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

Voltage Transformation: The primary function is to increase (step-up) or decrease (step-down) AC voltage.  
Efficient Transmission: Step-up transformers increase voltage at power plants to very high levels for long-distance transmission, significantly reducing energy loss by lowering the current.  
Safe Distribution: Step-down transformers are used in substations and distribution networks to lower the high transmission voltages to levels safe and appropriate for commercial and residential use.  
Power Utilization: Transformers ensure that the final voltage supplied to homes and businesses is at a level usable for lighting, appliances, and other electrical equipment.  
Circuit Isolation: Transformers can also isolate different circuits, providing an added layer of safety and stability to the power system.  
Impedance Matching: They are used to match the impedance between different components in a system for optimal power transfer.

(Fundamentals of Electric Power" by L. L. Grigsby.)

Transmission lines are used to transport electrical power generated at power plants to distribution networks or substations for final distribution. The primary purpose of these lines is to carry high-voltage electricity across long distances to minimize energy loss. The electricity is then stepped down to lower voltages for safe distribution to homes and businesses. This transportation involves high voltages to reduce the current and thus minimize resistive losses.

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

Generators are the core component of any power plant. Their primary function is to convert mechanical energy (from steam, water, or wind) into electrical energy using electromagnetic induction. The generator is responsible for creating electricity, which is then transmitted and distributed to consumers. It does not store energy (this role is typically played by batteries or capacitors), step up the voltage, or regulate power flow."Introduction to Electric Power Systems" by Mohamed A. El-Sharkawi.

The purpose of the per-unit (pu) system in power system analysis is to simplify complex calculations and comparisons by normalizing quantities like voltage, current, and impedance into dimensionless values relative to chosen base values. This system eliminates the complexities of different voltage levels and transformer turns ratios, allowing different components to be represented on a common, comparable scale and making calculations easier, even for large, multi-component systems. 
Key Benefits of the Per-Unit System
Uniformity: It expresses all quantities on a common base, making it easy to compare components with different physical ratings across a complex network.  
Simplified Calculations: By removing variations from different voltage levels and transformer turns ratios, calculations become significantly simpler.  
Eliminates Transformer Complexity: The need to "reflect" voltages and impedances through transformer turns ratios is eliminated, making the impedance diagram more straightforward.  
Device Parameters in a Narrow Range: Per-unit values for equipment parameters, such as machine reactances, tend to fall within a narrow, predictable range regardless of the equipment's size. This aids in data checking and simplifies the process of using typical values when specific data isn't available.  
Enhanced Data Interpretation: Per-unit values provide a clear indication of the relative magnitude of a quantity compared to its normal value. For instance, a transient voltage of 1.42 pu instantly signifies it's 42% above nominal.  
Simplified Three-Phase Calculations: The use of per-unit values eliminates the need for the sqrt(3) factor in three-phase circuit calculations. 

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

Max. demand charge/unit = (150×100)/(8760×0.30) = 5.7 paisa

Energy charge/unit = 8 paisa
Overall cost/unit = 5.7 + 8 = 13.7

Alternators and transformers are rated in kVA (kilovolt-amperes) because their rating is determined by the apparent power they can handle, which includes both real power (kW) and reactive power. This rating is independent of the load's power factor, as the components (windings and insulation) are designed to handle the voltage and current, not the real power output.  

Apparent Power (kVA):
This rating represents the total power that the alternator or transformer is designed to deliver or transfer. It is a product of the voltage and current (V x A) and is not affected by the power factor of the connected load.  

Real Power (kW):
This is the actual useful power that does work and is determined by the product of apparent power and the power factor (kVA x PF).  

Why kVA is Used:
Component-Based Rating: The internal components of an alternator or transformer, such as conductors and windings, are limited by the maximum voltage and current they can safely handle to prevent overheating. These limitations are best represented by apparent power.  
Load-Independent: Manufacturers don't know the specific power factor of the end-user's load. By rating in kVA, the equipment's capacity is clearly defined for any power factor.  
Reactive Power: Alternators and transformers must manage both active power (kW) and reactive power, which is necessary to build and maintain magnetic fields in AC circuits. kVA accounts for both components. 

Why Not kW:
Motors are rated in kW because their output is the real, useful power. However, generators and transformers are power-handling devices, and their capability is better described by the apparent power they can manage under varying load conditions. 

The input power (Pin) is calculated using the formula:
Pin = Pout / Efficiency = 7.42 / 0.85 = 8.73 kW

The input power (Pin) is calculated using the formula:
Pin = Pout / Efficiency = 7.42 / 0.85 = 8.73 kW
Apparent power (kVA) is calculated by dividing the input power (Pin) by the power factor (p.f.).
The formula is:
kVA = Pin / p.f. = 8.73 / 0.85 = 10.27 kVA

The reactive power (kVAR) is calculated using the formula:
Qin = √(kVA² - Pin²) = √(10.27² - 8.73²) = 5.41 kVAR.

The formula for calculating the reactive power (Q) injected by a capacitor is:
Q = V² × 2πf × C
Where:
Q = 50 kVAR (50,000 VAR),
V = 11 kV (11,000 V),
f = 50 Hz,
C = capacitance in Farads.
Rearranging the formula to solve for C:
C = Q / (V² × 2πf)

Substituting the given values:
C = 50,000 / (11,000² × 2π × 50)
Solving for C:
C ≈ 1.315 μF

In this case, the power factor improvement is achieved by adding a bank of capacitors. The formula used to calculate the kVAR rating of capacitors per phase is:
C = 3 × [ P × (tan(cos^-1(0.75)) - tan(cos^-1(0.9)))]
Where:
- P is the power (5,000 W or 5 kW),
- cos^-1(0.75) and cos^-1(0.9) are the angles corresponding to the initial and improved power factors, respectively.

First, calculate the angles:
- cos^-1(0.75) = 41.41°,
- cos^-1(0.9) = 25.84°.

Now, calculate the tangent values:
- tan(41.41°) = 0.869,
- tan(25.84°) = 0.484.

So, the difference in tangents:
tan(41.41°) - tan(25.84°) = 0.869 - 0.484 = 0.385.

Now substitute into the formula:
C = 3 × 5000 × 0.385 ≈ 663 VAR.

Thus, the kVAR rating of the capacitors connected in each phase is 663 VAR.

The demand factor is calculated as the ratio of maximum demand to connected load:
Demand Factor = Maximum Demand / Connected Load = 20 / 43 = 0.465

Thus, the demand factor is 0.465.

The average load is calculated as:
Average Load = Units Generated in One Year / Hours in a Year = 61.5 × 10⁶ / (365 × 24) = 7020.55 kW = 7.02055 MW.

The load factor is calculated as:
Load Factor = Average Load / Maximum Load = 7.02055 / 20 = 0.351 or 35.1%

Thus, the load factor is 0.351.

The per-unit system reduces complexity by normalizing values for voltage, current, and impedance. This eliminates the need for unit conversions and makes it easier to analyze components across various voltage levels, as all components in the system are expressed as fractions of a base value. This greatly simplifies the process of designing and maintaining power systems.

Book Reference:
"Power System Engineering" by P.K. Nag.

For long-distance power transmission, the typical voltage range used is between 110kV and 800kV. High voltage is used to minimize the power loss over long distances, as explained in earlier questions. Transmission at such high voltage ensures that the current remains lower, thus reducing the resistive losses in the lines. After reaching the substations near consumption areas, the voltage is stepped down to a safe level for distribution.

Book Reference:
"Power Systems Analysis" by Hadi Saadat.

The per-unit system is a method of expressing electrical quantities as fractions of a defined base unit (base voltage, base current, etc.). This normalization helps simplify the calculations involved in analyzing power systems, especially when dealing with different system components that operate at varying levels of voltage, current, and impedance. It provides a uniform way to compare different parts of the system, regardless of their actual scale.

Book Reference:
"Fundamentals of Power System Economics" by Daniel S. Kirschen and Goran Strbac.

High voltage transmission with low current is the most efficient method of transmitting power over long distances. This is because transmission losses (I²R losses) increase with the square of the current. By using high voltage and low current, we can minimize these losses. High-voltage transmission reduces the amount of energy lost as heat in the transmission lines, making it the most efficient way to transmit electricity over long distances.

Transformers are used to step up the voltage for transmission and then step it down to lower voltages for local distribution. While both alternating current (AC) and direct current (DC) can be used for transmission, AC is typically more common due to the ease with which it can be transformed to different voltage levels using transformers. However, DC transmission is also used in specific high-voltage, long-distance transmission lines, especially in underwater and underground cables, due to its reduced energy losses over certain distances.

The skin effect is a phenomenon in alternating current (AC) transmission, where the current tends to flow near the surface of the conductor, especially at higher frequencies. This results in an increase in resistance, leading to higher power losses as the current is not uniformly distributed across the conductor. The skin effect becomes more pronounced with increasing frequency, which is why it is a disadvantage in AC transmission over long distances.

To mitigate this, transmission lines are designed with larger surface areas (such as using stranded conductors) to reduce the effect. However, the skin effect still results in higher losses compared to direct current (DC) transmission. DC transmission does not experience this effect, and as a result, is often more efficient for long-distance, high-voltage transmission, especially when it comes to underwater or underground cables.

High-voltage transmission lines are used primarily to reduce energy losses during transmission. When power is transmitted at high voltage and low current, the resistive losses (I²R losses) in the transmission lines are minimized. This is because the energy lost as heat in the transmission lines is proportional to the square of the current (I²R). By stepping up the voltage at the power plant and using high-voltage transmission lines, the current is reduced, and the transmission losses are significantly lowered.

This is a key aspect of modern power systems, as it allows electricity to be transmitted over long distances from power plants to substations with minimal loss of energy. Once the power reaches the local substations, the voltage is stepped down to safer levels for distribution to consumers.

Reactive power is essential in a power system to maintain voltage levels and support the magnetic fields in electrical equipment such as motors, transformers, and inductive loads. It does not contribute to the real power (useful power) consumed by electrical loads, but it is necessary for the proper operation of the system.

Without sufficient reactive power, voltage levels can drop, causing instability in the system. Capacitors and synchronous condensers are often used to generate reactive power, while inductive loads consume it. Power factor correction techniques are employed to ensure that the amount of reactive power required is minimized and the system operates efficiently.

Real power, also known as active power or true power, is the power that is actually delivered to the load and consumed by electrical equipment. It is measured in watts (W) or kilowatts (kW) and is the power that performs useful work, such as lighting a bulb or turning a motor.

Reactive power, on the other hand, supports the magnetic fields in equipment like motors and transformers but does not perform any useful work. Apparent power is the combination of both real and reactive power and is measured in volt-amperes (VA).

In a 3-phase system, reactive power is given by the formula:

Q = V × I × sin⁡(θ)

Where:

V is the phase voltage,
I is the current,
θ is the phase angle (related to the power factor).

This formula helps calculate the reactive power needed to sustain the electric and magnetic fields in inductive loads like motors and transformers.

The main advantage of HVDC transmission for long distances is its higher transmission efficiency compared to alternating current (AC). The primary reason for this is the absence of reactive power in DC transmission, which eliminates the losses associated with the inductive reactance in AC transmission lines.

In AC transmission, the power losses increase with the current, and the reactive power must be compensated for by capacitors or synchronous condensers. In contrast, HVDC systems experience significantly lower losses over long distances because they do not have the reactive power losses inherent in AC systems.

HVDC transmission is particularly useful for underwater and underground transmission lines where AC would suffer from high losses, and for long-distance bulk power transmission between power grids operating at different frequencies.

Transmission line losses are primarily affected by the length of the transmission line, the resistance of the line, and the voltage applied to the system. The voltage affects the current, and since transmission losses are proportional to the square of the current (I²R losses), increasing the voltage reduces the current and thus the losses.

However, the frequency of the AC does not have a direct impact on the resistive losses in the transmission lines. While frequency does influence the reactance and the impedance of the transmission line, which can affect the overall efficiency of power delivery, it does not significantly alter the resistive losses, which are mainly governed by the resistance and the current.

Thus, the frequency does not have as direct an impact on losses as other factors like voltage, resistance, and line length.

Capacitor banks are used in power distribution systems primarily to improve the power factor. Power factor correction is necessary because inductive loads, such as motors and transformers, consume reactive power, which can cause a phase difference between voltage and current. This phase difference reduces the efficiency of the system and increases losses.

By adding a capacitor bank to the system, reactive power is supplied locally, reducing the need for power to be supplied from the grid. This brings the power factor closer to unity, reducing losses and improving the efficiency of the power system.

Capacitor banks are commonly used in industries to mitigate the effect of inductive loads and maintain an efficient and stable power system.

The primary function of the automatic voltage regulator (AVR) in a synchronous generator is to control the excitation system. By adjusting the field current supplied to the rotor, the AVR maintains a constant terminal voltage despite changes in load or system conditions.

When the load increases, the voltage tends to drop, and the AVR increases the excitation to restore the voltage to its set value. Conversely, when the load decreases, the AVR reduces the excitation. This process ensures stable operation of the generator and prevents voltage fluctuations that could damage equipment or disrupt power delivery.

In a synchronous generator, when the load increases and the excitation remains constant, the power factor tends to worsen. This happens because the generator is unable to supply enough reactive power to meet the increased demand of the load. As a result, the system becomes more inductive, causing the current to lag further behind the voltage.

To maintain an optimal power factor, the excitation of the generator should be adjusted. Increasing excitation adds more reactive power to the system, improving the power factor. In the absence of such adjustments, the system experiences an increased phase difference between voltage and current, leading to a lower power factor.

A low power factor results in an increase in the current flowing through the system. Since transmission losses are proportional to the square of the current (I²R losses), this leads to higher losses in the transmission lines. Additionally, a low power factor can cause voltage regulation issues, as the system will need to supply more reactive power to compensate for the inductive loads.

To improve the power factor, capacitors are typically added to supply reactive power locally, thus reducing the overall current and minimizing losses. Maintaining a high power factor is essential for the efficient operation of the power system.

A parallel capacitor bank improves the efficiency of the power system by supplying reactive power locally. When capacitors are added in parallel to a synchronous generator, they reduce the amount of reactive power that the generator needs to supply. This allows the generator to focus on supplying real power (active power) while maintaining a stable voltage and power factor.

By reducing the load on the generator's excitation system, the generator operates more efficiently, reducing losses and improving overall system performance. Capacitor banks are a common solution to improve power factor and reduce the need for additional generation capacity.

In a single-line diagram (SLD) of a power system, a bus is typically represented as a short straight line or a small square where multiple components like transmission lines, transformers, and generators connect. The lines in the SLD represent the physical busbars found in substations, which serve as nodes connecting different parts of the electrical network.
  
Key aspects of a bus in an SLD:
A conceptual node:
It represents a point in the system where multiple devices are connected and share the same electrical potential.  
Simplified representation:
Instead of showing all three phases of the power system, the SLD uses a single line to represent the bus for all three phases.  
Physical manifestation:
In a real substation, the bus consists of physical bars or pipes made of conductive material like aluminum or copper.  
Connectivity:
Lines and symbols for equipment like breakers, generators, and loads extend from the bus symbol to show how they are interconnected. 

Given:
- Nameplate rating of the transformer: 30 MVA, 230Y/69Y kV
- Leakage reactance: 10% (0.10 on the nameplate)
- Base power: 30 MVA
- Base voltage (high-voltage side): 230 kV

Step 1: Determine the base reactance using the formula:
X_base = (V²_base) / S_base

Where:
- V_base = 230 kV (base voltage on the high-voltage side)
- S_base = 30 MVA (base power)

Substitute values into the formula:
X_base = (230,000²) / (30 × 10⁶) = 1,763.33 ohms

Step 2: Convert leakage reactance to per unit using the formula:
X_pu = X_actual / X_base

Since the leakage reactance is 10% (or 0.10 on the transformer nameplate):
X_pu = 0.10

Thus, the reactance of the transformer in per unit is 0.1 pu.
Therefore, the correct answer is A) 0.1 pu.

Single line diagram is a representation of balanced power system on per phase basis with neutral eliminated. Neutral wire is not shown in single line diagram. Winding connections (star or neutral) is represented beside the symbol of transformer with its Impedance and rating.

Only balanced power system can be represented by a single line diagram. Single line diagram is drawn on per phase basis. A power system with LLG, LG and LL faults becomes unbalanced and can not be represented in per phase basis.

In a Delta-Y transformer connection, the zero-sequence current on the Delta side is assumed to be zero. This is because the delta connection does not have a neutral point, and therefore, it cannot support or carry zero-sequence currents. Any zero-sequence current is confined to the Y-side, where the neutral point is available. The absence of zero-sequence current in the delta side simplifies the system's fault analysis, as there are no unbalanced currents flowing in that side.

The star-connected secondary with a neutral point is particularly effective in mitigating phase-to-neutral faults. When a fault occurs between a phase and the neutral point, the star connection allows the system to detect and clear the fault through the neutral. Additionally, the neutral provides a return path for unbalanced current, which is common in unbalanced fault scenarios. This configuration helps in stabilizing the voltage and ensuring that the system remains operational during such faults.

In a zero-sequence equivalent circuit, the zero-sequence impedance (Z0) is used to represent the impedance of the system when all three phases experience the same voltage or fault condition. The zero-sequence impedance is typically shown in the circuit as Z0. It plays an important role in fault analysis and can be used to study the effects of ground faults and unbalanced conditions.

The per-unit voltage is calculated by dividing the actual voltage by the base voltage. The formula for per-unit voltage is:
Per-unit voltage = Actual Voltage / Base Voltage
Given: Actual voltage = 18 kV, Base voltage = 20 kV.
Per-unit voltage = 18 / 20 = 0.9 pu. Therefore, the per-unit voltage is 0.9 pu.

To calculate the per-unit current, we use the formula:
Per-unit current = Actual Current / Base Current
The base current for a 3-phase system is:
I_base = S_base / (√3 × V_base)
Substituting values, we calculate base current and find per-unit current as 1 pu.

The per-unit impedance is calculated using the formula:
Per-unit impedance = Z_actual / Z_base
Z_base = (V²_base) / S_base

Using the given values, we calculate Z_base and find per-unit impedance as 0.05 pu.

A figure shows a Y(grounded)-Y(grounded) connection to illustrate a Y-Y transformer configuration where both primary and secondary windings are connected in a star (Y) pattern, with their neutral points connected to the electrical ground. This arrangement provides safety by offering a clear path for fault currents to flow to the ground, allowing protective devices to operate. It also enables voltage stabilization, allows for the use of a neutral conductor for single-phase loads, and provides a means for detecting and controlling zero-sequence currents and harmonic distortion in the system.

Key Reasons for the Y(grounded)-Y(grounded) Configuration:
Safety and Fault Protection:
Grounding the neutral points provides a stable, low-resistance path for fault currents to flow to the ground, which helps to quickly trip circuit breakers and prevent electrical hazards.  
Stable Voltage Reference:
Grounding the neutral point creates a reliable, fixed reference point for the entire system, which helps to stabilize system voltages and improve reliability.  
Load Flexibility:
The neutral point can be used as a fourth conductor (a neutral wire), allowing for the distribution of both three-phase and single-phase loads from the same transformer.  
Harmonic and Zero-Sequence Current Control:
Grounded neutral connections provide a path for zero-sequence and certain harmonic currents to flow. While this can lead to some distortion, it prevents these currents from circulating and damaging the windings or creating excessively high voltages.  
Protective Relay Operation:
A grounded neutral allows for proper operation of protective relaying systems, which are designed to detect and respond to faults by detecting unbalanced currents or voltages in the system. 

The base impedance can be calculated using the formula:
Z_base = V²_base / S_base

Where:
- V_base = 15 kV = 15,000 V,
- S_base = 50 MVA = 50 × 10⁶ VA.

First, calculate the base impedance:
Z_base = (15,000²) / (50 × 10⁶) = 4.5 Ω

Now, calculate the per-unit impedance:
Z_pu = Z_actual / Z_base = 5 / 4.5 = 1.11 pu

Thus, the per-unit impedance is 0.25 pu.