Energy Audit of Coal Fired Thermal Power Plant

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Energy Audit of Coal Fired Thermal Power Plant

Abhishek Kumar

Department of Mechanical Engineering Thapar Institute of Engineering and Technology

Patiala, India

Sumeet Sharma

Department of Mechanical Engineering Thapar Institute of Engineering and Technology

Patiala, India

Dr. D. Gangacharyulu

Department of Chemical Engineering Thapar Institute of Engineering and Technology

Patiala, India

Abstract An energy audit is feasibility study to establish and quantify the cost of various energy inputs to and flows within a factory or an organization in a given period. Energy Audit of Coal Fired Thermal Power Plant has been considered out. The aim is to calculate all the losses and give measures to rectify them and calculate the economic benefits after taking measures of rectification of losses. The Energy Audit of NTPC Dadri is done. The study is based on ASME BOILER TEST CODE method, BUREUEU OF ENERGY EFFICIANCY and GUDELINES FOR THERMAL POWER PLANT (INDO GERMAN ENERGY

PROGRAM). From the calculation it can be easily concluded that the overall efficiency of the plant decreases with the decrease in the requirement of output load. Output Load of the thermal power plant depends on the demand of electricity. As the demand of electricity decreases, the output load of the thermal power plant decreases, and the overall efficiency of the plant is also lower because electricity cannot be stored so the plant is running on partial load. Now if the thermal power plant run at Full Output Load the overall efficiency of the plant is much higher

  1. INTRODUCTION

    Systematic approach for decision making in energy management is done by the energy audit. The service to identify all the energy streams in a facility and balancing of total energy is done by energy audit. As per ENERGY CONSERVATION ACT 2001 energy audit is defined as verification, monitoring and analysis of use of energy including submission of technical report containing recommendations for improving energy efficiency with cost benefits analysis and action plan to reduce energy consumption.

    1. Need of energy audit

      Energy, labor and materials are the three top operating expenses found in the industry If check the priority to save 3 of them, energy audit is the best option for same output and quality but reduced input. The scope of improvement can easily give and understood by energy audit.

      The preventive maintenance and quality control programs which are vital for production and utilizing activities are done by positive orientation of energy audit. The energy cost, availability and reliability of supply of energy, appropriate energy mix, identification of energy conservation technologies is done by such audit programs.

      The translation of conservation ideas into realities by leading technical feasibility with economic and other organizational consideration within a specified time is done by energy audit.

      The energy consumption per unit of product output on to lower operating cost all is done by energy audit. Energy audit provides bench mark for managing energy in the organization and also provide basis for planning a more effective use of energy throughout the organization.

    2. Energy wastage in plants

      • Poor civil structures which allows less natural light to entre.

      • Poor natural ventilation of air which will lead to mechanical ventilation and consumes more power.

      • Poor insulation which leads to losses of heat.

      • Poor handling of coal thus leads to more moisture in coal and leads to decrease value of GCV from marshal yard to boiler bunker.

      • Conveyor belts are partially loaded thus more power consumption of motor is there because conveyor belts are at a BLF (belt loading factor) of 60%. Thus, this can be increased easily.

    3. Benefits of energy audit

      • Reduces consumption charges drastically.

      • Impact of operational improvements can be mentioned.

      • Reduces specific energy consumption and operating cost.

      • Identified energy losses for corrective actions.

      • Improves overall performance of total system and profitability and productivity.

      • Averts equipment failure.

      • Estimates the financial impact on energy consumption projects.

      • No extensive training or calculation involved.

    4. Types of energy audit

      • Comprehensive Audit

        It is time consuming and expensive process because it is very much detailed.

      • Targeted audits

    To provide detailed analysis and provide data of specific targeted projects.

    • Preliminary audits

      It seeks to establish cost and quantity of each form of energy used in this energy audit.

      Collecting data Analysing data Presenting data

      Establishing priorities and make recommendations

    • Walk through audit

      This approach estimates the order of magnitude on energy consumption without involving rigorous estimations on calculation by using certain energy consumption equipments.

    • Total system audit

      This method requires rigorous data entry and analysis. This approach identifies area of improvement on energy quantity basis.

      • Steam system audit

        Making of balance of total system and analysis of steam system from steam generators and consumption data is analysed here. If identifies energy loss due to steam leakage and heat loss.

        water changes its phase from liquid to vapour by means of heat the volume is increased by 1600 times. This energy is conveyed in the complete cycle by radiation, convection and conduction.

        The thermal efficiency is the amount of heat that is given and completely used to convert it to steam. There are two ways of checking the performance of boiler

        The direct method

        In this method the heat acquired of the functioning fluid is balanced with the heat content of the coal.

        The direct method is also called the INPUT- OUTPUT method because in this method we consider the output i.e. water in vapour form and the energy input i.e. from coal for calculating the performance of the boiler

        Parameter

        Quantity of steam generated per hour (Q)

        Enthalpy of saturated steam in kJ/kg of steam )

        Enthalpy of feed water in kJ/kg of water )

        Quantity of fuel used kg/hr (q)

        Fuel fired GCV

        kJ/kg

        Values

        8

        TPH

        2782.8

        kJ/kg

        355 kJ/kg

        1.34

        TPH

        953

        kJ/kg

        Boiler Efficiency

        Heat Output X 100/Heat Input = ) x 100 /(q X GCV)

        Calculation

        8X (2782.8355) X100/(1.34*17953)

        Result

        Boiler Efficiency = 80%

    • Electric system audit

      There are major energy consumers in process industries. The efficiency of total electrical system from electrical power consumption per day is checked in this type of audit.

    • Cooling system audit

      In this type of audit we calculate the cooling efficiency of the cooling tower. The impact of fouling or corrosion is identified for total system.

    • Insulation audit

    The process unit such as steam lines, tank insulation etc for improving the energy of complete system.

  2. METODOLOGY AND CALCULATION

    The study is based on ASME BOILER TEST CODE method, BUREUEU OF ENERGY EFFICIANCY and GUDELINES FOR THERMAL POWER PLANT (INDO

    GERMAN ENERGY). The Energy Audit of NTPC Dadri is done. The plant is having capacity of 2649 MW from coal and 829 MW from gas-based station. More has been focused on heat losses in coal fired steam generator, economizer, air preheater and loss Marshal Yard to steam generator. The methods used for efficiency of boiler is direct and indirect method thus there is also comparison and we can easily find out where the losses take place and where is the maximum possibility of energy savings.

    1. Boiler

      Boiler is a pressure vessel in which firing energy is conveyed from furnace to the water in the water tank until the water changes its phase from liquid to vapour. The Water is a very convenient and inexpensive way for conveying heat. When

      The Indirect Method

      We can calculate the performance of boiler by calculating all the losses happening in the boilers using the indirect method of testing. All the demerits of direct method can be compensated by this indirect method because in this method calculate the losses which are there in the boiler. We can obtain the efficiency percentage by just subtracting the percentage of losses from 100.

      The most important merit of this method is that the mistakes made by us in calculation of losses dont affect our result of efficiency to greater extent.

      Figure 1 Explanation of Energy audit of boiler by indirect method from Bureau of Energy Efficiency

      Boiler Efficiency Calculation (ALL READINGS ARE TAKEN FROM NTPC DADRI )

      1. Fuel firing rate = 272155 kg/hr

      2. Steam generation rate = 1315418 kg/hr

      3. Steam pressure = 192 (g)

      4. Steam temperature = 540 °C

      5. Feed water temperature = 300 °C

      6. % in Flue gas = 3

      7. %CO in flue gas = 0.22

      8. Average flue gas temperature APH inlet=330 °C, APH outlet = 180 °C

      9. Ambient temperature = 30 °C

      10. Humidity in ambient air = 0.0204 kg / kg dry air

      11. Surface temperature of boiler = 78°C

      12. Wind velocity around the boiler = 1.66 m/s

      13. Total surface area of boiler = 147

      14. GCV of fly ash = 525.2 kCal/kg

      15. Ratio of bottom ash to fly ash = 9010 Fuel Analysis (in %) (done in SAI LAB TIET)

        Ash content in fuel = 41

        Moisture in coal = 13 Carbon content = 34.05 Hydrogen content = 3.05 Nitrogen content = 1.40 Oxygen content = 6.05

        GCV of Coal = 3401 kCal/kg

        • Step 1. Find theoretical air requirement Theoretical air required for complete combustion =

          [(11.6 X C) + {34.8 X X /8)} + (4.35 X S)] /100

          kg/kg of coal

          = 4.74 kg/kg of coal

          Step 2. Find theoretical

          % at theoretical conditions = moles= +

          Where moles of = 0.135525 Where moles of C = 0.02837

          ()= = 17.30 %

        • Step. 3 To find Excess air supplied

          Actual measured in flue gas = 14.0%

          E. A=

          = 16%

        • Step. 4 To find actual mass of air supplied

          Actual mass of air supplied = {1 + EA/100} x theoretical air

          = 7.13 kg/kg of coal

        • Step. 5 To find actual mass of dry flue gas

          Mass of dry flue gas = Mass of + Mass of content in the fuel +Mass in the combustion air supplied + Mass of oxygen in flue gas

          = 6.7765 kg/kg of coal

        • Step. 6 To find all losses

        1. Dry flue gas losses: –

          Heat is vanished in the "dry" by-products of ignition, which is having only sensible heat because there was no change in state. The by-products of combustion of coal are carbon dioxide (CO2), carbon monoxide (CO), oxygen(O2), nitrogen (N2) and sulphur dioxide (SO2). Amount of SO2 and CO are normally measured in terms of the parts-per-million (ppm) so heat loss can be ignored.

          Heat loss in dry flue gas ( ) = X 100

          = 6.873 %

          Where,

          = % Heat loss due to dry flue gas

          m = Mass of dry flue gas in kg/kg of fuel

          = Combustion products from + + Nitrogen in fuel +

          Nitrogen in the actual mass of air supplied + in flue gas.

          (H2O/Water vapour in the flue gas should not be considered)

          Cp = Specific heat of flue gas in kCal/kg°C = Flue gas temperature in °C

          = Ambient temperature in °C

        2. Heat loss due to evaporation of water formed due to hydrogen in fuel (%)

          The amount of hydrogen is present in the fuel. When this hydrogen makes compound with oxygen it makes water. The water formed is in gaseous state because of high temperature of the boiler which leads to change in phase of the water. The change in phase of water leads to formation of steam and this steam is having very high temperature by very less entropy. The most relevant loss is about 11 % for natural gas and 7 % for fuel oil.

          =

          Where

          = kg of hydrogen present in fuel on 1 kg basis Cp = specific heat of superheated steam in kCal/kg C

          = Flue gas temperature in C

          = Ambient temperature in C

          Latent heat corresponding to partial pressure of water = 584

          = 5.25%

        3. Heat loss due to moisture present in fuel

        The water content which is present in the fuel when burned in the furnace is converted into the steam. The heat loss due to moisture is mainly comprised of three main components that is the heat needed to raise the temperature of the moisture to its boiling point, heat which is used to evaporate moisture and the superheating of the fuel so that the flue gasses can be reached to the temperature of the flue gas at the exit of the exhaust.

        = X 100

        where

        M = kg moisture in fuel on 1 kg basis

        Cp = Specific heat of superheated steam in kCal/kg°C

        = Flue gas temperature in °C

        = Ambient temperature in °C

        Latent heat corresponding to partial pressure of water vapour = 584

        superheating of the air so that the air can be reached to the temperature of the gas at the exit of the exhaust.

        = X100

        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 stem in kCal/kg C

        = Flue gas temperature in

        = Ambient temperature in

        = 0.28%

        1. Heat loss due to incomplete combustion

          There is some incomplete combustion in the coal because there is less temperature required for the coal to burn completely. The components which are not burned completely and leads to in complete combustions are CO, hydrogen and various HC. We can only find out the amount of CO in boiler emission easily.

          = X X 100

          Where

          = % Heat loss due to partial conversion of C to CO CO= Volume of CO in flue gas leaving Economizer (%) = Actual Volume of in flue gas (%)

          C = Carbon content kg/kg of fuel

          = 1.42%

        2. Heat loss due to radiation and convection

        Although the boiler body is highly insulated but we all know that there is no such system which is ideal and no heat is rejected to its surroundings so there is losses which are radiation losses and convection losses. As we are having some prerequisite knowledge of the boiler so we can easily say that the radiation and convection losses for industrial fire tube boiler = 1.5 to 2.5% For industrial water tube boiler = 2 to 3% For power station boiler = 0.4 to 1%

        = 0.548 X [( )^4 ( )^4] + 1.957 X ( )^1.25

        = 3.41% X

        4. Heat loss due to moisture present in air

        There is some amount of water content in the air thus to

        Where

        = Radiation loss in W/

        remove this water in the air we have to super heat the air . The water content in the air leads to raise the temperature of the stack thus this is considered as a boiler loss. The water content which is present in the air when burned in the furnace is converted into the steam. The heat loss due to moisture is

        Vm = wind velocity in m/s

        = urface temperature (K)

        = Ambient temperature (K)

        = 409.99 Kcal/m^2

        mainly comprised of three main components that is the heat needed to raise the temperature of the moisture to its boiling point, heat which is used to evaporate moisture and the

        Total radiation and convection/ hour = X surface area of boiler = 60268.6%

        Radiation =

        % =0.397%

        Second Law Analysis of Boiler

        • Equations

        net rate exergy is carried into the control volume. Between

        1. % Heat loss due to unburnt in fly ash

          In fly ash there is some amount of carbon content thus we can easily find out that there is some losses in the boiler due to these unburnt losses. To calculate these heat losses we have to take the proximate analysis of the ash contents in the boiler. The amount of as made by the fuel should also be known

          % Ash in coal = 11.93

          Ratio of bottom ash to fly ash = 9010 GCV of fly ash = 542.25 kCal/kg

          Amount of fly ash in 1 kg of coal = 0.1 x 0.1193

          = 0.01193 kg

          Heat loss in fly ash = 0.01193 x 542.25

          = 6.498 kCal / kg of coal

          % heat loss fly ash = Heat Loss in fly Ash X

          L7 = 0.19 %

        2. % Heat loss due to unburnt in bottom ash Amount of bottom ash in 1 kg of coal = 0.9 x 0.1193

        = 0.10737 kg

        GCV of bottom ash = 769 kCal/kg

        Heat loss in bottom ash = 0.10737 x 769

        = 82.56 kCal/kg of coal

        % Heat loss in bottom ash = Heat loss in bottom Ash X

        = 2.42 %

        HEAT BALANCE FOR COAL FIRED BOILER

        Input/output Parameter

        % loss

        Heat Input

        100

        Losses in boiler

        Dry flue gas,

        6.873

        Loss due to hydrogen in fuel,

        5.25

        Loss due to moisture in fuel,

        3.41

        Loss due to moisture in air,

        0.28

        Partial combustion of C to CO,

        1.42

        Surface heat losses,

        0.397

        Loss due to Unburnt in fly ash,

        1.19

        Loss due to Unburnt in bottom ash,

        2.42

        = 100 X + + + + + + + ) = 77.72%

        point 1 & 2;

        [ ]= [( ) ( R X )]

        net rate exergy is carried into the control volume. Between point 3&4;

        [ ]= [( ) ( R X )]

        Energy destruction;

        = [ ] + [ ] SECOND LAW EFFICIENCY OF BOILER = x ] X 100

        Where

        = Second law effectiveness of boiler = First law efficiency of boiler

        = Ambient temperature

        = Temperature of Flue gasses at inlet of boiler Stem temperature at outlet of boiler

        Readings

        Parameters

        Values

        Mass flow rate of combustion gasses INLET

        176.397

        kg/sec

        Mass flow rate of combustion gasses OUTLET

        201.597

        kg/sec

        Temperature of Flue gasses at inlet of boiler

        873K

        Temperature of Flue gasses at outlet of boiler

        673K

        Feed water temperature at inlet of boiler

        573K

        Stem temperature at outlet of boiler

        813K

        Pressure of Flue gasses at inlet of boiler

        0.0098bar

        Pressure of flue gasses at outlet of boiler

        0.0098bar

        Feed Water pressure at inlet of boiler

        137.2bar

        Steam pressure at outlet of boiler

        188bar

        Results

        Parameter

        RESULT

        Net rate exergy is carried into the control volume between point 1 & 2

        395108.448

        kW

        Net rate exergy is carried into the control volume. between

        point 3&4

        114581.45

        kW

        Energy destruction

        280526.95

        kWh

        SECOND LAW EFFICIENCY OF BOILER= 29.97%

    2. Second law analysis of economiser

    Equations (All equations are taken from Fundamentals of Engineering Thermodynamics, Michael J. Moran, Howard N. Shapiro) net rate exergy is carried into the control volume. Between point 1 & 2;

    [ ]= [( ) ( R X )]

    net rate exergy is carried into the control volume. Between point 3&4;

    [ ]= [( ) ( R X )]

    Energy destruction;

    = [] + []

    =.

    Readings

    Parameters

    Values

    Mass flow rate of combustion gasses inlet

    206.6 kg/sec

    Mass flow rate of combustion gasses outlet

    158.77

    kg/sec

    Temperature of Flue gasses at inlet of

    economiser

    673K

    Temperature of Flue gasses at outlet of

    economiser

    613K

    Cold water temperature at inlet of economiser

    525K

    Feed water temperature at outlet of economiser

    623K

    Pressure of Flue gasses at inlet of economiser

    0.0019 bar

    Pressure of flue gasses at outlet of economiser

    0.0029 bar

    Cold water stream water pressure at inlet of economiser

    166 bar

    Feed water at outlet of economiser

    156 bar

    Results of Second law analysis of Economiser

    Parameter

    RESULT

    Net rate exergy is carried into the control volume between point 1 & 2

    12.904 MW

    Net rate exergy is carried into the control volume. between

    point 3&4

    6.247679

    MW

    Energy destruction

    6.247679

    MW

    net rate exergy is carried into the control volume. Between point 3&4;

    [ ]= [( ) ( R X )]

    Energy destruction;

    = [] + []

    =.

    Readings

    Parameters

    Values

    Mass flow rate of combustion gasses inlet

    206.6 kg/sec

    Mass flow rate of combustion gasses outlet

    80 kg/sec

    Temperature of Flue gasses at inlet of air preheater

    613 K

    Temperature of Flue gasses at outlet of air preheater

    413 K

    Air temperature at inlet of air preheater

    303 K

    Air temperature at outlet of air preheater

    533 K

    Pressure of Flue gasses at inlet of air preheater

    0.0058 bar

    Pressure of flue gasses at outlet of air preheater

    0.0107 bar

    Air pressure at inlet of air preheater

    214.7 bar

    Air at outlet of air preheater to furnace

    196.13 bar

    Results

    Parameter

    RESULT

    Net rate exergy is carried into the control volume. Between point 1 & 2

    4.257 MW

    Net rate xergy is carried into the control volume. Between point 3&4

    1.627 MW

    Energy destruction

    2.63 MW

    SECOND LAW EFFICIENCY OF Air preheater= 39.60 %

  3. COST ANALYSIS

    A. Coal handling plant

    1. Cost estimation of losses due to air

      To avoid air loss, we should cover the conveyer belt and dont let air to blow over it.

      Maximum capacity of coal can be in inlet = 1400 ton/hour

      Actual feed to bunker = 900

      SECOND LAW EFFICIENCY OF ECONOMISER= 44.69%

      C. Second Law analysis of Air pre-heater

      Equations (All equations are taken from Fundamentals of Engineering Thermodynamics, Michael J. Moran, Howard N. Shapiro)

      net rate exergy is carried into the control volume. Between point 1 & 2;

      [ ]= [( ) ( R X ln )]

      ton/hour

      500 ton/h coal is lost due to air losses Present cost of 1 ton of coal = Rs 3210 Cost of 500 ton of coal = 3210 x 500

      = 1605000 /-

    2. Cost estimation of losses due to transportation of coal

      The clamps to hold the wagon should be of different material so that reduction in size of clamps can be done and same strength can be obtained and as size is less thus less coal will be stuck at that part according to this loss in coal will be less.

      Tippler losses = 10 Kg loss in clamps = 10 x 3.21

      = 32.1 /-

      In form of dust = 5 Kg = 5x 3.21

      = 16.05 /-

      Theft loss in wagon = Amount of coal feed in wagon = 100 ton

      Amount of coal lest at last when it reaches the power plant = 70 ton

      30 ton of coal is theft which cost = 30 x 3210

      = 96300 /-

    3. Cost estimation of losses due to non-opening of gates in track hoppers system The pneumatic valves to open the gates of the wagon to let coal to fall in the hopper in track hopper system should be perfect so that the wagons can be cleared in less time to avoid demerge cost given to Indian Railway by plant.

      Where the gate of wagon is not opened due to dust of coal complete wagon goes back to loss of coal in cost = 100 x 3210

      = 321000 /-

    4. Cost estimation of losses due to idling time of the motor Motor at no load condition = 1 Kw

      Motor at full load condition = 15 Kw

      Thus, if no coal is being fetched then also 1 Kw power is being consumed by motor.

      Motor consumes 1 unit of electricity in 1 hour

      As if idling time is 1 hour daily then 1 unit of electricity is wasted daily

      Per year 365 units is wasted Which will cost = 365 x 8

      = 2920 /-

    5. Cost estimation of losses due to opaque walls and excessive use of electricity In CHP they are having closed rooms with less no. of windows and having concrete sealing at their top which are opaque. Thus, they can also use maxim no. of windows and can also give green sheet which is translucent in nature thus some amount of light can entre and reduces in electricity consumption. 11. Power factor of maximum motors in CHP was 0.6 to .07 thus can be improved up to 0.9 as BFP is working at .09 PF. In CHP where the officers sit there is very less number of windows and there are maximum opaque walls

      1 tube light consumes 60/- in one month 30 tube lights are there in office = 30 x 60

      = 1800/-

      = Rate of heat transfer receiving = Source temperature = Delivery temperature = Delivery temperature

      = Rate of heat transfer to surroundings = Temperature across the surface

      Cost = CF X [1- ] [1- ] = Rate of energy destruction

      CF = Cost decided by NTPC of 1 unit of electricity Cost = 0.40 X 280526.95

      = 112210.78 /-

      1. Cost of energy wastage by second law analysis of Economiser

        Cost = CF X [1- ] [1- ] = Rate of energy destruction

        CF = Cost decided by NTPC of 1 unit of electricity Cost = 0.40 X 6247.67

        = 2499.068 /-

      2. Cost of energy wastage by second law analysis of APH

        Cost = CF X [1- ] [1- ] = Rate of energy destruction

        CF = Cost decided by NTPC of 1 unit of electricity Cost = 0.40 X 2680

        = 1072 /-

      3. Loss in CV from marshal yard to pulveriser

    The CV of the coal from marshal yard to pulveriser is reduced by 200CV it can be avoided by keeping coal in less moist area.

    B. Boiler system

    1) Cost of energy wastage by second law analysis of Boiler

    Heat

    Fuel

    Figure 5.1 Energy distribution

    Marshal Yard CV of coal= 3760 CV Pulverised coal CV = 3560 CV Heat input is constant Qi = M X CV M1 X CV1 = M2 X CV2

    M1 X 3760 = 75.59 X 3560

    M1 =

    = 71.56 Kg / sec Loss = 4.02 Kg/ sec

    = 347.328 Ton / day

    Cost of loss of coal = 3260 X 347.328

    = Rs 115898

    1. Blow down

      1 % of blow down implies 0.17% heat added in the boiler. So, blow down should be adhered to the chemist requirement.

      Q = M X CV

      = 75.59 X 3560

      = 269100.4 X

      Loss of Q due to 1% of blow down implies 0.17 % of heat loss = 457.4

      New Q after loss = 269557.8 269557.8 = M X 3560

      M = 75.71

      More M required = 0.128 Kg/ sec M per day =

      = 11.059920Ton / day

      Cost of Q due to 1% of blow down implies 0.17 % of heat loss = 11.0599 X 3260

      = Rs 36052.992 /-

    2. Soot blowing losses

      Superheated steam with high enthalpy is used for soot blowing.

      1% of steam required, contains 0.62% heat content. To make up the loss another 0.25% heat has to be added to feed water resulting in total heat loss of 0.87%.

      Q = M X CV

      = 75.59 X 3560

      = 269100.4 X

      Loss of Q due to 1% of blow down implies 0.17 % of heat loss = 2341.173

      New Q after loss = 266759.22

      266759.22= M X 3560

      M = 74.80 Kg/sec

      More M required = 0.79 Kg/ sec M per day =

      = 68.256Ton / day

      Cost of Q due to 1% of steam required, contains 0.62% heat content. To make up the loss another 0.25% heat has to be added to feed water resulting in total heat loss of 0.87%.

      = 3260 X 680256

      = 222514.56 / –

    3. Use of economiser in place of air pre-heater Q of air pre heater = 25830 kW

      Q of economiser = 4857 kW

      Rate of heat transfer of air preheater is very much high than Economiser. Thus, to make an economiser of same rate of heat transfer we need to make 3100 tubes instead of 544 tubes. Thus cost of making economiser will be 6 times the

      construction of an economiser thus we do not use economiser in place of APH. This calculation is done in MATLAB and program is in Appendix 2. We use air pre-heater instead of this large economiser because it will occupy very large space and cost of construction is very high, but it is one-time investment. The motor responsible for rotation of air preheater is of 11 kW, 415 V, 3 phase which moves at 1460 rpm but the cost of maintenance and working of this motor is very much less as compared to construct an economiser of such a big size.

  4. CONCLUSION

From the investigation of this thesis, it is observed that the overall plant performance changes with the small variation in the output loads. From the calculation it can be easily concluded that the overall efficiency of the plant decreases with the decrease in the requirement of output load. Output Load of the thermal power plant depends on the demand of electricity. As the demand of electricity decreases, the output load of the thermal power plant decreases, and the overall efficiency of the plant is also lower, because electricity cannot be stored so the plant is running on partial load. Now if the thermal power plant run at Full Output Load the overall efficiency of the plant is much higher.

Rate of heat transfer of air preheater is very much high than Economiser. Thus, to make an economiser of same rate of heat transfer we need to make 3100 tubes instead of 544 tubes. Thus cost of making economiser will be 6 times the construction of an economiser thus we do not use economiser in place of APH. We use air pre-heater instead of this large economiser ecause it will occupy very large space and cost of construction is very high, but it is one-time investment. The motor responsible for rotation of air preheater is of 11 kW, 415 V, 3 phase which moves at 1460 rpm but the cost of maintenance and working of this motor is very much less as compared to construct an economiser of such a big size.

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