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

Line-‐to-‐Line FaultsTo represent a line-‐to-‐line fault through impedance Zf thehypothetical stubs on the three lines at the fault are connected asshown. Bus k is again the fault point P, and without any loss ofgenerality, the line-‐to-‐line fault is regarded as being on phases band c. Clearly:

I fb

I fc

a

Zf

k

kc

I fa I 0fa

kb

Ifb I

fc

Vkb Vkc Z I

f fb

2

Line‐To‐Line FaultThree‐phase generator with a fault through animpedance Zf between phase band c.

Ia=0

Zs

N

ZsZs

EaEb

Ec

Ib

Va

ZfIcVb

Vc

Assuming the generator is initially on no‐load, theboundary conditions at the fault point are:

Vb Vc Z fIb Ia 0Ib I c 0

3

Line‐To‐Line Fault

4

Substituting for Ia = 0, and Ic = ‐Ib, the symmetrical components ofthe currents from are:

From the above equation, we find that:

0aI 0 (10.68)

ba3

11 2I (a a )I

a

12 2I (a a)Ib

(10.69)

(10.70)

3

Line‐To‐Line FaultAlso, from (10.69) and (10.70), we note that:

1 2

aaI I

From (10.16), we have:

(10.71)

1 20

b

a aaa

V V 0 a2V 1 aV 2

a a a

V V V V

(10.16)

aa

Z f Ib

(a a)(V V )

V V (V 0 a2V 1 aV 2) (V 0 aV 1 a2V 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 V 1 and V 2 from (10.54) and noting I 2 I 1, 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 0I 0

a a

V 0 Z I

V 1 E Z 1I 1

a a a

Substituti ng for I b from (10.69), we get :

13I a

a a f(a a 2)(a 2 a)

E (Z 1 Z 2)I 1 Z (10.74) ba

11 2I (a a )I3 (10.69)

5

Line‐To‐Line Fault

aSince (a a2 )(a2 a) 3, solving for I 1 results in :

f

a

Ea

(Z 1 Z 2 Z )I 1

(10.75)

The fault current is

bI I (a 2 a)I 1

c aor

abI j 3I1(10.77) (10.78)

6

Line‐To‐Line FaultEq. (10.71) and (10.75) can be represented by connecting thepositive and negative – sequence networks as shown in thefollowing figure.

1 2

aaI I 1 aE

I f

a(Z 1 Z 2 Z )

7

Line-‐to-‐Line Faults

Thus, all the fault conditions are satisfied by connecting thepositive-‐ and negative-‐sequence networks in parallel throughimpedance Zf as was shown.

The zero-‐sequence network is inactive and does not enter into theline-‐to-‐line calculations. The equation for the positive-‐sequencecurrent in the fault can be determined directly from the circuit as:

8

Double Line-‐to-‐Ground Faults

Again it is clear, with fault taken on phases b and c, that therelations at fault bus k are:

I fa

I fb

I fc

a

Zf

I fa 0

VkbVkc Z fI fb I fc

k

k

b

k

c

9

Double Line‐To‐Ground FaultFigure 10.14 shows a three‐phase generator with a fault on phases band c through an impedance Zf to ground. Assuming the generatoris initially on no‐ load, the boundary conditions at the fault point are

Vb Vc Z f (Ib Ic) (10.79)

(10.80)I I 0 I1 I 2 0

a a a a

From (10.16), the phase voltages Vb and Vc are

10

Double Line‐To‐Ground Fault

0 2 1 2

aaa

V V 0 aV 1 a2V 2

Vb V a V aV

aac a

(10.81)

(10.82)

SinceVb Vc , from above we note that1 2

aaV V (10.83)

Substituting for the symmetrical components ofcurrent in (10.79), we get

V Z (I 0 a 2I 1 aI 2 I 0 aI 1 a 2I 2)(b ) f a a a a a a

Z (2 I 0 I 1 I 2)f a a a

f aI 0 3 Z

(10.84)

11

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 V 1

3Z I 0 V 0 (a 2 a)V 1

(10.85)

E Z 1I 1

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 Z 1I 1

(10.86)

Z 2aI 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 3Zf

Z (Z 3Z f )Z

I 1 (10.88)

12

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 3I 0

c abf

Figure 10.15 Sequence network connection for double line‐to‐ground fault

13

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 currentan flow, and the equations revert back to those for the line-‐to-‐linefault.

Again we observe that the sequence currents, once calculated, canbe treated as negative injections into the sequence networks atthe fault bus k and the sequence voltage changes at all buses of thesystem can then be calculated from the bus impedance matrices, aswe have done in all along.

14

15