design and development of virtual experimental set-up to determine efficiency of boiler

8/3/2019 Design and Development of Virtual Experimental Set-Up to Determine Efficiency of Boiler

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2011 Student Design Competition College Of Engineering,Pune (India)

Design and Development of Virtual Experimental Set-up to Determine Efficiency of Boiler 1

1. Title: Determination of Boiler Efficiency by Direct and Indirect Methods.

2. Prior Concept:

Steam, Steam generation, Boiler, Boiler operation

3. New Concepts:

Various fuels used in the Boiler, Efficiency calculation, Heat Balance Sheet

3.1 Concept structure:

Boiler Efficiency by Direct Method (Direct) =

Heat Input Heat Output

Fuel Supplied * Calorific Value of Fuel

Heat gained by the steam = ( + x* ) kJ/kg

Boiler Efficiency

Direct Method Indirect Method

Loss of heat due to dry flue gases

Heat Input Heat Output

Fuel Supplied * Calorific Value of Fuel

Heat gained by the steam = ( + x* )... kJ/kg

Loss due to moisture present in fuel (H2O)

Loss due to hydrogen in fuel (H2)

Loss due to moisture in air (H2O)

Loss due to carbon monoxide

Loss due to surface radiation, convection and other unaccounted

Unburnt losses in fly ash

Unburnt losses in bottom ash

Boiler Efficiency by Indirect Method (Indirect) = 100


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4. Learning Objectives:

4.1: Intellectual Skills:

a. To know the variation of efficiency by using various fuels.

b. Understand the steam generation process.

4.2: Motor Skills:

a. To determine enthalpy of feed water.

b. To determine enthalpy of wet steam.

c. To determine heat produced in the combustion chamber

d. To determine heat loss

5. Apparatus:

a. Boiler

b. Dryness fraction measuring device

c. Flue gas analyzer

6. Diagram:

Fig. 2.1 Packaged Steam Boiler


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7. Theory:

Thermal efficiency of boiler is defined as the percentage of heat input that is

effectively utilized to generate the steam. There are two methods of assessing boiler

efficiency.

A)The Direct Method: Where the energy gain of the working fluid (water and steam) is

compared with the energy content of the boiler fuel.

B) The Indirect Method: Where the efficiency is the difference between the losses and

the energy input. (Energy produced from the combustion of fuel is taken as 100%

energy input)

A. Direct Method

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. Parameters to be monitored for the

calculations of boiler efficiency by direct method are:

Quantity of steam generated per hour (Q) in kg/hr.

Quantity of fuel used per hour (q) in kg/hr.

The working pressure (in bar ) and superheat temperature (C), if any

The temperature of feed water (C)Type of fuel and gross calorific value of the fuel (GCV) in kJ/kg of fuel

Boiler Efficiency () =

=

Where,

Q Rate of water evaporation in kg/s

Enthalpy of feed water in kJ/kg at feed water temperature

h = Enthalpy or total heat of steam in kJ / kg of steam corresponding

to a given working pressure

= + x* . (For wet steam)

= + . (For dry wet steam)

= + . (Forsuper-heated steam)

Enthalpy of feed water in kJ / kg at steam pressure


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Enthalpy of wet steam kJ/kg at steam pressure

x Dryness fraction of wet steam

Advantages of direct method:

Plant people can quickly evaluate the efficiency of boilers Requires few parameters for computation

Needs few instruments for monitoring

Disadvantages of direct method:

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

Procedure:

1. Observe the setup and all the connected pressure and temperature measuring

equipments.

2. Open the valves supplying water and fuel to boiler

3. Wait till water level reaches required level

4. Check the fuel level in the fuel tank

5. Check the fuel feed pump

6. Check the water feed pump

7. Start the boiler

8. Open the outlet valve of header which then diverts the steam to heat exchangerheader.

9. When desired pressure is generated in system, start taking observations of various

instruments.

10.Measure the feed water temperature with the help of thermometer

11.Measure the temperature of output steam

12.Measure the pressure of output steam

13.Measure the dryness fraction of output steam with the help of calorimeter

14.Check the fuel used and get the calorific value for the same

15.Measure the fuel supply to the boiler16.Note all the readings in the table

17.Calculate the efficiency of boiler

18.Vary the type and quantity of fuel and take the readings

19.Vary the dryness fraction and take the readings

20.Change the opening of outlet valve and vary flow rate, this will change the pressure

of steam and again take the reading

21.Follow the above method three to four times and record the data in observation table


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Observation Table:

Sr. No. Particular Readings

1 Temperature of feed water (C)

2 Quantity of water supplied (kg/hr)3 Steam Generation Pressure (Bar)

4 Dryness Fraction (x)

5 Steam output temperature (C)

6 Mass of Steam Generated (kg/hr)

7 Fuel Type i.e. Coal / LDO/ FO etc.

8 Calorific Value of Fuel supplied (kJ/kg)

9 Fuel rate (kg/hr)

Calculations:

1. Enthalpy of Feed Water:

Enthalpy of feed water ( ) from steam table at a feed water temperature

Feed water temperature: ______________

Enthalpy ( ): ________________

2. Enthalpy of wet steam (hfg) from steam table at a steam pressure:

Steam Pressure: _________________

Enthalpy of feed water ( ): ________________

Enthalpy of wet steam ( ): _________________

Total Heat Content of Wet Steam:

h= Enthalpy of feed water + (Dryness Fraction * Enthalpy of Wet Steam)

h = ( +(x * ))

h= ___________________

3. Total heat content of output steam (Heat Output):

Total heat content of output steam = (Enthalpy of wet steam + Enthalpy of feed

water at steam pressureEnthalpy of feed water at feed water temperature)

= ( )

= ______________

4. Heat Produced During the Combustion of Fuel (q) (Heat Input):

q = (Mass flow rate of the fuel consumed * Gross Calorific Value of the fuel)

q = ____________________


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5. Efficiency of Boiler ( ) = * 100

= ______________

8. Results:

A. Efficiency of boiler by using:

a. Coal Lignite: _____________________

b. Coal Anthracite: _____________________

c. Coal Bituminous: _____________________d. Light Diesel Oil: _____________________

e. Furnace Oil: _____________________

B. Efficiency of boiler by varying dryness fraction:

a. Dryness Fraction x = 0.6 = _________________

b. Dryness Fraction x = 0.7 = _________________

c. Dryness Fraction x = 0.8 = _________________

d. Dryness Fraction x = 0.9 = _________________

e. Dryness Fraction x = 1 = _________________

Example

Find out the efficiency of the boiler by direct method with the data given below:

Type of boiler : Coal fired

Quantity of steam (dry) generated : 8 TPH = 8000 kg/hrSteam pressure (gauge): 10 bar

Steam temperature: 170C

Quantity of coal consumed : 1800 kg/hr

Feed water temperature : 85C

GCV of coal : 13395 kJ/kg

Enthalpy of wet steam at 10 bar pressure : 2776.2 kJ/kg

Enthalpy of feed water : 334.9 kJ/kg

Dryness Fraction: 0.93


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Required values from steam table:

at 85C = 355.9 kJ/kg

at 10 bar = 762.6 kJ/kg

at 10 bar = 2016.6 kJ/kg

Boiler Efficiency () =

Where,

For wet steam = + (x* ) = 762.6 + (0.93 * 2016.6) = 2638.038

=

= 75.70 %

Note: It should be noted that boiler may not generate 100% saturated dry steam, and there

may be some amount of wetness in the steam.


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B. Indirect Method:

Indirect method is also called as heat loss method. The efficiency can be arrived at,

by subtracting the heat loss fractions from 100. The standards do not include blow down

loss in the efficiency determination process. A detailed procedure for calculating boiler

efficiency by indirect method is as follows.

Fig. 2.2: Illustration of heat loss from the Boiler[iv]

The principle losses that occur in a boiler are:

Loss due to dry flue gas (sensible heat).... L1Loss due to hydrogen in fuel (H2).... L2

Loss due to moisture in fuel (H2O).. L3

Loss due to moisture in air (H2O) L4

Loss due to carbon monoxide (CO)...... L5

Loss due to surface radiation, convection and other unaccounted L6

Unburnt losses in fly ash (Carbon)... L7

Unburnt losses in bottom ash (Carbon) L8

Boiler Efficiency by indirect method = 100 - (L1 + L2 + L3 + L4 + L5 + L6 + L7 + L8)

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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b) Flow meter measurements for

1. Fuel

2. Steam

3. Feed water

4. Condensate water5. 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

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.

Theoretical (Stoichiometric) air fuel ratio and excess air supplied are to be determined

first for computing the boiler losses. The following relation can be used for the same.

1. Theoretical air required for combustion =

2. % Excess Air Supplied (EA) =

Note: Normally O2 measurement is recommended. If O2 measurement is not available

use of CO2 can do.

Or

% Excess Air Supplied (EA) = . from flue gas analysis


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Where:

(CO2 %)t = Theoretical CO2

(CO2 %)a = Actual CO2 % measured in flue gas

(CO2 %)t =

Moles of N2 =

Moles of C =

Actual mass of air supplied / kg of fuel (AAS) = * Theoretical air

The various losses associated with the operation of a boiler are discussed below:

1. Heat loss due to dry flue gas (L1):

This is the greatest boiler loss and can be calculated by the followingmethod:

L1 =

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 + Moisture in flue gases + O2 in flue gas + Mass of air supplied for

combustion of fuelCp = Specific heat of fuel gas in kJ/kg K

Tf = Flue gas temperature in C

Ta = Ambient temperature in C

2. Heat loss due to evaporation of water formed due to H2 in fuel (%) (L2):

The combustion of hydrogen causes a heat loss because the product of

combustion is water. The evaporation of water absorbs the heat in the form of LatentHeat.

L2 =

Where:

H2 = kg of hydrogen present in fuel (per kg of fuel)

Cp = Specific heat of superheated steam in kJ / kg K




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2676 = Latent heat corresponding to pressure of water vapour in kJ/kg

3. Heat loss due to moisture present in fuel (L3):

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 by following method:

L3 =

Where:

M = kg moisture in fuel on 1 kg basis

Cp = Specific heat of superheated steam in kJ/kg C



2676 = Latent heat corresponding to pressure of water vapour in kJ/kg

4. Heat loss due to moisture present in air (L4) :

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 psychometric charts and typical values are included

below:

Dry-Bulb Wet Bulb Relative Humidity Kilogram water

per Kilogram dry

TempC Temp

C (%) air (Humidity Factor)

20 20 100 0.016

20 14 50 0.008

30 22 50 0.014

40 30 50 0.024

L4 =

Where:


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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 kJ/kg K

Tf = Flue gas temperature in (C)

Ta = Ambient temperature in (C) (Dry bulb temperature)

5. Heat loss due to incomplete combustion (L5):

Products formed by incomplete combustion could be mixed with oxygen and burned

again with a further release of energy. Such products include CO, H 2, 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 = * * 100

Where:

L5 = % Heat loss due to partial conversion of C to CO

CO = Volume of CO in flue gas leaving economizer (%)

= Actual Volume of CO2 in flue gas (%)

C = Carbon content kg/kg of fuel

OrWhen CO is obtained in ppm during the flue gas analysis

CO formation (Mco) = CO (in ppm) * 10-6 * Mf *28

Mf = fuel consumption in kg/hr

L5 = Mco * 5744

5744 = heat loss due to partial combustion of carbon kJ/kg

6. Heat loss due to radiation and convection (L6):

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 water tube boiler = 2 to 3%

For power station boiler = 0.4 to 1%


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However it can be calculated if the surface area of boiler and its surface

temperatures are known as given below:

L6 = 0.548 * [ 4 ) 4] + 1.957 * (Ts Ta) 1.25 *

Where,L6 = Radiation loss in W/m

Vm = Wind velocity in m/s

Ts = Surface temperature (C)

Ta = Ambient temperature (C)

7.Heat loss due to unburnt in fly ash (%) (L7):

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.

L7 =

8.Heat loss due to unburnt in bottom ash (%)(L8):

L8 =

Ultimate Fuel Analysis:

ULTIMATE FUEL ANALYSIS

Contents Lignite Bituminous Anthracite LDO FO

% C 68.0 81.8 90.8 86.2 86.1

% O 21.2 9.1 1.9 0 0% H 4.3 5.1 2.6 11.5 11.3

% N 1.1 1.4 1.4 0 0

% S 0.4 0.6 0.7 1.5 3.5

Ash 3 1 1.5 0.02 0.1

Moisture 2 1 1.5 0.25 0.25

GCV 26700 36100 36200 43950 44788

(Source: http://en.citizendium.org/wiki/Coal)
http://en.citizendium.org/wiki/Coalhttp://en.citizendium.org/wiki/Coal


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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.

Input / Output Parameter kJ / 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) = 100( )

Indirect =


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Procedure:

1. Observe the setup and all connected pressure and temperature measuring

equipments.

2.

Open the valves supplying water and fuel to boiler3. Wait till water level reaches required level

4. Check the fuel level in the fuel tank

5. Check the fuel feed pump

6. Check the water feed pump

7. Start the boiler

8. Open the outlet valve of header which then diverts the steam to heat exchanger

header.

9. When desired pressure is generated in system, start taking observations of various

instruments.

10.Measure the feed water temperature with the help of thermometer and note the

readings

11.Measure the feed water rate

12.Measure the temperature of output steam

13.Measure the pressure of output steam

14.Measure the steam generation rate

15.Measure the dryness fraction of output steam with the help of calorimeter

16.Measure the fuel firing rate of Boiler

17.Observe the fuel used and get the calorific value for the same

18.Measure the flue gas temperature19.Measure the atmospheric temperature

20.Measure the surface temperature of Boiler

21.Measure the humidity in ambient air

22.Measure the wind velocity of surrounding air

23.Measure the total surface area of Boiler

24.Take ultimate analysis of fuel from the table

25.Note all the readings in the table

26.Calculate the efficiency of boiler

27.Vary the fuel and take the readings

28.Vary the dryness fraction and take the readings

29.Vary the content of fuel and observe the change in the efficiency of boiler

30.Observe the change in the efficiency of boiler by varying above parameters

31.Change the opening of outlet valve and vary flow rate, this will change the pressure

of steam and again take the reading

32.Follow the above method three to four times and record the data in observation table


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Observation Table:

Sr.No. Particular Readings / Observations

1 Fuel Firing Rate (kg/hr)2 Steam generation rate (kg/hr)

3 Steam pressure (bar)

4 Steam temperature (C)

5 Feed water temperature (C)

6 % CO2 in Flue gas

7 % CO in Flue gas

8 Average flue gas temperature (C)

9 Ambient temperature (C)

10 Humidity in ambient air (kg /kg dry air)

11 Surface temperature of boiler (C)

12 Wind velocity around the boiler (m/s)

13 Total surface area of boiler (m2)

14 GCV of Bottom ash (kJ/kg)

15 GCV of fly ash (kJ/kg)

16 Ratio of bottom ash to fly ash

Fuel Analysis (%)

1 Ash content in fuel

2 Moisture in coal

3 Carbon content

4 Nitrogen content

5 Oxygen content

6 GCV of Fuel (kJ/kg)


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Calculations:


= ___________________

2. % Excess Air Supplied (EA) = * 100

= ___________________

3. Actual mass of air supplied / kg of fuel (AAS) = * Theoretical air

= ______________ kg / kg of fuel

4. Mass of dry flue gas m = CO2 + SO2 + Moisture in flue gases + O2 in flue gas +

Mass of air supplied for combustion of fuel

m = +

= _____________ kg/kg of oil


L1 =

= _________ %


L2 =

= __________ %


L3 =

= __________________ %

8. Heat loss due to moisture present in air (L4

L4 =

= _____________ %


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L6 = 0.548 * [ 4 ) 4] + 1.957 * (Ts Ta) 1.25 *

=__________________ W/m2

=_______________ * 3.6 . ( 1 W = 3.6 kJ)

=__________________ kJ / m2

Total radiation and convection loss per hour =_________________W/m2

= ________________ kJ/m2

% Radiation and Convection loss =

L6 =________________ %

Note: Normally it is assumed this loss between 0.5 to 1 %

Boiler efficiency by Indirect Method = 100 (L1 + L2 + L3 + L4 +L5 +L6)

Indirect = _________________ %


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Summary of Heat Balance for the Boiler Using _______________:

Input / Output ParameterkJ / kg of

furnace oil% Loss

Heat Input 100Losses in Boiler

1. Dry flue gas, L1

2. Loss due to hydrogen in fuel, L2

3. Loss due to Moisture in fuel, L3

4. Loss due to Moisture in air, L4

5. Partial combustion of C to Co, CO2

6. Surface heat losses, L6

Efficiency of Boiler = 100(L1 + L2 + L3 + L4 + L5 + L6 )

= 100( )

Indirect= %

8. Results:

A. Efficiency of boiler by using

a. Coal Lignite: _____________________

b. Coal Anthracite: _____________________c. Coal Bituminous: _____________________

d. Light Diesel Oil: _____________________

e. Furnace Oil: _____________________

Example of Oil Fired Boiler Efficiency Calculation:

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 = 41858 kJ/kg

Fuel firing rate = 2648.125 kg/hr


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Surface Temperature of boiler = 80C

Surface area of boiler = 90 m2

Humidity = 0.025 kg/kg of dry air

Wind speed = 3.8 m/s

Flue gas analysis (%)

Flue gas temperature = 190C

Ambient temperature = 30C

Co2% in flue gas by volume = 10.8

O2% in flue gas by volume = 7.4

Solution:


=

= 13.92 kg/kg of oil

2. % Excess Air Supplied (EA) = * 100

=

= 54.4 %

3. Actual mass of air supplied / kg of fuel (AAS) = * Theoretical air

= * 13.92

= 21.49 kg / kg of fuel

4. Mass of dry flue gas m = CO2 + SO2 + Moisture in flue gases + O2 in flue gas + Mass of

air supplied for combustion of fuel

m = +

= +

= 21.36 kg/kg of oil


L1 =

=


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= 8.16 %


L2 =

=

= 7.77 %


L3 =

=

= 0.033 %

8. Heat loss due to moisture present in air (L4):

L4 =

=

= 0.42 %


L6 = 0.548 * [ 4

) 4] + 1.957 * (Ts

Ta) 1.25 *

= 0.548 * [ 4 ) 4] + 1.957 * (353 303) 1.25 *

= 1303 W/m2

= 1303 * 3.6

= 4690.8 kJ/m2

Total radiation and convection loss per hour = 4690.8 * 90 m2

= 422172 kJ

% Radiation and Convection loss =

=

L6 = 0.38 %


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Note: Normally it is assumed this loss between 0.5 to 1 %

Boiler efficiency by Indirect Method = 100 (L1 + L2 + L3 + L4 +L6)

= 100 (8.16 + 7.77+0.033 +0.42+0.38)

= 100 16.763

= 83.24 %

Summary of Heat Balance for the Boiler Using Furnace Oil:

Input / Output Parameter kJ / kg of

furnace oil

% Loss

Heat Input 41858 100

Losses in Boiler

1. Dry flue gas, L1 3417.6 8.16

2. Loss due to hydrogen in fuel, L2 3252.96 7.77

3. Loss due to Moisture in fuel, L3 14.184 0.033

4. Loss due to Moisture in air, L4 179.424 0.42

5. Partial combustion of C to Co, CO2 0 0

6. Surface heat losses, L6 159.422 0.38

Efficiency of Boiler = 100(L1 + L2 + L3 + L4 + L5 + L6 )= 100(8.16+7.77+0.033+0.42+0.38)

= 83.24 %


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D i d D l t f Vi t l E i t l S t t D t i Effi i f B il 23

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 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.

design and development of virtual experimental set-up to determine efficiency of boiler

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