chapter7 symmetrical faults

7/29/2019 Chapter7 Symmetrical Faults

1/44

7.1 SERIES R-L CIRCUIT TRANSIENTS

The closing of SW at t=0 represents to a first approximation a three-phase short circuit at the terminals of an unloaded synchronous

machine.

For simplicity, assume zero fault impedance; that is, the short circuit is

a solid or bolted fault.

L i(t)R

t=0)sin(2)( += tVte SW

Circuit

breaker


2/44

Ldi(t)Ri(t) 2 Vsin( t+ )

dt

+ = t 0 (7.1.1)

The solution to (7.1.1) is

The current is assumed to be zero before SW closes, and the source angle

determines the source voltage at t=0. Writing a KVL equation for the

circuit,

tT

ac dc2Vi(t) i (t) i (t) [sin( t ) sin( )e ]Z

= + = + (7.1.2)

ac

2Vi (t) sin( t )Z= + (7.1.3)

tT

dc

2V

i (t) sin( )eZ

= (7.1.4)


3/44

ac

2Vi (t) sin( t )

Z= +

tT

dc2Vi (t) sin( )eZ

=

ac dci(t) i (t) i (t)= + Asymmetrical fault current

ac fault current

(symmetrical or steady-state fault current)

Dc offset current

e(t) i(t) iac

(t)

idc

(t)

t


4/44

1 1L X

tan tanR R

= =

L X XT

R R 2 fR

= = =

2 2 2

Z R ( L) R X= + = +

2

s

RMS ac fault current isac

VI

Z=

ac

2Vi (t) sin( t )

Z= +

tT

dc

2Vi (t) sin( )e

Z

= dc

dc ac

When

Wh

i (t) 0

( ) i (t) In 22

e

= =

= =

A short circuit can happen at any instant during a cycle of ac source.

that is can have any value.

The largest fault current occurs when ( )

2

=


5/44

tT

aci(t) 2I [sin( t ) e ]2

= +

ac

VI

Z

= A

(7.1.8)

RMS value ofi(t) should found:

tT

2V

i(t) [sin( t ) sin( )e ]Z

= + Then becomes

( )2

= when

where


6/44

[ ] [ ]

[ ]

2 2

rms ac dc

2t2T

rms ac ac

2tT

rms ac

I (t) I I (t)

I (t) I 2I e

I (t) I 1 2e

= +

= +

= +

It is convenient to use T=X/(2fR) and t=/f, where is time in cycles, andwrite (7.1.10) as

(7.1.10)

rms acI ( ) K( A)I =

where(7.1.11)

(7.1.12)

The RMS asymmetrical fault Irms(t) with maximum dc offset is

4( X / R)

puK( ) 1 2e

= +


7/44

4( X / R)

per unitK( ) 1 2e

= +rms acI ( ) K( )I =

rms ac

ac

decreases from whI ( en

to when

) 3I =0

I is l rgea

rmsHigher ratios give higer valuesX / R Iof ( )


8/44

2 2

rms ac dcI (t) I i (t)= +

rms acI ( ) K( )I =

tT

dc

2Vi (t) sin( )e

Z

=

ac 2Vi (t) sin( t )Z= + ac

VIZ

=

with maximum dc offset

Component Instantaneous Current (A) rms Current (A)

Asymmetrical

(total)

dc offset

Symmetrical (ac)

TABLE 7.1

ac dci(t) i (t) i (t)= +


9/44

3 3

ac 2 2

20 10 20 10I 2.488

8.040(8) (0.8)kA = = =

+

EXAMPLE 7.1 Fault currents: R-L circuit with ac source

A bolted short circuit occurs in the series R-L circuit of Figure 7.1 with

V=20 kV, X=8 , R=0.8 , and with maximum dc offset.

The circuit breaker opens 3 cycles after fault inception.

Determine (a) the rms ac fault current, (b) the rms momentary current

at =0.5 cycle, which phases through the breaker before it opens, and (c)

the rms asymmetrical fault current that the breaker interrupts.

SOLUTION

a. From (7.1.9),


10/44

4 (3) /10K(3cycle) 1 2e 1.023 = + =

4 (0.5) /10K(0.5cycle) 1 2e 1.438 = + =

b. From (7.1.11) and (7.1.12) with (X/R) = 8/(0.8) =10 and = 0.5 cycle,

Imomentary=K(0.5cycle)Iac=(1.438)(2.488)=3.576 kA

c. From (7.1.11) and (7.1.12) with (X/R) =10 and = 3 cycles,

Irms(3cycles)=(1.023)(2.488)=2.544 kA


11/44

7.2Three-phase Short Circuit-Unloaded

Synchronous Machine

Ac fault current in one phase of an unloaded synchronous machine

during a three-phase short circuit (the dc-offset current is removed)

iac

(t)

t

2I

2I


12/44

Ac fault current in a synchronous machine can be modeled by the series R-L

circuit of Figure 7.1 if a time-varying inductance L(t) or reactance

X(t)=L(t) is used.In standard machine theory texts, the following reactances are defined :

d direct axis transient reactX cean =

d direct axis subtransient reactX n ea c =

d direct axis synchronous reac eX tanc=

d d dX X X

chapter7 symmetrical faults

Documents