Thursday, July 30, 2015


 I have been    conceptualizing the subject on analytical paradox since beginning my career.   Though they  have assigned me other wakes of the technical matters like what we have produced, on particular quality ,if not that has been  deviated what is the result etc.etc. and its effect on market reaction. As example what is the density of the product , if it is not then what is it? One day this density is very correct , other day the same density is not correct. Every product  has more or less variation in quality , so either way you are at fault and you are a product losing company money as well as your credibility and your pay package. 
I was thinking of application of probability of finding fault analytically with mathematics. Instead the probable variation of density. I started reading and started practicing probability on another aspect. As probability is the mathematics of assumptions and there is no correct and incorrect answer , every answer is write. Eventually any one can devise his own method  to find a minimum possible way to escape and excel. You may call it escape probability and excellent productivity.
Now one can ask why this paradox. When you have three asking proprietors and three asking managers  . You are really in a paradox .Simultaneously you do not know which manager is favorite for a particular director on a particular day , so pre judgement  is also a hard task for a hard core mathematician . But gradually i have mastered the method of my own probability function. The probability of escape and escaping paradox.
I have started gathering data. Even i have gathered data , believe it that simply  counting of footsteps of one particular manager while he used to return from a particular directors  chamber can be predicted if this day is his or others . Similarly  eye movement of others, following his posture,  following the style of opening his chamber door , slamming or gently closing the door. The sound of the chair, asking for the helping hand  is the probability  data of my technical analysis. Now it is clearing my judgement day by day.
Sincerely, leaving the product quality on heaven , i have been gathering this kinds of data to prepare and practicing  what to say on Monday ,Tuesday following days. Now i am the expert in this domain, leaving the product . The product is always good or best to the hard core salespersons and hardcore liars  .
I have written all the data in my secret diary , every day i plot them on my own method , not exactly on my own, i have best respect to the geniuses from whom i  have learned the basics and the higher order , but as i have told earlier that in probability forecasting and taking judgement by the past and comparing the present is the key.
But success has not come in a day , it has taken a lot of hard work. I have been spending countless sleepless nights to analysis . but when there is question of survival not only for me but others this  efforts  have given its fruit. The pitch or slight modulation of speech, slightest variation signals me, what  he wants  to listen today and what at that particular time. Even if it is fifteen minutes earlier or thirty minutes afterwards , the signals will be other. But you can ask how it become a data, how can you quantify that , how can you write it but the question is why i am going to tell you. You follow my path, you analyze of your own , you can invent your  own method .  
But now it is striking me , is not it possible to quantify it. So my secret diary will be a good book of my experience and knowledge. I am finding a wizard to associate with me.
Is a  great batsman is only a statistics, every one will say no. But is its batting style, no.  More stylish and technically perfect batsman has sunk in the history. The judgement he took which delivery is expected to come from a particular bowler on the next ball , is it quantifiable, truly it is yes. I know a batsman who can tell by the returning style or pattern, of  that  bowler to his starting point and body movement , what is he going to deliver to him.  Is it thought reading, yes, very much yes.
It is also a sense. By nature and by practicing , he has got mastery on it.
Is it came without hard work. No certainly not.
If any one knows how to quantify it or give reference, please call me. I have stuck on it.

     

Thursday, November 27, 2014

Saving Energy

SAVE ENERGY SAVE FUEL  

TIPS ON ENERGY SAVING

IN HOME APPLIANCES AND ELECTRICITY SAFETY

The Domestic Sector accounts for 30% of total energy consumption in the country. There is a

tremendous scope to conserve energy by adopting simple measures.

This information is a guide, which offers easy, practical solutions for saving energy in Home

Appliances. Please, take a few moments to read the valuable tips that will save energy & money

and ultimately help conserve our natural resources.

It would be useful to know which gadget consumes how much electricity. Economic use of home

appliances can help in reducing electricity bills.



By following these simple tips one can save energy to a large extent.

Lighting

Ø Turn off the lights when not in use

Ø Take advantage of daylight by using light-colored, loose-weave curtains on your windows

to allow daylight to penetrate the room. Also, decorate with lighter colors that reflect

daylight

Ø De-dust lighting fixtures to maintain illumination

Ø Use task lighting; instead of brightly lighting an entire room, focus the light where you need

it

Ø Compact fluorescent bulbs are four times more energy efficient than incandescent bulbs

and provide the same lighting

Ø Use electronic chokes in place of conventional copper chokes

Fans

Ø Replace conventional regulators with electronic regulators for ceiling fans

Ø Install exhaust fans at a higher elevation than ceiling fans

Electric iron

Ø Select iron boxes with automatic temperature cutoff

Ø Use appropriate regulator position for ironing

Ø Do not put more water on clothes while ironing

Ø Do not iron wet clothes

Kitchen Appliances

Ø Mixers

§ Avoid dry grinding in your food processors ( mixers and grinders) as it takes longer

time than liquid grinding

Ø Microwaves ovens

§ Consumes 50 % less energy than conventional electric / gas stoves

§ Do not bake large food items

§ Unless you're baking breads or pastries, you may not even need to preheat

§ Don't open the oven door too often to check food condition as each opening leads to a

temperature drop of 25°C

Ø Electric stove

§ Turn off electric stoves several minutes before the specified cooking time

§ Use flat-bottomed pans that make full contact with the cooking coil

Tips on Energy Savings Page 3 of 5

Ø Gas stove

§ When cooking on a gas burner, use moderate flame settings to conserve LPG

§ Remember that a blue flame means your gas stove is operating efficiently

§ Yellowish flame is an indicator that the burner needs cleaning



§ Use pressure cookers as much as possible

§ Use lids to cover the pans while cooking

§ Bring items taken out of refrigerators (like vegetables, milk etc) to room temperature

before placing on the gas stove for heating

Ø Use Solar Water Heater – a good replacement for a electric water heater

Electronic Devices

Ø Do not switch on the power when TV and Audio Systems are not in use i.e. idle operation

leads to an energy loss of 10 watts/device

Computers

Ø Turn off your home office equipment when not in use. A computer that runs 24 hours a

day, for instance, uses - more power than an energy-efficient refrigerator.

Ø If your computer must be left on, turn off the monitor; this device alone uses more than half

the system's energy.

Ø Setting computers, monitors, and copiers to use sleep-mode when not in use helps cut

energy costs by approximately 40%.

Ø Battery chargers, such as those for laptops, cell phones and digital cameras, draw power

whenever they are plugged in and are very inefficient. Pull the plug and save.

Ø Screen savers save computer screens, not energy. Start-ups and shutdowns do not use

any extra energy, nor are they hard on your computer components. In fact, shutting

computers down when you are finished using them actually reduces system wear - and

saves energy

Refrigerator

Ø Regularly defrost manual-defrost refrigerators and freezers; frost buildup increases the

amount of energy needed to keep the motor running.

Ø Leave enough space between your refrigerator and the walls so that air can easily

circulate around the refrigerator

Ø Don't keep your refrigerator or freezer too cold.

Ø Make sure your refrigerator door seals are airtight

Ø Cover liquids and wrap foods stored in the refrigerator. Uncovered foods release moisture

and make the compressor work harder.

Ø Do not open the doors of the refrigerators frequently

Ø Don't leave the fridge door open for longer than necessary, as cold air will escape.

Ø Use smaller cabinets for storing frequently used items

Ø Avoid putting hot or warm food straight into the fridge

Tips on Energy Savings Page 4 of 5

Washing machines

Ø Always wash only with full loads

Ø Use optimal quantity of water

Ø Use timer facility to save energy

Ø Use the correct amount of detergent

Ø Use hot water only for very dirty clothes

Ø Always use cold water in the rinse cycle

Ø Prefer natural drying over electric dryers

Air Conditioners

Ø Prefer air conditioners having automatic temperature cut off

Ø Keep regulators at “low cool” position

Ø Operate the ceiling fan in conjunction with your window air conditioner to spread the

cooled air more effectively throughout the room and operate the air conditioner at higher

temperature

Ø Seal the doors and windows properly

Ø Leave enough space between your air conditioner and the walls to allow better air

circulation

Ø A roof garden can reduce the load on Air Conditioner

Ø Use windows with sun films/curtains

Ø Set your thermostat as high as comfortably possible in the summer. The less difference

between the indoor and outdoor temperatures, the lower will be energy consumption.

Ø Don't set your thermostat at a colder setting than normal when you turn on your air

conditioner. It will not cool your home any faster and could result in excessive cooling.

Ø Don't place lamps or TV sets near your air-conditioning thermostat. The thermostat senses

heat from these appliances, which can cause the air conditioner to run longer than

necessary.

Ø Plant trees or shrubs to shade air-conditioning units but not to block the airflow. A unit

operating in the shade uses as much as 10% less electricity than the same one operating

in the sun.

Tips on Energy Savings Page 5 of 5

Electrical Safety Tips for Homes

Electrical Hazards

Ø Shocks

§ Electric Shock causes an involuntary grip which prolongs the period of contact.

§ More the period of contact, more the damage

§ Passage of current through the heart , stops the blood supply to the brain , resulting in

loss of consciousness and termination of breathing

§ When a person standing at a height receives an electrical shock , it is most likely that

he will fall

§ Personal sensitivity to electrical shock varies from person to person

Ø Burns

§ Whenever an electrical flash appears, and if a part of a body comes within flashing

distance, burns can be caused

§ Burns may be caused by short circuits as well, because a short circuit could create an

electrical fire
Fuels are NOT EVERLASTING. ENERGY SAVING  IS RETAINING NATIONAL AND INTERNATIONAL PROPERTY SAVING
Preventive Measures

Ø Allow only a qualified person to attend to your electrical repairs

Ø Service your electrical equipment at frequent intervals through a competent electrician

Ø In case of a short circuit or a fire, switch off the mains instantly Make sure that you have

easy access to switch off the supply source quickly, in case of an emergency

Ø Make sure your extension cords are free from cuts, improper insulation, or joints

Ø Ensure pins of your plugs are tight and not loose

Ø Use switches of the correct current rating and preferably with indicators to indicate whether

the switch is ON/OFF

Ø Use appliances with 3 pin plugs and connect them to 3 pin sockets

Ø Do not overload electrical outlets or use extension cords in place of additional outlets

Ø Switch off electrical appliances when not in use

Ø Provide proper earthing for the building/house
DULONG (1820) = (80.8 x C) + (344.6 x H) – (43.1 x O) + (25 x S)BOIE (1953) = (84 x C) + (277.7 x H) – (26.5 x O) + (15.0 x N) + (25 x S)
SEYLER (1938) = (123.9 x C) + (388.1 x H) + (25 x O2) - 4269NEAVEL (1986) = (81.05 x C) + (316.4 x H) – (29.9 x O) + (23.9 x S) - (3.5x Ash)
MOTT & SPOONER (1940) OXYGEN < 15% = (80.3 x C) + (339 x H) - (34.7 x O) + (22.5 x S)GIVEN (1986) = (78.3 x C) + (339.1 x H) – (33.0 x O) + (22.1 x S) + 152
MOTT & SPOONER (1940) OXYGEN > 15% = (80.3 x C) + (339 x H) - (36.6 x O) + (0.17 x O2) + 22.5 x S

NOTE: these formulas are not valid for coal blends. 







Design   Efficiency sizing calculation  for CFB BOILER                                                                                                                                                    

FIRST STEP  :::COMPOSITION OF FUEL TO BE USED IS VERY MUCH ESSENTIAL TO KNOW BEFORE GOING FOR SIZING CALCULATION OF BOILER AND BOILER COMPONENTS

EXAMPLE

Let the composition of Coal 
 Ash: 45% 
Carbon: C= 25.90% 
Hydrogen  H2 = 3.80%
Oxygen O2= 8.50%
Nitrogen N2= 1.80%
Sulphur S=00.00%
Moisture = 15%
HHV=(ARB) 14233.556 KJ/KG              

ARB  =AS RECEIVED BASIS

Amount of air required per kg of fuel

Theoretical dry air required

MDA=[ [ 11.53 *C+34.34 *(H-O/8)+ 4.34 *S +A*S]  ] 
PUTTING THE ABOVE VALUES FROM THE ABOVE MENTIONED ULTIMATE ANALYSIS  CONSIDERING 
A= 2.38 FOR S CAPTURE
A=0 FOR NO SULPHUR FUEL
.

MDA=  3.926  KG/KG of FUEL  
















Sunday, November 23, 2014

AVOGADROS LAW SIMPLIFIED

PLEASE TAKE SAY THREE EQUAL VOLUME  CONTAINERS -----  IT MAY BE  n NUMBERS OF             



FILL THIS CONTAINERS WITH DIFFERENT GASES    G1,G2,G3,.........,Gn

NOW WHAT I AM DOING

I AM MAKING PRESSURE AND TEMPERATURE OF ALL THE CONTAINERS EQUAL

now if we count number of  molecules in each containers 

we will find the numbers are same in each containers



other expression based on above hypothesis

molecular volume of gases under same pressure and temperature is same for gases.


NOW WE CONCENTRATE A  SINGLE CONTAINER AND DEDUCE SOMETHING WHICH WILL BE APPLICABLE TO ALL CONTAINERS.


mathematical expression

pv= MRT

V MOL = MRT/p = 8314 T/p                             T= DEGREE KELVIN
                                                                          p= N/M^2
                                                                           V= M^3






UNIVERSAL GAS CONSTANT AND MOL VOLUME OF A GAS AT STP

The UNIVERSAL GAS CONSTANT (MR) of any gas is
8314 J/KG MOL degree K.

MOL volume of a gas at STP

V mol= 0.08314 T/p  M^3    where T= 15 deg C  p=1.013254.

Characteristic Gas Constant 

CALORIFIC VALUE OF FUEL

YOU TAKE I KG OF FUEL EITHER SOLID OR LIQUID
YOU REQUIRE TO BURN IT FULLY SO ALL HEAT IS LIBERATED FROM THE FUEL.
THE AMOUNT OF HEAT LIBERATED IS ITS CALORIFIC VALUE.
ITS UNIT WILL BE KJ/KG (KILO JOULE/KILOGRAM)

FOR GASEOUS FUELS IT CAN EXPRESSED AS KJ/CUBIC METER (KJ/ M^3) OF GAS AT (STP) STANDARD TEMPERATURE AND PRESSURE.


NOW UTILIZATION OF THE HEAT

HYDROGEN BURNS TO FORM STEAM. ALSO IF THE FUEL CONTAINS MOISTURE . THIS TAKE AWAY SOME HEAT FROM THE FUEL WHILE FORMING STEAM.
WHEN THIS STEAM IS CONDENSED AGAIN AT ITS INITIAL TEMPERATURE AT WHICH TEMPERATURE YOU STARTED TO BURN THE FUEL. WITH THIS PROCESS  TOTAL HEAT OF THE FUEL  CAN BE DETERMINED.
THIS  HEAT VALUE IS  HIGHER CALORIC VALUE OF THE FUEL.

IN PRACTICE HEAT CARRIED AWAY BY STEAM IS NOT UTILIZED , BUT YOU HAVE TO KNOW THE NET HEAT VALUE GENERATED BY THE FUEL.
THIS NET HEAT VALUE IS ITS LOWER CALORIFIC VALUE

HEAT TAKEN AWAY BY STEAM IS THE MASS OF STEAM AND THE LATENT HEAT OF STEAM FORMATION AT THAT CONDITION.

COMMON PRACTICE IS TO TAKE THE CONDITION AT 15 DEGREE CENTIGRADE

LATENT HEAT AT SATURATION TEMPERATURE OF STEAM IS 2466 KJ/KG.

IF P IS THE AMOUNT OF STEAM

SO

LCV= HCV-Px2466 KJ/KG.


  

Friday, May 2, 2014

CFBC BOILER KNOWLEDGE



Viewers are cordially invited to CONTACT AT kaacconsultant@gmail.com  for knowledge and experience sharing.


steam turbine speed control governor

In a Steam Turbine system we control it's speed that is Turbine needs to rotate at fixed speed. Now if we want to add some additional load say if we want to raise its Power output, speed will tend to reduce as equivalent amount of heat energy is not available at that moment . Now at that condition we have to introduce more steam that is heat energy to increase that load maintained at the same speed. Now how that steam will be introduced in the turbine either someone has to operate the control valve or through some mechanized arrangement. that mechanized arrangement is its governing system. Which is either hydraulic or electro hydraulic . 
Now take the case of load increase as the speed of the turbine tends to decrease, in case of a fully hydraulic governing system , a pump which is mounted in the governing system through some gear arrangement at turbine main shaft receives oil and produces a fixed discharge pressure(This is your primary oil pressure) corresponding to the preset rotation per minute,  will also reduce as when the speed of the turbine reduced , the pump speed also reduced as it is coupled with gear arrangement. This reduced oil pressure will facilitate to introduce more secondary oil pressure in to the pilot valve which is through its hydraulic piston system will open steam entry valve to turbine , as more stem will enter into the turbine its speed will increase, now two phenomenon will occur simultaneously, primary oil pressure will rise as turbine speed has increased and one feedback mechanism fitted in the pilot valve will restrict pilot valve movement and controlled secondary oil will try to maintain the rated speed. 
Now it is the function of secondary oil and primary oil. 
Now how to protect the turbine if we want stop or trip the turbine when it is dangerous for the machine. This is the function of the trip oil. Trip oil generates from control oil which is through some mechanical device and through some solenoid valves are routed throughout the governing system. It has the mechanism to be drained rapidly so trip oil drains and blocks the oil passage of control oil. Mostly by spring arrangement steam stop valve(ESV) and control valve(regulating valve) stem stops to turbine. 
ANY HOW THIS ARE VERY BRIEF DESCRIPTION .
Now more sophisticated mechanism has been introduced which is electro hydraulic control . This system maintains turbine speed almost steady at all load condition with minimum variation.

Details on contact to kaacconsultant@gmail.com

Sunday, April 21, 2013

Finance for Managers

https://www.icapb2b.gr/b2b_web/CMSContent/FINANCIAL_RATIOS.pdf

Techno commercial managers are gaining advantage day to day because

Only technicalities of machines  will not be fruitful unless it is balanced with market competencies in this cut throat environment unless with a close watch on business profitability which reflects in financial indicators. Potential engineering managers at some points of time has to focus on commercial aspects on their carrier ladder as technical aspects goes on conception and monitoring parameters will be financial results.

https://www.icapb2b.gr/b2b_web/CMSContent/FINANCIAL_RATIOS.pdf

above link is a good guide for reference of financial performances.
      

Sunday, January 27, 2013

Sunday, March 18, 2012

BOILER OPERATION AND MAINTENANCE TRAINING SESSIONS . ANSWERS ARE GIVEN ON PAGE FOR THE QUERIES AT kaacconsultant@gmail.com


6 Do’s and Don’ts for Boiler
Operation
DO’S
• Maintain all instruments in good working
condition
• All equipment interlocks should always be in
line
• Maintain normal water level in steam drum
• Maintain water quality as per the recommended
limits. A table showing the DM water & drum
water quality is included at the end of this
section
• FD damper must be in smooth operating
condition
• Pressure raising from cold start must be done
as per the cold start up curve
• All the duct joints must be leak proof
• Use proper lubricant and maintain the schedule
as recommended by the manufacturers
• Operate the boiler within the recommended
operation limits
• Boiler, piping, ducts must be properly insulated
• Servicing of equipments should be done as per
the manufacturer’s schedule
• Maintain proper operation log sheets regularly
• Maintain the instrument air free from moisture
and oily matters and the pressure as
recommended
• Carry out regular cleaning of direct water level
gauge glasses of Boiler drum
• Use proper valve gland packing to avoid
leakage
• Use proper gaskets for ßange joints
• Operate the blowdown valves as per
recommendation
• In case of power failure close the steam stop
valve
• If the water level goes up above the limits
operate the intermittent blowdown valve
immediately and maintain the water level to
normal
• Maintain the feed water temperature at
Economiser inlet and ßue gas temperature at
Economiser outlet as recommended
• Use genuine spares
• Boiler surroundings and equipments must be
properly illuminated
DON’TS
• Don’t bypass any instruments and safety
interlocks
• Don’t use raw water as boiler feed water
• Don’t operate the boiler beyond the operation
limits
• Don’t leave the furnace door open while the
boiler is in operation
• Don’t mix up different lubricants
• Don’t alter the equipment maintenance
schedule
• Don’t leave the instrument control panel
unattended
• Don’t allow unauthorised persons to operate
the boiler and associated equipments.
• Do not dose chemicals into the boiler in batch
wise, they should be done on a continuous
basis
• If boiler is running under combustion control
manual mode, then while increasing load
air should be increased Þrst followed by Fuel



Saturday, March 17, 2012

ENERGY PERFORMANCE ASSESSMENT OF BOILERS


1. ENERGY PERFORMANCE ASSESSMENT OF BOILERS
1.1 Introduction
Performance of the boiler, like efficiency and evaporation ratio reduces with time, due to poor combustion, heat transfer fouling and poor operation and maintenance. Deterioration of fuel quality and water quality also leads to poor performance of boiler. Efficiency testing helps us to find out how far the boiler efficiency drifts away from the best efficiency. Any observed abnormal deviations could therefore be investigated to pinpoint the problem area for necessary corrective action. Hence it is necessary to find out the current level of efficiency for performance evaluation, which is a pre requisite for energy conservation action in industry.
1.2 Purpose of the Performance Test
• To find out the efficiency of the boiler
• To find out the Evaporation ratio
The purpose of the performance test is to determine actual performance and efficiency of the boiler and compare it with design values or norms. It is an indicator for tracking day-to-day and season-to-season variations in boiler efficiency and energy efficiency improvements
1.3 Performance Terms and Definitions
1. Boiler Efficiency, η = 100 x input Heatoutput
Heat
= 100)()(x  kCals input fuel in Heat kCals output steam inHeat
2. Evaporation Ratio = nconsumptiofuelofQuantitygenerationsteamofQuantity
1.4 Scope
The procedure describes routine test for both oil fired and solid fuel fired boilers using coal, agro residues etc. Only those observations and measurements need to be made which can be readily applied and is necessary to attain the purpose of the test.
Bureau of Energy Efficiency 1
1. Energy Performance Assessment of Boilers
1.5 Reference Standards
British standards, BS845: 1987
The British Standard BS845: 1987 describes the methods and conditions under which a boiler should be tested to determine its efficiency. For the testing to be done, the boiler should be operated under steady load conditions (generally full load) for a period of one hour after which readings would be taken during the next hour of steady operation to enable the efficiency to be calculated.
The efficiency of a boiler is quoted as the % of useful heat available, expressed as a percentage of the total energy potentially available by burning the fuel. This is expressed on the basis of gross calorific value (GCV) .
This deals with the complete heat balance and it has two parts:
�� Part One deals with standard boilers, where the indirect method is specified
�� Part Two deals with complex plant where there are many channels of heat flow. In this case, both the direct and indirect methods are applicable, in whole or in part.
ASME Standard: PTC-4-1 Power Test Code for Steam Generating Units
This consists of
�� Part One: Direct method (also called as Input -output method)
�� Part Two: Indirect method (also called as Heat loss method)
IS 8753: Indian Standard for Boiler Efficiency Testing
Most standards for computation of boiler efficiency, including IS 8753 and BS845 are designed for spot measurement of boiler efficiency. Invariably, all these standards do not include blow down as a loss in the efficiency determination process.
Basically Boiler efficiency can be tested by the following methods:
1) The Direct Method: Where the energy gain of the working fluid (water and steam) is compared with the energy content of the boiler fuel.
2) The Indirect Method: Where the efficiency is the difference between the losses and the energy input.
1.6 The Direct Method Testing
1.6.1 Description
This is also known as ‘input-output method’ due to the fact that it needs only the useful output (steam) and the heat input (i.e. fuel) for evaluating the efficiency. This efficiency can be evaluated using the formula: 100xInputHeatOutputHeatEfficiencyBoiler=
Bureau of Energy Efficiency 2
1. Energy Performance Assessment of Boilers
BoilerFuel Input 100%+ Air Steam OutputEfficiency = Heat addition to Steam x 100Gross Heat in Fuel Flue Gas Water 100 valuecalorific Gross x rate firing Fuelenthalpy) water feed enthalpy (steam x rate flow SteamxEfficiencyBoiler−=
1.6.2 Measurements Required for Direct Method Testing
Heat input
Both heat input and heat output must be measured. The measurement of heat input requires knowledge of the calorific value of the fuel and its flow rate in terms of mass or volume, according to the nature of the fuel.
For gaseous fuel: A gas meter of the approved type can be used and the measured volume should be corrected for temperature and pressure. A sample of gas can be collected for calorific value determination, but it is usually acceptable to use the calorific value declared by the gas suppliers.
For liquid fuel: Heavy fuel oil is very viscous, and this property varies sharply with temperature. The meter, which is usually installed on the combustion appliance, should be regarded as a rough indicator only and, for test purposes, a meter calibrated for the particular oil is to be used and over a realistic range of temperature should be installed. Even better is the use of an accurately calibrated day tank.
For solid fuel: The accurate measurement of the flow of coal or other solid fuel is very difficult. The measurement must be based on mass, which means that bulky apparatus must be set up on the boiler-house floor. Samples must be taken and bagged throughout the test, the bags sealed and sent to a laboratory for analysis and calorific value determination. In some more recent boiler houses, the problem has been alleviated by mounting the hoppers over the boilers on calibrated load cells, but these are yet uncommon.
Heat output
There are several methods, which can be used for measuring heat output. With steam boilers, an installed steam meter can be used to measure flow rate, but this must be
Bureau of Energy Efficiency 3
1. Energy Performance Assessment of Boilers
corrected for temperature and pressure. In earlier years, this approach was not favoured due to the change in accuracy of orifice or venturi meters with flow rate. It is now more viable with modern flow meters of the variable-orifice or vortex-shedding types.
The alternative with small boilers is to measure feed water, and this can be done by previously calibrating the feed tank and noting down the levels of water during the beginning and end of the trial. Care should be taken not to pump water during this period. Heat addition for conversion of feed water at inlet temperature to steam, is considered for heat output.
In case of boilers with intermittent blowdown, blowdown should be avoided during the trial period. In case of boilers with continuous blowdown, the heat loss due to blowdown should be calculated and added to the heat in steam.
1.6.3 Boiler Efficiency by Direct Method: Calculation and Example
Test Data and Calculation
Water consumption and coal consumption were measured in a coal-fired boiler at hourly intervals. Weighed quantities of coal were fed to the boiler during the trial period. Simultaneously water level difference was noted to calculate steam generation during the trial period. Blow down was avoided during the test. The measured data is given below.
Type of boiler: Coal fired Boiler
Heat output data
Quantity of steam generated (output) : 8 TPH
Steam pressure / temperature : 10 kg/cm2(g)/ 180 0C
Enthalpy of steam(dry & Saturated)
at 10 kg/cm2(g) pressure : 665 kCal/kg
Feed water temperature : 850 C
Enthalpy of feed water : 85 kCal/kg
Heat input data
Quantity of coal consumed (Input) : 1.6 TPH
GCV of coal : 4000 kCal/kg
Calculation 100)()(xGCVxqhHxQefficiencyBoiler−=η
Where Q = Quantity of steam generated per hour (kg/hr)
q = Quantity of fuel used per hour (kg/hr)
GCV = Gross calorific value of the fuel (kCal/kg)
H = Enthalpy of steam (kCal/kg)
h = Enthalpy of feed water (kCal/kg)
Bureau of Energy Efficiency 4
1. Energy Performance Assessment of Boilers
100/4000/10006.1)85665(/10008)(xkgkCalxTkgxTPHxTkgxTPHefficiencyBoiler−=η
= 72.5%
Evaporation Ratio = 8 Tonne of steam / 1.6 Tonne of coal
= 5
1.6.4 Merits and Demerits of Direct Method
Merits
�� Plant people can evaluate quickly the efficiency of boilers
�� Requires few parameters for computation
�� Needs few instruments for monitoring
Demerits
�� Does not give clues to the operator as to why efficiency of system is lower
�� Does not calculate various losses accountable for various efficiency levels
�� Evaporation ratio and efficiency may mislead, if the steam is highly wet due to water carryover
1.7 The Indirect Method Testing
1.7.1 Description
The efficiency can be measured easily by measuring all the losses occurring in the boilers using the principles to be described. The disadvantages of the direct method can be overcome by this method, which calculates the various heat losses associated with boiler. The efficiency can be arrived at, by subtracting the heat loss fractions from 100.An important advantage of this method is that the errors in measurement do not make significant change in efficiency.
Thus if boiler efficiency is 90% , an error of 1% in direct method will result in significant change in efficiency. i.e.90 + 0.9 = 89.1 to 90.9. In indirect method, 1% error in measurement of losses will result in
Efficiency = 100 – (10 + 0.1) = 90 + 0.1 = 89.9 to 90.1
Bureau of Energy Efficiency 5
1. Energy Performance Assessment of Boilers
The various heat losses occurring in the boiler are: BoilerFlue gas sampleSteam OutputEfficiency = 100 –(1+2+3+4+5+6+7+8) (by Indirect Method)AirFuel Input, 100%1. Dry Flue gas loss2. H2 loss3. Moisture in fuel4. Moisture in air5. CO loss7. Fly ash loss6. Surface loss8. Bottom ash lossWaterBlow down
The following losses are applicable to liquid, gas and solid fired boiler
L1- Loss due to dry flue gas (sensible heat)
L2- Loss due to hydrogen in fuel (H2)
L3- Loss due to moisture in fuel (H2O)
L4- Loss due to moisture in air (H2O)
L5- Loss due to carbon monoxide (CO)
L6- Loss due to surface radiation, convection and other unaccounted*.
*Losses which are insignificant and are difficult to measure.
The following losses are applicable to solid fuel fired boiler in addition to above
L7- Unburnt losses in fly ash (Carbon)
L8- Unburnt losses in bottom ash (Carbon)
Boiler Efficiency by indirect method = 100 – (L1+L2+L3+L4+L5+L6+L7+L8)
1.7.2 Measurements Required for Performance Assessment Testing
The following parameters need to be measured, as applicable for the computation of boiler efficiency and performance.
a) Flue gas analysis
1. Percentage of CO2 or O2 in flue gas
2. Percentage of CO in flue gas
3. Temperature of flue gas
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1. Energy Performance Assessment of Boilers
b) Flow meter measurements for
1. Fuel
2. Steam
3. Feed water
4. Condensate water
5. Combustion air
c) Temperature measurements for
1. Flue gas
2. Steam
3. Makeup water
4. Condensate return
5. Combustion air
6. Fuel
7. Boiler feed water
d) Pressure measurements for
1. Steam
2. Fuel
3. Combustion air, both primary and secondary
4. Draft
e) Water condition
1. Total dissolved solids (TDS)
2. pH
3. Blow down rate and quantity
The various parameters that were discussed above can be measured with the instruments that are given in Table 1.1.
Table 1.1 Typical Instruments used for Boiler Performance Assessment.
Instrument
Type
Measurements
Flue gas analyzer
Portable or fixed
% CO2 , O2 and CO
Temperature indicator
Thermocouple, liquid in glass
Fuel temperature, flue gas temperature, combustion air temperature, boiler surface temperature, steam temperature
Draft gauge
Manometer, differential pressure
Amount of draft used or available
TDS meter
Conductivity
Boiler water TDS, feed water TDS,
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1. Energy Performance Assessment of Boilers
make-up water TDS.
Flow meter
As applicable
Steam flow, water flow, fuel flow, air flow
1.7.3 Test Conditions and Precautions for Indirect Method Testing
A) The efficiency test does not account for:
�� Standby losses. Efficiency test is to be carried out, when the boiler is operating under a steady load. Therefore, the combustion efficiency test does not reveal standby losses, which occur between firing intervals
�� Blow down loss. The amount of energy wasted by blow down varies over a wide range.
�� Soot blower steam. The amount of steam used by soot blowers is variable that depends on the type of fuel.
�� Auxiliary equipment energy consumption. The combustion efficiency test does not account for the energy usage by auxiliary equipments, such as burners, fans, and pumps.
B) Preparations and pre conditions for testing
�� Burn the specified fuel(s) at the required rate.
�� Do the tests while the boiler is under steady load. Avoid testing during warming up of boilers from a cold condition
�� Obtain the charts /tables for the additional data.
�� Determination of general method of operation
�� Sampling and analysis of fuel and ash.
�� Ensure the accuracy of fuel and ash analysis in the laboratory.
�� Check the type of blow down and method of measurement
�� Ensure proper operation of all instruments.
�� Check for any air infiltration in the combustion zone.
C) Flue gas sampling location
It is suggested that the exit duct of the boiler be probed and traversed to find the location of the zone of maximum temperature. This is likely to coincide with the zone of maximum gas flow and is therefore a good sampling point for both temperature and gas analysis.
D) Options of flue gas analysis
Check the Oxygen Test with the Carbon Dioxide Test
If continuous-reading oxygen test equipment is installed in boiler plant, use oxygen reading. Occasionally use portable test equipment that checks for both oxygen and carbon dioxide. If the carbon dioxide test does not give the same results as the oxygen test, something is wrong. One (or both) of the tests could be erroneous, perhaps because of stale chemicals or drifting instrument calibration. Another possibility is that outside air is
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1. Energy Performance Assessment of Boilers
being picked up along with the flue gas. This occurs if the combustion gas area operates under negative pressure and there are leaks in the boiler casing.
Carbon Monoxide Test
The carbon monoxide content of flue gas is a good indicator of incomplete combustion with all types of fuels, as long as they contain carbon. Carbon monoxide in the flue gas is minimal with ordinary amounts of excess air, but it rises abruptly as soon as fuel combustion starts to be incomplete.
E) Planning for the testing
�� The testing is to be conducted for a duration of 4 to 8 hours in a normal production day.
�� Advanced planning is essential for the resource arrangement of manpower, fuel, water and instrument check etc and the same to be communicated to the boiler Supervisor and Production Department.
�� Sufficient quantity of fuel stock and water storage required for the test duration should be arranged so that a test is not disrupted due to non-availability of fuel and water.
�� Necessary sampling point and instruments are to be made available with working condition.
�� Lab Analysis should be carried out for fuel, flue gas and water in coordination with lab personnel.
�� The steam table, psychometric chart, calculator are to be arranged for computation of boiler efficiency.
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1. Energy Performance Assessment of Boilers
1.7.4 Boiler Efficiency by Indirect Method: Calculation Procedure and Formula
In order to calculate the boiler efficiency by indirect method, all the losses that occur in the boiler must be established. These losses are conveniently related to the amount of fuel burnt. In this way it is easy to compare the performance of various boilers with different ratings.
Conversion formula for proximate analysis to ultimate analysis
%C
=
0.97C+ 0.7(VM+0.1A) - M(0.6-0.01M)
%H2
=
0.036C + 0.086 (VM -0.1xA) - 0.0035M2 (1-0.02M)
%N2
=
2.10 -0.020 VM
where
C
=
% of fixed carbon
A
=
% of ash
VM
=
% of volatile matter
M
=
% of moisture
However it is suggested to get a ultimate analysis of the fuel fired periodically from a reputed laboratory.
Theoretical (stoichiometric) air fuel ratio and excess air supplied are to be determined first for computing the boiler losses. The formula is given below for the same.
a) Theoretical air required for combustion
=
100/)]35.4()}8/(8.34{)6.11[(22SxOHxCx+−+ kg/kg of fuel. [from fuel analysis]
Where C, H2, O2 and S are the percentage of carbon, hydrogen, oxygen and sulphur present in the fuel.
b) % Excess Air supplied (EA)
= 10021%%22xOO− [from flue gas analysis]
Normally O2 measurement is recommended. If O2 measurement is not available, use CO2 measurement ]%)(100[%)(]%)(%)[(79002222taatCOxCOCOCOx−− [from flue gas analysis]
Where, (CO2%)t
=
Theoretical CO2
(CO2%)a
=
Actual CO2% measured in flue gas
( CO2 )t
= CofMolesNofMolesCofMoles+2
Moles of N2
= 2222..NofWtMolfuelinNofWtNofwtMolairltheoriticainNofWt+
Moles of C
= CofWtMolecularfuelinCofWt
c) Actual mass of air supplied/ kg of fuel (AAS)
=
{1 + EA/100} x theoretical air
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1. Energy Performance Assessment of Boilers
The various losses associated with the operation of a boiler are discussed below with required formula.
1. Heat loss due to dry flue gas
This is the greatest boiler loss and can be calculated with the following formula:
L1
= 100)(xfuelofGCVTTxCxmafp−
Where,
L1
=
% Heat loss due to dry flue gas
m
=
Mass of dry flue gas in kg/kg of fuel
=
Combustion products from fuel: CO2 + SO2 + Nitrogen in fuel + Nitrogen in the actual mass of air supplied + O2 in flue gas. (H2O/Water vapour in the flue gas should not be considered)
Cp
=
Specific heat of flue gas in kCal/kg
Tf
=
Flue gas temperature in oC
Ta
=
Ambient temperature in oC
Note-1:
For Quick and simple calculation of boiler efficiency use the following.
A: Simple method can be used for determining the dry flue gas loss as given below.
a) Percentage heat loss due to dry flue gas = 100)(xfuelofGCVTTxCxmafp−
Total mass of flue gas (m)/kg of fuel = mass of actual air supplied/kg of fuel + 1 kg of
fuel
Note-2: Water vapour is produced from Hydrogen in fuel, moisture present in fuel and air during the combustion. The losses due to these components have not been included in the dry flue gas loss since they are separately calculated as a wet flue gas loss.
2. Heat loss due to evaporation of water formed due to H2 in fuel (%)
The combustion of hydrogen causes a heat loss because the product of combustion is water. This water is converted to steam and this carries away heat in the form of its latent heat.
L2
= 100fuelofGCV )} T-(T C {584 x H x 9afp2x+
Where
H2
=
kg of hydrogen present in fuel on 1 kg basis
Cp
=
Specific heat of superheated steam in kCal/kgoC
Tf
=
Flue gas temperature in oC
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1. Energy Performance Assessment of Boilers
Ta
=
Ambient temperature in oC
584
=
Latent heat corresponding to partial pressure of water vapour
3. Heat loss due to moisture present in fuel
Moisture entering the boiler with the fuel leaves as a superheated vapour. This moisture loss is made up of the sensible heat to bring the moisture to boiling point, the latent heat of evaporation of the moisture, and the superheat required to bring this steam to the temperature of the exhaust gas. This loss can be calculated with the following formula
L3
= 100fuelofGCV )} T-(T C {584 xMafpx+
where
M
=
kg of moisture in fuel in 1 kg basis
Cp
=
Specific heat of superheated steam in kCal/kgoC
Tf
=
Flue gas temperature in oC
Ta
=
Ambient temperature in oC
584
=
Latent heat corresponding to partial pressure of water vapour
4. Heat loss due to moisture present in air
Vapour in the form of humidity in the incoming air, is superheated as it passes through the boiler. Since this heat passes up the stack, it must be included as a boiler loss.
To relate this loss to the mass of coal burned, the moisture content of the combustion air and the amount of air supplied per unit mass of coal burned must be known.
The mass of vapour that air contains can be obtained from psychrometric charts and typical values are included below:
Dry-Bulb
Wet Bulb
Relative Humidity
Temp oC
TempoC
(%)
Kilogram water per Kilogram dry air (Humidity Factor)
20
20
100
0.016
20
14
50
0.008
30
22
50
0.014
40
30
50
0.024
L4
=100fuelofGCV ) T-(T xC afpxxfactorhumidityxAAS
where
AAS
=
Actual mass of air supplied per kg of fuel
Humidity factor
=
kg of water/kg of dry air
Cp
=
Specific heat of superheated steam in kCal/kgoC
Tf
=
Flue gas temperature in oC
Ta
=
Ambient temperature in oC (dry bulb)
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1. Energy Performance Assessment of Boilers
5. Heat loss due to incomplete combustion:
Products formed by incomplete combustion could be mixed with oxygen and burned again with a further release of energy. Such products include CO, H2, and various hydrocarbons and are generally found in the flue gas of the boilers. Carbon monoxide is the only gas whose concentration can be determined conveniently in a boiler plant test.
L5
= 1005744%%%2xfuelofGCVxCOCOCxCO+
L5
=
% Heat loss due to partial conversion of C to CO
CO
=
Volume of CO in flue gas leaving economizer (%)
CO2
=
Actual Volume of CO2 in flue gas (%)
C
=
Carbon content kg / kg of fuel
or
When CO is obtained in ppm during the flue gas analysis
CO formation (Mco)
=
CO (in ppm) x 10-6 x Mf x 28
Mf
=
Fuel consumption in kg/hr
L5
=
Mco x 5744*
* Heat loss due to partial combustion of carbon.
6. Heat loss due to radiation and convection:
The other heat losses from a boiler consist of the loss of heat by radiation and convection from the boiler casting into the surrounding boiler house.
Normally surface loss and other unaccounted losses is assumed based on the type and size of the boiler as given below
For industrial fire tube / packaged boiler = 1.5 to 2.5%
For industrial watertube boiler = 2 to 3%
For power station boiler = 0.4 to 1%
However it can be calculated if the surface area of boiler and its surface temperature are known as given below :
L6
=
0.548 x [ (Ts / 55.55)4 – (Ta / 55.55)4] + 1.957 x (Ts – Ta)1.25 x sq.rt of [(196.85 Vm + 68.9) / 68.9]
where
L6
=
Radiation loss in W/m2
Vm
=
Wind velocity in m/s
Ts
=
Surface temperature (K)
Ta
=
Ambient temperature (K)
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1. Energy Performance Assessment of Boilers
Heat loss due to unburned carbon in fly ash and bottom ash:
Small amounts of carbon will be left in the ash and this constitutes a loss of potential heat in the fuel. To assess these heat losses, samples of ash must be analyzed for carbon content. The quantity of ash produced per unit of fuel must also be known.
7. Heat loss due to unburnt in fly ash (%). 100fuelofGCVashfly of G.C.V xburnt fuel of kg / collectedash Total7xL=
8. Heat loss due to unburnt in bottom ash (%) 100fuelofGCVash bottom of G.C.V xburnt fuel of kg / collectedash Total8xL=
Heat Balance:
Having established the magnitude of all the losses mentioned above, a simple heat balance would give the efficiency of the boiler. The efficiency is the difference between the energy input to the boiler and the heat losses calculated.
Boiler Heat Balance:
Input/Output Parameter
kCal / kg of fuel
%
Heat Input in fuel
=
100
Various Heat losses in boiler
1. Dry flue gas loss
=
2. Loss due to hydrogen in fuel
3. Loss due to moisture in fuel
=
4. Loss due to moisture in air
=
5. Partial combustion of C to CO
=
6. Surface heat losses
=
7. Loss due to Unburnt in fly ash
=
8. Loss due to Unburnt in bottom ash
=
Total Losses
=
Boiler efficiency = 100 – (1+2+3+4+5+6+7+8)
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1. Energy Performance Assessment of Boilers
1.8 Example: Boiler Efficiency Calculation
1.8.1 For Coal fired Boiler
The following are the data collected for a boiler using coal as the fuel. Find out the boiler efficiency by indirect method.
Fuel firing rate
=
5599.17 kg/hr
Steam generation rate
=
21937.5 kg/hr
Steam pressure
=
43 kg/cm2(g)
Steam temperature
=
377 oC
Feed water temperature
=
96 oC
%CO2 in Flue gas
=
14
%CO in flue gas
=
0.55
Average flue gas temperature
=
190 oC
Ambient temperature
=
31 oC
Humidity in ambient air
=
0.0204 kg / kg dry air
Surface temperature of boiler
=
70 oC
Wind velocity around the boiler
=
3.5 m/s
Total surface area of boiler
=
90 m2
GCV of Bottom ash
=
800 kCal/kg
GCV of fly ash
=
452.5 kCal/kg
Ratio of bottom ash to fly ash
=
90:10
Fuel Analysis (in %)
Ash content in fuel
=
8.63
Moisture in coal
=
31.6
Carbon content
=
41.65
Hydrogen content
=
2.0413
Nitrogen content
=
1.6
Oxygen content
=
14.48
GCV of Coal
=
3501 kCal/kg
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1. Energy Performance Assessment of Boilers
Boiler efficiency by indirect method
Step – 1 Find theoretical air requirement
Theoretical air required for complete combustion
=
100/)]35.4()}8/(8.34{)6.11[(22SxOHxCx+−+
kg/kg of coal
=
[(11.6 x 41.65) + {34.8 x (2.0413 – 14.48/8)} + (4.35 x 0)] / 100
=
4.91 kg / kg of coal
Step – 2 Find theoretical CO2 %
% CO2 at theoretical condition
( CO2 )t
= CofMolesNofMolesCofMoles+2
Where,
Moles of N2
=2222..NofWtMolfuelinNofWtNofwtMolairltheoriticainNofWt+
Moles of N2
= 1356.028016.028100/7791.4=+x
Where moles of C
=
0.4165/12 = 0.0347
( CO2 )t
=0347.01332.00347.0+
( CO2 )t
=
20.37%
Step – 3 To find Excess air supplied
Actual CO2 measured in flue gas
=
14.0%
% Excess air supplied (EA)
=]%)(100[%)(]%)(%)[(79002222taatCOxCOCOCOx−−
=]37.20100[14]1437.20[7900−−xx
=
45.17 %
Step – 4 To find actual mass of air supplied
Actual mass of air supplied
=
{1 + EA/100} x theoretical air
=
{1 + 45.17/100} x 4.91
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1. Energy Performance Assessment of Boilers
=
7.13 kg/kg of coal
Step –5 To find actual mass of dry flue gas
Mass of dry flue gas
=
Mass of CO2 +Mass of N2 content in the fuel+ Mass of N2 in the combustion air supplied + Mass of oxygen in flue gas
Mass of dry flue gas
= 10023)91.413.7(1007713.7016.012444165.0xxx−+++
=
7.54 kg / kg of coal
Step – 6 To find all losses
1. % Heat loss in dry flue gas (L1)
=100)(xfuelofGCVTTxCxmafp−
=1003501)31190(23.054.7xxx−
L1
=
7.88 %
2. % Heat loss due to formation of water from H2 in fuel (L2)
=100fuelofGCV )} T-(T C {584 x H x 9afp2x+
=1003501 )} 31-(190 0.45 {584 x 0.02041 x 9x+
L2
=
3.44 %
3. % Heat loss due to moisture in fuel (L3)
=100fuelofGCV )} T-(T C {584 xMafpx+
=100fuelofGCV )} T-(T C {584 xMafpx+
L3
=
5.91 %
4. % Heat loss due to moisture in air (L4)
=100)(xfuelofGCVTTxCxhumidityxAASafp−
=1003501)31190(45.00204.013.7xxxx−
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1. Energy Performance Assessment of Boilers
L4
=
0.29 %
5. % Heat loss due to partial conversion of C to CO (L5)
=1005744%%%2xfuelofGCVxCOCOCxCO+
=100350157441455.04165.055.0xxx+
L5
=
2.58 %
6. Heat loss due to radiation and convection (L6)
=
0.548 x [ (343/55.55)4 – (304/55.55)4] + 1.957 x (343 - 304)1.25 x sq.rt of [(196.85 x 3.5 + 68.9) / 68.9]
=
633.3 w/m2
=
633.3 x 0.86
=
544.64 kCal / m2
Total radiation and convection loss per hour
=
544.64 x 90
=
49017.6 kCal
% radiation and convection loss
=
49017.6 x 100
3501 x 5599.17
L6
=
0.25 %
7. % Heat loss due to unburnt in fly ash
% Ash in coal
=
8.63
Ratio of bottom ash to fly ash
=
90:10
GCV of fly ash
=
452.5 kCal/kg
Amount of fly ash in 1 kg of coal
=
0.1 x 0.0863
=
0.00863 kg
Heat loss in fly ash
=
0.00863 x 452.5
=
3.905 kCal / kg of coal
% heat loss in fly ash
=
3.905 x 100 / 3501
L7
=
0.11 %
8. % Heat loss due to unburnt in bottom ash
GCV of bottom ash
=
800 kCal/kg
Amount of bottom ash in 1 kg of coal
=
0.9 x 0.0863
=
0.077 kg
Heat loss in bottom ash
=
0.077 x 800
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1. Energy Performance Assessment of Boilers
=
62.136 kCal/kg of coal
% Heat loss in bottom ash
=
62.136 x 100 / 3501
L8
=
1.77 %
Boiler efficiency by indirect method
=
100 – (L1+ L2+ L3+ L4+ L5+ L6+ L7+ L8)
=
100-(7.88 + 3.44+ 5.91+ 0.29+ 2.58+ 0.25+ 0.11+1.77)
=
100-22.23
=
77.77 %
Summary of Heat Balance for Coal Fired Boiler
Input/Output Parameter
kCal / kg of coal
% loss
Heat Input
=
3501
100
Losses in boiler
1. Dry flue gas, L1
=
275.88
7.88
2. Loss due to hydrogen in fuel, L2
=
120.43
3.44
3. Loss due to moisture in fuel, L3
=
206.91
5.91
4. Loss due to moisture in air, L4
=
10.15
0.29
5. Partial combustion of C to CO, L5
=
90.32
2.58
6. Surface heat losses, L6
=
8.75
0.25
7. Loss due to Unburnt in fly ash, L7
=
3.85
0.11
8. Loss due to Unburnt in bottom ash, L8
=
61.97
1.77
Boiler Efficiency = 100 – (L1 + L2+ L3+ L4+ L5+ L6+ L7+ L8) = 77.77 %
1.8.2 Efficiency for an oil fired boiler
The following are the data collected for a boiler using furnace oil as the fuel. Find out the boiler efficiency by indirect method.
Ultimate analysis (%)
Carbon
=
84
Hydrogen
=
12
Nitrogen
=
0.5
Oxygen
=
1.5
Sulphur
=
1.5
Moisture
=
0.5
GCV of fuel
=
10000 kCal/kg
Fuel firing rate
=
2648.125 kg/hr
Surface Temperature of boiler
=
80 oC
Surface area of boiler
=
90 m2
Humidity
=
0.025 kg/kg of dry air
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1. Energy Performance Assessment of Boilers
Wind speed
=
3.8 m/s
Flue gas analysis (%)
Flue gas temperature
=
190oC
Ambient temperature
=
30o C
Co2% in flue gas by volume
=
10.8
O2% in flue gas by volume
=
7.4
a) Theoretical air required
=
100/)]35.4()}8/(8.34{)6.11[(22SxOHxCx+−+ kg/kg of fuel. [from fuel analysis]
=
[(11.6 x 84) + [{34.8 x (12 – 1.5/8)} + (4.35 x 1.5)] / 100
=
13.92 kg/kg of oil
b) Excess Air supplied (EA)
=10021%%22xOO− [from flue gas analysis]
=1004.7214.7x−
=
54.4 %
c) Actual mass of air supplied/ kg of fuel (AAS)
=
{1 + EA/100} x theoretical air
=
{1 + 54.4/100} x 13.92
=
21.49 kg / kg of fuel
Mass of dry flue gas
=
Mass of (CO2 + SO2 + N2 + O2) in flue gas + N2 in air we are supplying.
Mass of dry flue gas
=1007749.21100234.7005.03264015.0124484.0xxxx++++
=
21.36 kg / kg of oil
% Heat loss in dry flue gas
=100)(xfuelofGCVTTxCxmafp−
=10010000)30190(23.036.21xxx−
L1
=
7.86 %
Heat loss due to evaporation of water due to H2 in fuel (%)
=100fuelofGCV )} T-(T C {584 x H x 9afp2x+
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1. Energy Performance Assessment of Boilers
=10010000 )} 30-(190 0.45 {584 x 0.12 x 9x+
L2
=
7.08 %
% Heat loss due to moisture in fuel
=100fuelofGCV )} T-(T C {584 xMafpx+
=10010000 )} 30-(190 0.45 {584 x0.005x+
L3
=
0.033%
% Heat loss due to moisture in air
=100fuelofGCV ) T-(T xC afpxxfactorhumidityxAAS
=10010000 ) 30-(190 x0.45 025.036.21xxx
L4
=
0.38 %
Radiation and convection loss (L6)
=
0.548 x [ (Ts / 55.55)4 – (Ta / 55.55)4] + 1.957 x (Ts – Ta)1.25 x sq.rt of [(196.85 Vm + 68.9) / 68.9]
=
0.548 x [ (353 / 55.55)4 – (303 / 55.55)4] + 1.957 x (353 – 303)1.25 x sq.rt of [(196.85 x 3.8 + 68.9) / 68.9]
=
1303 W/m2
=
1303 x 0.86
=
1120.58 kCal / m2
Total radiation and convection loss per hour
=
1120 .58 x 90 m2
=
100852.2 kCal
% Radiation and convection loss
=100125.2648100002.100852xx
L6
=
0.38 %
Normally it is assumed as 0.5 to 1 % for simplicity
Boiler efficiency by indirect method
=
100 – (L1 + L2+ L3+ L4+ L6)
=
100-(7.86 + 7.08 + 0.033 + 0.38 + 0.38)
=
100 – 15.73
=
84.27 %
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1. Energy Performance Assessment of Boilers
Summary of Heat Balance for the Boiler Using Furnace Oil
Input/Output Parameter
kCal / kg of furnace oil
%Loss
Heat Input
=
10000
100
Losses in boiler :
1. Dry flue gas, L1
=
786
7.86
2. Loss due to hydrogen in fuel, L2
=
708
7.08
3. Loss due to Moisture in fuel, L3
=
3.3
0.033
4. Loss due to Moisture in air, L4
=
38
0.38
5. Partial combustion of C to CO, L5
=
0
0
6. Surface heat losses, L6
=
38
0.38
Boiler Efficiency = 100 – (L1 + L2+ L3+ L4+ L6) = 84.27 %
Note:
For quick and simple calculation of boiler efficiency use the following .
A: Simple method can be used for determining the dry flue gas loss as given below.
a) Percentage heat loss due to dry flue gas = 100)(xfuelofGCVTTxCxmafp−
Total mass of flue gas (m) = mass of actual air supplied (ASS)+ mass of fuel supplied
= 21.49 + 1=22.49
%Dry flue gas loss = %27.810010000)30190(23.049.22=−xxx
1.9 Factors Affecting Boiler Performance
The various factors affecting the boiler performance are listed below:
�� Periodical cleaning of boilers
�� Periodical soot blowing
�� Proper water treatment programme and blow down control
�� Draft control
�� Excess air control
�� Percentage loading of boiler
�� Steam generation pressure and temperature
�� Boiler insulation
�� Quality of fuel
All these factors individually/combined, contribute to the performance of the boiler and reflected either in boiler efficiency or evaporation ratio. Based on the results obtained from the testing further improvements have to be carried out for maximizing the performance. The test can be repeated after modification or rectification of the problems and compared with standard norms. Energy auditor should carry out this test as a routine manner once in six months and report to the management for necessary action.
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1. Energy Performance Assessment of Boilers
1.10 Data Collection Format for Boiler Performance Assessment
Sheet 1 - Technical specification of boiler
1
Boiler ID code and Make
2
Year of Make
3
Boiler capacity rating
4
Type of Boiler
5
Type of fuel used
6
Maximum fuel flow rate
7
Efficiency by GCV
8
Steam generation pressure &superheat temperature
9
Heat transfer area in m2
10
Is there any waste heat recovery device installed
11
Type of draft
12
Chimney height in metre
Sheet 2 - Fuel analysis details
Fuel Fired
GCV of fuel
Specific gravity of fuel (Liquid)
Bulk density of fuel (Solid)
Proximate Analysis Date of Test:
1
Fixed carbon
%
2
Volatile matter
%
3
Ash
%
4
Moisture
%
Ultimate Analysis Date of Test:
1
Carbon
%
2
Hydrogen
%
3
Sulphur
%
4
Nitrogen
%
5
Ash
%
6
Moisture
%
7
Oxygen
%
Water Analysis Date of Test:
1
Feed water TDS
ppm
2
Blow down TDS
ppm
3
PH of feed water
4
PH of blow down
Flue gas Analysis Date of Test:
1
CO2
%
2
O2
%
3
CO
%
4
Flue gas temperature
OC
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Sheet 3 – Format sheet for boiler efficiency testing
Date: ………………… Boiler Code No. …………………
S.No
Time
Ambient air
Fuel
Feed water
Steam
Flue gas analysis
Surface Temp of boiler, oC
Drybulb Temp, oC
Wet Bulb Temp, oC
Flow Rate, kg/hr
Temp
oC
Flow rate, m3/hr
Temp
oC
Flow rate,
m3/hr
Pressure
kg/cm2g
Temp
oC
O2
%
CO2
%
CO
%
Temp
0C
1.
2.
3.
4.
5.
6.
7.
8.
Boiler Supervisor Energy Manager Energy Auditor
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1. Energy Performance Assessment of Boilers
1.11 Boiler Terminology
MCR: Steam boilers rated output is also usually defined as MCR (Maximum Continuous Rating). This is the maximum evaporation rate that can be sustained for 24 hours and may be less than a shorter duration maximum rating
Boiler Rating
Conventionally, boilers are specified by their capacity to hold water and the steam generation rate. Often, the capacity to generate steam is specified in terms of equivalent evaporation (kg of steam / hour at 100oC). Equivalent evaporation- “from and at” 100oC. The equivalent of the evaporation of 1 kg of water at 100oC to steam at 100oC.
Efficiency : In the boiler industry there are four common definitions of efficiency:
a. Combustion efficiency
Combustion efficiency is the effectiveness of the burner only and relates to its ability to completely burn the fuel. The boiler has little bearing on combustion efficiency. A well-designed burner will operate with as little as 15 to 20% excess air, while converting all combustibles in the fuel to useful energy.
b. Thermal efficiency
Thermal efficiency is the effectiveness of the heat transfer in a boiler. It does not take into account boiler radiation and convection losses – for example from the boiler shell water column piping etc.
c. Boiler efficiency
The term boiler efficiency is often substituted for combustion or thermal efficiency. True boiler efficiency is the measure of fuel to steam efficiency.
d. Fuel to steam efficiency
Fuel to steam efficiency is calculated using either of the two methods as prescribed by the ASME (American Society for Mechanical Engineers) power test code, PTC 4.1. The first method is input output method. The second method is heat loss method.
Boiler turndown
Boiler turndown is the ratio between full boiler output and the boiler output when operating at low fire. Typical boiler turndown is 4:1. The ability of the boiler to turndown reduces frequent on and off cycling. Fully modulating burners are typically designed to operate down to 25% of rated capacity. At a load that is 20% of the load capacity, the boiler will turn off and cycle frequently.
A boiler operating at low load conditions can cycle as frequently as 12 times per hour or 288 times per day. With each cycle, pre and post purge airflow removes heat from the boiler and sends it out the stack. Keeping the boiler on at low firing rates can eliminate the energy loss. Every time the boiler cycles off, it must go through a specific start-up sequence for safety
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1. Energy Performance Assessment of Boilers
assurance. It requires about a minute or two to place the boiler back on line. And if there is a sudden load demand the start up sequence cannot be accelerated. Keeping the boiler on line assures the quickest response to load changes. Frequent cycling also accelerates wear of boiler components. Maintenance increases and more importantly, the chance of component failure increases.
Boiler(s) capacity requirement is determined by many different type of load variations in the system. Boiler over sizing occurs when future expansion and safety factors are added to assure that the boiler is large enough for the application. If the boiler is oversized the ability of the boiler to handle minimum loads without cycling is reduced. Therefore capacity and turndown should be considered together for proper boiler selection to meet overall system load requirements.
Primary air: That part of the air supply to a combustion system which the fuel first encounters.
Secondary air: The second stage of admission of air to a combustion system, generally to complete combustion initiated by the primary air. It can be injected into the furnace of a boiler under relatively high pressure when firing solid fuels in order to create turbulence above the burning fuel to ensure good mixing with the gases produced in the combustion process and thereby complete combustion
Tertiary air: A third stage of admission of air to a combustion system, the reactions of which have largely been completed by secondary air. Tertiary air is rarely needed.
Stoichiometric: In combustion technology, stoichiometric air is that quantity of air, and no more, which is theoretically needed to burn completely a unit quantity of fuel. ‘Sub-stoichiometric’ refers to the partial combustion of fuel in a deficiency of air
Balanced draught: The condition achieved when the pressure of the gas in a furnace is the same as or slightly below that of the atmosphere in the enclosure or building housing it.
Gross calorific value (GCV): The amount of heat liberated by the complete combustion, under specified conditions, by a unit volume of a gas or of a unit mass of a solid or liquid fuel, in the determination of which the water produced by combustion of the fuel is assumed to be completely condensed and its latent and sensible heat made available.
Net calorific value (NCV): The amount of heat generated by the complete combustion, under specified conditions, by a unit volume of a gas or of a unit mass of a solid or liquid fuel, in the determination of which the water produced by the combustion of the fuel is assumed to remain as vapour.
Absolute pressure The sum of the gauge and the atmospheric pressure. For instance, if the steam gauge on the boiler shows 9 kg/cm2g the absolute pressure of the steam is 10 kg/cm2(a).
Atmospheric pressure The pressure due to the weight of the atmosphere. It is expressed in pounds per sq. in. or inches of mercury column or kg/cm2. Atmospheric pressure at sea level is 14.7 lbs./ sq. inch. or 30 inch mercury column or 760mm of mercury (mm Hg) or 101.325 kilo Pascal (kPa).
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1. Energy Performance Assessment of Boilers
Carbon monoxide (CO): Produced from any source that burns fuel with incomplete combustion, causes chest pain in heart patients, headaches and reduced mental alertness.
Blow down: The removal of some quantity of water from the boiler in order to achieve an acceptable concentration of dissolved and suspended solids in the boiler water.
Complete combustion: The complete oxidation of the fuel, regardless of whether it is accomplished with an excess amount of oxygen or air, or just the theoretical amount required for perfect combustion.
Perfect combustion: The complete oxidation of the fuel, with the exact theoretical (stoichiometric) amount of oxygen (air) required.
Saturated steam: It is the steam, whose temperature is equal to the boiling point corresponding to that pressure.
Wet Steam Saturated steam which contains moisture
Dry Steam Either saturated or superheated steam containing no moisture.
Superheated Steam Steam heated to a temperature above the boiling point or saturation temperature corresponding to its pressure
Oxygen trim sensor measures flue gas oxygen and a closed loop controller compares the actual oxygen level to the desired oxygen level. The air (or fuel) flow is trimmed by the controller until the oxygen level is corrected. The desired oxygen level for each firing rate must be entered into a characterized set point curve generator. Oxygen Trim maintains the lowest possible burner excess air level from low to high fire. Burners that don’t have Oxygen Trim must run with Extra Excess Air to allow safe operation during variations in weather, fuel, and linkage.
Heat transfer mediums
There are different types of heat transfer medium e.g. steam, hot water and thermal oil. Steam and Hot water are most common and it will be valuable to briefly examine these common heat transfer mediums and associated properties.
Thermic Fluid
Thermic Fluid is used as a heat transfer mechanism in some industrial process and heating applications. Thermic Fluid may be a vegetable or mineral based oil and the oil may be raised to a high temperature without the need for any pressurization. The relatively high flow and return temperatures may limit the potential for flue gas heat recovery unless some other system can absorb this heat usefully. Careful design and selection is required to achieve best energy efficiency.
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Hot water
Water is a fluid with medium density, high specific heat capacity, low viscosity and relatively low thermal conductivity. At relatively low temperature e.g. 70oC -90oC, hot water is useful for smaller heating installations.
Steam
When water is heated its temperature will rise. The heat added is called sensible heat and the heat content of the water is termed its enthalpy. The usual datum point used to calculate enthalpy is 0oC.
When the water reaches its boiling point, any further heat input will result in some proportion of the water changing from the liquid to the vapour state, i.e. changing to steam. The heat required for this change of state is termed the 'latent heat of evaporation' and is expressed in terms of a fixed mass of water. Where no change in temperature occurs during the change of state, the steam will exist in equilibrium with the water. This equilibrium state is termed 'saturation conditions'. Saturation conditions can occur at any pressure, although at each pressure there is only one discrete temperature at which saturation can occur.
If further heat is applied to the saturated steam the temperature will rise and the steam will become 'superheated'. Any increase in temperature above saturated conditions will be accompanied by a further rise in enthalpy.
Steam is useful heat transfer medium because, as a gas, it is compressible. At high pressure and consequently density, steam can carry large quantities of heat with relatively small volume.
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QUESTIONS
1)
Define boiler efficiency.
2)
Why boiler efficiency by indirect method is more useful than direct method?
3)
What instruments are required for indirect efficiency testing?
4)
What is the difference between dry flue gas loss and wet flue gas loss?
5)
Which is the best location for sampling flue gas analysis?
6)
Find out the efficiency by direct method from the data given below.
An oil fired package boiler was tested for 2 hours duration at steady state condition. The fuel and water consumption were 250 litres and 3500 litres respectively. The specific gravity of oil is 0.92. The saturated steam generation pressure is 7 kg/cm2(g). The boiler feed water temperature is 30o C. Determine the boiler efficiency and evaporation ratio.
7)
What is excess air? How to determine excess air if oxygen / carbon dioxide percentage is measured in the flue gas?
8)
As a means of performance evaluation, explain the difference between efficiency and evaporation ratio.
9)
Testing coal-fired boiler is more difficult than oil-fired boiler. Give reasons.
10)
What is controllable and uncontrollable losses in a boiler?
REFERENCE:
1. Energy audit Reports of National Productivity Council
2. Energy Hand book, Second edition, Von Nostrand Reinhold Company - Robert L.Loftness
3. Industrial boilers, Longman Scientific Technical 1999
www.boiler.com
www.eng-tips.com
www.worldenergy.org
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