R and do all the calculations based on X.

According to the electrical engineer who is a veteran (but never explored arc flash stuff), inductance of 20 feet of wire AWG 1 is insignificant at 60 hertz AC voltage. So you can practically ignore the inductance of wire. Now if you get the resistance of wire, it is only very small value, just 0.002 ohm. So to add it to say the existing 0.01536 ohm inductance of transformers, you use the formula

Z = sqrt (R^2 + X^2)
=sqrt (0.002^2 + 0.01536^2)
=0.01549 ohm.

You can see combined 0.01549 ohm is very close to the pure transformer impedance of 0.01536 ohm. So the 0.002 resistance of wire is practically insignificant.

The above is the argument of the EE. Now are you saying we must really add the inductance of the wire? But it is insignificant. If you believe it is significant. What is the conventional formula to get the inductance of 20 feet of wire awg 1? What value do you get so we can plug it in the formula above and see if it would increase the combined Z significantly to get 25% bolted short circuit current instead of just 1% as the EE calculated.

Please show if there is any error in the formulas above. Kindly focus on the above in your reply because I'm still reading the following today. If you agree the 20 feet awg 1 wire has insignificant inductance and even insignificant resistance, then the reasoning why the bolted short circuit current is 25% less after 20 feet awg 1 wire in the IEEE 1584 formulas compared to the naïve computation above which is just 1% is then due to your other arguments (detailed below). I just want clear clarification of this issue so I can share it to electrical engineer who never work with arc flash and never interested in it. Like I can tell him "You are right, resistance and inductance of 20 feet of awg 1 wire is insignificant compared to the transformer inductance, but he reason the bolted short circuit current is 25% lower instead of just 1% in the naïve computation is due to the contribution of the arcing current and arcing resistances and power transfer to it and heating up of air which lowers the threshold arcing voltage, and other arc flash dynamics". Thanks.

Quote:
A further issue is that arcing current is much lower than short circuit current. It is almost an adjustment unto itself relying on short circuit current, system voltage, and arc gap. The IEEE 1584-2002 model is given here:

https://www.ecmag.com/section/systems/a ... it-current

In reality there are a couple things going on. The first one arises from our simplification in assuming that arcing current is RMS when in fact it's not. It extinguishes and restrikes at a finite (nonzero) voltage. At low voltages it looks almost like a square wave whereas at higher voltages this effect almost disappears entirely. Thus the system voltage shows up in the empirical equation. Second for obvious reasons the arc gap has a pronounced (linear) effect on the arcing voltage in that it produces a voltage drop. Third the available fault current wll increase air temperature as it goes up which lowers the threshold arcing voltage which is in turn driving the arcing current (RMS perception).

Aside from this considering the Lee model which is that maximum power transfer occurs at the point where the arcing resistance (assumes arcs are purely resistive) equals the system resistance, which is a purely theoretical argument. So we get Iarc = 0.5*I-short circuit. This value cannot be less than this since maximum power transfer cannot exceed half the short circuit power transfer. With DC arcs it becomes more obvious that there are factors at play that decrease the arcing voltage (equivalently the arc resistance increases) to one of two stable points that are below the maximum power transfer point. This is an underlying effect on the AC arcs as well. You can see it very clearly in the "bimodal distribution" that shows up in IEEE 1584-2002 in the charts and tables. Sorry that this did not show up in the 2018 edition. I haven't seen the chart anywhere outside of the standard so I can't give you a link to what I mean.

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 Post subject: Re: Quantity of carbonized particles before Arc Flash InitiaPosted: Wed Mar 20, 2019 12:49 pm
 Plasma Level

Joined: Tue Oct 26, 2010 9:08 am
Posts: 2173
Location: North Carolina
OK, here is a link to what I believe to be the NEC table that I usually use, since the values match.

ftp://ftp.conagua.gob.mx/OCPBC/aps/Proy ... 202008.pdf

X-subL is 0.057 ohms per 1,000 feet so j0.000114 ohms for 20 feet; R is 0.16 ohms/1000 feet so for 20 feet we get 0.0032 ohms. In comparison to a transformer impedance of j0.01536 ohms, the reactance is pretty miniscule (1%) but the resistance is around 20% of the transformer impedance...not a lot but not insignificant.

The spreadsheet I believe you are referring to is the one that IEEE distributed on the 1584-2002 CD. You can find it here:

The Cooper EPR-1 document referenced in the spreadsheet contains quite a few multipliers to account for errors but it kind of looks like the IEEE 1584 spreadsheet doesn't take these into account. So transformer short circuit is simply (kVA*1000)/(1.732*V-line-line)*(100/%Z) which is pretty much the text book answer for the infinite bus assumption which basically assumes no impedance on the primary side. It also doesn't appear to be accounting for motor contribution which is done for traditional short circuit methods but not done here. In terms of arc flash motor short circuit contribution dies down in a couple cycles anyways so quite often especially in simplified calculations it gets ignored.

Then we get to the "f" and "M" factors you referred to. These in turn depend on a "C" factor. It took a little digging but Cooper reveals what the "C factors" are in another version of the calculation here:

http://www.cooperindustries.com/content ... rmulas.pdf

Basically C is equal to 1 over the impedance (Z) of the wire. Values are taken from the standard IEEE Color Books series on performing short circuit calculations. I didn't check very many but it looks about right to me.

The f and thus M factor is visible in a tab of the spreadsheet. I'm not 100% sure on the math of what they are doing but it sure looks like the right formula for calculating short circuit current given an upstream short circuit value (point to point method) and just the cable impedance.

Try plugging your values into the IEEE 1584 spreadsheet. I can't vouch for everything in it except to say that it has been verified by a number of people and I have never heard of anyone questioning the math in it. I don't have my copies of the IEEE Color Books with me this afternoon to try to dig further into the math behind it but suffice to say that IEEE 1584 seems to have taken their calculations directly from those published by Cooper (Bussmann at the time) which in turn appear to be a mix of the IEEE Gray and Buff books. At one time the IEEE Color Books series, particularly the Red Book, were all but required reading for electrical engineers. Short circuit calculations are detailed in those books.

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 Post subject: Re: Quantity of carbonized particles before Arc Flash InitiaPosted: Thu Mar 21, 2019 3:14 am

Joined: Wed Feb 20, 2019 3:06 am
Posts: 45
PaulEngr wrote:
OK, here is a link to what I believe to be the NEC table that I usually use, since the values match.

ftp://ftp.conagua.gob.mx/OCPBC/aps/Proy ... 202008.pdf

X-subL is 0.057 ohms per 1,000 feet so j0.000114 ohms for 20 feet; R is 0.16 ohms/1000 feet so for 20 feet we get 0.0032 ohms. In comparison to a transformer impedance of j0.01536 ohms, the reactance is pretty miniscule (1%) but the resistance is around 20% of the transformer impedance...not a lot but not insignificant.

I think you made an error above. 20/1000feet = 0.02 * 0.057 = 0.00114. You wrote "0.000114".

It's not far from the resistance of 0.0032 ohm so the inductive reactance is somewhat significant too.

Computing for Z at 20 feet is sqrt(0.0032^2 + 0.00114^2)= 0.003397 ohm

But let's just focus for now on 1000 feet as I'm verifying the NEC table manually (how it gets the values in the table).

In the NEC list for AWG 1 above, the inductive reactance (Xl) for 1000 feet has value of 0.046 ohm for PVC conduit
See:
ftp://ftp.conagua.gob.mx/OCPBC/aps/Proy ... 202008.pdf

1000 feet of awg 1 has calculated inductance of 687000 nH or 0.000687 H see: https://www.allaboutcircuits.com/tools/ ... alculator/

The inductive reactance is 2 pi f L... or 2 (3.141) (60) (0.000687) = 0.259 ohm http://www.66pacific.com/calculators/in ... lator.aspx

Why is the 0.259 ohm so far from the NEC listed value of 0.046 ohm?

How does NEC derive the 0.046ohm? Can you help derive 0.046 ohm? I've been trying it out for hours already.
[/quote]

Quote:

The spreadsheet I believe you are referring to is the one that IEEE distributed on the 1584-2002 CD. You can find it here:

The Cooper EPR-1 document referenced in the spreadsheet contains quite a few multipliers to account for errors but it kind of looks like the IEEE 1584 spreadsheet doesn't take these into account. So transformer short circuit is simply (kVA*1000)/(1.732*V-line-line)*(100/%Z) which is pretty much the text book answer for the infinite bus assumption which basically assumes no impedance on the primary side. It also doesn't appear to be accounting for motor contribution which is done for traditional short circuit methods but not done here. In terms of arc flash motor short circuit contribution dies down in a couple cycles anyways so quite often especially in simplified calculations it gets ignored.

Then we get to the "f" and "M" factors you referred to. These in turn depend on a "C" factor. It took a little digging but Cooper reveals what the "C factors" are in another version of the calculation here:

http://www.cooperindustries.com/content ... rmulas.pdf

Basically C is equal to 1 over the impedance (Z) of the wire. Values are taken from the standard IEEE Color Books series on performing short circuit calculations. I didn't check very many but it looks about right to me.

The f and thus M factor is visible in a tab of the spreadsheet. I'm not 100% sure on the math of what they are doing but it sure looks like the right formula for calculating short circuit current given an upstream short circuit value (point to point method) and just the cable impedance.

Try plugging your values into the IEEE 1584 spreadsheet. I can't vouch for everything in it except to say that it has been verified by a number of people and I have never heard of anyone questioning the math in it. I don't have my copies of the IEEE Color Books with me this afternoon to try to dig further into the math behind it but suffice to say that IEEE 1584 seems to have taken their calculations directly from those published by Cooper (Bussmann at the time) which in turn appear to be a mix of the IEEE Gray and Buff books. At one time the IEEE Color Books series, particularly the Red Book, were all but required reading for electrical engineers. Short circuit calculations are detailed in those books.

The IEEE 1584 spreadsheet is just using the simple FLA, f and M formula. The values entered was the same in other spreadsheest and one can easily compute it by hand. For example, the bottom matches the IEEE 1584 spreadsheet because they use the same formula:

The following spreadsheet is more intuitive as the formulas are written on the page:

So it boils down to how the formula f = 1.732 * L * I / N * C * E(l-n) and M = 1/ (1+1f) is derived. I need to know this because I'd like to know if the formula integrate already the following arc flash dynamics, such as arc flash added impedance, heating up lowering threshold voltage, arc gap causing voltage drop, etc. or is the above formula independent of these??

If they are already integrated. How did they do it?

This is my last questions. Lol.

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 Post subject: Re: Quantity of carbonized particles before Arc Flash InitiaPosted: Thu Mar 21, 2019 5:57 pm

Joined: Wed Feb 20, 2019 3:06 am
Posts: 45
PaulEngr wrote:
OK, here is a link to what I believe to be the NEC table that I usually use, since the values match.

ftp://ftp.conagua.gob.mx/OCPBC/aps/Proy ... 202008.pdf

X-subL is 0.057 ohms per 1,000 feet so j0.000114 ohms for 20 feet; R is 0.16 ohms/1000 feet so for 20 feet we get 0.0032 ohms. In comparison to a transformer impedance of j0.01536 ohms, the reactance is pretty miniscule (1%) but the resistance is around 20% of the transformer impedance...not a lot but not insignificant.

The spreadsheet I believe you are referring to is the one that IEEE distributed on the 1584-2002 CD. You can find it here:

The Cooper EPR-1 document referenced in the spreadsheet contains quite a few multipliers to account for errors but it kind of looks like the IEEE 1584 spreadsheet doesn't take these into account. So transformer short circuit is simply (kVA*1000)/(1.732*V-line-line)*(100/%Z) which is pretty much the text book answer for the infinite bus assumption which basically assumes no impedance on the primary side. It also doesn't appear to be accounting for motor contribution which is done for traditional short circuit methods but not done here. In terms of arc flash motor short circuit contribution dies down in a couple cycles anyways so quite often especially in simplified calculations it gets ignored.

Then we get to the "f" and "M" factors you referred to. These in turn depend on a "C" factor. It took a little digging but Cooper reveals what the "C factors" are in another version of the calculation here:

http://www.cooperindustries.com/content ... rmulas.pdf

Basically C is equal to 1 over the impedance (Z) of the wire. Values are taken from the standard IEEE Color Books series on performing short circuit calculations. I didn't check very many but it looks about right to me.

After 1 hour of looking at references and verifying if C is 1/Z. I got it.

from https://peritoselectricos.mx/wp-content ... cial-B.pdf

Table 65: at 1000 feet

Resistance is 0.129 ohm

Reactance is 0.0342

For 1 foot.. resistance is 0.129 ohm * 1/1000= 0.000129 ohm
reactance at 1 foot is 0.0000342

Impedance = sqrt (0.000129^2+0.0000342^2) = 0.0001335 ohm

1/Z = 1/0.0001335 = 7493 which is the C listed for AWG 1 Cooper nonmetallic 600v in the cooper pdf.

now going to the formula f = [1.732 * L * I] /[ N * C * E(l-l)]

since C = 1/Z. We could use 1/Z

This would make the equation f = [1.732 * L* Z* I] / [N* E(l-l)]

Since L is just length of conductors
N is number of conductors per phase
E(l-l) is just the voltage line to line
I is the short circuit current
Z is impedance
1.732 is just sqrt (3)

then it looks like the equation doesn't take into account the following arc flash dynamics:

2. heating up lowering threshold voltage
3. arc gap causing voltage drop

Right? if these are not included in the equations of f and M, then it means the actual bolted short circuit current is lower, right?

Quote:

The f and thus M factor is visible in a tab of the spreadsheet. I'm not 100% sure on the math of what they are doing but it sure looks like the right formula for calculating short circuit current given an upstream short circuit value (point to point method) and just the cable impedance.

Try plugging your values into the IEEE 1584 spreadsheet. I can't vouch for everything in it except to say that it has been verified by a number of people and I have never heard of anyone questioning the math in it. I don't have my copies of the IEEE Color Books with me this afternoon to try to dig further into the math behind it but suffice to say that IEEE 1584 seems to have taken their calculations directly from those published by Cooper (Bussmann at the time) which in turn appear to be a mix of the IEEE Gray and Buff books. At one time the IEEE Color Books series, particularly the Red Book, were all but required reading for electrical engineers. Short circuit calculations are detailed in those books.

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 Post subject: Re: Quantity of carbonized particles before Arc Flash InitiaPosted: Fri Mar 22, 2019 5:03 pm
 Plasma Level

Joined: Tue Oct 26, 2010 9:08 am
Posts: 2173
Location: North Carolina
Quote:
2. heating up lowering threshold voltage
3. arc gap causing voltage drop

Right? if these are not included in the equations of f and M, then it means the actual bolted short circuit current is lower, right?

Correct. The arc flash formulas START with the bolted fault short circuit current. The equations you are looking at are modelling the bolted fault short circuit current. The "point to point" method is only concerned with modelling short circuits. Arc flash is modeled from that data.

Most arc flash models (except the maximum power transfer type theoretical models) estimate arcing current from the short circuit current, arc gap, system voltage, and short circuit current (yes, nonseparable equations). They differ in how they are used. The Wilkins simplified as well as time domain models are essentially physics based. The various constants and parameters are derived from curve fitting to the IEEE 1584-2002 data set. The IEEE 1584-2002 and 2018 equations essentially just curve fit everything to exponential curves without regard for the underlying physics.

When it comes to arcing current system voltage plays a role in changing the duty cycle (restriking) of the arc. From an RMS point of view at low voltages it lowers the RMS arcing current significantly while above about 1 kV it has almost no effect at all. Arc gap is used as a proxy to estimate arc length. The ends (cathode and anode) of the arc have different physics but the majority of the arc is almost a constant voltage drop. This shows up in reducing the arcing voltage a little bit with the arc gap. Finally arc impedance is kind of an odd effect. The arc voltage only very slowly rises in a power arc so in effect the arc resistance (arcs are purely resistive) decreases as available short circuit current increases. This is modelled (nonseparable equation here) by using the short circuit current as a factor. These factors apply whether the equations are the Wilkins variety or the IEEE 1584-2002/2018 variety.

No arc flash model currently models the time-domain lowering of the arcing voltage threshold directly. It's a small effect that lasts only 1 or 2 cycles. Incident energy is modeled as linear with time in ALL models. Once we have calculated arcing current, we then look up the opening time of the overcurrent protective devices based on the arcing current to calculate arcing time. Then we calculate incident energy from those values and various other factors. There is no time dependent effect. Arcs are purely linear with time. This is actually very useful in practice. Without aid of a computer I can very accurate predict exactly what happens to incident energy if I know arcing current and I have time-current curves to work with. Often I can just look at the TCC charts and estimate the effect of changing device settings/sizes without ever doing a single calculation. It saves a lot of time when doing mitigation.

"Heating up" as you put it caused a lot of "noise" in the IEEE 1584-2002 data set. For the 2018 data set they cleaned it up by ignoring the first couple cycles and only collecting data on incident energy after the voltage and current measurements were stable waveforms. They did look at the phase angle when the arc was initiated but essentially the phase angle appears to be almost a random variable. There is research data that suggests that X/R has an effect but it is not currently taken into account by any model except maybe the Wilkins time domain model and even then probably indirectly because of the nature of the model.

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 Post subject: Re: Quantity of carbonized particles before Arc Flash InitiaPosted: Fri Mar 22, 2019 8:06 pm

Joined: Wed Feb 20, 2019 3:06 am
Posts: 45
PaulEngr wrote:
Quote:
2. heating up lowering threshold voltage
3. arc gap causing voltage drop

Right? if these are not included in the equations of f and M, then it means the actual bolted short circuit current is lower, right?

Correct. The arc flash formulas START with the bolted fault short circuit current. The equations you are looking at are modelling the bolted fault short circuit current. The "point to point" method is only concerned with modelling short circuits. Arc flash is modeled from that data.

But there is a problem. Using direct impedance computations from the table. The bolted short circuit current varies the ones with f and M (point to point method) by a wide margin.
Let's compute:

https://peritoselectricos.mx/wp-content ... cial-B.pdf

Table 65 for AWG 1 at 1000 feet for Nonmetallic conductors and Several 1/C Nonmagc conductor Data

Resistance is 0.129 ohm
Reactance is 0.0342

For 1 foot:
resistance is 0.129 ohm * 1/1000= 0.000129 ohm
reactance is 0.0342 * 1/1000 = 0.0000342

Impedance = sqrt (0.000129^2+0.0000342^2) = 0.0001335 ohm

1/Z = 1/0.0001335 = 7493 which is the C listed for AWG 1 Cooper nonmetallic 600v in the cooper pdf (see my last message for the table)

now going to the formula f = [1.732 * L * I] /[ N * C * E(l-l)]

entering the above data,
f is 0.347,
M = 0.742
Isc (bolted scc) = 11597 amp

This is same result in the IEEE 1584-2002 spreadsheet

Now if you will compute it directly by getting the impedances of the wires and transformers and dividing 240v by the total impedances to get the bolted short circuit current, the result doesn't match.

from the same impedance table above where the C=1/Z came from, and adding the transformer impedance to the wire inductive reactance side of it to get overall bolted impedance:

20 feet awg 1 impedance

For 20 foot:
resistance is 0.129 ohm * 20/1000= 0.000258 ohm
reactance is 0.0342 * 20/1000 = 0.0000684
Impedance = sqrt (0.000258^2+(0.0000684+0.01536)^2) = 0.01625 ohm
Transformer impedance of 0.01536 comes from 240^2/75000 * 0.02

240v/0.01625 = 14769 amp.

This differs from the IEEE f and M point to point formulas by wide margin (14769A vs 11597a).

I assume the X/R of the transformers is purely inductive. Even if you divided it among the resistance and inductive reactance. It won't lower the bolted short circuit current unless you assume X/R is purely resistive (which isn't true because the transformer impedance is most inductive reactance).

Since this addition of impedances and getting the bolted short circuit current by dividing the voltage by the total bolted impedance is more direct and accurate. Then is it possible the IEEE f and C formulas are just wrong??

Try to compute and compare the results too. Also how did IEEE exactly derive f and C? We need to know how they are derive to detect if there is error in the derivations.

Quote:
Most arc flash models (except the maximum power transfer type theoretical models) estimate arcing current from the short circuit current, arc gap, system voltage, and short circuit current (yes, nonseparable equations). They differ in how they are used. The Wilkins simplified as well as time domain models are essentially physics based. The various constants and parameters are derived from curve fitting to the IEEE 1584-2002 data set. The IEEE 1584-2002 and 2018 equations essentially just curve fit everything to exponential curves without regard for the underlying physics.

When it comes to arcing current system voltage plays a role in changing the duty cycle (restriking) of the arc. From an RMS point of view at low voltages it lowers the RMS arcing current significantly while above about 1 kV it has almost no effect at all. Arc gap is used as a proxy to estimate arc length. The ends (cathode and anode) of the arc have different physics but the majority of the arc is almost a constant voltage drop. This shows up in reducing the arcing voltage a little bit with the arc gap. Finally arc impedance is kind of an odd effect. The arc voltage only very slowly rises in a power arc so in effect the arc resistance (arcs are purely resistive) decreases as available short circuit current increases. This is modelled (nonseparable equation here) by using the short circuit current as a factor. These factors apply whether the equations are the Wilkins variety or the IEEE 1584-2002/2018 variety.

No arc flash model currently models the time-domain lowering of the arcing voltage threshold directly. It's a small effect that lasts only 1 or 2 cycles. Incident energy is modeled as linear with time in ALL models. Once we have calculated arcing current, we then look up the opening time of the overcurrent protective devices based on the arcing current to calculate arcing time. Then we calculate incident energy from those values and various other factors. There is no time dependent effect. Arcs are purely linear with time. This is actually very useful in practice. Without aid of a computer I can very accurate predict exactly what happens to incident energy if I know arcing current and I have time-current curves to work with. Often I can just look at the TCC charts and estimate the effect of changing device settings/sizes without ever doing a single calculation. It saves a lot of time when doing mitigation.

"Heating up" as you put it caused a lot of "noise" in the IEEE 1584-2002 data set. For the 2018 data set they cleaned it up by ignoring the first couple cycles and only collecting data on incident energy after the voltage and current measurements were stable waveforms. They did look at the phase angle when the arc was initiated but essentially the phase angle appears to be almost a random variable. There is research data that suggests that X/R has an effect but it is not currently taken into account by any model except maybe the Wilkins time domain model and even then probably indirectly because of the nature of the model.

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 Post subject: Re: Quantity of carbonized particles before Arc Flash InitiaPosted: Sat Mar 23, 2019 3:38 pm

Joined: Wed Feb 20, 2019 3:06 am
Posts: 45
Ommi wrote:
PaulEngr wrote:
Quote:
2. heating up lowering threshold voltage
3. arc gap causing voltage drop

Right? if these are not included in the equations of f and M, then it means the actual bolted short circuit current is lower, right?

Correct. The arc flash formulas START with the bolted fault short circuit current. The equations you are looking at are modelling the bolted fault short circuit current. The "point to point" method is only concerned with modelling short circuits. Arc flash is modeled from that data.

But there is a problem. Using direct impedance computations from the table. The bolted short circuit current varies the ones with f and M (point to point method) by a wide margin.
Let's compute:

https://peritoselectricos.mx/wp-content ... cial-B.pdf

Table 65 for AWG 1 at 1000 feet for Nonmetallic conductors and Several 1/C Nonmagc conductor Data

Resistance is 0.129 ohm
Reactance is 0.0342

For 1 foot:
resistance is 0.129 ohm * 1/1000= 0.000129 ohm
reactance is 0.0342 * 1/1000 = 0.0000342

Impedance = sqrt (0.000129^2+0.0000342^2) = 0.0001335 ohm

1/Z = 1/0.0001335 = 7493 which is the C listed for AWG 1 Cooper nonmetallic 600v in the cooper pdf (see my last message for the table)

now going to the formula f = [1.732 * L * I] /[ N * C * E(l-l)]

entering the above data,
f is 0.347,
M = 0.742
Isc (bolted scc) = 11597 amp

This is same result in the IEEE 1584-2002 spreadsheet

Now if you will compute it directly by getting the impedances of the wires and transformers and dividing 240v by the total impedances to get the bolted short circuit current, the result doesn't match.

from the same impedance table above where the C=1/Z came from, and adding the transformer impedance to the wire inductive reactance side of it to get overall bolted impedance:

20 feet awg 1 impedance

For 20 foot:
resistance is 0.129 ohm * 20/1000= 0.000258 ohm
reactance is 0.0342 * 20/1000 = 0.0000684
Impedance = sqrt (0.000258^2+(0.0000684+0.01536)^2) = 0.01625 ohm
Transformer impedance of 0.01536 comes from 240^2/75000 * 0.02

240v/0.01625 = 14769 amp.

This differs from the IEEE f and M point to point formulas by wide margin (14769A vs 11597a).

I assume the X/R of the transformers is purely inductive. Even if you divided it among the resistance and inductive reactance. It won't lower the bolted short circuit current unless you assume X/R is purely resistive (which isn't true because the transformer impedance is most inductive reactance).

Since this addition of impedances and getting the bolted short circuit current by dividing the voltage by the total bolted impedance is more direct and accurate. Then is it possible the IEEE f and C formulas are just wrong??

Try to compute and compare the results too. Also how did IEEE exactly derive f and C? We need to know how they are derive to detect if there is error in the derivations.

To add to the above computations and facts.

I corresponded with a veteran EE last week who helped me compute the bolted short circuit after 20 feet of awg 1 conductor. He was not familiar with arc flash so couldn't comment anything on the f and M formulas or anything about arc flash. He said though:

"Transformer resistive and reactive components

The transformer nameplate states the impedance as a percentage. Knowing only the percentage does not allow the resistive and reactive components, R and X, to be determined. Therefore the result of the above calculation is the combined resistance and reactance, Z. Often nameplates state X/R as well as percentage. That allows Z to be calculated when necessary. Another alternative is to find a listing of typical X/R for various transformer sizes. At this kVA level, X/R is likely more than 5 and therefore the impedance can be considered to be mostly inductive reactance.

If any added external impedance is mostly inductive, it can simply be added. If it is mostly resistive, it can be added using the square root of the sum of squares. In the case of adding the impedance of half of an additional winding, X/R is assumed to be the same for both components and can be simply added."

So to add the inductive reactance of transformers to the wiring conductors. You need to use the formula:

The inductive reactance must be added inside the X component. Now look at the IEEE bolted short formulas of f and M

The C is just 1/Z of the wiring conductor only. You are supposed to add the transformers inductive reactance inside the X component of C. What the formula f does is adding it outside! So there is a possibility it is wrong.

Why did they add it outside? It makes the bolted short circuit current after 20 feet of wiring conductors much lower than in reality.

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 Post subject: Re: Quantity of carbonized particles before Arc Flash InitiaPosted: Wed Mar 27, 2019 2:25 am

Joined: Wed Feb 20, 2019 3:06 am
Posts: 45
The Point to Point short circuit calculations method results vary even to another NEC method mentioned in :

https://iaeimagazine.org/magazine/2015/ ... t-current/

Here's it's more intuitive.

How can two results vary by over 30%?

How is the Point to Point f and M really derived?

Could they be wrong? Maybe they were created by an IEEE 1585-2002 Ad Hoc Committee who wants to make the bolted short circuit current smaller than really is so the overcurrent protection can have lower curve trip (to avoid the incident energy getting larger from slower bimetallic trip instead of instantaneous magnetic trip?)

Hope we can track who developed m and F and whether it is based on any calculations or just estimates.

Remember IEEE 1584-2002 was wrong that 5kA couldn't cause arc flash when in fact they used incomplete experimental setup without any barrier (representing breakers) present. So they could be wrong on other parts too.

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 Post subject: Re: Quantity of carbonized particles before Arc Flash InitiaPosted: Thu Mar 28, 2019 3:11 am

Joined: Wed Feb 20, 2019 3:06 am
Posts: 45
I am finally able to derive f and M and solve it by using different formula (direct method) but same results after trying different things in excel.

I adjusted the same formula I used before and it matched.

For 2 pairs of 20 feet AWG 1.. the conductors must be 40 feet instead of 20 feet, and using the same resistance and reactance from the IEEE table (where they got the C= 1/Z).

resistance is 0.129 ohm * 40/1000= 0.00516 ohm
reactance is 0.0342 * 40/1000 = 0.001368 ohm

Z = sqrt (0.00516^2 + 0.001368^2) = 0.005338 ohm

transformer impedance + conductor impedance = 0.01536 + 0.005338 = 0.020698 ohm

240/0.020698 ohm = 11595A bolted short circuit current... this result is identical to the one using M and f (shown in excel below)

(note Transformer impedance of 0.01536 comes from 240^2/75000 * 0.02)

So what they did is to add by brute force the impedance of the transformer and conductor...

Adding Z1 + Z2 is based on the assumption that X/R is the same for both. Is it?

If the X/R ratio is not equal, then the equation is not the one for it.
What do you think? Can you just add Z1 + Z2 if the X/R ratio is not the same for both?

I'm reviewing vector algebra and it doesn't seem to hold. Correct me where I comprehended wrong.

Identical results to:

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 Post subject: Re: Quantity of carbonized particles before Arc Flash InitiaPosted: Fri Mar 29, 2019 2:39 am

Joined: Wed Feb 20, 2019 3:06 am
Posts: 45
Anyway. Here is how exactly the formula of f and M are derived. I can't believe I missed it for a week. That's the problem when one didn't pay attention to math in college

After looking at many entries in excel and trying different data. I finally saw the pattern. It's simply:

They are simply two impedances where the Z of wire is divided to the Z of transformers. Why is shown at bottom. But first, remember C = 1/Z, then it's in 1 foot.. so you multiply it by the Length to get the total wiring impedance, now if there is say 2 conductors per phase, then it's divided by 2 simply because for parallel resistors, you get 1/2R for two resistors in parallel.

Now for the transformer impedance. It's simply R=V/I or Z = E(l-l) / I(l-l)

The reason the transformer impedance is below with wiring impedance above in the divide sign is the following.

The formula is just telling you that if you have a certain impedance at the short circuit current, Isc. Then if you add the wiring impedance to the transformer impedance, then the bolted short circuit current would be inversely proportional to the total over the partial Isc. Inversely proportional because the current goes down if impedances are high. Using the same data in last message, then it's simply

Zwire = Z * 2L/ n = 0.000133458 * 2 (20)/1
= 0.0053383ohm

Ztransformer = 240^2/75000 * 0.02 = 0.01536 ohm

What the M = 1/(1+f) formula does is simply getting the inverse proportional algebra...

That is

I(trans)/Ibolted= (0.01536 + 0.0053383) / 0.01536

given I(trans)sc of 15625A,

then solving for I (bolted scc) would give you 11595A. Exactly the same as the excel sheet in last message because this is all there is to it.

Adding Z1 + Z2 is based on the assumption that X/R is the same for both transformer and wiring. If X/R is not the same, then the bolted short circuit currents computed would be incorrect.

So my only questions now are. How often are X/R ratio of transformers and wiring the same? If seldom, then what formulas must you use instead?

Hope when mentioning the formulas of f and M, they can put disclaimer that it's only if the X/R of both transformers and wiring is the same. A veteran EE emphasized to me that's the only time you can add Z1 + Z2. Otherwise you need to use Z =sqrt(R^2 + X^2) where the components must be separately added.

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 Post subject: Re: Quantity of carbonized particles before Arc Flash InitiaPosted: Tue Apr 02, 2019 6:24 pm

Joined: Wed Feb 20, 2019 3:06 am
Posts: 45
After exhaustively exploring the derivations of the f and M formula and comparing it with complex numbers addition with different transformer setups. I came to the conclusion the difference is only less than 7%. In my exact setup with the 3 phase open delta transformers 20 feet from the service panel.

Isc (transformer Isc) = 6971A

Using f and M method (with 20 feet of awg 1): 5913A

Using complex numbers addition of each component (with 20 feet of awg 1): 6248A.

They are close. The f and M results are a bit lower. Just 5% difference. This is even when the transformers and voltage were increased up to Isc of 50,000A.

The reason I'm knit picking on them is because of IEEE-1584 2018 in which anything above 2000A can arc flash. And it's true. I can now confirm that at 6000A, it can arc flash when the panel was not empty (that is when it has breakers). Remember the IEEE-1584 2002 tests assume empty panel (what were they thinking?)

Now my two last questions before I move on from all these.

1. What are the arc flash standards used in say Europe or China or Russia? Do they all follow the IEEE 1584? In the US. arc flash at low voltage incidents are so low. Of course because it is 120 volts. But elsewhere where it is 240v. How is the arc flash incident statistics?

2. Until now I still haven't used my arc flash PPE. And remember it's only to open and close breakers for this panel:

With arc flash incident energy only critical at the arms part for 6000A maximum short circuit current and since the opening above is only very small. It's justified only arc flash jacket is required. Therefore I plan to cut my Oberon PPE suit so I can use it for example as cool jacket in malls or wear it elsewhere.

How best to cut it. Or better yet. since it's inherently flame resistant. I don't think more washing can damage it anyway. Because if I won't cut it. Then it will be used only once in say 3 years or so when the panel above needs to be turned off or repaired. And with only 6000A maximum short circuit current. I don't think the slags can reach the legs part with such small panel opening, isn't it.

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 Post subject: Re: Quantity of carbonized particles before Arc Flash InitiaPosted: Wed Apr 03, 2019 7:32 pm

Joined: Wed Feb 20, 2019 3:06 am
Posts: 45
Ommi wrote:
After exhaustively exploring the derivations of the f and M formula and comparing it with complex numbers addition with different transformer setups. I came to the conclusion the difference is only less than 7%. In my exact setup with the 3 phase open delta transformers 20 feet from the service panel.

Isc (transformer Isc) = 6971A

Using f and M method (with 20 feet of awg 1): 5913A

Using complex numbers addition of each component (with 20 feet of awg 1): 6248A.

They are close. The f and M results are a bit lower. Just 5% difference. This is even when the transformers and voltage were increased up to Isc of 50,000A.

The reason I'm knit picking on them is because of IEEE-1584 2018 in which anything above 2000A can arc flash. And it's true. I can now confirm that at 6000A, it can arc flash when the panel was not empty (that is when it has breakers). Remember the IEEE-1584 2002 tests assume empty panel (what were they thinking?)

Now my two last questions before I move on from all these.

1. What are the arc flash standards used in say Europe or China or Russia? Do they all follow the IEEE 1584? In the US. arc flash at low voltage incidents are so low. Of course because it is 120 volts. But elsewhere where it is 240v. How is the arc flash incident statistics?

2. Until now I still haven't used my arc flash PPE. And remember it's only to open and close breakers for this panel:

With arc flash incident energy only critical at the arms part for 6000A maximum short circuit current and since the opening above is only very small. It's justified only arc flash jacket is required. Therefore I plan to cut my Oberon PPE suit so I can use it for example as cool jacket in malls or wear it elsewhere.

How best to cut it. Or better yet. since it's inherently flame resistant. I don't think more washing can damage it anyway. Because if I won't cut it. Then it will be used only once in say 3 years or so when the panel above needs to be turned off or repaired. And with only 6000A maximum short circuit current. I don't think the slags can reach the legs part with such small panel opening, isn't it.

On second thought. I won't cut it anymore. Last night before I slept. I tried a free software that can only register up to 208v. I hope the creator of this web site can create one too. I entered 2 secs and the incident energy at 305mm away is 25.9cal/cm2 and arch flash boundary is 1.662 meters (1.2cal/cm2). It horrified me (see attachment).

Also I read "Investigation of Factors Affecting the Sustainability of Arcs Below 250 V
Michael J. Lang, Member, IEEE, and Kenneth Jones, Member, IEEE"

"The testing discussed in this paper shows that sustained arcs are possible at 208 V even at relatively low fault currents but are dependent on several factors including voltage variations, conductor material, the configuration of conductors, and the presence of insulating barriers. The challenge to industry is to advance the research identified in the references, do additional testing on a variety of low-voltage equipment, and incorporate those findings into improved standards. These test strategies must consider all practical locations within the equipment where arcs may occur; within all equipment is the possibility for different electrode orientations."

And I read "Effect of Insulating Barriers in Arc Flash Testing Robert Wilkins, Member, IEEE, Mike Lang, Member, IEEE, and Malcolm Allison, Member, IEEE "

"With an insulating barrier arc lengths and voltages are lower, the arcing currents are higher, and the maximum energy density, which is found on the middle calorimeter, is much higher than the one measured with the standard IEEE 1584 setup. Self sustaining arcs can also be produced at 208 V with relatively low levels of bolted-fault current."

I hope IEEE can create experiment where the 208v 6kA 12mm gap can be sustained for 10 secs. I want to see if it would equal the incident energy of major high voltage switchgear meters away.

Whatever, I guess I need to get Oberon ballistic PPE suit. Our PPE is only based on thermal and not adequate.

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