Steam Power Plants MCQ Quiz - Objective Question with Answer for Steam Power Plants - Download Free PDF
Last updated on May 8, 2025
Latest Steam Power Plants MCQ Objective Questions
Steam Power Plants Question 1:
Why do modern steam power plants use a regenerative feedwater heating system?
Answer (Detailed Solution Below)
Steam Power Plants Question 1 Detailed Solution
Use of a regenerative feedwater heating system
Modern steam power plants use regenerative feedwater heating systems primarily to improve thermal efficiency. This is because:
- By preheating the feedwater using steam extracted from intermediate stages of the turbine, the energy required in the boiler to convert water into steam is reduced.
- This reduces fuel consumption for the same power output, leading to better efficiency based on the Rankine cycle with regeneration.
- Regeneration increases the average temperature at which heat is added in the cycle, which directly boosts the cycle efficiency (per Carnot's principle).
- With improved efficiency, lower emissions per unit of electricity generated are achieved, making the plant more environmentally friendly.
Steam Power Plants Question 2:
Which of the following is NOT a part of the thermal station?
Answer (Detailed Solution Below)
Steam Power Plants Question 2 Detailed Solution
The correct answer is option 4.
A moderator is not a part of a thermal power station; it is used in nuclear power plants to slow down neutrons for efficient fission.
Components of Steam Power Plant
1.) Coal and ash handling plant
- The coal from the storage plant is delivered to the coal handling plant where it is pulverized (i.e., crushed into small pieces) to increase its surface exposure, thus promoting rapid combustion.
- The pulverized coal is fed to the boiler by belt conveyors. The coal is burnt in the boiler, and the ash produced after the complete combustion of coal is removed to the ash handling plant and then delivered to the ash storage plant for disposal.
2.) Boiler
- The heat of the combustion of coal in the boiler is utilized to convert water into steam at high temperatures and pressure.
- The flue gases from the boiler make their journey through the superheater, economizer, and air pre-heater and are finally exhausted to the atmosphere through the chimney.
3.) Superheater
- The steam produced in the boiler is wet and is passed through a superheater where it is dried and superheated by the flue gases on their way to the chimney.
4.) Economiser
- The feed water is fed to the economizer before supplying to the boiler.
- The economizer extracts a part of the heat of flue gases to increase the feed water temperature.
5.) Air preheater
- An air preheater increases the temperature of the air supplied for coal burning by deriving heat from flue gases.
6.) Steam turbine
- The heat energy of steam when passing over the blades of the turbine is converted into mechanical energy.
7.) Alternator
- The steam turbine is coupled to an alternator. The alternator converts the mechanical energy of the turbine into electrical energy.
Steam Power Plants Question 3:
Electrostatic precipitator is an important device in which of the following power plants?
Answer (Detailed Solution Below)
Steam Power Plants Question 3 Detailed Solution
Electrostatic precipitator
- An Electrostatic Precipitator (ESP) is an important device used in thermal power plants, particularly in coal-fired power plants.
- Thermal power plants, especially those that burn coal or oil, produce large amounts of particulate matter (PM), including fly ash, dust, and other pollutants. An ESP removes these particulates from flue gas before it is released into the atmosphere, reducing air pollution.
Working Principle:
The ESP works on the principle of electrostatic attraction:
- Ionization: High-voltage electrodes create an electric field that ionizes the particles in the flue gas.
- Charging of Particles: The dust and smoke particles become negatively charged due to the corona discharge.
- Collection of Particles: The charged particles are attracted to positively charged collecting plates, where they stick and accumulate.
- Cleaning Process: The collected dust is periodically removed by rapping or washing the plates.
Steam Power Plants Question 4:
In a thermal power plant, feed water from feed water pump enters the boiler through the __________.
Answer (Detailed Solution Below)
Steam Power Plants Question 4 Detailed Solution
The correct answer is option 1.
Components of Steam Power Plant
1.) Coal and ash handling plant
- The coal from the storage plant is delivered to the coal handling plant where it is pulverized (i.e., crushed into small pieces) to increase its surface exposure, thus promoting rapid combustion.
- The pulverized coal is fed to the boiler by belt conveyors. The coal is burnt in the boiler, and the ash produced after the complete combustion of coal is removed to the ash handling plant and then delivered to the ash storage plant for disposal.
2.) Boiler
- The heat of the combustion of coal in the boiler is utilized to convert water into steam at high temperatures and pressure.
- The flue gases from the boiler make their journey through the superheater, economizer, and air pre-heater and are finally exhausted to the atmosphere through the chimney.
3.) Superheater
- The steam produced in the boiler is wet and is passed through a superheater where it is dried and superheated by the flue gases on their way to the chimney.
4.) Economiser
- The feed water is fed to the economizer before supplying to the boiler.
- The economizer extracts a part of the heat of flue gases to increase the feed water temperature.
5.) Air preheater
- An air preheater increases the temperature of the air supplied for coal burning by deriving heat from flue gases.
6.) Steam turbine
- The heat energy of steam when passing over the blades of the turbine is converted into mechanical energy.
7.) Alternator
- The steam turbine is coupled to an alternator. The alternator converts the mechanical energy of the turbine into electrical energy.
Steam Power Plants Question 5:
A diesel power station has fuel consumption of 0.28 kg per kWh, the calorific value of fuel being 10,000 kcal/kg. The overall efficiency of the station is:
Answer (Detailed Solution Below)
Steam Power Plants Question 5 Detailed Solution
Concept
The overall efficiency of the diesel power station is given by:
\(η={Electrical\space output\space in\space heat \space units \over Heat\space combustion }\)
Calculation
Heat produced by 0·28 kg of diesel = 10,000 × 0·28 = 2800 kcal
Heat equivalent of 1 kWh = 860 kcal
\(η={860\over 2800}=0.307\)
η = 30.7 %
Top Steam Power Plants MCQ Objective Questions
In thermal power plants, the function of economizer is to:
Answer (Detailed Solution Below)
Steam Power Plants Question 6 Detailed Solution
Download Solution PDFEconomizer:
It is also known as a feedwater heater. It is a device in which the waste heat of the flue gases is utilized for heating the feed water.
In economizer, feed water is preheated by using flue gases to improve overall efficiency and only sensible heat transfer is taking place so feed water is heated without converting it into steam. Therefore, the economizer is placed after the superheater and located in the feeding water circuit.
Functions of economizer:
- Reduce fuel consumption
- Preheating a fluid (feed-water in case of steam boiler)
- Increases the efficiency of the power plant
Following are the advantages:
- There is about 15 to 20% of coal saving.
- It increases the steam raising capacity of a boiler because it shortens the time required to convert water into steam.
- It prevents the formation of scale in boiler water tubes.
A steam power station has an overall efficiency of 25%, and 0.5 kg of coal is burnt per kWh of electrical energy generated. Determine the calorific value of fuel.
(Take heat equivalent of 1 kWh as 860 kcal)
Answer (Detailed Solution Below)
Steam Power Plants Question 7 Detailed Solution
Download Solution PDFConcept:
Calorific values of fuel: The calorific value of fuel is the quantity of heat produced by its combustion at constant pressure and under normal conditions. Calorific value mention in kcal / kg.
The overall efficiency of the steam power station: The overall efficiency of the steam power station is defined as the ratio of the power available at the generator terminal to the rate of energy released by the combustion of fuel. It is given by,
\({\eta _{overall}} = \frac{{EO}}{{HC}}\)
Where EO is the Electrical output in the heat unit.
HC is Heat combustion.
Calculation:
Given: Overall efficiency = 25 % , 0.5 kg coal burnt.
Let x cal /kg be the calorific value of the fuel.
Heat equivalent of 1 kWh = 860 kcal
\({\eta _{overall}} = \frac{{EO}}{{HC}}\)
\(0.25 = \frac{{860}}{{0.5x}}\)
\(x = \frac{{860000}}{{25 \times 5}}\)
\(x = 6880\;kcal/kg\)
In a steam power plant, the value of the reheat factor is of the order of:
Answer (Detailed Solution Below)
Steam Power Plants Question 8 Detailed Solution
Download Solution PDFExplanation:
Reheat factor:
- It is defined as the ratio of cumulative heat drop to the adiabatic heat drop in all stages of the turbine.
- The value of reheat factor depends on the type and efficiency of the turbine.
- The value of the reheat factor is of the order of 1.1 to 1.5
Reheat factor (R.F.) = \(\frac{{Cumulative\,\,heat\,\,drop\,\,(isentropic)}}{{Isentropic\,\,heat\,\,drop\,\,(overall)}}\)
The function of electrostatic precipitator in a coal-based thermal power plant is to:
Answer (Detailed Solution Below)
Steam Power Plants Question 9 Detailed Solution
Download Solution PDFElectrostatic precipitators (ESP):
- In electrostatic precipitators, the flue gas is made to pass through a highly ionised zone, where the particles get electrically charged and are separated out from the gas, with the help of electrostatic forces in the power full electric field.
- Electrostatic precipitator in a coal-based thermal power plant collects dust from flue gas.
- They are widely used in Thermal power plants, Pulp and Paper industries, Mining and Metallurgical Industries, Iron and Steel plants, Chemical industries, etc.
- The main two advantages of ESP are particles may be collected wet or dry and efficiency more than 99% can be achieved.
- ESP can handle both gases and mists for high volume flow and even very small particles can be removed.
- It uses very high voltage to generate a powerful electric field. Hence, power consumption is very high and subsequently, ESP is very costly as compared to other air pollution control devices.
The following data is available for a steam power station:
maximum demand = 25 MW
Load factor = 0.4
Coal consumption = 0.88 kg/kWh
Boiler efficiency = 85%
Turbine efficiency = 90%
Price of coal = Rs. 55 per tones
Find the thermal efficiency of the station.
Answer (Detailed Solution Below)
Steam Power Plants Question 10 Detailed Solution
Download Solution PDFConcept:
Thermal Efficiency:
- Thermal Efficiency (\(\eta_T\)) is the ratio of the heat utilized by a heat engine to the total heat units in the fuel consumed.
- It is also the product of Boiler efficiency and Turbine efficiency.
- \(\eta_T\) = Boiler efficiency × Turbine efficiency.
Load Factor:
- It is the ratio of Average load to maximum demand.
- Load Factor is always less than or equal to 1.
Calculation:
Given
Maximum demand = 25 MW
Load factor = 0.4
Coal consumption = 0.88 kg/kWh
Boiler efficiency = 85%
Turbine efficiency = 90%
Price of coal = Rs. 55 per tone
From concept,
\(\eta_T = 0.85 × 0.9 = 0. 765\)
= 76.5 % (In percentage)
A pulverized coal boiler generates approximately _______ fly ash and _______ bottom ash, respectively.
Answer (Detailed Solution Below)
Steam Power Plants Question 11 Detailed Solution
Download Solution PDF- In the coal thermal power plant, the steam is produced at high pressure in the steam boiler due to the burning of fuel (pulverized coal) in boiler furnaces.
- The pulverized coal boiler produces approximately 80% fly ash and 20% bottom ash.
Additional InformationCoal ash: A collection of residuals produced during the combustion of coal.
Fly ash: A light form of coal ash that floats into the exhaust stacks.
Bottom ash: The heavier portion of coal ash that settles on the ground in the boiler.
Important Points
- A typical pulverized coal boiler consists of the furnace, which is the lower boiler part, where the burners and the firing system are installed and where the combustion takes place
- In the convective section, where the convective heat exchanger surfaces are installed.
- The boiler membrane walls have usually the function of the boiler evaporator.
- The heat exchangers sections are placed in the convective part.
A diesel power station has fuel consumption 0.28 kg per kWh, the calorific value of fuel being 10000 kcal/kg. The overall efficiency of the station is
(Heat produced equivalent of 1 KWh = 860 KCal)
Answer (Detailed Solution Below)
Steam Power Plants Question 12 Detailed Solution
Download Solution PDFConcept
The overall efficiency of the diesel power station is given by:
\(η={Electrical\space output\space in\space heat \space units \over Heat\space combustion }\)
Calculation
Heat produced by 0·28 kg of diesel = 10,000 × 0·28 = 2800 kcal
Heat equivalent of 1 kWh = 860 kcal
\(η={860\over 2800}=0.307\)
η = 30.7 %
Which of the following feed water treatments is basically one type of thermal treatment?
Answer (Detailed Solution Below)
Steam Power Plants Question 13 Detailed Solution
Download Solution PDFExplanation:
Deaerator:
- Deaerator in steam power plant removes condensable gases from water and non-condensible from circulated water.
- If a large number of closed regenerators are used in the cycle, at least one should be an open type for this deaeration process.
- This deaeration process removes some steam with gases which is makeup feed water.0
A 100 MW steam station uses coal of calorific value 6400 kcal/kg. Thermal efficiency of the station is 30% and electrical efficiency is 92%. Find the overall efficiency of the plant.
Answer (Detailed Solution Below)
Steam Power Plants Question 14 Detailed Solution
Download Solution PDFConcept:
The overall efficiency of the power plant is given as
\(Overall\;Efficiency\;\; = \frac{{Electrical\;energy\;output}}{{chemical\;energy\;input}}\)
Overall efficiency is also expressed as
\({\eta _0} = \;\frac{{Thermal\;Energy}}{{Chemical\;Energy}} \times \frac{{Mechanical\;Energy}}{{Thermal\;Energy}} \times \frac{{Electrical\;Energy}}{{Mechanical\;energy}}\)
So, the overall efficiency of the power plant is equal to the multiplication of the efficiencies of the boiler, turbine, and generator.
Overall Efficiency = Boiler efficiency × turbine efficiency × generator efficiency
ηpower plant = ηboiler × ηturbine × ηgenerator
Where,
Boiler efficiency = Thermal efficiency
Turbine efficiency × Generator efficiency = Electrical efficiency
Calculation:
Given-
Thermal efficiency = 30%
Electrical Efficiency = 92%
Overall efficiency of the plant = (0.30) (0.92) = 0.276
Overall efficiency of the plant = 27.6%.
Which of the following generating power stations requires lot of time for starting?
Answer (Detailed Solution Below)
Steam Power Plants Question 15 Detailed Solution
Download Solution PDFSteam power station:
A generating station which converts heat energy of coal combustion into electrical energy is known
as a steam power station.
A steam power station basically works on the Rankine cycle. Steam is produced in the boiler by utilising the heat of coal combustion.
Advantages:
- The fuel (i.e., coal) used is quite cheap.
- Less initial cost as compared to other generating stations.
- It can be installed at any place irrespective of the existence of coal. The coal can be transported to the site of the plant by rail or road.
- It requires less space as compared to the hydroelectric power station.
- The cost of generation is lesser than that of the diesel power station.
Disadvantages:
- It pollutes the atmosphere due to the production of large amount of smoke and fumes.
- It is costlier in running cost as compared to hydroelectric plant.
- It requires lot of time for starting.