Arc Flash Hazard Analysis and Mitigation. J. C. Das
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Название: Arc Flash Hazard Analysis and Mitigation
Автор: J. C. Das
Издательство: John Wiley & Sons Limited
Жанр: Техническая литература
isbn: 9781119709794
isbn:
This gives Ia = 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 0.85Ia. Let us assume for this example that 30 cycles is applicable even for reduced 0.85Ia. Then, calculate normalized incident energy from Equation (1.12):
Here, K1 = −0.555 (arc in a box), K2 = −0.113, resistance grounded system, and G = 153 mm (Table 1.5). This gives log En = 1.074. Calculate incident energy in J/cm2 using 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/cm2 = 20.08 cal/cm2.
Calculate the arc flash boundary (see Equation 1.15):
Here, Cf = 1, En as calculated before is 11.858 J/cm2, EB = 5 J/cm2 by definition. This gives DB = 16,541 mm = 54.267 ft. This is rather a large distance at which a worker can get threshold of second-degree burns.
Example 1.2
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 K1 is −0.792. This gives:
E = 4.184(1)(6.871)(0.5/0.2)(6100.973/9100.973) = 48.70 J/cm2. This is equal to 11.60 cal/cm2.
Also, arc flash boundary reduces to:
This shows a considerable difference between arc in open air and a box.
Example 1.3
Repeat Example 1.2 using Lee’s equations.
The incident energy is given by Equation (1.7). Substituting the values:
This gives 126.6 cal/cm2, 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/cm2 with IEEE method and 126.6 cal/cm2 with Lee’s method; his equations do not consider system grounding, and do not calculate arc flash currents.
Example 1.4
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/cm2 = 46.80 cal/cm2
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/cm2
The arc flash boundary is:
Substituting, this gives: 20.9 in.
Example 1.5
Repeat Example 1.4 with rigorous calculations.
Table 1.8 equations do not specify the gap length and also the system grounding. As the equations are for the maximum incident energy, it can be assumed that the system is ungrounded. Also for Table 1.8 equations, a working distance of 610 mm (2 ft) is specified in 1584 Guide. Thus, there are qualifications and compromises in using these equations.
To start the rigorous calculations, a time–current plot of settings on 800-A LVPCB and 400-A MCCB is shown in Figure 1.10. The maximum time on the trip band is used, shown by dots in this figure. This figure is based on circuit breakers of a certain manufacturer. This shows that the LVPCB maximum ST delay band is used, as the approximate equations are based upon this assumption. The opening time of the low voltage circuit breakers is built in the time–current plots supplied by the manufacturer. Thus, we consider the maximum operating times as shown by bold dots in СКАЧАТЬ