IEEE 1584-2018 Arc Flash Calculator

⚠️ FOR EDUCATIONAL PURPOSES ONLY
Not intended for actual arc flash hazard analysis. Consult a qualified engineer for real-world applications.

Incident Energy Analysis with Time-Overcurrent Coordination

System Parameters
Real-time calculation active
kA
480 V (fixed)
inches
mm
seconds
VCBB Vertical Cond. in Box Below
Overcurrent Protection
A
A
cycles
*3 cycles is the assumption for instantaneous element delay
Trip Point Adjustment
Using curve intersection with arcing current
Time-Current Characteristic Curve
Phase TOC Curve (51/50)
Arcing Current
Operating Point (draggable)
Incident Energy Curves
Understanding Arc Flash Energy

Why Available Fault Current Is Not Always Indicative of Arc Flash Energy

This is one of the most common misunderstandings I see when people first start looking at arc flash studies. They will look at a one line. They will see a bus with 50 kA available. They assume it must be the "worst" bus on the site.

Sometimes it is. Often it is not.

The reason is simple. Arc flash energy is not a short circuit problem. It is an energy over time problem. The energy delivered by an arc flash event is proportional to current squared (I²) times time (t). Often, at lower voltages, time overcurrent curves are used to clear faults. These time overcurrent curves are inverse, to varying degrees. Some slope down hard and some have shallow slopes. In coordinating these curves with each other, and with the equipment damage curves they are trying to stay under to protect, different curve shapes are selected with different time dial settings.

This is where things get a little interesting. Time overcurrent plots are log log. This means for P = T×I², you get a downward sloping curve at fixed energy. Your cable damage curves all have this same slope because they are thermally limited by fault current energy. Now, your PPE energy boundaries for the different levels of PPE, 1 = 4 cal/cm², 2 = 8 cal/cm², 3 = 25 cal/cm², and 4 = 40 cal/cm², are all energies per unit area and will have downward sloping curves with the same slope.

The exception to this is that studies often include a 2 second escape time, which is the amount of time an employee needs to get out of the arc flash boundary. This is not a rule or part of the standard. It is up to the engineer's judgement to assess the situation. Without this, arc flash energy calculations can get unrealistically large. You could have a high impedance arc that never clears, which could provide a ridiculous amount of calculated energy, but was not as dangerous as the numbers might imply in practice.

In the toolkit linked above, I created a simple test case where someone can adjust the various parameters and see how to push a PPE level into a lower region with different overcurrent curves, escape times, breaker clearing times, and fault current levels. More fault current, with respect to arc flash calculations, is not always a bad thing. This is a calculation involving energy and time. Squeezing either the current or the time helps.

This is not the only way arc flash energies can be mitigated. Sometimes arc flash light sensors are used. Differential schemes can be put into place that operate instantly. Crowbars are sometimes dropped. Zone interlocking and directional schemes can be used to get away from time coordination. The approach above, adjusting overcurrent curves, is the first option if it is a solution, since it can be done by adjusting relay settings and does not involve additional equipment or added complications.

Arc Flash Results
Educational Use Only
Incident Energy (E)
0.0 cal/cm²
At working distance of 18"
Arcing Current (Iarc)
0.0 kA
Reduced: 0.0 kA (85%)
Clearing Time (t)
0.0 sec
0.0 cycles @ 60Hz
Personnel Exposure Time
2.0 sec
Used for energy calculation
Arc Flash Boundary
0.0 ft
Distance for 1.2 cal/cm²
Operating Point
0.0× pickup
Current: 0 A
PPE Category (NFPA 70E)
1
≤4
2
≤8
3
≤25
4
≤40
Calculation Breakdown
lg(Iarc)
lg(E) J/cm²
CF (enclosure)
VarCf (variation)
Iarc_min factor
IEEE 1584-2018 Model
Iarc = 10^(k₁+k₂lg(Ibf)+k₃lg(G)) × f(Ibf)
E = (12.552/50)×T×10^(polynomial)