In this slide given the information about symmetrical fault , its derivations , examples etc..

1. Name :- Smit Shah -140410109096
T.Y Electrical 2 Sem 6
Subject:-Electrical power system 2
Topic :-Line to Line & Double Line to
Ground Fault On Power System 1

2. Line-‐to-‐Line Faults
To represent a line-‐to-‐line fault through impedance Zf the
hypothetical stubs on the three lines at the fault are connected as
shown. Bus k is again the fault point P, and without any loss of
generality, the line-‐to-‐line fault is regarded as being on phases b
and c. Clearly:
Ifb
Ifc
a
Zf
k
k
c
I fa I  0
fa
k
b
I fb  I fc
Vkb Vkc  Z If fb
2

3. Line‐To‐Line Fault
Three‐phase generator with a fault through an
impedance Zf between phase b
and c.
Ia=0
Zs
N
ZsZs
Ea
Eb
Ec
Ib
Va
ZfIc
Vb
Vc
Assuming the generator is initially on no‐load, the
boundary conditions at the fault point are:
Vb  Vc  Z fIb Ia  0
Ib  I c 0
3

4. Line‐To‐Line Fault
4
Substituting for Ia = 0, and Ic = ‐Ib, the symmetrical components of
the currents from are:
From the above equation, we find that:
 0a
I0
(10.68)
ba
3
11 2
I  (a  a )I
a
12 2
I  (a  a)Ib
(10.69)
(10.70)
3

5. Line‐To‐Line Fault
Also, from (10.69) and (10.70), we note that:
1 2
aa
I  I
From (10.16), we have:
(10.71)
1 20
b
a aaa
V  V 0
 a2
V1
 aV 2
a a a
V  V V V
(10.16)
aa
 Zf Ib
(a  a)(V V )
V V  (V 0
 a2
V1
 aV 2
)  (V 0
 aV1
 a2
V 2
)
b c a a a a a a
2 1 2
(10.72)
0 1 2 2
aaacV  V  aV  a V
Substituti ng for V1
and V 2
from (10.54) and noting I 2
 I1
, we get :
a a a a
a f b(a  a)[Ea  (Z  Z )I ]  Z I2 1 2 1 (10.73)
(10.54)
2 2 2
aa
V 0
 0  Z 0
I 0
a a
V  0  Z I
V1
 E  Z1
I1
a a a
Substituti ng for I b from (10.69), we get :
1
3Ia
a a f
(a  a2
)(a2
 a)
E  (Z1
 Z 2
)I 1
 Z (10.74) ba
11 2I  (a  a )I
3 (10.69)
5

6. Line‐To‐Line Fault
aSince (a  a2
)(a2
 a)  3,solving for I1
results in :
f
a
Ea
(Z1
 Z 2
 Z )
I1

(10.75)
The fault current is
b
I  I  (a2
 a)I 1
c a
or abI   j 3I1(10.77) (10.78)
6

7. Line‐To‐Line Fault
Eq. (10.71) and (10.75) can be represented by connecting the
positive and negative – sequence networks as shown in the
following figure.
1 2
aa
I  I 1 aE
I 
f
a
(Z1
 Z 2
 Z )
7

8. Line-‐to-‐Line Faults
Thus, all the fault conditions are satisfied by connecting the
positive-‐ and negative-‐sequence networks in parallel through
impedance Zf as was shown.
The zero-‐sequence network is inactive and does not enter into the
line-‐to-‐line calculations. The equation for the positive-‐sequence
current in the fault can be determined directly from the circuit as:
8

9. Double Line-‐to-‐Ground Faults
Again it is clear, with fault taken on phases b and c, that the
relations at fault bus k are:
Ifa
Ifb
Ifc
a
Zf
I fa 0
Vkb  Vkc  Zf Ifb  I fc
k
k
b
k
c
9

10. Double Line‐To‐Ground Fault
Figure 10.14 shows a three‐phase generator with a fault on phases b
and c through an impedance Zf to ground. Assuming the generator
is initially on no‐ load, the boundary conditions at the fault point are
Vb Vc  Z f (Ib  Ic ) (10.79)
(10.80)
I  I0
 I1
 I2
 0
a a a a
From (10.16), the phase voltages Vb and Vc are
10

11. Double Line‐To‐Ground Fault
0 2 1 2
aaa
V V 0
 aV1
 a2
V 2
Vb V  a V aV
aac a
(10.81)
(10.82)
SinceVb  Vc , from above we note that
1 2
aaV V (10.83)
Substituting for the symmetrical components of
current in (10.79), we get
V  Z (I 0
 a2
I1
 aI 2
 I 0
 aI 1
 a2
I 2
)
(b) f a a a a a a
 Z (2I 0
 I1
 I 2
)
f a a a
f aI 0
 3Z
(10.84)
11

12. Substituti ng for V from (10.84) and for V 2
from (10.83) into (10.81), we have :b a
a a
f a a a
 V 0
V1
3Z I 0
 V 0
 (a2
 a)V1
(10.85)
E  Z1
I1
Substituti ng for the symmetrical components of voltage from (10.54) into (10.85)
aand solving for I 0
,we get :
0
0 a a
a
(Z  3Z )
I  
f
Also, substituting for the symmetrical components of voltage in (10.83), we obtain
E  Z1
I1
(10.86)
Z2a
I 2
  a a
Substituti ng for I 0
and I2
into (10.80) and solving for I 1
, we get:
a a a
(10.87)
Ea
a 2 0
1
Z 2
 Z 0
 3Z
f
Z (Z  3Zf )
Z 
I1
 (10.88)
12

13. Equation (10.86) - (10.88) can be represented by connecting the positive - sequence
impedance in series with the paralel combination of the negative - sequence
and zero - sequence networks as shown in the equivalent circuit of figure 10.15.
The value of I1
found from (10.86) is substituted in (10.86) and (10.87),a
and I0
and I2
are found. The phase current are then found from (10.8).
a a
Finally, the fault current is obtained from
(10.89)I  I  I  3I0
c abf
Figure 10.15 Sequence network connection for double line‐to‐ground
fault
13

14. Double Line-‐to-‐Ground Faults
For a bolted fault Zf is set equal to 0. When Zf = ∞, the zero-‐
sequence circuit becomes an open circuit, no zero-‐sequence current
an flow, and the equations revert back to those for the line-‐to-‐line
fault.
Again we observe that the sequence currents, once calculated, can
be treated as negative injections into the sequence networks at
the fault bus k and the sequence voltage changes at all buses of the
system can then be calculated from the bus impedance matrices, as
we have done in all along.
14

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