৪৯তম বিসিএস ⎯ তথ্য ও যোগাযোগ প্রযুক্তি (EEE) [ ৮৯২]

Reduces volume of conductor material: High-voltage transmission reduces the need for large amounts of conductor material for the same power transmission capacity. This is because power loss in the form of heat (I²R losses) decreases when voltage is increased. High voltage allows for transmitting the same amount of power using lower current, and since I²R losses are directly proportional to the square of the current, reducing the current reduces the losses. Consequently, the size of conductors can be smaller for the same power, reducing material requirements and the overall cost of construction.

Increases transmission efficiency:High-voltage transmission significantly increases the efficiency of power transmission. When voltage is increased, the current decreases for the same amount of power. The reduction in current minimizes resistive losses (I²R losses) in the transmission lines, making the process more efficient. High-voltage transmission helps ensure that more of the generated power reaches its destination with minimal loss, which is especially important in long-distance transmission.

Decreases percentage line drop: The line voltage drop is directly related to the current flowing through the line. By increasing the voltage, the current is decreased, which in turn reduces the percentage line voltage drop.
A smaller voltage drop results in more stable and efficient transmission because the voltage at the receiving end remains closer to the required value, reducing losses and improving the quality of power delivered.

{"Power System Analysis and Design" by J. Duncan Glover, Thomas Overbye, and M. S. Sarma}

The corona effect is the phenomenon where ionization of air occurs around a conductor when the voltage on the conductor exceeds a certain threshold. This leads to the creation of a visible blue glow around the conductor, accompanied by energy loss, noise, and possible damage to the conductor and surrounding materials.
 

A stranded conductor is made up of multiple smaller wires twisted together. This increases the effective surface area of the conductor compared to a solid conductor of the same cross-sectional area.
A larger surface area results in more ionization of air around the conductor, which enhances the corona effect.
In a stranded conductor, the individual strands often do not form a perfectly smooth surface, unlike a solid conductor. The small gaps and irregularities between the strands lead to local high electric fields. These localized high fields can increase the likelihood of corona discharge compared to a smooth, solid conductor surface.
In a solid conductor, the voltage stress is more uniformly distributed across the surface of the conductor. In a stranded conductor, however, the voltage is not as evenly distributed due to the nature of the structure, leading to higher local electric fields and a greater tendency for corona discharge.

("Electric Power Transmission and Distribution" by S. Sivanagaraju and C. S. Indulkar)

Resistance of a Conductor:

The resistance (R) of a transmission line is given by the formula:
 R=ρ×L/A

Where:

ρ is the resistivity of the material (a constant),
L is the length of the conductor,
A is the cross-sectional area of the conductor.

ffect of Length on Resistance:

As L (length) increases, the resistance R increases proportionally, assuming the material and cross-sectional area (A) remain constant. This is because the resistance is directly proportional to the length of the conductor.
In simple terms, the longer the transmission line, the greater the total resistance because the electrical current must travel through a longer path, leading to more opposition to the current flow.

("Electric Power Transmission and Distribution" by S. Sivanagaraju and C. S. Indulkar)

Why is the Capacitance Neglected in the Case of a Short Transmission Line?

Short Transmission Line: A short transmission line typically refers to a transmission line whose length is relatively short (usually less than 50 km). For such lines, the impedance (resistance and inductive reactance) is much more significant compared to the capacitance.

In short transmission lines, the voltage and current variations over the length of the line are relatively small, and the effect of capacitance is negligible. Therefore, the capacitance is typically ignored in the analysis of short lines to simplify the calculations.

Capacitance in Transmission Lines: The capacitance of a transmission line exists due to the electrostatic interaction between the conductors and the ground or between conductors themselves. However, in short lines, the capacitive reactance is high and the line's length is so short that the voltage drop and other effects of capacitance are minimal. Hence, resistance and inductive reactance dominate the behavior of the line."

"Electrical Power Systems" by C.L. Wadhwa"

Why the Capacitance Between One Conductor and a Neutral Point is 2C:

Capacitance in a Single-Phase, 2-Wire Transmission Line:

In a single-phase, 2-wire line, there are two conductors, and the capacitance between these two conductors is C. This capacitance is typically due to the electrostatic interaction between the two conductors.
Capacitance Between a Conductor and Neutral:

If we introduce a neutral point between the two conductors, the total capacitance between the conductors is distributed such that each conductor experiences a capacitance relative to the neutral.
In this case, the total capacitance between the two conductors is C, and the capacitance of each conductor with respect to the neutral will be half of the total. However, since both conductors are symmetrical, the total capacitance seen from one conductor to the neutral point is actually twice the individual capacitance between one conductor and the other.
Therefore, the capacitance between one conductor and the neutral will be 2C.

("Electrical Power Systems" by C.L. Wadhwa)

What is Skin Effect?
The skin effect refers to the phenomenon in alternating current (AC) systems where the current tends to flow more on the surface of the conductor as the frequency of the AC increases. This is due to the electromagnetic field created by the AC, which induces opposing currents within the conductor, forcing the majority of the current to flow near the surface.

Reduces effective area of conductor: Skin effect effectively reduces the effective cross-sectional area of the conductor that carries current. As the current concentrates near the surface, the core of the conductor is not fully utilized, reducing the area through which current flows.

Increases resistance of conductor:The current crowding near the surface of the conductor means that less of the conductor is actively conducting the current. This increases the resistance of the conductor because the current is confined to a smaller surface area. The resistance increases with the frequency of the AC due to the skin effect.

Causes greater power loss:Since the effective resistance of the conductor increases due to the skin effect, it results in higher power losses in the form of I²R losses. The greater the resistance, the higher the energy dissipation in the form of heat.("Electric Power Transmission and Distribution" by S. Sivanagaraju and C. S. Indulkar)

Factors Affecting the Skin Effect:

(i) Nature of Material: The material of the conductor affects the skin effect. Materials with higher conductivity (like copper and aluminum) will experience a different skin effect compared to materials with lower conductivity (like steel or iron).
The magnetic permeability of the material also plays a role. For example, ferromagnetic materials (such as iron) have higher permeability and therefore exhibit a more pronounced skin effect compared to non-ferromagnetic materials.

(ii) Diameter of Wire: The diameter of the conductor affects the skin effect. For larger diameter conductors, the depth of current penetration increases. This means that for larger conductors, the skin effect is more pronounced and the current is confined to a smaller outer layer. Conversely, for smaller diameter conductors, the current is more evenly distributed throughout the conductor.
The larger the conductor, the more pronounced the skin effect becomes, as the current tends to flow only along the outer surface at higher frequencies.

(iii) Frequency: The frequency of the alternating current is directly related to the skin effect. As the frequency of the current increases, the skin effect becomes stronger. At higher frequencies, the current is pushed more to the surface of the conductor, leaving the inner part of the conductor with little or no current flow.
This is due to the inductive reactance increasing with frequency, which forces the current to travel along the outer surface to minimize the total impedance.

("Electric Power Transmission and Distribution" by S. Sivanagaraju and C. S. Indulkar)

Given Data:
The following data is given for the 3-phase, 3-wire system:

1. Distance between conductors D₃₁ = 4m

2. Distance between conductors D₁₂ = D23 = 2m

3. Diameter of each conductor = 2.5 cm (0.025 m)

4. The conductors are transposed

Explanation of Calculation:
The equivalent equilateral spacing is the distance between the conductors arranged such that the system is equivalent to having three conductors placed at equal distances from each other in an equilateral triangle arrangement. This can be calculated using the following formula for three conductors in a horizontal plane (with unequal spacing):

Let D₁₂, D₂₃, and D₃₁ be the distances between the conductors.

The formula for the equivalent equilateral spacing is given by:

E = (D₁₂ × D_{23 ×} D₃₁)^(1/3)

Substituting the given values:

E = (2 × 2 × 4)^(1/3) = (16)^(1/3) ≈ 2.52 m or 252 cm.

How Capacitance Affects a Transmission Line:

(i) Affects Transmission Efficiency:The capacitance of a transmission line can impact transmission efficiency, especially in long-distance transmission. The line capacitance causes a charging current, which increases the total current flowing through the line. This additional current leads to increased I²R losses, thereby reducing the transmission efficiency.

(ii) Affects Power Factor:The charging current due to the line's capacitance also affects the power factor. Since the charging current leads the voltage by 90º (purely reactive), it introduces a reactive power component. This reactive power does not contribute to the active power delivered to the load but affects the overall power factor of the system.
A high reactive power increases the phase difference between voltage and current, reducing the power factor.

(iii) Causes Charging Current:Capacitance inherently causes a charging current in the transmission line, which is a reactive current. The magnitude of this charging current increases with the length of the transmission line and the operating voltage. This current, while not delivering power, must still be supplied by the source and can affect both the voltage profile along the line and the efficiency of the transmission system.

("Power Transmission and Distribution" by G. R. Nagpal)

How Does the Supply Frequency Affect the Skin Effect?
Increase in Frequency: As the frequency of the AC increases, the skin effect becomes more pronounced. Higher frequencies cause the current to concentrate more on the outer surface of the conductor, effectively reducing the effective cross-sectional area of the conductor carrying the current.

This happens because higher frequencies induce stronger opposing currents (eddy currents) within the conductor, pushing the main current to the surface.
Impact on Resistance and Losses:With an increase in frequency, the effective resistance of the conductor increases because the current is confined to a smaller surface area. This can lead to higher power losses in the form of I²R losses, especially in high-frequency AC transmission systems.

("Power Transmission and Distribution" by G. R. Nagpal)

In a 3-phase transmission line, the total capacitance between two conductors (let’s call them C__AB) is shared between the conductors and the neutral. The capacitance between any conductor and the neutral is half of the total capacitance between two conductors because each conductor contributes equally to the total system capacitance.

Capacitance Between Each Conductor and Neutral:
The capacitance between each conductor and the neutral (C__A-N or C__B-N) will be half of the total capacitance between the two conductors. This is because the system is symmetrical, and the total capacitance is divided equally.
C_AB = 4µF
C_AN or C_BN= 2 × C_AB= 2 × 4 = 8 µF 

The following data is given for the short single-phase transmission line:

1. Loop impedance: Z = (0.2 + j 1.2) Ω

2. The voltage regulation needs to be zero

Explanation of Calculation:
Voltage regulation is zero when the voltage at no load equals the voltage at full load. This happens when the load's reactive power compensates exactly for the reactive voltage drop across the line impedance.

The voltage regulation can be defined as:

Voltage Regulation = (|V_no load| - |V_full load|) / |V_full load| × 100

For voltage regulation to be zero, the voltage at full load must match the voltage at no load. This condition is met when the reactive power generated by the load perfectly compensates for the line’s reactive voltage drop.

The impedance is Z = 0.2 + j 1.2 Ω, and the power factor angle φ is calculated using the formula:

tan(φ) = 1.2 / 0.2 = 6

From this, we find that φ = tan^{-1}(6) ≈ 80.5°. The power factor angle required to cancel the reactive voltage drop is approximately 80.5°, and the corresponding power factor is 0.986 lead.

What are Bundled Conductors?
Bundled conductors refer to the use of multiple smaller conductors in parallel for each phase of a transmission line, instead of using a single large conductor. These conductors are arranged in a geometric pattern, often in the form of a bundle, to achieve specific electrical benefits.

Effects of Bundled Conductors:
Reduce Corona Loss:Corona loss occurs when the electric field around the conductor exceeds a critical value, ionizing the air around the conductor and causing energy to be lost in the form of heat and radiation.
Bundling conductors reduces the electric field intensity around each conductor because the conductors in the bundle are spaced further apart from each other compared to a single conductor. This reduces the likelihood of the electric field reaching the critical value needed for corona discharge, thereby reducing corona loss.

Reduce Line Inductance: The use of bundled conductors also affects the inductance of the transmission line. The total inductance of the line depends on the spacing between the conductors. By increasing the spacing between the conductors in a bundle (compared to a single conductor), the mutual inductance between the conductors decreases.
As a result, the overall inductance of the transmission line is reduced, improving the efficiency of power transmission.

"Power System Analysis" by John J. Grainger and William D. Stevenson

In power transmission systems, the classification of transmission lines is based on their length and impedance characteristics. A short transmission line typically refers to a line whose length is relatively short, and where the resistive and reactive components of the line are dominated by the series impedance. A short transmission line is generally defined as a line whose length is up to 50 km. This length ensures that the line inductance and capacitance do not significantly affect the performance of the line, and the voltage drop across the line is relatively simple to calculate using basic equations that focus primarily on series impedance (resistance and inductive reactance).

(Principals of Power System "V.K metha")

The resistance drop in a transmission line is given by the formula:

V_drop = I × R

Where:

V_drop = Voltage drop across the transmission line
I = Current flowing through the line
R = Resistance of the transmission line

Explanation:
The current in the transmission line is inversely proportional to the transmission voltage for a constant power, as given by the equation:

I = P / V

Where:

P = Power transmitted
V = Transmission voltage

When the transmission voltage increases, the current decreases. Since the resistance drop (V_drop) is proportional to the current (I), a decrease in current will lead to a decrease in the resistance drop. As a result, the percentage resistance drop also decreases.((Principals of Power System "V.K metha")

Given Data:
The voltage regulation of a transmission line is the difference between the no-load voltage and the full-load voltage, expressed as a percentage of the full-load voltage.

Explanation of Calculation:
For a short transmission line, the total impedance consists of resistance (R) and inductive reactance (X_L). The capacitance is neglected. The voltage regulation is calculated based on the voltage drop across the line's impedance.

At unity power factor , the current in the line is purely resistive, and the voltage drop is determined by the resistance.

The voltage regulation is calculated as:

Voltage Regulation = (|V_{no load}| - |V_{full load}|) / |V_{full load}| × 100

The voltage drop is given by:

V_drop = I × (R + jX_L)

At unity power factor, the voltage drop is due to the resistance (R), and the voltage regulation becomes:

Voltage Regulation = I × R

Steel is Used for Cross-Arms in Transmission Lines. Reasons:
Strength and Durability: Steel is commonly used for cross-arms in transmission lines due to its high strength and durability. Steel cross-arms can withstand the mechanical stresses and forces caused by the weight of the conductors, wind, and other environmental factors.
Load-Bearing Capacity: Transmission lines require cross-arms to support the weight of the conductors, insulators, and other components. Steel provides the necessary load-bearing capacity and can resist deformation over time.
Corrosion Resistance: Steel cross-arms are typically galvanized or coated with anti-corrosive materials to protect them from rust and corrosion. This ensures that they can last for a long time even in harsh outdoor environments.
Cost-Effectiveness: Compared to materials like aluminium or copper, steel is a more cost-effective option for constructing the structural components of transmission lines."Power Transmission and Distribution" by G. R. Nagpal

When transmitting power over a distance, the transmission voltage is chosen to minimize losses and maximize efficiency. High-voltage transmission reduces the current for the same amount of power, which in turn reduces the resistive losses (I²R losses) in the transmission line.

For the given scenario:

Power to be transmitted = 30 MW (30,000 kW)
Distance = 70 km
The power loss in the line is proportional to the square of the current (I2RI^2 RI2R), and since P=V×IP = V \times IP=V×I, for a given power, increasing the voltage reduces the current, which lowers the losses in the line.

Transmission Voltage Consideration:
At lower voltages (e.g., 11 kV, 33 kV), the current would need to be higher to transmit the same amount of power. This would lead to higher losses and inefficiencies.
66 kV is a typical transmission voltage for medium-distance power transmission (70 km in this case), as it strikes a balance between minimizing current and maintaining manageable equipment sizes and costs.
General Guidelines for Transmission Voltages:
Low voltage (e.g., 3.3 kV to 11 kV) is used for short distances and low power levels.
Medium voltage (e.g., 33 kV to 66 kV) is commonly used for medium-distance transmission, especially for 50 km to 100 km ranges.
High voltage (e.g., 110 kV, 220 kV, 400 kV) is used for long-distance and high power transmission (two hundreds of kilometers).
Thus, 66 kV is the optimal voltage for transmitting 30 MW of power over a 70 km distance, as it reduces transmission losses effectively while ensuring equipment efficiency. "V.K mehta"

In the analysis of medium transmission lines, localized capacitance methods are used to approximate the line’s capacitance by considering it to be concentrated at certain points along the line. These methods make it easier to model the behavior of the transmission line, especially in terms of voltage drops, power losses, and the effect of capacitance.

Localized capacitance methods assume that the capacitance of the line is concentrated at specific locations (like the ends of the line or in segments), which simplifies the analysis.

Localized Capacitance Methods for Medium Transmission Lines:
End Condenser Method:In this method, the capacitance of the line is assumed to be concentrated at the ends of the line. This simplifies the calculation of the voltage and current at the receiving and sending ends, especially for medium-length lines.
Nominal T Method:This method assumes that the capacitance is concentrated at the midpoint of the transmission line. It’s a simplification used for analyzing the performance of medium transmission lines.
Nominal π Method: Similar to the Nominal T method, the Nominal π method divides the line into segments, placing the capacitance at both the sending and receiving ends of the line. This method is also used to simplify the analysis of medium transmission lines. ((Principals of Power System "V.K metha"))

Distributed Constant Method (ঘ) – Not a Localized Capacitance Method:
The Distributed Constant Method considers that the line’s impedance and capacitance are distributed uniformly along the entire length of the line.
Unlike the localized capacitance methods, which treat capacitance as concentrated at specific points, the distributed constant method treats the line parameters as being continuously distributed over the entire line. This method is more accurate and is used for long transmission lines where the line effects such as capacitance and inductance need to be considered at every point along the line.

Understanding the End Condenser Method:

The End Condenser Method is a simplified approach used for analyzing medium transmission lines, especially in cases where the line’s capacitance plays a significant role. This method assumes that the total capacitance of the transmission line is concentrated at the ends of the line.

While this method helps simplify calculations by reducing the complexity of the line model, it introduces a key drawback in terms of accuracy.

Why Does the End Condenser Method Overestimate Capacitance?

Capacitance at the Ends: By assuming that the entire capacitance is located at the ends of the line, the method overestimates the actual capacitance effects, as the capacitance is not distributed evenly along the line in reality.

Line Length Consideration: For short to medium lines, this assumption can lead to an overestimation of the capacitance, which causes the method to overestimate the voltage drop and other effects related to capacitance.

Resulting Error: Typically, the overestimation leads to about a 10% error in the predicted line behavior, which is significant for accurate power transmission analysis.

cos φ_R = 0·8 ; sin φ_R = 0·6

IR = IR (cos φR − j sin φR) = 284 (0·8 − j 0·6) = 227 − j 170

Sending end current, I_S = I_R + I_C + = (227 − j 170) + j 92 = 227 − j 78

The phasor sum of currents:

I_S = sqrt{227² + 78²} ≈ 240 A

In the Nominal T method, the total line capacitance is assumed to be concentrated at the middle point of the transmission line. This simplification is made for analytical purposes in the study of transmission lines. Capacitance Distribution: In reality, the capacitance is distributed along the entire length of the transmission line. However, for simplicity, the Nominal T method models this distributed capacitance as if it were concentrated at the midpoint of the line.

This assumption makes the analysis more manageable, as it allows for easier calculation of the charging current and voltage drops across the transmission line. By considering the capacitance as concentrated at the middle, the Nominal T method effectively balances the complexity of real-world distribution with the practical needs of simplifying the model.

Reducing the capacitance to a single concentrated point (the middle) simplifies calculations, particularly when solving for the current and voltage in the transmission line. The assumption helps in estimating the effects of line capacitance on transmission line behavior, particularly for long lines, without having to account for the distributed nature of capacitance along the entire length.

"(Principals of Power System "V.K metha")

 
Given Data:
The following constants are given for the overhead transmission line:
Resistance/km/phase = 0.1 Ω
Inductive reactance/km/phase = 0.2 Ω
Capacitive susceptance/km/phase = 0.04 × 10⁻⁴ siemen
Transmission Line Length = 100 km
Load = 10,000 kW at 66 kV
p.f. = 0.8 lagging
Frequency = 50 Hz

We begin by calculating the total resistance, reactance, and capacitive susceptance per phase:
Total Resistance/phase, R = 0.1 × 100 = 10 Ω
Total Reactance/phase, X_L = 0.2 × 100 = 20 Ω
Capacitive Susceptance, Y = 0.04 × 10^−4 × 100 = 4 × 10^−4 S

The receiving end voltage per phase (VR) is given as:
Receiving End Voltage, V_R = 66,000 / √3 = 38,105 V

Now, the load current is calculated as:
Load Current, I_R = 10,000 × 10^3 / (3 × 66,000 × 0.8) = 109 A
Using the impedance per phase, Z = R + jX_L = 10 + j20 Ω,
Taking the receiving end voltage as the reference phasor, we have,
Load current, I_R = I_R (cos φ_R − j sin φ_R) = 109 (0.8 − j 0.6) = 87.2 − j 65.4 A

The voltage across the capacitor (V1) is:
V₁ = V_R + (I_R × Z/2) = 38,105 + (87.2 − j 65.4) × (5 + j 10) = 39,195 + j 545 V
Charging current (I_C) is calculated as:
I_C = j × Y × V1 = 4 × 10^−4 × (39,195 + j545) = −0.218 + j15.6 A

Finally, the sending end current is:
Sending End Current, I_S = I_R + I_C = (87.2 − j 65.4) + (−0.218 + j 15.6) = 87.0 − j 49.8 A
Sending End Current = 100 ∠29º47′ A

(Principals of Power System "V.K metha")

The line drop refers to the reduction in voltage as the electrical power travels from the sending end to the receiving end, primarily caused by the resistance and inductive reactance of the transmission line. In the Nominal π method, the capacitance at the sending end is not involved in creating this voltage drop, which is why it does not affect the line drop. The capacitance at both the sending and receiving ends affects the charging current. The charging current is a result of the line's capacitance and is responsible for energizing the line's electric field. In the Nominal π method, capacitance is split between the sending and receiving ends, and while the capacitance at the sending end does not affect the voltage drop, it contributes to the charging current that flows through the line.

In the Nominal π method, capacitance is considered in two places: half at the sending end and half at the receiving end. This capacitance leads to charging currents, which need to be added to the total current calculation. The charging current is important because it affects the total current that the line must supply from the sending end. However, this charging current does not influence the line's voltage drop or overall resistance/reactance characteristics.(Principals of Power System "V.K metha")

Applying lumped constants to long transmission lines results in serious errors in the performance calculations. For long lines, the line constants must be treated as uniformly distributed over the entire length for accurate calculations. For long transmission lines (length greater than 150 km), the line constants are treated as uniformly distributed along the length of the line to achieve accurate performance calculations.In the equivalent circuit of a 3-phase long transmission line, the resistance and inductive reactance are the series elements that represent the line's opposition to current flow.The leakage susceptance (B) represents the capacitive effect between the line and neutral, which causes charging currents along the transmission line.The leakage conductance (G) takes into account the energy losses occurring due to leakage over the insulators or the corona effect between the conductors.(Principals of Power System "V.K metha")

When a three-phase short circuit occurs:

1. All three phases are connected directly through a very low impedance (essentially just the line and system impedance).
2. Ohm’s Law: I=V/Z

Here, Z (impedance) is very small.
Therefore, the current becomes extremely large.
3. This is why short-circuit currents in power systems are much higher than normal operating currents, often tens or hundreds of times the rated current.
4. Such high currents can cause: 1. Damage to equipment   2. Tripping of protection devices (circuit breakers)

The per-unit reactance of a transformer changes when the base KVA changes according to the formula:
X_new==X_old​×S_new​​/S_old

Where:

X_old=5%=
S_old=1000 kVA
S_new=2000 kVA

​Xnew= 0.05 * 2000/1000 = 0.10

The short-circuit current for a symmetrical fault can be calculated using the formula:

I_sc​=​I_FL​​/X%×100 = 20/25 ×100 = 80A

A three-phase short circuit causes the maximum fault current because it involves all three phases directly shorted together (or to ground).
Other types of faults (L-L, L-L-G, or L-G) are less severe since they involve fewer phases and result in lower fault currents compared to the three-phase fault.("Power System Analysis" by John J. Grainger and William D. Stevenson)

In a balanced 3-phase system, the system voltages are equal in magnitude and displaced by exactly 120°.
Therefore, the system contains only positive sequence components.
That means:

Negative sequence current = 0 (because there’s no unbalance).
Zero sequence current = 0 (because the sum of three balanced phase currents is zero).

To calculate the percentage resistance of the line at the given base, we need to use the following formula:

Percentage resistance = (R_actual / R_base) * 100

Where: 
- R_actual = Actual resistance of the line = 1 ohm
- S_base = Base apparent power = 10,000 kVA = 10,000,000 VA
- V_base = Base voltage = 10 kV = 10,000 V

For a 3-phase system, the base resistance can be calculated using the formula:

R_base = (V_base^2) / S_base

⇒R_base = (10,000)^2 / 10,000,000
⇒R_base = 100,000,000 / 10,000,000 = 10 ohms

Now, we can use the formula to find the percentage resistance:

Percentage resistance = (1 / 10) * 100 = 10%

Impedance in Balanced Systems: When there are unbalanced conditions (such as line-to-line faults or unsymmetrical loads), the negative sequence impedance could become more important because the negative sequence currents would increase. However, even in these cases, the magnitude of the negative sequence impedance still remains equal to that of the positive sequence impedance unless the transmission line is explicitly designed with asymmetry (which is uncommon).
In a balanced three-phase system, the positive and negative sequence impedances of a transmission line are typically equal under normal operating conditions. To understand why, it's important to first know the concept of sequence components in a three-phase system. Sequence components help analyze unbalanced systems, where the system's voltages or currents deviate from the balanced state. There are three types of sequence components: positive sequence, negative sequence, and zero sequence.

Positive sequence components represent the balanced, normal operating condition of the system, where the voltages or currents are of equal magnitude and 120 degrees out of phase with each other. These components indicate the healthy, symmetric operation of the power system.

Negative sequence components arise when there is an unbalance in the system, which can be due to faults or asymmetry in the transmission lines or loads. These components are the reverse of the positive sequence components and indicate the system's deviation from balanced operation.

However, under ideal conditions—where the transmission line and the system are perfectly balanced, and there is no fault or asymmetry, then "the positive and negative sequence impedances will be the same". This symmetry occurs because both sequence components experience the same impedance properties of the transmission line. The line’s impedance doesn’t differentiate between the positive or negative sequence components unless there is some form of imbalance or fault.

Impedance in Unbalanced Systems: When there are unbalanced conditions (such as line-to-line faults or unsymmetrical loads), the negative sequence impedance could become more important because the negative sequence currents would increase. However, even in these cases, the magnitude of the negative sequence impedance still remains equal to that of the positive sequence impedance unless the transmission line is explicitly designed with asymmetry (which is uncommon). ("Electrical Power Systems" by C.L. Wadhwa)

Power Systems Analysis and Design by J. Duncan Glover , Mulukutla S. Sarma, Thomas Overbye

3-Phase Short-Circuit Fault: This is a fault where all three phases are shorted together. In such a case, the system becomes completely unbalanced, and the positive sequence component of the voltage is zero because there is no longer any "normal" phase sequence or balanced voltage. The voltage at the fault point drops to nearly zero due to the short-circuit, which disrupts the phase relationships completely.
Line-to-Line-to-Ground (L-L-G) Fault: In this fault, there is a short between two phases and the ground. It introduces both positive and negative sequence components but does not completely eliminate the positive sequence component.
Line-to-Line (L-L) Fault: In a line-to-line fault, the fault is between two phases, and although the system is unbalanced, the positive sequence component is still present, though it will be distorted.
Line-to-Ground (L-G) Fault: This fault occurs between one phase and the ground. The positive sequence component is also present, but the system is unbalanced.

When a synchronous generator stator has balanced three-phase positive-sequence currents under steady-state conditions, the rotor speed produced by these positive-sequence currents are rotated in the rotor rotating speed in the same direction as that of the rotor. Under this condition, a high value of magnetic flux penetrates the rotor and the positive-sequence impedance Z ⁽¹⁾   has a high value. Under steady-state conditions, these positive-sequence generator impedance is called the synchronous impedance.

When a synchronous generator has balanced three-phase negative-sequence currents, the net mmf produced by these currents rotates at synchronous speed so the net mmf is not stationary but rotates at twice synchronous speed. Under this condition, currents are induced in the rotor windings that prevent the magnetic flux from penetrating the rotor. As such, the negative-sequence impedance Z⁽²⁾ is less than the positive-sequence synchronous impedance.

When a synchronous generator has only zero-sequence currents, which are line (or phase) currents with equal magnitude and phase, then the net mmf produced by these currents is theoretically zero. The generator zero-sequence impedance Z⁽⁰⁾ is the harmonic sequence impedance that is due to leakage flux, and turns, and is smallest flux from windings and do not produce a perfectly sinusoidal mmf." (Power Systems Analysis and Design by J. Duncan Glover , Mulukutla S. Sarma, Thomas Overbye)

Figure 8.18(a) is a schematic representation of an ideal Y-Y transformer. The grounded through neutral impedances Z_n and Z_n' are resistances Y-B (or shunt) sequence currents associated with this ideal transformer.

When balanced positive-sequence currents are considered and negative-sequence currents are applied to the transformer, the neutral currents are zero and there are no voltage drops across the neutral impedances. Therefore, the ideal transformer, Figure 8.18(c) and (d), are the same as the per unit single-phase ideal transformer.

Zero-sequence currents have equal magnitudes and equal phase angles. When unit positive-sequence current I_a0 = I _B0= I_C0=I₀ are applied to the high-voltage winding of an ideal Y-Y transformer, the neutral current I_n = 3I₀ flows through the neutral impedance Z_n, with a voltage drop (3Z_n)I₀. Also, per-unit zero-sequence current I₀ flows in each low-voltage winding (from the impedances), and therefore phase a, through neutral impedance Z_n with a voltage drop (3I₀)Z_n. The per-unit zero-sequence network includes the impedances 3Z_n and 3Z_n, as shown in Figure 8.18(b).

Figure 8.18: Ideal Y-Y Transformer

Note that if Z_n' is one in the sequence network of an ideal transformer, grounded Z_n are zero, the per-unit zero-sequence network is as shown in the figure. For example, if the high-voltage winding has an open neutral, then I_n = 0, so maximum current I₀ = 0 for the low-voltage side. This can be shown in the zero-sequence network circuit in the one in 8.18(b) which corresponds to an open neutral circuit.

The per-unit sequence network characteristics of Y-Y transformers are indicated in the sequence networks of the ideal transformer. If the phase a,b and c winding have equal leakage impedence Z_H​=R_H​+jX_H​, then the series impedances are symmetrical with sequence networks, as shown in Figure 8.10, where Z_H0=Z_H1=Z_H2=Z_H​. Similarly, the leakage impedances of the low-voltage windings are symmetrical series impedances with Z_X0=Z_X1=Z_X2=Z_X​. These series leakage impedances are shown in per-unit in the sequence networks of Figure 8.19(a).

The shunt branches of the practical Y-Y transformer, which represent exciting current, are equivalent to the Y load of Figure 8.3. Each phase in Figure 8.3 represents a core loss resistor in parallel with a magnetizing inductance. Assuming these are the same for each phase, then the Y load is symmetrical, and the sequence networks are shown in Figure 8.4. These shunt branches are also shown in Figure 8.19(a). Note that (3Z_N)and (3Z_n) have already been included in the zero-sequence network. (Power Systems Analysis and Design by J. Duncan Glover , Mulukutla S. Sarma, Thomas Overbye)

In a balanced three-phase system, the vector sum of the positive sequence currents is zero. This is because the positive sequence currents are balanced and are 120 degrees apart, leading to their vector sum being zero.

A short-circuit can lead to:

1. Excessive Current :When a short-circuit occurs, it creates a path of low resistance between two points, typically across the terminals of a power source or electrical circuit. This allows a very large current to flow, far exceeding the normal operating current. In a normal circuit, the resistance (or impedance) limits the current. However, in a short-circuit, this resistance becomes minimal, leading to a surge in current that can be hundreds or thousands of times greater than the rated current. This excessive current can lead to damage to wires, insulation, electrical components, and other parts of the system, potentially leading to failures.

2. Fire or Explosion : The excessive current flowing through the system during a short-circuit causes heating due to the power dissipation (P = I²R, where "I" is current and "R" is resistance). This heat can cause insulation on wires and components to melt, creating potential for electrical fires. In high-voltage systems, a short-circuit can lead to an arc flash, where an electric arc forms between conductors, causing extreme heat and potentially resulting in an explosion. In some cases, the heat and pressure generated from a short-circuit could trigger an explosion, particularly in confined spaces or when flammable materials are present. Certain components, like transformers, might even explode due to the extreme pressure caused by gas buildup.

3. Low Voltage in the System: A short-circuit often leads to significant voltage drops throughout the network. The excessive current created by the short can pull down the voltage in the rest of the system, affecting other connected loads. In some systems, circuit breakers or fuses may operate to disconnect the faulted section, which can also cause a temporary reduction in voltage or power to parts of the network. The short-circuit might cause the system to become unstable, with fluctuations in voltage, affecting sensitive equipment and devices that depend on a stable power supply.

Unsymmetrical faults (such as line-to-ground faults, line-to-line faults, and double line-to-ground faults) cause an unbalanced condition in the system. In a balanced system, the three phases have equal magnitudes and are 120 degrees apart in phase. However, unsymmetrical faults disturb this balance, leading to unequal voltages and currents in the system, causing an unbalanced load. This unbalance can affect the operation of generators, transformers, and motors, leading to reduced efficiency and potential damage to electrical equipment.
Unsymmetrical faults often occur due to abnormal conditions such as physical damage to the transmission line (e.g., due to storms, fallen trees, or accidents), equipment failure, or insulation breakdown. These faults indicate an issue within the system, such as a fault in one or more phases, which needs to be identified and resolved to prevent further damage.

Unsymmetrical faults are statistically more frequent than symmetrical faults (such as three-phase faults). This is because one-phase faults (like line-to-ground) are more likely to occur due to natural events, human activities, or equipment failures.Symmetrical faults are rarer and often require a simultaneous failure in all three phases, whereas unsymmetrical faults involve failures in one or two phases and can be caused by localized issues like lightning strikes, tree branches, or damaged equipment.

Current-limiting reactors are used in power systems to limit the fault current during short circuits by introducing impedance into the circuit. The reactor is designed to have high reactance and low resistance.
 
The primary function of a current-limiting reactor is to limit the fault current by introducing reactive impedance (inductive reactance). The large reactance reduces the fault current by opposing the flow of current, helping protect the system from high fault currents.
The resistance is kept low to minimize energy dissipation in the form of heat. A high resistance would waste energy in the reactor itself, defeating the purpose of limiting fault current efficiently.

In a 3-phase, 4-wire unbalanced system, the neutral current represents the sum of the phase currents that do not cancel each other out. The zero-sequence current is the part of the current that flows in the neutral wire due to the unbalanced load in the system.

For an unbalanced 3-phase system, the zero-sequence current can be calculated using the formula:

I₀ = I_N/3 = 18/3 = 6A

A 3-phase, 4-wire system supplies loads that are unequally distributed in the three phases. 
The analysis of currents flowing in R, Y, and B lines shows that in the R line, the positive phase sequence 
component is 200∠0° A, and the negative phase sequence component is 100∠60° A. The total observed current 
flowing back to the supply in the neutral conductor is 300∠300° A. 

To determine the zero-phase sequence current in the R-line, we use the formula:

    I_N = I_0 + I_1 + I_2

Where:
    - I_N is the total current in the neutral conductor.
    - I_0 is the zero-phase sequence current.
    - I_1 is the positive-phase sequence current.
    - I_2 is the negative-phase sequence current.

We are given the following values:
    - Positive-sequence current in R-line (I_1): 200∠0° A
    - Negative-sequence current in R-line (I_2): 100∠60° A
    - Total neutral current (I_N): 300∠300° A

To find the zero-sequence current, we perform the following calculation:

    I_0 = I_N - I_1 - I_2
    I_0 = 300∠300° - 200∠0° - 100∠60°

Performing the complex subtraction gives:
    I_0 = 100∠300° A

Thus, the zero-phase sequence current in the R-line is 100∠300° A.(V. K Mehta)

In a delta connection, the phases are interconnected, meaning the sum of the currents in the three lines should always be zero in a balanced system. There is no direct path for the neutral current, and consequently, the zero-sequence component of the current is zero. If the fuse in one of the lines (B-line) is removed, this can cause an unbalanced condition.

However, as the system is delta-connected, there is no neutral conductor in the delta connection to carry any zero-sequence current. The zero-sequence current is typically the current that flows in the neutral conductor of a 3-phase, 4-wire system, which doesn’t exist in a delta connection.

Since there is no neutral in a delta system to carry the zero-sequence current, the zero-sequence current is 0A.

Given an equipment with a per unit impedance of 0.9 p.u. on a base of 20 MVA and 33 kV, we need to convert 
this to a new base of 50 MVA and 11 kV.

The formula to convert the per unit impedance to a new base is:

    Z_new = Z_old × (S_{base_new} / S_{base_old}) × (V_{base_old²} / V_{base_new}²)

Where:
    - Z_new is the per unit impedance on the new base.
    - Z_old is the per unit impedance on the old base (0.9 p.u. in this case).
    - S_{base_old} and S_{base_new} are the old and new power bases, respectively (20 MVA and 50 MVA).
    - V_{base_old} and V_{base_new} are the old and new voltage bases, respectively (33 kV and 11 kV).

Given:
     Z_old = 0.9 p.u.
    S_{base_old} = 20 MVA
    V_{base_old} = 33 kV
    S_{base_new} = 50 MVA
    V_{base_new} = 11 kV

Substitute these values into the formula:

    Z_new = 0.9 × (50 / 20) × (33^2 / 11^2)

    ⇒Z_new = 0.9 × 2.5 × 9 = 20.25 p.u.

In a power system, the sequence reactances (positive, negative, and zero sequence reactances) characterize how an element responds to different types of faults, including balanced and unbalanced faults.

Synchronous Generator: The reactances for a synchronous generator are typically not equal for all three sequences. The positive-sequence reactance (X₁) is usually different from the negative-sequence (X₂) and zero-sequence (X₀​) reactances. This would result in different values for each sequence component.
Synchronous Motor: Similar to synchronous generators, synchronous motors typically do not have equal sequence reactances. The reactances for the positive, negative, and zero-sequence components would usually differ.
Static Load: A static load (like a resistive load) doesn't have a defined reactance to handle sequence components in this way, so it wouldn't typically exhibit sequence reactances as described here.
Transmission Line: A transmission line can have equal positive and negative-sequence reactances but often a larger zero-sequence reactance. The given reactances (0.3, 0.3, and 0.8 p.u.) fit the characteristics of a transmission line, which typically has a larger zero-sequence reactance compared to the positive and negative sequence reactances.

Total MVA Rating:
Since there are four alternators, each rated at 5 MVA, the total MVA rating of the system is:
Total MVA=4×5MVA=20MVA
The reactance for each alternator is 20%, but when the alternators are working in parallel, the equivalent reactance is divided by the number of alternators. So:
Equvalent Reactance =20/4 = 5 %
The short-circuit MVA at the bus-bars is given by the formula:
Short-Circuit MVA=Total MVA×100​/ Equivalent Reactance
Short-circuit MVA at bus bar = 5 * 100/5 =20

In a line-to-ground fault, the current in the faulted phase causes a zero-sequence current to flow in the system. For a balanced 3-phase system with a line-to-ground fault, the zero-sequence current is related to the faulted phase current.

The current in the faulted phase is given as 100 A.
For a line-to-ground fault, the zero-sequence current is typically one-third of the fault current in the phase (since, in a 3-phase system, the zero-sequence current divides equally among the phases in the case of an unbalanced fault).

Thus, the zero-sequence current is: I0​=I_{faulted phase}​​/3 =3100A​=33.3A

In a star-connected system without neutral grounding, there is no path for the zero-sequence current to flow. The system does not have a neutral point that is grounded, meaning there is no way for a zero-sequence current to exist in the system. Zero-sequence current is defined as the current that would flow in a neutral conductor if the system had one. However, in the absence of a neutral grounding (i.e., the neutral is not grounded or is floating), there is no neutral current, and consequently, there is no zero-sequence current.

Zero-sequence currents require a path to return, typically through the neutral of the system. In the case of transformers, the flow of zero-sequence currents depends on the configuration of the primary and secondary windings.

Delta/Star (ক): In this configuration, the delta side (primary) doesn't have a neutral, so zero-sequence currents can't flow through it. However, the star side (secondary) has a neutral, allowing zero-sequence currents to flow if grounded. But there’s no direct path from the delta side for zero-sequence currents.

Delta/Delta (খ): In this case, neither side of the transformer has a neutral point, so no zero-sequence current can flow. Both windings are isolated from the neutral, and thus, no path exists for zero-sequence current.

Star/Grounded Star (গ): Zero-sequence currents can flow through the neutral of the star-wound transformer if the neutral is grounded. However, there is no obvious direct path for zero-sequence currents from the line to the transformer bank in this configuration unless the system is designed with proper grounding and neutral points.

Grounded Star/Delta (ঘ): In this configuration, the star-wound primary side is grounded, providing a path for zero-sequence currents to flow through the neutral. The delta-wound secondary side (transformer bank) doesn't directly affect the flow of zero-sequence currents but can carry the resulting current. This configuration allows zero-sequence currents to flow from the line to the transformer bank.