Hi:
Would you please help to get the expected value for E & En for 480V in-box switchgear? We were trying to use the IEEE 1584 formula
Log En = K1 + K2 + 1.081 * Log Ia + 0.0011 * G AND
E = 4.184 * Cf * En * (t/0.2) * (610^x/D^x).

I (bolted fault) = 40kA.

If we got Log En = 515, would that be correct? and what is En?

Thanks,
abbhas2009

Good to see somebody using the calculations long hand (manually). En is the normalized Incident Energy for an arc duration of 0.2 seconds and for a distance of 610mm. The second equation that you show is to adjust this normalized figure to the actual duration and the actual distance to the worker. I'm not sure where your figure of 515 comes from for Log En, seems very big to me. I suspect that you've not considered that the K factors are negative numbers or perhaps you've misplaced a decimal point somewhere. If you log onto the link below, Jim Phillips provides really easy step by step work sheets for the calculation of arcing current, incident energy and flash protection boundaries. (They are free plus lots of other very useful material as well)

http://www.brainfiller.com/newsletterArchive.php

What are the values of E and En?

Hi Mike,

Thank you, we had considered the minus, however, the voltage was incorrect (480 was corrected to 0.48). The new result for Log En, En and E is 0.77, 5.93 and 3.72 respectively.

When we changed the bolted fault current to 106kA, the results got higher a little up to 1.15, 14.27 and 8.98.

E: in J/cm^2

Have a nice day,
abbhas2009

You haven't shown the calculation of Ia (arcing current) from Ibf (bolted fault current) using IEEE-1584 Equation (1).

I can't get your numbers using either the arcing current or the belted fault current.

Values for En and E

Hi jghrist;

It is the same equation for IEEE 1584. We believe that the values are coming reasonable.

Regards,
abbhas2009