J. C. Das - Arc Flash Hazard Analysis and Mitigation

5 Calculate the incident energy, arc flash boundary, and the PPE level.

NFPA table 130.7(C)(9)(a) [17] relates the hazard risk category and PPE with respect to maintenance tasks to be carried out on energized equipment within arc flash boundary For example, if a worker is inserting or removing a starter “bucket” from MCC, the risk category is 3, but if he is just reading a panel meter while operating a meter switch, the risk category is zero. This table is applicable for specific three-phase short-circuit currents for specific durations as specified in the many foot-notes to the table. The IEEE equations are based upon an arc in an enclosure, with one side open, and it seems logical to relate the hazard level with respect to task activity; the following observations apply:

All commercially available computer programs calculate the arc flash hazard for a bus fault. It is opined that for safety and conservatism, HRC for a bus fault should be considered. For example, the HRC for a fault downstream of a low voltage or medium voltage MCC (a fault on the load side of the motor starter) can be zero, while for a fault on the bus, cleared by an upstream device, it can be 3 or 4. MCC as a whole is labeled for the bus fault—a fault on the bus can shatter the isolation partition in the MCC on arc blast pressure, unless the equipment is arc resistant.

It will be practically difficult to provide multiple arc flash hazard labels on the same equipment, and lay down procedures for various personal protective outfits based upon the maintenance task to be performed.

A rigorous calculation procedure for task-based HRC is not available, though guidelines in this table can be applied.

Multiple tasks are normally performed in one single instance.

The recommendations of the table can be misapplied (see Section 3.13).

The table is judgmental; there are no mathematical calculations.

The calculations, examples, and study cases in the book consider a bus fault, which gives maximum HRC on an equipment and ignore the nature of the task activity.

This example is for calculations as per 2002 Guide. It is included here for reference. The results are compared with 2018 methodology in Chapter 3.

Calculate the incident energy using IEEE equations for a 30-kA three-phase bolted fault current in a 13.8-kV metal-clad switchgear, resistance grounded. The arcing time is 30 cycles for the arc fault current through the protective device.

Using IEEE Equation (1.11), calculate arcing current:

This gives I a= 28.578 kA. Note that the arc flash time is calculated based upon the arcing fault current through the protective device and not the bolted fault current. A further calculation for arc flash time is required at 0.85 I a. Let us assume for this example that 30 cycles is applicable even for reduced 0.85 I a. Then, calculate normalized incident energy from Equation (1.12):

Here, K 1= −0.555 (arc in a box), K 2= −0.113, resistance grounded system, and G = 153 mm ( Table 1.5). This gives log E n= 1.074. Calculate incident energy in J/cm 2using Equation (1.13):

Here, t is the arcing time = 0.5 second, the distance x is given in Table 1.7, and for 15 kV, it is 0.973, and D , the working distance from Table 1.6, is 910 mm.

This gives E = 84.4 J/cm 2= 20.08 cal/cm 2.

Calculate the arc flash boundary (see Equation 1.15):

Here, C f= 1, E nas calculated before is 11.858 J/cm 2, E B= 5 J/cm 2by definition. This gives D B= 16,541 mm = 54.267 ft. This is rather a large distance at which a worker can get threshold of second-degree burns.

Repeat Example 1.1, but for arc in the open air. For high voltage applications, no distinction is made between open and box construction for calculation of arcing current. Therefore, arcing current = 28.578 kA, same as before.

In Equation (1.12), the factor K 1is −0.792. This gives:

E = 4.184(1)(6.871)(0.5/0.2)(610 0.973/910 0.973) = 48.70 J/cm 2. This is equal to 11.60 cal/cm 2.

Also, arc flash boundary reduces to:

This shows a considerable difference between arc in open air and a box.

Repeat Example 1.2using Lee’s equations.

The incident energy is given by Equation (1.7). Substituting the values:

This gives 126.6 cal/cm 2, and from Equation (1.6), distance in feet for a curable burn is 30.82 ft. There is a vast difference in these calculations: 11.6 cal/cm 2with IEEE method and 126.6 cal/cm 2with Lee’s method; his equations do not consider system grounding, and do not calculate arc flash currents.

Calculate the incident energy and arc flash boundary based upon the IEEE simplified equations for a 480-V power circuit breaker (low voltage power circuit breaker [LVPCB]) of 800 A rating provided with electronic trip programmer, LS ( long time and short time). Also, calculate similarly for a 400-A molded case circuit breaker (MCCB) with thermal magnetic trip. The bolted fault current is 36.8 kA.

800-A LVPCB

The incident energy is given by the equation:

Substituting, incident energy is = 195.04 J/cm 2= 46.80 cal/cm 2

The arc flash boundary is given by the equation:

Substituting

Arc flash boundary = 173 in.

400-A MCCB

The incident energy equation is:

Which gives 1.80 cal/cm 2

The arc flash boundary is:

Substituting, this gives: 20.9 in.