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Incident Energy Study
Christian Brothers University
MAESC 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.
Arc Flash Test – 20kA on a 480V bus
1.
2.
3.
4.
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



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
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.
Site Specifications




Transformers
Conductors
Circuit Breakers
Bus Voltages
Transformers


Main transformer – Y-Y 2000KVA,
23KV/480V, Z=5.8%
16 sub-transformers - D-Y, 480V/208V,
Z = 5%
Conductors



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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
Bus Voltages
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Main Bus 480VLL
Lighting Panels 208VLL
Sample Schematics
23KV/480V
2000 KVA
Fault #1
Isc=41.4
KA
Z = 5.8%
Electric Panel
600 / 3
Load 1
2600 / 3
600 / 3
Load 2
700 / 3
30KVA
Isc=1.67
KA
480V/208V
Load 3
Fault #2
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 the
short circuit current on the secondary
transformer.
Calculate Incident Energy
Eqn (1):
l
g
(
I
)

K

.
6
6
2
l
g
(
I
)

.
0
9
6
6
V

.
0
0
0
5
2
6
G

.
5
5
8
8
V
(
l
g
I
)

.
0
0
3
0
4
G
(
l
g
I
)
a
b
f
b
f
b
f
Where;
lg is the log10
Ia is the arcing current (kA)
K is –0.097 for box configurations
Ibf is the bolted fault current for three-phase faults (kA)
V is the bus voltage (kV)
G is the gap between conductors (mm)
Calories/cm^2=35.616
Incident Energy Calculations Cont.
Eqn (2):
l
g
I
a


.
0
9
7

.
6
6
2
l
g
(
4
1
.
4
5
)

(
.
0
9
6
6

4
8
0
)

(
.
0
0
0
5
2
6

2
5
)

(
.
5
5
8
8

4
8
0
)
(
l
g
4
1
.
4
5
)

.
0
0
3
0
4
(
l
g
4
1
.
4
5
)
lgIa1.3425
Ia = 22.08 kA
Eqn (3):
lg En = K1 + K2 + 1.081 lg Ia + .0011G
where,
En is the incident energy (J/cm2) normalized for time and distance
K1 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.
Incident Energy Calculations Cont.
The normalized energy must now be converted to J/cm2.
t 610 x
Eqn(4) : E  4.184C f En 
.2 D x
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 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)
Incident Energy Calculations Cont
1.641

 .04  610

  12.96 J
Eqn(5) : E  (4.184)(1.5)(6.49)

2
1.641 
cm
.
2
460



E = 12.96 J/cm2
Converting to cal/cm2 we have
E = 3.095 cal/cm2
Arc Flash Suits
Fault to Fault Calculations
ra  xa
L
(1) : Zwire 

n
5280
2
2
VLL
(2) : ILN 
Zwire  Ztrans
Vreal
(3) : Z sec trans  2
a Ireal
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
Z sec  11.57 m
.1006 2  .432 2 360'
(1) : Zwire 

3
5280'
Zwire  10.08m
Ztot  10.08m  Z sec
Ztot  10.08m  11.57 m
Ztot  21.652m
480V
( 2) : ILN 
21.652m
ILN  22.17 KA
Current Progress
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Calculate short circuit currents at all 20
panels.
Calculate corresponding energy levels.
Determine protective rating of clothing.
QUESTIONS
?
Possible Design Solution:

Modify electrical components of the
facility i.e. decreasing bus voltage and
decreasing arc duration.
Sample Calculations Cont.
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Calculate short circuit current on
primary side of transformer
Per unit analysis used
V  23KV ; Sbase  2MVA; Zp.u  .058
base
Vreal  23KV; Sreal  2MVA
Vreal
23KV
Vp.u . 

 1p.u.
Vbase
23KV
Sample Calculations Cont-2
Vp.u .
1
Ip.u . 

 17.24 p.u .
Zp.u . .058
Sbase
2 MVA
Ibase 

 50.2 A
Vbase 3 23KV 3
Ireal  Ibase  Ip.u.  17.24  50.2 A  865.52 A
Is.c  865.52 A
Short Circuit Current of Secondary
Sbase
2MVA
Ibase 

 2.406 kA
Vbase sec 3 480 3
Ireal  Ip.u.  Ibase  17.24  2.406kA  41.45kA
Sbase
2 MVA
Ibase 

 2.406 kA
Vbase sec 3 480 3
Ireal  Ip.u .  Ibase  17.24  2.406kA  41.45kA
Vbase 23kV
Zbase 

 458.167
Ibase 50.2 A
Zreal  Zp.u.  Zbase  458.16  .058  26.57
26.57 26.57
Z sec 

 .01158
2
2
a
47.9