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Incident Energy Study
Christian Brothers UniversityMAESC CONFERENCE - 2005
Jermichael Beaver, Bruce Luong, David Temple, and John Ventura
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.
What is Arc Flash? An arc flash is
basically a short circuit current through the air.
Temperatures >5000o F
At least 5-10 people are injured every day e.g. facial burns, brain damage, hearing loss.
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
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
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
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.
Transformers
Main transformer – Y-Y 2000KVA, 23KV/480V, Z=5.8%
16 sub-transformers - -Y, 480V/208V, Z = 5%
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
Circuit Breakers
Manufacturer – SquareD Breaker name – Px-2500 Opening time – 5 cycles Arc gap – 2 inches
Load 1 Load 2
Load 3
23KV/480V
2000 KVA
Z = 5.8%
30KVA
480V/208V
Sample Schematics
2600 / 3
700 / 3600 / 3600 / 3
Electric Panel
Fault #1
Fault #2
Isc=41.4 KA
Isc=1.67 KA
Determine 3-phase short circuit currents
Per unit analysis used in calculations ISC = short circuit current
Isc = 41.455 KA at main panel
Sample Calculations Cont.
Calculate short circuit current on secondary side of transformer
Per unit analysis used to determine theshort circuit current on the secondary transformer.
Calculate Incident Energy
Calories/cm^2=35.616
Where;lg is the log10Ia is the arcing current (kA)
K is –0.097 for box configurationsIbf is the bolted fault current for three-phase faults (kA)
V is the bus voltage (kV)G is the gap between conductors (mm)
)(lg00304.)(lg5588.000526.0966.)lg(662.)lg( bfbfbfa IGIVGVIKI Eqn (1):
Incident Energy Calculations Cont.
)45.41(lg00304.)45.41)(lg4805588(.)25000526.()4800966(.)45.41lg(662.097.lg Ia
Ia = 22.08 kA
lg En = K1 + K2 + 1.081 lg Ia + .0011G
where,En is the incident energy (J/cm2) normalized for time and distanceK1 is –0.555 for box configurations
K2 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.
334425.1lg aI
Eqn (2):
Eqn (3):
Incident Energy Calculations Cont.
The normalized energy must now be converted to J/cm2.
x
x
nf D
tECEEqn
610
2.184.4:)4(
where,E is the incident energy (J/cm2)Cf is the calculation factor (1.5 voltages at or below 1kV)
En is the normalized energy
t is the arcing time in secondsD 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)
Incident Energy Calculations Cont
2641.1
641.1
96.12460
610
2.
04.)49.6)(5.1)(184.4(:)5(
cmJEEqn
E = 12.96 J/cm2
Converting to cal/cm2 we haveE = 3.095 cal/cm2
Fault to Fault Calculations
real
realtrans
transwire
LLLN
aawire
Ia
VZ
ZZ
VI
L
n
xrZ
2sec
22
:)3(
:)2(
5280:)1(
L = distance from transformer in feet
ra = Resistance/mile
xa= Inductance/mile
n = # of conductors/phase
VLL = Line-Line Voltage
Zwire = wire impedance
Zsectrans = transformer impedance
ILN = Fault Current
Fault to Fault Calculations
KAI
I
LN
LN
tot
tot
tot
wire
wire
m
V
mZ
mmZ
ZmZ
mZ
Z
mZ
17.22652.21
480:)2(
652.21
57.1108.10
08.10
08.10'5280
'360
3
432.1006.:)1(
57.11
sec
22
sec
Current Progress
Calculate short circuit currents at all 20 panels.
Calculate corresponding energy levels.
Determine protective rating of clothing.
Possible Design Solution:
Modify electrical components of the facility i.e. decreasing bus voltage and decreasing arc duration.
Sample Calculations Cont.
Calculate short circuit current on primary side of transformer
Per unit analysis used
058.;2;23 . upbase ZSV MVAKVbase
.123
23
2MVAS23KV;V
...
realreal
upupKV
KV
Vbase
VrealV
Sample Calculations Cont-2
AI
AAIII
AKV
MVA
V
SI
Z
VI
cs
upbasereal
base
basebase
upup
upup
52.865
52.8652.5024.17.
2.50323
2
3
24.17058.
1
.
.
....
....
Short Circuit Current of Secondary
kAkAIII
kAMVA
V
SI
baseupreal
base
basebase
45.41406.224.17
406.23480
2
3..
sec