This document provides tables and calculations related to various parameters at different points in a thermal power plant. It includes calculations of enthalpy, entropy, and mass flow rates at the boiler inlet and outlet, superheater inlet and outlet, turbine inlets and outlets, reheater inlet and outlet, and condenser inlet and outlet. Properties are obtained from steam tables using interpolation, with enthalpy and entropy expressed in units of MW for the entire flow. The tables show energy transfers and property changes throughout the plant.
This document discusses the working cycles of thermal power plants, specifically the Rankine, regenerative, and reheat cycles. It begins with an introduction to the Rayalaseema Thermal Power Plant located in Kadapa, India, including its layout, raw materials (coal, furnace oil, diesel, water), combustion process, and operational data from 1995-2013. It then provides details on the Rankine, regenerative, and reheat thermodynamic cycles used in thermal power generation, including diagrams, components, efficiencies, and factors that increase efficiency. The document contains extensive tables of thermodynamic properties and calculations for the plant components as well as graphs comparing exergy destruction and efficiencies.
This document discusses heat engines, refrigerators, and the second law of thermodynamics. It contains the following key points:
1. A heat engine transforms heat partly into work by a working substance undergoing a cyclic process, but is not 100% efficient due to the second law of thermodynamics.
2. The Carnot cycle involves two reversible isothermal processes and two reversible adiabatic processes, and represents the maximum possible efficiency for a heat engine.
3. Refrigerators and heat pumps move heat from a cold to a hot reservoir, but also require work input. Their performance is measured by the coefficient of performance.
1. A gas is compressed from 6.0 L to 4.0 L at a constant temperature of 5.0 atm. The work done on the gas is +30 liter atm.
2. A system's internal energy increases by 80 J when 50 J of work is done on it. The heat change of the system is +30 J.
3. A refrigerator operates in a cyclic and reversible manner between 0°C inside and 27°C outside. If it absorbs 334.72x103 J from the inside, the heat rejected outside is 340 J.
The document describes a dynamic Otto cycle analysis that models the performance of spark ignition internal combustion engines. It makes three modifications to the static air-standard Otto cycle analysis: 1) using representative specific heats and heat ratios for each process, 2) relating heat release during combustion to engine parameters, and 3) including an equation for volumetric efficiency as a function of engine speed. This dynamic analysis predicts engine power and torque curves with reasonable accuracy by modeling the mass of air and temperatures at each state as engine speed varies. Sample results are shown for a Nissan Maxima engine that agree well with manufacturer specifications.
1. Exergy, also known as available energy, is the maximum useful work possible during a process that brings a system into equilibrium with its surroundings. It quantifies the potential of a system to cause a change as it achieves equilibrium.
2. For any thermodynamic system, the energy supplied can be broken down into available (exergy) and unavailable (anergy) components. Exergy is the maximum work possible through reversible processes, while anergy is the energy that cannot be converted to work even through reversible processes.
3. Exergy can be defined and calculated for closed and open systems. For closed systems, the sum of expansion work and work from an ideal heat engine represents the maximum available work. For
The document contains examples of problems related to applied thermodynamics and heat engines. It includes 6 examples that cover topics like determining interface temperatures, heat transfer in heat exchangers, radiation from blackbodies, compression of gases, heating of water, and heat transfer over flat plates. The examples provide calculations and step-by-step workings to arrive at the solutions.
This document provides an overview of thermodynamic concepts related to steady flow processes involving steam and water in common power plant devices. It describes the basic functions and analysis of a steam boiler, steam turbine, steam condenser, and mixing chamber. The analysis involves applying the continuity equation and the steady flow energy equation to determine heat transfer rates and work output. Sample problems are provided to illustrate the use of thermodynamic property tables and calculations for each device.
This document discusses the thermal design of a simple boiler. It presents the calculation procedures for boiler design, focusing on heat transfer modes, heat and mass balances, and a worked example. The key points are:
- Heat transfer in boilers occurs via conduction, convection, and radiation. Conduction is not considered in simple calculations.
- Heat and mass balance equations relate the heat input from fuel to the heat output via steam as well as accounting for air and flue gas flows.
- A worked example calculates furnace conditions like flue gas temperature for a methane-fueled boiler, assuming radiation is the only heat transfer mode in the furnace. Tube bank calculations then determine the exit gas
This document discusses the working cycles of thermal power plants, specifically the Rankine, regenerative, and reheat cycles. It begins with an introduction to the Rayalaseema Thermal Power Plant located in Kadapa, India, including its layout, raw materials (coal, furnace oil, diesel, water), combustion process, and operational data from 1995-2013. It then provides details on the Rankine, regenerative, and reheat thermodynamic cycles used in thermal power generation, including diagrams, components, efficiencies, and factors that increase efficiency. The document contains extensive tables of thermodynamic properties and calculations for the plant components as well as graphs comparing exergy destruction and efficiencies.
This document discusses heat engines, refrigerators, and the second law of thermodynamics. It contains the following key points:
1. A heat engine transforms heat partly into work by a working substance undergoing a cyclic process, but is not 100% efficient due to the second law of thermodynamics.
2. The Carnot cycle involves two reversible isothermal processes and two reversible adiabatic processes, and represents the maximum possible efficiency for a heat engine.
3. Refrigerators and heat pumps move heat from a cold to a hot reservoir, but also require work input. Their performance is measured by the coefficient of performance.
1. A gas is compressed from 6.0 L to 4.0 L at a constant temperature of 5.0 atm. The work done on the gas is +30 liter atm.
2. A system's internal energy increases by 80 J when 50 J of work is done on it. The heat change of the system is +30 J.
3. A refrigerator operates in a cyclic and reversible manner between 0°C inside and 27°C outside. If it absorbs 334.72x103 J from the inside, the heat rejected outside is 340 J.
The document describes a dynamic Otto cycle analysis that models the performance of spark ignition internal combustion engines. It makes three modifications to the static air-standard Otto cycle analysis: 1) using representative specific heats and heat ratios for each process, 2) relating heat release during combustion to engine parameters, and 3) including an equation for volumetric efficiency as a function of engine speed. This dynamic analysis predicts engine power and torque curves with reasonable accuracy by modeling the mass of air and temperatures at each state as engine speed varies. Sample results are shown for a Nissan Maxima engine that agree well with manufacturer specifications.
1. Exergy, also known as available energy, is the maximum useful work possible during a process that brings a system into equilibrium with its surroundings. It quantifies the potential of a system to cause a change as it achieves equilibrium.
2. For any thermodynamic system, the energy supplied can be broken down into available (exergy) and unavailable (anergy) components. Exergy is the maximum work possible through reversible processes, while anergy is the energy that cannot be converted to work even through reversible processes.
3. Exergy can be defined and calculated for closed and open systems. For closed systems, the sum of expansion work and work from an ideal heat engine represents the maximum available work. For
The document contains examples of problems related to applied thermodynamics and heat engines. It includes 6 examples that cover topics like determining interface temperatures, heat transfer in heat exchangers, radiation from blackbodies, compression of gases, heating of water, and heat transfer over flat plates. The examples provide calculations and step-by-step workings to arrive at the solutions.
This document provides an overview of thermodynamic concepts related to steady flow processes involving steam and water in common power plant devices. It describes the basic functions and analysis of a steam boiler, steam turbine, steam condenser, and mixing chamber. The analysis involves applying the continuity equation and the steady flow energy equation to determine heat transfer rates and work output. Sample problems are provided to illustrate the use of thermodynamic property tables and calculations for each device.
This document discusses the thermal design of a simple boiler. It presents the calculation procedures for boiler design, focusing on heat transfer modes, heat and mass balances, and a worked example. The key points are:
- Heat transfer in boilers occurs via conduction, convection, and radiation. Conduction is not considered in simple calculations.
- Heat and mass balance equations relate the heat input from fuel to the heat output via steam as well as accounting for air and flue gas flows.
- A worked example calculates furnace conditions like flue gas temperature for a methane-fueled boiler, assuming radiation is the only heat transfer mode in the furnace. Tube bank calculations then determine the exit gas
The document discusses energy and exergy analysis of a diesel engine powered cogeneration plant (DEPC). It first provides background on cogeneration and its benefits. It then describes the components and working of the DEPC plant, which has a total installed capacity of 25.32 MW electricity and 8.1 tons/hr of steam. The plant uses three diesel engines fueled by heavy fuel oil. The document presents the thermodynamic analysis of the plant using concepts of energy and exergy balances. It finds that the plant has a first law efficiency of 40.4% and second law (exergetic) efficiency of 40.6%. The largest exergy destruction of 46% occurs in the diesel engines.
The document contains 7 examples of thermodynamics calculations involving concepts like steam tables, work, heat, ideal gases, refrigeration cycles, and processes involving gases. The examples calculate things like inlet and outlet steam pressures, net work done by systems, heat transferred in cycles, minimum heat rejection rate, refrigeration power requirement, work done by compressing an ideal gas, and net work in a sequence of gas processes.
Simulation of the effects of turbine exhaust recirculationZin Eddine Dadach
For an effective carbon capture by an amine mixture, the molar percentage of CO2 in the flue gas should be at least equal to 10%. Moreover, in order to reduce technical problems due to amine oxidative degradation, the molar percentage of O2 in the flue gas should be limited to 5%. One possible option for increasing the concentration of CO2 and decreasing the amount of O2 in the flue gas from power plants using natural gas is recirculation of a part of the flue gas.
The document contains 10 examples solving thermodynamics problems involving concepts like the first law of thermodynamics, ideal gas law, steady state conditions, heat transfer, work done, efficiency and refrigeration cycles. The last example involves a reversible heat engine and refrigerator operating between different temperature reservoirs. It is determined that the heat rejection to the 40°C reservoir is 5539 kJ.
This document summarizes key concepts from a chapter on thermodynamics including:
1) Thermodynamics deals with heat and work in systems and their surroundings. A system is the focus, while the surroundings are everything else in the environment.
2) The first law of thermodynamics states that the change in a system's internal energy equals the heat added minus the work done.
3) There are four main types of thermal processes - isobaric, isochoric, isothermal, and adiabatic - which involve constant pressure, volume, temperature, or no heat transfer respectively.
4) The second law of thermodynamics involves entropy and the principle that heat cannot spontaneously flow from cold
Ch 3 energy transfer by work, heat and massabfisho
The document discusses energy transfer through heat, work, and mass. It defines key concepts like the first law of thermodynamics, heat transfer, work, power, and various types of work including boundary work, shaft work, spring work, and more. It provides equations to calculate work, heat transfer, and power for different processes. It includes several examples calculating work, heat transfer, and analyzing processes on P-V diagrams for closed systems operating in various thermodynamic cycles and processes.
This document discusses various thermodynamic power cycles including:
- The Carnot cycle, which is the most efficient but impractical cycle.
- Rankine cycles, which are more practical vapor power cycles that use steam as the working fluid.
- Simple Rankine cycles involve heating water to steam then expanding it in a turbine before condensing it back to water.
- Rankine cycles with superheated steam, which increase efficiency by heating steam above its saturation temperature.
- The efficiencies of different cycles are calculated and compared in examples. Superheated steam cycles have higher efficiencies than simple Rankine cycles due to higher average temperatures.
- A heat engine absorbs heat (Q1) from a hot body and uses it to do work (W) on the surroundings, rejecting some heat (Q2) to a cold body.
- Common working substances inside heat engines include air and steam, which undergo heating and cooling to produce work in a cyclic process.
- The efficiency (η) of a heat engine is defined as the ratio of work output to heat input, and is highest when the minimum amount of heat (Q2) is rejected to the cold body.
- The Carnot cycle operates between two temperature reservoirs and represents the most efficient heat engine possible according to the second law of thermodynamics.
The document discusses thermal power cycles. It begins by explaining that a thermal power plant involves heating water to create steam that spins a turbine and generates electricity.
The basic energy flow in a thermal power plant is: chemical energy is converted to mechanical energy by the steam turbine, which is then converted to electrical energy. Various fuel sources can be used.
It then discusses the key power cycles used in thermal plants like the Carnot, Rankine, Diesel, Otto, and Brayton cycles. It also covers the laws of thermodynamics and important thermodynamic processes.
The Rankine cycle most closely models actual steam power plants. It involves pumping water, boiling it to create steam, expanding the steam
This document provides lecture notes on gas power cycles from a mechanical engineering course. It covers the ideal gas relation, specific heats of gases, and the first law of thermodynamics as they relate to closed systems. Formulas are presented for the ideal gas law in various forms, as well as specific heat capacities for air. The notes also define important terms used in analyzing thermodynamic cycles, such as pressure ratio, compression ratio, and cutoff ratio. An example problem is included to demonstrate calculating maximum pressure, net specific work, and thermal efficiency for a diesel engine cycle.
This document provides a mid-term review covering three topics: 1) energy analysis of closed systems, 2) mass and energy analysis of control volumes, and 3) the second law of thermodynamics. For the first topic, it provides examples of energy balance calculations for constant pressure processes in closed systems. For the second topic, it discusses the energy balance equation for control volumes and provides examples of its application to turbines, compressors, and throttling valves. For the third topic, it defines thermal efficiency and the coefficient of performance and discusses heat engines, refrigerators, and heat pumps.
1) Heat engines operate in a thermodynamic cycle to absorb heat (Q>0) from a hot reservoir and use it to do mechanical work (W>0), dumping excess heat (QC<0) into a cold reservoir.
2) The efficiency of a heat engine is defined as the work done (W) divided by the total heat absorbed (QH), and is always between 0 and 1.
3) Refrigerators and heat pumps operate using the same thermodynamic cycles as heat engines but with the goal of maximizing heat transfer (QC for refrigerators, QH for heat pumps) for a given expenditure of work (W). Their performance is quantified by coefficients of performance K that can be greater than 1
Thermodynamic Chapter 4 Second Law Of ThermodynamicsMuhammad Surahman
This document provides an overview of the second law of thermodynamics. It discusses how the second law establishes conditions for equilibrium and determines theoretical performance limits. The document defines key concepts like thermal efficiency, the Carnot cycle, and entropy. It presents examples calculating efficiency and heat transfer for systems like power plants, refrigerators, and heat pumps operating between different temperature levels.
This document provides information about gas turbine and steam power plant cycles. It describes the Brayton cycle used in gas turbines and the Rankine cycle used in steam power plants. It discusses components, processes, thermal efficiencies and improvements to the cycles such as regeneration, intercooling and reheating. Examples are provided to calculate efficiency, work and heat inputs/outputs for simple and improved cycles.
This document contains multiple problems involving ideal gas processes. The first problem describes a steady flow compressor handling nitrogen with known intake conditions and discharge pressure. It asks to determine the final temperature and work for two process types. The second problem involves air in a cylinder being compressed in a polytropic process with known initial and final pressures and temperatures. It asks to determine the work and heat transfer. The third problem describes a gas turbine expanding helium polytropically and asks to determine the final pressure, power produced, heat loss, and entropy change.
The document discusses the second law of thermodynamics. It explains that the first law has limitations and does not determine whether a process is possible. The second law states that heat cannot spontaneously flow from a cold body to a hot body. It presents two common statements of the second law by Kelvin and Clausius. The document also discusses entropy, ideal gas entropy changes, and the reversible Carnot cycle.
first law of thermodynamics and second lawnaphis ahamad
This document discusses the first law of thermodynamics and conservation of energy. It explains that the first law states that energy cannot be created or destroyed, only transformed between different forms. The total energy in a closed system remains constant. The document provides several examples of applying the first law to closed systems, control volumes, and various thermodynamic processes like isochoric, isobaric, and polytropic processes. It also discusses other concepts like the conservation of mass, work done by fluids, and applying energy balances to devices like nozzles, turbines, and heat exchangers.
Exergy analysis of magnetic refrigerationManoj maurya
The document discusses magnetic refrigeration and exergy analysis of magnetic refrigeration cycles. It explains the magnetocaloric effect and how magnetic fields can be used to achieve cooling via adiabatic demagnetization. Key equations presented include those relating the magnetocaloric effect to changes in magnetic field and temperature. The document also summarizes the reversible Brayton refrigeration cycle used in magnetic refrigeration and equations for the exergy efficiency and exergy destroyed. Major breakthroughs in 1997 that accelerated progress in the field are noted. In closing, the summary states that magnetic refrigeration provides an effective and efficient cooling method and has improved significantly since initial work in the 1920s-1930s.
The document discusses second law efficiencies and exergy change of systems. It defines second law efficiency for various devices like heat engines, work-producing devices, refrigerators, and general processes. Second law efficiency compares the actual performance of a device to its theoretical maximum performance under reversible conditions. The exergy change of a closed system depends on the change in internal energy, pressure-volume work, and entropy between initial and final states, accounting for environmental properties. Exergy can represent the useful or recoverable work of a system.
Energy and Exergy Analysis of a Country Sectors - Advanced ThermodynamicsMostafa Ghadamyari
This document outlines sectoral energy and exergy analysis procedures. It discusses how energy, exergy, and entropy concepts apply to thermodynamic systems and can be used to analyze macrosystems like societies. The document then provides an example of analyzing Turkey's residential/commercial sector, showing the steps to estimate energy and exergy efficiencies for applications like space heating and cooking. Finally, it discusses analyzing Turkey's industrial sector, dividing energy usage into process categories and examining industries like iron-steel production.
The document discusses energy and exergy analysis of a diesel engine powered cogeneration plant (DEPC). It first provides background on cogeneration and its benefits. It then describes the components and working of the DEPC plant, which has a total installed capacity of 25.32 MW electricity and 8.1 tons/hr of steam. The plant uses three diesel engines fueled by heavy fuel oil. The document presents the thermodynamic analysis of the plant using concepts of energy and exergy balances. It finds that the plant has a first law efficiency of 40.4% and second law (exergetic) efficiency of 40.6%. The largest exergy destruction of 46% occurs in the diesel engines.
The document contains 7 examples of thermodynamics calculations involving concepts like steam tables, work, heat, ideal gases, refrigeration cycles, and processes involving gases. The examples calculate things like inlet and outlet steam pressures, net work done by systems, heat transferred in cycles, minimum heat rejection rate, refrigeration power requirement, work done by compressing an ideal gas, and net work in a sequence of gas processes.
Simulation of the effects of turbine exhaust recirculationZin Eddine Dadach
For an effective carbon capture by an amine mixture, the molar percentage of CO2 in the flue gas should be at least equal to 10%. Moreover, in order to reduce technical problems due to amine oxidative degradation, the molar percentage of O2 in the flue gas should be limited to 5%. One possible option for increasing the concentration of CO2 and decreasing the amount of O2 in the flue gas from power plants using natural gas is recirculation of a part of the flue gas.
The document contains 10 examples solving thermodynamics problems involving concepts like the first law of thermodynamics, ideal gas law, steady state conditions, heat transfer, work done, efficiency and refrigeration cycles. The last example involves a reversible heat engine and refrigerator operating between different temperature reservoirs. It is determined that the heat rejection to the 40°C reservoir is 5539 kJ.
This document summarizes key concepts from a chapter on thermodynamics including:
1) Thermodynamics deals with heat and work in systems and their surroundings. A system is the focus, while the surroundings are everything else in the environment.
2) The first law of thermodynamics states that the change in a system's internal energy equals the heat added minus the work done.
3) There are four main types of thermal processes - isobaric, isochoric, isothermal, and adiabatic - which involve constant pressure, volume, temperature, or no heat transfer respectively.
4) The second law of thermodynamics involves entropy and the principle that heat cannot spontaneously flow from cold
Ch 3 energy transfer by work, heat and massabfisho
The document discusses energy transfer through heat, work, and mass. It defines key concepts like the first law of thermodynamics, heat transfer, work, power, and various types of work including boundary work, shaft work, spring work, and more. It provides equations to calculate work, heat transfer, and power for different processes. It includes several examples calculating work, heat transfer, and analyzing processes on P-V diagrams for closed systems operating in various thermodynamic cycles and processes.
This document discusses various thermodynamic power cycles including:
- The Carnot cycle, which is the most efficient but impractical cycle.
- Rankine cycles, which are more practical vapor power cycles that use steam as the working fluid.
- Simple Rankine cycles involve heating water to steam then expanding it in a turbine before condensing it back to water.
- Rankine cycles with superheated steam, which increase efficiency by heating steam above its saturation temperature.
- The efficiencies of different cycles are calculated and compared in examples. Superheated steam cycles have higher efficiencies than simple Rankine cycles due to higher average temperatures.
- A heat engine absorbs heat (Q1) from a hot body and uses it to do work (W) on the surroundings, rejecting some heat (Q2) to a cold body.
- Common working substances inside heat engines include air and steam, which undergo heating and cooling to produce work in a cyclic process.
- The efficiency (η) of a heat engine is defined as the ratio of work output to heat input, and is highest when the minimum amount of heat (Q2) is rejected to the cold body.
- The Carnot cycle operates between two temperature reservoirs and represents the most efficient heat engine possible according to the second law of thermodynamics.
The document discusses thermal power cycles. It begins by explaining that a thermal power plant involves heating water to create steam that spins a turbine and generates electricity.
The basic energy flow in a thermal power plant is: chemical energy is converted to mechanical energy by the steam turbine, which is then converted to electrical energy. Various fuel sources can be used.
It then discusses the key power cycles used in thermal plants like the Carnot, Rankine, Diesel, Otto, and Brayton cycles. It also covers the laws of thermodynamics and important thermodynamic processes.
The Rankine cycle most closely models actual steam power plants. It involves pumping water, boiling it to create steam, expanding the steam
This document provides lecture notes on gas power cycles from a mechanical engineering course. It covers the ideal gas relation, specific heats of gases, and the first law of thermodynamics as they relate to closed systems. Formulas are presented for the ideal gas law in various forms, as well as specific heat capacities for air. The notes also define important terms used in analyzing thermodynamic cycles, such as pressure ratio, compression ratio, and cutoff ratio. An example problem is included to demonstrate calculating maximum pressure, net specific work, and thermal efficiency for a diesel engine cycle.
This document provides a mid-term review covering three topics: 1) energy analysis of closed systems, 2) mass and energy analysis of control volumes, and 3) the second law of thermodynamics. For the first topic, it provides examples of energy balance calculations for constant pressure processes in closed systems. For the second topic, it discusses the energy balance equation for control volumes and provides examples of its application to turbines, compressors, and throttling valves. For the third topic, it defines thermal efficiency and the coefficient of performance and discusses heat engines, refrigerators, and heat pumps.
1) Heat engines operate in a thermodynamic cycle to absorb heat (Q>0) from a hot reservoir and use it to do mechanical work (W>0), dumping excess heat (QC<0) into a cold reservoir.
2) The efficiency of a heat engine is defined as the work done (W) divided by the total heat absorbed (QH), and is always between 0 and 1.
3) Refrigerators and heat pumps operate using the same thermodynamic cycles as heat engines but with the goal of maximizing heat transfer (QC for refrigerators, QH for heat pumps) for a given expenditure of work (W). Their performance is quantified by coefficients of performance K that can be greater than 1
Thermodynamic Chapter 4 Second Law Of ThermodynamicsMuhammad Surahman
This document provides an overview of the second law of thermodynamics. It discusses how the second law establishes conditions for equilibrium and determines theoretical performance limits. The document defines key concepts like thermal efficiency, the Carnot cycle, and entropy. It presents examples calculating efficiency and heat transfer for systems like power plants, refrigerators, and heat pumps operating between different temperature levels.
This document provides information about gas turbine and steam power plant cycles. It describes the Brayton cycle used in gas turbines and the Rankine cycle used in steam power plants. It discusses components, processes, thermal efficiencies and improvements to the cycles such as regeneration, intercooling and reheating. Examples are provided to calculate efficiency, work and heat inputs/outputs for simple and improved cycles.
This document contains multiple problems involving ideal gas processes. The first problem describes a steady flow compressor handling nitrogen with known intake conditions and discharge pressure. It asks to determine the final temperature and work for two process types. The second problem involves air in a cylinder being compressed in a polytropic process with known initial and final pressures and temperatures. It asks to determine the work and heat transfer. The third problem describes a gas turbine expanding helium polytropically and asks to determine the final pressure, power produced, heat loss, and entropy change.
The document discusses the second law of thermodynamics. It explains that the first law has limitations and does not determine whether a process is possible. The second law states that heat cannot spontaneously flow from a cold body to a hot body. It presents two common statements of the second law by Kelvin and Clausius. The document also discusses entropy, ideal gas entropy changes, and the reversible Carnot cycle.
first law of thermodynamics and second lawnaphis ahamad
This document discusses the first law of thermodynamics and conservation of energy. It explains that the first law states that energy cannot be created or destroyed, only transformed between different forms. The total energy in a closed system remains constant. The document provides several examples of applying the first law to closed systems, control volumes, and various thermodynamic processes like isochoric, isobaric, and polytropic processes. It also discusses other concepts like the conservation of mass, work done by fluids, and applying energy balances to devices like nozzles, turbines, and heat exchangers.
Exergy analysis of magnetic refrigerationManoj maurya
The document discusses magnetic refrigeration and exergy analysis of magnetic refrigeration cycles. It explains the magnetocaloric effect and how magnetic fields can be used to achieve cooling via adiabatic demagnetization. Key equations presented include those relating the magnetocaloric effect to changes in magnetic field and temperature. The document also summarizes the reversible Brayton refrigeration cycle used in magnetic refrigeration and equations for the exergy efficiency and exergy destroyed. Major breakthroughs in 1997 that accelerated progress in the field are noted. In closing, the summary states that magnetic refrigeration provides an effective and efficient cooling method and has improved significantly since initial work in the 1920s-1930s.
The document discusses second law efficiencies and exergy change of systems. It defines second law efficiency for various devices like heat engines, work-producing devices, refrigerators, and general processes. Second law efficiency compares the actual performance of a device to its theoretical maximum performance under reversible conditions. The exergy change of a closed system depends on the change in internal energy, pressure-volume work, and entropy between initial and final states, accounting for environmental properties. Exergy can represent the useful or recoverable work of a system.
Energy and Exergy Analysis of a Country Sectors - Advanced ThermodynamicsMostafa Ghadamyari
This document outlines sectoral energy and exergy analysis procedures. It discusses how energy, exergy, and entropy concepts apply to thermodynamic systems and can be used to analyze macrosystems like societies. The document then provides an example of analyzing Turkey's residential/commercial sector, showing the steps to estimate energy and exergy efficiencies for applications like space heating and cooking. Finally, it discusses analyzing Turkey's industrial sector, dividing energy usage into process categories and examining industries like iron-steel production.
This document discusses magnetic refrigeration, which provides cooling through the magnetocaloric effect. It begins by introducing magnetic refrigeration and the magnetocaloric effect. It then covers the basic principles and mechanism of magnetic refrigeration, including the thermodynamic cycle and components required. Potential magnetocaloric materials are discussed. Applications for magnetic refrigeration include household appliances, buildings, transportation, food storage, and electronics cooling. Benefits include higher efficiency and lower environmental impact compared to traditional refrigeration. Further research is still needed to improve temperature changes and develop stronger permanent magnets for widespread commercial use.
1. The document presents solutions to two problems involving a Rankine cycle.
2. For the first problem, the summary includes the pump work, turbine work, net work output, thermal efficiency, quality of steam entering the condenser, specific steam consumption, and quality of steam entering the condenser at a reduced pressure.
3. For the second problem, the summary provides the condition of steam entering the turbine, turbine work per unit mass of steam, actual pump work per unit mass of water, net work output and thermal efficiency of the cycle, and quality of steam entering the condenser.
The document contains solutions to several problems involving thermodynamic cycles and processes. It calculates things like:
- Water flow and power output for a hydroelectric plant given head, efficiency and installed capacity.
- Power output, heat loss and efficiencies for a turbine given inputs like flow rate, head, and efficiencies.
- Exhaust properties like temperature, enthalpy and quality for steam turbines given inlet conditions and pressures.
- Parameters like work, efficiency and temperatures for Rankine cycles and gas turbine cycles.
Thermodynamics Assignment 02 contains calculations for various cycles of a steam power plant operating between 40 bar and 0.04 bar:
1) Carnot, simple Rankine, and modified Rankine cycles are analyzed. The modified Rankine cycle with superheat has the highest efficiency of 40.86% and lowest SSC of 2.4820 kg/kWh.
2) "Metallurgical limit" refers to the maximum safe pressures and temperatures a power plant's components can withstand without damage.
3) Implementing reheating in the Rankine cycle increases efficiency to 41.05% and lowers SSC to 2.4663 kg/kWh by utilizing the steam's initial high temperature again
Solution Manual for Physical Chemistry – Robert AlbertyHenningEnoksen
https://www.book4me.xyz/solution-manual-physical-chemistry-alberty/
Solution Manual for Physical Chemistry - 6th Edition
Author(s) : Robert A. Alberty
This solution manual include all chapters of textbook (1 to 21).
GAS POWER CYCLES PRESENTATION FOR STUDENT UNIVERSITYssuser5a6db81
This document summarizes the key concepts and equations related to gas power cycles, including ideal cycles like Otto, Diesel, and Brayton cycles that are used in internal combustion engines. It discusses the assumptions and processes involved in each cycle. It also covers concepts like regeneration, intercooling, and reheat that are used to improve upon ideal cycles. The document provides definitions of important terms and examples of applying the cycle equations.
An air refrigeration system operates between pressures of 16 bar and 1 bar to produce 33.5 kW of refrigeration.
The document provides calculations to determine:
(a) The mass flow rate of air circulated per minute is 15.3 kg/min
(b) The piston displacement of the compressor is 11.768 m3/min and of the expander is 6.025 m3/min
(c) The net work is 40.5 kW
(d) The COP is 0.827
The document then provides similar calculations for other refrigeration systems operating under different conditions and specifications.
- The document describes an evaporator that concentrates a 20% sodium hydroxide solution to 50% using steam at a temperature of 126.45°C.
- Key calculations include determining vapor and liquid temperatures and enthalpies, solving material and energy balances, and using the heat transfer equation to calculate the required vapor flow.
- Solving the system of equations gives the feed rate F to the evaporator as 11,700 kg/h.
This document contains information about a steam power plant, including:
1. A block diagram of the plant showing the high pressure turbine, intermediate pressure turbine, low pressure turbine, and condenser.
2. Heat balance diagrams for the high pressure and low pressure turbines showing steam flow, pressures, temperatures, and enthalpies at different points.
3. Key performance parameters like the heat rate which is the amount of heat energy needed to produce a given amount of electrical energy. The actual and guaranteed heat rates are provided.
4. Formulas for calculating the heat rate, specific steam consumption, and boiler efficiency.
This document describes experiments conducted to test the feasibility of using a thermoelectric cooler powered by a bicycle dynamo to heat and cool drinks. It was found that the power output of 60W from cycling at 15km/h was not sufficient to cool a drink to the desired temperature within an optimum time. Additional experiments showed heating 350ml of water would take a very long time and cooling was hindered by lack of forced convection. Insulation would be needed for practical application of this thermoelectric cooler system.
1) The document discusses the processes involved in a Carnot cycle for an ideal gas, including isothermal expansion and compression and adiabatic processes.
2) It examines the efficiencies of Carnot engines and refrigerators, noting that engines are more efficient when the temperature difference is large, while refrigerators are more efficient when the temperature difference is small.
3) It then shows how assuming the heat engine statement of the second law is false would allow using a refrigerator to violate the refrigerator statement of the second law by creating a perpetual motion machine.
Lecture 3-4: Exergy, Heating and Cooling, Solar Thermalcdtpv
The document discusses various topics related to energy and thermodynamics, including:
1) Different systems that can store 1 kW hr of energy and why we pay more for some forms.
2) The concept of exergy, which quantifies how useful energy is, and how it is lost as heat is converted to less useful forms.
3) Examples of heat engines and refrigerators/heat pumps, and calculations of their maximum possible efficiencies.
4) Ways of improving energy efficiency for heating, such as reducing heat loss, increasing heat pump coefficients of performance, and using combined heat and power systems.
Jean-Paul Gibson: Analysis Of An Open Feedwater Heater SystemJean-Paul Gibson
People often look surprised when I tell them that Thermodynamics was probably one of my favorite classes in school. This was the final project for my Thermo II class and was intended to be done in groups. I wanted to challenge myself to a near unnecessary limit by completing this entire project myself. The purpose of this project was to determine the optimum operating pressure for the feedwater heater for a power cycle (from a selected problem in the textbook). However the primary focus, and most difficult task, was writing a program in C++ to calculate multiple values for a selected problem in our textbook. I quite literally had to teach myself how to program in C++ well beyond the basics I had learned in an intro class. This program had to be capable of cross referencing multiple reference tables in the back of the book, each having dozens of values which had to be manually typed in. Which reference tables it pulled values from was dependent on the user inputted turbine efficiency and feedwater heater pressure. It easily took me over a month to debug and test this program to ensure it was always going to the correct reference tables to pull data when outputting the calculations it performed.
Update on 7/2/2016: I did attempt to make another program that would allow the user to type in any feedwater heater pressure rather than make a selection from values there was a reference table for in the back of the back. The program used a method of interpolation to obtain all the data needed to perform all the calculations that were outputted. I don't remember the last time I attempted to run that version of the program, but I seem to recall getting impossible answers outputted. This indicated to me that a more complicated method of interpolating values was required, or I just couldn't get it to work correctly. I ultimately decided against pursuing that version of the program any further. Just getting it work correctly with a predefined list of pressures to select from was difficult enough.
In a Brayton cycle with regeneration:
1) Without regeneration, the cycle efficiency is 19.9%.
2) With a regenerator that is 75% effective, the cycle efficiency increases to 28.37%.
3) This is a 42.56% increase in efficiency due to regeneration.
For an ideal Brayton cycle with an air inlet of 1 atm and 300K compressed to 6 atm and a maximum temperature of 1100K:
1) The thermal efficiency of the cycle is 40.1%
2) The work ratio is 0.545
3) The power output is 40.1 MW
4) The exergy flow rate of the exhaust gas is 20.
Fox and McDonalds Introduction to Fluid Mechanics 9th Edition Pritchard Solut...KirkMcdowells
Full download : https://alibabadownload.com/product/fox-and-mcdonalds-introduction-to-fluid-mechanics-9th-edition-pritchard-solutions-manual/ Fox and McDonalds Introduction to Fluid Mechanics 9th Edition Pritchard Solutions Manual
This document provides an overview of gas turbine power cycles and the Joule cycle. It begins by revising gas expansions and compressions, then introduces the basic Joule cycle which consists of four ideal processes - isentropic compression, constant pressure heating, isentropic expansion, and constant pressure cooling. The document discusses efficiency calculations and worked examples for the ideal Joule cycle. It then examines the effects of friction, providing diagrams and equations to model non-isentropic compression and expansion processes. The document concludes by noting some variants in practical gas turbine engines from the basic Joule cycle model.
The document discusses exergy, which is a measure of work potential. It provides explanations of concepts such as reversible work, irreversibility, and second-law efficiency. Several examples are given to illustrate how to calculate exergy, irreversibility, and second-law efficiency for systems involving heat transfer, work, and multiple states.
The document presents a preliminary design for an energy recovery system in a ramjet engine. It proposes introducing a moving element to bleed some compressed air from the ramjet into a turbine. This would generate electricity or other energy conservation. Calculations are shown comparing thrust produced with and without air bleeding. The goal is to extract thrust from the turbine to compensate for thrust lost from air bleeding, reducing battery weight and improving missile efficiency through energy recovery.
This document provides information about a lecture on reheat and intercooling in gas turbine systems. It includes:
- An explanation of the concepts and purposes of using reheat and intercooling in gas turbines.
- An example problem calculating efficiency and mass flow rate for a gas turbine cycle with reheat and intercooling.
- Diagrams of gas turbine cycles with the different enhancements labeled.
Gas Power Cycles in Chemical Engineering Thermodynamics.pptHafizMudaserAhmad
This document describes several gas power cycles including ideal cycles like the Otto cycle, Diesel cycle, and Brayton cycle as well as actual engine cycles. It provides details on the assumptions and processes of each ideal cycle. The Otto cycle involves four internally reversible processes including isentropic compression and expansion with constant-volume heat addition and rejection. The Diesel cycle similarly involves isentropic compression and expansion but with constant-pressure rather than constant-volume heat addition. The Brayton cycle models gas turbine engines with isentropic compression and expansion and constant-pressure heat transfer. Regeneration and intercooling are discussed as ways to improve upon actual engine cycles.
Gas Power Cycles in Chemical Engineering Thermodynamics.ppt
PROJECT DOCUMENT ON EXERGY final
1. Exergy Analysis of Thermal Power Plant
Department of Mechanical Engg, SVIST, Kadapa. Page 56
CHAPTER-6
6.0 TABLES AND CALCULATIONS
Enthalpy (H): The sum total of internal energy (U) and (PV) is called Enthalpy.
Denoted by “H”
H= U+ PV
Where H = total enthalpy J,
U = total internal energy
P = absolute pressure
V = Specific Volume
Entropy: The disorderness of molecules of the system of molecules is called Entropy.
Denoted by “S”
Unit of Measure
Enthalpy H = h×m×10-3
=
𝑘𝐽
𝑘𝑔
x
𝑘𝑔
𝑠
=
𝑘𝐽
𝑆
×103
= kW x 103
= Mw
Entropy S = S x m x 10-3
=
𝑘𝐽
𝑘𝑔−𝐾
×
𝑘𝑔
𝑆
=
𝑘𝐽
𝑆−𝐾
=
𝐾𝑊
𝐾
×10-3
S = Mw / K
In this chapter, calculations related to various tables have been shown and the various directly
measured and derived parameters have been tabulated.
An interpolation has been done extensively in order to obtain properties which were not readily
available in the steam tables.
2. Exergy Analysis of Thermal Power Plant
Department of Mechanical Engg, SVIST, Kadapa. Page 57
6.1 Boiler Inlet:
At boiler inlet
Pressure P = 157.8 bar, Temperature T = 574 K, mass flow rate, m= 175 kg/s
From steam tables
Specific volume
v = 0.001694
m3
𝑘𝑔
Pressure difference
dP = 157.8 – 157.8 = 0
Temperature difference
dT = 0k
At 574k, hf =13450.4
𝑘𝐽
𝑘𝑔
Enthalpy in Mw, H = 1345.4x175x10-3 =235.445 Mw
At 574 k, sf =3.271.
Entropy in Mw S = sf x mx10-3
= 3.271 x 175 x 10-3= 0.5724
𝑀𝑤
𝐾
6.2 BOILER OUTLET
PRESSURE P=146.05 bar, TEMPERATURE T = 613 K,
mass flow rate = 175
𝑘𝑔
𝑠
.
Form steam tables
Enthalpy h=2626.2
𝑘𝐽
𝑘𝑔
Entropy s = 5.3429
𝑘𝐽
𝑘𝑔−𝐾
Enthalpy in Mw H = hxmx10-3
H = 2626.2 x 175 x 10-3 = 459.58 Mw
Enthalpy in Mw S = sxmx10-3
= 5.3429 x 175 x 10-3 = 0.9349 Mw/K
3. Exergy Analysis of Thermal Power Plant
Department of Mechanical Engg, SVIST, Kadapa. Page 58
6.3 SUPER HEATER INLET
PRESSURE P = 146.05 bar, TEMPERATURE T = 613 K,
mass flow rate mass flow rate = 177.78
𝑘𝑔
𝑠
From steam tables
Enthalpy h = 2626.2
𝑘𝐽
𝑘𝑔
Entropy s = 5.3429
𝑘𝐽
𝐾𝑔−𝐾
Enthalpy in Mw
H = h x m x 10-3
= 2626.2 x 177.78 x 10-3 = 466.88 Mw
Entropy in Mw
S = s x m x 10-3
= 5.3429 x 177.78 x 10-3 = 0.9497
𝑀𝑤
𝐾
6.4 SUPERHEATER OUTLET
PRESSURE P = 146.05 bar, TEMPERATURE T = 540 c, mass flow rate m= 177.18
𝑘𝑔
𝑠
From steam tables
At 140.05 bar Enthalpy
h = 3472.12+
3482.62−3472.12
10
x (150-146.05) [By interpolation] = 3476.26
𝑘𝐽
𝑘𝑔
Entropy values
Pressure 500 o
c 600 o
c 540 o
c
140 bar 6.394 6.714 6.5872
150 bar 6.349 6.676 6.5452
At 146.05 bar, interpolation values at 540o c
Entropy s = 6.5452 +
6.5872 −6.5452
10
x (150 – 146.05) = 6.56179
𝑘𝐽
𝑘𝑔−𝐾
Enthalpy in Mw H = hxmx10-3
= 3476.26x177.78x10-3 = 618.009 Mw
Entropy in Mw S = sxmx10-3
= 6.56179x177.78x10-3 = 1.1665
𝑀𝑤
𝐾
4. Exergy Analysis of Thermal Power Plant
Department of Mechanical Engg, SVIST, Kadapa. Page 59
6.5 High Pressure Turbine [HPT] INLET
PRESSURE P = 146.05 bar, TEMPERATURE T = 540 o c, mass flow rate m= 177.18
𝑘𝑔
𝑠
For above values
Enthalpy h = 3476.26
𝑘𝐽
𝑘𝑔
Entropy s = 6.56179
𝑘𝐽
𝑘𝑔−𝐾
Enthalpy in Mw H = hxmx10-3
= 3476.26x177.78x10-3 = 618.009 Mw
Entropy in Mw S = sxmx10-3
= 6.56179x177.78x10-3= 1.1665
𝑀𝑤
𝐾
6.6 High Pressure Turbine [HPT] OUTLET
PRESSURE P=35.30 bar, TEMPERATURE T = 630 K =330 o c,
mass flow rate m= 161.11
𝑘𝑔
𝑠
From steam tables, enthalpy values
Pressure 300 o
c 350 o
c
34 bar 2982.2 3108.7
36 bar 2975.6 3108.7
At 34 bar, 330o c
Enthalpy h = 2982.2+
3108 .7−2982.2
50
x20 = 3032.8 kJ/kg
At 36 bar, 330 o c
h= 2975.6+
3104.2−2975.6
50
x20 = 3027.04 kJ/kg
At 35.3 bar, 330 o c
h = 3027.04+
3032.8−3027 .04
20
x (36-35.3) = 3029.056 kJ/kg
Entropy values
Pressure 300 o
c 350 o
c
34 bar 6.467 6.679
36 bar 6.432 6.647
5. Exergy Analysis of Thermal Power Plant
Department of Mechanical Engg, SVIST, Kadapa. Page 60
At 34 bar, 330o c
s = 6.467+
6.679 −6.467
50
x 20 = 6.5518 kJ/kg- K
At 36 bar, 330 o c
s = 6.432+
6.647 −6.432
50
x 20 = 6.518
At 35.30 bar, 330 o c
s = 6.518+
6.5518−6.518
2
x(36 − 35.3) = 6.52983 kJ/kg
Enthalpy in Mw
H = h xm x10-3
=3029.056x161.11x10-3 = 488.011Mw
Entropy in Mw
S = sxmx10-3
=6.52983x161.11x10-3 = 1.05202 Mw/K
6.7 REHEATER INLET
PRESSURE P=35.30 bar, TEMPERATURE T = 630 K, =330 o c,
mass flow rate m= 161.11
𝑘𝑔
𝑠
For above values from steam tables from steam tables from steam tables
Enthalpy h = 3029.056 kJ/kg
Entropy s = 6.52983 kJ/kg
Enthalpy in Mw H = hxmx10-3
=3029.056x161.11x10-3 = 488.011Mw
Entropy
S = sxmx10-3
=6.52983x161.11x10-3 = 1.05202 Mw/K
6.8 REHEATER OUTLET
PRESSURE P=34.32 bar, TEMPERATURE T=813 K = 540 oc ,
mass flow rate m= 161.11 kg/s
6. Exergy Analysis of Thermal Power Plant
Department of Mechanical Engg, SVIST, Kadapa. Page 61
From steam tables, enthalpy values
Pressure 500 oc 600 oc
34 bar 3451.7 3677.7
36 bar 3449.5 3676.1
At 34 bar, 540 o c
Enthalpy h = 3451.7 +
36677 .7−3451 .7
100
(600-540) = 3587.3 kJ/kg
At 36 bar, 540 o c
h = 3449.5 +
3679.6.1−3449.5
100
x 60 = 3585.46 kJ/kg
At 34.32bar, 540 o c
h = 3585.4+
3676.1−3585 .4
2.0
(36-34.32) = 3587.0056 kJ/kg
Entropy values
Pressure 540 o c 550 o c
34 bar 7.712 7.447
36 bar 7.144 7.420
At 34bar, 540 o c
Entropy s =7.712+
7.447−7.172
100
x 100 =7.337kJ/kg-K
At 36 bar, 540 o c
s =7.144+
7.420 −7.144
100
∗ 100 = 7.3096 kJ/kg - K
At 34.32 bar, 540 o c
s = 7.3096 +
7.337 −7.3096
2.0
(36-34.32) = 7.3326 kJ/kg – K
Enthalpy in Mw H = hxmx10-3
= 3587.0056-161.11x10-3 = 577.902 Mw
Entropy in Mw S = sxmx10-3
= 7.3326x16111x10-3
= 1.1813 Mw/K
6.9 Intermediate Pressure Turbine [IPT] INLET
PRESSURE P=34.32 bar, TEMPERATURE T=813 K = 540 oc,
7. Exergy Analysis of Thermal Power Plant
Department of Mechanical Engg, SVIST, Kadapa. Page 62
mass flow rate m= 161.11 kg/s
For above values
h = 3587.0056 kJ/kg
s = 7.3326 kJ/kg - k
Enhalpy in MW
H = hxmx10-3
= 3587.0056-161.11x10-3= 577.902 Mw
Entropy in Mw
S = sxmx10-3
= 7.3326x16111x10-3 = 1.1813 Mw/K
6.10 Intermediate Pressure Turbine [IPT] TURBINE OUTLET
PRESSURE P = 6.87 bar, TEMPERATURE T 613 k, mass flow rate m=152.78 kg/s.
From the steam tables, enthalpy values
Pressure 300 o c 350 o c
6bar 3062.3 3166.2
7bar 3059.8 3164.3
At 6bar, 340 o c
Enthalpy h = 3062.3+
3166 .2−3062.3
50
x10
= 3083.08 kJ/kg
At 7 bar, 340 o c
h = 3059.8+
3164.3−3059.8
50
x10
= 3080.7 kJ/kg
At 6.87 bar, 340 o c
h =3080.7+
3083.08−3080.7
1.0
∗ (7 − 6.87)
= 3081.0094 kJ/kg
8. Exergy Analysis of Thermal Power Plant
Department of Mechanical Engg, SVIST, Kadapa. Page 63
Entropy values
Pressure 300 o c 350 o c
6 bar 7.374 7.548
7 bar 7.300 7.475
At 6 bar, 340 o c
Entropy s = 7.374+
7.548−7.374
50
x10
= 7.48088 KJ/Kg – k
At 7 bar, 340 o c
Entropy s =7.300+
7.475−7.300
500
x 10
= 7.335 kJ/kg-K
At 6.87 bar, 340 o c
s = 7.335+
7.4088−7.335
1.0
x0.13= 7.344594 kJ/kg - K
Enthalpy in Mw
H = hxmx10-3
= 3081.0094x152.8x10-3 = 470.716 Mw
Entropy in Mw
S = sxmx10-3
= 7.3445x152.78x10-3 = 1.12209 Mw/K
6.11 Low Pressure Turbine [
LPT] INLET
PRESSURE P = 6.87 bar, TEMPERATURE T 613 K, m=152.78 kg/s
For above values from steam tables from steam tables from steam tables
Enthalpy h = 3081.0094 kJ/kg
Entropy s = 7.344594 kJ/kg - K
Enthalpy in Mw
H = hxmx10-3
= 3081.0094x152.8x10-3 = 470.716 Mw
Entropy in Mw S = sxmx10-3
= 7.3445x152.78x10-3 = 1.12209 Mw/K
9. Exergy Analysis of Thermal Power Plant
Department of Mechanical Engg, SVIST, Kadapa. Page 64
6.12 Low Pressure Turbine [LPT] OUTLET
PRESSURE P = 0.0873 bar, TEMPERATURE T 322 K = 49 o c, mass flow rate m=125 kg/s
From steam tables, enthalpy values
0.085 bar 2579.2
0.09 2581.1
At 0.0873 bar,
Enthalpy h = 2580.226 kJ/kg
Entropy s = 8.1988 kJ/kg-k
Enthalpy in Mw
H = hxmx10-3
=2580.226x125x10-3 = 2580.226 Mw
Entropy in Mw
S = sxmx10-3
= 8.1988x125x10-3 =1.024 Mw/K
6.13 CONDENSER INLET
P=0.0892 bar, T= 320 K = 47 o c, M=144.44kg/s
For above values from steam tables from steam tables from steam tables
Enthalpy h = 2579.504 kJ/kg
Entropy s = 8.192 kJ/kg-K
Enthalpy H = hxmx10-3
= 2579.504x144.44 x10-3 = 372.48 Mw
Entropy
S = s x m x10-3
=8.192x144.44x 10-3 = 1.1813 Mw/K
6.14 CONDENSER OUTLET
PRESSURE P= 0.0873 bar, T=316K = 43 o c, M=144.44kg/s
For above values from steam tables from steam tables from steam tables
Enthalpy h = hf - Cpw ( Tsat- T)
= 181.184-4.187(43.28-43) = 179.9948 kJ/kg
10. Exergy Analysis of Thermal Power Plant
Department of Mechanical Engg, SVIST, Kadapa. Page 65
Entropy s = sf - Cpw ln (
Tsat
𝑇
)
= 0.61556-4.187 ln (
43.28
43
) = 0.61179 kJ/kg K
Enthalpy in Mw
H = hxmx10-3
= 179.99x144x.44x10-3 = 25.998Mw
Entropy S = sxmx10-3
= 0.61556x144.44x10-3 = 0.0883Mw/K
6.15 Condensate extraction pump [CEP] INLET
PRESSURE P= 0.0873 bar, T=316K = 43 o c, M=144.44kg/s
For above values from steam tables from steam tables from steam tables
Enthalpy h = hf - Cpw ( Tsa t- T)
= 181.184-4.187(43.28-43)
= 179.9948 kJ/kg
Entropy s = sf - Cpw ln (
Tsat
𝑇
)
= 0.61556-4.187 ln (
43.28
43
)
= 0.61179 kJ/kg K
Enthalpy in Mw H = hxmx10-3
= 179.99x144x.44x10-3= 25.998Mw
Entropy inMw
S = sxmx10-3
= 0.61556x144.44x10-3= 0.0883 Mw/K
6.16 Condensate extraction pump [CEP] OUTLET
PRESSURE P= 18.63 bar, Temperature T= 320 K = 47 o c, mass flow rate m=144.44kg/s
At 18.63bar,
Specific Volume v=0.00117148 m3/kg
H = 25.99844+ (vdPx102+Cpwdt) x 10-3 ×m
Pressure difference
dP = 18.63-0.0873 = 18.5427 bar
11. Exergy Analysis of Thermal Power Plant
Department of Mechanical Engg, SVIST, Kadapa. Page 66
Temperature difference
dT =320-316 K = 4 K
Enthalpy in Mw
H = 25.99844+ (vdPx102+Cpwdt)x10-3 × m
H = 25.99844+ (0.00117148x28.7312x102+4.187x4) ×144.4= 28.7312 Mw
Entropy in Mw
S = 0.0883 + (
mCpwdT∗10−3
Tprev
)
= 0.0883 + (
144 .44∗4.187 ∗4∗10−3
316
) = 0.096022 Mw/K
6.17 Ejector Inlet
PRESSURE P= 18.63 bar, Temperature T= 320 K = 47 o c, mass flow rate m=144.44 kg/s
At 18.63bar,
Specific Volume v=0.00117148 m3/kg
Enthalpy in Mw
H = 25.99844+ (vdPx102+Cpwdt)10-3 ×m
Pressure difference
dP = 18.63-0.0873= 18.5427.
Temperature difference
dT =320-316 k= 4 K
Enthalpy in Mw H = 25.99844+ (vdPx102+Cpwdt) x10-3x m
H = 25.99844+ (0.00117148x28.7312x102+4.187x4)×144.4 = 28.7312 Mw
Entropy in Mw
S = 0.0883 + (
mCpwdT∗10−3
Tprev
)
S = 0.0883+ (
144 .44∗4.187∗4∗10−3
316
) = 0.096022 Mw/K
6.18 Ejector Outlet
PRESSURE P= 17.65 bar, Temperature T=325K=52o c, m=144.44kg/s
At 17.65 bar
Specific Volume v=0.0011674 m3/kg
Pressure difference
dP = 17.65-18.63 = -0.98 bar
12. Exergy Analysis of Thermal Power Plant
Department of Mechanical Engg, SVIST, Kadapa. Page 67
Temperature difference
dT = 325-320 = 5K
Enthalpy in Mw
H = 28.7312 + (vdPx102+CpwdT)10-3xm
= 28.7312 + (0.0011674x102+4.187x5)10-3x144.44 =31.7385 MW
Entropy in Mw
S = 0.096022 + (
mCpwdT∗10−3
Tprev
)
S = 0.096022 + (
144 .44∗4.187∗5∗10−3
320
) = 0.10547 Mw/K
6.19 Gland Steam Condenser [GSC] Inlet
PRESSURE P= 17.65 bar, T=325K=52 o c, m=144.44kg/s
At 17.65 bar
v=0.0011674 m3/kg
dP = 17.65-18.63 = -0.98 bar
dT = 325-320 = 5K
H = 28.7312 + (vdPx102+CpwdT)10-3x m
= 28.7312 + (0.0011674x102+4.187x5)10-3x144.44 =31.7385 Mw
S = 0.096022 + (
mCpwdT∗10−3
Tprev
)
S = 0.096022 + (
144 .44∗4.187∗5∗10−3
320
) = 0.10547 Mw/K
6.20 Gland Steam Condenser [GSC] outlet
PRESSURE P = 16.67bar, T=327 k = 54 c m=144.4kg/s
At 16.67 bar
Specific Volume
v= 0.00116232 m3/kg
Pressure difference
dP = 16.67-17.65 = -0.98 bar
Temperature difference
dT = 327-325 o c = 2K
H = 31.7385+ (vdPx102+Cpwdt) x10-3 ×m
H = 31.7385+ (0.00116232x10.98x102+4.187x2)144.4x10-3 = 32.9315Mw.
13. Exergy Analysis of Thermal Power Plant
Department of Mechanical Engg, SVIST, Kadapa. Page 68
S = 0.0883+ (
mCpwdT∗10−3
Tprev
)
S = 0.10547+ (
144 .44∗4.187∗2∗10−3
325
) = 0.10919 Mw/K
6.21 Low Pressure Heater 1 [LPH] INLET
PRESSURE P=15.69bar, T=327K = 54 o c, m=144.44kg/s
For above values
Specific volume v =0.001157m3/kg
Pressure difference
dP = 15.69-16.67 =-0.98bar
Temperature difference
dT = 327-327 = 0 K
Enthalpy in Mw
H = 32.9315 + (vdPx102+Cpwdt) x10-3 ×m
H=32.9315 + (0.001157x-0.98x102+4.187x0) 144.44x10-3 =32.91511Mw
Entropy in Mw
S = 01.0919 + (
mCpwdT∗10−3
Tprev
)
S = 0.10919 + (
144 .44∗4.187∗0∗10−3
327
) = 0.10919 Mw/K
6.22 Low Pressure Heater [LPH1] OUTLET
PRESSURE P=14.71 bar, Temperature T=346K, mass flow rate m=144.44kg/s
For above values from steam tables
Specific volume v =0.00115274 m3/kg
Pressure difference
dP =14.71-15.69 =-0.98bar
Temperature difference
dT = 346-327 =19K
Enthalpy in Mw
H =32.91511+ (vdPx102+Cpwdt) x10-3 ×m
H= 32.91511+ (0.001152x-0.98x102+4.187x19) x144.44x10-3 = 44.3894Mw
Entropy in Mw
S = 0.10919+ (
mCpwdT∗10−3
Tprev
)
14. Exergy Analysis of Thermal Power Plant
Department of Mechanical Engg, SVIST, Kadapa. Page 69
S = 0.10919+ (
144 .44∗.187∗19∗10−3
327
) =0.14432 Mw/K
6.23 Low Pressure Heater [LPH] 2 INLET
PRESSURE P=14.71 bar, Temperature T=346K, mass flow rate m=144.44kg/s
For above values from steam tables
Specific volume v =0.00115274 m3/kg
Pressure difference
dP =14.71-15.69 =-0.98bar
Temperature difference
dT = 346-327 =19K
Enthalpy in Mw
H= 32.91511+(0.001152x-0.98x102+4.187x19)x144.44x10-3
= 44.3894Mw
Entropy in Mw
S = 0.10919+ (
144 .44∗.187∗19∗10−3
327
)=0.14432 Mw/K
6.24 Low Pressure Heater [LPH] 2 OUT LET
PRESSURE P=13.73bar, Temperature T= 368 K, mass flow rate m=144.44 kg/s
For above values from steam tables
Specific volume
V = 0.11476m m3/kg
Pressure difference
dP = 13.73-14.71 = -0.98 bar
Temperature difference
dT = 368-346 = 22K
Enthalpy in Mw
H = 44.3894+ (vdPx102+Cpwdt) x10-3 ×m
H = 44.3894+ (0.0011476x102-0.98+4.187x22) x10-3x144.44 = 57.678Mw
Entropy in Mw
S = 0.14432+ (
mCpwdT∗10−3
Tprev
)
S = 0.14432 + (
144 .44∗4.187∗22∗10−3
346
) = 0.1827 Mw/K
15. Exergy Analysis of Thermal Power Plant
Department of Mechanical Engg, SVIST, Kadapa. Page 70
6.25 Low Pressure Heater [LPH3] INLET
PRESSURE P = 14.71 bar, Temperature T = 346K, mass flow rate m = 144.44kg/s
For above values from steam tables
Specific Volume
v=0.0011524 m3/kg
Pressure difference
dP = 14.71-15.69 = -0.98bar,
Temperature difference
dT = 368-346 =22K
H = 44.3894+ (vdPx102+Cpwdt) x10-3 ×m
H = 44.3894 + (0.00114796x102x-0.98+4.187x22)10-3x144.44 = 57.678Mw
S = 0.14432+ (
mCpwdT∗10−3
Tprev
)
S = 0.14432 +
144 .44∗4.187∗22∗10−3
346
= 0.1827 Mw/K
6.26 Low Pressure Heater [LPH3] OUTLET
PRESSURE P = 6.37 bar, Temperature T = 398 K =125 o c, mass flow rate m =144.44kg/s
For above values
Specific Volume
v = 0.0011023m3/kg
Pressure difference
dP = 6.37-13.73 =-7.36bar
Temperature difference
dT =398-368 = 20 K
H = 57.678+ (vdPx102+Cpwdt) x10-3 ×m
H = 57.678 + (0.0011023x-7.36x102+4.187x20)10-3x144.44 = 69.656Mw
S = 0.1827+ (
mCpwdT∗10−3
Tprev
)
S = 0.1827 + (
144 .44∗4.187∗20∗10 −3
368
) = 0.21556 Mw/K
6.27 DEAERATOR INLET
PRESSURE P = 6.37 bar, Temperature T = 398 K =125 o c, mass flow rate m =177.78 kg/s
For above values from steam tables
16. Exergy Analysis of Thermal Power Plant
Department of Mechanical Engg, SVIST, Kadapa. Page 71
Specific Volume
v = 0.0011023 m3/kg
Enthalpy
h =676.75 kJ/kg
Entropy
s =1.9458 kJ/kg – K
Enthalpy in Mw
H = hxmfx10-3
= 676.65x177.78x10-3 = 120.31Mw
Entropy in Mw
S = sxmx10-3
= 1.9458x177.78x10-3 = 0.3459 Mw/K
6.28 DEAERATOR OUTLET
PRESSURE P = 6.28bar, T = 433 K, m= 177.78 kg/s
For above values from steam tables
Specific Volume
v = 0.0011032 m3/kg
Pressure difference
dP = 6.28-6.27 =-0.09bar
Temperature difference
dT =433-396 = 37K
H =120.31+ (vdPx102+Cpwdt) x10-3×m
H = 120.31+ (0.0011032x-0.09-100+4.182x37)10-3x177.78 = 147.849 Mw
S = 0.3459+(
mCpwdT∗10−3
Tprev
)
S = 0.3459+ (
177.78∗4.187∗37∗10 −3
396
) =0.4154 Mw/K
6.29 BOOSTER PUMP INLET
PRESSURE P = 6.28bar, Temperature T = 433 K, m mass flow rate = 177.78 kg/s
For above values from steam tables
Specific Volume v = 0.0011032 m3/kg
dP = 6.28-6.27 =-0.09bar
17. Exergy Analysis of Thermal Power Plant
Department of Mechanical Engg, SVIST, Kadapa. Page 72
dT =433-396 = 37K
Enthalpy H = 120.31+ (0.0011032x-0.09-100+4.182x37)10-3x177.78= 147.849 Mw
Entropy S = 0.3459+ (
177.78∗4.187∗37 ∗10−3
396
)=0.4154 Mw/K
6.30 BOOSTER PUMP OUTLET
PRESSURE P = 14.71 bar, T = 440 K, m= 177.78 kg/s
For above values from steam tables
Specific Volume
v =0.0011527 m3/kg
Pressure difference
dP = 14.1-6.28 = 8.34 bar
Temperature difference
dT = 440-433 = 7K
H = 147.89+ (vdPx102+Cpwdt)x10-3×m
H = 147.89 + (0.0011527x8.34x102+4.187x7)177.78x10-3= 153.72 Mw
S = 0.415449+ (
mCpwdT∗10−3
Tprev
)
S = 0.415449+ (
177 .78∗4.187∗7∗10−3
433
) = 0.42748 Mw/K
6.31 Boiler Feed Pump [BFP] INLET
PRESSURE P = 14.71 bar, T = 440K, m= 177.78 kg/s
For above values from steam tables
Specific Volume v =0.0011527 m3/kg
Pressure difference
dP = 14.1-6.28 = 8.34 bar
Temperature difference
dT = 440-433 = 7 K
H = 147.89+ (vdPx102+Cpwdt) x10-3×m
H = 147.89 + (0.0011527x8.34x102+4.187x7)177.78x10-3 = 153.72Mw
S = 0.415449+(
mCpwdT∗10−3
Tprev
)
S = 0.415449+ (
177 .78∗4.187∗7∗10−3
433
) = 0.42748 Mw/K
6.32 Boiler Feed Pump [BFP] OUTLET
18. Exergy Analysis of Thermal Power Plant
Department of Mechanical Engg, SVIST, Kadapa. Page 73
PRESSURE P = 176.52 bar, Temperature T = 443K, mass flow rate m=177.78 kg/s
For above values from steam tables
Specific Volume v = 0.001828 m3/kg
Pressure difference
dP = 176.52-14.71= 161.81 bar
Temperature difference
dT = 443-440 = 3K
H =153.72+ (vdPx102+Cpwdt)x10-3×m
H = 153.72 + (0.001828x161.81x102×4.187x3)177.78x10-3 =161.2237Mw
S = 0.42748+ (
mCpwdT∗10−3
Tprev
)
S = 0.42748 + (
177 .78∗4.187∗10−3
440
) = 0.43255 Mw/K
6.33 HPH 5 INLET
PRESSURE P = 175.54 bar, Temperature T = 443 o c, mass flow rate m = 177.78 kg/s
For above values from steam tables
Specific Volume
v =0.001836 m3/kg
Pressure difference
dP = 175.54-176.52 = -0.98 bar
Temperature difference
dT = 443-443 = 0K
H =161.2237+ (vdPx102+Cpwdt) x10-3×m
H = 161.2237+ (0.001836x-0.98x102+4.187x0)177.78x 10-3 = 161.188 Mw
S = 0.43255+ (
mCpwdT∗10−3
Tprev
)
S = 0.43255 + (
4.187 ∗177 .78∗0∗10−3
443
) = 0.43255 Mw/K
6.34 HPH5 OUT LET
PRESSURE P = 173.58 bar, T Temperature =480K, mass flow rate m = 177.78kg/s
At above values from steam tables
Specific Volume
v = 0.001779 m3/kg
19. Exergy Analysis of Thermal Power Plant
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Pressure difference
dP = 173.58-175.54 =-1.96 bar
Temperature difference
dT = 480-443 =37 K
H = 161.188+ (vdPx102+Cpwdt) x10-3 ×m
H = 161.188 + (0.001779x-1.96x102+4.187x37)177.78x10-3 = 188.674 Mw
S = 0.43255+ (
mCpwdT∗10−3
Tprev
)
S = 0.43255 +(
177.78∗4.187 ∗37∗10−3
443
) = 0.49472 Mw/K
6.35 HPH6 INLET
PRESSURE P = 171.62 bar, Temperature T =480K, mass flow rate m = 177.78kg/s
At above values from steam tables,
Specific Volume v = 0.001792m3/kg
Pressure difference
dP = 171.62-173.58 =-1.96 bar
Temperature difference
dT = 480-480 = 0K
H = 188.66+ (vdPx102+Cpwdt) x10-3 ×m
H = 188.66 + (0.001792x-1.96x102+4.187x37)177.78x10-3 = 188.6624 Mw
S = 0.49472+ (
mCpwdT∗10−3
Tprev
)
S = 0.49472 +(
177.78∗4.187 ∗0∗10−3
480
) = 0.49472 Mw/K
6.36 HPH6 OUTLET
PRESSURE P = 171.62 bar, Temperature T=518K, mass flow rate m=177.78 kg/s
At above values from steam tables
Specific volume v = 0.001792m3/kg
Pressure difference
dP = 171.62-171.62 =0 bar
Temperature difference
dT = 518-480 = 38 K
H = 188.6624+ (vdPx102+Cpwdt) x10-3 ×m
20. Exergy Analysis of Thermal Power Plant
Department of Mechanical Engg, SVIST, Kadapa. Page 75
H = 188.6624 + (0.001792 x 0 x102+40187x38)177.78x10-3 = 216.948 Mw
S = 0.4942+ (
mCpwdT∗10−3
Tprev
)
S = 0.4942+ (
177.78∗4.187∗38∗10 −3
480
) = 0.55364 Mw/K
6.37 FRS INLET
PRESSURE P = 171.62 bar, T Temperature = 518K, mass flow rate m=177.78 kg/s
At above values from steam tables
Specific volume
v = 0.001792m3/kg
Pressure difference
dP = 171.62-171.62 =0 bar
Temperature difference
dT = 518-480 = 38 K
H = 188.6624+ (vdPx102+Cpwdt)x10-3 ×m
H = 188.6624 + (0.001792 x 0 x102+40187x38)177.78x10-3 = 216.948 Mw
S = 0.4942+ (
mCpwdT∗10−3
Tprev
)
S = 0.4942+ (
177.78∗4.187∗38∗10 −3
480
) = 0.55364 Mw/K
6.38 FRS OUTLET
PRESSURE P = 163.77 bar, T=518K, m=175 kg/s
At above values from steam tables
Specific Volume v = 0.001717m3/kg
Pressure difference
dP = 163.77-171.62 =-7.85 bar
Temperature difference
dT = 518-518 = 0 K
H = 216.948+ (vdPx102+Cpwdt) x10-3 ×m
H = 216.948+ (0.001717x-7.87x102+4.187x0) x175x10-3 =216.712 Mw
S = 0.55364+ (
mCpwdT∗10−3
Tprev
)
S = 0.55364+ (
175 ∗4.187 ∗0∗10−3
518
) = 0.55364 Mw/K
21. Exergy Analysis of Thermal Power Plant
Department of Mechanical Engg, SVIST, Kadapa. Page 76
6.39 ECONOMISER INLET
PRESSURE P = 163.77 bar, T, Temperature =518K, m mass flow rate =175 kg/s
At above values from steam tables
Specific volume, v = 0.001717m3/kg
Pressure difference
dP = 163.77-171.62 =-7.85 bar
Temperature difference
dT = 518-518 = 0 K
H = 216.948+ (vdP102+Cpwdt) x10-3×m
H = 216.948+ (0.001717x-7.87x102+4.187x0) x175x10-3 =216.712 Mw
S = 0.55364+ (
mCpwdT∗10−3
Tprev
)
S = 0.55364+ (
175 ∗4.187 ∗0∗10−3
518
) = 0.55364 Mw/K
6.40 ECONOMISER OUTLET
PRESSURE P = 157.89 bar, Temperature T=574K, mass flow rate m=175 kg/s
At above values from steam tables,
Specific volume v = 0.001694 m3/kg
Pressure difference dP = 157.89-163.77 = -5.88 bar
Pressure Temperature dT = 574-518= 56 K
H = 216.948+ (vdPx102+Cpwdt) x10-3 ×m
H = 216.948+ (0.001694x-5.88x102+4.187x56) x175x10-3 =257.57 Mw
S = 0.55364+ (
mCpwdT∗10−3
Tprev
)
S = 0.55364+ (
175 ∗4.187 ∗56 ∗10−3
518
) = 0.6328 Mw/K
23. Exergy Analysis of Thermal Power Plant
Department of Mechanical Engg, SVIST, Kadapa. Page 78
38. FRS Outlet 163.77 518 175 216.712 0.55364
39. Economiser Inlet 163.77 518 175 216.712 0.55364
40. Economiser Outlet 157.89 574 175 257.57 0.6328
24. Exergy Analysis of Thermal Power Plant
Department of Mechanical Engg, SVIST, Kadapa. Page 79
6.2 Thermodynamic properties of steam at extractions
Extraction
The principle of regeneration can be practically utilized by extracting steam from turbine at
several locations and supply it to the regenerative heater. The most advantageous condensate
heating temperature is selected depending on the throttle conditions and this determines the
number of heaters to be used. Figure shows the layout of condensing steam power plant in which
a surface condenser is used to condense all the steam that is not extracted for feed water heating.
The turbine is double extracting and boiler is equipped with a super heater.
Extraction 1 at Low Pressure Turbine
Pressure P=0.216 bar, Temperature T=346 k=73oc, m=5.56 kg/s
At above values from steam tables
h = hg =2612.42 kJ/kg
Enthalpy at extraction1
H = h×m×10-3
= 2612.42×5.56×10-3 =14.5250 MW
s = sg =7.8824 kJ/kg-K
Entropy at extraction1
S = s×m×10-3
= 7.8824×5.56×10-3 =0.0438 MW/K
Extraction 2 at Low Pressure Turbine
Pressure P = 0.858 bar, Temperature T = 380 K, m= 6.94 kg/s
At above values,
h =18.515 kJ
s =7.411 kJ/kg-K
Enthalpy at extraction2 ,
H = h×m× 10-3
= 2668×6.94×10-3= 18.515 MW
Entropy at extraction2,
S = s×m×10-3
= 7.4118×6.94×10-3 =0.0514 MW/K
25. Exergy Analysis of Thermal Power Plant
Department of Mechanical Engg, SVIST, Kadapa. Page 80
Extraction 3 at Low Pressure Turbine
Pressure P = 2.37 bar, Temperature T =473K , m=6.94 kg/s
At above values,
h =2713.17 kJ/kg
Enthalpy at extraction3
H = h×m×10-3
= 2713.17×6.94×10-3= 18.8293 MW
s = 7.0702 kJ/kg-K
Entropy at extraction3
S = s×m×10-3
S = 7.072×6.94×10-3= 0.0490 MW/K
Extraction 4 at Intermediate Pressure Turbine
Pressure P = 6.87 bar, Temperature T=613 K, m = 8.33 kg/s
At above values,
h = 2761.58 kJ/kg
Enthalpy at exraction4
H = h×m×10-3
= 2761.58×8.33×10-3= 23.0039 MW
Entropy at extraction4
S = s×m×10-3
= 6.7115×8.33×10-3 = 0.0559 MW/K
Extraction 5 at High Pressure Turbine
Pressure P= 16.7 bar, Temperature T= 706k, m = 8.33 kg/s
At above values,
h = 2792.6 kJ/kg
Enthalpy at extraction5
H = h×m×10-3
=2793.28×8.33×10-3=23.26MW
s = 6.4026 kJ/kg-K
Entropy at extraction5
S = s×m×10-3 = 6.4026×8.33×10-3 = 0.0533 MW/K
26. Exergy Analysis of Thermal Power Plant
Department of Mechanical Engg, SVIST, Kadapa. Page 81
Extraction 6 at High Pressure Turbine
Pressure P= 39.23 bar, Temperature T= 616 K = 343 C, m= 16.67 kg/s
For above values,
h =2800.685 kJ/kg
Enthalpy at extraction6
H = h×m×10-3
= 2800.685×16.67×10-3 =46.687 MW
s =6.0767 kJ/kg-K
Entropy at extraction6
S = s×m×10-3
= 6.0767×16.67×10-3 = 0.1013 MW/K
6.2.1 TABULATED VALUES OF THERMODYNAMIC EXTRACTIONS
Pressure
(bar)
Temperature
(K)
Mass
(Kg/S)
Enthalpy
(h)
KJ/Kg
Enthalpy
(H)
MW
Entropy
(s)
KJ/Kg-k
Entropy
(S)
Mw/k
Extraction
1(LPT)
0.216 346 5.56 2612.42 14.5250 7.8824 0.0438
Extraction
2(LPT)
0.858 380 6.94 2668 18.515 7.4118 0.0514
Extraction
3(LPT)
2.37 473 6.94 2713.17 18.8293 7.0702 0.0490
Extraction
4(IPT)
6.87 613 8.33 2761.58 23.0039 6.7115 0.0559
Extraction
5 (HPT)
16.70 706 8.33 2793.28 23.26 6.4026 0.0533
Extraction
6 (HPT)
39.23 616 16.67 2800.68 46.687 6.0767 0.10129
27. Exergy Analysis of Thermal Power Plant
Department of Mechanical Engg, SVIST, Kadapa. Page 82
6.3 TABULATED VALUES OF TURBINES
6.3.1 HIGH PRESSURETURBINE (HPT)
HPT h (kJ/kg) v (m3
/kg) m (kg/s)
Inlet 3476.26 0.023959 177.78
Extraction 6 2880.68 0.0507564 161.11
Outlet 3029.056 0.071115 161.11
Work done by HPT = ( h in – h out ) × mass
= (3476.26-3029.056)×177.78 = 79.503 MW
6.3.2 INTERMEDIATE PRESSURETURBINE (IPT)
IPT h (kJ/kg) v (m3
/kg) m(kg/s)
Inlet 3587.0056 0.1097 161.11
Extraction 4 2793.28 0.118692 152.78
Outlet 3081.0094 16.68792 152.78
Work done by IPT = ( hin – hout ) × mass
Work done by the IPT = (3587.00-3081.00)×161.11
= 81.51 MW
6.3.3 LOW PRESSURETURBINE (LPT)
LPT h (kJ/kg) v (m3
/kg) m (kg/s)
Inlet 3081.0094 3865.93 144.44
Extraction 3 2713.17 0.75549 137.50
Extraction 2 2668 1.95474 130.56
Extraction 1 2612.42 7.11948 125
Outlet 2580.26 16.68792 125
Work done by LPT = ( hin – hout ) × mass
= (3081-2580.26)×144.4
28. Exergy Analysis of Thermal Power Plant
Department of Mechanical Engg, SVIST, Kadapa. Page 83
= 72.328 MW
CHAPTER-7
7.0 EXERGY AND ENERGYANALYSIS ON THE COMPONENTS
7.1 EXERGY ANALYSIS:
7.1.1 Exergybalance of High pressure turbine:
The exergy balance for the high pressure turbine is given by :
10→HPT inlet
11→HPT ext
14→HPT outlet
Ψ = H-To.S
Work done by high pressure turbine,
W HPT = ˙ m10(Ψ10 − Ψ11) + (˙ m10 −˙ m11) (Ψ11 − Ψ14) – To× ˙Sgen
This gives:
T0 ×˙Sgen = ˙ m10(Ψ10 − Ψ11) + (˙ m10 −˙ m11) ( Ψ11 − Ψ14) –WHPT
˙ m10(Ψ10 − Ψ11) + (˙ m10 −˙ m11) ( Ψ11 − Ψ14) = Exergy input
And the entropy generation rate is:
˙Sgen = ˙ m10(s11 − s10) + (˙ m10 −˙ m11)(s14 − s11)
Irreversibility destroyed = exergy loss is:
˙I destroyed = To. ˙Sgen = To[˙ m10(s11 − s10) + (˙ m10 −˙ m11)(s14 − s11)]
The second law efficiency is:
ήII,HPT = 1 − (˙Idestroyed ÷ ˙m10(Ψ10 −Ψ11) + (˙ m10 −˙ m11)( Ψ11 −Ψ14))
= WHpT ÷ ˙ m10(Ψ10 −Ψ11) + (˙ m10 −˙ m11)(Ψ11 − Ψ14)
=1-(Exergy loss/Exergy input)
29. Exergy Analysis of Thermal Power Plant
Department of Mechanical Engg, SVIST, Kadapa. Page 84
= WHPT/Exergy input
Data:
Exergy of high pressure turbine inlet
Ψ10 = 618.009 − (301×1.1665) = 266.89 kW
Exergy of high pressure turbine extraction
Ψ11 = 23.26 − (301×0.0533) = 7.2167 MW
Exergy of high pressure turbine outlet
Ψ14 = 488.011 − (301×1.0520) = 171.359 MW
To = 301K
By substituting in the above equations we get:
˙Sgen = 177.78 (6.56179−6.4026) + (177.78 − 8.33) (6.529 − 6.4026)
=28.3 + 21.418 = 49.718 kW/K
To ˙Sgen = 301×49.718 = 14965.26 kW =14.96 MW
Work done by high pressure turbine
W HPT = 177(266.89 −7.2167) + (177.78 − 8.33) (171.359 − 7.216) – 14965.26
= 73978.749 −14965.26 = 59013.489 kW
Second law efficiency of high pressure turbine,
ήII,HPT = 59013.489 ÷ 73978.74 = 0.7977×100 = 79.77 %
7.1.2 Exergybalance of Intermediate pressure turbine:
The exergy balance for the Intermediate pressure turbine is given by:
15→IPT inlet
19→IPT outlet
23→IPT ext (deaerator)
W IPT = ˙ m15 (Ψ15 −Ψ19) +˙ m23 ( Ψ19 −Ψ23) – To ×˙Sgen
This gives:
30. Exergy Analysis of Thermal Power Plant
Department of Mechanical Engg, SVIST, Kadapa. Page 85
To× ˙Sgen = .m15 (Ψ15 −Ψ19) +˙ m23( Ψ19 −Ψ23) –WIPT
.m15 (Ψ 15 −Ψ 19) +˙ m23 ( Ψ 19 −Ψ 23) = Exergy input
and the entropy generation rate is:
˙Sgen = ˙ m23(s23 − s19) + ˙ m15(s19 − s15)
Irreversibility = exergy loss is:
˙I destroyed = To .˙Sgen
= To[˙m23(s23 − s19) + ˙ m15(s19 − s15)]
The second law efficiency is:
ήII,IPT = 1 – (˙Idestroyed ÷ ˙ m15 (Ψ15 −Ψ19) +˙ m 23( Ψ 19 –Ψ 23))
= W IPT ÷ ˙ m 15 (Ψ 15 –Ψ 19) +˙ m 23( Ψ 19 –Ψ 23)
= 1-(Exergy loss/Exergy input)
= W IPT /Exerg input
Data:
Exergy of intermediate pressure turbine inlet
Ψ 15=577.902−(301×1.1813)=222.33 MW
Exergy of intermediate pressure turbine outlet,
Ψ 19 =470.716−(301×1.122)=132.994 MW
Exergy of intermediate pressure turbine extraction,
Ψ 23 =23.2039−(301×0.0559)=6.378 MW
Entropy generated, ˙Sgen=8.33(7.34−6.71)+161.11(7.34−7.3326)
=7.189 kW/K
To ˙Sgen=301×7.189=2163.95 Kw =2.16 mW
Work done by intermediate pressureturbine,
W IPT =161.11(222.33−132.994)+8.33(132.994−6.378)−2163.95
31. Exergy Analysis of Thermal Power Plant
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=15448.278−2163.95=13284.328 kW
Second law efficiency of intermediate pressure turbine,
ήII,IPT=13284.328÷15448.278 =0.8599×100 =85.99%
7.1.3 Exergybalance of Low pressure turbine:
The exergy balance for the Low pressure turbine is given by :
1→LPTinlet
2→LPToutlet
3→LPText3
4→LPT ext2
5→LPText1
W LPT = .m 1 (Ψ1−Ψ3)+(.m1−.m3) ( Ψ3−Ψ4)+(.m1−.m3−.m4 )+(.m1−.m3−.m4−.m5)
( Ψ5−Ψ2) − To×˙Sgen
This gives:
To×˙Sgen = .m1(Ψ1−Ψ3)+(.m1−.m3) ( Ψ3−Ψ4)+(.m1−.m3−.m4)+(.m1−.m3−.m4−.m5)
( Ψ5−Ψ2) −W LPT
and the entropy generation rate is
˙Sgen = .m1(s2−s1)+.m3(s3−s2)+.m4(s4−s2)+.m5(s5−s2)
Irreversibility destroyed = exergy loss is:
˙ I destroyed = To ˙Sgen
= To(.m 1(s2−s1)+.m3(s3−s2)+.m4(s4−s2)+.m5(s5−s2))
The second law efficiency is:
ήII,LPT = 1 –( ˙Idestroyed ÷ .m1(Ψ1−Ψ3)+(.m1−.m3)(Ψ3−Ψ4)
+(.m1−.m3−.m4)+(.m1−.m3−.m4−.m5) ( Ψ5−Ψ2)
32. Exergy Analysis of Thermal Power Plant
Department of Mechanical Engg, SVIST, Kadapa. Page 87
= W LPT ÷ .m1(Ψ1−Ψ3)+(.m1−.m3)(Ψ3−Ψ4)
+(.m1−.m3−.m4)+(.m1−.m3−.m4−.m5) ( Ψ5−Ψ2)
= 1-(Exergy loss/Exergy input)
= W LPT /Exergy input
Data:
Mass of low pressure turbine inlet,
.m1=144.4 kg/s
Exergy of low pressure turbine inlet,
Ψ1→445.02−(301×1.0608)=125.71 MW
Mass of low pressure turbine outlet,
.m2=125 kg/s
Exergy of low pressure turbine outlet,
Ψ2→322.53−(301×1.024)=14.306 MW
Mass of low pressure turbine extraction3,
.m3=6.49 kg/s
Exergy of low pressure turbine extraction3,
Ψ3→18.829−(301×0.049)=4.08 MW
Mass of low pressure turbine extraction2,
.m4=6.94 kg/s
Exergy of low pressure turbine extraction2,
Ψ4→18.515−(301×0.0513)=3.073 MW
Mass of low pressure turbine extraction1,
.m5=5.56 kg/s
Exergy of low pressure turbine extraction1,
Ψ5→14.525−(301×0.0438)=1.3412 MW
33. Exergy Analysis of Thermal Power Plant
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Entropy generated,
˙Sgen=144.4(7.68−7.45)+6.94(7.68−7.421)+6.94(7.68−7.65)+5.56(7.88−7.68)
=31.4 kW/K
To ˙Sgen=301×31.4=9451.4 kW
Work done by low pressure turbine,
WLPT =144.4(125.7−14.3)+6.94(14.3−4.85)+6.94(14.3−4.12)+5.56(14.3−2.4)
=16288.526−9451.4=6837.126 kW
Efficiency of low pressure turbine,
ήII,LPT=6837.126÷16288.52 =0.4197×100 =41.97%
7.1.4 Exergybalance of condenser:
The exergy balance for the condenser is given by :
Ψ1→Condenserinlet
Ψ30→CEP inlet/condenser outlet
Ψw=˙ m30(Ψ30 – Ψ1) − ∑n
k=1 (1 – (To ÷Tk) )Qk − To× ˙Sgen
0= ˙ m30(Ψ30 – Ψ1) − ∑ n
k=1 (1 – (To ÷Tk) )Qk − To ×˙Sgen
This gives:
To ×˙Sgen = ˙ m30(Ψ30 –Ψ1) − ∑n
k=1 (1 – (To÷ Tk))Qk
˙ m30(Ψ30 –Ψ1) =Exergy input
Irreversibility destroyed= exergy loss is:
˙Idestroyed=To ˙Sgen = [{˙ m30(h30 – h1)} − To{˙ m30(s30 – s1)}] − ∑n
k=1 (1 –
(To÷Tk))Qk
The second law efficiency is:
ήII,Condenser = 1 – (˙Idestroyed ÷ ˙ m30(Ψ30 – Ψ1))
34. Exergy Analysis of Thermal Power Plant
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= 1-(Exergy loss/Exergy input)
Data:
Exergy of condenserinlet,
Ψ1 =25.998−(301×0.0883)=−0.5803 MW
Exergy of condenseroutlet,
Ψ30 =372.48−(301×1.183)=16.397 MW
To ˙Sgen= 4.05×1000 kW
Ψw=144.4(16.397+0.5803)−(6.5051×1000)−4.05×1000=0
Second law efficiency,
ήII,Condenser= 1− (4.05×103 ÷2.45×103 ) = 0.6530×100 =65.30%
7.1.5 Exergybalance of super heater:
m.
g(Ψ gi−Ψ go)+m.
s(Ψ si−Ψ so)−E.
Qsh= I.
SH
m.
g(Ψ gi−Ψ go)+m.
s(Ψ si−Ψ so)−(1−To÷Tk)Qk=I.
SH
m.
g(Ψ gi−Ψ go) = Exergy input
Irreversibility of superheater = exergy loss
I.
SH=243.05(958.95−68.47)+177.78(266.89−18.02)−(111.18×1000)
=216430.92+15265.96−(111.18×1000) =120516.88 kW =120.516 MW
To.s.
gen=m.
s(hsi−hso)−To(m.
s(ssi−sso))+∑(1−(To÷Tk))Qk
=177.78(2626.2−3476.2)−301(177.78(5.34−6.56))+111180
=−151123.66−301(−216.89)+111180=25340.71 kW =25.34 MW
Exergy input= 243.05.(958.95-68.47)=243.05.890.47=216430.92 kW
ή II=1−(Exergy loss ÷ Exergy input)
35. Exergy Analysis of Thermal Power Plant
Department of Mechanical Engg, SVIST, Kadapa. Page 90
= 1−(120516.881 ÷ 216430.92) =0.443×100 =44.3%
7.1.6 Exergy analysis of boiler:
m.
g(Ψgi−Ψgo)+m.
b(Ψbi−Ψbo)−E.
b= I.
destroyed
m.
g(Ψgi−Ψgo)+m.
b(Ψbi−Ψbo)−(1−(To ÷ Tk))Qk= I.
destroyed
m.
g(Ψgi−Ψgo) = Exergy input =243.05*890.47=216430.92 kW
I.
destroyed = 243.05(958.95−68.47)+175(178.17−63.717)−163770
=216430.92+20029.27−163770
=72690.19 kW =72.69 MW= exergy loss
To
.s.
gen = mw (hbi−hbo) −To(mw(sbi−sbo))+∑(1−(To ÷ Tk))Qk
=175(1345.4−301 (175(−5.3+3.2))+(16.377×1000)
=−224140+110617.5+163770
=50247.5 kW =50.24 MW
ή II = 1−(Exergy loss ÷ Exergyinput)
= 1−(72690.19 ÷ 216430.92)
=0.6641×100 =66.41%
7.2 ENERGY ANALYSIS
7.2.1 Energybalance of High PressureTurbine:
The energy balance for the high pressure turbine is given by :
10→HPT inlet
11→HPT ext
14→HPT outlet
WHPT = ˙ m 10(h10 – h11) + (˙ m10 −˙ m11)(h11 – h14) − Energy loss
This gives :
36. Exergy Analysis of Thermal Power Plant
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Energy loss = ˙ m10(h10 – h11) + (˙ m10 −˙ m11)(h11 – h14) –WHPT
˙ m10(h10 – h11) + (˙ m10 −˙ m11)(h11 – h14) = Energy input
Energy input= 177.78(3476.26−2983.41)+(177.78−161.11)(2983.41−2793.28)
=90788.33 kW
WHPT = 90788.33− Energy loss
79.505×1000 = 90788.33 –Energy loss
Energy loss = 11283.33 kW
The first law efficiency of high pressure turbine is:
ηI,HPT = 1 – (Energy loss ÷ ( ˙ m10(h10 − h11) + (˙ m10 −˙ m11)(h11 − h14))
= = WHPT ÷ (˙ m10(h10 − h11) + (˙ m10 −˙ m11)(h11 − h14))
=1-(Energy loss/Energy input)
=WHPT/Energy input
= 1−(11283.33 ÷ 90788.33)
= 0.875×100 =87.5%
7.2.2 Energybalance of Intermediate Pressure Turbine:
The energy balance for the intermediate pressure turbine is given by :
15→IPT inlet
19→IPT outlet
23→IPT ext (deaerator)
WIPT = m.
15(h15−h19)+m.
23(h19−h23) − Energy loss
This gives :
Energy loss = m.
15(h15−h19)+m.
23(h19−h23) –WIPT
Energy input = m.
15(h15−h19)+m.
23(h19−h23) =84182.42 kW
WIPT = 161.11(3587.00−3081) + 8.3(3081−2761.58) – Energy loss
37. Exergy Analysis of Thermal Power Plant
Department of Mechanical Engg, SVIST, Kadapa. Page 92
(81.51×1000) = 84182.42−Energyloss
Energy loss = 2672.42 kW
The first law efficiency of intermediate pressure turbine is :
ηI,IPT = WIPT ÷ m.
15(h15−h19)+m.
23(h19−h23)
= WIPT /Energy input
= (81.51×103) ÷ 84182.42 =0.968×100 =96.8%
7.2.3 Energybalance of Low Pressure Turbine:
The energy balance for the low pressure turbine is given by:
1→LPT inlet
2→LPT outlet
3→LPT ext3
4→LPText2
5→LPT ext1
WLPT = m.
1 (h1−h2) +m.
3 (h2−h3) +m.
4 (h2−h4)+m.
5(h2−h5)− Energy loss
This gives:
Energy loss = m.
1(h1−h2)+m.
3(h2−h3)+m.
4(h2−h4)+m.
5(h2−h5)−WLPT
Energy input = m.
1(h1−h2)+m.
3(h2−h3)+m.
4(h2−h4)+m.
5(h2−h5) =74014.62 kW
WLPT =144.4(3081−2580.26)+6.94(2580.26−2713.17)+6.94(−2580.26+2668)+5.56(2612−2580.26)
−Energy loss
72.32×10
3
= 74014.62 − Energy loss
Energy loss = 1694.62 kW
The first law efficiency is :
ηI,LPT = 1− ( Energy loss ÷ m.
1(h1−h2)+m.
3(h2−h3)+m.
4(h2−h4)+m.
5(h2−h5) )
38. Exergy Analysis of Thermal Power Plant
Department of Mechanical Engg, SVIST, Kadapa. Page 93
= 1-(Energy loss/ Energy input )
=1− (1694.62 ÷ 74014.62)=0.977×100 =97.7%
7.2.4 Energybalance of condenser:
The energy balance for the condenseris given by :
2→condenserinlet
3→CEP inlet /condenser outlet
0 = ˙m30(h30 – h1) – Qk − Energy loss
This gives :
Energy loss = ˙m30(h30 – h1) − Qk
= 144.4(2579.5−179.99)−(307.74×103)
= 38845.22 kW
Energy input = ˙m30(h30 – h1) =144.4(2579.5-179.99)= 346585.22 kW
The first law efficiency is :
ηI,Condenser = 1 – (Energy loss ÷ ˙m30(h30 − h1))
= 1-(Energy loss/Energy input)
=1−(38845.22 ÷ 346585.22)=0.8879×100 =88.79%
7.2.5 Energybalance of super heater:
The energy balance for the super heater are :
Wsup = Qk − m.
g (hgi−hgo) − m.
s (hsi−hso) – Energy loss
O = Qk − m.
g (hgi−hgo) − m.
s (hsi−hso) – Energy loss
Energy loss = Qk − m.
g (hgi−hgo) − m.
s (hsi−hso)
Energy loss=(148.908×103)−243.05(1111.969−509.789) −(177.78(2626.2−3476.26))
=148908−292189.849−(−151123.66)
=148908−292189.849+151123.66
39. Exergy Analysis of Thermal Power Plant
Department of Mechanical Engg, SVIST, Kadapa. Page 94
=300031.66−292189.8 =7841.86 kW=7.841 MW
Energy input = 148908 kW
First law efficiency of condenser
ηI= 1− (Energy loss ÷ Energy input )
= 1− (7841.86 ÷ 148908)
= 0.947×100 =94.7%
7.2.6 Energy balance of boiler:
The energy balance for the combustion/ boiler is given by :
0 = Qk −˙mw(h10 – h9) −˙ms(h15 – h14)] − Energy loss
where mw is the mass flow rate of water, ms is the mass flow rate of steam
combustion which gives:
Energy loss = Qk −˙mw(h10 – h9) −˙ms(h15 – h14)
Energy loss = (203.04×10
3
)−177.78(3476.26−2626.2)−161.11(587.7−3029.056)
=241014.017−(203.04×10
3
) =37974.017 kW
Energy input = 203.04×1000 kW
The first law efficiency of boiler is defined as
ηI,Boiler =( Energy output ÷ Energy input)
= 1 – (Energy loss ÷ Energy input)
= 1−(37974.017÷(203.04×103))
=0.842×100 =84.2%
40. Exergy Analysis of Thermal Power Plant
Department of Mechanical Engg, SVIST, Kadapa. Page 95
7.3 TABLES OF THE EXERGY, ENERGY EFFICIENCIES AND
LOSSES
7.3.1 First law and Second law Efficiency table:
Component
name
First law
efficiency(ηI)
Second law efficiency(ήII)
HPT 87.5 % 79.77%
IPT 96.8% 85.99%
LPT 97.4% 41.97%
Superheater 94.79% 44.3%
Condenser 88.79% 65.3%
Boiler 84.85% 66.41%
7.3.2 Energyand Exergylosses table:
Component name Energy loss(mW) Exergy loss(mW)
HPT 11.283 14.96
IPT 2.16 2.67
LPT 1.69 9.45
SUPERHEATER 7.84 25.34
CONDENSER 38.84 4.050
BOILER 37.97 72.6
41. Exergy Analysis of Thermal Power Plant
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42. Exergy Analysis of Thermal Power Plant
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CHAPTER-8
8.0 COMPARISON GRAPHS BETWEEN EXERGY AND ENERGY
8.1 EXERGY DESTRUCTION GRAPH
X-axis –components, Y-axis- Exergy loss in %
8.2 TURBINE EXERGYEFFICIENCYAND DESTRUCTIONGRAPH
X-axis- Turbines, Y-axis-Exergy loss in %
72.6
25.34
14.96
2.16
9.45
4.05
0
10
20
30
40
50
60
70
80
Exergy destruction
Series 1
Series 2
0
10
20
30
40
50
60
70
80
90
100
HP turbine IP turbine LP TURBINE
Exergy
efficiency
Turbine Exergy Efficiency &
distruction
Turbine Exergy
43. Exergy Analysis of Thermal Power Plant
Department of Mechanical Engg, SVIST, Kadapa. Page 98
8.3 EXERGY Vs ENERGYEFFICIENCYGRAPH
X-axis- Exergyefficiency of components, Y-axis- Energyefficiency of
components
8.4 COMPARISON CHARTS
1
X-axis-Energy loss of components, Y-axis-Exergy loss of components
0
10
20
30
40
50
60
70
80
90
100
Exergy
Exergy
Exergy v/s energy efiiciency
Series 1
Series 2
Comparison charts
44. Exergy Analysis of Thermal Power Plant
Department of Mechanical Engg, SVIST, Kadapa. Page 99
CHAPTER-9
9.0 CONCLUSION
Now-a-days there are few methods to measure the performance of a
power plant. Some researchers use the conservation of mass and conservation of energy
principles (first law of thermodynamics), however the evaluation is not actually complete. The
exergy analysis based on the second law of thermodynamics should be included in order to do a
complete analysis, which can also access accurately the utilization of energy. This method
provides the information which are useful for engineers or managers to know about the power
plant performance. The information obtained on the result of the analysis will form a basis for
the energy manager or operation engineer to make decisions on how he should operate the plant
in order to save cost and energy usage.
This project has presented the results of an exergy analysis performed
on 210 MW power plant. The analysis was applied on the unit with running load of 210 MW.
Exergy destruction on the plant components are also presented and energy losses are discussed.
The results of the exergy indicate that boiler produces highest exergy destruction of 72 MW.
Comparing the 3 turbine stages, the results of the analysis indicate that HPT produces highest
exergy destruction than IPT and LPT.
The exergy destroyed in the turbines, super-heaters are small compare
to exergy destroyed in the boiler. It is apparent from the analysis,72% of the total exergy
destruction occurs in the boiler. This large exergy loss is mainly due to the combustion reaction
and to the large temperature difference during heat transfer between the combustion gas and
steam. The factors that contribute to high amount of irreversibilities are tubes fouling, defective
burners, fuel quality, inefficient soot blowers, valves steam traps and air heaters fouling.
Inspections of this equipment need to be carried out during the boiler outage. Other factors like
heat loss, incomplete combustion and exhaust losses. This study pinpoints that boiler requires
necessary modification to reduce its exergy destruction, there by performance can be improved.
The exergy losses in the turbines are due to the frictional effects and
pressure drops across the turbine blades as well as the pressure and heat losses to the
45. Exergy Analysis of Thermal Power Plant
Department of Mechanical Engg, SVIST, Kadapa. Page 100
surroundings. The HPT, IPT, LPT constitutes a combined 28% of the total exergy destruction
which indicates a need for reducing its irreversibilities. Other factors that may contribute to the
irreversibilities are most likely due to the throttling losses at the turbine governor valves, silica
deposited at the nozzles and blades. Amongst the three turbines, HPT produces the highest
exergy destruction. Overhauling of the turbine may be needed to check the real causes for
improving the plant performance. All this information complemented by the engineers intuition
and judgement, can assist in the improvement of efficiency and the reduction in generation cost.
9.1 RECOMMENDATIONS FOR FURTHER STUDIES
An exergetic- economic analysis of the plant and of different potential options available
for the plant improvement.
The measuring devices are necessary to measure the different values and to proceed
further analysis.
46. Exergy Analysis of Thermal Power Plant
Department of Mechanical Engg, SVIST, Kadapa. Page 101
CHAPTER-10
10.0 BIBILOGRAPHY
Cengel Y.A. Boles M.A.,”Thermodynamics :An Engineering Approach ,”2nd
edition,Mc.Graw Hill,1994.
Kotas T.J., “ The exergy method of Thermal plant Analysis, ” Krieger Publishing
Co.,1995.
S.C.Kaushik, V.Siva Reddy, S.K.Tyagi,”Renewable and Sustainable Energy Reviews”15
(2011)”
T.Ganapathy, N.Alayamurthy, R.P.Gowkhar and K.Murugesan “ Journal of engg science
and Tech Review 2 (2009)”
Mali Sanjay.D, Dr.Mehta NS, “International journal of advanced engg research and
studies E-ISSN2249-8974”
I.Satyanarayana,A.V.S.S.K.S.Gupta and Dr.k.Govinda Rajulu, “International Journal of
Engineering (IJE)”
A.Hepbasli, “Renewable and sustainable energy reviews 12(2008).”
Vundala Siva Reddy, Shubash Chandra Kaushik, Sudhir kumar Tyagi, Naraya Lal
Panwar, “Smart grid and renewable energy,2010.”
Ravi Prakash Kurkiya, Sharad Chaudhary, “International journal of scientific and energy
research volume 3,2012.”
Vosough Amir, “2nd international conference on Mecanical,2012.”
P.K.Nag, “Engineering Thermodynamics, 4th edition, Mc.Graw Hill, 1995.”
Sam Cooper, Energy and exergy analysis. People.bath.ac.uk/en8c.
A.GALOVIC, M.ZIVIC, M.kokanovic, “ Analysis of exergy destruction of condenser-
1987”
A.Rashad, A.El Maihy, “13th International conference on Aerospaces and Aviation
Technology.”
www.bhel.com (bhel maintenance manuals)
www.suzlon.com and www.scribd.com
47. Exergy Analysis of Thermal Power Plant
Department of Mechanical Engg, SVIST, Kadapa. Page 102
www.plantmaintenance.com and www.apgenco.gov.in/