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

Ring distribution system

• A ring distribution system is an electrical power distribution network where feeders connect in a closed loop, providing multiple paths for power delivery.
• This design ensures that if one feeder fails, power can continue to flow through an alternative path, maintaining reliability and preventing power outages.
• There are fewer voltage fluctuations at the consumer's terminal.

The most common internal fault in transformers is winding inter-turn short circuits.

Winding inter-turn short circuits

An inter-turn short circuit in a transformer refers to a short between adjacent turns of a winding due to insulation failure. Unlike phase-to-phase or winding-to-core faults, these occur within the same winding, making them harder to detect early.

Causes of Inter-Turn Short Circuit in Transformers:

• Insulation degradation due to thermal aging
• Overvoltages / surges
• Mechanical vibrations or shocks during transport or operation
• Moisture ingress leading to partial discharge
• Manufacturing defects
• High inrush or fault currents causing turn displacement

Explanation:

Fault Analysis in Power Systems

In power systems, faults are abnormal conditions that disrupt the normal flow of electrical current. Fault analysis is crucial to understand the type of fault, its behavior, and its impact on the system. Faults are typically categorized as single line to ground faults, double line to ground faults, line to line faults, and three-phase faults. Each type of fault has unique characteristics that influence the symmetrical components of the system, which are represented by the components Ia0 (zero-sequence current), Ia1 (positive-sequence current), and Ia2 (negative-sequence current).

The question asks us to determine the type of fault in which all three components—Ia0, Ia1, and Ia2—are equal. The correct answer is Option 1: Single line to ground fault. Below is a detailed explanation:

Single Line to Ground Fault:

A single line to ground (SLG) fault is the most common type of fault in power systems and occurs when one phase of the system comes into contact with the ground or a grounded component. This fault can be represented by the following conditions:

• One phase current becomes significantly larger than the currents in the other two phases.
• The fault current flows through the ground or the grounded component.
• The system becomes unbalanced, leading to the presence of zero-sequence, positive-sequence, and negative-sequence components.

In an SLG fault, the following condition holds:

Ia0 = Ia1 = Ia2

This equality occurs because the fault creates a condition where the faulted phase current splits equally among the three symmetrical components. The symmetrical component method is commonly used to analyze unbalanced faults, and for an SLG fault, all three components—zero-sequence, positive-sequence, and negative-sequence currents—are equal in magnitude and phase.

Mathematical Representation:

Assume an SLG fault occurs on phase a. The fault current Ia can be expressed in terms of symmetrical components:

Ia = Ia0 + Ia1 + Ia2

For an SLG fault, the relationship between the symmetrical components is such that:

Ia0 = Ia1 = Ia2

Hence, the faulted phase current Ia becomes:

Ia = 3 × Ia0 (since Ia0 = Ia1 = Ia2)

Meanwhile, the currents in the other two phases (Ib and Ic) remain zero under ideal conditions. This characteristic makes SLG faults unique and relatively straightforward to analyze using symmetrical components.

Impact of SLG Faults:

• SLG faults are highly unbalanced and can cause significant damage to equipment if not cleared promptly.
• They are the most frequent type of fault, accounting for approximately 70-80% of all faults in power systems.
• Protective relays and circuit breakers are designed to detect and clear SLG faults to prevent further damage.

Additional Information

To further understand the analysis, let’s evaluate the other fault types:

Option 2: Double Line to Ground Fault:

A double line to ground (DLG) fault occurs when two phases come into contact with the ground or a grounded component. In this case, the symmetrical components are not equal. The zero-sequence component (Ia0) depends on the grounding impedance, while the positive-sequence (Ia1) and negative-sequence (Ia2) components are influenced by the fault location and the system's impedance. Therefore, Ia0 ≠ Ia1 ≠ Ia2 for a DLG fault.

Option 3: Line to Line Fault:

A line to line (LL) fault occurs when two phases come into contact with each other. In this type of fault, there is no zero-sequence current (Ia0 = 0) because the fault does not involve the ground. The positive-sequence (Ia1) and negative-sequence (Ia2) components are present and are generally not equal. Thus, Ia0 ≠ Ia1 ≠ Ia2 for an LL fault.

Option 4: Three Phase Fault:

A three-phase fault is a balanced fault in which all three phases short-circuit together, either with or without ground involvement. In this case, only the positive-sequence component (Ia1) is present, while the zero-sequence (Ia0) and negative-sequence (Ia2) components are zero. Thus, Ia0 = 0 and Ia2 = 0, and only Ia1 exists, making this fault type entirely different from the SLG fault.

Conclusion:

The correct answer is Option 1: Single line to ground fault, as it is the only fault type where all three symmetrical components—Ia0, Ia1, and Ia2—are equal. Understanding the behavior of symmetrical components for different fault types is essential for fault analysis and the design of protective systems in power networks.

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.

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

Electrical faults in three-phase power system mainly classified into two types, namely open and short circuit faults.

Open circuit faults: These faults occur due to the failure of one or more conductors.  The most common causes of these faults include joint failures of cables and overhead lines, and failure of one or more phase of circuit breaker and also due to melting of a fuse or conductor in one or more phases.

Type of open circuit faults:

• Single phase open circuit fault
• Two phase open circuit fault
• Three phase open circuit fault

Open circuit faults are also called as series faults. These are unsymmetrical or unbalanced type of faults except three phase open fault.

Important Points

Shunt faults:

The shunt fault involves short circuit between conductor and ground or short circuit between two or more conductors.

The shunt faults are characterized by:

• Increase in current
• Fall in voltage
• Fall in frequency

Shunt faults are classified as follows:

• Single line to Ground fault (LG fault)
• Line to Line fault (LL fault)
• Double line to Ground fault (LLG fault)
• Three phase faults

Unsymmetrical Fault:

The fault gives rise to unsymmetrical current, i.e., current differing in magnitude and phases in the three phases of the power system are known as the unsymmetrical fault. It is also defined as the fault which involves one or two phases such as LG, LL, LLG fault. The unsymmetrical makes the system unbalanced. 

Single Line to Line Ground (SLG): 

• The single line of ground fault occurs when one conductor falls to the ground or contact the neutral conductor. 
• The 70 – 80 percent of the fault in the power system is the single line-to-ground fault.
• All sequence networks are connected in series.

​Ia0 = Ia1 = Ia2

I_F = 3Ia0 = 3Ia1 =3 Ia2

\({I_F} = \frac{{3{E_a}}}{{{Z_1} + {Z_2} + {Z_3}}}\)

Line to Line Fault (LL):  

• A line-to-line fault occurs when two conductors are short-circuited. The major cause of this type of fault is the heavy wind.
• The heavy wind swinging the line conductors which may touch together and hence cause short-circuit. 
• The percentage of such types of faults is approximately 15 – 20%.
• Positive and negative sequence connected in series opposition

Ia0 =0 and Ia1 = - Ia2

I_F = √3 Ia1 = √3 Ia2

\({I_F} = \frac{{√ 3 {E_a}}}{{{Z_1} + {Z_2}}}\)

Double Line to line Ground Fault (LLG): 

• In double line-to-ground fault, the two lines come in contact with each other along with the ground. 
• The probability of such types of faults is nearly 10%. 
• All sequence networks are connected in parallel.

Ia0 + Ia1 + Ia2 = 0

Va0 = Va1 = Va2

IF = 3Ia0

a - 3, b - 1, c - 2

Additional Information

Frequency of occurrence:

• Among the given faults, LG or line to ground fault is most common and occurs frequently.
• The order of frequency of occurrence is given below.
   

LG > LL > LLG > LLLG

Severity of faults:

• Among the given faults, LLLG or 3 phase faults are most severe. LG or line to ground fault is least severe.
• Line to line fault is more severe than line to ground fault while double line to ground fault is one level severe than LL.
• The order of severity of faults given below.
   

LLLG > LLG > LL > LG

Concept:

Short circuit MVA = \( \frac{{\;\ MVA}_{base}}{{{X_{pu}}}}\)

Where,

MVA_base = Full load or base MVA

X_pu = Per unit reactance

Calculation:

Given that, MVA_base = 500 kVA = 0.5 MVA, X_pu= 5% = 0.05

Since the transformer is connected to an infinite bus, the p.u. the reactance of the circuit will be 0.05 i.e., the p.u. reactance offered by the transformer.

∴ short circuit MVA can be calculated as

Short circuit MVA = \( \frac{{\ 0.5}}{{{0.05}}} =10 ~MVA\)

Concept:

In a single line to ground fault, all the sequence components of fault currents are equal.

I_a1 = I_a2 = I_a0

In a double line to ground fault, the sum of all sequence components of fault currents is zero.

I_a1 + I_a2 + I_a0 = 0

Where, Ia1 = Positive sequence component of the current

Ia2 = Negative sequence component of the current

Ia0 = Zero sequence component of the current

Calculation:

Given

I_positive = j 1.5 p.u. 

I_negative = - j 0.5 p.u. 

I_zero = - j 1 p.u.

Ia1 = j 1.5, Ia2 = -j 0.5, Ia0 = -j 1.0

⇒ Ia1 + Ia2 + Ia0 = j 1.5 – j 0.5 – j 1.0 = 0

Therefore, the fault is double line to ground fault.

Concept:

\({Z_{pu}} = {Z_{\rm{\Omega }}}\frac{{{{\left( {MVA} \right)}_b}}}{{{{\left( {k{V_b}} \right)}^2}}}\)

Z_pu = per unit impedance

Z_Ω = Impedance in Ω

(MVA)_b = base MVA

(kV)_b = base voltage

Calculation:

Base MVA = 100 kVA = 0.1 MVA

Base Voltage = 5 kV

Reactance in ohms = 5 Ω

\({Z_{pu}} = 5 \times \frac{{0.1}}{{{5^2}}} = 0.02\)

The correct answer is option 2): (200 A )

Concept:

The short circuit current is given by

I_sc = I × \(100 \over percentage\: of \: reactance\)

Where 

I is the full load current 

Calculation:

Isc = I × \(100 \over percentage\: of \: reactance\)

= 40 × \(100 \over 20\)

 = 200 A 

The different type of faults in power systems are:

• Single line to ground fault (LG)
• Line to line fault (LL)
• Double line to ground fault (LLG)
• Three-phase faults (LLL or LLLG)

Frequency of occurrence:

• Among the given faults, LG or line to ground fault is most common and occurs frequently.
• The order of frequency of occurrence is given below.

LG > LL > LLG > LLL

Severity of faults:

• Among the given faults, LLLG or 3 phase faults are most severe. LG or line to ground fault is least severe.
• Line to line fault is more severe than line to ground fault while double line to ground fault is one level severe than LL.
• The order of severity of faults is given below.

LLL > LLG > LL > LG

Concept:

The short-circuit MVA is given by:

SCMVA = \( {1\over X_{eq}}\times(MVA)_{base}\)

Calculation:

Given, (MVA)_base = 5 MVA

X_eq = \(20\over 4\) = 5% = 0.05 pu

SCMVA = \( {1\over 0.05}\times5\)

SCMVA = 100 MVA

Positive sequence component: It has three vectors of equal magnitude but displaced in phase from each other by 120° and has the same phase sequence as the original vectors. It specifies that the current is flowing through the source to load.

Negative sequence component: It has three vectors of equal magnitude but displaced in phase from each other by 120° and has the phase sequence opposite to the original vectors. It specifies that the current is flowing from load to source.

Zero sequence component: It has three vectors of equal magnitude and also are in phase with each other. It specifies that the current is flowing from source to ground.

• Positive sequence components exist in both balanced and unbalanced conditions.
• Negative sequence components exist in unbalanced conditions only.
• Zero sequence components exist in the unbalanced conditions involving ground.

In double line fault or line to line fault, there is no ground. Therefore, the zero sequence current is zero.

Concept

The fault current in different types of faults are:

1.) Short circuit fault

\(I_{3\phi}={E_a\over X_1}\)

2.) Single line to ground fault

\(I_{LG}={3E_a\over X_1+X_2+X_0}\)

3.) Line-to-line fault

\(I_{LL}={\sqrt{3}E_a\over X_1+X_2}\)

where, X₁ = Positive sequence reactance

X₂ = Negative sequence reactance

X_o = Zero sequence reactance

E_a = Generator voltage

Generally, the fault current is calculated at no-load condition, hence E_a = 1 pu

Calculation

Given, X1 = 0.12 pu

X2 = 0.12 pu

Xo = 0.06 pu

\(I_{3\phi}={1\over 0.12}=8.33 \space pu\)

\(I_{LG}={3\over 0.12+0.12+0.06}=10\space pu\)

\({I_{3\phi}\over I_{LG}}={8.33\over 10}=0.833\)