I was at a manufacturing facility recently and saw labels on the fused disconnect switchgear line up with the numbers shown in the subject. This line up is one riser cable to overhead conductor to utility recloser. Recloser is within 200ft. I am trying to get a copy of the arc flash study to see what the inputs were as that seems awful extreme especially at 34.5kV.

I did play around with a model without any recloser action, limiting the arc flash to 2 sec, using typical values I have gotten from this utility in the past for available fault current and get a value less than 20% of the above value. It probably will be even less with the recloser modeled.

So, I am curious to see with the vast experience of the people on this forum if anyone seen values this high at this voltage before? Was it realistic or were the result of some modeling methods?

I wonder how old the label is?

Years ago before people become more comfortable with using the 2 second cut off for exceptionally long device clearing times, many would use whatever the time current curve indicated. Sounds like that might be the case. Some would even use 1000 seconds if there was not an intersection of current and the TCC. One such group rationalized that this was the limit of known data i.e. Time current curves only go to 1000 seconds.

So, out of curiosity, we (I know I'm not alone) would be interested in knowing the clearing time if you can find it.
Happy Hunting!

I have seen excessively large numbers, when the 1000 sec method, mentioned by Jim, was used.

I think my record was:
Volts: 13800
ARCING FAULT (kA): 0.4
TRIP/ DELAY TIME (sec): 1000
BREAKER OPENING TIME (sec): 0.083
ARC FLASH BOUNDARY (in): 15,6231
WORKING DISTANCE (in): 36
INCIDENT ENERGY (cal/cm2): 4136

Modelling 34.5 kV is problematic. You are limited to ArcPro, Duke heat flux, and Lee. There is no public test data. Lee is 300%+ of experimental data at 13.5 kV and error increases as the voltage increases. OSHA recently stated as part of the 1910.269 subpart V rule that Duke and Lee were just plain not recommended. However, parts of the Duke and Arcpro models are not published so competitors (SKM, ETAP, Easypower) are stuck implementing Lee. My experience with it is that it produces silly results. Lee assumes the circuit is purely inductive, the arc is purely resistive, that the arc power is maximum thus arc resistance equals circuit impedance, and that conversion to radiative energy is 100% efficient. Arc voltage is thus dependent on circuit parameters and has nothing to do with physics such as arc length. So the only good thing about it is knowing the incident energy is much lower. Unfortunately many sites blindly accept Lee results without question.

If I find further info I will post it. The study was done in 2012 by a major firm but I have seen some questionable results from them in the past.

I ran the scenario using ArcPro, assumed the same utility fault currents with a 2 sec time cutoff, arc gap of 6 inches, distance to arc 36 in and end up with ~26 cal/cm2 with a 6.5 multiplier for in a box fault.

Will post what the actual values used when and if I can get the report.

Hello,
I have received a copy of the report and it contains what I suspected:

1. Actual utility fault current or X/R not used but an assumed value of 40,000A at 8 X/R
2. Trip time was run out to 2 seconds so no credit taken for any utility recloser operation
3. IEEE 1584 equations used.

I would analyze the fuse switchgear line up using ArcPro as this is more consistent with a utility type setup and is the recommended method in the new OSHA 1910.269 regulations.

Using the ArcPro with a 6.5 multiplier for 3 phase in a box, 40kA, 2 secs, the result is 422 cal/cm^2 @ 36 inches. Still high but less than the 1645 cal/cm^2 that is on the label.

I have requested from the utility the actual fault current and recloser settings and will post an update when I receive that information.

As a frequent flyer when it comes to purchasing "utility" equipment...

There are basically 2 "fault classes" of utility grade gear: 12.5 kA, and 20-25 kA at over 10 kV. 12.5 kA is pretty much an industry norm. Since I have large potential fault currents here (dual 100 MVA incoming transformers), the fault current on the site near the main substation gets to around 20 kA @ 22.9 kV. The number of possible vendors for equipment with a 20-25 kA fault rating is considerably narrower and the price goes up substantially. Beyond that range even fuses, even as backup fuses, start to become rare. That being said, I really, really doubt 40 kA as a fault current is realistic at over 10 kV. For a 480 V system, sure. But for 34.5 kV, I'm very dubious.

Also with respect to using IEEE 1584 equations, what is generally known as "IEEE 1584" is the empirical equation. It is only acceptable up to 15 kV since it's a curve fit and the highest voltages actually tested for the curve fitting data were 13.5 kV. At best, the report could be using the Lee method which is quoted in IEEE 1584.

Assumption is the mother of all screw-ups, as they say. The 40KA at 34.5KV translates into approx 2.5GW power, roughly three (3) times the rated capacity of The Three Mile Island Nuclear Generating Station in the USA. Typical example of SISO.

arcad - very nicely put! I really hate to have to point this stuff out when I am in the field as I am sure they paid a nice amount for the study. I have seen similar things from this company.

40 kA at a generator is not totally unreasonable. Typical capacities for large generators are 300-500 MW with a 12-24 kV output voltage. Step up transformers are usually 69 (rare these days), 115, 230, or sometimes high transmission voltages. Its all simple economics. At 500 MW with a 12 kV output thats 24 kA. So if multiple generators feed one large step up transformer (possible but I have no idea why anyone would do this...transformers are not usually that large), it might turn into a purely fuse protected system. If transmission was done at 35.4 kV (again, unlikely), it is possible. But the equipment costs and the need for way more than quad heavy conductors would dictate a different configuration. The arrangement I have is fed directly from Duke's 230 kV transmission system and fault levels are nowhere near 40 kA. In the 80's (CP&L then) it was upgraded from 115 kV. The only place I get over 20 kA is with the tie closed on two 100 MVA transformers at 22.9 kV with a 50 MW cogen at full power. This drives breakers to the limits of VCB and the older ones are SF6. The 20 kA rating is theoretically possible but I can't imagine anything other than a transient condition with both utility feeders and the generator and all breakers closed. Less than a mile away on overhead lines, fault current is under 12.5 kA. Not to say 40 kA fault currents cannot exist at 35.4 kV, but it just does not seem likely even within a utility network except very close to transmission transformers.

I don't think you applied the assumed X/R of 8 when you did your rough calculation. I think you'll find it alters things considerably.

Update:
I received the actual available fault current from the utility:

3phase: 1,442A with X/R of 2.7

Quite a bit different from the 40,000A and X/R of 8 that the firm/engineer assumed.

Yes indeed, with fault impedance factored in, even more powerful source is needed to push the 40KA current through the impedance.

Yes, but ArcPro does not use X/R in calculating the incident energy.

I use short circuit software factoring in equipment X/R in short circuit current analysis as the ratio has an impact on calculated short circuit current. Please note that a lot of industry accepted calculation methods and software programs ignore the equipment X/R ratios in short circuit analysis and apply correction factors to presumably compensate for the error. Such approach introduces up to 15% uncertainty in a single step of adding two impedances alone while applying different correction factors does not quantifies or corrects the error but only amplifies it.

The calculated fault X/R ratio is a measure of current and voltage displacement when the circuit is shorted, and it is related to power factor by the PF = cos(atan[X/R]) equation. In the above example, the initially assumed value of 40,000A at 8 X/R means the fault current is lagging the voltage by 83 degree when circuit is shorted. The actual 1,442A available fault current with X/R of 2.7 from the utility means the fault current is lagging the voltage by 70 degree. Introducing the arc resistance would increase power factor by reducing overall X/R as the electiric arc has no inductance. I agree with you that neither ArcPro nor IEEE 1584 explicitly use the fault X/R in arc flash analysis.

So True - yet people blindly believe having a computer program makes them an expert. Sometimes good software makes inexperienced people dangerous.

Yes but that is why experienced people can spot when something does not seem logical or reasonable. It is not the first time I have had to inform someone that the arc flash study they had done is not valid. This is the third time I have seen these poor studies by this one company.

The X/R refers to the source impedance, not the fault impedance (mostly resistive). Much less real prime mover power is needed to push the 40 kA through the mostly reactive source impedance.

I believe the discussion was about fault current and fault X/R, or the current and X/R at the point of fault. The source X/R is normally close to 0. The fault X/R factors in all the impedance (active and reactive) in the current path starting with the source all along to the point of fault. I don't see how you get more than 40kA fault current from 40kA active source. The picture below shows that current through an impedance connected between points A and B will be less than the source current:



Would you mind sharing calculations proving what you say?

Example 1: An inductor stores energy in its magnetic field. Energy is available in the magnetic field within the coils. Generally speaking, this doesn't last long. Transformer transients are significant but short lived.
Example 2: A motor has stored kinetic energy (inertia, Wk^2) which will contribute energy during a fault. Although the energy source is actually mechanical in nature, it can be interpreted electrically. This lasts a lot longer in general but the power (short circuit current) is often not very significant.

The end result is that for a few cycles, we will be putting more energy (kA) into the system than will flow from the source alone. These are transient (and subtransient) effects that rapidly decay but nevertheless for very short interrupting times they are a significant source of energy.

Otherwise, your interpretation is extremely valid.

I can't tell you how many times in particular with high torque/power hydraulic units someone fails to recognize that if I have say a 10 HP hydraulic pump, I can't get 20 HP out of the system except with stored energy sources (accumulators). And the analogy is the same here...the stored energy source contributes to the total power available until the stored energy is depleted.