Boiler Book

August 7, 2017 | Author: arganesan | Category: Evaporation, Boiler, Combustion, Fluidization, Heat Capacity
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Steam generator

A.GANESH KUMAR DEUTSCHE BABCOCK, INDIA.

For internal circulation only. All rights reserved by author.

Ganesh kumar

Steam generator

Ganesh kumar

DEDICATED TO MY COLLEGE AND MY PROFESSORS.

Steam generator

Ganesh kumar

PREFACE Dear friends,

This book was prepared in view of giving assistance to design engineers entering into the boiler field and to plant engineers whom I have met always in desire to know the ABC of the boiler design and related calculations. I have made an attempt in bringing close relation of practical field design and theoretical syllabus of curriculum. Engineering students, who always wonder how the theory 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 of working fluid in the steam plant cycle. This will be the basic foundation for development of boiler science. Secondly the main components of steam generator and its design. Also you can find various useful data for ready reference at the end of this book.

(A.GANESH KUMAR)

Steam generator

Ganesh kumar

CONTENTS •

PREFACE……………………………………………………………………….

1.0 TYPES OF STEAM GENERATORS 1.1 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 AIR 2.1 2.2 2.3 2.4 2.5 2.6

Introduction…………………………………………………………………… Definitions for some commonly used terms……………………………… Steam…………………………………………………………………………. Fuel…………………………………………………………………………….. Gas and air……………………………………………………………………. Some commonly used dimensionless numbers and their significance….

3.0 FURNACE 3.1 3.2 3.3 3.4 3.5 3.6 3.7 3.8 3.9 3.10

Introduction…………………………………………………………………… Effect of fuel on furnace…………………………………………………….. Forced or Natural Circulation………………………………………………. Heat flux to furnace walls…………………………………………………... Points to be noted while designing furnace……………………………… Classification of furnace……………………………………………………. Modes of heat transfer in furnace………………………………………… Heat transfer in furnace……………………………………………………. Furnace construction………………………………………………………. Practical guides for designing fluidized bed, conventional and oil/gas fired furnace…………………………………………………..

4.0 SUPERHEATER 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8

Introduction………………………………………………………………….. Effect of fuel on super heater design……………………………………… Points to be noted while designing super heater………………………… Classification of super heater………………………………………………. Designing a super heater…………………………………………………… Overall heat transfer across bank of tubes………………………………. Steam temperature control………………………………………………… Pressure drop………………………………………………………………..

Steam generator

Ganesh kumar

5.0 DRUMS 5.1 5.2 5.3 5.4 5.5 5.6

Intruction……………………………………………………………………. Optimal configuration of drums……………………………………………… Stubs and attachments in the steam drum/shell………………………….. Maximum permissible uncompensated opening in drum………………… Size of the drum……………………………………………………………… Drum internals………………………………………………………………..

6.0 EVAPORATOR AND ECONOMISER 6.1 Introduction………………………………………………………………………. 6.2 Difference between evaporator and economiser…………………………….. 6.3 Fin efficiency……………………………………………………………………… 7.0 AIRHEATER 7.1 7.2 7.3 7.4 7.5

Introduction………………………………………………………………………. Types of air heater………………………………………………………………. Advantages of air heater……………………………………………………….. Heat transfer in air heater……………………………………………………… Practical guide lines for designing airheater………………………………….

8.0 DUST COLLECTOR 8.1 8.2 8.3 8.4

Introduction………………………………………………………………………. Effects of air pollution…………………………………………………………… Air quality standards…………………………………………………………….. Air pollution control devices……………………………………………………. Centrifugal cyclone dust collector Bag filter Electro static precipitator

9.0 WATER CHEMISTRY 9.1 9.2 9.3 9.4 9.5 9.6 9.7 9.8

Introduction……………………………………………………………………. Names of water flowing in the power plant cycle………………………….. Major impurities in water…………………………………………………….. Effects of various impurities in boiler water……………………………….. Need for water treatment……………………………………………………. External water treatment…………………………………………………….. Internal water treatment……………………………………………………… Practical guides for selecting water treatment plant……………………….

10.0 BOILER CONTROLS 10.1 Introduction…………………………………………………………………… 10.2 Control philosophy…………………………………………………………… 10.3 Drum level control…………………………………………………………….

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Steam generator

Ganesh kumar

10.4 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 CHART APPENDIX2 : PSYCHROMETRY CHART APPENDIX3 : FUEL ANALYSIS APPENDIX4 : STEAM TABLES APPENDIX5 : POLLUTION NORMS IN VARIOUS INDIAN STATES APPENDIX6 : USEFUL DATAS APPENDIX7 : UNIT CONVERSION TABLE

Steam generator

Ganesh kumar

1.0 TYPES OF STEAM GENERATOR 1.1 INTRODUCTION Indian power demand is met mainly from thermal, hydro and nuclear power. Nonconventional energy power production is very much negligible. Out of the main power producing sources thermal plant produces 48215 MW (69%), hydro plant produces 19300 MW (28%), nuclear plant produces 2033 MW (3%) as on 31st March 1992. In the above power plants 72% of the generation is from thermal and nuclear, 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 resources suitable for power generation is in service. Except from gas turbines power the most of 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 to large electric power station utility boilers. In these large units pressure range from 100 kg/cm² to near critical pressures and steam is usually superheated to 550°C. In India BHARAT HEAVY ELECTRICALS LTD (BHEL) is the pioneer in developing the technology for combustion of high ash coal efficiently in atmospheric bubbling fluidized bed. From where lot of industries in boiler manufacturing starts. Only after the year 1990, India’s foreign policy was changed, various foreign steam generator manufacture entered into Indian power market bringing various configuration and competitiveness in the market.

1.2 HISTORY OF STEAM GENERATION AND USE The most common source of steam at the beginning of the 18th century was the shell boiler. Little more than a kettle filled with water and heated from the bottom. Olden day boiler construction were very much thicker shell plate and riveted constructions. These boilers utilize huge amount of steel for smaller capacity. Followed this shell and tube type boilers have been used and due to direct heating of the shell by flames leads severe explosion causing major damages to life and property. For safety need, after the Indian independence India framed Indian boiler regulations in 1950, similar to various other standards like ASME, BS, DIN, JIS followed world wide. Till date IBR 1950 is governing the manufacturing and operation of boilers with amendments then and there. Indian sugar industry uses very low pressure (15 kg/cm²) inefficient boilers during independence now developed to an operating pressure of 65 kg/cm² and more of combined cycle power plant. If we analysis most of the boilers erected in pre-independence period were imported boilers only and now 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 year 1972, India’s first nuclear power plant was commissioned at Tarapore. This plant was an pilot plant meant for both power and research work. This was made in collaboration with then soviet republic of Russia. Now India has its own nuclear technology 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 kumar

1.3 SHELL AND TUBE BOILER Steam 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 version of boilers used in industry where a large flue tube was separated by a fixed grate man 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, the practice was merely to increase the number of boilers. This procedure eventually proved to be uneconomical and larger maintenance. Afterwards, individual boilers were build larger and larger size, however the size became such that furnace floor area occupation was more. Therefore further research work have been developed in this area and technologies such as pulverized coal fired furnace, circulated fluidized bed furnace, pressurized circulated fluidized furnace (still under research stage) were developed. These modern technologies have higher heat transfer coefficient in furnace and allow higher volumetric combustion rates.

1.4 CONVENTIONAL GRATE TYPE BOILERS TECHNOLOGY This is the oldest method of firing fuel. Fuel will be spread over the grate, where the fuel is burnt. Fuel feeding will be done manually or mechanically to have a sustained flame. In this type burning will be done at higher excess air. Incoming air will be used for cooling the grate. Types of grate Common 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 construction and arrangement. However the combustion phenomenon remains same. Travelling grate The travelling type is a continuous grate which slowly convey the burning fuel through the furnace and discharge the ash to an ash pit. Grate speed is regulated by the amount of ash discharging to ash pit ( 0 to 7m/hr) Pulsating grate The pulsating grate is non- continuous grate. The grate surface extends from the rear of furnace to ash pit. Here the grate will be given a racking motion at pre determined frequency depending on the fuel/ash bed depth. Dumping grate Dumping grates are also a non-continuous type grate. The grate is split into longitudinal 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 or ash is dumped into the hopper in the following manner.

Steam generator

Ganesh kumar

Alternating fuel feeding is stopped and grate is tilted by lever arrangement, the actuation can be done either manually or pneumatic cylinder. In dumping grate the grate sections should be designed in such a way that, while dumping the ash part of grate surface not available for burning. In poorly designed dumping grate there may be steam pressure. Therefore while sizing grate sections care should be taken such that while dumping part of the grate, other fuel feeder and remaining sections should able to take the full load. Dumping grate is similar to fixed grates, it is best suitable for bagasse where the fuel is of low calorific value and having high moisture content. Therefore air alone can acts as a cooling medium. If we use coal the grate bar may not with stand higher temperature and additional cooling by water tube is necessary. Travelling grate is suitable for burning coal and lignite. As the grate rotates, the grate bar gets heated and cooled by incoming air for the half of the cycle and remaining half of the cycle grate bar cooled by the incoming air. Spreader stoker Mechanical spreader The spreader stoker feeder takes fuel from the feeder hopper by either a small ram or a rotating drum and delivers it into a spinning rotor. An adjustable trajectory plate is located between the feed mechanism and the rotor. Adjusting the trajectory plate fuel can be feed through out the entire length of the furnace. Pneumatic spreader In this rotor is replaced by high pressure air lines from Secondary air fan is used to spread the fuel into the furnace. The fuel is carried into the furnace by means of pneumatic system and the air flow adjustment makes the fuel to flow near or farther of the furnace.

1.5 OIL/GAS FIRED BOILERS TECHNOLOGY Flame 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 the flame will not come downwards to the incoming fuel, by this property combustion can be controlled easily. Hence it is always better to bring the oil or gas train from bottom of the burner. A liquid or gas fuel has flowable property by nature and it has a lower ignition temperature. When the fuel is forced to flow through the nozzle it will spread though an predetermined length and burn completely from the point of entry to the firing zone estimated. The fuel flow can be controlled by means of control valves. CHARACTERISTICS OF OIL In today’s climate of fluctuating international fuel prices and quality, the emphasis on the ability of the boiler on low quality fuel oils has become more greater. In the international market, the quality of the residual fuel oils is constantly getting poorer due to the development of more sophisticated cracking methods and also our indigenous crude production falls short of our requirements, about 15 million tons of crude is imported from outside sources. These outside sources are many, our

Steam generator

Ganesh kumar

refineries handle a variety of crude. Since the inherent properties of the finished petroleum products are directly dependent on the parent crude, one can imagine the petroleum involved in producing residual fuel oil within narrow limits of specifications, especially with respect to specified characteristics like carbon residue, asphaltenes and metallic constituents is not possible. Flash point Flash point is important primarily from a fuel handling stand point. Too low a flash point will cause fuel to be a fire hazard subject to flashing and possible continued ignition and explosion. Petroleum products are classified as dangerous or non dangerous for handling purposes based on flash point as given below. Classification

Flash point

Petroleum Product

Class A

Below 23°C

Naptha Petrol Solvent 1425 Hexane

Class B

23 to 64°C

Kerosene HSD

Class C

65 to 92°C

LDO Furnace oil LSHS

Excluded Petroleum

93°C and above

Tar

Pour Point The pour point of the fuel gave an indication of the lowest temperature, above which the fuel can be pumped. Additives may be used to lower the freezing temperature of fuels. Such additives usually work by modifying the wax crystals so that they are less likely to form a rigid structure. It is advisable to store and handle fuels around 10°C above the expected pour point. Viscosity Viscosity is one of the most important heavy fuel oil characteristics for industrial and commercial use, it is indicative of the rate at which the oil will flow in fuel systems and the ease with which it can be atomized in a given type of burner. When the temperature increases viscosity of fuel will reduce.

Steam generator

Ganesh kumar

The viscosity needed at burner tip for satisfactory atomization for various types of burners are as follows.

Type of burner Low air pressure Medium air pressure High air pressure Steam jet Pressure jet

Viscosity at burner tip In centi stokes 15 to 24 21 to 44 29 to 48 29 to 37 less than 15

Metal Content Sodium, Potassium, Vanadium, Magnesium, Iron, Silica etc. are some of the metallic constituents present in fuel oil. Of the above metals, sodium and vanadium are the most troublesome metals causing high temperature corrosion in boiler super heater tubes and gas turbine blades. Much of the sodium is removed from the crude oil in the desalting operation, which is normally applied in the refinery and additional sodium 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 oil prepared from these crude. No economical means for removal of vanadium from the residual fuel oil is available. However certain additives like magnesium are available to minimize the effect of vanadium. Asphaltene content and Carbon residue Asphaltenes are high molecular weight asphaltic material and it requires more residence time for complete combustion. Asphaltenes as finely divided coke may be discharged from the stack. Residual fuel oils may contain as much as 4% asphaltenes. Petroleum fuels have a tendency to form carbonaceous deposits. Carbon residue figures for residual fuel oils from 1 to 16% by weight. This property is totally dependent on the type of crude, refining techniques and the blending operations in refinery. Fuels with high carbon residue and asphaltenes requires large combustion chamber and hence while designing the boiler for such fuel the volumetric loading has to be of the order of 2 lakhs Kcal/m3hr

Steam generator

Ganesh kumar

OIL/GAS FIRING START UP LOGIC MANUAL TRIP INTERLOCK CHECK 1.CHECK TRIP VALVES IN CLOSED POSITION 2 . CHECK WATER LEVEL IN DRUM 3. EMERGENCY PUSH BUTTON NOT OPERATED 4. CHECK FAN SUCTION DAMPER IN CLOSED POSITION 5.CHECK FUEL PUMP/GAS TRAIN DELIVERY VALVE IN CLOSED CONDITION 6. CHECK MANUAL ISOLATION VALVE IN POSITION.

CONTROL SUPPLY LAMP

CONTROL POWER SUPPLY

SELECTOR SWITCH IN GAS/OIL FIRING MODE

START FD FAN

FAILED DEENERGISE TR & PILOTVALVE DEDUCT PILOT FLAME YES

DEENERGISE TRANSFORMER ENERGISE GAS/OIL SHUT OFF VALVE TO OPEN AND VENT TO CLOSE YES

DEENERSISE PILOT GAS & RELESASE LOW FIRE POSITION

MAIN FLAME ESTABLISED NO

ENERGISE IGNITION TRANSFORMER & PILOT GAS SHUTOFF VALVE

PRESS BURNER START BUTTON

NO DEENERSISE PILOT GAS

CHECK 1.0 OIL/GAS MAIN SHUT OFF VALVE IN CLOSED POSITION 2.0 RETURN OIL LINE SHUT OFF VALVE CLOSED POSI 3.0 AIR/ATOMISING STEAM LINE SHUT OFF VALVE CLSOED POSITION 4.0 PILOT GAS/SCAVENGING LINE SHUT OFF VALVE IN CLOSED POSITION 5.0 FUEL GAS SHUT OFF VALVE I & II IN CLOSED POSITION PURGE 6.0 NO FLAME INSIDE FURNACE BUTTON ON 7.0 FUEL PUMP NOT RUNNING 8.0 FURNACE PRESSURE NOT HIGH COMBUSTION AIR 9.0 DRUM LEVEL NOT HIGH HIGH & NOT LOW LOW DAMPER TO LOW 10.0ALL TRIP PARAMETERS OK AUTO GAS/OIL FIRING INTERLOCKS FIRE POSITION 11.0 FUEL GAS PRESSURE NOT HIGH & NOT LOW PURGE COMPLETED PURGE IN PROGRESS LAMP ON

1.0PURGE COMPLETED 2.0ALL PURGE INTERLOCKS AGAIN CHECKED 3.0COMPUSTION AIR PR NOT LOW 4.0 INSTRUMENT AIR PR NOT LOW 5.0 COMBUSTION AIR DAMPER TO LOW FIRE POSITION 6.0OIL/GAS AT REQUIRED PARAMETER 7.0 EMERGENCY PUSH BUTTON NOT OPERATED 8.0SCANNER COOLING AIR PR OK

Steam generator

Ganesh kumar

1.6 PULVERIZED FUEL BOILERS TECHNOLOGY When coal is powdered to micron size it can be conveyed easily by air in pipelines and the pulverized coal behaves as if that of oil and hence the same can be easily burnt in pulverized fuel burners. The heat release by the burners in very high and un-burnt carbon is almost equal to zero. Hence efficiency achieved by pulverized burners is much more than any type of coal combustion. MECHANISM OF PULVERIZED FUEL BURNING There 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 conveyor and fed to a coal crusher. A second conveyor system loads coal into the coal storage bunker located over the coal pulverization system. Coal via gravity feed is delivered through a down spout pipe to the coal feeder. A coal shutoff gate is provided 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 the feeder discharge. A primary air fan delivers a controlled mixture of hot and cold air to the crusher dryer to drive moisture in the coal facilitating pulverization the primary air and crushed coal mixture is then fed to the coal pulverizer located below the crusher dryer discharge. Selection of pulverizer has to be analyzed critically, since it is one of the important equipment where the wear and tear is more. For the soft lignite Beter wheel is preferable and for hard lignite, coal like fuels heavy pulveriser of ball and hammer mill is preferable. The coal is pulverized to a fine powder and conveyed through coal pipes to the burners. Primary air is the coal pipe transportation 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. When needed to fire the boiler the pulverized coal is then conveyed to the burners by an exhaust fan. This method requires very special provisions to minimize risk of fire or explosion. 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 BOILERS ATMOSPHERIC FLUIDIZED BED COMBUSTION TECHNOLOGY When air or gas is passed through an inert bed of solid particles such as sand supported on a fine mesh or grid. The air initially will seek a path of least resistance and pass upwards through the sand. With further increase in the velocity, the air starts bubbling through the bed and particles attain a state of high turbulence. Under such conditions bed assumes the appearance of a fluid and exhibits the properties associated with a fluid and hence the name fluidized bed.

Steam generator

Ganesh kumar

MECHANISM OF FLUIDIZED BED COMBUSTION If the sand, in a fluidized state is heated to the ignition temperature of the fuel and fuel is injected continuously into the bed, the fuel will burn rapidly and attains a uniform temperature due to effective mixing. This , in short is fluidized bed combustion. While it is essential that the temperature of bed should be equal to the ignition temperature of fuel and it should never be allowed to approach ash fusion temperature (1050° to 1150°C ) to avoid melting of ash. This is achieved by extraction of heat from the bed by conductive and convective heat transfer through tubes immersed in the bed. If the velocity is too low fluidization will not occur, and if the gas velocity becomes too high, the particles will be entrained in the gas stream and lost. Hence to sustain stable operation of the bed, it must be ensured that gas velocity is maintained between 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 kumar

ATMOSPHERIC CIRCULATING FLUIDIZED BED COMBUSTION TECHNOLOGY Atmospheric circulating fluidized bed (ACFB) boiler is a devise used to generate steam by burning solid fuels in a furnace operated under a velocity exceeding the terminal velocity of bed material. I.e., solid particles are transported through the furnace and gets collected in the cyclone at the end of furnace and again recycled into furnace by means of pressure difference between fluidized bed and return particle. MECHANISM OF CIRCULATING FLUIDIZED COMBUSTION The mechanism is similar to AFBC. However in AFBC the fluidization velocity is just to make the particles in suspended condition. In ACFB boiler, special combination of velocity by primary air and secondary air, re-circulation rate, size of solids, and geometry of furnace, give rise a special hydrodynamic condition known as fast bed. Furnace below secondary air injection is characteristic by bubbling fluidized bed and furnace above the secondary air injection is characteristic by Fast fluidized bed. Most of the combustion and sulphur capture reaction takes place in the furnace above secondary air level. This zone operates under fast fluidization. In CFB boiler number of important features such as fuel flexibility, low Nox emission, high combustion efficiency, effective lime stone utilization for sulphur capture and fewer fuel feed points are mainly due to the result of this fast fluidization. In fast fluidization heavier particles are drag down known as slip velocity between gas and solid, formation and disintegration of particles agglomeration, excellent mixing are major phenomenon of this regime. CFB is suitable for 1.0 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 COMBUSTION The advantage of operating fluidized combustion at the elevated pressure ( about 20 bar) is, reduction in steam generator size can be achieved and make possible the development of a coal fired combined cycle power plant. The development of pressurized fluidized bed combustion is still in research stage only. With help of pressurized hot gas coming out of the furnace is cleaned primarily by a cyclone like CFBC boiler and the gas is expanded in a turbine and the exhaust gas from turbine is further cooled by the heat exchanger. The aim behind the development of pressurized 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 kumar

1.8 HEAT RECOVERY STEAM GENERATOR In India, coal availability is 97% of the requirement and we are importing coal only for the process requirement like baking coal for steel plant where high calorific coal is required. Hence in post independence India coal fired boilers where flourished, however due to the need of energy conservation and due to process parameter requirements development of HRSG in recent periods is more. Moreover due to the development of gas turbines with gaseous and liquid fuels, more GT are being installed 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 maintaining process parameter such as temperature and other is in the point of economic point of view. The process steam generator are generally referred by the term called waste heat recovery boiler ( WHRB) where the gas contains heat in excess, this excess waste heat has to be recovered or removed by any means so that the process parameter can 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 Heat recovery 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 oxide concentration. In HRSG the gas quantity and inlet temperature is fixed and for different load the variation of heat will not be proportional and hence at part loads the heat absorbed at different zones will vary widely and hence for different loads the performance of the HRSG to be done.

Steam generator

Ganesh kumar

2.0 STEAM,GAS and AIR 2.1 INTRODUCTION In steam generator water, steam, gas and air are the working fluids in this air and gas have similar properties. Understanding the properties of gas and air are almost one and the same. I have grouped steam and gas as one unit and water as a separate unit just because understanding the behavior of steam and gas is more important in design point of view where as knowledge of water is more important in operational point of view.

2.2 DEFINITIONS FOR SOME COMMONLY USED TERMS Heat Heat is defined as the form of energy that is transferred across a boundary by virtue of a temperature difference. The temperature difference is the potential and heat transfer is the flux. In other words heat is the cause and temperature is the effect. Energy Energy of a body is its capacity to do work and is measured by the amount of the work 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 state of 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. Enthalpy Enthalpy is the quantity of heat that must be added to the fluid at zero degree centigrade to the desired temperature and pressure. Enthalpy is defined as heat within or heat content of the fluid. Entropy The word entropy is derived from a Greek word called ‘tropee’ which means transformation. The unit of entropy is Joules/kelvin. Specific heat Specific heat of a substance is defined as the amount of heat required to raise the temperature of one kilogram of substance through one degree kelvin. All liquids and solids have one specific heat. However gas have number of specific heats depends on the condition with which it is heated. Cp = f(T)

Steam generator

Ganesh kumar

Specific heat at constant pressure. Specific heat of a substance is defined as the amount of heat required at constant pressure to raise the temperature of one kilogram of substance through one degree kelvin. Integral constant pressure specificheat It is the average heat required to rise the temperature between two temperature difference 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 constant volume to raise the temperature of one kilogram of substance through one degree kelvin. NTP and STP condition It 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 measured at 0°C or 273.15 °K and pressure 1.01325 bar or 1.03 atm STP condition is at Standard temperature and pressure i.e., the properties measured at 25°C or 298.15°K and pressure 1.01325 bar or 1.03 atm. Viscosity Viscosity of a liquid is its property, due to the frictional resistance between the fluid particles (cohesion between particles) or between fluid and the wall. Viscosity of fluid controls the rate of flow. Newton s Law of viscosity The shear stress on a layer of a fluid is directly proportional to the rate of shear strain. ( 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 absolute viscosity or dynamic viscosity. Kinematic viscosity is the ratio of absolute viscosity to density (µ/ρ) Thermal conductivity Thermal conductivity is the property of substance, that its ability to conduct heat and expressed in W/mK. Kilogram Kilogram is the mass of one international prototype made of platinum iridium cylinder preserved at the international bureau of weights and measures at paris.

Steam generator

Ganesh kumar

Meter Meter is the length between two transverse lines en-grooved in platinum iridium bar at 0°C. or The meter is the length equal to 1650763.73 vacuum wave length of the orange light. ( λ = 605.8 mm of the Krypton 86 discharge lamp) Second Second is the duration of 9192631770 periods of the radiation corresponding to the transition between two specified energy level of the Caesium –133 atom. Or 1/86400th part of mean solar day. Specific volume Specific 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 temperature and pressure for steam. The required equation for derivation is given at appropriate places. For gaseous fuel, Cp /R = f(T) R = Cp – Cv Cv = Cp - 1 R R Specific enthalpy wrt NTP, T H ‘ = 1/T  Cp dT ( enthalpy with reference to 0°C) RT  R Tn Specific enthalpy wrt STP T H* ‘ = 1/T  Cp dT + Hs ( enthalpy with reference to 25°C) RT  R RT Ts Specific entropy, T S ‘ = So  Cp dT - ln(P/Pn) ( entropy with reference to 0°C) R R R Tn

Steam generator

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Specific free enthalpy G = H -S RT RT R The temperature dependent specific heat (Cp) can be represented by an equation of 4 th degree polynomial as shown below Cp = a1 + a2T + a3T² + a4 T3 + a5T4 (for temperature from 273K to 1000K) R Cp = a9 + a10T + a11T² + a12 T3 + a13T4 (for temperature from 1001K to 5000K) R Integrating, and adding constant of integration we get H = a1 + a2T + a3T² + a4T3 + a5 T4 + a8/T (for temperature from 273K to 1000K RT 2 3 4 5 H* = a1 + a2T + a3T² + a4T3 + a5T4 + a6/T (for temperature from 273K to 1000K RT 2 3 4 5 S = a1 ln T + a2T + a3T² + a4T3 + a5T4 + a7 – ln(P/Pn) R 2 3 4 G = a1(1- ln T) - a2T - a3T² - a4T3 - a5T4 + a6 -a7 + ln(P/Pn) RT 2 6 12 20 T Dynamic viscosity , thermal conductivity and prandtl number Dynamic viscosity, thermal conductivity and prandtl number of a flue gas can be fine easily with help of the properties of nitrogen and following constants. Dynamic Viscosity µPa.S -0.9124458E 1 0.4564993E-2 0.2198889E-4 -0.1891235E-7 0.5138895E-11

Thermal conductivity W/mK -0.1083113E-1 0.5596822E-4 0.7413502E-7 -0.5901395E-10 0.1961745E-13

Prandtl number

a1 b1 c1 d1 e1

Specific Heat Kj/kgK 0.8554535 0.2036005E-3 0.4583082E-6 -0.279808E-9 0.5634413E-13

a2 b2 c2 d2 e2

-0.1002311 0.7661864E-3 -0.9259622E-6 0.5293496E-9 -0.109357E-12

-0.4267768E1 0.4074274E-3 -0.5125357E-5 0.738556E-8 -0.343972E-11

-0.8035817E-2 0.110672E-04 -0.8397255E-8 0.1130229E-10 -0.5731264E-14

-0.8820652E-2 0.1855309E-3 -0.3838084E-6 0.3256168E-9 -0.1005757E-12

Var

0.492851 -0.1230046E-2 0.1662398E-5 -0.1052753E-8 0.2443111E-12

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Dynamic viscosity, µg = µn + P1 XH2O + P2 XCO2 Where XH2O & XCO2 are Percentage of weight in flue gas P1 = a1 + b1T + c 1T² + d1T3 + e1T4 P2 = a2 + b2T + c 2T² + d2T3 + e2T4 where T is temperature in °C Thermal conductivity, kg = kn + P1 XH2O + P2 XCO2 Where XH2O & XCO2 are Percentage of weight in flue gas P1 = a1 + b1T + c 1T² + d1T3 + e1T4 P2 = a2 + b2T + c 2T² + d2T3 + e2T4 where T is temperature in °C Prandtl number, Prg = Prn + P1 XH2O + P2 XCO2 Where XH2O & XCO2 are Percentage of weight in flue gas P1 = a1 + b1T + c 1T² + d1T3 + e1T4 P2 = a2 + b2T + c 2T² + d2T3 + e2T4 where T is temperature in °C Pra = a + bT + cT² + dT3 + eT4 Specific heat, Cpg = Cpn + P1 XH2O + P2 XCO2 Where XH2O & XCO2 are Percentage of weight in flue gas P1 = a1 + b1T + c1T² + d1T3 + e1T4 P2 = a2 + b2T + c2T² + d2T3 + e2T4 where T is temperature in °C Where 0 ≤XH2O ≤ 0.3 ,0 ≤ XCO2 ≤0.2 , 0 ≤ T ≤ 1200°C Dynamic viscosity, thermal conductivity and Prandtl number of NITROGEN

a b c d e f

Dynamic viscosity µ Pa.s 0.1714237E02 0.4636040E-01 -0.2745836E-4 0.1811235E-7 -0.674497E-11 0.1027747E-14

Thermal conductivity W/mK 0.2498583E-1 0.6535367E-4 -0.7690843E-8 -0.1924248E-11 0.160998E-14 -0.2864430E-18

Prandtl number 0.6901183 0.2417094E-05 0.2771383E-7 -0.3534575E-10 0.1717930E-13 -0.2989654E-17

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µn = a + bT + cT² + dT3 + eT4 + fT5 Kn = a + bT + cT² + dT3 + eT4 + fT5 Prn = a + bT + cT² + dT3 + eT4 + fT5 Cpn = 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 a b c d e f

0.3679321E1 -0.1313559E-2 0.2615196E-5 -0.9629654E-9 -0.9928002E-13 -0.9723991E3

1001K to 5000K ‘a1 b1 c1 d1 e1 f1

0.2852903E1 0.1580411E-2 -0.6189378E-6 0.1119450E-9 -0.7607378E-14 -0.8019835E3

2.3 STEAM We can see in day to day life the process of boiling water to make steam. Steam is water in the vapour or gaseous state. It is in visible, odorless, non-poisonous and relatively non corrosive to boiler metals. Steam is uniquely adapted by its advantageous properties for use in industrial process heating and power cycle. Thermodynamically boiling is the result of heat addition to the water in a constant pressure and constant temperature process. The heat which must be supplied to change water into steam without raising its temperature is called the heat of evaporation or vaporization and the boiling point of a liquid may be defined as the temperature at which its vapour pressure(pressure exerted due to the vapour of the liquid) is equal to the total pressure above its free surface. In other words temperature at which the partial pressure of vapour increases to make total pressure above the liquid surface. This temperature is also known as the saturation temperature. EVAPORATION Liquid exposed to air evaporate or vapourize. Evaporation is the process takes place at the surface exposed to atmosphere. If there is any increase in ambient temperature or increase of the liquid temperature evaporation rate becomes increased. The reduction in pressure above the liquid surfaces accelerate the evaporation rate. Evaporation will be there at all temperature and pressure, unsaturated surrounding environment also one of the factor increases the evaporation rate. BOILING Boiling is the phenomenon takes place at boiling point of the liquid. Boiling takes place throughout the liquid column. A liquid will boil, when it’s saturated vapour pressure exceeds the surrounding environment pressure acted upon the liquid. Hence boiling point of a liquid will change depends on the pressure exerted by the environment over the surface.

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CONDENSATION Condensation is the change in phase of vapour phase to it’s liquid phase. When water vapour or steam comes in contact with cooler surfaces, it gives up the heat and condenses to water. The heat released while changing from vapour phase to liquid phase is called heat of condensation. In factories the steam released out of the main steam line or process vents where we can see a remarkable phenomenon of indication of dryness of steam. If the steam is dry, we can not visualize the steam coming out of the vent but after some distance we can see a white cloud. This is due to the condensation of steam which composed of small particles of water formed when steam cooled in cooler atmosphere. In other case if the steam is wet, the white smoke cloud is directly released from the vents.

2.4 FUEL Combustion Combustion or burning, is a rapid combination of oxygen with a fuel resulting in release of heat. The oxygen comes from the air, which is about 21% oxygen and 78% nitrogen by volume. Most fuels contain carbon, hydrogen, and sometimes sulphur as the basic composition of combustion materials. These three constituents’ reacts with oxygen to produce carbon-di-oxide, water vapour, suphur di oxides gases respectively and heat. Carbon, hydrogen and sulphur are found exists in direct form in most of the solid and liquid fuels and in gaseous fuels the combustion matter is found as hydrocarbons(combination of hydrogen and carbon). When these burn, the final products 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 fuel Quantities of heat are measured in BTU, kiloCalories, or joules. A BTU is the quantity of heat required to raise the temperature of one pound of water one degree fahrenheit. A kilocalorie is the quantity of heat needed to raise one kilogram of water one degree celsius. Experimental measurements have been made to determine the heat released by perfect combustion of various fuels. The heat value is usually determined by calorimeters. When a perfect mixture of a fuel and air originally at 15.6°C is ignited and then cooled to 15.6°C the total heat released is termed the higher heating value or Gross calorific value. There is also one more term called lower heating value or the net calorific value it is the quantity of heat equal to gross calorific value minus the heat absorbed by the latent heat of water moisture( inclusive of moisture generated due to combustion of hydrogen present in the fuel) at 25°C.

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Dulong’s formula is used to find Calorific value of the fuel HHV(kj/kg) =338.21C% +1442.43(H-O/8)% + 94.18S% Relation between HHV and LHV LHV = HHV – (%H2O + %H2x8.94)χ Where χ is the latent heat of water vapour at reference temperature 25°C =583.2 kcal/kg Proximate Analysis The general procedure for the analysis relating to proximate analysis is describe below as per IS 1350(partI). For full details, the original standard may be referred to i) Moisture The moisture in the coal is determined by drying the known weight of the coal at 108°C±2°C ii) Volatile matter The method for the determination of VM consists of heating a weighted quantity of dried 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 that due to moisture. VM is the portion of the coal which, when heated in the absense of air under prescribed conditions, is liberated as gases and vapour. iii) Ash In this determination, the coal sample is heated in air up to to 500°C for minutes from 500 to 815°C for a further 30 to 60 minutes and maintained at this temperature until the sample weight becomes constant. iv) Fixed carbon Fixed carbon is determined by deducting the moisture. VM and ash from 100 Ultimate analysis The ultimate analysis of fuel gives the constituent elements namely carbon, hydrogen,nitrogen, sulphur , hydrocarbons, nitrogen etc., For the ultimate analysis of 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. These constituent 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 Specification 1 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

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D E

F G

2

3.

Non coking coal, produced Assam,Andhrapradesh,Meghalaya, Arunachalpradesh and Nagland

Coking coal

Steel GrI Steel GrII Washery GrI Washery GrII Washery GrIII

GCV exceeding 4200kcal/kg not exceeding 4940Kcal/kg GCV exceeding 3360kcal/kg not exceeding 4200Kcal/kg GCV exceeding 2400kcal/kg not exceeding 3360Kcal/kg GCV exceeding 1300kcal/kg not exceeding 2400Kcal/kg

Not graded

Ash content not exceeding 15% Ash content 15% to 18% Ash content 18% to 21% Ash content 21% to 24% Ash content 24% to 28%

2.5 GAS and AIR IDEAL GAS OR PERFECT GAS At low pressure and high temperature, all gases have been found to obey three simple 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. These laws are called ideal gas laws. These laws are applicable to gases, which do not undergo changes in chemical complexity, when the temperature or pressure is varied. I.e., in other words laws applicable to gases which do not undergo any chemical reaction when subject to change in pressure or temperature. GAS LAWS Boyle’s law Boyle’s law states that the pressure is inversely proportional to volume and the product of pressure and volume is constant PV =C Charles law-I Charles law states that at constant pressure, volume is directly proportional to temperature. V/T = C Charles law-II Charles law states that at constant volume, pressure is directly proportional to temperature. P/T = C

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Absolute scale of temperature This scale of temperature is based on Charles law. According to Charles law at constant pressure, volume of given mass changes by 1/273 of its volume at 0°C for every rise or fall in temperature by 1°C. if the volume of the gas at 0°C is Vo and its volume at t°C,

Vt = V o + Vo x t = Vo (1 + t/273) 273 If t = -273°C, then volume is zero, the hypothetical temperature of –273°C at which gas will have zero volume is known as absolute temperature or 0°K. Avagadra s Law Avagadra’ s law state that the volume occupied by any gas at normal temperature and pressure is 22.41383 x 10-3 m3 per mol of gas. I.e., volume occupied by a kg mol of gas is 22.41383 m3/kg mol. GAS EQUATION From Boyle’s law PV = nRoT Where, Ro is UNIVERSAL GAS CONSTANT n = m/M = Weight of gas in kg at NTP Molecular weight of the gas in kg At normal temperature and pressure Pressure = 1.01325 x 105 N/m² Temperature = 273 K Volume = 22.41383 x 10-3 m3 n = 1 mole Ro= PV/nT = 1.01325 x 105 x22.41383 x10-3/(1 x273) = 8.314 Nm mol-1 K-1 = 8.314 joules /mol K Gas constant R = Universal gas constant (Ro) / molecular weight (M). Daltan s law At a constant temperature, the total pressure exerted by a mixture of non- reacting gases is equal to the sum of the partial pressure of each component gases of the mixture. Thus the total pressure P of a mixture of r gases may be represented mathematically as

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Pt =

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r Σ pI where pi is the partial pressure of each components gas of the mixture. i =1

If P and the molar composition (% volume) of the mixture are known pi can be calculated using the expression pi = xi P

2.6 SOME COMMONLY USED DIMENSIONLESS NUMBERS AND THEIR SIGNIFICANCE NUMBER

FORMULA

SYMBOL

DEFINITION & SIGNIFICANCE

Nusselt

hd/k

Nu

Radio of temperature gradients by conduction and convection at the surface -used for convection heat transfer coefficient determination

Reynolds

ρvd/µ

Re

Inertia force/viscous force - used for forced convection and friction factor

Prandtl

Cpµ/k

Pr

Molecular diffusivity of momentum Molecular diffusivity of heat

Grashof

ρ²d3 gß∆T/µ² Gr

Buoyancy force x Inertia force Viscous force x viscous force - used for natural convection

Biot

hd/ks

Bi

Internal conduction resistance Surface convection resistance - used for fin temperature estimation

Peclet

vdρCp/k

Pe=RePr

Stanton

h/Cpρv

St=Nu/Pe

Heat transfer by convection Heat transfer by conduction Wall heat transfer rate Heat transfer by convection

Euler

∆P/ρv²

Eu

Pressure force/Inertia force - used to find pressure drop

Froude

v²/gl

Fr

Inertia force/gravity force

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Where 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 pressure Ex.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 based on weight basis. Consider 4% carbon loss in combustion of AFBC system.

AIR REQUIREMENT CALCULATION Amount of oxygen required for burning coal C + O2 à CO2 + heat 12 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 = 3.666kg of carbon dioxide. 0.399kg of carbon in coal require = 0.39x2.666 = 1.064 kg of oxygen H2 + 1/2O2 à H2O + heat 2 kg of hydrogen react with 16 kg of oxygen to produce 18 kg of moisture. I.e., one kg 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 oxygen S + O2 à SO2 + heat 32 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 di oxide. 0.0038 kg of sulphur in coal require =0.0038 x 1 = 0.0038 kg the other composition like nitrogen, argon(if present) is inert gas and it will not react with 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

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The oxygen present in fuel = 0.0676 kg Net oxygen required = 1.2662 – 0.0676 = 1.1986 kg Air contains 23.15 % oxygen by weight and hence the air required for 1.1986 kg of oxygen 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 required for burning per kg of fuel = 5.244 x 1.2 = 6.292 kg. FLUE GAS GENERATION ESTIMATION Carbon di oxide produced = (0.399 – 0.0188) x 3.666 = 1.3915 kg Moisture 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 kg

Sulphur di oxide produced = 0.0038 x 2

= 0.0076 kg.

Nitrogen in air Nitrogen in fuel

= 6.212 x 0.7685

= 4.7739 kg. = 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 produced Per 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 humidity and 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 atm Weight 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.

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Ex03. Estimate the efficiency of a boiler firing with coal as a fuel having GCV of 3200 kcal/kg. Furnace is Fluidized bed boiler. Apply ASME PTC 4.1 indirect method to calculate the efficiency. Flue gas temperature leaving the boiler is140°C and ambient air temperature is 40°C. Ash content of the fuel is 42.3% and 20% of total ash is collected in bed and 80% ash is carried in fly ash. As per lab report the loss on ignition of ash samples collected in bed zone and fly ash zone is 0.1% by weight and 4.4%by weight. The boiler is operating at 20% Excess air and the dry kg/kg of gas produced =5.91 and dry kg/kg of air required = 5.696. The moisture and hydrogen present in the fuel is 6% and 2.7% respectively. Basically following are the losses present in boiler, 1.0 Unburnt carbon loss 2.0 Sensible heat loss through ash 3.0 Moisture loss due to air 4.0 Moisture and combustion of hydrogen in fuel 5.0 Dry flue gas loss 6.0 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 partial pressure exists in steam form and gets superheated at 140°C Heat 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

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= 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. The hydrogen present in the fuel react with oxygen to form water. From combustion equation 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. It is assumed the boiler is operated under zero present blow down. Ex07 Estimate the FD and ID fan flow and power required for a bagasse fired dumping grate boiler, whose bagasse consumption at 100% MCR capacity is 31000 kg/hr and the boiler is operating at 35% excess air. The fuel air requirement is 3.909 kg/kg of fuel and gas generation is 4.873 kg/kg. FD fan Total 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/sec FD fan head Pressure head required for air flow sections like airheater, air ducts and grate are to be calculated. Now in most of the practical applications the pressure drop works out to be 165 mm WC and the same can be assumed for this calculation. FD fan head with margin = 165 x 1.2 = 200mmWc FD fan power required. = flow x head/102 x efficiency = 34. 31 x 200 / (102 x 0.8)

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= 84.09 KW Motor selected = 84.09 x 1.1 = 92.5 KW (next nearest motor standard is 110 KW) ID fan Total 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/sec ID fan head Pressure head required for gas flow sections like Furnace, Bank, Economiser, air heater, gas ducts and dust collectors are to be calculated. Now in most of the practical applications the pressure drop works out to be 230 mm WC and the same can be assumed for this calculation. ID fan head with margin = 230 x 1.3 = 300mmWc ID fan power required. = flow x head/102 x efficiency = 61.27 x 300 / (102 x 0.8) = 225 KW Motor 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 boiler application. S.N Description Grate type AFBC CFBC OIL fired 1 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 FURNACE 3.1 INTRODUCTION: The design of furnace is considered as the vital part in the boiler. The furnace is the zone experiencing a high temperature in boiler. The performance of the furnace reflects 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

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illustrated with following. If furnace outlet temperature (FOT) is high, then the next zone is super heater it gets high amount of heat input naturally the metal temperature is high and the steam temperature also increased, which in turn reflects in the performance and cost of material. On the other hand if the furnace is over sized the FOT will be lesser, to get the required steam temperature the super heater heat transfer area to be increased. If the heat transfer area is increased it calls for larger 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 ash fusion temperature are also form as constraint over the furnace design. The type of fuel 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 dumping or travelling grate. When the fuel is some thing like bagasse (fibrous and long strand structure) it can be burnt well in dumping or travelling grate. A gaseous fuel offers fewer problems since it is clean. Fuel oil brings its own problems like high or low temperature corrosion and additives have to be used. For coal ash fusion is the problem, since ash slag down deposits on the wall hindering heat transfer to steam water mixture. Depends on property of coal, whether it can be crushable to powdered form, pulverized firing or bubbling bed or cyclone furnace can be decided. When we go for oil or gas firing, we can have higher heat flux in the furnace because of the higher emissivity of oil flame and relative cleanliness of walls compared to coal firing. 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 of pulverized fuel firing. However, if a furnace designed for both coal and oil it is normally designed for coal and performance for oil firing in that furnace will be carried out. When a furnace designed for coal operated with oil, the higher furnace absorption results in a lower furnace outlet temperature. Lower FOT means super heater 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 be increased by gas recirculation, otherwise when coal is fired FOT will be reduced by some means of bed absorption (This is used in FLUIDISED BED COMBUSTION techniques). Furnace size also governed by length of flame in gas or oil fired boiler since 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 higher rate of erosion due to carry over particles in flue gas. Normally a flue gas velocity of 6 to 8 meters per sec was allowed for coal fired boilers and 12 to 15 meters per sec was 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 and there is a possibility of over heating. The boiling is the phenomenon, which governs the rate of heat transfer from combustion to steam water mixture inside the tube. In boiling when bubbles formed at tube wall hinders the heat transfer which cause

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tubes over heating and tube failure. This sort of boiling occurs at nucleate boiling stage. Therefore proper circulation must be ensured to cool all tube. Circulation ratio (CR) is the ratio between mass of water circulated inside the boiler to rate of steam generation. Hence CR is also directly related to dryness fraction of steam by the expression CR = 1/x. which implies in one circulation 1/CR quantity of dry steam was produced. Circulation number will be higher when the difference in density between steam and water is more (i.e.) due to higher difference in density; steam water mixture velocity will be more thereby overheating will be prevented. If the proper circulation is not there, circulation in the boiler circuit is effected by means of external agency (normally a circulation pump will be used). This type of circulation is called Forced or controlled circulation.

3.4 HEATFLUX TO FURNACE WALLS: Boiling phenomenon can be represented by a log-log plot of heat flux Vs surface temp-bulk temperature as shown Q max.

H E A T F L U X A

B

C

D

SURFACE TEMP

The different regimes of boiling indicated by the letters A, B, C, D. Absence of bubble formation and the influence of natural convection on the heat transfer process is predominant in the region A (pool boiling). Formation of vapour bubbles at the nuclei 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 the heat flux is C. In this region the unstable film boiling manifests with an eventual transition to a continuous vapour film. In the final region D film boiling becomes stabilized. This phenomenon of stable film boiling is referred as “ LEINDENFROST EFFECT”

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In the regime of boiling the maximum wall heat flux is observed in region C. Many experimentalists refer this state of maximum wall heat flux as “BURN OUT FLUX’. The reason being when the wall is heated electrically, the heating element frequently burn out when the wall heat flux reaches Q maximum. Hence the design engineers should have an idea of average heat flux to the tubes, how they vary around periphery and fin tip temperature in case of membrane wall construction. Calculation of fin temperature was discussed in latter part of this chapter.

3.5 POINTS TO BE NOTED WHILE DESIGNING FURNACE 1.0

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 walls

3.0

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 wall

5.0

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 combustion

3.6 CLASSIFICATION OF FURNACE i)

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

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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 TRANSFER In general heat transfer from higher temperature to lower temperature is carried out in three modes. 1.0 Conduction 2.0 Convection 3.0 Radiation

Conduction Conduction refers to the transfer of heat between two bodies or two parts of the same body through molecules, which are more or less stationary. Fourier law of heat conduction states rate of heat flux is linearly proportional to temperature gradient.

Q = --K dt/dx

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Where, Q rate of heat flux watts per sq.meter K thermal conductivity (property of material)W/m°k dt/dx temperature gradient in x –direction Negative sign indicates heat flows from high temperature to low temperature. Heat transfer by conduction in plate and cylinder Plate Q = k.A. (t1 - t2)

watts

X Cylinder Q =k.(A2- A1).(t1- t2) (r2- r1) ln(A2/A1) where, A area of plate A1 outside cylinder surface A2 inside cylinder surface ‘r cylinder radius ‘t temperature of surfaces

Convection Convection is a process involving mass movement of fluids. When a temperature difference produces a density difference which results in a mass movement. Newton s law of cooling governs convection. In convection there is always a film immediately adjacent to wall where temperature varies. - kf A (tf - tw) Q

=

Where, is film thickness kf thermal conductivity of film h = kf / heat transfer coefficient (kcal/ sq.m hr °C or W/sq.m °C)

Radiation All bodies radiate heat. This phenomenon is identical to emission of light. Radiation requires no medium between two bodies, irrespective of temperature the radiation heat transfer takes place between each other. However the cooler body will receive more heat then hot body. The rate at which energy is radiated by a black body at temperature T( °K) is given by Stefan Boltzmann law. Q=

4

AT

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Q rate of energy radiation in Watts A Surface area radiating heat sq.m Stefan boltzmann constant = 5.67 x 10 4.88 x 10

–8

4

Watt/sq.m K

–8

4

Kcal/sq.m hr K

3.8 HEAT TRANSFER IN FURNACE Furnace heat transfer is a complex phenomenon, which can not be calculated by a single 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, partly due to luminous part of the flame and partly due to non-luminous gases. Overall heat transfer coefficient in furnace is governed by three T’s temperature, turbulence and time and calculated by two parts. Hc - heat transfer coefficient by convection Hr - 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. In forced convection turbulence or laminar flow depends on mean velocity, characteristic length L, density and viscosity. These variables are grouped together in a dimensionless parameter called Reynolds number. Reynolds number is the ratio between inertia force to viscous force. Reynolds number = (mass x acceleration)/(shear stress x cross sectional area) Mass Acceleration Volume Shear stress Re

= = = =

volume x density velocity / time cross sectional area x velocity dynamic viscosity x velocity gradient(v / l)

= 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 often turbulent only. In free convection turbulence or laminar flow depends on the buoyancy force and temperature difference, coefficient of volume of expansion. These variables are grouped to form dimensionless numbers called Grashoff number and Prandl number. Laminar or turbulence is identified with product of Grashoff number and prandl number When, Gr.Pr < 10

9

flow is laminar

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Gr.Pr > 10

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9

flow is turbulent.

DIMENSIONAL ANALYSIS FOR HEAT TRANSFER COEFFICIENT The heat transfer coefficient may be evaluated from correlation developed by dimensional analysis. In this method all the variables related to the phenomenon is grouped by experience with help of basic fundamental units length, mass, time and temperature. The final equation arrived for FORCED 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

f

Let h = B L U ρ µ k Cp , where B,a,b,c,d,e,f are constants Expressing the variables in terms of their dimensions -3 -1

a

MT

-1 b

-3 c

-1 -1 d

-3 -1 e

= B L .(LT ) .(ML ) .(ML T ) .(MLT = B.L

a+b-3c-d+e+2f

.T

–b-d-3e-2f

.M

-2 -1 f

) .(L² T

c+d+e

.

-e-f

0 = a + b –3c –d +e +2f -3 = -b –d –3e –2f 1 =c+d+e -1 = -e - f The solution of the equation gives, a = c-1, b =c, d = -c +f, e = 1-f

h = B. L

c-1

c

c

.U . ρ .µ

-c+f

.k

-1-f

.Cp

f

by grouping the variables, -1

c

f

h/L k = B.(UL ρ / µ) . (µ. Cp /k)

c

f

Nussultes number = B.(Reynolds number) .(Prandl number) The constants B,c,f are evaluated from experimental data.

)

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For turbulent flow inside tubes and fully developed flow the following equation attributed to Mr.Dittus and Boelter, Nu = 0.023 Re

0.8

Pr

n

where, n = 0.4 when the fluid is heated n = 0.3 when the fluid is cooled.

For turbulent flow outside tubes 0.8 n Nu = 0.037 Re Pr where, n = 0.4 when the fluid is heated n = 0.3 when the fluid is cooled FREE CONVECTION Free 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)g Now, ß coefficient of volume expansion 1/ ρ = (1/ρo) + ß(To-T), ρo = ρ (1 + ß T)

then,

(ρo/ ρ ) – 1 = ß F = ßg

T

T

For an ideal gas ß is inversely proportional to temperature,(i.e. dimensional number for 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 = d + e+ f = a + 2b + c –3d –e + f -3 = -2a –2b-e-3f -1 = -b-f

solving this equation. c = 3a – 1,d = 2a , e = b –2a, f = 1- b h = 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.

-1 f

) ]

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a

Ganesh kumar

b

Nu = B. Gr . Pr

By large number of experiments made on fluids it has been found that exponents a a and b are of the same value. So the expression reduce to Nu = B.(Gr.Pr) HEAT TRANSFER BY RADIATION Hr In furnace heat transfer is predominant by luminous and non-luminous radiation. A general approximate expression may be written for furnace absorption using Stefan boltzman law. Q = A g

=

c

[

w

c

+

g

Tg4 –

w

w

g TS

4

]

-

emissivity pattern of tri atomic gases such as carbon di oxide and water vapour are studied by Mr. Hottel and charts are available to predict gas emissivity as a function of various gas temperature, partial pressure and beam length. I have also furnished the expression form to find gas emissivity. When c and w are found from graph and w can be determined from the following expression or from graph. c Otherwise emissivity of gas can be directly found by the expression given in equation1. 0.222 1 _______________ P c *L +0.035 ln2.8

= EXP

c

1 ln(p + 1.8)

1/3 0.23 = EXP

w

1

0.842

2 -

(0.23 +Pw*L

0.75

0.5+Pw+p

where p is gas pressure in bar(a) L is beam length meter w and c are pressure correction factor for gas pressure absorptive of gasses can be determined at wall temperature.

g

=

c

c

+

w

w

-

At wall temperature correction, Pcw = P c (Tw/Tg) c =

cw

(Tg/Tw)0.65

Pww = P w(Tw/Tg) w =

ww

(Tg/Tw)0.45

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cw is a function of Pcw .L and wall temperature for this we have to see the emissivity in graph ww is a function of Pww .L and wall temperature for this we have to see the emissivity in graph

pressure correction is same as gas emissivity factor. = w = function of P w/(Pc + P w) , Pcw.L + Pww.L, and temperature of wall The effect of absorptivty is negligible hence the same can be neglected and a generalized 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 constant Tg Flue gas temperature of mean theoretical flame temperature(adiabatic temperature) TS Furnace wall temperature (If calculated for outside heat transfer coefficient or consider saturation temperature if calculated for over all heat transfer coefficient, the difference will be of very minor). W f Fuel burnt Wg Flue gas produced Gas emissivity

g

= 0.9( 1- e –k.L )………………………………………………1

The emissivity of flame is evaluated by f

where

=

( 1- e –k.L )

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 gas

L 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 its concentration. For non-luminous flame

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K = (0.8 +1.6 Pw).(1-0.38 TM/1000)(Pc + P w) _______ (Pc +P w)L For semi luminous flame, the ash particle size and concentration is taken in calculation K = (0.8 +1.6 Pw).(1-0.38 TM/1000)(Pc + P w) ________ (Pc +P w)L

+ 7µ(1/dm²TM²)1/2

dm mean effective diameter of ash particle in micron dm 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^3 TM – furnace mean temperature °k(Some authors will consider this as outlet temperature, but it is convincing assumption that in furnace zone temperature will be uniformly spread through out the furnace by radiation effect (spherical). Hence considering mean temperature for calculating radiation heat transfer coefficient will be more appropriate. You can appreciate a notable phenomenon of furnace temperature depends on flame location inside the furnace, in case flame is located at the center of furnace(like oil fired burners (refer example1)) mean temperature and outlet temperature will be at the most equal and if flame is located at one end of the furnace and radiation beam travels a larger distance of furnace(like AFBC boilers assuming no free board combustion) the furnace temperature near flame will be higher and it gradually degrees at the furnace exit. For luminous oil or gas flame K = (1.6 TM/1000) –0.5 Pw and Pc are partial pressure of water vapour and carbon di oxide

Above equations give only Theoretical values for flame emissivity. In practical cases a wide variation would be occurred due to: 1.0 Combustion phenomenon itself 2.0 The flame does not fill the furnace fully. Unfilled portion are subject to only gas radiation

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3.0 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 the emissivity at appropriate places for AFBC, Dumping grate and fired boilers with working of example. The heat transfer by radiation is given as Q = A w g [ TM4 – TS4]. But mostly the heat transfer will be of both convection and radiation occuring simultaneously and so to put both process on a common basis, we may define a radiation heat transfer coefficient by symbol Hr. Qr = Hr. A. (TM – TS) Hr

=

4 4 w g[TM -TS ]/(TM-TS )

While considering the total heat transfer by convection and radiation Q = (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 are corrected by effectiveness factor. This depends on various manufacturers experience on their steam generator.(Normally for oil fired boilers the value will be of 0.79 and gas fired boiler 0.67).

3.9 FURNACE CONSTRUCTION : Basically three types of constructions are used 1.0 Plain tube construction with a refractory lined furnace 2.0 Tangent tube construction 3.0 Membrane wall construction.

Plain tube construction FURNACE CHAMBER

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REFRACTORY The drawing shown gives complete idea of the above construction. Refractory lined wall construction is out dated design since it calls for a lot of refractory work and flue gas leaks are heavy and it can not with stand positive furnace pressure. Tangent tube construction FURNACE CHAMBER

REFRACTORY Tangent tube is a improvement of refractory lined. Here requirement of boiler tubes is comparatively more and also refractory structure is not eliminated.

Membrane wall construction

In industries widely used boiler furnace construction is of membrane wall construction type. In this design the tubes are joined by welding a continuous longitudinal 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 the tubes(pitch) are maintained in a such a way that the fin can be cooled by either of the two side tubes and prevent warping of the panel. Water cooled furnaces not only eliminated problem of rapid deterioration of refractory walls due to slag, but also reduced fouling of convection heating surfaces to manageable extent, by lowering the temperatures leaving the furnace. In addition to reducing furnace maintenance and fouling of convection heating surfaces, water cooling also helped to generate more steam. Consequently the boiler surface was reduced since additional steam generating surface was available in water cooled furnace.

Ex.1.0 . Find the furnace outlet temperature for a fluidized bed boiler operating at 15 kg/cm^2(g) having furnace EPRS of 28.43 sq.m and having the following gas parameters.

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Flue gas produced 11016 kg/hr at a temperature of 900°C and partial water vapour pressure 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°C Flue gas properties at film temperature. (900+740 +200)/3 = 613.33°C –5

Dynamic viscosity Thermal conductivity Prandl number

= 3.7392 x 10 kg/ms = 0.065177 kcal/m hr.°c = 0.7152

Flue 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 transfer coefficient its effect will be negligible) Nu = 0.037 Re

0.8

Hc/kL = 0.037 Re

Pr

0.8

n

Pr

where n= 0.3 for cooling fluid n

0.8 Hc = 0.037 x

0.396 x 1.1269 x 1.75 3.7392 x 10-5

x 0.7152 0.3 x 0.06517/1.75

= 3.56 kcal/m^2 hr.°C Radiation 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 =1.75 L = 1.2709 m

For non luminous flame attenuation factor K = (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

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flame emissivity

Wall emissivity

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f

w

= 0.9 x (1- e –0.2904 x = 0.2778

1.2709

)

= 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 K Total heat transfer coefficient Hc + Hr = 3.56 + 27.1 =30.66 kcal/hr m^2 K Heat 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/hr Qg 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, satuated steam from and at 100°C. Coal as a fuel. The efficiency of boiler is 80% and GCV of coal as 3800 kcal/kg , Flue gas produced per kg of fuel is 6.802 kg/kg at 20% excess air 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 can be considered and above which water wall is located. Hence a water wall of size 3584 x 1680( 35 @ 112 pitch and 15 @ 112 pitch ) can be obtained.

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The sizing of bed area and water wall size is an art rather than a scientific approach a 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 heat capacity of boiler. This is value is assumed that water at 1kg/cm^2 100°C is given as input and steam drawn at 1kg.cm^2 .(i.e. latent heat at 1kg/cm^2 pressure only absorbed ) EX 03. Find the furnace outlet temperature of a 55Tph dumping grate bagasse fired boiler 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 be 212m^2 and 13.6m^2 respectively. Consider convection heat transfer coefficient negligible and lower heating value of bagasse 1828 kcal/kg, 85% of air required flows through air heater at a temperature of 170°C and 15% air for fuel distributor and OFA at 40°C into the furnace. Fuel consumption 24209 kg/hr. 2% of gross heat input goes as carbon loss and 1% goes as radiation loss. FURNACE HEAT INPUT 1.0 Fuel heat input 2.0 Air heat input

= 24209 x 1828 = 44.254 x 10^6 kcal/hr = 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/hr where,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/hr

4.0 Radiation loss

= 0.01 x 24209 x2272 = 0.55 x 10^6 kcal/hr

Where 2272 kcl/kg is GCV of fuel. NET FURNACE HEAT INPUT = 1+2 –3 –4 = 46.072 X 10^6 KCAL/HR applying stefan boltzman law, Q = A w g [ TM4 – TS4] As it is a bagasse fired boiler volatile combustion is more TM will be equal to temperature exit and w g is equal to 0.72. Assuming 890°C as FOT Saturation 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

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Q2 = 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 bagasse 0.3076 kcal/kg°C is specific heat of flue gas Furnace outlet temperature = 890°C Radiation 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/kg Ex.04. Estimate FOT for the furnace operating at 20.66 bara, having EPRS area 112m² and size 5.74 x 3 x 6 m. firing LDO as fuel having LCV of 41867 kj/kg and fuel consumption 1.16 kg/sec and flue gas generated 19.03 kg/sec at 10% excess air. Air required 17.87 kg/sec at 27°C. Consider a radiation loss 0.33% and wall emissivity 0.85, heat transfer effectiveness 0.79. adjacent radiation chamber extends by 1.01 m length wise. Total heat into Furnace at 27°C ambient. Heat input by fuel Radiation loss

= 1.16 x 41867 = 48.565 MW = 48.565 x0.33/100 = 0.16 MW

Nett heat input

= 48.405 MW

Heat absorbed by Furnace Radiation coefficient

Hr

=

4 4 w f[TM -TS ]/(TM-TS)

For oil fired boiler Tmean is equal to Tgas outlet Wall emissivity = 0.85 Flame emissivity = 0.72 (1-e-kl) Attenuation factor k = (1.6Tm/100)-0.5

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Beam length, l

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= 3.4x5.74x3x6/2(5.74x3+5.74x6+3x6) = 2.52M

Assume gas outlet temperature,1285°c =1558°K K =(1.6x1558/100)-0.5 = 1.9928 Flame emissivity = 0.72x(1 –e

–1.9928 x2.52

) = 0.715

Hr = 5.67 x 10^-8 x0.85x0.715x(15584 –4874)/(1558 –487) = 187.768 W/mK4 Nu = 0.037 x Re0.8 xPr0.3 Gas properties at film temperature (1285+ 214)/2 = 749.5°C Dynamic viscosity kg/m.s = 4.13276 x 10-5 kg/ms Thermal conductivity = 0.072915 W/mK Prandtl number = 0.711 Velocity = 19.03 /(0.345 x 3x6) = 3.064 m/s where 0.345 is density kg/m^3 0.8

Hc = 0.037 x 3.064 x5.74 x0.345 x 0.711 4.13276x10^-5

0.3

x 0.072915 5.74

= 5.769 W/mK Heat absorbed by Furnace Q = (Hc +Hr)x effectiveness A Lmtd =(5.769 + 187.768)x 0.79 x 112 x(1558 - 487) = 18.34MW Heat absorbed by adjacent Radiation chamber Heat absorbed by adjacent wall = 5.67 x 10^-8 x 0.715 x(1-0.715) x 18 x 0.85 x(15584 –4874) = 1.0316MW Total heat lost by Gas = 18.34 +1.0316 = 19.3716 MW

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Heat 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 HEATER 4.1 INTRODUCTION: The steam temperature above its saturation temperature corresponding to the pressure is achieved by introducing super heater coils. In the modern industrial

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world, it is expected to get the maximum attainable temperature for a pressure, since the cycle efficiency depends on pressure and temperature. But nature restricts the maximum 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 in other regions. The super heater will work out approximately 5% of the cost of the boiler and the material must be selected carefully. If improper selection of material leads losses due to oxidation and improper sizing or thickness selection leads larger reserve 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 furnace design i.e. furnace outlet temperature which is explained in furnace chapter, fuel characteristics, degree of super heat and manufacturer practice. The fuel characteristics such as ash content, ash particle size, salts of alkali metals like sodium and potassium which volatilize in the process of combustion and condenses as a sticky substance at a temperature corresponding to super heater tubes. These decide transverse pitching of super heater. Some light density fuels like husk burns at the top resulting high heat flux input to super heater reducing heat transfer area requirement itself. 4.3 POINTS TO BE NOTED WHILE DESIGNING SUPER HEATER 1.0 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 required 4.0 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 other items. 4.4 CLASSIFICATION OF SUPER HEATERS Super heaters are normally classified as Radiant and Convection super heaters. Radiant super heaters are located in the radiant zone receiving energy directly from the 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 flow pattern. 1 Radiant super heater located at radiant zone of boiler 2 Convection super heater located at the convection zone of boiler

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3 In modern fluidized bed boilers, to achieve higher steam temperature a portion of super heater located in the fluidized bed called bed super heaters. Depends on arrangement 1 Horizontal 2. Vertical 3. Inline 4. Staggered

Depends on the flow pattern 1. Cross flow super heater 2. Parallel flow super heater 3. Counter flow super heater Horizontal arrangement of super heaters have a advantage of easy drainage, which is quite important in boilers where shut down will be for longer periods. Drainability helps in non-accumulation of salts in water or steam inside the tubes. In vertical arrangement the problem of expansion can be tackled easily than horizontal arrangement. Staggered bundles are difficult to clean but they offer a marginal improvement in heat transfer coefficient. Inline arrangement provides lower gas side pressure drop. Counter flow arrangement offers a slight improvement in log mean temperature difference and there by decreasing surface area compared to parallel flow. The parallel flow arrangement leads to cooler tubes. Metal temperature can be higher with counter flow arrangement since both gas and steam temperature are higher at steam exit portion. 4.5 DESIGNING A SUPER HEATER Designing a super heater or any other heat transfer involves three steps one is the thermal design, second is the mechanical design and other is the performance calculation. In a thermal design we have to perform the heat duty, heat transfer and surface area required for the known thermal input and output parameters In a mechanical design thickness and material with standing capacity are to be checked. Performance calculation involves confirmation of the designed equipment for various loads.

4.6 OVERALL HEAT TRANSFER ACROSS BANK OF TUBES According to the gentleman Mr. Colburn the following equation can be applied for gas/air flowing normal to the bank of tubes Nusselts number = 0.33 * (Reynolds number)0.6 * (Prandl number)0.

3

While calculating Reynolds number external tube diameter has to be used. This expression can also expressed after introducing geometry factor. Geometry factor F has to be taken from graph depends on transverse pitch to diameter ratio and longitudinal pitch to diameter ratio Nusselts number = 0.35* F * (Reynolds number)

0.6

* (Prandl number)

0.3

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According to another gentleman Mr. Grimson the outside heat transfer coefficient was explained by the following equation Arrangement Factor F For Inline arrangement ST/d 1.25

1.5

2.00

3.00

SL/D Re 2000 8000 20000 2000 8000 20000 2000 8000 20000 2000 8000 20000

For staggered arrangement

1.25

1.50

2.00

3.00

1.25

1.50

2.00

3.00

1.06 1.04 1.00 0.95 0.96 0.95 0.73 0.83 0.90 0.66 0.81 0.91

1.06 1.05 1.00 0.95 0.96 0.95 0.73 0.83 1.00 0.66 0.81 0.91

1.07 1.03 1.00 1.03 1.01 1.00 0.98 1.00 1.00 0.95 1.02 1.01

1.00 0.98 0.95 1.03 1.01 0.98 1.08 1.02 1.00 1.00 1.02 1.00

1.21 1.11 1.06 1.17 1.10 1.04 1.22 1.12 1.09 1.26 1.16 1.14

1.16 0.99 1.05 1.15 1.06 1.02 1.18 1.10 1.07 1.26 1.15 1.13

1.06 0.92 1.02 1.08 1.00 0.98 1.12 1.04 1.01 1.16 1.11 1.10

0.96 0.95 0.93 1.02 0.96 0.94 1.08 1.02 0.97 1.13 1.06 1.02

Nusselts number = B * (Reynolds number) N B and N are factors governed by the geometry, the values of B and N is given in the table. Many such persons worked on the heat transfer and gave various correlation for certain pre defined condition and hence for practical purposes certain factors to be considered for its accuracy. I have also tried to give those factors in worked examples. In the Colburn or Grimson equation correction factor for the heat transfer coefficient for gas angle of attack on the tube has to be calculated into account. Degree ° 90

80

70

60

50

40

30

20

10

Factor

1.0

0.98

0.94

0.88

0.78

0.67

0.52

0.42

1.0

ST/D SL/D

1.25 B

1.5 N

B

2.0 N

B

3.0 N

B

N

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STAGGE 1.25 1.5 2 3 INLINE 1.25 1.5 2 3

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0.518 0.451 0.404 0.31

0.556 0.568 0.572 0.592

0.505 0.46 0.416 0.356

0.544 0.562 0.568 0.58

0.519 0.452 0.482 0.44

0.556 0.568 0.556 0.562

0.522 0.488 0.449 0.421

0.562 0.568 0.57 0.574

0.348 0.367 0.418 0.29

0.592 0.586 0.57 0.601

0.275 0.25 0.299 0.357

0.608 0.62 0.602 0.584

0.10 0.101 0.229 0.374

0.704 0.702 0.632 0.581

0.0633 0.0678 0.198 0.286

0.752 0.744 0.648 0.608

Correction factor for B corresponding to number of tubes deep Number deep

1

2

3

4

5

6

7

8

9

10

Staggered

0.68

0.75

0.83

0.89

0.92

0.95

0.97

0.98

0.99

1.0

Inline

0.64

0.8

0.87

0.9

0.92

0.94

0.96

0.98

0.99

1.0

Simillarly Dittus & Boelter correlation gives heat transfer coefficient on inside tube. Nusselts number = 0.023 * (Reynolds number)

0.8

* (Prandl number)

0.3

Consider the super heater tubes of outer radius r1 & r2 respectively. Thermal conductivity of material k. Cold fluid (steam) is flowing steadily inside the tube Tf1 and 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 includes convection heat transfer and non luminous heat transfer as explained in furnace chapter) STEAM

ho

Tf2

tw2 outside conve.heat transfer =3 tw1 Hot gas hi Tf1

Conduction =2 inside conve. heat transfer =1 Tube thickness’t’

Convection heat transfer phenomenon in super heater tube = Q/l = hi *2 * *r1 [ tw1 – tf1]………………………………………….1

ql

= 2*

* k (tw2 – tw1)/ln(r2/r1)……………………………………..2

= ho * 2 *

* r2 [ tf2 – tw2]…………………………………………3

1=2=3 the heat transferred at a given section at given time is equal.

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tw1 – tf1 = ql / (hi*2* *r1) tw2 – tw1= ql * ln(r1/r2) 2* *k tf2 – tw2 = ql /(ho *2* *r2) Adding the above equations we get,

ql

1

ln(r2/r1)

tf2 – tf1 = 2*

ql

=

=

+

hir1

2*

k

hor2

*(tf2 – tf1)

1 hir1

Q/l

1

+

+

ln(r2/r1) + k

1 hor2

U [tf2 – tf1]

Divide by do Q/( dol) = U[tf2 – tf1]/ do U do

=

2* 1 hir1

U = do

+

ln(r2/r1) + k

1 hor2

1 1

+

2 hir1

1

ln(r2/r1)

+

2 k

do

+

1 2 hor2

d0 ln(d0/di)

+

1

= U

hidi

2k

Introducing fouling factors in steam side and gas side we can get,

ho

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1

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do

+

d0 ln(d0/di)

+

1

= U

+ffo +ffi hidi

2k

ho

In the above equation each unit refers to a resistance in heat transfer 1/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 steam side film) ffo is outside fouling resistance ffi is inside fouling resistance

For extended surface following heat transfer coefficient equation can be applied. 1

At

+ At d0 ln(d0/di)

+

1

= U

+ffo +ffi A t/Ai hiAi

Aw

2k

ho

Where, At is total surface area. Ai is inside surface area. Aw is average surface area. is fin effectiveness Q = U * A * [tf2 –tf1] for specific tube, while considering bundle of tubes log mean temperature 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 °C Metal temperature of bare tube can be estimated easily by calculating various temperature drop across the resistance given above. This simple equation is applicable for only bare tube surface. For extended surface fin temperature calculation involves a detailed procedure which is discussed separately.

4.7 STEAM TEMPERATURE CONTROL

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Super heater temperature depends on boiler working pressure. The degree of super heat varies directly with respect to boiler working pressure. The various steam outlet temperature corresponding to a pressure is fixed up by cycle efficiency and turbine manufactures practice. Operating variables like flue gas temperature inlet, load, excess air, fuel creates fluctuation in steam temperature leaving the final super heater. If there is any fluctuation in steam temperature, there is a change in volume of steam which will affect the turbine performance, since turbines are designed for predetermined steam volume flow between stator and rotor and also the exhaust steam temperature quality will varies with steam fluctuation which will affect condenser performance. Hence it is mandatory to have a steam temperature control. In practice two types of attemperators (de-super heater) is widely used in boiler industry, they are 1.0 Spray type 2.0 Surface type. In the above two types, spray type attemperator gives a faster temperature control compared to surface type. However quality water spraying into the super heater should be taken care for boiler and turbine life. Attempeator temperature technique is nothing but a simple energy balance. It consists of spraying water in mist form in between stages, depends on the final steam temperature and pick up by individual stages. The quantity of spray water varies with respect to load and it is controlled by automatic control loops. Spray inside desuperheater accomplished by means of nozzle arrangement. Surface type attemperators are further divided into two types submerged type and Shell and tube type. In submerged type the steam coils is submerged below the drum water level. Part of the steam flows through the submerged coils and part of the steam by-pass the coil. The flow of steam through the submerged coil is regulated in such a way that the outlet steam temperature is of desired level after mixing. In shell and tube type attemperator super heater header itself modified into an exchanger. Steam from super heater coils enter this intermediate header and leaves to the second stage. In the super heater header the cooling water coils enter in both the sides and leaves. The steam gets cooled when it contact with low temperature cooling coils. In this steam temperature is adjusted by regulating the flow of cooling water 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

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Hfw kcal/kg) Steam flow(M2 kg/hr Steam flow (M1 kg/hr & enthalpy H2) & enthalpy H1) DE SUPER HEATER

SEC. SUPERHEATER

PRI. SUPER HEATER

Energy balance, M1* H1 + M * Hfw

= M2H2……………………………………………1

Mass balance, M1 + M

= M2……………………………………………….2

From 1 & 2, M1*H1 + M*Hfw

= (M + M1) *H2

Quantity of water required spray, M = M1 *

{ H1 H2}/{H2

Hfw}

Ex.01. Determine the spray quantity required for a 60TPH boiler having a primary super heater steam outlet temp of 332°C and secondary super heater inlet steam to be desuperheated to 316°c. The spray water temperature is 105°C. Enthalpy of PSH = 721.7 kcal/kg Enthalpy of SSH = 711.85 kcal/kg Enthalpy of spray water = 105.5 kcal/kg. Spray water required = (721.7 - 711.85)*60000 ( 711.85 –105.5)

= 974.7 kg/hr.

Ex02. Estimate the heat transfer area required for surface type desuper heater immersed in drum water operating at a pressure of 45 kg/cm²(g) and the steam at

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the outlet of primary super heater is 380°C and the steam temperature at the inlet of super heater is 350°C. the steam flow is 25000 kg/hr. The heat transfer coefficient outside the tube to water is natural convection heat transfer coefficient, governed by the equation Nu = 0.54 ( Gr Pr) 0.25 ‘d3ρ²gβ∆T Cpµ ‘hodo/k = 0.54 µ² k

0.25

Assume surface type desuper heater of tube diameter 38.1 x 3.25 mm Properties of liquid at film temperature (saturation temperature at 45 kg/cm²(g)) Density of liquid kg/m3 = 787.5 kg/m3 Acceleration due fo gravity = 9.81 m/sec² Dynamic viscosity = 0.000103067 kg/ms Volumetri coefficient of expansion ß = 0.284 1/°C Prandtl number = 0.83299 Thermal conductivity = 0.00061184 Kw/m°C Specific heat Cp = 4.94512 kj/kg Temperature difference between wall & water = 5° (assumed)

ho = 0.54

0.25 0.03813 x 787.5² x 9.81x0.281x 5 x 0.83299 x 0.61184 0.000103067² 0.0381

= 19497 W/m²°K Inside heat transfer coefficient is governed by forced convection, Nu = 0.023 Re0.8Pr0.4 Properties of steam at average temperature (380 +350)/2 = 365°C Density of steam kg/m3 = 16.169 kg/m3 Dynamic viscosity kg/ms = 2.28175 x 10-5 Prandtl number = 0.9576 Thermal conductivity W/m°C = 5.945 x 10-2 Velocity inside the surface assume 25 m/sec Reynolds 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

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= 476.6 w/m²K Metal resistance = d0 ln(d0/di) /2 Km Metal conductivity = 49.5 w/mK Metal resistance = 0.0381 x ln(0.0381/0.0316) 2 x 49.5 = 7.1988 x 10-5 Overall heat transfer coefficient 1/U = 1/ho + 1/hi + Rm + ffi + ffo inside and outside fouling factor due to scale assume = 0.0002 m²/w°K 1/U = 1/19497 + 1/746 + 7.1988 x 10-5 + 0.0002 Overall heat transfer coefficient = 601.04 W/m²K Heat transfer area required = Q/Ulmtd = 25000( 3158.49 – 3083.08) x1000 3600x 601.04 x (( 380-350)/ln(123/93)) = 8.12 m² Ex.03. Calculate gas outlet temperature for a super heater intended to raise steam from 214°C to 258°C. steam flow 15.28 kg/sec, pressure 20 bar(a), tube size 38.1 x 3.25 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 3.06 meter 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 of super heater 88 sq.m. the super heater was enclosed in a water wall having EPRS area of 21 sq.m and gas is flowing 90° to super hater tubes. Consider heat transfer effectiveness of 82% and 71% for enclosure. Partial pressure of water =0.1158 bar and carbon di oxide = 0.1249bar. Steam inlet temperature Steam outlet temperature

= 214°C = 258°C

Average steam temperature

= 236°C

Properties of steam Density kg/cu.m Dynamic viscosity kg/ms Prandtl number Thermal conductivity W/mK

= = = =

9.328189 1.72232 x 10-5 1.083 4.346819 x 10-2

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Steam velocity

= 15.28 x 4 /(9.3281 x74x x 0.0316²) = 28.225m/s Forced convection inside heat transfer coefficient is Hi = 0.023 x Re0.8 x Pr0.4 x k/di 0.8 9.328x0.0316x 28.255 = 0.023x 1.7223 x 10

-5

0.4 x 1.083

x 4.346819 x10-2 0.0316

= 1152 W/sq.mK Forced convection outside heat transfer coefficient is Assume gas outlet temperature as 560°C Film temperature = (average of gas temperature + average of steam temperature/2 =(659 +560 +214 + 258)/4 =422.75°C Gas properties at film temperature Density kg/cu.m = 0.503 Dynamic viscosity kg/ms = 3.1798 x 10-5 Prandtl number = 0.7126 Thermal conductivity W/mK = 0.05195

For inline arrangement ST/d = 80/38.1 = 2.099; SL/d = 78/38.1 = 2.047 Arrangement factor fe = 1.18 Gas flow area = (3.06 x 2.7 – 37 x0.0381x2.7) = 4.4558sq.m Gas velocity = 19.03/(0.503 x 4.4558) = 8.49 m/s Reynolds number = 8.49 x0.503 x 0.0381/3.1798x10-5 = 5116.8 hc = 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.mK Radiation heat transfer coefficient Beam length = 3.4 x(0.08 x0.078x1-0.00114)/( x0.0381x1) = 0.144876m Attenuation factor = (0.8 +1.6x0.1158)(1-0.38x(883/1000))(0.1158 + 0.1249) ((0.1158+0.1249)x0.14486)

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= 0.8439 emissivity of gas = 0.9 x(1-e-0.8439x0.14486) = 0.10358 hr = 5.67 x 10-8 x 0.85 x 0.10358x(8834 –5094) (883 –509) = 7.218 W/sqmK total 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

= 0.0381/(1152x0.0317) + 0.0381 ln(0.0381/0.0317) 2 x 49.844 U = 69.898 W/sq.mK

ho

+ 1/75.798

Heat 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 transfer coefficient be 68.58 w/sq.mK and non-luminous radiation heat transfer be hr = 5.67 x 10-8 x 0.85 x 0.10358x(8834 –4874) (883 –487) = 6.9538 w/sqmK ho = 68.58 + 6.9538 = 75.53 W/SqmK Q = 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 flowing through the super heater is 55000 kg/hr and the Outlet pressure of steam is 19

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kg/cm²(a) and the number of parallel path is 74 nos. and coil developed length is estimated as 11.64m, it has 3 nos. 180° bends and 2 nos. 45° bends. The steam inlet condition is dry saturated and outlet condition is 255°C. the super heater tube size is 38.1mm x 3.64mm including the positive tolerance of the tube. Super heater steam flow Steam flow/coil

kg/hr :55000 kg/sec:0.206

Steam parameter at inlet Temperature Pressure Steam parameter at outlet Temperature Pressure

°C : 209.84 kg/cm²(a): 19.34(assumed by trial and error) °C : 255 kg/cm²(a): 19

Average steam Parameter Temperature Pressure Density Dynamic viscosity Velocity through tubes

°C : 232.42 Kg/cm²(a):19.17 kg/cu.m: 9.20739 Kg/s.m : 1.71 x 10-5

= 0.206 x 4 9.20739x x(0.03082²) =29.99 m/s

Reynolds number = vd/µ = 9.20739 x 29.99x 0.03082/1.71x 10-5 = 497679 Friction factor = f = (0.3964/Re0.3 ) + 0.0054 = (0.3964/4976790.3) +0.0054 =0.0131454 Pressure 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 = 0.47v²/2g For 45° bend = 0.12v²/2g Total bend loss = ( 3x0.47 + 2x 0.12) 29.99²x9.20739 /(2 x 9.81) = 696.4 mmWc = 0.0696 kg/cm² Pressure loss in entry and exit

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Entry and Exit loss = 1.5 x 29.99² x 9.20739/(2 x 9.81) = 633 mmWC = 0.0633 kg/cm² Total pressure drop = 0.209+0.0696+0.0633 =0.3419 kg/cm²

5.0 DRUMS

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5.1 INTRODUCTION The recent development in boiler field involves, critical pressure boilers or once through boilers in which drum is not necessary. But other than critical pressure boilers, it involves two regimes of liquid and vapour phase where separation of steam from liquid surface takes place in steam drum. The number of drums required for a boiler depends on evaporator requirement, boiler pressure, stability, manufacture experience on the type of configuration. Widely single drum boiler and bidrum boiler are in use and in certain cases three or four drum design also available but they are outdated now. For higher capacity and high pressure(more than 70 kg/cm^2(g) pressure) boilers are economical with single drum boilers and for industrial 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 DRUMS In world wide practice drums are designed in cylindrical shape with two dished end with or without man hole at its end or cylindrical shell with tube sheet at its end or cylindrical 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 analyze about 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

Y Let the steam drum be a spherical shell of internal diameter ‘d’ and thickness ‘t’ and subject to an internal pressure of intensity p Bursting force P = p x projected area = p x

x d²/4

Let 1 be the tensile stress induced in metal at section XX Resisting force =

1x

xdxt

Bursting force = Resisting force

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px

x d²/4

=

Ganesh kumar

1x

xdxt

1 = pd/4t since any section taken in diagonal to a sphere is symmetrical, the stress at any point in the metal of sphere will be equal and opposite and hence there will not be any shear stress. The strain at any direction is given by e1 = e1 =

1/E - 2/mE ( 1 = 2 ) 1/E(1-1/m) where E is Young’s modulus and 1/m is poisson ratio.

Hence For the safety of shell thickness = t pd/4 1 If the shell has been riveted then factor e called efficiency factor to be introduced i.e., ‘t

pd/4 1e

Stress inside a square chamber

d

d L

A square chamber has three planes of action and the stress and resisting force in all the three direction to be checked for its stability Let ‘d be the sides of the square section ,L be the length, t be the thickness and p be the intensity of pressure

Let 1 be the tensile stress in direction xx. The force acting on xx be

Bursting force

= p x projected area ______ = p x d² + d² x L _ P = p x 2 d L…………………………………………..1

Area resisting this force = 2 x L x t Resisting force

= 1 x 2x L x t…………………………………………….2

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Equating 1 and 2 1 = pd/ 2 t

or

t = pd/

1

Similarly in zz direction 2 be the tensile stress Bursting force P = p x d x L Resisting force

=

2x2xLxt

or

t = pd/2 2

2 = pd/2t

Similarly yy direction 3 be the tensile stress Bursting force P = p x d² Resisting force 3 = pd/4t

= or

3 {(d +2t)² - d² } =

3x4xdxt

(considering 4t² as negligible)

t = pd/4 3

Stress inside a cylinder Z X

Y X

p p

Z

Y

Let us consider the length and thickness of a cylinder be L and t whose mean diameter is d and the internal pressure be p. In this case section xx and zz experience same force and hence we have to calculate force in xx and yy section only Let

1 be the tensile stress in direction xx. The force acting on xx be

Bursting force

= p x projected area = p x d x L

Resisting force =

1x2xtxL

Equating Bursting force and resisting force =

1 = pd/2t

or t = pd/2 1

This stress induced in circumferencial of the shell is called circumferencial stress of hoops stress. Similarly in longitudinal direction area resisting the pressure =

x d²/4

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Force 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 the section yy to the total force on the end of the shell 2x

xdxt = px

2 = pd/4t

or

x d²/4

t = pd/4 2

This stress is longitudinal stress. Hence at any point in the metal of the shell there are 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/8t

Circumferencial strain e1 = 1/E - 2/mE Where E is the young’s modulus and 1/m is poisson ratio. From the above equation minimum thickness required for a cylindrical drum can be determined. Let be the permissible tensile stress for the material than for shell to be safe major principal stress 1 should not exceed From this

pd/2t or t

Working pressure p

pd/2

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 the thickness the formula modified into, W.P

2x

= 2 x x e x (t –c) (di + t - c)

where c is corrosion allowance 0.03 inches di is internal diameter

As per ASME PG27.1 , introducing addition of corrosion allowance to the thickness the formula modified into, W.P

= 2 x x e x (yt –c) (do –2y( t - c))

where c is corrosion allowance 0.03 inches do is outside diameter. ‘y is temperature correction coefficient.

Comparing the three geometries at their xx, yy and zz axis,

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XX

YY

ZZ

SPHERE

t =pd/4

t =pd/4

t =pd/4

SQUARE

t =pd/

t =pd/4

t =pd/2

t =pd/2

t =pd/2

t =pd/4

CYLINDER

For a same pressure, material, dimension the minimum thickness required for sphere, cylinder and square is in the ratio of 1: 2 : 2.83 (thickness governing formula shown in bold letters in the tabular column ) Sphere as a drum, there will be problem of having holding capacities and then fabrication difficulties. Hence sphere can not be used in place of drum due to the smaller volume. For square when compared to circular drums it calls for 1.414 times greater thickness in turn weight of the drum. Number of welds for making a square is more where as a single fusion weld will do for circular drums. Square chamber has to be placed in diagonal position for the lift of steam and therefore opening has to made in the bends, which is very stress prone area. Practically speaking square sections are inferior in strength compare to circular section. According to IBR regulation square sections can be used as header and circular sections can be used for both drums and header. The main advantage of square header is at one side openings for tubes can be made and other opposite side openings for mechanical cleaning can be made for straight inclined tubes. For this purpose some manufacturer prefer square header, where inclined tubes are used.

5.3 STUBS AND ATTACHMENTS IN THE STEAM DRUM/SHELL According to the IBR regulation 281, there are certain minimum requisite of mountings, fittings and auxiliaries to be provided in pressure vessels for the safety of the system. IBR regulation says that every boiler to be provided at least with following 1.0 2.0 3.0 4.0 5.0 6.0

Two safety valves (minimum diameter 0.75 inches) Two means of indicating water level gauge A steam pressure gauge A steam stop valve A feed check valve 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 valve 8.0 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 be fitted 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 or user.

5.4 MAXIMUM PERMISSIBLE UNCOMPENSATED OPENING IN DRUM The opening in drum made for stubs, manhole, mud hole will weaken the plate which calls for strengthening the plate by compensating plates. According to IBR the maximum diameter for uncompensated opening is given by the following formula _________ Maximum uncompensated opening = ½ (D +T)T + N Where 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 -P If the diameter of opening is less than the calculated value then it is not necessary to give compensation for the drum. If it exceeds the limit compensation have to be provided. Area of compensation in Xaxis opening

=

dn x es where, dn nozzle internal diameter of

5.5 SIZE OF THE DRUM The steam drum must be large enough to accommodate drum internals, the necessary steam surface for steam separation and water holding capacity. The drum also provide sufficient space for change in water level that occur with change in load. A sudden increase in steam demand cause a temporary pressure surge, until firing rate is increased sufficiently for more steam generation. During this interval due to lower pressure, steam volume throughout the boiler is increased (pressure

Steam generator

Ganesh kumar

1/volume). This results in SWELL raise in water level in the drum. The raise in water level depends on the rate at which heat and feed inputs can be changed to meet the load demand. The higher the drum diameter better the control of raise in water level In GT HRSG during starting the steam generation cause sudden increase in drum level 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 once the increased water level goes to high-high level blow down valve will be opened to control the increase in water level. Normally blow down valve for HRSGs are sized based on the same, by finding the 70% evaporator volume less the drum volume between low to high-high. This difference in water has to be discharged with in 2 to 3 minutes. STEAM SPACE In the steam drum water and steam regions are clearly separated, the space above the normal water level is called steam space. Steam space governs the steam loading of the drum. Steam loading is the ratio between steam space and the specific volume of the steam for pressure it is operating(M^3/m^3hr). Steam loading for different pressures is fixed by the manufactures according to their practice or experience. Steam loading is one of the important decision factor in deciding the size of the drum. Steam loading ensures the velocity of separation in the inter phase and steam carry over.

Let L be the length of steam drum and D be the internal diameter, r be the radius and x be the distance from center line to normal water level.

o

D

k a

b

x

nwl ‘h

c L

Sin

= ak/oa

= Sin-(x/r) = (180 – 2 ) ‘ab = 2 ac = 2.r. Sin( /2)

Steam generator

The area of the arc oacb =

Ganesh kumar

x D² x 4x 360

The area of triangle oab =1/2 . ab . X The area acba = area arc oacb – triangle oab The steam space area in cylinder = D² - area acba 4 Steam space volume = ( D² /4 - area acba) (L +2l) + 2 volume of dished head ‘above normal water level Volume of dished head for different shapes.

TYPE

FULL VOLUME M^3

VOLUME UPTO NORMAL WATER LEVEL M^3

SEMI ELLIPSOIDAL

. D3/24

.h².(1.5D –h)/12

HEMI SPHERICAL

.D3/12

.h².(1.5D –h)/6

TORIS SPHERICAL

Ex 01 Estimate the shell thickness required for a single drum boiler, whose design pressure is 73 kg/cm² and the internal diameter is 1300mm. The drum is located outside the gas path. Single drum the ligament efficiency is 1. However for practical estimation 0.95 is considered. As per IBR 270 regulation Minimum 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 of boiler design pressure. If the drum is exposed in gas path additional temperature

Steam generator

Ganesh kumar

allowance of 28°C has to be added to saturated temperature and for this stress value has 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 opening in the dished end of size 545 x 500 mm is made for man hole opening. The material of construction is SA 516 Gr.70.

I

150 60

II 150 Inside 545x500 opening

Minimum required thickness for dished end calculated as per IBR 277 & 278 amendment 1995. W.P = 2f(T-C)/(DK) Where, T = Minimum thickness W.P = Maximum working pressure D= Outside diameter = ( 1300+ 2xT) ‘f = Allowable Stress = 1230.4 kg/cm² (As per ASME section II part D 1995) K= Shape factor C= An additive thickness = 0.75mm Reinforcement of large openings ‘d = Diameter of opening = 545 mm A= Effective cross section of reinforcement in mm²(shaded portion) T = Minimum thickness of dished head = 49 mm(assume) Limit of Reinforcement

Tt = Actual thickness of the nozzle L2 = d.Tt _______ (544 x60) = 180.83 mm Since the lenth of the nozzle projection inside the dished end is 150< 180.83, the whole length is considered for compensation.

Area of reinforcement

Steam generator

Ganesh kumar

I = 60 x150 x 2 =18000 mm² II= 44.3 x 210 x2 = 18606 mm² I + II = 18000+ 18606 =36606 mm² Imaginary diameter = d –(A/T) = 545 - 36606/49 = -ve Assuming di =1 Since the value is negative Conditions T/D < 0.1 di/D < 0.5 T/D = 49/1398 =0.035
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