In the given scenario, if the protecting circuit breaker opened in 1 second, what would be the approximate actual incident energy at 18 inches?

• A. 2.92 cal/cm2
• B. 65.4 cal/cm2
• C. 30 cal/cm2
• D. 0.05 cal/cm2

Answer: (B)

The main idea here is that arc flash incident energy at a given distance depends on how long the arc lasts and how far you are from it. Energy scales with I^2t (the current squared times the duration) and falls off roughly with the square of the distance from the arc. If the protective device takes about 1 second to clear the fault, the arc can be present for that entire second, so the energy delivered to anything at 18 inches remains quite high.

Using standard arc flash energy estimation methods (IEEE 1584/NFPA 70E) for a typical high-current fault with a 1-second clearing time, the incident energy at 18 inches ends up in the ballpark of tens of cal/cm^2, and 65.4 cal/cm^2 is a representative value that matches the given scenario. That’s why this option is the best choice.

The other numbers would imply conditions that aren’t stated: much faster clearing or a greater distance, or a much lower fault current. With a 1-second clearing time at 18 inches, those lower values don’t align with the expected energy level.