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

Problem Statement: A 30 MVA, 13.2 KV synchronous generator has a solidly grounded neutral. Its positive, negative, and zero sequence impedances are 0.30 pu, 0.40 pu, and 0.05 pu respectively. We need to determine the value of reactance that must be placed in the generator neutral so that the fault current for a line-to-ground fault of zero fault impedance does not exceed the rated line current.

Solution:

To solve this problem, we will calculate the required reactance step by step:

Step 1: Determine the base current

The base current for the system can be calculated using the formula:

I_base = (S_base × 10⁶) / (√3 × V_base)

Here:

• S_base = 30 MVA = 30 × 10⁶ VA
• V_base = 13.2 kV = 13.2 × 10³ V

Substituting the values:

I_base = (30 × 10⁶) / (√3 × 13.2 × 10³)

I_base = 1311.78 A

So, the base current of the system is 1311.78 A.

Step 2: Determine the per unit fault current for a line-to-ground fault

In the case of a line-to-ground fault, the equivalent impedance is given by:

Z_LG = Z₁ + Z₂ + (Z₀ || Z_N)

Where:

• Z₁ = Positive sequence impedance = 0.30 pu
• Z₂ = Negative sequence impedance = 0.40 pu
• Z₀ = Zero sequence impedance = 0.05 pu
• Z_N = Neutral reactance to be determined

The fault current is inversely proportional to the equivalent impedance. The fault current in per unit (I_f) for a line-to-ground fault can be expressed as:

I_f = 1 / Z_LG

We are tasked with limiting the fault current to the rated line current (i.e., 1 pu). Hence, the equivalent impedance must satisfy:

Z_LG = 1 pu

Substituting the values into the equation for Z_LG:

1 = 0.30 + 0.40 + (0.05 || Z_N)

Step 3: Solve for Z_N

The parallel combination of Z₀ and Z_N is given by:

(0.05 || Z_N) = (Z₀ × Z_N) / (Z₀ + Z_N)

Substituting this into the equation for Z_LG:

1 = 0.30 + 0.40 + [(0.05 × Z_N) / (0.05 + Z_N)]

Rearranging:

0.30 + 0.40 = 0.70

1 - 0.70 = (0.05 × Z_N) / (0.05 + Z_N)

0.30 = (0.05 × Z_N) / (0.05 + Z_N)

Cross-multiplying:

0.30 × (0.05 + Z_N) = 0.05 × Z_N

Expanding and simplifying:

0.015 + 0.30 × Z_N = 0.05 × Z_N

0.30 × Z_N - 0.05 × Z_N = 0.015

0.25 × Z_N = 0.015

Solving for Z_N:

Z_N = 0.015 / 0.25 = 0.06 pu

Step 4: Convert Z_N to ohms

To convert the per unit reactance into ohms, we use the base impedance:

Z_base = (V_base)² / S_base

Substituting the values:

Z_base = (13.2 × 10³)² / (30 × 10⁶)

Z_base = 5.808 ohms

Now, the neutral reactance in ohms is:

X_N = Z_N × Z_base

X_N = 0.06 × 5.808

X_N = 0.34848 ohms

The correct answer is option 4): (Double line to ground fault)

Concept:

A double-line-to-ground (DLG) fault is typically modeled as a short-circuit connection from two phases (i.e., Phases A and B) to a common point, with a fault resistance from the common point to the ground.

The two lines contact each other along with the ground. The probability of such types of faults is nearly 10 %.

All sequence networks are connected in parallel.

The given fault diagram is a Double line to ground fault

I_a0 + I_a1 + I_a2 = 0

V_a0 = V_a1 = V_a2

I_F = 3I_a0

Additional InformationThree Phase Line to Ground Fault:

The 3-phase to-ground faults are faults in which all the phases (A, B and C) are shorted together and they are grounded.

Three Phase Line to Line Fault:

The three-phase faults occur when having A, B and C phases are shorted together but the ground is not involved.

Line-to-ground fault:

A short circuit between one line and the ground, is very often caused by physical contact, for example, due to lightning or other storm damage. In transmission line faults, roughly 65% - 70% are asymmetric line-to-ground faults.

Line to line fault:

A line to line fault occurs when two conductors are short circuited

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

The impedance at a point of fault is,

\({X_{1eq}} = \left( {j0.35} \right)\parallel \left( {j0.65} \right)\) 

= j0.2275 pu

Negative sequence reactance is the same as positive sequence reactance

\({X_{2eq}} = {X_{1eq}} = j0.2275\;pu\) 

The zero sequence reactance network is,

Zero sequence impedance

Z_0eq = j0.25 ∥ j0.75 = j0.1875 pu

For a double line to fault at bus 1 is,

\({I_{a1}} = \frac{1}{{j0.2275 + j\frac{{\left( {0.2275} \right)\left( {0.1875} \right)}}{{0.2275 + 0.1875}}}} = - j3.02767\;pu\) 

\({I_{a0}} = \frac{{1 - \left( {j0.2275} \right)\left( { - j3.02767} \right)}}{{j0.1875}} = j1.65975\;pu\) 

I_f = 3I_a0 = j4.979 pu

We have the base current \({I_B} = \frac{{30 \times {{10}^3}}}{{\sqrt 3 \times 13.8}} = 1255.109\;A\)

The positive sequence current, \({I_p} = 4270\;A\)

Per unit current, \({I_{p.u.}} = \frac{{4270}}{{1255.109}} = 3.402\;p.u.\)

Fault impedance = 0.1 pu

X_n = 3z_f + 0.05 + 3x_n = 0.35 + 3x_n

\(\begin{array}{l} {I_{g1}} = \frac{{{E_a}}}{{{X_1} + \left( {\left. {{X_2}} \right|} \right.\left| {\left. {{X_0}} \right)} \right.\;\;}}\\ {X_0}\; = \;3{X_n}\; + \;0.35\\ 3.402 = \frac{{1.0}}{{0.15 + \frac{{0.15 \times \left( {3{X_n} + 0.35} \right)}}{{0.15 + 3{X_n} + 0.35}}}} \end{array}\)

We have the base current \({I_B} = \frac{{30 \times {{10}^3}}}{{\sqrt 3 \times 13.8}} = 1255.109\;A\)

The positive sequence current, \({I_p} = 4270\;A\)

Per unit current, \({I_{p.u.}} = \frac{{4270}}{{1255.109}} = 3.402\;p.u.\)

Fault impedance = 0.1 pu

X_n = 3z_f + 0.05 + 3x_n = 0.35 + 3x_n

\(\begin{array}{l} {I_{g1}} = \frac{{{E_a}}}{{{X_1} + \left( {\left. {{X_2}} \right|} \right.\left| {\left. {{X_0}} \right)} \right.\;\;}}\\ {X_0}\; = \;3{X_n}\; + \;0.35\\ 3.402 = \frac{{1.0}}{{0.15 + \frac{{0.15 \times \left( {3{X_n} + 0.35} \right)}}{{0.15 + 3{X_n} + 0.35}}}} \end{array}\)

The correct answer is option 4): (Double line to ground fault)

Concept:

A double-line-to-ground (DLG) fault is typically modeled as a short-circuit connection from two phases (i.e., Phases A and B) to a common point, with a fault resistance from the common point to the ground.

The two lines contact each other along with the ground. The probability of such types of faults is nearly 10 %.

All sequence networks are connected in parallel.

The given fault diagram is a Double line to ground fault

I_a0 + I_a1 + I_a2 = 0

V_a0 = V_a1 = V_a2

I_F = 3I_a0

Additional InformationThree Phase Line to Ground Fault:

The 3-phase to-ground faults are faults in which all the phases (A, B and C) are shorted together and they are grounded.

Three Phase Line to Line Fault:

The three-phase faults occur when having A, B and C phases are shorted together but the ground is not involved.

Line-to-ground fault:

A short circuit between one line and the ground, is very often caused by physical contact, for example, due to lightning or other storm damage. In transmission line faults, roughly 65% - 70% are asymmetric line-to-ground faults.

Line to line fault:

A line to line fault occurs when two conductors are short circuited

Problem Statement: A 30 MVA, 13.2 KV synchronous generator has a solidly grounded neutral. Its positive, negative, and zero sequence impedances are 0.30 pu, 0.40 pu, and 0.05 pu respectively. We need to determine the value of reactance that must be placed in the generator neutral so that the fault current for a line-to-ground fault of zero fault impedance does not exceed the rated line current.

Solution:

To solve this problem, we will calculate the required reactance step by step:

Step 1: Determine the base current

The base current for the system can be calculated using the formula:

I_base = (S_base × 10⁶) / (√3 × V_base)

Here:

• S_base = 30 MVA = 30 × 10⁶ VA
• V_base = 13.2 kV = 13.2 × 10³ V

Substituting the values:

I_base = (30 × 10⁶) / (√3 × 13.2 × 10³)

I_base = 1311.78 A

So, the base current of the system is 1311.78 A.

Step 2: Determine the per unit fault current for a line-to-ground fault

In the case of a line-to-ground fault, the equivalent impedance is given by:

Z_LG = Z₁ + Z₂ + (Z₀ || Z_N)

Where:

• Z₁ = Positive sequence impedance = 0.30 pu
• Z₂ = Negative sequence impedance = 0.40 pu
• Z₀ = Zero sequence impedance = 0.05 pu
• Z_N = Neutral reactance to be determined

The fault current is inversely proportional to the equivalent impedance. The fault current in per unit (I_f) for a line-to-ground fault can be expressed as:

I_f = 1 / Z_LG

We are tasked with limiting the fault current to the rated line current (i.e., 1 pu). Hence, the equivalent impedance must satisfy:

Z_LG = 1 pu

Substituting the values into the equation for Z_LG:

1 = 0.30 + 0.40 + (0.05 || Z_N)

Step 3: Solve for Z_N

The parallel combination of Z₀ and Z_N is given by:

(0.05 || Z_N) = (Z₀ × Z_N) / (Z₀ + Z_N)

Substituting this into the equation for Z_LG:

1 = 0.30 + 0.40 + [(0.05 × Z_N) / (0.05 + Z_N)]

Rearranging:

0.30 + 0.40 = 0.70

1 - 0.70 = (0.05 × Z_N) / (0.05 + Z_N)

0.30 = (0.05 × Z_N) / (0.05 + Z_N)

Cross-multiplying:

0.30 × (0.05 + Z_N) = 0.05 × Z_N

Expanding and simplifying:

0.015 + 0.30 × Z_N = 0.05 × Z_N

0.30 × Z_N - 0.05 × Z_N = 0.015

0.25 × Z_N = 0.015

Solving for Z_N:

Z_N = 0.015 / 0.25 = 0.06 pu

Step 4: Convert Z_N to ohms

To convert the per unit reactance into ohms, we use the base impedance:

Z_base = (V_base)² / S_base

Substituting the values:

Z_base = (13.2 × 10³)² / (30 × 10⁶)

Z_base = 5.808 ohms

Now, the neutral reactance in ohms is:

X_N = Z_N × Z_base

X_N = 0.06 × 5.808

X_N = 0.34848 ohms

We have the base current \({I_B} = \frac{{30 \times {{10}^3}}}{{\sqrt 3 \times 13.8}} = 1255.109\;A\)

The positive sequence current, \({I_p} = 4270\;A\)

Per unit current, \({I_{p.u.}} = \frac{{4270}}{{1255.109}} = 3.402\;p.u.\)

Fault impedance = 0.1 pu

X_n = 3z_f + 0.05 + 3x_n = 0.35 + 3x_n

\(\begin{array}{l} {I_{g1}} = \frac{{{E_a}}}{{{X_1} + \left( {\left. {{X_2}} \right|} \right.\left| {\left. {{X_0}} \right)} \right.\;\;}}\\ {X_0}\; = \;3{X_n}\; + \;0.35\\ 3.402 = \frac{{1.0}}{{0.15 + \frac{{0.15 \times \left( {3{X_n} + 0.35} \right)}}{{0.15 + 3{X_n} + 0.35}}}} \end{array}\)

The correct answer is option 4): (Double line to ground fault)

Concept:

A double-line-to-ground (DLG) fault is typically modeled as a short-circuit connection from two phases (i.e., Phases A and B) to a common point, with a fault resistance from the common point to the ground.

The two lines contact each other along with the ground. The probability of such types of faults is nearly 10 %.

All sequence networks are connected in parallel.

The given fault diagram is a Double line to ground fault

I_a0 + I_a1 + I_a2 = 0

V_a0 = V_a1 = V_a2

I_F = 3I_a0

Additional InformationThree Phase Line to Ground Fault:

The 3-phase to-ground faults are faults in which all the phases (A, B and C) are shorted together and they are grounded.

Three Phase Line to Line Fault:

The three-phase faults occur when having A, B and C phases are shorted together but the ground is not involved.

Line-to-ground fault:

A short circuit between one line and the ground, is very often caused by physical contact, for example, due to lightning or other storm damage. In transmission line faults, roughly 65% - 70% are asymmetric line-to-ground faults.

Line to line fault:

A line to line fault occurs when two conductors are short circuited

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

The impedance at a point of fault is,

\({X_{1eq}} = \left( {j0.35} \right)\parallel \left( {j0.65} \right)\) 

= j0.2275 pu

Negative sequence reactance is the same as positive sequence reactance

\({X_{2eq}} = {X_{1eq}} = j0.2275\;pu\) 

The zero sequence reactance network is,

Zero sequence impedance

Z_0eq = j0.25 ∥ j0.75 = j0.1875 pu

For a double line to fault at bus 1 is,

\({I_{a1}} = \frac{1}{{j0.2275 + j\frac{{\left( {0.2275} \right)\left( {0.1875} \right)}}{{0.2275 + 0.1875}}}} = - j3.02767\;pu\) 

\({I_{a0}} = \frac{{1 - \left( {j0.2275} \right)\left( { - j3.02767} \right)}}{{j0.1875}} = j1.65975\;pu\) 

I_f = 3I_a0 = j4.979 pu

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

\({I_{{a_1}}} = \frac{E}{{{Z_1} + \left( {{Z_2}\parallel {Z_0}} \right)}} = \frac{1}{{j0.25 + \left( {j0.35\parallel j0.10} \right)}} = j3.05\;pu\)

\({I_{{a_0}}} = - \frac{{{Z_2}}}{{{Z_0} + {Z_2}}}{I_{{a_1}}} = - j2.372\;pu\)

Similarly \({I_{{a_2}}} = - \frac{{{Z_0}}}{{{Z_0} + {Z_2}}}{I_{{a_1}}} = - j0.677\;pu\)

We know that \(\left[ {\begin{array}{*{20}{c}} {{I_a}}\\ {{I_b}}\\ {{I_c}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 1&1&1\\ 1&{{\alpha ^2}}&\alpha \\ 1&\alpha &{{\alpha ^2}} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{I_{{a_0}}}}\\ {{I_{{a_1}}}}\\ {{I_{{a_2}}}} \end{array}} \right]\)

\({I_b} = {I_{{a_0}}} + {\alpha ^2}{I_{{a_1}}} + \alpha {I_{{a_2}}}\)

α = 1∠120°

α² = 1∠-120°

⇒ I_b = 4.804 pu   

Since the star point is isolated from ground LLG fault is just like LL fault.

\({I_b} = - {I_c} = \frac{{ - j\;\sqrt 3 \times 1}}{{j\;0.35 + j\;0.25}} = - 2.887\;pu\)

First find rated current of generator

\({I_g}rated\; = \frac{{30000}}{{\sqrt 3 \times 13.2}} = 1312.16\;AmP\)

Taking rated voltage and MVA as base

\(Base\;Impedance = \frac{{{{\left( {13.2} \right)}^2}}}{{30}} = 5.88\;{\rm{\Omega }}\)

Z₁ = j0.30 PU, Z₂ = J 0.40 PU Z₀ = J 0

Double line to ground fault

I_F = 3I_a0

\( = \frac{{ - 3{E_a}}}{{{Z_0} + 3{Z_f}}} + \frac{{{Z_1}}}{{{Z_0} + 3{Z_f}}}.\frac{{3{E_a}}}{{{Z_1} + \frac{{{Z_2}\left( {{Z_0} + 3{Z_f}} \right)}}{{\left( {{Z_2} + {Z_0} + 3{Z_f}} \right)}}}}\)

\({I_F} = \frac{{{E_a}}}{{{Z_0} + 3{Z_f}}}( - 1 + \frac{{{Z_1}\left( {{Z_2} + {Z_0} + 3{Z_f}} \right)}}{{{Z_1}\left( {{Z_2} + {Z_0} + {Z_f}} \right) + {Z_2}\left( {{Z_0} + 3{Z_f}} \right)}}\)

\({I_F} = \frac{{ - 3{Z_2}{E_a}}}{{{Z_1}\left( {{Z_2} + {Z_0} + 3{Z_f}} \right) + {Z_2}\left( {{Z_0} + 3{Z_f}} \right)}}\)

\({I_F} = \frac{{ - 3{Z_2}{E_a}}}{{{Z_1}{Z_2} + \left( {{Z_0} + 32F} \right)\left( {{Z_1} + {Z_2}} \right)}}\)

\(\therefore \left| {\frac{{3 \times 1 \times J\;0.4}}{{J\;0.3 \times J\;0.4 + J\;\left( {0.3 + 0.4} \right)\left( {J\;0.05 + J\;3{X_n}} \right)}}} \right| = 1\)

∴ 0.12 + 0.7 (0.05 + 3X_n) = 1.2

X_n = 0.5 PU = 0.5 × 5.88 = 2.94 Ω 

Problem Statement: A 30 MVA, 13.2 KV synchronous generator has a solidly grounded neutral. Its positive, negative, and zero sequence impedances are 0.30 pu, 0.40 pu, and 0.05 pu respectively. We need to determine the value of reactance that must be placed in the generator neutral so that the fault current for a line-to-ground fault of zero fault impedance does not exceed the rated line current.

Solution:

To solve this problem, we will calculate the required reactance step by step:

Step 1: Determine the base current

The base current for the system can be calculated using the formula:

I_base = (S_base × 10⁶) / (√3 × V_base)

Here:

• S_base = 30 MVA = 30 × 10⁶ VA
• V_base = 13.2 kV = 13.2 × 10³ V

Substituting the values:

I_base = (30 × 10⁶) / (√3 × 13.2 × 10³)

I_base = 1311.78 A

So, the base current of the system is 1311.78 A.

Step 2: Determine the per unit fault current for a line-to-ground fault

In the case of a line-to-ground fault, the equivalent impedance is given by:

Z_LG = Z₁ + Z₂ + (Z₀ || Z_N)

Where:

• Z₁ = Positive sequence impedance = 0.30 pu
• Z₂ = Negative sequence impedance = 0.40 pu
• Z₀ = Zero sequence impedance = 0.05 pu
• Z_N = Neutral reactance to be determined

The fault current is inversely proportional to the equivalent impedance. The fault current in per unit (I_f) for a line-to-ground fault can be expressed as:

I_f = 1 / Z_LG

We are tasked with limiting the fault current to the rated line current (i.e., 1 pu). Hence, the equivalent impedance must satisfy:

Z_LG = 1 pu

Substituting the values into the equation for Z_LG:

1 = 0.30 + 0.40 + (0.05 || Z_N)

Step 3: Solve for Z_N

The parallel combination of Z₀ and Z_N is given by:

(0.05 || Z_N) = (Z₀ × Z_N) / (Z₀ + Z_N)

Substituting this into the equation for Z_LG:

1 = 0.30 + 0.40 + [(0.05 × Z_N) / (0.05 + Z_N)]

Rearranging:

0.30 + 0.40 = 0.70

1 - 0.70 = (0.05 × Z_N) / (0.05 + Z_N)

0.30 = (0.05 × Z_N) / (0.05 + Z_N)

Cross-multiplying:

0.30 × (0.05 + Z_N) = 0.05 × Z_N

Expanding and simplifying:

0.015 + 0.30 × Z_N = 0.05 × Z_N

0.30 × Z_N - 0.05 × Z_N = 0.015

0.25 × Z_N = 0.015

Solving for Z_N:

Z_N = 0.015 / 0.25 = 0.06 pu

Step 4: Convert Z_N to ohms

To convert the per unit reactance into ohms, we use the base impedance:

Z_base = (V_base)² / S_base

Substituting the values:

Z_base = (13.2 × 10³)² / (30 × 10⁶)

Z_base = 5.808 ohms

Now, the neutral reactance in ohms is:

X_N = Z_N × Z_base

X_N = 0.06 × 5.808

X_N = 0.34848 ohms