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.

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 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 - -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

Bus Voltages

Main Bus 480VLL

Lighting Panels 208VLL

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

Arc Flash Suits

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.

QUESTIONS?

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

01158.9.47

57.2657.26sec

57.26058.16.458..

167.4582.50

23

45.41406.224.17

406.23480

2

3

22

..

sec

aZ

ZbaseuZpZrealA

kV

Ibase

VbaseZ

kAkAIII

kAMVA

V

SI

base

baseupreal

base

basebase