Steam & Gas Turbines and Power Plant Engineering [7 ed.] 8185444358, 9788185444352

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Steam & Gas Turbines And Power Plant Engineering (7th Revised and Enlarged Edition, SI Units) (For B.E./8.Tech., I.E.5. and 1.A.5. Examinations)

·R. YADAV· Ph.D., F.I.E., M.I.S.T.E.

Emeritus Professor of Mechanical Engineering, M.N.N.I.T. Allahabad-211 004 Former Prof. & Head, Mech. Engg. Deptt., M.N.R.E.C. (now M.N.N.I.T.), Allahabad Former Principal, R.I.T. (now N.I.T.) Jamshedpur

AND SANJAY B.E., M.E.,

Asstt. Prof. of Mechanical Engg., N.I.T., Jamshedpur-14 AND

RAJAY · B. Tech. Senior Engineer, B.H.E.L., Hardwar

Central Publishing House 18-C, Sarojini Naidu Marg, Allahabad

© Copyright 2004 Publisher & Author

First Edition, 1972 Sixtli Revised (SI) Edition, 1988 Seventh Revised and Enlarged (SI) Edition, 2004

Printed by

Halcyon Press, Allahabad

Published by

Central Publishing House, Allahabad

ACKNOWLEDGEMENTS 1. Sincere acknowledgements and thanks are due to All India Council for Technical Education (AICTE), New Delhi for providing financial help and opportunity by giving Emeritus Fellowship to revise and enlarge the book from "Steam and Ga■ Turbine■" to "Steam & Ga■ Turbine■ And Power Plant Engineering". 2. Sincere acknowledgements and thanks are due to Pequot Publishing Inc. USA for giving the permission to copy the data and informations from the Journal "Ga■ Turbine World".

Pasir Gudang 400-MW combined cycle plant

Preface to the First Edition This book is intended to serve the Steam and Gas Turbines courses followed in Engineering Colleges. As there is no book on this subject in M.K.S. system, the present volume is expected to fill this gap. I have attempted in this book to present the subject matter in a simple, lucid and precise manner. The students difficulties in solving problems have been kept in view and hints have been given where they are likely to go wrong. A novel feature of this book is that at appropriate places a large number of problems of both Indian and foreign Universities of viva-voce, theoretical and numerical questions with answers have been added. I hope that this will add to the usefulness of the book for the students whose difficulties have been my foremost thought. I also take the opportunity to express my gratitude to Dr. Jagdish Lal, Principal, M.N.R. Engg. College, Alld, Dr. V.P. Vasandani, Head Mechanical Engg. Deptt., Punjab Engg. College, Chandigarh, Dr. B.K. Gupta, Head, Mechanical Engg. Deptt., M.N.R. ·ngg. College, Allahabad, Shree P.N. Maskara, Reader in Mechanical Engg. Deptt., M.N.R. Engg. College, Allahabad for their encouragement. The author will be happy to receive suggestions for the improvement of the work and for any acts of errors and omissions tha may have cropped up in the book .

15 August, 1972 Allahabad

R. Yadav

Preface to the 7 th SI, Revised and Enlarged Edition !he present form of the book is the outgrowth of thorough revision and enlargement of 6th Edition of Steam and Gas Turbines which served the students and teaching communities satisfactorily for about 30 years. Syllabus of most of the Universities have been revised and as a result I got a large number of suggestions to include Power Plant Engineering. The revised book contains 30 Chapters covered in about 1100 pages in royal size. It gives the latest informations in Power Plant Engineering based on Research Papers and Companies Journals. The main emphasis ih the teaching of Power Plant Engineering is on the working principle of main and auxiliary systems, basic design principles, operation and maintenance. Emphasis on these aspects have been given adequately. A large number of typical problems have been solved. The diagrammatic representations are. unique describing systems and processes. A new concept of case study has been included in the text. Questions based on computer software have been included in the excercise portion of the book. The depth, richness and beauty of this book are such that it may serve many courses such as Steam & Gas Turbines, Power Plant Engineering, Energy Conversion, and Turbomachines. Whi 1 e revising and enlarging the scope of the' book, the authors have consulted a large number of books, research papers and journals which are listed in the References. Sincere acknowledge is made to all authors listed in the References. The authors will appreciate if any mistakes and errors found in the tex(are brought to their notice. Finally, the authors wish to thank Meera Yadav, wife of the main author and mother of last two co-authors who has been a consistent source of encouragement and pi I lar of strength for the authors and endured certain hard­ ships due to preoccupation of this venture.

Lord Krishna Janmashtamj, 20 August, 2003 Allahabad

R. Yadav Sanjay Rajay.

Contents

1.

2.

Power Plant Types and Economics

1--55

Rankine and Binary-Vapor Cycles

56-82

A Brief History of Turbines, Turbomachines, Turbines, Pumps, Compresson, Blowers and Fans, A Case Study of Turbines, Types of Power Plant, Hydel Power Plant, Steam Engine Power Plant, Diesel, Petrol ·and Gas Engine Power Plants, Steam ·Turbine ·Power Plant, Gas Turbine Power Plant, Com­ bined Gas and Steam Turbines· Power Plant and Co-generation, Ran�ne Organic Power Plant, Binary-Vapor Cycle Power Plant, Free-Piston Gas Gen­ erator Turbine Plant, Nuclear Power Plant, Solar Energy Power Plant, Central Receiver Thermal Power Plant, Distributed Collector Solar Thermal Power Pl�nt, Tidal Wave Power Plant, Geo-thermal Power Plant, Vapor­ Dominated System, Liquid Dominated Hot-water ·system, Wind Energy Power Plant, Biomass Energy Power Plant, Ocean Thermal Energy Power Plant (O'I'EC), Wave Power Plant, Magneto-Hydro-Dynamic ,Power Plant (M.H.D.), Solar Cells Power Plant, Electric Storage Batteries Power Plant, Fuel Cells Power Plant, Thermoelectric Power Plant, Thermionic Power Plant, Electro-Gas-Dynamic Power Plant (E.G.D.), Hydrogen Fuel Combus­ tion Turbines Based Power Plant (Indirect Type), Load-Duration Curves· and Definitions, Load Curve, Connected Load, Demand Factor, Average Load, Load Factor, Capacity Factor or Plant Factor, Reserve Factor, Diversity Factor, Plant Use Factor, Construction of Load-Duration Curves, Effect of Variable Load on Power Plant Design and Operation, Location of Power _ Plant, Power Plant Economics, Construction Cost (Cc), Fixed Costs (FC), In­ terest Rates (I), Depreciation Rates, Fuel Cost (Cr), Present-Worth (PW) Concept, Plant Net Heat Rate (PNHR), Incremental Heat Rate OHR) and Plant Net Efficiency (11,J/ant), Economic Scheduling Principle For Load Di'stribution, Effect of Load Factor on Cost per kWh, Tariff for Electrical Energy, Indian Energy Scenario, Steam and Gas Turbine Design Procedure Simple Rankine Cycle, Working Principle, Thermodynamic Analysis, Limita­ tions of Carnot Cycle As Vapor Cycle, "Principles of Increasing the Thermal Efficiency, Effect of Operating Conditions on Efficiency, Performance Curves, Methods of Increasing the Thermal Efficiency, Deviation of Actual Cycle from Theoretical Cycle. (Internally Irreversible cycle), Externally Irreversible Rankine Cycle, Efficiencies, Requirements of an Ideal Working Fluid, Critical Temperature and Pressure of Water, Binary-Vapor Cycle, Advantages of Mer­ cury, Disadvantages of Mercury

Regcncrativa Feed Heating, Reheating, Reheating-Regenerative 83-179 and Water-extraction Cycles Basic Principle of Regenerative Feed Heating, Most Ideal Regenerative Feed Heating Cycle, Regenerative Feed Heating Cycles and their Representation of Ideal Process on T-s and h-s Diagrams, Representation of Actual process On T-s and h-s Diagrams in Regenerative Cycle, Other Types of Regenerative Feed Heating Arrangements, Surface Heaters with One Drain Pump, Plain Surface Heaters with One Drain Pump and One Drain Cooler, Surface Heaters With Drain Fed to Hot-Well, Optimum Feed Water Temperature and Saving in Heat Rate, Gain in Thermal Efficiency Due to Regenerative Feed Heating, Advantages and Disadvantages of Regenerative Feed Heating Cycle over Simple Rankine Cycle, Feed Heaters, Direct Contact (DC) Heaters, Surface Heaters, Deaerators, Effect of Flow of Wet Steam in Nozzles and Blades, Velocity Diagram for Dry Steam and Water Particles, Correction to Condition Curve for Wetness, Erosion and Corrosion of Blades, Prevention of Erosion and Corrosion, Reheating of Steam, Practical Reheating and Non-reheating Cycles, Advantages of Reheating, Disadvantages, Reheat-Regenerative Feed Heating Cycle, Ideal Reheat-Regenerative Feed Heating Cycle, Actual Reheat-Regenerative Feed Heating Cycle, Regenerative Water-Extraction Cycle, Practical Feed Heating Arrangements, Feed Heating System for a 120 MW Unit, Feed Heating System for a 200 MW Unit, Feed Heating System for a 350 MW Unit, Feed Heating System for a 500 MW Unit, Feed Heating System. for a 660 MW Unit, Advantages and Disadvantages of Turbine Driven Boiler Feed Pump 180-209 Back-Pressure, Pass-out and Mixed Pressure Turbine Cycles Back-Pressure Turbines, Pass-out Turbine, Process of Pass-out Turbine with Single Extraction, Partial Extraction, Throttle Governing, Full Extraction Nozzle Control Governing, No Extraction, Essence of Low-Pressure Turbines, Working of Mixed-Pressure Turbine, Heat Accumulator Fuels, Combustion, Combustion Equipment and Fuel Handling 210-276 Systems Types of Fuels for Steam Generators, Coal, Fuel oil, Natural and Petroleum Gas, Coal-oil and Coal•Water Mixture, Synthetic Fuels, Coal Gasification, Underground Coal Gasification, Coal Liquefaction, Industrial Wastes and Byproducts, Biomass, Combustion Reactions, Stoichiometric Air-Fuel Ratio, Combustion Equation, Dew Point Temperature of Exhaust Gases in Boiler, Heating Value of a Fuel (Coal), Optimum Excess Air in Boiler, Mass Balance of a Boiler Furnace of Steam Generator, Energy Balance in a Steam Generator, Draught System of a Boiler, Natural Draught, Mechanical Draught, Forced Draught, Induced Draught, Balance Draught, Pressure and Flow Measurement in Gas Flow Path, FD, ID and other Fans, Heat of Combustion, Open System, Closed System; Heating Values, Theoretical (or Adiabatic) Flame Temperature, Free Energy of Formation, Types of Boiler Furnace, Kinetics of Combustion Reactions, Mechanism of Solid Fuel Combustion, Combustion Equipments for Coal, Fuel Bed Combustion Furnace, Overfeed Fuel Bed, Underfeed Fuel Bed, Mechanical Stokers for Fuel Bed Combustion, Travelling Grate Stoker, Chain Grate Stoker, Spreader Stoker, Vibrating Grate Stoker, Underfeed Stoker, Pulverized Coal Firing System, Advantages and Disadvantages of Pulverized Coal Firing System, Crushers, Pulverizers,

Ball-Tube Mill, Ball and Race Mill, Hammer Beaters, Performance of Pulverizers, The Pulverized Coal System, Bin or Storage System, Direct Firing System, Semi-Direct Firing System, Design Aspect of Pulverized Coal Fired Furnace. Dry-bottom Furnace, Wet-Bottom Furnace, Pulverized Coal Burners Circular and Slot Burners, Straight-Flow Burners, Turbulent or Vortex Burners, Distributed Mixing Burner (DMB), Multifuel Burners, Burner Arrangement, Cyclone Furnace, Fluidized Bed Combustion, Regimes of Fluidization, Terminal Velocity of Particles, Fluidized Bed Combustion of Solid Fuels, Circulating Fluidized Bed (CFB) Combustion System, Pressurized Fluidized Bed Combustion (PRBC), Coal Gasifier, Fuel Oil Combustion System, Mechamism of Fuel Oil Combustion, Oil Burners, Gas Burner, Combined Gas-Fuel Oil Burners 6. Steam Generators, Ash Handling Systems And Feed Water Treatment 277-357 Classification of Steam Generator, Construction and Working of Fire-tube Boilers, Horizontal-Return Tubular Boiler, Oil-fired Package (Nestler) Boiler, Construction and Working of Water-tube Boilers, Babcock and Wilcox Water Tube Boiler, Stirling Bent Water Tube Boiler, Heat Absorption in Water-Tube Boilers, Circulation in Downcomer-Riser Circuit and their Sizing, Sizing of Downcomers, Sizing of Risers, Steam Drum and its Internals, Modern and High Pressure Water Tube Boilers, Radiant Type Power Boiler With Natural Circulation, Boiler Water Wall, LaMont Boiler, Benson Boiler, Loeffler Boiler, Velox Boiler, Once-through or Monotube Boiler (Critical Pressure Boiler), Fluidized Bed Boilers, Bubbling Fluidized Bed Boiler (BFB), Cyclone Separators, Non-mechanical Valves, Pressurized Fluidized Bed Combustion (PFBC) Boilers, Pressurized Bubbling Fluidized Bed Combustion (PBFBC) Boiler, Pressurized Circulating Fluidized Bed Combustion (PCFBC) Boiler, Advantages and Disadvantages of Fluidized Bed Combustion (FBC) Boilers, Advantages of Pressurized Bubbling Fluidized Bed Combustion (PBFBC) Boiler, Advantages of Pressurized Circulating Fluidized Bed Combustion (PCFBC) Boilers, Boiler Mountings and Accessories, Economisers, Superheaters and Reheaters, Air Preheater, Steam Generator Control, Feed Water and Drum-Level Control, Steam Pressure Control, Steam Temperature Control, Ash Collection and Its Disposal, Electrostatic Precipitator, Fabric Filters and Baghouses, Ash Handling System, ?oiler Makeup and Feed Water Treatment, Pretreatment, Demineralisation, Condensate Polishing, Deaeration, Internal Treatment, Boiler Blowdown, Evaporators 7.

Nozzles and Diffusers 358-413 Definitions and Applications, Types of Nozzles, Types of Diffusers, Equation of Continuity, Sonic Velocity, Mach Number and Stagnation Properties, The Steady Flow Energy Equation in Nozzles, The Momentum Equation for the Flow through Steam Nozzle, Nozzle Efficiency, The Effect of Friction on the Velocity of Steam Leaving the Nozzle, Adiabatic Flow Through Diffusers, Diffuser Efficiency, Total Pressure Loss Coefficient and Pressure Recovery Coefficient, Diffuser Efficiency, Total Pressure Loss Coefficient (), Pressure Recovery Coefficient (Ce), Shape of Nozzle for Uniform Pressure Drop, Mass of Discharge through Nozzle, Throat Pressure for Maximum Discharge or Existence of a Critical Pressure in Nozzle flow or Choked Flow, Physical Explanation of Critical Pressure, Critical Pressure Ratio for Adiabatic and

iv Frictionless Expansion of Steam from a Given Initial Velocity, Critical Properties-Choking in Isentropic Flow Based on Stagnation Values, Critical Pressure, Critical Velocity, Critical Temperature, General Relationship between Area, Velocity and Pressure in Nozzle Flow, Effect of Friction on Exponent of Expansion and Critical Pressure Ratio, Critical Pressure Ratio in a Frictionally Resisted Expansion from a Given Initial Velocity, Design of Nozzles, Design of Diffusers, Supersaturated Flow in. Nozzles, Effect of Variation of Back Pressure, Parameters Affecting the Performance of Nozzle, Experimental Methods to Determine Velocity Coefficient, Experimental Results 414-429 8. Steam Turbines Types Principle of Operation of Steam Turbine, Comparison of Steam Engines and Turbines, Classification of Steam Turbine, The Simple Impulse Turbine, Compounding of Impulse Turbine, Pressure Compounded Impulse Turbine, Simple Velocity-Compounded Impulse Turbine, Pressure-Velocity-Compounded Impulse Turbine, Impulse-Reaction Turbine, Combination Turbines, Overview of Turbine, Difference between Impulse and Reaction Turbines 430-474 9. Flow of Steam Through Impulse Turbine Blades \.„ Velocity Diagrams for Impulse Turbines, Combination of Vector Diagram, The Effect of Blade Friction on Velocity Diagram, Forces on the Blades and Workdone by Blades, Force, Work and Power, Blade or Diagram Efficiency, Axial Thrust or End Thrust on the Rotor, Gross Stage Efficiency, Energy Converted to Heat by Blade Friction, Influence of Ratio of Blade Velocity to Steam Irelocity on Blade Efficiency in a Single Stage Impulse Turbine, Efficiency of Multi-Stage Impulse Turbine with Single-row Wheel, Velocity Diag-am for Three-Row Velocity-Compounded Wheel, Most Economical Ratio of Blade Velocity to Steam Velocity for a Two-row Velocity Compounded Impulse Wheel, Impulse Blade Sections, Choice of Blade Angles , Inlet Blade Angle Outlet Blade Angle, (132),Blade Heights in Velocity-Compounded and Pressure Compounded Impulse Turbine, Advantages of Velocity Compounded Impulse Turbine, Disadvantage of Velocity- Compounded Impulse Turbine, Twisted Blading, Breadth of Blading 0. Flow of Steam Through Impulse-Reaction Turbine Blades 475-515 Velocity Diagram And Work done, Degree of Reaction, Impulse-reaction Turbine with Similar Blade Section and Half-degree Reaction (Parsons Turbine), Degree of Reaction, Gross Stage Efficiency and Optimum Value of p, Blade Efficiency, Comparsion of Enthalpy Drops in Various Stage Impulse-reaction Turbines and Impulse Turbines, Height of Impulse-Reaction Turbine Blading, Parallel Exhaust and Casing : General, First Stage Nozzles and Blades, Last Stage Blade Height, Parallel Exhaust : Number of Last Stages, Casing Arrangement, Effect of Working Steam on the Stage Efficiency of Parsons Turbine, Operation of Impulse Blading with Varying Heat Drop or Variable Speed, Impulse-Reaction Turbine Blade Section 11. Energy Losses in Steam Turbines 516.534 V List of Energy Losses, Regulating Valve Losses, Losses in Nozzles, Moving Blade Losses, Losses Due to Trailing *Edge Wake, Impingement Losses, Losses Due to Leakage of Steam Through the Annular Space Between the Stator and Shrouding, Blade Frictional Losses, Losses Due to the Turning of the Steam Jet in the Blades, Losses Due to Shrouding, Disc Friction Losses,

V

Blade Windage Losses or Partial Admission Losses, Clearance Losses, Impulse Turbines, Leakage Losses in Impulse- Reaction Turbines, Losses Due to Wetness of Steam, Carry Over Losses, Exhaust Piping Losses or Leaving Velocity Losses, Radiation and Conduction Losses, Mechanical Losses, Losses Due to Steam Leakage Through the End Seals 12. State Point Locus, Reheat Factor and Design Procedure 535-559 Stage Efficiency of Impulse Turbines, State Point Locus of an Impulse Turbine, State point Locus for Multi-stage Steam Turbine, Reheat Factor, Internal and Other Efficiencies, Increase in Isentropic Heat Drop in a Stage Due to Friction in Preceeding Stage, Correction For Terminal Velocity, Reheat Factor for an Expansion with a Uniform Adiabatic Index and a Constant Stage Efficiency, Correction of Reheat Factor for Finite Number of Stages, Design Procedure of Impulse Turbines, Design Procedure for impulse-Reaction Turbines, Calculation of Axial Thrust \Al Governing and Performance of Steam Turbines 560-591 Need of Governing,,Hydromechanical Speed Governing Loop, Speed Governors, Relays, Throttle Governing Employing Mechanical Speed Governor, Analysis of Impulse Turbine, Throttle Governing Employing Electrical Speed Governor, Throttle Governing in Reaction Turbine, Nozzle Control Governing, Condition Curve with Nozzle Control Governing, Comparison of Throttle and Nozzle Control Governing, By Pass Governing, By Pass Governing of Reaction Turbines, Speeder Gear, Anticipatory Gear, Governing of Reheat Turbines, Direct Digital Control (DDC), Governing Characteristics, Hydraulic System and Fluids 14. Steam Turbine Auxiliary Systems 592-611 Turbine Protective Devices, Possible Hazards, Emergency Steam Stop Valve Operating Gear, Tripping Devices, Overspeed Trip, Emergency Hand Trip, Remote Trip Gear, Oil Failure Trips, Thrust-bearing Wear Trip, Low Vacuum Trip, Low Boiler Pressure Trip, Unloading Gears, Low Vacuum Unloading Gear, Low Steam Pressure Unloading Gear, Lubricating System, Oil Pumps, Oil Relief Valve, Oil Coolers, Tanks, Strainers and Filters, Lubricating Oil, Generator Hydrogen S .als, Turbine Greasing, Glands and Sealing Systems, Essence of Glands and Sealing System, Labyrinth Gland and its Forms, Requirements of Gland , Gland Sealing System, Carbon Ring Gland, Water-Sealed Glands, I ange Heating Systems Construction, Stress Analysis, Operation and Maintenance of Steam Turbines 612-648 Construction of Nozzles and Diaphragm, Design Requirements of Nozzles, Construction of Convergent Nozzles, First Stage Convergent Nozzle, Built-up Nozzle, Diaphragm Nozzles, Construction of Convergent-Divergent Nozzles, De-Laval Nozzle, Cast-in Type, Built-up Nozzle, Construction of Turbine Blades, Production of Blades, Long Blades (L.P. Stage Blading), Hollow Blades, Shrouding, Blade Materials, Turbine Blade Attachment To the Rotor, de Laval Blade Attachment, Inverted-T Attachment, Serrated Blade Root or Annular Fir-Tree Root, Straddle (Single or Multiple Fork) Attachment, Modified Straddle Attachment., Side Entry Blades Attachment, Attachment of Shrouding Strip, Parsons End-tightened Blading, Parsons Integral Blades, Vibration of Blades, Types of Vibration and Remedies, Rotor Construction, Types of Rotors, Rotor Materials, Approximate Determination of Spindle Di-

vi ameter, Balancing of Rotors, Static Balance, Dynamic Balance, Critical Speeds of Rotors, Construction of Cylinders (Casing), Cylinders and. Diaphragms, Thickness of Cylinder, Steam Chests and Valves , Steam Chests, Steam Valves, Couplings, Bearings, Turning Gear, Stresses in Turbine Blades And Rotors, Centrifugal Stress in Turbine Blades of Uniform Cross-Section, Bending Stress in Symmetrical Impulse Blades of Uniform Cross-Section, Stress in Thin Rotor Rotating Ring, Stresses in Drum Type Turbine Rotors, Stresses in Rotating Discs of Variable Thickness, Continuous Disc of Constant Thickness, Steam Turbine Operation, Maintenance of Steam Turbines, Aims and Objectives of Maintenance, Steam Turbine Overhaul 649-679 16. Condensers And Cooling Towers The Function of a Condenser, Cooling System, Elements of a Water Cooled Condensing and Cooling System, Types of Condensers, Direct Contact Condensers, Spray Condensers, Barometric Condenser, Jet Condenser, Surface Condensers, Design Aspects of Surface Condensers, Non-condensable Gases (Air) Removal or Deaeration, Circulating Water System, Once-through Cooling System, Closed-loop Cooling System, Combination Cooling System, Cooling Towers, Wet Cooling Towers, Mechanical Draught Cooling Towers, Natural Draught Cooling Towers, The Water Distribution System, The Fill, Drift and Drift Eliminators, Dry Cooling Towers, Direct Dry-Cooling Towers, Indirect Dry-Cooling Towers, Wet Cooling Tower Thermodynamic Analysis 680-746 17. Gas Turbines—Shaft Power Cycles Classification, Application and Modern Developments, Simple Open Cycle Gas Turbine (Constant Pressure Heat Addition) or Air Standard Brayton (or Joule) Cycle, Actual Brayton Cycle, Polytropic or Small Stage Efficiency, The Cycle Air Rate, Work Ratio and Specific Fuel Consumption, Optimum Pressure Ratio for Maximum Specific Output in Actual Simple Gas Turbine Cycle, Optimum Pressure Ratio for Maximum Cycle Thermal Efficiency, Means of Improving the Efficiency and the Specific Output of Simple Cycle, Open Cycle Gas Turbine with Regeneration (Recuperation) : Ideal Cycle, Actual gas Turbine Cycle With Regeneration, Open Gas Turbine Cycle with Reheat : Ideal Cycle, Actual Cycle Gas Turbine with Reheat, Open Gas Turbine Cycle with Intercooling : Ideal Cycle, Actual Cycle Gas Turbine with Interco°ling, Open Gas Turbine Cycle with Reheat and Regeneration : Ideal Cycle, Actual Gas Turbine Cycle with Reheat and Regeneration, Open Gas Turbine Cycle with Intercooling and Regeneration : Ideal Cycle, Actual Gas Turbine Cycle worth Intercooling and Regeneration, Open Gas Turbine Cycle with Reheat and Inter-cooling : Ideal Cycle, Actual Gas Turbine Cycle with Reheat and Intercooling, Open Gas Turbine Cycle (OGTC) with Intercooling, Reheat and Regeneration : Ideal Cycle, Actual Gas Turbine Cycle with Regeneration, Reheat and Intercooling, Effect of Various Modifications, Effect of Regeneration. Intercooling and Reheating on Efficiency, Effect of Operating Variable on Thermal Efficiency, Effect of Operating Variables on Air Rate, Effect of Operating Variables on Work Ratio, Water Injection, Closed Cycle Gas Turbine, The Advantage of Closed Cycle Gas Turbine Over the Open Cycle, Disadvantages of Closed Cycle as Compared to Open Cycle, Semi-closed Cycle Gas Turbines, Advantages and Disadvantages of Gas Turbines over Steam Turbine Power Plants, Advantages and Disadvantages of Gas Turbines over Diesel or Petrol Engines

VII

747-819 18. Compressors Centrifugal Compressors, Velocity Diagram, EulermWork and Effect of Blade Shape, Velocity Diagrams, Slip Factor and its Effect on Work Input, Workdone and Pressure Rise in a Centrifugal Compressor, Dimensionless Parameters of Centrifugal Compressors, Prewhirl, Flow Through Impeller and Design Aspects, Velocity Distribution in the Vane to Vane Plane, Velocity Distribution in Meridional Plane, Main Flow Pattern, Impeller Design Aspects, The Diffuser System, Vaneless Diffusers, Vaned Diffuser, Volute Casing, Classification, Methods of Designing Volutes, Principle of Constant Moment of Momentum, The Principle of Constant Mean Velocity of Flow, Volute Angle, Base Circle Diameter, Volute Width, Losses in Centrifugal Compressor, Surging, Axial Flow Compressors, Velocity Triangle and Workdone, Pressure Rise and Aerodynamic Forces in Flow without and with Friction through Compressor Cascade, Isentropic Flow, With Friction, Cascade or Diffuser Efficiency, Dimensionless Parameters for Axial Flow Compressors, Factors Affecting Stage Pressure Ratio, Tip Speed, Axial Velocity, Fluid Deflection in the Rotor Blades, Losses in Axial-Compressor Stage, Choking Flow, Stalling, Three Dimensional Flow in Axial Flow Compressor, Theory of Radial Equilibrium, Aerodynamic Design Process of Axial Flow Compressor, Aerodynamic Design Example, Rotational Speed and Annulus Dimensions, Number of Stages, Stage-by-Stage Design. Design of Stage-1, Comments about Annulus Shape, Variation of Air Angles from Root to Tip, Blade Design and Construction of Shape, Off-design Performance, Characteristics of Centrifugal and Axial Flow Compressor, Compressor Materials and Manufacture, Rotor bladings, Rotor, Stator bladings, Casings, Manufacture, Case Study, Operational Problems, Comparison between Centrifugal and Axial Flow Compressors Combustion Chambers 820-826 Requirements of Combustion Chamber, Types of Combustion Chambers, The Combustion Process, Case Study, Factors Affecting Combustion Chamber Performance, Flame Tube Cooling, Gas Turbine Emissions, Methods for Reducing Emissions . Gas Turbines 827-848 Velocity Diagram and Workdone by Gas Turbine, Turbine Blade Cooling, Sources of Losses Due to Cooling of Turbine Blades, Cooling Flow Requirements, Convection Cooling, Film Cooling , Transpiration Cooling, Turbine Blade Materials, Protective Coatings, Blade Attachments, Turbine Blade Manufacture, Case Study, Losses in Turbine Blades, Performance of Turbines, Radial Flow Turbines, Matching of Turbine. Components or Equilibrium Running Diagram 21. Gas Turbine Auxiliary Systems, Operation and Maintenance 849-859 Starting and Ignition Systems, Types of Starter, Ignition System, Lubrication System, Elements of Lubrication System, Working Principle of a Lubrication System, Fuel System and Controls, Fuel System, Control of Gas Turbines, Hydro-mechanical Speed Governing System (Prime Control), Shut Down Control, Modulating Controls, Operation, Usintenance and Trouble Shootings, /Operation, Maintenance, Trouble Shooting. .22. Jet Propulsion 860-894 Classification of Jet Propulsion Engines, The Turbojet Engine, Thrust, Thrust

VIII Power, Propulsive Efficiency and Thermal Efficiency, Advantages and Disadvantages of Jet Propulsion over the Other System, Performance of Jet Propulsions, Intake and Propelling Nozzle, Optimization of the Turbojet Cycle, The Turbofan Engine, Optimization of Turbofan Cycle, Turbojet Engine with Afterburner, Turboprop, Ram Jet, Pulse Jet Engine, Rocket Engines, Basic Theory of Operation of Rocket Engines, Solid Propellant Rockets, Liquid Propellant Rockets 33. Combined, Co-generation and Mixed Cycle Power Plants 895-924 Introduction, Classification Combined Gas/Steam, Mixed and Co-generation Cycle, Early History, Combined Cycle Power Plants in India, Various Configurations of Combined Cycle Power Plants, Mixed Cycle, Elementary Thermodynamics of Unfired Combined Cycle, Thermodynamic Analysis of Combined Cycle and Co- generation Plants, Ideal Cycle Analysis, Actual Cycle Analysis, Advantages of Combined Cycle Power Generation, Exergy Analysis of Combined Cycles, Performance Curves of Combined And Steam Injected Cycles, Coal Based Combined Cycle Plants, Repowering, Other Types of Combined cycles \ /24. Nuclear Power Plant 925-959 The Atomic Structure, Comparison Between Chemical and Nuclear Equations, Nuclear Binding Energy, Nuclear Fusion and Fission, Fusion, Fission, Energy From Fission and Fuel Burnup, Radioactivity, Decay Rates and HalfLives, Neutron Energies and Scattering, Thermal Neutrons, Nuclear Cross-Sections, Neutron Flux and Reaction Rates, The Variation of Neutron Cross-section With Neutron Energy, Main Production of the Reactor, Principal Components of a Nuclear Reactor, Classification of Nuclear Reactor, Pressurized Water Reactor (PWR) Power Plant, Construction Details, Working Fluid Loop, Pressurizer, Case Study-A Typical PWR Power Plant, Advantages, Disadvantages, Boiling Water Reactor (BWR) Power Plant, Thermodynamic Analysis, The Current BWR System, Case Study:-A Typical BWR Power Plant, Advantages, Disadvantages, Gas Cooled Reactor (GCR) Power Plant, case Study:-A Typical U.K. Advanced GCR Power Plant, Case Study:A Typical-Temperature Gas-Cooled Reactor (HTGR), HTGR Direct Cycle-Gas Turbine Plant, Advantages, Disadvantages, Heavy Water Reactor (HWR) Power Plant, Advantages, Disadvantages, Liquid Metal Fast Breeder Reactor (LMFBR) Power Plant, Fission Reactions, Working Principal, Advantages, Disadvantages, Fusion Reactor Power Plant, Fusion Reactions, Working of a Conceptual Fusion Reactor, Indian Nuclear Power Plants

,F5' Hydro-electric Power Plant

960-1019 Advantages and Disadvantages of Hydroelectric Power Plants, Advantages, Disadvantages, Optimization of Hydro-thermal Mix Load Sharing, Site Selection for a Hydro-eleCtric Plant, Hydrology, Hydrological Cycle, Run-off, Hydrographs, Mass Curve, Applications of Hydrographs, Pondage and Storage, Main Elements of a Hydroelectric Power Plants, Catchment Area, Reservoir, Dam, Classifications of Dam, Selection of Site of Dam, Choice of Dams, Trash Rock, Forebay, Spillways, Overall (or Solid Gravity) Spillway, Chute or Trough Spillway, Side Channel Spillway, Saddle Spillway, Shaft Spillway, Siphon Spillway, Conduit (Canal, Flume, Tunnel, Pipeline Penstock, etc, Surge Tank, Draft Tubes, Power house, Classification of Hydroelectrical Power Plants, High Head Power Plants, Medium Head Power

Plants, Low Head Power Plants, Base Load Plants, Peak Load Plants, Hydel Plant With Storage Reservoir, Run-off River Hydel Plants with or Without Pondage, Pump Storage Hydel Plants, Mini and Micro-Hydel Plants, Hydraulic Turbines and Their Classifications, Ranges of Water Turbine Design Parameters, Theory of Pelton Turbines, Construction Details and Velocity Diagram, Workdone and Efficiency, Erosion and Cavitation, Degree of Reaction for Water Turbines, Theory of Francis Water Turbines, Construction Detail, Velocity Diagram, Workdone and Efficiency, Theory of Propeller and Kaplan Water Turbines, Construction Details, Velocity Triangle, Workdone and Efficiency, Theory of Deriaz Water Turbine, Theory of Bulb Water Turbine, Comparison of Water Turbines, Specific Speed, Scale Ratio, Unit Speed, Unit Power and Unit Discharge, Cavitation Problem in Water Turbines, Cavitation Factor, Methods to Avoid Cavitation, Governing of Hydraulic Turbines, Governing System of Impulse Water Turbines, Governing System of Reaction Water Turbines, Performance of Water Turbines, Constant Head Characteristic Curves, Constant Speed Characteristics Curves, Constant Efficiency Curves, Factors Governing the Selection of Water Turbines 26. Diesel Engine Power Plants 1020-1038 Applications, Advantages and Disadvantages, Working and Classification of Diesel Plants, General Layout of a Diesel Engine Power. Plant, Principal Parts of a Diesel Engine, Air Intake System, Exhaust System, Fuel Handling System, Fuel Injection System, Common Rail Injection System, Individual Pump Injection System, Distributor System, Fuel Injection Bosch Pump, Fuel Injector, Types of Nozzles, Cooling System, Open Cooling System, Natural Circulation System, Forced Circulation Cooling System, Lubrication System, Mist Lubrication System, Splash System, Pressure Feed System, Splash and Pressure Feed Lubrication System, Dry Sump Lubrication System, Starting System, Supercharging, Combustion in Diesel (CI) Engine, Knock in CI Engines, Performance Characteristics Energy Storage 1039-1059 Need of Energy Storage, Energy Storage System, Electrical Storage, Thermal Storage, Pump Hydro, Above Ground Pumped Hydro Storage System, Underground Pumped Hydro Storage System, Compressed-Air Energy Storage (CAES), Adiabatic Storage System, Hybrid Air Energy Storage System (HAES), Energy Storage by Flywheels, Electrochemical Energy Storage, Battery Storage, Fuel Cells, Magnetic Energy Storage, Thermal Energy Storage, Sensible Heat Energy Storage, Pressurized Water Sensible Energy Storage System, Variable Pressure Accumulator, Expansion Accumulator, Displacement Accumulator, Latent Heat Energy Storage, Chemical-reaction Energy Storage, Hydrogen Energy and Its Storage, Compressed Gas, Chemical Compound, Liquid Hydrogen, Metallic Hydrides

\YA.

Major Electrical Equipments in Power Plants 1060-1074 Layout of Electrical Equipments, Single Bus-Bar System, Duplicate Bus-Bar System, Ring and Bridging Bus-Bar System, Generators, Stator, Rotor, Exciter, Power Generated, Generator on Load, Voltage Regulation, Generator Cooling, Cooling Methods, Open System Using Air, Closed System Using Air, Advantages and Disadvantages of Closed System over Open System, Hydrogen Cooling, Switchgear, Indoor Switchgear Installation, Unit Type Switchgear Installation, Outdoor Switchgear Installations, Power Trans-

former, Working Principle of a Transformer, Main Elements of a Power Transformer, Circuit Breaker, Function, Principle of Circuit Breaker, Classification of Circuit Breakers, Oil circuit Breakers, Air Circuit Breakers, Protection of Electrical System, Types of Relays, Short Circuit in Electrical Installations and Limiting Methods, Limiting of Short Circuit Currents, Earthing of a Power System, Control Room, Voltage Regulation, Transmission of Electrical Power 29. Power Plant Instruments 1075-1081 Important Instruments used in Power Plants, Measurement of Water Purity, Measurement of Dissolved Solids in Water, Dissolved Oxygen Recorders (Katharometer), Measurement of pH (Potential of Hydrogen) Value of Water, Gas Analysis, 02, CO2 and CO Measurement, Measurement of CO2 by Thermal Conductivity Method, Magnetic Wind Method Measurement for 02 in Fuel Gases, Measurement of Carbon Monoxide, Measurement of Smoke and Dust, Light Extinction Type, Measurement in Nuclear Power Plants 30. Air Pollution Caused by Power Generation And Its Control 1082-1089 Effect of Air Pollution, Green House Effect, Acid Precipitation, Human Health, Basic Types of Systems for Air Quality Control, Fuel-Gas Desulfurization (FGD) System, The Wet-Gas Desulfurization System, The Dry Fuel-Gas Desulfurization System, Single Alkali Scrubbing, NO Removal, Thermal Pollution, Pollution from Nuclear Power Plants, Radioactivity Release, Radioactive Waste, High Efficiency Engines and Turbines (HEET) Technology With Near Zero Emissions •

References

•

Appendices

•

Index

1 Power Plant Types and Economics

Energy is the driving force of civilization. Until recent centuries one of the factors limiting population growth was the constancy of the energy supply per capita only slightly more than the food supply. Only after breakthrough in fossil fuels, civilization was released from the constraints of subsistence agriculture and home industry and began its breathtaking growth in both population and per capita standard of living [18]. Like per capita income, per capita energy consumption is now an another index to know the standard of living of the people of a particular country. It is worth mentioning that with 6 percent of world's population, U.S.A. consumes about 33 percent of the world's energy production [40]. The electric power generation in India by March 1994 is around 76,718 MW and India energy consumption is about 12 percent of global average. Energy in the form of electricity is easy to produce, transport, use and control. At present, 70 percent of energy production in the world is from fossil fuels. Hydro-electric, nuclear power, Diesel engine and other energy resources such as solar energy, solar cells. geothermal, biomass, etc. contribute only 30 percent of energy production. Since the deposits of fossil fuels are getting exhausted day by day, scientists and engineers are engaged in searching out new energy resources, especially non-conventional and renewable and developing energy converting devices for the new resources. The increasing awareness of the exponential growth of energy consumption, the rapid depletion of our natural resources, the lagging development of new energy resources and technologies and the growing public and institutional demands for energy and material conservation and environmental protection have forced the various countries to look forward to a more rational plan for the energy economy. As result, many government and private agencies have initiated broad studies of projected energy consumption and the possibility of conserving energy while enhancing the quality of environment. Among the various energy converting devices, steam and gas turbines are extensively used as these use fossil fuels. These devices convert thermal energy as a result of combustion of fuels into mechanical work and are termed as prime movers. In steam turbines, the working fluid is steam whereas in gas turbines, it is a mixture of air and fuel. The steam turbine is supposed to be the most important prime mover on the earth for generating electricity and gas turbine is supposed to the best in the field of jet engines and also for generating electricity. There are many other applications of these two devices which are depicted in the later stages of the book.

2

Steam & Gas Turbines And Power Plant Engineering

In any type of turbine, either thermal or hydel, there is no reciprocating part as in the case of steam, diesel or petrol engines The principle of operation of turbine is based on the Newton's Second-Law of Motion which states that the force is proportional to the rate of change of momentum. The turbine consists of a rotor over which many blades or vanes are attached which changes the direction of the working fluid. The working fluid (steam/gas/water) having achieved high velocity by means of nozzles or fixed blades due to decrease in pressure energy or head energy passes through the moving blade channels and is forced to change its direction due to blade curvature which involves a change in momentum and therefore a "driving force". If a generator is mounted on the sliaft supported on bearings to which the rotor is mounted, electricity will be generated. So we see that a turbine is free from unbalanced forces due to absence of reciprocating masses unlike as in steam engines. Since there are no mating parts the turbine is free from internal lubrication. A steam turbine thus has many advantages over the steam engines. About three-fourths of the electrical power generated in the world is produced by steam turbo-generator employing steam turbines. This is the age of nuclear reactors and they are gaining popularity in the field of power generation, even then, steam turbine will continue to be used because the reactor is only a source of producing heat which will generate steam and finally it will be expanded in steam turbines. Due to radiation hazards from nuclear reactors, the momentum for more nuclear power plants have lost. Now, hydrogen as a fuel obtained from electrolysis process is supposed to be most clean energy and a 400 MW based on hydrogen fuel utilising steam and gas turbine working on Rankine cycle is coming up in Japan by 2020. In solar and geothermal power plants, steam turbines are the prime movers, Thus the steam turbine is irreplaceable. The modern steam turbines can produce much power as the modern boiler is capable of developing steam pressure around 200 bar. Even steam pressure may be obtained at more than critical pressure in a supercritical boiler. So in the present set-up, we see many steam turbines of 1000 MW unit capacity working all over the world. Modern gas turbines are capable of producing 320 MW from a single unit. The purposes of turbine technology are to extract the maximum quantity of energy from the working fluid, to convert it into useful work with maximum efficiency, by means of a plant having maximum reliability, minimum cost, minimum supervision and minimum starting time. This book deals with the theory of steam, gas turbines and power plant engineering. It covers a wide range of power plants namemly steam turbine, gas turbine, nuclear, hydel and diesel engine power plants. Detailed analysis of steam and gas turbines, has been done. Production of electricity has seen its exponential growth. The rate of change of electricity production per year at the same fractional rate is expressed as dE dt

Ei

Where E is electricity production, i is the production rate per year and t is the time in year. After integration, one obtains In(E/ Ed = i (t — to) or

E = E0 (r-id

(1.2)

Power Plant Types And Economics

3

Where E0 is the electricity production in the base year to. The exponential behaviour gives rise to the concept of doubling time, which means the time required for electricity production to double. To find doubling time, the above equation (1.2) is expressed as EZ , 02 - i i ) E1 = e (1.3) Where E1 and E2 are the power generation in time t1 and t2 respectively. If td denotes the double time, and E2/E1 = 2, the eq. (1.3) becomes /n 2 = itd or

td

—

0.693 (1.4)

If i = 5%, the doubling time td = 13.86 years. It has been observed that there is linear relationship between the demand of electricity and gross national product (GNP). Higher GNP means higher growth which means higher demand of electricity. 1.1. A Brief History of Turbine The first turbine historically recorded, worked on the reaction principle. It was the "Hero turbine" shown in Fig. 1.1 (a) developed around 150 B. C. by Hero of Alexandria. The steam generated in the boiler, flowing through hollow trunnions enters a hollow spherical receiver. As the pressure in the receiver increases, the steam issues tangentially from the nozzles at the end of two opposite arms, the reaction of the steam leaving the nozzles rotates the sphere about its axis. Many centuries later, in the year 1629, Giovanni Branca built a steam turbine based on impulse principle. It consisted of a steam boiler, the lid of which was made in the shape of man, a long tube acting as nozzle, a horizontal wheel with blades, a shaft and a toothed gear transmission for driving a pounding mill. During 1806-13, a Russian inventor, Moving nozzle .. i Polikarp Zalesov built a number of models of steam turbines. Spherical Hollow turnion In the year 1890, Swedish engineer, Car receiver 41 1 Gustav Patrik da-Laval (1845-1913) built a single disc steam turbine of 3.5 kW capacity working on the principle of impulse. After Boiler this, steam turbines were commercialised and many improvements were made. Later on, Parsons developed a fifty percent reaction turbine which revolutionised the turbine Fig. 1.1 (a). Hero's Turbine industry.

flit/

1.2. A Case Study of Turbines Fig. 1.1 (b). shows a general arrangement of a 120 MW General Electric steam turbine. It is a three cylinder reheat turbine designed to operate at 3,000 rpm under the following full-load steam conditions : Fig. 1.1 (c) shows an industrial gas turbine. Initial steam pressure and temperature = 102 bar and 537°C. Reheat steam temperature = 537°C

„wiloseeql: k.tclionrith

) Tty' Oh.

I

111

:r4=fit.sr= ruall 71 Wimp iffisTA

Fig. 1.1(b). General Arrangement of G.E.C. 120 MW Steam Turbine.

r

..

- - .1 /47 "5-74•N.71.-0116.!"-

LP

Steam & Gas Turbines And Power Plant Engineering

Air

4

-15 kM2ICCEI

L

•

Turbine

Fig. 1.1 (c). Inaustrial Gas Turbine.

r 1 1101-4011111,411T74-

•

lit ,/1

CD

Flue gas

-4-

IP

/

LP

LP

Flue gas

Stack

Cooling tower(s) A

AA AA 217_ IL

Fig. 2.1 (b). Layout of Pulverized Coal Fired Power Plant.

Circulating water pump(s)

Makeup

i Circulat ng water system

Induced draught fan Scrubber solids and I )." fly ash to landfill ---,I

I

Filter

Many auxiliary subsystems

Subsystems

(i) Coal handling system (2) Steam generation system (3) Post combustion clean-up system (4) Power conversion system (5) Circulating cooling water system

Major Systems

A : Air A + V : Air + Vapor

Post-combustion cleanup system —)

Particulate removal n* system NZN7

SO2 removal system

I 'Bus bar

Generator

Turbines HP/IP/LP)

Reserve pile

Extraction steam "aL. Deaerator

HP

Stockout

Active pile \

c'nveYor

Unloading Sto Zeyor cknuvrec

Train unloading

Make up water Feedwater heater(s) Condensate pump (s) Power conversion system

HP BF

>

veyor

Reclaim

roeslof

Forced draught fan(s) Primary air fan(s)

1 Air heater

V

Reheater

HP exhaust steam

Super heated steam

Reheated steam

Steam generation system

Steam generator

0-

Downcorner

Feeder(s) ❑

Unit silos

sooxant

comes!' Iv)

Crusher building

NS

Transfer building

Coal handling system

Rankine And Binary Vapor Cycles

58

Steam & Gas Turbines And Power Plant Engineering

shown in Figs. 2.1 and 2.2 respectively. There are four fluid circuits in a steam power plant (Fig. 2.1 a). These are as follows— (ii) Cooling water circuit (closed or open) (i) Steam circuit (closed) (iii) Cooling air circuit (open) (iv) Combustion gas circuit (open) A steam power plant consists of four fundamental elements-(a) Boiler, (b) Steam turbine, (c) Condenser, and (d) Feed pump. There are many other elements, such as coal handling system, cooling tower, feed heaters, ash handling system, etc., some of them are shown in Fig. 2.1 (b). (a) Boiler :- In the boiler plant, the working fluid, water, receives heat due to the combustion of fuel and is converted into steam at constant pressure. Its efficiency is about 90%. (b) Steam turbine. In the steam turbine, the steam from the boiler expands (i.e. steam does work by reducing its pressure, temperature and heat content) and thus performs mechanical work. It is a rotodynamic machine. Its isentropic efficiency is about 86 to 88%. (c) Condenser. In the condenser, the exhaust steam from the turbine gives up heat on condensation to the cooling water which can not be converted into work and must be rejected to restore the initial condition of the working fluid. The condenser enables the exhaust steam to be used as the working fluid of the boiler again and again. It also increases the output of the turbine due to vacuum created inside the condenser. About 50% of the heat energy input is rejected in the condenser. The performance of condenser is judged by condenser and vacuum efficiencies. (d) Feed Pump. The feed pump (centrifugal type), pumps the feed water coming out of from the condenser to the boiler at the desired pressure. It is either motor or turbine driven. It consumes about 2 to 2.5% of the power output. 2.1.1. Working Principle. In the case of coal-fired boiler, the coal is supplied from the coal storage to the boiler through the coal handling plant. The atmospheric air is fed to the boiler through an air preheater (heat recovery device) where air is heated by the flue gases coming out as a waste heat. The heated air enters the boiler and thus increases its efficiency. As a result of combustion water supplied in the boiler at desired pressure gets converted into steam and ash and flue gases are formed. The ash is removed by the ash handling and disposal system whereas the flue gases passes through the air preheater, dust collector and finally chimney to the atmosphere. The steam so generated passes through the superheater tubes and gets converted into superheated steam. This superheated steam enters to the turbine through the steam stop valve (SSV.) and governor valve (G.V.). The stop valve is used for starting and stopping the turbine whereas the governor valve maintains the speed of the turbine sensibly constant irrespective of the load. Expansion of steam takes place in the turbine because steam does work in the turbine. Alternator converts the mechanical energy produced by turbine into electrical energy which, is fed to the transformer, circuit breaker and finally to the bus bar. The exhaust steam from the turbine is condensed in the condenser (shell and tube type heat exchanger) due to exchange of heat with cooling water. Condenser is equipped with a vacuum pump to extract any air which may be present due to leakage thronc.;11 joints and gases released upon condensation. The condensate is extracted by a condensate extraction pump and led to 1.p., feed heater (direct contact type) where feed water is heated with steam bled from the turbine. The heated feed water. is pumped back to the boiler through H.P. feed heater (surface

59

Rankine And Binary Vapor Cycles type).

The cooling water is supplied to the condenser by atOrculating water pump through a closed circuit. The heated water is cooled in a cooling tower. Some quantity of cooling water in the form of water vapor is carried away by the air hence make-up cooling water to the condenser is supplied from the river or lake or ocean through a filter. If the source of cooling water is an ocean, then there is a need of desalination plant. If the source of cooling water (i.e. river or lake, etc.) is very vast, the cooling tower can be dispensed with and the hot water is led to the river or lake as the case may be, as an open system. Due to leakage of steam from the turbine, some quantity of steam gets lost. Hence make-up water, well treated through a water treatment plant is generally added up in the well of condenser. The coal and ash handling systems are shown as simplified form in Fig. 2.1 (b) and will be discussed in detail in subsquent chapters. 2.1.2. Thermodynamic Analysis. The above processes are shown in p-v, T-s and h-s diagrams of Rankine Cycle. The various processes are :Process 3-4. The water from the hot-well or the surge tank which is at low pressure is pumped into the boiler at high pressure p1. Here pumping process 3-4 is isentropic.

—► s

V

Fig. 2.2(a). p-v Diagram of Rankine cycle. v,

Fig. 2.2(b). T-s Diagram of Rankine cycle ,--

P2

►s Fig.. .2(c). h-s Diagram of Rankine cycle.

60

Steam & Gas Turbines And Power Plant Engin Jering

Process 4-5. As the water enters the boiler, water is first heated up to the saturation temperature or evaporation temperature T1 called sensible heating and during this process the state point moves along curve 4-5. The heat supplied during this process is hi5 —hp and is represented by the area L-3-4-5-M, i.e. sensible heat of water. Process 5-1. At constant pressure p1 and temperature Ti, water is completely evaporated into steam. The heat supplied in this process is equal to h1 —h./5 and is represented by M-5-1-N i.e. latent heat of vaporization. The state point 1 shows the dry and saturated condition of steam. Process 1-2 shows isentropic expansion of steam in the turbine from pressure pi to p2. Process 2-3. At constant pressure /22 and temperature T2, the exhaust steam from the turbine is condensed in the condenser. It is possible that the steam leaving the boiler may be dry and saturated, wet or superheated and so the corresponding cycle on the T-s and h-s diagrams are 1-2-3-4-1, 1 '-T-3-4-5-1 or 1"-2"-3-4-1" respectively. The thermal efficiency of the Rankine cycle can be obtained by applying the First Law of Thermodynamics separately for each process. Let us assume that 1 kg of the working fluid flows through the various elements of the cycle. From First Law of Thermodynamics for flow process, 8q — Sw = de = dh + AKE + APE Since AKE and APE are usually negligible, hence (2.1) 8q — 8w = dh For boiler process 4-1. Process 4-1 is the combined process of 4-5 and 5-1 in the boiler. In this process, the work done is zero because no work is done in the boiler. The heat supplied to the fluid is given by q bolter = ji4

dh = hi —hp, = (171-12J3)— (hr4hi3)

Here, hp is the sensible heat of water (fluid) in the boiler at state 4.

For turbine process 1-2. In this process, heat supplied is zero because the process is isentropic and hence W turbine = j2. — 1

dh = 2

dh= h l —h2

For condenser process 2-3. In this process, no work is done and so from the First Law of Thermodynamics = h2 hf2 q condenser = h2 Here hp is the enthalpy of water of state 3 corresponding to pressure p2 i.e., hp = For feed pump process 3-4. In this process, no heat is supplied because the process is isentropic; hence from First Law pump = 1114 — 113 = vi.2 (p — p2) Here, vfl = specific volume of water in m3/kg corresponding to the final pressure of p2 , kN/m2 , pi = initial pressure of steam in kN/m2, p2 = final pressure of steam in kN/m2 Heat supplied = aboiler = h1 — hf3 — vfl (p1 —p2)

Rankine And Binary Vapor Cycles

So net work done is No =

= W turbine — W pump

(hi — h2 ) — (1714 — hf3) = (hi — h2 ) — vi2 (p — p2 )

Thermal efficiency = or

61

Net work Heat Supplied

.th

wnet (1 boiler

(hi — h2) — (1114 — ht.3) (rldRankine (h — hi3)— (hft — 11.13 ) 1

(2.2)

Compared to turbine work, pump work (w)pump is very-very small and it may be neglected because the specific volume of water is very small. Hence, we have 0 so h = 11 pump ' f4 J3 Therefore, (1-6 )Rankine =

h2 )

—h)

hi — h2 h —h

(2.3) I J3 1 f2 The thermal efficiency of the Rankine Cycle may be also expressed as (in the from of areas). Referring to Fig. 2.2 area 1 2 3 4 5 1 ("lth)Rankine area L 3 5 1 NL It is to be noted that if the condensate is pumped back to the boiler from the state point 3' instead of 3, then it forms the Carnot cycle. So 1-2-3'- 5-1 is Carnot cycle. We know that Carrot cycle has more efficiency than Rankine cycle and this is also obvious T from the Fig. 2.2 with the same temperature limits. The reason is that the average temperature at which heat is added in the Rankine cycle lies between T4 and 7'5 (or T1) and is thus less than the constant temperature T1 at which heat is added to the Carnot cycle. Though, the efficiency of the Carnot cycle is more than Rankine, still we take the Rankine cycle as the theoretical cycle for the steam power plant because of two main facts as Fig. 2.2 (d). Rankine Cycle with given below : Supercritical Boiler Pressure The Rankine cycle working on supercritical boiler pressure of steam is shown in Fig. 2.2 (d). 2.1.3. Limitations of Carnot Cycle As Vapor Cycle. (a) In the Carnot cycle, the condensate is pumped to boiler from state point 3' ( Fig. 2.2) where the condensate is in the form of water and its vapor to state point 5 where the water is saturated. It is very difficult to build a pump which can suck a mixture of water and its vapor at state point 3' and deliver saturated liquid at state point. 5. 'Practically, it is very easy to completely condense the vapor upto state point 3 and handle only liquid in the pump as is done in the Rankine cycle. (b) In1he Carnot cycle, superheating of steam is a great problem. Superheating has to

62

Steam & Gas Turbines And Power Plant Engineering

be done at constant temperature along path 1-6 but during this process pressure drops. This means that the heat will transfer to the vapor in the expansion process. In actual practice this is very difficult to achieve. But in the Rankine cycle, the vapor is superheated at constant pressure from 1 to 1" without any difficulty. From the above facts it is clear that the Rankine cycle is possible in actual practice and not Carnot cycle so the former is taken as theoretical cycle for the steam power plant.

2.2. Principles of Increasing the Thermal Efficiency. In the Carnot cycle, the heat addition is at the constant temperature but in Rankine cycle temperature of the working fluid increases during heat addition at constant pressure process, hence a concept of average temperature of heat addition is introduced which gives the same heat supplied as given by the variable temperature heat addition in Rankine cycle (Fig. 2.3). Thus,

qA = hi —hf4 = T. x (A).

Tave (s2 — s3)

R = T2(s2 — s3) '•' (1111Nankine

I—

R "IaA

= 1

T2 = f (T.) only as T2 is fixed

Tave

•

Thus higher the average temperature of heat additions the higher will be the Rankine efficiency. Based on the concept of average temperature of heat addition in Rankine cycle and the outcome of Carnot cycle, following two principles are established to increase the thermal efficiency of Rankine cycle., (a) By increasing the tWerage temperature at which heat is added to the cycle. (b) By decreasing the average temperature at which heat is rejected from the cycle.

2.3. Effect of Operating Conditions on Efficiency (a) Effect of Superheat. Rankine cycle 1-2-3-4-1 and 1'-2'-3-4-1' using dry and saturated steam and superheated steam respectively are shown in Fig. 2.4. Comparing these two cycles, it is obvious from the figure that superheat cycle delivers more work and the

T 2

s Fig. 2.3. Concept of Average-Temperature of Heat Addition.

-111.

S

L

Fig. 2.4. Effect of Superheat on Rankine Cycle.

Rankine And Binary Vapor Cycles

63

excess work is represented by area 1-1'2'-2-1. But it also takes in more heat and this is represented by area 1-1'-M-L-1. The result is that the efficiency of Rankine cycle using superheated steam is more than that of the dry and saturated steam. From the thermodynamic point of view also, this result could have been anticipated. Due to the superheating of steam the average temperature of heat addition to the cycle increases, so there should be an increase in the thermal efficiency compared to the cycle using dry and saturated steam. (b) Effect of Maximum Pressure. In Fig. 2.5, two cycles 1-2-3-4-1 and 1'-2'-3-4'-1' have the same maximum temperature T1 ' = T1 but different maximum pressures p'1 > pi . The condenser pressure is the same in both cases. It is obvious from the figure that due to increase in maximum pressure from pressure pi to p'i , the net work increases by the area . shown by the horizontal hatching line and decreases by the area shown by cross hatching. After drawing the diagram true to scale, it has been found that these areas are nearly equal, so the net work is nearly the same, but the heat rejected decreases by the area 2- 2'-L-M. The thermal efficiency is given by Heat rejected qn —— a• A — Heat added Since the heat rejection is reduced in the case of increasing maximum pressure, so thermal efficiency also increases. From the Second law it can be anticipated that the thermal efficiency increases with increase in maximum pressure because the average temperature of heat addition increases with increase of maximum pressure. (c) Effect of Exhaust Pressure. In Fig. 2.6, two cycles 1-2-3-4-1 and 1-2'-3'-4'-1 have the same maximum pressure and temperature but different exhaust pressures p2 . `c0

a) 44 Dry saturated To a) 6 40 a) 12

36

'\b'°'6

vapor l'ne, , i/ ,' ,1

Back pressure =0.5 cm of Hg abs I 1 I 32 150 250 350 450 550 650 750 850 900 Throttle temp (°C) Fig. 2.7. Effect of Throttle Temperature.

in exhaust pressure because the average temperature of heat rejection decreases with decrease of exhaust pressure. 2.3.1. Performance Curves. Fig. 2.7 shows effect of initial i.e. throttle temperatures and initial pressures on the Rankine thermal efficiency. At a particular value of initial pressure except very high pressure, the thermal efficiency increases approximately linearly as the temperature increases. At a very high pressure in the beginning the increase in thermal efficiency is very fast at the start and after that it increases slowly. The effect of back pressure on thermal efficiency is shown in Fig. 2.8. As the back pressure decreases the thermal efficiency increases, It is also obvious that as the initial pressure increases the thermal efficiency increases. • Fig. 2.9. shows the variation of cycle efficiency and specific steam consumption with boiler pressure and temperature.

•

45

■

g 40

175 bar - 540°C .ar. #1 7 " 540° 2 bar o mmimmo . 36 bar - 4 5°C

1111111111.1..-I 35 1

al

a) 30

25

15 bar - 200°C

20 15 0

0.1 0.2 03 04 05 06 07 08 09 10

Cycle exhaust pressure Fig. 2.8. Effect of Back and Initial Pressure.

Rankine And Binary Vapor Cycles

5

0.4

4

p- 0.3 t

3 1

0.2 — Boiler pressure 2 Li 30 bar 1 05, 0.1 Cond. press, 0.04 bar — 0 i 0 u) 200 400 600 800

Superheat temperature °C

0.5

11 0.4

2—

— 4

SS c

0.3

— 3

0.2

Superheat temperature 600°C

0.1 —

Condenser pressure 0.04 bar

2 — 1

S.S.C. kg/kWh

0.5

65

0

0 0

50

150

100

200

250

Boiler pressure, bar

(a)

(b)

Fig. 2.9. Variation of Cycle Efficiency and SSC with Boiler Pressure and Temperature

2.4. Methods of Increasing the Thermal Efficiency. Based on above two principles, the methods of increasing the thermal efficiency are as follows :(i) By increasing the superheating temperature of steam (ii) By increasing the maximum pressure of steam. (iii) By reducing the exhaust pressure of steam. (iv) By regenerative feed heating. (v) By reheating of steam. (vi) By water extraction. (vii) By using binary-vapor. Generally, a modern steam turbine power plant is a combination of all the above except (vii). In some case (vii) is also used. 2.5. Deviation of Actual Cycle from Theoretical Cycle. (Internally Irrevessible cycle) The actual steam turbine power plant cycle deviates from the theoretical i.e. ideal cycle because losses occure in various components like boiler, turbine, pump, condenser

Feed pump Fig. 2.10. Losses in Rankine Cycle

Steam & Gas Turbines And Power Plant Engineering

66

and piping. Internal irreversibility of Rankine cycle is caused by fluid friction, throttling and mixing. These losses are shown in Fig. 2.10. We will now discuss each of them briefly. (a) Turbine Losses, The losses in turbine arises mainly due to; (i) friction resulting from the flow of the working fluid through the turbine. (ii) heat loss from the turbine to-the surrounding. Throttle pressure loss

Throttle pressure

loss

Pi. pi

t

s

(b)

(a)

Fig. 2.11. Effect of Friction on Rankine Cycle. (iii) leakage loss. The effect of friction is to increase the entropy and it can be visualised from Fig. 2.11. Process 1-2 represents isentropic expansion while process 1-2' represents the actual expansion. The isentropic turbine work =h1 — h2 So turbine efficiency =

—

actual turbine work =h1 — h2'

and

Actual work in the turbine hi Isentropic work in the turbine h1 — h2

The turbine efficiency ranges from 85% to 88%. The items (ii) is negligible. Leakage loss is around 0.5% which may be neglected. (b) Condenser Loises. The losses in the condenser is small. These include the loss in pressure (from p2 to p3) and the cooling of the condensate below the saturation temperature. The degree of undercooling varies from 4 to 6°C. (c) Pump Losses. The losses in pumps are similar to losses in turbines and friction is mainly responsible for these. Heat loss may be neglected because it is very low. The pump efficiency is defined as Isentropic work supplied hp h/3 Tipump — Actual work supplied h ' —h /4

13

(2.5)

Generally, pump work is neglected in comparison to turbine, so pump losses do not come in the picture. The overall pump efficiency (rlp) which is a product of hydraulic efficiency, mechanical efficiency and volumetric efficiency ranges from 80% to 90%..

Rankine And Binary Vapor Cycles

67

(d) Boiler Losses. The losses in the boiler include the heat loss to surrounding and heat carried away by the flue gases. Boiler efficiency ranges from 88% to 92%. Boiler losses cannot be depicted in T-s and h-s charts but it affects the overall thermal efficiency of the plant. Due to inefficiency of boiler, fuel requirement will .be more. (e) Piping Losses. There are mainly two types of losses in steam pipes. Firstly, there is a pressure drop of steam (from p/' to pt') due to friction between the pipe wall and steam, Secondly, there is a heat loss from the pipe to surroundings. Due to these two losses the state of steam entering the turbine is at a lower pressure and temperature and greater entropy as compared to the state leaving the boiler. So the effect is that the net work of the cycle is reduced and thus the efficiency is lower. (f) Regulating valve losses. When the steam passes through regulating valves, there is a pressure drop (from pi ' to pi ) in the steam at constant enthalpy (hi' = hi) due to throttling action. As a result the net work of the cycle is reduced and thus the thermal efficiency drops. 2.6. Externally Irreversible Rankine Cycle It is the temperature differences between the combustion gases and the working fluid on the source side and the temperature differences between the condensing working fluid and the condenser cooling water on the sink side cause external irreversibility in the Rankine cycle (Fig. 2.12). The products of combustion i.e. flue gases get cooled from a to d and the working fluid (water) temperature rises from 4 to 1 in counter-flow heat exchange4r (steam generator). The minimum temperature difference between the two fluids are c-5 and a-1, and the points where these a occurs are called pinch points. Generally, a-1 Flue gass tq a-1 : pinch i is known as approach temperature differpoint ence. The pinch point plays an important c c-5 : pinch point part in deciding the steam generator efficiency. Too small a pinch point difference causes a lower external thermal irreversibility and an increase in surface area means more expensive steam generator, while a too large pinch point differences results in small and inexpensive steam generator resulting in reduced plant efficiency. This suggests that an optimum pinch point is desirable taking care efficiency and cost. In the condenser side, line e-f represents the rise in temperature of cooling water Fig. 2.12. External Irreversibility in while condensation of steam takes place Rankine Cycle. along 2-3. S

2.7. Efficiencies

(i)

Isentropic cycle efficiency =

= (iiih)•

Rankine —

,sen qA

where Mii.ser; =isentropic heat drop in the turbine.

qA = Heat supplied in the cycle. = (h, — h,) = nit x (LCV),

(2.6)

Steam & Gas Turbines And Power Plant Engineering

68

h.) = mass flow rate of fuel of fuel I (LCV)f = lower colorific value of fuel (ii)

Thermal efficiency of the turbine plant= ti th—

where Ahcoupa (iii)

=

Thermal efficiency ratio = coupa

nr

qA

(2.7)

actual coupling work considering all type of losses upto coupling. —

lth

—

r11

or

Ohcoupa

Ashisen .

_

6hcoupa

qA

Ah,sen

ilth X lmech

(2.8)

So we see that the thermal efficiency ratio is the ratio of actual coupling work to the isentropic heat drop based on per kg of steam. (iv)'Overall thermal efficiency (n od. The overall thermal efficiency takes care of all the losses in various elements of steam power plant. It is given by [wrisen X lt X rim — Wp,isentilp1

loth —

(q/1)a

1generator

where ri = overall pump efficiency = rih x ri m x rly p x (LCV)f X 1boiler (qA) =[(h1 — hf2)— w hi ]= p pi . Neglecting pump work and where qAci = Mf x (LCV)f X TI bode r loth

th) Rankine X 1 boiler X 1 Curb X 1 mech x 1 generator

The overall thermal efficiency varies from 35 to 40%. whet Net work (v) Work ratio — Turbine work wT (vi) Specific steam consumption (s.s.c) : It is defined as the steam consumption per power output. It is expressed as 3600 s.s.c. = , kg/kWh wnet s.s.c. varies from 3 to 5 kg/kWh. (vii) Heat Rate (HR) : It is defined as the amount of heat supplied per power output. Thus, 3600 . qA 3600 ; kJ/kWh Tith

Wnet

2.8. Requirements of an Ideal Working Fluid. • There are several vapors which have physical properties suitable for working fluid. They are steam, mercury vapor, sulphur dioxide, diphenyl oxide and certain hydro-carbons. The following are the requirements for an ideal working fluid :(i) Ample amount should be available at low cost.

Rankine And Binary Vapor Cycles

69

(ii) Critical temperature should be higher than metallurgical limits. (iii) Reasonable saturation pressure at maximum temperature of the cycle from metallurgical point of view of boiler. (iv) Steep saturated vapor line to minimise moisture problem in expansion of steam in the turbine. (v) It should wet the boiler surfaces enveloping it and should be chemically stable at the maximum temperature of the boiler. (vi) Saturation pressure at minimum temperature of the cycle should be higher than atmospheric, otherwise the maintenance of the condenser will be costly. (vii) Low liquid specific heat so that most of the heat is added at the maximum temperature. (viii) Considerable decrease in volume upon condensation. (ix) Non-toxic and non-corrosive. (x) Freezing point should be much below the normal atmospheric pressure. Among all types of working fluids, water satisfies the maximum requirements. Its general abundancy at low cost is of prime importance due to which it is selected as the working fluid in steam turbine power plant. 2.9. Critical Temperature and Pressure of Water. At all temperatures above the critical, it is impossible to liquify water vapor by using pressure, no matter how great the pressure is employed. The critical temperature of water is 374°C. At this temperature the vapor is called critical pressure and its value is 221.2 bar. Problem 2.1. Dry and saturated steam at pressure 11 bar is supplied to a turbine and expanded isentropically to pressure 0.07 bar. Calculate the following-(a) Heat supplied, (b) Total change of entropy, (c) Heat rejected (d) Theoretical thermal efficiency. (e) Also calculate the overall, actual output and thermal efficiency. if rig = 80%, n„, = 95%, ug= 96%, rip= 80%. Solution. Refer to Fig. 2.1(a). and Fig. 2.13. From h-s chart : h1 = 2781kJ/kg (at 11 bar), h2 = 2035 kJ/kg. (at 0.07 bar) From steam table :

= 163 kJ/kg (at 0.07 bar)

Pump work = vi2 (p1 —p2) = 0.001(11— 0.07)x 102 = 1.093 kJ/kg (a) heat supplied = qA = (hi — hf3 )— pump work =(2781-163)-1.093 = 2616.9 kJ/kg.K (b) Total change in entropy = s2 — s3 = — (6.554 — 0.56) = — 5.994 kJ/kg (c) Heat rejected=h2 — = 2035-163=1872kJ/kg.

Ans Ans. Ans.

(d) Theoretical thermal efficiency _Turbine work — Pump work (h1 — h2 ) — wp — Heat Supplied 2616.9 —

2781 — 2035 — 1.093 — 28.46% 2616.9

Actual output = (Wrsen x 11 / x rim — wp/rld = (2781 — 2035) x 0.8 x 0.95 — 1.093/0.8 = 565.59 kJ/kg

Ans.

Ans.

Steam & Gas Turbines And Power Plant Engineering

70

— ms

—mss (b)

s

(a) Fig. 2.13.

Overall actual thermal efficiency —

Hisen W

wpirl ni

x ri gen

A

_ [(2781-2035) 0.8 x 0.95 — 1.093/0.81 x 0.96 = 20.75% 2616.9

Ans.

Problem 2.2. A steam turbine receives steam at pressure 20 bar and superheated to 88.6°C. The exhaust pressure is 0.07 bar and the expansion of steam takes place isentropically. Using steam table only, calculate the following. (a) Heat supplied, assuming that the feed pump supplies water to the boiler at 20 bar. (b) Heat rejected (c) Net work done (d) work done by the turbine (e) Thermal efficiency (0 Theoretical steam consumption. If the actual steam consumption is 5 kg/kWh what is the efficiency ratio of the turbine. and actual power output at turbine coupling

-

Solution. Refer to Fig. 2.14. From steam table, ti at 20 bar=212.4°C

So initial temperature of steam=212.4+88.6 = 301°C, h1 =3025 kJ/kg

Q^

s

S

(b)

(a) Fig. 2.14.

Rankine And Binary Vapor Cycles Now si = s2 = x2 si. +(1—x) so g2 or x2 =0.8674 Again ,

or

71

6.768=0.7718 x2 +(1—x2 ) 0.559

h2 =h/2 +x2 hfg = 163 + 0.8674 x 2409 = 2252.56 kJ/kg

From steam table, hp = 163 kJ/kg Pump work = vfl (pi —p2 )102 = 0.001(20-0.07)102 = 1.993 kJ/kg (a) Heat supplied =hi — hf3 — pump work =3025-163-1.993 =2860.807 kJ/kg

Ans.

(b) Heat rejected = h2 — h./3 = 2252.56 — 163 = 2089.56 kJ/kg (c) Net work done = Turbine work—pump work = (hi — h2 ) — 1.993 = 3025-2252.56-1.993 = 771.147 kJ/kg (d) Work done by the turbine = hi — h2 = 3025-2101.52 = 772.42 kJ/kg. (e) Thermal efficiency =

Net work done 771.147 26.95% Heat supplied — 2860.807 —

(f) Theoretical steam consumption — Efficiency ratio =

Ans. Ans. Ans.

360 30 60 — 4.668 kg/kWh Net workdone — 771.147 Ans.

(Ah)coupa (Ah),sen

Theoretical steam consumption in kg/kWh 4.668 — 93.36% 5 Actual steam consumption in kg/kWh

Ans.

Actual power output at turbine coupling wneta = risen X lir —

kJ/kg = 772.44 x 0.9396 — 1.993 = 723.79

Ans.

Problem 2.3. A steam turbine plant is supplied with steam at pressure 17 bar and superheated to 100°C. The exhaust pressure is 0.06 bar. The temperature of the condensate in the hot-well is actually 33°C(vf = 0.001 m3/kg.). If the measured steam condensate is 5 kg/kWh, and if the generator efficiency is 96%, what is the absolute thermal efficiency of the whole boiler and turbine plant? Solution : Refer to Fig. 2.14. From steam table : is at 17 bar = 204.3°C Initial temperature of steam = ti = 204.3+100 = 304.3°C From h—s chart : hi = 3039 kJ/kg (at 17 bar and 304.3.3°C), h2 = 2110 kJ/kg (at 0.06 bar), l;j3 =.138.5 kJ/kg (at 33°C) Pump work = vfl (p1 —p2 ) 102 = 0.001 (17 — 0.06) x 102 = 1.694 kJ/kg Heat supplied = h l — hj3 — pump work = 3039 —138.5 — 1.694 = 2898.8 kJ/kg Net work done = hi — h2 — pump work = 3039 2110 —1.694 = 927.3 kJ/kg 3600 Theoretical steam consumption —927.3

3.882 kg/kWh

Steam & Gas Turbines And Power Plant Engineering

72

Efficiency ratio = rl r

— (6,h)coupa Oh).

Theoretical steam consumption in kg/kWh 3.882 — 77.64% 5 Actual steam consumption in kg/kWh 927.3 Thermal efficiency — Net Work — — 31.989%. Heat Supplied 2898.8 .*. Absolute thermal efficiency of the whole plant = TIrh x rir x rigen = 0.31989 x 0.7764 x 0.96 = 23.83%

Ans.

Problem 2.4. The following are the mean observations made during test on a steam turbine—Duration of test=2 hours, Power developed (turbine coupling)=4500 kW, Total mass of steam condensed=50,000 kg, Steam pressure at stop valve=14 bar, Steam temperature at stop valve=300°C, Barometric height=76 cm of Hg, Vacuum =72.5 cm of Hg. Calculate the thermal efficiency of the turbine and also its efficiency ratio. Assume that the condensate is returned to the boiler plant at the exhaust saturation temperature. Also calculate the fuel cost per kWh generated in steam power plant if rib = 0.92, .L.C.V. of coal = 29000 kJ./kg, cost of coal = Rs. 1800.00 per ton and 5% of steam generated is used to cover losses and auxiliary power. Solution : Refer to Fig. 2.14. Absolute exhaust pressure =67 — 72.5 =3.5 cm of Hg. P2 —

3.5x1.013 76 — 0.04665 bar

From Mollier chart : h1=303'8 kJ/1a

S

(a)

(b) Fig. 2.15

Neglecting the pump work, Heat supplied to turbine A = hi — h./3a = 3039 — 192 = 2847 kJ/kg Heat supplied to turbine B = hi — hi.3b = 3039 — 101 = 2938 kJ/kg Work done by turbine A = hi — h2a= 3039 — 2177.7 = 861.3 kJ/kg Work done by turbine B = hi — h2b= 3039 — 2025 = 1014 kJ/kg (a) Steam consumption of turbine A

3600 86 1 3

4.179 kg/kWh

Ans.

3600

Steam consumption of turbine B = — — 3.55 kg/kWh 1014 (b) %Steam consumption reduction in turbine B = 4.1.7 ;:35 — 17.71% . 8 1 .3 (c) %Theoretical thermal efficiency of turbine A = 6 30.25% 2847

21 90 31 48

Theoretical thermal efficiency of turbine B = —

34.51%

Ans. Ans. Ans. Ans.

Steam & Gas Turbines And Power Plant Engineering

74

(d) % increase in thermal effi ciency in B =

34.51 — 30.25 30.25

— 14.08%

Ans.

(e) Actual % increase in overall thermal efficiency will remain the same as same losses will be in each turbine.

2,10. Binary-Vapor Cycle. The main aim of the steam turbine designer is to achieve higher thermal efficiency of the turbine. We have already seen that the thermal efficiency of the Rankine cycle is increased by (a) increasing the average temperature at which heat is added to the cycle and (b) decreasing the average temperature at which heat is rejected by the cycle. Metallurgical problems in the boiler and steam turbine limit the maximum temperature but at a lower pressure. At present this maximum temperature limit is about 650°C. The cooling medium in the condenser is a limit to the minimum temperature of the cycle. This temperature is about 35°C, i.e., the temperature of natural water. If steam is used as the working fluid in Rankine cycle operating over the above temperature range, i.e., from 650°C to 35°C the difficulties arise. These are (i)

Since the critical temperature of water is 374°C so for obtaining maximum temperature of 650°C, the steam has to be superheated and this is only possible at very high pressure. The metallurgical problems do not allow high pressure and high temperature, and also, the average temperature of heat addition reduces thus reducing efficiency. This high temperature is therefore desirable at lower pressure.

(ii)

High initial pressure causes excessive moisture in the low pressure stages.

(iii)

As the pressure increases the latent heat of vaporization decreases.

Water also has disadvantages at the minimum temperature of the cycle. The condenser operates at vacuum because the saturation pressure corresponding to 35°C is 0.05622 bar which is much lower than atmospheric pressure. So it is essential to minimise air leakage in the condenser. It is also desirable that the back pressure is somewhat above the atmospheric pressure. We observe, therefore, that all the requirements for an ideal working fluid are not met by water or any single working fluid. Two working fluids are therefore used in the cycle so that most of the requirements are fulfilled. Such a cycle is called Binary-Vapor cycle (also called combined cycle) and in this mercury and steam are commonly used as working fluids. The following are the advantages and disadvantages of mercury as a binary vapor cycle fluid :-

2.10.1. Advantages of Mercury (i)

Moderate vapor pressure at high temperature. At 540°C, the saturation pressure is less than 14 bar.

(ii)

Very stable.

(iii) Liquid mercury has a high density which is advantageous in the separation of the liquid from vapor in boiler and in feeding back to the boiler under hydrostatic head rather than under the pumping head. (iv) Specific heat of the liquid is only 0.13, so the liquid line is steep, making the cycle closer to Carnot cycle. (v)

Specific enthalpy of mercury vapor is low thus resulting in lower jet velocities in mercury turbines.

(vi) Thermal efficiency is higher than Rankine steam cycle.

Rankine And Binary Vapor Cycles

75

2.10.2. Disadvantages of Mercury (i) High cost and limited supply. (ii) Toxic in nature. (iii) Pervasive, thus tends to leak through joints, cracks, etc. (iv) Latent heat is very low, requiring a large amount of mercury for the same heat utilisation. (v) The whole boiler operates under high temperature and not just the superheater. (vi) It does not wet surfaces, which cause poor heat transfer. This can of course be overcome by chemical treatment. Hence, it is clear that neither mercury nor any other vapor can replace steam in all the ranges of working. However, mercury is really superior in the high temperature range. The schematic diagram of a binary vapor (mercury-steam) cycle is shown in Fig. 2.16 and its corresponding T-s Diagram is shown in Fig. 2.17. In the binary-vapor (mercury-steam) cycle there are two different circuits, one for

Steam superheater and mercury boiler

Steam cond nser

Fig. 2.16. Binary-Vapor Cycle

Mercury cycle 9.o

Mercury line

vapor

sat.

Steam sat. line Qti

S

Fig. 2.17. T-s Diagram for Binary-Vapor Cycle.

76

Steam & Gas Turbines And Power Plant Engineering

steam and another for mercury. Mercury vapor, dry and saturated, enters the mercury turbine at state point a and expands to b. The exhaust mercury vapor is condensed in mercury condenser to state point c. The condensate is pumped back to the boiler by the mercury pump. The heat rejected in the mercury condenser is used to vaporize feed water fed back by the feed water pump into steam at state point 5. Thus, we can say that the mercury condenser also acts as a boiler for feed water. The heat transfer from the exhaust mercury vapor to feed water is possible because there is a large difference in temperature between these two. The saturated steam coming out from the mercury condenser is superheated to state point 1 in the superheater and expanded to state point 2 in the steam turbine and condensed in the condenser to state point 3. Thus 1-2-3-4-5-1 and a.-b- c-d-a show the steam and mercury cycle respectively. To calculate the thermal efficiency of the binary vapor cycle the process is as follows: Let mm = mass of mercury circulated per kg of steam. Assume 1 kg of steam entering the steam turbine. In mercury condenser, heat given by mercury-vapor is received by feed water. Thus

mm(hb - hma) = 1 x(h5 - ho) or

(h - h ) m - 5 J4 m (hb - hmd

(2.9)

Net work per kg of steam flow in the cycle =(Mercury turbine work-mercury pump work)+ (Steam turbine work-water pump work)

= mm(ha - hb) - mm(hmd- hmc) + (hi - h2) - (ho - hr2) kJ/kg of steam Neglecting pump work, the net work wnet =

mm(ha - hb)+(hl - h2) kJ/kg of steam

Heat supplied per kg of steam in the cycle = qA

= mm(ha - hmd)+ (hi - h5)

•••• m m(ha - hmc) + (h1 - h5) neglecting pump work Thermal efficiency = ilth

(2.10)

mm(ha - hb)+ (hi - h2) o mm a _ kJ+ (hi - h5)

(2.11)

(2.12) The thermal efficiency of the binary-vapor cycle is more than the Rankine steam cycle. Frass [16] proposed a potassium steam binary-vapor cycle that operates with a turbine inlet temperature of 830°C to 890°C and a condenser temperature of about 600°C. This topping cycle is superimposed on a conventional steam cycle with a turbine inlet temperature of 590°C by transforming the waste heat rejected from the condensing potassium vapor to the boiling water of steam cycle. This cycle is similar to the mercury-steam binary vapor cycle (employed in seven plants in U.S.A). However, the operating temperature of these plants was limited to 500°C. But the potassium-steam binary cycle employing higher temperature made the mercury-vapor cycle obsolete. The results obtained by Frass indicates that potassium-steam binary vapor cycle would give an overall thermal efficiency of about 54% as compared to only 40% from a conventional steam cycle. Binary-vapor cycle is also called combined cycle. Problem 2.6. In a binary-vapor cycle, the steam cycle operates between pressure of 30 bar and 0.07 bar and uses a superheat temperature of 35°C. The mercury cycle works

Rankine And Binary Vapor Cycles

77

between the pressure of 12.68 bar and 0.07 bar, the mercury vapor, entering the turbine, being in a dry and saturated condition. Compute the efficiency of the combined cycle assuming expansion in both cycle as isentropic. Data for mercury :p bar

Saturation

hfm kJ/kg

h g kJ/kg

sfm kJ/kg

s g kJ/kgK

0.07 12.68

236.5 537.5

32.395 71.9796

326.667 360.704

0.08548 0.145798

0.662906 0.50185

Solution. Refer to Fig. 2.18.

S

Fig. 2.18. Binary — Vapor Cycle Neglecting Pump Work Mercury cycle : sa = sb = 0.50185 = sfin +x (sg — sfin) = 0.08548 + (0.662906 — 0.08548) x

0.50185 — 0.08548 — 0.72 0.662906 — 0.08548

Enthaply at b = hb = him + x(hg — hid= 32.395 + 0.721 (326.667 — 32.395) = 244.56 kJ/kg of mercury Isentropic work done by mercury turbine = ha — hb = 326.667-224.56 = 102.107 kJ/kg of mercury Heat rejected by mercury in mercury condenser = (hb — hme) = 244.56 — 32.395 =212.165 kJ/kg of mercury Heat supplied in mercury boiler = ha — hmc = ha — hfm = 326.667 — 32.394 = 294.27 kJ/kg of mercury For steam cycle : h1 = 3114 kJ/kg of steam, h2 = 2090 kJ/kg of steam, h4 = 2794.5 kJ/kg of steam Since nothing is mentioned about the heating of feed water, we assume that the feed

78

Steam & Gas Turbines And Power Plant Engineering

water comes out from the mercury condenser as dry and saturated steam and superheated to 35°C in super-heater. Isentropic work done by steam turbine = hi — h2 = 3114 — 2090 = 1024 kJ/kg of steam Heat to be supplied in steam boiler, i.e., superheater h I — h4 = 3114 — 2794.5 = 319.5 kJ/kg of steam Heat rejected by mercury/kg of steam = Heat received by water/kg of steam moi(hb — hod = (h4 — h13)xl or

mm=

2794.5 — 161.76 — 12.4 kg of mercury/kg of steam 212.165

Total work done in the cycle per kg of steam = work done in the steam turbine/kg of steam+ Work done in mercury turbine/kg of steam = 1024 + 102.107x12.4 = 2290.126 kJ/kg of steam Heat supplied in the cycle = Heat supplied in mercury boiler/kg of steam + heat supplied in superheater/kg of steam = (ha — hadmm + (h1 — h4) = 294.27x12.4 + 319.5 = 3968.44 kJ/kg of steam Work done

2290.126 _

nth — Heat supplied — 3968.44

Ans.

Problem 2.7. In a binary-vapor cycle, mercury vapor dry and saturated enters the mercury turbine at 10 bar and exhausts at 0.1 bar into the mercury condenser-steam boiler unit. Steam comes out from this unit at 40 bar, dry and saturated, and expands in the steam turbine to 0.06 bar. Assuming that the efficiency ratio for mercury and steam turbine are 0.75 and 0.8 respectively, Calculate—(a) Amount of mercury circulating per kg of steam, (b) Thermal efficiency of the cycle, (c) Workdone by mercury and steam turbine p bar

is °C

kJ/kg

10 0.1

515.5 249.6

72.105 34.485

kJ/kg

sfmkJ/kgK

sg kJ/kgK

362.406 332.975

0.14755 0.089

0.5158 0.6604

fin

Solution:— Refer to Fig. 2.19. Processes a-b'-c-a and 1-2'-3-1 show the actual mercury and steam cycle respectively. For mercury cycle : so = Sb = 0.5158 = 0.089 + x (0.6604 — 0.098) x—

0.5158 — 0.089 — 0.7469 0.6604 — 0.089

Enthalpy at b = hb = hatf +x(hg — hod hi) = 34.485 +0.7469 (332.975 — 34.485) = 257.429 kJ/kg Here ha = 362.406 kJ/kg Hence isentropic work done = 362.406-257.429 = 104.977 kJ/kg of mercury

Rankine And Binary Vapor cycle.

79

S

Fig. 2.19. Binary-Vapor Cycle Actual work done by mercury turbine = ha —

= 104.977x0.7409 = 77.777 kJ/kg

= 362.406 — 77.777 = 284.629 kJ/kg of mercury Heat supplied in the mercury boiler per kg of mercury ha - h = 362.406 — 34.485 = 327.92 kJ/kg Heat rejected by mercury in mercury-condenser hb' - hmc = 283.999 — 34.485 = 250.144 kJ/kg For Steam Cycle : From. Mollier chart : hi = 2792.5 kJ/kg, h2 = 1876 kJ/kg Actual work done in the cycle = (hi — h2)tir = hi — h2 = (2792.5 — 1876)x0.8 = 733.2 kJ/kg of steam Heat supplied to feed water = hi — hfi = 2792.5-150 = 2642.5 kJ/kg of steam Heat rejected by mercury in condensing = Heat received by feed water mm(hb — ha,d = 1 x (h1-17J3) (a)

m

2642.5 — — 10.575 kJ/kg of steam m 250.144

Ans.

Total work done in the cycle = 78.406 x 10.575 + 733.2 = 1576.06 kJ/kg of steam 1576.06 Total workdone — 45.44% (b) h Heat supplied in mercury boiler 327.92x10.575 Ans. (c) Work done by mercury turbine = 77.777x10.575 = 822.49 kJ/kg steam Work done by steam turbine = 733.2 kJ/kg of steam EXERCISE Viva Voce and Theoretical 2.1.

Ans.

What is the reason that the Carrot cycle is not a practical cycle for steam turbine plant even though its efficiency is higher?

2.2. What is the effect of maximum pressure, superheater and back pressure on the

Steam & Gas Turbines And Power Plant Engineering

80

thermal efficiency of steam turbine power plant? 2.3. Why is the water used as working fluid in steam turbine plant? 2.4.

Is it possible to operate the steam turbine with steam having temperature and pressure higher than critical?

2.5. What are the principles and methods for increasing the thermal efficiency of Rankhie cycle? 2.6. What do you mean by the critical temperature and pressure of water? 2.7. What do you mean by binary-vapor cycle and in what case is it required? 2.8. What are the requirements of an ideal working fluid in steam turbine? 2.9. What are the advantages and disadvantages of mercury as working fluid? Numerical 2.10. Steam is supplied to a turbine at pressure 11 bar with a dryness fraction of 80%. The steam is expanded isentropically to pressure 0.07 bar. Calculate the following : (a) Heat supplied (b) Total change in entropy (c) Heat rejected (d) Theoreticalthermal efficiency. Ans : (a) 2213.326 kJ/kg, (b) 5.12 kJ/kg-K (c) 1589.779 kJ/kg (d) 28.3%. 2.11. Two steam turbine A and B operate under the following conditions : Initial pressure : 17 bar, exhaust pressure, 0.046 bar. In turbine A, the steam in initially dry and saturated, whereas in turbine B it is initially superheated by 94.5°C. Calculate —(a) Theoretical steam consumption, kg/kWh, (b) Theoretical thermal efficiency of the two turbines, (c) Percentage reduction in steam consumption due to superheating, (d) Percentage increase in thermal efficiency due to superheating. Ans. : (a) 4.28, 3.73 (b) 31.9% (c) 12% (d) 3.18%. 2.12 Two steam turbines operate between the same temperature limits. i.e., 427°C and 26.5°C. In turbine A the initial pressure is 13.5 bar and in B it is 68 bar. Calculate the theoretical thermal efficiencies of turbine A and B and the value of dryness fraction Ans.: A = 34.56%, 0.86 & B = 40.82%, 0.758. after isentropic expansion. 2.13. Steam at 13.5 bar and superheated to a temperature of 317°C expands isentropically to 0.07 bar. Find the pressure at which the steam is just dry and saturated, the final dryness fraction and the isentropic heat drop. Ans : 2.8 bar, 0.835, 882 kJ/kg 2.14. Calculate the fuel cost per kWh generated in a steam power plant under_the following conditions—Initial pressure=20 ba, Initial superheat = 160°C, Exhaust pressure =0.05 bar, Turbine efficiency ratio = 0.78, Boiler efficiency = 0.82, Alternator efficiency = 0.96, C.V. of coal = 29260 kJ/kg, Cost of coal = Rs. 1500 per ton. Assume (a) that the feed temperature to the boiler is equal to the saturation temperature at 0.05 bar. (b) That 6% of the steam generated is used to cover losses and auxiliary power. 2.15. A binary-vapor cycle operates under the following conditions :Pressure in the mercury boiler=17.76 bar, Pressure in the mercury condenser=0.28 bar, The initial steam pressure to steam=50 bar, Steam condenser pressure=0.072 bar, Initial condition of mercury and steam is dry and saturated. Calculate—(a) mass of mercury per kg of steam (b) Thermal efficiency of the cycle.

81

Rankine And Binary Vapor Cycles

p

h fm

h g

bar

kJ/kg

kJ/1s Fig. 3.37

Fig. 3.38

124

Steam & Gas Turbines And Power Plant Engineering At Hot-well : (1 — mi)hf2 +

= hf3

(b)

Putting the value of hi.3 from (b) into (a), we get + (1or

+

h„ — h,, m= ' ' — 1 h1' — h f2

= hn + milt;

417 — 163 2452.5 — 163

— 0.11094 kg/kg of entering steam Ans.

Workdone by the engine and turbine both = (hi) — h1') + (1 — mi) (h1 ' — h2') = 337.5 + 0.889 x 264.38 = 572.53 kJ/kg Heat supplied = 130 — hn = 2790 — 417 = 2373 kJ/kg Thermal efficiency

—572.53 — 24.12% 2373

Ans.

Problem 3.11. Dry and saturated steam enters a steam turbine at 40 bar and exhausts at 0.07 bar. It is planned to use a regenerative feed heating system employing three heaters. (a) Design suitable extraction points and fmd amounts of bled steam and feed water enthalpies at inlet and outlet of heaters. (b) Find efficiency of the regenerative cycle and compare with that of Rankine efficiency. Solution.. Refer to Fig 3.39 and Fig. 3.40. Let us assume regenerative feed heating arrangement with three heaters as shown in Fig. 3.32. Let m1, m2 and m3 kg/kg of steam amount of steam bled off from three heaters respectively. The theoretical design approach is to divide the total temperature interval between the boiler and condenser (or hot—well) into (n + 1) parts for n number of heaters tmployed. (a) Saturation temperature at 40 bar = 250.3°C, Saturation temperature at 0.07 bar =39°C Hence total temperature interval =250.3-39 = 211.3°C —211.3 Temperature interval per heater — 52.825°C

0

\ \LC) AC% 71 )431

S

Fig. 3.39

Fig. 3.40

Regenerative Feed Heating, Reheating

and Water-extraction Cycles

125

Feed water temperature at inlet to heater No. 3= Condenser temperature =39°C = 14 Feed water temperature at exit from heater No. 3 and inlet to heater No. 2 39 +52.825 = 91.825°C =t 3 Feed water temperature at exit from heater No. 2 and inlet to heater No. 1. =91.825 + 52.821 = 144.65°C =t2 Feed water temperature at exit from heater No. 1 =144.65 + 52.825 = 197.475°C=ti In ideal case, the feed water temperature at heater outlets should be same as the extraction temperature. Hence for isentropic expansion of steam in the turbine, the extraction pressure corresponding to heater outlet temperature can be obtained from steam table and similarly the extraction steam enthalpies from Mollier chart.: ho = 2801 kJ/kg First heater, corresponding to ti : pi =14.75 bar, hi =2610 kJ/kg at pi Second heater, corresponding to ti : p2 =4.14 bar, h2 =2400 kJ/kg at p2 Third heater, corresponding to t2 : p2 =0.75 bar, h3 =2160 kJ/kg at p3, h4 =1855 kJ/kg at p4 Hence the extraction points are at pressure 14.75, 4.14 and 0.75 bar

Ans.

Let the enthalpy of feed water be ho, hp and hp at the three extraction points.

At t 4 =39 °C,

hf4 = 163 kJ/kg,

At (3 = 91 825°C,

At t2 = 144.65°C,

hp = 606.67 kJ/kg

At ti = 197.475°C, hfi = 841.25 kJ/kg

hf3 = 384 kJ/kg

Ans. The amount of bled steam may now be calculated from the analysis of individual heaters. Writing the enthalpy balance at heaters : At Heater No. 1: m 1 h 1 +hJ2 =hJ1 +m 1 hJ1 • m

I

—

h

/1/2

' hi— hfi

841.25 — 606.67 2610 — 841.25

— 0.132 kg/kg of steam Ans.

At Heater No. 2 : m2h2 + hf3 + mihri = hp + (mi + m2)hi2 m2 -

—

— m1 (hfl h2 — hJ2

— lin) (606.67 — 584) — 0.132 x (841.25 — 506.67) •

2400 — 606.67

= 0.10689 kg/kg of entering steam

Ans.

At Heater No. 3 : m3h3 + hJ4 + (mi + m2 )hfi = hi3 + (mi + mi + m2 + m3 )hfl d — (mi + m2) (hp — hp) (hp — hi or

m3 =

h3 - hf3

— (384 — 163) — 0.2388(606.67 — 384) = 0.09449 kg/kg entering steam 2160 — 384

(b) Heat supplied = gA = ho - hn = 2801 — 841.25 = 1959.75 kJ/kg

Ans.

126

Steam & Gas Turbines And Power Plant Engineering

Neglecting pump work, work done in cycle = Turbine Work w =qA — qR =

— hi) + (1 — mi) (hi — h2) + (1 — m i — m2) (h2 — h2) + (1 —m1 — m2) (h3 — h4)

R=

(1 — Em) (h4 — hfid = 0.667(1855 — 163) = 1128.56 kJ/kg

w = qA — qR = 1959.75 — 1128.56 = 831.19 kJ/kg w 831.19 Thermal efficiency = (t1d )Reg — 59.75 — 42.41% qA 19 2801 — 1855 — 35.81% ho ha dr)Rank — h h — 2801 — 163 o f4

Rankine efficiency =

Ans.

Ans.

Problem 3.12. Steam is supplied to a turbine at 20 bar and 380°C. The mass flow of steam is 46,000 kg/h. The condenser pressure is 0.05 bar. Steam is extracted at pressure 3.5 bar for feed heating and there is 5% pressure loss in extraction piping. Condensate leaves the condenser and heater as saturated liquid and feed water leaves the heater 5°C cooler than the extraction steam drain which is pumped in to the feed line after the heater by a draM pump. The expansion efficiency upto the extraction stage is 81% while the overall efficiency is 0.84. The blade height of the last stage is 40 cm and the pitch diameter is 150 cm. The direction of flow from last stage of turbine is axial. Calculate (a) Amount of steam bled off., (b) Percentage of the possible temperature rise of feed water in heater., (c) Power developed., (d) Percentage reduction in output due to extraction., (e) Extraction and non-extraction heat rates., (j) Extraction and non-extraction steam rates., (g) Extraction and non-extraction residual or leaving losses. Solution. Refer to Fig. 3.41 and Fig. 3.42. Here pressure pi' = 3.5 x 0.95 = 3.325 bar denotes the pressure in the heater. The expansion of steam in L.P. stage starts from point I'. From Mollier chart, ho = 3225 kJ/kg, hi =2790 kJ/kg, h3 =2160 kJ/kg hi) — hi' = 0.81(3225 — 2790) or hi' = 3225 — 352.35 = 2872.65 kJ/kg h2 =2250 kJ/kg Since the overall efficiency ofexpansion is 0.84, thus — hi') + (hi ' — h2') 0.84 = or

h2'

= 2340.4

(3225 — 2872.65) + (2872.63 — h2')

h0 — h3

3225 — 2160

kJ/kg

From steam table, : hn = 575.95 kJ/kg at 3.325 bar and ti = 137°C ti — 5 = 137 — 5 = 132°C = temp. of feed water. Corresponding to 132°C, h; =554.3 kJ/kg Thus hfl =138 kJ/kg, t2

=

32.9°C

(a) Actual temperature rise in the heater = 132 — 32.9 = 99.1°C Possible temperature rise = 137 — 32.9 =104.1°C

Regenerative Feed Heating, Reheating

and Water-extraction Cycles

127

20 bar 1.3" 1'&

3.5 bar 3.3 bar

s

Fig. 3.42

Fig. 3.41

% Possible temperature rise —

99.1 x 100 = 95.19% 104.1

Ans.

(b) Enthalpy balance at heater : mihi' + (1 — ml)hp = mt hfl + (1 — mi)hp or

(hp' — hp) m — 1 (hit — — hfi + hp') 554.33 — 138 = 0.1534 keiiitbf 46ring steam (2872.65 — 138 — 575.95 + 554.33)

(c) Power developed =

Ans.

— h11) + (1 — m1) (h1' — h21)

= (3225 — 2872.65) + 0.8466(2872.65 — 2330.4) = 811.41 kJ/kg 4600 = 811.41 x — — 10368 kW 3600 (d) Non-extraction output =

Ans.

h3) x 0.84 =(3225 — 2160) x 0.85 = 894.6 kJ/kg

Hence, percentage reduction in output due to extraction _894.6 — 811.41 — 9.299% 894.6

Ans.

(c) Heat supplied in extraction cycle = ho — enthalpy of mixture after heater. Now enthalpy of mixture = him = (1 — mi)hp' + mihp = 0.8466 x 554.33 + 0.1534 x 575.95 = 642.68 kJ/kg Heat supplied in extraction cycle = qA = 3225 — 642.68 = 2582.32 kJ/kg Extraction heat rate =

qA ?it Power

2582.32 x 46000 — 11457 kJ/kWh 10368

Heat supplied in non-extraction cycle = h0 — hp = 3225 — 138 = 3087 kJ/kg x 46000 Non-extraction output = 894.6— 11431 kWh 3600

Ans.

Steam & Gas Turbines And Power Plant Engineering

128

114346000 x - 12422.5 kJ/kWh Non-extraction heat rate = 3087 000 (f)Extraction steam rate = 46 - 4.4367 kJ/kWh 10368 6000 Non-extraction steam rate = 4 - 4.024 kg/kWh 11431

Ans. Ans. Ans.

(g) Extraction cycle :From the continuity equation, itD/Cf2 = my Quantity of steam passing through the last stage = (1 - mi) in = (1 - m1)46000 = 38943.6 kg/h Specific volume of exhaust steam =x x vs = 0.91 x 28.2 = 25.662 m3/kg 38943.6 x 25.662 3600 38943.6 x 25.662 C= C - 147.27 m/s 12 2 3600 x x 1.5 x 0.4

Hence, 71 x 0.15 x 0.4 x Ci2 -

Leaving loss =

C2 2 x 1000

147.272- 10.84 kJ/kg 2 x 1000

Non-extraction cycle :- TC x 0.15 x 0.4 x Cf2 C=C2

12

46000 x 0.91 x 28.2 TC X 1.5 x 0.4 x 3600

Ans.

46000 x 0.91 x 28.2 3600

- 173.95 m/s

c22

- 173 '952 - 15.12 kJ/kg 2 x 1000 2 x 1000 Ans. Problem 3.13. Steam enters a turbine at 60 bar and 600°C. Steam is bled off at 7 bar for regenerative feed heating and the remaining steam is condensed in condenser to condenser temperature 30°C. Calculate (a) the amount of bled steam (b) cycle net work and (c) the ideal thermal efficiency of cycle. For an ideal turbine and with same states, determine (d) ideal turbine work (e) ideal efficiency (j) steam rate in kg/kWh. Solution. Refer to Fig. 3.43 and 3.44. In the absence of specific information thz feed arrangement shown in Fig. 3.37 is assumed. Assuming no undercooling, the condenser pressure corresponding to 30°C = 0.04242 bar. Since cycle net work and turbine work are separately required, so pump work has to be considered in cycle work. From mollier diagram and steam table : ho = 3657 kJ/kg, h1 = 3045 kJ/kg, hn 697 kJ/kg„ h2 = 2170 kJ/kg, h12 = 125.7 kJ/kg. The total pump work may be approximated to feed pump work i.e. work required in pumping from p1 to po Leaving loss =

vfi(po - p2) x 102 =0.0010044 (60 - 0.0242) x 102 = 6.02 kJ/kg Writing enthalpy balance equation : p

Regenerative Feed Heating, Reheating

and Water-extraction Cycles

129.

1 kg Steam ho Turbine 2

mh

hf1 1 kg

®

BFP

-mss

m1 h11

Hot-well Fig. 3.43

Fig. 3.44

and At Hot-well : (1 — mi)hf2 + m1hn = hfi At Heater : mh I 1+h13 = hfl +m1 h mihi + mihn + (1 — in" = hn + mihfi

Ans.

Cycle net work =w.— Ew = (ho — h1) + (1 — mi) (hi — h2) — 6.02 p = (3657 — 3045) + 0.8044 x (3045 — 2170) 6.02 = 1309.83 kJ/kg

Ans.

1

(b)

h„ — h,

697 —125.7 — 0.1956 kg/kg of entering steam 3045 —125.7

(a) m

h — hf2

—

Since after the heater there is no pump, so it will not effect heat supplied. Hence supplied = h0 — hn = 3657 — 697 = 2960 kJ/kg (c) Cycle % —

wnet qA

1309.83 44.25% 2960

Ans.

— h1)+(1 — m1) (hi — h2)

(d) Ideal turbine work =

=(3657 — 3045)+0.8044 (3045-2170) = 1315.83 kJ/kg T

(e) Ideal efficiency w = 1315.85 2960 qA

Ans.

44.45%

(f) Steam rate = 3600— 2.735 kg/kWh 1385.85

Ans. Ans.

(B) Reheating 3.12. Effect of Flow of Wet Steam in Nozzles and Blades. Practically, in the L.P. stages of the turbine the flowing steam will no longer remain dry and saturated. The steam will be in wet condition i.e, there will be suspended water

130

Steam & Gas Turbines And Power Plant Engineering

particles in the steam. Uptil now we have considered steam as a homogeneous fluid and that the steam and the suspended water particles are moving in the same direction and with the same velocity. But in actual practice this is not true. The direction and velocity of wet steam is different than that of dry steam due to difference in specific volumes. The water particles produce a very detrimental effect on the nozzles and the blades and ultimately on the performance while passing through it. Let us consider the expansion of wet steam through a nozzle. Now, a large number of suspended water droplets of varying size will be carried by the flowing steam. There will be exceedingly minute water particles and also larger droplets due to joining of these smaller particles. Let us consider two transverse section at distance Sx apart as shown in Fig. 3.45. Let p and p — Sp be the uniform pressure at section 1-1 and 2-2 respectively. Let A and B having area SA represent the elementary mass of the steam and water particles respectively .B having a prismatic shape instead of natural spherical shape; also assume that all the particles have the same velocity at any moment at a section considered. Let ()sand p„,be the density of steam and water particles in kg/m3 respectively. Elementary mass of steam A = NSA . Sx and elementary mass of water particles B = p„, . SA . ox Accelerating force on the elementary steam mass A = Sp . SA; N Accelerating force on the elementary water particles = Sp . OA; N We see that accelerating force in both case is equal but not the mass. From Newton's Second Law of motion, Accelerating force F= massxacceleration (a) Thus accelerating force on elementary steam mass = Sp . SA = ps SA . Sx . Sp or Acceleration of steam particles = a = — . 1, m/s2 Sx Sp Acceleration of water particles = a = — p S .1 ;

(3.28)

2 111/S

Pw

2

Fig 3.45. Effect of Wet Steam.

(3.29)

Regenerative Feed Heating, Reheating

and Water-extraction Cycles

as ps, vs Dividing (3.28) by (3.29), we get, — = — = aw Ps V

131

(3.30)

Let us compare this ratio at two different pressures. as

—

2.364

— 2.2818 x 103 = 2281.88

At 0.7 bar ,

a

At 0.05 bar,

as = 28.2— 28031.8 aw 0.001006

0.001036

It is thus clear that acceleration of steam is 2281.45 times that of water particles at 0.7 bar and 28031.8 times at pressure 0.05 bar thus a great difference exists at different pressures. This also shows that velocity of steam and water particles at any pressure is not the same. The water particle is accelerated at a much smaller rate than the steam, and soon lags behind. This is so because of the fact that the specific volume of water particles is very little than that of steam at the same pressure. Now, due to the difference in velocity of steam and water particles, there is a certain relative velocity of steam past the lagging particles which introduces a new accelerating force. If Cs and Cm be the velocity of steam and moisture at any point, then (Cs—Cm) will be the relative velocity. The accelerating force, F on the water particles due to relative velocity (Cs — Cm) only, is given by F = y . ps (Cs — Cm )2 . 8s ; N

(3.31)

where Ss —.external surface of the droplet and y = a number, function of Reynolds number. Though, we have assumed the prismatic shape of the water droplets for simplicity, but if the droplets are of spherical type, then external surface area 5, = TC d3 where d is the diameter of sphere in meter and volume is 7C d3/6 Hence, the mass of water particles = mn, pw Ird3/6 We can thus find the acceleration of the water particles caused by relative velocity from Newton's Second Law of motion. Accelerating force = mass x acceleration pw d3 y ps (C — Cm)2 Ss — 6 . a., Y. or

. Ps) (C — C.)2 m/s2 =6 H aw d pH, s

(3.32)

It is obvious from the above expression that the acceleration of the water particles will be least for larger droplets and also least in the low pressure stages where ps is very small and pw is very -very large. It is also clear from the above analysis that in the stages of a turbine where the steam is wet, the velocity of water particles is much less than the expanding steam and also that this velocity difference is maximum when the steam pressure is lowest for the particles of larger diameter. The ratio of moisture velocity to steam velocity varies from 0.1 to 0.3. In

132

Steam & Gas Turbines And Power Plant Engineering

addition to this; the moisture has a detrimental mechanical effect on the blades. 3.13. Velocity Diagram for Dry Steam and Water Particles. Fig 3.46 shows the velocity diagram for an impulse turbine for dry steam and water particles. Let Cm/ = absolute velocity of moisture at exit from nozzle, m/s, represented by AG. C si= absolute velocity of dry steam at exit from nozzle, m/s, represented by AB. C.1 ni = c (3.33) si cm! = relative velocity of moisture at inlet to the blade, m/s, represented by DG. Cr s I = relative velocity of moisture at inlet to blades, m/s, represented by DB. Crm 2 = relative velocity of moisture at outlet from blades m/s represented by DH. Crs2= relative velocity of steam at outlet from blades, m/s represented by DF. Cm 2 = absolute velocity of moisture at exit from blade, m/s represented by AH

C

F

cw$ Fig. 3.46. Velocity Diagram for Dry Steam and Water Particles Cs2= absolute velocity of steam at exit from blade, m/s, represented by AF.

C,„„,= velocity of whirl for moisture. CH, s = velocity of whirl for steam. Crm2

n22 = C

rs2

(3.34)

The main assumptions in drawing the velocity diagram are (a) that the water particles issue from the nozzle in the same direction as the steam and (b) that the relative velocity direction of the dry steam and water particles are the same at the blade outlet. ABD and AGD are the inlet velocity diagrams for the steam and water particles respectively and ADF and ADH are the outlet velocity diagrams for the steam and water particles respectively. We see that the direction of relative velocity C.1 of water particles is not the same as that of steam and thus it strikes the leading face of the moving blades instead of gliding off as shown in Fig. 3.47. Consequently the spherical water droplets are flattened out and broken up into smaller droplets which are then carried away be the rapidly flowing steam. The relative velocity of water particles Crm2 at the blade outlet is shown by DH which is very-very low compared to that of steam Crs2 shown by DF. Thus, the direction Cm2 is not

Regenerative Feed Heating, Reheating

and Water-extraction Cycles

133

the same as that of Cs2 and it strikes the fixed blades as shown in Fig. 3.47 causing impact. Due to the striking of moisture at the in- Crm1 let of the moving blades the momentum of Moving the incoming moisture particles is practically blades destroyed at inlet and hence the momentum of dry steam is only available. To calculate the value of Crm2, we assume that x is the dryness fraction of steam, then x. Ci.si will be momentum per kg of the stuff i.e. of the steam and moisture. The momentum at outGuide vanes let, assuming homogeneous flow K.x.Crsi where K is Crs2/Crsi. If we equate the momentum of the steam and moisture at outlet from the blade channel, then we can find out Fig 3.47. Diagram Showing Oblique Impact of Water Drops on Moving Blades Crm2 and C rs2 and Nozzle Guide Vanes. x.Crs2 + (1 — Crm 2 =X . K.Crsi or

and

Crs 2 [X + (1 —x)n2 ] = X . K.Crs I X K.Crs I Cr s 2 — X + (1 — x) n 2

(3.35)

n2 . X . K Crs Cr in 2 = X +(1 —x)n2

(3.36)

Equations (3.35) and (3.36) give the value of Crs2 and Crm2 and hence Cs2 and Cm2 can be drawn. The tangential force on the blades per kg of the stuff (i.e. steam and moisture) per second is given by.

Ft = x Cos + (1— x)Cw. Here the sign of C. is negative because Cm2 is greater than Cmi .The workdone per kg of stuff is given by W T = [X Cw s

V• •

(3.37)

12 = p2 = 44.6 K = 0.85 n2 = n

5

10

15

Percentage wetness Fig 3.48. Reduction in Blade Efficiency due to water particles in steam.

'134

Steam & Gas Turbines And Power Plant Engineering So we see that work done decreases due to the presence of water particles. The blade efficiency —

Work done per kg of stuff Incoming energy per kg of stuff

The incoming energy per kg of stuff=

x C.!1 (1 — x) Cm2 2

2

2 u[x Cs,s + (1 — x)

Therefore, the blade efficiency =

(3.38) Fig. 3.48. shows the reduction in blade efficiency due to moisture of steam. We see that for a fixed value of rti blade efficiency decreases as the moisture content increases. The stage efficiency of wet steam, given by Dr. K. Baumann is 115 =x .1ls

.(3.39)

Where x is the dryness fraction and is is the stage efficiency of dry steam. 3.14. Correcticin to Condition Curve for Wetnesi. The condition curve of the wet steam would not be same as that of dry steam and so it needs. correction. If AhciandtAhc2 are the cumulative heat drop in the superheat part and wet part of expansion, then

Ahd =

hi — h,

hc 42

and Ah 2 = „

where is is the constant stage efficiency.

15

The work done in a wet stage is x . Ah,sen where x is the dryness fraction of steam and Ahise„ is isentropic heat drop in that stage. Fig. 3.49 shows the values of E Ahisen and x plotted against saturation temperature for wet stages. It is obvious that the relations are linear: Hence, from this relationship, internal work done in the turbine is given by. -

Ah,2 =115 oho

x.• Ahc2)

Where x„, is the mean dryness fraction in all wet stages and A h,1 and A h,2 are the internal workdone in dry stage and wet stage respectively. In the Fig. 3.50 the dotted line shoivs the actual condition line in the wet stage because the internal work done in wet stage is less than the dry stage. So modified condition curve will give, for a constant dry stage efficiency, slightly higher reheat factor and slightly greater specific volumes in L.P. 0. 2 .c C'Y sat' ration !t ph., i

ca

2"

0

steam tvnperature (°C) Fig.-3.

Correction In Condition for Wetness.

3. 50. Relation of Eihisen x for wet steam. '1-

Regenerative Feed Heating, Reheating

and Water-extraction Cycles

135

stages. Problem 3.14. The following particulars refer to a low- pressure stage of an impulse steam turbine : Nozzle angle = 24°, Dryness fraction at nozzle outlet = 0.9, Steam velocity at nozzle outlet = 470 m/s, Mean velocity of water particles at nozzle outlet = 70 m/s, Mean blade velocity = 220 m/s, Ratio of relative velocity at outlet to relative velocity at inlet with homogeneous flow = 0.87, Steam flow = 75 kg/s If the blades are equiangular, calculate the power developed in the blades and the blade efficiency ignoring carry-over effects : (a) Assuming homogeneous flow, and (b) Assuming non-homogeneous flow Assume that there is no rebound at the inlet edge of the blades, that the momentum of the water particles referred to the relative velocity is reduced to zero at impact, and that the relative velocity of water particles at outlet is 0.15 times that of the dry steam. Solution. Refer to Fig. 3.51 and 3.52. Given, al= 240, C Si =470 m/s, x = 0.9, C., =70 m/s,u =220 m/s,

Crm 2

— 0.15 K =0.87 m = 65 kg/s, R1 =132

""rs2

(a) Assuming homogeneous flow. With the given value of u, a1 and C51 the inlet velocity triangle ABD can be drawn. This gives pi = 44° = P2 . With 132 = 44° and C „.2 = 0.87 C,.51, the outlet velocity triangle can be completed. It given CH, = CE = 370 m/s m Power developed — =

.0 1000

Blade efficiency = 11b =

65 x 370 x 220 1000

5291 kW

. u 2x 370 x 220 _ 2. = (Cs (470)2

Ans.

Ans.

(b) Assuming the non-homogeneous flow the inlet velocity triangle ABD for steam will remain the same. Taking AG=Cm =70 m/s, the inlet velocity triangle AGD can be drawn for moisture particles. The relative velocity of steam at outlet of blades can be calculated by the equation. x K.Crsi Crs2

X + (1 — x)n2

0.9 x 0.87 x 285 0.9 x 0.1 x 0.15

243.8 m/s

From the diagram, Pi = 42°. Now with the help of u , P2

= 42° and C„2 the outlet

Cw = 370m/s

C

D u 44° (Jo.'

S2 A

B

F Scale : 1cm = 100 m/s Fig 3.51

B

Scale : 1cm = 50 m/s Fig. 3.52

136

Steam & Gas Turbines And Power Plant Engineering

velocity triangle ADF can be drawn. We have,

Crm2 =n2. Crs2 = 0.15 x 243.8 = 36 .58 m/s

Hence, the outlet velocity triangle of the ADH can be drawn, C, = CE = 390 m/s = HG = — 126 m/s considering the sign Work done per sec = m {x . cs+(1—x 220 Power developed =65 (0.9x390-0.1x126) 1:13 — 4839.12 kW 2u Ix . CH, s + (1 — x ) Cu.} _ 2x 220 { 0.9 x 390 — 0.1 x 126} _ xC120

—x)C2.1

0.9 x 4702 + 0.1 x702

Ans.

0 74.7 /0 Ans.

3.15. Erosion and Corrosion of Blades. The presence of water particles in steam not only reduces the nozzle and blade efficiency but also brings about a serious erosion and corrosion of blades. Erosion of turbine blading is the wasting away or grooving of the back of the inlet edge of the blade, and only occurs where the moisture in steam exceeds about 10% and the blade speed is high.

We have seen that the water particles strike the leading surface of the blades instead of gliding smoothly to produce an impact on the blades, and if this impact is sufficiently heavy, severe local stresses develop in the blade material, causing the surface metal to fail and flake off. The water particles also corrode the metal surface. In the last one or two stages of the turbine where steam is the wettest, the erosion is more likely to occur. The erosion problem is the mostly affected by the two factors (i) Wetness steam (ii) peripheral velocity. (i) Wetness of steam. It has been found that a nominal wetness fraction exceeding 10 per cent produces severe erosion of the blades and the fmal wetness depends on the initial temperature and pressure of steam, the stage efficiency and the exhaust pressure. (ii) Peripheral velocity. It has been found that the intensity of pressure produced by the impact of water droplets does not depend on the size of the drops but is directly proportional to the velocity of the impact and this velocity of impact depends upon the blade speed. If the blade tip speed is very high, severe stresses are set-up by the impact of water droplets. The erosion is worst towards the tips of low pressure blades because centrifugal force tends to concentrate the water particles in the outer annulus, the the tip speed is greater than the root speed.

3.15.1. Prevention of Erosion and Corrosion. The following are the methods to combat the erosion and corrosion problems in steam turbines :— (i) Raising initial temperature. Raising the initial temperature of steam so that the wetness never exceeds 10%, even in the last stage of the low-pressure turbine. (ii) Reheating. Remove the whole of the steam before the moisture content in the expanding steam approaches 10% and reheat it in a reheater so that the wetness at the exhaust does not exceed 10%. (iii) Water extraction. By passing of whole of the steam through a separator where the steam is appreciably wet. (iv) Shielding leading edge. The leading edge of the blade may be shielded of hard material like tungston.

Regenerative Feed Heating, Reheating

and Water-extraction Cycles

137

In practice, generally (1) and (ii) are preferred but (i) is limited by the metallurgical condition. Sometimes (iii) and (iv) are also used. 3.16. Reheating of Steam. By reheating of steam erosion and corrosion dculties due to the presence' of water particles in the steam are avoided because of the fact that the steam becomes dry after reheating. This is done by taking out the whole of the steam from the turbine at a suitable point (i.e. where the steam tends to become wet or before reaching 10% wetness and it is reheated i.e. a further supply of heat is given to it at constant pressure by the flue gases or

live steam until it is superheated to near about the same initial temperature, after which the steam is re-admitted to the turbine and expanded to the condenser pressure. Generally, the expansion of steam is carried out in several stages and the steam is reheated by addition of heat between the stages as suitable points. Any power plant using reheating of steam also employs regenerative feed heating. The reheater may be incorporated in the walls of the main boiler or it may be a separately fired superheater. Double reheating may be employed. A schematic diagram of a theoretical single stage reheat cycle and its corresponding representation of ideal reheating process on T-s and h-s chart are shown in Figs. 3.53 & and 3.54. In Fig. 3.54 (a), 6-1 shows the formation of steam in the boiler. Steam at pressure pi and temperature ti represented as state point 1 enters the turbine and is expanded isentropically to a certain pressure /32 where the state point is 2 and temperature t2. At this point the whole of the steam is taken out and is reheated in a reheater to temperature t3(t3 t1) and is represented by state point 3. After this the steam is re-admitted to the turbine in which it is expanded isentropically to condenser pressure P. Assuming 1 kg of steam flowing through the cycle let us compare the thermal efficiency with and without reheating. With Reheating, Neglecting Pump Work Heat supplied = (hi - 1114) + (h3 - h2) kJ/kg. where h./4 =11.13 = enthalpy of water at exhaust pressure p3 (bar) Heat rejected = qR = h4 - hi4 kJ/kg. Work done by the turbine = qA - qR = (h1 - hi4)+ (h3 - h2 ) - (h4 -1:14) Reheater

Steam out

W P

BFP Fig. 3.53. Reheat Cycle

Steam & Gas Turbines And Power Plant Engineering

138

= (h1 — h2) + (h3 — hi)

(3.39)

Also, Wm = wT = (h1 — h2) + (h3 — h4)

(3.40)

Hence, the theoretical thermal efficiency of the reheat cycle is (h1 — h2) + (h3 — OldReh

— hid + (h3 — h2)

(3.41)

Without Reheating : wnet = wT = (hi —

(3.42)

Heat supplied = (hi — h14)

(3.43)

Increase in work done due to reheating

T

wp

S

S

Fig. 3.54 (a). Reheated Turbine Process on T—s Chart

Fig. 3.54 (b). Reheated Turbine Process on h—s Chart.

So, the theoretical thermal efficiency without reheating i.e., Rankine is (h, — h7) OrdRan - (h _ h ) i f4

(3.44)

We see from the equation (3.40) and (3.42) that the work done by the turbine is more in the case of reheating as compared to without reheating and it is obvious from the T— s hatched diagram. With Pump Work : If pump work has to be considered, then the result will slightly differ. Since pump work is very small (2 to 2.5% of turbine output), it is generally neglected. Considering pump work, lib with reheating is (r1111 )Reh

(hI — h2) + (h3 - h4) - W p _ f4 h) (h3 — h ) — 2 wp

(3.45)

where w = Vb (pi — Pb) x 102 kJ/kg and pi` and pb are iplet and back pressure in bar. p 3.17. Practical Reheating and Non-reheating Cycles. (I) Practical Reheating Cycle. Let us consider a practical single stage reheating cycle.

and Water-extraction Cycles

Regenerative Feed Heating, Reheating Pressure loss in valves Po

p

l

139

Pressure loss in reheater P2 / P2' 3

h

4

S

Fig. 3.55. Practical h-s Chart for Single Stage Reheating The corresponding h—s diagram is shown in Fig. 3.55. Let po, p1 and p3 be the stop-valve, nozzle box and exhaust pressure respectively. Let the steam condition before stop valve be represented by 0 then process 0-1 shows the throttling of steam in the stop valve. Process 1-2' shows the actual expansion line before the steam is taken out for reheating. At pressure p2 where the steam tends to become wet as represented by state point 2' the whole of steam is extracted and reheated to temperature t3 . Actually due to resistance in the steam pipe and reheater tubes there is a pressure loss and the steam then re—enters the turbine at a pressure p2 and represented by the state point 3. This drop in pressure is assumed to occur at constant enthalpy and is represented by process 2'-2". Therefore 2'-3 shows the reheating of steam and 3-4' the subsequent expansion of steam to the exhaust pressure p3 . If there is no reheating of steam, then the actual expansion line is 1-5' where the actual dryness fraction is x5', but due to reheating the new dryness fraction is x4' which is greater than x5' . Neglecting the inlet and outlet kinetic energy of steam we can find the total internal work in both cases (a) ignoring the effect of wetness and (b) allowing the effect of wetness. (a) Ignoring First the Effect of Wetness— The total internal work done in the cycle is given by Ah, = (hi — h21) + (h3 — h41) Heat supplied in the cycle = (h1 — hf4)+(h3 — h2') where h14 is the enthalpy of water at pressure p3 The actual thermal efficiency of reheating cycle is given by (hi — h2') + (h3 — w p (r11h)Reh = (h l — h.f4) + (h3 - h21) - Wp If pump work is neglected, wp 0 (h) Allowing the Effect of Wetness : Let us assume before reheating.

(3.46)

Steam & Gas Turbines And Power Plant Engineering

140

Ahcl ' Ahc2 =cumulative heat drop down to and below saturation line. x2' =fmal dryness fraction. Let after reheating : Aho , Ahc4 =cumulative heat drop down to and below saturation line. Let is be the dry stage efficiency and it is uniform for all stage then the total internal work is given by Ahi = is [Ahci +

1 + x'2 \ 2 Ahc2

1 +x'4 L 2 ane4

Ahc3

Ahi (hl — hi 4) (h3 — h2') (3.47) (ii) None-reheating Cycle. When reheating is not done, then this is simple Rankine cycle. Processes 1-5 and 1-5' show the isentropic and actual expansion line respectively, The actual fmal dryness fraction in this case is x5'. (11 th)Reh

(a) First Ignoring the Effect of Wetness in the L. P. Stages. The internal work done is given by (h1 — h51) and the internal thermal efficiency is hi — h 5'

hi — h5'

(r1th) Rank

h 1 — hf h I — hf4 (b) Allowing the Effect of Wetness:— Let Ahct = cumulative heat drop down to saturation line.

(3.48)

Ahc2 = cumulative heat drop from saturation line down to exhaust pressure. x5' = actual dryness fraction at p3. +

So the actual workdone = Ahr = i I Ahci + and the internal thermal efficiency =

5) Ah c2 C l 2x ,

th)Rank

—

(h1 — h14)

(3.49)

(3.50)

3.18. Advartages of Reheating (I) Erosion and corrosion problems in the steam turbine are avoided. (ii) The output of the turbine is increased. (iii) The thermal efficiency is increased. A gain of 4 to 7% of thermal efficiency over an equivalent non-reheat cycle is possible. Fig. 3.56 shows the gain in thermal efficiency due to single stage reheating. (iv) Final dryness fraction is increased. (y) Nozzle and blade efficiencies are increased. 3.19. Disadvantages (0 Relative to the expenditure incurred in reheating the thermal efficiency does not increase appreciably.

and Water-extraction Cycles

Regenerative Feed Heating, Reheating

1

1 1 I I 1 1 1 — p : 135 bar, ti= 450°C

10 9 w 7.

1

p : 0.035 bar, r. temp. = t3= 450°C Optimum reheating pressure 14.6 bar

I 8 7

Approximate actual gain (allowing for wetness of steam)

6 .

1

1

1

1

141

5 Optimum reheating pressure 4.5 bar Thermodynamic_ gain I

I

I

I

I

I

l

0 10 20 30 40 50 60 70 Percentage of heat drop at which reheating takes place Fig. 3.56. Curves Showing the Gain in Thermal efficiency due to Single Stage

Reheating at Various Pressures.

(ii) Maintenance is more. Note. It is to be noted that there is a particular pressure in the expansion at which it is most economical to reheat the steam. If the steam is reheated early in its expansion then the additional quantity of heat supplied will be small, consequently the gain in thermal efficiency will be small. If the reheating is done at a fairly low pressure, then, although a large amount of additional heat is supplied, the steam will have a high degree of superheat as is obvious from the Mollier diagram, thus a large proportion of the heat supplied in the reheating process will be thrown to waste in the condenser. So reheating should be done at the optimum pressure. In high pressure steam turbine, as is obvious from the Mollier diagram the initial enthalpy content at any superheating temperature is less, corresponding to moderate or low pressure at that temperature. During expansion, up to certain H.P. stages only, the steam will be wet, so it is essential to have reheating arrangement in H.P. turbine than the turbine working with moderate or low pressure. The clue for the solution of problems on reheating is to know the properties of steam at various states represented on h— s chart by possible methods. Problem 3.15. Steam is supplied to a turbine at a pressure of 32 bar and a temperature of 410°C. It expands isentropically to a pressure of 0.08 bar. What is the dryness fraction at the end of expansion and the thermal efficiency of the cycle ? Calculate the modified exhaust condition and thermal efficiency if the steam is reheated at 5.5 bar to a temperature of 395°C and then expanded isentropically to a pressure of 0.08 bar. Solution. Refer to Fig. 3.57 and 3.58. From Mollier chart : h1 = 3250 kJ/kg, h2 = 2175 kJ/kg, x2 = 0.83 Ant Workdone = h1 — h2 = 3260 — 2175 = 1085 kJ/kg

Steam & Gas Turbines And Power Plant Engineering

142

From the steam table the enthalpy of water at 0.08 bar =

= h.f2 — 174 kJ/kg Heat supplied = hi — hf2 = 3260 — 174 = 3086 kJ/kg

Thermal efficiency Work done 1085 — 35.15% - Heat supplied - 3086

Ans.

(b) In the modified case the reheating of steam takes place at 5.5 bar to a temperature of 395°C (Fig. 3.58). From Mollier chart hi = 3260 kJ/kg, h2 = 2830 kJ/kg , h3 = 3265 kJ/kg, h4 = 2420 kJ/kg ,

S

Fig. 3.57

Work done = (h2 — h2) + (h3 — h4) = (3260 — 2830) +(3265 — 2420) =1275 kJ/kg Heat supplied = (hi — hi4) + (h3 — h2) = (3260 — 174) + (3265 — 2830) =3521 kJ/kg 1275 Thermal efficiency = (7-1th-)Reh = 3521 — 36.21%

Ans.

Exhaust condition of steam = x4 = 0.935

Ans.

The results show that the workdone, the dryness fraction and thermal efficiency in-

s

Fig. 3.58 crease as a result of reheating of steam. Problem 3.16. A steam turbine is divided into two sections, H.P., and L.P. with a reheater interposed in between the two sections. The steam on its way to the turbine at 30 bar abs. and 500°C passes through a reheater where it gives up heat at constant pressure to heat the steam flowing from the H.P. turbine to the L.P. section. The steam then enters the H.P. turbine at 30 bar abs. and 380°C. The steam leaves the H.P. turbine at 7 bar and the L.P. turbine at 0.07 bar abs. Assuming no loss of pressure between the two sections of the turbine and an internal efficiency of 0.8 for both the sections, determine the steam condition at entrance to the L.P. section and the thermal efficiency of the plant. Solution. Refer to Fig. 3.59 and 3.60. Given, pi = 30 bar , p3 = 7 bar, 11 = 500°C, p4 = 7 bar , p2 = 30 bar, p5 = 0.07 bar, t2 = 380°C , s, =0.8

and Water-extraction Cycles

Regenerative Feed Heating, Reheating

143

-C

5 From boiler 1 kg/s

1

Fig. 3.60

Fig. 3.59

Assume 1 kg of steam flowing through the turbine. From the Mollier chart : hi =3456 kJ/kg, h2 =3180 kJ/kg, h3 =2832 kJ/kg Since the internal efficiency of the turbine is 0.8, so actual work in H.P. turbine h2 — h3' = rl. (h2 — h3) = 0.8(3180 — 2832) = 278.4 kJ/kg h3' = 3180 — 278.4 = 2901.6 kJ/kg Assume the efficiency of reheater be 100% i.e. the heat given by the heating steam must be equal to the heat taken by the steam coming from H.P. turbine. hi — h2 = h4 — h3' .or (3456 — 3180) = h4 — 2901.6 or

h4 = 3177.6 kJ/kg

With the value of h4 and pressure line 7 bar the point 4 can be located. Hence, the condition of steam at inlet to L.P. turbine are 7 bar and 360°C Ans. Again, from Mollier chart, — h5 = 3177.6— 2330 = 847.6 kJ/kg and ki — h5' = 0.8 x 847.6 = 678.08 kJ/kg. Total work done from the plant = (h2 — h3') + (h4 — h51) = 287.4 +678.08 = 956.48 kJ/kg Heat supplied = hi — hj5 = 3456 — 163 = 3293 kJ/kg Thermal efficiency of the plant = (ri - •th- Reh

956.48 3293

29.04%

Ans.

Problem 3.17. Steam is supplied to a turbine at a pressure of 52 bar and 420°C. It is expanded in a H.P. turbine to 5.5 bar, the internal efficiency ratio of the H.P. turbine being 0.8. The steam is reheated at constant pressure by saturated live steam and is returned to L.P. turbine at 5.5 bar and 250°C. It is then expanded to 0.035 bar with the same internal efficiency ratio. Calculate the mass of steam generated by the boiler per kWh out-put and the heat consumption of the turbine per kWh if the mechanical efficiency of the turbine is 99% and the alternator efficiency is 96%. Assume that the heating steam gives up heat by condensing only. Solution. Refer to Fig. 3.61 and 3.62.: Given pi = 52 bar, ti = 420°C, p2 = 5.5

Steam & Gas Turbines And Power Plant Engineering

144

—mss Fig. 3.62

Fig. 3.61

bar, 1, = 0.8 , p4 = 0.035 bar, p3 = 5.5 bar , rim = 0.99 , t3 = 250°C , rlai = 0.96 From Mollier chart : h1 = 3245 kJ/kg, h3 = 2900 kJ/kg, h2 = 2710 kJ/kg, h4 = 2122kJ/kg ActualworkdoneinH.P.turbine= hi — h2' — h2) = 0.8 (3245 — 2710) = 428 kJ/kg

=

Actual work done in L.P. turbine = h3 — h4' =

(h3 — h4) = 0.8 x (2900 — 2122) = 622.4 kJ/kg.

Alternator output = {(hi — h21) + (h3 —

m x la/

= (428 + 622.4) x 0.99 x 0.96 = 998.3 kJ/kg If Mi be the rate of steam consumption per kWh through the turbine, then = 3600 m — 3.606 kg/kWh. 98.3 1 Let M2 amount of steam be used for reheating purpose. According to question the saturated steam at 52 bar gives its latent heat only to reheat the expanding steam: Assume the efficiency of heater is 100%. h21 = 3245 — 428 = 2817 kJ/kg. Enthalpy balance at reheater : in1(h3 — h'2) = in2(h — hi) = in2L = in2

.

Latentheatat52bar = hfg = 1625.4 kJ/kg in 2 =

h2')

hfg

3.606(2988 — 2817) _ 0.295 kg/kwh 1625.4

Hence, the mass of steam generated by the boiler per kWh output is

= ini + in2 = 3.606 + 0.295 = 3.901 kg/kWh

Ans.

Heat consumption = ini(hi — hi-4)+ i12 hfg = 3.606(3245 — 112) + 0.295 x 1625.4 = 11777.09 kJ/kWh

Ans.

Regenerative Feed Heating, Reheating

and Water-extraction Cycles

145

Problem 3.18. Steam is supplied to a turbine equipped with two stage reheating at pressure 200 bar and temperature 500°C. At pressure 35 bar and temperature 280°C, whole of the steam passes through the first reheater and comes out at a pressure 34 bar and temperature 480°C to expand in the next stages. Again, at pressure 3 bar and temperature 180° whole of the steam is taken out and made to pass through the second reheater. The steam comes out from the second reheater at 2.5 bar and temperature 480°C and expands in the L.P. stage to pressure 0.04 bar and dry and saturated. Calculate— (a) Ideal and actual heat supplied to turbine, (b) Ideal and actual work done by turbine, (c) Ideal and actual efficiency of turbine, (d) Ratio of actual to ideal work of turbine, (e) Ideal cycle efficiency for the same state points. Solution. Refer to Fig. 3.63. Pressure and temperature give actual condition.

=ms

—ms s

s

Fig. 3.64

Fig. 3.63

From Mollier chart and steam table : hi = 3245 kJ/kg, h2 = 2884 kJ/kg, h4 = 2778 kJ/kg, h3 = 3405 kJ/kg, h3' = 2925 kJ/kg , h4 = 2778 kJ/kg, h4' = 2820 kJ/kg, h5 = 3448 kJ/kg, h6 = 2520 kJ/kg, h6' = 2550 kJ/kg , hj6 = 121.41 kJ/kg (a) Ideal heat supplied will be obtained by neglecting the losses in turbine. Ideal heat supplied = (h1 — hf6) + (h3 — h2) + (h5 + h4) = (3245 — 121.41) + (3405 — 2884) + (3448 — 2778) = (4314.59 kJ/kg

Ans.

Actual heat supplied = (hi — hf6) + (h3 — h2') + (h5 — h41) = (3245 — 121.41) + (3405 — 2925) + (3448 — 2820) = 4231.59 kJ/kg

Ans.

(b) Ideal work done = Ideal heat supplied — Ideal heat rejected. = 4314.59 — (2520 — 121.41)=1916 kJ/kg

Ans.

Actual work done =4231.59 — (2550 — 121.41 = 1803 kJ/kg

Ans.

(c) Ideal thermal efficiency = 1916 4314.59

Ans.

44.4%

Actual thermal efficiency = 1803 — 42.60% 4231.59

Steam & Gas Turbines And Power Plant Engineering

146

1,16 d) Ratio of actual to ideal work = 1803

0.941

Ans.

(e) It includes pump work also. Whenever the turbine efficiency and cycle efficiency are required separately, it means pump work has to be considered in the cycle efficiency. w, = vb(p p2) x 102, where vb at 0.04 bar =0.001007 m3 / kg. Hence, pump work = wpm 0.001007(200 — 0.04)102 20.13 kJ/kg (Wnet)cycle = (Wndlurbine— W p

=1916 — 20.13 =1895.87 kJ/kg

Cycle heat supplied= (a.2)turbine— w , = 4314.59 — 20.13 = 4294.46 kJ/kg Hence, ideal thermal efficiency of the cycle = 412 8, 9% 4 87 6 — 44.14%

Ans.

Problem 3.19. Steam is supplied to a turbine at 100 lytr anu 470°C. The steam while in expansion is taken out at 12 bar, 210°C for reheat inz comes out from reheater at 11 bar and 470°C and expands to 0.07 bar. The steam rate is 3.1 kg/kWh, the generator efficiency is 0.94 and the heat loss through the turbine casing is 2% of the throttle enthalpy. Calculate the turbine internal and thermal efficiency and the approximate condition of the exhaust steam. Solution. Refer to Fig. 3.64. From Mollier chart : hi =3300 kJ/kg, h2 =2766 kJ/kg, h2' =2842 kJ/kg, h3 =3417 kJ/kg, h4 =2367 kJ/kg , Writing the energy balance of the turbine as a system. Hence hi + h3 = h2' + + woo+ gloss or

Whet

=

Turf ine net work =

h2') + (h2 h4') gloss Wnet

3600 = 3.1 x 0.94

(i) 1235.41 kJ/kg

Here, a gloss = 0.02 x hi From equation (1), we have = h4' = hi + h3 — h2 — Whet — gloss to 3300 + 3417 — 2842 — 1235.41 — 0.02 x 3300 = 2573.59 kJ/kg As pressure 0.07 bar and enthalpy 2573.59 kJ/kg, t4' =57°C But at 0.07 bar saturation temperature is 39°C Tence, exhaust steam is superheated by =57 — 39 =18°C hi — h2' Internal efficiency upto extraction point =

h1 — h2

3300 — 2842 — 85.76% 3300 — 2766

Ans.

Ans.

(wodoomoi (hi — h2') + (h3 — h4) tAall efficiency of the turbine —

(wndideal

1235.41 — 77.99% 3300— 2766) + (3417 — 2367)

(h— h2) + (h3 — h4) Ans.

and Water-extraction Cycles

Regenerative Feed Heating, Reheating Thermal efficiency =

147

1235.41 (hi — ho) + (h3 — h21)

Work done Heat supplied

1235.41 — 32.9% (3300 — 163.27) + (3417 — 2842)

Ans.

Problem 3.20. A steam turbine plant operates between the pressure of 180 and 0.07 bar, the initial steam temperature being 430°C. During the expansion the steam is extracted at 26 bar and reheated to 430°C. Due to friction there is a drop of pressure in the reheater. Find the pressure drop in the reheater for which the gain in thermal efficiency due to reheating just vanishes. Assume isentropic expansion throughout. Solution . Refer to Fig. 3.65 and 3.66. Let us first consider the case of without reheating. From h — s chart and steam tablehi = 3035 kJ/kg, h2 %4 1806 kJ/kg, ho =163.27 kJ/kg Workdone by the turbine = hi — h2 = 3035 — 1806 = 1219 kJ/kg Heat supplied = h— hf2 = 3035 — 163.27 = 2871.73 kJ/kg Thermal efficiency =

1219 2871.73 — 42.44%

Now, considering the reheating arrangement (Fig. 3.66) : h2 =2630 kJ/kg The work done by the turbine = (hi — h2) + (h3 — h4) Heat supplied = (ht — Ito) + (h3 — h4) , here h14 =enthalpy of water at 0.07 bar Thermal efficiency = 01/dReh

(hi — h2) + (h3 — h4) — (h i — ho) + (113 h2)

Since due to loss of pressure in the reheater the gain vanishes, so the thermal efficiency in both case is equal. (3035 — 2630) + (h3 — h4) 0.4244 —

(3035 — 163.27) + (h3 — 2630)

h4 -0'5751 h3 = 302.34

(a)

This above equation (a) will be solved by error and trial as there is one equation and

Fig. 3.65

Fig. 3.66

Steam & Gas Turbines And Power Plant Engineering

148

two unknowns h3 and Pressure of reheater exhaust

h3

h4

h4

- 0.5751

20bar

3310

2250

346

23 bar

3300

2210

312.17

h3

Thus, approximately the pressure of steam leaving the reheater is 23 bar The pressure drop in reheater (approximately) = 26 — 23 = 3 bar

Ans.

Problem 3.22. Compare the theoretical heat consumption in kJ/kWh and the fmal wetness fraction of the exhaust steam for the following cycle in which steam is supplied to a turbine at 67 bar and exhaust at 0.035. bar (a) Initial steam temperature 480°C. Overall efficiency ratio 0.82. Alternator efficiency 0.96, (b) Initial steam temperature 480°C. Steam reheated to 480°C. Steam pressure entering reheater 7 bar, steam pressure entering turbine from reheater 6 bar. Overall efficiency ratio of each turbine 0.82. Alternator efficiency 0.96. Solution. Refer to Fig. 3.67. Case (a) Given : p1 = 67 bar,010 = 0.82 , ti = 480°C, rlat = 0.96, p2 = 0.035 bar From Mollier Chart : hi = 3362 kJ/kg, h2 = 2025 kJ/kg, hi — h2 = 1337 kJ/kg Since the mechanical loss is not mentioned, so the overall efficiency will be the internal efficiency. h2 — h2' = 0.82 x.1337 = 1096.34 kJ/kg or

/12

=

3362 — 1096.34 = 2265.66 kJ/kg

With the value of h2' and pressure line 0.035 bar the point 2' can be located on Mollier chart. x2' = 0.88 Hence the wetness fraction =1 — 0.88 =0.12 Ans. Alternator output =1096.34 x 0.96 = 1052.48 kJ/kg.

Reheater Fig. 3.67

Regenerative Feed Heating, Reheating

and Water-extraction Cycles

14')

—ms s

s

Fig. 3.68

Fig. 3.69

If in be the steam consumption per kwh, then M x 1052.48 = 3600 . 3600 or m = — 3.42 kg/kWh 1052.48 Heat supplied per kg of steam = hi — ha = 3362 — 111.85 = 3250.15 kJ/kg .

Heat consumption = 3250.15 x 3.42 = 11115.5 kJ/KW

Atr,

Refer to Fig. 3.68 and 3.69. Case (b) Given : pi = 67 bar, ti = 480°C, p2 = 7 bar, p3 = 6 bar, 13 = 480°C, p4 = 0.035 bar, io = 0.82 Form Mollier chart, we have, hi — h2 = 585 kJ/kg hi — h2' = 0.82 x 585 = 479.7 kJ/kg or

h2' = 3360 — 479.7 = 2880.3 kJ/kg , h3 = 3450 kJ/kg , h4 =2387 kJ/kg h3 — h4' = 0.82(3450 — 2387) = 871.66 kJ/kg.

h4

1

=

3450 — 871.66 = 2578.34kJ/kg

Thuspoint4' can be located on the pressure line 0.035 bar, 141 = 45°C Ans.

Degree of superheat = 45 — 26.694 = 18.30°C Alternator output = [(hi — h21) + (h2 — h4')]

= (479.7 + 871.66) 0.96

= 1297.3 kJ/kg If in be the steam consumption per kWh, then

• m =

3600— 2.774 kg/kWh

1297.3

Heat supplied = (hi — hfl) + (h3 — h2') kJ/kg = (3360 — 111.85) + (3450 — 28803) = 3817.85 kJ/kg Hence, heat consumption per kWh = 2.774 x 3817.85 =10590.71 kJ/kWh

Ans.

Problem 3.23. A turbine receives steam at a pressure of 34 bar and superheated by

150

Steam & Gas Turbines And Power Plant Engineering

150°C. It exhausts steam at pressure 0.07 bar. Find the thermal efficiency of the cycle. In a modification of the cycle, the steam is extracted from the turbine at a point where it becomes just dry and saturated and is reheated to the same initial temperature. After further isentropic expansion the steam is extracted at the point where it becomes just dry and saturated, and is reheated to such an extent that it becomes dry and saturated after expansion to 0.07 bar. It is then exhausted and condensed. Find the reheating pressures and the final superheat of the steam as it leaves the second reheater. Also, calculate the thermal efficiency of the modified cycle. Solution. Refer to Fig. 3.70. Process 1-7 shows the isentropic expansion from 34 bar to 0.07 bar. From steam table : tt = t3 = 150 + 240.88 = 390.88°C From Mollier chart : hi = 3200 kJ/kg, h7 = 2120 kJ/kg Work done = hi — 1/7 = 3200 — 2120 = 1080 kJ/kg h = 163.38 kJ/kg (Corresponding to 0.07 bar) Heat supplied = hi — hfl = 3200 — 163.38 = 3036.62 kJ/kg Thermal efficiency = 1080 3036.62

35.56s Ans.

In modified cycle, processes 2-3 and 4-5 show reheating of steam in first and second reheater respectively. • From Monier chart : h2 = 2743 kJ/kg , h5 = 2818 kJ/kg , h3 =3248 kJ/kg , h6 = 2571 kJ/kg , h4 =2617 kJ/kg, The reheating pressure and the temperature of steam leaving the second reheater is 0.25 bar and 150°C Ans. Work done = (hi — h2) + (h3 — h4) + (h5 — h6) = (3200 — 2743) + (3248 — 2617) + (2818 — 2571) = 3742.62kJ/kg Heatsupplied= (hi — hid+ (h3 — h2) + (h5 — h4) = (3200 — 163.38) + (3248 + 2743) + (2818 — 2617) = 3742.62kJ/kg

s

Fig. 3.70

Regenerative Feed Heating, Reheating

and Water-extraction Cycles

Thermal efficiency of modified cycle = (-q th)Reh

-

1335 3742.62

-

35.67%

151

AIn.

Problem 3.24. A high pressure turbine running at 3000 r.p.m. is geared to the same low speed shaft as a low pressure turbine running at 2000 r.p.m. The two turbines devfow equal power and the total useful power developed is 1365 kW. Steam enters the H.P. turbine at 40 bar and 460°C and after that. it is reheated to 410°C and then enters to L.P. turbine. The efficiency ratio of II.P ...P. turbine are 0.76 and 0.82 respectively. The final exhaust pressure is 0.05 bar. C.11r.: the following : (a) Pressure of steam entering to L.P. turbine, (b) Condition of eteam at L.P. turbiac exhaust, (c) Volume of exhaust steam entering to condenser, (d) Exit velocity of steonassumed axial, for a mean blade speed of 280 m/s. of the last row of blading wt•v• blade height to mean diameter ratio is 0.12. Solution. Refer to Fig. 3.71. Process 1 — 2' and 3 — 4' show the actual expa. steam and process 2' — 3 shows the reheating of steam. From chart, ' 3370 Since power developed in both turbine is equal. so, or

(h1 — h2)0.76 = (h3 — h4)0.82

c:

h1

(3376 — h:.).07

— h4' (/73 — h4)0

(a) By error and trial at pressure ! .4 1^ the Shove equation is Now, h2 =2590 kJ/kg.

•3b xJ/kg, h4 =2575 kJ/kg

h3 — h4 =3315 — 2575 =740 kJ/kg or h3 — h4' =740 x 0.82 =606.E 1..j/kg h4 ' =3315 — 606.8 =2708.2 kJ/kg (b) Condition of steam is superheated. The temperature of steam is 100°C •

Ans.

(c) From Mollier chart or steam table : Specific volume of exhaust steam =34.4 m3/k4 Actual heat drop in the H.P. turbine = (3370 — 2590)0.76 = 592.8 kJ/kg Power developed by H.P. turbine —

13965

kW

If in be the consumption of steam in kg/v, then

13965— 11.778 kg/s. 2 x 592.8

Volume entering to condenser =11.778 x 24.4 = 287.38 m3/s

Fig. 3.71

Ans.

152

Steam & Gas Turbines And Power Plant Engineering

(c0

u = 280 —

or

D—

D x 2000

ION 600

280 x 60 it x 2000

60

— 2.673 m

Hence,bladeheight= / = 0.12 x 2.673 = 0.32m = 32cm From the continuity equation at outlet of blade, itD1C1 = inv or

C= I

11.55 x 34.4 — 146.13 m/s. x 2.673 x 0.32

Ans.

(C) Reheat-Regenerative Feed Heating Cycle 3.20. Ideal Reheat-Regenerative Feed Heating Cycle All the modern steam turbine power plants are equipped with reheat and regenerative feed heating arrangement. The advantages of each system has been discussed separately in the previous article. The purposes of combined reheat—regenerative feed heating arrangement are (i) to increase thermal efficiency (ii) to increase output (iii) to avoid erosion and (vi) to avoid corrosion. Fig. 3.72 shows the diagrammatic arrangement of one stage reheating and two stage regenerative feed heating. Its corresponding representations of ideal processes are shown on T—s and h—s diagram in Fig. 3.73. Let 1 kg of steam enters the turbine. The state point 0 represents initial condition of steam entering the turbine i.e. Po is the nozzle box pressure. At state point 1, m1 kg of steam is extracted for regenerative feed heating and the remaining (1 — m i ) kg passes through a reheater where the steam is reheated to state point a, near about the initial temperature. The reheating process occurs at constant pressure pi . At point a steam is re-admitted in the turbine and expands to pressure p2. At state point 2, again m2 kg of steam is blade-off for the second heater for feed heating and the remaining quantity of steam (1 — mi — m2) expands to the back pressure p3. Considering pump work : Ideal heat supplied = qA =

h fi) + (ha — h1 ) (1 — mi)—w kJ/kg p Ideal work done in kJ/kg of entering steam.

SSV TV

0

Reheater (1--m,) kg

mi+m1 hf2

Fig. 3.72. Reheat-Regenerative Feed Heating Cycle

(3.51)

Regenerative Feed Heating, Reheating Po

153

and Water-extraction Cycles

Pi

S

S

Fig. 3.73 (b). Representation of Reheat— Regenerative cycle on h—s Chart.

Fig. 3.73 (a). Representation of Reheat— Regenerative Cycle on T—s Chart.

wT = (1'10 — h 1) + (1 — m i ) (h — h2) + (1 — m 1 — m2) (h2 — h3)

(3.52)

The value of m 1and m 2 will be calculated by writing the enthalpy balance at heater No. 1 and 2 as done in previous article. If pump work has to be considered then, the net work will be turbine work minus pump work i.e., W

But were

net

= W 7urbine

pump

—

pump

= V b(p —pb) x 102 kJ/kg

vb = specific volume of water at back pressure pb (m3/kg).

p1 = initial pressure (bar), Hence, thermal efficiency = rim =

Pb = back pressure, bar.

Net work done Heat supplied

W net

(3.53)

3.21. Actual Reheat-Regenerative Feed Heating Cycle. Fig. 3.74 shows the actual reheat—regenerative cycle on T—s and h—s diagrams of the system shown in Fig. 3.72. In the actual cycle, process 0-1' shows the actual expansion of steam before extraction. In the reheater, there is pressure loss and so steam from the reheater enters the turbine at pressure pa and state point a. Process a-2' shows the actual expansion of steam before m2 kg of steam extracted at pressure /32 and process 2'-3' shows the actual expansion in the L.P. stages. Actual heat supplied = ( Aa = (ho

1n) + (ha — hi s) (1 — m1) — w pa kJ/kg

Work done by turbine = wra = (ho h11)

(3.54)

(ha — 112') 1- (1 — m1 — m2) (h2' h3') kJ/kg (3.55)

where ho — h1 ' = 1, (ho — h1), ha — h2' =

(ha — h2) and (h2' — h3') = it(h2' — h3)

The values of m1 and m2 will be calculated by energy balance as discussed above. The actual thermal efficiency is given by

Steam & Gas Turbines And Power Plant Engineering

154

Pressure p0 Po loss 0

T

Pressure loss 7

Tf1 m2) kg P3

Fig. 3.74 (a). Actual T—s Representation of Reheat—Regenerative Cycle. Po

PO 4-4

Pressure loss

Pressure Pi loss

Pe 1 - m1) kg P2

(1- m1-m2) kg P3

S Fig. 3.74 (b). Actual h-s Representation of Reheat-Regenerative cycle.

wneta 1tha =

Aa

w

where

W

ned

= W Ta — W pa and w = Pa ri p

(3.56)

Generally, ri = overall puihp efficiency = 85 to 90%. If mechanical, generator and p boiler efficiencies are given, these have to be considered. Problem 3.25. A steam power plant equipped with regenerative as well as reheat arrangement is supplied with steam to the H.P. turbine at 80 bar and 470°C. For feed heating, a part of steam is extracted at 7 bar and the remainder of steam is reheated to 350°C in a reheater and then expanded in L.P. turbine drown to 0.035 bar. Determine the following — (a) Amount of.steam blade off for feed heating, (b) Amount of steam in L.P. turbine, (c) Heat supplied in boiler and reheater, (d) Output of the turbine, (e) Cycle efficiency Solution. Refer to Fig. 3.75 and 3.76. From Mollier Chart : ho = 3330 kJ/kg, h1 = 2720 kJ/kg, ha = 3170 kJ/kg, h2 = 2256 kJ/kg,

Regenerative Feed Heating, Reheating

and Water-extraction Cycles

155

Reheater

S

Fig. 3.76

Fig. 3.75

Let m1 kg/kg of entering steam be the amount of steam blade off for feed heating. Writing enthalpy balance equation at heater : mihi + (1 — m1)hfl = lift —

(a)

I

—

11

lift h 1 — hfl

697 — 111.85 - 0.2243 kg/kg of entering steam - 2720 — 111.85

(b) Amount of steam in L.P. turbine

Ans.

= 1 — mi = 1 — 0.2243 = 0.7757 kg/kg of steam

(c) Heat supplied in boiler = h0 — lift = 3330 — 697 =2633 kJ/kg

Ans.

Heat supplied in reheater = (1 — mi ) (ha — hi ) = (3170 — 2720) x 0.7757 =348.75 kJ/kg.

Ans.

Total heat supplied = 2633 +348.75 = 2981.75 kJ/kg

Ans.

(d) Turbine output = (hi) — hi) + (1 — m1) (ha — h2) = (3330 — 2720) +0.7757(3170 — 2236) = 1333.85 kJ/kg. (e) Thermal efficiency =

Output Heat supplied

2193831'8755 — 44.73%

Ans. Ans.

Problem 3.26. A steam turbine equipped with single-stage reheating and regenerative feed heating operates under the following conditions—Initial steam pressure = 100 bar, Initial steam temperature = 530°C, Pressure at which steam is heated = 20 bar, Reheat temperature at inlet to I.P. turbine = 510°C, Feed temperature at outlet from final feed heater = 215°C, Exhaust pressure = 0.055 bar, Steam bled off before reheater = 4% steam entering turbine, Total steam blade off = 21% of steam entering turbine, Internal efficiency of H.P. turbine = 0.83, Internal efficiency of I.P. and L.P. turbines = 0.88, Efficiency of alternator = 0.98. Neglecting heat losses and pressure losses due to reheating, calculate—(a) The heat rate in kJ/kWh, (b) The total volume of steam passing through the exhaust in m3/s if the electrical output is 100,000 kW. Solution. Refer to Fig. 3.77 and 3.78. In the problem, it is not mentioned at what pressure and at what point 17% of the entering steam is extracted, though for the solution it is not needed at all.

Steam & Gas Turbines And Power Plant Engineering

156

h1 = 2987 kJ/kg

From Mollier chart : h0 = 3450 kJ/kg , Since internal efficiency is 0.83, hence,

ho — hi' = 0.83(3450 — 2987) = 384.29 kJ/kg or

hi ' = 3450 — 384.29 = 3065.71 kJ/kg

From Mollier chart, ha = 3493 kJ/kg and from table, lift = 920.63 kJ/kg, h = 144.91 kJ/kg fi Work done can not be calculated from the turbine work equation because condition of extraction of 17% entering steam is not mentioned. We can also find the work done from the difference of heat supplied and heat rejected. Heat supplied = (ho — hf4) + (ha — 170(1 —

1)

= (3450 — 320.53) + (3493 — 3065.71)0.96 = 2939.56 kJ/kg From Mollier chart, h3 = 2287 kJ/kg ha — h3' = 11j (ha — h3) = 0.88(3499 — 2287) = 1061.28 kJ/kg h' 3 = 3493 — 1061.28 = 2131.72 kJ/kg Heat rejected = (1 — m i — 2) (h3' — hi3) = 0.79(2431.72 — 144.91) = 1806.57 kJ/kg Internal work done by the turbine = Heat supplied—Heat rejected =2939.56 —1806.57 =1132.99 kJ/kg —

1132.99— 0.3147 kWh/kg 3600

Heat rate Heat supplied/kg of steam 2939.56 _ 9340.83 kJ/kWh — 0.3147 kWh/kg of steam (b) Electrical output =10,000 kW Turbine output = 100,000— 102040.81 kW. 0.98 Reheater

0.04 kg

0.17 kg

Fig. 3.77

Ans.

and Water-extraction Cycles

Regenerative Feed Heating, Reheating

157

s Fig. 3.78 If M be the mass flow rate in kg/s, then m—

102040.81 — 90.12 kg/s 1132.19

Mass of exhaust steam = (1 — mi — m2) M = 0.97 x 90.12 =71.19 kg/s From Mollier chart, specific volume of steam at 3' = v3' = 25 m3/kg Volume flow rate through exhaust = 25 x 71.19 = 1779.75 m3/s

Ans.

Problem 3.27. Steam enters a turbine at 80 bar and temperature 400°C. At pressure 30 bar some quantity of steam is bled off for regenerative feed heating. The remaining steam expands to pressure 20 bar from where steam is passed through a reheater and comes out at 19 bar and 410°C. After that it expands to condenser pressure 0.04 bar. In actual turbine, the steam enters the reheater at 250°C, the feed water enters the boiler at 225°C. The combined steam rate is 4.5 kg/kWh. The generator efficiency is 0.95. The actual quantity of steam bled-off is 0.26 of the throttle flow i.e. entering steam. Determine, for the ideal turbine--(a) Percentage of steam bled off (b) Thermal efficiency. For actual turbine : (c) Shaft work (i.e. brake work) (d) The brake and electrical thermal efficiency. Solution. Refer to Fig. 3.79 and 3.80. From Mollier chart, ho = 3142 kJ/kg , hi = 2908 kJ/kg, h2 = 2814 kJ/kg, h3 = 3274 kJ/kg, h2' = 2908 kJ/kg, h4 = 2176 kJ/kg From steam tables : hf4 = 121.41 kJ/kg, h'fi = 966.89 kJ/kg at 250°C, hf1 = 1008.4 kJ/kg, nil = 0.26 Ideal Turbine : (a) m ih + (1 — m i)hf 4 = or

hi-1 — hp' 1008.4 — 121.41 m — — 31.83% I h 1 — hf 4 2908 — 121.41

Ans.

(b) Heat supplied = qA = (ho — h f1)+ (1 — m1) (h3 — h2) = (3142 — 1008.4) + 0.6817(3274 — 2814) = 2447.18 kJ/kg Heat rejected = qR = (1 — Mi)(h4 — hf4)= 0.6817 x (2116 — 121.41) = 1400.6 kJ/kg

Steam & Gas Turbines And Power Plant Engineering

158

1 kg

.0

h'11 h f 1 = 225°C h f4

Fig. 3.79

Fig. 3.80.

Work done by turbine = qA — qR = 2447.18 — 14006 = 1046.58 kJ/kg Hence, thermal efficiency = 1046.58 _ 42.76% 2447.18 Actual or brake work = wTa =

3600 steam rate x

(d) Actual heat supplied = qAa =

i

gen

—

Ans. 3600 — 842.1 kJ/kg 4.5 x 0.95

Ans.

3 — h2) (1 — m1 1)

= (3142 — 966.89) + (3274 — 2908) x 0.74 = 2445.95 kJ/kg Hence, the brake thermal efficiency = brake =

Wm ri

-

842.1 2445.95 = 34.42%

Ans.

Therefore, the electrical thermal efficiency is given by = ribrake X rigen = 0.3442 x 0.95 = 32.69% Problem 3.28. A 32 MW turbo-generator receives steam at 40 bar and 390°C. The steam while in expansion at 10 bar and 230°C is taken out for regenerative feed heating and reheating. A part of steam is taken to regenerative heater and the remaining steam is reheated at 9.5 bar. to 360°C and then expands to 0.04 bar with 80% efficiency. Again, second heater is used in between, such that it heats the feed water through half the temperature range between the condenser temperature and the first feed heater water discharge temperature. The actual steam rate is 4 kg/kWh. Find—(a) The extraction pressure for second heater, (b) Ideal cycle steam rate, (c) cycle efficiency, (d) Actual efficiency of the pint Solution. Refer to Fig. 3.81 and 3.82. From Mollier chart : = 3190 kJ/kg, hi' = 2866 kJ/kg = hi", ha = 3184 kJ/kg In the absence of specified feed heating arrangement, let us assume the arrangement shown in the Fig. 3.81. Condenser temperature = 28.983°C at 0.04 bar

Regenerative Feed Heating, Reheating

and Water-extraction Cycles

159

Reheater

Fig. 3.81

S Fig. 3.82

First heater discharge temperature = 179.88°C at 10 bar Temperature range = 179.88 — 28.983 = 150.897°C Half of this temperature range = .

150.897 — 75.448°C 2

Exit temperature of feed from second heater = to = 28.893 + 75.448 = 104.43°C

(a) Hence, pressure corresponding to 104.43°C = 1.1205 bar (b) Now from Mollier chart : h2 = 2688 kJ/kg , h21 = ha — (ha — h2) rli = 3184 — (3184 — 2688) x 0.8 = 2787.2 kJ/kg or

h3 = 2301 kJ/kg h2' — h3' = (h2' — h3) rig = (2787.2 — 23b1) x 0.8 = 388.96 kJ/kg

From steam table : hj3 = 121.41 kJ/kg at 0.04 bar, 11/2 = 440 kJ/kg at 1.1205 bar, hji = 162.61 kJ/kg at 10 bar

Steam & Gas Turbines And Power Plant Engineering

160

Writing enthalpy balance equation At Heater No. 1 : m i hi ' + hn = hf + mi hf m

I

=

— hp 762.61 — 440 — 0.1533 kg/kg of entering steam h 1 ' — hi, 2866 — 762.61

At Heater No. 2 : m2h21 + (1 — mi — m2) hi3 = (1 — mi )hr2 or

m2 =

— m i) p — hp-) _ 1 — 0.1533) (440 — 121.41) — 0.0975 kg/kg h2 ' (2787.2 — 121.41)

Work done = (ho — hi') + (1 — m 1) (h — h21) + (1 —m1— m 2)(h21 — h3') = (3190 — 2866) + 0.8467(3184 — 2787.2) + 0.7491(2787.2 — 2398.24) = 951.18 kJ/kg Hence, steam rate = 3600 — 3.784 kg/kWh 951.18 (c) Heat supplied = (ho — h/i) + (1 —

Ans.

1 )(ha — h1')

= (3190 — 762.61) + 0.8467 (3184 — 2860) = 2696.64 kJ/kg Cycle efficiency =

951.187 lc 1704 — 2696.64

Ans.

(d) Since actual steam rate = 4 kg/kWh, so Heat supplied/kWh = 4 x 2696.64 = 10786.56 kJ Actual thermal efficiency = 3600 — 33.37% 10786.56

Ans.

Problem 3.29. A steam turbine plant developing 120 MW of electrical output is equipped with reheating and regenerative feed heating arrangement consisting of two feed heaters—one surface type on h.p. side and other direct contact type on l.p. side. The steam conditions before the steam stop valve are 100 bar and 530°C. A pressure drop of 5 bar takes place due to throttling in valves. Steam exhausts from the h.p. turbine at 25 bar. A small quantity of steam is bled off at 25 bar for h.p. surface heater for feed heating and the remaining is reheated in a reheater to 550°C and the steam enters at 22 bar in l.p. turbine for further expansion. Another small quantity of steam is bled off at pressure 6 bar for the l.p. heater and the rest of steam expands up to the back pressure of 0.05 bar. The drain from the h.p. heater is led to the 1.p. heater and the combined feed from the l.p. heater is pumped to the high-pressure feed heater and finally to the boiler with the help of boiler feed pump. The component efficiencies are : turbine efficiency 85%, overall pump efficiency 90%, generator efficiency 96%, boiler efficiency 90% and mechanical efficiency 95%. It may be assumed that the feed-water is heated up to the saturation temperature at the prevailing pressure feed heater. Work out the following : (a) Sketch the feed heating system and show the process on T-s and h-s diagrams. (b) Amounts of steam bled off. (c) Overall thermal efficiency of turbo-alternator considering pump work. (d) Specific steam consumption in kg/kWh. and mass of steam in kg/h

Regenerative Feed Heating, Reheating

161

and Water-extraction Cycles

(e) Specific cooling water consumption in condenser in kg/kWh if temperature rise is 10°C. (f) fuel cost per kWh generated in steam power plant, if (LCV)/ = 29000 kJ/kg and cost of coal is Rs 1800.00 per ton and 5% of steam generated is used to cover losses and auxiliary power. Solution. Refer to Fig. 3.83. and 3.84. From Mollier diagram : ho = 3500 kJ/kg = h' , hi = 3050 kJ/kg , h2 = 3165 kJ/kg, h3 = 2310 kJ/kg, ho = 3575 kJ/kg hi ' =

0

-r11(h'0 — h1 ) = 3500 — 0.85(3500 — 3080) = 3143 kJ/kg

100 bar, 550°C Reheater Valve 95 bar

(1—m,) ha Pa

CIR

Fig. 3.83.

pressure loss in valve

Po pc:

pressure loss in valve 1 ka

1' Ai Po

—ms s

pressure Po po loss in reheater Pi Pa

S

(a)

(b) Fig. 3.84.

Steam & Gas Turbines And Power Plant Engineering

162

h2' = ha - it(ha - h2) = 3575 - 0.85(3575 - 3165) = 3226.5 kJ/kg a - h3) = 3575 - 0.85(3575 - 2310) = 2499.75 kJ/kg From steam table : hfi = 961.8 kJ/kg, vfi = 1.4521 m3/kg, hf2 = 670.5 kJ/kg h3 = ha -

h.13 = 137.83 kJ/kg, of = 1.0053 x 10-3 m3/kg Energy balance at heater No. 1 : m1h1' +ha = hp-hp - 961.8 - 670.5 (b) or m1 = h , 3143 - 961.8 nJI I

mihfi

0.1335 kg/kg of entering steam Ans.

Energy balance at heater No. 2 : m2h2' + (1 - mi - m2)h.f3 + mi hfi = hn or

m2

(hf2 - hf3) -

-

-

h2' - hfi

(670.5 - 137.83) - 0.1335(961.8 - 137.83) _ 0.136 kg/kg of steam 3226.5 - 137.83

Ans.

w.0 = (ho - ht')+(1- m1)(ha - h2')+(1- mi - m2)(hi - h3') = (3500 - 3143) + (1 - 0.1335) (3575 - 3226.5) + (1- 0.1335 - 0.136) (3226.5 - 2499.75) = 1192.96 kJ/kg

E v tt v„.3 (p0 - p3) =

1.0053 x 10-3 (100 - 0.05) x 102

10.04 kJ/kg

(qA)a = (ho - 9+ (1 - mi ) (ha - hi') = (3500 - 961.8) + 0.8665(3575 - 3575 - 3143) = 2912.52 kJ/kg (GO)a l oth

(6

or

nap.

Actual generator output where (qA)a= -Al TI• boiler Actual heat supplied

(q) a la

m

TIP

1 IWla g

:J r g %oiler 9A

( gi )e,

0.05) x 0.96 (1192.96 x 0.95 10.9 1077.58 2912.52

2912.52

36.987%

Ans.

(d) Specific steam consumption (ssc) based on G.O. is ssc =

3600

-

W neto 1 g

3600

-

W

Fro% - ri

3600 - 3.341 kg/kWh 1077..58

)Tig

"P

Mass of steam = ms = 3.341 x 120 x 103 = 400920 kg/h (e) Specific cooling water consumption (scwc) is

Ans. Ans.

Regenerative Feed Heating, Reheating

and Water-extraction Cycles

163

= (scwc) (kg/kWh) cpw (kJ/kgK)Ac (K) = (scc) (kg/kWh) qR (kJ/kg) (ssc) (1 — mi — m2) (h3' — hi3) Or

scwc Cr , /

Atw

3.341 x 0.7305(2499.75 — 137.83) _ 137.9 kwkwh 4.18 x 10 (f)

Mf (kg/kWh) x LCV (kJ/kg) x 1 n boiler = ssc (kg/kWh) x

or

hi x 29000 x 0.9 = 3.341 x 2912.52 x 1.05

or

Ans. 0 (kJ/kg) x 1.05

= 0.391 kg/kWh

Ans.

Fuel cost = 0.391 x 1800/1000 = Rs 0.7038/kWh

3.22. Regenerative Water-Extraction Cycle Erosion and corrosion problems in the L.P. stage of turbines are also avoided by the partial or complete extraction of suspended water particles from the steam during its flow through the wet stages. The extracted water particles are used to heat the feed-water hence the name regenerative water-extraction cycle. The larger droplets of water are concentrated in the outer half of the blades because of the fact that due to inertia, the water droplets move radially outwards during their passage through the moving blades. So a certain quantity of moisture is flung off the blade, probably at the outlet. For the purpose of trapping this water, various type of water-catching devices are used. Here the water is cascaded from stage to stage; the advantage of this is that there is a partial recovery of the sensible heat. Now a days complete extraction of suspended water particles is done at a certain definite point. Arrangement is made such that water only is extracted. The cascaded water is driven by pump into the feed pipe to heat the feed water. Fig. 3.85 shows such a regenerative two point extraction system and its corresponding h — s diagram shown in Fig. 3.86. Process 0-1 shows the throttling in the stop valve. Let pi be the pressure in the nozzle

.c Condenser

he 1 kg

Pump hf4

hf5 3 kg x, kg 2 Fig. 3.85. Diagrammatic Arrangement of Regenerative Water- Extraction Cycle.

Fig. 3.86. h-s Diagram for Water Extraction Cycle.

Steam & Gas Turbines And Power Plant Engineering

164

box of first turbine. Process 1-2' shows the actual expansion of steam before first point extraction of water particles. Complete extraction is assumed to take place at pressure p2 and p3. Let the dryness faction at 2' be x2', so the water extracted will be (1 — x2) kg per kg of steam entering the turbine and the steam passing through the turbine will be x2' kg per kg of steam entering the turbine. The actual expansion line in the second stage turbine is 2" 3' and x3' is the dryness fraction at state point 3'. At the second point water extracted is (1 — x31)x2' kg per kg of steam entering the turbine and x2 x3' the amount of steam entering the third stage of the turbine. If we neglect the water extraction pumps work, then by equating the heat entering and the heat leaving at the junction points gives. At junction 2 : x2 hf5 = x2' . x3' . + x2' (1 — x3')hj3 or

11 = x3' hi4 + (1 - x3')hi3 .5

At junction 1 : hf,6 = x2'(/1.5) x (1 — x2')h/2 Putting the value of hfi . h./6 can be calculated Heat supplied = 170 — h./6 = hi — hj6 Internal work done = (hi — h2) + x2'(h2" — h3') + x2ix3'(h3" — h4') So the internal thermal efficiency is (hi — 122) + x21(h2" - h3') + x x 3' (h3" - h4') 11 uh =

110 - J5

(3.57)

Problem 3.30. Steam enters a turbine employed with regenerative water-extraction at 50 bar and temperature 400°C and expands up to pressure 3 bar from where whole of the steam is passed through a separator. The steam comes out from the separator as dry and saturated and expands up to pressure 0.5 bar from where again passes through a separator. From separator steam comes out as dry and saturated and expands upto condenser pressure 0.05 bar. The internal efficiency ratio of the turbine is 0.9. Extracted water particles in the separator is used for feed heating. Compare the work done and thermal efficiency with and without regenerative water extraction. Solution. Refer to Fig. 3.85. for block diagram and Fig. 3.87 for h-s chart. From Moilier diagram : hi = 3220 kJ/kg h2 = 2595 kJ/kg, hf2 = 561.43 kJ/kg, h2" = 2730 kJ/kg, h3 = 2437 kJ/kg, h3" = 2640 kJ/kg, hi3 = 340.56 kJ/kg, h4 = 2322 kJ/kg, /IA = 137.77 kJ/kg Now h1 or

- h ' = 0.9(3220 — 2595 = 562.5 kJ/kg

2

h2' = 3220 — 562.5 = 2657.5 kJ/kg

•

Again, (h2" — h3') = 0.9(2730 — 2437) = 263.7 kJ/kg • Again, or

x2' = 0.97 or h3' = 2466.3 kJ/kg

x ' = 0 924 (h3" — h4') = 0.9(2640 — 2322) = 286.2 kJ/kg h4' = 2640 — 286.2 = 2353.8 kJ/kg

x4' = 0'918

165

and Water-extraction Cycles

Regenerative Feed Heating, Reheating

►s Fig. 3.87. Work done by the turbine using water-extraction = (hi — h'2) + x21(h"2 — h3') + x2' x3'(h"3 — h'4) = 1562.5 + 0.97 x 263.7 + 0.97 x 0.924 x 286.4 = 1074.8 kJ/kg

Ans.

We have from enthalpy balance equation. = x2'[x3' hi4 x (1 — X3')/1.13] (1 — X21)17/2 = 0.97[0.924 x 137.77 + 0.076 x 340.56] + 0.03 x 561.43 = 165.32 kJ/kg Heat supplied in water extraction cycle = hi — hf6 = 3220 — 137.77 = 3082.22 kJ/kg Thermal efficiency with water-extraction. =

1074.8 3082.22 — 34.87%

Ans.

Work done without water-extraction = (hi —

x 0.9 = (3220 — 2050) x 0.9 = 1053 kJ/kg

Ans.

Heat supplied = hi — hfii = 3220 — 137.77 = 3082.23 kJ/kg Thermal efficiency without water-extraction = 1053 — 34.16% 3882.23

Ans.

3.23. Practical Feed Heating Arrangements There are many ways to arrange the various types of heater. Figs. 3.88 to 3.92 show the feed heaters arrangement in various capacities of power plants. As mentioned earlier, feed water de-aeration is an important feature of feed system design. A de-aerator (Direct contact heater) combined with storage tank is arranged in an elevated position on the suction of the fed pump, or in the case of units with split pumping systems on the suctions of the booster feed pump, thus ensuring a positive suction head under all conditions of operation. It is to be noted that the store of de-aerated feed water supplies the boilers demands, particularly during period of changing load on the unit. The water level in the de-aerator is restored by make-up water admitted at the condenser. It is common practice to use h.p. heaters of the surface type, their drains cascaded from heater to heater and their combined drains lifted into the de-aerator by the pressure differential between the first h.p. heater and the de-aerator. So far the 1.p. heaters

Steam & Gas Turbines And Power Plant Engineering

166

are concerned, they are either of the surface type with their combined or of the direct contact type. This arrangement is a subject to variation only when the number of de-superheaters and drain coolers fitted are many. For example, the feed system shown in Fig. 3.90 has the .maximum number of drain coolers, and drains from LP3 is cooled by vapor condenser when the evaporator is not in service. Evaporators are used for make up water needed in the boiler. With larger units it has become the common practice to arrange the h.p. heaters in twin banks or increase the dc. heaters and reduce h.p. heaters. The power absorbed by the boiler feed pump is considerable about 2.5 to 3% of the turbine output for a steam pressure of 155 bar. In small power unit; the feed water pump may be electrically driven, but in larger power plant using reheater, it is now customary to modify the bled steam cycle to incorporate a turbine, powered by steam from the main turbine h.p. cylinder exhaust, to drive the main feed pump. This system gives a gain in thermal and overall efficiency. Table. 3.1 shows the economic selection for feed heating for a number of unit ratings and steam conditions. The average temperature rise across each heater is in the range 28 to 33°C. Table : 3.1. Feed Heating Stages Rating MW 30 60 100 120 200 275 300 350 500 550 660

Turbine Pressure bar

Stop Valve Reheat Temperature Temp. °C °C

Final Feed Temp.

Feed Heating Stages

41 103

454

—

566

—

166 204

4 5

103 103 162

524

224 224 238

6

538 566

510 538 538

158

566

566

252

7 or 8

°C

6 6 or 7

3.24. Feed Heating System for a 120 MW Unit Fig. 3.88 shows a typical feed heating arrangement for a 120 MW steam turbine unit with reheat. The steam conditions at the turbine stop valve are 103 bar and 538°C. The reheat temperature is also 538°C. There are\six stages of feed heating. The overall thermal efficiency obtained is 35%. Following are the salient features of this system :— (a) There are three low pressure and three high pressure heaters-all plain surface type except de-aerator. The maximum temperature rise (38°C) in feed water takes place across low pressure heater No. 1 and the minimum (22°C) across the

Fig. 3.88120 MW Feed Heating System. 280°C _feed 224'C pumps --®—

To A boiler

NO. 5

r N 5 •

tootC

NO 4

Deaerator water tank

Boostel feed pumps

feed line

Turbine

?

Double flow

-r

Extraction tank

AEP

28

Condenser

Surplus V extraction valve tai;:1r b

Turbine

we 4 Iv

Reheater

Drain No. 1 cooler No. 2 100°C NO 69'C ■ 81*C

—__Egonclense

o Deaerator Vent

-o-

125'C

Steam/ from boiler

Live tnp steam tro. 3 LP evaporator

RFW tanks

Emergency

3 L.P. surface heaters 3 H.P. surface heater

160'C

RFW main

Heater by pass Heater automatic by-Pass RFW Reserve feed water

--4/— Orifice plate in drain Line

sap,( uoporma-imem pue

Boiler

Regenerative FeedHeating, Reheating

168

Steam & Gas Turbines And Power Plant Engineering de-aerator. (b) Air pumps have been used to drive the air from the condenser. (c) The feed pump arrangement is of split type. The boiler feed pump has been installed after the h.p. heaters and the booster feed pump has been arranged before h.p. heaters, and enough suction head for the boiler feed pump. Both types of pump are electrically driven and in duplicate. (d) The high pressure cylinder is of double casing so bled steam cannot be taken along cylinder except at outlet. (e) The cascade drains on the h.p. heaters return to the de-aerator via an orifice plate. (f) The feed water from all the heaters may be by-passed manually and if the water level in h.p. heaters gets too high, it may be automatically by-passed.

3.25. Feed Heating System for a 200 MW Unit Fig. 3.89 shows a typical feed heating system for a 200 MW steam turbine unit with reheat. The steam conditions at the turbine stop valve are 162 bar and 566°C. The reheat temperature is 538°C. There are six stages of feed heating producing a final feed temperature of 243°C and giving an overall thermal efficiency of 36.9 percent. Following are the salient features of this system:— (i) There are three low pressure and three high pressure heaters. Except de-aerator all heaters are of surface type. All h.p. heaters are having separate de-superheater and drain cooler. (ii) Number 5 de-superheater is hotter than No. 6 de-superheater and so it is the last heater in the system before the boiler feed pumps. (iii) The cascade drains from all h.p. heaters pass through orifice plates and then into separate drains coolers where more heat is taken up by the feed water. 3.26. Feed Heating System for a 350 MW Unit Fig. 3.90 shows a typical feed heating system of a 350 MW steam turbine unit with reheat. The steam conditions at the turbine stop valve are 165 bar and 568°C. The reheat temperature is also 568°C. Seven stages of feed heating is employed to give a final feed water temperature of 252°C and an overall efficiency of 36.5 percent. Following are the salient features of this system— (i) There are four low pressure and three high pressure heaters. Except de-aerator, all heaters are of surface type. The three l.p. heaters are of plain surface type except the heater No. 1 which has a drain cooler. All the high pressure heaters are of surface type with integral de-superheater and draincooler. (ii) The condensate from the condenser passes through the hydrogen cooler, the generator stator cooler and the low pressure drain cooler for the recovery of the some of the waste heat. (iii) A unit evaporator is used in the system and the steam and water vapor from the evaporator outlet passes through a vapor condenser when it is condensed by the cooler feed water on its way to the de-aerator. The distillate is taken out and is used as a make up water to the boiler. In preparing this distillate the feed water gains a little or more heat. (iv) The 17c-,Per feed pump is driven by a separate steam turbine meant for this purpose only. :27.,e to larger size of the pump, steam driven pump is more economical.

Fig. 3.89 200 MW Feed HeatingSystem. 243°C

-

238°C !235°C 252°C

• •

99"C

L.P.2

12

4°C

32°C

188°C 1 I

166°91-v165°c

130°C

4

Booster 4-pump

.E.P

ondenser

Exhaust

Vent condenser

8e

129°C

Drain L P.1 cooler

A.E.P.

3

Deaerator

Drain cooler Drain cooler H.P. 4 5 6 H.P.5

191°C T

H.P. 6

3 LP surface heaters 3 HP surface heaters 2 DC 2 DS

Orific plate in drain line Heater by pass - -- Heater automatic by pass D.S.: Desuperheater

Steam

D.S. 5 D.S.I 6 Boiler feed 4 pump

Feed water to boiler

I Exhaust

•

--Trr:r5 • • •••

H.P.

Reheatter

Turbine

Double flow

Boiler

Regenerative Feed Heating,Reheating and Water-extractionC ycles

€0Condenser extraction pump

AEP

Fig. 3.90350MW Feed Heating System. D.0

93°C 2

Distillate pump

7 70°C 2

Stator cooler

H cooler

Gland steam condenser

12346 1

Booster pump

Evaporator

L.P.

A

111°C

3 171eC

Raw water

Steam

Turbine

Reheater WV

flow

Double

Deaerator

252°C Boiler feed pump

Bled turbine—,

4 L.P. Surface heaters 3 H.P. Surface heaters D.C. Drain cooler D.S. Desuperheater A.E.P. Air extraction pump

. 140°C

7

Feed water

Steam & Gas Turbines And Power Plant Engineering

Regenerative Feed Heating, Reheating

and Water-extraction Cycles

171

This turbine receives required steam from the outlet of h.p. turbine. For the heater No. 5, steam is bled from the feed pump steam turbine. The exhaust steam from this turbine returns to the de-aerator. (v) The final h.p. heater No. 7 is split in two to give additional surface area so that the maximum amount of heat can be given to the feed water before going to the boiler feed pump. Another. advantage of the twin arrangement is that one heater may be in service even if the other has a leak and has to be isolated for maintenance. 3.27. Feed Heating System For a 500 MW Unit Fig. 3.91 shows a typical feed heating system for a 500 MW steam turbine unit with reheat. The steam conditions at the turbine stop valve are —158 bar and 538°C. The reheat temperature is also 538°C. Eight stages of feed heating is employed to give a final feed water temperature of 254°C. The salient features of this system are as follows :— (i) There are four low pressure direct contact heaters and one de-aerator. The direct contact heaters also act as a de-aerator to some an extent. The direct contact heaters 1 and 2, 3 and 4 are positioned physically on top of each others so that the water from the upper heater to pass by gravity into the lower heater without using an extraction pump. By this there is a saving of two extraction pumps. This cannot be done with the high pressure heaters. (ii) There are three high pressure surface type heaters arranged in parallel banks for the same reasons as stated in article for 350 MW. (iii) The boiler feed pump is driven by a separate steam turbine meant for this purpose only similar to that in 350 MW unit. This turbine receives required quantity of steam from the outlet of h.p. turbine. (iv) The boiler feed pump is positioned before the high pressure heaters. 3.28. Feed Heating System for a 660 MW Unit Fig. 3.92 shows a typical feed heating system for a 660 MW steam turbine unit with reheat. The steam conditions at the turbine stop valve are 158 bar and 538°C. The reheat temperature is also 538°C. Eight stages of feed heating is employed to give a final feed water temperature of 254°C. The salient features of this system are as follow :— (i) There are five direct contact Lp. heaters in two cascades one deaerator and two h.p. surface type feed heaters. (iii) There is a single bank of h.p. heater instead of two. (iii) Similar to 350 and 500 MW, boiler feed pump is driven by a separate steam turbine meant for this purpose only. For more details on feed heaters references [2, 6] may be consulted. 3.29. Advantages and Disadvantages of Turbine Driven Boiler Feed Pump The boiler feed pump consumes a considerable amount of power, of the order of 2.5 to 3% of the turbine output for a steam pressure of 160 bar. With reheat-regenerative cycle, it is now customary to drive the boiler feed pump by a bled turbine which is fed with steam from h.p. turbine exhaust. This bled steam turbine is also known as boiler feed pump turbine. Some quantity of steam is bled off from the pump turbine to h.p. heater and the exhaust steam either is led to some feed water or to condenser. Following are the advan-

Fig. 3.91. 500 MW Feed HeatingSystem. Moisture extraction and gland steam condensers

Condensate ®— ()extraction pump AEP

Condenser

♦• V • • 1 2 3 45 •

LP turbine

Steam

H2 and stator coolers

• 8

Double W

DC : drain cooler DS : desuper heater AEP : air extraction pump Reheater

4

2 LP4

5 LP heaters 3 in two rows HP surface heaters

LP2

L. P:

LP3

Boiler

3

-► P

S HP7

DC

DC

Bled turbine

5

DS

HP6

DC

DC

Boiler • feed pump

Deaerator

Feed water

DS

DS

DS

DS

HP8

DC

DC

Steam& Gas Turbines And Power PlantEngineering

Regenerative Feed Heating, Reheating

Reheater— Double flow LP

Steam

•

Deaerator 8

LP6

Turbine = HP 234

Feed water

Boiler

v LP1

V 8

173

and Water-extraction Cycles

V

LP4 0.0

BFP

Condensate

4

Bled • turbine

AEP C.E.P.

Generator coolers GC

6 L.P. heater 2 H.P. surface heaters

DS HP7

HP 8

Fig. 3.92. 660 MW Feed Heating System tages are

disadvantages of turbine driven boiler feed pump. Advantages : (i) There is a gain in thermal and overall efficiency over the electrically driven pump. This is because of the fact that the extraction of steam from the feed pump turbine requires an increased steam flow through the early stage of that turbine and the use of longer blades results in higher stage efficiency. There is an absence of generator losses and transformer iron losses. (2) The speed of the pump driven by turbine is infinitely variable. Disadvantages : When the feed pump turbine is unavailable, two 50% capacity electrically driven pumps are used for starting and stand-by purpose. Under this condition the feed heaters which were getting steam from pump turbine will not be in use and the boiler reheater circuit will be running at excessive capacity. Though, there are methods to deal with such situation but they are uneconomical. EXERCISE Viva Voce and Theoretical 3.1

What is the thermodynamic principle of regenerative feed heating ?

3.2

Is it possible theoretically to achieve the efficiency of Rankine cycle with that of Carnot ? State the conditions for this.

3.3

What is the effect of the number of heaters on the thermal efficiency ? Discuss the law of diminishing return.

3.4

How many heaters are used in practical regenerative cycle ?

3.5

What are the advantages and disadvantages of regenerative feed heating cycle over the simple Rankine cycle ?

3.6

Explain how the regenerative cycle efficiency many approach the Carnot cycle efficiency.

3.7

Discuss the saving in Heat Rate from regenerative heating. How are the number of

174

Steam & Gas Turbines And Power Plant Engineering heaters related to the increase in feed water temperature for maximum saving?

3.8. What are the effects of moisture contents on the performance of steam turbines ? Ans. Erosion, corrosion, lesser blade efficiency, lesser nozzle efficiency, lesser thermal efficiency, lesser power developed, 3.9. Show the impacts of water droplets on blade surface by drawing velocity triangle. 3.10. What do you mean by reheating of steam ? 3.11. Is it essential to reheat at a optimum pressure ? If yes, why ? 3.12. It is true that modern plants are equipped with reheat and regenerative arrangement ? Why ? 3.13. Discuss the advantages and disadvantages of reheating in steam power plants and explain why it becomes more necessary to adopt a reheat cycle with high steam pressure. 3.14. Draw a schematic diagram of a steam turbine plant equipped with reheat-regenerative arrangement. Draw the temperature distribution in various direct and indirect contact feed heaters. Give a feed heating arrangement of 120, 200, 500 and 600 MW turbines. 3.15. Derive the expressions for work done and blade efficiency when the steam contains water particles. Numerical 3.16

A steam turbine operates under the following conditions—Initial pressure of steam =26.5 bar, Initial temperature of steam = 377.5°C, Exhaust pressure = 0.03 bar The expansion process is isentropic and at a stage of turbine where the pressure is 2.7 bar, a connection is made to a surface feed heater in which the feed-water is heated by bled steam to a temperature of 127°C. The condensed steam from-the feed heater is cooled in a drain cooler before entering the feed heater. The cooled drain water combines with condensate in the well of condenser. Assuming no heat losses in the system, calculate the following—(a) The mass of steam used for feed heating•per kg of steam entering the turbine, (b) Thermal efficiency of the cycle. If the turbine efficiency is 85% recalculate (a) & (b) Ans. (a) 0.17 kg(b) 38.6%

3.17 Steam is supplied to a turbine at a pressure of 30 bar and superheated to 93°C. The turbine exhaust pressure is 0.06 bar. The main condensate is heated regeneratively in two stages by steam bled from the turbine at 4.7 bar and 0.8 bar respectively. Calculate the mass of steam bled off at each pressure per kg of steam entering the turbine and the theoretical efficiency of the cycle. The following assumptions are to be made— (a) That the expansion of the steam in the turbine is isentropic, (b) That the condensate is heated to the saturation temperature in each of the heaters., (c) That the drain water from the H.P. heater passes through a trap without loss of heat into the steam space of L.P. heater, (d) That the combined drains from the L.P. heater are cooled in a drain cooler to the condenser temperature. Ans. : mi =0.17 kg, m2 = 0.0824 kg, 36.9% 3.18 A steam turbine plant operating between pressures of 40 bar with 72°C superheat and 0.04 bar, employs regenerative feed heating. The feed-water leaving the hotwell passes through two heaters in succession. The steam for the heaters is bled from the turbine at 6.0 bar and 1.4 bar, and the condensed steam leaving the heaters is pumped into the feed-pipe at points immediately after each heater. As-

Regenerative Feed Heating, Reheating

and Water-extraction Cycles

175

suming that the feed-water leaves each heater at the turbine is 0.75, find the masses of steam bled and the thermal efficiency of the turbine. Ans. :m i =0.101 kg, m2 =0.1 23 kg, i=30.23%.

3.19 The stream is supplied to a turbine at 17 bar with 4.5°C superheat, and the condenser pressure is 0.07 bar. For regenerative feed heating steam is bled off at 3.4 and 0.7 bar. It may be assumed that the temperature of the feed is raised to that of the bled steam and that the condensate from each heater is returned, by means of separate drain pumps, to the feed water at points immediately upstream of each heater. The condensate from each heater may be as being at the same temperature as the feed entering the heater. The stage efficiency between 17 bar and 3.4 bar is 0.7 and in the other two stages 0.65. Calculate (a) the mass of stream bled for each heater per kg of steam supplied to the turbine; (b) the total work done per kg of steam supplied to the turbine, (c) the overall thermal efficiency of the cycle. Ans. : m i =0.09 kg, m 2 =0.08 kg, w=5.1109 kJ/kg 1 =22.9%

3.20 Steam is bled off from two stages of a turbine for feed heating purpose. The steam conditions at throttle is 40 bar and 370°C and at the exhaust to the condenser is 0.07 bar and 0.9 dry. Steam is extracted at 2 bar and 0.4 bar. Assume that in each heater the feed water is heated to saturation temperature and the bled steam in the heater is pumped into the feed line immediately ahead of each heater, the condition line is straight and no heat loss to surrounding. Find-(a) Steam bled off per kg of entering steam. (b) Thermal efficiency of the plant. . Ans.: (a) m 1=0'.0782 kg, m 2 =0.06 kg, (b) 1 =29.4%

3.21 A regenerative steam cycle operates between the limits of 82 bar and 427°C and 2.54 cm of Hg. If only one extraction from the turbine occurs at 10 bar what percentage of total flow to the turbine is ideally extracted. Ans. 23.6% 3.22 A regenerative cycle operates between the limits of 82 bar, 429°C and 2.54 cm of Hg absolute. (a) Assuming the efficiency of turbine and pump are 100% determine the optimum extraction pressure if one heater is used. (b) Recalculate optimum pressure if turbine efficiency is 84%. Ans. (a)6.4 bar 3.23 In a regenerative cycle, steam leaves the boiler and enters the turbine at 17.5 bar and 300°C. Steam leaves the turbine at 0.07 bar and 0.9 dry. There are two stages of steam extractions at 4 bar and 0.8 bar. The low pressure heater releases its drain to condenser and the condensate from the high pressure heater is pumped directly into the boiler feed line. Calculate (a) steam bled off from each heater, (b) thermal efficiency, (c)total power developed by the turbine rotor if steam flow rate is 16000 kg/h, (d) percentage gain in thermal efficiency due to bled heating, (e) increase in steam consumption rate with regeneration, (f) percentage increase in boiler capacity for given output, (g) percentage reduction in condenser capacity. Ans. (a)m 1 =0.0872 kg, m2 =0.09417 kg; (b) 26%; (c) 2820 kW; (d) 6.53%; (e) 0.544 kg/kWh; (I) 11%; (g) 9.1%

3.24 In a regenerative steam cycle employing two closed feed heaters the steam is supplied to the turbine at 40 bar and 500 °C and is exhausted to the condenser at 0.035 bar. The intermediate bled pressures are obtained such that the saturation temperature intervals are approximately equal, giving pressure of 10 bar and 1.1 bar. Find the amount of steam bled at each stage, the work output of the plant in kJ/kg boiler steam and thermal efficiency of the plant assuming isentropic expansion.

176

Steam & Gas Turbines And Power Plant Engineering Ans. mi =0.147 kg/kg, m2 =0.124 kg/kg of steam, w =1134.5 kJ/kg, fith=42.3%

3.25 A power plant operates on the regenerative cycle. Steam is supplied to the turbine at 80 bar and 400°C. It is exhausted to condenser at 0.05 bar. The plant has two feed beaters where steam extracted from the turbine is directly mixed with the feed water. Steam is extracted from the turbine at a pressure of 10 bar and 1.5 bar. Determine the mass of steam extracted at each of these pressures per kg mass flow at the turbine inlet, the heat rejected in the condenser per kg mass flow at the turbine inlet, the cycle efficiency and the gain in cycle efficiency over a Rankine cycle operated under the same initial and final conditions. Assume expansion of steam is isentropic. 3.26. Develop a software for the selection of feed heating system. 3.27. Two steam turbine receives steam at an initial pressure of 100 bar and temperature of 540°C. In each turbine the steam expands in H.P. turbine up to 8 bar with efficiency ratio of 80 per cent. In turbine No. A, the steam expands in L.P., turbine to final pressure of 0.035 bar" with efficiency ratio of 0.78. In turbine No. B, the steam is reheated after its expansion through the H.P. turbine and returned to the L.P. turbine at 6.8 bar and 540°C after which it expands to 0.035 bar with an efficiency ratio of 0.84. Compare for the two cycles : (a) Thermal efficiency, (b) Total steam consumption for a full load output of 60,000 kW, (c) Total volume of exhaust steam per second for a full load output of 60,000 kW. Ans. Turbine No. A-35.4%; 181500 kg/h; Turbine No. B:— 37.8%; 142000 kg/h, 3.28. Steam enters a turbine employed with reheating arrangement at 90 bar and temperature 500°C. The steam is reheated at 10.5 bar and exhausts at 0.07 bar. The net work developed by the turbine is 1554 kJ/kg. The thermal efficiency is 41%. Ans. 453°C Calculate the temperature of steam coming out from the heater. 3.29. Steam enters a turbine employing reheat-regenerative arrangement at 90 bar and 500°C. At pressure 10 bar some steam is bled off for regenerative feed heating and the remaining is reheated in a reheater to 500°C. Then the steam re-enters the turbine and expands to.0.07 bar. For ideal cycle, calculate (a) heat supplied (b) heat rejected (c) net work and thermal efficiency. For an ideal turbine calculate (d) net work (e) thermal efficiency and steam rate in kg/kWh. Ans. (a) 3134 kJ/kg (b) 1745 kJ/kg (c) 1385 kJ/kg (d) 44.39% (c) 3134 kJ/kg (1) 1396 kJ/kg (g) 2.55 kg/kWh. 3.30. Steam enters a turbine at 27 bar and 310°C and expands isentropically to 5.5 bar. The steam is then reheated at constant pressure to 260°C and then throttled to 2.7 bar before admission to a low pressure steam turbine, where it expands to 0.07 bar with an overall efficiency of 78%. Determine (a) the heat supplied in the reheater per kg of steam; (b) the Rankine cycle efficiency of the L.P. turbine : (c) the final dryness fraction of steam; (d) the power of the L.P. turbine for a steam flow of 1090 kg/h Ans. (a) 295 kJ/kg (b) 21.2% (c) x = 0.87 (d) 140.5 kW 3.31. In a lovi-pressure turbine of a two-cylinder machine, the steam is initially dry and saturated with an enthalpy hi. The final pressure is p2 (liquid enthalpy lift and latent heat hfg2• The overall efficiency ratio of the stage would be ri if they are using dry steam, but with wet steam having a mean dryness fraction x. , the efficiency ratio is x.. Show that the final dryness fraction of the steam atp2 is

Regenerative Feed Heating, Reheating

—

2(h — hm2)

and Water-extraction Cycles

177

risen where Ah is isentropic heat drop. nen

2L2 -I- tl . Ahisen

3.32. Feed water at 26.6°C is pumped to a pressure of 68 bar and is then heated to the saturation temperature. It is evaporated at pressure 68 bar and superheated to a total temperature of 400°C. The steam is then expanded isentropically to a pressure of 6 bar at which pressure it is reheated to 370°C. It is then expanded isentropically to back pressure 0.035 bar. Find the (a) gross and (b) net theoretical thermal efficiency of the cycle. Recalculate (a) and (b) if turbine efficiency is 85%. Ans. (a) 41% neglecting pump work (b) 40.7% including pump work. 3.33 Steam enters a H. P. steam turbine at a pressure of 81 bar and temperature of 430°C. The back pressure is 0.035 bar. What is the internal cycle efficiency (a) neglecting the effects of wetness, (b) allowing a standard wetness correction of 1% wetness ? The dry stage efficiency is 80%. In the alternative arrangement, steam is exhausted from the H.P. turbine at 13.5 bar reheated and returned to the I.P. turbine at 12 bar and 430°C. It is expanded to the same back pressure and with the same efficiency. What is the internal thermal efficiency of the reheating cycle (a) neglecting the effects of wetness (b) allowing the same wetness correction as before. Ans. Non-Reheat Cycle. (a) 34.87% (b) 33.12% Reheating Cycle:—(a) 35.86% (b) 35.7% 3.34. It is planned to employ a single stage reheating and regenerative feed heating with two heaters in a turbine which receives steam at 90 bar and temperature 500°C and exhausts at 0.07 bar. The internal efficiency of the turbine is 80%. Design suitable points for reheating and regenerative feed heating and find the power developed and the thermal efficiency. Hints. The reheating should be at such a pressure so that the wetness at the exhaust should not exceed 10% or maximum 12%. 3.35. An ideal reheat-regenerative cycle is operated with steam entering the high pressure turbine at 21 bar and 600°C and it is expanded to 7 bar. Part of steam is extracted to heat the feed water at 7 bar and the remaining steam is reheated to 600°C and returned to low pressure turbine where it is expanded to 0.035 bar. Find the kg of steam which must be supplied per hour by the steam generated for an output of 40 MW by the turbine. 3.36. An ideal reheat-regenerative cycle operates with throttle conditions at the turbine inlet of 140 bar and 600°C, and reheat at 7 bar to 540°C. A closed feed water heater operates at 14 bar, and the drain from the closed heater is trapped back into an open feed water which operates at 2 bar. The condenser pressure is 0.07 bar. Calculate (a) the efficiency of the cycle, and (b) the required mass flow rate through the steam generator for a turbine output of equivalent to 100,000 kW. Ans. (a) 47.9%. 3.37. A steam power generating station operates on the reheat—regenerative cycle, with part of the steam extracted for heating the feed water and the rest of the steam is reheated before being expanded to the back pressure in the low-pressure turbine. Steam enters the high-pressure turbine at 7 bar, and the rest is superheated to 485°C before entering the low pressure turbine where it is expanded to 0.07 bar. Pump work in low pressure pump is 0.625 kJ/kg. Pump work in the high pressure pump is 8.3 kJ/kg. Assume isentropic expansion in each turbine. Calculate (a)

178

Steam & Gas Turbines And Power Plant Engineering

mass of steam extracted for heating per kg of steam entering the turbine, and (b) thermal efficiency of the cycle. 3.38. Consider a combined reheat—regenerative steam turbine cycle where the inlet conditions to the high pressure turbine are 140 bar and 600°C and reheat is at 14 bar with a temperature at the low pressure turbine inlet of 540°C. Closed feed water heaters operate at pressure of 14 and 3.5 bar with a trap outlet of 3.5 bar connected to the condenser drain. There is just one feed water pump located at the inlet to the 3.5 bar heater. Calculate (a) the enthalpy change of the feed water as it passes through the heater, (b) the ratio of the mass of steam bled at 140 bar to the mass of Ans. (a) 242.6 kJ/kg, (b) 0.118 the steam entering the high pressure heater. 3.39. A regenerative—reheat plant operates between the limits of 540°C and 105 bar and 2.54 cm of Hg. Reheat to 485°C occurs after the steam has expanded to 14 bar in the turbine. Three regenerative feed heaters are used. The first is a closed type heater; it is supplied with steam extracted at a pressure of 28 bar (condensate from this heater is returned to the condenser). The remaining two are direct contact type heaters operate at a pressure of 3.5 bar and 1.4 bar. (a) Draw a diagram showing the plant components and indicating flow directions. Show all necessary pumps and cooling water circuits, (b) Sketch the cycle on T—s diagram, (c) Calculate the ideal cycle thermal efficiency, (d) Calculate overall thermal efficiency if it = 86% , = 95% , 31g = 96% and rib = 90%. 3.40. In a 100 MW (generator output) commercial steam power plant operating on a reheat and regenerative cycle steam leaves the boiler-super-heater at 77 bar and 400°C and enters the high pressure turbine after a loss of 2 bar in the supply line. Steam exhausts from the high pressure turbine at 17 bar. It is passed through a reheater and then supplied to the low pressure turbine at 16.5 bar and 400°C. Steam exhausts from the low pressure turbine at 0.035 bar to condenser. 12% and 15% quantities of the flows to the high pressure and low pressure turbines are expected' at 26 bar and 5.2 bar respectively and supplied to open-type heaters for feed water heating. The condenser cooling water enters at 15°C and leaves at 23.3°C. Component efficiencies are : generator 94%, boiler (including superheater and reheater) 85%, high pressure turbine, 90%, low pressure turbine 85%., Mechanical efficiency, 95%. Workout the following— (a) Sketch the actual cycle on T—s coordinates, (b) Calculate the thermal efficiency of the actual cycle, (c) Calculate the overall plant efficiency, (d) Calculate the steam consumption rate in kg/kWh and kg/h, (e) Calculate the condenser cooling water flow rate in kg/kWh and in kg/h. 3.41. A steam power plant generates 200 MW. The superheat outlet pressure of the boiler is to be 170 bar and the temperature 600°C. After expansion through the first stage turbine to a pressure of 40 bar, 15% of the steam is extracted for feed heating. The remainder is reheated to 600°C and is then expanded through the second turbine stage to a condenser pressure of 0.035 bar. For preliminary calculations it is assumed that the actual cycle will have an efficiency ratio of 70% and that the generator and mechanical each is 95%. Determine the maximum continuous rating of the boiler in kg/h. Ans. 633 x 103 kg/h, 3.42. A 660 MW steam turbine is equipped with reheat and regenerative feed heating system as shown in Fig. 3.92. The steam Conditions at the turbine stop valve are

Regenerative Feed Heating, Reheating

and Water-extraction Cycles

179

158 bar and 538°C. The reheat temperature is also 538°C. The back pressure is 0.035 bar. The turbine internal, alternator and mechanical efficiency are 80%, 95% and 96% respectively. The pressure loss in inlet valves and reheater pipings may be assumed each as 2% of entry pressure. Select or design suitable reheating and extraction pressures and estimate the amount of steam bled off, the thermal efficiency, the steam consumption and specific team consumption. Assume that the drain from the surface heater (integral with drain cooler) is cooled upto the saturation temperature of the next low pressure heater. Also assume 1% leakage loss of entering steam in the turbine. Make suitable assumptions if any needed. Estimate overall thermal efficiency and ssc. 3.43. Design a feed heating system for the following power plant units— a) 120 MW, b) 200 MW, c) 350 MW, d) 500 MW, e) 660 MW, f) 1000 MW. Assume suitable throttle and condenser conditions and other values. to suit real conditions. 3.44. Develop a software for the design of feed heating system for a steam power plant unit developing power from 120 MW to 1000 MW. 3.45. A steam turbine power plant developing 200 MW of electrical output is equipped with reheating and regenerative feed heating arrangement consisting of three feed heaters-one surface type on h.p. side and other two direct contact type on l.p. side. The steam conditions before the steam stop valve are 162 bar and 566°C. A pressure drop of 6 bar takes place due to throttling in valves. Steam exhaust from the h.p. turbine at 30 bar. A small quantity of steam is bled off at 30 bar for h.p. surface heater for feed heating and the remaining is reheated in a reheater to 538°C and the steam enters at 26 bar in l.p. turbine expansion. Another small quantity of steam is bled off at 15 bar and 5 bar for first and second 1.p. feed heaters, and the rest of steam expands up to back pressure of 0.05 bar. The drain from h.p. heater is led to the I.p. heater and the combined feed from l.p. heater pumped to the high pressure feed heater and finally to the boiler with the help of a boiler feed pump. The component efficiencies are:— turbine efficiency 85%, Overall pump efficiency 90% and mechanical efficiency 96%, boiler efficiency 90% and alternator efficiency 95%. It may be assumed that the feed water is heated up to the saturation temperature of the prevailing pressure in feed heater. Workout the following— (a) Sketch the feed heating system and show the process on T-s and h-s diagram, (b) Amount of steam bled off, (c) Overall thermal efficiency of turbo-alternator considering pump work, (d) Specific steam consumption in kg/kWh, and amount of steam in kg/h, (e) Specific cooling water consumption in condenser in kg/kWh if temperature rise in 10°C, (0 Fuel cost per kWh if the lower calorific value of fuel (coal) is 19,000 kJ/kg and the price of fuel is Rs 1800 per ton, by assuming that 5% of steam generated is used to cover losses and auxiliary power.

4 Back-Pressure, Pass-out and Mixed Pressure Turbine Cycles

There is a dual demand of "power" and "steam" for heating and process work in many industries, such as paper making, chemical, textile, dyeing, sugar, refining, carpet making, brewing, etc. For this type of work previously the steam was generated at some particular pressure for power and at some different pressure for process work separately. Steam was generated in a separate boiler at a pressure which could give the desired heating temperature. In this type of system much of the heat of the steam that is used for power work is carried away by the cooling water and that is wasted. This loss can be saved by utilising the exhaust steam in process work by a suitable selection of initial and exhaust pressure of steam and thus by this way power and process work are achieved at the same time. Since all the steam is condensed in the process work and no steam is condensed in a separate condenser, the thermal efficiency of such a combined power and heating plant is very high and may approach to unity. Reciprocating engines and steam turbines are both available for such plant but the turbines are most suitable because of its many advantages over the reciprocating engines. For combined power and process work, two types of turbine, namely "Back Pressure Turbine" and Pass-out Turbine" are employed. Combined electrical power and heat can be obtained by gas turbine plant also in which the steam is generated in HRSG by the turbine exhaust heat energy. This is called co-generation. Now, co-generation plants are being used world over. Mixed-pressure turbines receive steam at two different pressure-one live steam from boiler and other waste steam of other source having some energy of utilisation. This chapter deals with, the thermodynamic analysis of back-pressure, pass-out and mixed pressure turbine cycles. 4.1. Back-Pressure Turbines. This type of turbine is employed where both power and process work are required. The expansion of steam from an economical 'initial pressure down to heating pressure generates the required power and the heat carried away by the exhaust steam from the turbine is used for heating or process work. Such type of turbine with its cycle is diagrammatically shown in Fig. 4.1. Steam at suitable initial pressure enters the turbine A and exhausts at such a pressure as it is suitable for process work. Generally, the exhaust steam

Back-Pressure, Pass-out and Mixed Pressure Turbine Cycles

181

Heating surface

Fig. 4.1. Back-Pressure Turbine for Power and Heating Purpose. from the turbine is in superheated state and in most of the process works it is not suitable, the reason being firstly, the imposibility to control its temperature and, secondly, the rate of heat transfer from superheated steam to the heating surface is lower than that of the saturated steam. Due to these reasons, a desuperheater B is sometimes used. In this desuperheater, a jet of water, thermostatically controlled, is sprayed and due to contact with superheated steam spray water is evaporated and the steam is cooled and thus becomes saturated steam. Saturated steam from the desuperheater B enters the heater C and is entirely condensed by giving heat to the heating surface. Depending upon the conditions, the condensed steam may or may not be returned to the boiler. Generally, it is not possible that the steam required for power generation will always be equal to that required for process work. To avoid variation in exhaust pressure and therefore, of the steam saturation temperature, some methods of controlling the exhaust steam pressure must be employed. According to circumstances, the method of control is adopted. If the turbine is only power unit, then the centrifugal governor is employed and the quantity of available exhaust steam is controlled by the load on the turbine. If the available exhaust steam is too small, live steam may be passed through the valve E into the desuperheater. If the quantity of exhaust steam is in excess of what is strictly necessary for the process work, then the excess steam may be blown into a feed tank or to atmosphere by Work done regulating the valve F. in turbibe Generally, valves E and F which maintain the exhausts pressure constant Heat available are controlled automatically. Sometimes for process work the back pressure turbine operates in parallel with other machines, in that case heat load controls its output. There is also a pressure regulator which controls the supply of steam to the turbine and consequently constant exhaust press sure will be maintained. Representation of thermodynamic Fig. 4.2. Representation of Back-Pressure process of back pressure turbine is Turbine Process on h-s Chart.

Steam & Gas Turbines And Power Plant Engineering

182

shown on h-s chart in Fig. 4.2. The boiler pressure is po and the initial state point is 0. pi is the inlet pressure to the turbine and accordingly the state point is 1. Process 0-1 shows throttling IA the throttle valve. Let p2 be the heating steam pressure or exhaust pressure of steam from back pressure turbine. The process 1-2' shows the actual expansion of steam in the turbine and the process 2' to /92 shows subsequent condensation of steam in the heaters. Therefore, (h2' - hi2). is the heat given by the steam to the heater per kg of steam where at pressure p2. Let us assume :Si = flow of steam through the turbine, (kg/h), H= heat required in heaters, (kJ/h.), Ki= internal power of the turbine (kW), Heat required in heaters = H = M (ha' Flow of steam = M -

- h f2)

H

(h2' - hp)

H = Heat required in heater may also be expressed in kW instead of kJ/h Internal power = Ki

-

M (hi - h2') 3600

Putting the value of St in above equation, we get

H (hi - h2') Kt - 3600 h21 - h fl Therefore,

(4.1)

h,' - h, H = 3600 K H-) ' h1 - 2 1

(4.2)

If H is in kW, eqs. (4.1) and (4.2) will be without the number 3600. Problem 4.1. In a factory where L.P. steam is required for heating purpose and electrical energy is required for power purposes, it is proposed to install a back-pressure turbine to operate under the following conditions :Initial steam pressure = 17.6 bar, Initial steam temperature = 316°C Exhaust pressure = 1.4 bar, Efficiency ratio of turbine = 0.7 What is the total power available in kW, if the generator efficiency is 94% and the amount of heat required per hour equals 52.7 million kJ. It may be assumed that the condensate drains from the heater return to the boiler at the condensing temperature. Solution. Refer to Fig. 4.3. From Mollier chart : hi = 3029 kJ/kg (at 17.6 bar, 316°C), h2 = 2531 kJ (at 1.4 bar). From steam table: hi2 = 457.2 kJ/kg (enthalpy of condensed steam i.e. water at 1.4 bar) Since efficiency ratio is 0.7, hence hi - h2' = 0.7 (hi - h2) = 0.7(3029 - 253) = 348.6 kJ/kg

h2' = 3029 - 348.6 = 2680.4 kJ/kg Thd heat required for heating work = H. 52.7 x 106 kJ/h Here, heat available for heating work = h2' - hfZ = 2680.4 - 457.2 = 2223.2 kJ/kg. The power developed in the turbine is

Back-Pressure, Pass-out and Mixed Pressure Turbine Cycles

183

S

Fig. 4.3 H K—

hi — h2'

3600 h ' - hfl 2

52.7 348.6 x 106 — 2295.39 kW. 3600 x 2223.2

Since the generator efficiency is 0.94, hence power developed by generator = 2295.39 x 0.94 = 2157.66 kW.

Ans.

Problem 4.2. Steam is supplied to a back pressure turbine at a pressure of 27 bar and superheated by 70°C. The exhaust steam from the turbine is utilised in heater at a pressure of 2.7 bar. If the amount of heat required per hour is 31.63 million kJ and if the efficiency ratio of turbine is 0.75, calculate the power generated and the hourly heat consumption of the plant assuming that the condensate from the heater is returned to the boiler at 130°C and that there is no heat losses. Assume a boiler efficiency of 80 per cent. Estimate the hourly heat consumption of a plant consisting of a condensing turbine generating the same power supplied with steam at the same pressure and superheat exhausting at 0.07 bar and having an efficiency ratio of 0.75 and separate low pressure boiler generating dry and saturated steam at 2.7 bar from feed water at 130°C and supplying the same quantity of heat. Assume again an efficiency of 80% for the high pressure and low pressure boilers. Solution. Refer to Fig. 4.4. From steam table : t1 = 228.07 + 70 = 298.07°C From the Mollier chart : h1 = 2995 kJ/kg, h2 = 2543 kJ/kg Since there is no mechanical losses, the efficiency ratio of turbine will be the internal efficiency of turbine. • or

h1 — h21 = (2995 — 2543) x 0.75 = 339 kJ/kg h21= 2995-339=2656 kJ/kg. A;f (hi — h21)

Internal power developed =

3600

Heat supplied at the heater = h21 — hf.2 so total heat required = H=

(h2' - hi)

184

Steam & Gas Turbines And Power Plant Engineering

S

►s

Fig. 4.4.

Therefore,

Fig. 4.5

H (h1 — hi')

Ki

31.63 x 106 x 339 — 1411.45 kW. 3600 02 — hi2) — 3600(2556 — 545.77)

Total heat required = 31.63 x 106 = M (h2' - hfl) Aii •

—

31.63 x 106 (2156 — 545.77)

1.49888x 104 kg/h.

Now, heat consumption oftheplant = hi — h/2 = 2925 — 545.77 = 2449.23 kJ/kg. Actual heat consumption for a plant =

2449.23 — 3061.53 kJ/kg 0.8

= 3061.53 x 1.4988 x 104 kJ/h = 4588.89 x 104 kJ/h

Ans.

In the second part of the problem, there are two separate boiler-one H.P. boiler for turbine and another a L. P. boiler for heater. From Mollier chart : h2 = 2044 kJ/kg

h 1 — h2' = (2995 — 2044) x 0.75 = 711.75 kJ/kg Power to be developed by the turbine = 1411.45 kW Steam flow rate through the turbine =

1411.55 x 3600 = 7.139 x 103 kg/h 711.7

From steam table ••hn = 163.38 kg (corresponding to 0.07 bar) Heat consumption of turbine =hi — hn = 2995 — 163.38 = 2831.62 kJ/kg Heat consumption per hour = 7.139 x 102 x 2831.62 = 20.2149 x 105 kJ/h Actual heat consumption by the boiler —

20.2149 x 105, — 25.268 x 105 kJ/h 0.8

Hence, total heat consumption of the whole plant = (2.5268 + 39.45) x 106 = 41.976 x 106 kJ/h

Ans.

Back-Pressure, Pass-out and Mixed Pressure Turbine Cycles

185

Problem 4.3. A back pressure turbine is supplied with steam at 14 bar and temperature 260°C. The exhaust pressure is 1.4 bar and the internal efficiency ratio being 0.75. If the coupling power is 1000 kW and mechanical losses amount to 20 kW, calculate the total amount of heat available assuming that the boiler condensate temperature is 109°C. Calculate the coupling power and the heat available when the quantity of steam is reduced to one half of that at 1000 kW coupling power. Assume that the turbine is governed by throttling, that the internal efficiency ratio, exhaust pressure, mechanical losses remain the same as before.

Fig. 4.6 Solution. Refer to Fig. 4.6. From Mollier chart and steam table : h1 = 2950 kJ/kg (at 14 bar), h2 = 2515 kJ/kg (at 1.4 bar), hfl = hfl = 461.32 kJ/kg (at 110°C) risen = hi

- h = 2950

2

— 2515 = 435 kJ/kg.

Since the internal efficiency ratio is 0.75, so the internal work done h1 — h2 = 0.75(h1 — h2) = 0.75(2950 — 2515) = 32625 kJ/kg h2' = 2950 — 326.25 = 2623.75 kJ/kg. Internal power = coupling power + mechanical losses = 1000 + 20 = 1020 kW H (hi — h21 ) We know that, Ki —

1020 =

H—

3600 h2' —

H 326.25 3600 2623.75 — 461.32

H x 326.25 860 x 2162.43

1020 K.

hl - h2:

- 24.338

x 106 kJ/h.

Steam consumption when the coupling power is 1000 kW

Ans.

186

Steam & Gas Turbines And Power Plant Engineering

m

3600 K. h l— h

3600 x 1000 — 11255.17 kg/h 326.25

Whe the steam consumption is half of that 1000 kW coupling power and heat 41'11255 .17 availabl, = 2 — 5627.58 kg/h. M' Under this new condition, the nozzle box pressure = p ' = — p = x 14 = 7 bar 1 M• 1 2 From Mollier chart under new condition; h1' = h1 = 2950 kJ/kg, h3 = 2627.55 kJ/kg Ahisen = ht' — h3 = 2950 — 2627.55 = 322.45 kJ/kg. We know that,

K: = - XI' Ah'hen 11,

M

(Refer Chap. 13)

435 — 283.53 kW K = 1020 x 0.75 x 21 x 322.45 Coupling power = Internal power—Losses = 283.53 — 20 = 263.53 kW

Atm:-

Now, h1 ' — h3' = 0.75(h3' — h3) = 0.75 x 322.45 = 241.83 kJ/kg h3' = 2950 — 241.83 = 2708.17 kJ/kg. So, heat available= H' =

03'

= 5627.78(2708.17-461.32) =12.664x 106 kg/h.

Ans.

4.2 Pass-out Turbine. The pass-out turbine is used in place of the back-pressure turbine when the power available from the back-pressure turbine through which the whole of the heating steam flows is less than that required in the factory. Such type of turbine is shown in Fig. 4.7. In this turbine, live steam enters and expands through the H.P. stage before the extraction branch for the heating or process work. Continuously, required quantity of steam is extracted from this extraction branch for heating purpose. The remaining steam passes through a pressure control valve into the L.P. stages of turbine and the exhaust steam from it is condensed in the condenser. The control arrangement is very essential and it should keep (i) speed of the turbine constant (ii) pressure of the heating steam sensibly constant irrespective of power and heating loads. The control arrangement consists of (a) a speed governor of the centrifugal type which controls the admission of H.P. steam to the turbine, and (b) a pressure regulator which controls the admission of steam to L.P. stage and is also responsive to changes of pressure of the extracted heating steam. So any change either in power load or heating load will operate both types of controls. Let us suppose, for example, that there is sudden increase in heat load. The result will be the falling in heating steam pressure which wiii reduce the. lift of the pressure control valve. This reduction in the lift will reduce the flow of steam through the L.P. stages which will reduce the power developed, leading to a fall of speed of the turbine. As the speed of the turbine falls the H.P. throttle opens, thus greater quantity of

Back-Pressure, Pass-out and Mixed Pressure Turbine Cycles

Oil

187

Pressure control valve

pi or pi'

H.P. steam Pass-out steam M 2 V

Steam to condenser

Fig. 4.7. Pass-out Turbine.

steam should pass through the L.P. stages and the heaters. Thus we see that a change in the heating load causes both the controls to operate. Now suppose that the generator load is increased and as a result the speed of the turbine will decrease. As the speed of the turbine decreases the speed governor will increase the lift of the H.P. throttle valve; in this way more steam will enter the turbine. As a results of more steam entering to the turbine, the pressure in the heating steam pipe will now increase and this pressure rise will actuate the oil relay in such a manner as to increase the lift of the pressure control valve allowing more steam to flow through L.P. stages. Thus, by this way the increase in generator load is compensated and the speed of the turbine is maintained constant. In this case also, we observe that an increase

Governor

Pressure regulator oil Lever 1

Speed Live steam

Extraction Fig. 4.8. Speed Governor for Pass-out Turbine.

188

Steam & Gas Turbines And Power Plant Engineering

in power load is compensated and the speed of the turbine is maintained constant. In this case also, we observe that an increase in power load operates both the control valves. Fig. 4.8. shows diagrammatic arrangement of speed governor and pressure regulator On pass-out turbine. In this figure, only one extraction point is shown but it is possible to extract steam from two or more points. 4.3. Process of Pass-out Turbine with Single Extraction. Fig. 4.9 and Fig. 4.10 show the h-s diagram when ideal nozzle control governing and

S

Fig. 4.9. h-s Diagram for Pass-out Turbine with Ideal Nozzle (Control Governing).

throttle valves are employed for both speed and pressure control respectively. Let, M1 = quantity of steam entering turbine (kg/h), 11;12 = quantity of steam extracted for heating (kg/h), i13 = quantity of steam entering L.P. stages (kg/h), = quantity of steam entering turbine (kg/h) at part load. A' = quantity of steam entering L.P. stages (kg/h) at part load, H = quantity of heat required for heating (kJ/h), K = total internal power (kW), K2 = amount of power developed by heating steam in expanding through

KA 3

Valve Steam for process work

S

Fig. 4.10. h-s diagram for pass-out Turbine with Throttle Governing.

Back-Pressure, Pass-out and Mixed Pressure Turbine C9'cles

189

both controls. Thus, it is obvious that with the nozzle control governing (p2 —p2") will be much less and thus it may be neglected (Refer Chap. 13). 11.1 ' Pi = M I Pi

143' P2 = M3'Pz

and

4.3.1. Case 1 (a). Partial Extraction.—(a) Ideal Nozzle-control Governing : (Refer to Chap. 13) In the Fig. 4.9, process 1-2' shows the actual expansion of steam in H.P. stages. Process 2' —2" shows the throttling of steam in the pressure control valve and process 2" —3' shows the actual expansion of steam in L.P. stages. (h2' — hp) will be the heat available per kg of steam for heater where hp is taken at pressure p2. H= U2 (h21 — hfl) kJ/h or M — 2 h _h 2 f2

(4.3)

Power developed K2 by the heating steam = K2 — Putting the valve of M2, we have, IC

11;1 2 3600

I, 1 "2'1

H (hi h2' ) 3600 02' — hi2

(4.4)

Power developed by the remainder of steam M3 = K3 = K — K2. In expanding through H.P. and L.P. stages the work done per kg of steam is given by (hi — h2') + (h2" — h3') = hi —h3' Now K3 = K — K2 —

300

(hi — h3')

(4.5)

We have, M i = U2 + U3 3600(K — K2)

H

or

or

1

M

—

h2' —

(hi — h3')

3600 K t

-hl-h3

, + A 2(h2i

3600 K —H + hi — h3' h2'

(h2' — h3' ) h I — h3'

(4.6)

h3')

(4.7) The same equation (4.7) may also be found another way. The total power developed. = K = power developed in H.P. turbine+power developed in L.P. turbine

But

M i(hi — h2')

M3(h2' — h31)

3600

3600

A;12 = A.4 1

A;12 — h2')

K— or

3600

— M2) (h2' — h3') 3600

• (h2' — h31) +M hi — h3' 2 (h1 — h3') 3600 K

1

(4.7)

Steam & Gas Turbines And Power Plant Engineering

190

4.3.2.Case (i)(b). Throttle Governing. In Fig. 4.10, process 1-1' shows the actual expansion of steam in H. P. stages. Process 2' —2" shows the throttling of steam by the pressure regulator valve and process 2" —3' shows the actual expansion of steam in the L.P. stage. Here, pi' and p2' are not known and that can be calculated. Generally, two alternative methods are used to get this. Firstly, let us assume pi ' and p2" and M1' and M3' will be calculated by the method of previous article. After this, assumed pressures are corrected and accordingly M2 and M3 are found out. Secondly, a series of values of M and M3 are taken and correspondingly calculation for heat and power load is done. 4.3.3. Case (ii). Full Extraction Nozzle Control Governing. When the extraction of steam is full, it means that no stream is passing through the L. P. stages and this turbine is equivalent to back pressure turbine. (h2 — h21) is the internal work done per kg of steam Thus, we have, K =

i

M1

h2 )

36100 3600

M

360 0 (h2 - h2')

(4.8)

In heating process, the heat rejected per kg of steam = h2' — hj2 Thus, the total heat available = H = M1 (h2' — hf2) 3600 K(h21 — or

H—

(h1 — h21)

h.'`)

kJ/h (4.9)

H(h1 — h2') or

K—

3600(h2' — h' f2

kW (4.10)

Equation (4.8) shows that the steam flow id is the largest. But to remove the heat generated due to disc friction and blade windage in L. P. stages, a small amount of steam is allowed to flow through it. 4.3.4. Case (iii). No Extraction. When there is no extraction of steam, then the turbine will operate as a normal condensing plant. Here M3 = M1. Naturally, the pressure control valve will be fully opened. The steam consumption M 1 is given by M

1

3600 K kg/h hl — h3'

(4.11)

Equation (4.11) shows that M1 is the lowest steam consumption. Unless some form of nozzle control governing is employed at the L. P. end there will an appreciable loss due to wire drawing at pressure control valve resulting in loss of power. On account of this the specified power output for no extraction should be kept as low as possible. Problem 4.4. Steam is supplied to a pass-out turbine at 27 bar and 270°C. The pass-out steam pressure is 4 bar and the exhaust pressure is constant at 0.07 bar. The total power required is 4410 kW and the heat load is 52 million kJ/h. What is the total steam consumption of the turbine if the efficiency ratio of H. P. and L. P. groups of stages is 0.75 ? If the total steam consumption is reduced to 32200 kg/h and the heat load remains the same, calculate the new power output. Assume that there is ideal nozzle control governing at H. P. turbine inlet but throttle governing at inlet to L.P. stages. Solution. Ref. to Fig .4.11 and 4.12. From Mollier chart: hi = 2928 kJ/kg, h2 = 2560 kJ/kg,

Back-Pressure, Pass-out and Mixed Pressure Turbine Cycles

191

►S

Fig. 4.11 No Extraction Case.

Fig. 4.12

h1 — h2' = (2928 — 2560) x 0.75 = 216 kJ/kg or h2' = 2928 —276 = 2622 kJ/kg From Mollier chart : h3 = 2075 kJ/kg , 1112 = 604.67 kJ/kg at 4 bar

h2 ' — h3' = (2652 — 2075) x 0.75 = 432.75 kJ/kg Heat available for process work =112 —1112 = 2652 — 604.67 = 2047.33 kJ/h. Heat required for process work = 52 x 106 = U2 x 2047.33 2

x 106 — 25.398 x 103 kg/h. - 52 2047.33

Now power produced by H. P. turbine =

11 — h21) = 276 /1.41 kJ/h.

Power produced by L.P. turbine = ir3(h2' — h31) = 432.75 kJ/h. 276M1 + 432.75/43 Total power produced = But

=4410 kW

M3 = kr i — A;f2 = /41 — 25.398 x 103 4410 —

or

3600

276k1

4- (i1

1 — 25.398 x 103 )432.75 3600

ki = 3.790756 x 104 kg/h.

Since the heat load is the same, so

Ans.

M2 =1421 = 25.398 x 103 kg/h.

New value of steam entering the H. P. turbine, kl 1' = 32200 kg/h

M3' = 32200 — 25.398 x 103 = 6802 kg/h. Since governing at entrance of H.P.'turbine is by nozzle control governing, so pi =pii = 27 bar

Steam & Gas Turbines And Power Plant Engineering

192

Power produced by H. P. turbine =

A/ l (hi - h21) 276 x 32200 2468.66 kW 3600 3600

Since steam consumption at L.P. turbine is controlled by throttle governing, so 143'

4 x 6802

P2" =P2 A;13 - (37.9 - 25.398) x 10'

- 2.176 bar

h2' = h2" = 2653 kJ/kg, h3' = 2168 kJ/kg, thus h2" - h3' = (2653 - 2168) x 0.75 = 363.75 kJ/kg. .•'

h " - h3' = (2653 a

- 2168) x 0.75 = 363.75 kJ/kg. (

Power produced by L.P. turbine -

M31

- h3I) - 6802 x 363.75 687.28 kW 3600 3600

Total power developed by turbine = 2468.66 + 687.28 = 3155.94 kW.

Ans.

Problem 4.5. A factory has an average load of 1000 kW and requires 15.8 x 106 kJ/h for process work. The heat is to be supplied by saturated steam which condenses in heaters in at 2 bar. It is recommended to install a pass-out turbine to operate under the following conditions :- Initial pressure = 17 bar, Initial superheat = 50°C, Exhaust pressure = 0.05 bar. The turbine efficiency ratio for the pressure ranges 17 to 2 and 2 to 0.05 bar is 0.76 and the boiler efficiency may be taken as 0.8. If the calorific value of coal is 29300 kJ/kg, calculate the daily coal consumption of the factory allowing 10% for the auxialiaries. Solution. Refer to Fig. 4.13. From steam table : ti = 204.31 + 50 = 254.31°C From Mollier chart : hi = 2920 kJ, h2 = 2518 kJ/kg •• •

h2 - h21 = (2920

- 2518) x 0.76 = = 305.52 kJ/kg

h2' = 2920 - 305.52 = 2614.48 kJ/kg Also, h2' - h3 = (2614.48 - 2100) x 0.76 = 391 kJ/kg Let M 1, M2 and /43. kg/h of steam flowing through H.P, heater and L.P. turbine respectively, thus M 3 = M 1 - M 2 Heat available for heating purpose is = (h2' - hi2) M2, kJ/h 15.8 x 106 = /42(2614.48 - 504.17) or

M

2

= 15.8 x 106 - 7.487 x 103 kJ/h 2110.31

Power produced by H. P. turbine = (hi - h21) M1 = 305.52 Mi kJ/h Power produced by L. P. turbine = k3(h21 - h2) = 391 1113 kJ/h Total power produced = 305.52 M1 + 391 tl;f3 As we know that, /43 =

-

=

1 - 7.487 x 103

Back-Pressure, Pass-out and Mixed Pressure Turbine Cycles

193

—ms s Fig. 4.13

Fig. 4.14 305.92

Hence, total power developed = 1000 — or

of

+391k — 391 x 7.487 x 103 3600

9.366 x 103 kg/h

Heat supplied by the boiler =

— il;f2h/2 — A;f3 h/3

= 9.366 x 103 x 2920 — 7.487 x 103 x 502.16 — (9.366 — 7.487) x 103 x 137.77 = 23.315 x 106 kJ/h Actual heat consumption of boiler =

23.315 x 106 — 29.14375 x 106 kJ/h. 0.8

Hence, coal consumption — 29.14375 x 106 — 994.667 kg/h 29300 = 994.667 x 24 = 23872 kg/day Allowing 10% for auxiliary, net coal consumption is = 1.1 x 23872 = 26259.2 kg/day = 26.259 tons/day

Ans.

Problem 4.6. A pass-out turbine developing 1000 kW and which has a heat load varying from nil to 19 million kJ/h operator under the following conditions—Initial steam pressure = 20 bar, Initial steam superheat = 115°C, Pass-out pressure = 3 bar, Exhaust pressure = 0.07 bar. If the efficiency ratio of H.P. and L.P. turbine are 0.7 under all condition of operations, calculate the maximum and minimum hourly steam consumption. Assume that heating steam gives up heat by cooling and condensing only, and control is affected by nozzle-control governing with no pressure drop at the control valves. Solution. Refer to Fig. 4.14. From steam table, t1 = 115 + 24.37 = 327.37°C From Mollier chart : hi = 3083 kJ/kg , h2 = 2677 kJ/kg hi — h2 = 0.7(h1 — h2) = 0.7(3083 — 2677) = 284.2 kJ/kg h = 3083 — 284.2 = 2798.8 kJ/kg

Steam & Gas Turbines And Power Plant Engineering

194

Locating the point 2' with the help of enthalpy h2' = 2798.8 and pressure 3 bar, we have h3 = 2225 kJ/kg .•.

h2' — h3' = 0.7(h2' — h3) = 0.7(2798.8 — 2225) = 401.66 kJ/kg.

••

h3 ' = 2798.8 — 401.66 = 2337.14 kJ/kg.

The maximum amount of steam consumption will be needed when the heat load requirement is maximum, i.e. 19 x 106 kJ/h. Let M kg of steam required for heat load, so M.. amount will be passing through the H.P. turbine and M..—M through L. P. turbine. The power required under this condition is 1000 kW. Amount of steam required for heat load 19 x 106 19 x 106 — _ 8.492 x 103 kg/h h2' — hJ2 2798.3 — 561.43 Also

K1 —

mx— (h — + 3600 1 h2)

k max 1000 — — 3600 (284.2) +

MaX

— max 3600

(h2 ' — h3') 8.492 x 10)401.66 3600

= 1.0222 x 104 kg/h

Ans.

Minimum steam consumption will be required when the heat load is zero, i.e. the same amount of steam will be flowing through the H. P. and L. P. stages. The internal power developed when heat load zero is given by

iff in. • M. — 1000 — a [284,2 + 401.66] — h + (h2' = — h3')] [(ht K. 3600 or

M

min

=

5248.88 kg/h

Ans.

Problem 4.7. The results obtained during a test of a pass-out turbine were—Initial steam pressure = 30 bar, Initial steam temperature = 370°C, Steam pressure before first stage nozzles = 24 bar, Pressure at which steam is extracted for heating i.e. pass-out pressure = 5 bar, Pressure before L.P. first stage nozzles = 3.5 bar, Exhaust pressure = 0.07 bar, Efficiency ratio of H.P. and L.P. turbine = 0.8, Temperature of condensed heating steam = 95°C, Electrical power developed = 7800 kW, Discharge from surface condenser = 2.72 x 104 kg/h, Alternator efficiency = 0.96, Turbine mechanical losses = 60 kW. Calculate the steam consumption of the turbine and the total amount of heat available. Solution. Refer to Fig. 4.15 and 4.16. From Mollier chart : hi = hi' = 3159 kJ/kg, h2 = 2786 kJ/kg hi' — h2' =

— h2) = 0.8(315 — 2786) = 298.4 kJ/kg

h2 ' = 3159 — 298.4 = 2860.8 kJ/leg ' Now, h'2' = h2' = 2860.8 kJ/kg

and

h3 = 2247 kJ/kg

Jig' — h3' = 0.8(h2" — h3) = 0.8(2860.8 — 2247) = 491 kJ/kg

Back-Pressure, Pass-out and Mixed Pressure Turbine Cycls

195

M3 For heating

Fig. 4.15

Fig. 4.16

li;f 1 be the total steam consumption in the turbine and out of which M2 quantity of steam is extracted for heating and kf3 goes to L.P. turbine. Here, /43 = 2.72 x 104 kg/h Let

Internal power developed by 1113 kg/h of steam in H. P. and L.P. stages 143[01' — h2') (h2" h3')] 2.72 x 104(298.4 + 491) — 5964.35 3600 3600 Electrical power developed by turbine = 7800 kW Power supplied to alternator = + 71: 1(6) — 8125 IcW Internal power developed by turbine = 8125 + 60 = 8185 kW Internal power developed in turbine by .4%/2 steam in H. P. stage = 8185 — 5964.35 kW = 2220.65 kW 2(h1' — h2')

2220.65 =

3600

/42(298.4) 3600

A.4

2220.65 x 3600 — 26790.68 kg/h 298.4 Hence total steam consumption 2

= M2 + M3 = 26790.68 + 2.72 x 104 = 53990.68 kg/h

Ans.

Heat available = M2(h2' — hfl), here hfl = 397.99 kJ/kg corresponding to 95°C Heat available = 26790.68(2860.8 — 397.99). = 65.98 x 106 kJ/kg

Arts. Problem 4.8. A factory has an average load of 1200 kW and requires 19 million kJ/h to be supplied by steam which condenses in heaters at 2 bar for process work. Steam for power work is to be generated at 17 bar and superheated by 60°C. There are two following arrangements for this work. Compare the fuel consumption for the two following cases. (a) The steam expanding the turbine to a fmal pressure of 0.06 bar and the heating

Steam & Gas Turbines And Power Plant Engineering

196

steam in separately generated at 2 bar. (b) The supply of heating steam is extracted from the turbine at 2 bar; the steam not required for heating purpose is being expanded to 0.06 bar. It may be assumed that the boiler efficiency is 0.76 in each case, that the turbine efficiency ratio for any range of pressure is 0.78 and that the calorific value of coal is 29709 kJ/kg. Both speed and pressure are regulated by nozzle control. Solution. Refer to Fig. 4.17 and 4.18. Case (a):— From Mollier chart : hi = 2945 kJ/kg, h2 = 2062 kJ/kg hi — h21 = (2945 — 2062) x 0.78 = 688.74 kJ/kg x 688.74 Power developed by turbine = 1200 —

M

I

'

3600

— 1200 x 3600 6.27 x 103 kg/h 688.74

h = 151.5 kJ/kg (corresponding to 0.06 bar )

Heat given to turbine by boiler =(h1 — l f2) /41 = (2945 — 151.53) 6.27 x 103 kJ/h = 17.51 kJ/h. Heat supplied to the heater by another boiler = 19 x 106 kJ/h. Hence total heat to be supplied by the boilers = (17.51 + 19)106 = 36.51 x 106 6.51 x 106 Actual heat to be supplied = 3 — 48.039 x 106 kJ/h 0.76 Since calorific value CV = 29706 kJ/kg, so Coal consumption =

=

48.039 x 106

29709

— 1617.148 kg/h

Ans.

Fig. 4.18

Case (b) : From Mollier chart : hi = 2945 kJ/kg , h2 = 2535 kJ/kg hi — h2 = (2945 — 2535) x 0.78 = 319.8 kJ/kg

.0

S

Fig. 4.17

Fig. 4.18

Back-Pressure, Pass-out and Mixed Pressure Turbine Cycles

197

h21 = 2945 — 319.8 = 2625.2 kJ/kg From Mollier chart : h3 = 2130 kJ/kg

h — h31 = (2625.2 — 2130) x 0.78 = 386.25 kJ/kg Let /41, M2 and M3 be the amount of steam in kg/h flowing through H. P., heater and L. P. turbine. Required heat load = 19 x 106 = 11.12 (h2' — hf2)

M — 2

19 x 106

2121.03

— 8.957 x 103 kg/h

(h1 — h2') + Af3(h2' — h3') Power developed = 1200 = But

3600

11;i3 = ii;i1 — A./2 = M t — 8.957 x 103 x 319.8 +(M1 8.957 — x 103)386.25 Power developed = 1200 —

or

Mi = 11.018 x 103 kg/h

3600 and

/43 = 2.061 x 103 kg/h

Heat supplied to turbine and heater = M1 h1— M 2 /In — M3 hfi = (11.018 x 2945 — 8.957 x 504.17 — 2.061 x 151.5)x 103 = 27619.91x 103 kJ/h Actual heat supplied = Coal consumption —

17619.91 x 103

0.76

3 = 36341.97 x 10 kJ/h

36341.97 x 103 — 1223.647 kg/h 29709

4.4. Essence of Low-Pressure Turbines. There is a wastage of energy associated with the exhaust steam coming out from the non-condensing steam engine. This exhaust steam is actually capable of doing work by further expansion because its temperature is about 100°C or slightly more which is well above the temperature of the available condensing water. The complete expansion of this exhaust steam is possible but it is more economical to expand in a turbine than in engine with low pressure cylinder. This fact can be visualised from the p —v diagram shown in Fig. 4.19. Due to the limitation of the size of the cylinder the steam is released at pressure p2 and after that the steam expands to the pressure p3 in the exhaust space. So the hypothetical indicator diagram is ABCDEA. Ifpv" = constant be the laws of expansion in both steam engine and turbine; then the work done by the engine represented by area ABCDEA. is

pivi —p2v2 =pivi +

n-1

p3v2

If we employ a steam turbine utilising the exhaust steam from the engine which is at low pressure (thus name low-pressure turbine) the work done by the turbine is represented by the area CDFC is

Steam & Gas Turbines And Power Plant Engineering

198

pd' = Constant

/

P2 E

3

1 7

Gain due to complete expansion Gain due to higher vacuum

D 1

f."

0 VI V2

V 3

V 4 -IP V

Fig. 4.19. Gain Due to Higher Vacuum. P2v2 — P3V3

— p3(v2 — v3)

n—1

The same above work done may be possible expanding exhaust steam in a LP. cylinder with greater difficulty but the most economical vacuum is less than that is possible in steam turbine. This is because of the fact that if a higher vacuum is employed in steam engine, then we have to balance a greater amount of sensible heat against relatively small increase in work done. Therefore, the most economical vacuum in steam engine is comparatively low, about 66 cm of Hg and this is limited by engine cylinder. But in steam turbine there is no limit of the vacuum; it is limited by the temperature of available cooling water, some practical consideration and cost. Thus, work done by the L. P. engine is less than the turbine. In this way, the L.P. turbine has greater advantages over the L. P. reciprocating engine. When the L.P. reciprocating engine is running continuously it is essential to install a L.P. turbine and suitable condensing plant. It must be noted however, that this turbine will have a small number of stages. 4.5. Working of Mixed-Pressure Turbine. It has been seen that the utilisation of exhaust steam in a L. P. turbine is advantageous. There are so many reciprocating non-condensing steam engine installations, for example, colliery winding engine, a haulag engine, rolling mill engine, etc. in which the supply of exhaust steam is intermittent. Here, it is essential to provide the installation a heat accummulator wherein the surplus exhaust steam (or its latent heat) may be stored during the time that the engine is operating, and from which the turbine may draw a supply of exhaust steam during the time that the engine is standing. A diagrammatic arrangement of such a plant is shown in Fig. 4.20; In some cases where, it may be necessary to run the turbine during long periods 'when the engine is idle or when the quantity of exhaust steam is not quit sufficient to produce the power demanded, it is essential to design the turbine in such a way that live steam (i.e. high pressure steam) may be admitted to the turbine at an earlier stage than the first L. P. stage. So we can say that this turbine consists of two turbines, one H.P. and the other L. P.

Back-Pressure, Pass-out and Mixed Pressure Turbine Cycles

199

Separator Heat accumulator Engine L.P steam Rolling mill

H.P. steam Mixed-pressure turbine Fig. 4.20. Diagrammatic Arrangement of Steam Engine, Heat Accummulator and Mixed-pressure Turbine.

turbine but placed generally in the same casing. As it obvious from the figure that the steam coming out from last stage H.P. turbine enters the first set of nozzles of the L.P. , turbine. If the engine is also operating then the L.P. steam is getting mixed with the steam coming out from the last stage of H.P. turbine and after that enters the first set of nozzles of the L.P. turbine, thus the name "mixed-pressure turbine". Both H.P. steam and L.P. steam are controlled by centrifugal governor to maintain the speed of the turbine fairly constant irrespective of load. So there are three possible conditions of operation :— (i) Power developed entirely by L.P. steam, (ii) Power developed entirely by H.P. steam, (iii) Power developed partly by H.P. steam and partly by L.P. steam, thus the name mixed-pressure turbine. Conditions of Operation : (i) Power Developed Entirely by L. P. Steam : Let, pi = pressure of L. P. steam assumed dry and saturated. M

2

= steam flow through turbine, kg/h.

K = internal power of turbine in kW (i.e. coupling power + mechanical losses) pea = exhaust pressure when operating with L. P. steam, = Actual internal work done per kg of steam between the pressure p2 and pee as shown in Fig. 4.21. Now /*2 X Ahn = 3600 K

(4.12)

Since Aha is very small, A.12 will be larger, thus the nozzles and blades are made sufficiently large to accommodate flow rate. (ii) Power Developed Entirely by H. P. steam : Let, pi = pressure of the H. P. steam. Ah fl = actual internal work done per kg of steam between pressures pi and pet as

Steam & Gas Turbines And Power Plant Engineering

200

—ms

s

Fig. 4.21. h-s Representation for L.P. and H.P. Operation shown in Fig. 4.21. Therefore, M i x & 11 = 3600 K

u _ 3600 K

or

I

Alz,1

(4.13)

It is to be noted here that the value of M I will be much less than i/2 and the pressure of steam at the first set of L. P. nozzles will be no longer p2 but p2' so that the same power has to be developed by L. P. stages. (Refer the chapter 13) M1 „ r2 =1.2 if4. 2 (iii) Power Developed by H.P. as well as L.P. Steam i.e. Mixed-pressure Operation: Fig. 4.22 shows the representation of various processes on h-s chart for a mixed pressure turbine. Mixed-pressure operation is needed when the power required is not met by L.P. steam. Let,. M4 = amount of L. P. steam (less than M2 when power developed entirely by L. P. steam) in kg/h thus H. P. steam is needed, M3 = amount of H. P. steam, kg/h Since it is possible that the pressure of exit steam /92" from the last stage of H.P. turbine may not be the same as L. P. steam pressure p2, say for example p2" is less than low pressure p2, then the L. P. steam has to be throttled to the pressure p2". H.P. steam is also throttled if required to maintain the speed constant. Now, the total flow through L. P. turbine is (A!3 -FV4), so the mixed pressure p2" will be P2 — P2

(M3 + k4) A,' 2

(4.14)

Here, the L. P. steam is just dry and saturated at pressure p2 and is throttled to pressure

Back-Pressure, Pass-out and Mixed Pressure Turbine Cycles

—ms

201

s

Fig. 4.22. h-s Diagram for Mixed- Pressure Operation. p2" so that the mixed pressure is the same. The final state point is 2' . 3' is the actual condition of H.P. steam in the mixing chamber. Hence, the mixture condition of steam will be 4 and 5' is the actual condition of exhaust steam from mixed pressure turbine. At inlet to the L. P. stage, M4 kg of steam with the enthalpy value h2' are mixed with with enthalpy h3', resulting in the enthalpy h4 which is calculated by

M3 kg

h4(M 3 + M 4) = M 3 h3 ' + M 4 h2 ' or

h4 -

M 3 h3' M 4 h21 (4.15)

IV' 3 + 1/;14

Once the point 4 is fixed on h—s chart, the condition curve 4-5' can be drawn. Total internal power in kW [11)1 3 (h2 — h3') + (11;f 3 + 11;1 4) (h4 — h5')] 3600

(4.16)

Problem 4.9. A reciprocating steam engine develops 1102 kW with a steam consumption of 14.53 kg/kWh. The engine exhausts at 1.3 bar. Calculate the power available from an exhaust steam turbine utilizing all the exhaust from the above engine assuming that the steam enters the turbine at 1.06 bar, dry and saturated, that a separator in the engine exhaust pipe removes 590 kg/h of water, that the vacuum at the turbine exhaust branch is 73.5 cm (76 cm barometer) and the turbine thermal efficiency ratio is 70%. Solution. Refer to Fig. 4.23. Steam consumption by steam engine = 14.53 x 1102 = 16000 kg/h. Condensed steam removed by separator = 590 kg/h, Therefore, steam consumption by the turbine itif 2 = 16000 — 590 = 15410 kg/h. Condenser pressure = 76 — 73.5 = 2.5 cm of Hg = 2.5 x 1.013 = 0.0333 bar Isentropic heat drop = h2 — h3 = 481 kJ/kg Actual power developed —

t1;i 2 X Alicia 14510 x 481 — 2058.94 kW — 3600 3600

Ans.

Steam & Gas Turbines And Power Plant Engineering

202

Fig. 4.23

Fig. 4.24

Problem 4.10. Following particulars relate to a mixed-pressure turbine— Output at turbine coupling = 800 kW, Pressure of H. P. steam, 70.5°C superheat = 12.7 bar, Pressure of L.P. steam, dry and saturated = 1.27 bar, Exhaust pressure = 0.04 bar. If the power absorbed by bearing friction, governor drive, etc. amounts to 15 kW and that the internal efficiency ratio is 65% for both H. P. and L. P. operation, calculate the steam consumption in kg/kWh for both methods of operation. Also, find the steam pressure in the L. P. steam belt during operation with H.P. steam. Solution. Refer to Fig. 4.24. ti = 70.5 + 190.5 = 261°C Total internal power developed = 180 + 15 = 815 kW H.P. operation :— From Mollier chart, : Aha = hi — h4 = 849 kJ/kg. Actualinternalwork= Ahd = hi — hi4 = 849 x 0.65 = 551.85 kJ/kg Al xAh i Total internal power developed = K = 815 — or

11;11 = 3600. K

3600 551.85

3600

kW

6.523 kg/kWh Ans.

L. P. operation :— Ahisen = h2 - h3 = 456 kJ/kg Actual internal work done = Aha = 456 x 0.65 = 296.4 kJ/kg ki2 X

Again, we have, K — or

M2 K

Aha

3600

kW

860 _ 3600 12.145 kg/kWh Aha - 296.4 —

Ans.

Problem. 4.11. 6800 kg/h of live steam is supplied to a mixed pressure turbine, the pressure and temperature of the steam after governor valve being 14.12. bar and 260°C.

Back-Pressure, Pass-out and Mixed Pressure Turbine Cycles

203

s Fig. 4.25

The turbine is also supplied with 9050 kg/h of dry and saturated steam at 1.4 bar. The exhaust pressure is 0.035 bar. If the internal efficiency ratio of H. P. and L. P. stages are 0.76 and 0.7 respectively and the mechanical losses are 25 kW, calculate the output of turbine coupling. Solution. Refer to Fig. 4.25. In the absence of sufficient data, it is assumed that exit pressure from the last stage of H. P. turbine is the same as entering steam. Amount of H. P. stem = /1.4 3 = 6800 kg/h, Amount of L. P. steam = A.4 4 = 9050 kg/h From Mollier diagram : hi = 2950 kJ/h Since internal efficiency ratio of H. P. turbine is 0.76, hence hi — h3' = 0.76(h1 — h3) or h3' = 2623 kJ/kg h2 = 2686 kJ/kg Hence h4 =

M 3 h3'Mh 4 2 11./3 + M44

6800 x 2623 + 9050 x 2686 _ 2658.97 kJ/kg. 6800 + 9050

Again hi — h5' = 0.7 (h4 h5) or h51 = 2171 kJ/kg. 11;1 3 Internalwork done —,

- h31)

(A;1 3 ± /V4)

4 -h5')

3600

6800 (2950 — 2623) + 15850 (2658.97 — 2171) _ 2766 kW 3600 Coupling power = Internal power—Mechanical losses = 2766 — 25 = 2741 kW

Ans.

Problem 4.12. Dry and saturated steam at 1.41 bar is supplied to a 2,000 kW mixedpressure turbine and develops its full output when the exhaust pressure is 0.07 bar and the efficiency ratio of the turbine is 0.75. There is no throttling loss. Calculate the power output of the turbine when supplied with 6800 kg/h of high pressure steam at 14.1 bar and 315°C and 4210 kg/h, dry and saturated steam at 1.41 bar, the exhaust pressure remaining at 0.07 bar. Assume that the low-pressure steam is admitted through a throttle valve and

Steam & Gas Turbines And Power Plant Engineering

204

20°G

-ms s

► S

Fig. 4.26

Fig. 4.27

there is no throttling of the high-pressure steam and that the efficiency ratios of both high pressure and low-pressure sections of the turbine are 0.75. Solution. Refer to Fig. 4.26 and 4.27. L. P. operation : Isentropic heat drop = h2 — he = 431 = 323.25 kJ/kg Actual internal work done = Ah,2 = h2 — he = 0.75 x 431 = 323.25 kJ/kg We know that, K— or

A.2 4

X Ahi2 3600

' 3600 K 3600 x 2000 — 22273.78 kg/h 2 Ah 323.25 i

M 4 = 4210 kg/h

Here if43 = 6800 kg/h, Now, p2" —

A;14)

P2 (A.113

3

1.41 x 11010 — 0.6969 bar 22273.78

From Mollier diagram : hi = 3075 kJ/kg, h2 = h2' = 2680 kJ/kg hi — h31 = 0.75 (hi — h3) or h3' = 2640 kJ/kg h4 =

if3xh3'+/144xh2 6800 x 2640 + 4210 x 2680 — 2657.58 kJ/kg 11010 XI3 + 4

h5' = h4 — 0.75 (h4 — h5) = 2401 kJ/kg (h 1 — h3') + Internal power developed —

3 + M4) (h4 - h5t) 3600

— 6800 (3075 — 2640) + 11010(2657.58 — 2401) _ 1606.37 kW 3600

Ans.

Back-Pressure, Pass-out and Mixed Pressure Turbine Cycles

205

4.6. Heat Accumulator. As the name implies the function of heat accumulator is to store the surplus exhaust steam from a non-condensing steam engine during the time that the engine is operating and supply to turbine when the engine is standing. This is also called L. P. heat accumulator. Fig. 4.28 shows a typical steam consumption curve for a non-condensing engine. At point a, the wind commences and steam enters the cylinders at full pressure. At A the steam is cut off by cut-off control and at B and steam is shut off entirely, allowing the engine to coast until the wind is stopped at C. At D, again, the next wind commences and more or less there is a repeatition of the curve. (Note that the area of the diagram represents the quantity of steam consumed by the engine and the average height represents average steam consumption). Average height of the curve will also represent the turbine steam consumption if the turbine uses whole of the steam available. As is obvious from the curve, that between a and b, the amount of exhaust steam is greater than that required by the turbine and between b and a the amount is less than required. Generally, the heat accumulator consists of one or more large horizontal cylindrical vessel, about 60 percent full of water. Fig. 4.2 shows that the surplus exhaust steam is led through a pipe into the tank and allowed to pass through a large number of sub-merged nozzles or orifices. As the steam comes out from the nozzles it has a direct contact with water and thus condenses. The latent heat of steam is absorbed by the whole water mass as sensible heat, the temperature of the water rises, and the pressure of the steam in the space above the water also rises. This happens from a to b. When the turbine begins to take more steam than supplied by the engine, there is a slight decrease of pressure in the steam space of the accumulator resulting in evaporation of water. In this way, the sensible heat of whole water mass decrease resulting in fall in temperature and pressure. This is called regeneration and this regeneration period happens from b to a'. At a', again, the quantity of exhaust steam is greater than required by the turbine. The inclined hatched area as Steam pressure in accumulator

Regeneration

Storage

lb

P. Time (sec)

Fig. 4.28. Time-consumption Diagram or a Non-Condensing Engine.

Steam & Gas Turbines And Power Plant Engineering

206

shown in Fig. 4.28 shows the quantity of steam to be condensed in the accumulator during the storage period and it also shows the quantity of steam re-evaporated. The quantity of water stored in the accumulator controls the variation of steam saturation temperature and thus the variation of steam pressure. For regeneration period : Let, II = lowest saturation temperature, (°C.), t2 = highest saturation temperature, (°C), M1 = minimum quantity of water in accumulator, (kg.), M2 = maximum quantity of water in accumulator, kg, hfgi = latent heat corresponding to t1. (KJ/kg), hfg2 = latent heat corresponding to h. (kJ/kg) At any instant, let M, t and L be the values of quantity of water, saturation temperature and latent heat and let the whole water mass cools by a very small amount Si, hence producing the small quantity of dry and saturated steam SM. Therefore, Mcp . St = L. 8M Since the latent heat is a function of temperature, so it may be written as L = a - bt where a and b are positive constants MC St = (a - bt) SM. or

SM _ C . St M

a - bt dM M

Integrating, M2

cp = b

112 c p dt Ji a - bt a - bt2

or

loge

or

M c hig1 loge to m b log io lo

loge a - bti

ig2

(4.17)

If (M2 — M1) amount of steam will be stored in the accumulator, and let this amount be m kg. then M2 M2 — MI M _ -1+ M M1 Mt1 or

m logig (1 + — A1 -) = -t, logig r_

(4.18) i fg2 It is to be noted that m is found from the steam-consumption curve. Therefore, maximum quantity of water M2 = MI + (4.19) Problem 4.13. Find the minimum water capacity of a low-pressure heat accumulator when the mass of steam to be stored is 450 kg per cycle. The maximum and minimum steam pressures in the accumulator are 1.6 and 1.2 bar respectively.. Solution. The condition of steam in the accumulator may be assumed dry and saturated. At 1.6 bar, t = 112.73°C; hfgi = 2218 kJ/kg, At 1.2 bar, t =104.27°C; hfg2 = 2243 kJ/kg Let us assume, L = a - bt Thus, two equations are possible :-a - 112.73 b = 530.5 and a - 104.27 b = 536.1

207

Back-Pressure, Pass-oat and Mixed Pressure Turbine Cycles

Solution of the above two equations give, b = 2.955 h ) We know that, login (1 + IA = b logio iiigi fg2

2243 450) 4.18 - 0.00564 Putting the values, we get login (1 + , , - ,.955 logio 2218 'i ' Or

1+

450

mi

= 1.01139 or MI = 39600 kg

Hence the least capacity = 39600 + 450 = 40050 kg

Ans.

EXERCISE Viva-voce and Theoretical Questions 4.1. Is it true that back pressure and pass-out turbines are advantageous for power and process work than generating steam for both work separately ? Explain. 4.2. In which circumstance is the desuperheater used in back pressure ? 4.3.

Explain briefly the following- (a) Back pressure turbine, (b) Pass-out turbine.

4.4.

Prove that the steam flow through a multi-stage turbine is M = C .4 Pi - P2 v l v2 Where p1 and p2 are the steam pressures before the first stage nozzles and at exhaust, respectively, and v1 and v2 are the corresponding specific volume.

4.5.

What is the reason that the L. P. steam turbine is more efficient than L. P. steam engine with the condition that both of them utilize the exhaust steam from engine ?

'4.6. When is mixed pressure turbine needed ? 4.7. What is the function of heat accumulator ? 4.8. Write notes on the following-(a) Mixed-pressure turbine, (b) Heat accumulator 4.9

Derive the expression 1 Li a_ login ( 1 + — = -f logo L , m1 2

4.15. Find the theoretical steam consumption in kg/kWh of L. P. reciprocati g engine and a L. P. turbine operating with initially dry and saturated steam at 1.4 bar when the exhaust pressure is 0.17 bar the release pressure in the engine is 0.56 bar. Find also the steam consumption of the turbine when the exhaust pressure is reduced to 0.035 bar. Assume n = 1.135 in both case. Ans. 14.31: 11.33: 7.18 kg/kWh 4.10. The steam is supplied to a back-pressure turbine at full load before the first stage nozzle at pressure 27 bar and temperature 372°C. The exhaust pressure is 13.5 bar. Under this condition the efficiency ratio is 0.8, and the steam flow 6.95 kg/s At part load the steam is throttled from the above initial condition to a pressure of 20 bar, the back pressure being the same. If the efficiency ratio is 0.80 calculate the steam flow and the internal power developed. Ans. 4.44 kg/s, 3871W. 4.11. Following particulars relate to a pass-out turbine : Initial steam pressure = 20 bar, Initial steam temperature = 370°C, Pass-out steam pressure = 2.6 bar, Exhaust

208

Steam & Gas Turbines And Power Plant Engineering

steam pressure = 0.07 bar. When the turbine is developing its normal full load and giving out the normal quantity of heating steam, the total consumption is 1.8 x 104 kg/h and the mass of steam extracted is 7.24 x 102 kg/h. If the total, quantity of steam supplied is 1.18 x 104 kg/h and the amount of heat extracted is 14.64 million kJ/h calculate the power available. The internal efficiency ratio of H. P. and L. P. turbine group stages is 0.72. The heating steam is condensed but the resulting condensate is not cooled below the saturation temperature. The admission of H.P. steam is affected by nozzle control but the pass-out Ans. 1766 kW. turbine is regulated by a simple throttle valve. 4.12. A factory requires 2205 kW for its electric motors for driving machinery and 836,000 kJ per min. for process steam, at a temperature of 140°C. Two alternative schemes are considered. (a) The power is taken from the main electricity grid with an overall thermal efficiency of, from fuel to motors, of 20%. The process steam is raised in a special boiler with an overall efficiency of 80%. Alternatively, (b) The power is generated in the factory steam power plant. The initial steam conditions are 28 bar and 375°C and the condenser temperature is 27°C. The efficiency from turbine rotor output to electric motor outputs is 70%, the internal isentropic efficiency of the turbine it 80%, the boiler efficiency is 85%. Process steam is obtained by bleeding off the required amount of steam from the same intermediate stage of the turbine (Assume a straight line condition curve on the Monier chart). Estimate the relative merits of the two schemes from the point of view of fuel economy. 4.13. Develop a software for the design of back-pressure and Pass-out turbine. 4.14. Develop a Software for the analysis of back-pressure and pass-out turbine. 4.16. Cyclic variation of steam consumption rate of a reciprocating engine is given by m = 1.82 + 1.36 sin (-P—) 0.07 where p is the pressure in bar and t is the time in seconds. The total time of the cycle is 120 sec. The exhaust steam is utilised in a L.P. turbine which used all the available steam at a steady rate. A heat accumulator is used to store the surplus exhaust steam. If the lower pressure is 1.08 bar and the higher pressure is 1.45 bar find the mass of the steam to be stored and regenerated each cycle and the total mass of water to be stored in the accumulator. Ans. 52 kg, 3150 kg. 4.17. In a rolling-mill engine, for the first five seconds of rolling the steam consumption increases uniformly from zero to 6.33 kg/s and then diminishes at a uniform rate to 2.02 kg/s at the end of the 30 seconds from the commencement of rolling. The engine then comes to rest and the interval between the successive rolls is one minute. The L.P. turbine is supplied with steam at 1.4 bar dry and saturated from the exhaust of rolling-mill engine. The turbine exhausts at 0.07 bar and the efficiency ratio'N: 0.64. Calculate the power developed by the turbine. Find the total mass of water to be \stored in a heat accumulator to be used in conjuction with the exhaust turbine and operating under the following conditionsMaximum steam pressure = 1.54 bar, Minimum steam pressure = 1.26 bar Assume that the whole of the water in the accumulator is at the same temperature at the instant. Ans. 573.3 kW; 5600 kg

Back-Pressure, Pass-out and Mixed Pressure Turbine Cycles 4.18.

209

In a collery winding engine, during first 20 sec of winding from rest, the rate of steam consumption increases uniformly from zero to a maximum and then remains constant for a further 20 sec. When steam is shut off and engine is allowed to coast the total time of winding and stoppage is 2 min and during each wind the engine consumes 273 kg of steam. The exhaust steam from the winding engine passes to L.P. turbine and a heat accumulator. It is assumed that the turbine may utilize the whole of the exhaust steam at a steady rate. Find the minimum mass of water which must be stored in the heat accumulator if the maximum and minimum Ans. 15300 kg. engine exhaust pressure are 1.4 and 1.12 bar.

5 Fuels, Combustion, Combustion Equipment and Fuel Handling Systems

Steam power plants use fossil fuels. The choice of fuels, their preparation and feeding, and their methods of firing deserve special attention. The increasing worldwide demand for energy has focused attention on fuels, their availability and environmental effects. The world energy reserves is given in Table-5.1. Table-5.1. World Energy Reserves. Fuel

Type

Energy ( x 1015 MJ)

Fossil

Coal, oil and gas

32 .6

Fissile

Uranium and thorium

600

Fusile

Deuterium

1010

This chapter' deals with combustion of fossil fuels, their combustion equipments and fuel handling systems for steam generators. 5.1. Types of Fuels for Steam Generators. The primary fuels used in steam generators (boilers) are coal, fuel oil and natural gas that are preserved in the sedimentary rocks. Besides these, industrial wastes like blast furnace gas, coke oven gas, refinery gas, sugar factory refuse (bagasse), saw mill wood. dust, rice husk, etc. are also used as boiler fuels as a boost to one of the primary fuels. The calorific value of industrial waste is low. 5.1.1. Coal. Coal is the primary energy source and available in India in large quantity. As per geological order of formation, coal may be of the following types : (i) Peat, (ii) Lignite, (iii) Subbituminuos, (iv) Bituminuos, (v) Subanthracite and (vi) Anthracite, with increasing percentage of carbon. After anthracite, the formation of graphite takes place. Anthracite is often difficult to burn as it contains more than 86% fixed carbon (in amorphous form), and less volatile matter. Bituminous coal is the largest group containing 46

•

Fuels, Combustion, Combustion Equipment and Fuel Handling Systems

211

to 86% of fixed carbon and 20 to 40% of volatile matter which helps in burning. Lignite is the lowest grade of coal containing moisture as higher as 30% and high volatile matter. Peat contains about 90% of the moisture and not attractive at all as fuels for utility boilers. The analysis of coal is done either by proximate or ultimate analysis. The properties of coal are described by swelling index (an index of caking of coal), grindability (power required to grind the coal to a specified particle size), weatherability (stock piling ability for a long period). Sulphur content (should be as minimum as possible), heating value (low and high), ash softening temperature), and spontaneous combustion. The heating value of coal is HHV = 33.83 C+ 144.45 (H — 8) + 9.385 ; MJ/kg LHV = HHV — 2.395 mW ; MJ/kg

(5.1) (5.2)

where HHV and LHV are high and low heating values respectively. 5.1.2. Fuel oil. Liquid fuels are an excellent energy source as it is easy to handle, store and burn. They are primarily a mixture of hydrocarbon compounds. They are obtained by refining the crude oil [C (83-87%), H2 (11-16%), and N2 (0-77%), S (0-4%)] which is believed to have been formed during past geological ages from decayed marine life, both vegetables and animals. In the refining process, it is distilled into a number of fractions lighter and heavier. The lighter fractions (having lower boiling point) like gasoline, aviation fuel, kerosene, light diesel oil, heavy diesel oil, lubrication oi) and so on are principally transportation and machine fuels. The heavier fractions (having,higher boiling point) are used for boiler fuels and chemical production. The physical properties of oil are described by specif G gravity, viscosity, pour point, flash point, heating value, etc. 5.1.3. Natural and Petroleum Gas. The formation of natural gas took place millions of years ago from decaying vegetable matter generally alongwith petroleum. The oil is taken out by drilling oil wells into the geologic formation containing the gas'and trapped oil. The best method of transporting the natural gas is through pipelines. At present in India, there is one HBJ (Hazira- Bijaipur-Jagdishpur) pipeline. Many new pipelines will be coming up from Godavari basin. It is the cleanest fuel among all fossil fuels. The calorific value of fuel varies from 33.5 to 40 MJ/m3. It is-most suited for utility boilers, power gas turbine and combined cycle power plants. Since the major constituent of all natural gas is methane (92 to 94%) whose critical temperature is —83°C, so cryogenic temperatures are required to keep as a liquid at moderate pressure (-100°C at 36 bar). That is why liquid natural gas (LNG) is transported by special tankers and stored in spherical vessels to be used when needed. For automobiles, compressed natural gas (CNG) is gaining ground as a alterate fuels. Liquid petroleum gas (LPG) belongs to propane, propylene, butane, butylene, etc. group is liquefied under moderate pressures at normal temperatures. At present, it is widely used as domestic fuels. It is also used as a supplement to natural gas. 5.1.4. Coal-oil and Coal-Water Mixture. Liquid fuels using mixtures of fine coal in oil is known as colloid fuel. Now-a-days it is known as coal-oil mixtures (COM). It has been observed that combustion of colloidal fuels upto 40% coal concentration with 90% of the coal powder having size 61 µm is comparable to coal or oil fuel. It is cheaper than oil and having less ash content than coal. Coal-water mixture (CWM) is superior to COM. It is often preferred to replace nil If

Steam & Gas Turbines And Power Plant Engineering

212

contains 70 to 80% of coal mass. It is prepared in the form of slurry and fired like oil. 5.1.5. Synthetic Fuels. Synthetic fuels are gaining ground. They are called synfuels and are in the form of gaseous and liquid fuels produced largely from coal in an economical and environmentally acceptable manner either through coal gasification or coal liquedification. These fuels are used as utility boilers but also as domestic, industrial and transportation purposes. 5.1.5.1. Coal Gasification. In the past, there had been widespread use of coal gas produced by destructing distillation of coal for illuminating and cooking purpose. If the coal is heated in the absence of air, coal is carbonized to coke by removing its volatiles and using them as a byproduct gas. In the past, this gas was distributed in urban areas as town gas. The coke so produced is burnt in big beds with less than the stoichiometric quantity of air to yield•the producer gas. 2C + (02 + 3.76 N2) —> 2C0 + 3.76 N2 coke air producer gas

(5.3)

Further, C + H2O CO H2 (5.4) coke steam water gas The modern gasification techniques are based on the producer and water gas reactions and are continuous and efficient processes. Fig. 5.1 shows the basic steps in coal gasification process. The coal is first ground into a sandlike powder and then preheated and dried (by air and steam) to reduce caking during conversion. For a caking coal a high-temperature pretreatment is often used by steam to give the coal particles a thin coating of oxygen to prevent sticking. Three gas mixtures classified according to heating values are produced by coal gasification namely low, medium and high heating gases. The process illustrated by the lower Air Steam

Steam

Jr

.1

Gasification CO, CO2

Gas

CH4, H2

Gas Shift conversion

Pipe line Methanation CO, CO2 (nickel catalyst) Gas removed CH 4

quality gas (HHV)

0 Cu

Tar and dust removal

a o. Cu

Air

Sulphur removal

Steam

0

Gasification Gas CO2' CO2 CH2, H2 N2

Gas clean Up

Gas

SO2 removal

LHV gas synthesis gas CO, CO2, H2, CH4

Tar and dust removal

Sulphur removal

Fig. 5.1. Coal Gasification Process.

Fuels, Combustion, Combustion Equipment and Fuel Handling Systems

213

flow sheet (Fig. 5.2) yields a low heating value gas while the upper flow sheet a high heating value gas, called pipeline gas with properties close to natural gas (CO +3H2 —> CH4 + H2O) having heating value 38 MJ/m3. There exists another method of gasification, called hydrogasification in which the fluidized coal is gasified directly with hydrogen - rich steam to a methane-rich gas that requires very little additional shifting. The overall reaction is expressed as CH08 + 0.55 H2O + 1.15 H2 --> 0.575 CH4 + 0.425 CO2 (5.5) The conversion efficiency of this method is higher than the above mentioned method. Now low heating value gas finds an attractive application as fuel in combined cycle power.plant. There are various commercial coal gasification processes namely Lurgi, Winklr, Kopper-Totzek, Synthane processes, etc. Fig. 5.2 shows Synthane coal gasification process developed by US Burease of Mines. In this process, dry crushed coal is fed from a pressurized hopper into a preheater which is also fed with steam and oxygen. The reaction of coal with oxygen and steam takes place at 400°C, 7 MPa to devolatilize the coal and prevent caking. The coal mixture flows to the gasifier where it is partially gasified in a dense phase at 800°C and later at 1000°C, 7 MPa. Removal of unreacted char and ash is accomplished from the bottom of the vessel. The gas passes through a cyclone separator which removes dust and tar. For further cleaning the gas passes through shift converter, purifier clean up vessel and nickel catalyst vessel and finally comes out gas pipeline gas (CH4) of high heating value. Treated coal

9

e, Cyclone separator

0 0 ,..,te0 000 ry o 0 06 0

Coal Stear50

0, Gasifier 0 b O ° o Coal

I Oxygen

Stear3 o Oxygen )

D 0 0 c7, 0 0

o

Gas

Dust and tar

0'

0 6 o '

Char and ash

Methanation Nickel catalyst Pipeline gas (CH4) of high heating value

Fig. 5.2. Synthane Coal Gasification Process.

214

Steam & Gas Turbines And Power Plant Engineering

5.1.5.2. Underground Coal Gasification. Underground coal gasification can be carried out economically in those deposites of coals which could not be mined competitively by other processes. The gasification of coal is carried out in situ and the gas so produced is conveyed to the surface for utilisation in various requirements. Two deep drilled holes are made-one for supplying air and other for product gas (CO, CO2, CH4, H2). With initial ignition combustion takes place at the bottom of air hole and the combustion zone proceeds towards the product gas hole forming carbon dioxide (C + 02 —> CO2). While in reduction zone (ahead of combustion zone), carbon monoxide is formed (CO2 + C -4 2C0). Due to the moisture in the coal CO and H2 are formed (C + H2O —> CO + H2). The heating value of product gas is low, about 5.5 MJ/m3. The situ coal gasification is attractive due to (i) capacity to extract energy from inaccessible reserve of coals, (ii) reduction of mining personnel and equipments, (iii) safer and minimum occupational hazards, (iv) fuel production, (v) easy to handle SO2 formed. 5.1.5.3. Coal Liquefaction. Due to shortage of world petroleum supplied and rapidly increasing cost of oil, the interest of producing liquid fuel from coal has revived. In South Africa, major oil demand is met by coal Liquefaction by Fischer- Tropsch process. Coal has a ratio of hydrogen atoms to carbon atoms of only 0.8 : 1, while in petroleum this ratio 1.75; 1. That is why the conversion of coal into a liquid fuel requires the addition of hydrogen to the coal. There are three basic processes of coal Liquefaction namely (i) hydrogenation, (ii) Catalytic conversion and (iii) hydropysolysis. (i) Hydrogenation Process. In this process, coal and catalyst are suspended as a slurry, which reacts with hydrogen at a high pressure and moderate temperature to form liquid hydrocarbon. (ii) Catalytic Conversion Process. In this process, a synthesis gas is produced from the coal as described earlier. The hydrogen and carbon monoxide of the gas are then combined in the presence of catalyst to form a liquid hydrocarbon. (iii) Hydropysolysis Process. This process involves heating the coal beyond 450°C. The fraction of coal volatilized greatly upon heating exceeds the volatile matter in the coal. The hydrogen pulverized coal is flash pyrolyzed and upto 50% of the coal can be' liquefied. Fischer-Tropsch Process. This is an old method but still used in South Africa. This process first produces a mixture of CO and H2 from coal and steam followed by catalytic reactions at about 150°C and 150 bar which yield a range of hydrocarbons from gaseous methane to higher liquid hydrocarbons. These hydrocarbons are then separated with methane going as pipeline gas and the rest going to different liquid fuels. 5.1.6. Industrial Wastes and Byproducts. There are various types of industrial wastes which are combustibles in nature and therefore are receiving increased attention to utilise them as fuels for steam generators serving two purposes — one is disposal and other is reduction in use of oil. Such industrial wastes include refinery gas, coke-oven gas, regenerator gas, blast furnace gas, liquid wastes, wood wastes, sugarcane (waste called bagasse), solid wastes, municipal vastes, refuse-derived fuels (RDF), etc. Refinery gas is a high heating value gas which is produced during the conversion of crude oil to gasoline and other refinery products. It is often blended with low heating value gas prior t% combustion. Generation of coke-oven gas is accomplished during the production of coke from raw coal in a coke oven. It is a mixture hydrogen (50%), methane (33%) and other gases and its heating value ranges fro 6.2 to 21.3 MJ/m3. Regenerator gas is produced in refinery by catalytic cracking process at about 540°C and has low heating

Fuels, Combustion, Combustion Equipment and Fuel Handling Systems

215

value. It is burned at the site. Blast furnace gas is produced during reduction process of iron oxide to iron in blast furnace. This gas contains about 30% CO and high dust loading and thus requires cleaning for burning. There are large number of liquid wastes such as solvents, waste oils and oil sludges, phenols, tars, combuitible chemicals, greases and fats. These liquid wastes have different heating values and other properties. Solid wastes include wood waste and sugar factory refuse. Sugarcane waste (bagasse) is the residue of sugarcane after juice extraction. Its heating value is 8.4 to 9.77 MJ/kg Boilers installed in sugar industry use bagasse to produce steam. Solid wastes generated by industrial and domestic processes in the form of garbage are often used as boilers fuel in incinerators, particularly in cities either to generate electricity or to use as process heat. Use of unserviceable types because of high heating value (32 MJ/kg) is becoming popul4 as a fuel for boiler. It is very common in USA. 5.1.7. Biomass. Biomass is an organic matter produced by plants both in land and water and includes forest crops, the crops grown in energy forms and animal manure. It is renewable resource of energy as the plant life renews and adds to itself every year unlike fossile fuels which take millions of year to form and nonrenewal. In biomass, it is the solar energy stored by way of photosynthesis as given below(5.6) 6H20 + 602 6 CO2 + 12 H2O -----p C6111206 + Air

Soil

Chlorophyll

glucose or starch

ejected by air transpiration

In other words, biomass (also called biofuels) is the products of photosynthesis. These are bulky and contain large amount of water, so it is not economical to transport then over long distances rather it is utilised by direct burning in region close to the source. They can also be converted to a variety of gaseous, liquid or solid fuels. The following are the various types of land crops. (i) ,Sugar crops, such as sugarcane refuse or bagasse. (ii) Herbaceous crops, such as non-woody plants. (iii) Forest crops, such as eucalyptus, sycamore, sweetgum, cultural hybrid poplar, etc. Animal and human wastes are the indirect land crops from which methane and ethylene can be produced while their manure values are retained. Aquatic crops which include seadwoods, marine algae and various kelps are grown fresh, sea and brackish water. There are three types of bioconversion process (i) Direct combustion such as wood waste and bagasse (ii) Thermochemical conversion (iii) Biochemical conversion In thermochemical conversion process there are two stages, gasification followed by liquefaction. Gasification is performtd by heating with limited oxygen and liquefaction yields methanol and ethanol. In Biochemical conversion process, there are two routes namely anaerobic digestion and fermentation. Anaeration digestion is the bacterial decomposition of organic matter (biomass) in the absence of air to produce gaseous mixture (biogas) of methane and carbon dioxide in an approximately 2:1 volume ratio. CO2 so produced can be converted to synthetic natural gas (SNG) and the sludge formed due to anaeration digestion serves as manure. In fermentation process, the breakdown of compir molecules in organic compounds

216

Steam & Gas Turbines And Power Plant Engineering

takes place with the help of a ferment such as yeast, bacteria, enzymes, etc. Grains and , sugar crops are converted by fermentation into ethanol which is mixed with gasoline to produce gasohol fuels for automobiles. In non-fertile areas, sycamore and eucalyptus trees are grown to produce energy by combustion. The heating values of some waste bioproducts is given in Table 5.2. Table 5.2. Heating Value of Some Waste Bioproduots. Waste Bioproducts

Heating value (MJ/kg dry)

Wood chips Bagasse Cereal straw

186-20.9 9.5 16-17

Organic refuse

13.2

5.2. Combustion Reactions. Combustion is defined as the high temperature oxidation of the combustible elements in coal and fuel oil are the presence of carbon, hydrogen and sulphur contents. The basic chemical equations for complete combustion of fuels are expressed as C + 02 -p CO2'' 2H20 + 02 -+ 2H20; S+02 -4 SO2 (5.7) In the case of insufficient oxygen, combustion will be incomplete forming CO. Thus 2C + 02 —> 2C0

(5.8)

It is worth to note that for burning a fuel completely, four basic conditions must be fulfilled, namely (i) supply of enough air for complete combustion, (ii) creation of enough turbulence for thorough mixing of air and fuel, (iii) maintaining a furnace temperature sufficiently high enough to ignite the incoming mixture of air and fuel, and (iv) providing a furnace volume sufficiently large enough to allow time for complete combustion. In other words, combustion is governed by a four letter word, "MATT" where M stands for sufficient mixture turbulence, A stands for proper air-fuel ratio, T stands for temperature and other T is for time-enough for combustion. The analysis of coal is performed either by proximate (volume basis) analysis or by ultimate (mass basis) analysis. 5.2.1. Stoichiometric Air-Fuel Ratio. The ultimate analysis of fuel (coal) shows the following components on mass basis : carbon (C), hydrogen (H), oxygen (0), nitrogen (N), sulphur (S), moisture (M) and ash (A). Therefore C+H+0+N+S+M+A=1.0 The mass of oxygen needed for oxidation process are calculated as follows— (i) Similarly (ii) C+ = CO2 2H2 + 02 = 2H20 02 12 kg 32 kg 44 kg 4 kg 32 kg 36 kg 1 kg 2.67 kg 3.67 kg 1 kg 8 kg • 9 kg C kg 2.67C kg 3.67C kg H kg 8H kg 9H kg (iii) S + 02 32 kg 32 kg

=

SO2 2 kg

(5.9)

Fuels, Combustion, Combustion Equipment and Fuel Handling Systems

217

5kg 5kg 25 kg Therefore, mass of oxygen required for complete combustion of 1 kg of fuel m02 = 2.67C + 8H + S — 0 where 0 is the oxygen in the fuel if any. As we know that air contains 23.2% oxygen by mass, therefore, theorectically air required for complete combustion of lkg of fuel 8 mar in 02 2.67C — = Air-fuel ratio — 0.232 — 0.232 + 0.232 mf or

S — 8 + 0.232

A/F = , = 11.5C + 34.5 (H — ) + 4.3S m 8 I

(5.10)

where C, H, 0 and S stands for mass fraction of carbon, hydrogen, oxygen and sulphur It is a well known fact that complete combustion of fuel cannot be achieved without supplying excess air than stoichiometric. The percentage of excess air is given by % excess air — maa

—

ma' x 100

m as

(5.11)

where maa is the actual air supplied for complete combustion of 1 kg of fuel. Due to excess air, a term dilation coefficient may be defined m d = maa

(5.12)

al

For large utility boilers, the percentage of excess air varies from 15 to 30%. Natural gas in which methane constitutes about 94% is also one of the fuel for utility

boilers. The combustion of methane is expressed as (5.13)

CH4 + 202 CO2 + 2H20

Since for each mole of oxygen taking part in combustion reaction, there are 0.79/0.21 or 3.76 moles of nitrogen, so for combustion of methane with 150% excess air CH4 + 2(1.5)02 + 2(3.76)(1.5)N2

CO2 + 2H20 + 02 + 7.5(1.5)N2 '

The actual air required for 1 kg of fuel is given by 3.04 N2 Cab maa aa =

CO + 2

N

CO 0.768

(5.14)

where Cab is the fraction of carbon in fuel which has been burnt to CO2 and CO. and N is the nitrogen present in fuel (%) if any nitrogen is absent, N = 0 The analysis of combustion products is carried out by (i) Orsat analyser (ii) Haldane apparatus, (iii) Infra-red gas analyser, and (iv) gas chromatograph. 5.2.2 Combustion Equation. In order to find the combustion equation based on the ultimate analysis of fuel and volumetric analysis of combustion products, let us consider the following example. C = 62%, H = 4%, S = 3.0%, 0 = 4%, N = 2%, H = 4% and A = 21% The exhaust gas has the following volumetric analysis,

218

'Steam & Gas Turbines And Power Plant Engineering

CO2 + SO2 = 13%, CO = 2%, 02 = 4%, and N2 = 83% Let a moles of oxygen be supplied for 100 kg fuels thus combustion equation for above example may be written as 62 4 3 4 —S+— —C+- 1 2 + 32 3202 +a02 +3.76aN2 12 2 —> b CO2 + dC0 + eS02 + f02 + gN2 + H20

(5.15)

By equating the coefficients of C, H2, S, 02 and N2 the constants are evaluated. For example, let us consider the combustion of propane gas with stoichiometric air C3H8 + 502;+ 5(3.76)N2 —› 3CO2 -I- 4H20 + 18.8N2 (5.16) With 80% theoretical air, the above combustion equation becomes with addition of formation of carbon monoxide due to incomplete combustion C3H8 + 5(0.8)02 + 5(3.76)(0.8)N2. —> aCO2 + bCO2 + 4H20 + 15.04 N2 (5.17) carbon balance gives, 3 = a + b oxygen balance gives, 8 = a + 2b + 4 By solving :, a = 2, b = 1, hence combustion equation is C31-18 + 402 + 15.04 N2 —) 2C0 + CO2 + 41120 + 15.04 N2 5.2.3. Dew Point Temperature of Exhaust Gases in Boiler. The partial pressure of water vapor in the mixture of gases of combustion product having total pressure of 1 atm is given by PH20

=

20

•p=

2 o • (atm)

=xH o atm Z

(5.18)

where, xH o = the mole fraction of water vapor formed 2

nH 0 2

nH 0 2

Ntotai

nCO2 + nC0 + no2 + nS02 + nfi2 + nH2O

(5.19) The saturation temperature at the partial pressure of water vapor is called the dew point temperature (dpt) as shown in Fig. 5.3. The flue gases coming out from superheaters/reheater is further cooled in economiser- and air preheater in order to minimise the exhaust gas losses through chimney. It is essential that the exhaust gases should not be

T

-ms s Fig. 5.3 Dew Point Temperature of Exhaust Flue Gases.

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cooled below its dew point temperature. If it is cooled below the dpt, the water vapor in exhaust gases condenses into liquid droplets which combines with SO2 or SO3 to form acid (H2SO4). The acid so formed corrodes the metal surfaces of the duct or stack while flowing through it. That is why the temperature of flue gases leaving the preheater is kept around 160°C. 5.2.4. Heating Value of a Fuel (Coal). The heating value of coal may be calculated based on the heat of reaction of its constituents which is given in Table-5.3. Table-5.3. Heats of Reaction of Coal Constituents. Constituents

Formula and State

Products of combustion and state

Heat of reaction (kJ/kg mole)

Carbon (coke)

C (s)

CO2 (g)

— 407 x 103

Carbon

C-(s)

CO2 (g)

— 397 x 1.03

Carbon monoxide

CO (g)

CO2 (g)

— 283 x 103

Hydrogen

H2 (s)

H2O (1)

— 286 x 103

Sulphur

S (s)

SO2 (g)

— 291 x 103

The heat released by combustion of carbon (C) constituent in a 1 kg of coal is kg . m x j — 33917 kJ/kg C (kg) x 407 x 103 kg . kmole) 12k gole

g'

291 x 103 S — 90945 kJ/kg 32 It is worth to note available hydrogen for combustion is the total hydrogen less than 0 that required to combine with oxygen in the coal, (H — — , and the heat released by it 8 Similarly, heat released by sulphur (S) —

=

0) 286 x 103 2 8

143 x 103 H — 2 kJ/kg 8

Thus, the higher heating value (HHV) for complete combustion of 1 kg of coal is expressed as 0 HHV = 33.91 C + 143 (H — 8) + 9.094S MJ/kg (5.20) The above equation is very close to Dulong's formula given as HHV = 33.83 C + 144.45 H — 8) + 9.38S MJ/kg

C

. LHV = HHV-2.395 mw MJ/kg

(5.21) (5.22)

5.2.5. Optimum Excess Air in. Boiler. In order to maintain optimum combustion efficiency in boiler, the proper control of right amount of excess air is essential. The amounts of CO2 and 02 in combustion gases are the indexes of excess air and which is a function of fuel. Fig. 5.4 shows the variation of CO2 in flue gases with fuels and excess air while the variation of 02 in flue gases with excess air is given in Fig. 5. 5 Boiler losses are

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Steam & Gas Turbines And Power Plant Engineering

—*Excess air

Excess air Fig. 5.5. 02 Variation in Flue Gases.

Fig. 5.4. CO2 Variation in Flue Gases.

Incomplete combustion loss Optimum zone of excess air Exhaust loss Loos due to exhaust gas

t•—• Optimum excess air Excess air Fig. 5.6. Optimum Excess Air For Maximum Combustion Efficiency

estimated for different steam outputs and the excess air is then adjusted by controlling air supply to show the optimum value of CO2 or 02. By the method shown in Fig. 5.6 the optimum value of excess air for best combustion efficiency is determined and instruments are used to control this. The excess air is also calculated by equation proposed by Skrotzki and Vopat (1960). 02 — 0.5 CO % excess air x 100 0.264 /42 — (02 — 0.5 CO)

(5.23)

Where 02, CO, N2 are the volumetric percentage in the dry flue gases 5.3. Mass Balance of a Boiler Furnace of Steam Generator. For the analysis of boiler, mass balance is needed. Fig. 5.7 shows the mass entering rtif = 1 kg/s =C+H+0+S+N +M+A

Mdfg = CO2 + CO + P2 + N2 + SO2

Boiler furnace

-

rhiizo

9H + M + Ya ma

-

Mref.= A + C - Cab

Ma (02 + N2) —

Fig. 5.7. Mass Balance of a Boiler Furnace.

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Fuels, Combustion, Combustion Equipment and Fuel Handling Systems

and leaving the boiler furnace. 1 kg of coal with amount of air (ma) needed for combustion ) and refuse (mref ) are coming out. are entering while dry flue gases (Mdf. ), steam (il H0 2 Thus, tilf (1 kg coal/s) + ma = lildfg + ' fill 0 + thref or

(C +H + 0+ S +N+ M.+ A)+

or

ma = mdfg + 8 (H - 8)

=thdfg + 9H +M+ A +C- Cab

Cab — N - S ; kg/kg-fuel

(5.24)

Mass of dry flue gases, tindfg is expressed as thdfg

Cab (44 CO2 + 28 CO + 28 N2 + 32 02) 12 (CO2 + CO)

Cab [11 CO2 + 7

+ 8 02 + 7 (100 - CO2 - CO - 02)]

3(CO2 + CO) Cab [4 CO2 + 02 + 700]

3(CO2 + CO)

kg/kg fuel (5.25)

Using perfect gas law, the volume of flue gases (wet) per kg coal is expressed as [tit, Ifift o T 101.325 V = =g "-+ x 22.4 —g- x A Mdfg 18 273 Pg

m3/kg (5.26)

where pg = gas pressure, (kPa) and Mdfg = molecular weight of dry flue gas The mass of refuse (mref ) per kg coal is expressed as = .A mref

Aref

where, A = ash content in 1 kg of fuel = Aref Cref Aref = mass fraction of ash in the refuse. A Mass of unliunit carbon is refuse per kg coal = Clef mref = Cref Aref The fraction of carbon burnt to CO2 = Cab = C - Cre f

A Aref

•

(5.27)

5.4. Energy Balance in a Steam Generator. The fuel and air are supplied to the steam generator and due to complete combustion, the fuel releases its heating value. The heat so released is absorbed by feed water supplied by the boiler feed pump and it is converted into steam While passing through economiser (ECO), evaporator (EVA) and superheater (SH) and reheater (RH). Fig. 5.8 shows the control volume of a boiler furnace in which the various masses and energies entering and leaving are depicted. Based on 1 kg of fuel (coal), the useful energy and various energy losses may be computed which will determine the efficiency of a steam generator. (i) Useful energy is the energy absorbed by feed water

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222

RH steam out RH steam in

h

(1 -

rildfg, tg.

hg

mH2o mref

1

h i , ti , pi

111, / enf Feed water inlet

SH steam outlet

Fig. 5.8. Energy Balance in a Steam Generator. tit

= Q. =

=

tilf

[(h2 — hi) + (1 — Ma) (h4 — h3)]

kJ/kg (5.28)

where of = fuel burning rate, (kg/s),'ms = steam generation rate, (kg/s), Illst = mass fraction of inlet steam bled from hp turbine before reheating. (ii) Energy loss due to dry exhaust gases Qerh = Q2 = mdfgcPdig (tg — ta)

kJ/kg (5.29)

where thug = mass of dfg produced per kg of fuel, -c- = average specific heat of dfg, pdf, (kJ/kgK), tg = exhaust gas temperature °C, to = ambient temperature (°C) (iii) Energy loss due to unburnt carbon Q3 = (C — Cab)

Qua

407 x 103 12

— 33917 (C — Cab) kJ/kg

(5.30)

where C and Cab are unburnout and burnout carbon in thy gas per kg of fuel (iv) Energy loss due to incomplete combustion is caused by the formation of CO. The loss of energy per kg of C oxidized to CO = Coca —

283 x 103 — 2358.3 kJ/kg of carbon 12

Thus, energy loss due to incomplete combustion Qincomb =

Q4 = 2358.3 ( id X dfg kg carbo mdfg kg fuel)

Fuels, Combustion, Combustion Equipment and Fuel Handling Systems

•

x [44CO2

223

(g3) 12 (kg carbon) 28C0 x 28 (kg CO) + 28C0 + 3202 + 28N2 kg dfg

= 10100 rn

28 CO + 28C0 + 3202 + 28N2 2

dfg [44CO

kJ kg

(5.31)

(v) Energy loss due to moisture in fuel = Qmoist = Q5 = M[c (100 — tf ) + hfg + cpg (1g — 100)1

kJ kg

kJ

= M [4.18 (100 — 9+ 2256.8+ 2.09 (tg — 100)] T

(5.32)

where tf = temperature of fuel entering the fuel,( °C), tg = temperature of exhaust gas (°C), M = mass of moisture per kg fuel (vi) Energy loss due to hydrogen in fuel which converts into steam kJ QH = Q6 = 9H [4.187 (100 — tf.) + 2256.8 4- 2.09 (tg — 100)] kg (5.33) (vii) Energy loss due to moisture coming with air supplied kJ Qam = Q7 = ya ma c (t — ta) g kg pr where ya = specific humidity of air, kg moisture/kg dry air

(5.34)

P = average specific heat of superheated water vapor in air (viii) Energy loss due to ash and slag (refuse) Q = ref

Q8 = (A + C — Cab)Z_p (t,f urnace — ta) kJ/kg rer

(5.35)

where c average specific heat of refuse (ash) Prd

(ix) Energy loss due to convection and radiation from the boiler surface = Qar = Q9 = (ha + ha) As (Ts — ta)/In f ; kJ/kg

(5.36)

where he = convective heat transfer coefficient (W/m2K), hr = radiative heat transfer coefficient (W/m2K), A, = total boiler surface area exposed to ambient air (m2), is = temperature of wall surface of boiler (°C), ta = ambient temperature (°C) Energy released by complete combustion of 1 kg of fuel = high heating value (HHV) Energy utilized in heating the feed water = HHV — E energy losses = Q1; kJ/kg (5.37) 'The efficiency of steam generator

listeam gas = lboiler —

Energy utilized Qi Energy released HHV

pits (h2 — h1) + (1 — tiasi ) (h4 — h3)1 riaf x HHV '

HHV — E energy losses HHV

(5.38)

Based on the useful energy and energy losses, heat balance sheet of boiler in bar chart or graph or table form may be prepared. A tabular form of 'heat balance sheet is shown below (Table 5.4) :-

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224

Table 5.4. Heat Balance sheet per kg of dry fuel Heat (kJ/kg fuel)

Item

Heat (%)

Heat input : Heat supplied Expenditure (i) Heat consumed to generate steam (ii) (iv)

100

Total

100

5.5. Draught System of a Boiler Draught (or draft) is defined as small pressure difference causing flow of air and gases (in and out) through the boiler. It is essential to supply a sufficient quantity of air to

support combustion in boilers and to remove the products of combustion. This is done by producing draught. The functions of draught producing equipment are (i) to create a differential pressure sufficient to allow the desired volume of air flow into the furnace' at the required velocity (ii) to overcome the resistance of passage in boiler, and (iii) to discharge gases i.e., products of combustion at a sufficient height in order to avoid pollution to atmosphere. The amount of draught necessary for a certain boiler depends on the following :— (i) rate at which combustion takes place (ii) characteristics of fuel and its depth on the grate (iii) design of combustion chamber and the method of burning the fuel. (iv) resistances in flue gases circuits offered by baffles, tubes, superheaters, reheater, evaporator, economiser, preheater, etc. There are mainly two ways of producing draught : (a) natural (chimney draught) and (b) artificial (mechanical) draft. The artificial draught is produced by steam jet or by mechanical fans. Mechanical (fan) draught is of three types : (i) induced, (ii) balanced, and,(iii) forced. In large output boiler balance (forced and induced) draught is common. 53.1. Natural draught. The natural draught is produced by a chimney or stack and is caused by the density difference between the atmospheric air and the hot gas in the stack. For a chimney of height H meters (Fig. 5.9) the draught or pressure difference, Ap produced is expressed as •

AP = gH (pa -pg) ; NIm2

(5.39)

where pa = density of atmospheric (cold) air (kg/m3), pg = average density of hot gas in the chimney (kg/m3), g = acceleration due to gravity = 9.81 m/s2 Using perfect gas laws, we have [ Pa

Pg

= - H Ra T a Rg T g

(5.40)

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pa = pg = pat.= 101.325 kN/m2, Ra = Rg = 0.287 kJ/kgK

where

Ta = ambient (cold) air temperature (K), Tg = average hot gas temperature in the chimney, K =

(T0 + TH) 2 , To = hot gas temperature at inlet to the chimney K, TH = hot gas

temperature at exit from chimney, K Thus,

Ap —

g HP [1 1 — Ra Ta Tg

N/m2 (5.41)

From the above equation, it is evident that the natural draught produced is inversely proportional to the average hot gas temperature for a given H and pa. The draught produced increases with the increase of Tg but the boiler efficiency goes down due to increase in stack loss. Therefore, an optimum value of Tg is selected. The natural draught created by stack or chimney is not sufficient for modern boiler due to high fuel burning rate and large heat transfer surface areas, so mechanical draught is used The natural draught produced, Ap given by above expressions is the theoretical draught. Stack introduces pressure losses of their own caused by wall friction and the pressure equivalent to the kinetic energy of the gases leaving the stack. Thus, the actual draught produced is less than the theoretical value, Ap and is given by the following expression C2 Apaa = Ap — pg

(it+ D

; N/m2

(5.42)

where f is the friction factor, H is the stack height (m), D is the stack inside diameter (m), Cs is the stack exit velocity (m/s), .15g is the average gas density is stack (kg/m3). The exit velocity of flue gases at stack exit results in a plume rise AH above the actual. Virtual source _

t

Stack

----—± Gas

Draught ---------

1

I

Cold column gas

)1-(1

A

Hot column gas

Gases to atmosphere

Wind velocity

---

Concentration profile

He

H

Stack

Gas 1 Exhaust fuel gas

Fig.

iL_ Reference L_ level

5.9. Natural (Chimney) Draught.

Fig. 5.10. Dispersion of Flue Gases at Stack Exit.

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Steam & Gas Turbines And Power Plant Engineering

stack as shown in Fig. 510. Dispersion of the flue gases into the atmosphere is caused by the movement of the flue gases horizontally as well as vertically and their dilution by the atmosphere. The gases bend in the direction of wind flow. The plum height, AH is given by imperial equation prepared by Carron and Moses as given below0.5 C .D AH = 2.62 — 0.029 s C., (5.43) where Qc is the heat emission (W) and C., is the wind velocity at stack exit (m/s). 5.5.2. Mechanical Draught'. Centrifugal or axial flow fans are used to produce mechanical draught. There are three types of mechanical draught : (i) induced draught produced by induced draught (ID) fan, (ii) forced draught produced by forced draught (FD) fan and (iii) balance draught produced by the combination FD and ID fans. When either one is used alone, it should overcome the total air and gas pressure losses within the steam generator. 5.5.2.1. Forced Draught. It is worth to note that forced draught fans are installed at inlet to the air preheater and so they handle cold air. The power input is given by Power input = friav dp = m f (A/F)x v x AP/TIFD;"1 (5.4 ) where Mf is the mass flow rate of fuel (kg/s), A/F is the air-fuel ratio, v is specific volume of inlet air (m3/kg), Ap is the pressure developed by fan to overcome resistances in flow path (kN/m2) and TIFD is the overall efficiency of FD fan. Since FD fans handle cold air, so they consume less power and have less maintenance problems. In general, in view of good reliability, two forced draught fans operate in parallel, each capable of undertaking 60% of full load air flow in the event of breakdown of one of them. In many large steam generators and almost in all marine applications, FD fans alone are used and they maintain the entire system upto the stack under positive gauge pressure. As a result, the furnace has to be gas tight. The height of the stack for forced draught fan alone is shorter in height just to meet the pollution regulation. 5.5.2.2. Induced Draught. Induced draught fans are normally located at the foot of stack and so they handle hot combustion gases. As a result, the power requirements are greater than that of FD fans.. The maintenance problems due to handling of corrosive combustion gases and fly ash are increased. Induced draught fans are seldom used alone. They discharge gases essentially at atmospheric pressure and place the system under negative gauge pressure in upstream side. The power input is given by Power input = f rog vg Ap = rinf [(1 + A/F) vg x 4/D — ] , kW

(5.45) 11D where vg is the specific volume of hot gas (m3/kg), Apo is the pressure head developed by ID fan (kN/m2) and ri/D is the overall efficiency of ID fan. 5.5.2.3. Balance Draught. As the name implies, balance draught system uses both FD and ID fans. The FD fans located at the inlet of preheater push atmospheric air through the air preheater, dampers, various air ducts and burners into the furnace and ID fans located at the foot of stack such out the flue gases through the heat transfer surfaces in the superheaters, reheaters, economiser, gas-side air preheater and dust collectors and discharge into the stack as shown in Fig. 5.11. The natural driving pressure is generated by the stack due to its height. When both FD and ID fans are used, the furnace is said to

Fuels, Combustion, Combustion Equipment and Fuel Handling Systems

227

Dampers APH

Windbox and burners

Air , ducts I

Atmospheric pressure SH

Furnace

RH

1 1— ECO

APH 0 8

Fig. 5.11 Balance Draught System. operate at approximately atmospheric pressure. In real situation, it is maintained at slightly negative gauge pressure ensuring leakage if any would be invariably inward. Due to the advantages of balance draught system, designers adopt this system for modem boilers. 5.5.3. Pressure and Flow Measurement in Gas Flow Path. The stagnation or total pressure (p0) at a point in gas flow path of a steam generator is sum of static pressure (p) and dynamic head (pd). Thus, Po = P +Pd = P.+ Pg C2I2

(5.46)

The static pressure is measured by tapping wall connected to a water tube manometer while dynamic pressure is measured by a pilot tube connected to a water tube manometer. Static and dynamic pressure may be simultaneously measured by a three hole probe. Since the pressure in boiler gas flow path is slightly more than atmospheric or less than atmospheric, so it is expressed in mm of water. One meter head of water =p=pgh = 103 x 9.81 x 1 = 9810 N/m2 The dynamic pressure head in terms of mm of water, is expressed as

Pd

Pg C2 in3 p C2 2 = 9810 x 2 x — 19.62 min of H O

If the area of the duct is A, the volume flow rate of gas in the duct is given by

1/2

V = A [19.62p/pg]

= 4.43 A 1.57p; 1 m3IS

(5.47)

where pd is in mm of water and pg is in m3/kg. 5.5.4. FD, ID and other Fans. There are two types of fans—centrifugal and axial. Centrifugal fans are cheaper and rugged than axial flow fans. The blades (or vanes) of centrifugal fan may be backward, radial and forward swept. Generally, FD fans are centrifugal type with backward swept blades. Since ID fans handle dust-laden flue gases, so they are operated at low speeds and therefore, centrifugal fan with forward swept blades or radial blades may be used. For controlling the output of fans, two methods are used. They are damper control and variable-speed control. Dampers are generally placed at the output of the fan. Variable-

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228

speed control has less power consumption and it is the most efficient method of fan control but the control equipment is costly. The effect of speed on the fan performance is as follows— Volume flow oc N where N is the speed in rpm. Pressure head oc N2 and Power input oc N3 The types of drives used for fan are :—(i) Variable-speed steam turbine, (ii) hydraulic coupling, (iii) variable-speed dc motor (iv) multijpeed ac motor and (v) electronically adjustable motor drive Over and above of mechanical draught fans, two other types of fans are used in power plants. One is primary air fan to supply •air to dry and transport pulverized coal to the furnace and the second is gas recirculation fan for steam temperature control system. Fans are noisy in operation, so special care is taken to dampen the noise. 5.6. Heat of Combustion. The heat of combustion from reactive system is obtained by applying the steady flow energy equation considering the chemical energy which is ignored in non-reactive system. 5.6.1. Open System. Fig. 5.12 shows a chemically reactive system. Applying steady flow energy equation ignoring kinetic and potential energy, we have HR + Qcv = H + W P

(5.48)

where HR and H are the enthalpies of reactants and products respectively evaluated for their constituents at their respective pressure and temperature. Q, is the net heat added to the control volume (which is usually negative, as the combustion heat is rejected to the surroundings) and Wcv is the net workdone by the control volume. Since the reactants and products are usually composed of several constituents, so the above equation may be expressed as mass basis

E R

(mh)+ Qom, =

E

(mh)+ Wcv

p

(5.49)

For example, to define the enthalpies value, consider the combustion of ethane (C2H6) with oxygen C2 116 + 3.502 —> 2CO2 + 31120 (5.50) It is worth to note that the enthalpies of various reactants and products are those that start at the same datum of composition, temperature and pressure, which are arbitrarily chosen as the elemental substances at 25°C and 1 atm respectively. For example, C2H6 is formed from elemental carbon C and hydrogen H2, CO2 from C and 02 on so on. These are exothermic reactions, that when they begin at 25°C and are cooled back to 25°C after the reaction takes place, yield 2817.2 kJ/kg and 8946.8 kJ/kg of product respectively (Fig. 5.12). In other words, in steady flow at 25°C, formation reactions are

and

2C + 3H2 —> C2H6 + 2817.3 kJ/kg C2H6

(5.51)

C + 02 --+ CO2 + 8946.8 kJ/kg CO2

(5.52)

The quantities 2817.3 and 8946.8 kJ/kg leave'the system and hence are negative. They are called enthalpies of formation (hp. The value of hf° for various substances at 25°C and 1 atm are given in Appendix in the last of this book. In general, chemical equations such as eq. (5.49) are balanced in terms of moles and not masses and since m = tiM where n is the number of moles and M is molecular weight,

Fuels, Combustion, Combustion Equipment and Fuel Handling Systems

229

C.V.

wcv r C

25°C 1 atm C2H6

25°C 1 ahm

Reactants

Products

Qcv

Q v

C.V.

= - 2817.3 kJ/kg C2H6 Fig. 5.13. Formation of Ethane in an Open Reactive System.

Fig. 5.12. An Open Reactive System

so eq. (5.49) takes the form

E

(n M

R

=

E

(n M Wc,

p

(5.53)

where h is in kJ/kg mole. At any other temperature than 25°C and 1 atm hT ,p = h °f

(5.54)

5.6.2. Closed System. In many situation burning takes place in closed system such as cylinder or bomb. The combustion equations for fuels burning in a closed system may be obtained by writing the energy balance for closed system. Thus UR + Q = U + W

(5.55)

where W is the non-flow work For gases, H = U + pV = U + nRoT where Ro is universal constant, hence

E

(n M hf — n Ro + Q =

E

(n M hf — n Ro l) (5.56)

5.7. Heating Values. There are two types of heating values-higher heating value (HHV) and lower heating value (LHi 9. The former is the heat released when water vapor is the products due to the combustion of hydrogen in the fuel condenses, the later when it stays in the vapor state. Based on the analysis of open system, the heating value (HV) is given by putting Wc, = 0 and replacing Qc,, by HV for heating value HV = [E (n M hf ) —

E

(n M hf )1 T i

(5.57)

where T1 indicates the standard temperature. This heating value is sometimes called the enthalpy of combustion and depends oft Ti ; whether the water vapor formed condenses giving HHV otherwise LHV. It also depends upon whether the reactants fuels were in liquid or vapor state, as it takes a certain amount of energy to vaporize a liquid fuel such as octant.

Steam & Gas Turbines And Power Plant Engineering

230

Based on the closed system, Mini. is sometimes called the internal energy of combustion. The heating values of fuels are given in Appendix. 5.8. Theoretical (or Adiabatic) Flame Temperature. A fuel burning with no heat exchange with the surroundings and no work done will result in the theoretical or adiabatic flame temperature. It is greater for stoichiometric than either a lean or rich mixture because, as above, a lean mixture has a dilution effect, whereas a rich mixture results in incomplete combustion. It will also be greater if the fuel is burned in oxygen than in air because of the dilution effect of nitrogen. At the adiabatic flame temperature, Q,,,= 0 and We, = 0, so (n M h ) =

(n

Al II )

(5.58)

Adiabatic flame temperature so calculated are higher than actual due to dissociation of some of the products takes place at higher temperature. However, the effect Of dissociation is lower for lean mixtures because the excess oxygen tends to drive the reactions towards completion, and for higher furnace pressure. That is why, the effect of dissociation is not significant in steam generator furnace combustion which occurs with excess air at about 1500°C. 5.9. Free Energy of. Formation. Gibbs function is defined as G = H —TS is called free energy. For equilibrium in a system at constant temperature and pressure, DG.fp < 0. For a process at constant T and p, G always decreases and becomes a minimum at the state of equilibrium. For driving force in a chemical reaction at constant temperature AG = AH — T AS

(5.59)

The free energy of the chemical elements at 1 atm and 25°C is arbitrarily assigned zero value. Thus, gf° is the free energy of formation of a compound when one mole of the compound is formed directly from its constituents elements, both the reactants and the products being at 25°C and partial pressure of 1 atm. 5.10. Types of Boiler Furnace. The factors such as temperature, concentration, preparation, distribution of reactants and mechanical turbulence greatly affect the rate of chemical reaction in the combustion

Flue gases Hot gas

Coal '0

r).p, • - cs - 0.,

Fuel bed

I)

Grate Air

Fig. 5.14. Grate-Fired Furnace

Furnace wall

Fuel and air mixture

_

421 I

Fig. 5.15. Chamber Type or Flame Furnace

Fuels, Combustion, Combustion Equipment and Fuel Handling Systems

231

process. In boiler, the firing systems introduce the fuel and air for combustion, mix these reactants ignite the combustible mixture, arid distribute the flame envelop and the products of combustion. There are two types of boiler furnaces as given below— (i) Grate-fired furnaces and (ii) Chamber type or flame furnaces Fig. 5.14 shows the grate-fired furnace. These are suitable for burning solid fuels such as coal, wood, chips, bagasse, urban waste, etc. The chamber type or flame furnace is shown in Fig. 5.15 which is used for firing pulverized coal, fuel oil and gas. A mixture of fuel and air is delivered into the furnace chamber for combustion where it burns and the combustion gases go upward as shown in Fig. 5.15 5.11. Kinetics of Combustion Reactions. The phenomena of combustion reactions is very complex. It is determined by number of physical and chemical factors. The physical factors consists of the process of mixing of fuel and air, the size of fuel particles and the surface area exposed to reaction. The chemical factors include the temperature and concentration of reactants. The kinetics of chemical reactions are determined by the complex fields of velocities, temperatures, and concentration. The combustion reactions are accomplished with exothermic type of evolution of heat. The chemical reactions can proceed in either direction - forward or backward and they are called reversible reactions. Reaction rates (r) are governed by the law of mass action and are expressed as r = K c74 c'; = KpApB where cA and cB are the concentration reactants of A and B. Here PA and pB are the partial pressure of components in gas mixtures. The reaction rate constant (k) is expressed as K = v „—E/Ro T (5.61) 1'0 where K0 is constant, E is the activation energy, kJ/kg mole, Re is the universal gas constant (kJ/kg mole) and T is the absolute temperature. From the nature of the curve, it is clear that combustion is possible only during a certain concentration limit. 02 CO

Gas film Coke particle Ash layer CO & CO2

8f

1 4 4__ 8f--11

(a)

(b)

Fig. 5.16. Effect of Variation of Concentration of Gases Near the Burning of Carbon Surface (a) At Low Temperature, (b) At High Temperature

Pressure distribution

P02 02

Reaction surface

Fig. 5.17. Burning of A Coke Particle

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Steam & Gas Turbines And Power Plant Engineering

5.12. Mechanism of Solid Fuel Combustion. Initiation of solid fuel (pulverized coal) combustion starts first with the stage of thermal preparation which includes the evaporation of residual moisture and distillation of volatiles. Volatiles are evolved in the temperature range of 400-600°C and this occurs within 0.2 to 0.5 seconds. A high yield of volatile produces enough heat to ignite coke particles. The final stage of combustion is the combustion of coke particles at a temperature above 800-1000°C. The burning of a coke particles takes up 1/2 to 2/3 of the total combustion time (1-2.5 seconds). The carbon-oxygen reaction at 1200°C and 1700°C are as follows • (5.62) 4C + 302 = 2C0 + 2 CO2 at 1200°C 3C + 202 = 2C0 + CO2 at 1700°C

(5.63)

At 1700°C, the ratio of CO/CO2 is around 2. The effect of variation of gases near the burning carbon surface at moderate and high temperature are shown in Fig. 5.16. The burning of a coal particle and its various zones around it is shown in Fig. 5.17. 5.13. Combustion Equipments for Coal. There are two ways to feed the coal into the furnace—(i) in lump pieces and (ii) in powder form. Based on this, the following types of furnaces may be used for the combustion of coal. (i) Fuel bed furnaces (coarse particles) (ii) Pulverized coal furnaces (fine particles) (iii) Cyclone furnaces (crushed particles) (iv) Fluidized bed furnaces (crushed small particles). Construction and working principle of each system is discussed separately. 5.14. Fuel Bed Combustion Furnace. In the case of fuel bed combustion furnace utilising coarse particles of coal a grate is used at the furnace bottom to hold a bed of fuel as shown in Fig. 5.14. In general, there are two ways to feed the coal on to the grate as given below— (i) Overfeeding and (ii) Underfeeding. 5.14.1. Overfeed Fuel Bed. Fig. 5.18, shows an overfeed fuel bed section. It receives fresh coal on its top surface and is characterized by the following five zones from top to the bottom. (i) a layer of fresh or green coal —fresh coal zone (ii) a layer of coal losing moisture — drying zone. (iii) a coking layer of coal losing its volatile content-distillation zone. (iv) a layer of incandescent coke where the fixed carbon is consumed—combustion zone. The ignition plane lies between green coal and incandescent coke. (v) a layer of ash progressively getting cooler — ash zone. The heat is released due to combustion in the combustion zone and as a result the heat transfer occurs upward and downward by conduction, while the air flow from below will tend to carry away the heat upward by convection. The primary air supplied from the bottom gets warmed up as it flows through the ash layer and when it passes through the incandescent coke layer at about 1200°C, the exothermic reaction (C+02 = CO2) taking place initially provides the sufficient heat release for continuing the combustion process till all the oxygen is consumed. In the event of thick incandescent layer, the CO2 may

Fuels, Combustion, Combustion Equipment and Fuel Handling Systems

Flame

II1

1

233

CO2 + 02 + N2 + H2O Secondary air + H2O

I I I fit (i'iv;,f

VM + CO + CO2 + N2 + H2

Green coal feed

VM (Volatile matter)

040:0

0ert% 0 Ota 0-

Incandescent coke

V Ash

CO + CO2 + N2 + H2

Yo%t. 9143 cf3"'

Grate

Primary air + H2O

Air Fig. 5.18. Over feed Fuel Bed Section.

partly or full reduced to CO (CO2 + C = 2C0). Further, water gas reaction (H20 + C = H2 + CO) may also take place with the moisture from air. These reactions are of endothermic nature resulting in considerable temperature drop of bed and gas stream. As the hot gas stream passes through the distillation zone where the volatile matter is added and then through the drying zone where the moisture is picked up and finally comes out from the fuel bed containing N2, CO2, CO, H2, volatile matter and water vapor. Some of the contents such as CO, volatile matter and hydrogen are not desirable, and if the combustion of these components is to be completed before the gases leave the furnace the following requirements have to be met :— (i) A hot ignition surface in the range of 1000°C to 1300°C is to be provided by using fire brick-lined arch which stores up the heat and remains at a high temperature. (ii) Sufficient fresh air or secondary air or overfire air is to be supplied. (iii) Turbulence is to be created in the secondary air by feeding at right angles above the fresh coal zone to the up flowing gas stream from the fuel bed for thorough mixing of air and gases (Fig. 5.18). It is not desirable to feed secondary air along with primary air as it will produce more carbon monoxide.. It is worth to note that the carbon contents of coal combined with hydrogen in the volatile matter leaves the fuel bed as part of the hydrocarbon gases, the molecules of which breakdown or disintegrates at high temperature of the furnace releasing some free

t

A Flame Ash

CO2 + 02 + N2 + H2O Secondary air + H2O VM + CO + CO2 + N2 + H2

Incandescent coke i

Green coal ,6*

Tuyere it

•

o.•

Imo • W.

Primary air + H2O

+ Coal feed

Fig. 5.19. Underfeed Fuel Bed Section.

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Steam & Gas Turbines And Power Plant Engineering

carbon in the form of finally divided particles suspended in the gas stream. For burning these suspended carbon particles, it is essential that coal feeding apparatus must be designed properly and if the well designed furnace is not operated properly, it may appear At the chimney top as black smoke indicating energy wastage and inefficient combustion. 5.14.2. Underfeed Fuel Bed. Fig. 5.19 shows the underfeed fuel bed showing the various zones. In this case coal is fed from underneath the grate between two tuyers by a screw conveyer or retort. The primary air passing through the holes in the tuyers goes through the various zones. The ash zone is above the incandescent coke zone. The secon-

Coal distribution block Furnace brick wall Fig. 5.20. Contour of Fuel-Bed and Ash Dumps for Single-Retort Stokes.

dary air fed in the mid of flame zone which help in burning the unburnt carbon and other coal constituents Fig. 5.20 shows the single retort stoker showing contour of fuel bed and as, h dumps. .75.15. Mechanical Stokers for Fuel Bed Combustion As mentioned earlier, grate is used to hold the fuel in fuel bed combustion system. In small steam generator, the grate is stationary and coal is fed manually by shovels. But the bigger units demands more uniform operating conditions, higher burning rate and greater efficiencies and these are accomplished by moving grate or stokers. There 'are various types of stokers as given below— (i) Travelling grate stoker (ii) Chain grate stoker (iii) Spreader stoker

(iv) Vibrating grate stoker

(v) Underfed stoker Advantages : The advantages of mechanical stoker are—(i) No necessity of coal preparation, (ii) free from danger of explosion, (iii) capital and maintenance costs are less, (iv) suitable for medium capacity unit and (iv) firing system reliable. 5.15.1 Travelling Grate Stoker. Fig. 5.21 shows the schematic diagram of a travelling grate stoker. The grate surface consists of a series of cast-iron bars joined together by links to form an endless belt running over two sets of sprocket wheels with a wide surface as per need. The depth of the coal on the grate is regulated by a coal gate provided at the rear of the ccal hopper which can be raised or lowered as per need. The burning rate of the coal is controlled by the simultaneous adjustment of grate speed, fuel bed thickness, and air flow sothat nothing but ash remains on the grate by the time it reacheftheitarportiorr

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235

Adjustable coal gate Coal feed Furnace wall Hopper Rear ignition arch

SA :—> secondary air Front ignition arch

Primary air (PA) inlets controlled by dampers

To ash pit

Front sprocket driven by a variable speed motor

Rear drive sprocket

Fig. 5.21. Travelling Grate Stoker.

of the furnace. The ash falls into the ash pit as the grate turns on the rear sprocket to make the return trip for the next feed. As soon as the raw or green coal on the grate enters the furnace, the surface coal gets ignited from the heat of the furnace flame and from the radiant heat rays reflected back by the ignition arch provided over the furnace. With the burning of coal, the coal bed becomes thinner toward the furnace rear. The primary air supplied by FD fan from under the grate is controlled with the help of dampers and the supply is reduced slowly as required. The secondary air is supplied perpendicular to the main flow to complete combustion. 5.15.2. Chain Grate Stoker. The chain grate stoker is similar to travelling grate stoker except that it is made of a series of cast iron links connected by bars or pins to form an Coal

Furnace wall Furnace hot gas Fuel falling trajectory

Air Feeder Spreader Lever

qqqqq Dumping grate Wind box and ash pit Fig. 5.22. Spreader Stoker.

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Steam & Gas Turbines And Power Plant Engineering

endless chain. It can burn a wide range of coal. Overfeed stokers are suited for industrial power plants with steady demand. It is simple in construction and initial cost is low. 5.15.3. Spreader Stoker. Fig. 5.22 shows the schematic of a spreader stoker. It consists of a rotating feeder (a drum fitted with short blades) and spreader installed at the bottom of hopper one over the other. Coal from the hopper is fed by a rotating feeder to the spreader or distributor below, which projects the coal particles in a continuous stream on to the grate holding an ignited fuel bed, The fine particles of coal burn in suspension where coarse particles bwn on the grate. The speed of the feeder is a function of load (steam output) of the boiler. The secondary air is admitted above the fuel blade to promote the turbulence and to complete combustion. The grate may be stationary or travelling as per the capacity of the boiler. High capacity boilers need travelling grate. The grate made of CI bars is connected to a lever through links underneath the grate. In order to allow the ash to fall below the ash pit, the lever is moved back and forth which makes the bar rock about the pivot. A variety of coal is burnt over this grate. However, it is necessary to size the fuel property for spreader stokers. Its size should be around 32 mm or less. Spreader stoker is applied to a wide range of boiler sizes 155 MW to 265 MW. 5.15.4. Vibrating Grate Stoker. In order to distribute the fuel bed properly, the stoker is vibrated which shakes the fuel bed intermittently. The frequency and amplitude of vibration depend on the load of boiler. The fuel bed is made inclined in order to move the fuel by gravity towards the rear of the boiler. In order to prevent slagging, the grate is water cooled. 5.15.5. Underfeed Stoker. Fig. 5.23 shows a underfeed retort type stoker. This stoker consists of either single or multiple retort in the form of troughs into Which coal is pushed by ram or screws. The primary air is fed by FD fan into the fuel bed through tuyres. The volatile matter is distilled off the coal and burns above the incandescent fuel bed. It is worth to note that coal forms a thick sloping pile (15 to 25°) over the whole surface of the grate and the carbonization occurs above and proceeds downwards. Ram pushes the fuel bed forward till the refuse or ash is discharged to the ashpit. The fuel bed may be as thick as 0.6 m. The underfeed stoker gives higher thermal efficiency as compared to chain grate stoker. The part-load efficiency is better with multiple retort. The grate is self cleaning. The initial cost is high. The use of any type of stoker .is limited. It becomes impractical and uneconomical beyond a certain size of the boilers due to floor area and boiler house cost. Modern boilers use pulverized cool combustion system which grows vertically upward

Feed hopper

Direction of penetration of heat

Retort

Ram

Front wall of boiler

O Air supply

Clinker crusher

Fig. 5.23. Retort Type Underfeed Stoker.

Fuels, Combustion, Combustion Equipment and Fuel Handling Systems

237

- 5.16. Pulverized Coal Firing System. The commercial development of methods for firing coal is treated as a landmark in the history of steam generation. It made possible the construction and reliable operation of large and efficient steam generators and power plants. John Anderson and his associates used first the pulverized coal firing in Lakside Station, Wisconsin (USA) in 1920. Since then it has become almost universal in central utility stations using coal as fuel. In pulverized coal firing system, coal is first ground to dustlike size and powdered coal is then carried in a stream of primary air to be fed through burners into the furnace of boiler. With the heating' of coal particles in high temperature flames in the furnace, the volatile matter is distilled off resulting in the formation of minute sponge-like masses of fixed carbon and ash. After the mixing of volatile gases with air, they get ignited and burn quickly releasing heat energy. Thus a pulverized coal system consists of pulverizing, delivery and burning equipment. In order to burn pulverized coal successfully in a furnace, the two following requirements must be satisfied. (i) The existence of large quantities of very fine particles of coal, usually those that would pass a 200 mesh 'screen, to ensure ready ignition because of their large surface to volume ratio. (ii) The existence of a minimum quantity (0.9%) of coarser particles to ensure high combustion efficiency. These coarser particles should be able to be retained by a 50 mesh screen. The following is a typical screen analysis of a high volatile bituminous coal sample pulverized to 80% 200 mesh (0.074 mm openings). 99.5% - 50 mesh (0.297 mm openings) 96.5% - 100 mesh 80% - 200 mesh (0.074 mm openings) The coarser particles which would be retained over 50 mesh screen is just 0.5 per cent. The above size of pulverized coal represents a surface area of roughly 150,000 mm2/g with 97% of the surface in the 200 mesh portion. It is worth to note that the greater surface area per unit mass of-coal provides faster combustion reaction as more carbon particles are exposed to oxygen and heat. This results in less requirement of excess air, reduction in dry exhaust loss and increase in steam generator efficiency. The proper burning of fuel requires the supply of correct proportion of air, thorough mixing of fuel and air, high temperature and adequate time to complete combustion reaction. As a result of combustion ash is formed. A part of ash falls to the furnace bottom and rest is carried in gas stream asfly ash which is separated before exhausting through chimney. 5.16.1. Advantages and Disadvantages of Pulverized Coal Firing System. Advantages. The following are the advantages of pulNierized coal firing system. i) Greater surface area of coal per unit mass allows faster combustion reaction.. ii) Excess air requirement is low. iii) The ability to use preheated air is high to reduce exhaust losses. iv) The boiler efficiency is high. v) It can burn wide varieties of coal. vi) The response to load changes is very fast enabling almost constant steam pressure

238

Steam & Gas Turbines And Power Plant Engineering

under wide load conditions. vii) It can burn combination of gas and oil with ease. viii) It is able to release large amount of heat enabling to generate steam above 2000 t/h in one boiler. ix) Pressure losses are less and so draught. x) Fly ash emerging out has the ability to be used in brick preparation. xi) The system is free from clinker and slagging troubles. xii) The boiler can be started from cold very rapidly and efficiently which is an important factor during emergency. xiii) There is no moving parts in the furnace subjected to high temperature resulting in more system life. xiv) Practically there is no ash handling problem. xv) The furnace volume required is considerably less enabling minimum travel of flame length. Disadvantages. The following are the disadvantages of pulverized coal firing system. i) The capital cost of the pulverized system is considerably high. ii) Additional power is required for pulverizing coal. iii) Fly ash removal equipments such as electrostatic precipitator is costly. iv) Due to high flame temperature, the refractory lined furnaces are inadequate, so water cooled walls are essential. v) The storage of powder coal requires special attention against fire hazards. vi), -"Special starting up equipments are needed. • /5.16.2. Crushers. To prepare-coal for pulverization, the following types of crushers are used— (1) Ring crusher or granulator, (ii) Hammer mill crusher, (iii) Bradford breaker Fig. 5.24 and Fig. 525 show a ring and hammer type coal crusher. The coal is fed at the top and is crushed by the action of rings that pivot off centre on a rotor or by swinging hammers attached to it. Adjustable screen bars determine the maximum size of the discharge coal. Ring crushers and hammermills are used off or on plant site. They discharge a large amount of fines suitable for further pulverization but not for cyclone-furnace firing. Bradford breaker is shown in Fig. 5.26. It is used for large capacity work. It comprises

aft de. •

I

r.

Clean out opening

.Access

Adjustable plate Crushed s Screen (bars) door

co

Fig. 5.24. A Ring Type Coal Crusher.

Fuels, Combustion, Combustion Equipment and Fuel Handling Systems

239

Raw coal

Crushed coal Fig. 5:25. A Hammermill Coal Crusher.

Casing Perforated plate

Fig. 5.26. A Bradford Breaker. of a large cylinder consisting of perforated steel or screen plates to which lifting shelves are attached on the inside. The cylinder rotates slowly at about 20 rpm and receives feed at one end. The coal is lifted by the shelves, and the breaking action is accomplished by the repeated dropping of the coal until its size permits it to discharge through perforations made. the size of perforations determine the size of crushed coal. The crushing force due to gravity limits the quantity of fines. The main advantage of Bradford breakers is to reject foreign patter and to produce relatively uniform size coals. 3:16.3. Pulverizers. Pulverizing process consists of three stages namely (i) feeding, (ii) drying and (iii) grinding. The feeding system must automatically control the fuel-feed rate according to the boiler demand and the air rates required for drying and transporting pulverized fuel to the burner (primary air). For pulverization, coal has to be dry and dusty. Sitic,e-coals have varying quantities of moisture and in order that lower rank coals can be used, dryers are an integral part of pulverizing equipment. A part of the air (primary air) from steam-generator air preheater is forced into the pulverizer at 350°C or more by primary air fan where it is mixed with the coal as it is being circulated and ground. The third stage is the grinding and its equipment is called pulverizer or grinding mill. Grinding

240

Steam & Gas Turbines And Power Plant Engineering To burners

Coal Hot primary air

Coarser particle fed back III

,.., Coal + (...' ball 1 k) C' P .3 a .01 ci...e.....0 pd.4 , r. k.R kz• -4 c, 13 0 ....31: • 13.0..30.4t, •.f.t.D.4 i-sy< Drum rotated at 18 to 25 rpm

Classifier

Fig. 5.27. Ball-tube Mill.

is performed by impact, attrition, crushing or combination of these. There are several types of pulverizers, classified by speed. They are— (i) low speed (below 75 rpm) : the ball-tube mill, (ii) medium speed (75 to 225 rpm) : the ball-and-race and roll-and race mill, (iii) high speed (above 225 rpm) : the impact or hammer mill, and the attrition mill. 5.163.1. Ball-Tube Mill. Fig. 5.27 shows the schematic of a ball-tube mill. It is one of the oldest in the market. Basically, it is a hollow cylinder with conical ends and heavy-cast wear resistance liners less than half-filled with forged steel balls of mixed size. Pulverization is accomplished by attrition and impact as the balls and coals ascends and falls with cylinder rotation. The primary air is supplied over the charge which carries the pulverized coal to classifiers. The system suffers from the drawback of more power consumption due to heavier construction and poor air circulation. Now efficient system is used. 5.16.3.2. Ball-and Race Mill. Fig. 5.28 shows a widely used ball-and-race type pulverizers. It operates on the principle of crushing and attrition. Pulverization takes place Coal and air discharge Openings (individual burner lines) Signer pipe shutoff

valve

Classifer

Raw coal feeder Windbo 1

Spring for adiusting load Stationary ring

Air seat Driving ring

Pyrites trap

Grinding ball

Driving ring Fig. 5.28. Ball-and-Race coal Pulverizer.

Fuels, Combustion, Combustion Equipment and Fuel Handling Systems

241

between two surfaces, one rolling on top of the other. The rolling elements may be balls or ring-shaped rolls that roll between two races. The balls are between a top stationary race or ring and a rotating bottom ring, which is driven by vertical shaft of the pulverizer. The primary air supplied to the pulverizer causes coal feed to circulate between the grinding elements and when it becomes fine enough, it becomes suspended in the air and is carried to .the classifier. In order to achieve most efficient grinding of various coals, the grinding pressure is varied by externally adjustable springs on top of the stationary ring. The ball-and-race pulverizer has ball circle diameters ranging from 42 cm to 20 cm and capacities between 2 to 20 tons/h. It is suitable for direct-firing systems. Boul Mill. Fig. 5.29. shows boul type grinding mill in which coal is fed to a rotating bowl and thrown outward by centrifugal force to the rotating ring where it is pulverized by

Spring loaded spinning rollers

Coal feed

To classifier

To classif e

Rotating bowl

Fig. 5.29. Boul Mill.

rolling under spring-loaded rollers. It is suitable for direct firing system and widely used. 5.16.3.3. Hammer Beaters. Hammer beaters are high speed pulverizers that revolve in a chamber equipped with high-wear resistant liners. They are suitable for low-rank coals with high moisture content and use fly gas for drying but they are not widely used for pulverized coal systems. As mentioned above the classifier receives the pulverized coal. It is usually a cyclone with adjustable inlet vanes and is located at the pulverizer exit. The function of the classifier is to separate the oversized coal and returns it to the grinders to maintain the proper fineness for the particular application and coal used. It is the inlet vanes that is used for varying the gas suspension velocity in the classifier. 5.17:Performance of Pulverizers. The main factors affecting the performance of pulverizer are (i) the grindability of the coal, (ii) the surface moisture, on the coal and (iii) the fineness of the grind needed. It is worth to note that the power consumption of the pulverizer should be low to minimize the operating cost. The grindability of a coal is measured in terms of hardgrove index. A low index number designates coal hard and difficult to grind, while a high index indicates soft coals which are easy to grind. The fineness of the pulverized coal needed for successful burning depends largely on the ratio of volatile matter (VM) to fixed carbon. Low VM coals should be ground finer as carbon burns slower than volatile matter. Moisture makes fine coal particles adhere to each other defeating the very purpose of grinding.

242

Steam & Gas Turbines And Power Plant Engineering

5.18. The Pulverized Coal System. A total pulverized-coal system consists of pulverizing, delivery and burning equipments. The system must be capable of both continuous operation and rapid change as required by load demands. The following are the three main pulverized-coal system— (i) Bin or storage system, (ii) Direct firing system, (iii) Semidirect firing system. 5.18.1. Bin or Storage System, It is essentially a batch system by which the pulverized coal is prepared away from the furnace. The resulting pulverized-coal-primary-air mixture goes to a cyclone separator and fabric bag filter that separate and exhaust the moisture laden air to the atmosphere and discharge the pulverized coal to storage bin as shown in Fig. 5.30. From there, the coal is pneumatically conveyed through pipelines to Cold (tempering) air from forced draught fan Tempering air Coal dam•er

Hot air from boiler air heater Boiler front wall Burner windbox

Pulverized fue and air piping

• Pulverizer Basement floor (a)

Coal Raw-coal bunker

Raw-coal feeder P Hot air Motor •

P Pulverizer (b)

o boiler furnace

Exhauster

Fig. 5.30. Bin or Storage System.

utilisation bins near the face to fed to burners. This system was widely used before the development of direct firing system. Since the drying, storing and transporting are carried out in many stages, so the bin system is prone to fire hazard. However, it is still in use in old pl .18.2. Direct Firing System. Pulverized coal direct firing system shown in Fig. 5.31, is extensively used in modern power plants. In comparison to bin system, this system offers greater simplicity and hence greater safety, lower space requirements, lower capital and operating costs and greater plant cleanliness. It continuously processes the coal from the storage receiving bunker through a feeder, pulverizer. and nrimary-air fan to the

Fuels, Combustion, Combustion Equipment and Fuel Handling Systems

243

Cyclone separator

Coal tripper and conveyor

Bag filter

Pulverized coal bin

Coal feeder Pulverizer Air heater fan

To feed hopper

Fig. 5.31. Pulverized Coal Direct Firing System.

furnace burners. The fuel flow suitable to load demand is accomplished by a combination of controls on the feeder and on the primary air fan in order to give air-fuel ratio as per load. The control operating range is only about 3 to 1. It is worth to note that large steam generators are provided with more than one pulverizer system, each feeding a number of burners enabling wide control range. APH . Air preheater Secondary air

Air Coal from waggon tippler

Tempering air-* Primary ai

Roller conveyor

10 to 25 mm size

Raw•coa bunker

80% through_ 200 mesh (88 gm)

Exhaust fan

Raw•coal feeder

Furnace wall

Coarser particles fed back Prima air

Burner

Windbox

(a)

Motor Pulverizer ik Hot air

Exhiiister fan (b) Fig. 5.32. Semi-Direct Pulverized Coal Firing System.

f.f..0

Steam & Gas Turbines And Power Plant Engineering

244

5.18.3. Semi-Direct Firing System. Fig. 5.32 shows a semi-direct firing system. A cyclone separator between the pulverizer and furnace separates the conveying medium from the coal. The coal particles fall on by gravity from the cyclone to feed the coal particles through the burners into the furnace. 5.19. Design Aspect of Pulverized Coal Fired Furnace. The shape, type and design of the furnace plays an important role in achieving the higher steam generator efficiency. Pulverized coal bums in suspension and the heat released by the combustion is transferred to the water wall tubes lined around the the furnace by radiation mechanism. The temperature of the gas leaving the furnace is such which does not cause clinkering to the subsequent heating surfaces. The water wall tubes receive water from the drum via downcomers tubes and discharge the water- tube mixture to the drum. The shape of pulverized coal fired furnace is characterized geometrically by its linear dimensions such as front width (w), depth (d) and height (hf) as shown in Fig. 5.33. The cross-sectional area of the furnace is at = w x d, through which hot furnace gas passes at a speed ranging from 7 to 12 m/s. The heat release rate per unit bed cross-section is expressed as HHV q — f' ,kW/m2 A a (5.64) where mf = mass of fuel (kg/s) HHV = high heating value (kJ/kg) qA is supposed to be the most important characteristics of the furnace and depends on kind of fuel and method of burning. It ranges from 3500 kW/m2 to 6400 kW/m2 for brown coal. For meeting a certain power of a furnace (Qf), the cross-sectional area (ad is expressed as of =

Qf qA

, m2 (5.65)

In general, w = 9.5 to 31 m, d = 6 to 10.5 m are taken. Sometimes, w is expressed in terms of steaming capacity (Ms) is tonns per hour as w = 0.674K

(5.66)

-Flame length

n.,. _.. Burneru, Flame

Coal P.A. + Coal Good mixing SA Air fuel mixture -0. SA

—0.

SA : Secondary air PA : Primary air

(a) Circular burners

(b) Slotted burners

Fig. 5.36. Pulverized Coal Burner (a) Circular and (b) Slot

creates a proper turbulent environment for thorough mixing and sucking hot furnace gases for preheating. (ii) In addition to maintaining stable ignition of the fuel-air mixture, it should also control the flame shape and travel in the furnace as the rate of flame propagation control the combustion. The coal-air mixture must move away from the burners as a rate equal to the (iii) flame-front travel in order to prevent flash back into the burner. (iv) A proper control of secondary-air quantity is highly desirable as too much air can ti re.and prevents its heating to ignition temperature. cool the iLAtu Based on cross-section, the burners for pulverized coal are classified as a) circular and b) slot (rectangular) type as shown in Fig. 5.36. Circular and Slot Burners. The circular burner consists of a cylindrical tube through which the pulverized coal and primary air mixture is fed into the furnace. The secondary air is delivered through a separate tube. The slot type is similar to circular type except the cross-section. Based on kind offlow of the two streams, the burners are classified as (i) Straight-flow burner (external mixing of air-fuel mixture and secondary air in the furnace space) (ii) Turbulent or vortex burner (internal mixing of air-fuel mixture and secondary air before entering the furnace). ----5.20.2.Straight-Flow Burners. Fig. 5.37 shows straight flow burners for pulverized coal in which, air-fuel mixture and secondary air are blown in as parallel jets and the intermixing takes place_in the furnace space,_not_in the ,burners, by properly arranging burners on the furnace walls. In high capacity boiler, burner assembly consisting of many burners are used. (Fig. 5.38). The range of the primary air velocity at the burner outlet varies from 20 to 28 m/s, while the range of secondary air velocity is 35 to 45 m/s.

248

Steam & Gas Turbines And Power Plant Engineering

View along X

O O

3'

(b) with central channel for hot air. (a) with tilting nozzle at the exit of fuel-air mixture. 1- supply of fuel-air mixture; 2- supply of hot air; 3- fuel-air mixture outlet; 4- hot air outlet; 5- suction of furnace gases Fig. 5.37. Straight Flow Burners. (a) Titling Nozzle (b) Central Channel for Hot Air.

4

3

4— Mixing

Furnace wall 2 1

SA

Furnace wall

Mixture

Fig. 5.38. Assembly of Three Burners. (1) Supply of Fuel-air Mixture to Burners (2) Secondary Air Supply, (3) Pipe for Mounting the Fuel-Oil Burners with Electric Gas Igniter, (4) Tilting Air Pipes.

Fig. 5.39. Two Scroll Turbulent Burner. 1- Scroll for Fuel-air Mixture, 2-Secondary Air Scroll, 3-Annular Channel for Discharging Fuel-air Mixture, 4-Ditto-for Secondary Air, 5-Main Fuel-oil Burner, 6-Furnace wall, AB-Boundary of Fuel-air Mixture Ignition, C-Suction of Gases to Flame Root.

5.20.3. Turbulent or Vortex Burners. Fig. 5.39 shows a turbulent or vortex burners. In such burners, a whirling rotary motion is imparted to the coal-air mixture by entering it tangentially to a central nozzle and also to the secondary air that flows through the annular space surrounding the nozzle. The mixing of two streams takes place as soon as they enter the furnace. Sometimes, it is called two scroll burner, where two scrolls are provided for whirling the dust-air mixtures and secondary air. Fig. 5.40 depicts a straight scroll type turbulent (vortex) burner. In this case, the dust air mixture is supplied through a central straight flow nozzle and spread by a deflector and the secondary air is whirled in a scroll. Vane type vortex burner is also sometimes used in which the whirling of the dust-air mixture and the secondary air is effected by axial and

249

Fuels, Combustion, Combustion Equipment and Fuel Handling Systems

tangential vanes. Such burner generally uses horizontal firing of coal. 520.4. Distributed Mixing Burner (DMB). Fig. 5.41 and 5.42 show the distributed mixing burner (DMB) and its mixing concept. Here, the total air supplied is distrib- PriMa air + coal uted in such a way that complete combustion of fuel is achieved with minimum excess air. Basically, the total input is divided in three zones. In the first zone, stoichiometric air-fuel ratio is maintained at a very rich condition when devolatization starts. In the second zone,

Furnace wall

Suction

CA

g 4- 01

gases

Mixing SA

Fig. 5.40. Straight Scroll Burner.

— Inner secondary air Outer secondary air

Throat

windbox

Tertiary air Fig. 5.41. Schematic of DMB. Outer secondary air

Tertiary air Flames

Inner secondary air Coal and primary air

I • • • •• •

>

leery fuel rich+ Progressive air 14Final air addition zone (average addition zone zone for burnout stoichiometry (overall (overall 40% stoichiometry stoichiometry 70%) 120%) Fig. 5.42. Mixing Concept of DMB.

250

Steam & Gas Turbines And Power Plant Engineering

supplementary secondary air is then added to the rich products so as to delay the volatilized fuel-nitrogen compounds. The third zone is the tertiary air zone which provides the oxygen necessary for complete combustion. The third zone is also called the oxidizing zone around the rich primary zone prevents corrosion and limits the interactions between adjacent burners. DMB burner is supposed to be very efficient one. 5.20.5. Low NOi Burners. It is the oxidation of both atmospheric nitrogen and the nitrogen chemically combined with the fuel which produces NOx. The mechanism of NOx formation suggests that the nitrogen available in primary combustion process is to be controlled by fuel-rich operation through reduction of primary air and subsequently to control the overall gas temperature, the combustion is completed with fuel-lean mixture. Fig. 5.43 shows the effect of varying quantities of over-fire on NOx production for various values of percentage of excess air. Fig. 5.44 depicts a two stage venturi furnace for reducing NOx emissions. Two separate furnaces are used for burning coal. In the primary furnace, coal is burnt at substoichiometric conditions resulting in incomplete combustion. As a result NO production is less. The secondary air is added in secondary furnace to complete combustion. The combined effect will be the reduction of NO formation. The other method of NO reduction is the re-circulation of flue gases which reduces the bulk flame temper Lure and flame oxygen concentration. - 5.20.6. Multifuel Burners. In many situation steam generators use simultaneously or alternately a variety of fuels. Multi-fuel burns are capable to accomplish this job. Fig. 5.45 shows a multi-fuel burners to burn pulverized coal, oil and natural gas. The b 1rner has an enlarged diameter in central channel consisting of main fuel oil burner and whirling device of air. Natural gas is fired through feeder tubes. 5.20.7. Burner Arrangement. The arrangement of burners on the furnace plays an -important role on the performance of combustion as it ensures complete combustion in the flame care itself. Based on the type of burners and their characteristics, burners are Convective pass —1

Secondary furnace To stock

% excess air Ox z 10 c

2724

20 16

Second stage air

2 20 2 30 40

0 20 40 60 80 100 Overfire air damper % open Fig. 5.43. Effect of Overfire Air on NOx Production.

Coal and primary air _.. Fig.. 5.44. Two-Stage Venturi Furnace for Low NOx Production.

251

Fuels, Combustion, Combustion Equipment and Fuel Handling Systems

Multi-purpose oil lighter

Windbox Inner air Outer air register register

B

000

Water cooled throat

* 111111 71 Central t

4° .°;

11111

channel el es Spin vanes

(4 1 7 ! 11 1 1dl Fig. 5.45. Multi-Fuel Burners.

B = Burners (a)

(b)

(c)

Fig. 5.46. Arrangement of Turbulent Burners. (a) Front Firing, (b) Opposed Firing, (c) Side Firing (double), (d) Double-side Fring

arranged optimally on the furnace. (i) Turbulent Burner Arrangement. The arrangement of turbulent burners on the furnace is as follows (Fig. 5.46) (a) Front firing (b) Double front or Opposed firing (d) Double-side firing (c) Side firing The diameter of the port (Dr) is taken as the basis in the arrangement. Burners are arrange (2.2 to. 3) Dp from the side walls of the furnace. This arrangement helps in preventing the flames from interacting prematurely and touching the walls. In the case of single wall firing the opposite water wall has a high heat absorption (10 to 20% above the average value) which may clinker the wall. To prevent clinkering the furnace depth in dry bottom furnace must not be less than d = (6 to 7) D.In high capacity boilers, arrangement of all required number of burners is not possible in single wall, so opposite arrangement is employed ensuring more uniform heat absorption by water walls. (ii) Straight Flow Burner Arrangement. They are arranged on the furnace wall as follows (Fig. 5.47) (a) Opposite displaced (offset) firing, (b) Corner firing with encountering flame jets, (c) Corner firing with tangential jets, (d) Vertical firing (a) The opposite displaced (offset) firing arrangement [Fig. 5.47 (a)] produced intensive turbulization of flames in the main combustion zone as a result of large velocity gradients between adjacent jets moving in opposite direction. This is suitable for peat and brown coal. (b) The corner firing with encountering flame jets arrangement [Fig. 5.47 (b)] produces a highly turbulized flame core but if the flame moves away from the centre to sides walls there is a possibility of clinkering of the walls. (c). Corner firing with tangential jets arrangement [Fig. 5.47 (c)] is widely used in pulverized coal fired boilers. It directs the mixture of fuel and air tangentially to an imaginary circle 1-2.5 m in diameter in the furnace centre. Both fuel and air are projected horizontally from the corners of the furnace along lines tangent to a vertical cylinder at the centre of the furnace. At the meeting point of two streams, intensive mixing occurs which

Steam & Gas Turbines And Power Plant Engineering

252

Main fuels,

Igniter

nozzle N4,

Secondary air dampers

if B = Burners

(a)

(e)

(c)

(b)

—± Hot gases Upper front (or rear wall) Arch

Hot gases

NA4

Tertiary air —> admission

High pressure jet air Primary air and pulverized coal Secondary air 4— Arch

Tertiary air 13CO2 + 11.5H20 + 18.75(3.76)N2 15 18.75 (b) The percentage of excess air = 22. x 100 = 18.13% 18.75

Ans. Airs.

Ans.

The actual air supplied per mole of C13 H23 is expressed as 22.15 x 32 - 3055.17 kg 0.232 (c) A/F ratio =

3055.17

= 17.06

Ans. Problem 5.2. The following data relate to a steam generator-Coal analysis : Carbon 62%, hydrogen 3.0%, oxygen 3.0%, nitrogen .2.0%, sulphur 1.0%, moisture 4.0%, ash 25.0 HHV of coal = 24.3 MJ/kg Dry flue gas analysis : CO2 12.5%, CO 1.5%, 02 8.5%, N2 77% Unburnt carbon : nil, Exhaust gas temperature = 180°C, Unaccounted energy loss = 2.5% of HHV, Steam generation rate = 200 t/h, Steam condition at boiler outlet = 120 bar, 500°C, Feed water inlet temperature = 170°C Heat of reaction for CO and CO2 = 33.083 MJ/kg carbon and 9.5 MJ/kg carbon cp for dry flue gases = 1.05 kJ/kgK; Ambient temperature = 30°C Calculate-(a) the amount of dry flue gas produced per kg fuel; (b) the dry exhaust loss and the incomplete combustion loss per kg fuel, (c) the boiler efficiency; (d) the fuel burning rate; (e) the percentage of excess air used and (f) prepare the heat balance sheet of steam generator. Solution. (a) The mass of dry flue gas produced per kg fuel is expressed as (12 x 13 + 1 x 23)

266

Steam & Gas Turbines And Power Plant Engineering

mdfg

-

Cab (44CO2 28C0 + 3202 + 28N2) 12 (CO2 + CO)

0.62 (44 x 0.125 + 28 x 0.015 + 32 x 0.085 + 28 x 0.77) 12 (0.125 + 0.015) =

18.724 - 11.14 kg 1.68

Ans.

(b) Energy loss due to dry exhaust gas is given by dfg C (1 - la) = 11.14 x 1.05 (180 - 30) pg g

=

Ans.

= 1754.55 kJ/kg = 1.7545 MJ/kg Energy loss due to incomplete combustion is expressed as 28 CO 12 = mdfg 44CO2 + 28C0 + 3202 + 28N2 OIR,C0 HR,CO2, 28 = 11.14

28 x 0.015 12 [33083 - 9500] 28 44x0.125 + 28x0.015 + 32x0.085 + 28x0.77

11.15 x 28 x 0.015 x 10107 - 1580.4 kJ/kg fuel 29.92

Ans.

= 1.5804 MJ/kg fuel (c) Unaccounted energy loss = 0.025 x 24.3 = 0.6075 MJ/kg fuel Total energy loss = 1.7545 + 1.5804 + 0.6075 = 3.9424 MJ/kg fuel HHV - E Losses x 100 = 24.3 - 3.9424 x 100 = 83.77% HHV • 24.3 (d) The energy balance gives : mf x HHV x boiler = ms (h1 hf) Boiler efficiency =

or

inf

= M s (hl - 19 HHV x Tl boiler

Ans.

200 x 103 (3347 - 719.2) _ 25818.2 kgm = 7.17 kgis 24.3 x 103 x 0.8377 Ans.

(e) Theoretical air required per kg of fuel = mai = 11.5C + 34.5

- i°) + 4.3 S

0.03 = 11.5 x 0.62 + 34.5 (0.03 - -) + 4.3 x 0.01 = 8.078 kg/kg of fuel 8 Actual air supplied to the furnace when nitrogen is present in the fuel 3.04 N2 Cab M N " CO2 + CO 0.768 3.04 x 0.77 x 0.62 1 - - x 0.02 = 10.34 kg/kg of fuel 0.768 (0.125 + 0.015) Percentage of excess air used =

m°° - Mab M at

10.34 - 8.078 x 100=28% 8.078

Ans.

Fuels, Combustion, Combustion Equipment and Fuel Handling Systems

267

(0 The heat balance sheet is given by Item Heat input Expenditure

Heat supplied (i) Heat consumed to generate steam (ii) Energy loss in dry exhaust gas (iii) Energy loss due to incomplete combustion (iv) Unaccounted energy loss

Total

Heat (M]/kg fuel) 24.3 20.356.

Heat (%)

1.7545

7.22

1.5804

6.505

0.6075

2.505

24.3

100

100 83.77

Problem 5.3. During a test of a boiler trial, the following data were obtained. Coal analysis : C = 85.2%, H = 4.8%, ash = 8.2%, gross calorific value of dry coal = 35295 kJ/kg, moisture content = 1.8%, coal consumption = 18 t/h, boiler room temperature = 30°C, feed water temperature = 150°C. Steam pressure = 140 bar, steam temperature = 500°C, steam raised = 200 t/h. The analysis of dry flue gases by volume gave : CO2 -7- 9.4%, 02 = 1 1. 1%, N2 = 79.5%, the temperature of gas in stack = 180°C, mean specific heat of dry flue gas = 1.005 kJ/kgK and temperature of atmospheric air = 30°C. Prepare a complete heat balance sheet per kg of dry coal. 200 = 11.111 kg Solution. Steam generated per kg of fuel = 18 Heat consumed in steam generation = ms (hi — hf) = 11.111 (3321 — 632.2) = 29872.5 kJ/kg filel Actual air supplied = maa —

3.04 N2 Cab 3.04 x 0.795 x 0.852 _ 21.905 kg/kg fuel 0.094 + 0 CO2 + CO

Theoretical (minimum) air required = mat = 11.5 C + 34.5 (H —

+ 4.35

= 11.5 x 0.852 + 34.5 (0.08 — 0) + 4.3 x 0 = 12.558 kg/kg fuel Excess air = 21.905 — 12.558 = 9.347 kg/kg fuel Total mass of flue gases = 21.905 + 1 = 22.905 kg/kg coal Mass of water vapor in product = 0.018 + 0.048 x 9 = 0.45 kg/kg coal Mass of dry products excluding excess air is Mdfg = 22.905 — 0.45 — 9.347 = 13.108 kg/kg coal

Heat loss in dry flue gas products = Mfgd Cpg (1g — = 13.108 x 1.005 (180 — 30) = 1976.03 kJ/kg fuel Heat loss due to moisture in fuel = M[c

pw (100 —

hfg + cps,. (tg — 100)]

268

Steam & Gas Turbines And Power Plant Engineering = 0.018 [4.18 (100 — 30) + 2256.8 + 2.09 (180 —100)] = 48.89 kJ/kg fuel Heat loss due to hydrogen in fuel which converts into steam 9H

[cpw (100 — ti) + h fg+ c ps,. (tg — 100)]

= 9 x 0.098 [4.18 (100 —30) + 2256.8 + 2.09 (180 —100)] = 1299.97 kg/kg fuel The heat balance sheet is given below. Heat balance sheet per kg of fuel Item Heat input Expenditure

Heat (kJ/kg fuel) 35295

Heat (%)

Heat consumed in steam generation (ii) Heat loss in dry flue gas products (iii) Heat loss in excess air (iv) Heat loss due to moisture in fuel (v) Heat loss due to hydrogen in fuel and converted into steam

29872.5

84.63

1976.03

5.559

1409 48.89

0.1385 3.683

(vi) Unaccounted heat loss (by difference)

688.61

1.9975

352995

100

Heat supplied (i)

100

1299.97

Total

Problem 5.4. An oil fired steam generator uses fuel whose analysis by mass gives : carbon 85%, hydrogen 9%, sulphur 3%, oxygen 1.5%, and remainder uncombustible. The analysis of dry flue gases by volume gives : combined (CO2 + SO2) 16.5%, 02 2%, and there is no traces of CO or S03. Evaluate per kg of fuel. (a) mass of air supplied, (b) percentage excess air supplied in kg mass of dry flue gas forced, and (d) mass of water vapor formed due to combustion. Solution. Since the fuel analysis is given on mass basis, so let us consider 100 kg of fuel. Let x moles of oxygen are supplied for combustion. The combustion equation on mole basis is given as 85 8.5 3 C+ H2 + S+ 12 2

32

02 + X02 + 3.76 x N2 —)

a CO2 + b SO2 + CO2 + d N2 + e H20 Equating coefficients from the equation Carbon : 85 = a or a = 7.08

Hydrogen :

1.5 Oxygen : — 4- x=a+b+c+ e/2 32

Sulphur :

Nitrogen : 3.76 x = d

3

8.5

= e ore = 4.25

= b orb = 0.25

Fuels, Combustion, Combustion Equipment and Fuel Handling Systems

269

Based on volumetric analysis of dry flue gases a+b - 0.165 (i) a+b+c+d

and

a+b+c+d

a+b c

Dividing eq. (i) by eq. (ii), we have :

- 0.02 (ii)

0.165 - 8.25 0.02

a+b 7.08 + 0.25 - 0.888 8.25 8.25 From oxygen balance, we have x = 7.08 + 0.25 + 0.888 + 4.25/2 - 1.5/32 = 10.296 moles Thus, the mass of oxygen supplied = 10.296 x 32 = 329.47 kg

or

c=

The mass of actual air supplied for 100 kg fuel - 329'47 - 1420.12 kg 0.232 (a) Therefore, the mass of actual air supplied per kg of fuel 1420.12 - 14.2012 100

Ans.

0 (b) Theoretical air required is = mat = 11.5 C + 34.5 CH - - + 4.3 S 8 -0.01 5 ) + 4.3 x 0.03 = 12.944 kg = 11.5 x 0.85 + 34.5 (0.09 8 Percentage of excess air - M aa M at

Mal

- 14.2012 12.944 x 100 = 9.7% 12.944

Ans.

(c) The mass of dry flue gases formed from 100 kg of fuel = a CO2 + b SO2 -1- CO2 + dN2 = 7.08 x 44 + 0.25 x 64 + 0.888 x 32 + 3.76 x 10.296 x 28 = 1439.89 kg The mass of dry flue gas per kg of fuel = e H2O = 4.25 x 18100 = 0.765 kg

Ans. boiler uses a propane gas. The analysis of dry products of combustion Problem 5.5. A gives CO2 11.6%, 02 2.8%, CO 0.6%. Calculate the percentage of excess air used in the combustion. Solution. Let x moles of propane reacts with a mole of oxygen supplied and thus the combustion equation becomes x C3H8+ a02 + 3.76aN2 -› 11.6 CO2 + 0.6 CO + 2.8 02 + 85 N2 + bH2 0 Equating moles, we have : Nitrogen : 3.76a = 85 or a = 22.60 Carbon : 3x = 11.6 + 0.6 or x = 4.066 b Oxygen : 22.60 = 11.6 + 0.3 + 2.8 + 2.8 + - or b = 15.8 Hydrogen : 8 x 4.066 = 2 x 15.8 (Approx.) Thus the combustion equation becomes 4.066 C3H8 + 22.602 + 3.76 x 22.6 N2 -> 11.6 CO2+0.6 C0+2.8 02+85 N2+15.8 H2O

Steam & Gas Turbines And Power Plant Engineering

270

For one mole of C3 Fl8, the above equation becomes C3 H8 + 5.558 02 + 20.899 N 2 2.852 CO2 + 0.147 CO + 0.688 02 + 2.09 + 3.885 The stoichiometric propane combustion equation is given by C3H8 + 502 + 5 (3.76) N2 --> 3CO2 + 4H20 +.18.8 N2 The excess air supplied =

.

(5.558 - 5) x 100 = 11.16% 5

Ans.

Problem 5.6. A steam generator generates 100 t/h steam at 100 bar and 500°C. The calorific value of fuel used by steam generator is 41 MJ/kg with an overall efficiency of 80%. For efficient combustion 16 kg of air per kg of fuel is required for which a draught of 22 mm of water gauge is required at the base of stack. The flue gases leave the steam genertor at 310°C. The average temperature of the gases in the stack may be taken as 290°C and the atmospheric temperature is 30°C. Calculate the height of the stack and the diameter at its base. The velocity of the gases at the stack exit may be assumed negligible. Solution. The draught produced is expressed by AP = g h (Pa P = gH 14 T T L

Where Ap = draught in N/m2

or

g H 1.013 x 102 1 22 x 10-3 (m) g (1 s,2)10- 3 (— m2 = 0.287 303

or

H-

1 563

22 x 10-3 x 103 x 0.287 x 303 x 563 - 40.89 m 1.013 x 102 (563 - 303)

Ans.

The overall efficiency of steam generator is given by ins(h1 - hf) boiler —

or

inf x CV -

100 x 103(3372 - 632.2) riif x 41 x 103

b.8

th = 8353 kg/h = 2.32 kg/s

ina = 16M 16 x 2.32 = 37.12 kg/s 1= The mass flow rate of flue gases = tilfg = (16 + 1) x 2.32 = 39.44 kg/s From continuity equation applied at base of stack tia = p A C = Lc D2 (2gH)I A fg g RT 4 1.00 x 102 u- (2 x 9.81 x 40.89)1/2 0.287 x 583 4

or

39.44 =

or

D = 1.71 m

Ans.

Problem 5.7. A FD fan is used by a boiler to supply the air. Consumption of coal is 20 t/h. The coal analysis gives : C 77.0, H 3.0, 0 2.5, S 0.8. The excess air used is 35%. The plenum chamber pressure is 200mm water gauge. The mechanical efficiency of the fan is assumed to be 70%. The room temperature is 30°C. Calculate—(a) the amount of air to be handled by the fan, (b) the motor capacity of the fan, If an ID fan is used to produce

Fuels, Combustion, Combustion Equipment and Fuel Handling Systems

271

draught of the same magnitude having gas temperature 179°C and fan efficiency 60% and (c) the fan motor capacity of ID fan. Solution. (a) The mass of theoretical air required is given by mat =

0 11.5 C+34.5 (H---)+4.3 S 8

= 11.5 x 0.77 + 34.5 (0.03 —

0.025 13 ) + 4.3 x 0.008 = 9.816 kg of air/kg of fuel

Since the excess air 35% thus the actual air supplied is given by maa = 9.816 x 1.35 = 13.25 kg of air/kg of fuel

Ans.

(b) The fan (FD) motor capacity is given by M aa • tiliV air

Pmotor

Ivan

0.287 x 303 1.0135 x 102

_ RT

But air

AP

P

(Pmotor)FD

0.858 m3/kg

13.25 x 20 x 103 x 0.858 x 200 x 10-3 x 9.81 x 103 0.7 x 3600

= 177 x 103 W = 177 kW

Ans.

(c) The FD fan handles flue gas. Thus, M fg = lil f + a = 20 x 103 + 13.25 x 20 x 103 = 285 x 103 kg/h, = 79.16 kg/s

The specific volume of fuel gas at the inlet of FD fan is + 273 170 x 273 v, = v — 0.858 — 1.254 m3/kg ° t + 273 30 + 273 g

Thus FD fan motor capacity = (Peaoter)

JFD =

v„

"

— 324.62 x 103 W = 324.62 kW

rifan

Ans.

Comment : The power input to FD fan is more than FD for producing the same draught. Problem 5.8. A steam generator uses a stack of 250m high and 5 m diameter at its exit. The mass of the gases leaving the stack is 1200 kg/s at 120°C. into the atmospheric air at 10°C. The prevailing wing velocity is 59 km/h. and the atmospheric condition is in its neutral stability. Determine the height of plum (AH) based on correlation of Carson and Moses. Solution. The hight of plum (All) may be calculated by the correlation of Carson and Moses as given below : Q1/2 Cs D • AH = 2.62 — 0.029 Cw Cs, Where Cs = stack gas exit velocity (m/s), D, = stack diameter at exit (m), CW = wind velocity (m/s), Qe = heat emission from plume (W), Further, Qe = m c (Ts — Ta) = 1200 x 1.005 (120 — 10) = 132660 kW g p

C

= 50 x 103/3600 = 13.88 m/s

Steam & Gas Turbines And Power Plant Engineering

272

m C = s

p .A g

ATI = 2.62

-

12 - 133.54 m/s 1.01385 x 0.287 x (393)

132.661/2 133.54 x 5 - 0.029 - 34.97 m 13.88 13.88

Ans.

Problem 5.9. A steam power plant empley a fluidized bed combustion system and its output 150 MW at 90% efficiency. The calorific value of the coal used is 26 MJ/kg with a sulphur content of 3.8%. In orde to limit SO2 emissions, limestone is fed to it at a calcium-sulphur ratio of 3.0. The limestone used contains 88% CaCO3. Calculate (a) the coal burning rate and (b) mass flow rate of limestone. to be fed with coal. Solution (a) The coal burning rate is expressed as

in f

150 0.9 x 26

- 6.41 kg/s = 23076 kg/h = 23.076 t/h

(b) The flow rate of sulphur with coal = =

2307 6 32 x 0.038

Ans.

- 27.4 kmoiih

Since the culcium-sulphur ratio is 3.0, thus the flow rate of calcium is = 27.4 x 3.0 = 82.2 kmol/h The mass flow rate of CaCO3 required = 82.2 M = 82.2 x 82.2 x .100 = 82.20 kg/h 220 Thus the mass flow of limestene = -8-- - 9340.9 kg/h = 9.3409 t/h 0.88 Ans. Problem 5.10. A fluidized bed combustor uses high volatile coal with calorific value 24 MJ/kg. The design of the bed and the combustion system allows the release of only 70% of calorific value in the bed and remainder in the above bed zone from which the combustion products leave at 850°C. The inlet temperature of air is 30°C, the bed temperature is 850°C and the air-fuel ratio by mass is 14:1. The specific heat of gas leaving the bed surface is 1.05 kg/kgK. If the burning rate of coal is 20t/h, calculate the rate of heat removal (a) from the bed and (b) from the above-bed zone of the combustion system. Solution. (a) The heat removed from the fluidized bed is given by = heat released-heat carried away by the products of gases = 0.7x24000-(14+1)x 1.05 x (850!- 30) 3885 kJ/kg Thus, the rate of heat removal from the bed is given by = 3885 x 20 x 103 - 21583.3 kW = 21.5833 MW 3600

Ans.

(b) The rate of heat removal from the above-bed zone is -

20 x 103 x 0.3 x 24000 = 56.6 x 103 kW = 56.6 MW 3600

Ans.

Problem 5.11. A fluidized bed operates at atmospheric pressure having bed temperature of 850°C. The fuel used has a calorific value of 26 MJ/kg. The excess air supplied is 30% The heat release rate is 120 MW. Assume density of air at 850°C is 0.32 kg/m3. Calculate the planform area required for two cases namely if (a) the firing rate is 2 MW/m2 and (b) the fluidized velocity is 3 m/s. 120 Solution. (a) The planform area = - = 60 m2 2 Ans.

Fuels, Combustion, Combustion Equipment and Fuel Handling Systems (b) The fuel burning rate - in, =

273

120 x 103 - 4.615 kg/s 26 x 103

The air flow rate = 'ha = 14 x 1.3 x 4.651 = 83.993 kg/s The planform area = Ma/pa U = 83.993/0.32 x 3 = 87.49 m2

Ans.

Problem 5.12. In a fluidized bed boiler, the fuel particles of mean size of 430 pm is used under the ambient condition. The gas density is 1.22 kg/m3 and its viscosity is 1.82 x 10-5 kg/m.s. The density of the loosely packed bed is 1650 kg/m3 while that of solid is 2800 kg/m3. Calculate (a) the voidage of the bed, and (b) the minimum fluidization velocity. Solution. The voidage at the minimum fluidization velocity is given by Pb

, = 1

1650 2800

0.41 Ans. Ps (b) The minimum fluidization velocity is expressed by Archimedes number (Ar). C

mf

=1—

1.242800 - 1.22)9.81 x (430 x 10-6)3 Ar - Pg(Ps -2P dgdp _ (1.82 x 10- 5)2

11

Ar is related to Remf as follows — C1 = [27.22 + 00408 x 804015 - 27.2 = 5.748 Remf = [ 12 + C2 . U Cf p . d But Remf inj g P — 5.478 5.478 x 1.82 x 10- 5 — 0.19 m/s 1.22 x 430 x 10— Ans. Problem 5.13. A fluidized bed of coal particles has a mass of 3000 kg/with a density of 2700 kg/m3. The mean particle size is 850 pm having sphericity of particles is 0.84. Calculate the total surface area of the particles in the fluidized bed. Solution. The surface to volume ratio of bed is given by or

U mf

A =

=

N nd3 6 PP = where N is the number N wod3 d p\

For non-spherical particles, the surface-area of the particles is expresed as 6V 6xm 6 x 3000 - 9337 m2 A= s = s d (i) dpps 0.84 x 850 x 10-6 x 2700 EXCERSISES Viva-Voce and Theoretical Questions 5.1. What are types of fuels for steam generators ? 5.2. How was coal formed ? What are the various types of coal ? 5.3. How was natural gas and petroleum formed ? 5.4. What are synthetic fuels ? • 5.5. How is coal gasification achieved ?

Ans.

274

Steam & Gas Turbines And Power Plant Engineering

5.6. What do you mean by coal liquefaction ? 5.7. How is lower heating value of fuel different from the higher heating value ? 5.8. What is the main constituents of natural gas ? 5.9. What are the major constituents of coal ? 5.10. What is compressed natural gas (CNG) ? Mention its applications. 5.11. What is LPG ? Mention its constituents and application. 5.12. What is town gas ? Mention its applications. 5.13. What are the various forms of biomass ? Mention various bioconversion process. 5.14. What do you mean by industrial wastes and byproducts ? 5.15. Derive the expression for stoichiometric air-fuel ratio. 5.16. Discuss the energy balance in a steam generator. 5.17. What do you understand by draught of a boiler ? 5.18. Derive the expression for draught. 5.19. What are the various types of draughts ? 5.20. What are the various locations of ID and FD fans for producing draught. 5.21. What do you mean by balance draught ? 5.22. What is the approximate height of stack in natural draught ? 5.23. Mention the type of fans used in mechanical draught system. 5.24. What do you mean by heat of reactants and products ? 5.25. What is the adiabatic flame temperature ? 5.26. What do you mean by Gibb's free energy of formation ? 5.27. Mention the types of boiler furnace. 5.28. What are the various combustion equipments for coal ? 5.29. Discuss the working principle of overfeed and underfeed fuel bed systems. 5.30. What are the various mechanical stokers for fuel bed combustion ? 5.31. Discuss the working principle of travelling grate stoker, chain grate stoker, spreader stoker, vibrating grate stoker and underfeed stoker. 5.32. What do you mean by pulverized coal firing system ? 5.33. List the advantages and disadvantages of pulverized coal firing system. 5.34. Discuss the working principle of various types of crushers. 5.35. What are the various types of pulverized coal system ? 5.36. With the aid of simple sketch discuss the working principle of bin or storage system, direct firing system and semi-direct firing system. 5.37. Discuss the design aspects of pulverized coal fired furnace. 5.38. Mention the various types of pulverized coal furnace. 5.39. Discuss the working principle of dry-bottom and wet-bottom furnace. 5.40. Mention the various types of pulverized coal burners. 5.41. What are the requirements of pulverized coal burners ?

Fuels, Combustion, Combustion Equipment and Fuel Handling Systems

275

5.42. With the help of sketch discuss the working principle of circular c/s, rectangular c/s, distributed mixing burners, low NO. burners and combined burners. 5.43. What are the various arrangement of burners ? 5.44. What is tangential or corner firing ? What are merits and demerits ? 5.45. What is a cyclone furnace ? Where is it used ? Mention its advantages and disadvantages. 4.46. What is a fluidized bed ? Define minimum fluidization velocity. What are the various types of fluidized bed ? 4.47. Discuss the various regimes of fluidization. 4.48. Define voidage and sphericity of a particle. 5.49. Discuss the working principle of bubbling and circulating fluidized bed combustion. 5.50. What is the advantage of using limestone or dolomite as the bed material in a fluidized bed ? • 5.51. Mention the/optimum combustion temperature and the approximate calcium sulphur molar ratio for maximum absorption of sulphur oxides. 5.52. What is a pressurized fluidized bed combustion (PFBC) ? 5.53. Explain the sequence of events of burning of a fuel particle in a fluidized bed. 5.54. With the help of sketch discuss the working principle of fluidized bed combustion of solid fuels. 5.55. What are the various types of coal gasifiers ? 5.56. Discuss the working principle of fixed bed counter flout and entrained parallel flow. 5.57. Discuss the working principle of integrated gasification combined cycle. 5.58: With the help of a sketch, discuss the working principle of fuel oil combustion system. 5.59. Discuss the mechanism of fuel oil combustion. 5.60. Sketch the various types of oil burners and describe their merits and demerits. 5.61. Sketch the gas burner and explain its working. 5.62. With the help of sketch discuss the combined gas-fuel oil burners. Numericals 5.63. A boiler consumes coal having following composition : carbon 85.5%, hydrogen 5.0%, oxygen 4.0%, and incombustibles 5.5%. The ashes discharged from the furnace contained 0.015 kg of unburnt carbon per kg of coal supplied to the furnace. The analysis of dry flue gas gives : CO2 14.5%, and CO 1.3% by volume. Determine the mass of air supplied and the complete volumetric composition of the flue gas. Ans. 13.05 kg/kg-coal, by volume : CO2 13.77%, CO 1.23%, 02 3.0%, N2 76.66%,H20 5.34% 5.64. A boiler uses coal whose analysis by mass is as follows— C 73%, ash 12%, moisture 15%. After burning the residue contains 18% C by mass. The analysis of dry flue gases by volume gives : CO2 11.8%, CO 1.3%, 02 5.5%. Determine the percentage of carbon in the coal which was really used in combustion and actual air used. Ans. 97.05%, 13.46. 5.65. A steam generator employing a pulverized coal firing system relate to the follow-

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ing data : Evaporate rate = 200 t/h, boiler pressure = 160 bar, superheater outlet temperature = 570°C, economiser inlet temperature of feed water = 180°C, overall boiler efficiency = 90%, calbrific value of coal = 23.5 MJ/kg, heat release rate in furnace =.190 kW/m3. The ratio of width : length : height of furnace = 1:2:3. Calculate—(a) the dimensions of the furnace, (b) the fuel burning rate, and (c) the percentage of each heating surface of boiler. '5.66. The following data relate to a boiler trial : coal analysis by mass = C 80%, H 4%, 0 5%, N 1%, M (moisture) 4%, ash (A) 6%, ,HHV of coal = 26 MJ/kg. Refuse analysis : C 10%, A 90%. Dry flue gas analysis : CO2 13%, CO 1%, 02 6%, N2 80% Flue gas temperature = 180°C. Air data : dry bulb temperature 20°C, wet bulb temperature 15°C. Prepare the heat energy balance sheet. 5.67. The following data relate to a boiler trial. Coal analysis (by mass) : C 76%, H 5%, O 3%, N 2%,S 4%, M 1%, A 9%. Refuse analysis : A 92%, C 8%. Air (ambient) has 1.8% moisture. Determine per kg of coal (a) actual air supplied, (b) theoretical air required and (c) the percentage of excess air. 5.68. A fuel oil to be used in a boiler plant composed only of carbon, hydrogen, and sulphur. The volumetric flow of gas analysis on a dry basis is CO2 11.7%, CO 0.44%, 0 4.002%, SO2 0.176%, and N2 83.682%, Calculate (a) the fuel mass composition, (b) the A/F ratio by mass, (c) the excess air used, (d) the dew point of the flue gases if the pressure is 1.5 bar. 5.69. The following data relate to a boiler trial : Coal analysis : C 84%, H 5%, M 2.5% and A 8.5%. The coal consumption is 16 t/h, boiler room temperature = 30°C, feed water temperature = 160°C, Steam pressure = 150 bar, steam temperature 600°C, steam raised = 200 t/h, dry flue gas analysis : CO2 10%, 02 18%, N2 78%. The temperature of the gas in the stack = 180°C. Mean specific heat of dry flue gases 1.005 kJ/kgK and temperature of atmospheric air =. 30°C. Prepare a complete heat balance sheet per kg of fuel. 5.70. A boiler used a fluidized bed combustion system having an output of 50MW operating at 80% efficiency with a coal heating value of 26 MJ/kg with a sulphur content of 4%. The system requires a limestone to be fed to the bed at a calcium sulphur ratio of 3.2 to limit SO2 emission. The limestone used contains 86% of CaCO3. Calculate the consumption rate of limestone. 5.71. A pulverized coal fed boiler uses coal of mean size 4801.tm. The air density is 1.22 kg/m3 and the viscosity is 1.88 X kg/m.s. The bulk density of locally packed bed is 1700 kg/m3. The density of the solid in 2800 kg/m3. Determine (a) the voidage of the bed, (b) the minimum fluidization velocity and (c) the sphericity of particles. 5.72 Develop a software.for the design of all types of combustion system.

6 Steam Generators, Ash Handling Systems And Feed Water Treatment

The function of a steam generator (i.e. boiler) is to generate steam at a desired pressure and temperature by transfering heat produced by burning of fuel in a furnace to water to change it into steam. According to ASME Code, a steam generating unit is defined as follows— "A combination of apparatus for producing, furnishing or recovering heat together with the apparatus for transforming the heat so made available to water which could be heated, vaporised and superheated to steam form"

Steam generators are used in both fossil-fuel and nuclear-fuel electric generating stations. In this chapter, only fossil fuel steam generators will be discussed and nuclear fuel steam generators will be discussed in the chapter of Nuclear Power Plant. A steam generator is very complex, integration of furnace, drum, a evaporator, superheater, reheater, economiser and air preheater along with various mountings and auxiliaries such as safety valves, pulverizers, burners, fans, stokers, dust collectors and precipitators, ash-handling system, chimney (or stack), etc. Superheaters, reheaters, economisers and preheaters are called accessories of boiler. The water is supplied to the boiler by a feed pump (multi-stage centrifugal pump) at a pressure -slightly higher than the pressure at which steam is generated. Heating takes place at constant pressure and there is only a conversIon of liquid phase of water into steam in the boiler. This chapter deals with steam genetator and their ash handling systems and feed water treatment. 6.1. Classification of Steam Generator Steam generatots are classified on the basis of many parameters such as tube contents, method of firing, pressure of steam, method of circulation of water, nature of service, position and number of drums, gas passage, heat source, once through boiler, fluid used, \ boiler shell material, etc. (a) Based on the contents of tubes, steam generators or boilers are classified as— i) Fire-tubes boilers ii) Water-tube boilers (i) Fire-tube boilers. In the fire-tube boile'rs, the hot gases (i.e. flue gases) pass

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278

Stack Steam F-lue, or ,-Chimney t gasT

Water level

Stack Flue gas Or Steam Chimney

Boiler drum -y

Water -_-_' m om

°

Flue gases

Air

Furnace bed—, Fig. 6.1. Fire-tube Boiler,

Header flue gases...., Air

%

-

r— Furnace bed Fig. 6.2 Water-tube boiler.

through tubes and water surrounds them. Fig. 6.1 shows the principle of operation of fire-tube boilers. The products of combustion (hot gases) leaving the furnace pass through fire tubes which are surrounded by water. Heat from hot flue gases is transferred to water which is converted into steam. The spent flue gases are then discharged to atmosphere through the chimney (stack). The examples of fire-tube boilers are Cochran, Lancashire, Cornish and locomotive boilers. ii) Water-tube Boiler. In water-tube boilers, water flows inside the tubes and the hot flue gases flow outside the tubes. Fig. 6.2 shows the principle of operation of water tube boilers. A bank of water tubes containing water is connected with a steam-water drum by means of two sets of headers. The hot flue gases from the furnaces pass over the tubes and are discharged through the chimney. The water thus absorbs heat from the hot gases and evaporates in the form of steam. The steam thus formed gets accumulated in the steam space of drum from where it may taken into superheater to superheat steam. The examples of the water-tube boilers are Babcock and Wilcox boiler, stirling boiler, power boilers for utility, etc. (b) Based on nature of service, steam generators or boilers are classified as follows.r— (i) Utility boilers (ii) Industrial boilers (iii) Marine boilers

(iv) Locomotive boilers (now obsolete)

(i) Utility Boilers. These boilers or steam generators are used by utilities for electricpower generating plants. Based on whether the pressure of steam is below or above the critical pressure (pc = 221.2 bar), they can be either subcritical or supercritical boilers. The subcritical boilers are invariably water tube-drum type and they usually operate at between 130 bar and 180 bar and steam is superheated between 540°C-560°C with one or two stage reheating, whereas supercritical boilers are drumless once-through type and operate at 240 bar or higher and steam is superheated between 570°C to 600°C with one or no stage reheating. The steam capacities of modern utility boilers range from 120 to 1300 MW with power plant unit output ranging from 120 to 1300 MW. So far the fuel is concerned, pulverized coal is mainly used. However, LNG and natural gas are also being used. Now circulating fluidized bed steam generators are also gaining popularity. (ii) Industrial Boilers. These boilers are used in process industries like sugar, carpet,

'

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279

paper, jute and so on, and institutions like hospitals, commercial and residential buildings. The sizes of industrial boilers are small. In general, the fuel used is coal or oil and it may be pulverized or fluidized or fired with the help of stoker. They can be heat recovery type using the waste of industries. Now, oil-fired boiler is gaining popularity. The pressure generated varies from 3 to 100 bar with steam capacities vary up to 110 kg/s. They supply dry and saturated steam or wet steam or even hot water. Superheated steam is not common with industrial boilers. (iii) Marine Boilers. They are used by marine ships and ocean liners and in general the prime mover is steam turbine. They are usually oil-fired. They operate at pressure between 50 to 70 bar and steam is superheated up to 540°C. (iv) Locomotive Boilers. They were fire-tube type and used in locomotive which are now obsolete. At normal load, boiler had an evaporating capacity of 8500 kg/h at 14 bar and 370°C temperature and the rate of firing coal was 1585 kg/h. 6.2. Construction and Working of Fire-tube Boilers. Fire-tube boilers are now not used in utility power plants but they are used in industrial boilers, locomotives (now obsolete) and river launches (now obsolete). They operate at pressure equal or less than 18 bar with steam capacity upto 6.5 kg/s. Fire-tube boilers are suitable for small steam requirements. The advantages are low first cost, reliability in operation, quick response to load changes, need of only unskilled labour, less draught required, relatively inexpensive, etc. A fire-tube boilers may be externally fired (e.g. locomotive boiler, Lancashire boiler, horizontally return tubular (HRT) boilers, etc.) or it may internally, fired (Scotoch marine boilers, package boilers, etc.). The major shortcomings of the fire-tube boilers is the definite size and pressure limitation inherent in its basic design. In other words, the maximum size of the unit and the maximum operating pressure are limited. The tensile stress on the drum wall is a function of the drum diameter, and the internal pressure. The thickness of the drum is given by the well known stress formula for thin cylinder, t = pd/2a where p is the gauge pressure (N/m2), 6 = tensile stress (N/m2), d = internal diameter of the shell (m), and t = thickness of wall (m). Thus with high steam pressures and capacities, the thickness would become very large. Further, they are also subjected to large scale deposit and susceptible to boiler explosions, and become very costly. 6.2.1 Horizontal-Return Tubular Boiler. The Horizontal Return Tubular (H.R.T.) boilers are eery common in U.S.A. Fig. 6.3 shows a H.R.T. boiler. Its simplicity and cheapness are the main advantages. They produce pressures up to approximately 17.5 bar with capacities from 4500 to 7000 kg/h. They are not suitable for large power plants because of their small capacities, limited steam pressures and low rates of steaming capacity. Construction. The boiler consists of a horizontal cylindrical shell closed at each end by a flat tube sheet or head. A large number of fire tubes which are usually 75 to 100 mm in diameter are supported between the two end covers are expanded into the cover at each end, thus serving not only the flues through which the hot combustion gases flow but also act as tie rods to hold the flat end covers in place against the steam pressure in the boiler. The flat surface of the heads above the tubes are braced to the boiler shell by diagonal braces. The normal water level is such that all the fire tubes remain immersed in water. The water level in the boiler is indicated by a water column which is connected to the

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Steam & Gas Turbines And Power Plant Engineering

Flue gas

Mounting block for steam stop alve

Main hole door aL7

Mounting block for safety valve Diagonal braces

14- Hot gas

Fire tubes longitudinal

T

.G"--r31out off connection

Hot gas /r ,, Bridge aL

Fire door Ash pit

Fire-brick wall

F6.3. Horizontal Return Tubular Boiler.

boiler by two pipes, one above and the other below the water level. The inside surface of the furnace wall is lined with fire bricks. A brick wall bridge is provided to deflect the hot combustion gases, A fire door is provided along with a single retort under feed stoker. It is an externally fired boiler. A blow-off line is connected to the bottom of the drum at the rear. Valves in this line are opened periodically and some of the boiler water is blown down to a sewer, thus cleaning out the impurities carried to the boiler by the feed water. In small boilers of this type, chemicals are added to the feed water to prevent scale forming impusities in the feed water from precipitating on the heating surfaces as an adherent scale. If the boiler produces dry steam all these impurities remain in the boiler and have to be periodically removed by blowing down. The boiler shell is provided with suitable openings for the attachment of spring loaded safety valves, feed water inlet, steam stop valve, manhole, etc. Working. Since the furnace grate is below the front portion of the water shell the hot gases of combustion heat the bottom of the shell and are deflected by the brick wall bridge to travel its full length as shown by the arrow in Fig. 6.3 to the rear of the boiler. From this position the hot combustion gases pass into the horizontal tubes and return to the front of the boiler where they pass through the smoke box and are finally discharged to the chimney. The damper provided in the smoke box regulates the quantity of hot combustion gases. Thus the hot gases complete two passes, one to the back and other from back to the front. The water circulation in the boiler is natural circulation caused by convection currents. In one type of arrangement shown in Fig. 6.4(a) the fire tubes are staggered and the circulation of water takes,place down the shell walls and then up; as shown by the arrows.

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281

Pan type separator

Perforated pipe se• arator Tubes

Baffles

Feed water

Feed water

Sludge

sludge

(b)

(a)

Fig. 6.4. (a) and (b). Water Circulation in H.R.T.

Boiler.

Fig. 6.4(b) shows the other type of arrangement in which internal baffles are provided to help circulation of water. The arrows show the circulation of water. 6.2.2. Oil-fired Package (Nestler) Boiler. It is an oil fired, fire tube horizontal boiler having working pressure, temperature and steam capacity equal to 11.8 bar gauge, 238°C and 545 kg/h respectively. Water is fed to the boiler through a feed check valve with the help of a multi-stage centrifugal pump run by an electric motor. When the boiler is in operation, a Duplex feed pump (reciprocating type) run by a steam engine using superheated steam feeds the water to the boiler to save the first class energy. The light diesel oil from the tank is pumped by a fuel pump (gear type) and it passes through an oil heater to

N=18 number of tubes III pass Spring loaded safety vale Stea stop valve— II pas 9—r R N-24

Flue gas to chimney V)1(et steam Superheated steam r

O

4

Watecall______ gauge Fusible—' pluge Fuel 4 _. Fuel Pum6?—ineater

Steam for steam engine

•

Superheater tubes

j

Motor-p

Ca.

Hot Fuel gas-----—* - , -

1 El ‘:1

Air

1 Combustion Centrifugal chamber /Flue pump Fue nozzle L motor and swirler Fig. 6.5. Nestler Oil Fired Boiler.

Pump Water

4

Mat-6r :dank

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Steam & Gas Turbines And Power Plant Engineering

decrease its viscosity for better atomizations and is fmally injected in the combustion chamber of boiler by an injector. The air is supplied to the combustion chamber with the help of a centrifugal blower through swirler for rapid and proper combustion. The hot flue gases formed as a result of combustion make three passes as shown in Fig. 6.5 and fmally escape through the chimney. The number of fire tubes employed in 88. The wet steam formed as a result of heat taken by water from flue gases is allowed to pass through the superheater tubes and fmally becomes superheated. Various mounting such as steam stop valve, spring loaded safety valve, water level indicators, pressure gauge, man hole, fusible plug, etc. are installed over the boiler for its safe and reliable operations. 6.3. Construction and Working of Water-tube Boilers. The limitation of fired-tube boilers forced to develop water tube boilers to permit the increase in boiler pressure and capacity with reasonable stress. Water-tube boilers are classified on the basis of circulation of water—natural or forced circulation, tube layout— horizontal straight tube or bent tube, or vertical tube, layout of drum—longitudinal drum or cross drum, number of drum—one, two, three, four or drumless and pressure of steam—subcritical or supercritical boilers. They are invariably used in utility plants. 6.3.1. Babcock and Wilcox Water Tube Boiler. It is a longitudinal drum, externally fired, water tube, natural circulation type of stationary boiler. Since the boiler is externally fired it is suitable to all types of fuels and for hand and stoker firing. Evaporative capacity in such boilers ranges from 20,000 to 40,000 kg/h of steam and operating pressures from 11.5 to 17.5 bar are quite common. ,But the operating pressures may be as high as 42 bar. It is suitable for small size thermal power plants and other industrial works. Construction. Fig. 6.6 shows the details of a Babcock Wilcox water tube boiler. It consists of a high pressure drum mounted at the top. The drum of the boiler is made of welded steer-Or-fin& course joined by longitudinal butt strap. The heads of the drum are forged by hydraulp i ress and are dished to a radius equal to the diameter of the drum. From each end of the drum connections are made with -the upper header and downtake header. A large number-of watertubes connects the uptake and downtake headers. The watertiiiies_are_Lnclinea 5 to 15 d4rees to promote water circulation. The water tubes are straight, solid drawn steel tubes about 10 cm in diameter and are expanded into the bored holes of the headers to ensure proper fixing. The headers have a serpentinejs_inusDidal) form. This serpentine form of headegeriille rs arran wger_hib_es _such that they are staggered and this exposes the con _iipktclleating_gurrace to flue gases_The heating surface of thenit uis the outer surface of the tubes and half of the cylindrical surface of the water drum which is exposed to flue gases. A mud box is attached to the bottom of the rear header (i.e., downtake header). Any foreign matter held in suspension in the water gets collected in the mud box due to gravity and it can be blown off from time to time. A hand hole is also provided for cleaning the mud box. To clean the outside of the water tubes and to remove the sot, a door is provided so that access to the interior of boiler can be made. The whole of the assembly of water tubes is hung along with the drum from a steel girder frame by steel rods called slings in a room made of masonry work, lined with fire bricks. The sling passes around each of the drum and is thus entirely independent of the brickwork of the setting. The boiler can expand of contract freely without straining the brickwork setting. Below the uptake header the furnace of the boiler is arranged. The coal is fed to the chain grate stoker through the fire door. The chain speed is so adjusted that by the time the

Steam Generators, Ash Handling Systems And Feed Water Treatment

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coal reaches the other end of the grate, its combustion has been complete. There is bridge wall deflector which deflects the combustion gases upwards. Baffles are arranged across the water tubes to act as deflectors for the flue gases and to provide them with gas passes. Here two baffles are arranged which provide three passes of the flue gases. A chimney is provided for the exit .of the gases. A damper is placed at the inlet of the chimney to regulate the draught. There are superheating tubes for producing superheated steam. Connections are provided for other mountings and accessories. Working. The hot combustion gases caused by burning of fuel on the grate rise and are deflected upwards by the bridge wall deflector and pass over to the front portion of the lete the first pass. With the of way the complete water tubes and haffles they deflect downwards and complete the second pass. Again, with the provision of baffles they rise upwards and complete the third pass and, fmally, come out through the chimp. The flow path of the combustion gases is shown by the arrows outside the tubes. During their travel they give heat to water and steam is formed. The circulation of water in the boiler is natural circulation set up by convective currents. Feed water is supplied by a feed water inlet pipe. The hottest water and steam rise from the tubes to the uptake fader and then through the risers enters the boiler drum. Superheated steam for use

Fig. 6.6. Babcock and Wilcox Water Tube Boiler. 1. Drum, 2. Pressure gauge, 3. Water gauge, 4. Safety valve, 5. Feed water inlet, 6. Man hole, 7. Headers, 8. Down corner, 9. None-retum valve, 10. Anti priming pipe, 11. Superheater, 12. Baffles, 13. Water tubes, 14. Fire grate, 15. Fire door, 16. Ash pot, 17. Clean out doors, 18 Riser, 19. Blow-off pipe, 20. Chimney, 21. Damper.

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Steam & Gas Turbines And Power Plant Engineering

The steam vapors escape through the water to the upper half of the drum. The cold water flows from the drum to the rear header and thus the cycle is completed. The down— ,comers arelocatedoutside the furnacedo enhance-circulation) To get superheated steam, the steam accumulated in the steam space is allowed to enter into the superheater tubes which are placed above the water tubes. The flue gases passing over the flue tubes produced superheated steam. The steam thus superheated is finally supplied to the user through a steam and stop valve. It to be noted that when the steam has to be raised from the cold boiler, the superheater-tubes are filled with water to the drum water level. The superheaters remain flooded until the steam reaches the working pressure. The superheater is then drained and steam is allowed to enter in it for superheating purposes. 6.3.2. Stirling Bent Water Tube Boiler. In 1888, Allan Stirling invented the bent water tube boiler which has the following features— (a) It consists of steam water drums, more than two in number. This improves the evaporation rates of water and the unit can meet the demand of load variations with less pressure drops. (b) The tubes connecting the drums are bent instead of being straight. The tubes are bent for the following reasons— (i) The bent tube allows free expansion and contraction and thus prevents the undue thermal stresses which are often met with in straight tubes. Bent tubes enter the drum in an aporoximately radial directim which (ii) permits easy connection with drums. (iii) It permits great flexibility in design, particularly in regards to drum arrangement. (iv) Tube replacement becomes easier. Construction and working. Fig. 6.7 shows the details of a stirling bent water tube boiler. It consists of four drums. The drums 1,2 and 3 are arranged in the upper part of the setting called steam drums containing water and steam and the drum 4 which is arranged in the lower part is called mud or water drum. The drums 1, 2 and 3 are connected with drum 4 by a series of bent tubes, arranged to give ample space for superheater tubes and with alternate wide and narrow space to permit removal of tubes without difficulty. The ends of the steam drums are dished and the drum has a man hole in one end. The water and steam spaces of the steam drums are connected by tubes. The tubes which connect the water space of adjacent steam drums are called circulating tubes and these help in the equalization of the water level in the drum. The tubes which connect the steam space of the steam drums are called equalizing tubes and these serve to equalize the steam pressure. The middle drum is fitted with a water gauge to show the water level in the steam drums. The steam drums are usually suspended by a steel structure and the mud drum is supported by tubes to permit free expansion and contraction of tubes. ' The feed water enters the rear drum 1 which is farthest from the fire and passes downward to the rear bank of tubes into the mud drum. Any foreign matter suspended in water settles down in the mud drum which is removed by blow off cock at regular intervals and only pure water passes upwards through the bank or the bent tubes to the front drum 3. The major portion of evaporation takes place in the front tubes and drums because the furnace is below them. The water from the front drum 3 passes through the water circulating tubes into the middle drum 2 and then downward through the middle bank of tubes to the mud drum and then retraces its course. The rear steam drum 3 is

Steam Generators, Ash Handling Systems. And Feed Water Treatment

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10

110 6

Flue gaS

9 14

1*

vr Hot gas

t t t t t t t ft

•LL.: I

Fig. 6.7. Stirling Bent Tube Boiler. 1, 2, 3. Steam drums, 4. Mud drum, 5. Equalizing tubes, 6. Water header, 7. Baffles, 8. Superheat tubes, 9. Inlet super heater, 10. Outlet superheater tubes, 11. Grate, 12. Fire brick arch, 13. Fire door, 14. Cleaning doors, 15. Steam stop.valve, 16. Safety valve, 17. Blow7off cock.

provided with safety valves to the main steam outlet and a dry pipe to ensure dry steam at the boiler outlet. The middle drum thus acts as a dryer and there is little moisture in the steam leaving the middle drum, and this is effectively removed by the baffle arrangement in the rear drum from which it passes to the dry pipe into superheater tubes. The steam from the drum is led into the superheater tubes which are bent and zig-zag and placed in the front portion of the boiler. The steam passing through the superheater tubes gets further heat due to the flow of hot gases over it. The steam thus superheated passes to the pipe leading to the steam stop valve from where it passes to its destination. The furnace is provided below the front portion of the drum. A cocking arch is sprung over the front of the grate with the grate directly under the arch when the boiler is fired by band stocking using solid fuels. The large triangular space above this arch (between the boiler front and the front bank of water tubes) serves as a combustion chamber. The fire brick arch gets incandescent hot due to fire which helps in combustion by heating the air required for combustion, ignites the gases distilled from coal and prevents the chilling of the furnace when the door is opened and cold air rushes in. The furnace gases are directed by fire brick baffles to past from the grate along the first bank of tubes; then down the middle bank and up the rear bank and fmally discharged to the chimney. Thus we see that the flue gases make three passes. Gas flow is generally in the opposite direction of the feed water. With a view to recover some of the heat from the flue gases the feed water is first circulated through an economiser before entering the drum 1.

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Steam & Gas Turbines And Power Plant Engineering

The entire boiler is surrounded by four walls. Doors are provided in front; side and rear walls for access to the interior of the setting and for inspection cleaning and repairs. There is also a provision of sooting door in the side wall and opposite to the middle of each bank of tubes, through which a jet of steam can be introduced to clean the outside of the tubes. The steaming capacity of the Stirling boiler is as high as 50,000 kg/h and pressure is as high as 60 bar and superheating temperature is upto 150°C. 6.3.3. Heat Absorption in Water-Tube Boilers. While designing the water-tube Superheated steam to h.p. turbine

To IP/LP turbine ___.

From h.p. turbine

Fig. 6.8 (a). Heat Absorption in a Water-tube Boiler. Superheated steam temperature

___

___ 4

6 .

Feed water inlet temp.

-I_ 1

Fig. 6.8 (b). T-s Representation of Heating Process in a Water- Tube Boiler.

Steam Generators, Ash Handling Systems And Feed Water Treatment

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boilers, it is essential to estimate the percentage of total heat absorbed in their various types of heat exchangers. In a water-tube boiler, feed water is heated in four types of heat exchangers namely economizer, evaporator (downcome-riser circuit), Superheater and reheaters (Fig. 6.8). It is to be noted that theoretically heating takes place at constant pressure. For each kg of water, the heat absorbed by different heat exchangers. g- eca = h2 —h1 (sensible heating) qeva = h3 — h2 (evaporation) q sup = /14 — h3 (superheating)

qreh

h6 — h5 (reheating)

The percentage of total heat absorbed in the economizer, evaporator, superheater and reheaters are given by (h2 — ht) x 100 (lea) x 100 % economiser = gm + eva + qsh + q reh [04 — h1) + (h6 — h5)] % evaporator =

eva

(h3 — h2) x 100

gee° + qe va + qsh + q reh [(h4 — hi) + (h6 — h5)]

% superheater =

% reheat =

x 100

gsup x 100

(h4 — h3) x 100

gee° + gem qsh + q reh [(h4 — h1) (h6 — h5)] reh X 100

qeco + qeva qsh + q reh

(h6 — h x 100 [(h4—hi)+(h6—h5)]

The use of feed water heaters is a must in steam power plant. The use of large number of heaters (upto ten) means a smaller economiser, a high pressure means a small evaporator surfaces (risers) because the latent heat of vaporisation decreases rapidly with pressure. In other words, a modern high pressure steam generator requires more superheating and reheating surface and lesser surface for evaporator and economiser than older units. It is worth to note that the fraction of total heat absorbed in superheaters alone in a modern boiler amounts to about 60%. 6.3.4. Circulation in Downcomer-Riser Circuit and their Sizing. In water-tube boiler, Superheater circulation means the flow of water and steam Economiser within the boiler circuit. Adequate circulation must be provided to transfer the heat from the flue gases of the furnace to the water. If the Heat circulation is caused by density difference, in rsoner Riser the boiler circuit, it is called natural circulatube tion. If it is caused by a pump, it is called forced or controlled circulation. H V Heat Heade Fig. 6.9 shows a simple downcomer-riser Pm circuit connecting a drum and a header. The Furnace downcommer is outside the furnace, so it \is inwall Downcomer sulated, while riser is inside the furnace. The Hot gas number of downcomer tubes is less as compared to the riser while the size of downcomer Fig. 6.9. Natural Circulation System.

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tubes in larger than that of riser. From the drum nearly saturated water falls by gravity through the downcomer into the bottom header and from this bottom header water flows along the riser where it partially boils with the formation of bubbles and then back into boiler drum. In other words, the flow is in two phase (steam-water mixture) in the riser. The density of steam-water mixture in the riser is less than that of saturated water in the downcomer and as result of this density difference a circulation current is set up within the downcomer-riser circuit. The steam so collected in the drum is taken out of the drum to the superheater where it is superheated. The feed water from the economiser enters the drum. The available pressure head for natural circulation is expressed as Op = gH (p0 —

(6.5)

Where H is the height of riser (i.e. furnace), pp is the density of saturated water in the downcomer and pm is the mean density of steam-water mixture in the riser. Though, the density of water-steam mixture (two phase) varies along the height of riser, however, for simplicity, the•mean density, pm may be taken as the arithmetic mean of the densities at the bottom and top of the riser. Thus Pm —

Pbottom + Ptop

2

(6.6)

Here, • pbottom = PD But

1

and

Plop— V

lop

•

vlop = of + xtop. V fg

where xlop is the quality of mixture (dryness fraction) at the top of the riser. The burning of fuel and the heat release rate decides the maximum height of the furnace (H). From eq. (6.5), for a certain height H of the furnace, we have AP cc (PD — Pm)

or

Ap cc

lvD

—

vm

(6.7)

The equ. (6.7) suggests that higher is the density difference, more will be the pressure

p

vg

V

Sat. liq line ‘. f

(a)

tc Sat. yap line

(b)

Fig. 6.10. Differential Specific Volume and Density Vs. Pressure.

Steam Generators, Ash Handling Systems And Feed Water Treatment

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head available for natural circulation. But the density difference decreases with the increase of pressure of steam as obvious from p-v and p-p diagram shown in Fig. 6.10. At critical pressure, pg = pf and thus there will be no circulation. The heat transfer takes place from the gas produced by burning fuel in the furnace to the water flowing through the risers which are installed (embedded) all around four walls of the furnace and thus act as a cooling tubes or water walls. It is essential that the rate at which heat is released must be absorbed at the same rate and for this reason, adequate circulation must be provided in the circuit. In the event of inadequate circulation, the rate at which heat is carried away will be less than the rate of heat release and thus the difference will be stored in the metal of the riser tubes leading to their overheating and rupturing when the tube temperature exceeds the metal melting point. The location of downcomers depends on the boiler pressure. As a general practice, if the boiler pressure is more than 30 bar, downcomers are placed outside the furnace in order to obtain more density difference for more natural circulation. If the boiler pressure is less than 30 bar, both downcomers and risers are placed inside the furnace with the risers installed in the hoter zone. The tubes which are heated less will act as downcomers while the tubes which are heated more will act as risers. In the event of boiler pressure exceeding 180 bar, the density difference (pp — Pm) becomes very small, as obvious from Fig. 6.10, which means poor and unreliable natural circulation and thus force circulation becomes essential. The forced circulation is achieved by using a pump installed in the downcomers which circulates saturated water through all the risers installed inside the furnace as shown in Fig. 6.11. For the analysis of process of steam generation, a term called "Circulation ratio" (CR) is used which is defined as CR —

Flow rate of saturatedwater in downcomers Flow rate of steam released from the drum Superheater Economiser uperheated steam

Feed water

Heat Boiler drum (BD) Forced circulation pump Downcome

Saturated water

Steam water mixture Riser Heat

urnace wall Hot gas

Fig. 6.11. Forced Circulation System.

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290

Saturated water

Bubbles QI Heat Surface blanketed by bubbles

— 02 = AU = mcAt Fig. 6.12. Nucleate and Film Boiling in a Riser Tube.

Mi.+ M

mg

mg

g

1 m Amf + m g

1 xiop

(6.8)

Where in is the mass of saturated water flowing through the downcomer-riser circuit during a certain time, ing is the mass of steam released during the same time, t is the mass of the saturated liquid water (fluid) at the riser exit and Xtop is the quality of mixture at the riser top. It is worth to note that circulation ratio varies from one riser to another due to varying heat absorption owning to placement position of risers. The riser tubes located opposite to the burners are subjected to more thermal loading and thus generate more steam which means less circulation. While the riser tubes located in the corner are subjected to less thermal loading (i.e. relatively cooler) which means higher circulation ratio. As a design practice, the circulation ratio in any tube should not be less than 6 otherwise the tube will get overheated and fail prematurely. Further, circulation ratio should not be higher than 25 in any tube for its efficient utilisation in the steam generation process. It has been observed that too much steaming (i.e. steam formation) is not desirable in the riser tube and the circulation ratio has to be maintained above 6 in any riser tube. From the point of view of heat transfer considerations a wetted surface is always desired and nucleate boiling should always persist in riser tubes. During boiling heat transfer, the initiation of bubbles start on the heated surface of the riser tubes. With the increase in heat transfer to the risers, the formation of bubbles may increase and filially it may form a stable film. The thermal conductivity of a vapor film is much less than that of a liquid film, so it offers a large thermal resistance, almost blanketing the surface where it forms and as a result the heat absorbed and carried away (Q2) will be much less than the heat transferred to the wall (Q1) (Fig. 6.12). The difference of heat (Q1 — Q2) will be stored in the metal of the riser tubes which will enhance the temperature of metal, If the temperature of metal exceeds its melting point, the tube will rupture and the leakage of water and steam will start. 6.3.4.1. Sizing of Downcomers. From the point view of design considerations, downcomers are fewer in number and bigger in diameter, whereas risers are more in number and smaller in diameter. In downcomers, water falls by gravity. Bigger the diameter, less the pressure drop due to friction since pressure drop is inversely proportional to tube

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diameter which is expressed as L C2 (6.9.) 6P = fT3 PT Therefore, the downcomers are made bigger in diameter which may vary from 150 mm to 200 mm or even higher. For estimating the number of downcomers continuity equation is used which is given as

inf = A pf C = (n 4 D2) piC

(6.10)

Where ri,j- = mass flow rate of saturated water in the drum (kg/s), n = number of downcomer tubes, D = diameter of downcomer tubes (m), pf = density of saturated water (kg/m3), C = average circulation velocity of water in downcomers (m/s) The value of circulation velocity in downcomer (C) depends on the circulation ratio in tubes and varies from 0.4 to 1.4 m/s depending on the boiler capacity. Thus the number of tubes estimated from eq. (6.10) can be 6, 8 or even high as 20 depending upon boiler capacity. 6.3.4.2. Sizing of Risers. Since risers absorb heat from the furnace, so for the same total cross-sectional area, the smaller the diameter, the larger the surface area exposed to hot gas for heat transfer. From this point of view, risers are of small diameter and as a practice, the diameter varies from 62.5 to 76.5 mm. For estimating the number of tubes, continuity equation is used where the determination of density of water-steam mixture (two phase) is a tedious and complex proposition. For efficient heat transfer it is essential that nucleate boiling should occur in each riser tube and film boiling is avoided. Departure from nucleate boiling and on-set of film boiling occur due to too much steaming and less circulation in risers. Vapor The recent trend is to provide internal twisters and bubbles springs in riser tubes placed in high absorbing areas of the furnace which break the vapor film and retard the onset of film boiling. Tubes ribbed helically on their inside surface are often used now as ribbing creates centrifugal action which washes away the vapor film formed. Fig. 6.13. Two-phase Flow in a A term "void fraction (a)" or volumetric qualRiser Tube. ity is used to analyse the two phase flow which is defined as Volume of vapor a— Volume of liquid + Volume of vapor The specific volume of a two phase mixture is given by v = (1 — x)v + xv v + xv Is Is The density of a two-phase mixture with void fraction a is given by p = (1 — x) pf + a p pf + a p g g Consider a two-phase flow in a riser tube where

Af

is the cross-section occupied by

292

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saturated liquid and Ag is the cross-section occupied by vapor bubbles (Fig. 6.13). Thus the void fraction for such a flow is expressed as A 1_a A g _Ig — or a— A a A +A (6.11) I g In the movement of a two-phase mixture, the vapor, because of bouyancy has a tendency to slip past the liqi4id i.e. the velocity of vapor is higher than the liquid. To account this, a slip ratio (S) is defined as K Velocity of vapor S=_ — Velocity of liquid C

(6.12)

The value of S at any cross-section of the riser is more than unity. A P g AgCg =— g Cg mv in m g

in

Since quality = x =

in . v .x

or

C — g

A

g

(6.13)

g

(1 — x) inv

Similarly,

=

' 1 The ratio of Cg and C1 is expressed from eqs. (6.13) and (6.14) C g C I S

A

xv . A g I Ag (1—x) v

(6.14)

(6.15)

— j[(2) _11-1 ,K) = (x 1 x Ag vi.

From eq. (6.11), we have =( x 1—a 2 1g (vf.) —x)(( a ) (6.16) (11—x) From the eq. (6.16) it is obvious that as the boiler pressure increases, the slip ratio decreases. Experimentally, it has been observed that the slip ratio varies between 1 and 10. The value of S = 1 occurs at high pressure where. liquid and vapor densities approach each other. After re-arranging the eq. (6.16), we have 1 1 _ a= + 1 x 22 1 1 x As + .S 1 x (x v (6.17) g V, where (I)= S V (6.18) S=

g

Similarly,

1

x— 1+

1 —a

a

(6.19)

Steam Generators, Ash Handling Systems And Feed Water Treatment

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The quality distribution along the height is obtained by assuming a reasonable value of S (say between 1 and 2). The mixture density distribution may be obtained from P = (1 — ct) Pf + a Pg

(6.20)

The average mixture density in the riser of height H is given by Pm = — H

PJ Wdz

(6.21)

o

where z stands for axial distance from the bottom of the riser. For uniform axial heating, El Wakil (87) has suggested the following distribution for the mean density in the riser Pf-

Pm = Pf

Pg {

[ 1 • a—

1—

1—

—

In

1 - ae (1 — 4))

(6.22)

where ae = void fraction at the riser exit It is worth to mention that the total pressure head produced due to density difference (given by eq. 6.5) is balanced by the pressure losses in the downcomer-riser circuit. AP = gH

Pm) = APD + APR + AP bends + AP header

(6.23)

where, suffixes D, R, bends, header placed on ep indicate the pressure losses in downcomers, risers, bends and header respectively. From the heat transfer analysis, we have Q=UAs AT = UOr DIV)

AT.

(6.24)

Non-heated riser Upper header

Downcomer*

Heating Saturated water Risers heated by hot gases Lower header

Fig. 6.14. Natural Circulation in a Boiler.

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Steam & Gas Turbines And Power Plant Engineering

where U is the overall heat transfer coefficient and AT„, is the log-man temperature difference. By knowing the value of U, D and AT,,, the number of riser tubes (N) can be estimated from eq. (6.24). The value of U depends on many factors such as heat transfer coefficients (inside and outside), thermophysical properties, geometry of tubes, flow behaviour inside and outside tubes, etc. The circulation circuit in a natural circulation boiler showing for simplicity three downcomers, a lower header from which water enters the riser tubes, nine heated risers, an upper collecting drums and four pipes for delivery of the steam is shown in Fig. 6\14. As mentioned the circulation rate depends upon the thermal loading of the circuit canemed. The steam formation rate will depend upon configurations and dimensions, density of water and steam and thermal loading of the circuit. 6.3.5. Steam Drum and its Internals. The purpose of steam drum is to receive the mixture of water-steam, foam and sludge from the risers and to separate the steam from the mixture before the steam leaves the drum for superheaters. Any moisture carried with steam to the superheater tubes contains dissolved salts. During superheating process water evaporates and the salts remain deposited on the inside surface of the tubes in the form of scale which is difficult to remove. The scale so formed reduces the rate of heat absorption which in turn leads to the overheating of the tubes and finally failure through ruptures. The superheater tubes are subjected to the highest steam pressure and temperature on the inside and the maximum gas temperature on the outside, and so they are made of the costliest materials which can withstand high temperature and pressure. This suggests that utmost care should be taken so that no damage is done to superheaters due to moisture carried away with steam. Further, some of the impurities in the steam may be vaporized silica which may cause turbine blade deposits resulting in poor turbine performance due to disturbed flow. The another requirement of steam drum is that no vapor bubble should flow along with saturated water from the drum to the downcomers as it will reduce the density difference and the pressure head for natural circulation. Further, the bubbles due to its own characteristics tends to flow upward and thus it will impede the flow in the downcomers resulting in reduction in circulation. This suggests that the steam drum has to insure moisture-free steam going to the superheater and bubble-free water going to the downcomer as shown in Fig. 6.15. Though, adequate treatment of feed water along with make-up water is done prior to feeding Moisture-free Feed water but still it carries impurities in the form of total vapor to SH from dissolved solids (TDS) which are generally execonomiser pressed in ppm (parts per million) by mass. One • (EGO) ppm means one part of impurity in millions parts tea m ' goiter Twa teo of water by mass. Since whole circulation andrum feeding process are continuous and feeding procBubble-free ess are continuous, so solid content of water goes Riser (R) water on increasing. In order to maintain a certain ppm Downcomer in the drum, continuous or intermittent blowdown (D) 1Hot gas is essential. Continuous blowdown is more cornmon in utility boilers. Hot pressurized water is used to blowdown solids precipitated at the bottom by opening the blowdown valve at the drum Fig. 6.15. Separation of Two-phase bottom. The ppm of drum water depends on presMixture in Steam Drum.

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sure and varies from 1000 to 2000. In order to help the precipitation process, trisodium phosphate or tannin is injected periodically in the drum. For continuous blowdown, the total solids entering will be equal to the total solids leaving. Thus infr x ppm = BD(ppm)drum + (1—x) ms (pPM)canyover where

is the rate of feedwater entering the drum, BD is the necessary blowdown rate, .fw ms is the rate of steam flow from the drum and x is the quality of steam leaving the drum. From the above discussion it follows that steam drum has to perform many functions Baffles

Steam to SH

Steam to SH From riser

fi 'QV Prirt1VV\ •••••••••• ••••• ••••

From riser

Screen

From riser

To downcomer

From riser To downcomer

To downcomer (b)

(a) Steam to SH

Scrubber

Scrubber

Cyclone

Cyclone

From riser

From riser From riser

From riser Blowdown

To downcomer (c)

Fig. 6.16. Mecharism of Separation in Drum (a) Gravity (b) Baffle and Screen, (c) Cyclone and Scrubber.

•296

Steam & Gas Turbines And Power Plant Engineering

as given below— (i) It stores water and steam in sufficient quantity to meet the varying load requirements. (ii) It helps in circulation of water. (iii) It separates steam from water-steam mixture discharged by risers and sends moisture free steam to superheaters. (iv) It supplies bubble free water to downcomers. (v) It provides enough surface area for liquid vapor disengagement. (vi) It permits the blowdown to remove precipitated dissolved solids. (vii) It maintains a certain desired ppm of dissolved solids in the drum by phosphate injection and blowdown. Mechanism of Separation. The method of separation of steam from steam-water mixture discharged by the risers is based on the boiler pressure. If the boiler pressure is below 20 bar, gravity separation is simply used with the condition that sufficient disengaging surface must be provided as shown in Fig. 6.16(a). Water being heavier separates out by gravity and steam moves up as pf» pg. But at high pressures, separating forces due to gravity is small. This is because of the fact that as pressure increases, the density difference (pf — pg) decreases and so the separating force. At 100 bar, the separating force is about one-third of at 20 bar so it is difficult to achieve adequate separation. Though, an improvement in separating force may be achieved by increasing the size of the drum but the construction of drum is a costly affair, so it is not economical. This suggests that a positive mechanical means of steam-water separation becomes a necessicity. There are various types of mechanical separators such as baffles, screens and cyclones which are housed inside the drum and known as drum internals. Fig. 6.16 (b) shows the baffler plates which act as a primary separators. These baffles plates change or reverse the steam flow direction and act as impact plates that cause water to drain off. In other words, baffle plates assist in gravity separation. Screens (Fig. 6.16(b)) made of wire mesh act as a secondary separators where individual wires attract and intercept the fine droplets, just as fabric filters attract dust from gases. The water accumulated on the screen wires drops and then fall by gravity back to the main body of water. In this way, we can say that screen also assist in gravity separation. Fig. 6.16(c) shows the cyclone separators which utilises the effect of centrifugal forces for separation of two-phase mixture. The mixture is entered tangentially to direct the water downward and to make the steam flow upward. After coming out from cylcone, the steam then goes through the zig-zag path in corrugated plates, called the scrubber or dryer positioned on the upper portion of the drum which removes even the last traces of moisture from steam. Finally, the perforated plates or screen under the drum exit perform the fmal drying action of steam. Fig. 6.17 shows the drum internals of a controlled (forced) circulation boiler. In many cases, the separation process also involves a washing of the steam by passing it through a rain of the relatively pure feed water having very low ppm before passing through the scrubber which reduces the ppm of moisture carry-over. The other advantage of washing of the steam lies in the reduction of silica concentration in steam as silica vapor is absorbed in the feed water. If there is too much moisture carry-over because of a high water level or a high steam rate, such a boiler drum is said to be priming. Priming phenomenon is induced by erratic feedwater control and rapid changes in steaming rate. This suggests that in order to reducing priming a certain water level in the drum should be maintained to give adequate steam space.

Steam Generators, Ash Handling Systems And Feed Water Treatment

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Foaming is another phenomenon encountered in the delivery of impure steam resulting from the formation of bubbles on the water surface. This is due to the presence of saponification agents in the water drum such as oil, certain dissolved salts, and high alkalinity. Some of the foam can be drawn off the water surface and wasted by blowing down through a connection made on the drum at the normal water level. The design of the shell of steam drum is based on the theory of pressure vessel. The size of the steam drum in the utility boilers having diameter as high as 5 m, length more than 30 m and weight a few hundred tons largely depends upon the space requirements of all the internals, steaming rate and minimum water level. The steam drums being a pressure vessel are made in cylindrical sections, called courses, which are welded together, and two nearly hemispherical heads which are welded to the ends. It has to be safeguarded against the explosion. 6.4. Modern and High Pressure Water Tube Boilers Modern pulverized coal-fired boilers stand up vertically integrated with a furnace, a steam drum, evaporators, (downcomers and risers circuits) superheaters, reheaters, economizers, air preheater, tube mills, forced draught and induced draught fans, coal feeders and burners, windbox, various control systems, ductwork and all accessories and mountings for safe operation of boiler. The principal structural members consist of vertical structural steel columns, a top support grid and beams (i.e. backstays) meant to provide stiffness to the furnace, combustion air and flue gas flow enclosures etc. Burning of the fuel is initiated by igniters and warmup burner for directing the flow of combustion air and flue gases, a duct work system is used. In order to remove ash or slug deposits, soot blowers are used. Modern water tube boilers include both natural and forced circulation boilers while high pressure boilers in general include boilers only with forced circulation. The modern Steam and water mixture rom user

Steam outlet Steam and water mixture rom riser Drum shell

Feedwater from economiser

To downcomers Fig. 6.17. Drum Internals of a Controlled (Forced) Circulation Boiler.

298

Steam & Gas Turbines And Power Plant Engineering

trend in central power station is to generate and use steam at high pressure and efficiency. The modern boilers used for power generation are for steam capacities ranging from 4 to 1300 kg/s and above with a pressure up to 200 bar and maximum steam temperature about 600°C having furnace height varying from 32 to 62 m. Now, it is possible to generate steam above the critical pressure of water (221.2 bar). But these supercritical boilers are different than subcritical boilers in both design and operation because of the fact that properties of steam in the critical range is different. The high pressure boilers are characterised by the following— (i) Forced circulation of water— In high pressure boilers, circulation of water is forced instead of natural. Therefore, a pump is used to force the water in the boiler. (ii) Arrangement of drums and tubing. In order to avoid any large resistance to flow of water, these boilers have a parallel set of tubes arrangement. They have a very small steam separating drum or may be entirely free of drum. (iii) Imported method of heating. The following methods are used to improve heating— (a) Superheated steam is used to heat the water. (b) Saving of latent heat by evaporation of water at pressure above critical. (c) The heat flow through the tube walls may be increased with the use of hot gases travelling with supersonic velocity. Advantages of High Pressure Boilers. The following are the advantages of high pressure boiler. (i) Smaller bore and therefore lighter tubes make the unit more compact. The space and weight requirements are minimized which reduces the erection time and cost. (ii) Reduction in the number of drums. (iii) There is greater freedom for disposing of the heating surface and hence greater evaporation for a given size. (iv) Lighter structure for a given output. (v) The boiler is capable of meeting rapid changes of load without the use of complicated or delicate control devices. (vi) All the parts are heated uniformly which eliminate the danger of overheating and setting up thermal stresses. (vii) Due to uniform temperature of parts the differential expansions are minimized, this reduces the leakage of gas and air. (viii) There is complete elimination of high head which is needed for natural circulation. (ix) Due to high circulation velocity the tendency to form scales is eliminated to a large extent. (x) If a external supply of power is available, very rapid start from cold state is possible. Hence the boiler is suitable for carrying peak loads. It is also useful for standby purposes in hydraulic station. The system is slightly complicated and certain percentage of the power is consumed by the circulating pump. Examples of high pressure boilers are Lamont, Benson, Velox Loeffler, Monotube, Radiant type (Natural circulation), etc. Three design concepts of water tube boilers are shown in Fig. 6.18. Type A is a boiler with natural circulation used up to 180 bar. Heat transfer to the water tubes around the walls is mostly by radiation from the fuel flame and less by convection from the flue gases. Type B is the forced circulation by a special pump and suitable up to pressure of

Steam Generators, Ash Handling Systems And Feed Water Treatment

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A

Fig. 6.18. Boiler Concept based on Circulation. (A) Natural Circulation, (B) Forced Circulation, (C) Once-through boiler. 1. Boiler drum, 2. Steam separator, 3. Boiler feed pump, 4, Circulation pump, 5. Economiser, 6. Evaporator, 7. Superheated. 8. Furnace. 200 bar. Type C is once-through boiler, there is no drum, so no circulation, and no separation of water and stream and they operate above critical pressure (221.2 bar). 6.4.1. Radiant Type Power Boiler With Natural Circulation. Most of the steam power plants are of the radiant type, and have moderately rated, water cooled (water wall) combustion chamber in which the heat is transferred to the evaporating surface by direct radiation as shown in Fig. 6.19. These boilers employ natural circulation and may use pulverized coal or employ cyclone furnace or fuel oil burning equipment. Some radiant boilers are stoker fired. The boiler shown in Fig. 6.19 employs a pulverizing mill. The pulverized fuel is fed into the combustion chambers with the help of burners. There are three zones of heat transfer (Fig. 6.19). In the zone I, heat transfer is predominantly by radiation as the flame is this zone is diffused yellow-flame which radiates much more than the premixed flue flame. With the upward movement of burnt gases and the addition of secondary air, the effect of radiation gets reduced and the convection becomes predominant as the flame (hot gases) changes from diffused to premixed. In the upper part of zone I, the heat transfer takes place by radiation as well as convection provided suitable heat transfer surface is introduced into the path. This space is marked by (R+C). The heat transfer in the zones II and III takes place mainly by convection and they are identified as high and low temperature zones respectively. The height of furnace varies from 42 m to 70 m. Water at about 170 to 260°C depending on boiler pressure from high pressure feed heater enters the economizer and leaves saturated liquid or as a two-phase mixture of low quality and then enters the steam drum at midpoint along lengthwise. The downcomers are insulated and are situated outside the furnace through which water flows and feeds to the bottom headers through which riser tubes situated inside the furnace and lined to the furnace wall are connected. The water in the riser tubes receives heat mainly by radiation (zone I) from the combustion gases and boils. The circulation is set up due to the difference in densities between fluid in the risers and downcomers. Steam is separated from the bubbling water in the drum and goes to the convective or primary superheater (CSH) where heat is absorbed by convective mode (zone II). From the convective superheater steam is fed to the radiant superheater (RSH) installed at the top of furnace, where heat is absorbed by radiation. Stearn after leaving the radiant superheater goes to the desuperheaters where water of high purity is sprayed on to the steam to bring down the steam temperature to its

Wind box

▪5'

CJ

eA Bumers

RSH

SA •:i PA 1:1 7 zz z 77,777777 ( PA) < Tube mills

Coal jCoa

op

Downcomers (D)—*

Blowdown

Drum internals

to 2. Primary air+coal

0

0

•

n ▪.

•

Boiler drum (BD)

Heated water

-0` Water

ai

(

Steam from HP turbine

Steam to IP turbine

Stack

160 °C

T

Ash

Ash

Slurry pump

Fuel gas

To site of disposal

-6E- FW from HP Heater Electrostatic APH FD fan precipitator t . =r-=-zAir (ESP) ID fan

Heated water 6 r. Economiser (ECO)

CSH

,,Reheater (RH)

Baffle'

(CZ)

Risers (R)

(

® Main stop valve

Cinker grinder Ash

)J,)

I(IF Furnace

1300 °C

Desuperheater

—*Superheated steam to H.P. turbine

CSH : Convective superheater RSH : Radiant superheater PSH :Pendent superheater RZ :Radiant zone CZ : Convective zone PA : Primary air SA : Secondary air FW : Feedwater FD : Forced draught

0

Steam &GasTu rbinesAnd Power Plant Engineering

Steam Generators, Ash Handling Systems And Feed Water Treatment

301

desired value if it exceeds the latter. After leaving the desuperheater, steam fmally goes to the pendent superheater (PSH) for further superheating before it leaves through the main stop valve to the high pressure (HP) turbine. In general, both RSH and PSH are often termed as secondary superheaters. The steam exhausting from HP turbine is fed to the reheater where it is reheated to the desired temperature. After leaving the reheater, steam is fed to the IP turbine for further expansion. The location of superheaters may vary as per design of the furnace and flue gas path. In many cases desuperheaters are not provided. Forced draught (FD) fan supplies the air for combustion which is preheated in a heat exchanger called air preheater (APH) by the outgoing flue gases. Air heated to the desired temperature in APH which varies from 250°C to 400°C depending upon the type of coal burnt, is split into two streams— primary and secondary. Hot primary air constituting about 15 to 30% of total air is fed to the pulverizer mill to dry the coal for better grinding and also to help push the coal particles through the burners into the furnace. The hot secondary air is fed into the windbox and where it gets distributed to the burners to help combustion. Based on the quality, the fuel burns with great intensity to some 1350°C or even higher. The combustion gases (flue) transfers their energy to the riser tubes, then to superheaters, reheaters, economizer and preheater and fmally leaves at about 160°C and goes to

Boiler drum

1

11111111

Ea;,..v,garAir. i'111111111111 -11111111,

Secondary superheater section

•

liPI

I Primary :superheater section

Coal Hopper

Economizer

wwV!! -ii Pulverizers

Risers

Fig. 6.20. Modem Radiant Power Boiler with Natural Circulation Using Pulverized —Coal Showing Coal Hopper and Structure.

302

Steam & Gas Turbines And Power Plant Engineering

atmosphere through stack. Dust collectors and electrostatic precipitators (ESP) collect the entrained solid particles (mostly ash) from the flue gas stream before it enters into the stack. An induced draught (ID) fan is used to draw the flue gases from the furnace and sends them through

Boiler drum 21

1. Feed water inlet 2. Economisers 3. Bottom bank 4. Top bank 5. reheater bottom bank 6. Reheater top bank 7. Primary superheater outlet lebs 8. Reheat outlet legs 9. Risers 10. Upflow cage rear outlet 11. Reheat outlet manifold 12. Platen superheater 13. Final superheater 14. Reheat final loop 15. Inlet boxes 16. Rear screen 17. Final superheater outlet manifold 18. Front wall 19. Rear wall 20. Side wall bottom boxes 21. Boiler drum 22. Downcomers Fig. 6.21.(a). A 660 MW CEGB (UK) Steam Generator.

303

Steam Generators, Ash Handling Systems And Feed Water Treatment Steam to IP turbine

Primer reheater vertical

Primary superheate rf vertical

Final reheater

Steam A to HP

turbine

"Ai Final

vvvv\A Platen super * super heater heater

A

E

co a)

co

(.9

co a)

V E

c

Primary super heater

Primary reheater

RI

0 2 To drum Feed wat

Supe heater economiser

Reheater economiser

Combustion gases Steam from drum

To drum Feed water U-

Air heater

Gas to stack

Fig. 6.21.(b). Flow Path of Flue Gas/Water/Steam in 660 ME CEGB (UK) Boiler.

the stack to the atmosphere. The flue gas leaving the air preheater at about 160°C represents the availability loss for the plant but it is accepted and reasonable due to the following reasons— (i) It is essential that the flue gas leaving the stack should be kept well above the dew point temperature (DPT) of the water vapor present in the flue gas (which is equal to saturation temperature at the partial pressure of water vapor) in order to prevent the condensation which would form acids due to presence of SO2 or SO3 and corrode the metal components in its path, i.e. stack. (ii) Enough buoyancy must be present in the flue gases in order to rise in a high plume above the stack for proper atmospheric dispersion. The clinkers from the dry bottom furnace fall to the clinker grinder which are broken into pieces and them carried by water in a hydraulic sluice wherein also falls the ash from the dust collectors and the precipitator and finally it is led to the site of disposal with the ,help of slurry pump.

Steam & Gas Turbines And Power Plant Engineering

304

An another modern radiant power boiler with natural circulation and single drum using pulverized coal is shown in Fig. 6.20. Fig. 6.21 (a). shows the cross-section of 660 MW CEGB (UK) steam generator and the flow path of flue gases/water/steam through the various elements of boiler. The flow path is shown in Fig. 6.21(b). The development of boiler has seen many changes. In 1952, the height of furnace was about'32 m and steaming rate was 35 kg/s while in 1974, these were 61.4 m and 560 kg/s respectively. At present, these values have crossed 70 m and 1300 kg/s respectively. 6.4.2. Boiler Water Wall. The design of boiler water wall plays an important role in the overall design of water tube boiler. It consists of a number of riser or evaporator or \ water tubes spaced all around the walls of the furnace and carries away the heat released by burning of fuel in the furnace. Fig. 6.22 shows four ways of laying the riser tubes on the wall namely (a) tangent tubes touching the refractory wall, (b) tubes embedded in the refractory, (c) studded tubes and (d) membrane tubes. The membrane water wall is in current use. It consists of tubes spaced on centres having centre to centre distance varying from 1.2 to 1.3 times the tube diameter while tube diameter varies from 68 to 76.2 mm. The tubes are connected by membrane or bars welded to them at their centre lines. The membranes give two advantages as they act as a fins to increase the heat transfer as well as to strengthen the furnace construction. In order to reduce heat loss, insulation is provided on the outer side of the wall and then metal logging is provided to protect the insulation. Since the riser tubes are located on the four sides of furnace wall, so the water wall absorbs mainly by radiation. The radiant energy is expressed as QR = a A eF (74 — 7',4v) g where T and Tw are the gas and wall temperature respectively, a is Steffan Boltzman constant, F is the shape factor and a is the emissivity which depends upon the constituents of combustion gases (CO2, H2O, SO2, etc.) and their partial pressures. 6.4.3. LaMont Boiler. This is a high pressure, forced circulation, water tube type boiler invented by LaMont in 1925. Fig. 6.23 shows the details of a LaMont high pressure boiler. The feed water from the hot-well is supplied through an economizer to a separating and storing drum that contains a feed regulator which controls the speed of the feed pump.

Centerline wall tubes Refractory Membrane bar Refractory wall

Refractory wall

Metal lagging

Metal lagging

Insulation

Risers

Cylindrical studs

Hot gas

Hot gas (a)

Insulation

Hot gas (b)

(c)

Fig. 6.22. Types of Water Wall.

(d)

Steam Generators, Ash Handling Systems And Feed Water Treatment

Air preheater Feed watv>

305

7 i 11111111111_, '

Ai Economiser

Ai Storage and separetra drum Down corner

Super heater m Main steam Evaporator (convective) Preheated air Evaporator (radiant)

Circulating pump Distributing header Fig. 6.23. LaMont Boiler.

Since the economizer is placed in the boiler at a place from where hot combustion gases pass, the economizer supplies sensible heat to the feed water from the boiler drum flows by gravity to a circulating pump, which discharges into a distributing header. Water from the distributing header flows through long small diameter boiler tubes located in the wall and roof of the furnace to the drum where the steam is separated and water returns to the pump. Orifice located at inlet to each circuit on the distributing header correctly proportions the water among the many parallel circuits, so that, each receives its proper share. The circulating pump rises the water pressure to about 3.5 bar above the drum pressure to overcome the resistance to the flow- controlling orifices and the long circuit of small diameter tubing. At normal load, the quantity of water circulated in each tube is about 3.8 times the steam evaporated in the long circuit of small tubes. This avoids the overheating of the tubes. Since three circuit tubes are placed in the combustion chamber through which hot gases are passing upwards, the steam is generated in them. The mixture water and steam from these tubes enter the boiler drum where the moisture is separated from the drum. Now the steam is led to the superheater tubes in which the steam gets superheated. The superheated steam goes to ultimate destination through the steam stop valve. It is essential to maintain a constant level of water in the drum. This is possible by supplying feed water equivalent to the steam quantity blown-off continuously. The pump consumes about 0.5 to 0.6% of the boiler output which is supplied by the power unit using the steam from boiler. For economic combustion forced air preheated by the flue gases is used. 6.4.4. Benson Boiler. The benson high pressure boiler introduced by Mark Benson in 1923 is a forced circulation water tube boiler. The main feature of this boiler is the absence of the steam separating drum. The entire process of heating steam formation and superheating is done is a single continuous tube but to increase efficiency, many parallel

306

Steam & Gas Turbines And Power Plant Engineering

circuits are used. The efficiency of this boiler is as high as 90%. Fig. 6.24. shows the details of a Benss.•`%%% Fan son boiler. It consist of (i) Air preheater in , Air • which air is preheated for economical 1111111111111111Pro 4— it preheater combustion, (ii) Economizer in which Economiser sensible heat is imparted to feed water, Feed (iii) Radiant surface of tubes in which rapump diant heat is supplied by combustion, (iv) Evaporative Evaporative surface where the major surface quantity of water is evaporated, (v) SuperThrottle valve heater tubes in which steam is superheated and led to the steam turbine. Superheater Main steam It is an established fact that the rate of heat transmission from flue gas to water is Radiant seriously impaired by the presence of surface steam bubbles in contact with the tube, and that, the release of these bubbles causes the water circulation to pulsate. It was pointed out by Mark Benson' that if the boiler pressure was raised to the critical pressure, the steam and water would Fig. 6.24. Benson Boiler. have the same density and, therefore no bubbles would be fonned, thus, eliminating the trouble. But to raise the pressure above critical, a large amount of power is consumed by the feed pump which lowers the efficiency of the plant. Hence by using pressure slightly lower than critical improved efficiency is possible. At normal load, this boiler has an operating pressure of 210 bar with 13500 kg/h evaporating capacity at 405°C superheating temperature. Bension boiler is very mean to supercrisical boiler. The Benson boiler has the following advantages— (i) There is no presure limitations and it may be as high as supercritical. (ii) Absence of circulating pump and downcomers. (iii) Absence of drum hence the cost is less. (iv) Lighter in weight, higher specific output with high safety factor. (v) Evaporation is quick. (vi) Self-contained unit and can be easily erected. The disadvantages are— (i) Evaporation process is accompanied by formation of salt and solids in the tubes. Special arrangements are required to remove this. (ii) On evaporative surface, there is chance of corrosion of the tubes. (iii) Overheating of the tubes in case of insufficient water supply. (iv) Since the storage capacity is small, it requires close coordination between steam, feed water and fuel input. 6.4.5. Loeffler Boiler. This boiler is a forced circulation, indirect heating type of boiler. In this, water is evaporated solely by means of superheated steam i.e., steam is used as the heat carrying and heat absorbing media.

Steam Generators, Ash Handling Systems And Feed Water Treatment

307

To/chimney Fan A

4—Air

Oil fuel or pulverised fuel

Feed pump 4— Economise Feed

tank Convection surface Radian surface

0)

Superheater coils

Steam circulating pump

>

Main steam

-------------------------------------------------Drain

Evopor

r::d :

Feed

6.25. Loeffler Boiler.

Fig. 6.25 shows the details of a Loeffler boiler. It consists of an evaporator drum which may be placed at any convenient point outside the furnace setting. The feed water pump is placed where the feed water passes through as economizer on the way to the drum. Since the economizer is placed in the path of the outgoing hot gases; the economizer gives sensible heat to water. The steam circulating pump extracts steam from the evaporator drum and forces it to pass through the radiant and convective surface of the superheater tubes placed in the path of the hot combustion gases. From the superheater tube, a big portion (about 3/4) of the superheated steam is trapped off for external use while the remainder (1/4) passes on to the evaporator drum where it gives its superheat to the water contained in this drum that generates an amount of steam equal to that trapped off. Nozzles are provided which distribute steam throughout the water contained in the drum. The nozzles are of special design which avoid priming and noise so that the boiler can carry a higher salt concentration than any other type of boiler. This makes it very suitable for marine transport and for power generation. The air, preheated through the air preheater placed in the path of the outgoing hot gases is used for economical combustion. The operating pressure of this boiler is 140 bar and the steaming capacity is 95000 kg/h. 6.4.6. Velox Boiler. The velox boiler is a high pressure, forced circulation, pressurized or forced combustion boiler but with the limitation of firing with oil or gas under pressure with little or no ash problem. The evaporative capacity is limited to 10200 kg/h of steam. Fig.. 6.26 shows the details of a Velox boiler. Air is compressed to about 2.5 bar in an axial compressor (which, at the time of start is driven by a motor) before being supplied to an oil fuel furnace. The purpose of this compression of air is to secure a high velocity of

308

Steam & Gas Turbines And Power Plant Engineering.

Oil burner

1

4

Compressor

5

10 9

1g

Counter flow economiser

Li Feed Gas

Feed ring main Circulating pump

Mud drain

Fig. 6.26. Velox Boiler.

gas and also at the same time release of a great amount of heat. The fuel and the compressed air are injected downwards into a vertical combustion chamber which is surrounded by hollow evaporator tubes on reaching the bottom of the combustion chamber the products of combustion are deflected upwards into the evaporator tubes by means of a spiral water coil. The evaporator tubes consist of an outer annulus through which 10 to 20 times the .water evaporated is circulated at a high velocity. The core of the lower half of the evaporator tube or element is occupied by a central pipe which supplied water to the outer annulus, while the upper half is occupied by a central pipe which supplies water to the outer annulus, while the upper hill is occupied by U-type superheater tubes. In the space between the inner pipes and outer annulus, the flue gaS rushes at a speed of about 250 m/s. There is provision of a ring main (4) which collects the steam and water, and discharges it tangentially into the separating chamber (5). This forms a forced vortex, which, by centrifugal loading on the water particles, allows steam release, without priming about two hundred times as great as in boilers of normal design. The dry steam then passes up the central tube (2) the superheater ring main, which distributes it to the various superheat elements. The muddrum collects the separated water which is extracted by means of a circulating pump. This circulating pump also creates a high water velocity through the evaporator tubes. The flue gases, after passing through the superheater enters an exhaust gas turbineS that drives the compressor. The exhaust from the tuibine passes through the counter flow feed heater where the feed water is preheated and which is discharged tangentially into the separating drum. This boiler is very compact steam generating plant of great flexibility. It is capable of quick starting and its theimal efficiency is Lout 50 to 95%.

Steam Generators, Ash Handling Systems And Feed Water Treatment

309

6.4.7. Once-throught or Monotube Boiler (Supercritical Pressure Boiler). The oncethrough or monotube boiler as the name implies has only one tube almost about 2 km long and there is no steam drum. It is the only type suited to supercritical pressure operation, i.e. above 221.2 bar as there is no latent heat of vaporization involved, and water on being heated at constant pressure (theoretically) suddenly flashes into vapor and steam is further heated to desired temperature in the superheater. Water is fed to the boiler by boiler feed pump at inlet through a number of parallel paths and comes out at outlet as superheated steam. LaMont may be called as monotube boilers. Subcritical boilers may also be monotube boilers. Many super-critical pressure once-through boilers were built either with single or double reheat having steam pressure 240-345 bar and steam temperature 540 to 650°C. A simplified view of CE Sulzer once-through boiler with triflux reheater where steam is reheated partly by steam and partly by the gas, is shown in Fig. 6.27. The main advantages of once-through boiler are the absence of drum, and best suited to quick starts rapid load changes. The main disadvantage is the poor quality boiler water. Advantages : The following are the advantages of supercritical boilers over sub-critical. (i) Large heat transfer coefficient, (ii) Stable pressure level, (iii) Higher thermal efficiency,

0 0 Steam 6

N

Steam

HPT

LPT 8

—10

4

0 Feed water

2" Fuel Air

Hot gas

1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

Economiser Evaporator Superheater Triflux reheater Final superheater Final reheater High pressure turbine (HPT) Low pressure turbine (LPT) Condenser Water separator

Fig. 6.27. Sulzer Once-through Boiler (Steam Generator).

310

Steam & Gas Turbines And Power Plant Engineering

(iv) Minimum erosion and corrosion, (v) the easy to generate peak load and (vi) Ease in operation 6.5. Fluidized Bed Boilers. The efforts to minimize the pollution resulting in from conventional steam power plants and to have efficient use of low grade fuels led to the discovery of fluidized bed boiler in which steam is generated by utilising the technique of fluidized bed combustion. The detailed analysis of fluidized bed combustion has been discussed in the chapter. Solid (coal) particles when subjected to a stream of air may be either (i) unaffected if the air velocity is below a certain value or (ii) air borne if the air velocity is above a certain value. Between the limits of velocity, a bed of small solid particles of fairly uniform size will not become fully air-borne but will have their reduced effective density and will move about in the bed as if bed is fluid. By proper regulation of air velocity the solid particles can be burnt in the combustion chamber almost as easily as if they are fluid. The critical velocity is called the minimum fluidization velocity at which the bed is said to be incipiently fluidized. With the further increase in velocity, the particles are buoyed up and imparted a voilently turbulent fluid like motion. The mixing is rapid and equilibrium between gas and particles is rapidly established which is called a fluidized bed. There are two types of fluidized bed boilers. They are— (i) Bubbling fluidized bed (BFB) boilers (ii) Circulating fluidized bed (CFB) boilers Both boilers may be operating at atmospheric pressure or pressurized condition. 6.5.1. Bubbling Fluidized Bed Boiler (BFB). Fig. 6.28 shows a schematic of a bubbling fluidized bed boiler working under atmospheric condition. Crushed coal having the size 6 to 20 mm is injected into the fluidized bed along with limestone through the feed hopper just above an air-distribution grid at the bottom of the bed. In some cases, coal and limestone are subjected into the lower section of the furnace from bottom. The air flows upwards through the grid from the air plenum into the bed, which now becomes the furnace where combustion of the swirling mixture takes place. The products of combusFlue gas and dust to particulate removal system Drum Furnace water walls Submerged tube bank Coal and limestone feed hopper\ — ,„ 4Transport air \i supply N NLAir . fan

L

Convective tube bank

RH Primary cyclone

Hot gas •

SH

r

Cyclone fines ecycle feed line

Fluidized bed

-3. •

z stack to

zone) Plenum chamber Grid plate

ECO

Bed drain tube

Fig. 6.28. Bubbling Fluidized Bed (BFB) Boiler.

u \/

I Ash‘r>

31 I

Steam Generators, Ash Handling Systems And Feed Water Treatment

tion leaving the bed contain a large proportion of unburned particles that are collected in a cyclone separator, which separates these particles from the gas by imparting centrifugal acceleration into the mixture. These separated unburned particles are returned back to the fluidized bed to complete their combustion. The boiler water tubes are located in the furnace. Finer solid residues (ash and spent sorbents) generated during combustion and desulphurization leave the furnace escaping through the cyclones, but they are collected by a bog-house or electrostatic- precipit4tor. The most important advantage of fluidized bed combustion lies in the concurrent removal of the sulphur dioxide that results normally from the combustion of sulphur content of the coal. Desulphurization takes 'place by the addition of limestone directly to the bed together with the crushed coal. The combustion temperature is low (800- 900°C). Due to low combustipn temperature, inferior grades of coal can be used without slagging,problem and there is no formation of NON. The other advantage of low combustion temperature is the use of cheaper alloy material which results in economical construction. Since there is no pulverizer, so further economy is achieved. Further, a fluidized.bed combustor can be designed to incorporate the boiler within the bed, resulting in volumetric heat transfer rates that are 10 to 15 times higher and surface heat transfer rates that are 2 to 3 times higher than a conventional boiler. All these make the boiler more compact. Economisers, superheater, reheaters, preheaters, bag-house or electrostatic precipitator, ID fan, stack, etc are used like conventional boiler. 6.5.2. Circulating Fluidized Bed (CFB) Boilers. Fig. 6.29. shows a second generation CFB boiler. The whole system is divided into two-sections namely solid fuel circulation loop and back- pass loop. The first section comprises of furnace or fast fluidized bed, cyclone (gas-solid separator), solid recycle device, (loop seal or L-valve) and external heat exchanger (optional). Similar to pulverized coal-fired (PC) boiler, the furnace enclosure is generally made of water tubes. In the a by-pass section, the remaining heat from the flue To SH Cyclone

Drum From ECO

Freheater euluerheater

Cross tubes

Water walls

46

Back pass Stand )43 IL/* pipe e_l_Voonomiser

Refractory

1Loop sea

Coal limestoneNN l e III! i

II I

Grid Fl u Fluiding (Primary) air

Air

Stack

__+ Bag Air — house heater

1

Y.YrY

External heat exchange

Fig. 6.29. Circulating Fluidized Bed (CFB) Boiler.

Ash

312

Steam & Gas Turbines And Power Plant Engineering

gases is absorbed by the superheaters, reheaters, economisers and air- preheater surfaces. In the first section, the furnace walls are lined with refractory upto the level of secondary air entry and beyond this furnace walls are generally cooled by evaporative superheater or reheater surface. The cyclone dust collector and the non-mechanical valves are also lined with refractory. Coal/limestone are injected into the lower section of the furnace through hopper system. There happens to be a loop furnace in which the recycled unburned coal particles are burned. The fluidizing (primary) air is fed to the furnace through an air distributor or grate and the secondary air is admitted at some height above the grate to complete combustion. With the injection of primary and secondary air the bed is fluidized, combustion takes place resulting in lower temperature of the gas (800 to 900°C). The ash from the bed falls down. The cyclone separator, the course sorbent (limestone) and coal particles which are recycled back near the base of the furnace while the finer solid residues (ash and spent sorbents) after passing over superheaters, reheaters economisers, air preheaters get captured by bag house or electronic precipitator and go finally to atmosphere via stack. The flow regimes are (i) turbulent fluidized bed, (ii) fast fluidized bed, (iii) swirl flow, (iv) moving packed bed, (v) bubbling fluidized bed, (vi) pneumatic transport. The first commercial CFB boiler in the world went into service at Pihlava, Finland in 1979. After that many CFO boiler were commissioned. The recent commissioned CFB boilers are 165 MWe Point Aconi Power Station at Cape Breton is Nova Scotia, Canada and 250 MWe Power unit at Lyon in France. 6.5.3. Cyclone Separators. As the name implies, the function of cyclone separator is to separate solid particles from the gas by utilizing the centrifugal force created by a spinning gas stream. Fig. 6.30 shows a standard tangential inlet vertical cyclone separator in which the gas flow is forced to follow the curved geometry of the cyclone, while the Dirty gas containing solid particles and dust

Clean gas outlet

Dirty gas containing solid particles and dust

Body

Inner cylinder Inner vortex

Outer vortex Top view of vertical cyclone on the left (b)

Solid particles and dust • (a) Fig. 6.30. Vertical Cyclone Separator with Tangential Inlet.

Steam Generators, Ash Handling Systems And Feed Water Treatment

313

inertia of the particles in the flow causes them to move toward the outer wall, where they collide with wall, fall down along the wall by gravity and are collected at the bottom end. The centrifugal force Fe on a particle of mass inp =Mr Cl/r while gravity force is giVen by, Fg = 9.81 Mr, thus the centrifugal force is about 20 times the gravity force. As a result of strong centrifugal force, the particles in the spinning gas stream move progressively closer to the outer wall as they flow through the cyclone and the gas stream executes several complete turns. The particles laden gas comes out from the furnace and enters tangentially to the cyclone. The centrifugal force forces the particles to the wall where they fall down along the wall by gravity force. It is worth to note that at the bottom of the cyclone the gas flow reverses to form the inner core that leaves to the top of the unit. The performance of cyclone 'separator is described by a parameter called collection efficiency which is defined as the ratio of the amount of particles separated to the amount of gas-solid mixture. The collection efficiency increase with increasing particle size, particle density, inlet gas velocity, cyclone body length, number of gas revolutions and smoothness of cyclone wall surface. Further, the collection efficiency goes on increasing by decreasing cyclone diameter, gas outlet duct diameter and gas inlet area. Cyclones used in CFB boilers are either steam or water cooled, due to relatively high temperature. Aeration

V-shaped 4— fluidized

bed

L-valve

Fluidizin gas

Aeration Fluidizing gas V-valve

Reverse seal

J-valve

(a)

Aeration gas

4_...Aeration gas

Fluidizing gas

Fluidizing gas Loop seal (fluoseal)

Seal pot (b)

Fig. 6.31. Schematic of Non-mechanical valves (a) Controllable, (b) Automatic.

314

Steam & Gas Turbines And Power Plant Engineering

6.5.4. Non-mechanical valves. Non-mechanical valves, extensively used in FBB are devices that allow the flow of solids between the return leg (stand pipe) and the furnace without any external mechanical force. There are various type of non-mechanical valves wig. 6.31) such as (i) L-valve, (ii) J-valve, (iii) reverse seal valve, (iv) loop valve and (v) seal pot. The movement of solids in these valves takes place by air which is fed at a short height above the exit of valve. The . L-valve comprises of a right-angled bent L-shaped pipe connecting the two vessels between which the solids are to be transferred. It is designed from pressure balance across the CFB loop. There is a difference in the density of bed. Even a small amount of air injected into the dense log moves solid to the diluted bed. The working principle of J-valve is similar to L-valve. It is the variation in the aeration rate in J-valve that controls the solid flow rate. A seal pot comprises of a bubbling fluidized bed with an overflow pipe and the standpipe of the cyclone. It is the higher pressure in. the seal pot that prevents the furnace air from entering the standpipe and by this way it serves as valve. There is minor• difference between the loop seal and the seal pot. The loop seal is smaller in size and here the solid movement is horizontally around a vertical partition separating the standpipe and overflow part of the seal pot and thus requires less air. From steam drum

To steam drum

Downcomer Risers

Combustor

Hot gas Ceramic filter

Super double omega reheater

.

In III no I CATWater wall I

Cyclone

Super double omega superheater

Hot clean gas to gas turbine

Ash Paste coal and sorbent feed heumatic Tr7 sorbent feed and trim

Secondary air

To ash system

Fig. 6.32. PFBC Hot Loop.

Steam Generators, Ash Handling. Systems And Feed Water Treatment

315

The L-valve, V-valve and J-valves are controllable while seal pot and loop seal are automatic. 6.5.5. Pressurized Fluidized Bed Combustion (PFBC) Boilers. In pressurized fluidized bed combustion (PFBC) boiler, combustion takes place in a pressurized environment created by supplying the compressed air by compressor of gas turbine unit in the furnace instead of atmospheric environment resulting in a compact furnace and improved combustion efficiency. Like atmospheric fluidized bed combustion (AFBC), the other advantage of PFBC is the capturing of sulphur and low NOx generation due to lesser operating temperature (900°C). Since the pressure of the gas is high (about 15 to 20 bar), gas turbine may be driven to develop power by using inexpensive coal. The furnace alongwith cyclones is housed in a pressure vessel and the compressed air from the compressor enters to the bottom of furnace via pressure vessel. The hot loop of PFBC system is shown in Fig. 6.32. The system may be used in turbocharged, combined and hybrid advance cycles. The following table (Table 6.1) depicts the values of important parameters in the major classes of FBCS. Table-6.1. Values of Important Parameters in Major Classes of FBCS. Pressurized (17 bar)

Atmospheric BFBC

CFBC

BFBC

CFBC

1.2

5

(i)

Gas velocity (m/s)

1.4-3.6

4-10.5

(ii)

Bed height (m)

1.1-1.4

(iii) Coal feed points/m2

0.5-1.1

36 3.5-4.2 . 0.006-0.008 0.26-6

0.2-0.6

(iv) Bed residence time (s)

0.5

5

4

8

Air

Combustion turbine

40

Hot clean gas

r

Pressure vessel

Cyclones Fluidized bed

Ash , - • Compressed air r):

I.-****GT Generator

i Reheat steam 'Steam

Exhaust gas

)10.-

Coal end clolornke";' ------ -

Solid rt5a waste ou U

Steam HP— LP Generator

;

Steam turbine Condenser Feedwater

ESP Economiser /heat recovery unit

Flue gas

) Fly ash 1,Stack ID Fan

Fig. 6.33. Pressurized Bubbling Fluidized Bed Boiler Serving Gas/Steam Turbine.

Steam & Gas Turbines And Power Plant Engineering

316

Cyclones Ceramic filter

Ash

Main steam

/ I

Reheat steam

Steam turbine Generator

I/

Boiler C

it Gas turbine Air

N CO 0) ,n

LP

▪ 0

2 co 0 Deaerator

Wet pest feed (WPF)

Gas © BFP

CWP

HP heatery

Ash

ANA I\ A

Stack

HRHE Fig. 6.34(a). Schematic of a 250 MWe PFBC System at Ohsaki, Japan.

Similar to AFBC, there are two types of pressurized fluidized bed combustor. These are— (i) Pressurized bubbling fluidized bed combustor (PBFBC) (ii) Pressurized circulating fluidized bed combustor (PFBC) 6.5.6. Pressurized Bubbling Fluidized Bed Combustion (PBFBC) Boiler. Fig. 6.33 shows a pressurized bubbling fluidized bed combustion (PBFBC) boiler where compressed air is supplied from the compressor of gas turbine unit and the granular solids are fluidized in bubbling fluidized mode. The gas velocity is fixed between the minimum bubbling velocity of coursesr particles and terminal velocity of finer particles. The water carrying tubes are embedded in the furnace bed which carries a part of heat generated in the furnace. The dirty gas is cleaned up in the two stage cyclones and then expanded through the gas turbine as shown in Fig. 6.33 which develops power. The furnace is housed in a pressure vessel in which primary air from the exit of compressor enters from the bottom of the furnace through the pressure vessel. The exhaust from the gas turbine is utilised in producing steam in a heat recovery steam generator which is superheated in fluidized bed. combustor and finally expands in steam turbine. The steam from the h.p. turbine is taken out . for reheating which .is readmitted in IP/LP steam turbine for further expansion. A 250 MWe PBFBC system commissioned at Ohsaki, Japan is shown in Fig. 6.34(a). The principle specification of this plant is given in Table 6.2.

Fig. 6.34(b). 350 Stack

Generatoweort_ 70 MW HPC

4-

LPGT

Air -0, LPC

4 7

Low pressure gas cooler • EP

z

High pressure gas cooler unit

HPGT

4

cn

0

411-11-t t

Tube

4--

Feed water

Mixer Lock hopper feeds coal and lime stone mixture Ash disposal

Il

Fuel slurry pump

oE c.)

)

DeaArano unit

)

Main steam

eed ter pump

IPST

,

le_

Feed water heater j

Condensate pump

Jondenser Sea Circulation pump

Steam turbine unit 7

Pregsure vessel

4-4

HPC : High pressure compressor LPST : Low pressure steam turbine HPGT : High pressure gas turbine

LPGT : Low pressure gas turbine IPST : Intermediate pressure steam turbine FPF BC unity

High temperature, High pressure air high pressure exhaust gas

Gas turbine unit

LPC : Low pressure compressor HPST : High pressure steam turbine

Steam Generators, Ash Handling SystemsAndFeed WaterTreatment

PFBC PlantatKartia, Japan.

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Steam & Gas Turbines And Power Plant Engineering

Table-6.2. Specifications of PBFBC Plant (250 MWe) at Ohsaki, Japan. Gross electrical output Boiler .

Gas turbine Steam turbine Emission control

Stack

250 MWe x2 (ST : 210 MWe, GT : 40 MWe) x2 Pressurized bubbling fluidized bed type Combustion regenerative once through type Steaming capacity : (520t/h) x 2 Bed temperature : 865°C Combustion pressure : 10 bar Cool and limestone feed system : Wet paste feed Single shaft, simple open cycle Tandem compound regenerative condensing type Steam condition : 166 bar, 566/593°C SO2 : In-site limestone desulfurization NOx ; NH3 injection SNCR and SCR Particulate : cyclone and ceramic filters Height : 200m

•

Fig. 6.34(b) shows a 350 MW PFBC plant commissioned at Kartia, Japan which is the biggest plant at present in the world. 6.5.7. Pressurized Circulating Fluidized Bed Combustion (PCFBC) Boiler. Fig. 6.35 shows a pressurized circulating fluidized bed combustion (PCFBC) boiler in which furnace is enclosed in a pressure vessel. In this case, the solid particles of coal and limestone in wet form is injected into the furnace and are kept in fast fluidized condition, they are mixed throughout the combustor, but there is very little mixing of gas in the axial direction. The gas flows upward with negligible back mixing. The water tubes are lined along the furnace of the wall vertically. It is worth to note that primary air, less than stoichiometric amount, enters the combustor through the grid at the bottom of the combustor through pressure vessel. The admission of secondary air takes place at some distance above the grid. On account of absence of back mixing of the air, substoichiometric condition prevails in the lower section of the bed and above this boiler tubes are located. Due to staging of air, NOx emission gets reduced and the chance of corrosion diminishes. Since air is added in stages, so desulphurisation is better. The circulation system of the boiler may be either once-through, natural circulation and forced circulation. In order to expand the high temperature and high pressure gas through the turbine two stage cyclones are used. The gas velocities through the cyclones at inlet are 30 to 42 m/s respectively. The gas is further cleaned by passing through a ceramic filter before admission to gas turbine. 6.5.8. Advantages.and Disadvantages of Fluidized Bed Combustion (FBC) Boilers. (A) Advantages The following are the advantages of fluidized bed combustion (FBC) boilers. (i) Fuel flexibility. The most attractive features of FBC boilers is the use of any cheap fuels without the support of any auxiliary fuel due io fluidized bed combustion. (ii) High combustion efficiency. As a result of superior mixing of solid fuels with air, availability of reaction space upto the top of furnace (even upto 40 m and further beyond into the hot cyclone) and long residence time of combustion, the combustion efficiency is

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319

R : Risers Pressure vessel ST

Exhaust steam

Fig. 6.35. Pressurized Circulating Fluidized Bed Combustion (PCFBC) Boiler.

very high (upto 99%). (iii) Concurrent removal of SO2. The presence of sulphur content in the coal produces SO2. Desulfurization is accomplished by the addition of limestone or dolomite which acts as sorbents directly to the bed together with the crushed coal. Limestone (CaCO3) absorbs the sulphur dioxide with the help of some oxygen from the excess air, according to CaCO3 + SO2 + 1/202 -*CaSO4 + CO2 Thus CaSO4 retains the sulphur in the bed in the solid form rather than allowing it to escape as gaseous SO2. It is worth to note that the rate of formation of CaSO4 is maximized at about 850°C and after 900°C, the sulphur capture gets reduced. That is why the temperature of FBC furnace is not alloWed to go beyond 900°C. The effect of pressure on sulphur to retention is appreciable. (iv) Low NOx emission. As a result of low combustion temperature (around 900°C) and the staged combustion process, NOx emission in FBC boiler is very low in the range of 50-150 ppm. (v) Simple fuel handling and feed system. Since the requirement of coal size in FBC boiler is coarse so there is no need of any conventional pulverizing mill and thus feed preparation is simple. The high degree of lateral solid mixing in turbulent zone at the bottom ensures uniform feed distribution within the bed. Further, the large height of the furnace prevents maldistribution of fuel.

320

Steam & Gas Turbines And Power Plant Engineering

(vi) Compact furnace. One of the main advantage of FBC boiler is the compactness of furnace due to high heat release per unit furnace cross-section (about 5 MW/m2). The high heat release and heat dispersion in the bed is possible due to intense gas-solid mixing and velocity (4 to 7 m/s). (vii) Good turndown and load following capability. As a result of relatively high fluidizing gas velocity and staged combustion ensure a fairly good turndown ratio (about 5:1) simply reducing the proportionate amount of fuel and air. (viii) High availability. FBC boilers offer high availability (above 90%). (B) Disadvantages The following are the disadvantages of FBC boilers. They are (i) Erosion of reactor walls, (ii) Attrition of fuel particles, (iii) Difficulty in immersing internals due to possible erosion and (iv) Complexities of hydfodynamics in furnace.. 6.5.9. Advantages of Pressurized Bubbling Fluidized Bed Combustion (PBFBC) Boiler. The following are the advantages of PBFBC boilers— (i) Reduced capital cost. The specific power output• gets increased which in turn reduces the capital cost. At a typical fluidizing velocity of 2 m/s, the furnace bed area of AFBC boiler at 1 bar is 2 m2/MW whereas it is 0.2 m2/MW at 10 bar. (ii) Increase in overall efficiency. The overall efficiency of power generation gets increased from 350/o for conventional power plants to more than 42% in the case of PBFBC plants. (iii) NOx emission reduces enormously. (iv) The combustion efficiency of PBFBC is higher than that of AFBC boiler. (v) Due to low fluidizing velocity (say 1 m/s) in PBFBC boiler, the gas residence time is high (about 5 seconds), whereas it is about 0.5 seconds in AFBC which results in better capturing of sulphate content. 6.5.10. Advantages of Pressurized Circulating Fluidized Bed Combustion (PCFBC) Boilers. The following are the advantages of PCFBC boilers— (i) Reduced space requirements. The pressure vessel enclosing the furnace is smaller in diameter due to higher fluidizing velocity, higher heat release per unit area and fewer components inside the pressure vessel. (ii) Less erosion and more accessibility. The erosion of the furnace is less due to low particle density fluidized bed with fine particles distributed thrOughout the furnace. The heat transfer surface in PBFBC is submerged whereas in PCFBC boiler, it is not submerged which results in less possibility of erosion. This also helps in better accessibility for inspection and maintenance of plate superheaters. (iii) Vigrous mixing. PCFBC boiler offers vigrous mixing of solid particles with air in the furnace which results in fewer feed points, thus a simpler and reliable furnace. (iv) Absence of material handling system. There is no need of bed material handling system to control the bed temperature. (v) Higher exit gas temperature. The exit gas temperature is higher which is very favourable for gas turbine performance. (vi) Lower NOx emissions. NOx emission is lower due to low temperature of combustion and enhanced staged 'combustion. (vii) Bettei load following. The following of load as per demand is better by varying the fuel and air flow and the proposition of primary and secondary air.

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(viii) Loss bed inventory, less total metal and refractory. Due to less bed inventory, less total metal and refractory, there is a shorter start-up and cool-down period and a quicker response to load changes. (ix) Use of dolomite. At higher pressure, the partial pressure of carbon dioxide in a bubbling bed is higher which requires higher temperature for calcination. Thus dolomite is more used in PBFBC boiler. 6.6. Boiler Mountings and Accessories. For safe and efficient operations, boilers are equipped with two categories of components such as and (b) Boiler accessories (a) Boiler mountings (a) Boiler Mountings. Boiler mountings are those machine components which are mounted over the body of the boiler itself for the safety of the boiler and for complete control of the process of steam generation. These mountings form an integral part of the boiler. According to the Indian boiler regulations, the following mountings should be fitted on the boiler— (i) Two safety valves, (ii) Two water level indicators, (iii) Pressure gauge, (iv) Fusible plug, (v) Steam stop valve, (vi) Feed check valve, (vii) Blow off cock, (viii) Inspector test gauge, (ix) Man and Mud hole, etc. Among the above mounting, safety valve, water level indicator and fusible plug are called safety fittings and the remaining are the control fittings. (b) Boiler Accessories. These are those machine components which are installed either inside or outside the boiler to increase the efficiency of the plant and or to help in the proper working of the plant. The following accessories are generally used in the boiler— (i) Air preheater, (ii) Economiser (iii) Superheater, (iv) Feed pump (v) Steam trap, (vi) Steam separator, (vii) Pressure reducing valve, etc. Only some of important accessories will be discussed here. 6.6.1. Economisers. The function of economiser is to heat the feed water coming out from high pressure feed heater upto nearly the saturation temperature corresponding to the boiler pressure by utilising the energy of hot flue gases leaving the superheater or reheater which is generally at a temperature varying from 370°C to 570°C. It is a heat exchanger, so its design will be based on the analysis of heat exchanger. The word economiser has been used to indicate that the energy leaving the superheater/reheater Economiser tubes

)

) )

)

Outlet header

• Mrw T

fw

V Inlet header (a)

Ao or L (b)

Fig. 6.36. Economiser Coil and Temperature Distribution.

Steam & Gas Turbines And Power Plant Engineering

322

which was going as a waste is utilized to heat feed water. Economisers are generally placed between the 0000000 last superheater- reheater, and the air preheaters. Tubes They have beep built with plain or extended surface 000000 tubes. The size of the economiser tubes are com°30000000 monly 45-70 mm in outside diameter. They are built 1 00000001 in vertical coils (sections) of continuous tubes con0000000 7, nected between inlet and outlet headers with each 0000000-t— section formed into several horizontal paths con0000000 nected by 180° vertical at a pitch of 45 to 50 mm spacings which depend upon the type of fuel and ash characteristics. Fig. 6.37. Section through Economiser An economiser coil with temperature distribuCoil (Fig. 6.36) Showing Height of tion is shown in Fig. 6.36. The heat transfer rate is Duct. expressed as

Q. = mg c g (To — Tg2) = tits, cos, (Tar — T fw) = Uo Ao ATin Ti —

Where AT. = log-mean temperature difference — T. —

(6.25)

AT°

In (AT,/ AT,

(6.26)

To — Ts.„ ATe = Tg2 — Tftv and

1 1 1 1 1 + + +— + Uo • h.(Ai/A0) hi(Ai/A0) Kw(AIni/A0) ho

(6.27)

Here, U0 = overall heat transfer coefficient, hsc ,h,,hfo and 170 are heat transfer coefficient due to scale formed on the inside the tubes, of water side, due to solids (ash and soots) on the outer surface of tubes and of the outside film (gas side) respectively. A . = n it Di I, Ao = mtDol where 1 is the length of one coil, n = number of coils, Aim = log-mean area = (A0 — Ayln (A0/4), x„, = thickness of tube, Kw = thermal conductivity of wall material. If h5 and hfo are negligible, then, 1= 1 x. 1 Uo kw ho

(6.28) The heat transfer coefficients, h, and ho may be calculated form the following standard equation Nu = 0.023 Re" Pe (6.29) Where n = 0.4 for heating, n = 0.3 for cooling, Nu = hd/k = Nussett number, Re = CD/ Wp Pr = (µ cr/k)f = Prandtle number From eqs. (6.25) and (6.28), 1.10 may be estimated and finally Ao may be found out where Ao = IT Do n 1. Using continuity equation at the exit of economiser where the water is saturated, we have mf~ .vf = AC = 7 114 D2,)Cf.

(6.30)

where vf is the specific volume of water and C is the velocity of water of exit. The high value of fluid velocity in the economiser tube leads to the lesser film thick-

Steam Generators, Ash Handling Systems And Feed Water Treatment

323

ness, i.e. more heat transfer coefficient but at the same time it needs more pumping power. These two conflicting requirements leads to the optimization of fluid velocity. On the gas-side the velocity of flue gases is about 10 to 12 m/s and on the water side, the water velocity is limited to 1 to 1.5 m/s. Thus, for the selected fluid velocity, the member of parallel coils (n) may be estimated for a given mass flow rate of saturated water. Since Ao is already known, the length of one coil may be computed. Fig. 6.37 shows the section along x-y of Fig. 6.36. This figure shows the pitch and height of the economiser duct. Let b is the width, c is the clearance on both sides of the gas duct then the number of turns, nt is expressed as n — ' b —2c The height of duct occupied by the economiser is expressed as Heco = nr .p

(6.31)

(6.32) where p is the vertical pitch or the centre to centre distance of the consecutive horizontal tubes forming the coil. 6.6.2. Superheaters and Reheaters. The function of a superheater is to superheat the saturated steam upto the desired temperature. It is essentially a surface heat exchanger generated located in the passagt of a hot flue gases. In some cases, superheaters may be placed in an independent fire furnace. With the use of superheater in the steam power plant, thermal efficiency increases as well as the moisture contents reduce in the last stage of turbine. About 40% of heat generated in modern steam generator is utilised by superheater. This indicates that a large surface area is needed by superheaters. Modern superheaters and reheaters operating at about 540°C, however are usually made of special high strength steels chosen for both strength and corrosion resistant. Below 460°C, carbon steel is sufficient. There are three types of superheaters namely convective (CSH), radiant (RSH) and combined (PSH) depending upon how heat is transferred from the gases to steam. As the name implies, convective superheaters are located in the convective zone, while radiant is in radiation zone and combined in convective radiation (pendant type) zone respectively. The radiant and combined superheaters are sometimes called secondary superheaters. Fig. 6.38 shows the heat absorption in three types of heaters. The heat transfer rate from the gas to steam in these three types of superheaters are

Fig. 6.38. Heat Absorption in Superheaters.

Steam & Gas Turbines And Power Plant Engineering

324

Superheater tubes Tgt mg

ms to RSH T

Mfg . From drum

g (a)

Fig. 6.39. Convective Superheaters and Temperature Distribution.

-CombinedSH

I

ro •sia E

.---------

Sx\

Go'

Rem E

ca

20

40 60 Steam flow (%)

80

100

Fig. 6.40. Steam-outlet Temperature Response of Superheaters. QCSH = QRs,H

g

c (T — 7' 2) = tir s (h2 — hi) = U0 A0 (AT) pg gi g

= c AT Ff w (71— 7;4v) = ins (h3 — h2) = U0 A0 07)

c (T — To) + (1—y) a AT fly (T'f1.— g pg g3 = ms. (h4 — h3) = U0 A0 (AT).

(6.33) (6.34)

=y

(6.35)

where Tf and T„, are the flame and tube wall temperature, Ff „, is the shape factor which respect to flame and wall and y is the fraction of energy shared by convection (say 50%). Fig. 6.39 shows the convective superheater coil and its temperature profiles. For radiant; it will be identical. The mass flow rate pa'ssing through the superheater is given by

'

— in . vg = s

D1C

S 4 (6.36) —Where Cs is the average velocity of steam in the superheater. For high, intermediate

Steam Generators, Ash Handling Systems And Feed Water Treatment

Gas flow

325

Gas flow

IM1 (a)

(c)

(b)

Fig. 6.41. (a) Pendent, (b) Inverted and (c) Horizontal Superheaters. Steam

Steam

(a)

(b)

Steam

(c)

Fig. 6.42. Types of Superheat Coils, (a) Sing e, (b) Double with One Tube, (c) Double with Two Tubes.

and low pressure, Cs are selected as 10, 12 and 22 m/s respectively. The erosion of superheaters tube is proportional to the square of gas velocity. This • suggests that the greater the ash content 'of the fuel, the less will be the allowable gas velocity in the system. But higher gas velocity yields higher heat transfer coefficient. So as a compromise the gas velocity for low-ash coals and high-ash coal should not exceed 12 and 8 m/s respectively. Superheaters and reheaters tubes are made of 50 to 75 mm outside diameter. The number of coils and length of tube are calculated using eqs. (6.33) to (6.36) similar to economiser. Fig. 6.40 shows the steam outlet temperature response of convective, radiant and combined (in series) superheaters. Itt convective superheater the steam receives greater heat transfer per unit mass flow and its temperature increases with load. The radiation heat transfer to radiant superheater is proportional (Tt — 74w) and since Ti is much greater than Tw, thus the heat transfer is essentially proportional to 71. Because Tf is not strongly dependent on load, the heat transfer per unit mass flow of steam decreases as the steam flow increases. Thus an increase in steam flow due to an increased load demand would result in reduction in exit steam temperature, the opposite effect of convective superheater (Fig. 6.41). Convective superheaters alone are used with low temperature steam generators. Radiant and convection superheaters and reheaters are used for high

Steam & Gas Turbines And Power Plant Engineering

326

temperature service. The radiant units and arranged in plant panels or platen sections with

wide spacings. Based on mechanical construction, there are three kinds of superheaters namely pendent, inverted and horizontal as shown in Fig. 6.41. Pendent type superheaters and reheaters are those that are hung from above. They have the advantage of firm structural support but the disadvantage of flow blockage by condensed steam after a cold shutdown, which necessitates slow restart to purge the water that accumulates in the bottom. Pendent type is generally used in radiant superheater. As the name implies inverted type units are supported from below, so they have proper drainage of the condensed steam but the structural rigidity is poor and that is why they are not commonly used. Horizontal-type units are usually supported horizontally in the vertical gas ducts parallel to the main furnace and receive the hot gases after a U-turn at the top. They don't view the flame directly and hence are mainly of the convection type. They have both proper drainage and good structural rigidity. As mentioned earlier, radiant and pendant superheaters may be made in the form of cells or platens. The coils may be single, double with one tube and double coil with two tubes as shown in Fig. 6.44.. Platens may be triangular and rectangular type as shown in Fig. 6.43. The function of reheaters is to reheat the steam upto the desired temperature. The design is similar to superheaters. ---- 6.6.3. Air Preheater. The function of air preheaters is to preheat the air before entering to the furnace by utilising some of the energy left in the flue gases before exhausting them to the atmosphere. They receive gases at 320 to 430°C. As mentioned earlier the flue gases are cooled not below 160°C to avoid gas condensation and corrosion problems and to allow for proper dispersion in the atmosphere. The air is heated from forced draught fan outlet temperature (slightly more than atmospheric) to 260°C to 345°C and even higher. Obviously, preheating the air saves fuel that would otherwise be used for Steam Steam

(a)

(b)

Fig. 6.43. (a) Triangular and (b) Rectangular Platens.

Steam Generators, Ash Handling Systems And Feed Water Treatment

By-pass damper

327

rG

mg Tg2 as out et

Cold air et Ta1

a

Air

Baffles By-pas 4 air duct

i Tubes

Heated air outlet

m' a Ta2 A

A

Expansion Me-- joint

Gas Gas inlet glTgl

Soot or cinder hopper Fig. 6.44. (a). A Tubular Counter-flow Preheater.

Ao or L

Fig. 6.44. (b). Temperature Profiles in APH.

that heating. For example, a typical fuel savings are 4 percent for a 110°C air temperature rise and about 11 percent for a 280°C air temperature rise in the preheater. There is an additional requirement of the preheated air for the operation of pulverizedcoal furnaces. Air in the range of 150 to 420°C is needed for drying the fuel. Air is also used for transporting it to the furnace and burning it there. Though, small stoker fire units don't require preheated air, but large stoker-fired bituminous coal steam generators have beneficial effect from preheated air, but only upto 180°C to prevent damage to the stoker moving parts due to high temperature. There are two general types of air preheaters namely recuperative and regenerative. In recuperative air preheaters, heat is transferred directly to the air across the heat exchanger surface. They are commonly tubular, although some plate type are still used. Tubular systems are essentially counter-flow shell and tube heat-exchanger in which the hot gases

Steam & Gas Turbines And Power Plant Engineering

328

Hot air outlet

Hot gas inlet

Basketed heating surface

Heating surface

Housing ----r,4( -----

Housing Rotor Radial sea (stationa

Rotation

Flow I seal (stationary)

Fuel-gas outlet

Flow

Air inlet

(a)

(b) Fig. 6.45. Regenerative Preheater (Ljungstrom) in Two Views. flow inside vertical or horizontal straight tubes and the air flow outside. Such a tubular preheater is shown in Fig. 6.44 involving baffles to maximize air contact with the hot tubes. The tubes are mechanically expanded into top and bottom tube sheets. A hollowtype expansion unit shown at the bottom sheet is used to provide the thermal expansion. A hopper provided below the tubes collects root and dust that deposit on the inside tube surfaces. There should not be any leakage from air (which is at high pressure) to the gas otherwise the power consumption of both forced and induced-draft fan will increase along with reduction in effectiveness of preheater. In order to suit particular steam generator spaces and duct layout, tubular preheaters are built in a variety of designs. They may use for one or more passes for both and air • gases in counter or crossflow, in vertical or horizontal arrangements. The size of tubes varies from 50 to 65 mm OD. Cost, cleaning requirements and fuel type determine the diameter to be used in a given situation. The design of the preheater is based on the heat exchanger analysis. The rate of heat transfer from gas to air is given by QAPH

=

crg (T — Tg2) = ina cpa (Ta2 —

Tal)= U0 A 0 ATn,

(6:37)

The value of U0 is calculated similar to that of economiser. Regenerative air preheaters are storage type heat exchangers in which heat is transferred from the hot flue gases, first to an intermediate heat storage medium called the matrix, then to the air. The most common type is the rotary air preheater, known as the Ljungstrom preheater. It consists of a rotor driven by an electric motor at a low speed of 1 to 3 rpm depending on diameter, through reduction gearing arrangement. The rotor consists of 12 to 24 radial members that form sectors as shown in Fig. 6.45. The matrix consisting of flat, corrugated, notched or undulated ribbing is filled in the sector. A stationary seal covers the equivalent of two opposite sectors and half of the remaining sectors are exposed at any instant to the hot gases. Rotary air preheaters are designed with vertical or horizontal shaft depending upon layout and ducting. The main advantage of rotary preheater is its compactness with a large heat transfer surface accom-

Steam Generators, Ash Handling Systems And Feed Water Treatment

329

Steam to SH A

Steam drum

Drum valve senor Normal water level Feedwater flow sensor

r) 0 •-• 0

0

X

Control valve

Feedwater

Steam flow sensor Trimming signal

Fig. 6.46. A Three-Element Feed Water Control System.

modated in a small volume. They are as popular as tubular preheater. The main disadvantage include (i) plugging of flow openings by fly ash, (ii) sealing of, gas to air path and vice-versa, and (iii) large pressure drop. 6.7. Steam Generator Control The purpose of steam generator control is to supply the steam flow required by the turbine at design pressure and temperature and according to the load.

Steam generator, and indeed total power plant control is rather broad subject that includes every important aspects of plant such as instrumentation, data processing and controls for combustion, steam flow, temperature, and pressure, drum level, burner sequencing, desulphurization precipitators, as handling, system integration, start-up and shut down and automation. This wide subject is not possible to be covered. The control system must act on the measurement of these parameters to control the plant. The current practice in control system is to employ closed loop control. The actual output of the system is measured and compared to some demand signal (set point). The error single (i.e. difference in output and demand) is used to control. Proportional control is the simplest either directly or inversely proportional to the error signal. Since the subject is very wide, only a few basic control systems that apply to steam generator such as feed water and drum control, steam pressure control, and steam-temperature control will be discussed here. 6.7.1. Feed Water and Drum-Level Control. The aim of feed water and driim-level control is to meet load demand by the turbine and at the same time maintain the level of water in the steam drum- within relatively narrow limits. It is common to maintain the normal drum level half-full; in reference to sight glass tube used outside the drum. In a particular situation of high steam consumption by the turbine, combined with low water supply would lower the water level in the drum. Fig 6.46 shows a three- element automatic control system, of which drum level is one element.

The drum level sensor senses the level of water and responds to the error signal

Steam & Gas Turbines And Power Plant Engineering

330

between actual drum and its set point, such as in the case of high steam consumption, and low feed water supply, and acts on the controller to increase the feed water valve opening to meet the steam flow demand. This action may be too slow and is supplemented by sensors for feed water and steam flow to meet the demand as soon as possible. The difference between the signals from these two sensors anticipates changes in drum level and sends a signal to the controller to actuate the valve in the proper direction. The zero difference in these two signals mean constant water level in the drum as per demand. 6.7.2. Steam Pressure Control. The purpose of steam pressure control system sometimes called the boiler master" is to maintain steam pressure by adjusting fuel and combustion air flows to meet the desired pressure. When pressure drops, the flows are increased. Three sensors namely steam flow, air flow and steam pressure are incorporated in the control system. The steam pressure sensor senses the steam pressure and acts directly on the fuel flow and air flow control such as the pulverized coal power drives and forced drought fan to affect the desired changes as per demand (Fig. 6.47.). A trimming signal obtained from the processing of fuel flow and air flow maintains the proper fuel-air ratio. Because it is often difficult to obtain accurate fuel flows, a steam flow sensor is sometimes substituted for the fuel-flow sensor. In general, about 5 seconds delay is permitted when changing coal flow and airflow to ensure the prevention of a momentary rich mixture and thus assure smoke-free combustion in the furnace. 6.7.3. Steam Temperature Control. For efficient power plant operation, it is necessary to control accurately the steam superheat temperature. The principal parameters affecting superheat temperature are furnace temperature, cleanliness of radiant and pendent superheaters, temperature of gases entering the convective superwater, cleanliness of convective superheater, mass flow rate of gases through the convective superheaters and

Steam flow sensor

Air flow sensor

Trimming signal Steam pressure sensor

Pulver zed coal power drive

Forced draught fan drive

Fig. 6.47. Steam Pressure Control System.

331

Steam Generators, Ash Handling Systems And Feed Water Treatment

Drum

To turbine Secondary superheater

Valve

•

L___

m1

Heat exchanger

m2 - m1

(attemperator)

mi

V

A

Primary superheater V

m1

Fig. 6.48. Principle of Surface Attemperator. variation of load on the unit which is the most important among all. The interrelationship among these variables is complex. A reduction in superheat temperature than the rated (design) value results in loss of plant efficiency. For example, a reduction of 20°C of superheat temperature results in above 1% increase in heat rate, whereas, a rise in steam temperature above rated value may result in overheating and failure of superheater and reheater tubes and turbine blades. As we know that steam leaving the drum is approximately saturated and its temperature corresponds to the boiler pressure and remains constant if the steam pressure controlSteam line Steam flow, ms

Spray nozzle Thermal sleeve

Water flow, mw

Spray water

m

ms S +MV )

(ril•

S

• M W

eTh

Venturi mixing Thermal sleeve

Total flow

1

Primary superheater

Secondary superheater

+

Fig. 6.49. A Spray Attemperator for Steam Temperature Control.

Fig. 6.50. Location of Attemperator.

332

Steam & Gas Turbines And Power Plant Engineering

ler is effective. The superheater and reheater temperature changes as per load and so it needs control. The following are the various methods to control the superheater/reheater temperature. (i) Combined radiant-convective superheaters. In this method, radiant and 'convective superheaters are arranged in series to yield a relatively flat final steam temperature over a wide load range (Fig. 6.40). This is possible as they have opposite steam temperature responses to load variation. (ii) Attemperation. Attemperation is the reduction of steam temperature. It is also called desuperheating. Attemperation is achieved by two methods. In the first method, a surface attemperator is used which removes heat from the steam in a heat exchanger. There are two types of surface attemperator namely shell type and drum type. The shell type, has a portion of the steam (th) taken out through tubes from between the primary and secondary superheaters by an automatic valve and diverts to a shell-and-tube heat exchanger (Fig. 6.48) containing some of the boiler water. The steam gives up some of its heat to that water and then remixes with the primary steam (M2) upon entering the secondary superheater. Temperature control is affected by controlling the amount of diverted steam. In drum type, heat exchange takes place between the diverted steam and boiler water within the main steam drum, which is made longer to accommodate the attemperator tains. In the second method, a spray or direct contact attemperator is used as shown in Fig. 6.49 which reduces the steam temperature by spraying low temperature wa:er from the boiler or economiser exit into the line between the primary and secondary sup6rheaters (Fig. 6.50). The spray nozzle injects water into throat of a mixing venturi, where the water mixes with high-velocity steam in the throat, vaporizes, and cools the steal 1. Further, the venturi and a thermal sleeve made of high chrome steel protect the main steam pipe from thermal shock caused by any unvaporized water droplets that might otherwise impact on the pipe leading to its failure. It is worth to note that water used must be of high purity to avoid adding deposits on the superheat tubes, pipes, and turbine blades leading to drop in performance and failure. It has been found operating satisfactorily all over the world. The main advantage of spray attemperator is to provide a rapid and sensitive means of temperature control. Fig. 6.51 shows the performance of a spray attemperator with primary Spray cooling (n - 0)

—560 Secondary S.H. (0 - p)

3

k

a)

E

Primary S.H. (m-n)

E as a) co

m

0

60 Load, (%)

100

Fig. 6.51. Attemperation with Oversized Primary and Secondary Superheaters.

333

Steam Generators, Ash Handling Systems And Feed Water Treatment

To stack A

Steam drum 0

4.

SH

APH SH

. Gas flow

c>_1%_ I X241 XIt s

Fig. 9.31. Profile Blades with Extended outlet Edge.

Fig. 9.32. Profile Blade with Unequal Angles.

The various methods of blade attachment are discussed in the chapter 15. 9.14. Choice of Blade angles 9.14.1. Inlet Blade Angle (a1) : To ensure the minimum shock loss at inlet of the impulse blade ; the flat edge at inlet HJ shown in Fig. 9.32 should make an angle 131 with the plane of the wheel because 131 is the relative direction of steam flow into the blade which is clear from the velocity diagram. The inlet blade angle is expressed as tan 13 — I

CI sin at

sin a l

C1 cos al — u

cos al — p

This condition is based on data at full load or at economical load in a large central steam turbine. At partial load the heat drop per stage will be smaller than at full load conditions, the greatest reduction of heat drop at partial load will be in the last one or two stages. Hence the jet velocity will be smaller than at full load. Let AB' be the jet velocity at partial load, so DB' will be the relative velocity at inlet. It is obvious from the Fig. 9.33 that inlet angle 131' is greater than angle Pi, so there will be shock in flow at inlet causing loss. When the turbine operates at partial load for a considerable period, the blade inlet angle may be greater than the diagram angle by the following amounts to ensure shockless entry In the last stage, 5 to 10 degree, In the last but one stage, 4 to 5 degrees, In the other stages, 2 to 3 degrees. 9.14.2. Outlet Blade Angle, 02) : For calculation of the outlet angle of the blade, the area required at outlet for the flow of steam is considered. Let us assume that S = mean circumferential pitch of the blades in cm. /2 = radial length of blade on the outlet side in cm. z = number of blades channel through which steam is flowing. With full admission or with partial admission and a nozzle arc , z =

Steam & Gas Turbines And Power Plant Engineering

464

(b) Fig. 9.33. Velocity Diagram for Part-load.

Fig. 9.34. Relation of Nozzle and Blade Height.

t = thickness on the blade at outlet edge in cm. So the mean effective width of the blade channel (Fig. 9.32) at outlet is d = S sin 132 - t Total area through active channel of blades = z . 12 (S sin 132 — t) sq. cm. Also, area required =

104 in.v2 —sq . cm. Cr2 104 .

Equating these two

sin P2 -1

— n12 [S . sin (32 -

C 104 .

or

.v

. V2

n . 12 . Cr2

+t (9.35)

It is passible that the calculated angle, P2 may not prove suitable or may not be accommodated by the standard blade sections, then it is preferable to use a blade section with somewhat smaller angle than with a slightly large angle because smaller angle will give some reaction in blades which is beneficial. From the equation (9.35), it is evident that if /2 is increased the angle 132 will be reduced and consequently an increased velocity of whirl and a gain in efficiency. The height /2 of a blade of constant height is selected according to a conditions at inlet as shown in Fig. 9.34 (a). In order to prevent the steam jets striking the shrouding strip or the blade root, the radial height of the blade at inlet is made slightly greater than that of the nozzles, and quite a small allowance should be sufficient for this. A. large allowance involves a certain amount of losses that may be avoided by employing a blade of increasing height at root and tip both in the direction of flow as shown in Fig. 9.34 (b). This practice is generally adopted. 9.15. Blade Heights in Velocity-Compounded and Pressure Compounded Impulse Turbine. In designing the blades, blade height plays an important role. Assume in kg of steam per second flowing through a two rows of moving blades of Velocity-Compounded

465

Flow of Steam Through Impulse Turbine Blades

Impulse turbine. Let /n, 1i ,1f. and /2 be the radial heights at the outlet side of the nozzle, first row of moving blades, fixed row and second row of moving blades respectively as shown in Fig. 9.35 (a). For simplification the specific volume v is assumed constant throughout the blade system though due to friction losses in the blade channels, the steam will be reheated and the specific volume v will be slightly increased (the increase is small). Let X = the effective length of arc over which the steam is flowing and 1 cc1 = nozzle angle. From the continuity equation, at the nozzle outlet, we have in .v=A .C= 1 C1

..II n sin 1a1 =

n•

1Cfl

(9.36)

If in be the nozzle thickness, so be the pitch of blades and for full admission X = TC D, and the number of nozzles (z„) = X / Sn = TED / Sn then from continuity equation X, ln — in .v= 1C1" S (Sn sin 1 — tn )

(9.37)

Let, S1 = mean circumferential pitch of blades in first row, Si-= mean circumferential . pitch of blades in fixed row, S2 = mean circumferential pitch of blades in second row. Let us assume that the arc covered by the steam as it passes through the blade system is constant. In this case, the number of the blade channels through which steam is flowing in the first row is A/S1 and the total area through the blade channels at outlet is (9.38)

Al = •T,..11 c (SI sin 1(32 — ti) Where t1 is the thickness of the outlet edge of the blade in the first row. Therefore, continuity equation at the outlet of the first row of, blade gives X /, .v = A .0 = i Cr2--=(S S 1 . sin 1 132 — t1 ) 1

(9.39)

From the equation (9.39) it is clear that angle 1132 may be calculated for a given value of /1 and if 1(32 is given, then 11 can be calculated. In some case, 1132 is equal to 1131 but in majority of cases 1(32 is less than I pi.

Stator Moving blade

Blade height I

Drum

Nozzle Blade tip or casing dia

Rotor \

(a)

(b)

Fig. 9.35. Blade Heights of Velocity Compounded Impulse Turbine.

466

Steam & Gas Turbines And Power Plant Engineering

For the fixed blade it may be shown that

ti? v = A .0 = 2Ct

X

.1f .(Sf . sin 2a, — tj.) (9.40)

where, t f is the thickness of the outlet edge of fixed or guide blades Similarly, for the second row of moving blades, continuity equation gives (Fig. 9.36).

in . v =A.C= 2Cr2 S

2

2 2

. sin

2

— t2) (9.41)

where t2 is the thickness of the outlet edge of the second row of moving blades. Eq. (9.41) is also applicable for pressure compounded impulse turbine. Generally, the ratio of the blade height in the last moving row to the nozzle height, has the ranges— Two row wheels, /2 / In = 1.8 to 2.5 and Three row wheels, /3//n = 2.5 to 3.5

Thus in general the blade height simple, velocity compounded or pressure comt,

Flow Tip

Height of blade = /

r

t2

Root

S sin 132 - t2 (b)

(a)

(c) Fig. 9.36. A General View of Moving Black. Channel. ,

Flow of Steam Through Impulse Turbine Blades

467

pounded impulse turbine will be calculated at the exit condition by using the continuity equation inv A.C. = Cr2 X n . I (S . sin p2 — 1) (9.42), But n S = nD and thus neglecting the blade thickness, in general (Fig. 9.35 b) inv = nD ICr2 sin 132 = 701 C (9.43) ./2 where D= mean diameter (m), 1= blade height (m), Cr2 = relative velocity at outlet (m/s), in = mass flow rate rate, (kg/s), v = specific volume (m3/kg) All the above equations are also valid for pressure compounded..

S

9.16. Advantages of Velocity Compounded Impulse Turbine. 80 1 Ro'w (i) Velocity-compounded impulse turbine requires a comparatively small 70 number of stages due to relatively large heat drop per stage. This is the reason 60 Rows of moying blade that in all modern large impulse-reaction turbine the initial stage is two-row veloc50 ity stage or single stage impulse turbine. 3 Rows 40 (ii) Due to small number of stages the initial cost is less. 30 (iii) The maximum stage efficiency and the optimum value of p decreases as 20 the number of moving blades rows in a wheel increase as shown in Fig. 9.37. to (iv) In a two or three row wheel, the steam temperature is sufficiently low, 0 01 02 03 04 05 07 hence a cast iron cylinder may be used, thus saving in material cost. Generally, two row wheel is more common. Fig. 9.37. Stage Efficiency Vs. p. 9.17. Disadvantage of VelocityCompounded Impulse Turbine. (i) The velocity-compounded impulse has low efficiency and high steam consumption. (ii) In a single row-wheel, the steam temperature is high so a cast iron cylinder cannot be used due to phenomenon of growth. Cast-steel cylinder is generally used which is costlier than cast iron. Problem 9.15 The nozzles in a stage of an impulse turbine are convergent type and formed by casting. The radial height of the nozzles is 15 cm and the mean diameter of the wheel is 1150 cm. The nozzle angle is 18°. The vanes are of uniform thickness amounting to 3 mm. The number of nozzles in the complete ring is 72. The specific volume of the steam at the nozzle outlet is 18 m3/kg. and the velocity of steam at nozzle outlet is 409 m/s. Calculate the mass flow rate of steam flowing through the nozzles. Calculate the power developed in the blades of the given stage, assuming that there is a single row of impulse blades having equal inlet and outlet angles, that the ratio of the relative outlet velocity to relative inlet velocity is 0.86, and that the rotor turns at 3,000 rpm.

468

Steam & Gas Turbines And Power Plant Engineering

CW — 370 m/s

EA

Cul *

Scale : 1 cm = 50 m/s B

Fig. 9.38 Solution. Refer to Fig. 9.38 Given : /„ = 15 cm, n = 72, D = 115 cm, v1 = 18 m3/kg, at = 18°, C1 = 400 m/s, t = 3 mm, N= 3,000 rpm, 131 = P2, ria =? and Power = ?, K =0.86 From continuity equation, Mxv = A x C The effective area at nozzles outlet = n (S . sin al — t)1„ where S = mean circumferential pitch of nozzles and t = vane thickness at outlet S—

TCD

n

—

7G X 11.5

72

— 5.02 cm

At nozzle outlet, inv1 = n . 1„(S sin al — t) . C1 in x 18

18„ = 72 x 100(5100 x sin

0.3 \ — 100 400

in = 2.988 kg/s

Ans.

itDN nx115x3000 — 180.641 m/s 60 100x60 With the known value of u , C1 and al the inlet velocity triangle ABD can be drawn. This gives : p, = 13°. 132 and CH = 235 m/s, The peripheral speed of blades = u =

..

Cr2 = 0.86x235 = 202 m/s

With the values of a2 , u , C,,2 the outlet velocity triangle DAF can be drawn. 2.98x180.641x370 In • U. u Cw 199.7kW 1000 1000 Ans. Problem 9.16. The nozzles in a two-row velocity-compounded impulse stage have 19 cm in height and extended over an arc of 76 cm. The effective angle at outlet is 16° and the specific volume of steam is 0.623 m3/kg. The mean diameter of blades is 72 cm and the rotational speed 3,000 rpm. The steam velocity at outlet from the nozzles is 558 m/s. The height of the second row of moving blades is 2.5 times radial height of the nozzle, i.e. 47.5 cm, and the blade heights increase in arithmatic progression, i.e. 1 = 28.5 cm, lf. =38 cm and 12 = 47.5 cm. Find the blade inlet and outlet angles, assuming that the blades have an axial width of 2.54 cm; that the thickness of the outlet edge 0.0508 cm and that blade velocity coefficient K = .0.86. The mean circumferential pitch of the blades may be Power developed —

Flow of Steam Through Impulse Turbine Blades

469

= 790 m/s 16° 20° GN 16.9°-

2Cw=219

20.9°

32.2° 2Cf1

19s )lt:441 2Cf2 , G‘ 2 Ctcl Scale : 1 cm = 80 m/s Fig. 9.39.

calculated by means of Briling's formula S = b/2 sin 213 Solution : Refer to Fig. 9.39. Given : /„ = 19 cm, /1 = 28.5 cm, X = 76 cm, /f =38 cm, t at = 16°, 12 =47.5 cm, v =0.623 m371cg , b =2.54 cm, D = 72 cm, t1 = t2 = 0.0508 cm, = 558 m/s, K= 0.86, N = 3,000 rpm, tat = = ?) 2P1 = 2P2 = ? First Moving Row :u—

TON 60

x 72 x 3000 — 117.5 m/s 100 x 60

We know that, inv = A . C = 1C1 . X . /„ sin loci or

1 131

my = x- 7 ICIIn sin ia i = 558x 19x0.276 100

— 29.3

With the value of u, i ai and I C I the inlet velocity triangle can be drawn. This gives = 20° and I Cri = 452 m/s. 1

Cr2 = 0.86x452 = 388 m/s

Mean pitch of blades = St =

We have

or or

sin 1 . 2

2 sin 2 181

S1' sin I1)2 = in X, 111C

2.54 3.54 2 sin 40° 2x0.645 — 1.975 cm t

1 ti 100 ill (/ C + — 29.3x 100 1 1 -a S 1

+

0.0508 1.975

l2 = 16.9°

Now the outlet velocity triangle for first row can be completed which gives 1

a2 = 23° and 1C2 = 272 m/s

Ans.

Steam & Gas Turbines And Power Plant Engineering

470 Fixed Row :- S f

2.54 = 2.54 — — 1.765 cm 2x0.72 2 sin 46°

100 0.0508 v 1 tf + + — (29.3) sin eat = ( ifrI)/f • C 38x234 1.765 2 1 Sf or

2

a = 25.9° 1

Ans.

Second row of Moving Blades :- 2C0 = 145 m/s and 2131 = 36° S — 2

2.54 2 sin 72°

2.54 2x0.95

1.34 cm

2Cr2 = 0.86x145 = 124.5 m/s sin 2132 — or

my

1 100 + 0.0508 _ + t2 293x 0.5328 /2. 2,Cr2 S2 1.34 * 47.5x124.5

2132 = 32.2°

Ans.

Now the velocity diagram is completed for second stage. Let

li n = 0.9 be the nozzle efficiency, then 11,

2 . rinu (l c + 2 C) = 2x0.9x117.5x1009 _ 65.7% C2 5582

Ans.

9.18 Twisted Blading We have seen that at part load, the approaching flow angle to the moving blade is larger than the inlet angle of blades. If the blades are parallel having uniform sectional area and angles throughout its active length, the shock losses would occur over the whole length of the blade at inlet This loss, may be minimised by the use of twisted bladings having angles that vary from root to tip. In this case, it is also possible to diminish the cross-section from the root outward. This decreases the stress at the root or if desired allows the use of a longer blade with the same stress. Thus, the length of tapered blade may be 25 to 35 percent of the blade ring mean diameter. In the larger size impulse turbine the blading of 1.p. turbines for handling larger volume are longer. As a result the peripheral velocity at the blade root will be much smaller than at the tip. Hence for shockless entry and better performance, the blade angles should vary along the blade height as shown in Fig. 10.12 (chapter 10) This results in twisting of the bladings. This is also true for impulse-reaction turbines since the peripheral velocity (u) is directly proportional to radius, so the blade angles vary accordingly. The velocity diagram at the root is impulse type while at the tip it is reaction type. 9.19. Breadth of Blading The optimum breadth of blading depends on the following parameters :(a) desired flow pattern, (b) loading of the blade, (c) angle of flow deflection [180— (131 + p2A, (d) relative velocity, (e) blade strength, (f) blade stiffness The latter two i.e. (e) and (f) closely related to the blade height. Since the deflection angle is the total angle turned through by the steam while travelling from entrance to exit through the blade passage, so a large deflection angle is usually desirable since it increases energy transfer and reduces the required number of stages, on the other hand, it accentu-

Flow of Steam Through Impulse Turbine Blades

471

ates the effects of centrifugal force in producing variations of velocity over the cross section, thereby increasing turbulence and decreasing the velocity coefficient K. A steam turbine blade is subjected to a combination of direct tension due to centrifugal forces and bending due to steam force, the former predominating. Increase in section dimension has little effect on blade centrifugal stress but does decrease the stress due to steam force. It has been found that the breakage of blades are primarily due to their repeated passage through fields of nonuniform discharge from the nozzle resulting in vibrations. If the period of pulsation closely approaches the natural vibration frequency of the blade, resonance will be created with large amplitude and stresses repeated with great frequency, often leading to speedy failure of blade. This may be prevented by adequate stiffness in the blades, meaning primarily the use of blades Sufficiently wide for their heights. In large impulse turbines, the breadths of blades should be within the range of 3.75 to 5 cm but at the high pressure end even up to 15 cm. It is to be noted that the total of the axial breadths .of blades and nozzle partitions constitute the primary factor in determining the span between bearings of a rotor. Any unnecessary increase of breadth may result in a reduction in the number of stage or efficiency. EXERCISES Viva-Voce and Theoretical 9.1.

What is the effect of blade friction on turbine performance ?

9.2.

What is the condition of maximum blade efficiency in single stage impulse turbine ?

9.3.

What is carry-over coefficient ?

9.4.

What are the different types of impulse blade sections. Give a sketch of them ?

9.5.

What are the advantages and disadvantages of velocity-compounded impulse turbines?

9.6.

Find expressions for the force, workdone, diagram efficiency, gross stage efficiency and axial thrust for an impulse turbine.

9.7.

Find the condition of maximum blade efficiency in a single stage impulse turbine.

9.8. What is the maximum workdone from a two row velocity compounded impulse wheel. 9.9.

Find an expression for the efficiency of multi-stage impulse turbine.

9.10. Find the optimum ratio of blade velocity to steam velocity for a two-row velocity compounded impulse wheel turbine. Numerical 9.11. A de Laval steam turbine is supplied with dry steam and work on a pressure range from 10 to 0.3 bar. The nozzle angle is 23° and the blade leaving angle is 30°, the mean blade speed is 300 m/s. If there is 10 per cent loss due to friction in the nozzle and if the steam velocity is reduced by 18% during its passage through the blades, find the end thrust on the shaft per power developed by the wheel. Ans. 0.268 N/kW 9.12. The first stage of an impulse turbine is compounded for velocity and has two rings of moving blades and one ring of fixed blades. The nozzle angle is 20°, and the leaving angles ofitthe blades are respectively : first moving 20°, fixed 25°, second moving 30°. The velocity of steam leaving the nozzle is 630 m/s, the blade speed is 130 m/s and the steam velocity is reduced by 10% during the•passage through each ring. Find the diagram efficiency under these conditions and power developed for a steam flow of 4 kg/s. Ans. ri b = 78%, Power = 617.4 kW 9.13. At a stage of an impulse turbine the wheel diameter is 100 cm and speed is 3,000

472

Steam & Gas Turbines And Power Plant Engineering

rpm. The blade speed is 0.34 of the steam speed, and the nozzle efficiency is 0.93, disc and vane friction is 3% of the stage heat drop. The blades inlet and exit angles are 290 and 25o respectively. The velocity coefficient for blading 0.78. Draw the velocity diagram and calculate (a) the work done per kg of steam (b) the stage efficiency. Ans. (b) 70% 9.14. For stage of impulse turbine with single acting wheel equiangular blades, nozzle angle 20° and the velocity coefficient for blades being 0.83, determine the maximum blade efficiency. Ans. 81% • 9.15. Dry and saturated steam is supplied to simple impulse turbine at 7 bar at the rate of 200 kg/h, exhausting at 0.45 bar. The efficiency of expansion is 85%. The nozzle angle is 20° and the blade tip outlet angle is 30°. The blade speed is 180 m/s. Blade velocity coefficient is 0.85 and 1.47 kW lost in disc friction and windage. Neglecting heat loss to surroundings, find—(a) net power at shaft coupling, (b) dryness fraction of steam in exhaust hood, assuming that it is leaving at a negligible velocity, (c) dryness fraction of steam at blade exit. Ans. (a) 13.14 kW, (b) x =0.87, (c) xb = 0.85 9.16. In a simple impulse turbine, the nozzle angle is 16° and the blade outlet angle is 25°. Inlet velocity of steam as it issues from nozzles is 720 m/s and the blade velocity is 180 m/s. The blade velocity coefficient may be taken as 0.75. The steam flow rate is 1 kg/s. Find—(a) Energy dissipated in blades due to friction, (b) Power developed, (c) Diagram efficiency. Ans. (a) 65.13 kJ/s (b) 158.098 kW (c) 61% 9.17. In a two-row velocity-compounded stage for an impulse turbine the initial velocity of steam from the nozzle is 550 m/s. The velocity of blade is 107 m/s. The nozzles have discharge angle 18° and the discharge angles of the three rows of blades in order, are 21.5°. 28° and 45°. Assuming loss due to friction to be 15% in each row of blades, find the work done by the steam per kg and the efficiency of the blading. Neglecting any increase of volume of steam, find the height required for each row blades, taking the height of nozzles as 17.75 cm. (Hints. Draw velocity diagram and find 1 Cfi 1C12 2Cn 2Ci2) I C, Ans. 1st moving blade height = 17.75x ' — 21.7 cm. 1 fl

C 1n Fixed „ „ = 17.75x— 26.3 cm. 2Cfi C 1n Second moving „ „ =17.75x— 31.7 cm. 2 ./2

Work done = 103005 J/kg, 1b = 69.9% 9.18 The first stage of a turbine is a two row velocity compounded impulse wheel. The steam velocity at inlet is 610 m/s and the mean blade velocity is 122 m/s. The nozzle angle is 16° and the exit angle for the first row of moving blades, fixed blades and second row of moving blades are 18°, 21° and 35° respectively. Draw the velocity diagram to suitable scale and calculate (a) diagram efficiency, (b) power developed per kg of steam flow 'per sec, (c) axial thrust at the wheel for mass flow of 1 kg/s. Take the blade velocity coefficient of 0.7 for all blades Ans. (a) 72.5%, (b) 146 kW, (c) 45.5 N

Flow of Steam Through Impulse Turbine Blades

473

9.19. In the nozzle of a single row impulse turbine steam at 5.45 bar with 0.91 dryness reaction is expanded to 3.4 bar at an efficiency of 0.90. The direction of steam at inlet and outlet is 20° with tangential direction and 10° from normal respectively. The ratio of blade speed to steam speed is 0.4. Turbine speed is 4000 rpm. and out put 18 kW. Friction reduces steam velocity by 15% across the blade. Calculate the mean diameter of the ring, (b) the blade inlet and outlet angle and (c) the output of steam nozzle. 9.20. (a) The steam velocity at inlet to the first stage of a turbine having two-row velocity compounded impulse wheel is 600 m/s and the mean blade velocity is 120 m/s. The nozzle angle is 16° and the exit angles for the first row of moving blades, the fixed blades, and the second row of moving blades are 18°, 21° and 35° respectively. Find the blade inlet angles for each row. Find also for each row of moving blades, the driving force and the axial thrust on the wheel for a mass flow rate of 1 kg/s of steam. Find the diagram efficiency for the wheel and the diagram power. What is the maximum possible diagram efficiency for a given inlet velocity and nozzle angle ? Take K = 0.9 for all blades. Ans. 1 13 1 =20°, 1 a1 = 24.5°, 2131 = 34.5°, 874 N, 292.2 N, 1F = 32 N, 2F = 9 N, 77.9%, 140 kW; 92.3%. a (b) In the above problem, the steam flow rate is 5 kg/s and the nozzle height is 2.5 cm. Neglecting the wall thickness between the nozzles, calculate the length of nozzle arc. Take specific. volume 0.375 m3/kg, If all the blade have a pitch of 25 mm, and exit tip thickness of 0.5 mm, calculate the blade height at exit from each row. Ails. 0.454 nr, 11 = 32.7 mm. I f = 41.5 mm, 12 =44.2 mm. 9.21. The exit velocity of steam from the nozzle of single wheel impulse turbine is 600 this. The nozzle angle is 20° to the plane of wheel. The speed is 3000 rpm and mean blade radius is 59 cm. The axial velocity of the steam at exit from the blades is 164 m/s and the blades are symmetrical. Calculate the following : (a) the blade angles, (b) the diagram work, (c) the diagram efficiency and (d) the blade velocity coefficient. Ans. 28° 28', 126.3 kJ/kg, 70%, 0.799. 9.22. (a) Steam at 40 bar and 400 °C is expanded in the nozzles to 15 bar and has a discharge velocity of 700 m/s in the first stage of an impulse turbine having two-row velocity compounded wheel. The inlet velocity to the first stage is negligible. The exit blade angles :-nozzle 18°, first row blade 21°, fixed blades 26.5°, second row blades 35°. Assume the blade velocity coefficient for all blades as 0.9. The mean diameter of the blading is 75 cm and speed of the turbine is 3000 rpm. By drawing the velocity diagram, calculate the following— (i) the diagram efficiency, (ii) the stage efficiency. (b) The mass flow rate is 4.5 kg/s and the nozzle height is 2.5 cm. Neglect the wall thickness of nozzle. The blades of the wheel has a pitch of 2.5 cm and the blade tip thickness at exit is 0.5 mm. Calculate the following—(iii) the length of arc occupied by the nozzles, (iv) the blade exit height for each row. Ans. (0 13.23 cm, (ii) 3.01 cm, (iii) 3.34 cm, (iv) 3.89 cm. 9.23. Steam.at 15 bar and 3000°C enters the nozzles of a impulse stage and discharges at 10 bar. The nozzle efficiency is 95% and the nozzle angle is 20°. The blade speed is that required for maximum work and the inlet angle of the blade is that required for entry of the steam without shock. The blade exit angle is 5° less than the inlet angle. Take the blade velocity coefficient as 0.9. If the mass flow rate of steam is

474

Steam & Gas Turbines And Power Plant Engineering

1350 kg/h, calculate the following— (a) the diagram power (b) the diagram effiAns. (a) 30.3 kW, (b) 86.3% ciency. 9.24. Based on the thermal analysis, show the advantages of a 4-stage pressure compounded impulse turbine over a single stage impulse turbine for following data— Inlet pressure = 42 bar, Inlet temperature = 400°C, Exhaust pressure = 2 bar, Power output = 2.5 MW, Speed of rotor = 8000 rpm Assume suitable data wherever necessary 9.25. The steam enters a single-stage impulse turbine at 380 m/s and the blade speed is 170 m/s. The steam flow rate is 2.2 kg/s and turbine develops 130 kW. Assume the blade velocity coefficient to be 0.8. For an axial discharge of steam find (i) nozzle angle (ii) blade angles and (iii) diagram efficiency 9.26. An impulse stage of a steam turbine has a mean diameter of 1.2 m. The speed of the rotor is 3000 rpm. The mass flow rate of steam is 20 kg/s. Steam is supplied to the stage at 15 bar and 300°C where it expands to 10 bar. Determine the efficiency and the power output of the stage if the nozzle efficiency is 0.9 and the blade velocity coefficient is 0.92. Assume acceleration from the rest for the steam expanding in the nozzle. Assume nozzle angle to be 25° 9.27. An impulse stage of a steam turbine is supplied with dry and saturated steam at 13 bar. The stage has a single row of moving blades running at 3600 rpm. The mean diameter of the blade disc is 0.9 m. The nozzle angle is 15° and the axial compo nent of absolute velocity leaving the nozzle is 93.42 m/s. The height of nozzle at exit is 100 mm. The nozzle efficiency is 0.9 and the blade velocity coefficier+t is 0.966. The exit angle of moving blades is 2° greater than that at the inlet. Determine (i) the blade inlet and outlet angle (ii) the isentropic heat drop in the stage (iii) stage efficiency and (iv) the power developed by the stage. 9.28. A set of steam nozzles in an impulse turbine stage is supplied with steam at 20 bar and 230°C. The mass flow rate of steam is 60 kg/s. The steam is expanded from rest to a back pressure of 14 bar with an efficiency of 98%. The mean diameter of the nozzle disc is 800 mm and the nozzle angle is 22°. Assuming that steam is admitted all round the periphery of the nozzle disc determine (i) whether a convergent or a convergent-divergent nozzle is used, (ii) the velocity of flow of steam at the exit of the nozzle, (iv) the area of the flow normal to the axis of the stage, (iv) the height of the nozzles. Assume adiabatic index forsuperheated steam to be 1.3. What would be mass flow rate when the back pressure is reduced to 9 bar keeping upstream condition same. 9.29 Design a two-row velocity compounded impulse wheel for following data:— Inlet steam pressure =30 bar, Inlet steam temperature = 400°C, Condenser pressure = 0.07 bar 9.30. Develop a software for the design of impulse turbine.

10 Flow of Steam Through Impulse-Reaction Turbine Blades

All modern steam power plants use impulse-reaction turbines as their blading efficiency is higher than that of impulse turbines. This chapter deals with the theory of impulse-reaction turbine which may be helpful in the design. The various types of rotors, blade roots and blade materials have been discussed in the chapter of "Construction, Operation and Maintenance". 10.1. Velocity Diagram And Work done It has been already discussed that in the impulse- reaction turbine, the pressure of steam at outlet from the moving blade is less than that of the inlet side i.e. steam expands while passing over the moving blades. Such a fall in steam pressure in the moving blades must be accompanied by an increase in the specific volume v2 and hence an increase in Ca. Increase in Cr2 and specific volume v2 are brought about by making the moving blades outlet angle actually smaller than the angle given by the following equation 104 . in . v2 sin (3 2 S n .1. 2 Cr2

t in order that this equation is satisfied.

It is also obvious from this relation that if (32 is reduced, keeping S, in, n and t constant, the value of v2 and Co must be varied to satisfy the equation. When pressure drops in the moving blade channels, Cr2 increases at a much faster rate than v2 and the equilibrium point is reached at which the equation of continuity is satisfied. This turbine blade section is very similar to aerofoil section, which has been discussed in Art. 10.7. The stage and velocity diagram of a impulse-reaction turbine is shown in the Fig. 10.1. (a) & (b). The force diagram is shown is Fig. 10.1 (c) & (d). Let AB = C1 be jet velocity at outlet from the nozzles or the fixed blades. AD = u, the blade velocity. Then DB will be the relative velocity at inlet to the blade. The angle 132 is made appreciably less than f31 . For pure impulse the relative velocity of the steam at outlet from the blade would be DG = K.0ri where K is the blade velocity coefficient. So the fmal absolute velocity would be AG. The change in velocity suffered by the steam would be represented by BG. And this change in velocity would give rise to the impulse force gb, which is equal to M (GB) and parallel to GB.

476

Steam & Gas Turbines And Power Plant Engineering

(b) Velocity diagram

(a)

force :T esaunta gelint tiaarI cfeorce: Axial (d) Fig. 10.1. Stage, Velocity and Force Diagram. Due to the expansion of steam in the moving blades, the steam velocity is increased from DG to GF. The increase in outlet relative velocity gives rise to a reaction force fg which is equal to in (FB). and parallel to FG. By combining the impulse force gb and reaction force fg, we get the resultant force fb which is equal to (FB). Component of the resultant force in tangential and axial direction gives the tangential force and axial force respectively. Like the impulse turbine, the tangential force is F = in C = where Cw= whirl velocity Work done = »r u. .u.C Power developed —

1000

W; kW

The same definition will be used for blade efficiency and gross stage efficiency as in the impulse turbine but the expressions are different. In reaction turbine, the velocity of flow per reaction pair is sensibly constant, so that the main end thrust is due to a difference in area and pressure on the rotor; this thrust is usually balanced by allowing low pressure steam to act on a dummy piston but a thrust bearing is also provided. 10.2. Degree of Reaction The degree of reaction, R is defined as the ratio of isentropic heat drop in the moving blades to the sum of the isentropic heat drops in the fixed and the moving blades i.e. in a stage. Mathematically, the degree of reaction is

Flow of Steam Through Impulse-Reaction Turbine Blades

477

(c) Fig. 10.2. Impulse-reaction Turbine

R=

isen)b

) + (hived b isen n

(10.1)

where (Ah ) and (Ahisewn 1 are the isentropic heat drop in moving blades and fixed blades respectively. To calculate the value of degree of reaction R the value of (Ah'sewn and (Ahisen)b should be known. In pure impulse turbine, degree of reaction is zero. In pure reaction turbine, R = 1. Let us consider a number of stage of an impulse-reaction turbine and let the mean diameter of all the stages be the same as shown in Fig. 10.2. Assume that the sections of the blades in each row are indentical and the velocity ratios in each row are the same. Due to uniform blade velocity, the absolute velocities C1 and C2 and the relative velocity, Crl and Cr2 will be the same for each pair of blades i.e., for each stage. ----Note that Cr2 is always greater than CH in reaction turbine/As steam expands in each row of blades, its specific volume increases and since there is a constant rate of steam flow through the blading the blade heights must be increased progressively as shown in the Fig. 10.2. Kinetic energy C1/2 in J per kg of steam comes out from each row of the moving blades and the steam with this energy is directed to pass over the fixed blades. Due to the losses at inlet, the available energy of steam for utilization in each row of fixed blade will be 4) C22 /2, where (I) is the carry-over coefficient. If Tin be the efficiency of the fixed blades when considered as nozzles, then .C2 Cz - (1)c 2 or (Ahisen)n = kJ/kg 2 in — 20hisedn

(10.2)

Steam & Gas Turbines And Power Plant Engineering

478

From the velocity triangle (Fig. 10.2) = U2 + C22 C2 2 r

2u . Cr2 COS 132 =

22(l p2 _ Z 2

p

cr

2

Cos p2)

. Putting the value of C22 in the eq. (10.2). we get

where p2 = r2

(Ahisen)n =

{C2, — C22 r (1 + p22 — 2(32 cos 132)}

(10.3)

211 n

The available energy for each row of moving blades at inlet is 4) cr2 /2. It we assume Ti n to be the efficiency of moving blades also for the sake of simplicity, then we have c2 r2 — cr21 — 2 (Misedb cr22

or

(Ahisen)b —

—

4,

C2

2 in

From the velocity diagram, C2i r = or

(10.4) + u2 — 2u Cl cos al

C2ri = C21(1 + p — 2p cos a l) , Wherep =

Putting the value of Cf2i in the expression for (Ah.)b in the equation 10.1, we get.

R—

C2r24 C21(1+p-2p cos al) I-4)Lr2( l+p2-2p2 cos 132)+0r2— • 0(1+p-2p cos al)

e

(10.5)

Workdone per kg. of steam = u . The velocity of whirl C,, = CD+DE = C1 cos ai —u+Cr2 cos P2 By putting the value of C,, in the above equation of work done per kg of steam, it becomes w = u (C1 cos a1—u+Cr2 cos (32) = ICkp cos al — p2) +p2.C2,,2.cos132} (10.6) The gross stage efficiency

= n•gs

Work done per kg of steam Isentropic heat drop in a stage

C21(p cos al — p2) + p2 C2r2 cos 132 cgs

( ) l(Ah isedn + -6th isen-

(10.7)

Due to the leakage of steam through the clearances between the fixed blades and the rotor and between the moving blades and the cylinder, the actual stage efficiency would be less than calculated from the equation (10.7). 10.3. Impulse-reaction Turbine with Similar Blade Section and Half-degree Reaction (Parson's Turbine). Consider a particular case of reaction turbine i.e. Parson's reaction turbine in which the degree of reaction is half and the section of blades is the same in both fixed and moving rows of blades (Fig. 10.3).

Flow of Steam Through Impulse-Reaction Turbine Blades

(a)

479

(b)

Fig. 10.3. Parson's Reaction Turbine

In most of the cases, the blades of the turbine are divided into groups, each group contains several rows of blades with the same mean diameter and the same blade height ; the results are, (i) increase in specific volume as the pressure falls, (ii) increase in the heat drop per stage compared to the preceeding stage and (iii) there is variation in the blade speed ratio from row to row, thus the degree of reaction is more than one-half. But in the Parson's reaction turbine, it is assumed that the blade section and the mean diameter of fixed as well as the moving blades are the same and the blade height is progressively so increased such that the velocity of steam at exit from each row of blades is uniform throughout the stage thus, the velocity triangle . at the inlet and outlet of moving blades will be similar.

The angle at -132, a2 =131 and velocity Cr2 = C1, and C2 = Cri (Fig. 10.3). 10.3.1. Degree of Reaction : For Parson's turbine we can prove that the degree of reaction is half. For the pure impulse working condition DG represents the velocity of steam from the moving blades which is equal to K.Cri Due to the expansion of steam in the moving blades, the relative velocity is increased to the value of Cr2 = C1 represented by DF. Gain in the kinetic energy in the moving blades is K2 e2 e2 ft, e2 CZr2 _ I rl rl where 4) = K2 = carry-over coefficient 2 2 If in be the efficiency of the moving blades as well as fixed blades, then the isentropic heat drop in each row of moving blades is (Ahisen)b —

C21 — C21 2 r kJ/kg (10.8)

From the velocity triangle. we know, Cr21 = Ci + u2 — 2u . C1 cos a t Since p = /.2 7 then, C21 r = Cal(1 + p2 — 2p cos ad 1 By putting the value of moving blades becomes

in equation (10.8) the isentropic heat drop in each row of

Steam & Gas. Turbines And Power Plant Engineering

480

C2 (6,h,Jb = 2 {1 — (1 + p2 — 2p cos al) } rin

(10.9)

Except for the first set of fixed blades, the gain in kinetic energy in each row of fixed blade is given by C2 _ 1

sr 2

1

rl

2 — 2 So the isentropic heat drop in fixed blades = (Ahisen)n =

Ct

2i — CrkJ/kg 2 nn

Thus the isentropic heat drop in the fixed and the moving blades is the same, i.e. (Ahisen)n = (Ah,sen)b, hence R = 0.5 10.3.2. Gross Stage Efficiency and Optimum Value of p The isentropic heat drop in a stage i.e. in a pair of blades consisting of fixed and moving is (Ahisen)n (61),Jb =

C21

il — (1 — p2 — 2p cos a1)}

(10.10)

n

Gross stage efficiency

=

n

Workdone per kg

Isentropie heat drop in a stage per kg

Work done per kg of steam = u . Cm, ; kJ/kg Cm, = C1 cos a1 — u + Cr2 cos 132 = Cl cos a —u+C1 cos a1 So work done per kg of steam = u . Cm, = u (CI cos at — u + CI cos al) =C1 (2p cos al — p2) li n (2p . cos at — p2)

(C21) (2p cos a t — p2) rigs —

(C2I /ri n) 11 —4) (1 + p2 — 2p cos al)

}

1 — (1+p2-2pcos ) (10.12)

11 n It may be put in another form also, ngs — { (1 7 4))/(2p cos at —132) I + (10.13) The above equation shows that the gross stage efficiency is maximum when 2 p cos at — p2 is a maximum where in and 4) are constant. For maximum value of gross stage efficiency dth s) — 0, which gives dp d(ri

— 0 — 2 cos a — 2p d dp or

p = port = cos at

(10.14)

481

Flow of Steam Through Impulse-Reaction Turbine Blades

2 cost Gross stage efficiency

100

1 -1:)S.2-- cT 1

I‘lulti-stage single row impulse uri)ine

90

50% Reaction

80

Multi stage impulse1 .. reaction turbine 1 i I

70 60 50

Simple impulse

It t a, = 20°, K = 0.9, fl = 0.94 ' 1 R = 9,5 I n 0

0.2

0.4

0.6

0.8

1.0

1.2

10 cos a1

C,

cos a1

2 Fig. 10.5. Blade Efficiency vs. p.

Fig. 10.4. Gross Stage Efficiency vs. p.

So we see that for maximum gross stage efficiency pop, is equal to cosa1 which is greater than pop, of multi-stage impulse turbine. Putting the value p = cos a1, in equation (10.12), we get 2

. COS al ) — gs max

1 — 4 (1 —

cos2a1)

(10.15)

Fig. 10.4 shows the variation of ri gs with p. The curve is flat and the optimum value of • p is 0.9 for at = 20 degree. 10.3.3. Blade Efficiency The diagram or blade efficiency = rib

— Work done per sec Energy input per sec

Workdone per kg = u CH, = C1(2p cos at — p2) and Energy input to fixed blades = C21 /2 Energy input to moving blades —

Total energy input —

ci 2

+

C2r2 — CZi 2

cr22 —r c21 c2 1 C21 — C-' 1 for Parson's turbine — + 2 2 2

C2, C2, C2, 2 (1+p2 — 2p cos ad = —' 2 [1 — p2 + 2p cos ad — CI — ±." = C‘i— --L C21 (2p cos ai— p2)

2 (2p cos ai— p2)

(CY2) [1—p2+2p cos al] — (1 — p2 + 2p cos al)

(10.16)

Since pope = cos at, hence 2 cos 2 a i (1dm= = 1 + COS2 a 1

(10.17)

Steam & Gas Turbines And Power Plant Engineering

,482

Fig 10.5 shows the variation of blade or diagram efficiency of impulse- reaction turbine with u/CI . For the sake of comparison the blade efficiency for impulse turbine is also depicted. It is obvious that the diagram efficiency for reaction turbine is higher and reasonably flat in the region of maximum value which is of great importance from the point of view of operation. 10.4. Comparison of Enthalpy Drops in Various Stage Impulse-reaction Turbines and Impulse Turbines. The optimum ratio of blade velocity to steam velocity for 50% impulse7reaction turbine is given by

— = cosa pops= C But Cl is the velocity of steam coming out from the nozzle or fixed fixed blade neglecting initial kinetic energy is given by CI — COSOC

44.72 q(Ahisen-)nozzle 2

or

4(Ah.nen) no;zle

Thus

4('6: h isen)50% reaction stage

(44.72cosa

(44.72ucosa i

(10.18) Eq. (10.18) gives the isentropic enthalpy drop in a single stage of 50% impulse-reaction turbine. The isentropic enthalpy drop in a single stage of impulse turbine and in a two-row Curtis stage impulse turbine for maximum efficiency are given as 2

q(6,hmen impulse—stage — 4 ( 44.72 cosy )

(10.19) 2

and

4(Ahisen-)Two—row Curtis

. Air Live steam supply Drain

0

Gland condenser

Fig. 14.4.2.

Gland Sealing System

14.4.3 and 14.4.4. In order to prevent the leakage of steam from labyrinth glands into the turbine house where it would condense on the walls and plant, all final gland elements such as A,B and D,E, etc. are maintained at approximately 0.034 bar below atmospheric by the use of a gland condenser vented to atmosphere via a blower. This small vacuum is sufficient to . draw air into the fmal gland section (Fig. 14.4.3 and 14.4.4) where it is mixed with steam leaking from the cylinder and by this way, a mixture of steam and air enters the gland condenser. The air is separated in the gland condenser and passed back to the atmosphere. From the h.p. and l.p. cylinder glands, where the pressure at the gland leak off points a few kN/m2 above atmospheric pressure (e.g. between B and C, etc.). is bled off to seal the gland of l.p. cylinders through cooler M, thus the outflowing steam prevents the ingress of air into the cylinder and condenser. The steam from h.p. and i.p. gland may be sufficient to seal the l.p. glands at high loads but at low loads a separate supply of sealing steam has, to be provided. This separate supply is usually taken from the main turbine supply and throttled through a reducing valve and cooled in a cooler N. Coolers are required to reduce the steam temperature acceptable to turbine shaft. These coolers may be either water spray type or tube and shell type heat exchangers. In order to reduce the power in external glands at the ends of the h.p. ana i.p. cylinders, the leaking steam is bled off and taken to an appropriate stage in the turbine or to a feed heater after each section. In other words, the leaking steam returns heat to the cycle, a pressure-controlled reducing valve is usually used to shut down automatically the sealing system. 14.4.5. Carbon Ring Gland. A compact and efficient form of gland may be made from carbon rings which are about 2.5 cm square in section. It consists of a number of segmental rings of graphite carbon and are enclosed in a suitable housing. It is self lubricating. Such a gland is shown in Fig. 14.4.5. The rings are held close to the shaft by a spring known as garter spring. If there is rubbing of ring to the shaft, the tension of the

Steam & Gas Turbines And Power Plant Engineering

608

spring should be such that rings should exert only a slightly pressure. The rotation of the ring gland is prevented by keys. Carbon ring glands are not used on large modern steam turbines as their maximum operating temperature and shaft speed is limited to about

Aires H.P. cylinder

Steam for sealing through / a cooler 'M'

CBlower

1 bar

Gland condenser

Steam and air Steam

I Drain

N

Steam

it

B

A

Shaft

H.P. final gland Fig. 14.4.3. H.P. Glands L.P. cylinder Steam from 'A'

Steam

Steam and air to gland condenser

ivavvorrA-- -livinrovvvenv..71 't.-474.77

•

r,"

E Fig. 14.4.4. L.P. Glands

Gland housing Carbon ring

(a) Carbon ring seal Fig. 14.4.5. Carban Ring Gland.

Steam Turbine Auxiliary Systems

609

Header tank

Impeller

Fig. 14.4.6. Water-Sealed Gland.

120°C and 5 m/s rubbing speed. 14.4.6. Water-Sealed Glands. As stated earlier, water-sealed gland completely prevent leakage. Fig. 14.4.6 shows a typical water-sealed gland. It consists of a shaftmounted impeller with a series of vanes or pockets machined in both faces which is enclosed within an annular chamber. When water enters the chamber,the impeller vanes force the water to rotate at a speed approximately equal to the• impeller speed. The difference in height h between the water levels across the impeller is approximately given by h—

P2)/ P CO

2

r

It is abvious from the above expression that the seal becomes relatively ineffective at low speeds. This suggests the use of auxiliary labyrinth gland in conjuction with large capacity air pumps to raise vacuum when starting. The water seal is ineffective on high pressure turbine, If water seal has to be used it is essential to use labyrinth seal in conjuction with it to break the pressure down to a figure that this seal can handle. Water is usually injected into the seal at approximately half speed. Since water contained in the annular chamber gets evaporated due to heat generated in seal, some make up water is added to the seal through a header tank. Glands are helpful in neutralising the axial thrust especially in single flow cylinders resulting in the use of smaller thrust bearing. The diameter of the gland is generally increased to take more axial thrust. Such glands are known as dummy or balanced pistons (Fig. 14.4.7). The net thrust on the dummy piston is given by Net thrust =

2 [P2 (D2 — 'q — P3 (D2

14.5. Flange Heating Systems The purpose of flange heating system is to control the heating rate of the casing

'2 wa)siCs &wall a 2u131 4'CV' !.4

Horizontal joint line

A V

A V •-itr )

A

V A

A

No,

Ary'

Steam tap off points on upper and bottom half of outer casings Cylinder bottom half outer casing flange

L.P.turbine

To. no. 2. L.P. heater bled steam pipe

0 0 0 0 A A V Inlet Outlet Inlet Outlet Outlet Section through joint bolt holes sh owing porting for flange heating steam 0

Cylinder top half outer easing flange

H.P. turnine

•

4

4—

To no. 4 deaerator blade steam pipe

'Ll.'171 uoism XuiwnCI ILI

C U

CTQ

to 00

c*

C

o-

PO.

—3

In

FD.

Steam Turbine Auxiliary Systems

61 1

flange, otherwise the differences in temperature between the rotor and casing will lead to differential expansion problems and differences in temperature across the cylinder will give rise to excessive stresses within the casing. Such problems arise during the warming especially from the cold condition where there is a different rate of heating of the rotor and the casing of the high and intermediate pressure cylinders; and inner and outer surfaces of cylinders particularly the flanges. In the flange heating system, steam is passed tlirough passages within the flanges. On leaving the flanges the steam is exhausted to a convenient point in the system as shown in Fig. 14.5. The heating system may be from an external source, such as the gland steam supply or it may be tapped from within the associated cylinder. EXERCISES 14.1. What do you mean by turbine protective devices ? Discuss each of them in brief 14.2. Mention the possible hazards in steam turbines. 14.3. List the various tripping devices. Discuss each of them in brief 14.4. What do you mean by unloading gears ? Discuss the working principle of various unloading gears. 14.5. What are the main elements of a lubricating system ? Discuss the working principle of such a system used in turbine. 14.6. Discuss the working principle of a gland sealing system. Sketch various types of labyrinth glands and discuss their working. 14.7. Discuss the working of flange heating system. 14.8. Mention the various types of pumps used in lubricating system. 14.9. Develop a software for the design of various protective devices installed on steam turbines. 14.10. Develop a software for the design of tripping devices. 14.11. Develop a software for the design of lubricating system. 14.12. Develop a software for the design of hydraulic control system for a modern turboalternator system. 14.13. Develop a software for the design of gland and sealing system for steam turbines.

15 Construction, Stress Analysis, Operation and Maintenance of Steam Turbines

The ultimate aim of any engineering science theory is to design and the ultimate aim of any design is to construct as per design so that the product maintains its high quality with minimum cost and requires minimum maintenance. This chapter deals with construction, stress analysis, operation and maintenance of steam turbine components. The construction of rotor will also be applicable for gas turbines. 15.1. Construction 15.1 (a). Construction of Nozzles and Diaphragm 15.1.1. Design Requirements of Nozzles. The following are the basic requirements for a nozzle— (a) The design of the nozzle should be such as to permit easy manufacturing and finishing and allow accurate channel sections to be obtained especially at the high pressure end of the turbine. (b) The inlet of the nozzle should be so designed as to utilise the carry-over energy from the previous stage to the largest possible extent. (c) There should not be any sudden change in the direction of flowing steam, especially at high velocity. (d) The shape and fmish of the nozzle should be designed so that the conversion of thermal energy into kinetic energy should take place with greatest possible efficiency. (e) In order to reduce friction, especially where the steam velocity is high, the wall surface should be as smooth as possible. 15.1.(a1). Construction of Convergent Nozzles. 15.1.2. First Stage Convergent Nozzle. A segment of convergent nozzle suitable for the first stage of an impulse turbine of medium power• is shown in Fig 15.1.1. This segment consists of six nozzles and comprises a casting. 'a' into which the nozzle guide vanes 'b' are embedded by 'casting in '. For casting of the guide vanes, they are first made from sheet metal of uniform thickness which is cut to shape and then curved in press.

Construction, Stress Analysis, Operation and Maintenance of Steam Turbines

613.

Endview

Steam chest of turbine section

(„0

I4s.1

Development on line AA Fig. 15.1.1. A Segment of Convergent Nozzle

With the correct shape and correct curvature at inlet these are embedded in a sand core to form the steam passage. Then molten metal is poured into the mould while the projecting edges of the guide vanes are surrounded; on freezing of the metal the vanes become firmly held and the casting is taken out. For this type of nozzle, rolled copper guide blades cast in gunmetal nozzle segments are used for saturated steam. But for superheated steam steel or alloy steel must be used. In steel or alloy-steel group materials such as low-carbon sheet steel, 3 to 5% nickel. stainless steel, iron and Hadfleld's Hecla A.T.V. steel may be used. In A.T.V. steel casting, special precautions are taken. 15.1.3. Built-up Nozzle. Built-up construction as shown in Fig.15.1.2. proVides an accurate nozzle segment. It consists of a number of vanes A, which are machined all over, and placed between curved angles B, likewise machined all over, The ends of segments are closed by pieces C of suitable shape. Through the rivets D the guide vanes are attached to the angles. A small spigot E is riveted over it. 15.1.4. Diaphragm Nozzles. A different type of built-up construction for high pressure diaphragms is shown in Fig 15.1.3. Each diaphragm contains a steel centre A, in halves to which several nozzles elements B are riveted as seen in the diagram. Each nozzle

}---4117c+ A Development on xx

Fig. 15.1.2. Built-up Nozzle.

614

Steam & Gas Turbines And Power Plant Engineering

(a) Fig. 15.1.3. Erste Brunner Type of Built-up Nozzle

Fig.15.1.4. Convergent-divergent de-laval Nozzle. elements are machined all over to a fine finish. 15.1 (a2). Construction of Convergent-Divergent Nozzles. 15.1.5. De-Laval Nozzle. It is a convergent-divergent type of nozzle made of gunmetal as shown in Fig.15.1.4. The nozzle is fitted with a valve arrangement C which opens or closes it to a steam chest A. It is mostly used in experimental type of impulse turbine. 15.1.6. Cast-in Type. Fig. 15.1.5 shows cast-in type of convergent- divergent nozzle. It is largely used in marine impulse turbine. Guide blades are rolled to the section shown by dotted lines in the upper part of the figure, and then cut to shape and cast into the nozzle segment in the usual manner. 15.1.7. Built-up Nozzle. Fig 15.1.6 shows a built-up type of convergent - divergent nozzle used in many steam turbines. It is made of steel to B.S.En 58B and is readily replaceable at any time. It consists of two parts-lower and upper forming rectangular cross-se.ction. It is machined from a bar by employing jigs and fixtures to the required shape. It is used in many impulse turbines.

Construction, Stress Ana► ysis, Operation and Maintenance of Steam Turbines

615

Section through AA

Fig. 15.1.5. Cast-in-Type ConvergentDivergent Nozzle.

Fig. 15.1.6. Built-up Type ConvergentDivergent Nozzle.

(II) Fig. 15.1.7. Three Segment Built-up Nozzle.

An another form of built-up nozzle is shown in Fig. 15.1.7. It consists of three partssegment strip 'a', in which the nozzle passages are machined, a covering segment 'b' and a wedge piece 'c'. The locking screws 'd' on the ring 'c' causes the strips 'a' and 'b' to be forced against the steam chest. The cap nut `e' covers the setscrew 'd' so that there should not be any leakage. It is used in so many turbines. 15.1.(b). Construction of Turbine Blades. 15.1.8. Production of Blades. Blades may be considered to be the heart of a turbine,

and all other members exist for the sake of the blades. Without blading there would be no power and the slightest fault in blading would mean a reduction in efficiency or lengthy and costly repairs. The following are some of the methods adopted for the production of blades. (i) Rolling. Sections are rolled to the finished size and used in conjunction with packing pieces. Blades manufactured by this method do not fail under combined bending and centrifugal force. (ii) Machining. Blades are also machined from rectangular bars. This method has more or less the same advantage as that of the first. Impulse bladings are manufactured by this technique. (iii) Forging. Blade and vane surfaces having airfoil sections are manufactured by specialist techniques. The simplest way is to determine the profiles required at

616

Steam & Gas Turbines And Power Plant Engineering

the hub and tip, and join them by straight, ruled lines. For more accuracy, a profile, at middle to each end separately is obtained. Once the geometry of the family of ruled lines is established they may be machined in turn by a milling machine, rest carefully for each line to generate the shape required in a master block from which the forging die may be copy-machined. This method ensures the accurate forging of blades to their finished size, requiring only fettling and polishing. The machining of the fir-tree root is often done by broaching, and electro-chemical machining may be used in some parts to avoid the conventional cutting processes. In advance method, computers are used to determine the blade shape required by aerodynamic and stress criteria. The computer may then instruct a numerically controlled milling machine to prepare the dies. (iv) Extrusion. Blades are sometimes extruded and the roots are left on for subsequent machining. This method is not reliable as rolled section, because of narrow limits imposed on the composition of the blade material. (v) Cold drawing. Blade are also cold drawn. 15.1.9. Long Blades (L.P. Stage Blading). The blades of low-pressure stage must be long to cater for the greatly increasing specific_volume of steam at the lower pressures. Irrespective of the design of previous stages, the final stage of 1.p. turbine employ little or no reaction at the root and up to about 65% reaction at the tip. This design allows the steam velocity to match the peripheral blade velocity of all radii. in order to dampen vibration long blades may be laced together in batches. It .is to be noted that the lacing holes are source of weakness and disturbs the flow path, so it should be avoided as for as possible by better designing against vibration. Sometimes an arched cover band may be used to brace the instead of lacing wires. L.P. stage bladings face an another problem of erosion of leading edges due to condensation droplets. In order to avoid this, satellite protection strips (extremely hard alloys of cobalt, chromium, tungston and carbon) are sometimes brazed to the leading edges. Due to centrifugal action much moisture can be extracted after leaving the moving blades, and provision is made in the cylinder to lead this water away. (Fig: 15.1.8). Collector annulus

Fig. 15.1.8. Moisture Drainage Bladings.

Fig. 15.1.9. Multi-exhaust System

Construction, Stress Analysis, Operation and Maintenance of Steam Turbines

617

Sometimes, multi-exhaust is used to reduce the length of 1.p. stage bladings (Fig.15.1.9). Invariably, double flow LP turbines are used. A long blade is not suitable because of the following reasons— (i) The blade speed varies from root to tip thus there is different blade angles, and if the steam is to flow on the blade without shock, the blade must be twisted. Generally, the discharge is axial. They are mainly impulse form at the roots and reaction form at the tips. The inlet angles of blades are varied to allow for change in blade speed. (ii) The space between the adjacent blades may increase so much from the root to the tip as could effect adversely the steam flow through the blades. At the tip speed of 330 m/s, the stress at the root of the blade is great. For this (iii) reason and from the point of view of stability, low pressure blades are not made longer than one-third the drum diameter, and even then the blade section is frequently tapered from the tip to the root. 15.1.10. Hollow Blades. Hollow blades satisfy the condition for ideal blades i.e. they give the most efficient control to the steam and are at the same time uniformly stressed. The hollow blades do not impose severe stresses in the rotor, and for that reason increased speed, leading to increased output is possible. 15.1.11. SAr_out _IILT. Stiffness against vibration and correct guidance to the steam is essential. To meet these conditions the outer ends of the blades are usually tied together by a perforated ribbon of metal known as shroud. In case where stress consideration is of primary importance, for example, the last row of the low-pressure blades, the shroud is omitted. In longer blades of L.P. turbine, lacing or binding wires are also silver soldered to connect bundles of blades together at various radii. 15.1.12. Blade Materials. Proper selection of blade material plays an important role in the blade design. The factors that influence the selection of blade materials are:(a) Method of manufacture. (i) Ease of machining. (ii) The ability to produce the blade section free from flaws. (iii) Ductility to allow of rolling to shape. (iv) The capacity for being brazed or welded. (v) Ease of forging easily.. (b) Conditions of operation. (i) Suitable tensile strength of material at high temperature. (ii) Resistance to creep. (iii) Resistance to corrosion and erosion specially in L.P. stages. (c) Cost. The commonly used blade materials are:Brass. Brass (70 to 72% Cu and 28 to 30% Zn) is suitable for temperature upto 230°C. It is resistant to corrosion and it may be cold drawn. This is rarely used now-a-days. Copper Nickel. This is an alloy containing about 80% Cu, 19% Ni and fraction of iron and manganese, etc. Nickel Brass. It is suitable for temperature range upto 230°C and 50% Cu,40% Zn and 10% Ni. It may be cold drawn. Manganese Copper; Its composition is 95 to 96% Cu,4 to 5% Mn and small amounts

618

Steam & Gas Turbines And Power Plant Engineering

of iron, carbon and lead. It is not suitable for high stress and temperature above 310°C. It may be cold drawn and cold rolled. Phosphor Bronze. This is copper-tin alloy with a small amount of phosphorous. Its composition is 86% Cu, 14% tin and 1% phosphorous for hard bronze and 92% Cu, 8% tin and traces of phosphorus for soft bronze. It is resistant to corrosion but not suitable for high temperature. It may be forged, rolled or cold drawn. Monel Metal. The composition of monel metal is 67% Ni and 28% Cu with a small amount of iron, carbon, manganese, etc. It is resistant to corrosion and is suitable for high temperatures. It is used for marine work. Mild Steel. In the past, it was used in many turbines but now is not used. It corrodes very soon with wet steam but it is very inexpensive. Nickel Steel. This alloy is more resistant to corrosion than in mild steel. It may be forged and machined but not welded. Generally, steel with 3 to 5% Ni is used. It is used in many turbines. Stainless Steel. It is an alloy of iron, chromium and carbon, containing 12 to 14% chromium and normal percentage of carbon. It is very resistant to corrosion and erosion. It is also very hard. Stainless steel is also very suitable for blading materials containing 0.1% carbon. It may be rolled, easily machined and welded. In some cases, FV 520B are used. Now-a-days it is used in many steam turbines. A.T. V. Steel. This is an alloy containing 36% Ni, 12% Cr, 0.25% C and the rest iron. It is highly resistant to corrosion and erosion. Its other qualities are that it can withstand high temperature. This steel is commonly used now a days in many turbines. CMV Steel. Most blading is made of 12% chromium-molybdenum vanadium steel which has good strength and creep properties. Niobium may be added to improve the creep properties. It is commonly used. 15.1.13. Turbine Blade Attachment To the Rotor. The attachment of the turbine blades to the rotor is the most critical aspect of steam turbine design. All the forces are transmitted through the attachment to the rotor. Especially, at the low pressure end of turbines of large output, the attachment has to bear relatively large forces due to high speed, the centrifugal force on the blade is many times its mass. Therefore, it becomes necessary to estimate the stresses in the attachment but sometimes it is difficult to get the exact value. There is always the possibility of stress concentration at the sharp corners. Therefore, selection of material is very important which can safeguard from this stress concentration and that is why the calculated stress is kept reasonably low. A careful study of the forms of attachment is also necessary because occasionally it influences the shape of the wheel rim and stresses in the disc. The form of the attachment should be such that the centrifugal force on the blade is transmitted to the disc in the simplest and most direct manner and it should give the security of attachment. The various forms are given below. (a) de Laval blade root attachment, (b) Inverted-T-attachment, (c) Serrated blade (Annular Fir-tree) root attachment, (d) Attachment for high-pressure Curtis wheel, (e) Straddle (single or multiple fork) attachment, ()J Modified straddle attachment, (g) Side entry blades attachment, (h) Shrouding strip attachment, (0 Parson's end-tightened blading, (j) Parson's integral blades Most of the above attachments are also used in gas turbine bladings. However, Annular Fir-tree or its modified versous are most common.

Construction, Stress Analysis, Operation and Maintenance of Steam Turbines

619

15.1.14. de Laval Blade Attachment. Fig 15.1.10 shows a de Laval blade attachment. To attach the blades to the wheel, a number of holes C are drilled in the rim of the wheel parallel to the axis and the slot for the blade tang is finished with the help of milling. The rectangular part B connects the working section of blade to the rectangular Section b-b tang. A is an integral shroudings provided at the tip of the blades to form a continuous, ring. To keep the blade tang free from bending stress, the centroid of the blade section is arranged in the same line as that of blade Fig. 15.1.10. de Laval Blade Attachment. tang. There is a negligible, bending moment due to impulse. To hold the blades in place a light caulking of the wheel rim is sufficient. • 15.1.15. Inverted-T Attachment. As the name implies the shape of the attachment is inverted-T type and its various forms are shown in Fig.15.1.11. Fig.15.1.11(a) shows a blade with a separate spacer. It is simple in construction. The blades are cut from bars or

ri

Section thro' ZZ (a)

Section thro' ZZ (c)

(b)

(d)

Fig. 15.1.11. Various Forms of Inverted T-Attachment

Steam & Gas Turbines And Power Plant Engineering

620

strip of blade section formed either by rolling drawing through dies or by machining. As little machining work is required so it is inexpensive. Fig.15.1.11(b) shows the spacers. A taper is provided at the leading edge of the spacer so that the blades should be set in radial position. S shows the length of spacer from the rotor. The main point to be noted is that the centroid of the section ZZ does not lie in the same radial line as the centroid of the blade section and so this section is subjected to tension as will as bending. In case of excessive stress across section Z-Z, the blade and spacer is made from one piece to withstand the stress. Fig 15.1.11(c) shows such a type of integral inverted T-attachment. It is an expensive affair compared to the separate spacers. In both cases, there will be stress concentration at the point A. To remove this stress concentration the blade tang is tapered as shown in Fig. 15.1.11(d). Since the tang is long, the depth of root is more, this is even more expensive affair because it requires relatively heavier rim and i-nore machining is needed. The resultant stress in the tang due to bending and direct tension is sometimes excessive when the root is constructed as shown in Fig 15.1.11(c) because of the long blades or high speed. This difficulty may be overcome by adoption the attachment shown in Fig. 15.1.12. The bending stress is practically eliminated by this. The type of wheel rim used in this type of attachment is shown in Fig.15.1.13(a). G is the centroid of the section A-B in one section for the wheel rim. The resultant tensile stress at B is F 3Fa = 2bd + bd2

(15.1)

where F/2bd is the uniform tensile stress produced by force F/2 and 3F.a/bd2 is the tensile stress produced at B due to bending moment F.a/2. In the case of excessive width of the rim, "side grip" type of attachment is used as shown in Fig. 15.1.13(b). This type of attachment reduces the bending moment which is given as

n

(b)

(a) Fig. 15.1.12. Invested T—Attachment Without Bending Stress.

Fig. 15.1.13. Side Grip Type Inverted T—Attachinent

621

Construction, Stress Analysis, Operation and Maintenance of Steam Turbines s--? I

(a) Fig.15.1.14. Serrated or Annular Fir-Tree Blade Root Attachment.

(b)

Fig.15.1.15. Single and Multiple Fork. Straddle Attachment.

F. a Bending moment — — — S . h . (15.2) 2 Invested—T attachments are used in many steam and gas turbines blades having shorter length. 15.1.16. Serrated Blade Root or Annular Fir-Tree Root. For fairly large centrifugal forces, serrated blade root attachment is used as shown in Fig.15.1.14. The blade and spacer is made in one piece to withstand large centrifugal stresses. Serrations of annular fir-tree are machined on both sides of the root and also in the blade groove. The angle of serrations is maintained at 45° to the groove. The depth of serrations is kept about 3/32 times the width of the groove. This type of attachment is most suited for high speed turbine blades. Inverted fir-tree is also used. Annular fir-tree root (serrated) attachments are extensively used in medium and large gas and steam turbines. 15.1.17. Straddle (Single or Multiple Fork) Attachment. In this attachment, the blade is forked as shown in Fig.15.1.15. Fig.15.1.15(a) shows the attachment for moderate speeds and short blades. In this attachment, the blade and spacer must be made from on piece of metal. The forked blade fits tightly on to a projecting tongue on the wheel as shown in the figure. The rivets are slightly countersunk and staggered. When the shear stress in the rivets is excessive, as with higher speeds or longer blades, the attachment shown in Fig. 15.1.15(b) is used, the rivets then being in quadruple shear. The straddle attachment enables a complete ring of blades to be employed as there is no necessity to cut any gap for the insertion of the blades. The other advantage is that one or more blades may be removed from the wheel if necessary without disturbing the remaining blades. 15.1.18. Modified Straddle Attachment. A modified form of straddle attachment is

622

Steam & Gas Turbines And Power Plant Engineering

Fig.15.1.16. Modified Straddle Attachment.

Fig. 15.1.17. Side Entry Blade Attachment.

used in the last two or three stages of turbines of large output. This is shown in Fig. 15.1.16. The periphery of the wheel is provided with a T-shaped section. As it is obvious from the figure the blade root is wide and provided with two small snugs which prevent the fork from opening out. The resultant moment across the section x.x-is F.a M — — —S . h . 2 (15.3) where S is the shear force induced in each prong. This type of attachment is really an inversion of that shown in Fig. 15.1.13(b). Suitable spacers, fitting tightly between the blades on either sides of the slot, are fastened to the rim as shown in the figure. 15.1.19. Side Entry Blades Attachment. This type of attachment is shown in Fig. 15.1.17 is suitable for very high peripheral speed of turbine. Slots are cut, one to each blade in the rim which is comparatively heavy. Depending upon the shape of the blade root profile, the slot may be either parallel or skew to the axis. a part of the pull is taken by the outermost teeth, so that the smaller area may be employed at 2 and similarly at 3. The advantages claimed are(a) maximum stress in the fastening is reduced by making the width of the wheel rim greater than the width of the blade. (b) possibility of eliminating the bending across section 1-1(c) complete ring of blade is possible (d) replacement of blades is simple. But it is an expensive attachment. 15.1.20. Attachment of Shrouding Strip. The radial acceleration in the shrouding strip is maximum because it is attached to the blades at the extreme radius. Consequently, the centrifugal force is often very large and the thickness of the shrouding strip must be chosen with reference to the bending stress, the shrouding being in the condition of continuous beam on several supports and carrying a uniformly distributed load. Fig.15.1.8 shows the various type of attachments. Fig.15.1.18(a) shows a round pin machined on the

623

Construction, Stress Analysis, Operation and Maintenance of Steam Turbines

El Section Clearance E

(a)

(b)

(c)

Fig.15.1.18. Attachment of Shrouding Strip. Fig.15.1.19. Earlies Parsons End-tightened Blading

end of the blade, the centre being as near to the.centroid as possible. Fig.15.1.18(b) shows the second,type in which the pipe for the shrouding is usually formed by milling. Sometimes because of thinner section the attachment is difficult and it becomes necessary to use two projections to thicken up blade end as shown in Fig.15.1.18(c). In order to avoid stress concentration, this section is gradually varied. 15.1.21. Parsons End-tightened Blading. This method of attachment was introduced by Parsons for reaction turbine. In this, the clearance which governs the leakage, past the blades is, axial instead of radial. In all the high and intermediate pressure stages these type of blades are used; only in L.P; stages thin tipped radial clearance blades are used. Fig.15.1.19. shows an earlier form of end-tightened blading. The deep spacers B and the a strip of shrouding strip C are silver soldered to the root and tip of blading respectively. Serrations are then milled on one side of the roots and shallow groove on the other.

Fig.I5.1.20. Recent Form of End-tightened Blading

Fig. 15. 1.21 Parsons Integral Blades.

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Steam & Gas Turbines And Power Plant Engineering

As is obvious from the figure that the rotor groove is serrated at D for the blade roots and E for the packing pieces F. To reduce the rubbing surface the shrouding strips are thinned down to a fine edge. As is obvious from the Fig. 15.1.19 the flow of steam through the clearance between the fixed shrouding strip and the bladings will disturb the main steam flow into the moving blades. This disturbance in flow is avoided by putting a second strip B shown in Fig. 1.1.20 which has a tapered edge serving to deflect the leakage of steam and consequently the disturbance is minimised. Now a days, the radial packing is also provided by means of a fm at the shrouding. 15.1.22. Parsons Integral Blades. Parson's integral blades are most suitable for long reaction turbine blading running at high speeds. As the name implies the blade and the spacer are made from one piece as shown in Fig. 15.1.21. The root of the blade is lozenge-shaped. These types of blades are rolled by a discontinuous process and made from a rectangular bar. The blade is pickled, polished and serrated at the root by machining. 15.1.23. Vibration of Blades. The vibrations of turbine blades are caused by the periodic nature of steam flow (i.e. the irregularities of steam flowing out from nozzles or guide blades behaving as a periodic external forces). Under normal operating conditions only forced vibrations are possible in steam turbines. The perturbation forces responsible for blade vibration are classified into two categories to facilitate tuning of blades out of dangerous regions of vibrations. They are as follows:(a) Forces having frequency equal to NZ where N is revolution per second and Z is the number of nozzles or guide blades along the periphery of the disc:In this case, the perturbations are caused by the partitions which constitute the guide passages. Due to friction the velocity of steam at the exit of nozzles or guide blades are not uniform but varies from a certain minimum value at the walls to the maximum value at the centre of blade passage. As soon as the moving blades come into contact with these non- uniform periodic forces, they begin to vibrate. if the natural frequency of these blades coincides with the frequency of the perturbation forces, resonance is set up which may lead to failure of blades. The frequency at which resonance may set up is expressed as f=NZ (b) Forces having frequencies equal to the rotor rps or its multiple, i.e. equal to KAI where K is an integer 1,2,3, etc:— In this case, the perturbation forces are caused by the following:— (i disturbances of the flow at the diaphragm joints (ii) local disturbances caused by the differences in the guide passage dimensions resulting from inaccuracies in manufacture. (iii) rotor vibration caused by improper balancing. The frequency at which resonance may occur in this case is given by f =K.N 15.1.23.1. Types of Vibration and Remedies: The following types of blade vibration are possible:—

Construction, Stress Analysis, Operation and Maintenance of Steam Turbines

.. 625

Tangential Vibrations—Vibrations in the plane of the disc. (ii)Axial Vibrations—Vibrations in a plane perpendicular to the direction of rotation. (iii) Torsional Vibrations—Twisting around the blade axis itself or the axis of a blade assembly. Remedies: Following steps are taken to reduce blade vibrations:— (iBy changing the number of blades in a. stage or increasing the spacing between blade rows. Use of different prime numbers of blade in each stage is also helpful. (ii)By altering the thickness of the blade airfoil all over or selectively. (iii)Introducing shrouds to provide a damping contact between adjacent blades. Wire lacing of blades has the same effect. (iv)Selecting appropriate material combinations. 15.1(c). Rotor Construction 15.1.24. Types of Rotors. There are five types of steam turbine rotor in common use. These are as follows (Fig.15.1.22) fig) The integral rotor (a) The built up rotor ',_(d) The solid drum rotor (c) The hollow drum rotor (e) e welded disc rotor Motors (a) and (b) are in use in large turbines which make use of bladings of the impulse type. In large turbines using 50% reaction bladings, rotors (c), (d), (e) and (a) are used. The built-up and welded discs rotors are mainly used for l.p. rotors of reaction turbines. (a) The built up rotor. (Fig 15.1.22a):— It consists of a forged steel shaft on which separate forged steel discs are shrunk and keyed. It is cheaper since the discs and shaft are relatively easy to forge and inspect for flaws and machining of these components can be carried out concurrently. Its shaft is machined with a series of stepped diameters ending with central collar. Each disc is heated and assembled on the shaft in turn, each being held in position by a form of circlip. Relative rotation is prevented either by keys, or by hub dowels known as buttons, which locate the hubs one to another to the central collar. The number of discs depend upon the number of stages which in turn depends upon the turbine output. (b) The integral rotor (Fig. 15.1.22b):— The shaft and wheels of this type of rotor are formed from one solid forging. Integral rotors are expensive and difficult to forge, and there is a high incidence of rejects. Over and above, a large amount of machining time and waste material are involved. Nevertheless, their advantages are such that they are invariably used for the h.p. rotors on modern reheat turbines, and sometimes for the i.p. and l.p. rotors as well. Followings are the advantages of integral rotors:— (i there is no chance of discs to become loose, particularly at the high temperature end where at times the wheels may be hot and the shaft cool as found in the built up rotor. (ii) This rotor is also free from the effect of creep which may cause the shrink fit of built-up rotor to disappear after a large number of running hours. (iii) The hoop stress is of lower magnitude as it contains a small hole meant for inspecting the forging.(iv) There is a saving in axial length and reduction in spindle diameter over the built-up type. (c) The hollow drum rotor (Fig. 15.1.2c):— This type of rotor promotes even temperature distribution because it is designed with the same thickness of material as the casing.

626

Steam & Gas Turbines And Power Plant Engineering

II Ily

/1

1 %

,

4

(b) Integral Rotor

(a) Buit-up Rotor

xN. / /11111011 .1p

IA

/• /

N I

(d) Solid Drum Rotor

(c) Hollow Drum Rotor

(e) Welded Disc Rotor • Fig. 15.1.22. Turbine Rotors

(d) The solid drum rotor (Fig.15.1.22d):— This type of rotors are suitable for cylinders where there are lower temperatures but large diameters, as in i.p. cylinders without reheat. (e) The welded disc rotor (Fig. 15.1.22e):— The last stage disc is the most heavily stressed part of the turbine and this is one of the main problem of 1.p. rotors. The centrifugal load of the large rotating blades set up a tensile stress in the rim of disc, and this stress increases with decreasing radius, its maximum value being at the bore of the hub. If the bore is exceedingly small, the hoop stress becomes very less but if there is no hole, the hoop stresses throughout the disc are theoretically halved. Since there is no central hole in welded disc rotor, so it is suitable for 1.p. rotors. It has two main advantages. It is less stressed and no need for large shaft forgings which are expensive and difficult to manufacture. The welding process and subsequent heat treatment should be performed with great attention. These five types are rotors are also used in gas turbines. 15.1.25. Rotor Materials. The selection of rotor materials depends mainly on the estimated stress in the rotor and the temperature to which the rotor is subjected. For rotors subjected to moderate stresses and temperature, plain carbon steel is used. It may be in two categories, steel No.1 and steel No.2. Steel No.1 .has the following composition:— C 0.25; Mr 0.58; Si 0.11; Ni 0.13 and safe working stress=10800 N/cm2 Steel No.2 has the following composition:— C 0.33; Mr 0.58; Sr 0.21; Ni 0.2

Construction, Stress Analysis, Operation and Maintenance of Steam Turbines

627

and safe working stress =1220 N/cm2 For rotors subjected to high stresses and temperature, h.p. and i.p. rotors are usually made of chromium-molybdenum-vanadium steel, a ferritic material suitable for wheelcase steam temperature up to 540°C.L.P. rotors are made either of 3% chromium molybdenum steel or 2.25% nickel chromium molybdenum steel. The English Electric 200 MW turbine has the following composition of rotor material H.P. and L.P. shafts C 0.25 max; Si 0.35 max; Mr 0.4 to 0.7; S 0.04 max; P.0.04 max;, Mo 0.5 to 0.85; Va 0.25; Ni 0.5 max; Cr 0.6 max. H.P. Disc C 0.35 to 0.4; Si 0.2 to 0.25; Ni 0.6 to 0.9; S 0.04max, P 0.04 max; Ni 0.3 max. L.P.Disc C 0.3 to 0.4; Si 0.35 max; Mr 0.6 to 0.9; S 0.04 max; P 00.4 max; Mo 0.5 to 0.7 max; Ni 0.5 max; Cr 3.0 to 3.5 max. In most of the impulse turbines, the rotor material is forged, heat treated with 3% Cr. M. alloy steel to 13.5. En 29. 15.1.26. Approximate Determination of Spindle Diameter. The approximate value of spindle diameter is given by d= . (15.4) where I 7 span between bearing centre in cm, N = speed of rotor in rpm, D = average mean diameter of disc, cm, d = mean diameter of middle portion of spindle(cm), k = constant In merchant turbines, k=8,900,000, in naval impulse and reaction turbine, Ic=1,2000,00. Where disc construction is much lighter, specially in land practice, k=15,3,00,000 It is to be noted that the rotor must be checked for the critical speed. 15.1.27. Thermal Stability of Rotors:Vs. great care is taken during manufacture of rotor to ensure that the forging is stable. A stable rotor is one whose physical properties do not change in service. There are three types of instability:— (a) permanent, (b) temporary and (c) transient (a) Permanent instability: This is due to assymetrical coefficients of expansion across a diameter. A close metallurgical control at the ingot stage may avoid it. (b) Temporary instability: It is due to locked-up stresses in the rotor. It is relieved by heating before and after machining in a special furnace in which the shaft is rotated. (c) Transient instability: It is due to differences in conductivity and emissivity. The use of turning gear overcomes this difficulty. 15.1.28. Balancing of Rotors*. From the theoretical point of view, a turbine (gas/steam/water) rotor is a balanced body but in actual practice, errors of balance are introduced by various causes such as: (i) Lack of homogeneity of the material. (ii) Slight errors in machining. (iii) Differences in the pitch of the blades and also in their individual masses. Articles 15.1.27, 15.1.28 and 15.1.29 are also valid for gas turbine rotors. *

628

Steam & Gas Turbines And Power Plant Engineering

Therefore, it is essential to test balance of a completed turbine rotor and make any adjustments necessary to ensure that the balance is as good as possible The purpose of balancing of a rotor is to reduce the amplitude of vibration on a tolerable level which can be taken to be about 0.0254 mm at the bearing pedestals of a 300 rpm machine. There are two types of balancing static and dynamic. Each rotor assembled with blades is balanced both statically and dynamically. 15.1.28.1. Static Balance. It means that the weight of the rotor is evenly disposed around the axis of the shaft. It is checked by rolling the rotor on horizontal knife-edge supports. 15.1.28.2. Dynamic Balance: It means that the moments of the out-of-balance weights along the axis about either bearing add up to zero. It is checked by spinning the rotor on risilient bearings, detecting the vibration and adding or subtracting weights until the vibration in negligible . Normally, rotors are balanced at low speeds say 400 rpm. The adjustment in weight is made in two planes, one at each end of the rotor by varying screwed plugs in tapped holes, or by removing metal from portions of a rim added for this purpose, or by fixing weights in a grooved by means of screws. Preference is given to subtraction of weights instead of addition, since there is a chance of coming the loose weights a drift. The modern balancing machine is a very sophisticated and accurate machine. However, a very small out-of balance force always remains. 15.1.29. Critical Speeds . of Rotors. Even with the utmost, care taken up in the construction and balancing of turbine shaft and disces i.e. rotor, due to some reason or another, the mass centre of the rotor does not coincide with the geometrical axis of the shaft. This means that after assembly the rotor has a certain amount of unbalance (imbalance). The distance between the mass- centre and geometrical axis of the shaft is known as the eccentriciy of the rotor. If the shaft is rotated, even a small eccentricity will give rise to a transverse force tending to deflect the shaft. The deflection of the shaft increases with the speed of the shaft. At a particular speed the deflection may be so great that the shaft be;:omes permanently bent:The speed at which the deflection of the shaft theoretically rends to infinity is known as the critical speed. In other words, the speed which numericult), coincides with the natural frequency of transverse vibrations of the shaft is known as the critical speed. ‘.y. This suggests that the operation of turbines at the critical speeds should be avoided otherwise resonance may occur leading to failure. From experience, it has been found that if the critical speed differs from the normal speed by 15 to 20%, safe working of the turbine is completely stored. Most of the manufactures use a normal speed either higher or lower than the critical by about 30 to 40%. Shafts having critical speeds less than their normal operating speed are known as flexible shafts. If the critical speeds are higher than the normal speeds, they are known as rigid shafts. Rotors of impulse turbines are made of both these type of shafts, flexible and rigid. Care has to be taken in the case of critical speed less than operating speed while starting the turbine so that the critical speed must be passed over quickly allowing hardly any time for the deflection to grow, which if allowed could result in a bent shaft and damaged bearings. For simply supported shafts loades at the centre, critical speed is given by

Construction. Stress Analysis, Operation and Maintenance of Steam Turbines 300 Ncr = 70 where yo= static deflection

629

(15.5)

The critical speed of the shaft with several discs is given by Ncr = 300 V [Em li

(15.6)

where m1, m2, m3, etc. be the masses of concentrated loads causing deflections y1, y2, y3...etc. 15.1(d). Construction of Cylinders (Casing) 15.1.30. Cylinders and Diaphragms. A steam turbine cylinder has to withstand the high pressure and temperature of steam; therefore, it falls essentially under the category of a pressure vessel. Its weight is being supported at each end. Following are the design requirements of a steam turbine cylinder: (i) It should withstand hoop stresses in the transverse plane. (ii) It should be very stiff in a longitudinal direction in order to maintain accurate clearances between the diaphragm or fixed blades and the rotor. Horizontal Joint Construction: A turbine cylinder has a complicated shape often varying in diameter along its length to incorporate the steam chambers for supply of steam to its various stages, to accommodate the rotor with increasing diameter at its 1.p. end and the exhaust pipings as well as extraction points or pass-out chambers. From the most ideal point of view the turbine cylinder should be truely circular and in one piece. But the ideal turbine cylinder is not possible in practice. Therefore, with all practical considerations in view it is essential to split the turbine casing horizontally to facilitate assembly and dismantling. Therefore, horizontal joint in some form or the other is essential as shown in Fig. 15.1.23. Since the horizontal joint is in two halves and if the rigid attachment is not provided at the horizontal joint, then an excess of temperature on the inside wall of the cylinder would cause expansion on the inside and thus a distortion shown in Fig. 15.1.23 by dotted lines in exaggerated form. In order to avoid this distortion and make it free from any steam leak on account of high internal pressure, a very stiff joint is essential. The gland housings, the horizontal flange and steam entry passages introduce the stress complexities in the cylinders. Furthermore, the external mass of the flanges may produce differential expansion during starting on account of warming up slowly than the remainder of the shell. This differential expansion may set up temperature stress and distortion. These difficulties are overcome by heating the flanges with the help of steam passing through passages during the starting period (see the article flange heating system). Parson's Clamped Joint Construction : This is suitable for high pressure and temperature reaction turbine. Here, the bolted joint is eliminated from the high pressure cylinder and the two halves are held by clamps as shown in Fig.15.1.24. The flanges are very narrow and deep. Double Shell Cylinder: When the turbine is using a very high pressure of steam, h.p. cylinder is generally made of double shell in which steam at exhaust pressure fills the space between the shell (Fig.15.1.25a). By this way, the differential pressure for whic,h each shell to be designed is reduced resulting reduced shell thickness and quicker warming of the turbine when starting without under temperature stress because of increased area of contact with the steam. The castings also became sounder. Sometimes, reverse flow bladings are used to minimise the

630

Steam & Gas Turbines And Power Plant Engineering

Fig. 15.1.23. Horizontal Joint Construction

Fig. 15.1.24. Parson's Clamped Cylinder differential expansion in h.p. cylinders (chap.4). This provides an opportunity to incorporate an additional bled steam tapping point at the point of reversal resulting in reduced net h.p. thrust. In general practice, most of the bled steam tappintg points are usually located in the i.p. and 1.p. cylinders and annular recesses are provided at extraction points. In reheated turbines, the steam enters the i.p. cylinder at high temperature but medium pressure so the design of the casing is a somewhat easier problem. However, a partial double shell (Fig. 15.1.25b) is sometimes adopted. The present practice is to drive the main boiler feed pump by means of bled steam turbine known as feed pump turbine. This turbine is fed with steam bled from h.p. cylinders exhaust. This arrangement comes out to be cheaper because it needs smaller size

Construction, Stress Analysis, Operation and Maintenance of Steam Turbines Inner shell

Outer shell

Partial inner shell

Outer shell

631

Carrier rings A/

•

Vii. s (a)

(b) Fig. 15.1.25. Double Shell Cylinder

•Stud tension

Mutal reaction

Hoop tension

Interface leakage pressure

Fig. 15.1.26. Force Arrangement of pipe leading back to reheater. Over and above, this arrangement simplifies the design of main i.p. cylinder as the extraction points which were supposed on i.p. cylinder are now provided on feed water pump turbine. Design of cylinder flanges requires a careful study. It must be thick for stiffness and, bringing the bolt centre line tangential to the shell and narrow for quick heating. Fig. 15.1.26 shows a force arrangement which may form the basis of design. It is obvious from this figure that flange faces are pressurised and the outer edge is only in contact. Cylinder Support : Fitting of various cylinders is also a difficult job. H. p. and i.p. cylinders c.re usually fitted with 'paw' supports which rest on the bearing pedestals (Fig. 15.1.27). It is the accurate positioning of these brackets which decides the concentricity of the rotor and cylinder. The presence of the condensers poses problems for the support of l.p. cylinders. There are various methods of supporting it. L.P. 'cylinders may be designed to form a stiff structure needing only end support. Alternatively, the gap in the foundation block for the condensers may be spanned by steel beams or by a reinforced concrete arch or by prestressed concrete beams. In order to withstand the compressive stress due to vacuum the condenser and I.p. exhaust casing of large turbine is rigidly connected. To ensure that the weight of the condenser is not carried by the turbine casing, condenser is supported on spring. In the case of large modern sets with three double-flow l.p. cylinders, four condenser

632

Steam & Gas Turbines And Power Plant Engineering Vertical guide Paw support Key

Casing expansion

Pedestal (cold)

Fig. 15.1.27. Cylinder Support

arrangements have been adopted in order to provide adequate rigidity of support for the turbine bearings. These are:(a) axial condenser, (b) bridge condenser (c) pannier condenser, (d) integral condenser. (a) Axial Condenser: In this case, the l.p. turbines are supported on two parallel concrete walls. The generator also rests on the extension of these two walls. Enough space has to be allowed beneath the generator for tube withdrawal. (b) Bridge Condenser: It bridges the gap between the i.p. turbine block and the generator. The condenser tubes are mounted transversely. (c) Pannier Condenser: The condensers are arranged parallel to the l.p. turbines. The tubes of the condensers are mounted axially but withdrawal takes place either side of the generator. The l.p. turbine is supported on a continuous concrete plinth and the basement depth is very much reduced. (d) Integral Condenser: It is a development of the pannier condenser. In this case, the condensers and l.p. turbine outer casings are combined in a single structure containing internal ribs to give it adequate stiffness. It also results a smaller basement depth. It is to be noted that the cylinders of a turbine are rigidly fixed together in the axial direction and a positive location with the foundations is provided at one point only, usually beneath one l.p. exhaust. This arrangement makes the turbine free to expand from this point, and at the governor end pedestal the movement may be 5 cm or more. In order to maintain transverse alignment, central sliding keys are used between the pedestal and the foundations. L.P. Exhaust Casings: The main design requirement of the l.p. exhaust casings is to give the maximum possible area of flow for the exhaust steam in order to keep the pressure drop to a minimum. In order to turn the huge volume of steam through without `bunching', curved vanes are employed to give diffusion action resulting in lower pressure at the blading exhaust annulus than that in the condenser. Insulation of Cylinder. It is essential to insulate the hot external surface of the cylinders in order to obtain nearly adiabatic expansion of steam and to protect the cylinder walls from severe temperature gradients which would cause cracking. The insulation is covered by outer covering known as clading to reflect a certain amount of heat and improves the appearance of the turbine like shape of a cylinder. It consists of planished steel, enamelled steel or aluminium sheet.

Construction, Stress Analysis, Operation and Maintenance of Steam Turbines

633

Cylinder Testing: Turbine cylinder is hydraulically tested after manufacture with a pressure of 50% in excess of the highest working pressure in the section. 15.1.31. Thickness of Cylinder For low pressure the thickness (t) of the cylinder without giving any allowance for diaphragm is given by p_2 t (15.7) 2a Here t and D are in cm, p is gauge pressure in N/cm2 and a is the allowable stress in N/cm2. With allowance for diaphragm' shown in Fig.15.1.28 aD a + - where 1 is in cm t= (15.8) 2a r2 For very high pressure rt

\ /-1 0. CY - p

(15.9)

where r1 and r2 are the internal and external radii. Where rigidity is essential and the diameter of cylinders is more than 76 cm, the thickness of L.P. end may be calculated from + 2.03 cm (15.10) 100 For relatively low pressure, the thickness of the flange should be from 1.5 to 1.75 times the thickness of the cylinder. For high pressure,the ratio may be 3 or more. Flange bolt diameter at the thread root is given by \ T- 4R min (15.11) gab t

where ab = tensile stress at the minimum bolt section, R = force exerted on bolt. 15.1.32. Cylinder Materials. Cast iron is the most suitable material for cylinders. But the use of cast iron above 240°C is not advisable due to phenomenon of growth. The typical cast iron for turbine work has the composition-C 3.3, Si 1.2, Mn 0.8; P 0.2 and S 0.1. Nickel cast iron containing 3.1.C, 1.2 Si, and 1.5 Ni adds strength to the cast iron and gives it good machinability and makes it less porous. Cast steel is suitable for moderate steam pressure and temperature range between 240°C and 433°C. The H.P. cylinder may be of cast steel and L.P. may be of oast iron. For steam temperature more than 420°C, alloy steel is used. This alloy steel contains 0.5 to 0.75 molybdenum and small amounts of nickel and chromium. In especial case, cylinders have been machined out of solid blocks of forged steel. The following is the composition of h.p and i.p.outer casings of the English Electric 200 MW turbine:C 0.15 max; Si 0.2 to 0.5; Mn 0.4 to 0.7; Ni 0.3 max; Cr 0.5 Max; Mo 0.5 to 0.7; 50.05 max; P 0.05 max;Va 0.20 min. The inner casing which is subjected to highest temperature has following composition:-C 0.16 max; Si 0.4 max; Mn 0.4 to 0.6; Ni 0.5 max;Cr 2.75 to 3.25; Mo 0.5 to 0.7;S 0.05 max; P 0.05 max; Va 0.7 to 0.9; W 0.4 to 0.6 15.1(e). Steam Chests and Valves 15.1.33. Steam Chests. Steam chest is fed with steam from the superheater outlet via the high pressure steam main. It normally houses the steam strainer, emergency stop valve

634

Steam & Gas Turbines And Power Plant Engineering

1.1

ll

Ll

w

11 II II zzzzi:WzzzziANA:>zwz); WfwA.

V'

Steam

11 11 11 wzz zi neeffzzeliTzzzz Mpze zezi, Z "0 oosimi,._

/

/ .1 z/ •

Ait li ./

Nozzle

r .A

(a)

Throttle valve

(c)

(b) Fig.15.1.28. Steam Chest.

and governing valves. It is steel casted or solid forged. Fig 15.1.28 (a) shows a steam chest suitable for nozzle control governing. The portion containing the governing valves embodies in the h.p. cylinder casting so that the channels from the governing valves to the nozzle groups may be cored. This type of casting is very difficult and therefore sometimes casted in two parts and welded together. Multi-valve integral steam chest is not suitable for temperatures above 480°C due to development of cracks caused by high local stress as a result of uneven temperature distribution. Fig 15.1.28(b) shows another design of steam chest suitable for high steam condition. It is very simple in construction and flexible enough to allow the turbine to expand freely without undue stress on the flange. In order to prevent solid particles of foreign matter from being carried into the turbine by the incoming steam and causing damage to the blading, the steam passes through a cylindrical strainer which surrounds the stop valve as shown in Fig.15.1.28(b). Fig.15.1.28(c) shows the third type of steam chest which provides better way of controlling overspeed by mounting a single governing valve chests on the cylinder. The disadvantage offered by this arrangement is the difficaltiqes in accommodating satisfactorily the relay gear. 15.1.34. Steam Valves. A steam turbine is equipped with the following valves:— (1) Emergency stop valves to cut-off the steam supply during periods of shut-down

Construction, Stress Analysis, Operation and Maintenance of Steam Turbines

635

and emergency. Fig. 15.1.29 shows various type of steam valves in use. (2) Governing valve to provide accurate control of flow rate of steam entering the turbine to control speed of the turbine. (3) Reheat emergency and intercept valves in the return path from the reheater to prevent overspeeding of the turbine following a trip. (a) Double-beat valve (Fig.15.1.29a): It has two seatings to balance the forces due to steam pressure. It is suitable for most pressures but not for high temperatures as differential expansion between the valve and cage may lead to the weeping problem in the seating. (b) Hollow double beat valve (Fig. 15.1.29b): This type of valve reduces the differential expansion problem of the double beat valve by leading the steam from one end seating to the centre of the valve. The thinner walls promote even heating and therefore the differential expansion is reduced. (c) Spherical valve (Fig.15.1.29c): This valve is a modern development for controlling high temperature steam. A large force is required to operate this valve as the pressure forces are not balanced being a single beat valve, with one seating. (d) Spherical valve with internal pilot valve (Fig. 15.1.29d): The internal pilot valve opens first, equalises the pressure and provides internal pilot valve opens first, equalises the pressure and provides initial fine control. (e) Cylinderical valve (Fig. 15.1.30e):- In this case, the steam pressure is prevented from acting on the back of the valve by a fine annular clearance. (/) Mushroom valve (Fig. 15.1.290:- It has a profiled skirt to give a more linear area/lift relationship. It is suitable for governing valve. (g) Flop valve (Fig. 15.1.29g):- This type of valve is suitable for reheat emergency valve, where the steam pressures are moderate and the specific volumes are large. This leads to larger valve diameter.

(a) Double Beat

(f) Mushroom

(b) Hollow Double Beat

(e) Cylinderical Fig.15.1.29.

(c) Modern Spherical

(d) Spherical with Internal Pilot

(g) Flap Valve

636

Steam & Gas Turbines And Power Plant Engineering

One of the main parameters in the design of the valve is the selection of steam velocity through valve. The diameters of valve opening are generally calculated to give maximum steam velocities of about 60 m/s for emergency valve and about 120 m/s for governing valves. In order to resist wear the mating annular faces of valves and their seats are nitrided or faced with stellite. Such wear is more due to erosion than the mechanical impact. The seating upon which any such valve closes is invariably part of a removable sleeve which is replaceabe when worn. 15.1.35. Couplings. The couplings are needed to couple two shafts for the following two main reasons— (i) The limiting length of the shaft which is possible to forge in one piece. (ii) Use of different materials for the various rotors in view of the various conditions of temperature and stress. The functions of coupling are the following— (a) to transmit torque from one shaft to another (b) to a allow relative angular misalignment (c) to transmit axial thrust from one shaft to another (d) to ensure axial location or to allow relative axial movement. There are three types of coupling. They are— (i) Flexible, (ii) Semi-flexible and (iii) Rigid.. JO/Flexible coupling- It absorbs small amounts of angular misalignment as ,ell as axial movement. Double flexible couplings can also accommodate eccentricity. Fig. 15.1.30 shows some designs of flexible couplings. The claw couplings is robust and slides easily when transmitting light load but on heavy load, it becomes axially rigid due to friction. For medium size load, Bibby couplings are suitable. The multitooth couplings transmit torque by external and internal gear teeth of involute form. Flexible couplings require continuous lubrication. A jet of oil feeding into an annular recess is led centrifugally to the coupling teeth through drilled passage ways (ilYSemi-flexible Coupling- It allows angular bending only and requires no lubrication. It consists of a bellows-piece having one or more convolutions as shown in Fig.15.1.31. It is interposed between the turbine and generator. fiii) Rigid Coupling- The use of flexible couplings becomes impracticable on the large turbines because high torque is to be transmitted. This requirement is met by the use of

(a) Claw Coupling

(b) Multi. Tooth Coupling Fig.15.1.30. Couplings.

(e) Biby Coupling

Construction, Stress Analysis, Operation and Maintenance of Steam Turbines

Fig. 15.1.31. Semi-flexible Coupling

637

Fig.15.1.32. Rigid Coupling

rigid couplings between the turbine cylinders, so that the turbine shaft behaves as one continuous rotor. Fig.15.1.32 shows such a rigid coupling in which a spigot locates the two half couplings and the flanged are joined by the numbered bolts. 15.1.36. Bearings. Two types of bearings are employed in the steam turbine. They are— (a) Journal bearings and (b) Thrust bearings (a) Journal Bearings. The function of the journal bearings is to support the shaft with the help of formation of oil wedge. Each section of the shaft is normally supported on two bearings although the advent of solid couplings now makes possible to use only one bearings between two cylinders in order to save length land bearing losses. Fig. 15.1.33 shows a plain white-metalled journal bearings which are invariably used because of their high loading capacity, reliability and absence of wear due to hydrodynamic formation of high pressure oil wedge between the white metal and the shaft which ensures no metallic contact between them as a result of hydrodynamic lubrication. The oil is continuously fed into the wedge by frictional drag and leaks away towards the sides of the bearing(Fig. 15.1.34) , Shim for adjusting alignment WWI/

Keep pin to prevent rotation

White metal Additional clearance for cooling oil Lubricating oil supply

Minimum film thickness

Oil guard

Oil wedge

Fig. 15.1.34. Formation of High Pressure Oil Wedge.

High pressure oil connection

Exit spillway for oil Fig. 15.1.33. Journal Bearing

638

Steam & Gas Turbines And Power Plant Engineering

In order to remove the heat produced in the bearing, the journal is continuously flushed with oil and for this purpose, a large clearances are allowed. Annular recesses are made at each side of the bearing which collect the side leakage of oil; and from these it drains into the pedestal and flows out through sight boxes. In order to prevent oil from dashing along the shaft and reaching the glands, oil-guards and oil throwers are provided on the bearing. The shell of the bearing is spilt horizontally and dowelled. There .are four pads fitted with shims on which the shell rests. A small adjustment to the alignment may be made by changing these shims. The pads may be spherical or cylindrical but the former is meant for self-alignment. The white metal lining may be thick and thin. Dovetails are used to attach the thick white metal lining to the shell. There are many drawbacks of this arrangement. (i) The white metal is subjected to fatigue due to its compressibility and changes in thickness. In order to safeguard against these, the thickness of expensive white metal should be considerable. (ii) Voids and stress-risers caused by dovetail corners make adhesion poor. On the contrary,thin uniform white metal lining adheres to the shell by surface tinning and now modern bearings employ this due to its obvious advantages over thick one. There is provision made in the bearing at its bottom to admit high pressure jacking oil to lift the journal when starting from the rest in order to prevent wear and reduce the starting torque of the turning motor. To ensure that the high pressure oil does not force the white metal away from the shell, the oil pipe is usually burried in the white metal. (b) Thrust Bearings. The function of Oil outlet thrust bearing is to provide a positive location for the rotor relative to the casings Adjusting and withstand thrust due to blade reaction spacers and steam pressure acting on unbalanced areas. Two types of thrust bearings are in common use namely multi-collar and Michell tilting-pad thrust bearings. The former is suitable for small turbines while the latter for large turbines. Oil inlets The Michell tilting-pad bearing use pads to thrust which are white-metalled and loosely fastened to removable half-rings and automatically tilt to the appropriate angle to build up a wedge of oil (Fig 15.1.35). In the case of rigidly coupled shafts Fig. 15.1.35. Michell Tilting-Pad Thrust Bearing only one thrust bearing of non- adjustable type is required where as flexlibly coupled shafts need one bearing to locate each shaft. 15.1.37. Turning Gear. There is an improper cooling of the rotor when the turbine comes to rest after running. This is because of the fact that the cooler steam being more dense tends to collect in the lower half of the rotor cools more quickly than the upper half. This improper cooling of the rotor causes the shaft to "hog" if special precautions are not taken. It is the use of turning gear which overcomes this problem. The purpose of turning gear is to keep the shaft turning slowly until cool when the turbine comes to rest. There are various types of turning gear. One type consists of an

Construction, Stress Analysis, Operation and Maintenance of Steam Turbines

639

electric motor driving train of reduction gearing, the last pinion of which meshes with a gear ring on one of the shaft couplings. As soon as the turbine comes to rest a gear lever is operated which slides the pinion to mesh along its splined shaft. At the same time, a motor driven flushing pump is started to provide lubricating oil to bearings and jacking pump is also started to support the shaft and thus to reduce the starting torque of the motor. The speed of rotation varies from 3 to 30 r.p.m but sometimes upto 100r.p.m is also selected to cool the shaft uniformly by creating turbulence within the cylinder. In the recent type of turning gear, the engagement is obtained automatically as the turbine shaft slows down and reaches barring speed. This renders full hydrodynamic lubrication in all bearings and is achieved by a synchro-self-shifting clutch. Other four types of turning gear are the following;— (a) It turns the shaft intermittently through 180°,thus saving electrical power. (b) It uses a water jet impinging on the 1.p. turbine blades. (c) It uses a special low-speed high torque electric motor mounted directly on the turbine shaft. (d) The generator itself provides the low speed drive of the shaft if the excitation is achieved by a special manner. This method is still under test. 15.2. Stresses in Turbine Blades And Rotors The ultimate aim of any engineering theory is to design the machine or machine parts. From the knowledge of previous articles based on thermodynamic and aerodynamic analysis the main dimensions such as inlet and out blade angles, blade height, pitch of the blades, etc. can be found out. It is also very essential to check the blades from the strength (stress) point of view. Turbine blades are subjected to centrifugal and bending stresses. The rotor is subjected to radial and tangential stresses. Now 3D Finite Element Method (FEM) based software is available for stress analysis and design of all mer, chanical parts such as rotors, blades, etc. The following analyses are valid for both steam and gas -_--/ ----turbines bladings and rotors. 18r _ t - - -- 15.2.1. Centrifugal Stress in Turbine Blades of Uniform Cross-Section. (A) Blades. Blades. Fig.15.2.1 shows a turbine blade having a constant cross-sectional area of A in m2, a mean diameter of D in m and a length of / in m. Now, let r, p = density of blade material, (kg/m3), N = rotational speed, co = angular velocity, (radians/s), F = internal force in blade at radius r, (N), F+ SF = Fig.15.2.I. Impulse Turbine internal force in blade at radius r + Sr,(1V). Blade Element Hence, volume of the blade element = A Sr m3 Centrifugal force on the blade element. =

mass x (linear velocity)2 — p A CO 2 r Sr r

By balancing the force F+ SF + pAco2r. Sr = F or

OF = — pAco2r8r

Steam & Gas Turbines And Power Plant Engineering

640

If

F = internal force at root, F2 = internal force at tip., due to centrifugal force on shrouding., N Then, integrating of the above equation between the limits r1 and r2 gives Awe ( 2 2) p A 032 D .1 — — F I —F2 = p A w 2 12 rdr — p 2 r2 r i

F2 The stress due to shrouding = a = — a

N/m2

A

The stress due to the inertia of the blade alone is

C

p.co2 .D.I 2

A

Nhil 2

(15.12)

For steel blades, p = 7970 kg/m3 a e—

7970 x (271N/60)2 D _ 43.7 D/ N2 2

The resultant centrifugal stress at the root of the blade is equal to the sum of the stresses a and as. It will be uniform stress if the centroids of the blade sections lie on a radial line. 15.2.2. Bending Stress in Symmetrical Impulse Blades of Uniform Cross-Section. We have seen that there is a centrifugal stress at the innermost blade section. In addition to this stress, there will always be a stress due to bending caused by impulse. Though, this bending stress is not large but its effect on maximum tensile stress in long blades has to be taken into account. To find this, the bending moment across the blade root must be found first. Blade heights are divided into a number of equal parts, say four equal parts as shown in Fig 15.2.2. The flow rate 1;11 ,1i:2,1'1;3 and th4 can be calculated through each part. For each part

Ui

14FL Ull U

(a)

(b) m4

(c) m3

ni2

• m1

M

(d) Fig.15.2.2. Bending Stress in Impulse Blades of Uniform Cross-Section

Construction, Stress Analysis, Operation and Maintenance of Steam Turbines

641

of the blade with peripheral velocity upu2 , etc. equal to that at the midheight of the part in question, the velocity diagrams are drawn. This gives the velocity changes C I ,C2, etc. If n is the total number of blades on the wheel, the impulse forces are tit 1 C1 in 2C2 , etc. and F2 — F= Therefore, the bending moment due to impulse forces are M1 = F1 11 and M2 = F2 /2 etc. Now, the resultant bending moment M across R-R is found out vectorially as shown in Fig. 15.2.2. Generally, the blades of an impulse turbine are profile blades with unequal angles. For these types of blades the axis X-X will be the axis of symmetry. (Fig. 15.2.3.) Let point G be the centroid of the blade section and axis Y-Y be perpendicular to X-X. Here X-X and Y-Y are the principal axis of the section. Assuming that the plane of resultant bending r ' moment is inclined at the angle 0 to XX, it can be resolved into two components Mx and My. As it is clear from the figure that Mx will produce tensile stresses in quadrants I and IV, compressive stresses in II, and III; while M will produce tensile stresses in III and IV. Therefore, we can conclude, that generally, the point A in quadrant I will be subFig. 15.2.3. Bending Moment Diagram jected greatly to tensile bending stress and on Impulse Blade. thus can be calculated by the equation Mx

ab

= /y

x

My .y I

(15.13)

The maximum tensile stress will also occur at A (Fig. 15.2.3) and is equal to 6max = 6s

-F6 -Fa c b

(15.14)

15.2.3. Stress in Thin Rotor Rotating Ring. Turbine rotors are designed on the basis of strength calculation. Therefore, correct estimation of rotor stresses is of great importance. The stresses in turbine rotors are due to the inertia of the rotor mass and this depends on the form of the rotor. Generally, the stresses developed are very high which limit the rotor design. Selection of type of turbine rotor depends on many factors, such as, use to which the turbine is put, type of turbine, speed, mass of rotor, stress developed, temperature, etc. A thin ring is one whose thickness in the radial direction is very small compared with the diameter. Let us consider such a thin ring rotating with angular velocity w radians per second about its central axis X—X. Let, t =radial thickness of the ring, (m), r =mean radius of ring, (m), u peripheral velocity, (m/s), p =density of material, (kg/m3), at, =stress in thin ring (tangential or hoop stress)-EN/m2), F =centrifugal force, (N), T = hoop tensions, (N),

Steam & Gas Turbines And Power Plant Engineering

642

. —•x

x—•

se

T Force diagram

Fig. 15.2.4.. Stress in Thin Rotating Ring Let us also assume that the length of the ring in axial direction is unity as shown in Fig.15.2.4.. The cross-sectional area of the small element shown by hatching = t m2. Volume of the element subtending the angle 60 at the centre, V = t.r.60 m3. Mass of the element, in = p .r.t. 60 kg Centrifugal force acting on the element = F = m co2 r = p.t 60 . o.)2 . r2 ; N As shown in the force diagram the element is in equilibrium under the action of the force F and the hoop tension T. The resultant of the two hoop tensions at subtended angle 60 with each other is T. 60 and the action is inwards. Therefore, T. 60 = p . t. . 60 . co2 r2 or The hoop stress = ira =

T = p. t. w2 . r2 .

_ n 032 T Area of ring

r2

N/m2

N

(15.l5)

Since wr = u since u is in m/s. = P u2,

N/m2

(15.16)

15.2.4. Stresses in Drum Type Turbine Rotors. Hoop stress in turbine rotor of drum type = stress due to rotor only+stress due to blade only r . Fb hoop

= p u2 +

A

N/m2

(15.17)

where Fb is the centrifugal force on blade 15.2.5. Stresses in Rotating Discs of Variable Thickness. Fig. 15.2.5. shows a steam turbine rotor with variable thickness. Let us consider a small element ABCD of a turbine wheel which is bounded by radial lines AB, CD and arcs AD,BC as shown in the figure. On the edges AD and BC of the element, the inertia forces due to the blades, the rim and the disc itself will produce direct stress known as radial stress. This stress induces certain

Construction, Stress Analysis, Operation and Maintenance of Steam Turbines

643

Disc

Fig. 15.2.5. Element of Rotor With Variable Thickness. direct stresses called hoop or tangential stress on the radial edges AB and CD of the small element. A torque T is also applied by the blades to the rim of wheel which produce a shear stress at = T/2it . r2 t on the faces AD and BC and a2 .= ai on the faces AB and CD. Note that shear stress is very small compared to the direct stress and in some cases if necessary it may be neglected. Let ar = radial stress., (N/m2), a, = hoop stress., (N/m2), r = radius of disc at any point, (m), t = thickness of disc at radius r, (m), µ = Poison's ratio, E = young's modulus, p = density of rotor material, (kg/m3). Starting from first principle the general equation for stress is 2 r3 (1 _ ) dt dt i jcly r [rldy 2 t dr dr t dr2 (15.18) p 0)

1

Equation (15.18) is a general equation of stress for a turbine rotor with variable thickness. Its derivation is out of scope of the book. 15.2.6. Continuous Disc of Constant Thickness. A continuous disc is one which has no hole at the centre, i.e, disc and shaft are made from a single piece by forging. Let us consider such a disc and assume that disc is not subjected to any edge loading i.e. the forces imposed due to masses attached to its periphery are absent. Using the expression (15.18), the stresses are given as follows. 3 +_,a Hence, at any radius r; ar = — 8

p

and a = 8 [(3 +1.1)R22 — (1 + 30r2]

w2

(R2 _ r2) 2

(15.19) (15.20)

For a disc having radial stress imposed at its periphery due to attached masses, the constant may be found as before, except that the applied stress at periphery will be equal to ar.

644

Steam & Gas Turbines And Power Plant Engineering

15.3. Steam Turbine Operation The following are the sequences of turbine operation. (A) Starting Sequence— (1) Application of control power illuminates all of the malfunction lights. This provides a check of the malfunction lights before starting the turbine. (2) Reset malfunction circuit by operating a reset switch. Malfunction lights go off and all control devices assume the condition for starting. (3) Inspect the governor mechanism, till all greese cups and oil where necessary. (4) Open the boiler stop valve to permit heating of the line and avoid condensation in the line. (5) Open header, separator, throttle and turbine casing drains. (6) Start auxiliary oil pump. This has to be stopped when the main oil pump starts delivering oil at normal pressure. (7) Adjust needle valve to secure required oil pressure(1.3 to 3 bar) for the bearings. (8) Start the circulating water pumps and dry vacuum pumps of the condenser. Operate the condensate extraction pumps as found necessary to remove water during the warming-up period. (9) Turn on the turbine steam or water seal. (10) Turn on the water to the generator oil cooler and other water requiring parts. (11) Keep open all the drains ahead of the throttle valve until all water of condensation has been removed. (12) Open the throttle or governor valve quickly to set the rotor in motion. (.13.) In order to check up whether the tripping mechanism operates properly or not and to prevent the turbine from accelerating too rapidly, operate the overspeed trip valve by using the hand lever as soon as turbine starts rolling. (14) Reset the emergency overspeed valve and before the turbine comes to rest, adjust the throttle so that the turbine will operate between 200 and 300 rpm. (15) While the rotor is in slow motion, observe any rubbing or mechanical difficulty by using a metal rod or listening device. (16) As soon as the temperature of the oil leaving the bearing reaches about 48 to 50°C, start the circulating water through the oil coolers to maintain bearing oil temperature. (17) Increase the speed gradually. Follow manufacture's instruction. (18) Adjust the water seal on the turbine and the atmospheric relief valve. (19) Once the machine comes under the control of governor test the emergency governor by opening the valve in the oil line to it. See that all valves controlled by this tripping mechanism close promptly. Reset open throttle valve and restore speed to normal. (20) Close all drains. (21) Open leak-off from h.p. side gland in order to flow any excess steam to the feed water heater or to one of the lower stage of the turbine. (22) Synchronize the generator and tie it in the line. (23) The speed is now under the control of governor. The turbine is now ready for load and is regulated from the turbine control panel.

Construction, Stress Analysis, Operation and Maintenance of Steam Turbines

645

(B) Sequence When the Turbine in Motion. (1) Apply the load gradually. (2) Check up the oil pressure going to the bearings and control gear. (3) Observe the oil bearing temperature. (4) Observe the turbine for any noise, vibration by watching the vibration and other indicators. (C) Sequence When Shutting the Turbine Down. (1) Gradually reduce the load to zero. (2) Start the auxiliary oil pump and make sure that oil will be supplied to bearings at proper pressure while the turbine is coming to a stop. (3) Trip the emergency valve. (4) Close the leak off from the h.p. glands. (5) Stop the supply of cooling water to the condenser. (6) Shut .down the condensing equipment and open drains on turbine pipings and casings. (7) Continue auxiliary oil pump in operation until the turbine rotor has stopped. (8) Operate turning gear to rotate rotor at about 3 to 30 rpm for sometime. During operation, it is good practice to keep a log sheet and record the hourly readings of the instruments. Some of the readings which might prove valuable are the following— Load on the generator throttle, steam pressure and temperature, exhaust pressure, temperature, of cooling water entering and leaving the cooler, bearing oil pressure and temperature, the throttle steam flow rate, speed,. frequency, vibration level. The main requirements of a steam turbine while in operation are the proper application of oil to the bearings and a continuous flow of cooling water. 15.4. Maintenance of Steam Turbines. 15.4.1. Aims and Objectives of Maintenance. The purpose of maintenance is to maintain the steam turbine in order to achieve as high a plant availability as possible at the minimum cost. The maintenance of large modem unit must be organised to minimise outages which affect availability. If the outages are unavoidable, the maintenance work should be planned so as to return plant to service in the minimum space of time. This may require round-the clock working on some activities. Plant efficiency is also a major aspect of maintenance team and it is the duty of maintenance department to keep a close liaison with the planning and operation departments in observing the plant performance. These observations will help in arranging the maintenance on plant short of its rated efficiency. It is essential to keep a close watch for faults which are inherent in the design of any equipment. These are generally known as 'type faults'. After careful study, its design may be improved with the help of manufacture. Maintenance of steam turbines used in power station may be classified broadly in three categories. The are as follows— (i) Rectification of defects- The deflects may or maynot be urgent. If immediate repairs are needed, these may be carried out on running units or the units should be shut down. If the repair is not urgent, it may be carried out in planed programme of future work.

646

Steam & Gas Turbines And Power Plant Engineering

(ii) Planned preventive maintenance or running units: (iii) Planned preventive maintenance on shut-down units, e.g. statutory surveys and major overhauls. Each category needs a different planning approaches and the maintenance is carried out daily, weekly, monthly and yearly by maintenance department. For rectifying the defects in a proper way, a "Defect Card" system is used which is in the form of standard pattern. The defects are posted in the defect cards and reported to the maintenance department for their rectification. (ii) Planned Preventive Maintenance on Running Units.A system of routine maintenance is adopted to prevent the defects and breakdowns. The planning of a such system needs careful consideration in order to arrive at the optimum levels of maintenance. The actual work content of each maintenance routine can only be determined after experience with the plant over a period of time. After carrying out the preventive maintenance, it is recorded on a Record Card' with full particulars of the items attended. (iii) Planned Preventive Maintenance on Shut-down Units- In order to minimise the maintenance costs of shut-down units, it is necessary to plan the work to be done carefully in advance sothat the duration of the plant shut-down is reduced to a minimum. This planning process should start some weeks or months before the shut-down is due to take place. A work list is prepared and based on this it is possible to programme the course of the overhaul; and the technique for critical path analysis is now generally used for this purpose. A record is maintained after doing the maintenance for future use. 15.4.2. Steam Turbine Overhaul Following are the predictable factors which influence the frequency of plant outage:— (i) Progressive deterioration in output on account of blade deposits. (ii) Erosion of 1.p. bladings. (iii) Retightening of nuts and bolts. (iv) Statutory inspection of associated pressure vessels and boiler plant and planned outage of the transmission system. .(v) Availability of manpower to carry out the planned overhual within the specific time. It is to be noted that the unscheduled turbine outage is a most serious aspect of maintenance due to extra costs of operating less efficient plant. When mechanical defects such as excessive shaft eccentricity, bearing vibration or excessive bearing temperature are noticed, unscheduled outage must follow. A common reason for this type of outage is rotor unbalance due to displacement and erosion of blade shrouding, and lacing wire and blade erosion. Deposits on the moving blades produces the same effect. Another common source of outage is the misalignment of shaft due to changes in the levels of bearings on account of foundation settlement on changes in pedestal support length. Following is the sequence of events of a typical turbine overhual: (1) Remove pipe works mounted on turbine and all top half cylinders and covers. (2) Remove coupling bolts add bearing covers leaving thrust bearing assembled. (3) Measure all releveant blade and gland clearances on the horizontal joint, with feeler or tapered guages.

Construction, Stress Analysis, Operation and Maintenance of Steam Turbines

64'/

(4) Remove the thrust bearings and then the turbine rotors to have a detailed examination of fixed and moving blades and diaphragms gland segments, casing, bearings, bolts, etc. concerning any type irregularities and damages. (5) Re-form all gland and baffle segment knife edges and restore all radial clearances. Examine the bedding of all bearings to journals and measure oil clearances. Check the adhesion of write metal of journal bearings. (6) After cleaning the blade deposits by water washing, chemical washing or blasting process, refit the rotors and measure clearances on the bottom points of glands and blades with lead or plastic strip. Measure similar clearances on the horizontal joint and compare for eccentricity of the shaft in the casing. 447) After cleaning the bearings, check the wear-down of bearing with the help of appropriate bridge gauge and feeler guages and compare the figures with those taken on the erection or previous overhaul. (8) Check the alignment of shafts by taking readings on all couplings and also record the level of all journals. (9) Refit top half cylinders and before bolting-up take a further set of top blade and gland clearances to confirm the concentricity of the rotor within the cylinder. (10) Remark the horizontal casing joints and refit all heavy parts. After this, take the readings for final coupling alignment and adjust it if necessary. (11) After refitting the coupling bolts, refit the thrust bearings and all other bearings covers. (12) Inspect turbine governor gear with stop, throttle and intercept valves. (13) Dismentle other accesories and mountings if necessary and inspect service and restall these. Clean or replace the oil filters and other items of lubricating system. Replace the (14) lubricating oil if necessary. (15) Inspect all the measuring instruments installed on the turbine and replace if necessary. EXERCISES Viva-Voce and Theoretical 15.1. What is the type of nozzle that is generally used in impulse turbine? 15.2. What is the method of manufacturing of the built-up nozzle? •15.3. What are the methods of manufacturing of turbine blades? 15.4. What is the reason that the blades of the L.P. stages are twisted? 15.5. Why are the long blades objectionable? 15.6. What factors influence the selection of blade materials? 15.7. Mention the suitable blade materials. 15.8. Mention the method of attachment of.turbine bladings to the rotor. 15.9. Which type of attachment is generally used in impulse turbine for moderate speed and small and medium power? 15.10. What are the methods of manufacturing the steam turbine rotors? Discuss type of rotors. 15.11. Mention the rotor materials. 15.12. What type of steam turbine casing is generally used in practice? Discuss them.

648

Steam & Gas Turbines And Power Plant Engineering

15.13. Mention the cylinder materials. 15.14. Discuss Inverted T-Attachment, serrated blade root attachment, Parson's integral blades and fir-tree. 15.15. What is the function of couplings? Mention their types? 15.16. What is the function of bearings? Mention the types of bearings used in steam turbine. 15.17. State the function of turning gear. 15.18. What do you mean by balancing of rotors? 15.19. Discus the critical speed of rotor. 15.20. Discuss the causes of vibration of turbine blades. 15.21. Discuss the sequence of steam turbine operation. 15.22. Write an essay an maintenance and operation of steam turbine. 15.23. Discuss the various type of couplings used in steam turbine. 15.24 Discuss the various types of steam valves used in practice. 15.25. Discuss the requirements of steam chest. 15.26. Derive the expression for stresses in rotor and bladings. 15.27 . Develop a software for the design of rotor and bladings.

16 Condensers And Cooling Towers

Condenser is one of the main elements of steam power plant. About 50 to 60% of the energy associated with steam is lost in condenser. This suggests that a comprehensive thermodynamic study of condensers in essential which will help in its designing. Cooling tower is one of the main elements of cooling system where the cooling water is to be circulated in cooling the steam in condenser. This chapter deals with the thermodynamic aspects of condensers, cooling system and cooling towers. 16.1. The Function of a Condenser The workdone and efficiency of a steam turbine plant are increased if the back or exhaust pressure of the turbine is reduced. This is so because of the fact that the average temperature at which heat is rejected in the cycle is reduced. It can be made possible by employing a condenser in which steam exhausts and get condensed. Fig. 16.1. shows the effect of reducing the back pressure from pb (atmospheric pressure) to pit,. The workdone is increased by hatched area due to reduction in back pressure. The back pressure can easily be reduced and maintained below atmospheric pressure by condensing it in a closed vessel as after condensing the steam volume gets reduced tremendously and vacuum is maintained and only latent heat is absorbed by cooling water. For example, one kg of steam at 0.1 bar occupies 19.9 in3 whereas after condensation it occupies only 0.001016 m3 of liquid. Further, little work is needed to discharge the condensate. From above discussion, it follows that the minimum condenser pressure is limited by the atmospheric condition. Thus condenser is defined as a closed vessel in which exhaust steam from steam turbine is condensed by cooling water and vacuum is maintained, resulting in an increase in workdone and efficiency of the steam power plant and use of condensate as the feed water to the boiler. Based on the above discussions, the advantages of a condenser incorporated in steam power plant are the following— (i) Improved workdone and efficiency due to low pressure (vacuum) of condenser. (ii) Recovery of the condensate to be fed to the boiler as a high quality feed water for reuse. (iii) Reduced steam consumption for the same power output due to increased workdone.

650

Steam & Gas Turbines And Power Plant Engineering

T pb = 1 atm

po Increase in workdone due to condenser

S

(a)

(b)

Fig. 16.1. Effect of Condenser Pressure on p-v and T-s diagrams. (iv) Reduced thermal stresses due to high temperature of feed water entering to boiler. (v) Economy in water softening plant as only make-up water is to be treated instead of full feed water. 16.2. Cooling system. The cooling system of condenser is one of the most important system of power_ plant. The cooling medium may be water or air. The cooling water system may be open or closed. The availability of water plays a major role in deciding the plant size. About 50 to 60% of the heat supplied in the cycle is rejected in the condenser. The cooling water requirement in open system is about 50 times the flow of steam in the condenser. Even with closed system using cooling towers the requirement of cooling water is considerably large, about 5 to 8 kg/kWh. This means that 2000 MW station will need at least 24 x 104 tons of water per day and out of this about 5% is evaporated in cooling tower, so make-up cooling water needed will be 12 x 103 tons per day. Since the cooling water takes the latent heat of steam, hence the temperature of water may increase upto 10°C. and the condensate temperature is the saturation temperature of steam at condenser pressure. In a closed system it is necessary to precool this in cooling tower. In open system (suitable where large reservoir is available), the water gets cooled in long pipe or open channel before entering to reservoir (lake, river, etc.). The circulating cooling water system and its type are discussed in subsequent chapter. 16.3. Elements of a Water Cooled Condensing and Cooling System Fig. 16.2 shows the main elements of a steam condensing unit (closed). The main elements are :(a) A closed vessel, condenser in which steam condenses. This is a heat exchanger. (b) A dry Air Extraction Pump-(AEP) whose purpose is to remove air (leaked into condenser through joints) and other non- condensable gases from the condenser in order to maintain vacuum constant. (c) A Condensate Extraction Pump (CEP) which extracts the condensed steam collected in hot-well of the condenser and pumps it to the feed water line. Sometimes, a wet air pump serves the purpose of CEP and AEP. (d) A Circulating Cooling Water Pump (CCWP) to circulate the cooling water in

651

Condensers and Cooling Towers

steam

Air + water vapor

CCWP : Circulating cooling water pump

Cooling tower

Make-up treated water Air extraction pump (AEP)

Feed to boiler through feed heaters

Condensate extraction pump (CEP)

Hot-well of condenser

Filter Make-up cooling water

Fig. 16.2. Main Elements of a Condensing Unit. condenser. (e) A Boiler Feed Water Pump (BFP) to pump the condensate to the boiler and is located below the deaerator (f) A cooling tower or spray pond to recool the circulating water in the condenser which is heated due to condensation of steam in a closed system. Cooling tower is essential where there is a scarcity of water. Cold and hot water flow in open channel with flow gates in open system. (g) A relieve valve to relieve the steam from the condenses when condenser is not in working order. Using this, the plant becomes non-condensing. (h) A makeup cooling water pump with screen at its inlet to supply make-up cooling water to the condenser as 2 to 5% of cooling water gets lost due to evaporation in cooling lower. 16.4. Types of Condensers. Broadly, condensers are classified in two categories. (i) Direct contact type, where the cooling water and steam directly meet and come out as a single stream. (ii) Surface condensers (Indirect type) where there is no mixing of cooling water and steam. It is a shell and tube type heat exchanger. The heat released upon condensation is transferred to circulating cooling water through the walls of the tubes. 16.5. Direct Contact Condensers They are classified into three categories (a) Spray condenses, (b) Barometric condenser and (c) Jet condenser 16.5.1. Spray Condensers. Fig. 16.3 shows the schematic of a spray condenser. The cooling water is sprayed into the steam and due to direct mixing, the steam gets condensed and mixed with cooling water. The exhaust steam from the turbine at state 2 mixes with cooling water at state 5 to produce saturated water at state 3, which is pumped to state 4. A

652

Steam & Gas Turbines And Power Plant Engineering Turbine exhaust

Dry cooling tower

2

4 To plant feedwater system

3 Condensate

Pump

s (b)

(a)

Fig. 16.3. Spray Condenser. part of the condensate (m4), equal to the turbine exhaust flow (m1 ) is sent back to the plant. The remaining condensate is cooled in a dry cooling tower to state 5 which is finally sprayed in the condenser on entering exhaust steam. The main drawback of direct contact spray condenser is the requirement of very high quality cooling water since a part of the condensate is used as feed water to the boiler. The mass and energy balances yield s

rn2

h2 — h 1 h3 — h5

as

m 2 = M4

(16.1)

16.5.2. Barometric Condenser. The schematic of barometric condenser is shown in Fig. 16.4(a). In this case the cooling water is made to fall in a series of baffles to expose large surface area for the steam fed from below to come in direct contact. Condensation of steam takes place and the mixture falls in a tail pipe to the hot well below. As a consequence of its static head, the tail pipe compresses the mixture to atmospheric pressure. Thus the height of tail pipe (H) is expressed as Path, Pcond + APf

= Pg H

where, p is the density of mixture and Apf is the pressure drop due to friction. For low value of Apf , H is around 9.5 m. Barometric condenser also suffers from the same drawback of spray condenser. Mixture is split and cooled in the same manner of spray condenser. 16.5.3. Jet Condenser. Fig. 16.4 (b) shows the schematic diagram of a jet condenser. Baffles of barometric type are replaced by cascade and to reduce the height of tail pipe, diffuser is provided. The diffuser acts on the same principle as the diverging section of a convergent-nozzle in subsonic flow. In this case also, the mixture is split and cooled in the same manner as in spray condenser. 16.6. Surface Condensers. Invariably, almost all the steam power plants employ surface condensers. They are essentially shell and tube type of heat exchanger where cooling water flows through tubes and exhaust steam fed into the shell surrounds the tubes. As a result of heat transfer from the steam to tube wall and then to cooling water, steam condenses outside the tubes. Fig. 16.5 shows the schematic of a surface condenser with two passes on the water side. The shell is made of steel with water boxes on each side. The right water box is divided to allow for two waterpasses. Tube sheets are provided at each end into which the

Condensers and Cooling Towers

653

Noncondensables to steam-jet air ejector Turbine exhaust Cooling water Baffles

-

i

Cooling water

Exhaust steam from turbine

Mixture

H

Tail pipe

Condensate

Hot well

Fig. 16.4(a). Barometric Condenser.

Fig. 16.4(b). Jet Condenser.

water tubes are rolled and also prevent leakage of circulating water into the steam. An expansion joint is provided which allows different rate of expansion between the tubes and shell. In order to provide support to the long tubes and to prevent tube vibration, vertical plates at intermediate.points between the tube sheets are provided. The hot well provided at the bottom acts as a reservoir of the condensate with a capacity equal to the total condensate flow' for a certain period of time (say five minutes) from where the condensate is extracted by a extraction pump and fed to the feed line. Condensers are normally designed with one, two or four cooling- water passes. It is the number of passes that determines the size and effectiveness of a condenser. Four passes are seldom used in utility plants. A single pass condenser is one in which cooling water flows through all the condenser tubes once from one end to another (Fig. 16.6). A Turbine exhaust Exhaust neck

Steam'done

Support plates

Shell Tube sheet Water in

- Water 4. out

r

Undivided water box

Tubes

Divided water box Hot-well

Condensate

Fig. 16.5. Schematic of a Two-Pass Surface Condenser.

654

Steam & Gas Turbines And Power Plant Engineering

single pass condenser with the same total number and size of tubes, i.e., the same heattransfer area, and with the same water velocity, requires twice as much water flow but results in half the water temperature rise and thus lower condenser pressure. From the above, it follows that such a single-pass condenser is good for plant thermal efficiency and reduces thermal pollution but requires more than twice the water and hence four times the pumping power. Two-pass condensers are more common in power plants. In order to facilitate the cleaning or repairing of tubes during operation in section of condenser, water boxes are often divided beyond the divisions required by the number of passes. A divided water box single pass condenser, for example, may have a partition in both inlet and outlet water boxes at opposite end of the condenser. Similarly, in the case of divided two-pass condensers the water boxes are divided into four quarters. Divided water boxes have duplicate inlet and outlet connections, each with its own circulating water circuit and the valves provided in the division plants permits backwashing by reversing water flow for cleaning purposes. The current design philosophy of condenser is to have a tube layout in the shape of funnel with most tubes and the largest tube passage area, where the steam enters from the

turbine. With the condensation of steam and consequent decrease in volume, there are fewer tubes and smaller passage areas. Steam is made to enter near the tube bundle or bundles from all sides toward a central air cooler for deaeration (below). This results in a low and balanced (to avoid cross-flow) pressure drop. An expansion joint is usually provided between the turbine exhaust and condenser steam inlet. The tube layout may -have triangular, square or irregular arrangement.

Large power plants, usually have two or more low pressure turbine sections in tandem. The condenser may be divided into corresponding sections or shells, situated below the low pressure turbine sections. In the event of same exhaust pressure in all sections, i.e., when the turbine exhaust ducts are not isolated from each other, a single pressure condenser is used. If the turbine exhaust ducts are isolated from each other, these individual condenser shell pressures will increase as it flows from shell to shell. Under this circumstances, multipressure surface condensers are used. It is worth to note that a multi-pressure condenser results in efficiency improvelExhaust steam Tubes sheet

Tubes

V

7

Cooling

water in Inlet water box

Hot-well

Cooling water out Ou let water box

Condensate pump

Fig. 16.6.. Single Pass Surface Condenser.

655

Condensers and Cooling Towers

ment because the average turbine back pressure is less compared with of a single-pressure condenser. Multi-pressure condensers are more commonly used in nuclear power plants. A typical modern two-pass surface condenser for a large steam power plant is shown in Fig. 16.7. Steam enters the tube bundles in two separate sections from the top, sides and bottom, and flows toward the centre of the tube nest in each section. At that point most of steam has condensed leaving air and other non- condensable gases which are cooled and removed by air extraction system. The side view shows the tube layout and baffles provided in the condenser. Fig. 16.8 shows a two-pass divided box surface condenser. The tube material selected has to withstand the temperature difference, pressure differential and load. The tube material can be (a) cupronickel (70% copper, 30% nickel), (b) aluminium brass (76% copper, 22% zinc and 2% aluminium, (c) aluminium bronze (95% copper and 5%.aluminium), (d) muntz metal (60% copper and 40% zinc), (e) admiralty alloy (71% copper, 28% zinc and 1% tin) and (f) stainless, etc. The outside diameter of the tubes is either 22 mm, 23 mm or 25.4 mm. The tube length for single pressure condenser varies from about 9 to 15 m while for multi-pressure condensers these ranges from about Shell expansion joint Back water box

Steam inlet

Air connection

AEI

Water inlet connection Front water box

_vim.s.u=s11=smizsgm A Tube sheet A

Hot well

EJ

Water outlet connection

Section B-B Condenser shell B

Tubes Section A-A

Baffle

Fig. 1677, A Typical Two-Pass Surface Condenser for large Steam Power Plant.

656

Steam & Gas Turbines And Power Plant Engineering Turbine connection

Circulating water connection

Expansion joint

Condenser tubes Tuble sheets

- Tube support plates

Fig. 16.8. A Two-Pass Divided Box Surface Condenser. 21 to 27m. Table 16.1 gives typical condenser dimensions for 100, 200 and 300 MW capacity. Table 16.1. Typical Condenser Dimensions (twi = 26.7°C, Vacuum = 5.08 cm) Capacity Steam flow Surface (MW) (kg/h) area (m2)

Number Tube of passes length (m)

Tube OD Turbine (mm) exhaust width

Turbine exhaust length

(m)

(m)

100

22.04 x 104

6967

2

8.534

22

4.267

4.267

200

41.2 x 104

13935

2

9.144

22

5.33

5.943

500

97 x 104

31586

2

. 9.144

22

5.486

6.096

16.7. Design Aspects of Surface Condensers. The calculations of heat transfer for determining the tubes and total surface area required by a surface condenser are rather complex. They require the knowledge of the total heat load on the condenser, the heat transfer mechanism and coefficients in various parts of the condenser. When the condenser is new, the outside surfaces are usually clean but quickly develop an oily film that changes condensation from dropwise to filmwise condensation. Thus the shell side heat transfer coefficients are conservatively based on the lower filmwise conden-

657

Condensers and Cooling Towers

sation mechanism. The shell side heat transfer coefficient depends upon the difference between the steam saturation temperature and tube-wall temperature, the relative position of the tubes, steam velocity and turbulence, the extent of noncondensables, and the existence of superheat steam if any. The circulating water side heat transfer coefficient depends upon its velocity, temperature and cleanliness of the inside surface. It is worth to note that the circulating water is obtained from natural body of water, so algae and other deposits accumulate on the inside surface of tubes which in turn affects the heat transfer. This necessitates the frequent cleaning of tubes. Owing to the large number of variables with many uncertainties, manufacturers have usually based their design in general proposed by the Heat Exchange Institute (HEI) Standards for Steam Surface Condensers. The method is based on the usual heat transfer equation for heat exchanger as given below— Q= U0 Ao ATni

(16.2)

Where Q = heat load on condenser (W), Uo = overall condenser heat transfer coefficient based on outside tube area (W/m2.K), Ao = total outside tube surface area (m2), AT,,, = log-mean temperature difference in condenser (°C) AT„, is expressed as (Fig. 16.9) ATi — ATe ATIn = ln(AT / )

(16.3)

where AT, = difference between saturation-steam temperature and inlet circulating water (°C). AT = difference between saturation-steam temperature and outlet circulating water temperature (°C), also called terminal temperature Ofference ('rID) HEI Method. The Overall heat transfer coefficient (U 0) is expressed empirically by U0 = C I .C2.C3.C41ZE 14,

(16.4)

Where C„ = circulating cold water velocity in tubes at inlet (m/s), C1 = dimensionless factor depending upon the tube outer diameter; considering heat transfer coefficient, C2 = dimensionless correction factor for circulating water inlet temperature, C3 = dimensionless correction factor for tube material and gauge, C4 = values of these factors are given in Table 162. Tsat

Tsat/ TTD c2 T

•ct

Ao Fig. 16.9. Temperature Distribution in a Condenser.

Steam & Gas Turbines And Power Plant Engineering

658

By knowing the values of 6,7',„ and Uo , Ao can be calculated for a known value of heat load, Q. Table 16.2. Factors (Cl, C2, C3, C4). Tube outer diameter (mm)

19

22

25.4

C1 (Cwin m/s, Uoin W/m2K)

2777

2705

2582

Water 1.66 4.44 7.22 10 temperature (°C)

12.77 15.55 21.11 26.66 32.22 37.7

0.57 0.64 0.72 0.79 0.86 0.92 1.00 1.04 1.08 1.10

C2 Tube material

304 stainless steel

Admiralty arsenic coper

Aluminium- Aluminium- 70-30 brass, muntz bronze 90-10 Cu-Ni Cu-Ni metal

C3

18 gauge

0.58

1.00

0.96

0.9

0.83

17 gauge

0.56

0.98

0.94

0.87

0.80

16 gauge

0.54

0.96

0.91

0.89

0.76

0.85

for clean tubes, less for algae covered or sludged tubes

C4

Conventional Method. In conventional method, the usual head transfer equations are used to calculate U.. For filmwise condensation the average heat transfer coefficient on steam side for a horizontal tube is given by Nusselt (h0 )Ss

=

[k.3 p2 g h ho = 0.725 -I f N

1/4

(16.5)

Where N = number of horizontal tubes in a vertical tier, = 0 Tsat Twaii , hfg = latent heat of condensation of steam, t.if = viscosity of condensate (fluid), p1= density of condensate (fluid), kf = thermal conductivity of condensate (fluid). do = outside tube diameter. It should be noted that Nusselt's equation for ho gives a conservative value for the condensing film coefficient for heat transfer. However, this value will also be influenced by many factors such as superheat, vapor velocity, turbulence, and the inside released gases and air leaked. The inside heat transfer coefficient on the water side is given by Mc Adams equation h. d. Nu =

k

= 0.023 R e0.8 - 0.4 (16.6)

where Re = Reynolds number due to flow of circulating water through tubes = Cd1 Pr = Prandtle number = c 1..t/k The overall heat transfer coefficient for a condenser is given by xwall 1 1 1 1 + + + Uo A0 h A hscale . A k A 7 74' wall m i 10 0 hscale =- heat transfer coefficient of scale formed where

(16.7)

Condensers and Cooling Towers

659

Xscale = wall thickness kwall = thermal conductivity of wall Aim = mean inside area including scale formed For simplicity, the tube wall resistance due to thin tube and good thermal conductivity may be neglected. Hence, 1 1 1 1 UO = ht hscale+ h0 (16.8) It is to be noted that he is much larger than hi and U0 mainly depends on water velocity as h. a 0.8 w 1 U u0

1 = A + B . c0.8 (16.9)

1

where A = — 1+ h0 hscale 1 amd B — k p0.4 0.023

f

cici

(16.10)

o-,.8 f

The rate of heat transfer from the condensing vapor to the cooling water is expressed as (16.12) c2 — Tel ) = U oA 0 ATm Where ins = mass flow rate of steam entering to condenser, inc is the mass flow rate of coolant and Tel and Ta are inlet and outlet temperature, of cooling water. Generally, (Te2 -Td) is around 10°C, AT, around 11 to 17°C and Are = TTD should not be less than 3°C. In practice, due to losses in condenser, the undercooling of the condensate (Tsar Tcondmmate) is around 4°C. From equation (16.12), the mass flow rate of coolant, i.e. water (6 c) is given by =

c

s(11 steam —

h condensate) = me pc

(T

tils(hsteam — hcondensate) C — T ) c2 ci

(16.13)

The outside surface area A0 is thus calculated from eq. (16.12) Ao —

iPl s(hsteam — h condensate)

— rm do 1 U0 AT (16.14) Where n = number of tubes, / = length of one tube (for a single pass condenser). Further, //lc =

pw CH,

(16.15) where pn, = density of water (103 kg/m3), and C,„„ is the velocity of water which varies from 1.8 to 2.5 m/s. Therefore, the length and number of tubes can be calculated from equations (16.14) and (16.15). Generally, tube length and diameter are selected so the estimation is made for number of tubes. In the case of large size power plant, the number of tubes may be as high as 50,000 or even more. The pressure drop in the condenser is composed of the pressure drop in the water box and (ii) the friction pressure drop in the tubes. The pressure drop in terms of head is

660

Steam & Gas Turbines And Power Plant Engineering

Ap = p gH 16.8. Non-condensable Gases (Air) Removal or Deaeration It is important to remove the non-condensable gases that gets accumulated in the condenser. The non-condensables are mostly air that leaks through the joints into the condenser as condenser operates much below atmospheric pressure. It also includes other gases caused by the decomposition of water into oxygen and hydrogen by thermal action and.by chemical reactions between water and materials of construction. The performance of condenser due to presence of noncondensables gets affected badly because of the following reasons :— (i) The total pressure of the condenser gets increased as it is the sum of the partial pressure of its constituents i.e. steam and air (p = ps + pa). An increase in condenser pressure lowers plant efficiency. (ii) They blanket the heat transfer surfaces such as outside the surface of. condenser tubes which reduces the condensing heat transfer coefficient and thus condenser effectiveness. (iii) The pressure of noncondensables results in various chemical activities such as corrosion especially by oxygen in steam- generator. The maximum acceptable oxygen concentration in the condensate leaving the condenser is 0.003 per cent by volume. The process of removing noncondensables is called deaeration. Good deaeration within a condenser requires time, turbulence, and good venting equipment. The cold condensate falling from the lower tubes must have sufficient falling height and scrumbling steam for reheat and deaeration. The scrubbing steam is provided by allowing some of the incoming steam to pass through an open flow area directly to the bottom tubes to reheat the condensate. This is done keeping in view that noncondensables are more easily released from a hotter than a colder liquid. Once the noncondensables are released they are cooled to reduce their volume before Steam and air

Steam and air

1

I

1

plus ~_• steam CW tubes in proper ;ayout ;symbolic)

Baffle Air cooler

Condensate

i 6. 10. Air Cooler Section (Side).

CW tubes in proper layout (symbolic)

Air cooler

Condensate

Fig. 16.11. Air Cooler Section (Central).

661

Condensers and Cooling Towers Noncondensables in Motor From auxiliary boiler for startup

First-stage ejector

From main steam

Intercondenser

Condensate in Drain Aftercondenser

4 Condensate out

4f

Drain

Fig. 16.12. A Two-Stage Steam Jet Air Ejector (SJAE). being pumped out of the condenser. For the convenience of noncondensable gases (air) removal, an air cooler section is provided in the condenser shell. This aircooler section may be provided either at the side below the centre position (Fig. 16.10) or in the centre (Fig. 16.11). In Fig. 16.10, most of the condensation is carried out on the main bank of the tubes and the air is drawn over another smaller bank (about 5 to 6%) which is shielded from the main bank by a baffle and is called air-cooler. Here, further condensation takes place at a lower temperature, and thus there is a saving in feed water as well as in air injection load. In the central air cooler section (Fig. 16.11), the air cooling tubes (about 5 to 6%) are in the centre of the condenser and the air is removed from this section. The incoming steam passes all round the bank of tubes and some is drawn upward to the centre. This air cooler section in baffled to separate the noncondensables from the main steam flow. The noncondensables flow toward the cold and of the condenser, where they connect to a vent duct that leads to the venting equipment. The venting equipment includes reciprocating compressor (called dry vacuum pump) and jet pump, called steam jet air ejector (SJAE), the latter is now invariably used. SJAE uses steam jet as their motive or driving flow. They are usually multistage units, usually two or three Fig. 16.12 shows a two-stage SJAE. It uses main steam at a reduced pressure that enters a driving-flow nozzle in the first-stage ejector, from which it exits with high velocity and momentum but at reduced pressure. This reduced pressure draws in the noncondensables from the condenser and as a result of momentum exchange, the gases are entrained by the steam jet. The combined flow of steam and gas is now compressed in the diffuser of the first stage ejector and discharged into a small intercondenser, where the steam is condensed by passing across cooling pipes similar to the main condenser. It is to be noted that cooling here, however, is accomplished by the main condenser condensate and is a part of feed water heating system, resulting in enhancement in efficiency of the plant. The condensed steam is drained and returned to a low pressure part of the cycle. The noncondensable and any remaining steam are then passed to the second stage ejector, where they are compressed and passed to an aftercondenser (also known as vent condenser). The steam if any get condensed and noncondensables (air) at higher pressure than atmospheric pressure is vented out.

662

Steam & Gas Turbines And Power Plant Engineering Low-pressure turbine

From steam plant

Circulating pump Condensate to plant

Surface of natural body of water

Fig. 16.13. Once-through Cooling Water System. 16.9. Circulating Water System. As mentioned earlier, the circulating water system is one of the most important system of steam power plant. It supplies cooling water to the turbine condensers and thus acts as a vehicle by which heat is rejected from the steam cycle to the environment. Circulating water system is broadly classified as (a) once- through, (b) closed loop and (c) combination system. 16.9.1. Once-through Cooling System. Fig. 16.13 shows once- through system. In this case, water is taken from a natural body of water such as a lake, river or ocean and pumped through the condenser where it is heated'and then discharged back to the source. There are mainly three methods of discharge namely (i) surface discharge, (ii) submerged discharge and (iii) diffuser discharge. • Once-through cooling system, is thermodynamically, the most efficient means of heat rejection. Due to scarcity of water or other environmental regulation, closed loop system is universally used. 16.9.2. Closed-loop Cooling System. Closed-loop water cooling is shown in Fig. 16.14. In this case, hot water coming out from the condenser is passed through a cooling device (such, as cooling towers, spray ponds, spray canals and cooling lakes) and is Hot humid air

Steam )

/-\

Cooling tower Atmospheric air

Condensate

Circulating pump

Makeup pump

Natural water Bleed Flow direction

Fig. 16.14. Closed Loop Cooling System.

Condensers and Cooling Towers

663

returned to the condenser with the help of a pump. A nearby natural body of water is still necessary to supply make-up water to replace the lost by evaporation during the cooling process (say in cooling tower) and to receive blowdown from it. 16.9.3. Combination Cooling System. In combination cooling system is such in which wet cooling tower is used and as per requirement either the cooling water coming out from cooling tower is led to the open mode or closed mode as per availability of water or scarcity of water. 16:10. Cooling Towers. The purpose of cooling tower is to cool the warmed water discharged from the condenser and feed the cooled water back to the condenser. By this way, the cooling water requirement get reduced to make-up water supply only. The cooling tower may be wet or thy type. 16.10.1. Wet Cooling Towers. Wet-cooling towAir + water I 4, Circulating ers cool the hot water by dissipating heat to the envivapor Is ✓ fan ronment through the mechanism of (i) addition of sensible heat to the air and (ii) evaporation of a portion of the recirculation water itself. When operated in Tci the open mode, there is a third mechanism (iii) addi- Hot water in tion of sensible heat to the natural body of water as a Packing result of terminal temperature difference. Wet cooling towers employ a hot-water distribution system that showers or sprays the water evenly k.../ Air in over a lattice work of closely set horizontal slats or T bars called fill, or packing (Fig. 16.15). Since the C2 Cold water splashes down from one fill level to the next by water out gravity, there is a thorough mixing of falling water with air moving through the fill. Outside air enters the Fig. 16.15. Packing or Fill tower via louvers in the form of horizontal slats on the in a Wet Cooling Tower. side of the tower. The slats are arranged in such way that they usually stop downward to keep the water in. The intimate mix between water and air results in the enhancement of heat and mass transfer (evaporation) which cools the water. The cold water gets collected in a concrete basin at the bottom of the tower from where it is pumped back to the condenser in closed system or returned to the water body in open system. The resulting hot and moist air leaves the tower at the top. Unsaturated air enters the cooling tower and as it comes in contact with the water spray. Water continues to evaporate till it becomes saturated. From this, it follows that the minimum temperature to which water can be cooled is the adiabatic saturation or wet bulb temperature (To) of the ambient air. Performance Parameters. A cooling tower is characterized by three performance parameters namely (a) approach, (b) range, and (c) cooling efficiency. The approach temperature (A) is defined as the difference between the exit temperature of cooling water and the wet bulb temperature of the ambient air. Thus A = T — Twb (16.16) Twb = the wet bulb temperature of air where = cooling water exit temperature from cooling tower This approach temperature (A) varies from 6 to 8°C.

664

Steam & Gas Turbines And Power Plant Engineering

The cooling range (R) is defined as the difference in temperature of the incoming warm water (I'd and existing cooler water (T 2). Thus, R = Ti — Te2 (16.17) This range R varies from 8 to 10°C The cooling efficiency is defined as the ratio of actual cooling of water to the maximum cooling possible. Thus licooling

=

Actual cooling Maximum cooling possible

Ti —T 2

T i — Twb (16.18) Classification of Wet Cooling Tower. Wet cooling tower are classified as either (i) mechanical draught or (ii) natural draught cooling tower. Each of these may be counter flow or cross-flow. 16.10.1.1 Mechanical Draught Cooling Towers. Fig. 16.16 shows the schematic of mechanical draught cooling tower. In this case, the air is moved by one or more mechanically driven fans. The fan could be of the forced-draught (FD) type or induced draught (ID) type. The FD fan (Fig. 16.16b) is mounted on the lower sides to force air into the tower while ID fan (Fig. 16.16a) is located on the top of the tower. Though, FD fan is thermodynamically superior as it handles cold air but it has shown some disadvantages because of air distribution problems, leakage, recirculation of the hot and moist exit air back to the tower and frost accumulation at fan inlets during winter operation. As a result, the majority of mechanical-draught cooling towers for utility application are therefore of the induced-draught (ID) type. In this case, air enters the sides of the tower through large openings at low velocity and passes through the fill. The fan located at the top of the tower exhausts the hot, humid air to the atmosphere. Induced-draught cooling towers are usually in the form of multi-cell with a number of fan stacks on the top and built in various arrangement like rectangular, octagonal, circular, etc. The fans are propeller type and driven by electric motor The blades of fans are usually made of cast aluminium, stainless steel or fibre glass to safeguard against corrosion. The advantages of mechanical-draught cooling towers include the assurance of movAir + water vapor outlet

Air t Water

Air + water vapor outlet

Counter flow 4— Drift eliminators —3. 4—

Hot water —> inlet Fan

Fan Air inlet

4-Air inlet

---* Air inlet

Air inlet

Cold water outlet (a) Induced draught

(b) Forced draught

Fig. 16.16. Mechanical Draught (Induced and forced) Counter-flow Cooling Tower.

665

Condensers and Cooling Towers

ing the required quantity of air at all loads and climatic conditions, low initial capital costs, and low physical profile. The main disadvantage include power consumption, operating and maintenance costs and greater noise generated from fans. 16.10.1.2. Natural Draught Cooling Towers. There is no fan used by natural draught cooling tower. They depend for air flow upon the natural driving pressure caused by the difference in density between the cool outside air and the hot, humid air inside (Fig. 16.17 and 16.18). The driving pressure differential is expressed as

APd = (130 Pi) gll where

(16.19)

H = height of tower above the fill (m)

pc, and p, = density of outside and inside air respectively (m3/kg) Since (Po — p) is relatively small, so H must be large to cause desired Apd and as a result natural draught cooling tower are therefore very tall. It is to be noted that the tower body, above the water distribution system and the fill, is an empty shell of circular cross-section but with vertical hyperbolic profile and due to this natural draught towers are called hyperbolic towers. The advantage of natural .draught hyperbolic cooling tower include its superior strength and greatest resistance to outside wind loading compared to other shape. The natural draught may be counter-flow or cross-flow type. In counter-flow the fill is inside whereas in cross-flow, the fill sits in a ring outside the tower outside the stilts. The choice between different type of cooling towers depends upon many factors, the most important of which are climatic and economic. Generally, mechanical draught towers are selected when the approach (A) temperature is low and when the broad range of water flow is expected. The latter is met easily because they are usually built as multi-cell units with a variable airflow fan which offers versatility and good response to changes in cooling demands and parameters. Natural draught cooling towers are selected under the following conditions— (i) in cool humid climates (low wet-bulb temperature and high relative humidity; (ii) when there is combination of low wet-bulb temperature and high condenser-water inlet and outlet temperatures and (iii) in cases of heavy winter loads. In general, they are Air + water vapor Air + wa er vapor 1\ 1 (1

Hot water in

Drift eliminators pc > pp > pB), with the result, two secondary flows are set up in the blade channels. Emperical relations are often used to take into account the three dimensional effects along the blade height. According to Carter, CDS = 0.018 C2

(18.59)

The above equation represents drag coefficients due to secondary flows produced by the boundary layers on the walls. Note that the secondary flow drag coefficient is independent of radial blade clearance.

Compressors

787

PD PB

PA > PC > PD PB

Pressure surface X (concave surface) Suction surface (convex surface)

Tip

-4-Secondary flow

Root

i Section at X-X

„A

Fig. 18.36. Secondary Flow.

18.19. Choking Flow

Pressure ratio

Theoretically, the mass flow rate becomes maximum when the pressure ratio is unity i.e. there is no compression. This generally occurs when the Mach number corresponding to the relative velocity at inlet C) becomes sonic. Under this condition. the mass flow Useful co C) C rate possible from a centrifugal or axial flow comoperating U) pressor becomes maximum which is known as chokrange 0 .c ing flow (Fig. 18.37). Choking means fixed mass flow 'rate regardless of pressure ratio, i.e. the characteristics Mass flow becomes vertical (Fig. 18.37). Fig. 18.37. Choking Flow. Problem 18.5. Calculate isentropic pressure rise and workdone by a cascade of axial flow compressor for the following data: a = 14°, u = 200.m/s, al = 45°, Ca = 186 m/s, p = 1 kg/ m3 2 Solution : The pressure rise through an axial flow compressor cascade is given by Op = Ca2(tan2ai — tan2a ) =

2

1 x 1862

2x

1 04

(tan2 45 — tan2 14) = 0.16235 bar Ans.

Steam & Gas Turbines And Power Plant Engineering

788

Workdone per kg of air in a ring of blades = u . Ca (tan al — tan a2) —

200 x 186 (tan 45 — tan 14) = 27.95 kW 1000

Ans.

Problem 18.6. In an eighteen stage axial flow compressor, the overall compressor pressure ratio achieved is 15:1 with an overall isentropic efficiency of 90%. The temperature and pressure at inlet is 20°C and 1 bar. The work is divided equally between the stages. The mean blade velocity is 175 m/s and 50% reaction design is used. The axial veloc1 ity through the compressor is —ms s constant and equal to 100 m/s. Calculate the power required Fig. 18.38 .and blade angles for mass flow rate of 300 kg/s. Solution : Refer to Fig. 18.38 for velocity diagram..In the case of 50% reaction, velocity triangles are identical. Thus, al = f32 and a2 = 131 Let suffix Z denotes the number of stages. - 1Yr

Toz = T 01(13 oz oi risen

or

or

29301.286 = 635.66K

Toz— T0, = 0'9 — T ' — T oz 01

635.66 — 293 Toz ' —293

To; = 673.73 K w ca = cp(TOZ'

TO1) = (C — Cwi )u . Z= Co (tan al — tan a2)u . Z

tan a 1 — tan a2

—

1.005 (673.73 — 293) x 1000

100 x 175 x 18

— 1.2147

(a)

75 we have, -1 = tan a1+ tan 13 = 100 = 1.75

(b)

a

Solving equations (a) and (b), we have, tan 131 = 131 = 62.2° = a2 Putting the value of tan 131 in equation (a) we get tan a1= 1.75 — 1.4823 = 0.2677

L2147 + 1.75 — 1.4823 2 Ans.

789

Compressors or

a1 = 132 = 16.65°

Ans. Power required by the compressor = in . c (To — T01) = 300 x 1.005 (673.73 — 293) = 11479 kW = 11.479 MW Ans. p Problem 18.7. An axial flow compressor is required to deliver 50 kg/s of air at a stagnation pressure of 5 bar. At the inlet to the first stage the stagnation pressure is 1 bar and the stagnation temperature is 300 K. The hub and tip diameters at this location are 0.436 m and 0.728 m. At the mean radius, which is constant through all stages of the compressor, the reaction is 0.5 and the absolute air angles at stator exit is 28.8° for all stages. The speed of the rotor is 8000 rpm. Determine the number of similar stages needed assuming that the polytropic efficiency is 0.89 and that the axial velocity at the mean radius is constant through the stages and is equal to 1.05 times the average axial velocity. Solution. Refer to Fig. 18.39. Data given : S2 mean = 0.5, in = 50 kg/s, a1 = 28.8°, poi = 1 bar, N = 8000 rpm, p02 = 5 bar, rip, = 0.89, DI, = 0.436 m, (Ca)p„ = Ca x 1.05, D, = 0.728 m, Z = ?, From continuity equation at the inlet, m = p1 A (Cal

\aye = P

(q

(C dace

1 X 105 P01 3 But p1 = — — RToi 287 x 300 — 1.161 kg/m

50 = 1.161 X T (0.7282 — 0.4362) (Ca)

\ Ave

or

(Ca) = 161m/s \ Jaye D — D, Dh + D, u= nDN ButD=D+/ —D+ ' " " 60 2 — h h h 2 7E(Dh + D,)N ic(0.728 + 0.436) x 8000 u— — — 243.78 m/s 2 x 60 2 x 60 lisen = r ist

50.4/1.4 — 1 T02 — T01 (P02/P01)"— "/Y — 1 — 0.862 T — — = /_ (n / n )(1. - 1)1r Tl ix. _ 1 — 50.4/1.4 x 0.89 _ 1 T02 01 .- 02 rOl•

Dt lb

Inlet Fig. 18.39

Outlet

790

Steam & Gas Turbines And Power Plant Engineering

PO2

1.4/0.4 6,71 = 1 -I-

Poi or

ATot] = 5 = [1 + 0.862 -31c3

n 01

AT0' = 202.9 K

The reaction = =

= 40 tan paiR -

1 - (13. tan a ms = 1-

a)ave 2) (tan pi + tan(32) 2u

(Caa ) ,e

2u '

(lanai + tan a2)

05 x 161 0.5 = 1 1. [tan28.8 + tana2] 2 x 243.78

or

tana2 = 0.893.

Since wm = cp (AT0')stage = U(Cw2 — Cwi ) or

(AT0' )stage = =

u (C) aa C P

ve

[tan a2 - tanai ].

243.78 x 1.05 x 1.05 x 161 [0.893 - 0.593] _ 14.1 K 1.005 x 105 AT0'

Number of stages - Z -

(AT0 )mage

-

202.9 - 14.34 :-.-:. 14 stages 14.1

Ans.

Problem 18.8. An axial flow compressor is required to deliver air at the rate of 50 kg/s and provide a total pressure ratio of 5:1, the inlet stagnation conditions being 288 K and 1 bar. The isentropic efficiency is 86%. The compressor shall have 10 stages with equal rise in total temperature in each stage. The axial velocity of flow is 150 m/s and the blade speed is kept at 200 m/s to minimise noise generation. The stage degree of reaction at mean blade height is 50%. Assuming workdone factor as 0.86, calculate all the fluid angles of the first stage. Also calculate the tip and hub diameter if hub-tip diameter ratio is 0.8. Determine the speed in rpm. Take R = 287 J/kgK and cp = 1.005 kJ/kgK. Solution. Refer to Fig. 18.40. PO2 Data given : Z = 10, in = 50 kg/s, u = 200 m/s, r = — = 5, S2 = 0.5 at mean blade P

p 01

height, p01 = 1 bar, X = 0.86, Toi = 288 K, DT' = 0.8, disc„ = image = 86%, 131 , P2, al , a2 = ?, (.6,To)mage = constant, Dh, D„ N = ?, Ca = 150 m/s, AT ' PO2 We have, — = 1+ isen T Poi oi or

1.4/0.4 = 5 = 1+0.86

AT0' = 195.6 K AT'

195

19.56 (ATOstage = — Z = 10 =

288

791

Compressors

Cwi --:1

E— Cm.'

Cm2

Cw2

L Inlet velocity triangle

--*s

Outlet velocity triangle

Fig. 18.40.

cp (ATAaage = u Ca (tan a2 — tanal)

ii),„age

or

1.005 x 19.56 x 103 = 0.86 x 200 x 150 (tana2 — tanai)

or

tana2

—

tana = 0.761

(1)

'C Reaction = = 1 — (1) tan am, = 1 —,2u (tan ai + tan a2) 150 (tan al + tan a2) 2 x 200

or

0.5 = 1 —

or

tan a i + tan a2

=

1.33

(2)

Solving equation (1) and (2), we have a = 16.1° and a2

=

46.26°

Since reaction is 50%, so a2 = 13; = 46.26°, i From continuty equation, in = p A C

= al = 16.1°

Ans. D2

it D1 - Do Ca = pD2 1 - —h 4 Ca D2 (

10 5 ) D2 (1 — 0.82) x 150 287 288 4

or

50 —

or

D = 0.987 m

Ans.

Dh = 0.8 x 0.987 = 0.789 m

Ans.

D — Dh Height of blade = lb =

2

• — 0.987 — 0.789 = 0.103 m

Ans.

Problem 18.9. In an experiment with two dimensional cascade of blades the total pressure measured downstream where the flow is uniform was found to be less than the inlet total pressure by 0.0065 bar. The pitch chord ratio of the blades was 0.7 and the inlet and outlet fluid angles were measured as 50.3° and 25.3° respectively from the axial

792

Steam & Gas Turbines And Power Plant Engineering

direction. The outlet velocity was measured as 150 m/s. Calculate coefficient of drag. Take density of air as 1.22 kg/m3. s COS3 an, Solution. The drag coefficient Co = 4 (c cos2a i where 4

and

-

Apo

,=

(1/2) p Co

0.0065 x 105

, - 0.0473

(l/) 1.22 x 1502

a l + a2 50.3 + 25.3 ani = — 37.8° — 2 2

CD —

0.0473 x 0.7 cos3 37.8 — 0.038 cos250.3

Ans.

18.20. Stalling

Stalling of stage of an axial flow compressor is defined as the aerodynamic stall or the break away of the flow from the suction side of the blade aerofoil. This break away of the flow may occur due to lesser mass flow rate than designed value or due to non-uniStall propagation formity in the blade profile or approaching flow. Thus, stalling is an ahead phenomenon of surging. Consider flow through a cascade as shown in Fig. 18.41. Suppose that some non- uniformity in the approaching flow or in a blade profile causes blade B to stall. The air now flows on to a blade A at an increased incident due to the blockage of channel AB. The blade A stalls as a result but the flow on to the blade C is now at a lower incidence and blade C may unstall. In this way, a stall cell may move along the Unstalling cascade opposite in the direction of blade motion or the direction of the lift of the blades. In other words, the phenomenon of reduction in lift force at higher Fig. 18.41. Stalling in Axial angles of incidence is known as stalling. Compressor. If the natural frequency of vibration of the blades coincides with that at which the stall cells pass a blade, the resonance and possible mechanical failure of the blades may occur. 18.21. Three Dimensional Flow in Axial Flow Compressor

The elementary theory concerning stage pressure and workdone assumes that the flow in the compressor annulus is two- dimensional as it ignores any effect due to radial movement of the fluid. This assumption is true for compressors with shorter blade height relative to mean diameter of the annulus, i.e. for higher hub-tip ratio (r/r,> 0.8). For lower hub- tip ratio, especially for front stages, the flow must have a radial component of velocity. An another cause of pressure of radial component of velocity is due to whirl component of velocity which increases the pressure with radius. To provide a balance between the pressure forces and the inertia forces, the flow will undergo some movement in the radial direction.

Compressors

793

With a longer blades i.e. low hub tip ratio, the blade speed will Casing Blade row vary appreciably from root to tip \i which will result in the variation of / velocity triangles and thus air angles approaching to blades. Further, l \_§treamthe change in pressure, and hence lines > Hub density, with radius will cause the fluid velocity vectors to change in magnitude and these too affect the Axis velocity triangles. All these suggests that air angles will vary appreFig. 18.42. Radial Equilibrium Flow Through ciably at root and tip with reference a Rotor Blade Row. to mean radius. It is essential that entry to the leading edge of blade at all radii must be shockless and for this the blade has to be twisted from root to tip to match the approaching air angles. To an observer travelling with fluid particle, a radial motion will continue until sufficient fluid is transported radially to change the pressure distribution to that necessary for an equilibrium. The flow in the annular passage in which there is no radial component of velocity, whose streamlines lie on circular, cylindrical surface and which is axisymmetric, is commonly known as radial equilibrium flow. The radial equilibrium method is widely used for three- dimensional design calculations in axial flow compressors and turbines. This method is based upon the following assumptions. (Fig. 18.42) (i) that any radial flow which may occur, is completed within a blade row, the flow outside the row (i.e. upstream and downstream) then being in radial equilibrium. 18.21.1. Theory of Radial Equilibrium. Consider a small element of fluid mass dm of unit depth and subtending an angle dB at the axis rotating about the axis with tangential or whirl velocity, at radius r as shown in Fig. 18.43. It is assumed that the fluid element is in radial equilibrium so that centrifugal forces balance the pressure forces. Thus

7

(p + dp) (r + dr)d8 — prdO — (p + — 21 dp)dr . dO = dm . C2w /r

(18.60)

The mass of fluid element dm = p r dO dr Rearranging the above by introducing the expression for dm and ignoring the terms of the second order of smallness, the above equation takes the form 1 clE p dr

C2w r

.-

(18.61) The above equation is known as radial equilibrium equation and the radial pressure distribution along the blade length (i.e. radius) can be determined if the swirl, velocity, Cw and density, p are known functions of radius. In fact, the radial component of velocity is being neglected in spaces between blade rows as it will be very small as compared to Ca or Cw. , The stagnation enthalpy = h0 = h + C2 —=h+— 21 (Ca2 + 2 The differentiation of above equation with radius results in the variation of enthalpy with radius

Steam & Gas Turbines And Power Plant Engineering

794

dr

p + dp 2

Fig. 18.43. Radial Equilibrium of Fluid Element.

dCa dCw dh o dh dr dr + Ca dr + C w dr

(18.62)

From the thermodynamic relation, we have, Tds = dh p

dh , ds+ 1 dp 1 di d dr 1 dr ds dT+ dr -- — p2 dr 4' Neglecting second-order terms, we have dh ds 1 c _12 •— = „, — + dr dr p dr

dh Substituting the value of — in eq. (18.62), we have dr

dr a`

dCa ds 1 c12 + +u + dr p dr a dr

dC v dr

Substituting the value ofd from radial equilibrium equation, we have dh0 ds dCa dC C2 = 2' — + C +C + w dr dr a dr w dr r ds dr

If entropy remains the same, then — = 0, thus dho dC + = c a + dr a dr w dr r

(18.63)

The above equation is known as Vortex energy equation. If stagnation enthalpy h0 remains constant, dho/ dr = 0, which is possible in the case of constant specific work at all radii, then , /

795

Compressors

dC dCH, Ca dr +Cw dr + r °

(18.64)

dCa If Ca remains constant along radius, then — = 0, thus 7r

dCw —dr Cw or

C .r = constant

(18.65) This is called free vortex condition. From the above derivation it follows that there are three following conditions naturally satisfy the radial equilibrium equations 18.61 and conducive for design. (i) free vortex variation of whirl velocity (ii) constant specific work (iii) constant axial velocity Any of the above three conditions may be used for design of bladings but each of them has some drawbacks. The expression for degree of reaction may help in discussing the limitation. (i) Free Vortex Design Condition. The degree of reaction is expressed as

Ca = 1— — 2u (t"1

+tana2) — 1—

Cwi + Cw2 2u

Since u = um . r/rm, where un, is the blade speed at the mean radius of annulus rm, so Cw2 r + C r 2

=1

2uni r /rm For free vortex flow, Cw . r= constant, so =1 —

constant r2

(18.66)

The above equation (18.66) suggests that the degree of reaction SI for free vortex design condition increases appreciably from root to tip of the blade. Since the blade speed is lower at the root section, so more fluid deflection is required for a given work input. This means a greater rate of diffusion is required at the root section which suggests that it is undesirable to have a low degree of reaction at the root section. (ii) Constant Specific Work Design Condition. The design condition of constant specific work may go hand in hand with an arbitary whirl velocity distribution. In order to obtain constant work input, u(C„2 — C,1) must remain constant across the annulus. The following assumptions are made for the distribution of whirl velocity

b aR"-- and Cwt = aR"+ 11— r

where a, b and n are constants and R is the radius ratio — and u = u . R. When n = -1 The whirl velocity distribution when n = -1 is given by

Steam & Gas Turbines And Power Plant Engineering

796

ab a b C = — - — and C = — + — wl R R w2 R R The above equations have the form of free vortex, C„ r = constant. To ensure radial equilibrium, the axial velocity, Ca has to be constant. Thus, the reaction takes the form by using (C„,1 r + Cw2 r)= constant = 2a r„, as follows 2a r171 a =1 2u R . r — 1 um R2 (18.67) When n = 1 The whirl velocity distribution becomes

C

b

b

= aR — — and Cw2 = aR + —

Using the radial equilibrium equation when ho = constant, it becomes C2

Ca dCa + Cw dCw + R dR = 0

After integration the reaction finally is given as 2a In R a =1+ Um

U

(18.68)

When n = 0 The whirl velocity distribution is given by

CW1 = a - — and Cw2 = a + R _R The radial velocity distribution is given by 2 ) = —2 [a2 In R-1+ ab] and (C2a1) — (Ca.) in R + cf1 ?) — ab (C2a2) (Ca2 m =-2 I a2 . Blading designed on this basis is known as exponential design. If (Ca2)m = S2 =

, + -a 2a u u R

al )no then (18.69)

771

For all three cases, at the mean radius (R=-1), the expression for a = u„,(1-0 m) holds good if 51„, is the reaction at mean radius: Thus substituting the values of a in the above three cases the reaction is tabulated as follows— n

S2

—1

1—

(1 — Om)

Free vortex

0

1 + [1 — - (1 — n„,) ?1 1 + (2 1n R-1) (1 — Om)

Exponential

1

of designing bladings

First power

797

Compressors

1.0 n

Exponential

0

1 0 -1

Free vortex

a)

rn a) 0

= 0.5 0 1 0.5

•

1.0 R = r/rm

1.5

Fig. 18.44. Radial Variation of Degree of Reaction for Om = 0.5.

Fig. 18.44 shows the radial variation of degree of reaction for mean value of reaction CZ = 0.5. It is obvious from the result that the free vortex design provides the greatest reduction in reaction, SI for low values of R while the first power (design, n=1) gives the least reduction in reaction. It is interesting to note that for all the methods of designing the bladings, there is a lower limit to R below which the degree of reaction becomes negative. This limiting value of R is for free vortex design. But as mentioned earlier, a hub-tip ratios as low as 0.4 are common at entry to the compressor of a jet engine. To satisfy this the reaction at mean radius has to be higher than 0.5. Since r/r, = (r/rm)(rni/ri) = R(rm /ri) = R[1+ (rr/rik2 so the variation of reaction in terms of r/r, for a specified value of hub-tip ratio (r/ri) may be found. Fig. 18.45 shows the variation of reaction with radius for free vortex design of r/r, = 0.4 1.0 It is obvious from the result that a stage of low hub-tip ratio must have a high value of C2„, to meet the satisfac- 0 tory conditions at the root of bladings. Another method of design is a constant reaction design. It results in a) twisted blading than the free vortex design. However, constant reaction design does not satisfy the radial equilibrium condition. 0 0 4 0.5 0.6 0 7 0.8 0.9 1.0 The following example gives the r/rt three dimensional design procedure for axial flow compressor. It will be obFig. 18.45. Variation of Reaction with Radius served that the process requires confor rr/r/ = 0.4 and Free Vortex Design. tinuous judgement by the designer.

Steam & Gas Turbines And Power Plant Engineering

798

18.22. Aerodynamic Design Process of Axial Flow Compressor The following is the design'process for an axial flow compressor— (i) Determination of rotational speed and annulus dimensions for aviation turbine. For power turbine, N = 3000 rpm or 3600 rpm. (ii) Determination of number of stages for an assumed polytropic efficiency. (iii) Determination of air angles for each stage at the mean radius of axial flow compressor (stage by stage design). (iv) Determination of variation of the air angles from root to tip. (v) Consideration of compressibility effects in the design. (vi) Selection of compressor blading from available cascade data. (vii) Checking on polytropic efficiency assumed using the cascade data. (viii) Estimation of off-design performance. (ix) Testing of compressor blgdings on test rig. The complete design procedure in sequential form has been given in chapter 1. All the stages stated above is not possible to discuss and out of scope of this book. 18.23. Aerodynamic Design Example Design an axial flow compressor to be used in a low cost turbo jet engine for the following data : Ambient conditions = 1.01 bar, 288 K, Total compressor pressure ratio =rp = 4.5, Air mass flow rate = ma = 22 kg/s Assume no inlet guide vanes (IGV) in axial flow of compressor 18.23.1. Rotational Speed and Annulus Dimensions. There is no equation available to calculate the rotational speed. Generally, a suitable value of rotational speed is selected. The best way is to select the blade tip speed, (u,), the axial velocity (Ca) and hub-tip ratio (yr) at the inlet to the first stage in the light of stress limit. The continuity equation gives the cross-sectional area. The experience gained over the years suggests that the blade tip speed, u, = 350 m/s will lead to acceptable stress and axial. velocity, Ca ranges from 150 to 200 m/s. Since there is no IGV, so there will not any whirl component of velocity at inlet. This will result in higher value of Mach number relative to blade at inlet and from aerodynamic consideration, this value should not be much more than unity. Thus, a modest value of Ca = 150 m/s is suitable. At inlet, the hub-tip ratio varies between 0.4 and 0.6. For a fixed blade speed, the rotational speed is a function of hub-tip ratio. Further, for a specified annulus area the tip radius will also obviously be a function of hub-tip ratio. The above discussion suggests that the designer has a wide variety of choice and it is the best judgement that will matter. The continuity equation at inlet to the first-stage gives 2

= pi A Cal = pi 7C

[1 —L-r-)] C t•

Assuming no loss in the intake system, Ta= Toi and Pa = P01, Cwl = 0, we can (Fig. 18.46) find out static values of properties at inlet, CI

= Cl = Ca = 150

m/s

799

Compressors

= 288

T1 = To — I T1

My-I)

P1 = -p01[T

1502 2x1.005x10 — 276.8 K

276.8 =1.01

P1

= 0.879 bar

288 ]

o

P I = RTI

mil A

0.879x102 — 1.106 kg/m3 0.287x276.8

Substituting the values in continuity equation, we have 2

22 = 1.106

— (1 1 IX

The peripheral speed = N =

rt

150

—

r

or

50 The following table is prepared which 27t r,

gives the values of rt and N for the various value of

rt

(m)

N (rps)

0.4

0.2241

248.56

0.45

0.230

242.19

0.5

0.2372

234.84

0.55

0.246

226.43

0.6

0.2568

216.91

r,./rt

r,

[ 1 — (r/rt)2] r

From the above table, we find that (rr/r,)= 0.5 gives a reasonable value of r, = 0.2372 m and N = 234.84 rps. Thus, the selected value of u, = 350 m/s seems to be reasonable. Let us select a round figure of rotational speed, N = 240 rps, which will slightly alter the tip speed. Thus, ut = 27c rt N = 27tx0.2372x240 = 357.68 m/s It is worth to note that the choice of (rr/rt)= 0.5 is arbitrary. It should be such that the compressor annulus dimension must match with that of turbine. It is pertinent to check the Mach number relative to the rotor tip at inlet to the compressor. Assume axial velocity constant across the annulus. Thus for no IGV, we have at the tip. zi2u Ca21= 357.682 + 1502 or Cot = 387..85 m/s

a = 4FirTI = 41.4x0.287x103x276.8 = 331.0 m/s

Steam & Gas Turbines And Power Plant Engineering

800

387.85 - 1.171 331.0 The Mach number relative to the rotor tip is slightly more than sonic in the first stage, so it will not create problem of shock losses and the selection of tip speed is reasonable and acceptable. Thus, the geometry of annulus of first stage is as follows at N = 240 rps r 0.5, ri = 23.72 cm, rr = 11.86 cm, rn, = 17.79 cm rr = C rit

a

(M11) ref

Assume ripe = 90%, The actual total temperature at the exit of compressor is given by 0-1)/n

(y-1)/yrip,

.9 = 464.26 K = 01 [Poz] = 288 (4.5)0.4"1.4x0.9 T02 = Toi pot Poi Assume that the axial velocity remains constant from inlet to outlet. Assume that air leaving the stator of the last stage has no swirl, so the static temperature, pressure and density at exit may be calculated [Po2

1502 Caa = 464.26 - 453 K T2' = TO2 - 2c 2x1.005x103 -

T, 2

P2 = pot T 02 p 2

Kr-1)

453 = 4.5 x1.01 464.26

3.5 - 4.17 bar

P2 = 4.17 x 102 _ 3.2 kg/m3 287 x 453

RT2'

Using the continuity equation at the exit of last stage we can find out the annulus area. Thus 22 - 0.0458 m2 m = p2' A2 Ca = 3.2xA2x150 = 22 or A2 = 3.2x150 Since the mean radius = rm = 0.1779 m, so the blade height is

A2 = 0.0458 = it D.1b2 = irx2 x 0.1779 x /b2 or

0.0458 - 0.0409 m 1/12 • itx2x0.1779 Thus the radii at exit from the last stator are

b 0.0209 4 rr = rni -- 0.1574 m 2 = 0.1779 4 r = r +-b = 0.1779 +0.0209 - 0.2024 m m2 Summarising the above, we have Inlet : rr = 0.1186 m and ri = 0.2372 m General :

N = 240 rps, ut = 357.68 m/s, Ca = 150

Ans.

Outlet :

r,= 0.1574 m, ri = 0.2024 m

Ans.

801

Compressors rm= 0.1779 m

Mean :

Ans.

18.23.2 Number of Stages. The number of stages Z may be estimated as Z = Overall actual total temperature rise_ AT0' Temperature rise in a stage (AT0')stage But

u Ca 0a/31 — tan(32) X u (Cw2 — Cw )

(PTO )stage —

p

p

Here, tanii i = C as there is no inlet guide vanes (Cw1 = 0) a

But

u = 27t rllt N = 27cx0.1779x240 = 268.26 m/s 268.26 t 131 — 1 50 or 131 = 67.54°

cr i

Ca

150

COS131

cos67.54

— 307.33 m/s

Applying de Haller criterion for maximum possible deflection C r2 — < 0.72. and taking Cr2 /C = 0.72, we have C = 0.72x307.33 = 221.27 m/s crl

cosi32

Ca

C r2

07'0 -)stage — Z —

150 or 221.27

132 = 52.57°

0.95x268.26x150(tan67.54—tan52.57) — 26.78 K 1.005x103

AT0'

464.26-288

(.6,To -)stage

26.78

— 6.58 7 stages

Ans.

The revised stage temperature rise = (AT0')stage —

464.26-288

— 25.18 K

As a thumb rule, the first and last stages are less loaded than the rest middle stages to avoid separation.

Thus, (AT0')is = (6,7'0')/s = 20 K and (AT0')es=25K 18.23.3. Stage-by-Stage Design. Design of Stage-1 (Fig. 18.46 ) The stage work = wstage = Xu (Cw2 — CWI) = (C AT0 -)stage p

Since Cwt = 0, so Cw2 — tanh

C_p

T0')- stage

—

1.005x103x20 _ 76.45 m/s 0.98x268.26

u u 268.26 or 131 = 67.54° Ca = 150 u — Cw2

tanP2 — Ca

268.26-75.45 or (32 =57.8° 150

Ans.

Ans.

Steam & Gas Turbines And Power Plant Engineering

802

To

S

Fig. 18.46. Stage and Overall Compression Process. tang

2

76.45 Cw = = Ca 268.26

or a2 =17.67°

Ans.

The deflection of rotor blades = pi — 132 = 9.74° is reasonable. The diffusion can be checked by de Haller criterion, Cr2 cos (3t cos67.54 — — cos57.8 — 0.79, a low rate of diffusion which is aceeptable. Crt cosi32 Now it is possible to calculate the pressure ratio per stage. Suffixes 1,2,3 etc. will indicate the stages (P02) .... [ 1 ÷ Ti s (Laci stage P01 Tot )

.5 1 + 0.9x20

=[

3.5

288

= 1.236

(pm), = 1.01 x 1.236 = 1.2495 bar and (7'02)1 = 288 + 20 = 308 K The reaction is given by

OnR

CI =

— 1—

(Al)stage or

SI = 1

Ca

2u

(C

(tana2 + Mucci) = 1 —

ya

+C

2u

76.45 — 0.8575 2x268.26

The degree of reaction is high in the first stage. However, this is necessary for low hub-tip ratio to avoid negative value of reaction at root radius. Design of Stage-2 : For second stage, stage temperature rise = (AT0')stage, 2 = 25 K. X = 0.93 and assume = 0.7 for second stage. From this data we can find out 131 and P2. We have X OTOstage,2— C

P

0.93 x268.26x150(tan(3i — tan132) uCa(tanfli — tan132) = 25 —

1.005x103

803

Compressors or

tan 131 — tan132 = 0.671

(1)

C

= or

2u

(tanPi + tan(3z) = 0.7 =

150 (tan131 + tan(3z) 2x268.26

(tanPi + tan(32) = 2.503

(2)

From (1) and (2), we have, 131 = 64.2° and 132 = 47.21° u = Mai + tan13 — 268.26 150 Ca 1

Further, or

Ans.

tana1+ tan131= 1.788 or a1= 12.64°

Ans.

Again, — = tana2 + tan12 = 1.788 or a2 = 45.65° Ca

Ans.

The whirl velocities are given by = C.,1 = Ca Mai = 150 tan 12.66 = 30.17 m/s Cwt = Ca tana2 = 150 tan 45.65 = 130.78 m/s Thus change in whirl velocity= AC., = Cw2 — Cw1 = 130.78 — 30.17 = 100.61 m/s The required change in whirl velocity is 100.61 m/s, compared to 78.45 m/s of the first stage. This higher value is due to the higher stage temperature rise and the lower workdone factor. The fluid deflection in the rotor blades = 131 —132 = 64.2 — 47.21 = 16.99 degrees The effect of degree of reaction on velocity diagram for first and second stage is given in Fig. 18.47. The de Haller number for the second stage rotor cos62'41 c°431 — 0.755 which is satisfactory. — cos47.21 cos132 The- de Haller number for the first stage stator taking (a1)2 = (al), Cl = (cosa2) '._ cos17.67 _ n 02

(C2 1

C2

1

cosa l

cos

''' '

1

This means that diffusion is small as a result of high degree of reaction due to absence of guide valnes. This shows guide vanes at the inlet should be given in the case of medium and high duty engines. Fig. 18.47 shows the velocity triangle for the second stage. The outlet pressure and temperature ( Fig. 18.29b) are given by 3.3 fro3) (1 + Tis (AT0')stase) (i 0.390x825) 0.33 = 1.28 iJo Or

T()I

(p 1 = (poilx 1.28 = (p0 ix 1.28 = 1.2495 x 1.28 = 1.5999 bar 03

Steam & Gas Turbines And Power Plant Engineering

804

(b) Scond stage

(a) First stage Fig.

18.47. Effect of Degree of Reaction on Velocity Diagram.

(T03) = (To21 (AT01)stage = 308 + 25 = 333 K At this stage (a1)3 will be for the third stage. It is worth to note that the degree of reaction is directly related to the stage of the velocity diagram. The reaction is also given by SI = — C,„,,,/u. The velocity triangle for the first and second stage shown in Fig. 18.47 yields C,„„,/u very small and thus the corresponding reaction is high and the velocity diagram is highly skewed. The high degree of reaction in the first stage is due to the decision of dispensing with inlet guide vanes and use a pure axial inlet velocity. The degree of reaction has reduced in the second stage but in other stage the requirement is 50% reaction. Design of Stage 3 For the third stage an attempt should be made to use 50 percent reaction design by assuming a stage temperature rise of 25 K and workdone factor of 0.88. Following the previous procedure, we have tanPi — tan = 2

tank, + tanf32 =

(ATOsfage p U Ca

25 X 1.005 x 103 — 0.7095 0.88 x 150 x 268.26

2u 0.5 x 2 x 268.26 c = — 1.788 150

Solution of above two equation gives (131)3 = 57.01 degrees and ((32)3 = 31.49 degrees de Haller number based on this result yields (cosPi 3 — (cos 57.01 _ 0.712 which seems to be reasonable. cos 31.44 3 cos(32 Now the pressure and temperature at the outlet of 3rd stage is

Ans.

Compressors

3.5 (P03) = + 1stage (ATO Pol 3

(T01)3

805

3.5 0.9 x 25 = 1.257 333 )

3 = 01)3 x 1.257 = (p02)2 x 1.257 = 1.5999 x 1.257 = 2.011 bar

(P 03)

(T03)3 = (T02)2 + 25

= 333 + 25 = 358 K

Since the reaction is 50 percent, so from the symmetry of the velocity diagram, (a1)3 = (132)3 = 31.49 degrees and (a2)2 = 01)3 = 57.01 degrees

Ans.

The whirl velocities (C.,1)3 = Ca tanai = 150 x tan 31.49 = 80.9 m/s (Cw2)3 = Catalla2 = 150 x 57.01 = 187.25 Design of Stages 4, 5 and 6 Stage 4, 5 and 6 may be designed following the previous procedure by assuming a workdone factor of 0.83 and 50% reaction and the results may be tabulated. Design of Final Stage 7 The fmal stage, 7th may be designed to give the required overall pressure ratio by assuming 50% reaction. 18.23.4. Comments about Annulus Shape. When the flow leaves the last stator blades it has whirl component of velocity whereas ideally the flow should be axial at entry to the combustion chamber. In general, the flow is straightened by incorporating vanes after the final compressor stage and these can form past of the necessary diffuser at entry to the combustion chamber. It is worth to note that all the design preliminary calculations have been carried out on the basis of a constant mean diameter of the compressor if a sketch is drawn on the basis of chosen scale of the compressor and turbine annuli, it results in an awkward shape of combustor and the required change in flow direction will cause additional pressure losses. The other satisfactory alternative is to design the compressor for a constant outer diameter which result in the increase in mean blade speed with stage number which makes possible to reduce the number of stages. Axial flow compressors designed on the basis of constant inner diameter, or constant mean diameter or constant outer diameter are found in service. Compressors having constant inner diameter are generally used in industrial units as it allow the use of same diameter of rotor disc resulting in the reduction of cost. Constant outer diameter compressors are commonly used in aircraft engines as they require minimum number of stages of compressors. 18.23.5. Variation of Air Angles from Root to Tip. Variation of air angles from root to tip plays an important rote in designing the blade shape. The following three methods of designing may be used to compare the variation of air angles by assuming 50% reaction at the mean radius. a) Free vortex, r C = constant b) Constant reaction, 0 = constant b

c) Exponential, Cl = a -- and Cw2 = a + -IL when n = 0

Steam & Gas Turbines And Power Plant Engineering

806

It has been observed on the outcome of design calculations that the three design methods considered above all have some advantages and the final choice of design would depend largely upon the previous experience of designer's team. The method of constant reaction offers less twisted bladings and hence attractive but in this case the radial equilibrium is ignored. Further, with the relatively low hub-tip ratio, the constant reaction is not the efficient method. The exponential design results in a substantial variation in axial velocity both across the annulus and through the stage. Further, exponential design yields appreciably lower rotor tip March number. Here, the calculation for free vortex will be given. First Stage In the first stage, there is no inlet guide vanes, so there is no whirl component, ie., al = 0 The inlet rotor blade angles at root, mean and tip are given as follows :u ir tariPir

a

u

271r, N 27c x 0.1186 x 240 ,i3ir = 67.54. —C — 150 a 2nr. N

tani31. —

— u„ 2nri N

tan131,

a

Ca

—

2Th X 0.1779 x 240 ; 150

= 67.54.

27c x 0.2372 x 240 „ = 74.72 . 1 50 ' R ll

For simplicity here blade angles have been given at only three locations while for designing of blades, angles are to be calculated at other locations also. In order to calculate the air angles 132 and a2 the radial variation of C.,2 is needed. Since there is a reduction of annulus area through the compressor, the blade height at exit from the rotor will be slightly less than inlet, so it is necessary to calculate the tip and root radius at exit from the rotor blades to find the variation of At the stator exit (say state 3 ) the absolute velocity may be taken as the inlet of rotor of second stage. Ca 150 a— — 153m/s C3 = (C1 )2 — cos 12.64 cos a3 (cosal)2 T3 = (T1)2 (T02)1 = 308 — 1532/(2 x 1.005 x 103) = 296.35 K (296.35 P3 = (P1)2 = (PO2) X 308

3.5

3.5

1.2495

196.35) 308

= 1.0917 bar

1.097 x 102 (P1)2 1.283 kg/m3 P3 = (P1)2 = R(T1)2 — 0.287 x 296.35 — A3 = (A1)2 —

22 inin — — — 0.1143 m3 P3 Ca3 (p1)2 Ca 1.283 x 150

The height of blade at the exit of stator = 1 —

A3 — 0.1143 — 0.1022 m 2itrIn 27t x 0.1779

Ans.

8U/

Compressors The radii at stator exit, rr= rm — — = 0.1779 — 2 r = r + — = 0.1779 + I in 2

0.1022

— 0.1268 m

Ans.

0.1022 0.1022 2 — 2 — 0.229 m

Ans.

Thus the radii at exit from the first stage rotor blades will be the mean of those (with negligible error) at rotor inlet and stator exit. Thus, at first stage rotor exit r—

0.1268 + 0.1186 0.229 + 0.2024 — 0.1227 m — 0.2157 m and rr— 2 2

Ans.

For free vortex flow, r Cwt = constant Thus applying the free vortex condition at root of tip, we have rr Cw2r = constant = r or

r m Cw2r= r

w

r

— 0.1779 x 76.45 = 110.84 m/s 0.1227

r

Cwv =

L

Cw =

0.1779 x 76.45 = 59.39 m/s 0.229

The air angles are given by C Cw2 tana2 = and tan(2,2 = (u — Cw2)/Ca a

Thus, tana2r

cw2r 110.84 ; a 2r = 40.51° — 150

cal tana2m —

a

Ans.

76.45 — 150 ' a2m = 30°

Ans.

CO I— 59.39 , a21 = 24° talla2i Ca 150 tani32r = (Ur Cw2rY Ca = (2nrrN —C,v2)/Ca — Or

Ans. 211 x 0.1227 x 240 — 110.84 150

I32r = 29.2°

2nr„,N —Cw2m tan(32m —

ca 27t riN—Cw2, tank = C

2TC X

0.1779 x 240 — 76.45 or 132m = 57.74° 150

27c x 0.2157 x 240 — 59.39

150

or (32i =

67.29°

Ans. Fig. 18.48.(a). shows the'radial variation of air angles of the first stage. It is obvious from the result that the deflection (PI - 02) is the largest at the root and decreases along the blade height, however, the blade is to be twisted along the height for shockless entry The degree of reactions are as follows. a

Steam & Gas Turbines And Power Plant Engineering

808

70

Pi 131, 67.29 57.24

a) 12 40.5 co) a) 30 a) 29.2

132 = 24°

rn co

c1 = 0

0

Root

Tp

Mean

Fig. 18.48.(a). First Stage, Radial Variation of Air Angles. Si r = 1 — 52~

—1

110.84 2r —1— — 0.769 s-/ 2u 2 x 240 Cw2, 2u

1

—

1—

Cw2m 2u

—1—

76.45 — 0.84 2 x 240

59." 240 — 0.876 2x

The degree of reaction is obviously high in the first stage The velocity diagrams of first stage at root, mean and tip are shown in Fig. 18.48.(b). Second to Last Stages Following the above variation of air angle for second stage to last stage are calculated and plotted. In all the cases free vortex with 50% reaction at the mean radius have been assumed for designing. 18.23.6. Blade Design and Construction of Shape. The correct geometry of the blade forms may be determined with the help of air angles distribution. The following are the design requirements of any particular blade row :(i) It should turn the air through the required angle; 0, — 132) in the case of rotor and (a2 — a3) ie (a2 — al) in the case of stator.

Root

Mean Fig. 18.48.(b). First Stage Velocity Diagram.

Tip

Compressors

809

(ii) Diffusion process should take place with optimum efficiency, i.e., with minimum loss of stagnation pressure. A large number of variables are involved in the geometry of compressor blade row, so the design becomes to a certain extent dependent on the particular preference and previous experience of the designer. Generally, correlated experimental results from wind tunnel tests are main source of informations, Fig. 18.49 shows the test results obtained for a wide range of geometrical forms of the cascade. In general, the design practice is to select a deflection which corresponds to a defmite proportion of the stalling deflection. This proportion is 0.8, so the selected or nominal deflection c = 0.8 es where es is the stalling deflection. Take the example of the design of first stage rotor which is based on free vortex design At r m= 0.1779 m, — 132m = 67.54 — 57.74 = 9.8°, a2 = 17.67° 5734°,e * = With s = 9.8° and a2 = 17.67°, the value of s/c is available from the Fig 18.50 of design deflection curve, s/c = 2.0 = 67.54°,132m

The blade height of the first stage rotor exit = / = r1 — tr = 0.2157 — 0.1227 = 0.093 m / The blade chord in general = c = 3 = 0.093 — 0.031 m 3 Thus, the pitch of blades = s = 2 x 0.031 = 0.062m 27crm _ 2n x 0.1779 The number of blades z=— — 18.02 r-r. 18. s 0.062 There is no need of corrected value of c and s after rounding off the value of z = 18.02

50

40 E.

c?

30

E'

to "

20 11.3"; g 0.075

40 cf, t.1.) 30 C O

20

10 a.)

0.050 < 10

2 0.025 a)

Minimum 0 —20 —15 —10 —5 0 5 Incidence i, degrees Fig. 18.49. Mean Deflection and Mean Stagnation Pressure Loss.

Twice Minimum 10

0 —10 0 10 20 30 40 50 60 70 Air outlet angle ce.2, degrees Fig. 18.50. Deflection Curves.

Steam & Gas Turbines And Power Plant Engineering

810

18. .. The shape of rotor blade angles will be based on the variation of 131 and 132 while the shape of stator blade angles will depend on the variation of al and a2. In fact, ignoring losses in recess, a2 will be the inlet angle of stator blades and al will be exit angles of stator blades. In choosing the blade angles at inlet, the angle of incidence has to be considered and at outlet the deviation angle is to be considered. Rotor outlet angle : 02' =132 + 8, similarly, Rotor inlet angle : 131' =131 + i Stator outlet angle : al' = al + 8

Stator inlet angle : a2' = a2 + i

Where i and 8 are incidence and deviation angle respectively. It is to be noted that, ideally, the mean direction of the air leaving the cascade would be that of the outlet angle of the blades, but in practice it is found that there is a deviation which is due to the reluctance of air to turn through the full angle required by the shape of blades. The deviation = 8 = m 0 4(7s/c) a P2 ( 22 + 0.1 — for rotor blades where m = 0.23 c 50 0

2

OC,

= 0.23 (-4-) + 0.1H for stator blades 50 a = distance of the point of maximum camber from the leading edge of the blade. 2 a = 1 for circular arc camber line Construction of Blade Shape . Assume circular arc camber-line for the construction of blade shape. The deviation angle is 57.74 ) (4(:) 0 = 0.393 0 50

8 = m0 -Nkri/c) = 0.23 x 1 + 0.1 x (

-p2i and 1311 = 01 -

Since 0 = or

0 = 011 - 02 + 8 =01' -02 + 0.3730

Assuming zero incidence (i = 0), 131' = (31

-132 =

0.6070 = or

0= 16.14° and 132' =

02 = 67.54

- 57.74

-e= 67.54 - 16.14 = 51.4°

The deviation angle = 8 = 0.393 x 16.14 = 6.34° The position of the blade chord is fixed relative to the axial direction by the stagger angle 4 which is expressed as , 4 = 131

20

Substituting the values, the stagger angle is 67.54 16. 14 - 59.4r 4= 2

Compressors

C

4.01 — 4.55 — 4.90 —

111 — 4.06 — 4.88 — 4.89

4.98 —

— 5.02

4.76 —

— 4.79

4.30 —

— 4.31

3.70 —

14 3.72

2.91 —

— 3.00

Centre or camber arc

Stator

Camber-line length L

Rotor

811

2.02 — — 2.15 1.05 — — 1.20 0.60 — — 0.68 0 0 X

Upper surface X%L

Lower surface Y%L

Fig. 18.51. Base Profile (C series) and Construction of Blade Shape.

Fig. 18.51 shows the method of construction of blade shape. The chord line AB is drawn equal to 0.031 m long at 57.47° to the axial direction. The lines AC and BD are drawn at an angle of p,' and 132' respectively. A circular arc is constructed tangential to these lines and having AB as chord. This arc is the camber line of the blade around which an aerofoil section as depicted in Fig. 18.51 RAF 27 profile called `C' series can be built up. Actual W The method described above Cs4 can be applied to a selected numStalled ber of points along the blade Ideal liwith al, p2 constant length. This is to be remembered that having arc fixed the pitch at the mean diameter by the choice of a particular number of blades, '--„ the pitch at all other radii is deStalled 8 termined. Similarly, the length of a) chord like previous case is deter1.1 is a) mined of other radii. By this E way, the whole blade shape can be constructed. 18.24. Off-design Performance Flow coefficient cp = Aerodynamic design of axial flow compressor meets the reFig. 18.52. Stage Characteristics.

Steam & Gas Turbines And Power Plant Engineering

812

quirements of a particular design point of mass flow, pressure ratio and efficiency. But any compressors will be required to operate at conditions different from the design point including engine starting, idling, reduced power, maximum power, acceleration and deceleration. This means that a compressor must be capable of satisfactory operation over a wide range of rotational speeds and inlet conditions. The temperature rise in a stage is given by uCa AT°stage' = (tani3i — tanf32) = — [14— Co (tutu, + tait32)] c c Or

C p

stage

u2

l 1—

(tana + tan(32) or = 1 — (I) K

(18.70)

c ATo ' where w — P st age is called temperature coefficient u2 Ca = — = is called flow coefficient and K = (tanai + tan(32) u The expression 18.70 gives the dotted line in Fig. 18.52. Considering the various losses, the actual curve for tv is plotted by firm line. Similarly, the stage efficiency is also plotted. 18.25. Characteristics of Centrifugal and Axial Flow Compressor Both centrifugal and axial flow compressors have well defined operating characteristics. They are normally plotted in dimensionless form. Rig testing is used to determine the characteristics. Fig. 18.53 shows a typical constant speed characteristic for one stage of a compressor in terms of pressure rise and flow coefficient. The effect of positive and negative incidence on the stage pressure rise is also depicted. Fig. 18.54 shows the characteristic of a centrifugal compressor at various speed in dimensionless from. The pressure curve is flatter in the case of centrifugal compressor. The maximum flow limit is set by choking at the eye, the critical flow limit causes the constant speed curves to turn down vertically. The minimum flow limit of the impeller is set by negative incidence stall of the inducer vanes, A typical working line may be located with 10 percent surge margin, the mass flow being 10 per cent greater than at surge. The constant efficiency curves are also superimposed on the curve. Fig. 18.55 shows the characteristics of axial flow compressors. At design point A (N/-470 7 7 = 100%) the flow conditions are correct for all stages through the compressor with incidence angles being ideal, the efficiency at its maximum and usually constant axial velocity. This corresponds to the majority of engine operation, At the same speed, pressure ratio could rise as far as B or fall to C. At B, the axial velocity is so low that it causes surge. But at C, the axial velocity increases progressively until the last stage encounters negative shock stall-which may cause fatigue failure of the blades. Therefore, it is necessary to avoid both extreme conditions B and C. At speed 110% and 120%, the characteristics become almost vertical below the working line and the inlet is chocked. At lower speed the characteristics is less steeper. Fig. 18.55 shows power and efficiency verses rate of flow for a particular speed. At a certain flow rate, the efficiency becomes maximum. The power consumed increases with the increase of rate of flow. This is valid for both centrifugal and axial flow compressors.

813

Compressors

A

(Ideal characteristic)

Incidence i =

Losses at high

Apo* deflection %pC2 angle

Effect of rl, and

Negative incidence

Losses of high Match number

Design point

Positive incidence

Incidence I Negative positive 0

4= Ca/u Fig. 18.53. Characteristics of a Compressor.

0

o.

i,i )ert

0

Impeller sugreline , i 2.5

Ce 0.2

‘,

Impeller stall Diffuser

100 110

stall 90 80 A 1 011A . dOP

- %WT., 6 i 50

. 70 I Ikr?"%idillp _ 44 PP: / A.11115.0. °V. 401P. r.". ...7° .•10021P101r

0.87 kEfficiency Tic -

0.88 0.2 0.4 0.6 0.8 1.0 Mass flow, m 4T0,/pol x 1.013/'I288)

1.2

1:4

Fig. 18.54. Characteristics of Centrifugal Compressor.

18.26. Compressor Materials and Manufacture 18.26.1. Rotor bladings. The following compressor blade materials are in common use : (a) Fibrous composites, (b) Aluminium, (c) Titanium, (d) Steel, (e) Nickel alloy The above listings are in increasing order of weight and their ability to withstand high temperature. The last three are the most common and a compressor rotor may use each in sequence to match the rate of temperature rise through the machine. Titanium is widely used in the front stages of aircraft engines for its toughness, strength and lightness, but is is prone to rapid fatigue failure. Steel is suitable above 700K. 18.26.2. Rotor. The rotor material includes steel for shafts and discs but aircraft engines may use titanium at the front stages and nickel alloy at the rear. The rotors are forged. All components are usually polished to remove highly stresses sharp edges and generously filled to avoid sharp internal stresses. 18.26.3. Stator bladings. Stator bladings may be of the same materials but steel is the most common. In this case, the blade stressing is different and involves no centrifugal loadings.

Steam & Gas Turbines And Power Plant Engineering

814

First stage positive shock stall

A

Detailed causes of compressor surge Surge line

4.0

Last stage positive stall

120%

Front stage in rotating First stage positive stall stall 1 RS.C, 2 R.S.CS

100

8 70%

1.0

choked

90% percentage of design n'IT01

3 R.S.C,

40%

ot• Inlet

Last stage negative shock stall

5(1A Laststage negative Stall drop stall below this line out line . Mass flow, ni4ToliToi

Fig. 18.55. Characteristics of Axial Flow Compressor.

Surge line

Efficiency at N,

/ / •—/--••• \

/ \ N2

Power consumed at N,

1\11>N2>N3>N4 Rate of Flow Fig. 18.56. Power and Efficiency vs. Rate of Flow of Axial Flow Compressor.

18.26.4. Casings. Casings may be of cast magnesium, aluminium, steel or iron or fabricated from titanium or steel. 18.26.5. Manufacture. Blades and vanes surfaces are manufactured by specialist techniques as they involve airfoil sections and may be twisted or curved or tapered along the height of blade. Generally, the profiles required at the hub and tip are determined and they are joined by straight ruled lines. If this "Single ruled" design philosophy is inadequate, a

Compressors

815

profile at mid-height may be chosen and the surface "double ruled" from the middle to each end separately is prepared. After establishing the geometry of the family of ruled lines they be machined in turn► by a milling machine, reset carefully for each line to generate the shape required in a master block from which forging die may be copy machined. This ensures the accurate forging of blades to their finished size, requiring only fettling and polishing. In the advanced method, computer instructs a numerically controlled machines to make forging dies ensuring accurate forging. 18.26.6. Case Study : In MS 7001F gas turbine, higher strength alloys have been applied in order to accommodate the increased compressor blade stresses. Custom 450 stainless steel has been selected for IGV and stages 0 through 8. A higher strength version of AISI 403 with calumbinm addition is the alloy of choice in stages 9 through the exit guide vanes. The MS 7001F rotor is of bolted disc and shaft constriction and consists of two major groups : the compressor and turbine. The compressor rotor is made up of 16 bladed disc plus a stub shaft on the forward end and a disc/cylinder on the aft end. The rotor assembly is bolted together by fifteen 76.2mm diameter 12 Cs bolts located, on a 940 mm diameter bolt circle. The blades are solidly retained in the wheel rims via dovelails at fixed in position via spaces. The forward stub shaft and the stage I through 15 discs are of NiCrMoV steel while the 16th and 17th stage forgings are of CrMoV steel. Stages 14 and 16 are hot spun at speed in excess of 5000 rpm. 18.27. Operational Problems The following are the operational problems. Performance Deficit. There is very likelihood that the axial compressor may fail to generate the required pressure ratio at any part of the operating speed range or it may lose efficiency. Once the compressor put in service, blade profiles get changed for various reasons : (i) Possible 'building up the dust or deposits in industrial atmosphere on the blade concave surface or some distance behind the leading edge of the concave surface will change the blade surface (ii) Salt deposits may form in the same way in marine environment (iii) Ice may form on the front stage blade in wet atmosphere near freezing point (iv) Sand in deserts or atmosphere dust and smoke may blunt the leading edge. Any changes in blade shape due to above reasons result in departures from the carefully chosen profile and cause a loss of performance. This loss can be restored by periodic cleaning. The increase of blade tip clearance also results in performance deterioration. This is caused due to erosion problems. Foreign. Object Damage. Injection of stones, small machines parts, hand, tools, tool kits, hats, ear muffs or head phones, bird, puddles etc, cause this type of damage. Operational Malfunctions. It means reduction in flow and pressure ratio at a giyen speed due to the following reasons : (a) aerodynamic stall (b) incorrect system matchirig (c) foreign object damage (d) component failure (e) intake flow distortion caused by surging. 18.28. Comparison between Centrifugal and Axial Flow Compressors. The following main features give the comparison between centrifugal and axial flow compressors. The comparison helps us to form some criterion for the choice between these two compressors.

Steam & Gas Turbines And Power Plant Engineering

816 Centrifugal (i) The flow is radial

(ii) The pressure ratio per stage is high about 5:1. Thus the unit is compact. In supersonic compressors, the pressure ratio per stage is about 10 but at the cost of efficiency. Operation is not so difficult and risky.

(iii) The isentropic efficiency is about 82%. (iv) Centrifugal compressors have a wide range of operation between surging and chocking limit. The head capacity curve is flat. The part load performance is better. (v) Centrifugal compressors have a larger frontal area and thus produces more drag force for the same mass flow rate and pressure ratio if used in aviation, (vi) When working with the contaminating fluids, the accumulation of deposits on the surface of flow passage do not adversely affect the performance. (vii) It needs low starting torque (viii) Its construction is simple, rigid and relatively cheap. At high altitude it is less sensitive to icing trouble (xi) Multistaging is slightly difficult and upto 400 bar delivery pressure is possible (x) It is.used in blowing engines in steel mills, low pressure refrigeration, big central air conditioning plants, fertiliser industry, supercharging, gas pumping in long distance pipe line, petrochemical industries. Previously, it was used in jet engines, small air craft and gas turbines.

Axial (i) The flow is axial i.e., parallel to the direction of the axis of machine. (ii) The pressure ratio per stage is low, about 1.2:1. This is due to absence of centrifugal action. To achieve the pressure ratio equal to that of per stage centrifugal, 10 to 12 stages are required. Thus, the unit is less compact and less rugged. The pressure ratio per stage in supersonic compressor is about 10 but the efficiency drops rapidly. The compression is achieved due to shock wave and the operation is risky. (iii) The isentropic efficiency is about 88%. (iv) Axial compressors have a narrow range of operation between surging and chocking limit. The part load performance is poor.

(v) Axial flow compressors have a small frontal area for the same mass flow rate and pressure ratio than that of centrifugal. This makes the axial flow compressors more suitable for jet engines. It is invariably used in aviation and power gas turbines. (vi) The accumulation of deposits on the surface of flow passages affect adversely the performance of axial flow compressors. (vii) It needs high starting torque (viii) Its construction is complex and costly. It is sensitive to icing troubles at high altitude (xi) It is most suitable for multistaging and upto 35 bar delivery. pressure in a single casing is possible (x) Due to higher efficiency and smaller frontal area, axial flow compressors are mostly used in jet engines. In power plant gas turbines, axial compressors are invariably used.

Compressors

817

EXCERCISES Viva-Voce and Theoretical 18.1. Describe the working principle of centrifugal compressor 18.2. Discuss the mechanism of pressure rise achieved in centrifugal compressor 18.3. Discuss the physical meaning of Euler's equation

[ C2 — C2 2 — 2 c2 — e 2

1

2 +

2

1

2 +

rl

2

2 r1•

18.4. Define slip factor of centrifugal compressor 18.5. Discuss the working principal of axial flow compressor with the help of T-s representation 18.6. What do you mean by prewhirl in centrifugal compressor ? 18.7. Derive an expression for CI, and CD with and without friction in the case of axial compressor cascade bladings. 18.8. Discuss the losses in axial compressor stage 18.9. Discuss surging, choking and stalling phenomena. 18.10. Discuss the characteristics of centrifugal and'axial flow compressor. 18.11. Discuss the operational problem of axial flow compressor 18.12. Give a comparison between centrifugal and axial flow compressor. 18.13. Following particulars relate to a centrifugal compressor :— Inlet diameter of impeller = 61.4 cm, Outlet diameter of impeder = 12.3 cm, Speed = 5000 rp, Velocity of flow — 61.6 m/s, Free air delivered = 1000 m3/min, Pressure ratio = 1.33, Index of compression =..L6 Assuming that all pressure rise takes place in the impeller, find the angle at which air from impeller enters the casing, breadth of the impeller blade at inlet and outlet. Ans.: pi = 21°, P2 = 17.2°, a3 = 26.6°, b1 = 14.05 cm, b2 = 6.04 cm. 18.14. The following particulars relate to a double sided centrifugal compressor :— Speed of compressor = 15,000 rpm, Mass flow rate = 10 kg/s, Total head pressure at inlet = 1.12 bar, Total temperature at inlet = 295 K, Isentropic efficiency = 0.8, Eye root diameter = 15 cm, Eye tip diameter = 20 cm, Impeller tip diameter = 50 cm. Assuming a slip factor of 0.9, a power input of 1.04 and axial entry, calculate total head pressure ratio and compressor power. Also calculate angles at impeller root and tip, assuming axial component of initial velocity to remain constant. Ans.: r = 3.134, power = 2585 kW, 31.1°, 50.5° P 18.15. A single-sided centrifugal compressor is to deliver 13.65 kg of air/s when operating at a pressure of 4:1 and a speed of 12,000 rpm. The total- head inlet conditions may be taken as 283K and 1.03 bar and an isentropic efficiency of 80%,. Calculate the overall diameter of the impeller. If the Mach number is not to exceed unity at the impeller tip, and 50% of the losses are assumed to occur in the impeller, calculate the minimum depth of the diffuser. Ans. 69 cm, 5.5 cm 18.16. All the stages of an axial flow compressor have the following conditions at the mean diameter; degree of reaction is half, blade peripheral speed 200 m/s, angles of

818

Steam & Gas Turbines And Power Plant Engineering absolute velocity at inlet to and exit from the rotor measured from the axial direction 15° and 45° respectively. A stage efficiency of 88% and workdone factor of 0.86 are to be taken as constant throughout the compressor. The combined efficiency of the intake and inlet guide vanes is 95% based on static conditions. If the ambient air pressure and temperature are 1 bar and 228K respectively find (a) the static temperature of the air at inlet to the first stage. (b) the static pressure Ans. 277K. 1.195, 3.81 bar of the air at exit from the tenth stage.

18.17. The first stage of an axial-flow compressor is to be designed for an axial velocity of 133 m/s, mass flow rate of air 22.7 kg/s. and for ambient condition of 1 bar and 10°C. At the mean radius both moving and fixed blade sections are to be suitable for relative air angles of 50° at inlet and 30° at out let. A row of guide vanes giving an outlet air angle of 30° is to be fitted before the stage. Draw the velocity triangles assuming the ratio of hub to tip diameter as 0.68, free vortex condition and ignoring end effects. Calculate (a) the compressor speed; (b) theoretical temperature rise across a stage, (c) the number of similar stages, which would theoretically be required to give a total head delivery pressure of 8.4 bar if Ans. (a) 8250 r.p.m; (b) 16°C (c) 13 the stage isentropic efficiency is 85%. 18.18. A single sided centrifugal compressor delivers 8.15 kg/s of air with a pressure ratio of 4.4 to 1 at 18000 r.p.m. The entry to the eye, for which the internal diameter is 12.7 cm, is axial and the mean axial velocity at the eye section is 148 m/s with no prewhirl. Static conditions at the eye section are 15°C and 1 bar respectively. The slip factor is 0.94 and the isentropic efficiency of compression is 78%. Neglecting the losses, calculate (a) the rise in temperature during compression if the change in kinetic energy is negligible; (b) the tip speed and impeller tip diameter; Ans. (a) 157°C (b) 493 m/s (c) 28.7 cm. (c) external diameter of the eye. 18.19. Design a centrifugal compressor for the following data. = 15 kg/s, r = 4.5, N = 16,000 rpm, p01 = 1.013 bar, T01 = 300K p 18.20. Design an axial flow compressor, for the following Data in = 500 kg/s, r = 15, N = 3000 rpm p 18.21. An axial flow compressor stage is to be designed fro a stagnation temperature rise of 22 K and axial velocitry of 155 m/s, which are constant from root to tip. The blade root, mean and tip velocities are 160, 210 and 250 m/s. The work done factor is 0.92. Calculate the stage air angles at root, mean and tip and the degree of reaction at mean and free vortex design. 18.22. The following data relate to an axial flow compressor— Overall pressure ratio = 4.0, Mass flow rate = 3 kg/s, Polytropic efficiency = 88%, Stagnation temperature rise per stage = 25 K, Absolute velocity approaching the last rotor = 165 m/s, Air angle approaching last stage measured from axial direciton = 20°, Work done factor = 0.83, Velocity diagram = symmetrical, Mean diameter of the last stage rotor = 18 cm, Ambient conditions = 1.01 bar, 288 K Calcualte the number of stages required, pressure ratio of first and last stage, rotational speed and the tength of last stage rotor blade at inlet to the stage. [Ans. 7, 1.273, 1.178, 414 rps, 1.325 cm] 18.23. The following data relate for the design of first stage of an axial flow compressor based on free vortex principal with no inlet guide vanes— N = 6000 rpm, Stagnation temperature rise per stage = 20 K, Hub-tip ratio = 0.6,

Compressors

819

Workdone factor = 0.93, Mach number relative to tip = 0.95, Isentropic efficiency of the stage = 0.89, Inlet velocity = 140 m/s, Ambient conditions = 1.01 bar and 288 K., Calculate—a) The mass flow rate entering the stage, b) The tip radius and corresponding rotor inlet and outlet angle, c) The stage stagnation pressure, d) The power required to drive compressor, e) The rotor angle at the root section. [Ans. a) 65.55 kg/s, b) 0.455 in, 64°, 56.5°, c) 1.2335, d) 1.318 MW, e) 50.82° and 18.33°] 18.24 The following date relate to an axial flow compressor— Mass flow rate = 50 kg/s, Inlet stagnation pressure & temperature = 100 kPa, 23°C, Oulet stagnation pressure = 500 kPa, Hub and tip diameter at inlet = 0.436 m and 0.728 m, The reaction at mean radius throughout = 0.5, Absolute air angles through all stages at stator exit = 28.8°, The speed of the rotor = 8000 rpm, Polytropic efficiency = 0.89, Axial velocity at mean radius = 1.05 x Average axial [Ans. 14 stages] velocity, Calculate the number of similar stages. 18.25. A multistage axial flow compressor compresses air from 1 bar and 300 K to 14 bar. Each stage is having 50% reaction and the mean blade speed is 275 m/s, flow coefficient 0.5 and stage loading factor 0.3 for all stages. Determine the flow angles, the number of stages required if the stage efficiency is 89%. 18.26. Develop a software for the design of axial flow compressor for power plant. List the design dimensions for the following data— rPo = 14, ina = 600 kg/s, N = 3000 rpm, poi = 1 bar, T01 = 300 K 18.27 Develop a software for the design of industrial centrifugal air compressor. List the design dimensions for the following data— r = 300, tha = 10 kg/s, N = 16,000 rpm,=1 p01 bar, '01 = 300 K

19 Combustion Chambers

The combustion chamber has the difficult task of burning a large quantities of fuel, supplied through the fuel burners, with extensive volume of air, supplied by the compressor, and releasing the heat in such a manner that the air is expanded and accelerated to give a smooth stream of uniformly heated gas at all conditions required by the turbine. Generally, the air-fuel ratio in an open gas turbines varies from 50:1 to 200:1 to get efficient combustion and to keep the turbine inlet temperature down to permissible limits. Aircraft and ships must carry the fuel required for their mission which has resulted in the universal use of liquid fuels. However, hydrogen has been experimented but the commercial use seems to be unlikely. Aircraft gas turbines operate over a wide range of inlet pressure and temperature within the flight envelop of Mach number and altitudr.% A typical subsonic aircraft will operate at a cruise altitude of 11 km, where the ambient pressure and temperature for ISA conditions are 0.227 bar and 216.9 K compared with the -ea level value of 1.013 bar and 288 K. Therefore, the combustor has to operate with a greatly reduced air density and mass flow at altitude, while using approximately, the same fuel air ratio as at sea level to maintain an appropriate value of turbine inlet temperature. Atmospheric condition changes quite rapidly during climb and descent and combustor chamber has to meet all operational requirements. Industrial gas turbines have a wider scope of fuel. Natural gas is the most preferred. 19.1. Requirements of Combustion Chamber The requirements of a combustion chamber are—(a) low pressure loss; (b) high combustion efficiency; (c) good flame stability; (d) low carbon deposit in combustion chamber, turbine and regenerator; (e) low weight and frontal area; (f) reliability and serviceability with reasonable life and (g) through mixing of cold air with the hot products of combustion. 19.2. Types of Combustion Chambers Types. In general, the types of combustion chamber used in open cycle are (a) tubular, (b) annular, (c) turbo-annular or cannular (d) industrial, (e) reversed and (f) cyclone. These are shown in Fig. 19.1. 19.3. The Combustion Process The combustion of a liquid fuel in the combustion chamber involves the following aspects:

Combustion Chambers

Engine casing

Liners

Engine casing

Fuel nozzles

IC. Combustion

Combustion chamber

chamber

Liner

(b)

(a) Liner

821

Combustion zone

(c)

.4--

Fuel nozzle

4,

Mixing zone

(0

(d)

Fig. 19.1. Forms of Combustion Chamber

(a) Breaking down of heavy hydrocarbons into lighter fractions. (b) Intimate mixing of molecules of these hydro-carbons with oxygen molecules. (c) Vaporization of droplets. (d) Chemical reaction themselves. Air from the compressor enters the combustion chamber at a velocity upto 170 m/s but because at this velocity, the air speed is far high for combustion; the first thing that the chamber must do to diffuse it. i.e. decelerate it and raise its static pressure. In other words, a region of low axial velocity has to be created in the chamber, so that the flame will

Swirler

Liner (Flame tube) Casing

Fuel Spray nozzle Gas to turbine

Air from compressor

Snout Diffuse

Primary zone 15 to 20%

Secondary zone 30%

Dilution zone 50 to 55%

Fig. 19.2. Construction of Combustion Chamber

822

Steam & Gas Turbines And Power Plant Engineering

remain alight throughout the range of the engine operating condition. This is achieved with the help of a diffuser as shown in Fig. 19.2. Since the overall air-fuel ratio in the combustion chamber is in the region of 100:1 while the stoichiometric ratio is approximately 15:1, the first essential requirement is that the air should be introduced in stages. In other words, the fuel must be burned with only a part of the air entering the chamber. This is achieved by means of a flame tube (combustion liner) that has various devices for—metering the air flow distribution along the chamber. There are three stages of introduction of air such as —primary, secondary and tertiary (dilution) as shown in Fig. 19.2. Around 15 to 20% of the air is introduced around the primary zone to provide the necessary high temperature for rapid combustion. The air-fuel ratio in this zone is around 15:1. Some 30% of the totzl air is then introduced through the holes in the flame tube in the secondary zone to complete the combustion. In order to achieve high combustion efficiency, this air must be injected carefully at the right points in the process, to avoid chilling the flame locally and drastically reducing the reaction rate in the neighbourhood. Finally, in the dilution zone, the remaining air is mixed with the products of combustion to cool them down from about 2000°C to the temperature required at inlet to the turbine, around 1400°C. Sufficient turbulence must be promoted so that the hot and cold streams are thoroughly mixed to give the desired outlet temperature distribution. It is essential to have a self-piloting flame (flame stabilisation) in the air stream of combustion chamber liner. The zonal method of introducing the air does not provide flame stabilization. Therefore, the third essential feature is to produce a recirculating flow pattern (i.e. toroidal vortex similar to a smoke ring) which directs some of the burning mixture in the primary zone back on to the incoming fuel and air. There are various method of flame stabilization. Fig. 19.2. shows one method of flame stabilization in which the fuel is injected in the same direction as the air stream and the primary air is introduced through twisted radial vanes, known as swirl vanes. This swirl vanes produce a vortex motion of the air which induces a region of low pressure along the axis of chamber. This vortex motion is sometimes enhanced by injecting the secondary air through short tangential chutes in the flame tubes, instead of through plain holes. The net result is the production of recirculating flow i.e. the burning gases tend to flow towards the region of low pressure, and some portion of them is swept around towards the jet of fuel as indicated by arrows. A spark initiated by a spark plug initiates the combustion and for this there is an ignition system. The liner is typically made of a nickel alloy which can operate up to about 1400K, or of cobalt alloy which may run above 1400K. 19.3.1. Case Study. In a particular gas turbine MS 7001F[51], the combustion system consists of 14 combustion chambers with 356 mm nominal diameter combustion liner. Transition pieces conduct the combustion gases to the first-stage nozzle. This particular design provides for extensive and effective film cooling of the inner wall as well as penetrations for combustion and dilution air and for cross-fire tube connections. These liners are constructed of Haselloy-X materials with the addition of HS-188 in the lower portion and the application of thermal barrier coating to the internal surface. These additions provide for improved high temperature strength and a reduction of metal temperatures and thermal gradients. A flow sleeve surrounds the liner to provide a controlled flow

823

Combustion Chambers

path for the combustion, dilution, and cooling air. The liner cap provides multifuel (six in number) nozzles facility. 19.4. Factors Affecting Combustion Chamber Performance. The factors which affect the performance of combustion chamber are as follows: (a) Pressure loss (b) Combustion efficiency (c) Outlet temperature distribution (d) Stability limits (e) Combustion intensity (a) Combustion chamber pressure loss. It is due to two distinct causes (i) skin friction and turbulence and (ii) the rise in temperature due to combustion. The pressure loss is around 2% of combustion entry pressure. From the experimental results, the overall pressure loss may be expressed by an expression as T Pressure loss factor (PLF) — m. 2 /2p

Am

—

K +K 1

2 T 01

— 1)

where A. = maximum cross-section area of the chamber and K1 and K2 = Constant The pressure loss factor (PLF) is shown in Fig. 19.3 which varies with temperatue ratio. (b) The combustion efficiency

(n.b)./t is defined as

Theoretical fuel for actual AT 1"6 — Actual fuel for actual AT The combustion efficiency is around 98%. (c) Outlet temperature distribution. A desired outlet temperature distribution helps in reducing the hot spot and thermal stresses in blades. (d) Combustion stability. It means smooth burning and the ability of the flame to remain alight over a wide operating range. For any particular combustion chamber there is

Pressure loss factor --).

60 50 A in co .o

40

A

30

Fundamental pressure loss

20

E 0

Cold loss

Stoichiometric

ocx`

20 0

Stable region

I

1

.—L

V

2 T 02 Temperature rat o

Fig. 19.3. Pressure Loss Factor.

Peak air velocity

—0

3

Air velocity, m/s

Fig. 19.4. Stability Loop.

824

Steam & Gas Turbines And Power Plant Engineering

both a rich and weak limit to the air-fuel ratio beyond which the flame is unstable (Fig. 19.4.) (e) Combustion intensity. It is defined as Heat release rate Combustion intensity — Combustion volume x Pressure It is expressed in kW/m3 atm. It is obvious that for a fixed value of heat release rate and pressure, a lower combustion volume i.e. smaller size of combustion chamber is needed to obtain high combustion intensity. 19.5. Flame Tube Cooling The increase in permissible turbine inlet temperature is the almost desire of the designer of gas turbine. This increase is limited by the turbine blade material, turbine blade cooling and flame tube cooling. The flame tube receives energy by convection from the hot gases and by radiation from the flame. It is cooled through convection by cooling air coming from compressor in the annulus and by radiation to the outer casing. But this simple cooling is not sufficient to maintain the tube wall a safe temperature. One of the common practice is the film cooling of flame tube as shown in Fig. 19.5(a) in which a narrow annular gap between the overlapping section of the flame tube is left so that a film of cooling air is swept along the inner surface. Corrugates, wigglestrip, spot welded to successive lengths of flame tube, provides adequate stiffness with annular gaps which donot vary too much with thermal expansion. Another method employs a ring of small holes with an internal splash ring to deflect the jet along the inner surface as shown in Fig. 19.5(b). The recent technique of flame tube cooling is transpiration cooling in which the cooling air enters a network of passages within the flame tube wall before exiting to form an insulating film of air.

(a)

(b)

Fig. 19.5. Film Cooling of Flame-tube of Combustor. 19.6. Gas Turbine Emissions Though the combustion in gas turbine is essentially a clean process due to steady flow process in which hydrocarbon fuel is burned with a large amount of excess air in order to maintain the turbine inlet temperature at an appropriate level. In recent years, the control of emissions is one of the important factor in the design of gas turbine. The exhaust of any gas turbine consists primarily CO2, CO, H2O, 02, N2 and unburned HC. The pollutants include oxides of nitrogen (NO), carbon monoxide (CO), unburned HC and SO2. NOx may react in the presence of sunlight to produce 'smog' which is similar to brownish cloud. NOx also causes acid rain in combination with moisture in the air and ozone depletion leading to increase in incidence of skin cancer. The unburned HC also contain carcinogens which is injurious to health. CO is well known fatal if inhaled in sufficient amount. A result of these harmful effects, the control of emissions is an highly

Combustion Chambers

825

important topic worldwide. Fig. 19.6 shows the dependence of Stoichiometric mixture emissions on air-fuel ratio. The formation of NOR is due to the flame temperature. This is theoretically a maximum at stoichiometric condition and will fall off at both rich and lean mixture. It is important to note that while NO could be reduced by well away from stoichiometric conditions this results in increasing formation of both CO and unburned HC. This is well known fact that the rate of formation of NO varies exponentially with flame temperature, so Rich +—I-3. Lean Air/fuel ratio the key to reducing NO is to reduce flame temperature. Further, an increase in resiFig. 19.6. Emissions As a Function dence time of the fluid in combustor which of A/F Ratio. correspondence to the increase in combustor volume provides favourable effect on reducing both CO and unburned HC. 19.6.1. Methods for Reducing Emissions. There are three major methods of minimizing emissions. They are— (i) Water or steam injection into the combustor (ii) Selective catalytic reduction (iii) Dry low NOR (i) Water or Steam Injection. The purpose of water injection into the combustion chamber is to decrease flame temperature. The amounts of water required are substantial and demineralised must be used to prevent corrosive deposits in the turbine. Though, there is an small increase in power output but it is acompanied by the decrease in thermal efficiency. Further by increasing the water/fuel ratio to decrease NOR, the emissions of CO and unburned HC increase. But every where water in sufficient quantity is not available. Steam injection into the combustion chambers reduces flame temperature which in turn reduces NOx emission. In combined cycle or co-generation plants, steam is generated in heat recovery steam generator (HRSG). The requirement of steam for steam injection is also very high. The drawback is the same as in water injection. (ii) Selective Catalytic Reduction (SCR). Selective catalytic reduction is used where extremely low limits of NO is specified (f \gCooling primary air " secondary air

ii

0 42 310 U

O

S

Fig. 22.2. Pressure, Temperature and Velocity Distribution in Turbojet.

Fig. 22.3. T—s Diagram of Turbojet.

nozzle. The basic cycle for the turbojet engine is the Joule or Braytons,ycle_ as shown in the Fig. 22.3. In process 0 to 1, the air entering from atmosphere is diffused isentropically from velocity Ca down to zero (i.e. C1 = 0). This means that the diffuser has an efficiency of 100 percent. This is termed as ram compression. Process 0 — 1' is the actual process. Process 1' to 2 shows the isentropic compression of air and 1' to 2' shows the actual compression of air. 2 to 3 shows the ideal addition of heat at constant pressure p2 = p3 and 2' to 3 shows the actual heat addition at constant pressure /32 = p3. Process 3 to 4 shows the isentropic expansion of gas in the turbine and 3 to 4' shows the actual expansion in the turbine. Process 4' to 5 shows the isentropic expansion of gas in the nozzle and process 4' to 5' shows the actual expansion of gas in the nozzle. Consider 1 kg of working fluid flowing through the system. The analysis is based on static values. However, the same analysis may be extended - based on stagnation values, by using suffix 0 in the enthalpy, temperature and pressure terminologies. Diffuser. Energy equation between states 0 and 1 gives C2a

T

C21

h+ got = 2- + ht + I of

where Cu = Velocity of catering air from atmosphere.

863

Jet Propulsion In an ideal diffuser, C1 = 0, goi = 0 and w01 = 0 C2a kJ/kg or So, enthalpy at state 1, h1 = h0 +

C2 Ti = T + 0 2

n

(22.1)

Process 0-1' shows the actual process in diffuser. h1 - h0 T1 -To Diffuser efficiency = rid = T ' - To h 1' - h0

C2

C2

h = ho+ — 2 rid or T1' = T°+ 2 . P rid Further, the pressure ratio is given by P

I

7)-0-

L

iYAY - 1)

CZ

rY - 1) 7 I

1 -1 1 + nd 2c T p

(22.2)

='[ 1 + ild 2

o

Ma

Generally, 1d varies from 0.9 to 0.93.

CZ C2 Compressor. Using energy equation, h1 +2+ q12+ wc = h2+ -1 2 •here q12 = or

w = h - h01 Neglecting Neglecting the02 change 01 in kinetic energy and potential energy, the ideal work consumed by compressor = h2 - hi = cp (T2 - T1) The actual compressor work = (w)ad = h2' hi =

!)

11

1; I (T) T1)

This compressor work has to be supplied by the turbine. Combustion Chamber. The ideal heat supplied =

= Cp(T 3

(.14

- T2)

The actual heat supplied = gaa = (1 + / a ) h3 - h2' qAa = Cpg (1+ inf hiza ) T3- p° . T2' Based on stagnation values, qac, = cpg(1 + »if /Ma) T03 - paT02' Turbine. Using energy equation between 3-4, we get h3

C23

Cq

+ q34 = + T + (C2

If

q34 = 0, then the turbine work,Iv , = (h3 - h4) +

3

- C2) 4 2

Neglecting the change in K.E., we have, w, = h3 - h4 = cp (7'3 - T4) Actual turbine work = wia = h3 - h4' = cp (T3 - T4') = c (T3 - T)rii p Based on stagnation values, woo = c (T03 - T04) rit p For the simplification, turbine work is equal to compressor work

h03- h04

Steam & Gas Turbines And Power Plant Engineering

864

(T2 — Tdc cp(T3 — T4) = C p (T3 — T4)111 — or

T4' = T3 — (T3 — T4)

= T3

p

Tic T2 — T1

ic Jet Nozzle. Using energy equation between states 4 and 5. r2 C24 5 h4+2 =5 h+ 2 C '2

5 In actual case, h4 ' + — 2 = h5' + 2

C

If C4' is very less compared to C5'2, so h4' = h5' + 25 Or Or

C5' = 4(2C (T4' — T5 1) p C5' = 4(2C i n (T4' — T5) p

(22.3) T4' — T5'

Here, lin is the nozzle efficiency which is defined as, rin —

Thermal efficiency

71 = •th —

(h3 — h' 5)— (h2' — h0) h3 — h2 '

T4 — T5 (T3 — T5') — (T2' — TO) T3 — T2'

(22.4) If the properties are given as stagnation values, then all above expressions will be in terms of stagnation values. Infact. in gas turbines and jet engines the velocities are appreciable and the analysis should be done for stagnation values. ,22.3. Thrust, Thrust Power, Propulsive Efficiency and Thermal Efficiency. 22.3.1 Thrust. Let Ca = vehicle velocity through the air, m/s. Since the atmospheric air is assumed to be at rest, the velocity of the air entering the engine, relative to the engine, is the velocity of the vehicle Ca . It is also called velocity of approach of air. Let C = velocity of the jet relative to the exist nozzle, m/s. 1

C

l 4. A in = (1 + Mai n I = mass of products leaving the nozzle for 1 kg of air l a)

--So, the thrust of the jet engine = (1 + Fuel/Air) Cj — Ca N/kg. of air/s

(22.5)

N/kg of air/s I'd VI , Neglecting the mass of fuel, T = (Ci — Ca) (22.5a) .1 It is assumed that the vehicle is flying in quiescent air. it-if If the exhaust gases are not expanded to atmospheric pressure (pa) in the propulsive t z.""' duct, the pressure pa in the plane of the exit will be greater than pa and there will be an additional pressure thrust exerted over the jet exit area Al equal to Ai (pa. — pa). Thus, the 7 net thrusts is then the sum of the momentum thrust and the pressure thrust, namely N

A.

(C — C )+ a

ma

—p) a

N/kg of aids

(22.5b) 22.3.2. Thrust Power, TP. It is the time rate of development of the useful work achieved by the engine, i.e. it is the product of the thrust times the flight velocity of the l`J

865

Jet Propulsion vehicle, or

Ci — Ca ] Ca

TP = TC, = [(1 +

W/kg

(22.6)

Neglecting the mass of the fuel, it reduces to C(C — C ) Ca ) , W/kg — a 1000 a kW/kg of air T.P. = (22.7) `22.3.3. Propulsive Power, PP. This represents the energy required to change the momentum of the mass flow of gas and may be expressed as the difference between the rate of kinetic energies of the entering air and exit gases, i.e.

P.P. = 4KE —

(1 + 2

W/kg

2

Neglecting mass of fuel, it reduces to PP — or

PP —

C2 — c2a 2 x 1000

C2 — C2a

(22.8)

W/kg

2

kW/kg

(22.9)

,./22.3.4. Propulsive Effkiency (71). It is defined as the ratio of thrust power (TP) to propulsive power (PP). TP

2[(1 + inf/rha)Cf — Ca] Ca

— PP [(1 th f/tha)C, C2a] Neglecting the mass of the fuel, 2(C, — Ca) Ca 2(Ci Ca)Ca (c„ cd (c„ —Ca) CZ —Ca

t PF 2Ca

c„

2 1 + Cf/Ca

(22.10)

(22.11)

Propulsive efficiency, tip is also called the Froude efficiency. It is a measure of the effectiveness with which the propulsive duct is being used for propelling the aircraft. t is Iobvious from the equation (22.11), that propulsive efficiency increases as the aeroplane velocity approaches jet velocity but the thrust decreasesL.-22.3.5. The Thermal Efficiency of the Plant (%). It is defined as the ratio of gain in = • E. K.E. by the jet to the heat supplied. fl `h

AK.E (thf./rha) Calorific value

+ nob.)

— c2c,

2 x (mt/ma) x Calorific value

gm)* (22.12)

Neglecting the mass of the fuel, 11th

(C.; — C2) — 2 x (inf/tha) x Calorific value

(22.13) 22.3.6. The Overall Efficiency (r1overall)* The overall efficiency is the ratio of the useful work done in overcoming drag to the energy in the fuel supplied. In other words, it is expressed as the product of thermal efficiency and propulsive efficiency. [(1 + — Ca ]Ca 11 overall = (flth rip) — (m/m a) Calorific value / (22.14)

866

Steam & Gas Turbines And Power Plant Engineering

Neglecting mass of the fuel, flove rail —

(C.

— Ca)Ca

I (111 filil

a

X

C. .

(22.15)

22.3.7. The Jet Efficiency, (II jet). It is defined as Final K E . in the jet

rife` = Isentropic heat drop in the jet pipe + carry over from turbine

(22.16)

22.4. Advantages and Disadvantages of Jet Propulsion over the Other System. The following are the advantages of jet propulsion. (i) Low specific weight.—The specific weight of jet propulsion is one-fourth to onehalf of the reciprocating engine. (ii) No unbalanced force.—There is no reciprocating parts and so jet propulsion is free from unbalanced forces. Greater reliability is thus achieved. (iii) Small frontal area.-'The frontal area of jet propulsion is less than one forth of the frontal area of the reciprocating engines which decreases nacelle drag greatly and hence makes available greater power, particularly at high loads. This reduces the air cooling problem. I (iv) No restriction in power output.—Compared to reciprocating engine the jet propulsion with greatly increased power output can be built because the power is not limited by detonation. The unit can operate over a large range of mixture strength. (v) High Speed—The speed of jet propulsion is not limited by propeller. High speed can be obtained. (vi) Neither lubrication nor radiators.—Jet propulsion requires neither internal lubrication nor radiators as reciprocating engines require. (vii) At high speed and at an altitude greater than 10,000 m, the efficiency of the jet is much higher than that of a propeller. Combustion (viii) and delivery of power are continuous, whilst peak and fluctuating pressure do not occure. (ix) There is no slip stream loss, the drag is reduced, and warm compressed air is available for cabin heating. , (x) The unit permits of a better position of the pilot whilst the absence of a propeller permits of a smaller under carriage. Disadvantages The disadvantage from which jet propulsion suffers are i) Particularly at low pressure, the thermal efficiency is lower. At low altitude and speed up to 148 m/s, the fuel consumption is 2 to 3 times that of a reciprocating engine. The plant is very noisy, materials costly and life short. iii) The compression ratio is not constant as in the reciprocating engines, but varies approximately with the square of the rotational speed. (iv) There are certain difficulties which are, ericountered in the running of the propulsive unit. 22.5. Performance of Jet Propulsions. Fig. 22.4 shows the comparison of propulsive efficiency among reciprocating, turbojet

Jet Propulsion

867

100 Reciprocating

Alcohol-oxygen rocket

80

0

Hydrogen-oxygen rocket ..7.) 60 tt) o Turbo

40 a. 2 20

1;040) U)

let:4 1:rtil:et Turbo

jet

Reciprocating propeller 190 Jet velocity, m/s

250

390

Vehicle speed

Fig. 22.4. Comparison of Propulsive Efficiency

Fig. 22.5. Comparison of Specific Fuel Consumption

and rocket. Fig. 22.5 shows the comparison of these for specific fuel consumption. Comparison of the relative performance characteristics of the various propulsions power is given below in the table 22.1. Table 22.1. Comparison of Relative Performance of Various Propulsion Power

Engine

Relative Relative Relative fuel Probable consumption weight of the frontal area best speed per power engine/power per power range m/s output output output

Reciprocaring

Turboprop

48.3 to 240

N

Eat

EM

CI)

97 to 290

0

240 to 570

:4 .. Turbo jet

\

A

.

d

.

r Ram jet

Pulse jet A

\

480 to 1/80

.N

0

ES

0

48.5 to 1180

r Liquid rocket

Above 965

868

Steam & Gas Turbines And Power Plant Engineering

Problem 22.1. In a jet propulsion unit, the total pressure and total temperature at intake to the compressor are 0.6 bar and 0°C. The speed of the propulsion unit is 190 m/s. The total temperature and total pressure of gases after the combustion entering the turbine are 750°C and 3.1 bar. The isentropic efficiencies of compressor and turbine are 85% and 80% respectively. The static back pressure of the propulsion nozzle is 0.52 bar and the efficiency of the nozzle based on total pressure drop available is 90%. Determine (a) power consumed by the compressor per kg of air (b) the air-fuel ratio if the calorific value of fuel is 41840kJ/kg of fuel (c) the total pressure of gas leaving the turbine (d) thrust per kg of air per second. Assume cp of gas = 1.1296 kJ/kgK, yg = 1.33, cra = 1.005, Ya = 1 .4 Solution. Refer to Fig. 22.6. Working in total pressures and temperatures, we have (y-lYy 0.286 pot = 436.65 K 7'02 = 01 = 273 0.6 P 01 r

(436.65 — 273) + 273) TO 1 = 0.85

(T02T

465.2K

or

T02'

(a)

iv. = cpa(T02' — Tod= 1.005(465.52 — 273) = 193.48 kW/kg

(b)

+ inf ) c T03 — ma pa To2' = i;if X C . V pg V — Cpg T03 iita C 41840 — 1.1296 x 1023 — 59.166:1 1.1296 x 1023 — 1.005 x 465.52 CpgT03 -- Cpa T02' thf

or

—

Ans.

a

Ans.

(b) Air-fuel ratio = 59.166:1 The power developed by the turbine is consumed by the compressor cps(To21 T01) = cpg(T03 — T04') (1 + riyina) Or 1.005(445.2 — 273) = 1.1296(1023 — T04') (1 + 1/59.166) or

T04'

=

1023 —

1.005(465.52 — 273) _ 854.56K 1.1296 x 1.0169

-C

S

Fig. 22.6

$

Fig. 22.7

Jet Propulsion

Also, To3 - TO4 =

(T03 - T04') •

869

or T04' = 1023 - 210.55 = 812.45K

it y/(y - 1)

1.33/0.33 T04 (. 12 45) = 1.2246 bar. = 3.18 Now, p04 = p03 1 . 7 1023 [ 03 Total pressure of gas leaving the turbine =p04 = 1.2246 bar

Ans.

For the first approximation, T04 and Am may be assumed as static value. P4 (y - 1)/y = 854.56 Now, T5 = T41(lI l

)_ 0.52 690.945K L2246°33/1 33

T5' = T4 - (T4' - T5)ri n = 854.56 - (854.56 - 690.945) x 0.9 = 707.3K 114' - h; = Cp(T4' - TO = 1.1269(854.56 - 707.3) = 166.34 kJ/kg C5' = C = 44.72 46Ahnozzle) = 44.724(1) = 576.76 m/s Ahnozzk =

Thrust = (1 + t.lif /Ma) C - C. = (1.0169 x 576.76 - 190) = 396.5 N/kg/s

Ans.

Problem 22.2. In a theoretical cycle for a jet-propulsion unit both the compression and the expansion are considered isentropic, and that heat is supplied at constant pressure. Show that the Thrust developed per kg of air per second when the velocity of approach is neglected is [2 c T (q - Or(r- 1)/Y - 0)1/2

P

P

where q is the ratio of the absolute temperatures after and before combustion, rp is the compression ratio and T is the absolute atmospheric temperature.

T3 P2 T = T, — Solution. Refer to Fig. 22.7. Here,— = p pr ,1 T =q i

2

Work consumed by compressor = cp(T2 - T1) and Workdone by the turbine = cp(T3 - T4) Net work available for KE in the jet = w = cp[(T3 - T4) - (T2 - T1)]

(22.16)

We know, T2 = TI r(Y P -1)4 and T4 = T3h4; - WY T3 = q T2 = q T1 r(Y - 1)4 and T4 = q T1

p

Putting the values in equation (22.16), we have w= Here,

1)/Y + TI ] P[q T I r(Y - " - q T1 - T1 rep( Ti = T, so w = cP. Tfq re ' 1)4 - g - re " I t f) + 1] P P

= CPT[(q - 1) (r(Y- 1)/Y - 1)] kJ/kg If C. = C4 be the outlet velocity from the jet and since the approach velocity Ca = 0, SO

C = 4(L Ah) and Here, Ah = w - 1)1]1'2 C. = [27{(g- 1)(r(Yp

Steam & Gas Turbines And Power Plant Engineering

870

Thrust = C-Co ; N/kg of air/s = [2c . T(q - 1) (r1:- 1Yr - 1) 1 / 2 p 1

Hence, proved

Problem 22.3. In a jet-propulsion unit, the air is compressed in a axial compressor from atmospheric pressure 1 bar and temperature 10°C to 3.4 bar. The fmal temperature is 1.14 times that for isentropic compression. The air is then led to a combustion chamber where the combustion takes place at constant pressure and the products of combustion at 475°C pass through a turbine which drives the compressor. The exhaust gases from the turbine are expanded in a nozzle down to atmospheric pressure. Assuming the velocity of approach is negligible, the gases are expanded isentropically in both turbines and the nozzle and values of R and y of gases are same as that of air, calculate —(a) the power required to drive the compressor per kg of air per second; (b) the air-fuel ratio if the calorific value of fuel is 43095 kJ/kg and (c) static thrust developed per kg of air per second. Solution. Refer to Fig. 22.8. Given : T1 = 283K, p2 = 3.4 bar, pi = 1 bar 72 = 748K (y -1)/y = 238 (3.4)0.286 = 401.59K

We know, T2 = Ti (it±'-2 Pt

Actual temperature rise = T2' - T1 = ( T2 - T1 ) x 1.14 or

T2' = (401.59 - 283) x 1.14 + 283 = 483 = 481.19K

wcp = c (T2' -• T1) = 1.005 (418.19 - 283) = 135.86 kJ/kg of air p (a) Hence, power required to drive the compressor = 135.86 kW/kg of air gila

= c(T3 - T21) = 1.005(748 - 418.19) = 331.459kJ/kg of air

in in x q,tic = -7° x 331.459 C. V = 43095 = -4' m

I

3 Turbine work Ahnozzle

—S Fig. 22.8

Fig. 22.9

Ans.

871

Jet Propulsion

m° 43095- 130 kg of air/kg of fuel in 331.459 f. Ans. (b) Air-fuel ratio = 130:1 ( 11 = 748 - 5271K Now, 7'5 = T3 (3.4) 286 P2 Since the power developed by the turbine is consumed by the compressor, hence Ahnazd, (h3 - h5) - (h2' - h1) = c (T3 - T5) - cp(T2' - Ti) p = 1.005(748 -257) -.135.86 = 86.245 kJ/kg of air

or

Velocity at exit of nozzle = C5 = i = 44.72- (Ahnazzle) = 44.724(86.245) = 415.36 m/s - C a= 1.0076 C.5 - 0 = 415.32 N/kg of air/s.

(c) Thrust = (1 + -137. /) 1 1a

•

Ans.

Problem 22.4. A jet propelled unit travels at 180 m/s in air at 0.65 bar and - 6°C. Air first enters a diffuser in which it is brought to rest relative to the unit and it is then compressed in a compressor through a pressure ratio of 5.8 and fed to a turbine at 925°C. The gas expands through the turbine and then through the nozzle to atmospheric pressure (i.e. 0.65 bar). The efficiencies of diffuser and nozzle are 0.9. The compressor and turbine efficiencies are 0.8. Pressure drop in the combustion chamber is 0.14 bar. Find the fuel-air ratio and the specific thrust of the unit. If the inlet cross-section of the diffuser is 0.1 m2, calculate the total thrust. Assume calorific value of fuel as 44141 kJ/kg of fuel. Solution. Refer to Fig. 22.9. For ideal diffuser i.e. for process 0-1, from energy equation C2 C2 T+ h1 = h o+ -2- or h - h = -2 2 -orT1 = 0 C2 /2cp (hi - ho) or

ht ' = h +

= + 11 a

T = 267 + and

Now,

1802 - 283.11K 2 x 10G0 x 1.0005

T = 267 + Ti

7o-

C2 C2a or T1 ' = T + 211d 4) 2cprid

1802 - 284.9K 2 x 1000 x 1.005 x 0.9

(y -1)%r (Pi) = 283.11 - 1.06 or pi = 0.65(1.06)36 = 0.797 bar = Po 267

Again, T2 = 7.1'(5.8)0286 = 284.9 x 1.655 = 471 K (T2 - TO And,

T2' = T1' +

- 284.9 + 11,

Assume Cpg

(471 - 284.9) - 491.67 K 0.9

=C =C pa p

Heat supplied = (ma + M1)cpT3 - MacpT2' = mj x C . V

Steam & Gas Turbines And Power Plant Engineering

872 or

ma

C . V - cp T3

44141 - 1.005 x 1198 - 60.48 1.005(1198 - 491.67) c (T3 - T2') p Air-fuel ratio = 60.48:1

Ans.

p3 = p2 -0.14 = 5.8 x 0.797 - 0.14 = 4.4826 bar In jet engines, the function of turbine is to drive the compressor and accessories. Here, assume that the turbine drives only compressor, so or 491.67 - 284.0 = 1198 - T4' or T4 ' = 991.23K cp(T2' = c (T3 - TO p - T) 198 - 991 23) 939.53K Also, T4 =1198 - 0 T - ' ' - 3 0.8 T3

I4 .

10

0.286

.. 3 H P4

3.5

-

1198 ( 1198 orp3- = H = 2.34 939.53 p4 939.53

4.4826 .n- 4 - 2.341 - 1.9148 bar 0.286 0.286 T4' (p4) 1.9148) 991.23 Now, -7,-- = = 1.362 or T - 727.77K 0.65 5 1.362 5 p5 And T5' = T4' - 0.9(T4' - Ti') = 91.23 - 0.9(997.23 - 727.77) = 754.11 K or

Velocity at the exit of nozzle = C = 44.724074' - h51) = 44.72'‘kcp(T4' - TO = 44.724(1.005(991.23 - 754.11) = 690.34 m/s 1 Specific thrust (1 + in f) x CI = + 60.48 690.3 = 701.75 N/kg of air/s 1 Ans. Volume of flowing air = 0.1 x 180 = 18 m3/s 0.65 x 102 Mass flow rate = 2-x 18 - 15.268 kg/s /2T = 0.287 x 267 Total thrust = 701.75 x 15.268 = 10714.31 N Ans. . Problem 22.5. For the consumption of 430 kg of petrol, a Flying Bomb had a range of 260 km, an average speed 174 m/s and a thrust of 2678 N. Assuming the calorific value of fuel 42676 kJ/kgK, the maximum temperature rise in the combustion chamber as 820°C, the diameter of the discharge nozzle as 0.304 m, the altitude of the flight as 606 m, and the c of exhaust gases as 1.05 kJ/kgK, calculate (a) The fuel-air ratio, (b) The approximate temperature of the exhaust gases and their relative velocity to the bomb, (c) The propulsive efficiency and overall efficiency of the power unit. Solution. Let in,be the mass of fuel flow rate and Ma be the man flow rate of air 42676 in f

a = 1 + m x 1.05 x 820 a

Air-fuel ratio = Ma/ h7/ - 0.01961 - 50.994:1

or

_Lilz =Fuel = 0.01961 Air ma

Ans. Let C be the velocity of flying bomb and C be the velocity of the jet relative to

873

Jet Propulsion machine. With this value of C, the air enters the machine at inlet, i.e.; Ca = C. Thrust =

Mr

Mr

Ma J

Ma

i

060 174 — 1494.25 s. Duration of flight = 260 430 1494.25 — 0.2877 kg/s Fuel consumption = in .7 = Air consumptions = ina = 0.287 x 50.994 = 14.67 kg/s Now, Thrust = 2678 = [(1 + 0.01961) C. —174] x 1.67 or

C. = 349.69 m/s

Ans.

(7c/4) (0.304)2 x 349.69 1.6968 m3/kg 14.67 + 0.2879 At an altitude of 606 metres, the barometric pressure from gas table is 0.964 bar. DV 0.964 x 1287 1.6968 569.93K Therefore, temperature of exhaust gases T = — R 0. Ans. Hence, propulsive efficiency, neglecting mass of fuel 2C 2C 2 x 174 — 66.45% C.j + Ca C.j + C 349.69 + 174 Ans. Specific volume of the exhaust gas = v =

Considering the amount of fuel 2[(1 + In( a)C — Ca ]C 11P = (1+mDIX" —Ca

Overall efficiency, rlOth

=

2[(1 + 0 01961) x 349.69 — 174)] 174 _67.29% (1 + 0.01961) x 349.69 — (174)2

[(1 + (Fuel/AOC./ — Cc,]Ca (Fuel)/Air) x C . V

[(1 + 0.01961) 349.69 — 174] x 174 _ 3.795% 1000 x 0.01961 x 42676 Problem 22.6. The following particulars relate to a jet engine which develops a thrust power 770 kW when flying at an altitude of 9120 m with a speed of 217.5 m/s :— Internal efficiency of turbine, 0.85; efficiency of compressor, 0.85; efficiency of jet tube 0.9: inlet pressure and temperature 0.306 bar and — 45.5°C; temperature of gas leaving combustion chamber, 670°C; pressure ratiu rp, 5; C. V. of fuel, 42676 kJ/kg; velocity in ducts constant at 197 m/s; for air, cp = 1.005, y = 1,4, R = 0.28; for combustion gases; cp = 1.087 kJ/kgK for gases during expansion, y = 1.33. Calculate (a) overall efficiency of the machine, (b) rate of air consumption, (c) the power

Ans.

s Fig. 22.10

874

Steam & Gas Turbines And Power Plant Engineering

developed by the turbine, (d) the outlet area of the jet tube, (e) specific fuel consumption in kg per kg thrust. Solution. Refer to Fig. 22.10 Let tn . and ina be kg of fuel and air flow rate .f C. V. ±-( = (1 + ±L 1 I )xcpg T3 -c pa T' th 2 a a (r)o.286 = 227.5 (5)0.286 = 360.48.48K T2 = '1 (T2 - T1) T2' = T1 + (lAa

or

- 227.5 + ric

(360 - 227.5) - 383.38K 0.85

in,. = 42676 --:-L -(1 + -.--L x 1.087 943 - 1.005 x 383.38 III a ma

1 -7-;L -*1 = 0.01446 = Fuel-air ratio or Air - fuel ratio = 0.01446 - 69.156:1 ina

The discharge velocity C. = C5 can not be determined from the thrust equation because the rate of air flow is unknown. It may be determined from the expression of jet efficiency. Here, C4' = 197 m/s C"/2 Final K . E . in the jet _ 1 Jel - Isentropic heat drop in jet pipe+ carry-over turbine c (T4' - 7's) + C4'2/2 pg Since the turbine's work is to drive the compressor only, so pa (T2' - T,) = c (1+ F/A) (T3 - T4') pg 1.005 x 155.88 or T4' = 943 800.93K 1.087 (1 + 0.01446) Let

r = expansion pressure ratio in turbine, i.e. rpt= p3/p4 Pt vi = expansion pr6sure ratio in jet tube, i.e. r = p4/ p5 id P3 r x r = P3 — x P4 = — A-. r = 5 PI PJ p4 P5 p5 P T4 = T -

3

(T3- TO _ 943 _ (943 - 800.93) _ 775.85K 0.85 1,

0.33/1.33 (r- 11)/y 03 T3 (1)3 (P3 P3 943 1 = —) Now, — = 1 or — = 1' = = 2.195 T4 p4 p4 P1 (775.85 P4 Hence,

Pi

=

p4 - 2.2779 p5 2.195

T4' 800.93 (r )0 .33,,1 ..,„,,, (2.2775)-,A2 = 653K PI Putting "all these values in jet efficiency equation, we have

Thus, T5

=

A

875

Jet Propulsion

CZ/2

0.9 -

1.087 x 1000 (800.93 - 653) + (197)26 or

C. = 569.53 m/s

(a) 0 ver all efficiency =

-

-

[(1 + FueVAir)Ci - Ca]Ca

° (Fuel/Air) x Calorific value [(1 + 0.01446) 569.53- 217.5] x 217.5 - 12.69% 1000 x 0.01446 x 42676

Ans.

•

C1 +

- Cal ma

Thrust power = 770 =

1000 770 x 1000 — 9.826 kg/s — [1.01446 x 569.53 - 217.5 ] 217.5

f _ (T3 - TO Power developed by the turbine = Ma (1 + =I- P M

a

P

8

= 9.826 x 1.01445 x 1.087(943 - 800.93) = 1539.37kW C2 - C4,2 Now,

Or

C pg

2

T5'

= T — 4'

(T4'

— T5')

- C4,2

569.532 - 1972

cpg' = 800.93 - 2 x 1000 x 1.087

669.57 K

The atmospheric pressure is 0.305 bar and let the exit pressure be equal to the atmospheric. 0.306x 102 0.15758 m3/kg Density of exhaust gases = p = -P— RT 0.20 x 669.57 Discharge of jet area = A x x p = ma(1 + Mf /ina) or

A - 9.826 x 1.01446 - 0.1111712 600 x 0.15758

Specific fuel consumption = 0.01446 x 9.826 x 3600 - 1.444 kg/1'hrust-h 1000 x 770/217.5

Ans. Ans.

22.6. Intake and Propelling Nozzle 22.6.1. Intake. The intake is a critical part of an aircraft engine installation. It has a significant effect on both engine efficiency and aircraft safety. The main requirement of the intake system is to minimise the pressure losses up to the compressor face and is ensure that the flow enters that compressor with a uniform pressure and velocity, at all flight conditions. It is worth to note that the non-uniform or distorted flow may cause compressor surge which can result in either engine flame-out or sever mechanical damage —ii-16-515dt de- vibration induced by unsteady aerodynamic effects. However, even with a 1 e, some flow distortion will be present during rapid maneuvering. well designed —TiTa-Further, the shape the intake system will depend upon the engine location on the e uov-e me hi-- O P-verrweve.._lv s eSIA43,,,,Icl nrut_IA. GO-rcAnc.t µ e•Are'-u (-4)L4g‘A..0,-..tot akk—v....---4.1-thm.1" lovemteAR,Or ilek:4-107

876

Steam & Gas Turbines And Power Plant Engineering

aircraft. Engines may be installed in the fuseleg, in pods (wing or rear fuselage mounted), or buried in the wing root. In all these installations the shape of the intake is different as the requirements are different. The design of intake involves a compromise between aerodynamic and structural requirements. As per the current practice, the air enters the first stage of compressor at an axial Mach number in the region of 0.4 to 0.5. The subsonic aircraft typically cruises at M = 0.8 to 0.85 while supersonic aircraft operates at M = 2 to 2.5. At take off, with zero forward speed, the engine will operate at maximum power and air flow. From the above discussion it is clear that the intake must satisfy a wide range of operating conditions and the design of the intake requires close collaboration between the aircraft and engine designers. 22.6.2. Propelling Nozzle. In the simple jet engine, a single nozzle down stream of Afterburner fuel the turbine is provided. The turbofan engine i may have two separate nozzles for the hot and cold streams or the flows may be mixed and leave from a single nozzle. Between the turbine exit and propelling nozzle, a jet pipe -(Fig. 22.11) is incorporated, the length of which is determined by the location of the engine in the aircraft. In the transition from Diffuser Nozzle Jet pipe the turbine annulus to circular jet pipe some --->i >l< l< increase in area serving as diffuser is proFig. 22.11. Propelling Nozzle System. vided to reduce the velocity cad hence the friction loss ;n the jet pipe. In the case of the requirement of thrust boosting, an after burner may be incorporated in the jet pipe {Fig. 22.11) Thus, depending on the location of the engine in the aircraft, and on whether reheat is to be incorporated for thrust boosting, the propelling nozzle will consists of some or all of the items shown in Fig. 22.11 The pertinent question before the designer comes whether a simple convergent nozzle is sufficient or whether a convergent-divergent nozzle should be employed. Even with the moderate cycle pressure ratio, the expansion pressure ratio (p04/pa) will be greater than the critical pressure ratio and thus convergent-divergent nozzle becomes necessary. But it should be noted that it is the thrust that is required and not maximum possible jet velocity. Further, the use- of a convergent-diveigent nozzle would result in significant increases in engine weight, length and diameter. This would then result in a penalty in aircraft weight and major installation difficulties. For expansion ratio upto 3, the thrust produced by a convergent nozzle is comparable with that produced by convergent-divergent nozzle. Over, and above, at operating pressure less than design value, a convergent-divergent nozzle of fixed proportion would certainly be less efficient because of the loss incurred by the formation of shock wave in the divergent portion. For the above mentioned reasons aircraft gas turbine normally employ a convergent propelling nozzle. There are other secondary advantages of convergent propelling nozzle, which are given below :— (1) Variable Area. It is essential when an after burner is incorporated. Fig. 22.12(a). shows the technique to achieve variable area by incorporating "iris" and "central plug". (ii) Thrust Reveser. This is achieved by providing "Clam" shut which reduces the

877

Jet Propulsion

'Clam' shut for reversal

Noise suppressor

b IAD

ly OPP (a)

(b)

Fig. 22.12. Provision of Variable Area, Thrust Reversal and Noise Separation.

length of runway required for landing. it is used almost universally in civil transport aircraft (Fig. 22.12(b) (iii) Noise Separation. It is the mixing of the high velocity of hot stream with the cold atmosphere and the intensity of noise decreases as the jet velocity is reduced. This is the reason that the jet noise of turbofan is less than that of simple turbojet. In any given situation, the noise level can be reduced by accelerating the mixing process and this is normally, achieved by increasing the surface area of the jet stream (Fig. 22.12(b).) This does not mean that convergent-divergent nozzles are never used. In supersonic aircraft (M = 2 to 3), convergent-divergent nozzles are used. 22.7. Optimization of the Turbojet Cycle The Following two independent parameters mainly affect the performance of turbojet engine cycle. a) Turbine inlet temperature and b) Compressor pressure ratio. However, the other two following parameters also affect. c) Cruising speed and d) Altitude In order to optimize the turbojet cycle it is necessary to carry out a series of design point calculations covering a suitable range of turbine inlet temperature and compressor pressure ratio using fixed polytropic efficiency for a certain subsonic cruise speed and an appreciable altitude. Fig. 22.13 shows a typical turbojet cycle performance. From the above result it is obvious that the specific thrust is strongly dependent on the value of turbine inlet temperature (T03). Further, in order to keep the engine as small as possible, for a given thrust, the utilisation of the highest possible temperature is desirable. At a constant pressure ratio, however, an increase in T03 will cause some increase in sfc. This trend is in contrast to the effect of T03 on shaft power cycle performance, where increasing T03 improves both specific power and sfc. In the case aircraft turbine, the gain in specific thrust with increasing temperature is invariably more important than the penalty in increased sfc, particularly at high flight speeds where small engine size is essential to reduce both weight and drag. Fig. 22.13. shows that the effect of increasing the pressure ratio (re) in order to reduce the sfc. It is obvious from the results that at a fixed value of T03, increasing the pressure ratio initially results in an increase in specific thrust but eventually leads to a decrease; and the optimum pressure ratio for maximum specific thrust increases as the value of T03 is increased.

878

Steam & Gas Turbines And Power Plant Engineering

Compressor ,ce-`6b 1800 5 pressure ratio ce" ,00- 1600 10 OQ 0.14 .. 15 c,03e 1400 25

0.16

0.12 - 1200

0.10

Ma = 0.8 alt. = 9000 m

0.08 500 600 700 800 900 1000 Specific thrust, Ns/kg Fig. 22.13. Turbojet Cycle Performance

In order to see the effect of cruising speed, a large number of results similar to shown in Fig. 22.13. will be plotted at higher cruising speed but at the same altitude. It will be found that for a given value of pressure ratio (re) and turbine inlet temperature T03 , sfc is increased but thrust is reduced with the increase in cruising speed. This is attributed due to the combination of an increase in inlet momentum drag and increase in compressor work consequent upon the rise of inlet temperature. Similarly, the results may be plotted for different altitudes while cruising speed remaining the same, will show that there will be an increase in specific thrust and a decrease in sfc with increasing altitudes, due to the fall in temperature and the resulting reduction in compressor work. It has also been noticed by increasing the design cruise speed, the optimum pressure ratio for maximum thrust is reduced which is due to the larger ram compression in the intake. The need for a higher jet velocity and the resulting higher temperature at the compressor inlet make the use of high turbine inlet temperature desirable for economic operation of specially supersonic aircraft. While dealing with the thermodynamic optimization of turbojet cycle, mechanical design considerations have to be taken into account. The choice of cycle parameters depends very much on the type of aircraft. Though, high value of T03 is thermodynamically desirable but this will require expensive alloys and cooled turbine blades leading to an increase in complexity and cost. Fig. 22.14. shows the relation between design considerations and performance, Demarcation line for cooled and uncooled blades has been given in the figure. A small business jet or trainer, for example needs simple, reliable engine for low cost. In such engines, sfc is not critical as flying time is less. Noise regulations have led to the displacement of the turbojet engine by the turbofan. Another matter interest is of the lifting engines for vertical take off and landing (VTOL) where the prime requirement was for maximum thrust per unit weight and volume. This resulted in high sfc but it was not critical due to low running times. But these requirements were possible by low pressure ratio and high turbine inlet temperature Now this type of engines are not used Previously, turbojets of high pressure ratio were used in early commercial aircraft and bombers because of the need for long range and hence low sfc. Due to

879

•Jet Propulsion

bte ssv•-..Decie vAG

sfc

ova

cc4 Lifting engine Increasing weight \ and cost

Business jet

Pressure ratio increasing —" teroPe' I I' increasing

ture

Long range subsonic

Specific thrust Fig. 22.14. Curves Showing Relations between Performance and Design Considerations.

restriction on noise, turbojet have now been superseded by turbofans for commercial subsonic aircraft. Now, the design and development of variable cycle engine are going on for supersonic aircraft which will be operating as turbofan during take-off and as turbojet of supersonic cruise conditions. It is worth to note that at different flight conditions both the thrust and sfc will vary, due to the change in air mass flow with forward speed. Due to change in intake conditions on account of attitudes the compressor pressure ratio and turbine inlet temperature will change even the engine were run at fixed rotational speed. The thrust decreases significantly with increasing altitude due to decrease in ambient pressure and density but sfc will decrease. Problem 22.7. The following data relate to a simple turbojet engine at the design point cruising at a speed of M = 0.8 and at an altitude of 10 km where pa = 0.2650 bar, Ta = 223.3 K and round speed, a = 299.5 m/s Conipressor pressure ratio = 10, Stagnation turbine inlet temperature = 1400 K, Isentropic 03 efficiency of intake = 93%, Isentropic efficiency of compressor = 87%, Isentropic efficiency of turbine = 90%, Isentropic efficiency of propelling nozzle = 95%, Mechanical transmission efficiency = 98%, Combustion efficiency = 98%, Combustion pressure loss = 4% of compressor delivery pressure, Lower calorific value of fuel = 43100 kJ/kg, cp , cpg and y for gas = 1.005, 1.148 kJ/kgK, 1.33 respectively —ms s Calculate the specific thrust, fuel-air ratio, specific fuel consumption and propulsive effiFig. 22.15 ciency. Solution. Refer to Fig. 22.15. The stagnation conditions at the intake are :— Caa = 223 3 + (0.8 x 299.5)2, 251.86 K 7'01 = Ta + 2c ' 2 x 1.005 x 10' P

Steam & Gas Turbines And Power Plant Engineering

880

Fr- 1) Pot _ [1

da

15

0.93 x 28.56

=[l

Pa -

223.3

2p a = 0.265 x 1.4818 = 0.3926 bar poi

- 1.4818

The stagnation conditions at the exit of compressor, PO2 = a X P01 = 10 x 0.3926 = 3.926 bar TO2 (POI

= (10)0.286= 1.931 or T02 = 1.931 x 251.86 = 486.341 K

T01 = P01

(T02 - T02) TO2' = TO1

-251.86 +

486.3 -251.86 -521.33 K 0.87

c g T03 — From the energy balance around combustor m (LC V) = 1 + m p f a or

pa Toe'

•-•'• x 43100 = 1 +-h/i x 1.148 x 1400 - 1.005 x 521.33 Ina

0.0261 ma Since the compressor work is equal to turbine work, hence (T02' — Tot ) h02' - hot (1 + iilf / (h03 or (1 + 4iL j cpg (T03 T04') - Pa = i rim 11,„ h°41) a lila) or

7-1- =

Or

(1 + 0.0261).x 1.148(1400 - T04') -

or

0.98 T04' = 1165.405 K and p03 =p02 x 0.96 = 3.926 x 0.96 = 3.768 bar (T03 -T04')

T04 =T03 T

PO3

- 1400

11

PO4 = (a 04

or

1.005(521.33 - 251.86)

It —1)

TO3

1400 - 1 165.405 - 1139.33 K 0.9

0133/033 ..

_ (1139.33) 1400 )

= 0.4358

p04 = 0.4358 x 3.768 = 1.642

1.642 The nozzle pressure ratio = =PO4 —= pa 0.265 6.196 The critical pressure ratio in propelling nozzle Pc = P5 PO4

PO4

=

PO4 1 ( 2 1/6.-1) =( — 2 )33/1133 = 0.54 or = 1.851 2.33 p 0.54 Y 1 a

-Pot 1104 Since — > — , hence the nozzle is choking Pa Pc

881

Jet Propulsion ) x 1165.405 = 1000.34 K = — T'2.33 +

-

The Critical temperature = Tc =

_ _ PO4 1.642 pc — p5 — 1.851 - 1:851 -.0.887 The density at the critical condition = Pc = P5

0.887 x 102 Pc 3 RT = 0.287 x 1000.34 - 0.3089 kg/m

Critical velocity C5 = C =(yRTc)112 = (1.33 x 0.276 x 1000.34 x 103)14 = 605.97 m/s As5 __ -A1=

in

m

1 - 0.0053 p5 C5 0.3089 x 605.97

The specific thrust = T =[(1 +

in

C1+: 7n1L (Pc — Pa) a = [1.0261 x 605.97 - 239.6] + 0.0053 x (0.887 - 0.265) x 105 = 711.84 The specific fuel consumption = (in ma) x 3600 0.0261 x 3600 sfc - 0.1319 kwh.N 711.84

Ans.

Ans.

2[(1 + filtha) C - Cal ./

Propulsive efficiency =ri = L p

[1 +

C:c_j2 -

2[1.0261 x 605.97 - 239.6] x 239.6 - 57.3% [1.0261 x 605.972 - 239.62]

Ans.

22.8. The Turbfan Engine L Lfr

t)1^17LAI-13 IA-At , The turbofan engine is an modified version of turbojet engine with an aim to reduce jet noise and to improve propulsive efficiency. 111 this turbine, a portion of total flow bypasses part of the compressor, combustion chamber, turbine and nozzle before being ejected through a separate nozzle (Fig. 22.16): This means that the thrust is made up of two components, the cold stream or fan thrust and the hot stream thrust. Though the arrangement shown in Fig. 22.16, is for separate exhaust for cold stream, but it is sometimes desirable to mix the two streams and eject them as a single jet having reduced velocity. In the analysis of turbofan engine, the term by-pass ratio_plays an important part. The by-pass_ratio is defined as the ratio of the flow through the by-pass duct (i.e. cold stream) ta_the-flow-at-entry_to_the_high-pressure compressor-(ire,hat-stream)-.-ThW the by-pass ratio (B) is in B= mh (22.17)

= me= and = thc thh B + 1 mh B +1 (22.18) The value of B varies from 0.3 to 8 or more If both streams are expanded to atmospheric pressure in the propelling nozzles the net

or

882

Steam & Gas Turbines And Power Plant Engineering

—14"- C.Jo 3

7 Cjh

._.—.— ._

Fig. 22.16. Twin-spool Turbofan Engine.

L

-->

7 1 /40%,./IVAZ ZZZZZAr7,01//17.0%

—*M•c,C jc mh,

- • - • • - - • -• - • - - • _ _ •

(a) Two-spool

SIPIM

Mh. Cjh

•- • -• - • - • -•_ •_ • _ • - - • -

(b) Two spool \ \

Cjc

mh, Cjh

Fig. 22.17. Configurations of Turbofan. thrust is expressed as T= (inc C + inh q— In Ca (22.19) o The conditions after, mixing will be found from enthalpy and momentum balance. In the two-spool configuration, the fan is driven by the l.p. turbine while h.p. compressor is driven by h.p. turbine (Fig. 22.16) The various configurations of turbofan engines is given in Fig. 22.17. The configuration shown in Fig. 22.17 (a) suffers from the fact that the later stages on the l.p. rotor, usually called "booster stages". contribute little because of their low blade speed. The configuration shown in Fig. 22.17 (b) is Tore attractive but it requires a high pressure ratio from h.p. compressor resulting in instability problems. The three spool arrangement shown in Fig. 22.17(c) is the most attractive concept as the pressure ratio in each compressor is modest. V2500 engines and Roll Royce Trent are examples of this. Fig. 22.17 (d) is two-spool geared system for smaller engines. An aft-fan configuration was used in some early turbofan. 22.9. Optimization of Turbofan Cycle 41i00,11-' " There are four independent thermodynamic parameters available for the designers of turbofans engines. These are Overall pre.ssure ratio, (b) Turbine inlet temperature, (c) By-pass ratio, (d) Fan

883

Jet Propulsion

Increasing bypass ratio

sfc

• •

sfc Optimum

Bypass ratio fixed

Specific thrust

Specific thrust

(a)

(b)

Fig. 22.18. Optimum Values of Turbofan Engine.

pressure ratio, However, the other two following parameters also affect. (e) Cruising speed, (f) Altitude Since there are many variables affecting the performance, so the optimization of the cycle is somewhat complex. At first place, let us consider the effect of fan pressure ratio while overall pressure ratio and by-pass ratio are fixed. With the low value of fan pressure ratio (FPR), the fan thrust will be small and the workdone by 1.p. turbine will also be small. This means that the little energy will be extracted from the hot stream and a large value of hot thrust will be available. As FPR is raised it is evident that the fan thrust will increase and thus there will be decrease in hot thrust. Fig. 22.18 shows a typical variation of specific thrust (1) and specific fuel consumption (sfc) with fan pressure ratio (FPR); for various values of turbine inlet temperature. It is obvious from the result that for any value of turbine inlet temperature there will be an optimum value of FPR; optimum values of FPR for minimum sfc and maximum specific thr.ust. Both coincide because of the fixed energy input. For plotting the optimized relation between sfc and specific thrust, a curve similar to shown in Fig. 22.18 is obtained by taking the sfc and specific thrust for each of these values of optimum FPR in turn. The above result if repeated for a series of by-pass ratio at the same overall pressure ratio, the family of curves as shown in Fig. 22.18. is obtained. This relation gives the optimum variation of sfc with specific thrust for the selected overall pressure ratio. Problem 22.8. The following data relate to a twin-spool turbofan engine where the fan is driven by the 1.p. turbine and the compressor by the h.p. turbine. Separate cold and hot streams nozzles are used in the turbofan. Overall pressure ratio = 30.0, Fan pressure ratio = 1.7, By-pass ratio, = 5.5, Stagnation turbine inlet temperature = 1600 K, Mechanical efficiency = 0.98, Combustion pressure loss = 2 bar, Mass flow rate of air = 220 kg/s, Lower calorific value of fuel 43000 kg/kg, cp,„ , cpg and y for gas = 1.005, 1.148 kg/kgK, 1.33, Polytropic efficiency for fan, compressor and turbine = 0.9, Ambient condition = 1.0 bar and 288 K, Nozzle

884

Steam & Gas Turbines And Power Plant Engineering

efficiency = 0.95 Calculate the specific thrust, air-fuel ratio, and specific fuel consumption at sea level condition Solution. Refer to Fig. 22.16. The values of (n-1)/n for polytropic compression and expansion are given by for compression : n •

n

= 0.9 x1 5.5 - 0.3175

y

-1 y

Under static condition assume Toi = T and poi = pa as nothing is mentioned about the ramming action. The actual condition will be directly available. (n-1)/n (P02)

oi

= 288 x (1.7)°•3173 = 341.81 K

Poi (n-1)/n

T03' = T02

= 341.81(17.647)°•3"3 = 850.35 K

n

2-02

Poz The cold nozzle pressure ratio is fan pressure ratio (FPR) = — = FPR = 1.7 Po 1.3/033 2 Pc The critical pressure for cold nozzle = — = Y = = 0.54 2.33 P02 • Y+1 or or

•P02 _ 1 • 0.54 = 1.851 pc PO2 — < PO2 —, so the nozzle is not choking. Thus p8 = pa Pa Pc

The cold thrust = Tc = inc

q = inc c.,

where C8 = CI = [ 2C (T02' - T81 1/2 pg

(Y-1YY

But Toz' - T8 ' = 1J Toe 1-N n 1-02

1/3.5

1 =0.95 x 341.81[1-H 1 = 45.67 K

Thus C8 = C1 = [2 x 1.005 x 103 x 45.6711/2 = 302.97 m/s in B 220 x 5.5 The cold mass flow rate = ins - 186.15 4/s B+1 5.5+1 The cold thrust = 7 'c = 186.15 x 302.97 = 56.3971N Heat balance around the combustor gives :(1 + m.-iII cpg T04 -cpa T03'= m!L i (LCV)f h ' or C

l + 4 x 1.148 x 1600 -1.005 x 850.35 = -I x 43000 Mh Mh

885

Jet Propulsion = 0.0238 mh Since h.p. turbine drives h.p. compressor, so cpa L(7'03' - TO2') in• (I. + cpg (T04 - T05')

or

or or

Ans.

1 (850.35 - 341.81) - 443.71 1.005 To4 — To5' = — 0.98 1.148 x 1.0238 x To; = 1600 -443.71 = 1156.29 K

Since 1.p. turbine drives fan, so Tos — T06' = (B+1)

C pg m

(T02

T01)

1.005 (341.81 - 288) = 321.44 K 1.148 x 0.98 • or Toh ' = 11656.29 -312.44 = 843.85 K n/n-1 1/0.225 1600 ) Poo (T04 ) — = 4.235 Now, •P06 may be found as follows : — - ,, 1156.29 pos 1 05 = (5.5+1) x

(

p03 — AP 30-2 p04 .n- - os = 4.235 - 4.235 - 4.235 - 6.611 bar , n/n-1 1/0.225 (1156.29) 1105 - 6.611 _ 1 63 P05 = (T05 = 4.055 or 10 T06 843.85 °6 = 4.055 4.055 ' PO6

or

PO6 1.63 The hot nozzle pressure ratio = — = — = 1.63 Pa 1 The critical pressure ratio in hot nozzle is given by 1.33/0.33 Pc ( 2 /Y-I ) = 0.54 P06 = 041) — 2.33

y

or

P06 Po6 Since — < — , hence nozzle is unchoked. Thus, p7 = pa Pa Pc (y-1)/y T06 - T7' = j 7' 06 [1 - (Pa V ) o6

= 0.95

Poo pc

=

1

0.54

= 1.851

0.33/1.33

x 843.85[1-H 1 ) 1.63

= 91.48K

- To)1/2 C7 = Cjh = 2cpg(Toh? [2 x 1.148 x 91.48 x 101= 458.29 [ in 220 .5.5+1 - 33.84 kg/s = B+1 = The hot gas thrust = Th = inh C7 = 33.84 x 458.29 = 15.5 kN The total thrust T = Tc + Th = 56.397 + 15.5 = 71.897 kN

Ans.

886

Steam & Gas Turbines And Power Plant Engineering The specific fuel consumption (sfc) is sfc =

(m/ih ) ih n x 3600

0.0238 x 33.84 x 3600 71.897 x 103

0.0403 kg/hN

Ans.

22.10. Turbojet tngine with Afterburner. Fig. 22.19 shows a schematic diagram of a turbojet engine equipped for a tail pipe burning or after burning. Tail pipe burning consists of introducing and burning fuel between the turbine and the exit nozzle which has the same effect as a reheater to the cycle. Its function is to increase the exit temperature which results in an increased jet velocity thus increasing the thrust. For turbojet engine with afterburner inlet velocities must be sufficiently low to support stable combustion and to avoid excessive pressure losses, a diffuser is provided between the turbine outlet and the tail—pipe burner inldt. To operate the engine as a straight turbojet, a variable area exit nozzle is used. As the turbojet with tail—pipe burning consists of a turbojet and essentially a ram jet, the engine is sometimes designated as a turbo-ram-jet engine. Tail-pipe burning is not only used to improve the take-off and high speed performance of an aeroplane but it may be treated as a distinct type of engine for flight at supersonic speed. 22.11. Turboprop.( tv-71-2,)' Fig. 22.20 shows a turboprop or propjet engine. It consists of a geared propeller attached to a turbojet engine. To add more power to the propeller shaft, the turbine is modified by adding additional stages. In this case, nearly 80 to 90 percent of the power

Combustion chamber

Compressor

Tail-pipe burner

Adjustable. exhaust nozzle

• 1!111J,,qtrimp.,!,-/a;14111811P,,,, ippo, AggioNlf1(1. •.::•:.:::

III

Diffuser Turbine

Turbine-discharge diffuser

Fig. 22.19. Turbojet Engine with After Burner.

Combustion chamber Reduction gear

Propeller

Compressor'

Turbine

Clutches to disengage either unit

Fig. 22.20. Dual Arrangement of Turboprop Engine.

Jet Propulsion

887

developed is used to drive the shaft and the remaining 10 to 20 per cent is used to obtain thrust from the jet. The control of the engine is governed by changing the propeller pitch and the quantity of fuel burned in the combustion chamber. The turboprop engine cycle is the same as that of theimhojet engine cycle except the turbine expansion process is greater. Regenerators, intercoolers and reheaters may be added to the turboprop cycle for improvement in power and efficiency. The turboprop engine combines in the merit of turbojet engines, i.e. low specific weight, small frontal area, simplicity and lower vibration and the merits of the propeller are high power for take off and climb, high propulsive efficiency at speeds below 800 'kg/h, high thrust per wilt frontal area and good fuel economy. 22.12. Ram Jet. Ram jet was invented by a French engineer, Rene Lorin in 1913. It is also called athodyd Lorin tube or flying stovepipe. It is a steady combustion or continuous flow engine and has the simplest construction of any propulsion engine. Fig. 22.21 shows a schematic diagram of a ram jet engine. It consists of an inlet diffuser, a combustion chamber and an exit nozzle or tailpipe, As the ram jet has no compressor hence the entire compression depends upon the ram compression. The ram pressure ratio increases very slowly in the subsonic speed range. That is why ram jet is boosted upto a speed of 290 km/h by a suitable means such as a turbojet or a rocket before the ram jet will produce any thrust and must be boosted to even higher speeds before the thrust produced exceeds the drag. After the boosting of ram jet, velocity of air passing through the diffuser decreases and hence pressure increases. This is called ram compression and a 'Pressure barrier" is created after the end of the diffuser. The fuel is injected through injection nozzles into the combustion chamber where it is ignited by means of a spark plug. The expansion of gases toward the diffuser entrance is restricted by the pressure barrier at the end of the diffuser and as a result the gases are constrained to expand through the tail pipe and out of the exit nozzle at a high velocity. As the ram jet engine has no turbine, the temperature of the gases of combustion is not limited to a relatively low value as in the turbojet engine. The air-fuel ratio is 15:1, the exhaust temperature ranges 1800°C to 2100°C. The jet action gives the necessary forward thrust to the engine. Fig. 22.22 shows the representation of ram jet cycle on T—s diagram. The basic characteristics of the ram jet engines are : (i) Simple in construction. -

Shock wave

Subsonic diffuser

Supersonic diffuser

Fuel injector

Igniter

Combustion chamber Fig. 22.21. Ram Jet.

Steam & Gas Turbines And Power Plant Engineering

(ii) No moving parts and hence free from unbalancing. (iii) Greater thrust per unit engine weight' than any other propulsion engine -at supersonic speed except rocket. (iv) The thrust per unit frontal area increases both with the efficiency and the air flow through the engine, therefore, much greater thrust per unit area is obtainable at high supersonic speeds. (v) The best performance of ram jet enS gine is obtained at flight speed of Fig. 22.22. T-s diagram of Ram Jet Cycle. 1700 km/h to 2200 km/h. (vi) For successful operation the diffuser has to be carefully designed so that the kinetic energy associated with high entrance velocity is efficiently converted into pressure. (vii) At low moderate speeds, the fuel consumption is too large. However, the fuel consumption decreases with flight speed. (viii)The performance is independent of fuel technology and a wide range of fuels can . be burned. 22.13. Pulse Jet Engine. The pulse jet engine (Fig. 22.23) is_somewhat similar to a ram jet engine with the exception that a mechanical valve (V-shaped non-return valve) is used to prevent the hot gases of combustion from going out the diffuser. Thus, pulse jet like ram jet develops thrust by a high velocity jet of exhaust gases without the aid of compressor or turbine. It was invented by the German inventor Paul Schmidt in 1930. Note that the turbojet and ram jet engine are continuous in operation and are based on the constant pressure heat addition Brayton cycle. The pulse jet is an intermittent combustion engine and it operates on a cycle similar to a reciprocating engine and may be better compared. The compression of incoming air is accomplished by ram effect in the diffuser section and the grid passages which are opened and closed by V-shaped non-return valve. The fuel is then injected into the combustion chamber by fuel atomizers or nozzles worked from the air pressure from the compressed air bottles. A spark plug is used to initiate combustion but once the engine is operating normally, the spark is turned off and residual flame in the combustion chamber is used for ignition. Due to combustion of mixture of air and fuel, the temperature and pressure of the combustion products increases. Since the combustion pressure is higher than the ram pressure hence the non-return valve gets closed and as a result of this the hot gases flow out of the tail pipe with a high velocity and in doing so, it gives a forward thrust to the engine. When the combustion products leave, the pressure in the combustions chamber drops end the high pressure air in the diffuser forces the valve to open and fresh air enters the engine. The alternating cycle of combustions, exhaust, induction, etc, are related to accoustic velocity at the temperature prevailing in the country. Since the temperature varies continu-

Jet Propulsion

Fuel injector Spark plug

Grid

889

Combustion chamber

Ignition bars Pressure equal both sidas Thrust action

Fig.. 22.23. Pulse Jet Engine.

ously, the actual process is very complicated. but a workable assumption is that the tube is acting similar to a quater wave length organ pipe. The advantages of the pulse jet engines are as follows : (i) Pulse jet engine is very inexpensive compared to turbo jet engine. (ii) fhe pulse jet produces static thrust and produces thrust in excess of drag at much lower speeds than a ram jet. The disadvantages of the pulse jet engine are : ' (i) apparent noise, (ii) serious limitation to mechanical valve arrangement, (iii) high rate of fuel consumption and low thermodynamic efficiency, (iv) the operating altitude is limited by air density consideration (v) severe vibration. 22.14. Rocket Engines. A rocket engine has its own oxidiser and does not depend on surrounding air. Hence, it can operate at any place even in vacuum. The fuel and the oxidiser are carried into the body of the unit which is to be propelled. The two substances together is termed as propellants. The propellants may be two types : (i) Solid propellants and and (ii) Liquid propellants The following are the requirements of an ideal rocket propellants : (i) high heat value, (ii) low toxicity and corrosiveness, (iii) highest possible density so that it occupies less space, (iv) reliable and smooth ignition and (v) stability and ease of handling and storing. Rocket propulsion at this time should not be regarded as a competitor of existing means for propelling airplanes but as a source of power for reaching objectives unattainable by other methods. Some of the applications of the rockets engines are : (i) artillery barrage rockets. (ii) anti-tank "Bazooka" rockets. (iii) all types of guided missiles. (iv) aircraft launched rockets. (v) jet-assisted take-off for airplanes. (vi) engines for long range, high speed guided missile and pilotless aircraft.

Steam & Gas Turbines And Power Plant Engineering

890

(vii) the main and auxiliary propulsion engines on transonic airplanes such as the D-558. It is the advances in rocket technology which landed the man on moon long back and sending a large number of satellites in space orbiting the earth and serving the mankind for communication, entertainment, video-conferencing, etc.

22.15. Basic Theory of Operation of Rocket Engines Fig. 22.24 shows a schematic diagram of a liquid hi-propellant rocket engine. It consists of an injection system, a combustion chamber and an exit nozzle. In the combustion chamber, the oxidizer and fuel burn which produces high pressure. The pressure thus produced depends on mass flow rate of propellants, the cross-sectional area of the nozzle throat and the chemical characteristics of the propellants. The products of combustion are ejected to the atmosphere at supersonic speed through the nozzle. In the nozzles, the pressure energy is converted into kinetic energy. The reaction of this high velocity jet emerging from the nozzle produces the thrust of the rocket engine. Thrust. The thrust produced is due to the resultant of the pressure forces acting upon the inner and outer surfaces of the rocket motor. Hence, Resultant internal forces = nipCi +pA

(22.20) Where nip = mass flow rate of propellant, (kg/s.), = Exit or jet velocity relative to nozzle, (m/s.), = Exit static pressure, (N/m2), = Exit area, m2 The resultant external forces acting on the rocket motor. = M .)

Where p0 = atmospheric pressure, N/m2 Thrust =T = Resultant of total pressure forces = inp Cf + A (pi —0

(22.21) The above equation shows the effect of atmospheric pressure on the thrust of a rocket engine, The lower is the atmospheric pressure the higher is the thrust. This means when po = 0 i.e. the rocket engine will produce maximum thrust in vacuum. Sometimes the thrust is expressed in terms of mass flow rate of propellants and an `effective jet velocity.' Thus T= C. (22.22) P le

The effective jet exit velocity is a hypothetical velocity and it is expressed as

Combustion chamber

Exit nozzle

A)1

Fuel

Oxidizer Fig. 22.24. Components of a Liquid Bi-propellant Rocket.

891

Jet Propulsion

A. C. = C + m

.—p I 0

m/s

Cie is an important parameter in rocket engine performance. Thrust Power TP; It :s defined as the thrust multiplied by the flight velocity, or TP = TCa = m C Co Watt (22.23) p ie Propulsive Efficiency : It is expressed as tiz pC. TP je a TP + K . E . loss (rn C Ca) + [in (Cie — Ca)2 /2 p ie p 2(Ca/C ) je — 1 + (Ca /Cie)2

11Pm

(22.24) Specific Impulse : This is also an important parameter in rocket engine performance and is defined as the thrust produced per unit mass flow rate of propellant consumption. T I = — = C ;N—s/kg sP tji le (22.25) 22.16. Solid Propellant Rockets. Fig. 22.25 shows the schematic diagram . of a solid propellant rocket. It differs from other engine in that the mass of fuel is stored and burned within the combustion chamber. There is no need of any fuel supply system. It consists of an seamless tube, usually made of steel closed solidly at one end. The open end holds the nozzle which may be a single or multi-orifice type. There are two types of solid propellant rockets—(i) restricted burning and unrestricted burning. In restricted burning rocket, the propellant is constrained to burn on only one surface as shown in Fig. 22.25 (a). Restricted burning is similar to cigarette burning. It is accomplished by pouring the charge when liquid so that on solidification it fits the chamber tight. This type of rocket is preferred when the unit is required to deliver a small thrust for a relatively longer duration. • In the unrestricted burning, the entire charge is free to burn at all surfaces at the same Combustion chamber

Propellant charge

Burns on end - surface only Exhaust nozzle

(a) Restricted buring

Propellant charge

Burns on outside and inside cylindrical surfaces

Ends prevented from burning by washers

(b) Unrestricted burning

Fig. 22.25. Details of Solid Propellant Rocket.

892

Steam & Gas Turbines And Power Plant Engineering

time. This type of rocket is suitable when a large thrust for a short period is needed. The solid propellant charge contains all the material necessary for combustion, i.e. fuel and oxidiser. The basic requirements of a solid propellants are : (i) Burning rate should be remain constant., (ii) Mechanical strength (compressive and impactive) should be adequate specially at low temperature,, (iii) Non-toxic in nature. (iv) Specific impulse should be high enough. 22.17. Liquid Propellant Rockets. As the name implies, liquid propellant rockets utilise liquid propellants which are stored in the containers outside the combustion chamber. It overcomes the undesirable effect of solid propellant such as short duration of thrust, inadequate cooling arrangement, etc. Fig. 22.26 shows a schematic diagram of two liquid bi — propellant rocket systems. It consists of rocket motor,' propellant system and controls. The rocket motor consists of an exit nozzle, a combustion chamber, propellant injectors and ignition system. The propellants in the liquid state are injected into a combustion chamber, burned and exhausted at a high velocity through the exit nozzle. Also, the liquid propellant is used to cool the rocket engine by circulation of fuel around the walls of the combustion chamber and around the nozzle. The following is the purpose of cooling motor : (i) It maintains metal strength by keeping combustion chamber walls and nozzle surface cooled, (ii) Heat losses are reduced by using the heat lost to motor walls to preheat the fuel, (iii) Less expensive on non-critical materials for combustion chambers may be used.

Oxidizer

Combustion Turbine chamber 463-* Exit nozzle

Fuel ( (

Oxidizer Exit nozzle Supply tank

Echaust Pumps

Fuel (b) Pump feed

Shut off valve Supply tank

Combustion chamber

gases Ci

Propellant control valve

Exit nozzle xhaust

Gases C Pressure regulator

Combustion chamber Fuel (a) Pressure feed

Propellant control valve

Fig. 22,26. Liquids Bi Propellant Rocket System.

Jet Propulsion

893

The propellant system employs either a pressure feed or a pump feed to transfer fuel from a storage tank to the combustion chamber. In the pressure feed system, the pressure exerted by the inlet gases is utilised for feeding while in pump' feed system, a pump is used to feed. The pumps are driven by relatively small gas turbines which may or may not have their own combustion chamber. The pressure feed system is simple, inexpensive and . reliable. Sometimes, mono-propellant is also used. The disadvantage of the mono- propellant system that the extra weight of the turbine fuel may be from 3 to 5% of the total weight. The selection of liquid propellant depends on (i) specific impulse, (ii) density impulse, (iii) availability and expense, (iv) ease for handling and (v) storage requirements. Table 22.2 gives the various oxidiser and fuel of liquid propellant. Table 22.2. List of Oxidiser and Propellant Oxidiser (i) Bi propellant Liquid oxygen

Fuel

II

II

Ethyl alcohol

II

II

Gasoline

Nitric acid

Liquid hydrogen

•

Aniline

EXERCISES Viva-Voce and Theoretical 22.1. What is the principle of jet propulsion ? 22.2. • Is it true that rocket is independent of atmospheric air ? Give reasons. 22.3. What do you mean by propulsive power and propulsive efficiency ? 22.4. Give the advantages and disadvantages of jet propulsion over other system. 22.5. Give the advantages of convergent propelling nozzle. 22.6. Show the performance of jet propulsion. 22.7 Discuss the adantages and disadvantages of solid and liquid propellant rockets. Numerical 22.8. What are the advantages and disadvantages of jet propulsion over the other systems ? Discusi the working principle of turbojet after-burner, pulse jet, rain jet, etc. 22.9. The atmospheric air in which a jet travels at the speed of 223 m/s is 0.562 bar. The gas enters the nozzle at 2 bar and 833K and leaves the nozzle exit at pressure 1 bar. Calculate the velocity of jet at nozzle exit, nozzle exit area, propelling thrust and thrust power if the nozzle efficiency is 0.95 and the mass flow 22.5 kg/s. Ans.533 m/s, 852.1 cm2, 12 kN, 2619 kW 22.10. Air enters at the rate of 900. kg/min into a compressor of jet aircraft travelling at 241 m/s. The air-fuel ratio is 60:1 and the compression pressure ratio is 6:1. The calorific value of fuel is 41820 kJ/kg. Neglecting all losses calculate the thrust, the specific fuel consumption and propulsive efficiency.

894

Steam & Gas Turbines And Power Plant Engineering .

Ans. 10104 N, 0.0898 kg/KW/thrust-h, 11.516 N 22.11. Atmospheric air enters at pressure 1.033 bar and temperature 15°C in a compressor of a turbojet engine. The isentropic efficiency of compressor is 0.8. Calculate the pressure at the turbine exhaust if gas enters the turbine at at 1056K and leaves the turbine at negligible velocity. The isentropic efficiency of turbine is 0.8. Assume c = 1.005 and y = 14 for for air and cp = 1.154 and y = 1.33 for gases. If the mass flow rate of air through the compressor is 22.5 kg/s and the nozzle efficiency is 0.94, Calculate the nozzle exit area and the thrust of the jet leaving Ans. 1.792 bar, 979 cm2, 11.516 lcN. nozzle when the jet is stationary. 22.12. An aircraft travels a 241 m/s at an altitude at which the pressure is 0.46 and the temperature is 17°8°C, the calorific value of fuel burnt being 41820kJ/kg. Assuming that the inlet duct (diffuser) converts the entire kinetic energy of the air into pressure energy, and that the isentropic efficiency of compressor is 80% when the pressure ratio is 5, calculate the air-fuel ratio if the maximum temperature must not exceed 874°C. Assume c as 1.005. Ans. 61.4: 22.13. A jet engine requires 1544 kW at 193 m/s. The fuel consumption is 0.316 kg per propulsive kW per hr and the C.V. of fuel is 42676 kJ/kg of fuel. The temperature rise in the combustion chamber is limited to 510°C. Calculate the air-fuel ratio, the velocity and the reaction of the jet, and the propulsive and thermal efficiency of the plant. Ans. 63.3: 1.910 m/s relative to m/c, 8338 N, 32.7%, 19.63%. 22:14. Develop a software for the analysis of jet engine, turboprop, turbofan, turbojet with after burner and rocket. 22.15. Develop a software for the design of jet engine, turboprop, turbofan, turbojet after burner and rocket. 22.16. The following data relate to a twin-spool turbofan engine, with the fan driven by the L.P. turbine and the compressor by the HP turbine having separate arrangement for cold and hot nozzles. Overall pressure ratio = 32, Fan pressure ratio = 1.75, By-pass ratio (inchhh) = 5.5, Turbine inlet temperature = 1600 K, Polytropic efficiency of fan, compressor and turbine = 0.91, Isentropic efficiency of each propelling nozzle = 0.95, Mechanical efficiency of each spool = 0.99, Combustion pressure loss = 1.65 bar, Total air mass flow rate = 250 kg/s, Calculate the thrust, Sfc for inlet condition of 0.8 bar and 255 K, 22.17. Develop a software for the thermodynamic analysis of jet propulsion systems. 22.18. Develop a software for the design of Jet Propulsion System.

23 Combined, Co-generation and Mixed Cycle Power Plants

Energy is a fundamental requirement of human life, and thus a driving force of civilization so it must be utilised very efficiently at all levels. Due to increased fuel prices and depletion of fossil fuels resources, high efficiency energy conversion systems are utmost desirable. There is a great potential for increased efficiency in energy converting devices by recovering waste heat to a greater extent. Development of combined cycle is the most efficient and effective effort in this direction. Combined cycle power plants are gaining increasing acceptance as alternatives to conventional or nuclear steam cycle due to high thermal efficiency as high as 60% utilising natural gas as fuel. This chapter deals with the analysis and performance of gas/steam combined cycle power plants. In India, about 4000 MW of combined cycle plant has been installed by 2000. In Europe, USA and Japan, a large number of combined cycle and co-generation plants have been installed. The plant efficiencies vary from 47% to 60% as per pressure ratio, turbine inlet temperature, configuration, component efficiencies, etc. Extensive research is going on to obtain efficiency above 60%. For more details, References [46 to 70] may be consulted. 23.1. Introduction The concept and development of combined cycle originated from the utilisation of waste heat of a thermal power plant. A combined cycle as the name implies is a combination of two cycles operating at different temperatures, each of which could operate independently. The heat rejected by the higher temperature cycle is recovered and used by a lower temperature cycle to produce additional power to realize an improved overall efficiency. For the combination, the separate cycles must operate on separate fluids. Combined cycles which have seen commercial service include Diesel-steam, Mercury-steam and Gas turbine-steam turbine. Development work is going on for steam-organic fluid, gas-organic fluid, liquid metal-steam and MHD-steam. The higher temperature cycle is called the Topping. Cycle (or topper) and the lower temperature cycle is the Bottoming Cycle (or bottomer). Topping cycle may include. Otto, Brayton and Rankine cycles. Bottoming cycle have all been Rankine cycle, in all the cases. In the gas/steam combined cycle power plant {Fig. 23.2) the .gas turbine working on Joule scycle is the topping cycle where steam cycle working on Rankine cycle is the

896

Steam & Gas Turbines And Power Plant Engineering

bottoming cycle. The gas/steam combined cycle efficiency ranges from 47 to 60% as compared to 35% to 40% of conventional steam or gas turbine power plant. The heat rejected by gas turbine is highly appreciable as the exhaust temperature varies from 450°C to 650°C depending upon pressure ratio and turbine inlet temperature. This energy goes as a waste if it in rejected to atmosphere. This waste heat energy may be utilised to produce steam in a heat recovery steam generator (HRSG) and may be expanded in steam turbine to develop additional power based on Rankine cycle. This combination is termed as gas/steam combined cycle power plant. The gas cycle and steam cycle may use separate generator or a single generator. It is desirable that the exhaust temperature of gas turbine should be above 570°C otherwise the steam cycle will be inefficient resulting in lower combined cycle plant efficiency. If the steam generated in HRSG is utilised for process heating the system is called Co-generation. Co-generation is defined as the simultaneous production of electrical power as well as heat energy from the same source. A Mixed Cycle utilizes the two working fluids in one power cycle (gas turbine) to improve the power output and thermal efficiency. Steam injected cycle is one of the example of Mixed cycle. The most suitable fuel for combined cycle is natural gas. However, plants using liquid and coal based gaseous fuels are also being established. In the past twenty years the development in power gas turbine is enormous. Previously, the power output from gas turbine was merely 1 MW to 30 MW but now simple open gas turbine in capable of developing 320 MW. Table 23.1. shows the specifications of latest combined cycle power plants manufactured by leading companies of the world. It depicts the total power, shared power, efficiency, number and type of gas turbines, and types of HRSG-used by various companies. Unfired HRSG is widely used in combined, mixed and co-generation cycles. 23.2. Classification Combined Gas/Steam, Mixed and Co-generation Cycle Fig. 23.1 shows the classification of combined gas/steam, mixed and co-generation cycle. Such a classification is for understanding the development process of such cycles. 23.3. Early History. The history of the combined cycle power plant goes back to the early part of this century. In 1925, Emmet described substantial development work on a mercury - steam plant. This presentation was based on the basic work of Charles Bradley of 1913. The early development of the gas/steam turbine plant came somewhat later and was described by Seippel and Beseuter (1960). They proposed seven possible combinations of gas/steam combined cycle. In 1978, Wunsch has given classificatio,n of combined cycle plant. • By the 1979s the so called 'recuperative plant', higher le. el gas turbine, exhausting to a heat recovery steam generator (HRSG), which supplies steam to the lower (steam turbine) cycle, with no supplementary heating of the exhaust had become well established, primarily by General Electric and Westinghouse in US and by Brown Boveri in Europe. In 1997, Wood gave a list of some 40 such plants of USA in the range of 15-20 MW with one exception of 63 MW using supplementary firing. Now a days gas/steam combined cycle are gaining popularity. Over 200 MW gas turbines, single units all over the world are working. Now research is going on for 400 MW single unit gas turbine. Specifications of some of the combined cycle plants are given

897

Combined, Co-generation and Mixed Cycle Power Plants Combined gas/steam cycle

Topping (gas) cycle

ISimple gas turbine cycle

Bottoming (steam) cycle

Complex IC-RH-RG gas turbine cycle

Exhaust in HRSG

Unfired HRSG

Fired HRSG

J Steam . for heating (Cogeneration) purpose

Steam for mixed cycle

For base load

For peak load

Fuel Steam for steam cycle

Same in GT and boiler

Different in GT and boiler

Steam cycle

Single pressure HRSG

Dual pressure HRSG

With feed heating in direct contact heater including deaerator

Steam bled from turbine

fla

Tripple pressure HRSG.

Without feed heating excluding deaerator

Steam from HRSG

Fig. 23.1 Classification of Combined Gas/Steam , Mixed and Co-generation Cycles.

383

1994 1994 1997

GE Power Systems S109FA

S209FA

S109H

480.0

786.9

390.8

772

— —

CC209FA

478

CC109 FA

CC2942

393

720

60.0

57.1

56.7

554

55.7

52.3

57.9

52.9

52.9

(%)

(MW)

480

Net plant efficiency

Net plant output

1998

1996

KA 26-1

BHEL

1993 1993

KA 13 E2-3

Year

KA 13 E2-2

POWER

ALSTOM

ABB

Model

6000

6305

6350

6502

6461

6883

6217

6805

6805

Heat rate (kJ/kWh)

50

50

50

50

50

50

50

50

50

—

508.2

254.1

251.7

503.4

308

251.0

477.9

318.6

Electrical GT frequency power (Hz) (MW)

—

289.2

141.8

13.70

279

177

142.0

249.0

167.0

ST power (MW)

1xMSS9001H

2x MS9001FA

1xMS9000FA

1xMS9001FA

2xMS9001FA

2xV94.2

1xGT26

3xGTBE2

2xGT13E2

No & Type of Gas turbine

Table 23.1. Specifications of Combined Cycle Power Plants [Courtsey—Gas Turbine World, 2000]

Triple pressure reheat

Triple pressure reheat

Triple pressure reheat

Triple pressure reheat

Triple pressure

Dual pressure

single shaft

Triple pressure HRSG,

Dual pressure HRSG

Dual pressure HRSG

Comments (Based on natural gas as fuel)

00 00 OC

Steam & Gas Turbines And Power Plant Engineering

1997 1997

MPCP2 (M 701G)

1997 1995 1995 1994

Westinghouse 2V942A

ISV943A 2V943A

2W501G

Siemens

509.2

1995

207 FA

Mitsubishi heavy Ind. MPCPI (M701G)

730.0

385.5 771.0

584.5

972.1

484.4

253.7

1995

Hitachi 107 FA

562.5

1999

S207 FB

Table-23.1. Continued

58

57.1 57.1

55.0

58.2

58.0

55.5

55.3

57.5

6210

6300 6300

6550

6187

6208

6490

6510

6250

60

_50 50

50

50

50

60

60

60

490.0

— 512.0

367

657.8

328.9

328

164

363.3

253.0

— 275.0

230

314.3

155.5

181.2

89.7

204.0

2xW502G

1xV943A 2xV943A

2xV942A

2xM701G

1xM701G

2xMS 7001FA

1xMS 7001FA

2xMS 70001 FB

Triple pressure, reheat

Triple pressure, reheat Triple pressure, reheat, single shaft

Dual pressure ,

—

—

—

—

Triple pressure reheat all single shaft.

000

Combined , Co-generationand Mixed Cycle Power Plants

Steam & Gas Turbines And Power Plant Engineering

900

in Table 230 Briesch et al. [46] has advocated to achieve combined cycle efficiency greater than 60%. Ref [46] to [70] may give an account of various advancement made in the combined cycle. 23.4. Combined Cycle Power Plants in India In India, NTPC initially planned three major combined cycle power plants totaling 1600 MW based on natural gas available from HBJ gas pipe line. This was a quite successful experience and NTPC continued to plan and build large size Gas/steam combined cycle power plants by 2004. Table 23.2 shows NTPC gas based projects. Table 23.2. NTPC Gas based Combined Gas/Steam Cycle Power Plants. Project

Location

Total capacity (MW)

Units commissioned (MW Fuel Source

Anta

Baran, Rajsthan

413

'3 x 88 GT + 1 x 149 ST

HBJ Pipe Line

Auraiya

Etawah, UP

652

4 x 112 GT+ 2 x 1025T

HBJ Pipe Line

Kawas

Surat, Gujrat

645

4 x 106 GT + 2 x 11057'

South Basin Gas field

Dadri

Ghaziabad, UP

814

4 x 131 GT + 2 x 146.5ST HBJ Pipe Line

Jhanor Gandhar

Bharuch, Gujrat 648

Kanyam Kulan

Allapozha, Kerala

400 (3.50)

Gandhar Gas Field

3 x 131 GT+ 1 x 255ST

Naptha from BPCL

Under Progress .

23.5. Various Configurations of Combined Cycle Power Plants. A large number of configurations of unfired combined cycle is possible however, only the main configuration will be discussed and from that many other configurations may be evolved. It has been already discussed in the chapter of Gas Turbines that the efficiency/specific work may lie increased by the addition of intercooling, reheating and recuperation arrangement in simple cycle at the cost of complexities of plants. For the better utilisation of exhaust gas energy in HRSG, steam may be generated at dual or triple pressure with reheat instead of single pressure arrangement. All these possibilities offer a wide scope for large number of configurations. (a) Simple gas cycle/steam cycle with single pressure (IP), dual pressure (2P) and triple pressure (3P) HRSG without and with reheating of steam. (b) Complex gas (Simple + intercooling + reheating + regeneration) /steam cycle with single pressure (1P), dual pressure (2P) and triple pressure (3P) HRSG without and with reheating of steam. Fig. 23.2 (a) shows the simple gas cycle with single pressure HRSG steam cycle (SGIP). The representation of processes of gas and steam cycle is shown in Fig. 23. 2 (b) while the T-Q diagram of HRSG is shown in Fig. 23.2 (c). Fig. 23.3 shows the simple gas/cycle with dual pressure HRSG with reheating of steam (SG2PR) along with its process representation on T-s diagram and T-Q diagram of

Combined, Co-generation and Mixed Cycle Power Plants

901

dual pressure HRSG. Fig. 23.4 shows the complex gas (simple + intercooling + reheating) cycle with triple pressure (3P) HRSG with reheating of steam (CG3PR) along with its representation on T-s diagrams. The temperature distributions in HRSG for the system shown in Fig. 23.2 and 23.3 are depicted in Fig. 23.4 and 23.5 respectively. After HRSG, if the steam cycle is absent in above case, and steam so generated may be used for process heating, the system will act as co-generation plants.

Exhaust gas to stack Eco Eva SH

Steam

m a Airin

Coolant

for Condenser heating (cogeneration) CEP

Fig. 23.2 (a). Simple Gas/steam Combined Cycle with Single Pressure HRSG. (SGIP)

Pressure loss

-- Tgas, in TSH

Pinch point

Q

Fig. 23.2 (b). T-s Representation of SGIP System.

Fig. 23.2 (c) T-Q Representation of Single Pressure HRSG (IP).

Steam & Gas Turbines And Power Plant Engineering

902

In all these configurations shown above, the gas turbines are air cooled bleeding air from compressor. The gas turbine may be steam cooled by the steam generated in HRSG or by bled steam from steam turbines. Extensive research is going on to cool the bladings by water. The cooling system includes open loop cooling or closed loop cooling. In open

Stack

It

Exhaust gas

RH me,hpi. Bled

Fuel 0 0 rris hp — HPST

A

Steam LPP

LPST

GT

0— C

HPP Condenser

A

Ai r

in

Deaerator

Coolant

CEP

(ms ms1)

Fig. 23.3(a). Simple Gas/Steam Combined Cycle with Dual Pressure HRSG with Reheating of Steam (SG2PR)

S

Fig. 23.3(b). T-s. Representation of SG2PR System

Fig. 23.3(c). T-Q Representation of Dual Pressure HRSG with Reheat (2PR).

Combined, Co-generation and Mixed Cycle Power Plants

Fuel

Fuel

U IIC

Reheater

x CC Y>NI CC Yj '\. HPC — HPT LPT — Coolant

0^ LPC

LAir in

1

4 V Exhaust gas S ack

Y4

HPD A SH

\A

A A

A NAA/

W RH

ms p MsiP

0

Mshp V5

— HPST

F

aerator

IPST

LPST -0

v LPP

. 10

IPP HPP

Feed water

Fig. 23.4(a). Complex Gas (Simple + Intercooling + Reheat) with Triple Pressure HRSG with Reheating of Steam (CG3PR)

—pi s

Fig. 23.4(b). T-s Representation of CG3PR System

Fig. 23.4.(c). Temperature Distribution in triple Pressure (3P) HRSG with Reheat

903

904

Steam & Gas Turbines And Power Plant Engineering

loop, the coolant mixes with gases in turbine whereas in closed loop there is no mixing of coolant and gases in turbine. The cooling techniques include convective, impingement, film and transpiration cooling. The intercooler used in complex system may be surface or evaporative type. The coolant air bled from compressor may be further cooled in a heat exchanger employing feed water going to HRSG. All these will contribute additional configurations. In practice two or three gas turbines coupled with respective generators at air entry side are installed in parallel, the exhaust of each gas turbine gas goes to single HRSG which generates the steam for expansion in steam turbine working on Rankine cycle or for heating purpose as co-generation plant or to be injected in combuster as mixed cycle. In order to save the capital cost steam and gas turbine are coupled to the single alternator. In the supplementary fired combined cycle power plant. a supplementary fuel firing system is provided in HRSG to produce more power from steam turbine plants. However, supplementary fired combined cycle is less efficient than unfired due to less utilisation of waste and supplementary heat and so it is seldom used. 23.6. Mixed Cycle. Since the bottoming cycle needs extensive cost, so efforts are going on to dispense with it and may use mixed cycle whose performance is slightly inferior or comparable to combined cycle. In these cycle, major part of steam generated is fed to the combustion chamber, get heated upto turbine inlet temperature (TIT) and expands in gas turbine however the rest part of steam generated is used for bled cooling. such system is either called Steam Injected Gas Turbine cycle (SIGTC) or integrated gas steam cycle (IGSC). in other words, in mixed cycle two fluid streams are used. Humidified Gas Turbine Cycle (HGTC) is another example of mixed cycle. Fig. 23.5 shows the configuration of a steam injected gas turbine cycle in which steam generated is fed into the combustion chamber. This increases the output and efficiency of steam injected cycle as compared to simple gas turbine but less than combined cycle. As a result of steam injection in combustor NOx emission is less. Some steam injected plants of small sizes are working in many parts of world.

T I

Steam Gas -->

Fuel

A

N

CC C

Steam

GT Deaerator

A Air in

A

0 —0 Pump Fig. 23.5. Steam Injected Gas Turbine Cycle.

905

Combined, Co-generation and Mixed Cycle Power Plants

23.7. Elementary Thermodynamics of Unfired Combined Cycle. Consider the case of unfired combined cycle (Fig. 23.6). The higher or topping cycle (H) receives heat QH, does WH work at a thermal efficiency of nil . It rejects heat QHR to a lower or bottoming cycle (L) which produces WL work output at a thermal efficiency of TIL . However, between the cycle there is a heat loss Qun (unused) to the surroundings. WH The thermal efficiency of the higher cycle = r1H = —0----H

The heat rejected from higher cycle = ---HR = QH (1 —111) The heat received by the lower cycle = QL = QHR — Q„„= QH[ 1 = QH (1 —11H)

un

(23.1)

Pun QH

Where Bun is the ratio of unused heat Qun to the heat supplied QH , i.e. B • un = QuniQH • The lower cycle (L) efficiency =

= WL L QL

The thermal efficiency of combined cycle plant (ri ucp ) is given by WH + WL 1HQH +11.{Q11(1 ilccp =

Q H

11H) 13unQH}

QH

(23.2) 1ccp — 11H + 1L -11H1L un • TIL The expression for combined cycle plant efficiency may also' be written in terms of a "boiler efficiency", (riB) of the heat exchange process between the two cycle (the heat recovery steam generator, HRSG for the gas/steam turbine plant.). Thus, TI B is defined as = QL =

QHR — Qun QunI QB Pun

=1

B QHR QHR

QHRIQB

—11H

Heat source TH QH

GTIeHWH QRH Qun

QL ST I f0 HWL QLR

Heat sink TL Fig. 23.6. Unfired Combined Cycle.

(23.3)

Steam & Gas Turbines And Power Plant Engineering

906

The combined cycle plant efficiency (rim,) may be written as liccp = TIH + (TILQL)1QB = 1111 +11L 1B (QHad = rih +

(23.4)

11B lit 11H

(23.5) = rill + (10)L 114110)L where (ri a)L = %TV is the overall efficiency of the lower plant, including the HRSG. For illustrating this elementary thermodynamic analysis for a practical operating station consider the Brown Boveri combined cycle plant which has the following data :Combined cycle output = 125 MW, Gas turbine output = 81 MW, Steam turbine output = 49 MW, Heat supplied based on = 276 MW, LCV of natural gas, Gas turbine efficiency = 0.29, Temperature of gas turbine exhaust = 491°C, Temperature of HRSG exhaust = 96°C, Energy lost in the exhaust = 34 MW, Steam power plant efficiency = 32% 0.12 34 = 0.12 and -riB = 1 - 0.83 dun-276 (1-0.29) Steam plant thermal efficiency = 0.32 so (rlo)L = 0.32 x 0.83 = 0.27 The combined plant efficiency = riccp = 11H + -1H 11L PUH• ilL = 0.29 + 0.32 - 0.29 x 0.32 - 0.12 x 0.32 = 0.48 = 48% Also, 71„p = 11H + (110)c. 71H (10)L = 0.29 + 0.27 - 0.29 x 0.27 = 0.48 = 48% The above results show that the combined cycle plant efficiency is much higher the individual topping and bottoming cycle plant. If the individual cycle efficiency is higher which is currently available e.g. 'nil = 39% and riL = 35%, and pun = 0.12, we have ri ccp = 0.39 + 0.39 - 0.35 x 0.35 - 0.12 x 0.35 = 56.15 = 56.15% 23.8. Thermodynamic Analysis of Combined Cycle and Co- generation Plants 23.8.1. Ideal Cycle Analysis. In ideal cycle analysis, no pressure losses and aerodynamic losses are considered and specific heat of air and gases are the same and constant. There is no cooling of gas turbine blading. Mass of fuel is neglected as he7,t input is due to heat transfer alone. Consider 1 kg/s of air entering to compressor. The analysis may be based on static or total values. Refer to Fig. 23.2 to 23.5. Gas Properties : pa = c = 1.005 kJ/kgK constant, All = cp Oland ina = Mg as ?hi= 0 pg Compressor : The compressor work = wc = how - hin = cp (Toui- 7',a) (23.6) Thus, for simple gas cycle (Fig. 23.2), the compressor work is given by we = h2 - hl = cp(T2 - Tl )

(23.7)

For complex gas cycle (Fig. 23.3), the compressor work is given by wc = (h2 - hi ) + (h4 - h3) = c p[(T2 - T1) + (T4 - 7'3)]

(23.8)

Intercooler For perfect intercooling,

T3 =

Combustion Chamber For a given turbine inlet temperature (TIT), the fuel flow rate per kg of air is found from the energy equation qA = m1 x (LHV)f = how- hin = c (Tout - 7n) (23.9) p

Combined, Co-generation and Mixed Cycle Power Plants

907

Thus, for simple gas cycle (Fig. 23.2) inf x (LHV )1. = (123 — h2) = cp(T3 — T2)

(23.10) For complex gas cycle (Fig. 23.3), the the fuel flow rate per kg of air is found as qA = fn (LHV)f = (h5 — h4) + (h7 — h6) = c [(T5 — T4) + (T7 — T6)] (23.11)

p .( and wt is neglected for calculating we For simplicity, inf

Gas Turbine : The turbine output = 11)1 = (how — h.) = cp (Tozo — Til,); kJ/kg For simple gas cycle (Fig. 23.2), w = (h3 — h4) = cp (T3 — T4) gi For complex gas cycle (Fig. 23.4), wgi = (h5 — /16) + (h7 — h8)

(23.12)

(23.13)

= c[(T5 — T6) + (T7 — T8)]

(23.14) HRSG : In general, It should be kept in mind that the temperature and pressure of steam depends upon the temperature of gas turbine exhaust entering into HRSG. The steam cycle efficiency is satisfactory only when the gas turbine exhaust temperature is above 570°C. For single pressure HRSG (IP) system, the throttle pressure varies from 40 bar to 50 bar while in dual pressure (2P) system (Fig. 23.3) the hp throttle pressure varies from 80 to 120 bar. In triple pressure HRSG system (Fig. 23.4), the hp throttle pressure varies from 120 bar to 166 bar. The throttle temperature is generally limited to 570°C, however in some case it may be taken up to 600°C. It is to be noted that the approach temperature difference has to be at least 20°C. The mass of steam generated per kg of air, ins (Fig. 20.4) is expressed as

ins Oheco+ N eva + d hsup + dhreh) = (hour — hin) where ins is the mass of steam generated in HRSG per kg of gas For simple (SG2P) and complex gas cycle (CG3P) the mass of steam generated in HRSG is given by ( Aheco + Ailey + 6'1 sup + h red = (114 h stack) s (23.15) ins [Aheco + Ahe, + Ahsup + Ah„d= [h8 — hsfack ] (23.16) where Ah = enthalpy rise in respective elements of HRSG. The stack temperature is limited by the possibilities of condensation of combustion products in stack and thus its minimum value is 78°C. Generally, stack temperature ranges from 85°C to 120°C. While calculating mass of steam (ins) generated, pitch point has to be satisfied. Steam Turbine : Assuming isentropic expansion in the turbine, the workdone by the steam turbine is given by for single pressure HRSG (Fig. 23.2) and

wst

(23.17) Where suffixes bb and ab stand for before bleeding of steam for deaerator and after bleeding of steam. Condenser : Under ideal condition the condensate temperature is equal to the saturation temperature corresponding to condenser pressure. The heat rejected by condenser is given by = ills (Ah)ssen, bb

qR =

s

(1115

sl)(hfg) cond ;

111 sl )(Ah)tsen,ab

kJ/kg of air

Feed Pump : The work input to the feed pump is given by

(23.18)

Steam & Gas Turbines And Power Plant Engineering

908

w = ines(hfio — hfi) = mcs "form ( boiler P con) p Where ins, = mass of condensed steam.

(23.19)

Overall Performance : wccp =-wgtna+wa net = (wgi— wc)+(wst— wp) kJ/kg of air (23., 0) ccp = Wccp/qA (First law efficiency)

(23.21)

The second law efficiency (exergetic efficiency) is given as ccp lex

(ex)f. where (e)1= exergy content of fuel input

(23.22)

Co-generation Performance Parameters : In general, mainly there parameters namely fuel utilisation efficiency (-0, power to heat ratio (Rph)and exergetic efficiency are used to predict the performance of co-generation plant. gt, net +qp Fuel utilisation efficiency, if = w qA

where gp.= process heat production = (hg)exii — (hg)stack, kJ/kg of air Power to heat ratio, RpH

=

Wgt net ' P

Exergetic efficiency = (wgt net + p)/ (ex)j, 23.8.2. Actual Cycle Analysis. In the actual cycle the fluid flowloses occur in every component of the combined cycle. The analysis of actual cycle is affected by introducing the concept of effectiveness, efficiency and pressure losses, etc. Gas Properties : As mentioned above the most suitable fuel for combined cycle power plants is natural gas. The lower calorific value of natural gas is around 42 MJ/kg. The composition of natural gas is 94% CH4, 3% C2H6, 1% N2 on mass basis. All gaseous mixtures at any point are treated as the mixture of air, water vapor and combustion products with excess air in their varying proportions. All nonreacting gases may be arbitrarily assigned zero thermodynamical enthalpy at the ambient temperature regardless of chemical composition. It actual practice, it is never constant and it is a function of temperature and very high pressure. The specific heats arid enthalpies of air and gases have been obtained in the form of pplynomials fits, as the function of temperature cp(T) = a + br + c72 + dT 3 h=

c (T)dT

(23.23) For approximation, the average value of specific heat of gases may be taken which is much higher than that of air i.e., cpg 1.15 to 1.2 kJ/kgK. Ta P

Compressor : The compressor, which is'axial flow in nature supplies compressed air to the combustion chamber/intercooler. Air is bled from the compressor for turbine blade cooling. For a compressor spool with specified inlet state and polytropic efficiency the actual overall temperature is expressed as

Combined, Co-generation and Mixed Cycle Power Plants

909

n-1

(pin

T2'

=

T

Pi

(23.24)

-1 -1 The polytropic index, n is given by the relation -n-- = LY- where rlpcmay be taken ria as 90 to 91%. Thus for polytropic stages (assuming large number of small stages) n- 1

Thus

Tout,

(19 out)

T m

pin

n

The number of stages for an axial flow compressor may be expressed as by assuming equal pressure in each stage: (1.12, to 1.2) or by assuming equal temperature drop. Thus in r The number of stages — z — (23.25) in(r)stage AT'

Alternatively, z =

(AT )stage where AT' = overall actual temperature rise and (AT' ).gage = temperature rise (20 K to 30 K) Using the concept of isentropic compressor efficiency, the overall exit temperature is expressed approximately as Tout Tin

11 c

Tout' — Tin

(23.26)

where Tout' = actual temperature of compressed air. It is worth to note that the concept of polytropic efficiency will give better representation of actual compression for any pressure ratio. The mass and enthalpies balance for the compressor work yields as thin — thou/

+

inci

wC = moth bout — E mci hci —.n hi.n Without considering cooling of turbine blading w = out bout ininh. out —min in

(23.27) (23.28a)

(23.28b) where Zinc, represents the total cooling flow requirements. After specifying the pressure at the desired state, the temperature and specific enthalpies may be determined at that states. After the turbine flow cooling calculations have been performed, one may determine the mass flows extracted from each bleed port from the compressor. If the cooling of the turbine blading is neglected, then E =0 Intercooler : In intercooler, pressure loss.occurs and the+ intercooling is imperfect. The actual temperature of air leaving the intercooler is related with effectiveness ATaci e

is

AT.

T2—T3

T2 —

(23:29)

Steam & Gas Turbines And Power Plant Engineering

910

If E is is known, the actual temperature leaving the intercooler may be calculated. Combustion Chamber : In combustor, the combustion does not take place under ideal condition but is accompanied by fuel throttling loss, mainflow pressure drop loss, incomplete combustion loss, and other losses. For a given turbine inlet temperature (TIT), pressure drop and combustion efficiency, the fuel flow is found from the mass and energy balance. mout

= tit 4in

(23.30)

f

(23.31) 1,c (LCV)f = M out hout — min hin Turbine : The number of stages in the turbine depends on the expansion ratio which varies from 2 to 2.5. The maximum number of turbine stages is generally 5 The turbine is cooled by the air bled from compressor. However, blades may also be cooled by steam generated in HRSG or by fresh water (under development). The concept of polytropic efficiency is used to calculate actual exhaust temperature. Alternatively, the concept of isentropic efficiency is used to calculate the actual exhaust temperature. For cooled turbine stages the mass and energy balance may be written as nlout = we =

M in

+

(23.32)

ci

hin+Erna ha

out

(23.33)

bout n-1

The actual temperature

(Po

Tout' Tin

n

(23.34)

pin

(Y — 1) 1pi where n — 1 n y Approximately, the concept of turbine isentropic efficiency may be used to calculate actual temperature at the exit of turbine. rl t

h'out

hin

"'out

— ho:

= h

Tout' — Tout

— T in

(23.35)

where Einci iepresents the sum of cooling flows to each blade row and are found from expressions given in Art 20.4 in the chapter of gas turbines. An additional coolant leakage flow is assumed as 0.4% of local gas flow. Generally, the air coolant required varies from 10% to 20% of entering air to the compressor depending on cooling medium techniques, turbine inlet temperature, compressor pressure ratio, etc. HRSG : As discussed above HRSG may be single pressure, dual pressure or triple pressure system. It may also incorporate reheater unit. The inefficiency of HRSG is taken care by efficiency or effectiveness of HRSG. In triple pressure HRSG with reheat, the recovery of exhaust gas energy is the maximum at the cost of complexity. The mass of steam generated per kg of air is expressed as s(Aheco where E hrsg

=

Ahem

Ahsup + Ah reh) = mg

— hgout) E hrsg

(23.36)

effectiveness of HRSG

Steam turbine : The internal losses in steam turbine are taken care by introducing the

Combined, Co-generation and Mixed Cycle Power Plants

911

concept of steam turbine isentropic efficiency. Thus for system shown in Fig. 23.2 ) ; kJ/kg of aiir (23.37) s(h 5 — h61)bh + (M s M1)(h6i hi t ab where ins is the mass flow rate per kg of air generated in HRSG and expanding through steam turbine. and nisi is the amount of bled steam for deaerator. Here, suffixes bb and ab represent bafore bleeding and after bleeding. Condenser : In real situation, there is an undercooling of the condensate due to inefficient heat transfer and pressure loss in condenser. The enthalpy of the condensate is calculated based on the degree of undercooling which generally varies from 4°C to 6°C. Thus enthalpy of the condensate is calculated grej = (Ms — sl)(h ex, act — hfcond)

=( ills st)(117'

(23.38)

h j8)

Pump : Due to inefficiency, more pump work has to be supplied w = ins (hfo — hf;)/ri p = m cs . V cs (Phrsg P condr p p where rip = overall pump efficiency (say 80 to 90%) Overall Performance of Combined Cycle Plant : Feed

ccp)act = (Wgt — Wdact + st (11 ccp) act =

cc/ q A ) 0a

dact

(23.39)

(23.40) (23.41)

23.9. Advantages of Combined Cycle Power Generation. The following are the advantages of combined cycle power plants. (i) High overall plant efficiency. The overall plant efficiency of combined cycle varies from 50 to 60% depending upon many factors which are much more than conventional steam power plant (35 to 40%). (ii) Less investment costs. Since 2/3 of the output is produced by gas turbine (GT) unit and only 1/3 by steam turbine (ST) plant, so the investment cost per kW are approximately 30% less than these of steam power plant. (iii) Greater operating flexibility. Due to gas turbine plant and simple steam power plant, there is a greater operating flexibility in terms of start-up and shut-down. The start up losses are less. (iv) Less Cooling water requirement. Since the steam power (bottoming) cycle contributes only about 1/3 of the total power output, so the cooling water requirement in condenser is greatly reduced to about 50% as compared to conventional steam power plant. (v) Phased installation. The installation of gas turbine plant (bottoming cycle) is much quicker, so it can start generating electricity while the installation of steam plant (bottoming cycle) may be under construction. This makes it possible to adjust the growth in demand for energy in a grid. Later on, combined cycle may be operated with coal gasification if price of natural gas or oil is increased considerably. (vi) Simplicity in operation. Due to gas turbine plant as topping cycle, the operation is very simple as in the era of state-of art technology, the operation of combined cycle plant is fully automatic. Even with less experience staff, the plant may be run. (vii) Less pollution. In general combined cycle uses natural gas as the fuel and since the efficiency is high, so the pollution level is low. The low NOx emission is one of the

Steam & Gas Turbines And Power Plant Engineering

912

attractive feature of combined cycle plant. It is worth to note that gas- fired combined cycle plants produce per kWh only about 40% of CO2 produced by coal-fired plant. Virtually, there is absence of fly ash. (viii) Possibility of co-generation. The combined cycle power plant is most suitable for co-generation. A part or full quantity of steam generated in HRSG may be used for process heating or a part of steam may be bled from LP turbine for process heating. 23.10. Exergy Analysis of Combined Cycles. The exergy of a system is the maximum work obtainable as the system comes to equilibrium with the surrounding. The higher the value of exergy, more is the work obtainable from the system. The traditional first-law cycle analysis based upon component performance characteristics coupled with energy balances can invariably lead to the correct final result. However, such analysis cannot locate and quantity the sources of losses which lead to that result. This is because the first law embodies no distinction between work and heat, no provision for quantifying the quality of heat and no accounting for the work lost, for example, in adiabatically throttling a pressurized gas. It is the second law which quantify the quality of energy and losses in a system. Energies of two systems may be quantitatively equal, but qualitatively they may be different. Exergy is a measure of energy quality and second law (exerget;c) efficiency is a measure of the perfectness of a thermal system. While energy of a system in any process remains constant, for example in throttling process a part of its exergy is always destroyed. The First law (i.e. thermal) efficiency is expressed as ' Wout W out Arj Q (23.42) "''rn , While second law (exergetic) efficiency is our rill

Ex,

our —

(23.43) Where AJ-I n is enthalpy of reaction while AGrn is Gibbs function of reaction and Ex in is the exergy input. In finding out Gibbs function for fuels, the ratio (4)), of chemical exergy, exs) to the net calorific value, (NCV)0 or (LCV)0 plays an important role. Kotas [94] have given expression for various fuels. The chemical Exergy for gaseous fuels of a given composition is given by f AGrn

Pk] e.,1:1 _ — Ah0 + To As0 + RT0 [x02 ln P02 — — k In p — P0 o Where the subscript k refers to the components of products of composition. Typical values for some industrial fuels are given in Table 23.3.

a

Table 23.3. Typical Values of (1) for Some Industrial Fuels. S.N.

Fuel

cb

(i) (ii) (iii)

Coke Coal Word

1.05 1.06 — 1.10 1.15 — 1.3

(23.44)

Combined, Co-generation and Mixed Cycle Power Plants Fuel oils and petrol

1.04 — 1.08

(v)

Natural gas

1.04 ± 0.5%

(vi)

Coal gas

1.0 ± 0.5%

(vii)

Blast furnace gas

0.98 + 1%

(viii)

Hydrogen

0.985

(ix)

Carbon monoxide

0.973

(iv)

913

For chemically reactive system, the specific exergy. ex,o = Agrn,o = the change in the standard Gibb's function of reaction (kJ/kg) and = — (NCV)0 = the change in the enthalpy of reaction (kJ/kg).

We have, (I) —

— AGr n, o — AHrn, o

(23.45)

and

AH = — inf (NCV)0 kW mo where inf.= fuel burning rate (kg/s)

(23.46)

Here, suffix r n represents reaction and suffix 0 stands for final dead or equilibrium condition (i.e. atmospheric condition) From eqs. (23.45) and (23.46), the value of AG n 0 can be evaluated. Further, AG, 0 =Off n, 0 — To • Sr n, o Here To . AS n, 0 = r n = Exergy loss in combustion or reaction Thus f2 n = To ASr n, 0 = (N C V)0+0[th f x(N C V)0] = th f x (NC V)0(0 — 1) (23.47) where (I) is given in Table 23.3 for various fuels. The exergy of fuel = (AGrno) =

tilf (NC 01 o

(23.48) The specific exergy of a flow stream for a given pressure pi and temperature Ti can be computed by the following equation ex (pi , Ti) = (hi — — To(si — so) kJ/kg (23.49) For an ideal gas, the term (s1 — so) can be expressed as cIT S I — SO = c (— 0PT

R 0P

= C In — R In -I31 P To

Po

The exergy of the stream in given by E = m [(hi — ho) — To(qi — So)] ; kW

(23.50) (23.51)

where in = mass flow rate (kg/s). In general, the specific exergy loss (f2) of a stream given by S2 = WillaX — W act = W rev — W act ; k W or

S2 = To As kJ/kg

The exergy balance for a stream gives the exergy loss. kJ/kg = exin exnu

(23.52) (23.53) (23.54)

Steam & Gas Turbines And Power Plant Engineering

914

The specific exergy loss or loss of availability or irreversibility in the various elements of combined cycle is given below (Fig. 23.7.) when working with 1 kg of air at inlet to the compressor. (i) Compressor : The specific exergy loss in the compressor without bleeding of air for cooling is given by T2' SIcomp = To(s 2' -s1) = To [c In — - R In P2] kJ/kg a p1 Pa Ti (23.55) (Y- 1)/y (P2 where T2 = T1 pi

and T2' = T1 + (T2 - Ty'ri c

where i c = isentropic compressor efficiency. Alternatively, T2' = T1 (p2/ pi)"- 1)/n = (pi / p)(Y -I)/Y (ii) Combustor : The exergy loss in the combustor is given by comb = ToRsp)3 - (Sr)2] kJ/kg where (s„)2 = (02 + (92

(23.56)

Here, the subscripts p, r, a and f stand for products, reactants, air and fuel respectively Thus, n comb = TO[(S p)3 (s,)2 — (92] = To {[(93 — (s )0 j+ (90 — Rso)2 — (so)o] — (s — [(.9 — (90] — (90 } p But

(s,)0 - [(9° + (s,0] = (As, do

Thus, C2 comb = To 1[(s 11)3 or

pq — [( 92 — (S. a) 01+ (As rn)o}

[ T3 T2 P3 P2 =T(1+in/in) c In — R In H [c In — - R In 1+(Asd o} CI combo r o Pg g po Pa T1 a po

(23.57) where To(Asrdo is given by the equation (23.47). Alternatively, fl comb may be expressed by taking exergy balance 1) comb = inf [Agri, + R T0 In 1 12 ]- [(I + in f hiio)(ex)3 — (ex)2] 0 (23.58) (iii) Gas Turbine. For uncooled gas turbine (GT), the exergy loss is given by il GT = (1 + 'illiii a) T o (54' -- s3) = (1 + iili/m)To [

In / - R 1n 1 -21 kJ/kg pg T3 g p3

[1 -1)/y where T4 = T3 — and T4'=T3 Tlgt (T3 -T4) P3

(23.59)

915

Combined, Co-generation and Mixed Cycle Power Plants n_1 D n = P4 Alternatively, T4' = T3H P3 P3

-1)10

where iv is the isentropic efficiency of turbine (iv) HRSG : The exergy loss in the HRSG is given by considering the two stream, i.e. water/steam and gas streams. HRSG = To[iii s(8 hrsg,out — hrsg,in)s/w + (1 + In f n (s steak — S hrsg,in>gas]

ino)[cpgIn TTstack R In

= To[ins (s 5 — s10) 4- (1 +

hrsg,in

P stock

(23.60)

g P hrsg,in

Here the feed water enters the HRSG at state 10, Ms is the mass of steam generated in HRSG per kg of gas (v) Exhaust Gas : The exhaust gas exergy loss is given by EXH = f ° (1 stack = (1 +

/) 8Q rei

)Cpg [(Tstack—

To) — To In ---1 Tri

(23.61)

Sometimes, exhaust gas exergy loss is included with HRSG. (vi) Steam Turbine : The exergy loss in the steam turbine is given by SI = ms T (sexu — sin) = ms T0 (s; — Ss)

(23.62) (vii) Condenser. The exergy loss in the condenser by considering the two streamsteam and cooling water. cond = ms To(S exit — S in)siecnn mw 0(8f:exit — Slin)cw =

s T0(S8 — S)steam

w T f,exit —

(23.63)

in)cw

(viii) Feed Pump : The exergy loss in the feed pump (FP) is given by

n FP = ms To (e exit

.

s in) = s T o(s

(23.64)

The total exergy loss = S2 = In component Fuel exergy input = (ex )1 = — Agro,0 = — (13 Ahmo

(23.65)

Total power generation = woot = Wgt,net

(23.67)

Exergetic or second law efficiency =

Wst,net

(23.66)

wog,/ (ex)f

(23.68)

23.11. Performance Curves of Combined And Steam Injected Cycles Fig. 23.7 shows the variation of gas turbine steam injected and combined cycle efficiencies with specific cycle work at various TIT and r for simple gas/simple steam combined cycle. At all TIT the cycle work deceasbs wait an increase in cycle pressure ratio but cycle efficiency first increases and then decreases resulting an optimum pressure ratio in all there cases at any TIT. Further it is obvious from the results of combined cycle that at any compressor pressure ratio (rix), the efficiency increases with TIT.

Steam & Gas Turbines And Power Plant Engineering

916

56 54 - Combined cyolc 52 ? 50 1000°C 48 g 46 E 44 DRIASI cycle 42 30 .1.13 40

e

38a) .... c 36 7 .S.1 34 32 7 30 28 26 -

le&

000

10 N.p =

—.,,,,_:::::..i.. --‘ ,

111.101.1%\

c

Steam injected cycle

TIT=c Utak

1000°C 10 1400°C

Simple cycle

rP = C

130 180 230 280 330 380 430 480 530 580 630 Specific work (w5) [kJ/kg air] Fig. 23.7. riccp vs.. wecp of Gas Turbine, Steam Injected and Combined cycle 75 70

65 60

11gc = 50% tlgc = 46%

t 55 a) 17) 50 cu c 45 5

tlgc =

42%'

llga = 38% tlgc = 34% Igo = 30%'

lac = 40% lac = 36%, L = .05 Tin = 34%, L = .05 lsc = 32%, L = .05 llsc = 28%, L = .05 llsc = 24%, L = .05

40

35 0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 Sizing ratio 1w • -st,netiwgt,net) Fig. 23.8 Combined cycle efficiency vs. Sizing Ratio

Fig. 23.8 shows the effect of gas turbine cycle efficiency, steam cycle efficiency and sizing ratio (,wst,net/w_gt,net) on the combined cycle efficiency. The higher plant efficiency is achieved at lower value of sizing ratio. Problem 23.1. A combined cycle power plant takes in air at 1 bar and 15°C. Th( compressor pressure ratio is 13 and the gas turbine inlet temperature. is 1400°C . The fue

Combined, Co-generation and Mixed Cycle Pcwer Plants

917

used in natural gas. The exhaust gas pressure of gas turbine is 1.08 bar. The stack temperature is 140°C. The HRSG is a single pressure system. The steam pressure and temperature are 50 bar and 500°C. The steam at 2 bar, is bled from the steam turbine for heating the feed water in the deaerator and the feed after the deaerator is pumped to the boiler. The condenser pressure is 0.05 bar. For ideal conditions calculate the. following :(a) compressor work, (b) gas turbine work, (c) mass of fuel (natural gas having lower calorific value of 42 MJ/kg), (d) gas turbine cycle efficiency, (e) steam turbine work, (f) steam turbine cycle efficiency, (g) combined cycle specific work, (h) combined cycle power plant efficiency. Also calculate the above by assuming the following actual conditions :— (i) compressor isentropic efficiency = 86%, (ii) combustion efficiency = 98%, (iii) gas turbine isentropic efficiency = 86%, (iv) HRSG effectiveness = 0.92, (v) steam turbine efficiency = 85%, (vi) mechanical efficiency for compressor and turbine each = 99% , (vii) generator efficiency = 98%, (viii) Cpa = 1.005 kJ/kgK, cpg = 1.15 kJ/kg K;, (ix) overall pump efficiency = 90%, Neglect the pressure losses and cooling requirements by gas turbine bladings. Solution. Refer to Fig. 23.2. For Ideal Conditions. The temperature at the end of compressor is given by - 1)/y T2 = T1 I pZ = 288(13)0.4/14 = 288 x 2.08 = 599.32 K. Pi (a) Compress work = = h2 — hi = cp(T2 — Ti) = 1.005(599.32 — 288) = 312.87 kJ/kg The gas turbine exhaust temperature is given by 0, - 1)/), 1.08 (P4 T4 = T3 ) = (1400 + 273) ( n 13 P 3

Ans. 0.4/1.4

= 821.84 K

(b) Gas turbine work = wgi = h3 — h4 = cp (T3 — T4) = 1.005(1673 — 821.84) = 855.41 kJ/kg

Ans.

Heat supplied = qA = h3 — h2 = cp(T3 — T2) = 1.005(1673 — 599.32) = 1079.04 kJ/kg (c) Mass of fuel required

1079.04 — — — 0.02569 kg/kg of air = f (LCV) 42 x 10-

(d) Gas turbine cycle efficiency = w 9,1

Ans.

w — w 855 41 — 312.87 — 49.44% 1079.04 9,1

From Mollier diagram, the steam properties are :— h5 = 3440 kJ/kg, h6 = 2650 kJ/kg, hf6 = 504.8 kJ/kg, 112 = 2140 kJ/kg, hfi = hw7 = 137.8 kJ/kg, The mass of steam generated in HRSG (ins) in kg/kg of air is given by (h4 — hick) = ins(h5 — hr6)

Steam & Gas Turbines And Power Plant Engineering

918 or or

c (Ti — Tstack) = s (113 — hid p cp(T4— T stack) 1.005(821.84 — 413) — 0.1399 kg/kg of air m— (3440 — 504.8) s (h5 — hid

(e) Steam turbine work = wst = "'s(h5 h6) + (ins — in 4)(h6 — h7) Energy balance on deaerator gives (ham— yins (504.8 — 1378) x 0.1399

— 0.0204 (2650 — 137.8 h6 — wst = ins(173 — h6) + s — msi )(h6 — h7) = 0.1399(3490 2650)+(0.1399-0.0204)(2650 — 2140) = 166.72 kJ/kg Feed pump work = MC,.vcond (-phrsg -pG_onGe,) ms1—

= 0.001005 (50 — 0.05) x 102 x (0.1399 — 0.0204) = 0.0594 kJ/kg (0 Steam turbine cycle efficiency = w, — w — W st

— wp

wstpet

166.72 — 0.594 — 40.45% 0.1399(3440 — 504.8)

ins(173 — hid

Ans.

(g) The combined cycle power plant efficiency is given by +W

iceP

sl,net

—

qA

533.54 + 166.12 64.84% — 1079.04

Ans.

For Actual Conditions (Ignoring blade cooling) From the concept of compressor isentropic efficiency, we have h2 — hi T2 — T1 rle

or

h21 — h 1 — T21 — T1 (7'2 — T1)

T21 =

— 288 + •

IC

T

599.32 — 288 — 650 K 0.86

(a) Actual compressor work = wca = —

—h1) r

c (T,' T = Pa 1m

1.005(650 — 288) — 371.23 kJ/kg of air 0.98

From the concept of gas turbine isentropic efficiency we have,

or

h3 — h4'

T3 — T4'

h3 — h4

T3 — T4

T4' = T3 —

(T3 — T4) = 1673 — 0.86(1673 — 821.84) = 941 K

Actual heat supplied = qAa = (1 + in/ ina) h3 — h2' = (1 + inf /tha)cpg T3 — Cpa7'2'

Ans.

Combined, Co-generation and Mixed Cycle Power Plants

919

in Thus qAa = (LCV)f .-: x 1= (1 + rintlik) Cpg T3 - Cpa T211 [ a cpgT3 pa T2' (b)i Mass of fuel required = Ma (LCV)f x cc - c T3 pg

1.15 x 1673 - 1.005 x 650 0.0323 kg/kg 42 x 103 x 0.98 -.1.15 x 1673

Ans.

The actual heat supplied = qAa = (1 inf /ina)cpg T3 - Cpa T2 = (1 + 0.0323) x 1.15 x 1673 - 1.005 x 650 = 1332.84 kJ/kg (c)Actual gas turbine work = (1 + »Vii24)(h3 - h4') = (1 + yi://ina)cpg(T3 - T4')

= (1 + 0.0323) x 1.15 (1673 - 941) = 868.99 kJ/kg

Ans.

(d)Actual gas turbine cycle efficiency = ri gca = W ig'net = W ga-W ca qAa gAa - 868.99 - 371.23 - 37.345% 1332.84

Ans.

The mass of steam generated (ins) in HRSG is given by (1 + inf.hha) or

in 8

hstack) E hrsg = ths(h5 hf6) (1 + 0.0323) x 1.15(941 - 413) x 0.92 - 0.1964 kg/kg of air (3440 - 504.8)

(e)Actual steam turbine work = wsia s(hs - h6') (Ms - insi )(h6' - h7') Here, ins1 is calculated by taking energy balance on deaerator . msl

hp)Ihs

(h6' - hp)

(504.8 - 137.8) x 0.1964 = 0.02733 kg/kg of gas (2775 - 137.83)

wsta, = 0.1964(3440 - 2775) + (0.1964 - 0.02733)(2775 - 2330) = 205.84 kJ/kg 0.67 Actual feed pump work th = "• cs(P hrs Pcond) - 0 9 - 0.728 kJ/kg .. (f) Actual steam cycle efficiency = rl

wsta -

pa

sca -ins(h5 - hi5)

205.84 - 0.728 - 30.81% (3440 - 50.48) x 0.1964

Ans.

(g) The combined cycle power plant efficiency is given by 'g

ar

W 041gt,net st,net X71 = wout = SSE gen Aa gyla Aa

[(866.99 - 371.23) + (205.84 - 0.728)] x 0.98 686.85 _ 51.53% = 1332.84 1332.84

Ans.

Steam & Gas Turbines And Power Plant Engineering

920

It is obvious that the combined cycle power plant efficiency is much higher than individual gas or steam cycle efficiency The gas turbine cooling has been neglected which will further alter the efficiency. Problem 23.2. A Combined cycle power plant takes in air at 1 bar and 15°C. The compressor pressure ratio is 23 and the gas turbine inlet temperature is 1700K. The exhaust gas pressure is 1.08 bar. The stack temperature is 110°C. The HRSG is a single pressure system. The steam pressure and temperature entering to steam turbine is 50 bar and 500°C. The steam at 2 bar is bled from the system turbine for heating the feed water in the deaerator and the feed after deaerator is pumped tc the boiler. The condenser pressure is 0.05 bar, Assume 1-1,, 11„, 6,,Lrsg, ri.gt, .st, /l m, 11./) and igen as 0.86, 0.98, 0.92, 0.86, 0.86, 0.99, 0.9 and 0.98 respectively. Take cpc, = 1.005 kJ/kg and cpg = 1.15 kJ/kg. Neglect pressure losses and cooling requirements for gas turbine bladings. The fuel used is natural gas giving calorific value equal to 42. MJ/kg. Calculate First law efficiency, second law efficiency and exergy losses in various elements of combined cycle. Solution. Refer to Fig. 23.2. Compressor. The isentropic temperature at the exit of compression T2 = Tl (p2/ p1 )(r-1)/Y = 288 (23)0A/1A = 705.43 K The actual temperature at the exit of compressor (T2— T1 — 288 + 705.43 — 288 T2 ' = T -F — 773.38K 0.86 11 c The exergy loss in the compressor for 1 kg of air is T' 2 P2 SI comp = To(s2' — s I) = To 7, — Ra ln T (i c ln — p1 1

9=

= 288 (1.005 In 7 288 — 0.287 In 26.744 kJ/kg 8 1 Combustor. For finding out exergy loss of combustion, the mass flow rate of fuel per kg of air (rnf/rna) is to be found out. f X

Or

ma or

LCV x i cc = ma[(1 + Tdc T3 — c , T2] pg pc a

[(LCV) X li cc — Cpg T3] = Cpg T3 — pa T2'

— [42000 x 0.98 — 1.15 x 1700] = 1.15 x 1700 — 1.005 x 773.38 a

[1.15 x 1700 — 1.99S x 773.38] — 0.3 kg/kg 42000 x 0.98 — 1.15 x 1700 ma The exergy of fuel = ev = mf (NCV)1.01) = 0.03 x 42000 x 1.045 = 1316.7 kJ/kg or

the

The exergy loss in combustion or reaction is given by = To(As)react = Mf (NCV)f (cD — 1) = 0.03 x 42000 (1.045 — 1) = 56.7 kJ/kg rn The exergy loss in the combustor is

921

Combined, Co-generation and Mixed Cycle Power Plants C2 = To [{(sp)3 - (s p) - {(sa )2 - ( sa ) + ( AS rea c )0)1 (As ) = To l In T3 - R In 1 } - c In + Ra Inreact 0 p cg P To g po Pa Ti

773.28 1700 23 + 0.287 In T + 56.7 .2861n - - 1.005 In = 288 [1.15 In 288 288 288 - 0 1 23] = 288 [2.041 - 0.896 - 0.9926 + 0.899] + 56.7 = 359.3 kJ/kg Gas Turbine. The exergy loss for uncooled gas turbine is TO [c pg in

CI GT = (1 + /ii/ ka) T o 41 s = (1 4. But

T4'= T3 (7 1)/Y

=

P4 Rg

7 31

1700(1.08/23)o.4/1 4 = 709.46 K

S2 GT = 1.03x288 1.15 In

1.08 848.1 - 0.86 In - = 22.24 kJ/kg 1 700 23

The gas turbine net power output is W = 0 + in / in ) c (T3 n pg

- pa (T2'

-T

= 1.03 x 1.15 (1700 - 848.1) - 1.005 (773.28 - 288) = 521.36 kJ/kg HRSG. For finding out exergy loss in HRSG, the amount of steam generated per kg of gas is to be calculated. Here, h5 = 3440, No (deaerator outlet) = 504.8 kJ/kg Energy balance yields (1 + in/ind (h4' - h steak ) hrsg

= 61s (I

1.03 x 1.15(848.1 - 413) x 0.92 (3440 - 504.8) The exergy loss in HRSG is given by or

Ill

=

hrsg

T

c[iii 3(S hrsg,out

- hno) -

0.1615 kg/kg gas

hrsg,m)s/w + ( 1 + rill1;1a) (S steak

= To ins(s5 - hd + (1 + inj /ina ) I cpg In

steak

Thrsg,tn

R g in

S hrsg en) gas]

;stack hrsg,en JJ

= 288 [ 0.1615(6.995 - 1.5302) + 1.03 x 1.15 In 8441: 1 - 0.286 In 11-it = 8.91 kJ/kg Exhaust Gas. The exhaust gas exergy loss is given by JJ ,T 0 ( To rstack] S2 Esh = J 1 - T.) 8q,,= (1 + Iii/hid ( Track - Ta) - To In 7' ,.th

To

413 = 1.03 [413 - 288) - 288 In -] = 21.81 kJ/kg 288 Steam Turbine. The exergy loss in steam turbine is given by Ci sr = in s To (1 7' -s 5 = 0.1763 x 288(7.6 - 6.995) = 30.71 kg/kg

922

Steam & Gas Turbines And Power Plant Engineering

Steam turbine net output = wsuiet = ins(h5 - h6') + (ins - msi) (h6' - h7') - wp The energy balance at deaerator given (h - h *i _ .16 .17 s - (504.8 - 137.83) x 0.1763 - 0.0245 kg/kg si h -h (2775 - 137.83) 6 f7 [0.1763(3440 - 2776) + (0.1763 - 0.0245) (2776 - 2330)] Thus W = = 184.79

kJ/kg Condenser. The exergy loss in the condenser is Ms T0 (sfl,- s7 ) + To (stout - SDI;)water cond = But mw (kg/kg of gas) is calculated from ins(h; - hjs) = in c or

in w

=

Thus SI cond

ins 0771- hid 0.1763(2330 - 137.83) 0.245 kg/kg 4.18 x 10 c xA pw = 0.1763 x 288 (0.4761 - 7.6) + 9.45 x 288(0.3672 - 0.2244)

= - 361.7 + 380.2 = 18.5 kJ/kg Feed Pump. The exergy loss in the condenser is S2 FP = ms To (sr' - sr) = 0.1763 x 288(0.5 - 0.4761) = 1.21 kJ/kg The exergy balance is-tabulated as follows Exergy input (kJ/kg) Power output(kJ/kg) 1316.7 wgt,net = 521.36 (39.59%) compressor (100%) wgt,net = 184.79(14.03%) Combustor GT HRSG Exhaust gases ST Condenser Pump Unaccounted Input = 1316.7 Output = 706.15

Second law efficiency =

wool e

Exergy losses (kJ/kg) = 26.74(2.03%) = 359.5(27.3%) = 22.24(1.689%) = 8.91 (0.676%) = 21.81(1.0656%) = 30.71(2.33%) = 18.5 (1.40%) = 1.21 (0.091%) = 120.93 (9.18%) 610.55

706.15 - 53.63% 1316.7

The exergy flow diagram is shown in \Fig. 23.9 23.12. Coal Based Combined Cycle Plants There are two dominant coal based combined cycle Technologies. They are

Ans.

Combined, Co-generation and Mixed Cycle Power Plants Ex (100%)

Exergy • loss

923

E'xergy

27.3%

39.59%

2.03% /Compressor

14.03% 1.689% 0.676% 1.656% 2.33%

1.4% Condenser 0.091%

Pump

V Unaccounted 9.8%

Fig. 23.9. Exergy Flow Diagram Based on Problem 23.2. (a) Pressurized fluidized bed combustion (PFBC) system which may be either a bubbling fluidized bed or a circulating fluidized bed (b) Integrated gasification combined cycle (IGEC) The schematic diagram of there plants have been discussed in chapter 6. 23.13. Repowering. The conversion of older steam power plant into combined cycle units are generally known as repowering. Boiler is replaced by gas turbine unit and HRSG. The building. Steam turbine, generator, condenser, water cooling system, and electrical equipments are used. By this way, the uneconomical steam power plant becomes economical 23.14. Other Types of Combined cycles. The following are the other possible combined cycle power plant. (a) M.H.D - steam power plant (b) Thermionic - steam power plant (c) Fuel-cells-gas turbine (Hybrid) power plant The first-two are still in the research stages while the third one, i.e. hybrid power plant is coming up as prototype (40MW unit) by 2007 in USA. EXERCISES 23.1. Define combined cycle and co-generation plant.

924

Steam & Gas Turbines And Power Plant Engineering

23.2. Derive an expression of combined cycle plant in terms of topping and bottoming cycle efficiencies and fraction of unutilised energy. 23.3. Following data relate to a combined cycle power plant based on simple gas and simple steam cycle. Power output (electrical) = 240 MW, Total pressure ratio = 30, Total turbine inlet temperature = 1650 K, Pressure of steam is bled from turbine = 2 bar, Condenser pressure = 0.05 bar, Stack temperature = 140°C, Polytropic compression efficiency = 90%, Polytropic turbine efficiency = 91%, Mechanical efficiency = 98%, Alternator efficiency = 98%, Combustion efficiency = 99%, Pressure loss in combustor = 3% of entry pressure, Turbine exhaust pressure = 1.08 bar , Ambient condition = 1 bar, 255 K, Effectiveness of HRSG = 0.9, Steam turbine efficiency = 0.86, Overall pump efficiency = 90%, Calculate the net gas turbine specific work, gas cycle efficiency, mass of fuel flow rate, air flow rate, steam turbine specific work, steam turbine efficiency, combined cycle specific work, combined cycle efficiency and exergy losses.. 23.4. Develop a software for predicting the performance of combined cycle power plant for various configurations.

24 Nuclear Power Plant

There is an economic necessity for nuclear power plants due to increasing price rise and depletion of fossil fuels. Now, the cost of electricity generated per kW from nuclear power plants is comparable with coal-fired power plants. But nuclear industry is facing many difficulties all over the world, primarily as a result of negative impact of the issues of nuclear safety (i.e. radiation hazard), waste disposal, weapons proliferation and the economics on the public and government. However, after so many hurdles, electricity generated by nuclear plant takes the share around 10% in west and USA. In India, the nuclear power plants contribute only about 1% of total electricity generated. The energy released in nuclear reaction is very large in comparison to the chemical reaction. The heat evolved by one kg of uranium is equivalent to the combustion of 1,13,350 quintals of coal or 30 x 106 litres of diesel oil. This chapter deals with the principle of nuclear energy and nuclear power plant systems. 24.1. The Atomic Structure In 1803, John Dalton postulated that all matter is composed of unit particles called atoms. This was simple but incomplete atomic hypothesis. An atom consists of a relatively heavy, positively charged nucleus and a number of much lighter negatively charged electrons that exists in various orbits around the nucleus. The nucleus consists of subparticles, called nucleons. Nucleons consist of neutrons which are electrically neutral, and the protons which are positively charged. It is worth to note that the electric charge on the proton is equal in magnitude but opposite in sign to that on the electron. Since the number of protons is equal to the number of electrons in orbit, so atom as a whole is electrically neutral. The most interesting aspect of this atomic structure is that one atom may be transferred into another by losing or acquiring some of the above subparticles. Such atomic reaction results in a change of mass (Am) and therefore, release (or absorb) large quantity of energy (E) which is given by famous Einstien Law AE = Am C2 where C is the velocity of light (3 x 108 m/s) Fig. 24.1 depicts the structure of three light atoms. Hydrogen has a nucleus composed of one proton, no neutrons, and one orbital electrons. It is worth to note that it is the only

926

Steam & Gas Turbines And Power Plant Engineering

(b) 0 = neutron e = proton • = electron

(c)

Fig. 24.1. Structure of Hydrogen, Deuterium and Helium Atoms. atom that has no neutrons. Deuterium has one proton and one neutron in its nucleus and one orbital electron while helium has two protons, two neutrons on nucleus and two orbital electrons. It is the nucleus that contains all the mass of the atom. The masses of the three primary atomic subparticles, which are supposed to be the building blocks of all atoms are given as follows— Neutron mass = m = 1.008665 amu Proton mass = m = 1.007277 amu Electron mass = me = 0.0005486 amu Where amu stands for atomic mass unit and 1 amu = 1.66 x 10-27 kg. Since the atoms contain these three primary atomic subparticles in varying number, so the atoms differ in their masses. The atoms with nuclei that have the same number of protons and have similar chemical and physical characteristics and dyer in their masses are called isotopes. For example, deuterium, frequently called heavy hydrogen, is an isotope of hydrogen. When combined with oxygen, ordinary hydrogen and deuterium form ordinary water and heavy water respectively. The number of protons in the nucleus is called ttie atomic number (z), while the total number of nucleons in the nucleus is termed as mass number (A). Thus, isotopes of the same element have the same atomic number. Nuclear symbols are generally written as XA where X is the usual chemical symbol. Thus, hydrogen nucleus is 1F1', deuterium is 1H2 and helium is 2He4. For particles containing no protons, the subscript indicates the magnitude and sign of the electric charge. For example, an electron is _le° and (sometimes e- or 131 and neutron is on'. There are many elements found in nature as mixture of isotopes of varying abundances such as naturally occurring uranium, called natural uranium is composed of 99.282 mass percent U238, 0.712 mass percent U235, and 0.066 mass percent U234 where the atomic number is deleted which is 92 in all three cases. Over and above three above primary atomic subparticles, the two other particles of importance are the positron and neutrino. The positron is a positively charged electron having the symbols +10, e+ or 13+ while neutrino is a tiny electrically neutral particle that is difficult to observe experimentally. Further, there are many other atomic subparticles. For example, it is the mesons which

Nuclear Power Plant

927

are unstable, positive, negative or neutral particles where mass lies between electron and proton. • It is to be noted that electrons that orbit in the outermost shell of an atom are called valence electrons, while the outermost shell is called the valence shell. 24.2. Comparison Between Chemical and Nuclear Equations. There are some similarities and dissimilarities between chemical and nuclear equations. Chemical reactions involve the combination or separation of whole atoms. Consider the reaction between carbon and oxygen which is (24.1)

C + 02 --> CO2

This equation is accompanied by the release of about 4 electron volts (ev). An electron volt is a unit of energy in common use of nuclear engineering. 1 ev = 1.6021 x 10-9 J = 4.44 x 10-26 kWh. In chemical reactions, each atom participates as a whole and retains its identity. The molecules change but the nuclei are unaffected. There are as many atoms of each participating elements in the products as in the reactants. It is to be noted that ,both chemical and nuclear reactions are exothermic or endothermic. In nuclear reactions, the reactant nuclei donot show up in the products, instead we may find either isotopes of the reactants or other nuclei. Further, in balancing nuclear equations it is essential to observe that the same, or equivalent, nucleus show up in the products as entered the reactions. As an example, if K, L, M and N are chemical symbols, the corresponding nuclear equations becomes KAI A2 wAA3 •KTA4

zl

z2L.

z3i"

a"

for balancing the above equation the following conditions must be satisfied Z1 + Z2 = Z3 + Z4 and

Al

+ A2 = A3 + A4

(24.2) (24.3) (24.4)

24.3 Nuclear Binding Energy. There exists a difference between theoretically calculated mass of nuclei (the sum of the masses of the protons and neutrons that comprise the nucleus) and the actual mass of the atomic nucleus. This difference in mass is called the mass defect (Am). This mass defect is found by adding up all the individual particles (neutron, proton and electron) masses and subtracting the actual mass (m) of the atom. Thus Am = nn

n + npp + nee — zMA

(24.5) where n refers to the number of particles, m the mass of particles and suffixes, n p and e refer to neutron, proton and electron respectively. The mass defect is converted to energy in a nuclear reaction as given by Einstein's Law AE = Am C2 ' (24.6) where E stands for energy (J), C stands for velocity of light = 3 x 108 m/s. The energy associated with mass defect is known as the binding energy of the nucleus. The energy equivalent of 1 g of mass is AE = (1 x 10-3) x (3 x 108)2 = 9 x 1013 J = 9 x 107 MJ Similarly, energy equivalent of 1 amu of mass is AE = (1.66 x 10r27) x (3 x 108)2 = 14.94 x 10-11 J = 9.3 x 108 eV = 931 MeV

928

Steam & Gas Turbines And Power Plant Engineering

24.4 Nuclear Fusion and Fission. Nuclear reactions are of three types namely fusion, fission and radioactivity. Infusion, two or more light nuclei fuse to form a heavier nucleus while in fission, a heavy nucleus is split into two or more lighter nuclei. In both, there is a decrease in mass resulting in exothermic energy given by above Einstein's Law. 24.4.1. Fusion. Energy is produced in the sun and stars by continuous fusion reactions. In this fusion process four nuclei of hydrogen fuse in a series of reactions and culminates in one nucleus of helium and two positrons as given below(24.7) 4 1 H i --> 2He4 + 2+1e° This results in a decrease in mass of about 0.0276 amu and thus release of 25.7 MeV. The heat produced in these reactions maintains temperature of the order of several million degrees in their cores and serves to trigger and sustain succeeding reactions. On the earth, it is difficult to have fusion reactions for meaningful use due to various reasons. In order to cause fusion, it is necessary to accelerate the positively charged nuclei to high kinetic energies, to overcome electrical repulsive forces, by raising their temperature to hundreds of millions of degree resulting in a plasma formation. Achieving such a high temperature is a difficult preposition. Further, plasma must be prevented from contacting the walls of the container and must be confined for a small period (about a second) at a minimum density. Fusion reactions are also called thermonuclear as very high temperature are required to trigger and sustain them. Many problems have to be solved before artificially made fusion reactor becomes a reality. 24.4.2. Fission. Fission is a practical preposition and can be caused by the neutron, which, being electrically neutral, can strike and fission the positively charged nucleus at high, moderate or low speeds without being repulsed and sustains chain reaction because two or three neutrons are usually released for each one absorbed in fusion. These keep the reactions going. A typical fission reaction is depicted in Fig. 24.2. It is worth to note that there are only a few fissionable isotopes. U235, Pu239 and U233 are fissionable by neutrons of all energies while U238 Th232 and pum are fissionable by high energy neutrons only. The fission reaction with 6235 is as followsxe140 + Sr94 + 2 nl U235 + n1 (24.8) 92 0 54 38 o The immediate (prompt) products of a fission reaction, such as Xe14° and Sr94 are called fission fragments while they, and their decay products are called fission products Fig. 24.3 shows the fission product data for U235 by thermal and fast neutrons and for U235 and Pu239 for thermal neutrons with their mass number.

Xenon nucleus /at:

0,51

Neutron lost by escape or consumed in nonfission reaction

Neutron for further (chain) fission Neutron

k•-.71 -J-) Uranium nucleus

::; Strontium nucleus 4:57

Fig. 24.2. A Typical Fission Reaction

929

Fission yield, per cent

Nuclear Power Plant 10

10

1.0

1.0

===============

immumoimi NI =1.111MINI4INV 1=1=1=111Id 111.1111.111•4011,•1•11l••• /JUNE

mim4 , -.4 .11. tIM•h. MI= MO WINN= WWI= I . •11•1111•11.•••••M IIIMM.,110•••

IMMIlinil== mom.. m -maidmi=====.01 -----------------

a. 0.1111111111114MI11I ========== == l'aM=ZMI= IM OOMIMMINIMM a) ========= ==.111MIMSO m.imam ME MENEMM EM 001 iiimmoRmirmommu c 0.01 agb 0 Ena 1MM •• MM WM LL MM ()COI ilimmmumm 0.001 iimmummummi-i ===== •Thermal neut on === == u 233 PU 239 .=1M1=IMMM 14 MeV neutrons 0.0001 ITIONIMMINININIMMan 00001UMMEMMEMMEMMEMEMMIN 0.1

70 80 90 100 110 120 130 140 150 160

70 80 90 100 110 120 130 140 150 183

Mass number (a)

Mass number (b)

Fig. 24.3. Fission Product Yield Data for (a) U235 by Thermal and 14 MeV neutrons and (b) U233 and Pu239 by Thermal Neutrons. The two fission fragments in eq (24.8) are not equal in size and are radioactive. The fission fragments, therefore, undergo several stages of 13 decay (converting neutrons into protons) until a stable product in formed in each case 54X

el

40

cs14O _

-13> 16s

55

P

and

Sr94 --> M

2min

13>

B-° 13 _,

56

66s

R

39Y94 -- —>

17 min

12.8 days

57

/jet°

40Zr94 (stable)

____13, 40 his

55ce

I40

(24.9)

The above series is known as fission chain. It is a well known fact that f3 decay is usually accompanied by y radiation so suitable shielding against y rays as well neutron must be provided in a reactor. . 24.4.3. Energy From Fission and Fuel Burnup. Energy released by fission reaction depends upon the type of reaction. The reaction given by eq. (24.8) yields 196 MeV. Consider the following reaction 92U235 + nl---> Ba137 + Kr92 + 2 onl 0 56 36 • (24.10) Taking mass balance, we have 235.0439 + 1.00867 —> 136.9061 + 96.9212 + 2 x 1.00867 or

236.0526 --> 235.8446 Am = 235.8446 — 236.0526 = —0.2080 amu

Thus, AE = 931 x (-0.2080) = —193.6 MeV It has been observed that on the average the fission of a U235 nucleus yields about 193 MeV. U233 and Pu239 also yield the similar value of energy release. It is to be noted that this amount in energy is prompt, i.e. released at the time of fission and more energy is produced because of (i) slow decay of fission fragments and (ii) the nonfission capture of excess neutrons in reaction that produce energy. As a result, the total energy produced per

Steam & Gas Turbines And Power Plant Engineering

930

fission reaction is about 200 MeV. The complete fission of 1 g of U235 nuclei produces 0.60225 x 1024 Energy releasedAogadro's v number x 200 eased = 235 X 200 MeV — 235.0439 U isotope mass = 0.513 x 1024 MeV = 2.276 x 1024 kWh = 0.948 MW-day In simple way, we can say that reactor burning 1 g of fissionable material generates nearly 1 MW-day of energy. It is to be noted that nuclear fuel is defined as all uranium, plutonium and thorium isotopes. Further, depending upon fuel type and enrichment (mass percent of fissionable fuel in all fuel), the energy released may vary from 103 to 105 MW-day/ton. 24.5. Radioactivity. Radioactivity is one of the important aspect of nuclear science. It provides an important source of energy for small power devices and a source of radiation for use in research, industry, medicine and a wide variety of applications and as well as an environment concern. Most of the naturally occurring isotopes are stable. These isotopes which are not stable are known as radioactive. The examples of radioactive are some naturally occurring isotopes of the heavy elements like thallium (Z = 81), lead (Z = 82) and bismuth (Z = 83) and all the isotopes of the heavier elements beginning with polonium (Z = 84). A few lower-mass naturally occurring isotopes are radioactive, such as K40, Rb87 and In115. In addition to above, several thousand artificially produced isotopes of all masses are radioactive. It is to be noted that natural and artificial radioactive isotopes, also known as radioisotopes have similar disintegration rate mechanisms. A continuously undergoing spontaneous (i.e. without outside help) disintegration of radioactive isotopes is called radioactivity. This is accompanied by the emission of one or more smaller particles from the parent nucleus. The resulting nucleus is known as daughter nucleus. The parent nucleus is said to decay into the daughter nucleus. The daughter nucleus may or may not be stable, and several successive decays may occur until a stable isotope is formed. The following nuclear reaction shows an example of radioactivity 49In15 50SnI15 + _1e (24.11) where the parent, I11115, is a naturally occurring radioisotope, and its daughter, Sn115, is stable. It is worth to note that radioactivity is always accompanied by a decrease in mass and thus is always exothermic. The energy liberated is in the form of kinetic energy and radiation. The emission from naturally occurring radioisotopes includes a , 13 or y particles or radiations. The artificial isotopes, in addition to the above, emit or undergo the following particles or reactions — positron; orbital electron absorption, called K capture; and neutron. In addition, neutrino emission is accompanied with 13 emission of either sign. (i) Alpha (a) decay : a particles are helium nuclei, each consisting of two protons and two neutrons and are commonly emitted by the heavier radioactive nuclei. The decay of Pu239 into fissionable U235 and a (He4) particles, is an example of this. 94

a PU239 —> 92U235 + 2He

(24.12)

931

Nuclear Power Plant

(ii) Beta (13) decay. It is commonly accompanied by the emission of neutrino (A) and y radiation. An example of 13 decay is equation (24.13) and the following pu2I4 82

Bi214

1- -1"

83

A

(24.13)

The penetrating power of p particles is small compared to y rays, however, it is larger than that of a particles. (iii) Gamma (y) radiation (decay). y particles are electromagnetic radiation of extremely short wavelength and very high frequency resulting in high energy. y rays originate from the nucleus while x rays from the atom. y wavelengths are, on an average, about one-tenth those of x-rays, though energy ranges overlap somewhat. There is no alternation of atomic or mass number due toy decay. (iv) Positron decay. Positron decay is caused when the radioactive nucleus contains an excess of protons. An example of this is the decay of 7Ni23 into OC13. Ni13 —) 6C13 + +l e°

(24.14) Since the daughter nucleus has one less proton than the parent, so one of the orbital electrons is released to maintain atom neutrality. As a result there is combination of emitted and orbital electron resulting in y radiation and known as annihilation process. eo +-1eo -+ y (24.15) +l 7

The energy emitted by y radiation is equivalent to the sum of their rest masses, i.e. 1.02 MeV. The reverse of annihilation process is known as pair production (i.e. positron- electron pair), an endothermic process that converts energy into mass. (v) K Capture. K capture is a process which captures an orbital electron from the orbit or shell nearest to the nucleus which possesses an excess of protons but not the threshold of 1.02 MeV necessary to emit positron (Fig. 24.4). The vacancy in K shell is filled by another electron falling from the higher orbit resulting in x-ray emission from the atom. K capture also effectively changes a proton into neutron. The example of K capture is as followscu64 29

Higher shell K shell

Fig. 24.4 K Capture Phenomena.

i64 28N

(24.16)

Here, Ni64 is stable (vi) Neutron Emission. It occurs when a neutron possesses an extremely high excitation energy. If the excitation energy of nucleus is more than 8 MeV, it could decay by the emission of neutron. For example, 54

Xe1"

54Xe136

onl

(24.17)

Here, the parent Xe137 is a fission product from the 0 decay of the fission fragment 1137, generally called precursor and daughter is an isotope of the parent. Generally, it happens rarely except nuclear reactor where it is the source of delayed fission neutrons. In reactor, the control of this neutrons are of utmost importance.

• Steam & Gas Turbines And Power Plant Engineering

932

24.6. Decay Rates and Half-Lives. In the event of a very large number of radioactive nuclei of the same kind, there happens to be a defmite statistical probability that a certain fraction will decay in a certain time. This decay rates increase exponentially with temperature and donot depend upon temperature, pressure or the physical and chemical states of matter. If N is the number of radioactive nuclei of one species at any time 0, the rate of decay is directly proportional to N. Thus,

Number of half lives Fig. 24.5. Radioactive Decay Rates Vs. Half-Life.

—dN = de

(24.18)

where, is a proportional factor called the decay constant and depends upon type of isotopes. After integration of the above equation 0

A, where N = No at = 0

-I 21 d1 = x. 0

(24.19)

Thus, N = No e xe

(24.20)

The rate of decay (—dN/N) is also called the activity (A) and commonly has dimension 51. In nuclear science, the common trend to represent decay rates is by using the concept of half-life (0/2). This is used as the time during which one-half of a number of radioactive species decay or one-half of their activity cases. Thus, N/No = A/A0 = 1/2 = elein

(24.21)

In2 0.6931 and 01/2 - A — A

(24.22) Eq. (24.22) suggests that half-life is inversely proportional to the decay constant. Fig. 24.5 shows the radioactive decay rates as a function of half-life. Table 24.1 shows the half-life of some of the radio-isotopes. Table 24.1. Half-Life of Some Isotopes.

Isotope

01/2

Activity

Isotope

0 1/2

Activity

Tritium (H3)

12.26 yr

13

Thorhim '232

1.41x1010 yr

a and y

Carbon 14

5730 yr

13

Thorium 233

22.1 min

Krypton 87

76 min

13

Protactinium 233

27 days

13 13 and y

933

Nuclear Power Plant Strotium 90

28.1 yr

Xenon 135

Uranium 233

1.65x105 yr

a and y

9.2 h

0 p and y

Uranium 235

7.1x108 yr

a and y

Barium 139

82.9 min

13 and y

Uranium 238

4.51x109 yr

a and y

Radium 223

11.43 days

a and y Neputnium 239

2.35 days

13 and y

x and y

2.44x104 yr

a and y

Radium 226 1600 yrs

Plutonium 239

From the table 24.1 it is observed that the readily fissionable isotopes, U233, U235 and PU238 have extremely long half-lives, so they can be stored practically indefinitely. U235 is found in nature but U233 and Pu239 are artificially produced from Th232 and U238 respectively. 24.7. Neutron Energies and Scattering. Since neutrons are essential in fission process, so it is essential to study neutron energies. The kinetic energy of a neutron KEn or simply En is expressed as E = mn C2/2 where mn is mass of neutron, C is the speed of neutron (24.23) Since inn = 1.008665 amu and if C is in cm/s. then

En = 1.008665 C2/(2 x 0.965 x 1018) = 5.227 x 10-19 C2; MeV As a result of fission reaction, the newly born fission neutrons have energies ranging between 0.075 to about 17 MeV. In the course of travelling through matter, neutrons collide with nuclei and are decelerated, mainly by the lighter nuclei resulting in loosing of some of their energy with each successive collision. This process is called scattering. Neutrons are classified into three categories namely (i) fast (greater than 105 eV), intermediate (1 to 105 eV) and slow (less than 1 eV). Newly born fission neutrons carry on an average about 2 per cent of a reactor fission energy which may be either prompt or delayed. Scattering may be classified in two types namely (i) inelastic scattering and (ii) elastic scattering. Inelastic scattering are those in which the momentum and total energy of the particles before and after collision are conserved. Elastic scattering are those in which both momentum and kinetic energy of the colliding particles are conserved. The number of collisions, n, required to slow down a neutron from initial energy En i to a final energy En f in elastic scattering in expressed by In En ] Enf n—

where = logarithmic energy decrement, (24.24)

For example, for U235 nucleus, the number of collisions (n) between 2 MeV and 0.025 eV is about 4480. 24.8. Thermal Neutrons. An effective scattering medium, called moderator is used to scatter or slow down the fission neutrons. The moderator has a small nuclei with high neutron-scattering and low neutron- absorption probabilities, such as H and D (in H2O and D20), C (graphite) and Be or BeO. In the process of slowing down, the neutrons achieve the lowest energy level that put them in thermal equilibrium with the molecules of the medium they are in. They become thermalized and are called thermal or slow neutrons.

Steam & Gas Turbines And Power Plant Engineering

934

The thermal neutrons possess a wide range of energies and corresponding speeds which are given by Maxwell distribution law. The most probable speed of thermal neutron is given by .5

=

2k7'\°

(24.25) -11 k = Boltzman constant = 8.617x10 MeV/K, T = absolute temperature (K)

Where

mn = mass of thermal neutron particle = 1.674 x 10-27kg The energy corresponding to the most probable speed is given by E = 1- m C2 kT (24.26) m 2 n m Eq. (24.26) suggests that the energy of thennalized particle is independent of mass and a function only of absolute temperature of medium. 24.9. Nuclear Cross-Sections The concept of nuclear cross-section arises when we are interested to know the beam intensity of neutrons. The actual cross-section area of a nucleus is obtained from its radius rn as given by (24.27) rc =roAi" Where r0 is a constant and having average value of 1.4 x 10-23 cm and A is the mass number. The average cross-sectional area of nuclei is about 10-24 cm2. It is worth to note that the probability of neutrons colliding or inte.acting with nuclei is proportional to an effective, rather than actual cross-section area of the nuclei in question. This effective cross-section area is known as microscopic cross- section or simply the cross-section of the reaction and designated as cr. It depends upon nucleus, type of reaction and neutron energy. In order to fmd out the intensity of a neutron beam striking target area A (Fig. 24.6) consider a number of nuclei in volume A Ax

- dl

x Fig. 24.6 Neutron Beam Striking Target Area.

which is equal to NAAx. If the beam intensities are I and I-AI at x and x + Ax respectively, then in the limit -dl

(NAdr) - Ndx A

After integrating from 10 (at x = 0) and I (at x = x), we have _ r dl j — = N 0

or

I = Io e-'1*

x=o (24.28)

935

Nuclear Power Plant

Neutrons have as many cross-sections as there are reactions. The total cross-section, at is given by a = as + ac + af + any other (24.29) Where as = microscopic cross-section for scattering, ac = microscopic cross-section for radiation capture and crf = microscopic cross-section for fission. The product aN is equal to the total cross-sections of all the nuclei present in a unit volume and is called macro-scopic cross-section represented as E = Na. Thus I =10

(24.30)

The value of N can be calculated for an element of atomic mass At, density p (g/cm3) N = p Avogadro's number ; nucleucm3 At (24.31) 24.10. Neutron Flux and Reaction Rates If n is the neutron density (neutrons/cm3) and C is the neutron velocity (cm/s), the product n C is called the neutron flux (y). Thus cp = nC (24.32) This cp represents the number of neutrons crossing a unit area from all directions per unit time. The reaction rate is expressed as Reaction rate =n C N a = 4:1E ; reaction/(s.cm3)

(24.33)

where a and E are the cross-sections of the particular reaction in question 24.11. The Variation of Neutron Cross-section With Neutron Energy. Fig. 24.7 shows the variation of neutron cross-section with neutron energy for U235. The variations are represented in three regions namely (i) The 1/v region, (ii) the reso582 barns at 0.0253 eV

.z...''::•,,,

k,,,, Fission ' -

U) 0

, Total Total

i Fission . ,../ i, ',1 kli ' ... ' i Radiat ve capture ...2 10 .,yv , f"----........_ „......, 0. . , Scattering I § 10 „...... 0 Fast neutron region — reg ion TResonance regio 2 V I I I I I I 01 10' 105 103 104 105 10 102 10-2 10-' 1.0 104 102

Neutron energy, eV Fig. 24.7. Neutron Cross-section for U-235.

Steam & Gas Turbines And Power Plant Engineering

936

nance region and (iii) the fast neutron region. (i) 1/v Region. It is the low energy range. The absorption cross- section for many, but not all, nuclei are inversely proportioned to the square root of the neutron energy, En.

c

j0.5

Ga =ac +af =C1

(24.34)

En= C

where C1 and C2 are constant and C is the neutron velocity. (ii) Resonance Region. In this region, most neutron absorbers exhibit one or more peaks occurring at definite neutron energies, called resonance peak .1J235 has many peaks. This effects the design of thermal reactor. (iii) Fast-neutron Region. In the fast-neutron region, the cross- section usually undergo a gradual decrease as neutron energies increase. The total cross-section = 0.125 A2/3 barns (1 barn = 10-24 cm2) . In the very high neutron energy range, the total cross-sections are rather low, usually less than 5 barns, each for the largest nuclei. Table 24.2 gives the microscopic cross-sections of uranium fuels. Table 24.2. Microscopic Cross-section for Uranium Fuels. Neutron

Fast, 1 MeV

Thermal, 0.0253 eV

Natural uranium

Nucleus

Micro-scopic cross-section, barns

U238

U235

as

5.30

6.6

6.6

ac

0.093

0.14

0.19

of

1.2

0.018

0.026

as

.6.0

8.0

8.0

ac

112

2.73

3.47

of

577

0

4.16

24.12. Main Production of the Reactor For power production nuclear reaction is carried out in a nuclear reactor. The main products of the reactor are— (i) Thermal energy, i.e. heat (ii) Neutrons (iii) Radioactive fission fragments (iv) Materials which are fissionable (v) Gamma quantas and other radiation (vi) Radio isotopes for commercial and military purposes. 24.13. Principal Components of a Nuclear Reactor. The nuclear reactor may be regarded as a substitute for the boiler fire box of steam power plant or combustion chamber of gas turbine plant. The heat produced in the nuclear reactor is by fission process whereas in steam and gas power plants, the heat is produced by combustion of fuel. The other cycle of operation and components required is the same either as steam plant if steam is generated by using the heat released due to

Nuclear Power Plant

937

fission or a gas turbine plant, if gas is heated by using the heat released Control rod due to fission. The steam or gas may be the working fluid in nuclear power plant. Each reactor consists of the folCoolant out lowing principal parts as shown in Biological Fig. 24.8. shield (1) Fuel Element. The nuclear fuels which are generally used in reactors are 92U235, 941}11239 and P P // P Reflector ,/ .0 // I, 92U2" Out of the three, the 92U2" ,/ /, // // is only naturally available upto Moderator /5 55 ve 0 0.7% in the uranium ore 239 and ,/ /, r 00 00 Fuel 011 b the remaining is 92U238. The other ,/ " // // 4 0 /' two fuels 94Pu239 and 92U233 are the a /0 Reactor byproduct and formed in the nuvessel clear reactor during fissioning proc0.:. ess from 92U2" and 90Th232 due to Coolant absorption of neutron without fisin sion. The selection of the shape of the fuels and their locations in the reacFig. 24.8. Principal Parts of a Nuclear Reactor. tor are made keeping in view of 'uniform heat production within the reactor. The fuel elements are designed taking into account the heat transfer, corrosion and structural strength. The cooling passage should be designed in such a way that there should not be any hot spot in the fuel element. (2) Moderator. It is a material used to slow down the neutrons from high kinetic energy (1MeV or 13200 km/s) to slow neutron (0.25 eV or 2200 m/s) in a fraction of a second. Further, a moderator is to increase the probability of reaction and to maintain the chain reaction due to slow neutrons. The slowing down of the neutrons is effectively done by the light elements such as H2, D2, N2, 02, C and Be. The graphite, heavy water or bergllium can be used as a moderator with natural uranium. In one of the arrangement, the moderator (solid) surrounds the fuel. (3) Reflector. In order to keep the critical size of the reactor and hence the amount of fissionable material as small as possible, it is important to conserve neutrons. This is possible by surrounding the reactor core with a material Which reflects escaping neutrons back into the core. This material is called reflector. The required properties of a good reflector are low absorption and high reflection for neutrons, high resistance to oxidation and irradiation as well as high radiation stability. Many times the materials used as moderator is also used as reflector. The H2O, D20 and carbon are also used as reflector. (4) Coolant. The main purpose of the coolant in the reactor is to transfer the heat produced in the reactor and to keep the fuel assembly at a safe temperature to avoid their melting and destruction. The same heat carried by the coolant is used in the heat exchanger for further utilization in the power generation either generating steam or using hot gas.

938

Steam & Gas Turbines And Power Plant Engineering

The following are the requirements of a good coolant— (i) It should not absorb the neutrons. (ii) It should have high chemical and radiation stability. (iii) (iv) (v) (vi)

It should be non-corrosive. It should have high boiling point (if liquid) and low melting point (if solid).

It should be non-oxidizing and non-toxic. It should have high density, low viscosity, high conductivity and high specific heat. The water, heavy water, gas (He, CO2), a metal in liquid form (Na) and organic liquids are used as coolant. (5) Control Rods. The control system controls the rate of energy generated. It starts, increase, decrease and stops the reaction. These rods may be shaped like the fuel rods themselves and are interspread throughout the core. Instead of containing fuel, they contain neutron absorber such as boron, cadmium or indium. The have high neutron-absorption cross-sections. (6) Biological Shield. The intensity of radiations and radioactive fragments from the reactor core is too high for the human body to tolerate. Therefore, it is necessary to surround the reactor with shielding material to prevent damage of human body due to radiation. (7) Reactor Vessel. The reactor vessel encloses the reactor core, reflector arid shield. It also provides coolant inlet and outlet passages. The reactor vessel has to withtand the pressure at 200 bar or above. The reactor core (fuel and moderator assembly) is generally placed at the bottom of the vessel. In order to obtain criticality or steady power three methods are used (i) building a heterogeneous reactor (ii) enriching the fuel or (ii) both. A heterogeneous reactor using natural uranium or low- enriched fuel, is one in which the fuel must be divided into separate fuel elements in the form of pins, rods, plates, hollow cylinders, pellets, spheres, etc. These are placed in the core, with space between them filled with moderator (Fig. 24.8). The second method of attaining criticality is to enrich the fuel by artificially increasing the percentage of U235 in it. Ordinary water, often incorrectly called light water, may be used as moderator, resulting in thermal heterogeneous reactor. 24.14. Classification of Nuclear Reactor. Nuclear reactors are classified on the basis of following— (A) Coolant. On the basis of coolant used the reactors are classified as (i) Gas cooled reactors, such as air, CO2 or helium. (ii) Liquid cooled reactors such as boiling water, pressurized water, liquid metal cooled reactors. The boiling water cooled reactor serves directly as a steam boiler. In the pressurized water cooled reactor, the water is under pressure in excess of saturation pressure at the maximum water temperature in the system, so that it cannot boil in the reactor. (B) Principal Product. On the basis of principal products, the reactors are classified as (i) Research reactors to produce neutrons (ii) Power reactor to produce heat (iii) Breeder reactor to produce fissionable materials

Nuclear Power Plant

939

(iv) Production reactor to produce isotopes. (C) Neutron Energies. On the basis of neutron energies, reactor are classified as (i) Fast fission reactors (ii) Intermediate or epithermal reactors (iii) Thermal (low energy) reactors (D) Fuel-Moderator Assembly. On the basis of fuel moderator assembly, reactors are classified as (i) Homogeneous reactors, in which the fuel is in the form of liquid such as uronyl sulphate dissolved in water. (ii) Heterogeneous reactors in which the fuel is in the form of solid (bar, rod, etc.) surrounded by moderator. This is the most common reactor. (E) Fuel State. On the basis of fuel state, reactors are classified as (i) Solid fuel reactors (ii) Liquid fuel reactors (iii) Gas fuel reactors (F) Fuel Materials. On the basis of fuel materials, reactors are classified as (i) Natural uranium with U235 reactors. (ii) Enriched uranium with more than 0.71 of U235 reactors. pu239 Pu241, or Pu 239 (artificial) reactors. 5 (iv) U233 (artificial) reactors. (G) Construction of Core. On the basis of construction of core, reactors are classified as (i) Spherical, (ii) Cylindrical, (iii) Octagonal, (iv) Cuboid, (v) Slab, (vi) Annulus reactor (H) Moderator. On the basis of moderator, reactors are classified as (i) Water, (ii) Heavy water, (iii) Graphite, (iv) Beryllium, (v) Hydrocarbon reactors. Table 24.3 gives the most common fission power reactor types in use commercially or under serious development world over. Table 24.3. The Most Common Fission-Power Reactor Types. Neutron energy Thermal (i)

Reactor Type

Coolant Pressurized water reactor H2O (PWR) (ii) Boiling water reactor H2O (BWR) (iii) Gas Cooled reactor (GCR) CO2, He (iv) Heavy water reactor D20 (HWR)

Fast

Moderator Fuel enrichment H2O

UO2, low

H2O

UO2 , low

Graphite

UC, low UO2, natural

D20

(I) Liquid-metal fast breader Na None reactor (LMFBR) (ii) Gas cooled fast reactor He None (GCFR) -—Table 24.4 gives the characteristics of typical power reactors.

pu23902+u238 02 pu23902+u23802

940

Steam & Gas Turbines And Power Plant Engineering

Table 24.4. Characteristics of Typical Power Reactors. Parameters

BWR

PWR

LMFBR

GCFR

Electric power (MWe)

1300

1050

1000

330

Thermal power (MWth)

3800

3000

2750

842

Specific power (KWth/kg)

33

26

575

50

Power density (KWth/m3)

100

60

300

10

Core height (m)

4.25

3.75

1.5

50

Core Diameter (m)

3.50

4.90

3.25

59

Coolant

H2O

H2O

Nal (liquid)

He

Pressure (bar)

155

72

8.0

48

Inlet temperature (°C)

280

275

330

400

Outlet temperature (°C)

310

285

500

770

Coolant flow rate (Mg/s)

20

12

.11

0.45

Average linear heat rate

22.5

20

30

—

24.15. Pressurized Water Reactor (PWR) Power Plant.

PWR reactor is currently the most extensively used type of power reactor in the world mainly due to the excellent properties of water as a moderator and coolant. The fuel used is alloy of uranium or UO2 The most stringent limitation on a PWR is the critical temperature of water, 374°C. This is the maximum possible exit temperature of coolant in the reactor but in practice, its value taken is around 310°C or less. In order to avoid the boiling of coolant in the reactor, the coolant pressure must be far greater than the saturation pressure corresponding to 310°C, i.e. 98.7 bar. In practice, the pressure of the coolant is maintained at about 155 bar and by this way the chance of boiling of coolant is prevented in the reactor. Fig. 24.9 shows a PWR power plant. It consists of two loop in series, the coolant loop, called the primary loop, and the water- steam or working fluid loop. The coolant picks up the heat released due to fission reaction in reactor and transfers it to the working fluid in Pressurizer Pressurized heated water Steam

Reactor Condenser

Coolant (pressurized water)

Pump

Feedwater heaters

Fig. 24.9. A Schematic of PWR Power Plant.

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941

Position indicator coil

Control rod drive mechanism Reactor vessel head Reactor vessel studs

Control rod port

Control rod guide bushing

Core hold down ring

Guide tube hold down plate Guide tube support plate

Head gasket Upper core support barrel Lower core support barrel

Control rod guide tube

Outlet nozzle Inlet nozzle for coolant Control rod drive shaft

Upper core support plate

Control rod shock absorber Baffle support flange

Control rod stop Control rod

Core baffle Inner thermal shield

Core barrel

Outer thermal shield

Fuel assemblies

Reactor vessel Lower core support plate Thermal shield support lug Control rod shroud

Fig. 24.10. Vertical Section through a PWR.

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Steam & Gas Turbines And Power Plant Engineering

the steam generator. The steam so generated is then used in a Rankine type cycle to generate electricity. 24.15.1. Construction Details. The construction details of PWR is shown in Fig. 24.10. The reactor vessel is in the form of a cylinder closed at each end by a hemispherical head. The upper head is bolted to the barrel, so that, for an access to the enterior of reactor, it can be removed. The size of the reactor vessel may be 3 m in external diameter, 4.5 m to 11 m in height and from 18 cm to 23 cm thick. It is lined with stainless steel. The fuel assembly which contains the fuel elements are mounted in a cage in the centre of reactor vessel. The cage consists of a core barrel of cylindrical shape closed at the top and bottom by horizontal core support plates. The length of fuel assembly is about 1.8 m to 2.8 m. It is mounted rigidly in a vertical position between the core support plates and can be removed from the reactor vessel. Commonly, the fuel may be an alloy of uranium which may be enriched by more than normal 0.7% of U-235 that is found in natural uranium or pillets of UO2. The fuel is encased in a stainless steel tubes that are seal- welded at the ends to produce a corrosionresistant gas tight cladding. Baffle plates are used to keep these stainless steel tubes in accurate parallel alignment. The baffle plates, themselves are mounted in a housing of rectangular cross-section that is constructed of some corrosion resistant alloy such as zircalloy 2. As the energy is released in the nuclear fuel within the small tubes, they are so spaced that cooling water flows around them in sufficient quantities so that they may be cooled properly and prevent an local boiling of water. Sometimes, fuel elements consist of flat bars of uranium alloy which are clad in stainless steel. Thermal shields are provided to absorb gamma rays and neutrons that are not reflected back into the core. Thus the thermal shields protect the walls of the pressure vessel from excessive radioactivity. They are constructed from steel plates in the form of thin walled cylinders and are supported from the reactor vessel wall and to extend for some distance below and above the core support plates. Since the thermal shield absorbs gamma and neutrons, it gets heated. In order to prevent them from overheating, these thermal shields are cooled. In order to the positions of core barrel thermal shields and fuel assemblies with respect to the wall of the reactor vessel, a horizontal section through a typical reactor may serve the purpose as shown in Fig. 24.11. Since control rod passes into the inside portion of the reactor, it is made of materials such as boron or cadmium which can absorb neutrons. Control rods are generally classified as regulating or skim rods and scram rods. The skim rods are moved in and out of the reactor core under close automatic control in such a manner as . to regulate the rate of energy release by controlling the number of neutrons available to produce fission. If the skim rods fail to control the rate of reaction, safety rods (called scram rods) can be inserted fully in a very short period of time. The arrangement of control rods is shown in Fig. 24.11. To control automatically the position of the control rods in the reactor core, a drive mechanism is provided outside the reactor vessel. A typical PWR contains about 200 fuel assemblies, each assembly being an array of rods. In a typical fuel assembly, there are about 264 fuel rods and 24 guide tubes for control rods. The coolant in this reactor is the pressurized water which is maintained with the help of pressurizer (dealt separately). It enters the reactor vessel through an inlet nozzle located

Nuclear Power Plant

943

Fuel assemblies Control rods

111=-111r 1 11112111C+

11

--L

Core barrel Thermal shield Wall or reactor vessel

Fig. 24.11. Horizontal Section through a PWR.

below the flanged head. The water flows downward along the wall of the reactor vessel through the thermal shields to the bottom of the vessel. After this, it flows upward through the fuel elements in the reactor core and goes out through the discharge nozzle which is located opposite to the inlet vessel. Baffles are provided so as to prevent the coolant flowing directly from the inlet to the outlet nozzle. It is to be noted that basic features of all types of reactors are similar to PWR. 24.15.2. Working Fluid Loop. The coolant leaving the reactor enters the steam generator which may be either shell and tube type with U-tube bundles or once through type, the former arrangement being more common. In the case of U-tube steam generator, the hot coolant enters at inlet channel head at the bottom, flows through the U-tube and reverses direction to an outlet at the bottom. Since the coolant temperature is less, so it can generate only saturated steam at moderate pressure (4.e. around 40 bar). In the once-through arrangement, the primary coolant enters at top, flows downward through tubes and exits at the bottom of main pumps while feed water flows upward (counter-flow) and gets converted in water in the condenser and is again recirculated in the steam generator. The first PWR power plant was built at shipping port USA in 1959 which generates 231 MW thermal output. The coolant pressure in the primary circuit is 14.1 bar and the coolant (water) outlet temperature is 282°C which generates dry and saturated steam in heat exchanger (i.e. steam generator) at 41 bar; 252°C. It develops 68 MW of electrical output at an thermal efficiency of 29.4%. The plant efficiency may be improved by using

944

Steam & Gas Turbines And Power Plant Engineering

an oil-fired superheater. 24.15.3. Pressurizer. As mentioned above, the coolant (water) in the PWR primary loop is maintained at a pressure (about 155 bar) much greater than the saturation pressure corresponding to the maximum coolant temperature (about 310°C) in the reactor to prevent bulk boiling. Since liquids are incompressible, small changes in volume due to changes in coolant temperature caused either due to load variation or sudden nuclear activity insertions results in severe or oscillatory pressure changes which are not at all desirable. In the event of increase of coolant pressure, some water will flash into steam which will affect the reactor performance and may lead to possible burn out of fuel element. In the event of decrease of coolant pressure, there may be a chance of cavitation. This suggests that a surge chamber often known as pressurizer may be incorporated in the primary loop that will accommodate the coolant volume changes while maintaining pressure within permissible limits. There are two types of pressurizers in common use- namely vapor pressurizers and gas pressurizer. PWR conveniently uses vapor pressurizers because their coolant, water can be vaporized. Gas- type pressurizers are commonly used in liquid metal cooled fast- breeder reactor. A vapor pressurizer (Fig. 24.12) is essentially a small boiler. In this case, liquid, the same as the primary coolant, is maintained by controlled electrical heating at a constant temperature and consequently a constant vapor pressure above its full surface. It is to be noted that this vapor pressure has to be the same as that of the primary coolant at the junction between the pressurizer and the hot leg of the primary loop. Thus the pressurizer temperature is higher than the primary coolant temperature because the latter is subcooled. The heaters are of the electric immersion type, located in the lower section of the pressurizer vessel. The bottom of the pressurizer is connected to the hot leg of the primary coolant system. A spray Spray nozzle is located at the top of presvalve surizer which is connected via conRelief trol valves to the cold leg of the primary coolant system after the Pressurizer pump. Under normal full-power oprelief tank eration, the pressurizer is about half Pressurize full of water and the top half is full of vapor. Electric In the case of positive surge, the heaters volume of the primary coolant increases and the vapor in the top half is compressed. Entry of the cooler primary coolant into the pressurizer condenses some of the vapor thus Heat limiting the pressure rise. Further, exchanger the spray valves are power-actuated and a cool spray (under pump pressure) enters the top, which helps condense vapor at a rapid rate and Feedwater Water limits the pressure rise. In order to Pump prevent the excessive cooling of the Fig. 24.12. Flow Diagram of a PWR Primary Loop spray piping and to maintain equal with a Vapor Type Pressurizer System. boron concentration in the primary

Nuclear Power Plant

945

coolant and pressurizer water, a small continuous spray is provided. Boron is used for chemical skim. In the case of negative surge the primary coolant volume gets decreased and the vapor in the pressurizer expands resulting in a momentary reduction in pressurizer pressure. The liquid in the pressurizer then partially flashes into vapor and assisted by further steam generation as a result of the automatic actuation in the •electric heater. By this way, the pressure is maintained above a minimum allowable limit. In order to protect the pressurizer against pressure surges that are beyond the capacity of the pressurizer, a power operated relief valve is attached to the top of pressurizer. 24.15.4. Case Study-A Typical PWR Power Plant. The working fluid system of a typical twin-reactor 100 MW (thermal) power plant consists of steam generator, steam turbine, generator, condenser, two condenser pumps, five stages of feed water heaters, two feed water pumps and auxiliary equipment. The steam turbine is composed of one doubleflow high-pressure element in tandem with two double flow, low pressure elements. There are four combination moisture separator-reheater assemblies between high and low pressure units. The wet steam from the exhaust of the high pressure elements enters each assembly of moisture-separator at one end, is distributed by internal manifolds, and rises through a wire mesh where the moisture is removed. After this, it flows over the tubes in the reheater where it is reheated by high pressure steam from the steam generators and after giving heat it gets condensed and is fed to the high pressure feed water heater. The reheated steam flows back to the low pressure turbine for further expansion. The generator coupled with steam turbines generates electricity. 24.15.5. Advantages. The following are the advantages of PWR— (i) The water which is used as coolant, moderator and reflector is cheap in first cost and available in plenty. (ii) Small number of control rods are needed. (iii) The secondary circuit may be optimized for higher efficiency. (iv) Fuel consumption is less. (iv) Safe and stable operation is possible. (vi) The reactor responds quickly for more power demand. (vii) The reactor is compact due'to use of enriched feed. 24.15.6. Disadvantages. The following are the disadvantages of PWR. (i) High capital cost due to high primary circuit pressure. (ii) The plant efficiency is low. (iii) Commissioning problems are more severe. (iv) Fuel suffers radiation damage. 24.16. Boiling Water Reactor (BWR) Power Plant. In BWR system steam is produced directly in the reactor core and thus it directly serves the purpose of steam generator. Here, the coolant (water) is in direct contact with the heat-producing nuclear fuel and boils in the same compartment in which fuel is located. It boils because the reactor pressure is maintained at about 70 bar less than half that in a PWR, with the nuclear fuel temperature roughly comparable. Since water and vapor coexit in core, the saturated steam is produced at about 283°C. The coolant in this case serves the triple function of coolant, moderator and working fluid. Fig. 24.-1-3-shows a simplest form of BWR Power Plant. It consists of a reactor, a

Steam & Gas Turbines And Power Plant Engineering

946

Saturated steam Saturated steam to turbine

gi

Water steam mixture

Chimney Feedwater Core

Turbine

Steam separator

ind

111111111111111

Recirculation water

Core

ll Downcomer recirculatio water (saturated)

111°

Feedwater d

Cooling Condenser Condensate pump

Feedwater pump (b)

(a)

Fig. 24.13. Schematic of a BWR Power Plant. (a) Internal and (b) External Recirculation.

steam turbine, a generator, a condenser, and associated equipments (air ejector, cooling system etc.) and a feed pump. The saturated liquid that separates from the vapor at the top of reactor or in a steam separator flows downward either internally within the reactor or externally outside the reactor via downcomer and mixes with the return condensate (Fig. 24.13). This recirculating coolant flows either naturally, caused by density differential between the liquid in the downcomer and the two- phase mixtures in the core or by the recirculating pump in the downcomer (not shown). The internal, forced recirculating type is most common in modern BWR. The performance of BWR is affected by recirculation ratio which is defined as the ratio of recirculation liquid to the saturated vapor produced. It is a function of the core average exit quality. Since boiling-water core exit qualities are low between 10 and 14 per cent, so recirculation ratio lies in the range of 6 to 10. This value of recirculation ratio is highly necessary to avoid large void fractions in the core which would materially lower the moderating powers of the coolant and result in low value of overall heat transfer coefficient or vapor blanketing and burnout of the fuel element. 24.16.1. Thermodynamic Analysis. For the analysis of BWR power plant, let us assume that a slightly subcooled liquid enters the core bottom of a rate of In; mass per unit. The liquid rises through the core and chimney if any. Here the chimney is an unheated section above the core which enhances natural recirculation. The resulting vapor separates and proceeds to the power plant at a mass flow rate of mg. Let saturated recirculation liquid flows via the downcomer at a mass flow rate of Mi.. Here, the recirculated liquid mixes with the relatively cold return feedwater and from the power plant to form the slightly subcooled inlet liquid in; (Fig. 24.12). The mass balance in the reactor core yields (24.35)

d =mg

and

g

f

=

The average exit quality of the entire core • m d . — and • - .171 g

f

rind + Mf

k is expressed as

(24.36)

(24.37)

Nuclear Power Plant lilt

The recirculation ratio (R) is expressed as, R =

mg The heat balance. results, m1 h. = mf

947 —

1

5_ c

(24.38)

(24.39) The condition of liquid entering the bottom is expressed by enthalpy of subcooling as given by hf

hd

h — hi Aliso = hf — hi = 5c- e (hf — hd) or ;ie — I h1-hd

(24.40)

or

(24.41)

degree of subcooling, =

=

t,

The total heat generated in the core Qi = thi [(121+ h e fg) _ h.] (24.42) Fig. 24.14 shows a BWR system supplying saturated steam directly to the turbine where it expands and does work. Throttle valve The exhaust wet steam from the turbine Saturated steam Turbine is condensed in the condenser and is Load pumped back to the reactor. A The advantages of the direct cycle Vapor are simplicity and its relatively low Feedwater capital cost. Because of the direct loop Condenser I arrangement, there are no heat exchang1 reactor ers. There is no need of any pressurizer, as there is large steam volume above Feedwater water. A major drawback of the direct pump cycle is that the reactor is not load following. Fig. 24.14. A Direct Cycle BWR Power Plant. 24.16.2. The Current BWR System. The current BWR designs are of the direct cycle, forced internal-recirculation-type. The reactor is about 7 m ID and 24 m high for a 100 MW plant. the fuel rods are similar to those of a PWR. They are Zircalloy-clad, 'about 4 in high in active length, and contain enriched UO2 pellets. The fuel assemblies are either 7 x 7 or 8 x 8 arrays and are enclosed in a Zircalloy-4 channel. The control rods are cruciform in shape and occupy the space between four channels. 24.16.3. Case Study:—A Typical BWR Power Plant. A typical BWR power plant produces 780 MW from 0.4 per cent moisture steam at 66.7 bar at turbine inlet condensing to 0.067 bar. The turbine is a tandem compound and double flow (with one HP section and two LP sections), four flows to the condenser and non-reheat. The cycle has a moisture separator between HP and LP turbines and five feed heaters. The throttle steam flow in 1210 kg/s. The thermal efficiency is 33.18 per cent. 24.16.4. Advantages. (i) The pressure vessel is less costly due to smaller pressure in the reactor vessel. (ii) The reactor does not require steam generator, pressurizer, circulating pump and connecting pipes, thus further reduces the cost. (iii) The metal surface temperature in the reactor is lower than PWR.

948

Steam & Gas Turbines And Power Plant Engineering

(iv) The reactor is more stable than PWR. (v) The plant efficiency is higher than BWR. 24.16.5. Disadvantages.

(i) It cannot meet sudden demand in load. (ii) The steam leaving the reactor is slightly radioactive. (iii) The power density is nearly 50% (33.6 kW/litre) of PWR, therefore the size of the vessel is large. (iv) The possibility of burn-out of fuel is more. 24.17. Gas Cooled Reactor (GCR) Power Plant.

GCR power plants have received great attention in UK and to a lesser extent in Germany, France, USSR and USA. This type of reactor uses natural or enriched uranium fuel, CO2 or helium as coolant and graphite or heavy water as a moderator. The main advantages with GCR lie in the fact that in general, (i) gases are safe, are (ii) relatively easy to handle, (iii) have low macroscopic neutron cross-section, (iv) are plentiful and cheap (except He) and (v) may be operated at high temperature without high pressurization. The main disadvantages include (i) the lower heat transfer and heattransport characteristics of gases, Gas Steam which require large contact surfaces Turbine and flow passage within the reactor Reactor and heat exchangers and (ii) their Heat high pumping requirements (beexchanger tween 8 to 20 per cent of plant Condgross power), which offers chalenser lenging problems of fluid flow, pressure drop, etc. Fig. 24.15 shows GCR power Feedwater Compressor plant.The gas coolant absorbs the heat released due to nuclear reacFig. 24.15. A GCR Power Plant. tion in the reactor and this heat is utilised to generate steam in a heat exchanger (i.e. generator) which expands in the steam turbine, get condensed in the condenser and fed back to steam generator. 24.17.1. Case Study:—A Typical U.K. Advanced GCR Power Plant. An example of AGCR program is Hinkley Point B nuclear power station. It has a twin reactor with a total

output of 1250 MW. The reactor uses enriched ceramic fuels clad in stainless steel, graphite as moderators and CO2 as coolant. The core is a 16 sided stack of graphite blocks, of which 308 form fuel channels and 81 are the control rods. The core is surrounded by an annular graphite reflector. The CO2 coolant enters at bottom at 310°C and 42.6 bar and leaves at top at 655°C and 40.4 bar. It then flows down 12 superheater-reheat boilers surrounding the core, within the cavity of a prestressed concrete reactor vessel leaving at 276°C and 40.2 bar. The coolant is then pumped back to the reactor by eight constant speed, electrically driven (28 MW) centrifugal circulators with variable-guide vane flow control. The steam is generated at 166.8 bar and 541°C. A plant thermal efficiency of 41.7 per cent is obtained. 24.17.2. Case Study:—A Typical-Temperature Gas-Cooled Reactor (HTGR). In

Nuclear Power Plant

949

Helium circulator

HPT a IPT Condenser PCRV

Steam generator

0

Boiler feed pump

Feedwater heaters

Condensate pump

Fig. 24.16. HTGR Power Plant Diagram.

USA, PWR and BWR power plants constitute the major share among nuclear plant. However, HTGR has been conceptualised and installed (Fig 24.16). In this concept, graphite is used for fuel particles coating, fuel structural material, moderator and coolant channel walls. In a 1160 MWe plant, the coolant is helium. The reactor (3000 MW thermal) is housed in a vertical prestressed concrete reactor vessel. It is 29.4 m in diameter, 27.6 m high, and includes the reactor core and reflector, helium and nuclear steam supply system delay-heat removal system, control rods and their drives, and facilities. The helium is circulated at an average pressure of about 48 bar. It enters the reactor core at the top at about 240°C, flows downward through the coolant holes in the graphite block and exits at 740°C. The steam generators are shell and tube type heat exchangers in which helium flows on the shell side. Superheated steam from the six steam generators, a total about 1000 kg/s, combines into a single header and enters the double flow HP steam turbine at 166 bar and 510°C. The exhaust steam from HP turbine is reheated and it leaves the reheater at 38 bar and 538°C, recombines and enters the double flow HP turbine and then to double flow LP turbine. The exhaust steam gets condensed and sent back to the boiler. 24.17.3. HTGR Direct Cycle-Gas Turbine Plant. This type of plant uses gas turbine for power production and coolant as helium gas. It consists a regenerator and multi-stage compression with intercooling (Fig. 24.17). The fuel used may be U-233 or Th-232. 24.17.4. Advantages. (i) It needs simpler fuel processing. (ii) The problem of limiting the fuel element temperature is not serious as in other reactors. (iii) There is no possibility of explosion due to CO2 as a coolant. (iv) There is no corrosion problem. (v) It gives better return economy due to low parasitic absorption. (vi) Graphite remains stable under irradiation at high temperature. 24.17.5. Dirqdvantages.

Steam & Gas Turbines And Power Plant Engineering

950

Intercooler 444 K

373 K

2 Stage compressor h = 0.85 726 K 12.5 bar Precooler 566 K Regenerator = 0.75 Fig. 24.17 A Schematic of Direct Cycle Gas Turbine Plant. (i)

Large vessel is required due to low heat transfer coefficient.

(ii)

The loading of fuel is more elaborate and costly.

(iii) The leakage of gas is a major problem if helium is used instead of CO2. (iv) Coolant circulation requires much power. (v)

The control is more complicated.

24.18. Heavy Water Reactor (HWR) Power Plant. The property of heavy water as very low neutron capturing cross- section makes it an excellent moderator and coolant and permits the use of natural uranium (0.7% U-235) as fuel. The advantages of HWR include good neutron economy resulting in good breeding ratios, high fuel burn up, and reduction in fuel cost. The disadvantages include the limita-

1. Zircaloy bearing pads 2. Zircaloy fuel sheath 3. Zircaloy end cap 4. Zircaloy end support plate 5. Uranium dioxide pellets 6. Canlub graphite interlayer 7. Inter element spacers 8. Pressure tube

Fig. 24.18. CANDU 28-Element Fuel Bundle.

Nuclear Power Plant

951

L.P. Turbine

it row.

8 4 kg/s

515 MW

40.3 0.81arIH C

1. Cylindrical steel vessel (calandria) 2. Pressure tubes (4) shown) 3. Heat exchangers 4. Moderator cooling circuit 5. Control rods

EN7„

L._ 249 °C

293 °C

Fig. 24.19. A Schematic of HWR Power Plant.

tions imposed by critical temperature like water. Due to economic incentive, Canada has developed this reactor for its maximum use as CANDU reactors. In this reactor the fuel elements are hermetically sealed Zircalloy-2 rods that contain compacted and sintered naturally enriched UO2 pellets. Twenty-eight such elements are attached mechanically at their ends to form a cylindrical fuel assembly (Fig. 24.18). Fig. 24.19 shows a diagram of HWR power plant. The heavy-water (D20) coolant enters the fuel channels at 249`C and leaves at 293°C and 90.7 bar via feeder pipes. Twelve U-shaped tube-in- shell heat exchangers generate a total of 814 kg/s of steam at 40.3 bar and 250,8°C from the ordinary-water working fluid. The steam expands in the turbine, does work, gets condensed in the condenser and returns back to the steam generator. 24.18.1. Advantages. The following are the advantages of HWR plant. The fuel need not be enriched. (ii) The reactor vessel may be built to withstand low pressure as compared to PWR and BWR. (iii) Control is very easy and so less costly. (iv) Good neutron economy is possible. (v) The reactor has high multiplication factor and low fuel consumption as heavy water is a very good moderator. (vi) High fuel burn-up is achieved. (vii) A shorter period is required for plant construction.. 24.18.2. Disadvantages. The following are the disadvantages. (i) The cost of heavy water is high.

Steam & Gas Turbines And Power Plant Engineering

952

(ii) The leakage is a major problem. (iii) Very standard of design, manufacture, inspection and maintenance are required. (v) The power density is considerably low (9.7 kW/litre) as compared to PWR and BWR. 24.19. Liquid Metal Fast Breeder Reactor (LMFBR) Power Plant. Fast reactors are those whose neutrons are not slowed down to thermal energies by a moderator. They are designed to create or breed new fissile material, while producing useful electric power. In fast- breeder reactor, the objective is to maintain a chain reaction with fast neutrons that have an average energy of about I MeV by fission in U235 and Pu239. It also must provide additional fast neutrons sufficient to convert U238 to PU239. 24.19.1. Fission Reactions. The typical fission reactions in fast reactors are given by 92

U235 + on' --> 56 Bain + 36 Kr97 + 20n"

(24.43)

92

U233 + on' --> 56 Ba' 36 +36Kr96 + 2 on'

(24.44)

Ba137 sr100 . + 0n1 56 35 011 (24.45) A fast neutron reaction with U238 results in a series of reactions that culminate in Pu239 and is given by 94Pu235

238

920 92

U239

1 + n0 ---->

239 92 U + y 24 min —> 13

(24.46) 239 + -1 e° 93Np

(24.47)

2-3 days

94 PU239 + -1e° (24.48) 13 This is a breeding reaction that converts fertile U238 into fissionable Pu239. In the analysis of fast breeder reactor the two following important fuel parameters are used A= the average number of neutrons emitted per fission and depends on fissile isotopes 930239

11

=

the fission factor, equal to the average number of neutrons emitted per neutron absorbed. It is strong function of neutron energy.

The conversion or breeding ratio is given by C = — 1 — L where L represents the losses.

(24.49)

Based on the values of ti and L, C can occupy three values which may be used to define reactor. If C is much less than unity, the reactor is called a burner. A reactor with low C is generally called a converter. One with high C but less than 1.0 is called an advanced converter. Breeding gain is defined as the gain in fissionable nuclei per fissionable nucleus consumed. Thus, G=C-1=i-2—L

(24.50)

Thus, Gin. = Cmax — 1 = — 2 (24.51) On economic importance to a breeder reactor is its doubling time. It is defined as the

953 •

Nuclear Power Plant

Na

Steam

Secondary heat exchanger (steam generator) Load

Reactor

Primary heat exchanger 111 t ibine eCondenser

Pump

. Feedwater pump heaters

Pump

>l< Intermediate loop

Primary loop

Working fluid

>1

Fig. 24.20. Schematic of a LMFBR Power Plant.

time required to produce as many new fissionable nuclei as the total number of fissionable nuclei that are both normally contained in the core and tied up in the reactor fuel cycle. In general, doubling time ranges between 10 to 20 years, the shorter the better. 24.19.2. Working Principal. Fig. 24.20 shows the schematic of a LMFBR power plant. Sodium and other liquid metals are used as a coolant but they suffer from high induced radioactives and are generally chemically active. This necessitates an intermediate cooling loop to be used between the primary radioactive coolant and the steam cycle. The intermediate coolant is usually also a liquid metal, often Na or NaK. This intermediate loop guards against reactions between the radioactive primary coolant and water. It

Control rod drive mechanism

Control rod Steam generator

Fuel rer'ion Blanket PRI sodium

Reactor loop

Pump

)1, Intermediate Power generation loop

Fig. 24.21. Loop Type LMFBR Plant.

loop

954

Steam & Gas Turbines And Power Plant Engineering

also ensures against its high pressure water or hydrogen entering the reactor. There are two types of primary loop concept : (i) the loop or pipe, type, (ii) the pool, tank or pot type. The loop type is the most common in USA and is shown in Fig. 24.21. In this case, the reactor vessel, heat exchangers liquid metal pumps, and other components of the primary system and their inter connecting piping are separated within a large building containing an inert atmosphere to preclude sodium fires in the case of a sodium leak. _ The pool type is more common in UK and is shown in Fig. 24.22. In this case, the entire primary system, including reactor, primary heat exchangers, and pumps, is submerged in a large tank filled with molten sodium. The tank is part of the primary coolant loops. The heat exchangers discharge coolant directly into the tank and the pumps receive coolant directly from it. It has been observed that the pool system has the edge in safety and economy while the loop system has the edge in the simple mechanical design. 24.19.3. Advantages. The following are the advantages of LMFBR (i) High plant efficiency at low cost power is possible due to high temperatures. (ii) The sodium as a coolant need not be pressurized. (iii) It is best suited to thermal reactor with slightly enriched fuel as the neutron absorption cross-section of sodium is low. (iv) The low cost graphite moderator can be used as it can retain its mechanical strength and purity at high temperature. (v) The reactor is more stable due to liquid metal cooled. (vi) The reactor size is comparatively small. 24.19.4. Disadvantages. The following are the disadvantages of LMFBR (i) The neutron economy is reduced. (ii) It is necessary to shield the primary and secondary cooling systems. Sodium level Heat exchanger 482 °C

Steam to turbine

Pump

Water from condenser Reactor loop

Intermediate Power generatioil 100p loop

Fig. 24.22. Pool Type LMFBR Power Plant.

Nuclear Power Plant

955

(iii) The leak of sodium is very dangerous. 24.20. Fusion Reactor Power Plant. As mentioned above the fusion reactor power plant is not yet commercialised due to very high temperature requirement resulting in plasma to initiate fusion process and to sustain it. 24.20.1. Fusion Reactions. Deuterium, a stable heavy isotope of hydrogen, present in natural water, is the main fuel for a fusion reactor. The following four reactions involving deuterium are given below— Energy per reaction Fusion Reaction I;\ 1

H2 + U2 I"

2

He3 + onl

+D —› He3 + n) 1He3 + 1pi T + p) (D + D 1H2 + if13 --> 2He4 + onl (iii)

3.2 MeV

(D

(ii) 1H2 + IH2 —›

(D

+ T -->

(iv) 1H2 + 2He3

2He4 + n) 2He4 + 1 /31

4.0 MeV 17.6 MeV 18.3 MeV

(D + He3 He4 + p) The symbols n, p, D and T stand for neutron, proton, deuterium (1H2) and tritium (1H3) respectively. Third fusion reaction occurs most easily between deuterium (D) and tritium (T) which is self-sustaining at temperature of 50 x 106 K releasing 17.6 MeV per reactions. The fourth fusion reactions requires a temperature equivalent to 10 K. 24.20.2. Working of a Conceptual Fusion Reactor. A futuristic deuterium-tritium fusion reactor is shown in Fig. 24.23. The plasma is contained inside an evacuated tube of about 4 m. The surrounding vacuum wall through which 14 MeV neutrons from the plasma pass, is maintained at 750°C. There happens to be two concentric regions outside this wall namely the lithium breeding moderator and the magnetic shield. Tritium is manufactured in the lithium blanket. The magnetic shield is maintained by large cryogenic superconducting magnets of 7 to 8 m in diameter. The binary vapor cycle is used in which potassium acts as topping cycle and steam as bottoming cycle. There is a recovery system for tritium. The following are the advantages of fusion power plant— (i) The fuel deuterium needed is almost in exhaustible. (ii) There is no production of radioactive waste. (iii) It is very safe to operate. (iv) The conversion efficiency is high as 60 per cent. (v) The rejection of heat per kW generated is low. If the large fusion power plant becomes successful, the energy crisis of the world will be over and a new era of high quality of standard of living of mankind will be achieved Mankind will be waiting for that day. 24.21. Indian Nuclear Power Plants. Among the various types of nuclear power plants available, India has selected Pressurized Heavy Water Reactor (PHWR) due to its several advantages suiting to her. A PHWR

956

Steam & Gas Turbines And Power Plant Engineering Magnet Magnet coil shield Tritium Tritium

Potassium turbine

Potassium turbine Potassium boiler Lithium blanket A


I

Steam & Gas Turbines And Power Plant Engineering

966

The time period may be hour, day, week or month. The unit discharge may be m3/s, km2-cm/h or day-s-m. Fig. 25.3. shows a hydrograph. In Fig. 25.3 (a) the discharge is decreasing with time (expressed in percentage of time) while in Fig. 25.3. (b), the discharge is first increasing with time (say onset of monsoon), reaches a maximum value and then decreases. The area under the curve show the water yield. Similarly, power duration curve may be plotted by representing power (=g 31 Q H) on ordinate. ' Fig. 25.4 shows an another typical flow duration curve showing the minimum (Qinin) and maximum (Qm ) flow rate as a percent of time. The power output available at Qmin (area OABC) is known as primary power and it will be available for all the times (i.e. 100% time). The additional output available at higher water flows is called secondary power (area BCDE). If the flow rate of Qn is a required for all the times (i.e. 100% times) as indicated by the area under the flow demand line DEF, then it would be possible to meet the uniform demand of the flow rate (or power) for all the times if storage equal to BEF is provided. Alternatively, a thermal power unit of BF capacity to work as a supplement to hydropower unit is to be installed. The curve also shows the natural flow sufficient to meet the flow demand, Qn is available for 53.5 per cent of time or 195 days in a year. 25.4.3.1. Mass Curve. A plot of cumulative values of water quantity (run-off) ex-

°Blowing water (say river) with respect to time for a specified time.

16 B ,, / /

14

/B

/11 / //

/ 0

12

/.// /'\ ../ ,/ tr,

E

co

0 0

/

/ /

10 /

0

0

/

/ j F/

/

,/' ,/

/'

/

/B /

C

/ /

/

/ H' / ,/ 4•',/ 1 / //

1? /6,

4,

/'

/ G/ E / / /

r /

Demand lines 100 cumec / • Parallel to AB / /. Parallel to GH 40 cumeo '

/ / D/ , / 2 ,,/ / / A" /

/

I'

,/

A

2

3 4 Time (years)

Fig. 25.5. Mass Curve.

5

6

Hydro-electric Power Plant

967

pressed in hectare-meter against time made from the records of mean monthly flows of stream is known as mass curve (Fig. 25.4). Mathematically, the mass curve is expressed as

V=

j Q(t)dt (24.5)

Where V is the volume run-off and Q(t) is the discharge in m3/s as a function of time. A typical mass curve is shown in Fig. 25.5. The slop of the curve at any point indicates the rate of flow at that particular time. This means thatif the curve is horizontal, the flow will be zero and if there is a high rate of flow, the curve will be steeper. The dry periods are indicated as concave depression on the mass curve. The straight line AB is obtained by joining the end points of the mass curve (Fig. 25.5). The slope of this straight line AB (tan 0 = V/t) represents the average discharge over the — total period. Two straight lines A'B' and A" B" are drawn parallel to AB and tangent to the mass curve at the lowest tangent point C and the highest tangent point D respectively. The vertical intercept A'A" represents the storage volume required to permit continuous release of water at this average discharge rate over the entire period. The rate of demand is an important parameter and is defined as the rate at which water is required for use in a power plant. The demand line represents the constant rate of demand. If the mass curve is steeper than the demand line some storage is possible and if it is flatter then water is withdrawn at a greater rate. Two demand lines are shown in Fig. 25.5. Many other problems can also be solved with the help of mass curve. The maximum uniform flow possible with a .given storage capacity can be determined from the mass curve by locating the lines with the smallest slope which will be tangent to the mass curve and has maximum departure from it equal to the assumed storage capacity. 25.4.3.2. Applications of Hydrographs. As discussed above hydrographs are presented in different forms. The following are the applications of hydrographs. (i) In preliminary studies of a hydro-project. (ii) In comparison of different streams. (iii) hi flood control studies. (iv) In evaluation of low level of flows. (v) In determining the primary and secondary power. (vi) In forecasting sedimentation. (vii) In design of drainage system. (viii) In forecasting average discharge and demand rate. 25.5. Pondage and Storage. Since the rain-fall and consequently run-off varies considerably over the years, so for harnessing a water power to meet a certain power demand, it is essential that some arrangement in the form of pondage and storage of water is required. Pondage defined as a regulating body of water in the form of a relatively small pond or reservoir provided at the plant. It is used to regulate the variable water flow to meet the

power demand. Fig. 25.6 shows the location of power house using pondage at a suitable site. A weir is provided across the river and water is directed through water channel to feed to pondage. From pondage water is fed to turbine through short penstocks.

968

Steam & Gas Turbines And Power Plant Engineering

Power channel

River

Weir

WS"

Intake Hydro-power house

Forebay to provide pondage

Tailrace Short penstocks (a) Power house with pondage Fig. 25.6. PoWer House with Pondage.

River

111

Hydro-electric power

River (b) Power house with storage Fig. 25.7. Power House with Storage.

Storage is defined as impounding of a considerable amount of excess run-off during season of surplus flow for use in dry seasons. This is made possible by constructing a dam across the river stream at a suitable site and building a big storage reservoir on the upstream side of the dam. The Water is taken to the turbine through the penstock on the downstream side of the dam so that sufficient head is available. Fig. 25.7 shows the location of a power house with storage. It is worth to note that the pondage increases the capacity of a river for a short-time such as week while storage increases the capacity of a river varying from 6 months to 2 years. 25.6. Main Elements of a Hydroelectric Power Plants. Fig. 25.8 shows the main elements of a typical hydroelectric power plant. They are— (i) Catchment area, (ii) Reservoir, (iii) Dam, (iv) Sluice Gate or valve, (v) .Penstock or

969

Hydro-eleetric Power Plant

4) „coco

Sluice gate Water carrying pipe (penstock) Shown in different plane

Transmitting tower

(17

Surge tank Anchor Transform

Transformer room Control room Generator Turbine Tail water level

Inlet valve Fig. 25.8. Main Elements of Hydroelectric Power

Plant.

conduits, (vi) Spillways, (vii) Surge tanks, (viii) Trash rock, (ix) Foreway, (x) Power House, (xi) Draft tube, (xii) Switch yard for transmission lines. 25.6.1. Catchment Area. Avast area upstream side of the river across which dam is construpted into which various streams of water in the form of run-off terminates into river is known as catchment area. The catchment area is characterised by many parameters such as size, shape, surface, orientation, altitude, topography, geological aspects, etc. The larger catchment area, steeper slope, and higher altitude result in greater total run- off water in the catchment area. 25.6.2. Reservoir. A reservoir is a storage of water either natural like a lake on a mountain or artificially built by erecting a dam across a.river. The efficient use of water resources lies in the storage of water during times of plenty for subsequent use in times of scarcity. Water stored in the reservoir is not only used for power generation, but also for irrigation, flood control, water supply mid navigation. This suggests that efficient management is a must. 25.6.3. Dam. A dam is a thick wall made of RCC, earth, masonry, etc. constructed across a river to store water at a high head. The design of the dam is such that it takes care all types of possible stresses acting upon it. 25.6.3.1. Classifications of Dam. Dams are classified on the basis of following— (i) Function..On the basis of functions, dams are classified as— (a) Storage dam, (b) Diversion dam and (c) Detention dam. (a) Storage Dams are constructed for storing water and using it subsequently as and when required for various uses such as hydropower, irrigation and water supply. (b) Diversion dams are constructed to raise the water level and to divert the river flow in another direction. (c) Detention dams are constructed primarily to store flood water. (ii) Shape. Based on shapes, the dams are classified as trapezoidal or arch dams to

970

Steam & Gas Turbines And Power Plant Engineering

suit the structural function. (iii) Construction material. On the basis of construction materials, dams can be classified as earth, rock pieces, stone masonry, concrete, RCC, etc. dams. Concrete, RCC, earthen and rockfill dams are most popular. (iv) Hydraulic design. On the basis of hydraulic design dams are classified as nonoverflow type and over flow type. In non-over flow type, water is not allowed to flow over the top of the dam while in overflow type, water flows over it respectively. (v) Structural design. On the basis of structural design, dams are classified as gravity dam, arch dam and buttress dam where water thrust is resisted by gravity, arch and buttresses action. 25.6.3.2. Selection of Site of Dam. Several factors should be taken into consideration in the selection of site of dam. They are as follows— (i) At neck formation of river (Fig. 25.7). (ii) At confluence of two rivers (Fig. 25.9). (iii) Geology of foundation. (iv) Hydraulic factors and river diversion during construction. (v) Availability of construction materials. (vi) Accessibility of materials by transport. (vii) Economics. (viii) Availability of know how. (ix) Sociological factors resulting from • submergence of area, displacing people and disturbing the ecology. (x) Safety consideration as failure of dam is a big catastrophe. The chances of failure if

River water

Mountaneous region

River

River

Water (a) Neck

(b) Confluence Fig. 25.9

Mountaneous region

Hydro-electric Power Plant

971

any may be during flood and earthquake and a dam has to be designed to withstand these shocks over and above the other stresses acting over the dam in normal situation. 25.6.3.3. Choice of Dams. (i) Gravity Dams. In gravity dams, the water thrust is resisted by gravity. They may be constructed either in stone masonry or concrete or RCC. The concrete or RCC dam can be constructed faster with better quality control. The cross-section of solid gravity dam is shown in Fig. 25.10. In a stable dam, the resultant of all the forces acting on•the dam will intersect the base at same point within the middle third of the base line. The water pressure P and the weight of the dam W are the principal forces acting on the dam and the resultant force R intersects the base at the required middle third. (ii) Buttresses Dam. Fig. 25.11 shows a buttresses dam. It has a flat deck laid on slopping buttresses. The weight of water P on the deck is transmitted as a thrust to the buttresses. It is also called hollow gravity type dam.

Headwater level

Tail water level

Middle third Fig. 25.10. Cross-Section of Solid Gravity Type Masonry Dam.

\\\\.\\\ _ \\\••••••••.\\\\\\\‘‘,, s•.,01,

Fig. 25.11. Buttress Type Masonry Dam with Flat Deck.

972

Steam & Gas Turbines And Power Plant Engineering

Fig. 25.12. Arch Type Masonry Dam.

Rock Fig. 25.13. A Cross-section of Earth Dam. (iii) Arch Dam. Fig. 25.12 shows an arch type masonry dam in which all forces are acting perpendicular to the surface so that they tend to compress the material in the arch. (iv) Earth Dam. Earth dam is constructed of earth fill with wide base. Generally, for small project, upto 70 m in height, earth dam is constructed. A large volume of earth material is required and it should be available in the vicinity. The construction of dam varies with height and the side slopes are flatter. Fig. 25.13 shows the cross-section of a earth dam. Stone pitching is done on the slopes. Though, it is cheaper than masonry dam, .but the seepage losses are more. The erosion by water overtapping the dam or seeping through it may cause serious damage to the dam. (v) Rock-fill Dam. Rock-fill dams are constructed from loose rock of all sizes and has a trapezoidal shape with a wide base. It must have a water tight section to reduce seepage as chances of seepage are more. It is suitable for mountaineous regions where plenty of all sizes of rocks are available. 25.6.4. Trash Rock. The water intakes from the dam or from the forebay are provided with trash rock to prevent the entry of the debris which might damage the wicket gates or

Hydro-electric Power Plant

973

turbine runners or chocking up the nozzles of impulse turbines. In the severe winter, ice may be deposited on trash rock and to prevent the icing trouble, it is heated electrically. Sometimes, air bubbling system is provided to overcome icing trouble. 25.6.5. Forebay. The purpcse of forebay is to serve as a regulating reservoir temporarily storing water when the load on the plant is reduced and provides water for initial increment of an increasing load while the water is the canal is being accelerated. In short, forebay is a naturally provided storage which is able to absorb the flow variations. The forebay is always provided with some type of outlet structure to direct water to the penstock depending upon local conditions. 25.6.6. Spillways. During the rainy season the water level in the reservoir basin rises which may endanger the stability of the dam structure. In order to relieve the reservoir of this excess water, a structure is provided in the body of a dam or close to it which allows the excess water to spill from the reservoir so that water level does not cross high flood level (HFL). Such as structure is known as spillway. There are various types of spillways as given below25.6.6.1. Overall (or Solid Gravity) Spillway. An overall spillway is shown in Fig. Crest

H.F.L.

Upper nappe

Lower nappe High flood level F.R.L. = Full reservoir level

Bucket

Fig.

25.14. Overall Spillway.

25.14. This provision is provided in concrete, RCC and masonry dam. Water spills and flows over the crest in the form of rolling sheet of water. The shape of the spillway is provided in such a way that it forms crest, upper nappe, lower nappe and fmally a bucket. The excess energy is destroyed by the bucket, 25.6.6.2. Chute or Trough Spillway. It is similar to overall spillway except chute or trough is provided at the end. The excess water spill and flows over the crest and shoots down a channel or trough to meet the river downstream of the dam. Though spillway is suitable to the situation when the valley is very much narrow which cannot accommodate the overall spillway. 25.6.6.3. Side Channel Spillway. In many situation when the valley is too narrow, even the trough spillway cannot be accommodated and it is not desirable to allow the flow of water over the dam, side channel spillway is provided (Fig. 25.15). After running parallel to downstream, the side channel either meets to the downstream or used as a canal. 25.6.6.4. Saddle Spillway. A saddle spillway is preferred only when any of above types of spillway are not favourable. A saddle spillway is a natural depression type of spillway and is provided on the periphery of the reservoir basin away from the dam (Fig.

974

Steam & Gas Turbines Arid Power Plant Engineering Dam

Spillway crest

Fig. 25.15. Side Channel Spillway. 25.16). 25.6.6.5. Shaft Spillway. The shape of the shaft spillway is like a funnel constructed adjoining the dam, the lower end of which is turned at right angle and then taken out below the dam horizontally. The excess water spills over the circular crest and then flows through the funnel below the dam and comes out horizontally to meet the downstream water. 25.6.6.6. Siphon Spillway. Fig. ,i5.17 shows a siphon spillway. At the fall reservoir level (FRL) a siphon is made in the main body of the dam. When water level rises above FRL, water spills through the siphon. An air vent is provided to drive entrapped air. 25.6.7. Conduit (Canal, Flume, Tunnel, Pipeline Penstock, etc. A conduit is an open or closed passage through which water flows from the head race to the turbine and water from the turbine goes to the tail race. There are various forms of conduits namely canals, flumes, tunnels, pipelines and penstocks. Canals and flumes are open, while tunnels, pipelines and penstocks are closed passages. A combination of conduits are shown in Fig. 25.18. A canal is an open waterway excavated in natural ground following its contour while allume is an open channel erected on a surface above the ground supported oil a trestle. In many situations there exists a Saddle spillway site

Top of dam

Free reservoir level

Profile of basin along its periphery •

Fig.

25.16. Saddle Spillway.

975

Hydro-electric Power Plant

Air vent

Crown

Full reservoir level

r%w

Lower limb

Step

Fig. 15.17. Siphon Spillway.

Reservoir 1 Reservoir 2

Max. elev.

Max. elev.

Regulating forebay Flume

Tunnel Intake

Fig. 25.18.

Penstock (to power station)

Sluice

Various Types of Conduits (Tunnel, Flume, Penstock, etc.).

ridge or mountain between the dam and powerhouse. To allow the water to' enter into the turbine, a closed channel called tunnel is excavated through the ridge or mountain from the reservoir to the powerhouse or to another reservoir as the case may be. A pipeline is a closed conduit supported on or above the surface of the ground. A penstock is a closed conduit for supplying water from reservoir/surge tank to the turbine. Canals and flumes are non- pressure conduits while tunnel, pipe and penstock are pressure conduits. In the event of short distance between the forebay and power house, separate penstock for each turbine is preferred while a single penstock is preferred to feed two or more turbines for moderate head and long distance. Penstocks may be constructed from steel, RCC, asbestos, cement cast steel, etc. and it may be supported or buried or embedded or exposed. The penstocks may be interconnected either by flexible couplings or expansion joints with rigid pipes having anchorage, semi-rigid pipes or flexible pipes. 25.6.8. Surge Tank. When the load on the generator decreases, there is a sudden increase of pressure in the penstock due to sudden decrease in the rate of water flow to the

Steam & Gas Turbines And Power Plant Engineering

976

Reservoir level o

Surge tank

Hectia-ce

Pen stock

Turbine Draft tube

Fig. 25.19(a)

Tailrace

Surge tank on ground level

Reservoir level Reservoir Hydraulic gradient line Tunnel

Penstock Power house (b1) Inclined surge tank

-

Tail race

Conduit

(b2) Fig. 25.19(b)

Riser Over flow —0. Water (c1)

Ports

Water (c2)

(c3)

Fig. 25.19(c)

•

Fig. 25.19. Surge Tank and Its Types (a) Simple Surge Tank (b) Arrangement and (c) Types.

Hydro-electric Power Plant

977

turbine when the gates admitting water to the turbines are suddenly closed owing to the action of governor. The sudden rise of pressure in the penstock above normal due to reduced load on generator is known as water hammer. When the load on the generator increases, the turbine gates open suddenly allowing more water to enter into turbine to meet the increased demand, water has to rush through the penstock and there is a tendency to cause a vacuum in the water system. The penstock must withstand the positive hammer caused by sudden decrease in load and no vacuum should be produced in the system when the gates suddenly open under increased load condition. A surge tank is a small reservoir introduced in the system between the dam and power house, preferably on the high ground to provide better regulation of water pressure by reducing the pressure swing during variable load conditions so that they are not transmitted to the penstock. Fig. 25.19(a) shows a surge tank placed in between reservoir and power house. At low load water flow into the turbine gets reduced suddenly, so water level rises in the surge tank. This produces a retarding head and decreases the velocity of water in the penstock. When the velocity of water in the penstock is reduced to the value demanded by the turbine, the level of water in the surge tank starts falling and fluctuates

0

1

4

o

o (a) Conical draft tube

(b) Trumpet shaped draft tube

(c) Elbow draft tube with circular section

f(D*I

r- D

(d) Elbow draft tube with variable shape section

(e) Hydrocone draft tube

D

f) Moody draft tube with low cone

(g) Moody draft tube with high cone

Fig. 25.20. Various Types of Draft Tubes.

Steam & Gas Turbines And Power Plant Engineering

978

up and down till its motion is damped out by friction. In the event of sudden rise in the load on the turbine, additional water is supplied from the surge tank. This lowers the water surface in the surge tank and thus producing an accelerating head which increases the flow of water in the penstock. When the discharge of water in the penstock corresponds to the turbine demand the water surface in the penstock ceases to fall down. By this way surge tank helps in stabilising the velocity and pressure in penstock and reduces the water hammer effect. Fig. 25.19(b) shows a typical arrangement of surge tank showing surge tank on ground level and inclined surge tank. The various types of surge tank such as conical, internal belt mouth spillway and differential types are shown in Fig. 25.19(c). In the differential surge tank, there is a central riser pipe having small ports at its lower end. It has the advantage of providing both the retarding and accelerating heads with the decrease and increase of load. A surge tank is located near to the power house as is feasible to reduce the length of penstock thereby reducing the water hammer effect. 25.6.9. Draft Tubes. The draft tube is a diverging discharge passage connecting the water turbine outlet with tailrace. It is shaped as diverging passage to decelerate the flow with minimum loss so that the remaining kinetic energy of water coming out of runner which would otherwise a waste as an exit loss in efficiently converted into head energy thereby increasing the total pressure difference on the runner. It also allows the inspection and maintenance of turbine. Fig. 25.20 shows the various types of draft tubes. The conical type is mainly used for low head turbine while elbow type is more common as it allows the water to come out horizontally in tailrace. 25.6.10. Power house. A power house comprises of two main parts, a sub-structure to support the hydraulic and electrical equipments and a superstructure to hous and protect these equipments. The turbines (prime movers) which rotate on vertical axis are placed just below the floor level while those rotating horizontally are placed on the ground floor. The generating units and exciters are placed on the ground floor. The elevation of water turbine with respect to tailrace is determined by the necessity of avoiding cavitation. It is always advantageous to locate the power house underground under certain topographical conditions where there is no convenient site for a conventional (on the surface) type. Many underground power houses are working world over for example Koyna power house in India and Kemano power house of Canada having generating capacity 1670 MW under a head of 785 m. The following equipments are installed in the power house (Fig. 25.21). (i) Hydraulic turbines (ii) Electrical generators (iii) Governors

(iv) Gate valves

(v) Relief valves

(vi) Water circulating pumps

(vii) Air duct

(viii) Switch board and instruments

(ix) Control room

(x) Storage batteries

(xi) Cranes 25.7. Classification of Hydroelectrical Power Plants. There are various ways to classify the hydroelectrical power plants. They are as follows— (a) Availability of head. Based on the availability of head, hydroelectric power plants

979

Hydro-electric Power Plant

are classified as (i) High head power plants (ii) Medium head power plants, (iii) Low head power plants (b) Nature of loads. On the basis of nature of loads, power plants are classified as— (i) Base load plants (ii) Peak load plants (c) Availability of water. On the basis of quantity of water available, power plants are classified as (i) Hydel plant with storage reservoir (ii) Run-off river hydel plant with and without pondage. (iii) Pump storage hydel plants (iv) Mini and micro hydel plants Various types of above hydel plants are discussed below in brief. 25.7.1. High Head Power Plants. The hydel power plants operating under a head of 100 m or above are called high head power plants. Fig. 25.22 shows a schematic of high head power plants. In such plants water is stored in lakes or high mountains during rainy season or when snow melts. Surplus water if any is discharged by a spillway. Tunnel is constructed through the mountain and a surge chambef is provided with the help of regulating valve, water is allowed to flow the turbine. Due to large head, Pelton wheel is used as prime mover. 25.7.2. Medium Head Power Plants. Such plants in large number are found world

Galleries for auxiliaries

•••••••••••••••••••• •••• •••••••, • •••••••, ,•••••••

••••,•••••••• ••••••,••••••• •,,••••• ••••••••,••••• ••,••••• •••• •••• •••• •••••••• ■ ••••

Third stage concretes Second stage concretes

. Penstock

Tailrace

Approach to manhole First stage concretes Fig. 25.21. A Sectional View of a Power House.

980

Steam & Gas Turbines And Power. Plant Engineering

Surge chamber

Headwork Spillway

Reservoir Valve house

Penstock

Pelton wheel power house -1Tailrace Fig. 25.22. High Head Power Plant Using Pelton Wheel.

over. They operate under heads varying from 30 m to 100 m. Francis turbines are invariably used as prime mover Fig. 25.23 shows a schematic of such a plant. In such cases, forebay before the penstock acts as reservoir as well as 'surge tank. In the forebay, trash rock is provided to prevent the entry of the debris so that water gates or turbine runners are not damaged. 24.7.3. Low Head Power Plants. Low head power plants are constructed where the head available is below 30 m. Under such a situation, a dam is constructed across a river and a sideway stream diverges from the river at the dam and a forebay is provided before the power house. After the power house, the water meets the original river. Since head is low, so Francis or Kaplan turbines are used as prime movers, Fig. 25.24. the management of such a plant. 25.7.4. Base Load Plants. Base load plants supply constant power to the grid without any interruption. 25.7.5. Peak Load Plants. The peak load plants are activated when the load is more Head race

Dam Forebay (or surge tank)

4 River (or reservoir)

Trash rock

\

Power house Penstock

Inlet valve

Francis turbine

Tailrace

Draft tube Fig. 25.23. A Medium Head Power Plant

981

Hydro-electric Power Plant Dam

River

Water.

Water

Tailrace Power house

Sluice gate Canal

Forebay

Fig. 25.24. A Low Head Power Plant.

than the nominal load. As discussed earlier, thermal and hydel plants and operated in tandem to meet the base and peak load during various seasons as per demand. 25.7.6. Hydel Plant With Storage Reservoir. This plant is found world over and more common in India. Water is stored in the reservoir during rainy season and electricity is generated throughout the year. In some cases, water is also used for irrigation purposes. 25.7.7. Run-off River Hydel Plants with or Without Pondage. These plants operate using the run-off water in the river and operate daily. It may or may not use pondage. If pondage is used it may take care of the peak load. 25.7.8. Pump Storage Hydel Plants. In such plants,'a small dam is constructed across the tailrace. Water after passing through the turbine is stored in the tailrace reservoir. During low load (say right time), the water is pumped back from the tailrace to the head water reservoir during excess electricity from the grid or from the nearby steam plant. The peak load is met by drawing excess water from the head reservoir and by this way about 70% of electricity consumed is reservoir and at the same time peak load demand is met. Fig. 25.25. shows the schematic of such a plant. The advantages of this plant lie in (a) substantial increase in peak load capacity at low cost, (b) high operating efficiency, (c) higher load factor and (d) independence of stream flow conditions. 25.7.9. Mini and Micro-Hydel Plants. Due to high capital cost, large gestation pe-

Dam Head water reservoir

Penstock

Powerhouse with pumps and turbines Head Tail water reservoir

Fig. 25.25. Pumped Storage Plants.

Steam & Gas Turbines And Power Plant Engineering

982

♦♦

Reservoir:::.

yYdraulic

h

gractie,,A lips

H (gross) Turbine

h (Net)

Penstock

••••••ti

a~Uac V.44

Fig. 25.26. Head on the turbine. riod, sociological problems, the large hydel plants are now less matter of attention. Now, more attention is given to mini and micro hydel plants. The natural water source in hilly terrain may be easily utilised to generate electricity using low head turbo-generator sets. There is a negligible disturbance of ecology. The Mini hydel plants operate under head of 9 to 20 m producing about 1 MW. From the latest survey, it has been estimated that the potential energy from mini and micro plants in India is about 22,000 MW. This shows the charm of such plants. 25.8. Hydraulic Turbines and Their Classifications. Hydraulic turbines are the oldest prime movers serving the mankind. They convert the head energy of water into shaft work and filially electrical energy through electric generators when turbines and generators are coupled on the same shaft. The oldest turbine is waterwheel. However, the modern turbines have gone through a lot of technological advances and as a result the size of the turbine are very large yielding high efficiency and large output. Hydraulic turbines are classified on the basis of the following— (a) Head and discharge of water available. The head of water available is the most important parameters in the selection of turbine. The gross head (H) is the difference in the elevation of water surface between the headrace and tailrace while the net head (hoe) is the difference between the gross head and head loss due to friction (hf) (Fig. 25.26). The water turbine works under a wide range of heads varying from 2 to 2000 m. Table 25.1. shows the classification of turbines with respect to head and the suitable turbine for that head. Turbines are also classified on the basis of discharge; low, medium and high discharge. It is to be noted that the distribution of head is a typical one. Table 25.1. A Typical Head Based Classification of Turbines. Head based turbine Low head turbine Medium head turbine

Head (m) 2-15 m 16-70 m

High head turbine

71-500 m

Very high head turbine

> 500 m

Discharge High discharge High or medium discharge Medium or low discharge low discharge

Turbine type Kaplan or Propeller Kaplan or Francis Francis or Pelton Pelton

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983

(b) Inventor of Turbine. On the basis of name of the inventor, water turbines are named as follows— (i) Pelton turbine. It is named after its inventor Lester Allen Pelton of USA. It is an impulse turbine and suitable for high head and low discharge. (ii) Francis turbine. It is named after James B. Francis who had invented it. It is a reaction turbine and suitable for medium head and medium discharge. (iii) Kaplan turbine. It is named after its inventor Dr. Victor Kaplan. It is an reaction turbine and used for low head and large discharge. (iv) Deriaz turbine. It is named after its inventor Deriaz, a Swiss engineer. It is a reversible turbine-pump and suitable upto 300 m. (c) Nature of working on blades. Turbines may be classified on the basis of energy conversion process when water impinges on blades. They may be impulse type or reaction type. The impulse type ,is one in which the whole head energy is converted into kinetic energy in nozzles and water comes out as a free jet and impinges on bucket i.e. moving blades. During this process water is always in contact with air and thus theoretically there is no head or pressure drop in bucket. Such a turbine is Pelton turbine. In reaction turbines, the entire flow takes place in a closed conduit. Part of head energy is converted into kinetic energy in nozzles and the runner converts both kinetic energy and head energy into mechanical energy. Many turbines such as Francis, propeller, Kaplan and Deriaz are examples of reaction turbines. (d) Direction of Flow. Based on the direction of flow, with respect to wheel, the water turbines are classified as axial, radial (inward), tangential, mixed and diagonal flow turbines. In axial flow, water flows along the axis of shaft while in radial flow, the direction of flow is perpendicular to the axis of shaft. Mixed is the combination of axial and radial. Tangential flow is tangent to wheel while diagonal flow is neither axial or nor radial but angular with respect to axis. Table 25.2 gives the flow directions of various turbines used in practice. Table 25.2. Flow Directions in Water Turbines.

Propeller and Kaplan turbines Francis turbines Pelton turbines

Flow direction Axial flow Radial inward flow or mixed flow Tangential flow

Deriaz turbines

Diagonal flow

Turbine types

(e) Orientation of axis of shaft. On the basis of orientation of axis of shaft, turbines are classified as vertical or horizontal turbine. Pelton turbines have horizontal shaft whereas Propeller, Kaplan, Francis and Deritz turbines have vertical shaft. (f) Specific speed. Specific speed of turbine is one of the most important parameter for the design of turbine. It is defined as the speed of a geometrically similar turbine which produces I KW power under 1 m head. Mathematically specific speed (Na) is expressed as N s— NH514 (25.6) Where N is speed in rpm, P is power developed in kW and H is head in metre. Although, technically, specific speed is not a dimensionless parameter but it exhibits the property of non-dimensional parameters. Table 25.3 shows a typical classification of water turbine based on specific speed. The specific speed varies from 4 to 1100 and the

Steam & Gas Turbines And Power Plant Engineering

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distribution has been made among slow, medium and fast speed for various types of turbines. Table 25.3. A Typical classification of Turbines Based on Specific Speed. Turbine Slow 4-20 60-150 300-450

Pelton Francis Kaplan

Specific speed Medium 21-35 151-250 451-700

Fast 36-70 251-400 701-1100

25.9. Ranges of Water Turbine Design Parameters. With the advancement of technology, water turbines have gone various changes with reference to their sizes, hydraulic design and efficiency. The present trend is towards bigger size. The turbine size and its number are chosen judiciously to obtain optimum working conditions as well as economy. Table 25.4. gives the ranges of water turbine design parameters. Table 25.4. Ranges of Water Turbine Design Parameters. Types of turbine

Maximum head (m)

Pelton Francis Kaplan

300-2000 30-500 2-70

Maximum power (MW) developed by the unit 250 750 230

Maximum wheel diameter of a runner (m) 5.5 10.0 10.0

Specific speed

4-70 60-400 300-1100

25.10. Theory of Pelton Turbines. 25.10.1. Construction Details and Velocity Diagram. A Pelton turbine consists of one nozzle or a set of nozzles and a pelton wheel. Fig. 25.27(a) shows the construction details of a pelton wheel. The runner consists of a large circular disc on the periphery of which a number of two-lobe ellipsoidal buckets are evenly mounted. Each bucket has a ridge or splitter in the middle which divides the jet into two equal stream as shown in Fig. 25.27(b). The symmetry of the bucket ensures zero momentum in the axial direction resulting in no axial thrust on the shaft bearing. The whole head energy is converted into kinetic energy in the nozzle and comes out, in the form of free jet. Pelton turbine is of impulse type. The nozzle directs the flow on the wheel and the flow approaching to the bucket is governed with the help of a spear valve which is under the control of governing system.. The water jet after impinging on the buckets is deflected through an angle of about 165° and flows axially in both directions thus avoiding the axial thrust. In general, there is a single nozzle feeding water to the turbine, but if the discharge is more upto six nozzles may be used, all symmetrically placed and causing rotation in the same direction. A four jet pelton wheel arrangement is shown in Fig. 25.28. Single jet pelton wheel has invariably horizontal shaft but multi-jet pelton wheel 'may have vertical

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Hydro-electric Power Plant

Side of bucket Fig. 25.27(a). Pelton Wheel and Nozzle.

Splitter

Ch

U

Bucket

Fig. 25.27(b). Shape of Bucket and Approaching Flow.

shaft. The specific speed of the multi- jet pelton wheel is given by N = Ns sinj

(25.7)

Where n is the number of jet and Ns is the specific speed of a single jet. The maximum number of nozzles used is 6 and the maximum specific speed of single jet turbine _wheel used is 6, so the maximum speed of multi jet pelton wheel will be about 70. In India, many pelton turbines are in operation such as at Koyna (475 m head, 4 jets),

Steam & Gas Turbines And Power Plant Engineering

986

Fig. 25.28. A Four Jet Pelton Wheel.

Sharavathi (570 m head, 4 jets). The world largest Pelton turbine is at Aurland-2, Brazil (840 m head, 6 jets) producing 243 MWe. The velocity diagram for pelton wheel is shown in Fig. 25.29. In Pelton wheel u1 = u2 = u. The bucket deflection angle is around 160° or 165° and as a result the water after leaving the bucket falls freely to the tailrace. The peripheral velocity (u) of the bucket is given by DN U = (Or (25.8) 60 where r is the bucket circle radius, D is the wheel diameter and N is the speed in rpm.

Bucket

Nozzle

8 = 165°

Cl >14

u

Crt Cr2 = U 1 2 =U Cw1 = C1

U

Outlet velocity triangle u2

Fig. 25.29. Velocity Diagram for Pelton Turbine.

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The ratio of wheel diameter to nozzle diameter (D/d) is an important size parameter for the design of pelton turbine. The velocity of jet issuing from the nozzle is given by C1 = Ic\ragri)

(25.9)

Where Kv is the coefficient of velocity and its value ranges from 0.97 to 0.99. 25.10.2. Workdone and Efficiency. The total energy transferred from the fluid to the wheel is given by Euler's equation. Thus E -

u l Cw1 - u2Cw2

U

- (C -C ) w1 w2

(25.10)

where the suffixes 1 and 2 refer to inlet and outlet conditions of runner, suffixes w, r and f represent whirl, relative and flow velocities respectively. From the outlet velocity diagram, we have Cw2 =

- Ca

cos (180 - ) = u + Cr2 cos°

(25.11)

where 0 is the bucket deflection angle (say 0 = 165°) The relative velocity of water at the .exit is reduced due to friction and thus given by (25.12) Crz = k Cr1 = k (C1-u) From equation (25.11) and (25.12), we have k(Ci - u) cos()

Cw2 =

(25.13)

Substituting the value of Cw2 in eq. (25.10) and since Cwt= C1 we have E = li [C - u - k (C1 - u) cos0 g

or

E -

(1 - k cosO)

(uc - u2)

(25.14)

The above equation is an equation of parabola. For a given value of k, cos 0 and C1 , there is a certain value of u, for which E will be maximum. Differentiating above eq. (25.14) w.r.t. u, and equate to zero, we have (1 - k cos0))

dE du =

Or

(C I - 2u) = 0

--- =0.5

Ct

(25.15)

Substituting the optimum value of u in equation (25.14), we have E — max

(1 - k cos0 C21 g ) 4

(25.16)

The blading or diagram efficiency (lb = id) is defined as the ratio of work output by a stage of the turbine to inlet kinetic energy. Thus, = rib

E /2g =

(1 k cog) )

2 \ 2g

(tic 11 k

or

lb = fid = 2(1 - k cos()) (p - p2)

Jc2 (25.17)

988

Steam & Gas Turbines And Power Plant Engineering where p = u/C,

The variation of 1b with u/C1 is plotted and is shown in Fig. 25.30. For maximum efficiency, in is differenti-

Ilmax

ated with w.r.t. p and equated to zero. drib

0 o

dp

1.0

0.5 p - u/C,

Thus 2(1 - k cos0 ) (1 - 2p) = 0

Fig. 25.30. Variation of rib with p. p , = 1/2 (25.18) op Thus we find that for maximum diagram or blade efficiency the ratio of u/C1 is just half. The same expression has been found with impulse steam turbine. or

Substituting the p , in eq. (25.17), we get (rib). = op cos (lb)max12k0

(25.19)

The diagram of blading efficiency formaximum work is given by max

(rldmax, work = c2 /2g =

1

For no blade friction, k = 1, thus

1 — kcos0 2 = (11b)max

max

cos0 -1 2

Generally, the value of k has been found to vary from 0.8 to 0.85. If k = 0.8 and 0 = 165°, (rib). = 0.886

(25.20)

Though, theoretically popt = (u/Cdopt = 0.5, but practically popt = 0.46 is used. The minimum number of buckets (z) on the wheel is given by an empirical equation. m z = — + 15 where m is the jet ratio (D/d). 2

(25.21)

25.10.3. Erosion and Cavitation. Erosion, and cavitation problems arise during operation of pelton turbine. The erosion problems occurs on the buckets due to erosive effects of flow and at the nozzle due to cavitation effect. Cavitation occurs at the exit of turbine or nozzles. In order to protect the buckets from wear and tear due to erosion and cavitation, chrome alloy steel or stainless steel is used as material. 25.11. Degree of Reaction for Water Turbines. Degree of reaction is defined as the ratio of energy transfer due to static pressure drop in the rotor to the total energy transfer to the turbine. R-

Energy transfer due static pressure drop in the rotor Total energy transfer to the turbine

(25.22) From Bernoulli's equation applied to the inlet of outlet of water turbine, we have

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989

C2

/0

- 2 2 12pg1 +5 = c+— + 2g pg 2g

(25.23)

where E is the total energy transferred from the fluid to the turbine. and p in the density of water. Thus [1 3

E—

p ) + (C2 C2) 2

2

1

pg

2g

(25.24) In the above equation, the first term on the R.H.S. represents the energy transfer due to static pressure drop and the second term represents the energy transfer due to drop in velocity head. For impulse turbine (i.e. Felton wheel), p1 = p2, there is no static pressure drop, so for impulse turbine C1- CZ - 2g (25.24a) PI P2 pg

(Reaction

(25.25)

If C I = C2, then E is expressed by eq (25.25). Eq. (25.25) is valid for pure reaction turbine. So, the degree of reaction (R) is expressed as R=

P2)/Pg

E

P

[(P 1

P2)/ Pg

p )/ IDA [ C1 2

C22)/2g]

(25.26)

Thus, R=0 for pure impulse and It=1 for pure reaction. Further, eq. (25.26) is modified as below R= 1 —

(cal - c22) 2gE

From Euler's equation, E = Thus, R = 1 —

(25.27) u2 c2)/g

C21 — C22 2(ui Cwi — u2 Cwt)

(25.28) The reaction of a water turbine may be calculated either from equation (25.26) or (25.27) or (25.28). 25.12. Theory of Francis Water Turbines. Francis turbines are widely used world over. It operates under the head varying from 30 to 500 m. The single unit may develop power as high as 750 MW. The specific speed ranges from 60 to 400. It is a reaction turbine. Formerly, its specific speed was limited to about 60 and it was radial inward flow type but at present they are the mixed flow type with radial entry and axial exit. 25.12.1. Construction Detail. Fig. 25.31 shows the schematic of a Francis turbine. It consists of a spiral casing, guide vanes (adjustable) and a runner. Water from the reservoir through the penstock enters the spiral or scroll casing which surrounds the runner. In order to keep the water velocity constant all along the flow path inside the spiral casing, its cross-sectional area diminishes uniformly along the circumference. The water enters the

0

990

s

Steam & Gas Turbines And Power Plant Engineering Shaft Crown

Pivot

Scroll casing

Guide vane o wicket gate Shroud ring Draft tube

Tailrace

Scroll casing

Fig. 25.31. A Schematic of a Francis Water Turbine.

adjustable guide vanes or wicket gates which are pivoted and can be turned suitably to control the flow and output. The purpose of the guide vanes is to impart a whirl component of velocity or angular momentum of the water before entering to the runner. The runner consists of number of curved blades numbering 12 to 22. These blades are welded to the shroud of the runner. The blade shape is such that it changes the direction of water flow from radial to axial direction. After leaving the runner, water enters a draft tube (either straight or elbow type) which increases the head by reducing the kinetic energy and fmally falls to the tailrace. It is worth to note that the pressure of water at the inlet of the runner is more than that at the outlet, so it is essential that the flow must take place in a closed conduit. The runner is always full of water and the draft tube is submerged into the tailrace. 25.12.2. Velocity Diagram, Workdone and Efficiency. Fig. 25.32 shows the velocity

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Hydro-electric Power Plant

diagram at inlet and outlet of the runner. The water comes out from the guide vanes at velocity C1 and with peripheral velocity u1, the water impinges the curved blade with velocity Cot at inlet and comes out with velocity Cr2. The discharge is axial, i.e. Cwt = 0, a2 = 90° and C2 = Cf2' From Euler's equation, the energy transfer to the turbine is E = (ui Cwt — u2 Cw2)/g

(25.29)

Since Cwt = 0, so E = ui Cwi /g But

Cwt = Cf.i cot a t where at is exit angle of guide blades u =Cc (cot a — cot (3t) where 13 is inlet blade angle fi Substituting the value of u1 and Cwt in eq. (25.29), we have E=

efi cot a (cot a — cot (3t)

(25.30)

Since it is a reaction turbine, so the total energy input is 2

Et.npui = E+Et

where Et = loss of kinetic energy — C2,2

2g

Thus, the blading or diagram efficiency is expressed as E1 lb E+E 1 E+E1 Empui 1 1 + 2 cot al(cot al — cot pi) CC cot a2 Degree of reaction R =

— 1—

(25.31) C2 w1

(25.32)

Hydraulic efficiency is expressed ai

C

Inlet velocity f.1 triangle

Cwt

Moving blade fl = Cf2 Cw2 = 0 Outlet velocity triangle

=C 2

Fig. 25.32. Velocity Diagram for a Francis Turbine.

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Steam & Gas Turbines And Power Plant Engineering

E

uiCwi

rlti H

gH

(25.33)

The overall efficiency of the turbine (1-1 0) is expressed as 1113

P pQgH

(25.34)

Where P is total power output from turbine, (w) p is the density of water (m3/kg), Q is discharge m3/s) and H is head (m). 25.13. Theory of Propeller and Kaplan Water Turbines. Propeller turbines are axial flow and reaction types machines. They are suitable for low heads (2 to 70 m) and high specific speed (300-1100) capable of handling large volume of water. 25.13.1. Construction Details. Fig. 25.33 shows a propeller turbine which consists of an axial-flow runner usually with four to six blades of aerofoil shape. The spiral casing and guide blades are similar to those in Francis turbines. The runner blades are fixed and non-adjustable. Kaplan turbine is a special type of a propeller turbine in which the individual runner Casing

Shaft

Guide vane

Runner blade

Runner hub

Runner blade

Fig. 25.33. A Schematic Diagram of a Propeller Turbine.

Hydro-electric Power Plant

993

blades are pivoted on the hub so that their inclination may be adjusted during operation as per load (Fig. 25.34). The blades are adjusted automatically rotating about the pivots with the help of a governor Vanes servo-mechanism. The varying demands mean varying flow rates which means varying approaching flow angle. For maximum efficiency, it is necessary when the flow enters the axial blades with zero incidence. This is possible only by using varying blade angle which is done in the case of Kaplan turbine. In other words, Kaplan turbines are high efficiency turbines at all loads. Fig. 25.35. shows a Kaplan turbine consisting of scroll casing, Runner iii Vane hub 111 adjustable guide blades and runner with attachment plovision adjustable blades. The draft tube is inclined type. The cross-section of typical low head concrete spiral case setting with Fig. 25.34. A schematic View of a Kaplan Kaplan turbine is shown in Fig. 25.36. Rotor with Adjust able Blades. 25.13.2. Velocity Triangle, Workdone and Efficiency. Fig. 25.37 shows the velocity triangle for propeller and Kaplan turbine. Since the peripheral velocity varies from root to tip, so the velocity triangles are different in symmetry. The expressions for workdone, and efficiencies are the same as of Francis turbine. 25.14. Theory of Deriatz Water Turbine. A schematic diagram of Deritz turbine is shown in Fig. 25.38. It is a diagonal turbine Water inlet from penstock

Main shaft

Scroll casing

Guide

ir 0. Runner blade

Tailrace = Flow I \ Boss

Draft tube

Fig. 25.35. A Schematic View of a Kaplan Turbine.

vane Guide wheel Link

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Steam & Gas Turbines And Power Plant Engineering

Crane

Motor generator set Runner Tailrace

Fig. 25.36. A Cross-section of a Power House with Kaplan Turbine.

and the flow over the runner is at an angle of 45° to the axis. Like Kaplan turbine, its runner has adjustable blades. The flow is diagonal or mixed. So for its suitability is concerned, it may be placed between two turbines Kaplan and Francis and is suitable for head up to 200 m. The number of blades on the runner varies from 10 to 12. The guide blades are inclined and

Inlet

Inlet velocity triangle

velocity triangle

Outlet velocity triangle

Propeller blade Outlet velocity triangle

(a) Hear the tip

(b) Near the hub

Fig. 25.37. Velocity Triangles for Propeller and Kaplan Turbines.

Hydro-electric Power Plant

995

Scroll casing

Shaft

Guide vane

4-- Water Water

Draft tube

Inclined vane

Fig. 25.38. Deriaz Water Turbine for Reversible Operation.

adjustable. This turbine is most suitable for reversible flow conditions when the turbine works as a pump in pump storage. power plants. The draft tube is straight conical type. The velocity triangles, the expression for workdone, reaction and efficiency are the same as that of Kaplan turbine. 25.15. Theory of Bulb Water Turbine.

Fig. 25.39 shows the schematic of a bulb or tubular turbine. It is suitable for low head in micro, mini and tidal power plants. It is an axial flow type. The turbo-generator is housed in an enclosed bulb-shaped casing, which is installed right in the middle of the flow passage. The bulb and the propeller form an integral unit followed by straight conical draft tube. 25.16. Comparison of Water Turbines

Table 25.5 gives a comparison of characteristics of Felton, Francis and Kaplan turbines.

Metal casing --Water —4

Runner vanes

maw orkir•-•• Draft tube

Fig. 25.39. A Schematic of a Bulb Turbine.

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Steam & Gas Turbines And Power Plant Engineering

Table 25.5. Comparison Between Turbines. S.N.

Parameters

Pelton turbine

Francis turbine

Kaplan/Propeller turbine

(i)

Flow, stage and impulse or reaction

Tangential, single stage, impulse type

Increased radial or mixed flow, single stage, reaction type

Axial flow, single stage, reaction type

(ii)

Maximum generating 250 MW capacity

750 MW

230 MW

(iii)

Jets/guide blades

1 to 6 jets, 2 for horizontal and 6 for vertical

Adjustable guide Adjustable guide blades blades

(iv)

Runner blades

Fixed

Fixed

Fixed for propeller and adjustable for Kaplan

(v)

Head

300-2000 m

30-500 m

2 to 70 m

(vi)

rpm

75-1000

93-1000

72 to 600

(vii)

Specific speed

4-70

60-400

300-1100

(viii)

Regulation mechanism

Guide vanes and re:ief valve

Guide vanes and runner vanes and relief valve better controlied

(ix)

Variation in operating head

Less controlled

Better controlled Better controlled

(x)

Size of runner, generator and power house

Larger

Smaller

Smaller

(xi)

Mechanical efficiency Decreases faster with time

Less decreases with time

Less decreases with time

(xii)

Hydraulic efficiency 85-90%

90-94%

85-93%

(xiii)

Uncleaned water

High wear with loss in efficiency

High wear with loss in efficiency

' Spear nozzle and deflector plate

Less wear

25.17. Specific Speed. In the development of hydraulic machines, first model is prepared for hydraulic test to predict the performance of prototype machine. The concept of specific speed helps in this regard. The specific speed of a turbine is defined as the speed of operation of a geometrically similar model of the turbine which produces 1 KW power when operating under 1 m head. The expression for specific speed is derived as follows—

Hydro-electric Power Plant

997 (25.35)

Power = p = 6QgH oc (QC).H where A is the cross-section flow area and C is the velocity of water

(25.36) P = D2 (2gH)I/2 Hoc D2 . H312 or where D stands for wheel diameter through which water flow axially and H stands for net head of water. The blade velocity = u cc C or

DN cc (2gH)1/2

(25.37)

or

DecH1/2 /N

(25.38)

From eq. (25 36) and eq. (25.38), we have P If

2 H512 oc H -- . H3/2 or N cc p1/2

N2

or N

CC

H514 P iz

H5/4

p1/2

or N = K p1/2 or K = N H5/4

P = 1 kW, H = 1 m, K = N = Ns = specific speed N P"2 N= —s=

H5/4

(25.39)

It is to be noted that the specific speed is not a non-dimensional number but it serves its purpose. The non-dimensional nuclear specific speed (Na) is expressed as N'is not used in practice. The ranges of specific speed for various water turbines are already given earlier. 25.18. Scale Ratio. The scale ratio is defined as the ratio of the diameters of the model turbine and the prototype turbine. We have seen that u cc C or

DN cc 47-7

Thus, the ratio of DN for model and protoype is given as D. N. 44-171: DN= PP

Or

D — = D

(25.40)

H N H N

P m

=scale ratio (25.41)

where suffixes m and p stand for model and prototype. 25.19. Unit Speed, Unit Power and Unit Discharge. The concept of unit speed (Nu), unit power (Pu) and unit discharge (Qu) are frequently used to express the operational characteristics of hydraulic turbine. Using this concept, the variation of speed, power and discharged for various turbines are found out. 25.19.1 Unit Speed. The unit speed (N) is defined as the speed of a geometrically similar turbine working under a head of 1 m. We have, u cc C or

DN cc 'sril

Steam & Gas Turbines And Power Plant Engineering

998 or

N oc 4-11 for geometrically similar turbine.

or

N = K \TY where K = constant H = 1 m, N = Nu = K,

If

Thus, Na =

(25.42)

25.19.2. Unit Power (Pu). It is defined as kW of power generated by a geometrically similar turbine working under a head of 1 m. -71) gH P = SQ g H = 8 (AC) gH = 8A .NWg1 we have, or

P oc H312 = KH312

If

H = I, P = Pa = K,

so,

Pu = H312

(25.43)

25.19.3. Unit discharge (Qu). It is the flow rate through the turbine working under a

head of I m.

We have,

Q = AC = A 42,1-1 cc

or

Q = K 4ii

If

H = 1, Q = Qu = K, thus

" = .411

Q

(25.44)

25.20. Cavitation Problem in Water Turbines. Cavitation is one of the undesirable problem in the operation of water turbines. The

formation of water vapor and air bubbles on the water surface due to reduction of pressure and sudden collapse are known as cavitation. In any part of the turbine if pressure drops below the vapor pressure at that temperature some of the liquid flashes into vapor. The bubbles formed during vaporization are carried by the water stream to higher pressure zones, where the bubbles condense into liquid forming a cavity or vacuum. The surrounding liquid rushes towards the cavity giving rise to a very high level pressure as high as 7000 atm. The rapid formation and collapsing of the bubbles cause the pitting of the metallic surface accompanied with vibration and noise. This phenomena is known as cavitation. It also reduces the hydraulic efficiency of turbine. Since the chance of reduction of pressure below atmospheric occurs only at the exit of runner blade of turbine or inlet of draft tube (eq. 25.45), so the cavitation occurs at the exit of runner. The energy balance at the exit of runner (er) and exit of draft tube (ed) yields

L)

per PO _ [h + 1C2er C _ 2g ) 8 8

(25.45)

where h is height between turbine exit and tailrace, hi. is the head loss between turbine and draft tube exit, pa is the atmospheric pressure exerting on the tailrace, per is the pressure of water at the exit of runner, Cer \and Ced are the water velocities at the exit of runner and draft tube respectively. 25.20.1. Cavitation Factor. Prof. D. Thowe of Germany suggested a cavitation factor to determine the zone where the turbines can work without any danger of cavitation.

Hydro-electric Power Plant

999

The critical value of cavitation factor is given by ac —

(H,— 11)— h H

(25.46)

where Ha = atmospheric pressure head (m), Hv = vapor pressure (m) at water temperature, H = working head of turbine (m), h = height of turbine outlet above tailrate level (m) The values of a, depend upon specific speed and turbine which are given in table 25.6. Table 25.6. Values of ac for various Values of Ns and Different Turbines. Francis turbines Ns 50 100 150 200 250 300 350

ac 0.04 0.05 0.07 0.1 0.14 0.2 0.27

Kaplan turbines Ns 300-450 450-550 550-600 650-700 700-800

ac 0.35-0.4 0.4-0.45 0.46-0.6 0.85 1.05

25.20.2. Methods to Avoid Cavitation. The following are the methods to avoid cavitation in water turbines. (i) Installation of turbine below tailrace level. The danger of cavitation increases in the case of low head and high speed propeller runner as the value of (C2er — C2C)/2g is considerably large in eq. 25.45. In order to keep the value of pressure at the exit of runner (p„) within the cavitation limitation, the value of h is made negative keeping the runner below tailrace level. For such installations, the turbines remain always under water. The main disadvantage of this method is the difficulty in inspecting and repairing the turbine. (ii) Cavitation free runner. By designing properly the cavitation free runner, the problem of cavitation is avoided. This is possible by selecting the proper shape of blade, blade angle and thickness of blade, etc. (iii) Use of material. The cavitation effect can be reduced by selecting a material which can resist better the cavitation- effect. Cast steel, stainless steel and alloy steel material for runner are in order of increasing resistant to cavitation. It has been observed that the welded part is more resistant to cavitation than casted. (iv) Polishing. By using polished surface the cavitation effect is minimised. The cast steel runners are coated with stainless. (v) Selection of specific speed. By selecting a runner of proper specific speed, it is possible to avoid cavitation. 25.21. Governing of Hydraulic Turbines The purpose of governing of hydraulic turbines is to maintain the speed of turbines fairly constant irrespective of the load. The generator coupled with hydraulic turbine is required to run at constant speed irrespective of the load so that the frequency of electricity generated remains fairly constant with a small permissible deviation (±2) otherwise

1000

Steam & Gas Turbines And Power Plant Engineering

electrical appliances will burn. The frequency or speed is expressed as N

_ 120f (25.47)

where f is the frequency of electricity generated, N is the speed of generator known as synchronous speed and p is the number of poles on the generator. When the load on the generator varies with the variation of consumer's demand and when input to the turbine remains the same, then the speed of the turbine or the frequency will vary due to the difference in power output and load which is not desired as all the electrical appliances and machines are designed to run on nominal frequency. Therefore, the speed of the turbine is always required to be maintained fairly constant irrespective of the load. This is made possible by using a governing system which regulates the quantity of water flowing through the turbine in proportion to the load so that the power output from the turbine is equal to the load and by this way the speed is maintained constant. 25.21.1. Governing System of Impulse Water Turbines. Fig. 25.40 shows the governing system of impulse water (Pelton) turbines. It consists of a centrifugal governor mounted on shaft which is geared with turbine shaft relay valve, servomotor, spear, deflector plate, oil sump, oil pump and linkages. In this case, water flow is regulated by joint action of the spear and the deflector plate. When the load on the generator drops, the speed of the turbine runner increases due to accelerating torque caused by the difference between turbine output and gerierator load. As a result the flyballs of centrifugal governor fly outward due to higher speed. The sleeve moves up, the portion of the lever to the right fulcrum moves down pushing the piston rod of relay or control valve downward. The relay consists of two valves - V1 and V2. With the downward movement of piston rod valve V1 closes and valve V2 opens. The oil is supplied to the relay valve at high pressure by gear pump. The high pressure oil enters to the left of servomotor and pushes the piston rod which partly closes the flow area, and hence the required (less) amount of water enters to the turbine so that the turbine output is equal to the generator load and by this way speed is maintained constant. When the load on the generator increases, the speed decreases, the valve V1 of relay opens while valve V2 closes. The high pressure oils enters through valve V1 opening to the right portion of servometer which pushes the. piston to the left hand side and by this the spear moves to the left and thus supply more water to the turbine resulting in turbine output equal to the load. By this way, the speed of the turbine is maintained fairly constant irrespective of the load. It is worth to note that the spear or needle valve is normally used for small load fluctuations. In the event of sudden fall of load the spear moves very rapidly to close the nozzle which results in water hammer problem in penstock. In order to avoid the water hammer problem, a deflector is introduced in the system (not shown). The purpose of deflector is to deflect some water from the jet advancing to the turbine runner is the event of sudden drop in load. The deflected water without doing any work falls into the tailrace. Thus, in the case of sudden rapid drop in load, the speed is maintained constant by the joint action of spear and deflector plate. 25.21.2. Governing System of Reaction Water Turbines. Fig. 25.41. shows the governing system of reaction water turbines. This system is similar to that of impulse turbine except that the motion of piston in servo-motor is used to partially close or open the guide vanes gates through which the water is supplied to the turbine instead of spear in

Hydro-electric Power Plant

1001

Centrifugal governor Fly ball

z Lever Sleeve --->

Piston rod of control valve Relay or control valve

To tu'bine main shaft

Piston rod of relay cylinder of servomotor Piston Oil sump

Water out

M Piston. Relay cylinder of servo-motor Water in

Spear

Nozzle

Fig. 25.40. Governing System of Impeller (Pelton) Turbine.

the nozzle. When the load is less than design on the generator, the centrifugal governor Senses the speed and valve V1 closes while V2 of relay valve opens which allows the high pressure oil to enter the left portion of servomotor. This pushes the piston forward and partially close the passage of, guide blades through rock and pinion arrangement via linkages. Reverse action takes place when the load is more. A compensating device is, however, added to prevent the governor from overshooting. The compensating device consists of ball-crank lever with a pivot when the servomotor piston moves to the right, the bell-crank lever EFG is rotated downward about F and the arm G is lowered. This pulls down the pivot A, which in turn, lowers the fulcrum B. By this way, the relay part "b" is partially or fully closed, restricting the piston motion of the servomotor to the right and by this way it prevents the governor from overshooting. In order to protect the system from water hammer a pressure relief valve is incorporated in the system (not shown). A sudden closure of the wicket gate of guide vanes due to sudden fall in load will open the relief valve due to sudden increase in pressure and protect the conduit from inertia effects of speeding water. The relief valve consists of a spear and is held by the fluid (oil) pressure to close the by pass of water from the spiral casing to the tailrace at design load. With the sudden decrease in load, a bell-crank lever opens the pilot valve of the pressure chamber so that the pressure on the spear is reduced which permits the spear to be lifted up and allows a portion of water to flow directly from the spiral casing to the tailrace through the bypass without passing through runner blades. Kaplan turbine is governed by the simultaneous control of adjustable guide vanes and runner vanes. Both are operated by separate servomotor and a control valve which are interconnected with those of guide vanes to ensure a proper inclination of runner vanes for a particular setting of guide vanes.

Steam & Gas Turbines And Power Plant Engineering

1002

Governor Fly ball Floating lever Water quantity regulating ring.

Motion from turbine shaft

Gear pump

Servo moto

Rack and pinion

Fig. 25.41. Governing of Reaction (Francis) Water Turbines.

25.22. Performance of Water Turbines. In real situation, turbines are supposed to work also at off- design conditions. In order to know the off-design performance of water turbines tests are conducted on prototype a model machine and the results are plotted in graphical forms called characterisdc curves. Generally, these curves are plotted in terms of unit quantities as given below— (a) constant head characteristic curves (b) constant speed characteristic curves (c) constant efficiency curves 25.22.1. Constant Head Characteristic Curves. In order to plot these curves, tests are conducted by maintaining a constant head and constant gate opening and the speed of the turbine is varied by varying the load on the turbine. By this way, a series of value of speeds (N) are obtained. For each value of N, the discharge and output of the turbine are measured. Similar tests are conducted for different gate opening. The values of Qu, Pu and Nu and overall efficiency (no) are computed for each gate opening. Such characteristics are shown in Fig. 25.42. for various types of turbines. Qu is independent of Nuu and it is only a function of gate opening as the discharge only varies with gate opening. For Kaplan turbine, Qu increases with Nu linearly for all positions of gate opening. Obviously, the higher Qu will be for full gate opening. But for Francis turbines, Qu decreases with the increase of Nu. The variation of Pu vs Nu and no vs Nu are parabolic in nature for all types of turbine. There exits a maximum value of Pu and no at a particular for any gate opening, however, the optimum value of Nu is different for different gate opening. 25.22.2. Constant Speed Characteristics Curves. In order to obtain constant speed characteristics curves say efficiency vs. load the constant speed is attained by regulating the gate opening, i.e. discharge as the load varies. The head may or may not remain constant. Fig. 25.43 shows the overall efficiency vs load for different turbine. It is obvious from the results that overall efficiency increases with load and attains its maximum value at full load, however, Kaplan and Pelton turbines maintain the high efficiency over a

Hydro-electric Power Plant

1003

longer range of part load (i.e. about 40% to 100%) as compared to other turbines. The variation of overall efficiency and power with discharge is shown in Fig. 25.4. The overall efficiency increases with discharge and it attains a maximum value while the power output increases with discharge linearly. 25.22.3. Constant Efficiency Curves. Constant efficiency or universal characteristics curves are shown in Fig. 25.45. They represent for all conditions of running. Iso-efficiency, and best performance curves and plotted between Nu and Qu for different gate opening. The inner most curve represents the highest efficiency while outer curves represents lower efficiency. A vertical line drawn at a specific value of Qu, it cuts the iso-efficiency curves at two points, giving two values of Nu for the same efficiency while it cuts at only one point at the best performance efficiency curves. Though, all the hydraulic design of water turbine is based on the optimum speed (Nupt) but it must also satisfy structurally the safety conditions at runaway speed (1.8 to 2.3 times N0) which is a maximum speed of the turbine under no load and no governing action. 25.23. Factors Governing the Selection of Water Turbines. The following factors govern the selection of water turbines. (a) Operating head. One of the parameter for the selection of water turbines is the operating head. As per present practice Propeller and Kaplan turbines are suitable upto 50 m head while for head varying from 50 to 400 m Francis turbines are suitable. For heads greater than 400 m, Pelton turbines are suitable. It is to be noted that the ranges of the head mentioned above is not hard and fast, it may vary depending upon other conditions to obtain maximum economy. (b) Specific speed. Specific speed plays an important role in the selection of turbine. It is always better to choose turbines of high specific speed as it means compact size of turbines, generator and power house, etc. The range specific speeds of Pelton, Francis and Kaplan are 6-60, 50-400 and 250-1100 respectively. (c) Height of installation. It is always advisable to install the turbines as high. above the tall water level as possible. This concept saves excavation cost for placing draft tube provided cavitation does not occur. (d) Performance characteristics 'of turbines. While selecting the turbine the performance characteristics of each turbine is to be studied carefully. If the turbine to be installed going to operate mostly at off-design conditions, the turbines should be selected such which gives better part-load performance. (e) Size of turbine. It is a well known fact that large size of the turbines offer better economy, so one should go for large size turbine. (f) Number of turbines. There is always a chance of breakdown, so one turbine does not serve the purpose for supplying constant electrical supply. It is always advisable to install at least two turbines sothat in the event of breakdown, one may supply the power. Problem 25.1. The quantity of water available for hydroelectric station is 300 m3/s under a head of 10 m. Assuming the speed of the turbine 200 rpm and its efficiency 85%, calculate the number of turbines required. Take specific speed 900. Solution. The power developed is given by

Steam & Gas Turbines And Power Plant Engineering

1004

Full gate opening

Unit discharge (Qu) ---).

0

3/4 gate opening 1/2 gate opening 1/4 gate opening 1/4 gate opening Unit speed (Nu) (a) For Pelton wheel

Full gate opening Full gate opening 3/4 gate opening

3/4 gate opening

1/4 gate opening Unit speed (Nu)

Unit speed (Nu)

0 Full gate opening

0 3/4 gate opening

Full gate opening 3/4 gate opening

a) 1/4 gate opening

U

0

(For Francis turbine) Unit speed (Nu)

Unit Discharge (Qu

1/2 gate opening

1/4 gate opening (For Francis turbine) Unit speed (Nu)

0

f

Full gate opening

Full gate opening

•••

3/4

0 O.

Unit speed (Nu) -->

3/4 gate opening

Unit speed (Nu) —+ (b) For reaction turbine

Fig. 25.42. Constant Head Characteristic Curves for Impulse and Reaction Turbines.

1005

Hydro-electric Power Plant 100 0 80 49

.-

60 0

F

6

(1) Kaplan turbine-Ns = 700 _ (2) Pelton wheel-Ns = 15 (3) Francis turbine- Ns = 350 (4) Propeller turbine-Ns = 600

8 40 •a w 20

0

I At constant speed test 0

20

40 60 Full load (%) (a)

1 80

100

100 Kaplan 80

a)

60

45

Constant speed test 11111111111 , 1

0

20 40 60 80 100 120 Full load (%) (b)

Power (P) --).

---)-Efficiency (q)

Fig. 25.43. (a) & (b). Overall Efficiency vs Load for Various Turbines at Constant Speed.

11

Power

Discharge (Q) Fig. 25.44. Variation of Overall Efficiency and Power with Discharge.

1006

Steam & Gas Turbines And Power Plant Engineering 1„-efficiency curves Best performance curve

111111111EMEINNIIIIMEIMINEME11111 1111111111111=1111111111•111111111111111111EMMIIII 111•111=MINECOMMINNOISIONIIMMIIII 70

-80

Nu

60

MEN1111105eaa*PZILia-Z-LiZclkikiElli

111•1111WWERIMEMERWRIMMI1`EIN MMEMINLIMigEEE'EsiltrillEREZMIll 40 MMENNINSS*SEELS'att.ziewroMPAIIIIIII 111•11116151 14a2=1;a:,-.-,=_=E-Ti.:114:Iallielin 50

30

minummiiiagiammunnm 9mm 11 mm 15 mm 20 mm 23 mm 27 mm

100

140

180

220

260

Qu

-'110

300

Fig. 25.45. Constant Efficiency Curves.

P

P g QH 110 103

103 x 9.81 x 300 x 10 x 0.85 — 25015.5 kW 103

The power developed by each turbine will be given by the expression of specific speed N

or

N '‘. 1T3 = pia

—

200 x 41'

= 64

10" 105/4

P = 642 7 4098.3 kW

The number of Kaplan turbines required n —

25015.5 — 6.10 P-1 6 4098.3

Ans.

Problem 25.2. The following data relate to a hydroelectric power plant— Available head = 30 m, Catchment area = 450 km2, Rainfall = 150 cm/year, Percentage of total rainfall utilised = 70%, Penstock efficiency = 90%, Turbine efficiency = 85%, Generator efficiency = 90%, Load factor = 45%. Calculate the power developed by the turbine and suggests suitable turbine for the plant. Solution. The quantity of water available per year in the reservoir = (450 x 106) m2 x 1.5(m) x 0.7 = 472.5 x 106 m 3 Quantity of water available per second for utilisation Q

472.5 x 106 = 14.98 m3/s 365 x 24 x 3600 x x

• Power developed = Pen

xpgQH g 103

= 0.95 x 0.85 x 0.9 x 103 x 9.81 x 14.98 x 30 kW = 3203.96 kW 103

Ans.

Hydro-electric Power Plant The load factor is expressed as, Load factor =

1007

Average load - 0.45 Peak load

3203.96 - 7119.9 kW 0.45 At least two turbines have to be provided. Peak load capacity -

The capacity of each unit = 7119.9 - 3955.5 kW 2 x 0.9

Ans. Problem 25.3. A Pelton turbine is to be designed for the following specifications : Power to be developed = 120 MW, Net head available = 500 m,. Speed = 200 rpm, Ratio of jet diameter to wheel diameter = 1/10, Hydraulic efficiency = 86%, Velocity coefficient = 0.98, Speed ratio (u/C1) = 0.46. Calculate (a) volume flow rate of water required, (b) number of jets, (c) diameter of each jet, (d) diameter of the wheel. Solution. The velocity at the exit of nozzle (jet) is CI =

= 0.98 42 x 9.81 x 500 = 99.04 m/s

Cv 4YR

Theperipheral velocity =u = p C1 = 0.46 x 99.04 = 45.55 m/s The overall efficiency = rjo = (a) or Q

120 x 106 0.86 p Q gH 103Q x 9.81 x 500 -

120 x 106 = 28.44 m3/s 103 x 0.86 x 9.81 x 500

Ans.

x 200 - 45.55 60

u- 7EDN 60

5x6 (d) or D - 45. 5 0 - 4.349 m x 200

Ans.

Since d/D = 1/10, thus (c)

d = diameter of each jet = 4.349 - 0.4344 m 0

(d) Number of jets C

d2

=

Ans.

28.44 x 4 _ 1.933 Fs 2 99.04 x (0.4344)2

t4

Ans.

Problem 25.4. A Pelton turbine with single jet develops 20 MW while working under a head of 600 m. The specific speed of the turbine is 20, the overall efficiency is 85%, and the coefficient of velocity is 0.98. Assuming the ratio of bucket speed to jet speed as 0.46 calculate (a) the volume flow rate. (b) the diameter of the wheel and (c) jet diameter. Take blade velocity coefficient as 0.98. N '4-13

Solution. The specific speed = N = H5/4 or

Ns H5/4 N

Ar/5

20 x 6005/4 419.95 rpm 428 x 103

The velocity of jet = C1 = Cv I2T1 = 0.98 /2 x 9.81 x 600 = 106.32 m/s The peripheral speed = u = 0.46 x 106.32 = 48.90 m/s

Steam & Gas Turbines And Power Plant Engineering

1008

TcxDx 419.95 — 48.9 60

DN 60

u—

48.9 x 60 — 2.223 m x 419.95

(b) or D —

=

The overall efficiency = (a) or Q =

Ans.

P pqQH

20 x 106 10' X 9.81 x 600 0 x 0.85

(c)

Q =A i C1 =

or

d1 =

20 x 106 — 0.85 103 x 9.81 Q x 600

— 3.997 m3/s

Ans.

di x 106.32 = 3.997 where d1 = jet diameter

x it 106.32

1/2

— 0.218

Ans.

Problem 25.5. A Pelton turbine employs four jets of each 60 mm diameter. These jets strike the bucket and each jet deflected by an angle of 165°. The speed of the bucket wheel is 50 m/s. Calculate (a) the velocity of jet for maximum efficiency and (b) maximum power and (c) blade efficiency. Solution. (a) For maximum efficiency, C1 = 24 = 2 x 50 = 100 m/s Ans. flow through each jet = (it/4) (12 C1 = (x/4) (0.06)2 x 100 = 0.2827 m3/s ,-,2 1 — kcos0 L'i mg ; W Maximum power developed = ( g 4. 1 — 0.98 x cos165 1002 x x 0.2827 x 103 x g/103 = 1375.76 kW 4 (Pmaxecrch jet — g For four jets, the maximum power developed \

(b)

Pmax = 4 x 1375.76 = 5503.04 kW —5.50304 MW

Ans.

1 —cos 165. = — kcos8 (c) The maximum blade efficiency = (rib)1 2 — 98.29% 2 max Ans. Problem 25.6. A Francis turbine operates under the following conditions : Outer diameter of runner = 2m, Inner diameter of runner = lm, Head = 200 m, Specific speed = 130, Power developed = 20 MW. The inclination of water jet with tangential direction = 12°. The water leaves the blades radially with no velocity of whirl. Hydraulic efficiency = 92%. Calculate the inlet and outlet blade angles. Solution. Refer to Fig. 25.46. The specific speed is given by Ns — or

14 —

N H5/4

— 130

130 x 200" — 691.98 rpm 420x 103

The peripheral velocity at inlet =

=

DN 60

n x2x 691.48 — 72.41 m/s 60

Hydro-electric Power Plant U

The hydraulic efficiency tih =

I

C -U C wt 2

W2

—

gH

Cwi —

1009

u CH, 1

1

gH

as CW2 = 0

0.92 x 9.81 x 200 — 24.92 m/s 72.41

C„ From velocity tr i4ngle,tan 1 = tan 12° = " Cwt or Cji = Cwt tanlr = 24.92 x tan12° = 5.296 m/s = C/2 tan

(3

u I — Cwl

I

5.296 or [31 = 6.36° 72.41 — 24.92

The peripheral velocity at outlet of runner = u2 = u1 tan132 —

=

72.41

Ans.

- 36.205 m/s

C„ — 5.296 — 0.1462 or P2 = 8.32° ' 36.205 u2 — Cw2

Ans.

Problem 25.7. The following data relate to a Francis turbine. Shaft power = 500 kW, Net head = 80 m, Speed = 600 rpm, Overall efficiency = 85%, Hydraulic efficiency = 92%, Flow ratio (Cfl/Ci) = 0.25, Breadth ratio (B/D) = 0.1, Outer diameter = 2 x Inner diameter, Cf = constant, Outer discharge = radial. The thickness of vanes occupies 7% of circumferential area of the runner. Calculate (a) the volume flow rate(b) the dia and width of wheel at inlet (c) the guide vane angle (d) the runner vane inlet and outlet angles. Solution. Refer to Fig. 25.47. The flow velocity at inlet is given by CI = 0.22 CI = 0.22 I2 x 9.8 80 = 8.715 m/s Shaft power P lo — Water power — pgQH (a) or Q =

500 x 103 103 x 9.81 x Q x 80

500 x 103 — 0.7445 m3/s 10' x 9.81 x 80 x 0.85

(b)

Q = A C = 0.93 7t Di Cfi

or

0.7495=0.93 itxD1 x 0.1 Di x 8.715

or

D1 =

0.85

Ans.

1/2

0.7495 (0.93 it x 0.1 x 8.715)

= 0.5425

[31 = 0.1 x 0.5425 = 0.05425 m = 5.425 cm (e)

DI 5425 D2 = 2 = 0' — 0.227225 m 2 7t Di N u— 1

60 •

—

TC

x 0.5425 x 600 — 17.043 m/s 60

Ans. Ans.

Steam & Gas Turbines And Power Plant Engineering

1010

C f1 U1

Moving blade

Moving blade

=0 = Cf2

= Cf2

Cf 2

Fig. 25.47

Fig. 25.46

1h or

ut Cwt — U2 Cw2 gH

Cw1 =

Ui Cwi

17.043 x Cwi

gH

9.81 x 80

— 0.92

0.92 x 9.81 x 80 — 42.364 m/s 17.043

n 8715 = • —.0.2057 From inletvelocitytriangle,wehavetana = 1 Cwt 42.364 or

a = 11.62°

(d)

tan[3, =

or

at = 18.992°

Cwt — u

Ans. 8.715 0.3441 42.364 — 17.043 —

From outlet velocity triangle, we have tanl32 — But

Ans. Cn u2

D2ui 17.043 u ——— — 8.521 m/s 2 Di 2 8.715 — 1.0227 or 02 = 45.644° tang = 8 521

Ans.

Problem 25.8. A hydro-power plant employs a inward flow reaction turbine. The peripheral velocity of the runner of the turbine is 40 m/s. The velocity of whirl of inflowing water is 35 m/s and the radial velocity of flow is 5 m/s. If the flow is 2 m3/s and the hydraulic efficiency is 90%, Calculate (a) the head on the runner, (b) the power generated by the turbine and (c) the angles of the vanes. Assume radial discharge. Solution. Refer to Fig. 25.46. Since the discharge is radial, thus Cwt = 0. The hydraulic efficiency is expressed as

1011

Hydro-electric Power Plant Cwi u 35 x 40 1 — 0.9 — gH 9.81 x H (a) or H—

35 x 40 = 158.56 m 9.81 x 0.9

Ans.

(b) Power generated =P=pQgHih —

x x 0.9 103 x 2 x 9.81158.56 103

Ans.

= 2799.85 kW = 2.79985 MW C„ 5 = — = 0.14285 (c) From velocity triangle, we have tam = I Cwi 35 or

al = 8.13° and = 180° — 8.13° = 171.87° tanfti =

—

u1 — C.,1

5

40 — 35

Ans.

— 1 or 131 = 45° Ans.

Problem 25.9. A hydroelectric power plant equipped with four units of vertical shaft Pelton turbine is coupled with 70,000 KVa, 3 phase, 50 hertz generators. The generators employ 10 pairs of poles. The gross design head is 550 m. The transmission efficiency of head race runner and penstock is 94 per cent. The power output from the four units is 280 MW with an efficiency of 92 per cent. The nozzle efficiency is 98%, and each turbine has four jets. Take u/C1 = 0.48, C,, = 0.98 and nozzle diameter is 28% bigger than jet diameter. The turning angle of buckets on wheel is 165°, the bucket friction factor is 0.98 and the discharge efficiency is 99.7%. Calculate— (a) the volume flow rate through turbine, (b) the jet diameter, (c) the nozzle tip diameter, (d) the pitch circle diameter of the wheel, (e) the specific speed, (0 the number of buckets on the wheel, (g) the workdone per kg of water on the wheel, the hydraulic efficiency. Solution. The synchronous speed of the generator is given by N = 120f 120 x 50 _

p

20

300 rpm

Net head available =ri•penstock x Gross head = 0.94 x 550 517 m Power developed per turbine =

Total power 280 4 in MW Number of turbines — — ""

In put water power for each turbine = 70 x 103 = 0.92 (a) or Q = Discharge —

p g QH 103 x 9.81 x Q x 517

103

=

70 x 103 x 103— 15 m3/s 0.92 x 103 x 9.81 x 517

15 Discharge per jet = T = 3.75 m3/s Velocity ofjet = C1=

= 0.98 I2 x 9.81 x 517 = 98.7 m/s

Discharge per jet = 3.75 =1(714) d2 Cl

103 Ans.

1012

Steam & Gas Turbines And Power Plant Engineering

/ (b) or d = diameter of each jet =

x 3'75 — = 0.2177 m 4 98.7

Ans.

(c) The nozzle tip diameter = 1.20 x 0.2199 = 0.26388 m

Ans.

7C

The pitch circle diameter of Pelton wheel (D) = u — D O But

u = 0.48 C1 = 0.98 x 98.7 = 47.376 m/s

or

u= 43.376 —

(d) or D —

N

1-cDx 300 60

47.376 x 60 — 3.016 icx 300

Ans.

300.4280 x 103 N .4T) — 64.392 (c) The specific speed = NS = Hsi4 — (517)5/4 The jet ratio —

D 3016 d = 0.2199

Ans.

11 7 c

(f) The number of buckets = z = 2d + 15 —

13.715

+ 15 = 21.85 22

Ans.

(C1 —u ) (1 k cos 0) u1 1 , kg.m/kg g (98.7 — 47.376) (1 — 0.98 cos 165) x 47.376 9.81 Workdone per kg The hydr au lic efficiency = ih lidischarge .head X ildischarge

The Workdone per kg = E =

487.27 x 0.997 = 93.966% 517

Ans.

Problem 25.10. An axial flow water turbine works under a gross head of 40 m. The mean diameter of the runner, is 2 m and rotates at 150 rpm. Water leaves the guide vanes at 30° to the tangential direction. The outlet angle of the runner vanes is 55°. The head lost in the casing and guide vanes is 5 per cent and the blade velocity coefficient of the runner is 0.92. Calculate (a) the blade angle, (b) the work output per kg of water in meter head and (c) the hydraulic efficiency. Solution. Refer to Fig. 25.46. The effective head = H = 0.94 x 40 = 37.6 m The velocity at the exit of guide vanes is given by C1 = 427-I g -- 42 x 9.81 x 37.6 = 27.16 m/s The peripheral velocity = u =

nDN 60

—

n x 2.0 x 150

60

— 15.7 m/s

From inlet velocity triangle, we have tan 13

I

(a) or R1 = 60°

CI sin ai CI cos ai — u

27.16 x sin 30° — 1.736 27.16 x cos 30° — 15.7 Ans.

Hydro-electric Power Plant

1013

The relative velocities at inlet and outlet of runner are CI sin at 27.16 x sin 30° 27.16 m/s Cr' — pi — sin 60° — and

C_r2 = k C = 0.82 x 27.16 = 22.27 m/s

Cwt = CI cos al = 27.16 cos 30° = 23.52 m/s Cwt = u — Cr2 cos 132 = 15.7 — 22.27 cos 55° = 2.92 m/s Workdone per kg of water in metre head of water is given by u (Cwi — C.,2) (b)

E—

15.7 (23 — 2.92) _ 32.96 m 9.81

g

Ans.

32.96 27 404 H 37.6 — Ans. Problem 25.11. The following data relate to a hydroelectric power plant employing Kaplan turbine. Power output = 30 MW, Head = 40 M, Speed ratio (u/C1) = 2, flow ratio (CJI/C1) = 0.65, Hub to tip diameter ratio = 0.3, Overall efficiency = 0.92. Calculate (a) the volume flow rate, (b) the speed, (c) the hub and tip diameter of the runner and (d) the specific speed. Solution. Power developed by the model turbine is (c) The hydraulic efficiency = rih —

(a)

P g Hilo P = m 103

103 x 9.81 x 0.6 x 5 x 0.88 — 25.898 kW 103

Ans.

The specific speed of the model and protoype has to be the same. Thus N N = m m — 300 H514 300 x 5574 (b) or m — V25.898

440.75

DN P Since DN oc -41 7 -1, so

DIn NM

=

D Nm H Scale ratio = D -\•1 P Np H in m Here H is not known, so it has to be calculated or

Nsp = P P H5/4 P

200 -4) x 103 - 300 H5/4 P

200 46(71 x 103 — 163.29 300

or

H5/4 P

or

H = (163.29)4/5 = 58.93 m

Ans.

1014

Steam & Gas Turbines And Power Plant Engineering

(c) Now scale ratio =

440 75 45r1r. 7.565 200 5

Ans.

The volume flow rate through prototype turbine is p

(d)

pgH

p

—

o

60 x 106 = 117.94 m3/s 103 x 9.81 x 58.93 x 0.88

Ans.

Problem 25.12. A test has been conducted on one-sixth scale model of a Francis turbine working under a head of 2 m. The results indicated that it could develop 10 kW at 600 rpm. Calculate (a) the speed and (b) power of a prototype turbine while working under a head of 50 m. Solution. The scale ratio is expressed as Dm

Hm N . Hp

P

Since the same turbine is used at both places, so Dm = D , thus

N. 600 _ FRI

11- or =-\1H p (a) or

NP =

600

N

N 10 —

= 346.4 rpm

Ans.

The volume flow rate through the turbine = Q = Cd i

D2 C or

or

Qp = D2p Cp _ \/:) _ --

=

I H .N — HP

D2 C

since Dm =p D

.4-i-

Qp — 10

I-- = 2.886 m3/s P =y3

(b) or Q

Ans.

Problem 25.13. The table below shows the run-off data of a river at a particular site : Table 25.7: Run-off data Month

Mean discharge (Millions of cu.m/ second)

Month •

Mean discharge (Millions of cu.m/ second)

January

40

July

90

February

35

August

110

March

30

September

120

April

05

October

70

May

15

November

50

June

60

December

40

1015

Hydro-electric Power Plant

140 — 120 — 100 — 80 — 60 — 40

1

20 0

J f MA MJJASOND Fig. 25.48. Hydrograph for Problem 25.13. (a) Draw the hydrograph and calculate the mean flow (b) Draw the flow-duration curve (c) Calculate the power developed for the available head of 200 m with the overall efficiency of 90 per cent. Assume 30 days in a month for simplicity. Solution. The mean discharge from the river 40 + 35 + 30 + 05 + 15 + 60 + 90 + 110 + 120 + 70 + 50 + 40 12

54.16 million m3/s

The hydrograph based on the above run-off data is given in Fig. 25.48. In order 'to draw the flow duration curve, it is necessary to find the lengths of time during which certain flow is available as given in the following table Table 25.8: Discharge-duration data. Discharge per month (Million m3/s) 0 10 20 30 40 50 60 70 80 90 100 110 120

Total number of months during which flow is available 12 ,11

10 9 8 6 5 4 3 3 2 2 1

Percentage time

100 91.666 83.332 75.00 66.66 50.00 41.666 33.333 29.00 25.00 16.66 16.66 8.33

1016

Steam & Gas Turbines And Power Plant Engineering

120 110 100 90 80

c 0 E

70 60

E Es) .c

50 40 30 20 10 10 20 30 40 50 60 70 80 90100 Time (%) Fig. 25.49. Flow Duration Curve.

Based on the above table, the flow duration curve plotted and shown in Fig. 25.49. Problem 25.14. Calculate the maximum height of the Francis turbine above the tailrace for the following data— Atmospheric pressure = 1.013 bar, Pressure at the exit of runner = 0.5 bar absolute, Velocity of water leaving the runner = 6 m/s, Velocity of water leaving the draft tube = 2 m/s, Head of water lost passing through the draft tube due to friction = 0.25 m of water head. Solution. The maximum height of the turbine above tailrace level is calculated by h

pa— pre C2re(C2 + de + h 8 2g 2g J.) (1.013 — 0.5) x 104 62 22 — + +0.25 2 x 9.81 2 x 9.81 103

= 5.13 — 1.834 + 0.2038 + 0.25 = 3.7498 m above tailrace

Ans.

Problem 25.15. The available discharge and head at a proposed site of hydroelectric power plant is 400 m3/s and 45 m respectively. The overall turbine efficiency is 90%. The generator coupled directly to the turbine has 24 poles and generates electricity at 50 cycles/s. Calculate the least number machines required if (a) a Francis turbine with specific speed of 350 is used and (b) a Kaplan turbine with specific speed of 700 is used. Solution. Since the generator is directly coupled to the turbine, the speed of the

1017

Hydro-electric Power Plant turbine must be equal to the synchronous speed of the generator. N=

120f 120 x 50 — 250 rpm p 24

The power developed =P=pgQHri o —

103 x 9.81 x 400 x 45 x 0.9 10

= 158922 kW = 158.22 MW The power capacity of each Francis turbine can be calculated by the following formula Ns

r\rii H 514 2

or P

(350 x 5"") — 26624.86 kW = 26.624 MW 250

(a) .'. Number of Francis turbine required = 156.22— 5.942 26.62486

6

Ans.

The power capacity of each Kaplan turbine is calculated from N., equation N 4-15 250 4T3 = H514 455/4 — 850 or

P—

(700 x 451 — 106499.4 kW = 106.4994 MW 2 50

(b) Number of Kaplan turbines required = 156.221.4856 -4 2 106.4994 —

Ans.

EXERCISES Viva-voce and Theoretical 25.1. What is the basic two parameters on which power output from water turbine depends ? 25.2. Discuss the advantages and disadvantages of hydroelectric power plants. 25.3. Explain the optimization procedure of hydro-thermal mix to meet the power demand of a particular region. 25.4. Discuss the parameters on which the selection of site for a hydraulic power plant depend. 25.5. Discuss hydrology and hydrological cycle. 25.6. What do you mean by hydrographs ? 25.7. What do you mean by mass curve ? How is it plotted ? 25.8. What are storage and pondage ? 25.9 List the essential elements of a hydroelectric power plant. 25.10. What do you mean by catchment area, reservoir and dam ? 25.11. Classify the dams and discuss each of them in brief. 25.12. What do you mean by gravity, buttress and arch dam ? 25.13. Discuss the parameters for the selection of site for dam. 25.14. What do you mean by spillway ? Discuss its various types ?

1018

Steam & Gas Turbines And Power Plant Engineering

25.15. What are the various types of conduits ? 25.16. What do you mean by surge tank ? What are the various types of surge tank ? 25.17. Who do you mean by water hammer problem ? How is it overcome ? 25.18. What do you mean by draft tube ? What are the various types of draft tube ? 25.19. Classify hydroelectric power plants and sketch each of them. 25.20. Explain the following—(i) headrace, (ii) tailrace, (iii) canal, (iv) Plume, (v) tunnel, (vi) pipeline, (vii) penstock 25.21. Explain with neat sketch a pumped storage plant. 25.22. What type of turbine would you recommend for the following heads ? (a) 1100 m, (b) 2 W, (c) 20 m 25.23. Classify hydraulic turbines. 25.24. With the neat sketch discuss the construction detail, velocity diagram, power out and efficiency of Pelton, Francis and Kaplan turbine. 25.25. On what basis you will select Pelton, Francis and Kaplan turbine for a particular site ? 25.26. With neat sketch, discuss the constructional aspects of Deritz and bulb turbines. 25.27. Define specific speed and derive the expression for it. 23.28. Define and derive the expressions for scale ratio unit speed, unit power and unit discharge. 25.29. Give a comparison for turbines. 2530. What is the purpose of governing of hydraulic turbines. 25.31. With neat sketch discuss the governing of impulse and reaction turbines. 25.32. Discuss the performance characteristics'of water turbines. 25.33. What is runaway speed ? How does it affect the turbine design ? 25.34. How is the type of turbine selected in a certain hydro unit ? 25.35. What do you mean by cavitation ? What are its effects and how can it be minimized ? 25.36. Derive the optimum speed ratio for maximum power and maximum blade efficiency of water turbine. 25.37. Explain the characteristic features of Francis, Pelton and Kaplan turbines. Numerical 25.38. The Pelton wheel diameter is 2 m and the deflecting angle of the bucket is 162°. The diameter of jet is 165 mm. The pressure behind the nozzle is 700 kPa. The speed of the bucket is 320 rpm. Neglecting friction, calculate the power developed by the turbine and the hydraulic efficiency: Ans. 352 kW, 0.62 25.39. A Pelton wheel develops 4500 kW at 400 rpm, working under a head of 360 m. There are two equal jets and the bucket angle is 170°. The pitch circle diameter of the bucket is 1.82 m. If the blade velocity coefficient is 0.85. Calculate (a) the blade efficiency and (b) the diameter of eac,h jet. Ans. (a) 0.91, (b) 0.1032 m 25.40. A Pelton wheel is to be designed to produce 12 MW under a head of 300 m at a speed of 500 rpm. The diameter of the jet is 1/9 of wheel diameter, calculate the number-and diameter of jets, diameter of the bucket wheel and the volume flow

Hydro-electric Power Plant

1019

rate. Take overall efficiency = 88%, C,, =.0.97 and speed ratio = 0.45. Ans. D = 1.32 m, d = 0.146 n = 4, Q = 4.62 m3/s

25.41. The runner of a Francis turbine works under a head of 150 m. The outer diameter is 2 m while the inner is 1 m. The specific speed is 200 and the turbine and generation 15 MW. If the hydraulic efficiency is 90%,.the nozzle angle is 12° and leaves the blades with no whirl, calculate the outlet and inlet angles. 25.42. A hydraulic power plant develops 100 MW. The head available is 40 m and the speed is 70 rpm. Calculate the number of turbine units required. 25.43. A hydroelectric power station is to be designed for a catchment area of 120 sq. km, run-off 70%, average rain is 130 cm. The head available is 400 m. If the overall efficiency is 85%, calculate the power developed. 25.44. The following table shows the mean discharge of a river at a particular location. Month January February March April May June

Discharge m3/s 150 350 450 750 900 950

Month July August September October November December

Discharge m3/s 1150 1350 1100 900 650 350

Draw (a) the hydrograph and (b) the flow duration curve. 25.45. The following data relate to a hydroelectric power plant— Available head = 180 m, Catchment area = 2100 sq. km, Annual average rainfall = 150 cm, Turbine efficiency = 85%, Generator efficiency = 95%, Percolation and evaporation losses = 15%,. Calculate the power developed and suggest the type of turbine to be used if the runner speed is to be kept below 250 rpm. 25.46. Develop a computer software for design of a hydroelectric power plant using Pelton, Francis and. Kaplan turbines. 25.47. Design a hydroelectric power plant for the following data— (a) H = 1200 m, P = 250 MW, (b) H = 400 m, P = 700 MW, (c) H = 60 m, P = 200 MW 25.48. Develop a software for the data given in Prob. 25.49.

26 Diesel Engine Power Plants

Diesel electric power plants 'available in the range of 1 MW to 50 MW capacity are used sometimes as central station for meeting small requirements and universally employed as standby plant to supplement thermal power station or hydroelectric station. either for starting condition from cold or to meet emergency situations. Enough literature is available in I.C. Engine books, here only systems and performance will be discussed. For more details, see the author's book titled "I.C. Engine and Air Pollution". 26.1. Applications. The following are the applications of diesel electric power plants — \ _,(i) Peak load plant. Diesel plants are suitable as peak load plants in combination with thermal or hydle plant. The peak load unit needs easy, starting and stopping and diesel plants serve this purpose. (ii) Standby unit. There are many situations in which standby units are needed such as the main unit fails or cannot cope with the demand. As an example, due to less rainfall in a particular year, the hydro-plant cannot meet the demand. Thus diesel units are installed as standby unit to supply power in parallel to generate the short-fall of power. (iii) Central Stations. Due to ease in installation, starting and stopping, diesel electric plants can be used as central station where the capacity requirement is small. (iv) Starting station. For starting large thermal power plant, auxiliaries such as FD and ID fans, BFP, circulating cooling water pump, etc. can be run by installing diesel electric power plant. (v) Energy unit. In many developing countries like India, power failure/interruption is a common features for hours. Under this circumstances, the power to vital units such as hospitals or industrial plant can be supplied by installing diesel electric plant. ,,,(vi) Mobile unit. In many remote areas, large civil construction work is carried out where there is no power. Diesel plants mounted on trailer are used to supply power to construction work. ✓(vii) Nursery station. In remote town where there is no supply of power by main grid, diesel electric plant can be installed. When the power becomes available by main grid, this plant can be shifted to other place. Such a diesel electric power plant is generally called nursery unit.

1021

Diesel Engine Power Plants

26.2. Advantages and Disadvantages. Advantages. The following are the advantages of diesel electric power plant— (i) Easy installation, (ii) Easily available in standard capacities, say 1 MW, 5 MW, (iii) Quick response for load changes, (iv) Less standby losses, (v) Ease in starting and stopping, (vi) Less space requirement, (vii) Less cooling water requirement per kW produced, (viii) Less capital cost, (ix) Less supervisory staff needed, (x) Reasonable thermal efficiency (about 32%), (xi) Better part load thermal efficiency, (xii) Less civil engineering work needed for installation, (xiii) No proper and specific site needed, (xiv) Fully capable of located at load centre, (xv) No ash handling problem, (xvi) Easy lubrication. Disadvantages. The following are the disadvantages of diesel electric power plant—(i) High operating cost, (ii) High maintenance cost, (iii) High lubrication cost, (iv) Limited capacity, (v) Noisy, (vi) Inability to meet overload, (vii) Emissions not suitable to human life. 26.3. Working and Classification of Diesel Plants. In a diesel (or compression ignition) engine, air is sucked through opening of inlet valve and is first compressed to high pressure through a compression ratio varying from 13 to 22 such that the temperature of the compressed air is more than the self- ignition temperature of the diesel fuel. The diesel fifel is sprayed in the cylinder at the end of compression in a very fine atomized form and combustion starts automatically resulting in very high pressure and temperature which forces the piston down for power or expansion stroke till the exhaust valve opens for exhausting the combustion products out of cylinder. It is worth to note that there is no spark plug for initiation of combustion. It is the high temperature of compressed air which initiates combustion and so the name is compression ignition engine also. Diesel engines may be classified as—(a) Four stroke, (b) Two stroke, (c) Horizontal engine, (d) Vertical engine, (e) Single cylinder, (0 Multi-cylinder, (g) Indirect ignition, (h) Exhaust to stack Muffler— Surge tanlz NI Fuel injection pump

Air filter Fuel day to Di sel engine

A

Exhaust line

•

Compressed air bottle

of water . 10 muffler1 Jacke water • p um p Generator 7 Oil 1

• uel Oilter

0

4

Cooling water

Air compressor

Foundation

Oil pump

Lubricating oil tank Overflow

06 Aux. oil pump Strainer

Transfer Fuel

pump

Oil ---4, ( 1 4_11 0:il cooler —0 -Cooing . Oil purifier tower Air —•

A

Heat Lt exchanger

Pump Li ) Cooling water

Fig. 26.1. General Layout of Diesel Plant.

Hot water

T

1022

Steam & Gas Turbines And Power Plant Engineering

Direct ignition, (i) Naturally aspirated, (j) Supercharged engine, (k) Air cooled, (1) Water cooled, etc. Four stroke multicylinder engines are generally preferred for standing plant. 26.4. General Layout of a Diesel Engine Power Plant. Fig. 26.1 shows the general layout of a diesel engine power plant. The engine and its various auxiliaries systems are depicted with their proper positions. Exhaust valve Inlet The flow path of air, fuel and gas are valve Injector Valve shown by arrows. The plant consists spring of the following— Exhaust Inlet (a) Engine, (b) Air intake system, port port (c) Exhaust system, (d) Fuel system, Cylinder (e) Fuel injection system, (f) Cooling head system, (g) Lubrication system, (h) Starting system. Cooling 26.4.1. Principal Parts of a Dief ins set Engine. Fig. 26.2. shows the Piston I Piston cross-section of an air cooled IC en• ii rings Wrist pin gine depicting . the principal parts. — Connecting • I Generally, for standby plant, water Cylinder rod cooled engine is preferred but where there is scarcity of water or in mobile ../ lie Crankcase power plant, air cooled engine is preferred. The principal parts are—(a) Crank pin Cylinder, (b) Cylinder head, (c) PisCrank ton, (d) Inlet valve, (e) Inlet port, (f) \ shaft Exhaust valve, (g) Valve spring, (h) Cooling fins, (i) Wrist pin, (j) Con*--"• 7 necting rod, (k) Crankcase, (1) Crank//, pin, (m) Crank, (n) Crankshaft, etc. ? Crank 26.4.2. Air Intake System. The „___,____._... 0, Lubricating _ oil sump .4 function of air intake system is to r A convey fresh air through louvres and air filter to the cylinder via intake Fig. 26.2. Main Elements of a Diesel Engine. manifold. In order to augment the power, supercharger is fitted in between the filter and engine and the supercharger is driven by the engine itself. Supercharging will be discussed separately in subsequent pages. 6.4.3. Exhaust System. The purpose of exhaust system is to discharge the engine A exhaust to the atmosphere with minimum noise. Fig. 26.3. shows the exhaust system. The exhaust manifold connects the engine cylinder exhaust outlets to the exhaust pipe which is provided with a muffler or silencer to dampen the fluctuating pressure of the exhaust line which in turn reduces most of the noise which may result if gases are discharged directly to the atmosphere. It is advisable to use flexible tubing system for exhaust pipe to take up the effects of expansion due to high temperature and to isolate the exhaust system from the engine vibration. Appreciable amount of heat from the engine exhaust goes as a waste. In order to utilise this, a heat recovery steam generator (HRSG) may be used to generate low pressure steam

Diesel Engine Power Plants

1023

Gases out

Silencer I

Expansion joint

Exhaust manifold

Cooling water out Diesel engine (Multi-cylinders)

Fig. 26.3. Diesel Engine Exhaust System.

for process work. Alternatively, it may also be used for air heating where exhaust passes through a heat exchanger. 26.4.4. Fuel Handling System. Fig. 26.4. shows the fuel handling system of a diesel engine power plant. The fuel oil may be delivered at the plant site by many means such as trucks, railway wagons or barges and oil tankers. With the help of unloading facility, the fuel oil is delivered to the main tanks from where oil is pumped to small service storage tank known as engine day tank through strainers. This day tank has the capacity to store oil equivalent to about 8 hours consumption. In order to reduce the pumping power input, oil is heated either by hot water or steam which reduces viscosity and so the power input. 26.4.5. Fuel Injection System. It is supposed to be the heart of diesel engine and its failure means stopping of the engine. The fuel injection system performs the following functions— (i) It filters the fuel ensuring oil free from dirt.

Full unloading line

Bulk storage tanks

Fuel oil pumps

Strainers Meters

Day tanks

To engines Fig. 26.4. Fuel Handling System.

1024

Steam & Gas Turbines And Power Plant Engineering

— Nozzle — High pressure pump R— Pressure relief valve M— Metering element — Filter S — Storage tank LP—Low pressure pump

Fig. 26.5. Common-rail Injection System.

(ii) It meters the correct quantity of the fuel to be injected in each cylinder. (iii) It times the injection process in relation to the crankshaft revolution. (iv) It regulates the fuel supply. (v) It atomizes finely the fuel oil for better mixing with the hot air leading to efficient combustion. (vi) It distributes, the atomized fuel properly in the combustion chamber. There are two ways to atomize the fuel. In one case air injection is used while in other method, pressure or mechanical or solid injection is used. Now, the air injection is obsolete and mechanical injection is invariably used. In mechanical or solid injection system, the fuel oil is forced to flow through spray nozzles at a pressure above 100 bar. There are three types of solid injection system namely— (a) Common rail injection system (b) Individual pump injection system (c) Distributor system. 26.4.5.1. Common Rail Injection System. As the name employs, a single pump supplies fuel under high pressure to a fuel header or common rail as shown in Fig. 26.5 from where the fuel goes to each of the nozzle located in the cylinder. The timing of injection is maintained by a mechanically operated valve and the amount of fuel is controlled by the push rod stroke. 26.4.5.2. Individual Pump Injection System. As the name employs, the system has an independent high pressure pump for each cylinder which meters, pumps and controls the timing of fuel injection as shown in Fig. 26.6. Each cylinder is provided with one injector and the pump and injector may be integrated as one unit. The fuel is brought to the individual pump from storage tank through course filter, low pressure pump and fine filter. The high pressure pump is equipped with control mechanism and at the proper time, a rocker arm actuates the plunger and thus injects the fuel into the cylinder. The amount of fuel-injected is regulated by the effective stroke of the plunger. It is the most popular fuel injection system in practice. 26.4.5.3. Distributor System. Fig. 26.7 shows the arrangement of distributor system. In this system, a metering and high pressure pump is used to pump the metered quantity,of fuel in the rotating distributor which distributes the fuel to the individual cylinder at the

1025

Diesel Engine Power Plants

N— Nozzle P— High pressure pump R— Pressure relief valve M— Metering element F — Filter S — Storage tank LP— Low pressure pump

Fig. 26.6. Individual Pump Injection System.

correct timing. The number of injection stroke per cycle for the pump is equal to the number of cylinders. The fuel is fed to the high pressure pump from storage tank through course filter, 1.p. pump and fine filter. Since the metering and timing of injection is accomplished by one plunger, equal amount of fuel is supplied to each cylinder at the same point in the cycle. 26.4.5.4. Fuel Injection Bosch Pump. Fig. 26.8 shows the construction details of individual Bosch injection pump. Here, all of the individual high pressure pumps are placed in one housing. High pressure lines start at the pump and terminate at the injector which is located in the cylinder head. Fig. 26.7. Distributor System. The pump is of the positive displacement type, the plungers of which are activated by individual cam (one cam per pump) which are arranged in line. At the proper moment the cam activates the plunger axially, which forces the fuel under high pressure to the injector. The amount of fuel injected is controlled by a horizontal movement of the rack, which in turn rotates the plunger and thus varies the effective stroke. 26.4.5.5. Fuel Injector. The liquid fuel in the injection system enters into the combustion chamber through the injector. Fuel injector employed in CI engine is of automatic type. It is mounted on the cylinder body at such a location which yields better performance. Quick and complete combustion is ensured by a well designed fuel injector. The fuel injector assembly consists of the following— (a) a needle or nozzle valve, (b) a compression ring, (c) a nozzle and (d) an injector body. Fig. 26.9 shows a cross-sectional view of a typical Bosch fuel injector. The fuel from the fuel pump is fed down to the nozzle mouth through long drilling passage. The fuel pressure acts-on the differential area of the nozzle valve which lifts against the spring

t

1026

Steam & Gas Turbines And Power Plant Engineering

force, and thus allows the fuel to enter Fuel to injector at high pressure into the combustion chamber via small orifice (holes) in the form of finally atDelivery valve omized spray. Once, the fuel from the spring Delivery valve delivery pump gets exhausted, the spring (in delivery pressure pushes the nozzle valve back on position) Fuel inlet its seat. Helix on Intake port plunger 26.4.5.6. Types of Nozzles. The deSpill port Fuel sump sign of nozzle is mainly based on the Plunger types of combustion chamber used enToothed suring proper and efficient combustion control Control sleeve of fuel. The types of nozzles used in rack Lug attached diesel engines are (Fig. 26.10)— (a) sinto plunger gle hole, (b) multi-hole, (c) Pintle type, Housing Plunger return (d) Pintaux type. spring 26.4.6. Cooling System. During combustion process the peak gas temperature in the cylinder of an internal To be jerked by cam combustion engine is of the order of 2500 K. Maximum metal temperature Fig. 26.8. Construction Details of Bosch Pump. for the inside of the combustion chamber space are limited to much lcrwer End cap Adjusting values than the gas temperature by a large Lock nut number of considerations and thus cooling for the cylinder head, cylinder and piston must therefore be provided. Necessity of engine Spring Leak off —A. cooling arises due to the following facts— (i) During combustion period, the heat Fuel inlet Spindle from fluxes to the chamber walls can reach as high fuel pump as 10 MW/m2. The flux varies substantially Nozzle Injector with location. The regions of the chamber cap body that are contacted by rapidly moving high Fuel temperature gases generally experience the Nozzle passage valve highest fluxes. In region of high heat flux, thermal stresses must be kept below levels Nozzle body that would cause fatigue cracking. So temperatures must be less than about 400°C for Fuel cast iron and 300°C for aluminium alloy for water cooled engines. For air cooled engines, Fig. 26.9. Bosch Fuel Injector. these values are 270°C and 200°C respectively. (ii) The gas side surface temperature of the cylinder wall is limited by the type of lubricating oil used and this temperature ranges from 160°C to 180°C. Beyond these temperature, the properties of lubricating oil deteriorates very rapidly and it might even evaporates and burn, damaging piston and cylinder surfaces. Piston seizure due to over= heating resulting from the failure of rubrication is quite common. (iii) The valves may be kept cool to avoid knock and preignition problems which result from overheated exhaust valves (true for S.I. engines).

1027

Diesel Engine Power Plants

Nozzle valve

Nozzle valve

Nozzle valve

Nozzle valve • r

Nozzle • body

Nozzle bod

0 I••

Nozzle body

)10/

1

Nozzle body

•IX X

/

At7

A

4•1!

Aroft

4, (a) Pintle type

(b) Single hole type

9 (c) Multi hole type

Auxiliary hole (d) Pintaux type

Fig. 26:10. Types of Nozzles. (iv) The volumetric and thermal efficiency and power output of the engines decrease with an increase in cylinder and head-temperature. Based on cooling medium two types of cooling systems are in general use. They are— (a) Air as direct cooling system. (b) Liquid or indirect cooling system. Air cooling is ❑sed in small engines and portable engines by providing fins on the cylinder. Big diesel engines are always liquid (water/special liquid) cooled. Liquid cooling system is further classified as— (i) Open cooling system (ii) Natural circulation (Thermo-system) (iii) Forced circulation system (iv) Evaporation cooling system 26.4.6.1. Open Cooling System. This system is applicable only where plenty of water is available. The water from the storage tank is directly supplied through an inlet valve to the engine cooling water jacket. The hot water coming out of the engine is not cooled for reuse but it is discharged. 26.4.6.2. Natural Circulation System. The system is closed one and designed so that the water may circulate naturally because of the difference in density of water at different temperatures. Fig. 26.11 shows a natural circulation cooling system. It consists of water jacket, radiator and a fan. When the water is heated. its density decreases and it tends to rise, while the colder molecules tend to sink. Circulation of water then is obtained as the water heated in the water jacket tends to rise and the water cooled in the radiator with the help of air passing over the radiator either by ram effect or by fan or jointly tends to sink. The direction of natural circulation which is slow is shown by arrows. 26.4.6.3. Forced Circulation Cooling System. Fig. 26.12 shows forced circulation cooling system which is closed one. The system consists of pump, water jacket in the cylinder, radiator, fan and a thermostat. The coolant (water or synthetic coolant) is circulated through the cylinder jacket with the help of a pump which is usually a centrifugal type, and driven by the engine. The function of thermostat which is fitted in the upper hose connection initially prevents the circulation of water below a certain temperature (usually upto 85°C) through the radiation so that water gets heated up quickly.

1028

Steam & Gas Turbines And Power Plant Engineering

Upper tank

Upper hose connection

Radiator core 4_

Air

Fan Lower tank

Fig. 26.11. Natural Circulation Cooling System.

Standby diesel power plants upto 200 kVA use this type of cooling. In the case of bigger plant, the hot water is cooled in a cooling tower and recirculated again. There is a need of small quantity of cooling make-up water. 26.4.7. Lubrication System. The purpose of lubrication system is to provide sufficient quantity of cool filtered oil to give positive and adequate lubrication to all the moving parts of the engine. The following are the function of lubrication— (i) To reduce friction which consequently reduces the power required to overcome the same. n) To reduce wear and tear between the rubbing and wearing surface. vAiii) To cool the.surfaces by carrying away heat generated by friction.

, , 7-- Radiator 1 1 Overflow; pipe' Thermostat

Fig. 26.12. Forced Circulation Cooling System.

Diesel Engine Power Plants

1029

(iv) To carry away the particles of worn metal and carbon. ,,_(v) To seal a space adjoining the surfaces such as piston moving in a cylinder. ,_(vi) To reduce the engine noise and to increase the engine life. (vii) To avoid corrosion and deposits. The lubrication system is classified as (a) Mist lubrication system, (b) Wet sump lubrication system (i) Splash system, (ii) Pressure feed system, (iii) Splash and Pressure feed system (c) Dry sump lubrication 26,4.7,1. Mist Lubrication System. In mist lubrication system, small quantity of lubricating oil is mixed in the fuel tank. It is used in two stroke engine. 26.4.7.2. Splash System. The application of this system is limited to only light duty, engines. As the name implies, a splasher or dipper is provided under each connecting rod cap which dips into the oil in the trough at every revolution of crankshaft and oil is , splashed all over the anterior of the crankcase. 26.4.7.3. Pressure Feed System. Fig. 26.13 shows the pressure feed lubrication system. The main elements of this system consists of oil in crankcase, strainer, pump, pressure regulator, filter, breather andoil gallaries. The oil is drawn from the sump through strainer which prevents foreign particles and is pumped with the help of gear pump submerged in the oil and driven by crankshaft to all the main bearings of the crankshaft through distributing channel. An oil hole is drilled in the crankshaft from the centre of each crankpin to the centre of an adjacent main journal, through which oil can pass from

Connecting rod • bearing

Piston Oil pressure gauge 4.:•••• Tappet and cam receiver oil Camshaft

Connecting rod bearing

End leakage

Gear Main bearing

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Ca(HCO3)2 Ca(HCO3)2 + SO2 + H2O —> CaS03 .21120 + 2CO2 CaS03 .21120 + 202 --> CaSO4 . 2H2 0

(30.1)

A spray tower is employed downstream of the particulate-removal system (i.e. electrostatic precipitator or fabric filler) as shown in Fig. 30.1. The flue gas is drawn into the spray tower by the main steam-generator induced draft fan where it flows in counter current fashion to the limestone-slurry spray. Any spray droplets entrained by the gas is removed by a mist eliminator located at the upper exit of the spray tower. The gas may • have to be slightly reheated before it enters the stack to improve atmospheric dispension. In the bottom of spray tower, the sprayed limestone slurry gets collected and is recirculated back to the spray nozzles by a pump. There is a system of feed and bleed in the system by which charges a fresh slurry, under pH control and discharges an equivalent

Air Pollution Caused by Power Generation And Its Control

Boiler

I

Bypass

1085

Stac

Dampec Lime 1 limestone

Coal

Pump Hot fan

To dryash handling system Slurry Recycle return system *-t? Thickener 9 Filter centrifuge Thickene Q overflow > • aliber. Pump tank Pump Cake

Spray tower Bleed

A_ Water g

11 Slacker/grinder Pump

Slurry tank

Fig. 30.1 Wet Flue Gas Desulfurization System with a Spray Tower.

amount from the circulating slurry. The preparation of fresh slurry is accomplished by mixing the lime-limestone with water in a slacker-grinder and stirred in a slurry tank. There is dewatering system in the form of thickness and fillers or centrifuges which receives the bled slurry where water is removed from the calcium-sulfur salts. The reclaimed water is utilised in making fresh slurry. The wet FGD system is mostly used in power plants. The following are the advantages of wet scrubber. (i) The SO2 removal efficiency is high (ii) The system offers good reliability (iii) It requires low flue gas energy (iv) It is capable of removing from the flue gases residual particulates that might have escaped the electrostatic precipitator or fabric filler. The main disadvantages of this wet FGD system include— (i) It builds up scale in the spray tower and the possibility of plugging. (ii) There is reheating of the flue gas. ' (iii) The larger gas pressure drop requires higher fan power. (iv) It offers higher capital and operating cost. (v) It offers costly disposal of waste material from wet- scrubbers which is waterlogged sludge. 30.3.2. The Dry Flue-Gas Desulfurization System. It is also known as dry scrubber. It utilises an aqueous slurry of lime, CaO, to capture flue gas SO2 by forming calcium sulfides and sulfates in spray absorber. Such a dry scrubber is shown in Fig. 30.2 which is placed before fabric filter. The preparation of slurry in this case is different than that of wet scrubber. In dry scrubber system, the slurry is atomized usually by a centrifugal atomizer, into a fine spray in the spray absorber that promotes the chemical absorption of SO2 and, because of the small spray particle size, is quickly dried by the hot flue gases themselves to a particulate suspension that is carried along with the desulfurized gas stream. The reaction particulates as well as those carried by the flue gases (fly ash) are

Steam & Gas Turbines And Power Plant Engineering

1086

Emergency shut-of valves

I

Lime storage silo

/Pneumatic transfer

truck

1

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(Watery Dilution control valve

Vibrating bin bottom Volumetric screw feeder Water Slacke Pressure Slurry con rol tank valve

Grits to.

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Slurry pumps

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