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Incident Energy Study Christian Brothers University MAESC CONFERENCE - 2005 Jermichael Beaver, Bruce Luong, David Temple, and John Ventura.

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Presentation on theme: "Incident Energy Study Christian Brothers University MAESC CONFERENCE - 2005 Jermichael Beaver, Bruce Luong, David Temple, and John Ventura."— Presentation transcript:

1 Incident Energy Study Christian Brothers University MAESC CONFERENCE - 2005 Jermichael Beaver, Bruce Luong, David Temple, and John Ventura

2 Background Everyday electricians across North America are injured as a result of arc explosions. Many of these injuries could have been prevented with proper training and adequate protective clothing. To reduce the likelihood of an arc explosion, Buckman Labs has decided to have the amount of incident energy exposure calculated to ensure that they are within safety standard regulations.

3 What is Arc Flash? An arc flash is basically a short circuit current through the air. Temperatures >5000 o F At least 5-10 people are injured every day e.g. facial burns, brain damage, hearing loss.

4 Arc Flash Test – 20kA on a 480V bus 1.2. 3. 4.

5 Objectives: Determine incident energy levels present at 20 service panels within Buckman Laboratories. Establish the rating of protective clothing worn at the plant Propose a corrective design if necessary

6 Calculation Guides NFPA (National Fire and Protection Agency) 70-E ASTM (American Society of Testing and Materials) F-1506 IEEE (Institute of Electronic and Electrical Engineers) 1584 OSHA (Occupational Safety and Health Association) 29CFR1910.269 Oberon

7 Selected Method Type IEEE 1584 (A guide to incident Arc Flash Calculations) – most extensive and comprehensive set of equations regarding arc-flash Based on lab testing on wide range of system conditions

8 Plan of Attack Gather site specifications Determine short circuit currents at the panel locations Obtain circuit breaker parameters e.g. opening times, arc gap Collect data concerning worker distance from panel.

9 Site Specifications Transformers Conductors Circuit Breakers Bus Voltages

10 Transformers Main transformer – Y-Y 2000KVA, 23KV/480V, Z=5.8% 16 sub-transformers -  -Y, 480V/208V, Z = 5%

11 Conductors 3-phase 4-wire system two to six conductors per phase Copper conductors used through out the plant Wire sizes: 300,000 cm to 600,000 cm

12 Circuit Breakers Manufacturer – SquareD Breaker name – Px-2500 Opening time – 5 cycles Arc gap – 2 inches

13 Bus Voltages Main Bus 480V LL Lighting Panels 208V LL

14 Load 1 Load 2 Load 3 23KV/480V 2000 KVA Z = 5.8% 30KVA 480V/208V Sample Schematics 2600 / 3 700 / 3600 / 3 Electric Panel Fault #1 Fault #2 Isc=41.4 KA Isc=1.67 KA

15 Determine 3-phase short circuit currents Per unit analysis used in calculations I SC = short circuit current Isc = 41.455 KA at main panel

16 Sample Calculations Cont. Calculate short circuit current on secondary side of transformer Per unit analysis used to determine the short circuit current on the secondary transformer.

17 Calculate Incident Energy Calories/cm^2=35.616 Where; lg is the log10 I a is the arcing current (kA) K is –0.097 for box configurations I bf is the bolted fault current for three-phase faults (kA) V is the bus voltage (kV) G is the gap between conductors (mm) Eqn (1):

18 Incident Energy Calculations Cont. I a = 22.08 kA lg E n = K 1 + K 2 + 1.081 lg I a +.0011G where, En is the incident energy (J/cm 2 ) normalized for time and distance K 1 is –0.555 for box configurations K 2 is –0.113 for grounded systems G is the gap between conductors (mm) lg En = -.555 + -.113 + 1.081 * lg(22.08) + (.0011)(25) En = 6.49 J. Eqn (2): Eqn (3):

19 Incident Energy Calculations Cont. The normalized energy must now be converted to J/cm 2. where, E is the incident energy (J/cm 2 ) C f is the calculation factor (1.5 voltages at or below 1kV) E n is the normalized energy t is the arcing time in seconds D is the distance from the possible arc point to the person (mm) x is the distance exponent (1.641 for voltages between.208 and 1 kV)

20 Incident Energy Calculations Cont E = 12.96 J/cm 2 Converting to cal/cm 2 we have E = 3.095 cal/cm 2

21 Arc Flash Suits

22 Fault to Fault Calculations L = distance from transformer in feet r a = Resistance/mile x a = Inductance/mile n = # of conductors/phase V LL = Line-Line Voltage Z wire = wire impedance Z sectrans = transformer impedance I LN = Fault Current

23 Fault to Fault Calculations

24 Current Progress Calculate short circuit currents at all 20 panels. Calculate corresponding energy levels. Determine protective rating of clothing.

25 QUESTIONS ?

26 Possible Design Solution: Modify electrical components of the facility i.e. decreasing bus voltage and decreasing arc duration.

27 Sample Calculations Cont. Calculate short circuit current on primary side of transformer Per unit analysis used

28 Sample Calculations Cont-2

29 Short Circuit Current of Secondary

30


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