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Past Questions — 2024
Future expansion in substation is possible for -
A)
underground substation
B)
indoor substation
C)
outdoor substation
D)
pole mounted substation
Future expansion refers to the ability to add more bays, transformers, or switchgear to a substation as the load demand increases over time.
1. Outdoor Substations: These are typically built on open land. Because they are not confined by walls or ceilings, expanding the facility usually only requires additional land or utilizing pre-allocated space within the yard. It is the most flexible and easiest type of substation to expand.
2. Indoor Substations: These are housed within a building. Expansion is severely limited by the physical dimensions of the building (walls and roof). To expand, one would either need to build a new extension to the building or construct a completely new building, which is costly and complex.
3. Underground Substations: These are the most constrained. They are limited by the excavated volume and the structural integrity of the surrounding earth/buildings. Expanding an underground facility is extremely difficult and expensive.
4. Pole Mounted Substations: These are small, distribution-level substations designed for a specific, limited capacity. They are not designed for 'expansion' in the industrial sense; if the load increases beyond the pole's capacity, the entire unit is typically replaced or a new substation is installed nearby.
Modern steam power plant works on which of the following cycle -
A)
carnot cycle
B)
rankine cycle
C)
otto cycle
D)
bell-coleman cycle
Step-by-step reasoning:
1. **Analysis of the Question**: The question asks for the thermodynamic cycle upon which modern steam power plants operate.
2. **Evaluating Option (b) Rankine Cycle**: The Rankine cycle is the idealized thermodynamic cycle for steam power plants. It consists of four main processes: pumping liquid water to high pressure (Pump), heating the water to create superheated steam (Boiler), expanding the steam to produce work (Turbine), and condensing the steam back into water (Condenser). This is the practical basis for all modern thermal power plants.
3. **Evaluating Option (a) Carnot Cycle**: While the Carnot cycle is the most efficient theoretical cycle, it is impossible to implement in a steam plant because it requires isothermal heat addition and removal, and the compression of a two-phase mixture (liquid-vapor) is mechanically impractical.
4. **Evaluating Option (c) Otto Cycle**: The Otto cycle is the theoretical cycle for spark-ignition internal combustion engines (petrol engines), not steam power plants.
5. **Evaluating Option (d) Bell-Coleman Cycle**: Also known as the Reverse Brayton cycle, this is used for air refrigeration systems, not for power generation in steam plants.
Conclusion: The Rankine cycle is the standard cycle for steam power plants.
Penstock in a hydroelectric power plant is -
A)
a pipe connected to runner outlet
B)
nozzle that release high pressure water on turbine blades
C)
a conduit connecting forebay and turbine
D)
a pipe connecting surge tank to dam
In a hydroelectric power plant, the penstock is a high-pressure steel or concrete pipe (conduit) that carries water from the reservoir or forebay (the storage area at the top) down to the turbine. Its primary purpose is to deliver water with high kinetic and potential energy to the turbine runner to generate electricity.
Analysis of options:
- Option (a) is incorrect because the pipe connected to the runner outlet is called the 'Draft Tube', which recovers pressure head.
- Option (b) is incorrect because the nozzle is a specific component (common in Pelton wheels) that converts pressure into a high-velocity jet, but the penstock is the entire conduit leading to that nozzle.
- Option (c) is correct as it accurately describes the penstock as the conduit connecting the forebay (or surge tank) to the turbine.
- Option (d) is incorrect because the surge tank is typically located between the main conduit and the penstock to relieve water hammer pressure; the penstock connects the surge tank to the turbine, not the surge tank to the dam.
A tariff system with fixed rate per unit of energy consumed.
A)
Simple Tariff
B)
Flat-rate Tariff
C)
Block-rate Tariff
D)
Two-part Tariff
Step-by-step reasoning:
1. **Definition of Flat-rate Tariff**: In a flat-rate tariff, the cost of energy is charged at a constant rate per unit (kWh) regardless of the total amount of energy consumed. The formula is: Total Charge = (Units Consumed) × (Fixed Rate per Unit).
2. **Analyzing the Question**: The question specifies a 'fixed rate per unit of energy consumed,' which aligns perfectly with the definition of a flat-rate tariff.
3. **Evaluating Other Options**:
- **Simple Tariff**: This is a general term and not a specific technical classification of a tariff structure in electrical engineering.
- **Block-rate Tariff**: In this system, energy is charged in blocks. The rate per unit decreases (or increases) as the consumption moves into higher blocks (e.g., first 100 units at $x, next 100 units at $y). It is not a single fixed rate.
- **Two-part Tariff**: This consists of two components: a fixed charge (based on maximum demand or connected load) and a running charge (based on actual energy consumed). It is not based solely on a fixed rate per unit.
Which of the following is used to maintain clearance between ground and live conductor in substations?
A)
Isolators
B)
Insulators
C)
Circuit breakers
D)
All of these
In a substation, the primary purpose of an insulator is to provide electrical isolation and mechanical support. Insulators are specifically designed to prevent the flow of current from live conductors to the grounded structures (such as steel towers, poles, or support frames), thereby maintaining the required safety clearance and preventing flashovers or short circuits to the ground.
Why other options are incorrect:
- a. Isolators: These are mechanical switches used to isolate a part of the circuit for maintenance purposes after the current has been interrupted by a circuit breaker. While they provide a visible gap, their primary function is switching/isolation, not the continuous maintenance of ground clearance for the conductor.
- c. Circuit breakers: These are protective devices designed to interrupt fault currents or normal load currents. They are used for switching and protection, not for providing structural insulation between the conductor and the ground.
- d. All of these: Since only insulators are specifically used for the purpose of maintaining electrical clearance between live parts and ground, this option is incorrect.
Usually distribution substation are -
A)
underground type
B)
indoor type
C)
outdoor type
D)
dry type
Distribution substations are the final stage of the power system before electricity is delivered to the end consumer. They typically step down voltage from a sub-transmission level (e.g., 33kV or 11kV) to a distribution level (e.g., 415V/230V).
1. Why 'Indoor Type' is correct: Distribution substations are often located in densely populated urban areas or residential neighborhoods. To ensure safety, protect equipment from environmental factors, and prevent unauthorized access or accidents in crowded areas, these substations are usually housed within a building or a secure enclosure, making them 'indoor type'.
2. Why 'Outdoor Type' is incorrect: Outdoor substations are typically used for high-voltage transmission substations where large clearances are required between conductors and equipment to prevent flashovers, and where land is readily available.
3. Why 'Underground Type' is incorrect: While some components (like cables) are underground, the entire substation (transformers, switchgear) is rarely fully underground due to cooling requirements and maintenance difficulties, although 'gas-insulated substations' (GIS) can be placed in basements.
4. Why 'Dry Type' is incorrect: 'Dry type' refers to a specific type of transformer cooling (air-cooled) rather than the overall classification of the substation layout.
In electrical work, the fuse should be placed -
A)
in earth wire
B)
in live or phase wire
C)
in lamp directly
D)
neutral wire
In electrical installations, a fuse is a safety device designed to protect the circuit from overcurrent or short circuits by melting and breaking the circuit. For maximum safety, the fuse must be placed in the live (phase) wire for the following reasons: 1. Isolation: If the fuse blows, the entire circuit downstream of the fuse becomes 'dead' (de-energized). This prevents the appliance or the wiring from remaining energized at a high potential. 2. Safety: If the fuse were placed in the neutral wire and it blew, the appliance would stop working, but the internal circuitry would still be connected to the live wire. If a person touched a fault in the appliance, they would provide a path to the ground, resulting in a severe electric shock. Therefore, placing the fuse in the phase wire ensures that the power source is disconnected immediately upon a fault.
Why other options are incorrect:
- Option (a): The earth wire is a safety path for leakage current; placing a fuse here would prevent the safety path from working, leading to dangerous chassis electrification.
- Option (c): Placing a fuse 'directly in a lamp' is not a standard wiring practice; the fuse must protect the entire branch circuit.
- Option (d): As explained, placing a fuse in the neutral wire leaves the circuit live even after the fuse has blown, creating a significant electrocution hazard.
Primary distribution substation voltage is -
A)
11kV
B)
33kV
C)
66kV
D)
220kV
In a typical electrical power distribution system, the distribution network is divided into two main stages: Primary Distribution and Secondary Distribution.
1. Primary Distribution: This is the stage where power is transmitted from the distribution substation to the distribution transformers. The standard voltage level for primary distribution in most electrical grids (especially in India and similar systems) is 11 kV. This voltage is used to carry power over a moderate distance to local neighborhoods or industrial areas.
2. Secondary Distribution: This is the final stage where the distribution transformer steps down the 11 kV primary voltage to a usable consumer voltage (typically 415V three-phase or 230V single-phase).
Analysis of options:
- Option (a) 11kV: Correct. This is the standard primary distribution voltage.
- Option (b) 33kV: This is typically used for sub-transmission or primary distribution in very large rural areas, but 11kV is the standard answer for 'primary distribution' in general technical exams.
- Option (c) 66kV: This is a sub-transmission voltage level.
- Option (d) 220kV: This is a high-voltage transmission level used for long-distance bulk power transport.
Distance relay is used for -
A)
Transmission line protection
B)
Transformer protection
C)
Bus bar protection
D)
Generator protection
A distance relay is a type of protection relay that operates based on the impedance of the line. It measures the ratio of voltage to current (V/I) at the relay location. Since the impedance of a transmission line is proportional to its length, the relay can determine the distance to the fault.
1. Why Option (a) is correct: Transmission lines are long conductors where faults can occur at various distances from the substation. Distance relays provide fast, selective protection by dividing the line into zones (Zone 1, Zone 2, etc.), ensuring that only the faulted section of the line is isolated.
2. Why other options are incorrect:
- Transformer protection: Typically uses Differential Relays (for internal faults) and Buchholz relays (for incipient faults).
- Bus bar protection: Typically uses Bus Differential Protection schemes because the bus bar is a single point of connection with no 'distance' or length to measure.
- Generator protection: Typically uses Differential Relays, Reverse Power Relays, and Overcurrent/Overvoltage relays to protect the machine windings and rotor.
Which is the most common voltage level of electrical lines that is running along the road sides on poles?
A)
66kV
B)
33kV
C)
11kV
D)
400 V
In a typical electrical power distribution system, power is transmitted from generating stations at very high voltages, stepped down at transmission substations, and then further stepped down at distribution substations. The lines running along roadsides on poles are typically 'Primary Distribution Lines'. In most standard power systems (especially in India and similar grids), the primary distribution voltage is 11kV. These lines carry power from the distribution substation to local distribution transformers (pole-mounted transformers), which then step the voltage down to the secondary distribution level (415V/230V) for end-user consumption.
Analysis of options:
- 66kV and 33kV: These are typically used for sub-transmission lines (connecting main substations to distribution substations) and are usually carried on larger steel towers or specialized high-tension poles, not the common roadside distribution poles.
- 400 V: This is the secondary distribution voltage (Low Tension - LT). While LT lines also run on poles, the 'most common' characteristic voltage associated with the primary distribution network seen on roadside poles is 11kV. In many exam contexts, when asked about the primary lines on poles, 11kV is the standard answer.
What type of relay is used in applications where high frequency switching is required?
A)
Electromechanical relay
B)
Reed relay
C)
Solid state relay
D)
Mercury relay
To determine the correct relay for high-frequency switching, we must analyze the switching mechanism of each option:
1. **Electromechanical Relays (EMR):** These rely on a physical armature moving to close or open contacts. Due to the inertia of the mechanical parts, they have a slow response time and a limited switching frequency. They are also subject to mechanical wear and tear.
2. **Reed Relays:** These are a type of electromechanical switch consisting of two ferromagnetic blades in a glass envelope. While faster than standard EMRs, they still involve physical movement and are not suitable for very high-frequency switching.
3. **Solid State Relays (SSR):** These use semiconductor devices (such as thyristors, TRIACs, or transistors) to switch the load. Because there are no moving parts, the switching speed is extremely fast (microseconds), allowing for very high-frequency operation. They also offer longer lifespans as there is no mechanical wear or contact arcing.
4. **Mercury Relays:** These use a pool of mercury to establish a connection. While they provide low contact bounce and are reliable, they are still mechanical in nature and cannot compete with the switching speeds of semiconductor-based SSRs.
Therefore, for applications requiring high-frequency switching, the Solid State Relay is the only viable choice.
Transmission efficiency of a transmission line increases with the -
A)
decrease in power factor and voltage
B)
increase in power factor and voltage
C)
increase in power factor but decrease in voltage
D)
increase in current
Transmission efficiency (η) is defined as the ratio of the power delivered at the receiving end to the power supplied at the sending end. Mathematically, η = (Power Delivered) / (Power Delivered + Power Losses). The primary source of loss in a transmission line is I²R loss (copper loss).
1. Effect of Voltage: For a constant amount of power (P = VI cosφ), increasing the transmission voltage (V) results in a decrease in the line current (I). Since power loss is proportional to the square of the current (I²R), increasing the voltage significantly reduces the losses, thereby increasing efficiency.
2. Effect of Power Factor: Power factor (cosφ) represents the ratio of real power to apparent power. For a fixed amount of real power to be delivered, a lower power factor requires a higher current to flow through the line (I = P / (√3 * V * cosφ)). By increasing the power factor (bringing it closer to unity), the current required to transmit the same amount of power decreases. This reduction in current leads to lower I²R losses and higher efficiency.
Therefore, both an increase in voltage and an increase in power factor lead to a reduction in line losses and an increase in transmission efficiency.
What environmental factors can affect the performance of circuit breakers?
A)
Humidity and dust buildup
B)
Exposure to direct sunlight
C)
Extremely high or low temperature
D)
Exposure to rain
Humidity and dust buildup are the most direct environmental factors that can degrade circuit breaker performance. Moisture can reduce insulation resistance and promote corrosion, while dust can contaminate contacts and moving parts, causing overheating, poor contact, or mechanical sticking. Option c is also an environmental factor in general, but the question asks for the factor that affects circuit breaker performance most characteristically in service conditions; humidity and dust are the standard answer. Option b (direct sunlight) may raise temperature slightly, but it is not a primary factor affecting breaker operation. Option d (rain) can be harmful if equipment is exposed, but circuit breakers are normally enclosed/protected, so rain is not the usual general factor cited compared with humidity and dust.
An insulator is so designed that it should fail only when -
A)
a line breakdown
B)
a flash over occurs
C)
it is punctured
D)
sag is less
In transmission line design, there are two primary ways an insulator can fail: puncture and flashover.
1. **Puncture**: This is the electrical breakdown of the insulating material itself. Once an insulator is punctured, it is permanently damaged and must be replaced. This is a catastrophic failure.
2. **Flashover**: This is an arc that forms through the air around the insulator surface. While a flashover causes a short circuit, it does not necessarily damage the physical structure of the insulator; once the fault is cleared, the insulator can often continue to function.
An ideal insulator is designed such that the **flashover voltage is lower than the puncture voltage**. This ensures that if the system is subjected to an overvoltage (like a lightning strike), the electricity will 'flash over' the surface rather than 'puncturing' through the material. Therefore, the insulator is designed to 'fail' (allow current to pass) via flashover first to protect the integrity of the insulator itself from permanent destruction.
- Option (a) 'line breakdown' is a general term for system failure, not a specific insulator design criterion.
- Option (c) 'it is punctured' is exactly what designers want to avoid, as it leads to permanent failure.
- Option (d) 'sag is less' refers to the mechanical tension of the conductor, not the electrical failure of the insulator.
Power factor can be improved by installing a device in parallel with load which takes -
A)
lagging reactive power
B)
leading reactive power
C)
apparent power
D)
leading apparent power
1. Understanding Power Factor: In most industrial electrical systems, the load is predominantly inductive (e.g., motors, transformers), which causes the current to lag behind the voltage. This results in a 'lagging power factor'.
2. The Problem: Inductive loads consume lagging reactive power (Q_L), which increases the total current drawn from the source for the same amount of real power, leading to higher losses and inefficiency.
3. The Solution: To improve (increase) the power factor, we need to neutralize the lagging reactive power. This is achieved by installing a device (typically a capacitor bank or a synchronous condenser) in parallel with the load.
4. Mechanism: Capacitors supply 'leading reactive power' (Q_C). Because leading reactive power is mathematically the opposite of lagging reactive power, the capacitor provides the reactive current required by the inductor locally, reducing the reactive power demand from the main supply.
5. Analysis of Options:
- (a) Lagging reactive power: Adding more lagging power would further decrease the power factor.
- (b) Leading reactive power: Correct. This cancels the lagging effect of inductive loads.
- (c) Apparent power: Apparent power (S) is the vector sum of real and reactive power; it is not something 'taken' by a device to specifically correct the phase angle.
- (d) Leading apparent power: This is a technically incorrect term; 'leading' or 'lagging' refers specifically to the reactive component, not the apparent power.
In a microphone, the acoustic energy is converted into -
A)
Vibration
B)
Pressure energy
C)
Magnetic impulses
D)
Thermal energy
A microphone is a transducer that converts acoustic energy (sound waves) into electrical energy. To understand why 'Magnetic impulses' is the most appropriate choice among the given options, we must look at the mechanism of common microphones (like the dynamic microphone). In a dynamic microphone, sound waves cause a diaphragm to vibrate, which moves a coil of wire within a magnetic field. According to Faraday's Law of Electromagnetic Induction, this movement induces an electromotive force (EMF) or electrical impulses. While the final output is 'electrical energy', the physical process of conversion in many standard microphones involves the creation of magnetic/electrical impulses.
Analyzing the other options:
- (a) Vibration: Acoustic energy is already a form of vibration; the microphone uses vibration as an intermediate step, not as the final converted form.
- (b) Pressure energy: Sound waves are essentially fluctuations in pressure; therefore, acoustic energy is already pressure energy.
- (d) Thermal energy: Converting sound to heat would be an energy loss, not the functional purpose of a microphone.
Therefore, in the context of these specific options, 'Magnetic impulses' refers to the electrical signals generated via magnetic induction.
Ferranti effect on long overhead lines is experienced when -
A)
the line is lightly loaded
B)
the power factor is unity
C)
corona effect is dominant
D)
the power factor is leading
The Ferranti effect is a phenomenon observed in long transmission lines where the voltage at the receiving end (Vr) becomes higher than the voltage at the sending end (Vs). This occurs due to the capacitive charging current of the line.
1. **Mechanism**: A transmission line acts as a distributed capacitor. When the line is open-circuited or lightly loaded, the charging current (which leads the voltage by 90 degrees) flows through the line's inductance. This creates a voltage rise along the line toward the receiving end.
2. **Condition**: The effect is most pronounced when the line is 'lightly loaded' or 'no-load' because the inductive voltage drop (due to load current) is not sufficient to counteract the capacitive voltage rise.
3. **Analysis of Options**:
- **Option (a)**: Correct. Light loading minimizes the voltage drop across the line inductance, allowing the capacitive charging effect to dominate.
- **Option (b)**: Incorrect. Unity power factor implies the load is purely resistive; while Ferranti effect can still occur if the load is light, unity PF is not the defining condition for the effect.
- **Option (c)**: Incorrect. Corona effect is a loss phenomenon related to air ionization and does not cause the receiving end voltage to rise above the sending end voltage.
- **Option (d)**: Incorrect. While a leading power factor load can exacerbate the voltage rise, the Ferranti effect is fundamentally a characteristic of the line's own capacitance under light/no-load conditions, regardless of the load's specific PF.
The height for light points and junction boxes from the floor level should be -
A)
3.0m to 4.0 m
B)
2.5m to 3.0m
C)
2.0m to 3.5m
D)
1.5m to 2.5m
In domestic electrical wiring standards, the mounting height of light points (such as ceiling roses or wall-mounted lamp holders) and junction boxes is designed to ensure both safety and accessibility while remaining out of reach of general accidental contact.
1. **Standard Practice**: For residential buildings, the standard ceiling height typically ranges from 2.7m to 3.0m. Therefore, light points and junction boxes are generally installed at a height of 2.5m to 3.0m from the finished floor level (FFL).
2. **Junction Boxes**: These are placed high up to keep wiring concealed and away from the living area to prevent tampering or accidental damage.
3. **Comparison with other options**:
- Option (a) 3.0m to 4.0m is too high for standard domestic rooms.
- Option (c) 2.0m to 3.5m is too broad and includes heights (2.0m) that are too low for standard light points.
- Option (d) 1.5m to 2.5m is the typical range for switchboards (usually 1.2m to 1.5m) and power sockets, not for light points and junction boxes.
At receiving end of secondary distribution substation voltage is -
A)
stepped up to 220 kV
B)
stepped down to 66 kV
C)
stepped down to 11 kV
D)
stepped down to 440 V
In a typical electrical power distribution system, the process occurs in stages:
1. Primary Distribution: Power is brought from the main substation to the distribution transformer at a voltage typically around 11 kV or 33 kV.
2. Secondary Distribution: This is the final stage of the distribution system. The distribution transformer (located at the secondary distribution substation) steps down the voltage from the primary distribution level (e.g., 11 kV) to a usable consumer level.
3. For three-phase systems, the standard secondary voltage is 415 V to 440 V (which provides 230 V to 240 V for single-phase loads).
Analysis of options:
- Option (a) 220 kV is a transmission voltage, not a distribution voltage.
- Option (b) 66 kV is a sub-transmission voltage.
- Option (c) 11 kV is the input (primary) voltage to the secondary distribution substation, not the voltage at the receiving end (output).
- Option (d) 440 V is the standard output voltage delivered to the end consumers in a secondary distribution network.
Distance relay measure -
A)
impedance of the protected line
B)
voltage at different points along a line
C)
current flowing through the line
D)
power factor of the line
A distance relay is a type of protective relay used primarily for the protection of transmission lines. It operates based on the measurement of the ratio of voltage (V) to current (I) at the relay location. According to Ohm's Law, V/I = Z, where Z is the impedance of the line. Since the impedance of a transmission line is proportional to its length (Z = z * L, where z is the impedance per unit length), measuring the impedance allows the relay to determine the 'distance' to the fault.
- Option (a) is correct because the fundamental operating principle of a distance relay is to measure the impedance of the line to locate the fault.
- Option (b) is incorrect because measuring voltage alone does not provide information about the fault location relative to the line length.
- Option (c) is incorrect because measuring current alone is the function of an overcurrent relay, not a distance relay.
- Option (d) is incorrect because the power factor is a characteristic of the load/fault but is not the primary quantity measured to determine the distance to a fault.
What type of circuit breaker is best for protecting high power appliances like air conditioners?
A)
Miniature Circuit Breaker
B)
Ground Fault Circuit Interrupter
C)
Double pole Breaker
D)
Single pole breaker
A double pole breaker is best for high-power appliances like air conditioners because such loads typically require a 240 V supply and need both live conductors to be disconnected simultaneously for safe isolation and overload protection. It is commonly used for heavy domestic appliances and higher-load circuits.
Why the others are incorrect:
- Single pole breaker: protects only one live conductor, usually for low-power 120 V/lighting circuits, not heavy 240 V appliances.
- Ground Fault Circuit Interrupter (GFCI): protects against earth leakage/shock, not primarily against overloads for high-power appliances.
- Miniature Circuit Breaker (MCB): is a general protective device, but the specific best choice for an air conditioner circuit is typically a double pole breaker because of the higher voltage and need to break both poles.
The components that connects incoming and outgoing electrical lines are called -
A)
Bus bars
B)
Isolator
C)
Insulators
D)
Circuit breakers
In an electrical substation, the bus bar is a metallic strip or bar (typically made of copper or aluminum) that serves as a common electrical node. Its primary function is to collect electrical power from incoming feeders (incoming lines) and distribute it to various outgoing feeders (outgoing lines).
Why other options are incorrect:
- b. Isolator: An isolator is a mechanical switch used to isolate a part of the circuit for maintenance purposes; it does not act as the central connection point for multiple lines.
- c. Insulators: Insulators are materials used to prevent the flow of current from conductors to the supporting structures (like poles or towers); they do not connect lines.
- d. Circuit breakers: A circuit breaker is a protective device designed to automatically interrupt current flow during a fault; while it is placed in the line, its purpose is protection, not the general interconnection of incoming and outgoing lines.
Buchholz relay is used for the protection of -
A)
transformer only
B)
transformer and transmission line
C)
transformer and alternator
D)
alternator only
A Buchholz relay is a gas-actuated safety device specifically designed for oil-immersed transformers. It is installed in the pipe connecting the main transformer tank to the conservator tank. Its primary function is to detect internal faults such as insulation failure, core heating, or winding shorts. These faults cause the transformer oil to decompose, releasing gases (incipient faults) or causing a sudden surge of oil (severe faults). The relay detects these gas accumulations or oil surges and triggers an alarm or trips the circuit breaker to protect the transformer.
Why other options are incorrect:
- Transmission lines (Option b) do not use oil-filled tanks and conservators; they are protected by distance relays, differential relays, or overcurrent relays.
- Alternators (Option c and d) are protected by different mechanisms such as stator earth fault protection, differential protection, and temperature sensors, as they do not utilize the specific oil-to-conservator piping system required for a Buchholz relay to operate.
The conductor vibration of an overhead line may cause -
A)
breaking of insulator discs
B)
excessive sag
C)
collapse of supporting structures
D)
sound pollution
Conductor vibration in overhead transmission lines is typically caused by wind (Aeolian vibrations). These high-frequency, low-amplitude vibrations create repetitive mechanical stress and fatigue at the points of support, specifically where the conductor is clamped to the insulator string. Over time, this mechanical fatigue can lead to the failure of the conductor strands or cause the insulator discs to break due to the constant oscillating stress.
Why other options are incorrect:
- b. Excessive sag: Sag is primarily a function of conductor weight, tension, and temperature, not high-frequency vibration.
- c. Collapse of supporting structures: While extreme wind loads (galloping) can damage towers, standard conductor vibration is localized to the conductor and its attachment points, not sufficient to collapse the entire structure.
- d. Sound pollution: While vibrations can produce a humming sound (Aeolian harp effect), the primary engineering concern and 'cause' of failure in a technical context is the mechanical degradation of components, not environmental noise.
Sag is provided in overhead lines so that -
A)
safe tension is not exceeded
B)
repair can be made
C)
conductor material is saved
D)
extra conductor is kept for future maintenance
In overhead transmission lines, 'sag' is the difference in level between the points of support and the lowest point of the conductor. It is essential to provide sag for the following reasons:
1. **Mechanical Stress Management**: If a conductor were stretched perfectly straight (zero sag), the tension in the conductor would theoretically become infinite. In reality, as the sag decreases, the tension increases exponentially.
2. **Temperature Variations**: Conductors expand in summer (increasing sag) and contract in winter. If there were no sag, the contraction during cold weather would increase the tension beyond the breaking strength of the material, leading to the conductor snapping or the supporting towers collapsing.
3. **Safety Factor**: By providing a calculated amount of sag, engineers ensure that the tension remains within the 'safe working tension' limits of the conductor material under all climatic conditions.
Why other options are incorrect:
- Option (b): Repairing a line is independent of the sag; in fact, excessive sag can make repairs more difficult due to ground clearance issues.
- Option (c): Providing sag actually requires *more* conductor material (the length of the arc is longer than a straight line), not less.
- Option (d): Sag is a structural requirement for tension management, not a reserve of material for future maintenance.