Can someone explain incident energy normalized, what is it and how it is obtained .Utilizing the Incident Energy at Working Distance -Empirically Derived Equation -NFPA 70E-D4.3,
En = incident energy normalized.
E= 4.184* Cf * En (t/0.2) (610/D) ^x

Thanks

What you are seeing is the IEEE 1584-2002 edition empirical equations. NFPA 70E just summarizes these equations and several others in the annex.

You jumped to the end. Need to start at the beginning of D4 and follow the whole thing.

The empirical equations are just curve fitted calculations without regard to the fundamental physics. They are curve fit to a database of about 300 lab tests. It is accurate to within about +/-15%.

First the arcing current is estimated given the system voltage, bolted fault current, and arc gap. You run the equation twice, once for the bolted fault current and again for 85% of that value. Above 1 kV, arcing current is the same as bolted fault current. Below 1 kV, it varies.

Next, you calculate the normalized incident energy given the arcing current and arc gap (again). This estimates incident energy at 610 mm and 0.2 seconds arcing time.

Finally you "denormalize" it by taking into account the actual opening time, distance, and equipment type, which is the equation you stated.

There is nothing magical about separating these. Normalized incident energy can easily be substituted into the equation you gave to have a one step process.

Here is a short article from Electrical Contractor Magazine showing the calculation process using the Normalized Incident Energy and correcting it to specific conditions - Time, Distance, Equipment Type (exponent) Hope it helps.

Incident Energy Calculations