Arc Flash PPE by the Tables

[david.mullins]

We have a system that runs from a 480 Hi-Wye Generator. I've gone through the PPE determination using the tables in 70E and it is pretty straight forward. My question is, being as OSHA and Cal OSHA require a PE to sign off on an arc flash analysis, is this also required when using the table method?

 

[jim dungar]

How did you determine the amount of fault current and the arc clearing time of the OCPD? Both are required in order to use the tables.
OSHA does not require a PE to perform the analysis, I do not know about California.

 

[david.mullins]

The fault current I got from the generator mfr who provides a very thorough datasheet specifically for this purpose. It is Full Load Current 421A, Steady State S.C. Current, 1263A, and max 3 ph. Symmetrical Short Circuit Current, 4048A. The generator has a GE spectra SF250 Frame 3ph breaker and our panel has a Schneider PowerPact L frame breaker. Both have a trip time of about 0.025 s. for overcurrent in that range.

 

[david.mullins]

Make that 0.015s on the Schneider.

 

[jim dungar]

Have you taken the impedance of the conductors, between the generator and your panel, into account?
The main device in the panel is usually ignored as it cannot protect itself.

 

[david.mullins]

Yes, but I do not see that this is necessary for the table method. That's the point of the method: all of the calculations don't need to be done if your situation is within the range of a any table criteria.

That said, it's Gexol 4c 1/0, 0.0234 Ohms in our 100' cable. All the Xn reactance values for this generator are on the order of 0.06 to 1.7 ohms. In a synchronous, bolted fault, 480V / 4048A = 0.119 Ohms. 480V / (0.119 + 0.0234) Ohms = 3370 Amps, which is still enough to trip the breakers instantly.

And the main breaker in the panel SCCA rating is 100kA so it can easily handle that.

 

[mayanees]

We have a system that runs from a 480 Hi-Wye Generator. I've gone through the PPE determination using the tables in 70E and it is pretty straight forward. My question is, being as OSHA and Cal OSHA require a PE to sign off on an arc flash analysis, is this also required when using the table method?

If Cal OSHA requires a PE for the incident energy (IE) calculations then it's likely required for the PPE Categories method, but I don't know that as my world is DE/MD/VA.
I can say that your PPE calc is probably conservative and I suspect your coming up with Category 2 which is good for 8 calories.
But I also don't agree with your numbers with the generator fla equal to 421 amps and your gen breaker is a 250.
Regardless, if I look at an SKM model where I have a 300 kW, 480V genset with a 400-amp fla, my IE at the generator is 4.24 cals and 0.82 calories after 80 feet of 500-amp conductor.

 

[jim dungar]

Yes, but I do not see that this is necessary for the table method. That's the point of the method: all of the calculations don't need to be done if your situation is within the range of a any table criteria.

This is a problem with the tables. The explanatory text provides criteria for the clearing time of the protective device. You cannot determine the clearing time without knowing the current at the point of the fault. For all intents the Table Method only saves the incident energy calculation.
My experience has been that small and medium size generators usually have long arc fault clearing times due to their relatively low fault current versus device clearing time.

 

[david.mullins]

But I also don't agree with your numbers with the generator fla equal to 421 amps and your gen breaker is a 250.

Thank you, Mayanees. The generator has two 250A breakers and we have one panel running off of each breaker. This analysis is for the panels, not the generator enclosure. No work will be performed in there when it's running.

Very interesting that 80 feet of 500-amp conductor reduces the IE by 80%. Can you share how that was calculated?

 

[paulengr]

Yes, but I do not see that this is necessary for the table method. That's the point of the method: all of the calculations don't need to be done if your situation is within the range of a any table criteria.

That said, it's Gexol 4c 1/0, 0.0234 Ohms in our 100' cable. All the Xn reactance values for this generator are on the order of 0.06 to 1.7 ohms. In a synchronous, bolted fault, 480V / 4048A = 0.119 Ohms. 480V / (0.119 + 0.0234) Ohms = 3370 Amps, which is still enough to trip the breakers instantly.

And the main breaker in the panel SCCA rating is 100kA so it can easily handle that.

The fallacy of the table method is it lists assumptions about opening time and maximum available short circuit current. You can’t use the tables unless you verify this data is true which means you have to do 95% of the work of the engineering method. Since you have the same data, you can just do the final step, plug the data into IEEE 1584 to calculate incident energy.

There are some other tables in development on the academic shed that promise to work better.

 

[paulengr]

This is a problem with the tables. The explanatory text provides criteria for the clearing time of the protective device. You cannot determine the clearing time without knowing the current at the point of the fault. For all intents the Table Method only saves the incident energy calculation.
My experience has been that small and medium size generators usually have long arc fault clearing times due to their relatively low fault current versus device clearing time.

A fallacy with arc flash is that lower current means lower arc flash. If the fault current falls into the “thermal” or inverse time region as the fault current decreases the opening time increases at a faster rate than the arc power is decreasing, resulting in an increase in arc energy. So lower fault current yields increased incident energy contrary to what you would expect to happen. This is a big problem with generators compared to utility power.

 

[mayanees]

Thank you, Mayanees. The generator has two 250A breakers and we have one panel running off of each breaker. This analysis is for the panels, not the generator enclosure. No work will be performed in there when it's running.

Very interesting that 80 feet of 500-amp conductor reduces the IE by 80%. Can you share how that was calculated?

Note from the spreadsheet that 1640 Arcing Fault amps takes 0.08 seconds to open the breaker. I did an example with 1400 feet of cable and the Incident Energy (IE) was 3.0 calories and it timed out at 2 seconds.
There's 3.7 cals IE at the generator, 0.2 cals midway, and 0.3 cals at the 80 foot mark.

 

[jim dungar]

Note from the spreadsheet that 1640 Arcing Fault amps takes 0.08 seconds to open the breaker. I did an example with 1400 feet of cable and the Incident Energy (IE) was 3.0 calories and it timed out at 2 seconds.
There's 3.7 cals IE at the generator, 0.2 cals midway, and 0.3 cals at the 80 foot mark.

But the OP said he was using (2) 250A devices and conductors.

 

[mayanees]

But the OP said he was using (2) 250A devices and conductors.

I modelled a typical 300 kW genset with a 500-amp breaker., and not his system.

 

[paulengr]

But the OP said he was using (2) 250A devices and conductors.

It doesn’t change anything. Look at the TCC. The current is in the instantaneous range. If you use a smaller breaker it’s going to trip at the same speed.

 

[jim dungar]

It doesn’t change anything. Look at the TCC. The current is in the instantaneous range. If you use a smaller breaker it’s going to trip at the same speed.

What current, the one through 500A conductors or the one through 250A conductors?

 

[mayanees]

But the OP said he was using (2) 250A devices and conductors.

I thought back to this and was going to post that if the OP clarifies the info I'll plot it. But I see that the OP gave everything needed! Note that I modified the typical generator in SKM to match the characteristics presented, like the fla and the steady-state fault current.
A picture of the situation follows. Note from the TCC that an INST setting of max allows the arcing fault current to get hung up in the characteristic curve of the 250-amp thermal magnetic breaker. If the INST were set any less the IE comes out to 0.16 calories. Otherwise, with the mag adjustment at max the IE is 1.2 calories.