Test: Line to Line & Double Line to Ground Fault - Electrical Engineering (EE) MCQ

Line to Line fault:

When two conductors of a 3-phase system are short-circuited line to line fault occurs.
I_R = 0
I_F = I_Y = -I_B

From equation (i), we found that zero sequence currents do not exist in the LL fault.

Concept:
In a double line fault, the fault current is given by

Where Z₁ is the positive sequence impedance
Z₂ is the negative sequence impedance

Calculation:
V = 6.6 kV
Z₁ = Z₂ = j5Ω
Z_f = 200 Ω
Z₀ = j2Ω
This is the case of the line to line fault.

I_a = 0 (∵ open circuited)
From the formula of a double line fault or L-L fault,

For line to line fault, there is no ground.
V_a0 = 0 and V_a1 = V_a2

In line to line fault,
fault current with fault impedance z_f is

fault current with zero fault impedance is,


⇒ z₁ + z₂ = k (z₁ + z₂ + z_f)
⇒ z₁ + z₂ - k z₁ - k z₂ = k z_f 

Voltage (E) = 11 kV
Phase voltage (E_ph) 
Three phase fault current = 5610 A

⇒ X₁ = 1.132 Ω
L-L fault current = 6760 A

⇒ X₂ = 0.495 Ω
L-G fault current = 8630 A

Concept:
Here star point is isolated so the flow of fault current is not possible in neutral or ground ( also zero sequence current is zero because zero-sequence current flow through the ground and here the ground is not available) so current is absent in neutral but current available in phase so this fault is considered as line to line  (L-L) fault.

Positive sequence current (i₁) for L-L fault is given by;
i₁ = v_oc/(x₁ + x₂)
Where;
v_oc = open circuit voltage, x₁ = Positive sequence reactance, x₂ = Negative sequence reactance
Fault current (i_f) for L-L fault is given by:
if = √3 × i₁

Calculation:
Given that:
v_oc = 1.0 pu, x₁ = j0.35, x₂ = j0.25 pu, x_o = 0.20pu, x_n = ∞ pu  (since Neutral is isolated from ground)
Here, v_oc = Open circuit voltage, x₁ = Positive sequence reactance, x₂ = Negative sequence reactance, x_o = Zero sequence reactance, x_n = Neutral reactance
Here L-L (line to line) fault happened so we will first calculate positive sequence current (i₁) then calculate fault current (i_f) by help of positive sequence current.
Calculation of Positive sequence current (i₁);
 i₁ = 1/(j0.35+j0.25)
i₁ = -j1.6666 pu
So fault current (i_f) ;
if = √3 × i₁
I_f = -j2.887 pu

For the given power system network, the positive sequence impedance network is given below.

The impedance at a point of fault is,
X_1eq = (j0.35)∥(j0.65) 
= j0.2275 pu
Negative sequence reactance is the same as positive sequence reactance
X_2eq = X1_eq = j0.2275pu 
The zero sequence reactance network is,

Zero sequence impedance
Z0_eq = j0.25 ∥ j0.75 = j0.1875 pu
For a double line to fault at bus 1 is,

We have the base current 

The positive sequence current, I_p = 4270 A

Fault impedance = 0.1 pu
X_n = 3z_f + 0.05 + 3x_n = 0.35 + 3x_n 

From above equation, we get  Xn = 1.07 p.u

The equivalent circuit for LLG fault can be drawn as:

Z₁ = j0.25 pu, E = 1 pu
Z₂ = j0.35 pu
Z₀ = j0.10 pu

α = 1∠120°
α₂ = 1∠-120°
⇒ I_b = 4.804 pu   

First find rated current of generator

Taking rated voltage and MVA as base

Z₁ = j0.30 PU, Z₂ = J 0.40 PU Z₀ = J 0
Double line to ground fault

∴ 0.12 + 0.7 (0.05 + 3X_n) = 1.2
X_n = 0.5 PU = 0.5 × 5.88 = 2.94 Ω