Short Circuit Study

(1)

Engineering Dependable Protection

Engineering Dependable Protection - Part I

"A Simple Approach to Short-Circuit Calculations"

T

Ta

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blle

e

o

of

f

C

Co

on

ntte

en

nttss

P

Pa

ag

ge

e

Basic C

Basic Consid

onsiderati

erations of Short-

ons of Short-Circ

Circuit Calc

uit Calculatio

ulations

ns ………

………

………3

…3

- Why

- Why Short-C

Short-Circu

ircuit Ca

it Calcula

lculations

tions ………

………

………3

- Gener

- General Comm

al Comments on Shor

ents on Short-Cir

t-Circuit Ca

cuit Calcula

lculations

tions ………

………

………3

…3

-

- Asym

Asymmetri

metrical

cal Compon

Components

ents ………

………

………3

3 -

- Interr

Interrupting Rating, Interru

upting Rating, Interrupting Capacit

pting Capacity, Short-C

y, Short-Circu

ircuit Currents

it Currents ………

………4

…………4

3

ØØ

Short-C

Short-Circu

ircuit Current Calculati

it Current Calculations, Procedur

ons, Procedures and

es and Method

Methodss ………

………5

-

- Ohmic

Ohmic Method

Method ………

………6

-

- Per

Per-Un

-Unit

it Met

Method

hod …………

………

………1

………11

1 - TRON

- TRON

®®

_Comput

_{Computer Software Pro}

_{er Software Procedur}

_{ceduree ………}

_………

_………16

- Poin

- Point-to-Po

t-to-Point Me

int Method

thod ………

………

………17

……17

- Comparis

- Comparison of

on of Resul

Results

ts ………

………

………19

……19

1

ØØ

Short-Circuit Calculation on 1

ØØ

T

Transfo

ransformer

rmer Syste

System, Pr

m, Procedu

ocedures a

res and Me

nd Methods

thods ………20

………20

- Per

- Per-Unit M

-Unit Method - Li

ethod - Line-to-

ne-to-Line F

Line Faults

aults ………

………

………21

…………21

- Per-Un

- Per-Unit Method - Line-to-

it Method - Line-to-Neutra

Neutral Faults

l Faults ………

………

………22

…………22

- Point-to-Po

- Point-to-Point Method - Line-to-Li

int Method - Line-to-Line Faults

ne Faults ………

………

………23

- Point-to-Point

- Point-to-Point Method -

Method - Line-to-Neutral F

Line-to-Neutral Faults

aults ………

………24

……24

- Compariso

- Comparison of

n of Resu

Results

lts ………

………

………24

…24

Data Section………25

- T

- Table 1

able 1 - T

- Transformer Impedance

ransformer Impedance Data

Data ………

………25

- T

- Table 2

able 2 - Current

- Current Trans

Transformer Reacta

former Reactance Data

nce Data ………

………25

- T

- Table 3 - D

able 3 - Disco

isconnecti

nnecting Switc

ng Switch Reac

h Reactance D

tance Data

ata ………

………

………25

…25

- Tab

- Table 4 - Circuit Bre

le 4 - Circuit Breaker Re

aker Reactan

actance Data

ce Data ………

………

………26

…………26

- Ta

- Table 5 - Insul

ble 5 - Insulated Con

ated Conductors I

ductors Impeda

mpedance Dat

nce Dataa ………

………

………26

…26

- Table

- Table 6 - “C” Values for PTP

6 - “C” Values for PTP Method Data

Method Data ………

………27

- T

- Table 7 - Buswa

able 7 - Busway Impedance Da

y Impedance Data

ta ………

………28

- T

- Table 8

able 8 - Asymmetric

- Asymmetrical Fac

al Factors

tors ………

………29

Selec

Selective

tive Coordi

Coordination

nation -

- EDP

EDP II

II ………

………

………29

29 Selec

(2)

Part 1

Part 1 A Simple Approach

A Simple Approach

T

To

o

Short Circuit

Calculations

(2004-1)

Protection

For An

Electrical

Distribution

System

(3)

Basic Considerations of Short-Circuit Calculations

Sources of short circuit current that are

Sources of short circuit current that are normally taken

normally taken

under consideration include:

-

- Utility

Utility Generation

Generation

-

- Local

Local Generation

Generation

-

- Synchronous

Synchronous Motors

Motors and

and

-

- Induction

Induction Motors

Motors

Capacitor discharge currents can normally be

neglected due to

neglected due to their short time

their short time duration.

duration. Certain IEEE

Certain IEEE

(Institu

(Institute of

te of Electrical and Electronic Engineers

Electrical and Electronic Engineers)

) publications

publications

detail how

detail how to calculate these

to calculate these currents if they

currents if they are substantial.

are substantial.

Asymmetrical Components

Short circuit current normally takes on

Short circuit current normally takes on an asymmetrical

an asymmetrical

characteristic during the first few cycles of duration. That is,

it is offset abou

it is offset about the zero axis, as indi

t the zero axis, as indicated

cated in Figure 1.

in Figure 1.

Figure 1

In Figure 2, note that the total short circuit current I

aa

is

the summation of two components - the

the summation of two components - the symmetrical RMS

symmetrical RMS

current I

SS

, and the DC component, I

DCDC

. The DC component

is a function of the stored energy within the system at the

initiation of the short circuit. It decays to zero after a few

cycles due to I

22

_{R losses in the system, at which point}

_{R losses in the system, at which point the}

_the

short circuit current is symmetrical about the zero axis. The

RMS value of the symmetrical co

RMS value of the symmetrical co mponent may be deter-

mponent may be

deter-mined using Ohm`s L

mined using Ohm`s L aw. T

aw. To determine

o determine the asymmetrical

the asymmetrical

component, it is necessar

component, it is necessar y to know the X/R ratio of the

y to know the X/R ratio of the

system. To obtain the X/R ratio, the total resistance and total

reactance of the circuit to the point of fault must be

determined. Maximum thermal and mechanical stress on

the equipment occurs during these first few cycles. It is

important to concentrate on what happens dur

important to concentrate on what happens dur ing the first

ing the first

half cycle after the initiation of

half cycle after the initiation of the fault.

the fault.

Why Short-Circuit Calculations

Several sections of the National Electrical Code relate

to

to proper overcurrent

proper overcurrent protection.

protection. Safe

Safe and

and reliable

reliable

application of overcurrent protective devices based on

these sections mandate that a short

these sections mandate that a short circuit study and a

circuit study and a

selective coordination study be conducted.

These sections include, among others:

110-9

110-9 Interrupting

Interrupting Rating

Rating

110-10

110-10 Component

Component Protection

Protection

230-65

230-65 Service

Service Entrance Eq

Entrance Equipment

uipment

240-1

240-1 Conductor

Conductor Protection

Protection

250-95

250-95 Equipment Gr

Equipment Grounding Conductor P

ounding Conductor Protection

rotection

517-17

517-17 Health Car

Health Care Facilities - Selec

e Facilities - Selective Coordinat

tive Coordination

ion

Compliance with these code sections can best be

accomplished by conducting a shor

accomplished by conducting a shor t circuit study and a

t circuit study and a

selective coordination study.

The protection for an electrical system should not only

be safe under all service conditions but, to insure continuity

of service, it s

of service, it should be selectively coordinated as

hould be selectively coordinated as well.

well. A

A

coordinated system is one where only the faulted circuit is

isolated without disturbing any other part of the system.

Overcurrent protection devices should also provide

short-circuit as well as overload protection for system

circuit as well as overload protection for system

components, such as bus, wire, motor controllers, etc.

T

To obtain reliable,

o obtain reliable, coordinated operation and assure

coordinated operation and assure

that system components are protected from damage, it is

necessary to first calculate the available fault current at

various critical points in the electrical

various critical points in the electrical system.

system.

Once the short-circuit levels are determined, the

engineer can specify proper interrupting rating

require-ments, selectively coordinate the system and provide

ments, selectively coordinate the system and provide

component protection.

General Comments on Short-Circuit Calculations

Short Circuit Calculations should be done at all

Short Circuit Calculations should be done at all critical

critical

points in the system.

These would include:

-

- Service

Service Entrance

Entrance

-

- Panel

Panel Boards

Boards

-

- Motor C

Motor Control

ontrol Centers

Centers

-

- Motor

Motor Starters

Starters

-

- Transfer

Transfer Switches

Switches

-

- Load

Load Centers

Centers

Normally

Normally, short

, short circuit studies involve calculating

circuit studies involve calculating a

a

bolted 3-phase fault

bolted 3-phase fault condition.

condition. This can be

This can be characterized

characterized

as all three phases “bolted” together to create a zero

impedance connection.

impedance connection. This establishes

This establishes a “worst

a “worst case”

case”

condition, that results in maximum thermal and mechanical

stress in the

stress in the system.

system. From this calculation, other types

From this calculation, other types of

of

fault conditions can be obtained.

C C U U R R R R EE N N TT TIME TIME

(4)

T

To accomplish this,

o accomplish this, study Figure 2, and refer to T

study Figure 2, and refer to Table 8.

able 8.

Figure 2 illustrates a worst case waveform that 1 phase

of the 3 phase system will assume during the first few

cycles after the fault initiation.

For this example, assume an RMS

For this example, assume an RMS symmetrical short

symmetrical short

circuit value of 50,000 amperes, at a 15% short circuit

power factor.

power factor. Locate the 15% P

Locate the 15% P.F

.F. in T

. in Table 8.

able 8. Said another

Said another

way

way,

, the X/R short circuit ratio of t

the X/R short circuit ratio of this circuit is 6.5912.

his circuit is 6.5912.

The key portions are:

- Symmetrical RMS Short Circuit Current =

- Symmetrical RMS Short Circuit Current = II

ss

- Instantaneous Peak Current =

- Instantaneous Peak Current = II

pp

- Asymmetrical RMS Short

- Asymmetrical RMS Short Circuit Current

Circuit Current

(worst case single phase) =

(worst case single phase) = II

aa

From Table 8, note the following relationships.

II

ss

= Symmetrical RMS Current

II

pp

=

= II

ss

x

x M

M

pp

(Column 3)

II

aa

=

= II

ss

x

x M

M

mm

(Column 4)

For this example, Figure 2,

II

ss

= 50,000 Amperes RMS

= 50,000 Amperes RMS Symmetrical

Symmetrical

II

pp

= 50,000 x 2.309 ( Column 3)

= 115,450 Amperes

II

aa

= 50,000 x 1.330 (Column 4)

= 66,500

= 66,500 Amperes RMS Asymmetrical

Amperes RMS Asymmetrical

With this basic understanding, proceed in the systems

analysis.

Basic Considerations of Short-Circuit Calculations

II

nterrupting Rating, Interrupting Capacity and Short-Circuit Currents

Interrupting Rating can

Interrupting Rating can be defined as

be defined as “the maximum

“the maximum

short-circuit current that a protective device can safely

clear

clear, under

, under specified test conditions.”

specified test conditions.”

Interrupting Capacity can be defined as “the actual

short circuit current that a protective device has been

tested to interrupt.”

The National Electrical Code requires adequate

interrupting ratings in Sections 110-9 and 230-65.

Section 110-9 Interrupting Rating.

Section 110-9 Interrupting Rating. Equipment intended to

Equipment intended to

break current at fault levels shall have an

break current at fault levels shall have an interrupting rating

interrupting rating

sufficient for the system voltage and the current which is

available at the line terminals of

available at the line terminals of the equipment.

the equipment.

Section 230-65. Available Short-Circuit Current.

Section 230-65. Available Short-Circuit Current. Service

Service

Equipment shall be suitable for the short circuit current

available at its supply terminals.

Low voltage fuses have their interrupting rating

expressed in terms of the symmetrical component of

short-circuit current, I

circuit current, I

SS

. They are given an RMS symmetrical

interrupting rating at a

interrupting rating at a specific power factor

specific power factor. This means

. This means

that the fuse can interrupt any asymmetrical current

associated with this rating. Thus only the symmetrical

component of short-circuit current need be considered to

determine the necessary interrupting rating of a low voltage

fuse. For U.L. listed low

fuse. For U.L. listed low voltage fuses, interrupting rating

voltage fuses, interrupting rating

equals its interrupting capacity.

Low voltage molded case circuit breakers also have

their interrupting rating expressed in terms of RMS

symmetrical amperes at a

symmetrical amperes at a specific power factor.

specific power factor. However,

However,

it is necessary to

it is necessary to determine a molded case

determine a molded case circuit breaker’

circuit breaker’s

s

interrupting capacity in order to

interrupting capacity in order to safely apply

safely apply it.

it. The reader

The reader

is directed to Buss bulletin PMCB II for an understanding of

this concept.

II_P_P== 115,450A 115,450A IIaa== 66,500A 66,500A IIss== 50,000A 50,000A

IIaa- Asymmetrical RMS Current- Asymmetrical RMS Current

IIDCDC- DC Component- DC Component

IIss- Symmetrical RMS Component- Symmetrical RMS Component

IIPP- Instantaneous Peak Current- Instantaneous Peak Current

Figure 2

C C U U R R R R EE N N TT IIaa IIDCDC TIME TIME IIss

(5)

Procedures and Methods

3Ø Short-Circuit Current Calculations,

Procedures and Methods

To determine the fault current at any point in the

system, first draw a one-line diagram showing all of

system, first draw a one-line diagram showing all of the

the

sources of short-circuit current feeding into the

sources of short-circuit current feeding into the fault, as well

fault, as well

as the impedances of the

as the impedances of the circuit components.

circuit components.

To begin the study, the system components, including

those of the utility system, are represented as impedances

in the diagram.

The impedance tables given in the Data Section

include three phase and single phase transformers, current

transforme

transforme rs, safety

rs, safety switches, circuit

switches, circuit breakers, cable,

breakers, cable, and

and

busway

busway.

. These tables can be used if information from the

These tables can be used if information from the

manufacturers is not readily available.

It must be understood that short circuit calculations are

performed without current limiting devices in

performed without current limiting devices in the system.

the system.

Calculations are done as though these devices are

replaced with copper bars, to determine the maximum

“available” short circuit

“available” short circuit current.

current. This is

This is necessary to

necessary to

project how the system and the current limiting devices will

perform.

Also, current limiting devices do not operate in

Also, current limiting devices do not operate in series

series

to produce a

to produce a “compounding” current lim

“compounding” current limiting effect.

iting effect. The

The

downstream, or load side, fuse will operate alone under a

short circuit condition if properly coordinated.

System A

3Ø Single Transformer System

M M Available Utility Available Utility S.C. MVA 100,000 S.C. MVA 100,000 1500 KVA Transformer 1500 KVA Transformer 480Y/277V, 480Y/277V, 3.5%Z, 3.45%X, .56%R 3.5%Z, 3.45%X, .56%R IIf.l.f.l.= 1804A= 1804A 25’ - 500kcmil 25’ - 500kcmil 6 Per Phase 6 Per Phase

Service Entrance Conductors Service Entrance Conductors in Steel Conduit in Steel Conduit 2000A Switch 2000A Switch KRP-C-2000SP Fuse KRP-C-2000SP Fuse Main Swb’d. Main Swb’d. 400A Switch 400A Switch LPS-RK-400SP Fuse LPS-RK-400SP Fuse 50’ - 500 kcmil 50’ - 500 kcmil

Feeder Cable in Steel Conduit Feeder Cable in Steel Conduit Fault X Fault X22 MCC No. 1 MCC No. 1 Motor Motor Fault X Fault X11

System B

33 Ø

Ø

Double Transformer System

1000 KVA Transformer, 1000 KVA Transformer, 480/277 Volts 3Ø 480/277 Volts 3Ø 3.45%X, .60%R 3.45%X, .60%R IIf.l.f.l.= 1203A= 1203A Available Utility Available Utility S.C. KVA 500,000 S.C. KVA 500,000 30’ - 500 kcmil 30’ - 500 kcmil 4 Per Phase 4 Per Phase Copper in PVC Conduit

Copper in PVC Conduit _{1600A Switch}_{1600A Switch} KRP-C-1500SP Fuse KRP-C-1500SP Fuse LPS-RK-350SP Fuse LPS-RK-350SP Fuse 400A Switch 400A Switch 225 KVA 225 KVA 208/120 Volts 3Ø 208/120 Volts 3Ø .998%X, .666%R .998%X, .666%R 20’ - 2/0 20’ - 2/0 2 Per Phase 2 Per Phase Copper in PVC Conduit Copper in PVC Conduit

In this example, assume 0% motor load. In this example, assume 0% motor load.

22

T

To begin the

o begin the analysis, consider the following system,

analysis, consider the following system,

supplied by a 1500

supplied by a 1500 KV

KVA, three phase transformer

A, three phase transformer having a

having a

full load current of

full load current of 1804 amperes at 480

1804 amperes at 480 volts.

volts. (See System

(See System

A, below)

A, below) Also, System B

Also, System B, for a double transformation, wil

, for a double transformation, willl

be studied.

To start, obtain the available short-circuit KVA, MVA, or

SCA from the local utility company.

The utility estimates that System A can

The utility estimates that System A can deliver a short-

deliver a

short-circuit of 100,00

circuit of 100,00 0 MVA

0 MVA at the primar

at the primar y of the transforme

y of the transforme r.

r.

System B can deliver a

short-System B can deliver a short- circuit of 500,000 KVA at the

circuit of 500,000 KVA at the

primary of the first transformer

primary of the first transformer. Since the X

. Since the X /R ratio of t

/R ratio of t he

he

utility system is usually quite high, only the reactance need

be considered.

With this available short-circuit information, begin to

make the necessary calculations to determine the fault

current at any point in

current at any point in the electrical system.

the electrical system.

Four basic methods will be presented in this text to

instruct the reader on short

instruct the reader on short circuit calculations.

circuit calculations.

These include :

- the ohmic method

- the per unit method

- the TRON

® ®

_{Computer Software method}

- the point to point method

11

22

Note: The above 1500KVA transformer serves 100% motor Note: The above 1500KVA transformer serves 100% motor load.load.

Fault X Fault X11

Fault X Fault X22

(6)

• •

Ohmic Method

33Ø

Ø Short Circuit Calculations,

Short Circuit Calculations,

Ohmic Method

Most circuit component impedances are given in ohms

except utility and transformer impedances which are found

by the following formulae* (Note that the transformer and

utility ohms are referred to the secondary KV by squaring

the secondary voltage.)

Step 1.

††

_X

_utility_utility_Ω_Ω

₌₌

1000 (KV

secondarysecondary

))

22

S.C. KVA

utilityutility

S

Stteep

p 22..

X

transtransΩΩ

==

(10)(%X**)(KV

secondarysecondary

))

22

KVA

transtrans

R

transtransΩΩ

==

(10)(%R**)(KV

secondarysecondary

))

22

KVA

transtrans

Step 3.

The impedance (in ohms) given for

The impedance (in ohms) given for current

current

transformers, large switches and large circuit breakers is

essentially all X.

S

Stteep

p 44..

X

cable and buscable and busΩΩ..

R

cable and buscable and busΩΩ..

Step 5.

T

Total

otal all

all X

X and all

and all R

R in system to point of fault.

in system to point of fault.

Step 6.

Determine impedance (in ohms) of the

Determine impedance (in ohms) of the system by:

system by:

ZZ

TT

==

√

(R

TT

))

22

+ (X

TT

))

22

Step 7.

Calculate short-circuit symmetrical RMS amperes

at the point of fault.

II

S.CS.C. sym. sym RMSRMS

==

EE

secondasecondaryry line-linline-linee

√

3 (Z

TT

))

Step 8.

Determine the motor

Determine the motor load.

load. Add up the

Add up the full load

full load

motor currents.

motor currents. The full load motor current in

The full load motor current in the system is

the system is

generally a percentage of the transformer full

generally a percentage of the transformer full load current,

load current,

depending upon the types of loads. The generally

accepted procedure assumes 50% motor load when both

motor and lighting loads are considered, such as supplied

by 4 wire, 208Y/120V and 480Y/277V volt 3-phase

systems.)

*For simplicity of calculations all

*For simplicity of calculations all ohmic values are single phase ohmic values are single phase distance one way, later compensated for in distance one way, later compensated for in the three phase short-circuit formula the three phase short-circuit formula by the factor,by the factor,

√

3.3. (See Step 7.)

(See Step 7.)

**UL Listed transformers 25 KVA and larger have a

**UL Listed transformers 25 KVA and larger have a ±±10% impedance tolera10% impedance tolerance. nce. Short circuit amperes can be affeShort circuit amperes can be affected by this tolerance.cted by this tolerance. †Only X is considered in this procedure since utility X/R ratios are usually quite high.

†Only X is considered in this procedure since utility X/R ratios are usually quite high. For more finite details obtain R of utility source.For more finite details obtain R of utility source. •

•AA more exact determinatimore exact determination depends upon the sub-transient reacton depends upon the sub-transient reactance of the motors in question and ance of the motors in question and associated circuit imassociated circuit impedances. pedances. A less conservativeA less conservative method would involve the total motor circuit

method would involve the total motor circuit impedance to a common bus (sometimes referred to as a “zero reactance bus”).impedance to a common bus (sometimes referred to as a “zero reactance bus”). ††Arithmeti

††Arithmetical addition results in conservative values of fault current. cal addition results in conservative values of fault current. More finite values involve vectorial addition of the currents.More finite values involve vectorial addition of the currents.

Step 9.

The symmetrical motor contribution can be

approximated by using an average multiplying factor

associated with the

associated with the motors in the sys

motors in the system.

tem. This factor varies

This factor varies

according to motor design and in this text may be chosen

as 4 times motor full load current for approximate

calculatio

calculatio n purposes.

n purposes. T

To solve

o solve for the

for the symmetrical motor

symmetrical motor

contribution:

••

_II

sym motor contrib

== (4)

(4) xx (I

(I

full load motorfull load motor

))

Step 10.

The total symmetrical short-circuit RMS current is

calculated as:

†† ††

_II

total

total S.C. syS.C. symm RMSRMS

== (I

(I

S.C. sym RMSS.C. sym RMS

)) ++ (I

(I

sym motor contribsym motor contrib

))

Step 11.

Determine X/R ratio of the system to the point of

fault.

X/R

ratioratio

==

X

totaltotal Ω Ω

R

totaltotalΩΩ

Step 12.

The asymmetrical factor corresponding to the X/R

ratio in Step 11 is found in Table 8, Column

ratio in Step 11 is found in Table 8, Column M

M

mm

. This

multiplier will provide the worst case asymmetry occurring

in the first 1/2 cycle. When the average 3-phase multiplier

is desired use column

is desired use column M

M

aa

..

Step 13

.

. Calculate the

Calculate the asymmetrical RMS

asymmetrical RMS short-circu

short-circu it

it

current.

II

S.C. asym RMSS.C. asym RMS

= (I

) x (Asym Factor)

Step 14.

The short-circuit current that the motor load can

contribute is an asymmetrical current usually approximated

as being equal to the

as being equal to the locked rotor current of

locked rotor current of the motor.

the motor.

As a close approximation with a margin of

As a close approximation with a margin of safety use:

safety use:

••

_II

_{asym motor contrib}_{asym motor contrib}

_{= (5)}

_{= (5) xx (I}

_(I

_{full load motor}_{full load motor}

₎₎

Step 15.

The total asymmetrical short-circuit RMS current is

calculated as:

(7)

M

M MM

Ohmic Method –

To Fault X

1

1 – System A

– System A

Available Utility Available Utility S.C. MVA 100,000 S.C. MVA 100,000 1500 KVA Transformer, 1500 KVA Transformer, 480V, 3Ø, 480V, 3Ø, 3.5%Z, 3.4 3.5%Z, 3.45%X5%X , 0.56%R, 0.56%R IIf.f.l.l. transtrans= 1804A= 1804A

25’ - 500 kcmil 25’ - 500 kcmil 6 Per Phase 6 Per Phase Service Entrance Service Entrance

Conductors in Steel Conduit Conductors in Steel Conduit 2000A Switch 2000A Switch KRP-C-2000SP Fuse KRP-C-2000SP Fuse Fault X Fault X11 Motor Contribution Motor Contribution

R

X

X ==

1000(.48)

22

== 0.0000023

0.0000023

—

00..00000000002233

100,000,000

X

X ==

(10) (3.45) (.48)

22

== 0.0053

0.0053

—

0.0053

1500

R

R ==

(10) (.56) (.48)

22

== 00..0000008866

00..0000008866

—

1500

X

X ==

25'

xx

0.0379

== 00..000000115588

—

00..000000115588

1000

6 1000

6 R

R ==

25'

xx

0.0244

== 00..000000110022

00..00000011002

2 —

—

1000

6 1000

6 X

X == 0.000050

0.000050

—

0.000050

Total R and X

Total R and X ==

00..000000996622

00..0000555511

ZZ

total pertotal per

==

√

(0.000962)

22

++ (0.00551)

(0.00551)

22

== 0.0056

0.0056

Ω

phase

II

==

480

480 = 49,489A

= 49,489A

√

3 (.0056)

II

= 4

= 4 xx 1804

1804 == 7216A

7216A

(100% motor load)

II

total S.C. sym RMStotal S.C. sym RMS

== 49,489

49,489 ++ 7216

7216 == 56,705A

56,705A

(fault X (fault X11))

X/R

ratioratio

== .00551

.00551 = 5.73

= 5.73

.000962

Asym Factor

Asym Factor == 1.294 (Table 8)

1.294 (Table 8)

II

== 1.294

1.294 xx 49,489

49,489 == 64,039A

64,039A

II

asym motor contribasym motor contrib

== 55 xx 1804

1804 == 9,020A

9,020A

(100% motor load)

II

total S.C. asym RMStotal S.C. asym RMS

== 64,039

64,039 ++ 9,020

9,020 == 73,059A

73,059A

(fault X

(fault X11))

O

Onnee--LLiinne D ie D iaaggr ar amm IImmppeeddaanncce D ie D iaaggrraamm

1

1 11

Note:

Note: See Ohmic Method Procedure for Formulas.See Ohmic Method Procedure for Formulas.

(Table 1.2)

(Table 3)

(Table 5)

(8)

R

X

X == 00..0000555511

—

00..0000555511

R

R == 00..000000996622

00..000000996622

—

X

X == ..0000000088

—

00..0000000088

X

X ==

50’

50’ xx .0379

.0379 == 00..0000118899

—

00..0000118899

1000

R

R ==

50’

50’ xx .0244

.0244 == 00..0000112222

00..0000112222

—

1000

Total R and X

Total R and X ==

00..000022118822

00..0000774488

ZZ

total pertotal per

==

√

(0.002182)

22

++ (0.00748)

(0.00748)

22

== 0.00778

0.00778

Ω

phase

II

==

480

480 == 35,621A

35,621A

√

3 (.00778)

II

== 44 xx 1804

1804 == 7216A

7216A

(100% motor load)

II

== 35,621

35,621 ++ 7,216

7,216 == 42,837A

42,837A

X/R

ratioratio

== .00748

.00748 == 3.43

3.43 .002182

.002182

Asym Factor

Asym Factor == 1.149 (Table 8)

1.149 (Table 8)

II

== 1.149

1.149 xx 35,621

35,621 == 40,929A

40,929A

II

== 55 xx 1804

1804 == 9,020A

9,020A

(100% motor load)

II

== 40,929

40,929 ++ 9,020

9,020 == 49,949A

49,949A

(fault X

(fault X22))

Note:

Note: See Ohmic Method Procedure for See Ohmic Method Procedure for Formulas. Actual motor contributionFormulas. Actual motor contribution will be somewhat smaller than calculated due to the impedance of the will be somewhat smaller than calculated due to the impedance of the feeder cable.

feeder cable.

Ohmic Method –

To Fault X

2

2 – System A

– System A

M M MM Adjusted Impedance Adjusted Impedance to Fault X to Fault X11 Fault X Fault X11 400A Switch 400A Switch LPS-RK-400SP Fuse LPS-RK-400SP Fuse 50’ - 500 kcmil 50’ - 500 kcmil Feeder Cable Feeder Cable in Steel Conduit in Steel Conduit Fault X Fault X22 Motor Contribution Motor Contribution One-Line Diagram

One-Line Diagram Impedance DiagramImpedance Diagram

22 22 11 11

(Table 3)

(Table 5)

(9)

To use the OHMIC Method through a second transformer,

the following steps apply:

Step 1a.

Summarize X and R values of

Summarize X and R values of all components on

all components on

primary side

primary side of transformer.

of transformer.

O

Onene-L-Liine ne DDiaiagrgramam IImmpepedadannce ce DDiaiaggraramm

Step 1b.

Reflect X and R values of

Reflect X and R values of all components to

all components to

secondary side of transformer

X

ss

== V

V

ss22

(X

pp

))

R

ss

== V

V

ss22

(R

pp

))

V

pp22

V

pp22

and proceed with steps 2 thru 15 from page 6.

Ohmic Method –

To Fault X

1

1 – System

– System

B

R

X

X ==

100 10000 (.4

(.48)

8)

22

== ..000000446611

—

..000000446611

500,000

X

X ==

(10) (3.45) (.48)

22

== ..0000779955

—

..0000779955

1000

R

R ==

(10) (.60) (.48)

22

== ..0000113388

..0000113388

—

1000

X

X ==

_11000000

30'

xx

.0303 == .000227

.0303

₄₄

.000227

—

..000227

000227

R

R ==

30'

xx

.0220

== ..000000116655

..000000116655

—

11000000

44 X

X == ..000000005500

—

..0000000055

TToottaal

l R

R aannd

d X

X ==

..000011554455

..000088668888

ZZ

total pertotal per

==

√

(.001545)

22

++ (.008688)

(.008688)

22

== .008824

.008824

Ω

phase phase

II

==

480

480 == 31,405A

31,405A

√

_{3 (.008824)}

X/R

ratioratio

== .008688

.008688 == 5.62

5.62 .001545

.001545

Asym Factor

Asym Factor == 1.285 (Table 8)

1.285 (Table 8)

II

== 31,405

31,405 xx 1.285

1.285 == 40,355A

40,355A

Available Utility Available Utility 500,000 S.C. KVA 500,000 S.C. KVA 1000KVA Transformer, 1000KVA Transformer, 480V, 3Ø, 480V, 3Ø, 3.45% X, .60% R 3.45% X, .60% R 30' - 500 kcmil 30' - 500 kcmil 4 Per Phase 4 Per Phase Copper in PVC Conduit Copper in PVC Conduit 1600A Switch 1600A Switch KRP-C-1500SP Fuse KRP-C-1500SP Fuse 11 11

(Table 1.2)

(Table 3)

(Table 5)

(10)

Ohmic Method –

To Fault X

2

2 – System B

– System B

Adjusted Impedance Adjusted Impedance to fault X to fault X11 400A Switch 400A Switch LPS-RK-350SP Fuse LPS-RK-350SP Fuse 20' - 2/0 20' - 2/0 2 Per Phase 2 Per Phase Copper in PVC Conduit Copper in PVC Conduit 225KVA Transformer, 225KVA Transformer, 208/120V, 208/120V, .998%X, .666%R .998%X, .666%R

R

X

X == ..000088668888

—

..000088668888

R

R == ..000011554455

..000011554455

—

X

X == ..0000000088

—

..0000000088

X

X ==

20'

20' xx .0327

.0327 == .000327

.000327

—

.000327

11000000

22 R

R ==

20'

20' xx .0812

.0812 == ..000000881122

..000000881122

—

11000000

22 TToottaal

l R

R aannd

d X

X ((448800V

V)

) ==

..000022335577

..000099009955

To Reflect X and R to secondary:

X

totaltotal

==

(208)

22

xx (.009095)

_(.009095)

== ..000011770088

—

..000011770088

(208V) (208V)

(480)

22

R

totaltotal

==

(208)

22

_{xx (.002357)}

_(.002357)

== ..000000444422

..000000444422

—

(208V) (208V)

(480)

22

X

X ==

(10) (.998) (.208)

22

== ..0000119922

—

..0000119922

225

225 R

R ==

(10) (.666) (.208)

22

== ..0000112288

..0000112288

—

225

225 TToottaal

l R

R aannd

d X

X ((220088V

V)

) ==

..000011772222

..000033662288

ZZ

total pertotal per

==

√

(.001722)

22

+ (.003628)

22

= .004015

Ω

phase

II

==

208

208 == 29,911A

29,911A

√

_{3 (.004015)}

X/R

ratioratio

== .003628

.003628 == 2.10

2.10 .001722

.001722

Asym Factor = 1.0491 (Table 8)

II

= 29,911

= 29,911 xx 1.0491

1.0491 == 31,380A

31,380A

One-Line Diagram

11 11 22 22

(Table 5)

(Table 1.2)

(11)

Per-Unit Method

33 øø

Short Circuit

Short Circuit Calculation Per-Unit

Calculation Per-Unit Method*

Method*

The per-unit method is generally used for calculating

short-circuit currents when the electrical system is more

complex.

After establishing a one-line diagram of the system,

proceed to the following calculations: **

Step 1.

††

_PUX

_utility_utility

₌₌

KVA

basebase

S.C. KVA

utilityutility

S

Stteep

p 22..

P

PU

UX

X

transtrans

==

(%X

••

_)(KVA

base base

))

(100)(KVA

transtrans

))

PUR

transtrans

==

(%R

••

_)(KVA

base base

))

(100)(KVA

transtrans

))

S

Stteep

p 33..

P

PU

UX

X

component (cable,component (cable,

==

(X

Ω

)(KVA

basebase

))

switches, CT, bus)

(1000)(KV)

22

S

Stteep

p 44..

P

PU

UR

R

component (cable,component (cable,

==

(R

Ω

)( KVA

basebase

))

switches, CT, bus)

(1000)(KV)

22

Step 5.

Next, total all per-unit

Next, total all per-unit X

X and all per-unit

and all per-unit R

R in system

in system

to point of fault.

Step 6.

Determine the per-unit impedance of the system by:

PUZ

totaltotal

==

√

(PUR

totaltotal

))

22

++ (PUX

(PUX

totaltotal

))

22

Step 7.

Calculate the symmetrical RMS short-circuit current

at the point of fault.

II

S.CS.C. sym. sym RMSRMS

==

KVA

basebase

√

3 (KV)(PUZ

totaltotal

))

Step 8

. Determine the

. Determine the motor load.

motor load. Add up

Add up the full load

the full load

motor currents.(Whenever motor and lighting loads are

considered, such as supplied by 4 wire, 208Y/120 and

480Y/277 volt

480Y/277 volt 3

3 phase systems,

phase systems, the generally

the generally accepted

accepted

procedure is to assume 50% motor load based on the full

load current rating

load current rating of the

of the transformer

transformer.)

.)

Step 9.

The symmetrical motor contribution can be

approximated by using an average multiplying factor

associated with the motors in the system. This factor varies

according to motor design and in this text may be chosen

as 4 tim

as 4 times motor f

es motor full load current

ull load current for

for approximate

approximate

calculation purposes. To solve for the symmetrical motor

contribution:

*** ***

_II

sym motor contrib

== (4)

(4) xx (I

(I

))

Step 10.

The total symmetrical short-circuit rms current is

calculated as:

•• ••

_II

tota

total l S.C. symS.C. sym RMSRMS

== (I

(I

)) ++ (I

(I

))

Step 11.

Determine X/R ratio of the system to the point of

fault.

X/R

ratioratio

==

PUX

totaltotal

PUR

totaltotal

Step 12.

From Table 8, Column

From Table 8, Column M

M

mm

, obtain the

, obtain the asymmetrical

asymmetrical

factor corresponding to the X/R ratio determined in Step

11. This multiplier will provide the worst case asymmetry

occurring in the first

occurring in the first 1/2 cycle.

1/2 cycle. When the average 3-ph

When the average 3-phase

ase

multiplier is desired use column

multiplier is desired use column M

M

aa

..

Step 13.

Step 13. The asymmetrical RMS short-circuit current can

The asymmetrical RMS short-circuit current can

be calculated as:

II

= (I

) x (Asym Factor)

Step 14.

Step 14. The short-circuit current that the motor load can

The short-circuit current that the motor load can

contribute is an

contribute is an asymmetrical current usually approximated

asymmetrical current usually approximated

as being equal to the locked rotor current of the motor.***

As a close approximation with a margin of

As a close approximation with a margin of safety use:

safety use:

*** ***

_II

asym motor contrib

== (5)

(5) xx (I

(I

))

Step 15.

Step 15. The total asymmetrical short-circuit RMS current

The total asymmetrical short-circuit RMS current

is calculated as:

•• ••

_II

totalS.C. asym RMS

== (I

(I

)) ++ (I

(I

))

** The base KVA used throughout this text will be 10,000 KVA.The base KVA used throughout this text will be 10,000 KVA. **

(12)

**

10,000 KVA Base

PUR

PUX

PUR

PUX

PUX ==

10,000

== 00..00000011

—

00..00000011

100,000,000

PUX

PUX ==

(3.45) (10,000)

== 00..22330000

—

00..22330000

(100) (1500)

PUR

PUR ==

(.56) (10,000)

== 00..00337733

00..00337733

—

(100) (1500)

(25')

xx

(.0379)

x (10,000)

PUX

PUX ==

((11000000))

((66))

== 0.00685

0.00685 —

—

00..0000668855

(1000) (.480)

22

(25')

xx

(.0244)

x (10,000)

((11000000))

((66))

PUR

PUR ==

(1000) (.480)

22

== 00..00004444

00..00004444

—

PUX =

(.00005) (10,000)

=

00..0000221177

—

00..0000221177

(1000) (.480)

22

Total PUR and PUX

Total PUR and PUX ==

0.0417

0.2391

PUZ

totaltotal

==

√

(0.0417)

22

++ (0.2391)

(0.2391)

22

== .2430

.2430

II

==

10,000

== 49,489A

49,489A

√

_{3 (.480)(.2430)}

II

== 44 xx 1804

1804 == 7,216A

7,216A

II

== 49,489

49,489 ++ 7,216

7,216 == 56,705A

56,705A

X/R

ratioratio

== .2391

.2391 == 5.73

5.73 .0417

.0417

Asym Factor = 1.294 (Table 8)

II

== 49,489 x 1.294

49,489 x 1.294 == 64,039A

64,039A

II

== 55 xx 1804

1804 == 9,020A

9,020A

(100% motor load)

II

total total S.C. asS.C. asymym RMSRMS

== 64,039

64,039 ++ 9,020

9,020 == 73,059A

73,059A

(fault X

(fault X11))

Per-Unit Method –

To Fault X

1

1 – System A

– System A

Available Utility Available Utility S.C. MVA 100,000 S.C. MVA 100,000 1500 KVA Transformer, 1500 KVA Transformer, 480V, 3Ø, 480V, 3Ø, 3.5%Z, 3.45%X, .56%R 3.5%Z, 3.45%X, .56%R IIf.l. transf.l. trans= 1804A= 1804A

25’ - 500kcmil 25’ - 500kcmil 6 Per Phase 6 Per Phase Service Entrance Service Entrance

Conductors in Steel Conduit Conductors in Steel Conduit

2000A Switch 2000A Switch KRP-C-2000SP Fuse KRP-C-2000SP Fuse M M M M Impedance Diagram Impedance Diagram One-Line Diagram One-Line Diagram Note:

Note: See Per Unit Method Procedure for Formulas.See Per Unit Method Procedure for Formulas. Actual motor contribution will be somewhat smaller than

Actual motor contribution will be somewhat smaller than calculatedcalculated due to impedance of the feeder cable.

due to impedance of the feeder cable. 1

(13)

M

M MM

Per-Unit Method –

To Fault X

2

2 – System A

– System A

Fault X Fault X11 400A Switch 400A Switch LPS-RK400SP Fuse LPS-RK400SP Fuse 50’ - 500kcmil 50’ - 500kcmil Feeder Cable in Feeder Cable in Steel Conduit Steel Conduit Motor Contribution Motor Contribution O

Onnee- L- Liinne De Diiaaggrraamm IImmppeeddaanncce De Diiaaggrraamm

10,000 KVA Base

PUR

PUX

PUR

PUX

PUX == ..22339911

—

..22339911

PUR

PUR == ..0044117

7 ..00441177

—

PUX

PUX ==

(.00008) (10,000)

== ..00003344

—

..00003344

(1000) (.480)

22

50’

50’ xx (.0379)

(.0379) xx (10,000)

(10,000)

PUX

PUX ==

1000

== ..0088222

2 —

—

..00882222

(1000) (.480)

22

50’

50’ xx (.024

(.0244) x (10,000)

4) x (10,000)

PUR

PUR ==

1000

== ..00552299

..00552299

—

(1000) (.480)

22

Total PUR and PUX

Total PUR and PUX ==

..00994466

..33224477

PUZ

totaltotal

==

√

(.0946)

22

++ (.3247)

(.3247)

22

== 0.3380

0.3380

II

==

10,000

== 35,621A

35,621A

√

_{3 (.480)(.3380)}

II

== 44 xx 1804

1804 == 7,216A

7,216A

II

total total S.C. symS.C. sym RMSRMS

== 35,621

35,621 ++ 7,216

7,216 == 42,837A

42,837A

X/R

ratioratio

==

.32477

== 3.43

3.43 .09465

.09465

Asym Factor

Asym Factor == 1.149 (Table 8)

1.149 (Table 8)

II

== 1.149

1.149 xx 35,621

35,621 == 40,929A

40,929A

II

== 55 xx 1804

1804 == 9,020A

9,020A

(100%motor load)

II

== 40,929

40,929 ++ 9,020

9,020 == 49,949A

49,949A

(fault X (fault X22)) Adjusted Impedance Adjusted Impedance to Fault X to Fault X11 1 1 11 22 22

(14)

10,000KVA Base

P

PU

UR

R

P

PU

UX

X

PUX

PUX ==

10,000 == ..0022

10,000

—

..0022

500,000

PUX

PUX ==

(3.45) (10,000)

== ..334455

—

..334455

(100) (1000)

PUR

PUR ==

(.6) (10,000)

== ..0066

..0066

—

(100) (1000)

(30')

xx

(.0303)

PUX

PUX ==

((11000000))

((44))

x (

x (10,000)

10,000)

== ..00009999

—

..00009999

(1000) (.48)

22

(30')

xx

(.0220)

PUR

PUR ==

((11000000))

((44))

xx (10

(10,00

,000)

0)

== ..00007722

..00007722

—

(1000) (.48)

22

PUX

PUX ==

(.00005) (10,000)

== ..00002222

—

..00002222

(1000) (.48)

22

TToottaal

l P

PU

UR

R aannd

d P

PU

UX

X ==

..00667722

..33777711

PUZ

totaltotal

==

√

(.0672)

22

++ (.3771)

(.3771)

22

== .383

.383

II

==

10,000

== 31,405A

31,405A

√

3 (.48)(.383)

X/R

ratioratio

== .3771

.3771 = 5.62

= 5.62

.0672

Asym Factor = 1.285 (Table 8)

II

S.C.asym RMSS.C.asym RMS

== 31,405

31,405 xx 1.285

1.285 == 40,355A

40,355A

Per-Unit Method –

To Fault X

1

1 – System B

– System B

Available Utility Available Utility S.C. KVA 500,000 S.C. KVA 500,000 1000 KVA Transformer, 1000 KVA Transformer, 480V, 3Ø 480V, 3Ø 3.45%X, .60%R 3.45%X, .60%R 30' - 500kcmil 30' - 500kcmil 4 Per 4 Per PhaPhasese

Copper in PVC Conduit Copper in PVC Conduit 1600A Switch 1600A Switch KRP-C-1500SP Fuse KRP-C-1500SP Fuse One-Line Diagram