Hydroelectric Power Station MCQ Quiz - Objective Question with Answer for Hydroelectric Power Station - Download Free PDF

Explanation:

Francis Turbine Components and Flow Regulation

Definition: A Francis turbine is a type of reaction turbine used for hydropower generation, designed to operate efficiently under a wide range of head and flow conditions. It converts the potential energy of water into mechanical energy, which is then used to generate electricity through a connected generator. A key feature of the Francis turbine is its ability to dynamically regulate flow and optimize power output under varying load conditions.

Correct Option Analysis:

The correct answer is:

Option 1: Guide Vanes

The guide vanes play a critical role in regulating the flow rate and optimizing the power output of a Francis turbine. They are adjustable components located just before the turbine runner. By dynamically altering their angle, guide vanes control the amount of water entering the runner and the direction of the flow. This ensures that the turbine operates at peak efficiency under varying load conditions.

Working Principle of Guide Vanes:

• Flow Regulation: The guide vanes adjust their position to regulate the flow of water entering the runner. When the load on the turbine decreases, the guide vanes close slightly to reduce the flow rate, and when the load increases, they open to allow more water to flow through.
• Flow Direction: The guide vanes direct the water flow at an optimal angle to the runner blades. This ensures that the kinetic energy of the water is effectively transferred to the runner, maximizing the turbine's efficiency.
• Dynamic Response: Modern Francis turbines are equipped with servo mechanisms that allow the guide vanes to respond quickly to changes in load or flow conditions, maintaining stable operation and consistent power output.

Advantages of Guide Vanes:

• Enable precise control over the turbine's power output by dynamically adjusting the water flow.
• Improve the overall efficiency of the turbine by ensuring optimal flow direction and velocity.
• Help in maintaining stable operation during fluctuating load conditions.
• Reduce the risk of cavitation and mechanical stress on the turbine components by controlling flow conditions.

Conclusion: The guide vanes are essential components in a Francis turbine that dynamically adjust to regulate the flow rate and optimize power output under varying load conditions. Their ability to control both the quantity and direction of water entering the runner ensures efficient and reliable operation of the turbine.

Important Information

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

Option 2: Spiral Casing

The spiral casing is the outer component of a Francis turbine that distributes the incoming water evenly to the guide vanes. It is designed to maintain a constant velocity of water around its circumference. However, it does not dynamically adjust to regulate the flow rate or optimize power output. Its primary function is to ensure uniform distribution of water, which is critical for the efficient operation of the turbine, but it does not have a direct role in dynamic regulation under varying load conditions.

Option 3: Draft Tube

The draft tube is a diverging passage located at the exit of the turbine runner. Its purpose is to recover the kinetic energy of the water leaving the runner and convert it into pressure energy, thereby increasing the overall efficiency of the turbine. While the draft tube plays a significant role in energy recovery, it does not dynamically adjust to regulate the flow rate or power output. Its function is passive rather than active in the context of flow regulation.

Option 4: Runner Blades

The runner blades are the primary moving components of a Francis turbine that convert the kinetic and potential energy of water into mechanical energy. While the shape and design of the runner blades are crucial for the turbine's efficiency, they do not dynamically adjust during operation. The runner blades are fixed in position and rely on the guide vanes to control the flow of water entering the runner.

Conclusion:

In a Francis turbine, the only component that dynamically adjusts to regulate the flow rate and optimize power output under varying load conditions is the guide vanes (Option 1). The spiral casing, draft tube, and runner blades, while essential for the turbine's operation, do not have the capability to dynamically adjust during operation. Understanding the distinct roles of these components is crucial for comprehending the operation and design of Francis turbines.

Hydropower Plant Categories

Definition: Hydropower plants are categorized based on their installed capacity, which is the maximum amount of electricity they can generate under ideal conditions. These categories help classify hydropower projects for planning, design, and regulatory purposes. The primary classifications are Small hydro, Mini hydro, Medium hydro, and Large hydro. Each category is defined by specific capacity ranges, which can vary slightly depending on the country or organization.

Correct Option Analysis:

The correct option is:

Option 1: Small hydro

Why is this correct?

In most international classifications, a hydropower plant with an installed capacity between 1 MW and 25 MW is categorized as a Small hydro project. Since the given hydropower plant has an installed capacity of 10 MW, it falls squarely within this range. Small hydro projects are typically used to supply power to local communities or industries and often have a minimal environmental impact compared to larger projects.

Characteristics of Small Hydro Projects:

• Capacity Range: Typically between 1 MW and 25 MW.
• Applications: Often used for rural electrification, decentralized power generation, or small industrial purposes.
• Environmental Impact: Generally lower than larger hydropower projects due to their smaller scale and less invasive construction requirements.
• Advantages:
  • Can be developed in remote areas to provide localized power solutions.
  • Lower initial investment and shorter construction times compared to large hydropower projects.
  • Can often operate without the need for large dams, preserving local ecosystems.
• Disadvantages:
  • Limited capacity means it may not be suitable for regions with high power demands.
  • May be less efficient in terms of energy production compared to larger hydropower plants.

Conclusion: Since the given hydropower plant has an installed capacity of 10 MW, it falls into the category of Small hydro. This classification is widely recognized and aligns with international standards for hydropower categorization.

Additional Information

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

Option 2: Mini hydro

Mini hydro projects typically have an installed capacity of less than 1 MW (or sometimes up to 2 MW, depending on specific definitions). These systems are smaller than small hydro projects and are often used for localized, off-grid power generation in rural or isolated areas. Since the given plant has a capacity of 10 MW, it does not fall under this category.

Option 3: Medium hydro

Medium hydro projects generally have capacities ranging from 25 MW to 100 MW. These plants are larger than small hydro projects and are used to supply power to larger grids or industrial facilities. The 10 MW capacity of the given plant is significantly below this range, so it does not qualify as a medium hydro project.

Option 4: Large hydro

Large hydro projects have an installed capacity exceeding 100 MW. These are massive infrastructure projects that often involve significant dam construction and reservoir creation. They are used for large-scale power generation and contribute to national or regional grids. Since the given plant has a capacity of only 10 MW, it does not fall under this category.

Conclusion:

The classification of hydropower plants is essential for understanding their scale, applications, and potential impacts. The given plant, with an installed capacity of 10 MW, clearly falls under the category of Small hydro, as defined by widely accepted standards. This classification highlights the plant's suitability for localized power generation with minimal environmental disruption compared to larger hydropower projects.

Explanation:

Hydroelectric Power Plant Efficiency and Net Head Calculation

Definition: A hydroelectric power plant converts the potential energy of water into electrical energy. The efficiency of the system depends on various factors, including the flow rate of water, the net head (height difference between water source and turbine), water density, acceleration due to gravity, and the overall efficiency of the system.

Problem Statement:

We are given the following parameters:

• Flow rate (Q) = 50 m³/s
• Electrical power output (P_out) = 8 MW = 8 × 10⁶ W
• Water density (ρ) = 1000 kg/m³
• Acceleration due to gravity (g) = 10 m/s²
• Overall efficiency (η) = 80% = 0.8

Objective: To calculate the required net head (H) for the hydroelectric power plant to achieve an overall efficiency of 80%.

Formula:

The overall efficiency (η) of the hydroelectric power plant is given by:

η = (P_out) / (P_input)

Where P_input is the power input due to the potential energy of water, which can be expressed as:

P_input = ρ × g × Q × H

Rearranging the formula to find H:

H = (P_out) / (η × ρ × g × Q)

Substituting the given values:

• P_out = 8 × 10⁶ W
• η = 0.8
• ρ = 1000 kg/m³
• g = 10 m/s²
• Q = 50 m³/s

H = (8 × 10⁶) / (0.8 × 1000 × 10 × 50)

H = (8 × 10⁶) / (400,000)

H = 20 m

Conclusion:

The required net head (H) for the hydroelectric power plant to achieve an overall efficiency of 80% is 20 meters.

Correct Option: Option 4

Additional Information

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

Option 1: 19.50 m

This option is incorrect because the calculated net head (H) using the formula does not match this value. It may arise from a miscalculation or incorrect assumptions in the problem-solving process.

Option 2: 22.15 m

This option is also incorrect for the same reasons as option 1. The value does not align with the mathematical calculation based on the given parameters.

Option 3: 18.75 m

This option is incorrect as well. It does not correspond to the calculated value of H using the formula provided.

Option 4: 20 m

This is the correct option, as demonstrated by the step-by-step calculation above. The net head required for the given efficiency is 20 meters.

Conclusion:

The formula for calculating the net head (H) in a hydroelectric power plant is straightforward when the efficiency, flow rate, water density, and acceleration due to gravity are known. By carefully substituting the given values into the formula, the correct net head of 20 meters is determined, making option 4 the correct choice. This calculation is essential for designing efficient hydroelectric power plants and ensuring optimal energy output.

• The Hydro-electric power plant has the least maintenance cost among all types of power generation plants.
• Hydro-electric power plants not only have low maintenance costs but also have the least operating cost.
•  Because Hydro-electric power plant has fewer moving equipment, a rigid structure then the operating and maintenance cost reduces.
• Another factor in operations is the fuel cost which is practically free for hydro stations.
• Where a typical coal-fired plant is has a coal handling unit, a DM water unit, a generating unit, a cooling water unit and an ash handling unit while a hydro plant has a generating unit only.

Explanation:

Surge Tank in Hydroelectric Power Plant

Definition: A surge tank is a protective device in hydroelectric power plants. It is a water storage reservoir connected to the penstock, which is the pipeline that delivers water from the reservoir to the turbines. The primary purpose of a surge tank is to mitigate the effects of water hammer and pressure fluctuations in the penstock, ensuring the safety and efficiency of the power plant.

Working Principle: In a hydroelectric power plant, water flows from a high elevation reservoir through the penstock to the turbines, where it generates electricity. When there are sudden changes in the water flow rate, such as the rapid closure or opening of a turbine valve, it can cause significant pressure surges, known as water hammer. This phenomenon can lead to damage in the penstock and other components. A surge tank acts as a buffer, absorbing sudden pressure changes and maintaining a steady flow of water.

Advantages:

• Reduces the risk of water hammer, protecting the penstock and other equipment from damage.
• Helps maintain a steady flow of water to the turbines, improving the efficiency and reliability of the power plant.
• Allows for better control of water flow during transient conditions, such as load changes or emergency shutdowns.

Disadvantages:

• Requires additional space and infrastructure, which can increase the overall cost of the hydroelectric power plant.
• Needs regular maintenance to ensure proper functioning and to prevent issues such as sediment buildup.

Applications: Surge tanks are commonly used in hydroelectric power plants with long penstocks and significant elevation differences between the reservoir and the turbines. They are essential in ensuring the safe and efficient operation of the plant by mitigating pressure surges and maintaining a steady water flow.

Correct Option Analysis:

The correct option is:

Option 4: Reduce water hammer effect in the penstock.

This option accurately describes the primary purpose of a surge tank in a hydroelectric power plant. The surge tank helps reduce the water hammer effect, which is the pressure surge caused by sudden changes in water flow within the penstock. By absorbing these pressure fluctuations, the surge tank protects the penstock and other components from potential damage, ensuring the smooth operation of the power plant.

Additional Information

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

Option 1: Increase the efficiency of the turbine.

While surge tanks contribute to the overall efficiency of the hydroelectric power plant by maintaining a steady water flow, their primary purpose is not directly related to increasing the turbine's efficiency. The main function of the surge tank is to mitigate pressure surges and protect the penstock and other components.

Option 2: Store water for future use.

This option is incorrect because the primary function of a surge tank is not to store water for future use. While the surge tank does hold some water, its main purpose is to act as a buffer to absorb pressure fluctuations in the penstock, not to serve as a reservoir for water storage.

Option 3: Generate additional power.

This option is incorrect as well. The surge tank does not generate additional power. Its function is to protect the penstock and other components from pressure surges, ensuring the safe and efficient operation of the hydroelectric power plant. The generation of power is solely the function of the turbines within the power plant.

In conclusion, the primary purpose of a surge tank in a hydroelectric power plant is to reduce the water hammer effect in the penstock. By absorbing sudden pressure changes, the surge tank protects the penstock and other components from potential damage, ensuring the safe and efficient operation of the power plant. While it contributes to the overall efficiency and reliability of the plant, its main function is to mitigate pressure surges, not to store water for future use or generate additional power.

Explanation:

Concept:

• Hydropower plants capture the energy of falling water to generate electricity.
• A turbine converts the potential energy of falling water into mechanical energy and the generator converts the mechanical energy from the turbine into electrical energy.

The power equation for a hydropower plant is

P = ρQgh

Where,

P = power generated (W), ρ = density of water (kg/m3), Q = volumetric flow rate of water (m3/s), h = head height (m)

Calculation:

Given:

Q = 100 m3/s, h = 75 m

P = ρQgh = 1000 × 100 × 9.81 × 75 = 73.5 × 10⁶ = 73.5 MW

Bernoulli’s equation:

Bernoulli’s equation is used to calculate the power output of the Hydropower plant.

Output power can be expressed as

\(P = 9.81\times 10^{-3} \times (\eta \ QWH )\) in KWatts

Where,

η = Plant efficiency 

W = Density of water = 1000 Kg / m³

H = Mean water head in meters

Q = Discharge of water in m³ / sec

Calculation:

Given that,

H = 324 m

Q = 1370 m³ / sec

W = 1000 Kg / m³

If plant efficiency is not given in the question, then we take it as 100%

∴ η = 1 (For calculating capacity of plant)

The output power is

P = 9.81 x 10^-3 η QWH kW

P = 9.81 x 10^-3 x 1 x 1000 x 324 x 1370

∴ P = 4.35 GW

Surge tank (or surge chamber) is a device introduced within a hydropower water conveyance system having a rather long pressure conduit to absorb the excess pressure rise in case of sudden valve closure.

The surge tank is located between the almost horizontal or slightly inclined conduit and steeply sloping penstock and is designed as a chamber excavated in the mountain

It also acts as small storage from which water may be supplied in case of a sudden valve opening of the turbine.

In case of a sudden opening of the turbine valve, there are chances of penstock collapse due to a negative pressure generation. If there is no surge tank.

Surge tanks are usually provided in high or medium - head hydroelectric power plants when there is a considerable distance between the water source and the power unit, necessitating a long penstock.

The main functions of the surge tank are:

1. When the load decreases, the water moves backward and gets stored in it.

2. When the load increases, an additional supply of water will be provided by the surge tank.

Explanation:

• Standby losses in a power plant are defined as the fuel wasted when no generation is happening.
• A hydropower plant has no standby loss as compared to a thermal power plant.
• In thermal plants, even if a generator is kept stand by still the boiler should run because it takes more than 24 hrs to restart a boiler if it is shut down so the energy generated by these boilers which are not being used is the standby loss.
• In the case of a hydropower plant, if the value is closed the water stops that remains on standby. When the valve is opened again, the water flows therefore, no standby losses occur in the hydropower plant.

Additional Information

Concept:

Potential energy, E = mgh

Total output energy, E'  = Potential energy x overall efficiency

\(\left[ {Density = \frac{{Mass}}{{Volume}}} \right]\)

Where,

M = Mass

g = gravitational acceleration

h = height

1 J = 1/ 3600 Wh

Calculation:

Given that

overall efficiency = 75%

Reservoir capacity (V) = 3.6 × 106 m3

Water Head (h) = 100 m

Now to calculate the potential energy, we know that

Density of water = 997 kg/m3

g = 9.8 m/s2

Put all these in formula of potential energy, we get

E = (D × V) × 9.8 × 100     

E = 997 × 3.6 × 106 × 9.8 × 100

E = 3517416 × 106 J

E =  (3517416 / 3600) × 103 kWh

E = 977060 kWh

Hence, total output energy can be calculated as

E' = 977060 × 0.75 

∴ E' ≈ 732795 kWh ≈ 732.7 MWh

The hydroelectric power plant has a high initial cost because of the construction of the dam.

The efficiency of hydroelectric plants is about 85%, hence highly efficient.

Hence, the correct answer is option 3.

Comparison between different power plants:

Note:

Hydro turbines, the oldest and the most commonly used renewable energy source, have the most efficient of all power conversion processes. The potential head of water is available right next to the turbine, so there are no energy conversion losses, only the mechanical and copper losses in the turbine and generator and the tail end loss. The efficiency is in the range of 85 to 90 %.

Wind turbines have an overall conversion efficiency of 30 % to 45 %. These two renewable sources, though efficient, are dependent on the availability of the energy source.

Solar thermal systems can achieve efficiency up to 20 %. The moving path of the sun and the weather conditions drastically alter the incident solar radiation. The efficiency on an annual basis, around 12 %, is considerably less than on a daily basis.

Geothermal systems, on the other hand, also use the Rankine cycle with steam temperatures at a saturation point. Since there is no other conversion loss, this plant can achieve efficiencies in the range of 35 %.

Concept

The hot reserve or reserve capacity of a power plant is given by:

Hot reserve = Plant capacity - Demand capacity

Calculation

Given, Plant capacity =  200 MVA

Demand capacity = 120 MVA

Hot reserve = 200 - 120

Hot reserve = 80 MVA

Additional Information The plant capacity factor is given by:

\(Plant \space capacity \space factor={Average \space demand\over Plant \space capacity}\)

The correct answer is option 2):(Draft tube)

Concept:

• The draft tube is used to increase the acting head on the water wheel in a hydro-electric power plant
• Draft Tube is a diverging tube fitted at the exit of the runner of the turbine and used to utilize the kinetic energy available with water at the exit of the runner. This draft tube at the end of the turbine increases the pressure of the exiting fluid at the expense of its velocity.
• The draft tube is used only in the reaction turbines as there is a change in pressure energy involved.
• It permits a negative or suction head to be established at the runner exit, thus making it possible to install the turbine above the tail race level without loss of head.
• It converts a large proportion of velocity energy rejected from the runner into useful pressure energy.

Additional Information

•  A penstock is a sluice or gate or intake structure that controls water flow, or an enclosed pipe that delivers water to hydro turbines and sewerage systems. 
• A Surge tank is a water storage device used as a pressure neutralizer in hydropower water conveyance systems in order to dampen excess pressure variance.
• A reservoir is an enlarged lake behind a dam. Such a dam may be either artificial, built to store fresh water or it may be a natural formation