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36954153 boiler-book 36954153 boiler-book Document Transcript

  • Steam generator Ganesh kumarA.GANESH KUMARDEUTSCHE BABCOCK, INDIA.For internal circulation only. All rights reserved by author.
  • Steam generator Ganesh kumarDEDICATED TO MY COLLEGE AND MY PROFESSORS.
  • Steam generator Ganesh kumar PREFACEDear friends,This book was prepared in view of giving assistance to designengineers entering into the boiler field and to plant engineers whomI have met always in desire to know the ABC of the boiler designand related calculations. I have made an attempt in bringing closerelation of practical field design and theoretical syllabus ofcurriculum. Engineering students, who always wonder how thetheory studying in curriculum will help them in real life of business.For them this book will give an inspiration.I have designed this book in two parts. First, the basic theory ofworking fluid in the steam plant cycle. This will be the basicfoundation for development of boiler science. Secondly the maincomponents of steam generator and its design. Also you can findvarious useful data for ready reference at the end of this book.(A.GANESH KUMAR) View slide
  • Steam generator Ganesh kumar CONTENTS• PREFACE……………………………………………………………………….1.0 TYPES OF STEAM GENERATORS1.1 Introduction…………………………………………………………………….1.2 History of steam generation and use………………………………………1.3 Shell and tube boiler………………………………………………………….1.4 Conventional grate type boiler……………………………………………….1.5 Oil/gas fired boiler…………………………………………………………….1.6 Pulverized fuel boiler………………………………………………………….1.7 Fluidized bed boiler……………………………………………………………1.8 Heat recovery steam generator………………………………………………1.9 Practical guide lines for selection of boiler………………………………….2.0 STEAM, GAS and AIR2.1 Introduction……………………………………………………………………2.2 Definitions for some commonly used terms………………………………2.3 Steam………………………………………………………………………….2.4 Fuel……………………………………………………………………………..2.5 Gas and air…………………………………………………………………….2.6 Some commonly used dimensionless numbers and their significance….3.0 FURNACE3.1 Introduction……………………………………………………………………3.2 Effect of fuel on furnace……………………………………………………..3.3 Forced or Natural Circulation……………………………………………….3.4 Heat flux to furnace walls…………………………………………………...3.5 Points to be noted while designing furnace………………………………3.6 Classification of furnace…………………………………………………….3.7 Modes of heat transfer in furnace…………………………………………3.8 Heat transfer in furnace…………………………………………………….3.9 Furnace construction……………………………………………………….3.10 Practical guides for designing fluidized bed, conventional and oil/gas fired furnace…………………………………………………..4.0 SUPERHEATER4.1 Introduction…………………………………………………………………..4.2 Effect of fuel on super heater design………………………………………4.3 Points to be noted while designing super heater…………………………4.4 Classification of super heater……………………………………………….4.5 Designing a super heater……………………………………………………4.6 Overall heat transfer across bank of tubes……………………………….4.7 Steam temperature control…………………………………………………4.8 Pressure drop……………………………………………………………….. View slide
  • Steam generator Ganesh kumar5.0 DRUMS5.1 Intruction……………………………………………………………………. Deleted: od5.2 Optimal configuration of drums………………………………………………5.3 Stubs and attachments in the steam drum/shell…………………………..5.4 Maximum permissible uncompensated opening in drum…………………5.5 Size of the drum………………………………………………………………5.6 Drum internals………………………………………………………………..6.0 EVAPORATOR AND ECONOMISER6.1 Introduction……………………………………………………………………….6.2 Difference between evaporator and economiser……………………………..6.3 Fin efficiency………………………………………………………………………7.0 AIRHEATER7.1 Introduction……………………………………………………………………….7.2 Types of air heater……………………………………………………………….7.3 Advantages of air heater………………………………………………………..7.4 Heat transfer in air heater………………………………………………………7.5 Practical guide lines for designing airheater………………………………….8.0 DUST COLLECTOR8.1 Introduction……………………………………………………………………….8.2 Effects of air pollution……………………………………………………………8.3 Air quality standards……………………………………………………………..8.4 Air pollution control devices……………………………………………………. Centrifugal cyclone dust collector Bag filter Electro static precipitator9.0 WATER CHEMISTRY9.1 Introduction…………………………………………………………………….9.2 Names of water flowing in the power plant cycle…………………………..9.3 Major impurities in water……………………………………………………..9.4 Effects of various impurities in boiler water………………………………..9.5 Need for water treatment…………………………………………………….9.6 External water treatment……………………………………………………..9.7 Internal water treatment………………………………………………………9.8 Practical guides for selecting water treatment plant……………………….10.0 BOILER CONTROLS10.1 Introduction……………………………………………………………………10.2 Control philosophy……………………………………………………………10.3 Drum level control…………………………………………………………….
  • Steam generator Ganesh kumar10.4 Super heater steam temperature control…………………………………..10.5 Furnace draft control………………………………………………………….10.6 Combustion control…………………………………………………………...10.7 Field instruments……………………………………………………………..10.8 Panel instruments……………………………………………………………APPENDIX 1 : MOLLIEAR CHARTAPPENDIX2 : PSYCHROMETRY CHARTAPPENDIX3 : FUEL ANALYSISAPPENDIX4 : STEAM TABLESAPPENDIX5 : POLLUTION NORMS IN VARIOUS INDIAN STATESAPPENDIX6 : USEFUL DATASAPPENDIX7 : UNIT CONVERSION TABLE
  • Steam generator Ganesh kumar 1.0 TYPES OF STEAM GENERATOR1.1 INTRODUCTIONIndian power demand is met mainly from thermal, hydro and nuclear power. Non-conventional energy power production is very much negligible. Out of the mainpower producing sources thermal plant produces 48215 MW (69%), hydro plantproduces 19300 MW (28%), nuclear plant produces 2033 MW (3%) as on 31stMarch 1992. In the above power plants 72% of the generation is from thermal andnuclear, where steam generation is one of the main activity. In the years to come,the demand of electricity is going on increasing and already most of water resourcessuitable for power generation is in service. Except from gas turbines power the mostof new electric capacity has to be met by utilizing steam.Steam boiler today range in size from those to dry the process material 500 kg/hr tolarge electric power station utility boilers. In these large units pressure range from100 kg/cm² to near critical pressures and steam is usually superheated to 550°C. InIndia BHARAT HEAVY ELECTRICALS LTD (BHEL) is the pioneer in developingthe technology for combustion of high ash coal efficiently in atmospheric bubblingfluidized bed. From where lot of industries in boiler manufacturing starts. Only afterthe year 1990, India’s foreign policy was changed, various foreign steam generatormanufacture entered into Indian power market bringing various configuration andcompetitiveness in the market.1.2 HISTORY OF STEAM GENERATION AND USEThe most common source of steam at the beginning of the 18th century was the shellboiler. Little more than a kettle filled with water and heated from the bottom. Oldenday boiler construction were very much thicker shell plate and riveted constructions.These boilers utilize huge amount of steel for smaller capacity. Followed this shelland tube type boilers have been used and due to direct heating of the shell byflames leads severe explosion causing major damages to life and property. Forsafety need, after the Indian independence India framed Indian boiler regulations in1950, similar to various other standards like ASME, BS, DIN, JIS followed worldwide. Till date IBR 1950 is governing the manufacturing and operation of boilers withamendments then and there. Indian sugar industry uses very low pressure (15kg/cm²) inefficient boilers during independence now developed to an operatingpressure of 65 kg/cm² and more of combined cycle power plant. If we analysis mostof the boilers erected in pre-independence period were imported boilers only andnow steam generators were manufactured in India to the world standards on budget,delivery and performance. In power industry India made a break through in the year1972, India’s first nuclear power plant was commissioned at Tarapore. This plantwas an pilot plant meant for both power and research work. This was made incollaboration with then soviet republic of Russia. Now India has its own nucleartechnology for designing nuclear power plant. Even though there is a development,Indian industry has to go a long way in power sectors.
  • Steam generator Ganesh kumar1.3 SHELL AND TUBE BOILERSteam was originally used to provide heat to the industrial process like drying,boiling. In small industry the people are not taken care in fuel consumption point,they have generated steam in crude manner. Shell and tube boilers are old versionof boilers used in industry where a large flue tube was separated by a fixed grateman power is used to throw husk and shells into the grate and firing was done.In early days, as individual electric generating stations increased in capacity, thepractice was merely to increase the number of boilers. This procedure eventuallyproved to be uneconomical and larger maintenance. Afterwards, individual boilerswere build larger and larger size, however the size became such that furnace floorarea occupation was more. Therefore further research work have been developed inthis area and technologies such as pulverized coal fired furnace, circulated fluidizedbed furnace, pressurized circulated fluidized furnace (still under research stage)were developed. These modern technologies have higher heat transfer coefficient infurnace and allow higher volumetric combustion rates.1.4 CONVENTIONAL GRATE TYPE BOILERSTECHNOLOGYThis is the oldest method of firing fuel. Fuel will be spread over the grate, where thefuel is burnt. Fuel feeding will be done manually or mechanically to have a sustainedflame. In this type burning will be done at higher excess air. Incoming air will beused for cooling the grate.Types of grateCommon types of grate that are used for fuel are fixed grate, pulsating grate,dumping grate, travelling grate. Each type of grate differ slightly in their constructionand arrangement. However the combustion phenomenon remains same.Travelling grateThe travelling type is a continuous grate which slowly convey the burning fuelthrough the furnace and discharge the ash to an ash pit. Grate speed is regulatedby the amount of ash discharging to ash pit ( 0 to 7m/hr)Pulsating grateThe pulsating grate is non- continuous grate. The grate surface extends from therear of furnace to ash pit. Here the grate will be given a racking motion at predetermined frequency depending on the fuel/ash bed depth.Dumping grateDumping grates are also a non-continuous type grate. The grate is split intolongitudinal sections, one for each feeder. Fuel is distributed on the grate and burns.When ash depth gets to a depth where air can not diffuse it , the grates are tilted orash is dumped into the hopper in the following manner.
  • Steam generator Ganesh kumarAlternating fuel feeding is stopped and grate is tilted by lever arrangement, theactuation can be done either manually or pneumatic cylinder.In dumping grate the grate sections should be designed in such a way that, whiledumping the ash part of grate surface not available for burning. In poorly designeddumping grate there may be steam pressure. Therefore while sizing grate sectionscare should be taken such that while dumping part of the grate, other fuel feeder andremaining sections should able to take the full load.Dumping grate is similar to fixed grates, it is best suitable for bagasse where the fuelis of low calorific value and having high moisture content. Therefore air alone canacts as a cooling medium. If we use coal the grate bar may not with stand highertemperature and additional cooling by water tube is necessary. Travelling grate issuitable for burning coal and lignite. As the grate rotates, the grate bar gets heatedand cooled by incoming air for the half of the cycle and remaining half of the cyclegrate bar cooled by the incoming air.Spreader stokerMechanical spreaderThe spreader stoker feeder takes fuel from the feeder hopper by either a small ramor a rotating drum and delivers it into a spinning rotor. An adjustable trajectory plateis located between the feed mechanism and the rotor. Adjusting the trajectory platefuel can be feed through out the entire length of the furnace.Pneumatic spreaderIn this rotor is replaced by high pressure air lines from Secondary air fan is used tospread the fuel into the furnace. The fuel is carried into the furnace by means ofpneumatic system and the air flow adjustment makes the fuel to flow near or fartherof the furnace.1.5 OIL/GAS FIRED BOILERSTECHNOLOGYFlame has a tendency to burn upward only. This forms the basic concept of burner.Whenever fresh fuel enters into the ignition zone it starts burning upwards and theflame will not come downwards to the incoming fuel, by this property combustioncan be controlled easily. Hence it is always better to bring the oil or gas train frombottom of the burner.A liquid or gas fuel has flowable property by nature and it has a lower ignitiontemperature. When the fuel is forced to flow through the nozzle it will spread thoughan predetermined length and burn completely from the point of entry to the firingzone estimated. The fuel flow can be controlled by means of control valves.CHARACTERISTICS OF OILIn today’s climate of fluctuating international fuel prices and quality, the emphasis onthe ability of the boiler on low quality fuel oils has become more greater. In theinternational market, the quality of the residual fuel oils is constantly getting poorerdue to the development of more sophisticated cracking methods and also ourindigenous crude production falls short of our requirements, about 15 million tons ofcrude is imported from outside sources. These outside sources are many, our
  • Steam generator Ganesh kumarrefineries handle a variety of crude. Since the inherent properties of the finishedpetroleum products are directly dependent on the parent crude, one can imagine thepetroleum involved in producing residual fuel oil within narrow limits of specifications,especially with respect to specified characteristics like carbon residue, asphaltenesand metallic constituents is not possible.Flash pointFlash point is important primarily from a fuel handling stand point. Too low a flashpoint will cause fuel to be a fire hazard subject to flashing and possible continuedignition and explosion. Petroleum products are classified as dangerous or nondangerous for handling purposes based on flash point as given below.Classification Flash point Petroleum ProductClass A Below 23°C Naptha Petrol Solvent 1425 HexaneClass B 23 to 64°C Kerosene HSDClass C 65 to 92°C LDO Furnace oil LSHSExcluded Petroleum 93°C and above TarPour PointThe pour point of the fuel gave an indication of the lowest temperature, above whichthe fuel can be pumped. Additives may be used to lower the freezing temperatureof fuels. Such additives usually work by modifying the wax crystals so that they areless likely to form a rigid structure. It is advisable to store and handle fuels around10°C above the expected pour point.ViscosityViscosity is one of the most important heavy fuel oil characteristics for industrial andcommercial use, it is indicative of the rate at which the oil will flow in fuel systemsand the ease with which it can be atomized in a given type of burner. When thetemperature increases viscosity of fuel will reduce.
  • Steam generator Ganesh kumarThe viscosity needed at burner tip for satisfactory atomization for various types ofburners are as follows.Type of burner Viscosity at burner tip In centi stokesLow air pressure 15 to 24Medium air pressure 21 to 44High air pressure 29 to 48Steam jet 29 to 37Pressure jet less than 15Metal ContentSodium, Potassium, Vanadium, Magnesium, Iron, Silica etc. are some of the metallicconstituents present in fuel oil. Of the above metals, sodium and vanadium are themost troublesome metals causing high temperature corrosion in boiler super heatertubes and gas turbine blades. Much of the sodium is removed from the crude oil inthe desalting operation, which is normally applied in the refinery and additionalsodium can be removed from the finished fuel oil by water washing and centrifuging.Vanadium is found in certain crude oils and is largely concentrated in fuel oilprepared from these crude. No economical means for removal of vanadium from theresidual fuel oil is available. However certain additives like magnesium are availableto minimize the effect of vanadium.Asphaltene content and Carbon residueAsphaltenes are high molecular weight asphaltic material and it requires moreresidence time for complete combustion. Asphaltenes as finely divided coke may bedischarged from the stack. Residual fuel oils may contain as much as 4%asphaltenes.Petroleum fuels have a tendency to form carbonaceous deposits. Carbon residuefigures for residual fuel oils from 1 to 16% by weight. This property is totallydependent on the type of crude, refining techniques and the blending operations inrefinery.Fuels with high carbon residue and asphaltenes requires large combustion chamberand hence while designing the boiler for such fuel the volumetric loading has to be ofthe order of 2 lakhs Kcal/m3hr
  • Steam generator Ganesh kumar1.6 PULVERIZED FUEL BOILERSTECHNOLOGYWhen coal is powdered to micron size it can be conveyed easily by air in pipelinesand the pulverized coal behaves as if that of oil and hence the same can be easilyburnt in pulverized fuel burners. The heat release by the burners in very high andun-burnt carbon is almost equal to zero. Hence efficiency achieved by pulverizedburners is much more than any type of coal combustion.MECHANISM OF PULVERIZED FUEL BURNINGThere are two systems of pulverized firing 1.0 direct firing 2.0 indirect firing.In the direct firing system, raw coal from the storage area is loaded on a conveyorand fed to a coal crusher. A second conveyor system loads coal into the coalstorage bunker located over the coal pulverization system. Coal via gravity feed isdelivered through a down spout pipe to the coal feeder. A coal shutoff gate isprovided prior to the coal feeder inlet to allow emptying the system down stream.The coal feeder meters the coal to the crusher dryer located directly below thefeeder discharge. A primary air fan delivers a controlled mixture of hot and cold airto the crusher dryer to drive moisture in the coal facilitating pulverization the primaryair and crushed coal mixture is then fed to the coal pulverizer located below thecrusher dryer discharge. Selection of pulverizer has to be analyzed critically, since itis one of the important equipment where the wear and tear is more. For the softlignite Beter wheel is preferable and for hard lignite, coal like fuels heavy pulveriserof ball and hammer mill is preferable. The coal is pulverized to a fine powder andconveyed through coal pipes to the burners. Primary air is the coal pipetransportation medium.The indirect firing system utilizes basically the same coal flow path to the pulverizer.After the classification of pulverized coal, it is delivered to a coal storage bin. Whenneeded to fire the boiler the pulverized coal is then conveyed to the burners by anexhaust fan. This method requires very special provisions to minimize risk of fire orexplosion. Of the two systems, the direct firing is more common.Neyveli lignite power corporation has pulverized boiler of direct firing system.1.7 FLUIDIZED BED BOILERSATMOSPHERIC FLUIDIZED BED COMBUSTIONTECHNOLOGYWhen air or gas is passed through an inert bed of solid particles such as sandsupported on a fine mesh or grid. The air initially will seek a path of least resistanceand pass upwards through the sand. With further increase in the velocity, the airstarts bubbling through the bed and particles attain a state of high turbulence. Undersuch conditions bed assumes the appearance of a fluid and exhibits the propertiesassociated with a fluid and hence the name fluidized bed.
  • Steam generator Ganesh kumarMECHANISM OF FLUIDIZED BED COMBUSTIONIf the sand, in a fluidized state is heated to the ignition temperature of the fuel andfuel is injected continuously into the bed, the fuel will burn rapidly and attains auniform temperature due to effective mixing. This , in short is fluidized bedcombustion.While it is essential that the temperature of bed should be equal to the ignitiontemperature of fuel and it should never be allowed to approach ash fusiontemperature (1050° to 1150°C ) to avoid melting of ash. This is achieved byextraction of heat from the bed by conductive and convective heat transfer throughtubes immersed in the bed.If the velocity is too low fluidization will not occur, and if the gas velocity becomes toohigh, the particles will be entrained in the gas stream and lost. Hence to sustainstable operation of the bed, it must be ensured that gas velocity is maintainedbetween minimum fluidization and particle entrainment velocity.Advantages of FBC.1.0 Considerable reduction in boiler size is possible due to high heat transfer rate over a small heat transfer area immersed in the bed.2.0 Low combustion temperature of the order of 800 to 950°C facilitates burning of fuel with low ash fusion temperature. Prevents Nox formation, reduces high temperature corrosion and erosion and minimize accumulation of harmful deposits due to low volatilization of alkali components.3.0 High sulphur coals can be burnt efficiently without generation of Sox by feeding lime stone continuously with fuel.4.0 The units can be designed to burn a variety of fuels including low grade coals like floatation slimes and washery rejects.5.0 High turbulence of the bed facilitates quick start up and shut down.6.0 Full automation of start up and operation using simple reliable equipment is possible.7.0 Inherent high thermal storage characteristics can easily absorb fluctuation in fuel feed rate.
  • Steam generator Ganesh kumarATMOSPHERIC CIRCULATING FLUIDIZED BED COMBUSTIONTECHNOLOGYAtmospheric circulating fluidized bed (ACFB) boiler is a devise used to generatesteam by burning solid fuels in a furnace operated under a velocity exceeding theterminal velocity of bed material. I.e., solid particles are transported through thefurnace and gets collected in the cyclone at the end of furnace and again recycledinto furnace by means of pressure difference between fluidized bed and returnparticle.MECHANISM OF CIRCULATING FLUIDIZED COMBUSTIONThe mechanism is similar to AFBC. However in AFBC the fluidization velocity is justto make the particles in suspended condition. In ACFB boiler, special combination ofvelocity by primary air and secondary air, re-circulation rate, size of solids, andgeometry of furnace, give rise a special hydrodynamic condition known as fast bed.Furnace below secondary air injection is characteristic by bubbling fluidized bed andfurnace above the secondary air injection is characteristic by Fast fluidized bed.Most of the combustion and sulphur capture reaction takes place in the furnaceabove secondary air level. This zone operates under fast fluidization. In CFB boilernumber of important features such as fuel flexibility, low Nox emission, highcombustion efficiency, effective lime stone utilization for sulphur capture and fewerfuel feed points are mainly due to the result of this fast fluidization.In fast fluidization heavier particles are drag down known as slip velocity betweengas and solid, formation and disintegration of particles agglomeration, excellentmixing are major phenomenon of this regime.CFB is suitable for1.0 Capacity of the boiler is large to medium.2.0 The boiler is required to fire a low grade fuel or highly fluctuating fuel quality.3.0 Sox and Nox control is important.PRESSURIZED FLUIDIZED BED COMBUSTIONThe advantage of operating fluidized combustion at the elevated pressure ( about 20bar) is, reduction in steam generator size can be achieved and make possible thedevelopment of a coal fired combined cycle power plant. The development ofpressurized fluidized bed combustion is still in research stage only. With help ofpressurized hot gas coming out of the furnace is cleaned primarily by a cyclone likeCFBC boiler and the gas is expanded in a turbine and the exhaust gas from turbineis further cooled by the heat exchanger. The aim behind the development ofpressurized fluidized bed are:1.0 To develop steam generator of smaller size for the higher capacity.2.0 To reduce the cost of generation of power per MW.3.0 To develop turbines which make use of solid fuels such as coal, lignite etc.,
  • Steam generator Ganesh kumar1.8 HEAT RECOVERY STEAM GENERATORIn India, coal availability is 97% of the requirement and we are importing coal only forthe process requirement like baking coal for steel plant where high calorific coal isrequired. Hence in post independence India coal fired boilers where flourished,however due to the need of energy conservation and due to process parameterrequirements development of HRSG in recent periods is more. Moreover due to thedevelopment of gas turbines with gaseous and liquid fuels, more GT are beinginstalled due to their lower gestation period and higher efficiency than Rankine cycle.As explained earlier HRSG can be classified into two types, one is for maintainingprocess parameter such as temperature and other is in the point of economic pointof view.The process steam generator are generally referred by the term called waste heatrecovery boiler ( WHRB) where the gas contains heat in excess, this excess wasteheat has to be recovered or removed by any means so that the process parametercan be maintained. ( e.g. Sulphuric acid plant, hydrogen plant, sponge iron plant,Kiln exhaust etc.,)The steam generator stands behind the gas turbine are usually referred as Heatrecovery steam generator.The HRSG or WHRB the design greatly vary with respect to the size of the plant,the gas flow, gas volumetric analysis, dust concentration and sulphur di oxideconcentration. In HRSG the gas quantity and inlet temperature is fixed and fordifferent load the variation of heat will not be proportional and hence at part loads theheat absorbed at different zones will vary widely and hence for different loads theperformance of the HRSG to be done.
  • Steam generator Ganesh kumar 2.0 STEAM,GAS and AIR2.1 INTRODUCTIONIn steam generator water, steam, gas and air are the working fluids in this air andgas have similar properties. Understanding the properties of gas and air are almostone and the same. I have grouped steam and gas as one unit and water as aseparate unit just because understanding the behavior of steam and gas is moreimportant in design point of view where as knowledge of water is more important inoperational point of view.2.2 DEFINITIONS FOR SOME COMMONLY USED TERMSHeatHeat is defined as the form of energy that is transferred across a boundary by virtueof a temperature difference. The temperature difference is the potential and heattransfer is the flux. In other words heat is the cause and temperature is the effect.EnergyEnergy of a body is its capacity to do work and is measured by the amount of thework that it can perform.Potential Energy( mgh = mass x gravitational force x datum level)Potential energy of a body is the energy it possesses by virtue of its position or stateof strain.Kinetic energy ( ½ mv² = ½ x mass x velocity²)Kinetic energy of a body is the energy possessed by it on account of its motion.EnthalpyEnthalpy is the quantity of heat that must be added to the fluid at zero degreecentigrade to the desired temperature and pressure. Enthalpy is defined as heatwithin or heat content of the fluid.EntropyThe word entropy is derived from a Greek word called ‘tropee’ which meanstransformation. The unit of entropy is Joules/kelvin.Specific heatSpecific heat of a substance is defined as the amount of heat required to raise thetemperature of one kilogram of substance through one degree kelvin. All liquids andsolids have one specific heat. However gas have number of specific heats dependson the condition with which it is heated. Cp = f(T)
  • Steam generator Ganesh kumarSpecific heat at constant pressure.Specific heat of a substance is defined as the amount of heat required at constantpressure to raise the temperature of one kilogram of substance through one degreekelvin.Integral constant pressure specificheatIt is the average heat required to rise the temperature between two temperaturedifference t1 and t2 i.e., Cp = ( H2 – H1)/(t2 –t1) H = f(Cp/T)Specific heat at constant volume.Specific heat of a substance is defined as the amount of heat required at constantvolume to raise the temperature of one kilogram of substance through one degreekelvin.NTP and STP conditionIt is customary to specify the gas or steam properties at NTP or STP condition,NTP condition is at Normal temperature and pressure, i.e., the properties measuredat 0°C or 273.15 °K and pressure 1.01325 bar or 1.03 atmSTP condition is at Standard temperature and pressure i.e., the properties measuredat 25°C or 298.15°K and pressure 1.01325 bar or 1.03 atm.ViscosityViscosity of a liquid is its property, due to the frictional resistance between the fluidparticles (cohesion between particles) or between fluid and the wall. Viscosity offluid controls the rate of flow.Newton s Law of viscosityThe shear stress on a layer of a fluid is directly proportional to the rate of shearstrain. ( Velocity gradient )τ α ν/l where τ is shear stress and ν is velocity , l is the distance or gap between layers.τ = µ ν/l where µ is the constant of proportionality and is known as absoluteviscosity or dynamic viscosity.Kinematic viscosity is the ratio of absolute viscosity to density (µ/ρ)Thermal conductivityThermal conductivity is the property of substance, that its ability to conduct heat andexpressed in W/mK.KilogramKilogram is the mass of one international prototype made of platinum iridium cylinderpreserved at the international bureau of weights and measures at paris.
  • Steam generator Ganesh kumarMeterMeter is the length between two transverse lines en-grooved in platinum iridium barat 0°C. or The meter is the length equal to 1650763.73 vacuum wave length of theorange light. ( λ = 605.8 mm of the Krypton 86 discharge lamp)SecondSecond is the duration of 9192631770 periods of the radiation corresponding to thetransition between two specified energy level of the Caesium –133 atom. Or1/86400th part of mean solar day.Specific volumeSpecific volume is the volume occupied per kg of steam or water or fluid.Specific volume is the inverse of density.For heat and mass transfer calculations, we have to know the above properties.The properties where mainly depends on the temperature for gases and temperatureand pressure for steam. The required equation for derivation is given at appropriateplaces.For gaseous fuel,Cp /R = f(T)R = Cp – CvCv = Cp - 1R RSpecific enthalpy wrt NTP, TH ‘ = 1/T  Cp dT ( enthalpy with reference to 0°C)RT  R TnSpecific enthalpy wrt STP TH* ‘ = 1/T  Cp dT + Hs ( enthalpy with reference to 25°C)RT  R RT TsSpecific entropy, TS ‘ = So  Cp dT - ln(P/Pn) ( entropy with reference to 0°C)R R R Tn
  • Steam generator Ganesh kumarSpecific free enthalpyG = H -SRT RT RThe temperature dependent specific heat (Cp) can be represented by an equation of4 th degree polynomial as shown belowCp = a1 + a2T + a3T² + a4 T3 + a5T4 (for temperature from 273K to 1000K)RCp = a9 + a10T + a11T² + a12 T3 + a13T4 (for temperature from 1001K to 5000K)RIntegrating, and adding constant of integration we getH = a1 + a2T + a3T² + a4T3 + a5 T4 + a8/T (for temperature from 273K to 1000KRT 2 3 4 5H* = a1 + a2T + a3T² + a4T3 + a5T4 + a6/T (for temperature from 273K to 1000KRT 2 3 4 5S = a1 ln T + a2T + a3T² + a4T3 + a5T4 + a7 – ln(P/Pn)R 2 3 4G = a1(1- ln T) - a2T - a3T² - a4T3 - a5T4 + a6 -a7 + ln(P/Pn)RT 2 6 12 20 TDynamic viscosity , thermal conductivity and prandtl numberDynamic viscosity, thermal conductivity and prandtl number of a flue gas can be fineeasily with help of the properties of nitrogen and following constants. Var Specific Dynamic Thermal Prandtl number Heat Viscosity conductivity Kj/kgK µPa.S W/mK a1 0.8554535 -0.9124458E 1 -0.1083113E-1 0.492851 b1 0.2036005E-3 0.4564993E-2 0.5596822E-4 -0.1230046E-2 c1 0.4583082E-6 0.2198889E-4 0.7413502E-7 0.1662398E-5 d1 -0.279808E-9 -0.1891235E-7 -0.5901395E-10 -0.1052753E-8 e1 0.5634413E-13 0.5138895E-11 0.1961745E-13 0.2443111E-12 a2 -0.1002311 -0.4267768E1 -0.8035817E-2 -0.8820652E-2 b2 0.7661864E-3 0.4074274E-3 0.110672E-04 0.1855309E-3 c2 -0.9259622E-6 -0.5125357E-5 -0.8397255E-8 -0.3838084E-6 d2 0.5293496E-9 0.738556E-8 0.1130229E-10 0.3256168E-9 e2 -0.109357E-12 -0.343972E-11 -0.5731264E-14 -0.1005757E-12
  • Steam generator Ganesh kumarDynamic viscosity,µg = µn + P1 XH2O + P2 XCO2Where XH2O & XCO2 are Percentage of weight in flue gasP1 = a1 + b1T + c 1T² + d1T3 + e1T4P2 = a2 + b2T + c 2T² + d2T3 + e2T4 where T is temperature in °CThermal conductivity,kg = kn + P1 XH2O + P2 XCO2Where XH2O & XCO2 are Percentage of weight in flue gasP1 = a1 + b1T + c 1T² + d1T3 + e1T4P2 = a2 + b2T + c 2T² + d2T3 + e2T4 where T is temperature in °CPrandtl number,Prg = Prn + P1 XH2O + P2 XCO2Where XH2O & XCO2 are Percentage of weight in flue gasP1 = a1 + b1T + c 1T² + d1T3 + e1T4P2 = a2 + b2T + c 2T² + d2T3 + e2T4 where T is temperature in °CPra = a + bT + cT² + dT3 + eT4Specific heat,Cpg = Cpn + P1 XH2O + P2 XCO2Where XH2O & XCO2 are Percentage of weight in flue gasP1 = a1 + b1T + c1T² + d1T3 + e1T4P2 = a2 + b2T + c2T² + d2T3 + e2T4 where T is temperature in °CWhere 0 ≤XH2O ≤ 0.3 ,0 ≤ XCO2 ≤0.2 , 0 ≤ T ≤ 1200°CDynamic viscosity, thermal conductivity and Prandtl number of NITROGEN Dynamic viscosity Thermal conductivity Prandtl number µ Pa.s W/mK a 0.1714237E02 0.2498583E-1 0.6901183 b 0.4636040E-01 0.6535367E-4 0.2417094E-05 c -0.2745836E-4 -0.7690843E-8 0.2771383E-7 d 0.1811235E-7 -0.1924248E-11 -0.3534575E-10 e -0.674497E-11 0.160998E-14 0.1717930E-13 f 0.1027747E-14 -0.2864430E-18 -0.2989654E-17
  • Steam generator Ganesh kumarµn = a + bT + cT² + dT3 + eT4 + fT5Kn = a + bT + cT² + dT3 + eT4 + fT5Prn = a + bT + cT² + dT3 + eT4 + fT5Cpn = a + bT + cT² + dT3 + eT4 + fT5 (for temp.273 K to 1000K)And Cpn = a1 + b1T + c1T² + d1T3 + e1T4 + f1T5 (for temp. 1001K to 5000K) 273 K to 1000K 1001K to 5000K a 0.3679321E1 ‘a1 0.2852903E1 b -0.1313559E-2 b1 0.1580411E-2 c 0.2615196E-5 c1 -0.6189378E-6 d -0.9629654E-9 d1 0.1119450E-9 e -0.9928002E-13 e1 -0.7607378E-14 f -0.9723991E3 f1 -0.8019835E32.3 STEAMWe can see in day to day life the process of boiling water to make steam. Steam iswater in the vapour or gaseous state. It is in visible, odorless, non-poisonous andrelatively non corrosive to boiler metals. Steam is uniquely adapted by itsadvantageous properties for use in industrial process heating and power cycle.Thermodynamically boiling is the result of heat addition to the water in a constantpressure and constant temperature process. The heat which must be supplied tochange water into steam without raising its temperature is called the heat ofevaporation or vaporization and the boiling point of a liquid may be defined as thetemperature at which its vapour pressure(pressure exerted due to the vapour of theliquid) is equal to the total pressure above its free surface. In other wordstemperature at which the partial pressure of vapour increases to make total pressureabove the liquid surface. This temperature is also known as the saturationtemperature.EVAPORATIONLiquid exposed to air evaporate or vapourize. Evaporation is the process takesplace at the surface exposed to atmosphere. If there is any increase in ambienttemperature or increase of the liquid temperature evaporation rate becomesincreased. The reduction in pressure above the liquid surfaces accelerate theevaporation rate. Evaporation will be there at all temperature and pressure,unsaturated surrounding environment also one of the factor increases theevaporation rate.BOILINGBoiling is the phenomenon takes place at boiling point of the liquid. Boiling takesplace throughout the liquid column. A liquid will boil, when it’s saturated vapourpressure exceeds the surrounding environment pressure acted upon the liquid.Hence boiling point of a liquid will change depends on the pressure exerted by theenvironment over the surface.
  • Steam generator Ganesh kumarCONDENSATIONCondensation is the change in phase of vapour phase to it’s liquid phase. Whenwater vapour or steam comes in contact with cooler surfaces, it gives up the heatand condenses to water. The heat released while changing from vapour phase toliquid phase is called heat of condensation. In factories the steam released out ofthe main steam line or process vents where we can see a remarkable phenomenonof indication of dryness of steam. If the steam is dry, we can not visualize the steamcoming out of the vent but after some distance we can see a white cloud. This isdue to the condensation of steam which composed of small particles of water formedwhen steam cooled in cooler atmosphere. In other case if the steam is wet, thewhite smoke cloud is directly released from the vents.2.4 FUELCombustionCombustion or burning, is a rapid combination of oxygen with a fuel resulting inrelease of heat. The oxygen comes from the air, which is about 21% oxygen and78% nitrogen by volume.Most fuels contain carbon, hydrogen, and sometimes sulphur as the basiccomposition of combustion materials. These three constituents’ reacts with oxygento produce carbon-di-oxide, water vapour, suphur di oxides gases respectively andheat.Carbon, hydrogen and sulphur are found exists in direct form in most of the solid andliquid fuels and in gaseous fuels the combustion matter is found ashydrocarbons(combination of hydrogen and carbon). When these burn, the finalproducts are carbon di oxide and water vapour unless there is a shortage of oxygen,in which case the products may contain carbon mono oxide, unburnt hydrocarbons,and free carbon.Heat value of fuelQuantities of heat are measured in BTU, kiloCalories, or joules. A BTU is thequantity of heat required to raise the temperature of one pound of water one degreefahrenheit. A kilocalorie is the quantity of heat needed to raise one kilogram of waterone degree celsius.Experimental measurements have been made to determine the heat released byperfect combustion of various fuels. The heat value is usually determined bycalorimeters. When a perfect mixture of a fuel and air originally at 15.6°C is ignitedand then cooled to 15.6°C the total heat released is termed the higher heating valueor Gross calorific value. There is also one more term called lower heating value orthe net calorific value it is the quantity of heat equal to gross calorific value minus theheat absorbed by the latent heat of water moisture( inclusive of moisture generateddue to combustion of hydrogen present in the fuel) at 25°C.
  • Steam generator Ganesh kumarDulong’s formula is used to find Calorific value of the fuelHHV(kj/kg) =338.21C% +1442.43(H-O/8)% + 94.18S%Relation between HHV and LHVLHV = HHV – (%H2O + %H2x8.94)χWhere χ is the latent heat of water vapour at reference temperature 25°C=583.2 kcal/kgProximate AnalysisThe general procedure for the analysis relating to proximate analysis is describebelow as per IS 1350(partI). For full details, the original standard may be referred toi) MoistureThe moisture in the coal is determined by drying the known weight of the coal at108°C±2°Cii) Volatile matterThe method for the determination of VM consists of heating a weighted quantity ofdried sample of coal at a temperature of 900°±10°C. for a period of seven minutes.Oxidation has to be avoided as far as possible. VM is the loss in weight less by thatdue to moisture. VM is the portion of the coal which, when heated in the absense ofair under prescribed conditions, is liberated as gases and vapour.iii) AshIn this determination, the coal sample is heated in air up to to 500°C for minutes from500 to 815°C for a further 30 to 60 minutes and maintained at this temperature untilthe sample weight becomes constant.iv) Fixed carbonFixed carbon is determined by deducting the moisture. VM and ash from 100Ultimate analysisThe ultimate analysis of fuel gives the constituent elements namely carbon,hydrogen,nitrogen, sulphur , hydrocarbons, nitrogen etc., For the ultimate analysisof the coal sample is burnt in a current of oxygen. As a result the carbon, hydrogen,sulphur oxidized to water, carbon di oxide and sulphur di oxide respectively. Theseconstituent are absorbed solvents to estimate the percentage of C,H2,S,N etc.,The classification of Indian coal on the basis of proximate analysis.S.n Description Grade Specification1 Non coking coal, produced A GCV exceeding 6200kcal/kg in all states other than Assam B GCV exceeding 5600Kcal/kg but Andhrapradesh,Meghalaya, not exceeding 6200Kcal/kg Arunachalpradesh and Nagland C GCV exceeding 4940kcal/kg not exceeding 5600Kcal/kg
  • Steam generator Ganesh kumar D GCV exceeding 4200kcal/kg not exceeding 4940Kcal/kg E GCV exceeding 3360kcal/kg not exceeding 4200Kcal/kg F GCV exceeding 2400kcal/kg not exceeding 3360Kcal/kg G GCV exceeding 1300kcal/kg not exceeding 2400Kcal/kg2 Non coking coal, produced Assam,Andhrapradesh,Meghalaya, Not graded Arunachalpradesh and Nagland3. Coking coal Steel GrI Ash content not exceeding 15% Steel GrII Ash content 15% to 18% Washery GrI Ash content 18% to 21% Washery GrII Ash content 21% to 24% Washery GrIII Ash content 24% to 28%2.5 GAS and AIRIDEAL GAS OR PERFECT GASAt low pressure and high temperature, all gases have been found to obey threesimple laws. These laws relate the volume of gas to the pressure and temperature.All gases, which obey these laws, are called ideal gases or perfect gases. Theselaws are called ideal gas laws. These laws are applicable to gases, which do notundergo changes in chemical complexity, when the temperature or pressure isvaried. I.e., in other words laws applicable to gases which do not undergo anychemical reaction when subject to change in pressure or temperature.GAS LAWSBoyle’s lawBoyle’s law states that the pressure is inversely proportional to volume and theproduct of pressure and volume is constant PV =CCharles law-ICharles law states that at constant pressure, volume is directly proportional totemperature. V/T = CCharles law-IICharles law states that at constant volume, pressure is directly proportional totemperature. P/T = C
  • Steam generator Ganesh kumarAbsolute scale of temperatureThis scale of temperature is based on Charles law. According to Charles law atconstant pressure, volume of given mass changes by 1/273 of its volume at 0°C forevery rise or fall in temperature by 1°C. if the volume of the gas at 0°C is Vo and itsvolume at t°C,Vt = V o + Vo x t = Vo (1 + t/273) 273If t = -273°C, then volume is zero, the hypothetical temperature of –273°C at whichgas will have zero volume is known as absolute temperature or 0°K.Avagadra s LawAvagadra’ s law state that the volume occupied by any gas at normal temperatureand pressure is 22.41383 x 10-3 m3 per mol of gas. I.e., volume occupied by a kg molof gas is 22.41383 m3/kg mol.GAS EQUATIONFrom Boyle’s law PV = nRoTWhere, Ro is UNIVERSAL GAS CONSTANT n = m/M = Weight of gas in kg at NTP Molecular weight of the gas in kgAt normal temperature and pressurePressure = 1.01325 x 105 N/m²Temperature = 273 KVolume = 22.41383 x 10-3 m3n = 1 moleRo= PV/nT = 1.01325 x 105 x22.41383 x10-3/(1 x273) = 8.314 Nm mol-1 K-1 = 8.314 joules /mol KGas constant R = Universal gas constant (Ro) / molecular weight (M).Daltan s lawAt a constant temperature, the total pressure exerted by a mixture of non- reactinggases is equal to the sum of the partial pressure of each component gases of themixture. Thus the total pressure P of a mixture of r gases may be representedmathematically as
  • Steam generator Ganesh kumar rPt = Σ pI where pi is the partial pressure of each components gas of the mixture. i =1If P and the molar composition (% volume) of the mixture are known pi can becalculated using the expression pi = xi P2.6 SOME COMMONLY USED DIMENSIONLESS NUMBERS AND THEIR SIGNIFICANCENUMBER FORMULA SYMBOL DEFINITION & SIGNIFICANCENusselt hd/k Nu Radio of temperature gradients by conduction and convection at the surface -used for convection heat transfer coefficient determinationReynolds ρvd/µ Re Inertia force/viscous force - used for forced convection and friction factorPrandtl Cpµ/k Pr Molecular diffusivity of momentum Molecular diffusivity of heatGrashof ρ²d3 gß∆T/µ² Gr Buoyancy force x Inertia force Viscous force x viscous force - used for natural convectionBiot hd/ks Bi Internal conduction resistance Surface convection resistance - used for fin temperature estimationPeclet vdρCp/k Pe=RePr Heat transfer by convection Heat transfer by conductionStanton h/Cpρv St=Nu/Pe Wall heat transfer rate Heat transfer by convectionEuler ∆P/ρv² Eu Pressure force/Inertia force - used to find pressure dropFroude v²/gl Fr Inertia force/gravity force
  • Steam generator Ganesh kumarWhere v is velocity‘ d is characteristic dimension Cp is specific heat ρ is density g is acceleration due to gravity h is convection heat transfer coefficient µ is dynamic viscosity ß is volumetric expansion coefficient T is temperature P is pressureEx.01. Estimate the air and flue gas produced per kg of the following coal analysis.Ultimate analysis: Carbon = 39.9%, Hydrogen = 2.48% , Sulphur = 0.38 %, Nitrogen= 0.67%, Oxygen = 6.76 %, Moisture =8% and Ash = 42%. The analysis is basedon weight basis. Consider 4% carbon loss in combustion of AFBC system.AIR REQUIREMENT CALCULATIONAmount of oxygen required for burning coalC + O2 à CO2 + heat12 kg of carbon react with 32 kg of oxygen to produce 44 kg of carbon di oxide. I.e.,one kg of carbon required 32/12 = 2.666 kg of oxygen and produce 44/12 = 3.666kgof carbon dioxide.0.399kg of carbon in coal require = 0.39x2.666 = 1.064 kg of oxygenH2 + 1/2O2 à H2O + heat2 kg of hydrogen react with 16 kg of oxygen to produce 18 kg of moisture. I.e., onekg of hydrogen requires 16/2 = 8 kg of oxygen and produce 18/2 = 9 kg of moisture.0.0248 kg of hydrogen in coal requires = 0.0248x8 = 0.1984 kg of oxygenS + O2 à SO2 + heat32 kg of sulphur require 32 kg of oxygen to produce 64 kg of sulphur di oxide. I.e.,one kg of sulphur require one kg of oxygen and produce 64/32 = 2 kg of sulphur dioxide.0.0038 kg of sulphur in coal require =0.0038 x 1 = 0.0038 kgthe other composition like nitrogen, argon(if present) is inert gas and it will not reactwith oxygen. Moisture is in saturated form and it does not require oxygen.The total oxygen required = 1.064 + 0.1984 +0.0038 = 1.2662 kg
  • Steam generator Ganesh kumarThe oxygen present in fuel = 0.0676 kgNet oxygen required = 1.2662 – 0.0676 = 1.1986 kgAir contains 23.15 % oxygen by weight and hence the air required for 1.1986 kg ofoxygen is = 1.1986/0.2315 = 5.176 kg of dry air.Amount of wet air required considering 60% Relative humidity = 5.176 x 1.013 =5.244 kg.Coal requires 20% excess air for combustion in AFBC system hence wet air requiredfor burning per kg of fuel = 5.244 x 1.2 = 6.292 kg.FLUE GAS GENERATION ESTIMATIONCarbon di oxide produced = (0.399 – 0.0188) x 3.666 = 1.3915 kgMoisture produced = (0.0248 x 9 ) = 0.2232 kg.Moisture in fuel = 0.08 kg.Moisture in air = 0.013 x 6.212 = 0.0807 kg.Total moisture in flue gas = 0.3839 kgSulphur di oxide produced = 0.0038 x 2 = 0.0076 kg.Nitrogen in air = 6.212 x 0.7685 = 4.7739 kg.Nitrogen in fuel = 0.0067 kg.Total nitrogen in the fuel = 4.7739 + 0.0067 = 4.7806 kg.Excess oxygen in gas = (6.212 – 5.176)x0.2315 = 0.2398 kg.Total Flue gas producedPer kg of fuel = 1.391 + 0.3839 + 0.0076 + 4.7806 + 0.2398 = 6.803 kg.Ex.02 Find the weight of water present in atmospheric air at 60% relative humidityand temperature 40°C.For 40°C, the saturation pressure of water is = 0.075226 atm (from steam tables)At 60% RH the partial pressure of water vapour is 0.6 x 0.075226 =0.045135 atmWeight of moisture present in air = 0.622 x Pw/(1.035 –Pw) = 0.622 x 0.045135 (1.035 – 0.045135) = 0.02836 kg/kg.
  • Steam generator Ganesh kumarEx03. Estimate the efficiency of a boiler firing with coal as a fuel having GCV of3200 kcal/kg. Furnace is Fluidized bed boiler. Apply ASME PTC 4.1 indirect methodto calculate the efficiency. Flue gas temperature leaving the boiler is140°C andambient air temperature is 40°C. Ash content of the fuel is 42.3% and 20% of totalash is collected in bed and 80% ash is carried in fly ash. As per lab report the loss onignition of ash samples collected in bed zone and fly ash zone is 0.1% by weight and4.4%by weight. The boiler is operating at 20% Excess air and the dry kg/kg of gasproduced =5.91 and dry kg/kg of air required = 5.696. The moisture and hydrogenpresent in the fuel is 6% and 2.7% respectively.Basically following are the losses present in boiler,1.0 Unburnt carbon loss2.0 Sensible heat loss through ash3.0 Moisture loss due to air4.0 Moisture and combustion of hydrogen in fuel5.0 Dry flue gas loss6.0 Radiation loss.Unburnt Carbon loss =4%Sensible heat loss in ash,Flyash = %Flyash x% of ash qty x sp.heat (Tgo – Tamb) x100/GCV= 0.8 x 0.423 x0.22(140-40) 100/3200=0.233%Bed ash= 0.2x0.423x0.22(900-40)100/3200=0.5%Sensible heat loss due to ash = 0.233+ 0.5 =0.733%Heat loss due to moisture in air= kg/kg of moist in air x kg/kg of dry air( Enthalpy of steam at Tgo in 0.013ata – Enthalpy of steam at Tamb in 0.013 ata)= 0.013 x 5.696 x( 660.33–615.25)100/3200=0.1043%Note: The above implies that the water vapour at ambient temperature at partialpressure exists in steam form and gets superheated at 140°CHeat loss due to moisture in fuel and combustion of hydrogen,=(%of moisture in fuel + % of hydrogen x8.94)(Enthalpy of steam –Tamb)100/3200= (0.06 + 0.027x8.94)(658.37 –40)100/3200
  • Steam generator Ganesh kumar= 5.824%Note: The above implies that the water moisture present in fuel is in liquid form,during combustion it will absorb latent heat and superheat from combustion. Thehydrogen present in the fuel react with oxygen to form water. From combustionequation of hydrogen it is found that 1 kg of hydrogen form 8.94 kg of water.Dry flue gas loss,= kg/kg of dry flue gas x (Enthalpy of gas at Tgo –Air enthalpy at Tamb)x100/3200=Kg/kg of dry flue gas x Spheat (Tgo –Tamb)100/3200=5.91 x 0.24 x(140 –40)100/3200 = 4.433%Radiation loss,From ABMA Chart the loss is estimated as =0.5%Note: In the indirect method Blow down losses will not be considered into account. Itis assumed the boiler is operated under zero present blow down.Ex07 Estimate the FD and ID fan flow and power required for a bagasse fireddumping grate boiler, whose bagasse consumption at 100% MCR capacity is 31000kg/hr and the boiler is operating at 35% excess air. The fuel air requirement is 3.909kg/kg of fuel and gas generation is 4.873 kg/kg.FD fanTotal air requirement = 31000 x 3.909 = 121179 kg/hr.Fan design flow with 15% margin = 121179 x 1.15/(3600 x1.128) = 34.31 m3/secFD fan headPressure head required for air flow sections like airheater, air ducts and grate are tobe calculated. Now in most of the practical applications the pressure drop works outto be 165 mm WC and the same can be assumed for this calculation.FD fan head with margin = 165 x 1.2 = 200mmWcFD fan power required.= flow x head/102 x efficiency= 34. 31 x 200 / (102 x 0.8)
  • Steam generator Ganesh kumar= 84.09 KWMotor selected = 84.09 x 1.1 = 92.5 KW (next nearest motor standard is 110 KW)ID fanTotal gas produced = 31000 x 4.873 = 151063 kg/hr.Fan design flow with 25% margin = 151063 x 1.25 x (273 +140)/(3600 x1.295x273) = 61.27 m3/secID fan headPressure head required for gas flow sections like Furnace, Bank, Economiser, airheater, gas ducts and dust collectors are to be calculated. Now in most of thepractical applications the pressure drop works out to be 230 mm WC and the samecan be assumed for this calculation.ID fan head with margin = 230 x 1.3 = 300mmWcID fan power required.= flow x head/102 x efficiency= 61.27 x 300 / (102 x 0.8)= 225 KWMotor selected = 225 x 1.1 = 247.7 KW (next nearest motor standard is 250 KW)Table showing percentage margin on flow and head required for different boilerapplication.S.N Description Grate type AFBC CFBC OIL fired1 FD Fan Flow 15% 25% 25% 15% Head 20% 25% 25% 20%2 ID Fan Flow 25% 25% 25% 20% Head 30% 25% 25% 20%3 SA/PA/OF fan Flow 10% 25% 25% Not Head 15% 25% 25% applicable 3.0 FURNACE3.1 INTRODUCTION:The design of furnace is considered as the vital part in the boiler. The furnace is thezone experiencing a high temperature in boiler. The performance of the furnacereflects or has an impact over other parts behind it such as super heater, evaporator,and air heaters. For instant, how the furnace design affects super heater can be
  • Steam generator Ganesh kumarillustrated with following. If furnace outlet temperature (FOT) is high, then the nextzone is super heater it gets high amount of heat input naturally the metaltemperature is high and the steam temperature also increased, which in turn reflectsin the performance and cost of material. On the other hand if the furnace is oversized the FOT will be lesser, to get the required steam temperature the super heaterheat transfer area to be increased. If the heat transfer area is increased it calls forlarger space and cost wise it becomes uneconomical.3.2 EFFECT OF FUEL ON FURNACE DESIGN:The type of fuel, form of fuel, heat content and the properties of the fuel such as ashfusion temperature are also form as constraint over the furnace design. The type offuel whether solid or liquid or gas and quantity decides how efficiently we can burn.Whether we can have a burner (for liquid & gases), solids bubbling bed or dumpingor travelling grate. When the fuel is some thing like bagasse (fibrous and long strandstructure) it can be burnt well in dumping or travelling grate.A gaseous fuel offers fewer problems since it is clean. Fuel oil brings its ownproblems like high or low temperature corrosion and additives have to be used. Forcoal ash fusion is the problem, since ash slag down deposits on the wall hinderingheat transfer to steam water mixture. Depends on property of coal, whether it can becrushable to powdered form, pulverized firing or bubbling bed or cyclone furnace canbe decided.When we go for oil or gas firing, we can have higher heat flux in the furnace becauseof the higher emissivity of oil flame and relative cleanliness of walls compared to coalfiring. There by size of furnace will be smaller for oil or gas fired steam generators.The volume of the furnace for oil fired boilers will be 60 to 65 percentage ofpulverized fuel firing. However, if a furnace designed for both coal and oil it isnormally designed for coal and performance for oil firing in that furnace will becarried out. When a furnace designed for coal operated with oil, the higher furnaceabsorption results in a lower furnace outlet temperature. Lower FOT means superheater pick up in super heater will be less and steam outlet temperature will be less.This is avoided by several techniques out of which, when oil is fired FOT will beincreased by gas recirculation, otherwise when coal is fired FOT will be reduced bysome means of bed absorption (This is used in FLUIDISED BED COMBUSTIONtechniques). Furnace size also governed by length of flame in gas or oil fired boilersince the flame should not impinge on the water walls and cause overheating.Likewise in coal fired boilers flue gas velocity should be optimized to prevent higherrate of erosion due to carry over particles in flue gas. Normally a flue gas velocity of6 to 8 meters per sec was allowed for coal fired boilers and 12 to 15 meters per secwas allowed for bagasse fired boilers.3.3 FORCED OR NATURAL CIRCULATION:Water wall is receiving radiation from flames and are exposed to high heat flux andthere is a possibility of over heating. The boiling is the phenomenon, which governsthe rate of heat transfer from combustion to steam water mixture inside the tube. Inboiling when bubbles formed at tube wall hinders the heat transfer which cause
  • Steam generator Ganesh kumartubes over heating and tube failure. This sort of boiling occurs at nucleate boilingstage. Therefore proper circulation must be ensured to cool all tube. Circulationratio (CR) is the ratio between mass of water circulated inside the boiler to rate ofsteam generation. Hence CR is also directly related to dryness fraction of steam bythe expression CR = 1/x. which implies in one circulation 1/CR quantity of dry steamwas produced. Circulation number will be higher when the difference in densitybetween steam and water is more (i.e.) due to higher difference in density; steamwater mixture velocity will be more thereby overheating will be prevented. If theproper circulation is not there, circulation in the boiler circuit is effected by means ofexternal agency (normally a circulation pump will be used). This type of circulation iscalled Forced or controlled circulation.3.4 HEATFLUX TO FURNACE WALLS:Boiling phenomenon can be represented by a log-log plot of heat flux Vs surfacetemp-bulk temperature as shown Q max. H E A T F L U X A B C D SURFACE TEMPThe different regimes of boiling indicated by the letters A, B, C, D. Absence ofbubble formation and the influence of natural convection on the heat transfer processis predominant in the region A (pool boiling). Formation of vapour bubbles at thenuclei with resulting agitation of liquid by the bubble characteristics at the region B(nucleate boiling). The most important perhaps the critical region with respect to theheat flux is C. In this region the unstable film boiling manifests with an eventualtransition to a continuous vapour film. In the final region D film boiling becomesstabilized. This phenomenon of stable film boiling is referred as “ LEINDENFROSTEFFECT”
  • Steam generator Ganesh kumarIn the regime of boiling the maximum wall heat flux is observed in region C. Manyexperimentalists refer this state of maximum wall heat flux as “BURN OUT FLUX’.The reason being when the wall is heated electrically, the heating element frequentlyburn out when the wall heat flux reaches Q maximum. Hence the design engineersshould have an idea of average heat flux to the tubes, how they vary aroundperiphery and fin tip temperature in case of membrane wall construction. Calculationof fin temperature was discussed in latter part of this chapter.3.5 POINTS TO BE NOTED WHILE DESIGNING FURNACE1.0 Optimal heat transfer area to reduce the gas temperature to a temperature required from the point of super heater.2.0 Sufficient height to ensure adequate circulation in the water walls3.0 Fins in the wall to be properly cooled, accordingly the pitch of water wall to be selected.4.0 Flames should not impinge on water wall5.0 Proper provision should be there to remove ash generated.6.0 Optimal furnace outlet temperature.7.0 Sufficient residence time inside the furnace for complete combustion3.6 CLASSIFICATION OF FURNACEi) According to ash removal a) Dry bottom: It consists of water walls or refractory walls enclosing the flame. Ash shall be removed dry from bottom. The fuel used has low heat flux and high ash fusion temperature. b) Wet bottom: Ash removed from bottom is of molten form. The fuel having high heat flux low ash fusion temperature is used. The flue gas generated here or clean and free from fly ash and hence erosion, fouling problems are minimized.ii) According to Type of combustion a)Conventional firing 1) Travelling grate 2) Dumping grate 3) Pulsating grate 4) Step grate 5) Fixed grate
  • Steam generator Ganesh kumar b)Bubbling Fluidized bed combustion c)Circulated Fluidized bed combustion d)Pulverized fuel combustion e) Cyclone furnace.iii) According to draft system a) Balance draft: In balanced draft both Forced draft and Induced draft fans are used so to maintain vacuum or zero pressure in furnace. There is no leakage of combustion product in the atmosphere. In the atmospheric pressure air leaks into furnace. This type of draft system is widely adapted in industries. b) Forced draft or pressurized draft: Considering economic aspect in oil or gas fired boilers Forced draft fan alone used. The furnace pressure will be of the order of 100 to 150 mm a water column. The furnace has to be designed to without leakage. Otherwise combustion product will leak into atmosphere. c) Induced draft: Induced draft fan is used for sucking the flue gas generated. The furnace pressure will be maintained below atmospheric pressure. d) Natural draft: There is no draft fan will be provided for this system. Natural draft generated due to chimney itself used for the boiler draft. Very small capacity steam generators will be of this type.3.7 MODES OF HEAT TRANSFERIn general heat transfer from higher temperature to lower temperature is carried outin three modes.1.0 Conduction2.0 Convection3.0 RadiationConductionConduction refers to the transfer of heat between two bodies or two parts of thesame body through molecules, which are more or less stationary. Fourier law ofheat conduction states rate of heat flux is linearly proportional to temperaturegradient. Q = --K dt/dx
  • Steam generator Ganesh kumarWhere,Q rate of heat flux watts per sq.meterK thermal conductivity (property of material)W/m°kdt/dx temperature gradient in x –direction Negative sign indicates heat flows from high temperature to low temperature.Heat transfer by conduction in plate and cylinderPlate Q = k.A. (t1 - t2) watts XCylinder Q =k.(A2- A1).(t1- t2) (r2- r1) ln(A2/A1)where,A area of plateA1 outside cylinder surfaceA2 inside cylinder surface‘r cylinder radius‘t temperature of surfacesConvectionConvection is a process involving mass movement of fluids. When a temperaturedifference produces a density difference which results in a mass movement.Newton s law of cooling governs convection. In convection there is always a filmimmediately adjacent to wall where temperature varies. - kf A (tf - tw) Q =Where, is film thicknesskf thermal conductivity of filmh = kf / heat transfer coefficient (kcal/ sq.m hr °C or W/sq.m °C)RadiationAll bodies radiate heat. This phenomenon is identical to emission of light. Radiationrequires no medium between two bodies, irrespective of temperature the radiationheat transfer takes place between each other. However the cooler body will receivemore heat then hot body. The rate at which energy is radiated by a black body attemperature T( °K) is given by Stefan Boltzmann law. 4 Q= AT
  • Steam generator Ganesh kumarQ rate of energy radiation in WattsA Surface area radiating heat sq.m –8 4 Stefan boltzmann constant = 5.67 x 10 Watt/sq.m K –8 4 4.88 x 10 Kcal/sq.m hr K3.8 HEAT TRANSFER IN FURNACEFurnace heat transfer is a complex phenomenon, which can not be calculated by asingle formula. It is the combination of above said three modes of heat transfer.However in a boiler furnace heat transfer is predominantly due to radiation, partlydue to luminous part of the flame and partly due to non-luminous gases. Overallheat transfer coefficient in furnace is governed by three T’s temperature, turbulenceand time and calculated by two parts.Hc - heat transfer coefficient by convectionHr - heat transfer coefficient by radiation.HEAT TRANSFER COEFFICIENT BY CONVECTION (Hc)Heat transfer by convection may carry out in turbulent or laminar flow of the fluid. Inforced convection turbulence or laminar flow depends on mean velocity,characteristic length L, density and viscosity. These variables are grouped togetherin a dimensionless parameter called Reynolds number. Reynolds number is theratio between inertia force to viscous force.Reynolds number = (mass x acceleration)/(shear stress x cross sectional area) Mass = volume x density Acceleration = velocity / time Volume = cross sectional area x velocity Shear stress = dynamic viscosity x velocity gradient(v / l) Re = density x velocity x characteristic length Dynamic viscosity.When Re > 2100 then flow is turbulence < 2100 then flow is laminar. In practical case the flow is most oftenturbulent only.In free convection turbulence or laminar flow depends on the buoyancy force andtemperature difference, coefficient of volume of expansion. These variables aregrouped to form dimensionless numbers called Grashoff number and Prandl number.Laminar or turbulence is identified with product of Grashoff number and prandlnumber 9When, Gr.Pr < 10 flow is laminar
  • Steam generator Ganesh kumar 9 Gr.Pr > 10 flow is turbulent.DIMENSIONAL ANALYSIS FOR HEAT TRANSFER COEFFICIENTThe heat transfer coefficient may be evaluated from correlation developed bydimensional analysis. In this method all the variables related to the phenomenon isgrouped by experience with help of basic fundamental units length, mass, time andtemperature.The final equation arrived forFORCED CONVECTION h = f(L,U, ρ,µ,k,Cp) ,where,L characteristic length (meters)U velocity (meters/second)ρdensity ( kilogram/ cub.meter)µ dynamic viscosity(kilogram/meter. Hour)k thermal conductivity (watts/meter°kelvin)Cp specific heat(watt/kilogram.°kelvin) a b c d e fLet h = B L U ρ µ k Cp , where B,a,b,c,d,e,f are constantsExpressing the variables in terms of their dimensions -3 -1 a -1 b -3 c -1 -1 d -3 -1 e -2 -1 fMT = B L .(LT ) .(ML ) .(ML T ) .(MLT ) .(L² T ) a+b-3c-d+e+2f –b-d-3e-2f c+d+e -e-f = B.L .T .M .0 = a + b –3c –d +e +2f -3 = -b –d –3e –2f 1 =c+d+e-1 = -e - fThe solution of the equation gives,a = c-1, b =c, d = -c +f, e = 1-f c-1 c c -c+f -1-f fh = B. L .U . ρ .µ .k .Cpby grouping the variables, -1 c fh/L k = B.(UL ρ / µ) . (µ. Cp /k) c fNussultes number = B.(Reynolds number) .(Prandl number)The constants B,c,f are evaluated from experimental data.
  • Steam generator Ganesh kumarFor turbulent flow inside tubes and fully developed flow the following equationattributed to Mr.Dittus and Boelter, 0.8 nNu = 0.023 Re Pr where, n = 0.4 when the fluid is heated n = 0.3 when the fluid is cooled.For turbulent flow outside tubes 0.8 nNu = 0.037 Re Pr where, n = 0.4 when the fluid is heated n = 0.3 when the fluid is cooledFREE CONVECTIONFree convection depends on buoyancy force F, which is defined by,Let a fluid at To with density ρo change to temperature T with density ρ then, F = (ρo –ρ)g/P = ((ρo/ρ) – 1)gNow,ß coefficient of volume expansionthen, 1/ ρ = (1/ρo) + ß(To-T), ρo = ρ (1 + ß T) (ρo/ ρ ) – 1 = ß T F = ßg TFor an ideal gas ß is inversely proportional to temperature,(i.e. dimensional numberfor ß is -1 and F is -1 * LT-2 ie LT-2)By dimensional analysis, h = B.(Fa.Cpb.Lc. ρ d.µe.k f)MT-3 -1 = B[ (LT-2)a.(L2 T-2 -1 b ) . Lc.(ML-3)d.(ML-1T-1)e.(MLT-3 -1 f ) ]1 = d + e+ f= a + 2b + c –3d –e + f-3 = -2a –2b-e-3f-1 = -b-fsolving this equation. c = 3a – 1,d = 2a , e = b –2a, f = 1- bh = B[ (gß T)a . Cpb. L 3a-1. ρ2a. µ b-2a. k1-b)]h = B[ (gß TL3 ρ2/ µ² )a . (µ.Cp/k)b] (k/L)hL/k = B. Gra. Prb.
  • Steam generator Ganesh kumar a bNu = B. Gr . PrBy large number of experiments made on fluids it has been found that exponents a aand b are of the same value. So the expression reduce to Nu = B.(Gr.Pr)HEAT TRANSFER BY RADIATION HrIn furnace heat transfer is predominant by luminous and non-luminous radiation. Ageneral approximate expression may be written for furnace absorption using Stefanboltzman law. Q = A w [ g Tg4 – g TS 4 ] g = c c + w w -emissivity pattern of tri atomic gases such as carbon di oxide and water vapour arestudied by Mr. Hottel and charts are available to predict gas emissivity as a functionof various gas temperature, partial pressure and beam length. I have also furnishedthe expression form to find gas emissivity. When c and w are found from graph c and w can be determined from the following expression or from graph.Otherwise emissivity of gas can be directly found by the expression given inequation1. 0.222 1 1 c = EXP _______________ - P c *L +0.035 ln2.8 ln(p + 1.8) 1/3 0.23 1 2 w = EXP 0.842 - (0.23 +Pw*L 0.75 0.5+Pw+pwhere p is gas pressure in bar(a) L is beam length meter w and c are pressure correction factor for gas pressureabsorptive of gasses can be determined at wall temperature. g = c c + w w -At wall temperature correction,Pcw = P c (Tw/Tg) Pww = P w(Tw/Tg) c = cw (Tg/Tw)0.65 w = ww (Tg/Tw)0.45
  • Steam generator Ganesh kumar cw is a function of Pcw .L and wall temperature for this we have to see the emissivityin graph ww is a function of Pww .L and wall temperature for this we have to see the emissivityin graphpressure correction is same as gas emissivity factor. = w = function of P w/(Pc + P w) , Pcw.L + Pww.L, and temperature of wallThe effect of absorptivty is negligible hence the same can be neglected and ageneralized form of Q = A w g [Tg4 –TS4] can be used.Heat absorption by energy balance method, Q = [ Wf . lower heat value – W g .gas exit enthalpy]Where,A effective projected area of heat transfer including wall opening w wall emissivity g gas emissivity Stefan boltzman constantTg Flue gas temperature of mean theoretical flame temperature(adiabatictemperature)TS Furnace wall temperature (If calculated for outside heat transfer coefficient orconsider saturation temperature if calculated for over all heat transfer coefficient, thedifference will be of very minor).W f Fuel burntWg Flue gas producedGas emissivity g = 0.9( 1- e –k.L )………………………………………………1The emissivity of flame is evaluated by f = ( 1- e –k.L )where is the characteristic flame filling volume. = 1.0 for non luminous flame(practical 0.9) of solid fuels. 0.90 for luminous and semi luminous flame of coal .lignite & husk(AFBC ) 0.85 for luminous and semi luminous flame of bagasse (conventional firing) 0.72 for luminous and semi luminous sooty flame of liquid fuels 0.62 for luminous and semi luminous flames of refinery gas fuel OR gas/oil mixture 0.50 for luminous and semi luminous flames of natural gasL beam length meters = 3.4* volume/surface area.For cuboid furnace chamber and bundle of tubes.K attenuation factor, which depends on fuel type and presence of ash and itsconcentration. For non-luminous flame
  • Steam generator Ganesh kumarK = (0.8 +1.6 Pw).(1-0.38 TM/1000)(Pc + P w) _______ (Pc +P w)LFor semi luminous flame, the ash particle size and concentration is taken incalculationK = (0.8 +1.6 Pw).(1-0.38 TM/1000)(Pc + P w) ________ + 7µ(1/dm²TM²)1/2 (Pc +P w)Ldm mean effective diameter of ash particle in microndm 13 for coal ground in ball mills 16 for coal ground in medium or high speed mill 20 for coal milled in hammer mill.µ - ash concentration in gm/Nm^3TM – furnace mean temperature °k(Some authors will consider this as outlettemperature, but it is convincing assumption that in furnace zone temperature will beuniformly spread through out the furnace by radiation effect (spherical). Henceconsidering mean temperature for calculating radiation heat transfer coefficient willbe more appropriate. You can appreciate a notable phenomenon of furnacetemperature depends on flame location inside the furnace, in case flame is located atthe center of furnace(like oil fired burners (refer example1)) mean temperature andoutlet temperature will be at the most equal and if flame is located at one end of thefurnace and radiation beam travels a larger distance of furnace(like AFBC boilersassuming no free board combustion) the furnace temperature near flame will behigher and it gradually degrees at the furnace exit.For luminous oil or gas flame K = (1.6 TM/1000) –0.5Pw and Pc are partial pressure of water vapour and carbon di oxideAbove equations give only Theoretical values for flame emissivity. In practical casesa wide variation would be occurred due to:1.0 Combustion phenomenon itself2.0 The flame does not fill the furnace fully. Unfilled portion are subject to only gas radiation
  • Steam generator Ganesh kumar3.0 The emissivity of radiation is far below the flame emissivity. Emissivity of gas radiation may be in the range 0.15 to 0.3. Therefore overall emissivity of flame reduces. Hence emissivity changes with respect to location.Due to the above fact I have tried to give the practical values and graphs for theemissivity at appropriate places for AFBC, Dumping grate and fired boilers withworking of example.The heat transfer by radiation is given as Q = A w g [ TM4 – TS4]. But mostly theheat transfer will be of both convection and radiation occuring simultaneously and soto put both process on a common basis, we may define a radiation heat transfercoefficient by symbol Hr. Qr = Hr. A. (TM – TS) 4 4 Hr = w g[TM -TS ]/(TM-TS )While considering the total heat transfer by convection and radiationQ = (Hc + Hr) A (TM –TS) for fired furnace where gas throughout furnace is same.Q = (Hc + Hr) A Lmtd for AFBC and Radiation chambers.By this equations we can get theoretical Hr value but in practice these values arecorrected by effectiveness factor. This depends on various manufacturersexperience on their steam generator.(Normally for oil fired boilers the value will be of0.79 and gas fired boiler 0.67).3.9 FURNACE CONSTRUCTION :Basically three types of constructions are used1.0 Plain tube construction with a refractory lined furnace2.0 Tangent tube construction3.0 Membrane wall construction.Plain tube construction FURNACE CHAMBER
  • Steam generator Ganesh kumar REFRACTORYThe drawing shown gives complete idea of the above construction. Refractory linedwall construction is out dated design since it calls for a lot of refractory work and fluegas leaks are heavy and it can not with stand positive furnace pressure.Tangent tube construction FURNACE CHAMBER REFRACTORYTangent tube is a improvement of refractory lined. Here requirement of boiler tubesis comparatively more and also refractory structure is not eliminated. Membrane wall constructionIn industries widely used boiler furnace construction is of membrane wallconstruction type. In this design the tubes are joined by welding a continuouslongitudinal strip forming a solid panel, which can be as large as transportable.Panels can be welded together on site to form the furnace. The gap between thetubes(pitch) are maintained in a such a way that the fin can be cooled by either ofthe two side tubes and prevent warping of the panel. Water cooled furnaces not onlyeliminated problem of rapid deterioration of refractory walls due to slag, but alsoreduced fouling of convection heating surfaces to manageable extent, by loweringthe temperatures leaving the furnace. In addition to reducing furnace maintenanceand fouling of convection heating surfaces, water cooling also helped to generatemore steam. Consequently the boiler surface was reduced since additional steamgenerating surface was available in water cooled furnace.Ex.1.0 . Find the furnace outlet temperature for a fluidized bed boiler operating at 15kg/cm^2(g) having furnace EPRS of 28.43 sq.m and having the following gasparameters.
  • Steam generator Ganesh kumarFlue gas produced 11016 kg/hr at a temperature of 900°C and partial water vapourpressure 0.15 ata , partial carbon di oxide pressure 0.14 ata .The furnace size is 2.424 x 2.828m and height of 1.75meters.Assume FOT 740°CFlue gas properties at film temperature. (900+740 +200)/3 = 613.33°C –5Dynamic viscosity = 3.7392 x 10 kg/msThermal conductivity = 0.065177 kcal/m hr.°cPrandl number = 0.7152Flue gas velocity at outlet = 11016 x (613.33 +273) 3600 x 273 x 1.286 x 2.424 x 2.828 = 1.1269 meter/sec.Convection heat transfer coefficient at gas side(Hc ) =(As steam side heat transfer coefficient is very high, in over all heat transfercoefficient its effect will be negligible) 0.8 nNu = 0.037 Re Pr where n= 0.3 for cooling fluid 0.8 nHc/kL = 0.037 Re Pr 0.8Hc = 0.037 x 0.396 x 1.1269 x 1.75 x 0.7152 0.3 x 0.06517/1.75 3.7392 x 10-5 = 3.56 kcal/m^2 hr.°CRadiation heat transfer coefficient (Hr)Beam length = 3.4 x(w x d x l)/2(l.w +l.d + w.d )Substituting w= 2.424,d = 2.828, l =1.75L = 1.2709 mFor non luminous flame attenuation factorK = (0.8 + 1.6x 0.15) x(1-0.00038x(820+273)) x (0.14 +0.15) _________________ (0.14 +0.15)1.2709 = 0.2904
  • Steam generator Ganesh kumarflame emissivity f = 0.9 x (1- e –0.2904 x 1.2709 ) = 0.2778Wall emissivity w = 0.9 (practically adopted for fluidized bed boilers)Radiation heat transfer coefficient Hr = 4.88 x 10-8 x 0.2778 x 0.9 x [(820+273)4 –(200 + 273)4] [820 –200] = 27.1 kcal/hr m^2 KTotal heat transfer coefficient Hc + Hr = 3.56 + 27.1 =30.66 kcal/hr m^2 KHeat transferred Qg = U A (lmtd) = 30.66 x 28.43 x[(900 - 740)/ln(700/540)]= 537419 kcal/hr.Heat lost by gas QL = Wg ( Hi – Ho) = 11016 (257.3 – 207.45) = 549147 kcal/hrQg not equal to QL try with 745°C.Ex 02. Evaluate the size of bed for a 10 tph boiler, operating at 14.5 ksc, satuatedsteam from and at 100°C. Coal as a fuel. The efficiency of boiler is 80% and GCV ofcoal as 3800 kcal/kg , Flue gas produced per kg of fuel is 6.802 kg/kg at 20% excessair operation.Heat output = 10000 x 540 = 5400000 kcal/hr.Heat input = 5400000/0.8 = 6750000 kcal/hr.Fuel input = 6750000/3800 =1776.3 kg/hr.Flue gas produced = 1776.3 x 6.802 = 12082.4 kg/hr.Bed area = (Flue gas qty x bed temp)/(velocity x density of gases) = 12082.4 x (900 +273)/(3600 x 273 x 1.295 x 2.8) = 3.977 m^2.Bed size arrived = 3200 x1250 mm x mm a refractory wall thickness of 370 mm canbe considered and above which water wall is located. Hence a water wall of size3584 x 1680( 35 @ 112 pitch and 15 @ 112 pitch ) can be obtained.
  • Steam generator Ganesh kumarThe sizing of bed area and water wall size is an art rather than a scientific approacha better configuration has to be arrived on the basis of experience.Note: From and at 100°C is the term used in boiler industry to specify the heatcapacity of boiler. This is value is assumed that water at 1kg/cm^2 100°C is given asinput and steam drawn at 1kg.cm^2 .(i.e. latent heat at 1kg/cm^2 pressure onlyabsorbed ) EX 03. Find the furnace outlet temperature of a 55Tph dumping grate bagasse firedboiler operating at 42 kg/cm^2 and 420°C super heater outlet at furnace exit plane.The effective projected area of furnace and superheater plane works out to be212m^2 and 13.6m^2 respectively. Consider convection heat transfer coefficientnegligible and lower heating value of bagasse 1828 kcal/kg, 85% of air requiredflows through air heater at a temperature of 170°C and 15% air for fuel distributorand OFA at 40°C into the furnace. Fuel consumption 24209 kg/hr. 2% of gross heatinput goes as carbon loss and 1% goes as radiation loss.FURNACE HEAT INPUT1.0 Fuel heat input = 24209 x 1828 = 44.254 x 10^6 kcal/hr2.0 Air heat input = 0.85 x 24209 x 3.909 x 0.24 x 170 + 0.15 x 24209 x 3.909 x 0.24 x 40 =3.418 x 10^6 kcal/hrwhere,3.909 is air required for burning one kg of bagasse at 35% excess air. 0.24 kcal/kg°c specific heat of air.3.0 Un burnt carbon loss = 0.02 x 24209 x2272 = 1.1 x 10^6 kcal/hr4.0 Radiation loss = 0.01 x 24209 x2272 = 0.55 x 10^6 kcal/hrWhere 2272 kcl/kg is GCV of fuel.NET FURNACE HEAT INPUT = 1+2 –3 –4 = 46.072 X 10^6 KCAL/HRapplying stefan boltzman law, Q = A w g [ TM4 – TS4]As it is a bagasse fired boiler volatile combustion is more TM will be equal totemperature exit and w g is equal to 0.72.Assuming 890°C as FOTSaturation temperature 263°c .Q1 = 212 x 0.72 x 4.88x10^-8 x ( 11634 – 5364) = 13.01 x 10^6 kcal/hr.superheater steam outlet 420°c
  • Steam generator Ganesh kumarQ2 = 13.6 x 0.72 x 4.88 x 10^-8 x( 11634 –6934) = 0.764 x 10^6 kcal/hr.Total heat absorbed by surface = Q1 + Q2 = 13.77x 10^6 kcal/hr.Heat lost by gas,Q = ( furnace heat input – gas flow x outlet enthalphy) = (46.072 x 10^6 - 24209 x4.873 x 890 x 0.3076 ) = 13.776 x 10^6 kcal/hr.where 4.873 is kg of flue gas produced per kg of bagasse0.3076 kcal/kg°C is specific heat of flue gasFurnace outlet temperature = 890°CRadiation heat pick up contribution to raise steam temperature,(it is assumed that 70% of heat absorbed will go to steam temperature raise)= 0.7 x 0.76 x10^6/55000 = 9.67 kcal/kgEx.04. Estimate FOT for the furnace operating at 20.66 bara, having EPRS area112m² and size 5.74 x 3 x 6 m. firing LDO as fuel having LCV of 41867 kj/kg andfuel consumption 1.16 kg/sec and flue gas generated 19.03 kg/sec at 10% excessair. Air required 17.87 kg/sec at 27°C. Consider a radiation loss 0.33% and wallemissivity 0.85, heat transfer effectiveness 0.79. adjacent radiation chamber extendsby 1.01 m length wise.Total heat into Furnace at 27°C ambient.Heat input by fuel = 1.16 x 41867 = 48.565 MWRadiation loss = 48.565 x0.33/100 = 0.16 MWNett heat input = 48.405 MWHeat absorbed by Furnace 4 4Radiation coefficient Hr = w f[TM -TS ]/(TM-TS)For oil fired boiler Tmean is equal to Tgas outletWall emissivity = 0.85Flame emissivity = 0.72 (1-e-kl)Attenuation factor k = (1.6Tm/100)-0.5
  • Steam generator Ganesh kumarBeam length, l = 3.4x5.74x3x6/2(5.74x3+5.74x6+3x6) = 2.52MAssume gas outlet temperature,1285°c =1558°KK =(1.6x1558/100)-0.5 = 1.9928 –1.9928 x2.52Flame emissivity = 0.72x(1 –e ) = 0.715Hr = 5.67 x 10^-8 x0.85x0.715x(15584 –4874)/(1558 –487) = 187.768 W/mK4Nu = 0.037 x Re0.8 xPr0.3Gas properties at film temperature (1285+ 214)/2 = 749.5°CDynamic viscosity kg/m.s = 4.13276 x 10-5 kg/msThermal conductivity = 0.072915 W/mKPrandtl number = 0.711Velocity = 19.03 /(0.345 x 3x6) = 3.064 m/s where 0.345 is density kg/m^3 0.8 0.3Hc = 0.037 x 3.064 x5.74 x0.345 x 0.711 x 0.072915 4.13276x10^-5 5.74 = 5.769 W/mKHeat absorbed by FurnaceQ = (Hc +Hr)x effectiveness A Lmtd =(5.769 + 187.768)x 0.79 x 112 x(1558 - 487) = 18.34MWHeat absorbed by adjacent Radiation chamberHeat absorbed by adjacent wall= 5.67 x 10^-8 x 0.715 x(1-0.715) x 18 x 0.85 x(15584 –4874)= 1.0316MWTotal heat lost by Gas= 18.34 +1.0316 = 19.3716 MW
  • Steam generator Ganesh kumarHeat balance (heat lost = heat gained)19.3716 = (48.405 – 19.03 x Enthalphy of leaving gas)Enthalphy of leaving gas = 1.526 Mj/kg For this gas temperature is 1255°C. 4.0 SUPER HEATER4.1 INTRODUCTION:The steam temperature above its saturation temperature corresponding to thepressure is achieved by introducing super heater coils. In the modern industrial
  • Steam generator Ganesh kumarworld, it is expected to get the maximum attainable temperature for a pressure, sincethe cycle efficiency depends on pressure and temperature. But nature restricts themaximum temperature with material availability and metallurgical limitations.Super heater is a critical section, where high metal working temperature involves,since working fluid in super heater is hot steam when compared saturated water inother regions. The super heater will work out approximately 5% of the cost of theboiler and the material must be selected carefully. If improper selection of materialleads losses due to oxidation and improper sizing or thickness selection leads largerreserve in thickness which is uneconomical and lesser thickness cause tube failure.4.2 EFFECT OF FUEL ON SUPER HEATER DESIGN:The mechanical arrangement of super heaters is governed by factors like furnacedesign i.e. furnace outlet temperature which is explained in furnace chapter, fuelcharacteristics, degree of super heat and manufacturer practice. The fuelcharacteristics such as ash content, ash particle size, salts of alkali metals likesodium and potassium which volatilize in the process of combustion and condensesas a sticky substance at a temperature corresponding to super heater tubes. Thesedecide transverse pitching of super heater. Some light density fuels like husk burnsat the top resulting high heat flux input to super heater reducing heat transfer arearequirement itself.4.3 POINTS TO BE NOTED WHILE DESIGNING SUPER HEATER1.0 The super heater surface required to give the desired steam temperature.2.0 The gas temperature zone in which the surface is to be located.3.0 The steam temperature required4.0 The type of steel and other material best suited to make the surface and its supports.5.0 The rate of steam flow through the tubes (mass velocity) which is limited by permissible pressure drop, which will exert a dominant control over the tube metal temperature.6.0 The arrangement of surface to meet the characteristics of the fuels anticipated with particular reference to the spacing of the tubes to prevent a) Erosion b) Accumulation of ash and slag or to provide space for the removal of such formation in the early stage.7.0 The mechanical design and type of super heater.A change in any of the first six items will call for counter balancing change in all otheritems.4.4 CLASSIFICATION OF SUPER HEATERSSuper heaters are normally classified as Radiant and Convection super heaters.Radiant super heaters are located in the radiant zone receiving energy directly fromthe flame in the furnace. Convective super heaters do not receive furnace radiation.A few super heaters receive energy partly from the flame are called semi radiant.The other classification of super heaters depends on location arrangement and flowpattern.1 Radiant super heater located at radiant zone of boiler2 Convection super heater located at the convection zone of boiler
  • Steam generator Ganesh kumar3 In modern fluidized bed boilers, to achieve higher steam temperature a portion ofsuper heater located in the fluidized bed called bed super heaters.Depends on arrangement1 Horizontal 2. Vertical 3. Inline 4. StaggeredDepends on the flow pattern1. Cross flow super heater2. Parallel flow super heater3. Counter flow super heaterHorizontal arrangement of super heaters have a advantage of easy drainage, whichis quite important in boilers where shut down will be for longer periods. Drainabilityhelps in non-accumulation of salts in water or steam inside the tubes.In vertical arrangement the problem of expansion can be tackled easily thanhorizontal arrangement.Staggered bundles are difficult to clean but they offer a marginal improvement inheat transfer coefficient. Inline arrangement provides lower gas side pressure drop.Counter flow arrangement offers a slight improvement in log mean temperaturedifference and there by decreasing surface area compared to parallel flow. Theparallel flow arrangement leads to cooler tubes. Metal temperature can be higherwith counter flow arrangement since both gas and steam temperature are higher atsteam exit portion.4.5 DESIGNING A SUPER HEATERDesigning a super heater or any other heat transfer involves three steps one is thethermal design, second is the mechanical design and other is the performancecalculation.In a thermal design we have to perform the heat duty, heat transfer and surface arearequired for the known thermal input and output parametersIn a mechanical design thickness and material with standing capacity are to bechecked.Performance calculation involves confirmation of the designed equipment for variousloads.4.6 OVERALL HEAT TRANSFER ACROSS BANK OF TUBESAccording to the gentleman Mr. Colburn the following equation can be applied forgas/air flowing normal to the bank of tubes 3Nusselts number = 0.33 * (Reynolds number)0.6 * (Prandl number)0.While calculating Reynolds number external tube diameter has to be used. Thisexpression can also expressed after introducing geometry factor. Geometry factor Fhas to be taken from graph depends on transverse pitch to diameter ratio andlongitudinal pitch to diameter ratio 0.6 0.3Nusselts number = 0.35* F * (Reynolds number) * (Prandl number)
  • Steam generator Ganesh kumarAccording to another gentleman Mr. Grimson the outside heat transfer coefficientwas explained by the following equationArrangement Factor F For Inline arrangement For staggered arrangementST/d SL/D 1.25 1.50 2.00 3.00 1.25 1.50 2.00 3.001.25 Re 2000 1.06 1.06 1.07 1.00 1.21 1.16 1.06 0.96 8000 1.04 1.05 1.03 0.98 1.11 0.99 0.92 0.95 20000 1.00 1.00 1.00 0.95 1.06 1.05 1.02 0.931.5 2000 0.95 0.95 1.03 1.03 1.17 1.15 1.08 1.02 8000 0.96 0.96 1.01 1.01 1.10 1.06 1.00 0.96 20000 0.95 0.95 1.00 0.98 1.04 1.02 0.98 0.942.00 2000 0.73 0.73 0.98 1.08 1.22 1.18 1.12 1.08 8000 0.83 0.83 1.00 1.02 1.12 1.10 1.04 1.02 20000 0.90 1.00 1.00 1.00 1.09 1.07 1.01 0.973.00 2000 0.66 0.66 0.95 1.00 1.26 1.26 1.16 1.13 8000 0.81 0.81 1.02 1.02 1.16 1.15 1.11 1.06 20000 0.91 0.91 1.01 1.00 1.14 1.13 1.10 1.02Nusselts number = B * (Reynolds number) NB and N are factors governed by the geometry, the values of B and N is given in thetable.Many such persons worked on the heat transfer and gave various correlation forcertain pre defined condition and hence for practical purposes certain factors to beconsidered for its accuracy. I have also tried to give those factors in workedexamples.In the Colburn or Grimson equation correction factor for the heat transfer coefficientfor gas angle of attack on the tube has to be calculated into account.Degree ° 90 80 70 60 50 40 30 20 10Factor 1.0 1.0 0.98 0.94 0.88 0.78 0.67 0.52 0.42 ST/D 1.25 1.5 2.0 3.0 SL/D B N B N B N B N
  • Steam generator Ganesh kumar STAGGE 1.25 0.518 0.556 0.505 0.544 0.519 0.556 0.522 0.562 1.5 0.451 0.568 0.46 0.562 0.452 0.568 0.488 0.568 2 0.404 0.572 0.416 0.568 0.482 0.556 0.449 0.57 3 0.31 0.592 0.356 0.58 0.44 0.562 0.421 0.574 INLINE 1.25 0.348 0.592 0.275 0.608 0.10 0.704 0.0633 0.752 1.5 0.367 0.586 0.25 0.62 0.101 0.702 0.0678 0.744 2 0.418 0.57 0.299 0.602 0.229 0.632 0.198 0.648 3 0.29 0.601 0.357 0.584 0.374 0.581 0.286 0.608Correction factor for B corresponding to number of tubes deepNumber deep 1 2 3 4 5 6 7 8 9 10Staggered 0.68 0.75 0.83 0.89 0.92 0.95 0.97 0.98 0.99 1.0Inline 0.64 0.8 0.87 0.9 0.92 0.94 0.96 0.98 0.99 1.0Simillarly Dittus & Boelter correlation gives heat transfer coefficient on inside tube. 0.8 0.3Nusselts number = 0.023 * (Reynolds number) * (Prandl number)Consider the super heater tubes of outer radius r1 & r2 respectively. Thermalconductivity of material k. Cold fluid (steam) is flowing steadily inside the tube Tf1and hot fluid Tf2 steadily outside the tube. Inner and outer wall temperature tw1 &tw2. Heat transfer coefficient of steam and gas sides be h1 and h2. (h2 includesconvection heat transfer and non luminous heat transfer as explained in furnacechapter) STEAM ho Tf2 tw2 outside conve.heat transfer =3 tw1Hot gas Conduction =2 hi inside conve. heat transfer =1 Tf1 Tube thickness’t’Convection heat transfer phenomenon in super heater tubeql = Q/l = hi *2 * *r1 [ tw1 – tf1]………………………………………….1 = 2* * k (tw2 – tw1)/ln(r2/r1)……………………………………..2 = ho * 2 * * r2 [ tf2 – tw2]…………………………………………31=2=3the heat transferred at a given section at given time is equal.
  • Steam generator Ganesh kumartw1 – tf1 = ql / (hi*2* *r1)tw2 – tw1= ql * ln(r1/r2) 2* *ktf2 – tw2 = ql /(ho *2* *r2)Adding the above equations we get, ql 1 ln(r2/r1) 1tf2 – tf1 = + + 2* hir1 k hor2ql = 2* *(tf2 – tf1) 1 + ln(r2/r1) + 1 hir1 k hor2Q/l = U [tf2 – tf1]Divide by doQ/( dol) = U[tf2 – tf1]/ doU = 2* do 1 + ln(r2/r1) + 1 hir1 k hor2U = 1 do 1 + ln(r2/r1) + 1 2 hir1 2 k 2 hor2 1 do + d0 ln(d0/di) + 1 = U hidi 2k hoIntroducing fouling factors in steam side and gas side we can get,
  • Steam generator Ganesh kumar 1 do + d0 ln(d0/di) + 1 = +ffo +ffi U hidi 2k hoIn the above equation each unit refers to a resistance in heat transfer1/ho is gas side heat transfer resistance.(this gives temperature drop in the film)doln(do/di)/2k is metal resistance(this gives temperature drop in metal side)do/hidi is steam side heat transfer resistance.(this gives temperature drop in steamside film)ffo is outside fouling resistanceffi is inside fouling resistanceFor extended surface following heat transfer coefficient equation can be applied.1 At + At d0 ln(d0/di) + 1 = +ffo +ffi A t/Ai U hiAi Aw 2k hoWhere, At is total surface area. Ai is inside surface area. Aw is average surface area. is fin effectivenessQ = U * A * [tf2 –tf1] for specific tube, while considering bundle of tubes log meantemperature has to be used.Q = U * A * lmtd.METAL TEMPERATURE CALCULATION :Q = U * do *(Tg – Ts) = hi * di * (Tw – Ts)Where Tw is metal temperature °CMetal temperature of bare tube can be estimated easily by calculating varioustemperature drop across the resistance given above. This simple equation isapplicable for only bare tube surface. For extended surface fin temperaturecalculation involves a detailed procedure which is discussed separately.4.7 STEAM TEMPERATURE CONTROL
  • Steam generator Ganesh kumarSuper heater temperature depends on boiler working pressure. The degree of superheat varies directly with respect to boiler working pressure. The various steam outlettemperature corresponding to a pressure is fixed up by cycle efficiency and turbinemanufactures practice. Operating variables like flue gas temperature inlet, load,excess air, fuel creates fluctuation in steam temperature leaving the final superheater. If there is any fluctuation in steam temperature, there is a change in volumeof steam which will affect the turbine performance, since turbines are designed forpredetermined steam volume flow between stator and rotor and also the exhauststeam temperature quality will varies with steam fluctuation which will affectcondenser performance. Hence it is mandatory to have a steam temperaturecontrol.In practice two types of attemperators (de-super heater) is widely used in boilerindustry, they are1.0 Spray type2.0 Surface type.In the above two types, spray type attemperator gives a faster temperature controlcompared to surface type. However quality water spraying into the super heatershould be taken care for boiler and turbine life. Attempeator temperature techniqueis nothing but a simple energy balance. It consists of spraying water in mist form inbetween stages, depends on the final steam temperature and pick up by individualstages. The quantity of spray water varies with respect to load and it is controlled byautomatic control loops. Spray inside desuperheater accomplished by means ofnozzle arrangement.Surface type attemperators are further divided into two types submerged type andShell and tube type. In submerged type the steam coils is submerged below thedrum water level. Part of the steam flows through the submerged coils and part ofthe steam by-pass the coil. The flow of steam through the submerged coil isregulated in such a way that the outlet steam temperature is of desired level aftermixing.In shell and tube type attemperator super heater header itself modified into anexchanger. Steam from super heater coils enter this intermediate header and leavesto the second stage. In the super heater header the cooling water coils enter in boththe sides and leaves. The steam gets cooled when it contact with low temperaturecooling coils. In this steam temperature is adjusted by regulating the flow of coolingwater flow inside the coils. Higher the water flow steam cooling will be more.The water which absorb heat usually mixed with feed water to avoid the heat loss.SPRAY TYPE ATTEMPERATOR Water qty (M KG/HR & enthalpy
  • Steam generator Ganesh kumar Hfw kcal/kg) Steam flow(M2 kg/hr Steam flow (M1 kg/hr & enthalpy H2) & enthalpy H1) DE SUPER HEATER SEC. SUPERHEATER PRI. SUPER HEATEREnergy balance,M1* H1 + M * Hfw = M2H2……………………………………………1Mass balance,M1 + M = M2……………………………………………….2From 1 & 2,M1*H1 + M*Hfw = (M + M1) *H2Quantity of water required spray,M = M1 * { H1 H2}/{H2 Hfw}Ex.01. Determine the spray quantity required for a 60TPH boiler having a primarysuper heater steam outlet temp of 332°C and secondary super heater inlet steam tobe desuperheated to 316°c. The spray water temperature is 105°C.Enthalpy of PSH = 721.7 kcal/kgEnthalpy of SSH = 711.85 kcal/kgEnthalpy of spray water = 105.5 kcal/kg.Spray water required = (721.7 - 711.85)*60000 = 974.7 kg/hr. ( 711.85 –105.5)Ex02. Estimate the heat transfer area required for surface type desuper heaterimmersed in drum water operating at a pressure of 45 kg/cm²(g) and the steam at
  • Steam generator Ganesh kumarthe outlet of primary super heater is 380°C and the steam temperature at the inlet ofsuper heater is 350°C. the steam flow is 25000 kg/hr.The heat transfer coefficient outside the tube to water is natural convection heattransfer coefficient, governed by the equationNu = 0.54 ( Gr Pr) 0.25 ‘d3ρ²gβ∆T Cpµ 0.25‘hodo/k = 0.54 µ² kAssume surface type desuper heater of tube diameter 38.1 x 3.25 mmProperties of liquid at film temperature (saturation temperature at 45 kg/cm²(g))Density of liquid kg/m3 = 787.5 kg/m3Acceleration due fo gravity = 9.81 m/sec²Dynamic viscosity = 0.000103067 kg/msVolumetri coefficient of expansion ß = 0.284 1/°CPrandtl number = 0.83299Thermal conductivity = 0.00061184 Kw/m°CSpecific heat Cp = 4.94512 kj/kgTemperature difference between wall & water = 5° (assumed) 0.25 0.03813 x 787.5² x 9.81x0.281x 5 x 0.83299 x 0.61184ho = 0.54 0.000103067² 0.0381 = 19497 W/m²°KInside heat transfer coefficient is governed by forced convection,Nu = 0.023 Re0.8Pr0.4Properties of steam at average temperature (380 +350)/2 = 365°CDensity of steam kg/m3 = 16.169 kg/m3Dynamic viscosity kg/ms = 2.28175 x 10-5Prandtl number = 0.9576Thermal conductivity W/m°C = 5.945 x 10-2Velocity inside the surface assume 25 m/secReynolds number = 16.169 x 25 x 0.0316/2 .28175 x 10-5 = 559812‘hi = 0.023 x 5598120.8 x 0.95760.4 x 0.0316/5.945 x 10-2
  • Steam generator Ganesh kumar = 476.6 w/m²KMetal resistance = d0 ln(d0/di) /2 KmMetal conductivity = 49.5 w/mKMetal resistance = 0.0381 x ln(0.0381/0.0316) 2 x 49.5 = 7.1988 x 10-5Overall heat transfer coefficient1/U = 1/ho + 1/hi + Rm + ffi + ffoinside and outside fouling factor due to scale assume = 0.0002 m²/w°K1/U = 1/19497 + 1/746 + 7.1988 x 10-5 + 0.0002Overall heat transfer coefficient = 601.04 W/m²KHeat transfer area required = Q/Ulmtd = 25000( 3158.49 – 3083.08) x1000 3600x 601.04 x (( 380-350)/ln(123/93)) = 8.12 m²Ex.03. Calculate gas outlet temperature for a super heater intended to raise steamfrom 214°C to 258°C. steam flow 15.28 kg/sec, pressure 20 bar(a), tube size 38.1 x3.25 mm thk and thermal conductivity 49.844 W/mK. The gas flow 19.03 kg/sec,inlet temperature is 659°C and the pattern of flow is counter, furnace width is 3.06meter and length 2.7meters, tube pitching width side 80mm and depth side 78mm,inline arrangement, number of tubes in steam side path is 74. Heat transfer area ofsuper heater 88 sq.m. the super heater was enclosed in a water wall having EPRSarea of 21 sq.m and gas is flowing 90° to super hater tubes. Consider heat transfereffectiveness of 82% and 71% for enclosure. Partial pressure of water =0.1158 barand carbon di oxide = 0.1249bar.Steam inlet temperature = 214°CSteam outlet temperature = 258°CAverage steam temperature = 236°CProperties of steamDensity kg/cu.m = 9.328189Dynamic viscosity kg/ms = 1.72232 x 10-5Prandtl number = 1.083Thermal conductivity W/mK = 4.346819 x 10-2
  • Steam generator Ganesh kumarSteam velocity = 15.28 x 4 /(9.3281 x74x x 0.0316²) = 28.225m/sForced convection inside heat transfer coefficient isHi = 0.023 x Re0.8 x Pr0.4 x k/di 0.8 9.328x0.0316x 28.255 0.4 = 0.023x x 1.083 x 4.346819 x10-2 -5 1.7223 x 10 0.0316 = 1152 W/sq.mKForced convection outside heat transfer coefficient isAssume gas outlet temperature as 560°CFilm temperature = (average of gas temperature + average of steam temperature/2 =(659 +560 +214 + 258)/4 =422.75°CGas properties at film temperatureDensity kg/cu.m = 0.503Dynamic viscosity kg/ms = 3.1798 x 10-5Prandtl number = 0.7126Thermal conductivity W/mK = 0.05195For inline arrangementST/d = 80/38.1 = 2.099; SL/d = 78/38.1 = 2.047Arrangement factor fe = 1.18Gas flow area = (3.06 x 2.7 – 37 x0.0381x2.7) = 4.4558sq.mGas velocity = 19.03/(0.503 x 4.4558) = 8.49 m/sReynolds number = 8.49 x0.503 x 0.0381/3.1798x10-5 = 5116.8hc = 0.287 x fex Re0.6 xPr0.364 x k/d = 0.287 x1.18x 5116.80.6 x 0.71260.364x0.05195/0.0381 = 68.58 W/sq.mKRadiation heat transfer coefficientBeam length = 3.4 x(0.08 x0.078x1-0.00114)/( x0.0381x1) = 0.144876mAttenuation factor = (0.8 +1.6x0.1158)(1-0.38x(883/1000))(0.1158 + 0.1249) ((0.1158+0.1249)x0.14486)
  • Steam generator Ganesh kumar = 0.8439emissivity of gas = 0.9 x(1-e-0.8439x0.14486) = 0.10358hr = 5.67 x 10-8 x 0.85 x 0.10358x(8834 –5094) (883 –509) = 7.218 W/sqmKtotal gas side heat transfer coefficient = hc + hr = 68.58 + 7.218 = 75.798 W/sqmK 1 do + d0 ln(d0/di) + 1 = U hidi 2k ho = 0.0381/(1152x0.0317) + 0.0381 ln(0.0381/0.0317) + 1/75.798 2 x 49.844 U = 69.898 W/sq.mKHeat transferred to super heater tubes,Q = U x effectiveness x Ax lmtd = 69.898 x 0.82 x 88 x 372 = 1.876 MW.Heat transferred to water wall encloser,Neglecting metal resistance and internal conductance the convection heat transfercoefficient be 68.58 w/sq.mK and non-luminous radiation heat transfer behr = 5.67 x 10-8 x 0.85 x 0.10358x(8834 –4874) (883 –487) = 6.9538 w/sqmKho = 68.58 + 6.9538 = 75.53 W/SqmKQ = 75.53 x 0.71 x 21 x 393 = 0.4426 MW.Total heat gained by super heater and encloser is = 1.8763 + 0.4426 = 2.319 MW.Total heat lost by gas = Gas flow x (enthalpy difference)= 19.03 x (752.63 – 629.46) /1000 = 2.34 MW.Ex 05 Find the steam side pressure drop for a superheater, the steam flowingthrough the super heater is 55000 kg/hr and the Outlet pressure of steam is 19
  • Steam generator Ganesh kumarkg/cm²(a) and the number of parallel path is 74 nos. and coil developed length isestimated as 11.64m, it has 3 nos. 180° bends and 2 nos. 45° bends. The steaminlet condition is dry saturated and outlet condition is 255°C. the super heater tubesize is 38.1mm x 3.64mm including the positive tolerance of the tube.Super heater steam flow kg/hr :55000Steam flow/coil kg/sec:0.206Steam parameter at inletTemperature °C : 209.84Pressure kg/cm²(a): 19.34(assumed by trial and error)Steam parameter at outletTemperature °C : 255Pressure kg/cm²(a): 19Average steam ParameterTemperature °C : 232.42Pressure Kg/cm²(a):19.17Density kg/cu.m: 9.20739Dynamic viscosity Kg/s.m : 1.71 x 10-5Velocity through tubes = 0.206 x 4 9.20739x x(0.03082²) =29.99 m/sReynolds number = vd/µ = 9.20739 x 29.99x 0.03082/1.71x 10-5 = 497679Friction factor = f = (0.3964/Re0.3 ) + 0.0054= (0.3964/4976790.3) +0.0054 =0.0131454Pressure loss in straight length = flv²/2g d=0.0131454 x 11.64 x 29.99²/(2x9.81x 0.03082) = 227.58 Mgc= 227.58 x 9.20739 = 2090 kg / m² or mmWc= 0.209 kg/cm².Pressure loss in Bends,For 180° bend = 0.47v²/2gFor 45° bend = 0.12v²/2gTotal bend loss = ( 3x0.47 + 2x 0.12) 29.99²x9.20739 /(2 x 9.81) = 696.4 mmWc = 0.0696 kg/cm²Pressure loss in entry and exit
  • Steam generator Ganesh kumarEntry and Exit loss = 1.5 x 29.99² x 9.20739/(2 x 9.81) = 633 mmWC = 0.0633 kg/cm²Total pressure drop = 0.209+0.0696+0.0633 =0.3419 kg/cm² 5.0 DRUMS
  • Steam generator Ganesh kumar5.1 INTRODUCTIONThe recent development in boiler field involves, critical pressure boilers or oncethrough boilers in which drum is not necessary. But other than critical pressureboilers, it involves two regimes of liquid and vapour phase where separation ofsteam from liquid surface takes place in steam drum. The number of drumsrequired for a boiler depends on evaporator requirement, boiler pressure, stability,manufacture experience on the type of configuration. Widely single drum boiler andbidrum boiler are in use and in certain cases three or four drum design also availablebut they are outdated now. For higher capacity and high pressure(more than 70kg/cm^2(g) pressure) boilers are economical with single drum boilers and forindustrial process steam application where bidrum boiler works out to be economical.In practice it was found that drum cost around 10% of the boiler pressure part cost.5.2 OPTIMAL CONFIGURATION OF DRUMSIn world wide practice drums are designed in cylindrical shape with two dished endwith or without man hole at its end or cylindrical shell with tube sheet at its end orcylindrical shell with water box but the uniqueness of the drum is cylindrical shape.We consider the following three basic configuration for drum and let us analyzeabout drum geometry.1.0 why not drum be sphere?2.0 why not drum be square?3.0 why it is a cylinder?Stress inside a sphere Sphere with internal diameter d Y X X YLet the steam drum be a spherical shell of internal diameter ‘d’ and thickness ‘t’ andsubject to an internal pressure of intensity pBursting force P = p x projected area = p x x d²/4Let 1 be the tensile stress induced in metal at section XXResisting force = 1x xdxtBursting force = Resisting force
  • Steam generator Ganesh kumarpx x d²/4 = 1x xdxt 1 = pd/4tsince any section taken in diagonal to a sphere is symmetrical, the stress at anypoint in the metal of sphere will be equal and opposite and hence there will not beany shear stress. The strain at any direction is given bye1 = 1/E - 2/mE ( 1 = 2 )e1 = 1/E(1-1/m) where E is Young’s modulus and 1/m is poisson ratio.HenceFor the safety of shell thickness = t pd/4 1If the shell has been riveted then factor e called efficiency factor to be introducedi.e., ‘t pd/4 1eStress inside a square chamber d d LA square chamber has three planes of action and the stress and resisting force in allthe three direction to be checked for its stabilityLet ‘d be the sides of the square section ,L be the length, t be the thickness and pbe the intensity of pressureLet 1 be the tensile stress in direction xx. The force acting on xx be = p x projected area ______ = p x d² + d² x L _Bursting force P = p x 2 d L…………………………………………..1Area resisting this force = 2 x L x tResisting force = 1 x 2x L x t…………………………………………….2
  • Steam generator Ganesh kumarEquating 1 and 2 1 = pd/ 2 t or t = pd/ 1Similarly in zz direction 2 be the tensile stressBursting force P = p x d x LResisting force = 2x2xLxt 2 = pd/2t or t = pd/2 2Similarly yy direction 3 be the tensile stressBursting force P = p x d²Resisting force = 3 {(d +2t)² - d² } = 3x4xdxt (considering 4t² as negligible) 3 = pd/4t or t = pd/4 3Stress inside a cylinder Z Y X X p p Z YLet us consider the length and thickness of a cylinder be L and t whose meandiameter is d and the internal pressure be p.In this case section xx and zz experience same force and hence we have tocalculate force in xx and yy section onlyLet 1 be the tensile stress in direction xx. The force acting on xx beBursting force = p x projected area = p x d x LResisting force = 1x2xtxLEquating Bursting force and resisting force= 1 = pd/2t or t = pd/2 1This stress induced in circumferencial of the shell is called circumferencial stress ofhoops stress.Similarly in longitudinal direction area resisting the pressure = x d²/4
  • Steam generator Ganesh kumarForce acting on the end of the shell P = p x x d²/4 2 be the tensile stress on the section yy, equating the resistance offered by thesection yy to the total force on the end of the shell 2x xdxt = px x d²/4 2 = pd/4t or t = pd/4 2This stress is longitudinal stress. Hence at any point in the metal of the shell thereare two principal stress namely hoop stress(pd/2t) and longitudinal stress(pd/4t)acting perpendicular and parallel to the axis of shell respectively.Greatest shear stress Qmax = ( 1 - 2)/2 = pd/8tCircumferencial strain e1 = 1/E - 2/mEWhere E is the young’s modulus and 1/m is poisson ratio.From the above equation minimum thickness required for a cylindrical drum can bedetermined. Let be the permissible tensile stress for the material than for shell tobe safe major principal stress 1 should not exceedFrom this pd/2t or t pd/2Working pressure p 2x x t / d introducing ligament efficiency ‘e =pitch –hole dia pitch p = 2 x x e x t / d where d is the mean diameter.As per IBR regulation 270, introducing addition of corrosion allowance to thethickness the formula modified into, W.P = 2 x x e x (t –c) where c is corrosion allowance 0.03 inches (di + t - c) di is internal diameterAs per ASME PG27.1 , introducing addition of corrosion allowance to the thicknessthe formula modified into, W.P = 2 x x e x (yt –c) where c is corrosion allowance 0.03 inches (do –2y( t - c)) do is outside diameter. ‘y is temperature correction coefficient.Comparing the three geometries at their xx, yy and zz axis,
  • Steam generator Ganesh kumar DESCRIPTION XX YY ZZ SPHERE t =pd/4 t =pd/4 t =pd/4 SQUARE t =pd/ t =pd/4 t =pd/2 CYLINDER t =pd/2 t =pd/2 t =pd/4For a same pressure, material, dimension the minimum thickness required forsphere, cylinder and square is in the ratio of 1: 2 : 2.83 (thickness governing formulashown in bold letters in the tabular column )Sphere as a drum, there will be problem of having holding capacities and thenfabrication difficulties. Hence sphere can not be used in place of drum due to thesmaller volume.For square when compared to circular drums it calls for 1.414 times greaterthickness in turn weight of the drum. Number of welds for making a square is morewhere as a single fusion weld will do for circular drums. Square chamber has to beplaced in diagonal position for the lift of steam and therefore opening has to made inthe bends, which is very stress prone area. Practically speaking square sections areinferior in strength compare to circular section. According to IBR regulation squaresections can be used as header and circular sections can be used for both drumsand header. The main advantage of square header is at one side openings for tubescan be made and other opposite side openings for mechanical cleaning can bemade for straight inclined tubes. For this purpose some manufacturer prefer squareheader, where inclined tubes are used.5.3 STUBS AND ATTACHMENTS IN THE STEAM DRUM/SHELLAccording to the IBR regulation 281, there are certain minimum requisite ofmountings, fittings and auxiliaries to be provided in pressure vessels for the safety ofthe system. IBR regulation says that every boiler to be provided at least withfollowing1.0 Two safety valves (minimum diameter 0.75 inches)2.0 Two means of indicating water level gauge3.0 A steam pressure gauge4.0 A steam stop valve5.0 A feed check valve6.0 One feed apparatus(pump) when the heating surface exceeds 200 sq.ft and two independent feed apparatus each such apparatus shall have a capacity of not
  • Steam generator Ganesh kumar less than the maximum continuous rating of the boiler.7.0 A blow down cock or valve8.0 A fusible pug when boiler has internal furnace.9.0 An attachment for inspector pressure gauge.10.0 A man hole where the size and construction feasible.In case of boilers fitted with integral super heaters an additional safety valve shall befitted at the end of the super heater outlet header.There are also other attachments not in the IBR regulation such as steam separator,air vent and pressure gauge stub more than one as desired by the manufacture oruser.5.4 MAXIMUM PERMISSIBLE UNCOMPENSATED OPENING INDRUMThe opening in drum made for stubs, manhole, mud hole will weaken the plate whichcalls for strengthening the plate by compensating plates. According to IBR themaximum diameter for uncompensated opening is given by the following formula _________Maximum uncompensated opening = ½ (D +T)T + NWhere D is internal diameter of the drum T is thickness of drum shell N = 3 where E does not exceed 0.5 = 6 (1- E) in case exceeds 0.5, E is required thickness of seamless un-pierced shell divided by thickness of shell(T).Thickness of un-pierced seamless shell, PD = --------------- + 0.03 where e is 1 for un-pierced shell 2Fe -PIf the diameter of opening is less than the calculated value then it is not necessary togive compensation for the drum. If it exceeds the limit compensation have to beprovided.Area of compensation in Xaxis = dn x es where, dn nozzle internal diameter ofopening5.5 SIZE OF THE DRUMThe steam drum must be large enough to accommodate drum internals, thenecessary steam surface for steam separation and water holding capacity. Thedrum also provide sufficient space for change in water level that occur with change inload. A sudden increase in steam demand cause a temporary pressure surge, untilfiring rate is increased sufficiently for more steam generation. During this intervaldue to lower pressure, steam volume throughout the boiler is increased (pressure
  • Steam generator Ganesh kumar1/volume). This results in SWELL raise in water level in the drum. The raise inwater level depends on the rate at which heat and feed inputs can be changed tomeet the load demand. The higher the drum diameter better the control of raise inwater levelIn GT HRSG during starting the steam generation cause sudden increase in drumlevel and it is estimated that 70% of the evaporator water gets raised into the drum.Hence the during start up the HRSGs are started with low water level and at oncethe increased water level goes to high-high level blow down valve will be opened tocontrol the increase in water level. Normally blow down valve for HRSGs are sizedbased on the same, by finding the 70% evaporator volume less the drum volumebetween low to high-high. This difference in water has to be discharged with in 2 to 3minutes.STEAM SPACEIn the steam drum water and steam regions are clearly separated, the space abovethe normal water level is called steam space. Steam space governs the steamloading of the drum. Steam loading is the ratio between steam space and thespecific volume of the steam for pressure it is operating(M^3/m^3hr). Steam loadingfor different pressures is fixed by the manufactures according to their practice orexperience. Steam loading is one of the important decision factor in deciding thesize of the drum. Steam loading ensures the velocity of separation in the inter phaseand steam carry over.Let L be the length of steam drum and D be the internal diameter, r be the radius andx be the distance from center line to normal water level. o D k a b x nwl ‘h c LSin = ak/oa = Sin-(x/r) = (180 – 2 )‘ab = 2 ac = 2.r. Sin( /2)
  • Steam generator Ganesh kumarThe area of the arc oacb = x D² x 4x 360The area of triangle oab =1/2 . ab . XThe area acba = area arc oacb – triangle oabThe steam space area in cylinder = D² - area acba 4Steam space volume = ( D² /4 - area acba) (L +2l) + 2 volume of dished head ‘above normal water levelVolume of dished head for different shapes. VOLUME UPTO NORMAL TYPE FULL VOLUME WATER LEVEL M^3 M^3 SEMI ELLIPSOIDAL . D3/24 .h².(1.5D –h)/12 HEMI SPHERICAL .D3/12 .h².(1.5D –h)/6 TORIS SPHERICALEx 01 Estimate the shell thickness required for a single drum boiler, whose designpressure is 73 kg/cm² and the internal diameter is 1300mm. The drum is locatedoutside the gas path.Single drum the ligament efficiency is 1. However for practical estimation 0.95 isconsidered.As per IBR 270 regulationMinimum thickness required T = PDi___ + C 2fE –P= 73 x 1300__________ + 0.762 (2x 1230.4 x 0.95 –73)= 42.66 mm(over this thinning allowance for rolling has to be considered)Note. Stress value 1230.4 kg/cm² is calculated for the saturated temperature ofboiler design pressure. If the drum is exposed in gas path additional temperature
  • Steam generator Ganesh kumarallowance of 28°C has to be added to saturated temperature and for this stress valuehas to be considered.Ex.02 Estimate the thickness required for the 2:1 semi ellipsoidal head as per IBR,whose design pressure is 73 kg/cm² and the inside diameter is 1300mm. A openingin the dished end of size 545 x 500 mm is made for man hole opening. The materialof construction is SA 516 Gr.70. I 150 60 II 150 Inside 545x500 openingMinimum required thickness for dished end calculated as per IBR 277 & 278amendment 1995.W.P = 2f(T-C)/(DK)Where,T = Minimum thicknessW.P = Maximum working pressureD= Outside diameter = ( 1300+ 2xT)‘f = Allowable Stress = 1230.4 kg/cm² (As per ASME section II part D 1995)K= Shape factorC= An additive thickness = 0.75mmReinforcement of large openings‘d = Diameter of opening = 545 mmA= Effective cross section of reinforcement in mm²(shaded portion)T = Minimum thickness of dished head = 49 mm(assume)Limit of ReinforcementTt = Actual thickness of the nozzleL2 = d.Tt _______ (544 x60) = 180.83 mmSince the lenth of the nozzle projection inside the dished end is 150< 180.83, thewhole length is considered for compensation.Area of reinforcement
  • Steam generator Ganesh kumarI = 60 x150 x 2 =18000 mm²II= 44.3 x 210 x2 = 18606 mm²I + II = 18000+ 18606 =36606 mm²Imaginary diameter = d –(A/T) = 545 - 36606/49 = -veAssuming di =1 Since the value is negativeConditionsT/D < 0.1 di/D < 0.5T/D = 49/1398=0.035<0.1di/D =1/1398 =0.0007<0.5 Both conditions are fullfilled.‘di/ DT = 1/ ( 1398x49) = 0.0038Ref. Fig 23-D The least valve of di/ DT =1The same is considered for calculationFor Semi elliposidal headH/D = 374/1398 = 0.267 (H is height of dished head for semi ellipsoidal ( (ID/4)+ T)K = 1.1 from fig 23-DHence Minimum thickness = W.P x Dx K + C 2f= 73 x 1398 x 1.1 + 0.75 = 47.08 2 x 1230.4= 49 mm selected is sufficient. (Over the selected thickness dishing thinningallowance has to be considered)Ex 03 Estimate Manhole nozzle thickness for the above problem.As Per IBR 338(a)Thickness = P.D____ +C 2f + P= 73 x 545_____ + 0 = 15.702 ( 2x 1230.4 +73)The nozzle thickness provided 60mm(Which is used in reinforcement calculation ofthe above problem)
  • Steam generator Ganesh kumarThe provided thickness has to be cross checked, by compensation calculation.As per IBR 279,Inside diameter of nozzle dn = 545 –2x60 = 425Actual nozzle thickness tn =60mmActual dished end thickness ts = 49 mmStress value for nozzle fn =1230.4 kg/cm²Stress value for shell fs = 1230.4 kg/cm²Equivalent thickness of nozzle(en)= Pdn/(2f-P) + C= 73 x 425______ + 0.75 =13.75 (2x 1230.4 –73)Equivalent thickness of dished end (es)= 47.08 mmAs per IBR three conditions,‘b1 = minimum of 2.5ts or 2.5tn‘b2= minimum of 2.5ts or 2.5tn of projection thicknessCi = maximum of (ts+76) or dn/2‘b1 = 2.5 x49 = 122.5 or 2.5 x60 =150b2 = 2.5x49 = 122.5 or 2.5x60 = 150 or projection 150mmCi = 425/2 = 212.5 or (49 +76 =125)The bolded values to be taken into account for calculationSectional area at X = dn xes= 425x 47.08 = 20009 mm²Sectional area at Y = 2(tn-en)b1(fn/fs) + 2tnb2(fn/fs) + 2(ts-es)Ci=2(60-13.75)122.5(1230.4/1230.4) + 2x60x122.5x(1230.4/1230.4) +2(49 – 47.08)212.5= 26574.25mm²Area X < Area YCompensation shall be considered adequate when the sectional area X to becompensated measured through the axis of the shell is less than the compensatingare Y. In other words area X should be less than Area Y for the design to besatisfied, if X area is greater than area Y additional compensation pad to beprovided or thickness of nozzle to be increased.
  • Steam generator Ganesh kumarIn our case condition is satisfied and hence there is no need foe externalcompensation pad.Ex 04 Estimate the thickness required for end plate of thickness or closing the manhole nozzle.Man hole opening (max) = 545 – 2x60 = 425mmMan hole opening (min) = 500 – 2x60 =380mmAs per IBR 342(b) , ____Minimum end plate thickness = {Opening dia x (PK/f)] + CC= 1mmFor cylindrical endsK= 0.19 for ends integral with or flanged and butt welded to the header. = 0.28 for ends directly strength welded to the headerFor rectangular ends,Opening dia = Minimum width between the walls of rectangular headerK= 0.32 for ends integral with or flanged and butt welded to the header. = 0.40 for ends directly strength welded to the headerIn our case Ellipsoidal header is assumed as if circular header/nozzle and checkedfor both major and minor axis. _____________= 425 x (73x0.19)/1230.4 +1 = 46.123 mmProvided thickness = 65mmEx 05 Find maximum allowable uncompensated opening can be made in the abovedrum.As per IBR 187, _________Uncompensated Opening d = 0.5 (Di+T)T+NDi = Internal diameter of the ShellT = Thickness of shellN = 3 where E does not exceed 0.5 = 3x (1-E) in other cases 0.5E = The required thickness of a seamless unpierced Shell divide by CalculatedthicknessDi =1300mmT = 42.66…………from problem 5.1Ts = 73 x 1300____ + 0.762 = 40.505 (2x 1230.4 –73)E= 40.505/42.66 =0.949
  • Steam generator Ganesh kumar _______N = 6x( (1-0.949) = 1.35 _____________________Max. Uncompensated Opening size = 0.5 (1300+42.66)42.66 + 1.35 =119.66mm
  • Steam generator Ganesh kumar 6.0 EVAPORATOR AND ECONOMISER6.1 INTRODUCTIONIn steam generator boiling phenomenon takes place in furnace and evaporatorsection only. Even though the terminology boiler is commonly used for whole steamgenerator, boiler is applicable to the furnace and evaporator only. The other unitssuch as super heater, economiser, air heater etc., are only the heat exchanger usedto improve the cycle efficiency.6.2 DIFFERENCE BETWEEN EVAPORATOR AND ECONOMISEREVAPORATOR ECONOMISERIt is a part of the boiler circuit Economiser is a preboiler unit and act as a heat recovery unit.Evaporator always precedes the economiser. Economiser follows the evap-Since working fluid involved in an evaporator is orator and working medium inboiler water and the temperature of the medium economiser is feed water.is nearer to saturation temperature and it can not Temperature of feed water iscool the flue gas beyond its working temperature. Well below its saturation temp. corresponding to the working pressure.Log mean temperature prevailed in evaporator Log mean temperature inIs less than economiser. Economiser is more due to the Temperature gradient.In evaporator both natural and forced circulation Economiser coils otherwisecan be provided which depends on design, called forced flow coil.circulation ratio and manufacturer’s practice. Economiser inlet header is connected to feed pump and forced circulation is ensured.Evaporator gets its working fluid (boiler water) Economiser gets its workingfrom steam drum. Fluid(feed water) from feed tank and delivers into steam drum.The temperature difference between the saturation The temperature differencetemperature and gas temperature leaving the between the saturationevaporator is called pinch point temperature and the water leaving the economiser is called approach point.
  • Steam generator Ganesh kumar6.3 FIN EFFICIENCYThere are several types of fin types such as helical solid fins, serrated helical fins,circular fins, rectangular (H type ) fins, peg pins, longitudinal fins are available due todevelopment in heat transfer study. Out of the various fin configuration longitudinalfin configuration is the simplest fin configuration for estimating the fin efficiency, fromwhich correlation can be made to get other fin type fin efficiency. a hf hb hi l=b l l=0 hfoAt any cross section as in figure let Tc be the constant temperature of the hot fluideverywhere surrounding the fin and let t be the temperature at any point in the finand variable. Let be the temperature difference driving heat from the fluid to the finat any point in its cross section. Then = Tc – tIf l is the height of the fin varying from 0 to b‘d /dl = -dt/dlThe heat within the fin which process through its cross section isQ = ka d /dl…………………………………………….ADifferentiating the equation‘dQ/dl = ka d² /dl²……………………………..1Where a is the cross sectional area of the fin. This is equal to the heat whichpassed into the fin through its sides from l=0 down to the darkened cross section.If P is the perimeter of the fin, the area of the sides is Pdl and the film coefficientfrom liquid to fin side, whether on fin surface or tube surface, is hf.‘dQ = hf Pdl or dQ/dl = hf ……………………..2
  • Steam generator Ganesh kumarEquation 1 and 2 are same and hence 1-2 =0kad² /dl² -- hf =0Rearranging,d² /dl² - hf /ka =0The direct solution of this equation is, 0.5 0.5 l(hfP/ka) = C1e + C2e-l(hfP/ka) ……………………..3 _____‘m = (hfP/ka)The general soution is = C1eml + C2e-mlApplying boundary layer concept,At the outer edge of the fin l =0, e = c1 + c2Assume no heat enters from outside edge of the fin, then‘d /dl =0 when =0 and C1 –C2 =0C1 = C2 = e/2Equation 3 becomes, e = (eml+ e-ml)/2 = e cosh(ml)Thus an expression has been obtained for the temperature difference betweenconstant fluid temperature and variable fin temperature in terms of the length of thefin.It is now necessary to obtain an expression for Q in terms of lDifferentiating the equation 2, we getd²Q/dl² = hfPd /dl………………………………….4 /dl = ‘d²Q/dl² hfPSubstituting in equation AQ = kad²Q/hfPdl²
  • Steam generator Ganesh kumard²Q/dl² = hfPQ/ka………………………………..5d²Q/dl² - hfPQ/ka = 0As before the solution is 0.5 0.5 l(hfP/ka)Q = C3e + C4e-l(hfP/ka) ……………………..6Q = C3eml + C4e-mlAt I =0,C3 + C4 =0 , C3 = -C4 and dQ/dl =0‘dQ/dl =hf e = mC3 –mC4 =0C3 = hf e/2m and C4 = -hf e/2mQ = hf e eml /2m - hf e eml /2mQ = hf e /m(eml - eml )/2Q = hf e /m sinh(ml)The ratio of heat load Q to the temperature difference at the fin base isQ/ = hf e sinh(ml) m e cosh(ml)Q/ = hfP tanh(ml) ……………………………………………………7 ‘mhf = Qm/ ptanh(ml)Let hf is the heat transfer coefficient of fin and the bare tube and the heat absorbedby fin through the heat transfer coefficient hf is getting transmitted into the tube bymeans of base heat transfer coefficient hb. The ratio of heat transfer coefficient hb tothe heat transfer coefficient hf is termed as fin efficiency.According to Fourier’s law of conduction,hb = Q/ lPη’ = hb/hf = Q P tanh(ml) Qm lP
  • Steam generator Ganesh kumar = tanh(ml) mlThis equation applies only to the fin and not to the bare portion of the tube betweenfins. The total heat removed from the annulus liquid and arriving at the tube insidediameter is a composite of the heat transferred by the fins to the tube outsidediameter and that transferred directly to the bare tube surface. These may becombined by means of the weighted efficiency.If the heat tranferred through the bare tube surface at the tube outside diameter isdesignated by Q0, thenQo = hf Ao where A0 is the bare tube surface at the outside diameter exclusive ofthe area beneath the bases of the fins. If there are N number of fins on the tubel.P.Nis all of the fins surface. The total heat transfer at the outside diameter is givenbyQ = Qb + Qo = hb.l.P.N. + hf.Ao=(hb.l.P.N.Ao/Ao + hf.l.P.N.Ao/P.l.N)=(hb/Ao + hf /P.l.N)l.P.N.Ao.=(l.P.N.tanh(ml) + Ao ) hf ………………………………………………..8 mlCalling hfo the composite value of hf to both the fin and bare tube surfaces whenreferred to the outside diameter of the tube, the fin effectiveness or weighted finefficiency is by definition η =hfo/hf. Combining equation 7 and 8η =hfo/hf = l.P.N η’ + Ao l.P.N +Ao = (η’.At + Ao /(At + Ao))inserting At –At, =( At –At + Ao + Atη’) /(At+ Ao) = (At+Ao -(1-η’)At)/(At+Ao)Fin effectiveness η = 1- (1-η’)At/A where At is Area of fin surfaceEx.01. Design a feed water heater for a 10 tph boiler whose exhaust gas flow is21355 kg/hr at an outlet temperature is 185°C and the desirable outlet temperature is140°C. The feed water is available at 60°C.I. Heat duty (gas side) = 21355 ( 48.26 – 36.09) =259890 kcal/hr
  • Steam generator Ganesh kumarWhere 48.26 kcal/kg enthalpy of gas at 185°C and 36.09 kcal/kg enthalpy of gas at140°C.II. Water outlet temperature to = 259890/10000 + 60 =85.98°C.III. For the heat exchanger, gas is flowing inside the tube internal heat transfercoefficient governs the overall heat transfer coefficient. The overall heat transfercoefficient is around 85% of inside heat transfer coefficient.Assume gas velocity inside the flue tubes 15 m/sec.Flow area required to maintain the velocity in tubes= 21355 (162.5 +273)/(3600x273x1.295x15) =0.48716m²Tube selected 50.8 x 3.25IV. Number of tubes = 0.48716 /( (0.0508 – 2x0.00325)² = 316 tubes.V. Heat transfer coefficientGas properties at average temperature 162.5°C.Thermal conductivity = 31.225 x 10-3 kcal/mhr°CKinematic viscosity = 28.577 x 10-6 m²/secPrandtl number = 0.6775Reynolds number = vd/γ = 15 x 0.0443 /28.577 x 10-6 = 23252.559HTC = 0.023 Re0.8 Pr0.3 K/d = 0.023 x (23252.559)0.8 x (0.6775)0.3 x (31.225 x10-3/0.0443) = 39.2 kcal/m²hr°CThe overall heat transfer coefficient = 0.85 x39.2 =33.32 kcal/hrm²°C.V Heat transfer area required = Q/(Ux Lmtd) = 259890/(33.32 x89.15) = 87.49m²VI Heat transfer length
  • Steam generator Ganesh kumar = 87.49/( x 0.0443) = 628 mNumber of tube required = 316Tube length = 628/316 = 1.98 mConsidering tube length for welding to the tube sheet clearance 10 mm on both sidesTube size chosen 2.0 m.Ex.02. Calculate gas outlet temperature for a evaporator intended to generatesteam flow 15.28 kg/sec, pressure 20 bar(a), tube size 38.1 x 3.25 mm thk andthermal conductivity 49.844 W/mK. The gas flow 19.03 kg/sec, inlet temperature is1180°C and the pattern of tube arrangement furnace width is 3.06 meter and length2.7meters, tube pitching width side 80mm and depth side 78mm, inline arrangement,number of tubes in width side 36 and depth side 20. Heat transfer area of evaporator246 sq.m. the evaportor was enclosed in a water wall having EPRS area of 26 sq.mand gas is flowing 90° to evaporator tubes. Consider heat transfer effectiveness of82% and 71% for enclosure. Partial pressure of water =0.1158 bar and carbon dioxide = 0.1249bar.Assume gas outlet temperature as 658°CHeat loss by the gas = 19.03 *( 1418.1 – 745.94) = 12791.2 KWInside heat transfer coefficient = 14000 W/m²°C (assume)Gas properties at average temperature = 919°CDensity kg/cu.m = 0.299Dynamic viscosity kg/ms = 4.573 x 10-5Prandtl number = 0.71Thermal conductivity W/mK = 0.083For inline arrangementST/d = 80/38.1 = 2.099; SL/d = 78/38.1 = 2.047Arrangement factor fe = 0.98 (from tables)Gas flow area = (3.06 x 2.7 – 36 x0.0381x2.7) = 4.558sq.mGas velocity = 19.03/(0.299 x 4.558) = 13.96 m/sReynolds number = 13.96 x0.299 x 0.0381/4.573x10-5 = 3478.5hc = 0.287 x fex Re0.6 xPr0.364 x k/d = 0.287 x0.98x 3478.50.6 x 0.710.364x0.083/0.0381 = 72.1 W/sq.mKRadiation heat transfer coefficientBeam length = 3.4 x(0.08 x0.078x1-0.00114)/( x0.0381x1) = 0.144876m
  • Steam generator Ganesh kumarAttenuation factor = (0.8 +1.6x0.1158)(1-0.38x(1192/1000))(0.1158 + 0.1249) ((0.1158+0.1249)x0.14486) = 0.6947emissivity of gas = 0.9 x(1-e-0.69470.14486) = 0.08617hr = 5.67 x 10-8 x 0.85 x 0.08617x(11924 –4854) (1192 –485) = 11.53 W/sqmKtotal gas side heat transfer coefficient = hc + hr =72.1+11.53 = 83.63 W/sqmK 1 do + d0 ln(d0/di) + 1 = U hidi 2k ho = 0.0381/(14000x0.0317) + 0.0381 ln(0.0381/0.0317) + 1/83.63 2 x 49.844 U = 82.55 W/sq.mKHeat transferred to evaporator tubes, Q = U x effectiveness x Ax lmtd = 82.55 x 0.82 x 246 x 673 = 11206 KW.Heat transferred to water wall encloser,Neglecting metal resistance and internal conductance the convection heat transfercoefficient be 82.55 w/sq.mKQ = 82.55 x 0.71 x 26 x 673 = 1025 KW.Total heat gained by evaporator and encloser is = 11206 + 1025 = 12231 KW.
  • Steam generator Ganesh kumar 7.0 AIRHEATER7.1 INTRODUCTIONAir heater is a heat recovery unit, which is employed to recover heat from lowertemperature levels usually 240°C and below. Air heater is used in most of theindustrial and utility boilers. The construction of this equipment being a simple andalso a non pressure part. Gas temperature drop 240°c to 170°C can be effectivelyachieved by means of air heater there by 4% boiler efficiency can be improved. Incases of low capacity , low pressure boilers manufactures prefers to have an airheater instead of economiser. This is to avoid problems arised due to economiserlike oxygen pitting, corrosion higher feed water temperature, necessity of havingdeaerator more over economiser is a pressure part. These are avoided by means ofair pre heater7.2 TYPES OF AIR HEATERThe basic classification of Air heater is based on their operating principle1.0 Recuperative2.0 RegenerativeRecuperative Heat Exchanger:In recuperative heat exchanger heating and cooling medium are separated bypartition and the heat is transferred from one fluid to other by means of conductionand convection and there is no moving part employed in recuperative heatexchangers. According to the construction it is again sub divided into tubular or platetype air heater.Tubular air heaterThis usually consists of large number of steel tubes of either welded or expandedinto the tube plate at the ends. Either gas or air may be designed to flow through thetube. Gas through the tube normally requires higher size tube and vertical flow toreduce fouling. Single or more passes on the gas side and multi pass to and fro onthe air side usually fits in with shorter tube length so as to facilitate maintenance osurfaces due to corrosion and fouling. In some cases instead of using of boiler fluegases separated external firing is used particularly during startingPlate air heaterThis comprise of parallel plates which provide alternate passage for gas and air. Thistype is simple and compact to that of tubular type. The narrow passes betweenplates make cleaning tedious but with shot cleaning method it is improved. Butreplacement is a major task and nowadays it is not used in industry or utility boilers.Regenerative Heat exchangersIn regenerative heat exchangers the heating fluid heat the closely packed matrix toraise its temperature which is again placed in cooling medium to transfer the heat to
  • Steam generator Ganesh kumarcooling fluid. In regenerative exchangers either matrix or the hoods are rotated toachieve this. This is sub divided by the names of inventors intoi) Ljungstrom typeii) Rothemuble typeIt is beyond the scope of this chapter to define regenerative heater.Recent technology involves usage of heat pipe in air heater.A heat pipe is a device with a high thermal conductance. It is based on the largelatent heat of vapourisation/condensation of working fluid inside the heat pipe.Hence all heat pipes are completely sealed one and the working fluid is the heattransfer element between the hot and cold fluid passing through its ends separatedby a partition plate. In heat pipe heat is absorbed in the evaporator portion of theheat pipe, and of vapourises the internal working fluid. The vapour is transferred tothe condenser portion on the other end where the fluid surrenders the heat to thecold surrounding and condenses. This fluid is then returned to the evaporator andthe cycle is repeated. Heat pipe contains a working fluid specifically selected foroperating temperature range and hence heat pipes are arranged in modulescontaining different working fluid for the range of temperature.Types of heat pipes differ according to the method used to transport the workingfluid.a) Thermo-syphon heat pipe.The name itself indicates heat pipes uses gravity, buoyancy and vapour pressureforces to transport the phase of the working fluid. It requires the heat source to bebelow the heat sink that is evaporator must be below the condenser. The vapourgenerated in the evaporator travels upward because of the buoyancy and expansionforces. In the condenser the vapour is condensed to a liquid and turns to theevaporator by gravity. The thermo syphon heat pipe can be made to functionefficiently with an orientation of approximately 10 to 90 degree from horizontal.b) Capillary action heat pipeThe working fluid is transferred through capillary action and it does not requiregravity to return the condensate to the evaporator section. This type of heat pipescontain an internal element known as wick, which spans the length of the pipe. Thewick may be made of gauze wire mesh or other materials that provide closelyspaced longitudinal paths for returning condensate. The path spacing is chosenbased on the working fluid to obtain high capillary forces and there are no orientationrestriction for this type of heat pipe.There should not be gap between heat source and sink in the heat pipe. If a gapexists then it can be spanned by an adiabatic(non heat transfer) section of heat pipe.This adiabatic region is created by externally insulating the appropriate section ofheat pipe.
  • Steam generator Ganesh kumarToluene and naphthalene are commonly used heat transfer fluids in air heaters. Themajor advantage of heat pipe design compared to conventional tubular heatexchanger commonly used are that it is isothermal and can be built with better sealsto reduce leakage. Each pipe is fixed only at the center support plate separating thehot flue gas and cold air. The ends of each pipe are free to expand and contractindependently. Since heat pipe has no hot and cold spot the possibility of cold endcorrosion is reduced. A COMPARISON OF VARIOUS AIR HEATER TYPESDESCRIPTION RECUPERATIVE REGENERATIVE HEAT PIPE (TUBULAR) (ROTARY)Size Large Compact CompactWeight High Medium MediumCorrosion Cold end corrosion Avoided Avoided Will be thereCleaning Water wash on load sootblowing/on load soot blowing Water wash water washPressure drop High Medium MediumCost of air heater Low High HighMaintenance Low High PeriodicEfficiency 65% 90% 90%Suitability Industrial outdated Utility boilers boilersDuct routing Length required Simple and Simple and More less costly less costly.Cleaning of Air heaterAir heater are in low temperature operating region, it is subject to metal temperaturebelow dew point, which is primary cause of condensation of acid or moisture fromflue gas. Minimum metal temperature occurs at cold end where most fouling andcorrosion occurs. If the air heater is not cleaned at frequent intervals the foulingincreases and velocity increases in other tubes which cause severe erosion of high
  • Steam generator Ganesh kumarvelocity tubes and corrosion of fouled tubes. These problems can be eliminated byfrequent cleaning of air heater.On load cleaningRecuperative type proven practical method is by shot rain and regenerative type canbe cleaned by fixed or moving soot blowers. Heat pipe which uses finned tubes canbe cleaned by soot blowers.Off load cleaningIf not provided with on load cleaning, air heater has to be cleaned at intervals duringshut down manually or mechanical methods. Large quantity of cold or warm waterhas to be used for flushing the air heater since small quantity of water will actually doharm by making deposits compact and hard. If deposit is severe soda ash solutionwill assist in dissolving it.LeakageLeakage in air heater can be checked during operation by analyzing the flue gas forcarbon di oxide drop across the air heater. This leakage is through the tube leakjoint corrosion, erosion holes etc., in case of tubular air heater. While through sealsin case of regenerative air heater.7.3 ADVANTAGES OF AIR HEATERIn addition to increase in boiler efficiency the other advantages that may result are1.0 Stability of combustion is improved by use of hot air.2.0 Intensified and improved combustion which result in faster load variation and fluctuation.3.0 Permitting to burn poor quality coal.4.0 High heat input to the furnace and hence high heat flux and high heat transfer rate.5.0 Less unburnt fuel particle in flue gas, thus combustion and boiler efficiency further improved.6.0 In case of FBC and pulverized coal combustion hot air can be used for drying the coal as well as for transporting the coal into furnace.7.0 This being a non pressure part shut downing of unit is not required due to corrosion of heat transfer surface or failure.8.0 High temperature of inlet cause reduction in carbon mono oxide.
  • Steam generator Ganesh kumar7.4 HEAT TRANSFER IN AIR HEATERThe operation region involves low temperatures, where heat transfer takes placepredominantly through convection and little conduction. Since the thickness of tubeis very less conduction heat transfer can be neglected.Convection heat transfer takes place in two modes1) Gas to metal heat transfer2) metal to air heat transfer.As the air heater is located in cold end of heat transfer side, non-luminous radiationwill be very negligible and hence the same can be eliminated. (Above 400°C nonluminous radiation heat transfer will be negligible).The over all heat transfer coefficient can be estimated as per the derivation given inthe super heater chapter. In that equation neglecting fouling resistance and metalresistance we will getU = ho x hidi/do (ho + hidi/do )7.5 PRACTICAL GUIDE LINES FOR DESIGNING AIR HEATER.1.0 Gas or air velocity inside tube has to be limited to maximum of 15 to 18 meters/sec and out side tube has to be limited to 6 to 7 meters/sec beyond which hysterisis will occur.2.0 The air or gas passes over the tubes, where the lesser number of tubes deep arranged this is to avoid higher pressure drop.3.0 The unit height of tube should not exceed 3.5 meters to avoid vibration problem and preferable height be 2 to 3m between tube sheets.4.0 In FBC and conventional boilers it is desirable to heat the air not more than 180°c, in order to have a better cooling for grate or Distributed plate.5.0 Normally a tube diameter 50.8 mm or 63.5 mm will be used.Ex 01 Design tubular air heater for 200 TPH steam generator having followingparameter,Gas flow kg/hr = 272160Air flow kg/hr =213156Air inlet temperature = 26.67°cGas Inlet temperature =271.3°cGas Outlet temperature =160°cLet the air heater be cross type and staggered arrangement and the boundarycondition of air heater be
  • Steam generator Ganesh kumarFlue gas inlet temperature °c TgiFlue gas outlet temperature °c TgoAir inlet temperature °cTaiAir outlet temperature °cTaoMass of gas flow kg/hr mgMass of air flow kg/hr maHeat duty Ql = mg (Hgi– Hgo) kcal/hr= 272160 ( 75.93-44.1244) = 8.656 x106 kcal/hrAir outlet temperature = {Ql/(cpo *ma)} + Tai=8.656 x106 + 26.67 = 194.47 0.242x213156Assuming optimal velocity of 15m/secGas flow area = flue gas flow * (Tgav+273) 3600* 273 * 1.218*15= 272160 x ( 215.65 +273)/(3600x273x1.218x15) = 7.4065 m²Assume 63.5 x 2.06 mm tube sizeNumber of tubes = GFA * 4/( * d²)= 7.4065 x4 / x 0.05938² = 2675Arrangement 88 x 30 =2640 tubes (velocity works out to be 15.198m/s)Height of air heater tube may be considered as h‘h = 5.5 m ( Intermediate support to be considered)Air flow area = h(num. of tubes in air path * pitch – num. of tubes in air path * od)= 5.5 (88x 0.080 – 88 x0.0635) = 7.986 m²Air velocity = ma * (Taav +273) 3600 *273 * AFA* density of air= 213156 x( 110.335 +273)/ (3600 x 273x 7.986 x 1.293)= 8.051 m/sProperties of gas at mean gas temperature,Specific heat = 0.2856 kcal/kg°CTh.conductivity = 0.0383 W/m°K =0.0329 kcal/mhr°CViscosity = 2.39 x 10-5 kg/ msecDensity of gas = 0.6804 kg/m3
  • Steam generator Ganesh kumarProperties of air at mean air temperature,Specific heat = 0.241 kcal/kg°CTh.conductivity = 0.03248 W/m°K = 0.0279 kcal/mhr°CViscosity = 2.2216 x 10-5 kg/ msecDensity = 0.9208 kg/m3Gas side heat transfer coefficient,Nu = 0.023 Re0.8 Pr0.3Reynolds number = 15.198 x 0.05942 x 0.6804/2.39 x 10-5 =25711Prandtl number =0.2856 x 2.39 x 10-5 x3600/0.0329 =0.7469‘hg = 0.023 x 257110.8 x 0.74690.3 x 0.0329/0.05942= 39.36 kcal/hrm²°CAir side heat transfer coefficient,Nu = 0.3 Re0.6Reynolds number = 8.016 x 0.0635 x 0.9208 /2.2216 x 10-5 = 21097‘ha = 0.3 x 21097 0.6 x 0.0279/0.0635 = 51.81 kcal/hrm²°C1/U = do/dihg + 1/ha + do ln(do/di) 2Km1/U = 0.0635 + 1/51.81 + 0.0635 x Ln(0.0635/0.05942) 0.05942 x 39.36 2 x 49.14U = 21.46 kcal/hrm²°CConsidering an effectiveness factor 0.8The Overall heat transfer coefficient is 21.46x0.8 = 17.168 kcal/m²hr°CAssuming Counter flowLmtd = (160-26.67)- (271.3 – 194.47) Ln[(160.-26.67)/(271.3 –194.47)]= 102.49°As the flow is not perfectly counter flow, correction factor in lmtd of 0.95 has to bemade.Corrected Lmtd = 0.95x 102.49 = 97.36 °Heat transferred = U A Lmtd
  • Steam generator Ganesh kumarHeat transfer area arranged in two blocksA = 8.656 x 106/17.168 x 97.36Required heat transfer area = 5178.65 m².Provided heat transfer area in two blocks = 2x x0.0635x5.5x88x30 = 5793 m²Gas side Pressure drop,‘f = 0.184/Re0.2f = 0.184/ 257110.2= 0.02414Pressure drop = 0.02414x5.5/0.05942 x 15.198² x 0.68/(2x9.81)=17.89 mmWcFor two Passes = 17.89 x2 = 35.78 mmWcEntry and exit loss = 1.5 x 2 x 15.198² x 0.68/(2x9.81) = 24.016 mmWcFor two passes gas sideTotal pressure drop = 35.78 + 24.016 = 59.796 mmWCAir side Pressure drop,‘f = 1.632/Re0.15f = 1.632/ 210970.15= 0.3665Number deep = 30 nos/passPressure drop = 0.3665 x 30 x 2 x 8.051² x 0.9208/(2x9.81) =66.89 mmWC
  • Steam generator Ganesh kumar 8.0 DUST COLLECTOR8.1 INTRODUCTIONThe advent of scientific and technological revolution has brought tremendousbenefits to mankind and also given rise to many problems related to ourenvironment. Rapid industrialization as associated with scientific exploitation ofnatural resources and this has resulted in contamination of three major constituentsof our environment – air, water and soil.Combustion process is the main source of elements that pollute the atmosphere. A500MW thermal power plant produces 2000 tones of ash every day, it also producesabout 50 MT of sulphur di oxide (considering sulphur content in coal as 0.5%).8.2 EFFECTS OF AIR POLLUTIONThe consequences of air pollution have to be gauged from the factors like1.0 First the amount of substances polluting the atmosphere are eventually accumulating2.0 The distribution of pollutants is not uniform and in certain places, their concentration is in-admissibly high3.0 Very small concentrations of certain substances can be highly dangerous.4.0 The damages caused by air pollution is enormous. It involves losses in all the spheres of the national economy.PARTICULATE EMISSIONSmog is closely linked to air pollution. Many cities in the world, particularly inLondon catastrophe of 1952, roughly 4000 people are known to have lost their lifedue to smog. The phenomenon of smog is a mixture of smoke and fog, has beenrelated to temperature inversion, the heavier cold air saturated with industrial andtransport emissions remains near the grounds compressed under a dome off lightwarm air. Hence it is necessary for industry to let the particulate emission to amaximum height possible. Pollution control board framed minimum height requiredfor letting out the smoke in atmosphere.CARBON MONO OXIDEThe unburnt carbon mono oxide in the presence of oxygen react and form carbon dioxide. This carbon mono oxide loss reduce efficiency of power cycle and also itaffects human population causing branchits problem.CARBON DI OXIDEThe effects of air pollution cannot be completed without a reference to the greenhouse effect. The dramatic increase in the carbon di oxide level in today’satmosphere is 27% higher than the year 1850. Scientists believe that the growingburden of carbon di oxide and other gases are the causes for change in earth’s
  • Steam generator Ganesh kumarclimate. ( Recent ELINO effect increase in pacific temperature cause shift inmonsoons) Carbon di oxide allows most of the solar radiation to penetrate theatmosphere but prevents part of the heat re-radiated by land and water bodies fromescaping into space. It is estimated that the earth’s mean temperature could rise 1.5to 4.5°C by the middle of next century if green house gases continue to increase atcurrent rate. This impact could be even greater at the poles. The rise in globaltemperature could result in shift in rain fall patterns, melting of the polar ice caps andraising of ocean levels.The wide ranging effects of atmospheric pollution under lines the need for controllingthe industrial and other emissions and heaving clean air in our atmosphere. Thecost of providing clean air may be more, but the cost of polluted air is much more.8.3 AIR QUALITY STANDARDSApart from particulate matter the other major air pollutants are gaseous pollutants. Ingeneral, gaseous pollution is classified into two categories, namely inorganic andorganic vapours. In the former class there are sulphur di oxide, nitrogen oxides,hydrogen sulphite, hydrogen chloride, hydrogen fluride, carbon mono oxide andammonia. In the latter are the hydrocarbons, mercaptans, keytones and esters.As per recent BIS standard the following limits have been recommended. Theambient air standards as propounded by EPA(Environmental protection agency) ofUSA and adopted by BIS are shown in the following table. INDUSTRIAL RESIDENCIAL SENSITIVE ENVIRONMENT ENVIRONMENT AREA POLLUTANTS PPM PPM PPM PARTICULATE 350 200 100 SULPHUR DI OXIDE 120 80 30 CARBON MONO OXIDE 5000 2000 1000 NITROGEN OXIDE 80 30 128.4 AIR POLLUTION CONTROL DEVICESAir pollution control equipment is broadly classified into two categories- one controlsparticulate matter and the other controls gaseous emissions. Various considerationare required to be made while selecting a control device for an industry. Mostlyindustries face problem with particulate matter collection and gaseous emission hasto be controlled by proper pre treatment and maintaining certain design parameterssuch as dew point. For particulate matters, some of the major factors are collectionefficiency of the device, initial cost operating and maintenance cost, space required
  • Steam generator Ganesh kumararrangement, material of construction. Important factors considered in thisconnection are as follows.1.0 Particulate characteristics such as particle size range, shape, density and physical, chemical properties such as agglomeration tendencies, corrosiveness, hydroscopic tendencies, stickiness, inflammability, toxicity, electrical conductivity and others.2.0 Gas characteristics such as temperature, pressure, humidity density, dew points of condensable components viscosity, electrical conductivity etc.,3.0 Process factors such as volumetric gas emission rate, particulate concentration, variability of material flow rates, collection efficiency allowable pressure drop, product quality requirement etc.4.0 Operational factors such as floor space, pressure, temperature, corrosion etc.In boiler applications bag filter, ESP, Wet scrubber and centrifugal mechanical dustcollector are quite widely used for fly ash removal. So let us try to understand howthese systems are operating and the merits and demerits of these systems. COLLECTION PRESSURE POWER EFFICIENCY IN DROP IN CONSUMED TYPE PERCENTAGE MM WC BHP PER100 WEIGHT M^3/MIN ELECTROSTATIC 80 TO 99.5% 2 TO 20 0.2 TO 0.5 PRECIPITATOR FABRIC FILTER 95 TO 99.5% 50 TO 150 0.8 TO 2.6 CYCLONE 50 T0 95% 25 TO 100 0.4 TO 2.2 SEPARATOR WET SCRUBBER 75 TO 99% 50 TO 400 0.8 TO 6CENTRIFUGAL CYCLONE DUST COLLECTORIt is the cheapest and most effective dust collector. In the multi cyclone device thereis a nest of cyclones in parallel having one header and dust hopper. Here thecentrifugal action removes the particulate from the gas stream. The larger theparticles, the more easily or they likely to be collected due to their proportionatehigher ratio of centrifugal force imparted to the particle of given mass. Dustcollection depends on the radius through which the gas is turned, the smaller theradius the higher the centrifugal force. Centrifugal action throw the heavy particles tothe side of the cyclone, where the dust can slide down to a hopper at the bottom.The cyclone works with two vortex. The cyclone possesses a high separation factor
  • Steam generator Ganesh kumargiven as the ratio of the radial velocity in the cyclones to the stokes velocity in thesetting chamber.Large separation factor requires high tangetical velocities and small diameters, bothof which result in a large pressure drop. In India the industrial application of bothlarge and small diameter cyclones are very common. The multi cyclone dustcollector can effectively remove dust particle of diameter above 50 microns andhaving a particle density of 600 kg/m^3. So it is not always possible to meet thestringent emission levels to PCB norms, it becomes essential to provide either a baghouse or ESP.BAG HOUSEBasically, bag house is a large metal box divided into two chambers of plenums, onefor dirty air and one for clean air. Rows of cloth bags form a partion or interfacebetween the plenums. A polluted gas stream is ducted into the dirty air plenum, andis exhausted into the atmosphere through a stack. Almost 100% of the particulatematter in the process effluent can be filtered out by bags if the system is designed,operated and maintained properly but bag house will not be suitable for fuels likebagasse where burning particles may enter into bag house causing burning of bags.When a new bag house is first started up with factory fresh bags, some stackemissions are usually visible. This is because the filtering medium (which is thebags made of fabric called Ryton and bag size is of 5m length and 0.3m diameter) isporous and allows a certain amount of very fine particulate matter to pass throughthe interstices between the fibers. After a short period of operation, a dust cakebuilds upon the surface of the bags and become actual filtering medium. The bags,in effect act primarily as a matrix to support the dust cake.The dust cake is desirable only upto a point, when that point is reached the bag mustbe cleaned properly other wise the pressure drop through the filter system willcontinue to increase. At high pressure drops, particles of dirt can be forced into thebag filters causing bags to become blinded. When this happens, air flow is restrictedand the bags may have to be replaced or removed and cleaned.Practical guides for proper maintenance of bag filter.Maintenance personnel must learn to recognize the cause of the difficulty, and tomend it either by in planed action or by contact with the manufacture. High pressuredrop across the system is a symptom for which there are many possible causes,example, 1. Difficulties with the bag cleaning mechanism 2. Low compressed air3. Loose bag tension are usually available the reasons for high pressure drop andcorrective action to be taken appropriately. I have furnished below some of thecheck points to be taken care while doing routine bag house inspection.----------------------------------------------------------------------------------------------------------------COMPONENTS CHECKLIST----------------------------------------------------------------------------------------------------------------Bags Worn, damaged bags, condensation on bags,
  • Steam generator Ganesh kumar improper bag tension, loose, damaged or improper bag connections.Differencial pressure Steadiness of pressure drop(should be read daily)Dust removal system Worn bearings, loose mountings, deformed parts, worn or loose drive mechanism, proper lubrication.Bag house structure Loose bolts, cracks in welds, corrosion, holes(housing, hopper)Solenoids, pulsing valves(reverse Proper operationPulse system)Compressed air system Proper operation, lubrication of compressor,(reverse pulse, plenum pulse) leaks in header, piping. Proper mounting.Dampers valves Proper operation and synchronization, leaking(Reverse flow, plenum pulse) cylinders, bad air connections, proper lubrication, damaged.Doors Worn, loose, damaged or missing seals, proper tight closing.----------------------------------------------------------------------------------------------------------------ELECTROSTATIC PRECIPITATORAn electrostatic precipitator (ESP) consists very basically of a precipitator chamberand an electrical unit. The precipitator chamber includes discharge and collectionelectrodes, an electronic cleaning system, gas distribution devices, and a precipitatorshell and hopper. The electrical unit is made up of a power supply, high voltagetransformers, rectifiers and precipitator bus sections.ESP is physical process by which a particulate suspended in a gas stream ischarged electrically and then, in the influence of an electrical field, is separated andremoved from gas stream. The system that does this consists of positively chargedcollecting plate in close proximity to a negative charged emitting electrode. A highvoltage charge is imposed on the electrode and the grounded collection surface.The dust particles pass between the electrode and grounded collection surface. Thedust particles pass between the electrodes, where negatively charged and divertedto the positively charged collecting plates.Periodically, the collected particles must be removed from the collecting surface.This is done by vibrating (usually by rapping) and or water washing the surface of thecollection plates to dislodge the dust. This dislodged dust drops into a dust removalsystem and is collected for disposal.Practical problems faced in ESP
  • Steam generator Ganesh kumarDischarge electrode failure is the primary cause of operational breakdown. After thisare rapper malfunctions, insulator failure, shorts caused by dust build up, hopperplugging and transformer- rectifier failures. Most of these problems occur whenpreventive measures are not used. For example discharge electrode failures can bereduced if the hoppers are properly discharged and cleaned to prevent grounding outand burning off the discharge electrodes.Fly ash build up on the collecting plates should normally be about 1/8 to ¼ inch thick.If the build up exceeds this thickness, the intensity of the plate rappers should beincreased. If the collecting plates are clean, this may be an indication of high gasvelocity or low operating voltage.Collecting plates should be checked for proper alignment and spacing. Hangers andspacers at the top and bottom should be adjusted so that they do not bind the platesor prevent proper rapping. It is necessary to check for corrosion.Hoppers should be checked periodically to be sure they empty properly and toinspect for corrosion, which is likely to be most severe at points where at pointswhere dust builds up. The heating system and insulation on the hoppers arechecked to prevent condensation.Insulator compartments and housings must be checked frequently, leakage ofcorrosive gases from the precipitators into this area can cause dirt deposit that resultin breakdown of electrical insulators. 9.0 WATER CHEMISTRY9.1 INTRODUCTIONAll system obeys law of conservation, energy can be neither created nor destroyedbut energy can be converted from one form to another. In the system of powercycle, water is basically most important raw material which acts as energytransferring element. H2O↓(w) + Heat à H2O ↑(s)As the raw water contains dissolved solids, suspended matters like mud, clay,calcium and magnesium salts free minerals and dissolved gases like oxygen andcarbon di oxide, hydrogen sulphite etc., it is required to be treated before it is fed intothe boiler to prevent boiler damages and to get required steam purity.Most of the boiler shut downs are arise due to uncontrolled water treatment, anunnoticed boiler shut down runs in loss of money to company. Therefore it isnecessary to have a controlled water quality for the boiler.Although the manufacturers supplies and recommends water quality, it is upto theplant personnel to meet the required water quality. It is not possible to suggest onetype of water treatment for all the industry as composition of different water supply
  • Steam generator Ganesh kumarvary greatly from place to place and from time to time. Therefore selecting watertreatment plant starts with analyzing the raw water. The water should be analyzedperiodically, if the supply or availability is seasonal, the treatment plant selected isbased on the worst analysis made by the study conducted in series for minimum ofone year. In high pressure boiler plant, engineer should not hesitate to stop theboiler if water treatment plant fails or water quality is poor.9.2 NAMES OF WATER FLOWING IN THE POWER PLANT CYCLEa) Raw water - Water as received from the source of supplyb) Treated water - Any treatment added.c) Softened water - Hardness removed waterd) Condensate - Condensed steame) Demineralised water or DM water - All ionisable solids removed by ION – Exchange treatmentf) Deaerated water - Oxygen content removed in deaeratorg) Feed water - Any mixture of the above to pass through feed pump.h) Boiler water - Water present in a boiler when steaming.i) Make up water - Added to make up for losses.SOURCES OF WATERSurface water: Ponds, Lakes, Reservoirs, Streams and rivers are the sources ofsurface water.Ground water: Wells, Mines and Springs are sources of ground water.9.3 MAJOR IMPURITIES IN WATERIn water the major dissolved impurities are bi-carbonate, sulphates, chlorides,nitrates of calcium and magnesium. Minerals have two components Radial or acidiccomponent and metallic component. In the water analysis as there are various solidsdissolved in various percentage, the concentration of dissolved salts impurities arereported in CaCo3 terms only.Temporary hardness are due to calcium, magnesium bi carbonate salts. These doesnot cause hard scales and these can be removed by boiling the water. Calcium andmagnesium bi carbonates are broken down by moderate heating(100°C) intorelatively insoluble mono carbonate and carbon di oxide as given in the reaction.
  • Steam generator Ganesh kumarThe less soluble salt like calcium carbonate precipitate and settle in the boiler waterwhich can be removed by blow down.Permanent hardness are due to sulphates, phosphates of calcium and magnesiumsalts. These form sticky hard scales and the amount of dissolved salts in water isthe measure of hardness. Among all, calcium is the principal scale forming agentand particularly calcium sulphate. Solubility of Ca SO4 decreases with increase inwater temperature.9.4 EFFECTS OF VARIOUS IMPURITIES IN BOILER WATERa) AcidityBoiler water should be always alkaline in nature. The alkalinity is to be maintainedfor the following reasons.1.0 To keep magnetite layer (Iron oxide) uniform, which protects the metal and minimize corrosion and to promote formation of magnetite layer as and when impaired.2.0 To keep sludge in floating condition.3.0 To keep silica in soluble condition preventing precipitation and carry over.4.0 To neutralize any acid generated in the system.Highly acidic water dissolves metal and mild acidic water hastens pitting and oxygencorrosion.b) AlkalinityAlkalinity should also be maintained as per recommended value based on drumpressure ratings. Higher values result in foaming leading to carry over and alsocaustic embrittlement. In highly concentrated alkali the protective film of magnetic isdissolved forming a mixture of ferrite and hypo ferrite irons. But at low concentrationthe layer of magnetite is porous. In the presence of pororus deposit hydroxyl ionstend to concentrate between metal and deposit leading to a typical corrosion patterncalled caustic gauging.c) HardnessMost of the calcium and magnesium salts present in the water contribute hardness.Ex: calcium carbonate, magnesium silicate etc., there salts form sludge and calciumsulphate, calcium silicate etc., form scales on the evaporating surface. Hardness ismeasured in terms of mg/lit or ppm as CaCo3.Alkaline hardnessHardness caused by the presence of bicarbonate, carbonate, hydroxide of calciumand magnesium (temporary hardness). Alkalinity is the concentration of alkalinesalts. Bicarbonate alkalinity due to bicarbonate salts, caustic alkalinity due tohydroxide salts (P or phenophthaline alkalinity).
  • Steam generator Ganesh kumarM- alkalinity is the total alkalinity which is a combined hardness of bicarbonate andhydroxide salts. Hence in an analysis P never exceeds M alkalinity.(if P= M thenalkalinity is due to hydroxide, if P=0 then alkalinity is due to bicarbonate andcarbonate)Total hardness(H)Total hardness is the concentration of calcium and magnesium salts. If totalalkalinity (M) exceeds total hardness(H) then the hardness due to bicarbonate,carbonate and hydroxide only the value(M-H gives sodium bicarbonate present inwater)Total alkalinity (M) less than total hardness (H) then the value (H – M then givescarbonate hardness) ie. Remaining is non alkaline hardness.(chlorides, nitrates,sulphate)If total hardness is equal to total alkalinity the only carbonate hardness is present.d) OxygenWater at room temperature always contain oxygen in dissolved form of air. Oxygenpromotes corrosion. The rate of wastage and thinning of tubes in highly acidic wateris enhanced heavily by the presence of oxygen. Oxygen is released from water onheating.e) Total dissolved solids (TDS) mg/lit of CaCo3All the salts dissolved in the water (sodium, calcium, magnesium ) are togetheraccounted against total dissolved solids. The effects of the total dissolved solids infeed water depends on the type of salt and the type of boiler. The effect are alsocomplex due to mutual action between the various salts. Generally, presence ofchlorides, iron salts accelerate corrosion. Presence of other salts increase scaleformation, priming (sweeping action of suspended solids towards steam outlet) andfoaming ( Persistent bubble formation). Priming and foaming cause carryover ofsolids and heavy fluctuation in water level. A high TDS in the boiler water will alsoincrease steam contamination. 0.1 mg/lit is the limit to TDS in steam used for TGsets. Approximately value of TDS in mg/lit can be obtained by measuring theelectrical conductivity in micro siemens/cm. It is found that TDS will beapproximately half that of conductivity.ConductivityPure water is a poor conductor of electricity, but water contain ionized impuritiessuch as salts, acids, alkali is conductive and it can be used to define purity of water.The conductivity produced by the presence of given concentration of impuritydepends upon nature of impurity and temperature of water. It is measured in termsof micro siemens per cm because the numerical value of the same is foundproportional to the concentration of dissolved solids in mg/lit or ppm.f) Turbidity (NTU)
  • Steam generator Ganesh kumarTurbidity is an indicator of undissolved solids. Turbid water may induce erosion.The undissolved solids also impair circulation of water on the heating surface whichin turn can reduce the cooling action of the water. The undissolved solids causeconsequent over heating of the surface. In practice suspended undissolved solidsare expressed in ppm and micro organic undissolved solids are expressed in NTU.g) Dissolved carbon di oxide.Dissolved carbon di oxide appears in the water in the form of carbonic acid. Thebicarbonate salts present in the water are often considered as bound carbon di oxidein the water. Carbon di oxide being a gas separated out of water on heating. Thecarbon di oxide released from water gets carried away along with steam and whensteam condenses the carbon di oxide redissolves in the condensate formingcarbonic acid. This acid corrodes pipe lines and other user equipment.h) PH value.It is not a constituent or impurity. The concentration of hydrogen ions produced byvarious chemicals dissolved in water. + _ H2O à H + OHIn pure water hydrogen ion concentration will be 0.0000001 gm/lit. of water or 10-7 ofhydrogen ions per liter. This is balanced by hydroxyl ions(same amount). If HCl orany acids dissolved in water it gives additional hydrogen ions causing increase inhydrogen ions 10-7 à 10-6 à 10-5 ie water becomes acidic. Hence for PH value of7 water is neutral and PH less than 7 is acidic greater than 7 in alkaline. Thecorrosion rate is foud to be very slow when PH range is 9.5 to 10.5 in boiler water.The protective magnetite (a corrosion product) is not distubed under the aboveregimes.Tube failure.1) Hydrogen damage/ EmbrittlementHydrogen damage is mainly due to low PH environment. Presence of carbonic acid,improper deaeration, condenser leakage. Hydrogen produced in the boiler waterdiffuses through underlying metal producing decarburisation and inter granular microfissuring of the structure. Brittle fracture occurs in the tubes, the failure will be ofthick edged fracture.2) Ductile gauging / pitting attack.Ductile corrosion is more probable when boiler water contains highly soluble alkalinetreatment chemicals such as sodium hydroxide. PH VS PHOSPHATE curve has tobe followed to limit free hydroxide. In this the metal is gradually eaten away andwhen tube is insufficient to with stand the pressure, tube fails by pinhole or burst. Itis predominantly seen in low sloped tubes such as roof tube, bed coils, idlecompartment with zero flow are attacked by this.
  • Steam generator Ganesh kumarj) IronIron will form sludge like layer and prevent heat transfer. Iron content in water will beremoved by oxidizing ferrous iron to ferric iron by air and catalyst, then filtered and insome cases special iron removing technique/ equipment will be employed.k) Organic matterThe soluble organic matter present in natural water are either fulvic or humic acidmaterial ( Fulvic fairly recent organic decay, Humic – old organic decay). Mainsource of organic matter in the boiler water is oil, grease with in the boiler, pipeline,pumps etc. It is difficult to predict the effects of organic matter in the water as itdepends on their nature. Oil and grease induce priming and foaming consequentlywater level fluctuations and carry over.Methods to get rid of the offending impurities in the water and to ensure thequality requirement is given below.Impurities Action to be taken Method EquipmentHardness Softening Base exchange SoftnerAcidity/PH Dosing Phosphate & Chemical Caustic dosing doserOxygen Scavenger/deaeration Sodium sulphite Deaearator Hydrazine chemical Deaeration doserTDS Reduce Blow down DM plant DemineralizeTurbidity Remove Filteration Clarifier/ Sand filterOrganics Preventing Boilout FilterationSilica Reduce Demineralize DM plant9.5 NEED FOR WATER TREATMENT
  • Steam generator Ganesh kumarHigh pressure boilers are normally designed close to the limiting conditions of heattransfer, tube metal temperature, circulation etc., to make the units compact andeconomical. Modern steam turbine rated for high capacities call for stringent steamquality to avoid damages. Without strict control over the impurities in steam cause,deposits over turbine blades and nozzles reducing the output. Thus successfuloperation of high pressure boiler and turbine unit requires a through understandingof all aspects of water treatment. In short the following are the reasons for need oftreatment.1.0 In an effort to optimize capacity and efficiency, deposit of any sort on tubes arevirtually intolerable.2.0 Optimization of tubes thickness can not even a moderate level of corrosion tgotake place.3.0 Blow down requirements are being cut to have energy conservation.4.0 High steam purity requirement calls stringent quality. Foreign particles like oil,organic matter, iron and copper ions contaminate the purity.TDS in steam = ( TDS in boiler water/ sodium in boiler water) x sodium in steamSteam purity is determined either by conductivity method or sodium method.9.6 EXTERNAL WATER TREATMENTCHLORINATION AND CLARIFICATIONRemoval of suspended solids or flocullants can be achieved by sedimentationcoagulants or salts of aluminium or iron added. Salts react with alkalinity to produceprecipitate that attract and absorb fine suspended solid and organic matter to formlarge floc particle which settle easily. This process is clarification and done inclarifier.Filter alumAl 2(SO4)3 + 3 Ca(HCO3)2 à 3 CaSO4 + 2 Al(OH)3 + 6 CO2↑Fe2(SO4)3 + 3 Ca(HCO3)2 à 3 CaSO4 + 2 Fe(OH)3 + 6 CO2↑Coagulation is the process by which the positively charged co-agulants attract thenegatively charged fine suspended matter and repulsive force is reduced. Foreffective coagulation temperature and PH conditions to be maintained properly.Floculation is the aggregation of particles under the influence of agitation. Chlorineaddition to destroy organic matter because organic matter tends to keep some ironand manganese in solution and aeration is preceded in the case of iron andmanganese presence. Aeration leads to oxidation of iron and manganese andremoval of carbon di oxide.FILTERATION
  • Steam generator Ganesh kumarRemoval of coagulant is carried by fiteration.Two type of filteration are01. Gravity filter.02. Pressurized filter.Filter bed are of stone gravel or coarse anthracite. Anthracite is non silcieous anddoes not add silica to water. Gravity filter are open type RCC tank and treatment iscarried 5 to 6 meter per hour and used for treatment of high amount of water.ION EXCHANGEDemineralization use of both cation and anion exchange for removal of dissolvedsolids. Ion exchange resins are hard spherical beads of 0.3 mm to 1.2mm diameter.It swells when it condact with water. Resin is to be kept moist and supplied in wetform. When resin gets dried the resin will be cracked and lose its ion exchangeproperties. Hence storing for long period is not advisable. In ion exchangeSAC,WAC,SBA, WBA,MB and degasser are arrangement in different combination toget the desired water quality based on the raw water analysis.Cat ion exchange and its regeneration.Ca HCO3 HCO3 Ca CaMg NO3 + HR à HNO3 + Mg R + HCl à HR + Mg ClNa SO4 H2SO4 Na NaAnion exchange and its regenerationHCO3 CO3 CO3HNO3 + ROH à R NO3 + NaOH à ROH + Na NO3H2SO4 SO4 SO4REVERSE OSMOSISRecent times due to the fast industrialization all natural resources are more pollutedwhich results in water level TDS increase to 1000 and above ppm of CaCO3. Touse thes high TDS water for boiler, TDS level has to be brought down and more overchemically treating this high TDS water will be highly uneconomical. Osmosis is thetechnique where in two solutions of low and high TDS water are separated by semipermiable membrane results movement of ions from high concentration to lowconcentration zone. Reverse osmosis is the reversal of osmosis process byapplying external force where in movement of ions will be from low concentration tohigh concentration zone.9.7 INTERNAL WATER TREATMENTLOW PRESSURE DOSING
  • Steam generator Ganesh kumarAfter deaeration, deaerated water gets collected in deaerated storage tank. Thisdeaerated water contains traces of oxygen due to the in complete mechanical/thermal deaeration. These traces of oxygen is scavenged by chemical deaeration.Chemicals like sodium sulphite or hydrazine is dosed to the suction header offeredpump by LP dosing pump. Residual hydrazine of 0.02 ppm has to be maintained infeed water.Hydrazine reactionN2H4 + O2 à N2 ↑ + H2OOne ppm of hydrazine react with one ppm of oxygenSodium sulphite reaction2Na2SO3 + O2 à 2Na2SO4Na2SO3 + H2O + HEAT à 2NaOH + SO2 at 900 to 1000 psiHence it is not recommended in high pressure boilers. It also can not be used withco-ordinated phosphate treatment since it affects balance of sodium and phosphate.Ex.02. Calculate the quantity of hydrazine required for the steam generator ofcapacity 65000 kg/hr. after deaerator. The oxygen percentage after deaerator is 0.02ppm and the desired level of oxygen in feed water is 0.007 ppm.The oxygen to be scavenged = 0.02 –0.007 = 0.013 ppmMinimum hydrazine required = 0.013 ppmPractically 2 times of required hydrazine will be dosed hence = 0.026 ppmResidual hydrazine to be present in water = 0.02 0 ppmTotal hydrazine required = 0.046 ppm or mg/kgHydrazine for 55.6 kg/sec of water = 0.046 x 65000 = 2990 mg/hr at 100% conc.Hydrazine available is at 60% conc. Therefore the required hydrazine of commercialgrade = 4983 mg/hr.HIGH PRESSURE DOSINGScale formation is limited by converting hardness salts to a free flowing sludge. Hpdosing is done by carbonate control or phosphate control by addition of sodium
  • Steam generator Ganesh kumarcarbonate or tri sodium phosphate respectively. Other chemicals such as sodiumhydroxide or calcium oxy phosphate can also be added. Best result of calciumcarbonate , magnesium hydroxide, calcium silicate, magnesium silicate will beavailable as free flowing sludge where caustic alkalinity is 10 to 15% of the dissolvedsolids. OH alkalinity leads to prevention of magnesium phosphate as bad scale byconverting magnesium to magnesium hydroxide. If there is no effective means ofblow down is available it is better not to concentrate on internal treatment. Trisodium phosphate is done in such a way that boiler blow down contain residualphosphate(PO4) of 7 ppm.( I.e., 4 ppm of Na3PO4.12H2O contain 1PPM PO4, hence28PPM tri sodium phosphate required, 0.028 kg tri sodium phosphate per 1000 kgof water blow down.)Ex.03. Calculate the tri sodium phosphate quantity required for the above specifiedboiler in example 01. Consider blow down of 1%.Blow down quantity = 0.01 x 65000 = 650 kg/hr0.028 kg of tri sodium phosphate required for 1000 kg of blow downtherefore required tri sodium phosphate = 0.028 x 650/1000 = 0.0182 kg/hr.BLOWDOWNIn spite of all treatment, boiler water will contain dissolved solids. In order to keepthe level of total dissolved solids in boiler within the limits and to remove any sludge,loose scales and corrosion products, a certain quantity of boiler water should beregularly drained. This process is known as blow down. The blow down can beintermittent say once a shift or continuous. The quantity of the water to be blowdown will depend on the dissolved solids entering the boiler through the feed waterand the maximum tolerable levels of these salts in the boiler water. Whiledetermining the dissolved solids content in the feed water it is necessary to takesinto account not only the original dissolved solids but also the dissolved solids addedin the form of dosing.Continuous blow downContinuous blow down is a continuous removal of boiler water controlled by aspecially designed adjustable valve or by an orifice plate. The installation of heatrecovery equipment in continuous system may be economically justified. Suspendedsolids may block or erode the adjustable valve or orifice plate and continuous blowdown is there fore limited to the control of dissolved solids. Additional manual blowdown is necessary to control suspended solids and prevent the build up of sludge.
  • Steam generator Ganesh kumarWhen boiler water from high pressure is suddenly reduced to low pressure blowdown water, water enthalpy will drop and steaming will occur. In utility boilersrecovering this blow down flash steam proves economical. The quantity of flashsteam generated is the product of ratio between difference in boiler water enthalpyand blow down water enthalpy to latent heat of blow down water and blow downquantity.Intermittent blow downIntermittent blow down may be effected by specially designed values either operatedmanually or automatically controlled by timers, feed water flow and conductivity. Itshould be noted that the most effective intermittent blow down is achieved byfrequent full open operation for short intervals rather than small extended infrequentoperation. It is not practicable to recover waste heat from infrequent intermittentblow down.Ex.04 Find the blow down quantity and flash steam produced for a 160tph boileroperating at a pressure of 120 bar(a) and the permissible silica content in feed waterand boiler water be 0.01ppm and 0.8ppm respectively. The flash steam blow downvessel is maintained at 4.5 bar(a).Blow down quantityContinuous blow down = S x 100 ( C –S) = 0.01 x 100/(0.8 –0.01) = 1.26 %Blow down quantity = 0.0126 x 160 = 2.025 tphFlash steam calculationEnthalpy of boiler water leaving the drum = 1491.2 kj/kgAt 120 bar(a)Enthalpy of water in blow down tank at = 623.14 kj/kg4.5 bar(a)Enthalpy of flashed steam in blow down = 2743.55 kj/kgTank at 4.5 bar(a)Latent heat of flashed steam at 4.5 bar(a) = 2120.41 kj/kgFlash steam percentage = enthalpy of boiler water – enthalpy of blow down water Latent heat of blow down water = (1491.2 – 623.14)x 100/2120.41 = 40.94%
  • Steam generator Ganesh kumarFlash steam quantity = 2.025 x 0.4094 =0.829 tph. 10.0 BOILER CONTROLS10.1 INTRODUCTIONIn modern industries automation is carried by means of micro processor chips (µp)made of silicon, the semi conductor which controls all the functions. The basicbuilding blocks of integrated chips are of diodes, gate circuit. Today’s boiler aredesigned with high heat flux and very high capacity, safety and automation isbecoming prime important. Automation of boiler is done primarily by relay systemand microprocessor system. Relay technology is the oldest one, even now adaptedfor small industrial boilers for the basic boiler inter locks like trips and alarms. Mostof the automation is carried by microprocessor based control.10.2 CONTROL PHILOSOPHYThe control in boiler is required for the following critical items1.0 Boiler water level has to be maintained always in the system in order to prevent starvation boiler and tube failure due to over heating.2.0 The steam temperature at the outlet of superheater has to be controlled within
  • Steam generator Ganesh kumar the limits other wise higher metal temperature leads metallurgical problem and also it affects turbine and condenser performance due to volumetric change and dryness fraction shift.3.0 The draft system has to be controlled according to the design of the boiler, i.e. if the boiler furnace designed for negative pressure and if subjected to positive pressure due failure of draft equipment back firing will take place, Which is hazardous to boiler and the people.4.0 In the view of boiler performance, the combustion of fuel takes place completely with minimum excess air. Hence air flow control with respect to fuel has to be done and fuel flow with respect to steam demand has to be achieved.5.0 Considering acid dew point corrosion to metals, the metal temperature at the back end of economiser tubes or air heater tubes has to be achieved. For this cooling medium temperature inlet temperature at exhaust gas exit temperature above the dew point has to be maintained.6.0 If there is any tripping of rotating equipment like fans, pumps due failure of the equipment or power the boiler has to be tripped automatically. For this interlock arrangement has to be made. Further to this, if equipment trips due to its own defect the stand by equipment has to be started automatically.In the above point numbers one to five involves variables with complex non linearand linear relation ship this is grouped as PID controls and point number six isinterlocks with change of two states (on or off) is called on/off control.InstrumentationIn boiler controls, as explained earlier the instrument can be broadly classified intofield instruments and panel instruments. The field instrument comprises mainly ofSensors, Transmitters, Actuators etc.,10.7 FIELD INSTRUMENTSSensorsSensors which has to sense the mechanical process parameters like pressure,temperature, flow etc., In boiler application the following are the major processparameters to be converted into electrical signal for measuring and controllingpurposes.1.0 Flow2.0 Pressure3.0 Temperature4.0 Level5.0 Flame6.0 Position of actuators
  • Steam generator Ganesh kumarTransmittersThe sensed or measured process parameter variable has to be transmitted from thefield to control panel. For this transmitters converts the process parameter intoelectrical current of 4 to 20 mA which will be transmitted through 2 core, coppercables.,1.0 FLOWThe flow measurement is made based on several principle, some of the commonlyused instruments are, differential pressure flowmeter, area flowmeter, positivedisplacement flowmeter, eletro magnetic flowmeter, ultrasonic flowmeter. The typeof flow meter selection is based on accuracy, magnitude of flow, type of fluid beingused like liquid or gas, properties of fluid like temperature, pressure, viscosity etc.,a Differential pressure flowmeterThis is also called constriction flowmeter. The flow sensors either orifice or nozzle orventuri etc., this type of flowmeters are widely used in industries for variousapplication.2.0 PRESSUREPressure is applied to the two sides of the twin diaphragm capsule. The pressure istransmitted from the twin diaphragm to the sensing diaphragm through the sealingliquid. Two fixed electrodes are placed symetrically on the left and right side of theinsulator and electrical capacitance is formed between these electrodes and sensingdiaphragm.3.0 TEMPERATUREa) Sebecks effectWhen junctions formed between two dissimilar metals kept in two differenttemperature, an electromotive force is induced in the system. It is called Sebeckseffect..4.0 LEVELIn boiler drum level maintenance is one of the critical parameter for safe boileroperation.5.0 FLAMEFlame or light is detected by photo electric cells. Presense of UV or IR rays isdetected by the Photo cells and produce an emf. This signal is used in burnerscanner and fire alarm application.6.0 POSITION FEED BACK
  • Steam generator Ganesh kumarThe position of an actuator has to be constantly feed backed to the remote center.In order to have a control over the process parameters. The actuator movement canbe either of linear or circular.b) Inductive SensorsThe actuator movement is transmitted to the rotating shaft of the positioner. Thecam is attached to the shaft is scanned by an inductive non contact sensors. Theangle of rotation is then converted to an electrical signal, as shown in the diagram.The main advantage of this type of sensors is there is no frictional contact betweenrotating part and sensor. The sensor senses the varying gap due to the cammovement and produces varying current level by induction. Sensors are suppliedwith 24 V DC for generation of 4 to 20 mA.(e.g.) The control valve actuators used in flow control and pressure control uses thistype of inductive principle.10.8 PANEL INSTRUMENTSThe signal given from field instrument is processed in panel and necessary controlsignal will be generated in the panel and the same is send to field for control action.Due to the industrial development various control techniques and automation hasbeen adapted in industry. The following Instrumentation system hierarchy gives thebasic concept behind the same. Management Distribution Concentration Monitor and Operation Control and Safety
  • Steam generator Ganesh kumar10.3 DRUM LEVEL CONTROLF SteamIEL Drum LT PT FT TTD 4 to 20 mA signalPA Σ √NE PVL SP internal Σ LICI CV ΣN SVST FICRM PVE I/P √NTF FCVIE Air supply FTLD Drum Feed waterThree element drum level control loop have one or more drum level, feed water flowand steam flow transmitters for achieving level control. The controller receives inputfrom the above three main transmitters and depends on the level set point theprocessed output of the controller will actuate the control valve in feed line either toclose or open as desired by the controller.
  • Steam generator Ganesh kumar10.4 SUPER HEATER STEAM TEMPERATURE CONTROLFI super heater steamE TE thermocoupleL 4 to 20 ‘mVD signalPA Controller mA signalN TICE I/P Air supplyF SCVIEL Air supplyD Spray water line to DSH10.5 FURNACE DRAFT CONTROLFI Furnace pressureE PTL 4 to 20 ‘mAD signalPA Controller mA signalN PICEL I/P Air supplyF Power cyliderI
  • Steam generator Ganesh kumarEL Air supplyD ID fan damperIn super heater steam temperature control the measured variable steam temperatureis compared with set point or desired temperature in the controller and accordinglythe spray control valve is regulated. Similar to the temperature control, in draftcontrol furnace pressure is measured by means of pressure transmitter andcontrolled by regulating the ID fan damper.