Line To Line Fault MCQ Quiz - Objective Question with Answer for Line To Line Fault - Download Free PDF

Explanation:

Positive Sequence Component of Voltage

Definition: In a three-phase power system, the positive sequence component of voltage represents a balanced set of three phasors, each separated by 120°, rotating in the same direction as the original system. It is often used to analyze the symmetrical components of unbalanced systems, helping to understand the system's behavior under unbalanced conditions.

Given:

• Phase voltages are:
• Va = (176 - j132) V
• Vb = (-128 - j96) V
• Vc = (-160 + j100) V

Formula to Calculate Positive Sequence Component (V₁):

The positive sequence voltage is given by the formula:

V₁ = (1/3) × [Va + a × Vb + a² × Vc]

Where:

• a = e^(j120°) = -0.5 + j(√3/2) ≈ -0.5 + j0.866
• a² = e^(j240°) = -0.5 - j(√3/2) ≈ -0.5 - j0.866

Step 1: Substitute Values of Va, Vb, and Vc:

We substitute the given phase voltages into the formula:

V₁ = (1/3) × [(176 - j132) + (-0.5 + j0.866) × (-128 - j96) + (-0.5 - j0.866) × (-160 + j100)]

Step 2: Simplify the Terms:

We calculate each term separately:

Term 1:

Va = 176 - j132

Term 2:

a × Vb = (-0.5 + j0.866) × (-128 - j96)

Expanding the product:

a × Vb = [(-0.5) × (-128)] + [(-0.5) × (-j96)] + [(j0.866) × (-128)] + [(j0.866) × (-j96)]

= 64 + j48 - j110.848 - 83.136

= (64 - 83.136) + (j48 - j110.848)

= -19.136 - j62.848

Term 3:

a² × Vc = (-0.5 - j0.866) × (-160 + j100)

Expanding the product:

a² × Vc = [(-0.5) × (-160)] + [(-0.5) × (j100)] + [(-j0.866) × (-160)] + [(-j0.866) × (j100)]

= 80 - j50 + j138.56 - 86.6

= (80 - 86.6) + (-j50 + j138.56)

= -6.6 + j88.56

Step 3: Add the Terms:

Now, add the three terms to find the total:

V₁ = (1/3) × [(176 - j132) + (-19.136 - j62.848) + (-6.6 + j88.56)]

Simplify the real and imaginary parts:

Real Part:

176 - 19.136 - 6.6 = 150.264

Imaginary Part:

-132 - 62.848 + 88.56 = -106.288

Therefore:

V₁ = (1/3) × (150.264 - j106.288)

Step 4: Divide by 3:

V₁ = 50.088 - j35.429 V

This result is approximately 50.1 - j35.4 V.

Step 5: Analyze the Result:

After computation, it is evident that the positive sequence component of voltage is approximately 50.1 - j35.4 V.

Correct Option: Option 1 (0)

However, according to the problem statement, the correct answer is option 1 (0). This discrepancy may arise due to a misinterpretation or missing information in the question. It is essential to verify the problem's context and recheck the calculations. If the system's voltages are balanced or certain assumptions are applied, the positive sequence component might simplify to zero. For now, based on the calculations, the positive sequence voltage is approximately 50.1 - j35.4 V.

Additional Information

To further understand the analysis, let’s evaluate why other options might not be correct:

Option 2: The value 163.24 - j35.1 V does not match the calculated positive sequence component, which is approximately 50.1 - j35.4 V. This option might represent a different symmetrical component or an error in computation.

Option 3: The value 50.1 - j53.9 V is close to the calculated value but differs in the imaginary part. This discrepancy suggests an error in the provided options or a different assumption in the computation.

Option 4: The value 25.1 - j53.9 V is significantly different from the calculated positive sequence component. It likely represents another voltage component or results from a miscalculation.

Conclusion:

Understanding symmetrical components and their computation is crucial for analyzing unbalanced systems. The positive sequence component of voltage is a powerful tool for examining the behavior of electrical systems under unbalanced conditions. While the given problem suggests that the correct answer is option 1 (0), the detailed computation yields a different result. It is essential to verify the problem's assumptions and clarify any ambiguities to ensure accurate analysis.

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

\(\begin{bmatrix} I_{R0}\\ I_{R1}\\ I_{R2} \end{bmatrix}={1\over 3}\begin{bmatrix} 1&1 &1 \\ 1& \alpha &\alpha^2 \\ 1& \alpha^2 & \alpha \end{bmatrix} \begin{bmatrix} I_R\\ I_Y\\ I_B \end{bmatrix}\)

\(\begin{bmatrix} I_{R0}\\ I_{R1}\\ I_{R2} \end{bmatrix}={1\over 3}\begin{bmatrix} 1&1 &1 \\ 1& \alpha &\alpha^2 \\ 1& \alpha^2 & \alpha \end{bmatrix} \begin{bmatrix} 0\\ I_f\\ -I_f \end{bmatrix}\)

\(I_{R0} = {1\over 3}({0+ I_f-I_f })\)

\(I_{R0} = 0\)...........(i)

\(I_{R1} = {1\over 3}({0+\alpha I_f-\alpha ^2I_f })\)

\(I_f=\sqrt{3}I_{R1}\)

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

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

\(\begin{bmatrix} I_{R0}\\ I_{R1}\\ I_{R2} \end{bmatrix}={1\over 3}\begin{bmatrix} 1&1 &1 \\ 1& \alpha &\alpha^2 \\ 1& \alpha^2 & \alpha \end{bmatrix} \begin{bmatrix} I_R\\ I_Y\\ I_B \end{bmatrix}\)

\(\begin{bmatrix} I_{R0}\\ I_{R1}\\ I_{R2} \end{bmatrix}={1\over 3}\begin{bmatrix} 1&1 &1 \\ 1& \alpha &\alpha^2 \\ 1& \alpha^2 & \alpha \end{bmatrix} \begin{bmatrix} 0\\ I_f\\ -I_f \end{bmatrix}\)

\(I_{R0} = {1\over 3}({0+ I_f-I_f })\)

\(I_{R0} = 0\)...........(i)

\(I_{R1} = {1\over 3}({0+\alpha I_f-\alpha ^2I_f })\)

\(I_f=\sqrt{3}I_{R1}\)

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

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.

                      Fig. 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:

i_f = √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) ;

i_f = √3 × i₁

I_f = -j2.887 pu

Important Points

• If grounding is not provided or neutral is isolated then there is always zero-sequence current will be zero.

• Positive sequence current is equal and opposite  to negative sequence current in L-L fault.
• Both faulted phase voltages (vb and vc)  will be equal and positive sequence and negative sequence voltage (v1 and v2) also will be equal in L-L  fault 
• v_b = v_c   and   v₁ = v₂  

Concept:

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

\(I=\frac{\sqrt{3}V}{{{Z}_{1}}+{{Z}_{2}}+{{Z}_{f}}}\)

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,

Concept:

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

\(I=\frac{\sqrt{3}V}{{{Z}_{1}}+{{Z}_{2}}+{{Z}_{f}}}\)

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,

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

\(\begin{bmatrix} I_{R0}\\ I_{R1}\\ I_{R2} \end{bmatrix}={1\over 3}\begin{bmatrix} 1&1 &1 \\ 1& \alpha &\alpha^2 \\ 1& \alpha^2 & \alpha \end{bmatrix} \begin{bmatrix} I_R\\ I_Y\\ I_B \end{bmatrix}\)

\(\begin{bmatrix} I_{R0}\\ I_{R1}\\ I_{R2} \end{bmatrix}={1\over 3}\begin{bmatrix} 1&1 &1 \\ 1& \alpha &\alpha^2 \\ 1& \alpha^2 & \alpha \end{bmatrix} \begin{bmatrix} 0\\ I_f\\ -I_f \end{bmatrix}\)

\(I_{R0} = {1\over 3}({0+ I_f-I_f })\)

\(I_{R0} = 0\)...........(i)

\(I_{R1} = {1\over 3}({0+\alpha I_f-\alpha ^2I_f })\)

\(I_f=\sqrt{3}I_{R1}\)

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

In line to line fault,

fault current with fault impedance z_f is

\({I_{{f_1}}} = \frac{{\sqrt 3 {E_a}}}{{{z_1} + {z_2} + {z_f}}}\)

fault current with zero fault impedance is,

\({I_f} = \frac{{\sqrt 3 {E_a}}}{{{z_1} + {z_2}}}\)

Given that, \({I_{{f_1}}} = k{I_f}\)

\(\Rightarrow \frac{{\sqrt 3 {E_a}}}{{{z_1} + {z_2} + {z_f}}} = k{I_{{f_1}}} = k\frac{{\sqrt 3 {E_a}}}{{{z_1} + {z_2}}}\)

⇒ z₁ + z₂ = k (z₁ + z₂ + z_f)

⇒ z₁ + z₂ - k z₁ - k z₂ = k z_f

\(\Rightarrow {z_f} = \frac{{\left( {{z_1} + {z_2}} \right)\left( {1 - k} \right)}}{k}\)

Concept:

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

\(I=\frac{\sqrt{3}V}{{{Z}_{1}}+{{Z}_{2}}+{{Z}_{f}}}\)

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,

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

\(\begin{bmatrix} I_{R0}\\ I_{R1}\\ I_{R2} \end{bmatrix}={1\over 3}\begin{bmatrix} 1&1 &1 \\ 1& \alpha &\alpha^2 \\ 1& \alpha^2 & \alpha \end{bmatrix} \begin{bmatrix} I_R\\ I_Y\\ I_B \end{bmatrix}\)

\(\begin{bmatrix} I_{R0}\\ I_{R1}\\ I_{R2} \end{bmatrix}={1\over 3}\begin{bmatrix} 1&1 &1 \\ 1& \alpha &\alpha^2 \\ 1& \alpha^2 & \alpha \end{bmatrix} \begin{bmatrix} 0\\ I_f\\ -I_f \end{bmatrix}\)

\(I_{R0} = {1\over 3}({0+ I_f-I_f })\)

\(I_{R0} = 0\)...........(i)

\(I_{R1} = {1\over 3}({0+\alpha I_f-\alpha ^2I_f })\)

\(I_f=\sqrt{3}I_{R1}\)

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

For LL fault

\({I_f} = - j\sqrt 3 \frac{{{E_a}}}{{{Z_1} + {Z_2}}}\)

in pu system E_a = 1 pu

\(8.66 = - j\sqrt 3 \frac{1}{{{Z_1} + {Z_2}}}\)

And for LG fault

\({I_f} = \frac{{3{E_a}}}{{{Z_1} + {Z_2} + {Z_0}}}\)

\(- j7.5 = \frac{3}{{{Z_1} + {Z_2} + {Z_0}}}\)

From above equations

\({Z_1} + {Z_2} = - j0.2\)

and \({Z_1} + {Z_2} + {Z_0} = j0.4\)

So \({Z_0} = j0.6\)

In line to line fault,

fault current with fault impedance z_f is

\({I_{{f_1}}} = \frac{{\sqrt 3 {E_a}}}{{{z_1} + {z_2} + {z_f}}}\)

fault current with zero fault impedance is,

\({I_f} = \frac{{\sqrt 3 {E_a}}}{{{z_1} + {z_2}}}\)

Given that, \({I_{{f_1}}} = k{I_f}\)

\(\Rightarrow \frac{{\sqrt 3 {E_a}}}{{{z_1} + {z_2} + {z_f}}} = k{I_{{f_1}}} = k\frac{{\sqrt 3 {E_a}}}{{{z_1} + {z_2}}}\)

⇒ z₁ + z₂ = k (z₁ + z₂ + z_f)

⇒ z₁ + z₂ - k z₁ - k z₂ = k z_f

\(\Rightarrow {z_f} = \frac{{\left( {{z_1} + {z_2}} \right)\left( {1 - k} \right)}}{k}\)

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.

                      Fig. 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:

i_f = √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) ;

i_f = √3 × i₁

I_f = -j2.887 pu

Important Points

• If grounding is not provided or neutral is isolated then there is always zero-sequence current will be zero.

• Positive sequence current is equal and opposite  to negative sequence current in L-L fault.
• Both faulted phase voltages (vb and vc)  will be equal and positive sequence and negative sequence voltage (v1 and v2) also will be equal in L-L  fault 
• v_b = v_c   and   v₁ = v₂  

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

\(\begin{bmatrix} I_{R0}\\ I_{R1}\\ I_{R2} \end{bmatrix}={1\over 3}\begin{bmatrix} 1&1 &1 \\ 1& \alpha &\alpha^2 \\ 1& \alpha^2 & \alpha \end{bmatrix} \begin{bmatrix} I_R\\ I_Y\\ I_B \end{bmatrix}\)

\(\begin{bmatrix} I_{R0}\\ I_{R1}\\ I_{R2} \end{bmatrix}={1\over 3}\begin{bmatrix} 1&1 &1 \\ 1& \alpha &\alpha^2 \\ 1& \alpha^2 & \alpha \end{bmatrix} \begin{bmatrix} 0\\ I_f\\ -I_f \end{bmatrix}\)

\(I_{R0} = {1\over 3}({0+ I_f-I_f })\)

\(I_{R0} = 0\)...........(i)

\(I_{R1} = {1\over 3}({0+\alpha I_f-\alpha ^2I_f })\)

\(I_f=\sqrt{3}I_{R1}\)

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

Voltage (E) = 11 kV

Phase voltage (E_ph) \(= \frac{{11}}{{\sqrt 3 }}kV = 6.35\;kV\)

Three phase fault current = 5610 A

\({I_{f\left( {3 - \phi } \right)}} = \frac{{{E_{ph}}}}{{{X_1}}}\) 

\(\Rightarrow 5610 = \frac{{6.35 \times {{10}^3}}}{{{X_1}}}\) 

⇒ X₁ = 1.132 Ω

L-L fault current = 6760 A

\({I_{f\left( {L - L} \right)}} = \frac{{\sqrt 3 {E_{ph}}}}{{{X_1} + {X_2}}}\) 

\(\Rightarrow 6760 = \frac{{\sqrt 3 \times 6.35 \times {{10}^3}}}{{1.132 + {X_2}}}\) 

⇒ X₂ = 0.495 Ω

L-G fault current = 8630 A

\({I_{f\left( {L - G} \right)}} = \frac{{3{E_{ph}}}}{{{X_1} + {X_2} + {X_0}}}\) 

\(\Rightarrow 8630 = \frac{{3 \times 6.35 \times {{10}^3}}}{{1.132 + 0.495 + {X_0}}}\) 

⇒ X₀ = 0.58 Ω

\({X_{base}} = \frac{{\left( {kV} \right)_{base}^2}}{{{{\left( {MVA} \right)}_{base}}}} = \frac{{{{\left( {11} \right)}^2}}}{{25}} = 4.84\;{\rm{\Omega }}\)