Stoichiometry of Combustion and Boiler Efficiency Calculations By Dr. S. A. Channiwala Professor, Mechanical Engineering Department S.V. National Institute of Technology Ichchhanath, Surat, Gujarat

CONTENTS INTRODUCTION STOICHIOMETRY OF COMBUSTION ESTIMATION OF BOILER EFFICIENCY BY DIRECT METHOD ESTIMATION OF BOILER EFFICIENCY BY INDIRECT METHOD OR LOSS METHOD AS PER BS-2885 BASIS OF THE METHOD ESTIMATION OF VARIOUS LOSSES ESTIMATION OF BOILER EFFICIENCY NUMERICAL EXAMPLES PACKAGE BOILER UKAI THERMAL POWER STATION PULVERISED FUEL FIRED BOILER GIPCL THERMAL POWER STATION CIRCULATING FLUIDISED BED BOILER WITH LIMESTONE ADDITION CONCLUSIONS

INTRODUCTION FUEL - A substance basically comprising of C, H, O, N and S, which on combustion liberates heat with minimum emissions. CALORIFIC VALUE : Energy content of fuel per unit mass or unit volume of fuel. UNITS : kJ/kg or kcals/kg for Solid & Liquids :kJ/nm3 or kcals/nm3 for Gaseous Fuels Gross calorific value or Higher Heating Value : Amount of heat energy liberated per unit mass or unit volume with H2O in its liquid state under standard conditions [25C, 1atm. pressure] Lower Calorific Value/Lower heating value : Amount of heat energy liberated per unit mass or unit volume with H2O in its vapour form under standard condition [25C, temp. 1 atm. pressure]

Qvh = GCVv=c = LCVv=c+ufg x mw ufg = hfg –p.Vfg at 25C = 2305 kJ/kg FOR CONSTANT VOLUME COMBUSTION : Qvh = GCVv=c = LCVv=c+ufg x mw ufg = hfg –p.Vfg at 25C = 2305 kJ/kg mw = MC+9H, kg/kg of fuel MC = Moisture, kg/kg of fuel H = Hydrogen, kg/kg of fuel FOR CONSTANT PRESSURE COMBUSTION : Qph = Qpl + hfg . mw GCVp=c = LCVp=c +hfg .mw hfg = 2442.3 kJ/kg at 25C mw = MC+9H RELATION BETWEEN CONSTANT PRESSURE & CONSTANT VOLUME COMBUSTION GCVp=c = GCVv=c +n.Ro.T n = np - nr = No. of Moles of Gaseous Product - No. of moles of Gaseous Reactants Ro = 8.314 kJ/kg mole- K T = Temp. in K=298 K

STOICHIOMETRY OF COMBUSTION DATA : Coal : Ultimate analysis : Proximate : Analysis C = 66.0 % FC = 50.19 % H = 4.1 % VM=30.11 % S = 1.7 % Ash = 7.7 % O =7.2 % MC = 12.0 % N= 1.3 % MC = 12.0 % Excess Air= 40 % Ash = 7.7 % Gross CV=27 MJ/kg. DETERMINE :- Theoretical Air Required Actual Air Composition of Flue Gas Density of Flue Gas at 0C & 25 C & at 160 C Adiabatic Flame Temperature SOLUTION :- 1. Combustion of Carbon :

2. Combustion Hydrogen : 3. Combustion of Sulphur : (a) Theoritical O2 Required :  Theoretical air required for complete combustion is : 8.839 kg/kg of fuel (b) Actual Air [40% excess] :- O2  N2  Air 2.8462 kg  9.5284 kg  12.3746 kg Air/Fuel = 12.3746 :1

Composition on mass basis % (c) Flue Gas Composition (With 40% Excess Air) : Constituent Mass kg/kg of fuel Composition on mass basis % Mol. Mass Moles Composition on volume/mole Basis % Wet Basis Dry basis Wet. Dry basis CO2 2.42 18.199 18.893 44 2.42/44= 0.055 12.253 13.042 {MC {H2O 0.12 0.369 - H2O]fg 0.489 3.677 0.00 18 0.2717 6.053 SO2 0.034 0.256 0.265 64 0.000531 0.118 0.126 O2]ex 0.8132* 6.115 6.349 32 0.02541 5.661 6.026 {N2}fuel {N2}air 0.013 9.5284 N2]fg 9.5414 71.753 74.492 28 0.34076 75.915 80.806  mwet 13.2976 kg/kg 100.00 0.448871wet  mdry 12.8086 kg/kg 0.421701dry *2.8462-2.033

(d) Flue Gas Density :-

1 kg.-mole of any gas  22.4 m3 at 0 & 1 atm. Check : We know that 1 kg.-mole of any gas  22.4 m3 at 0 & 1 atm.  0.448871  10.05471 m3 Adiabatic Flame Temperature :- GCV = LCV+mH2O. Levap 2700 0 = LCV+0.489 x 2305 LCV = 25872.86 kJ/kg 25872.86 = 13.2976 x 1.25 x (Tad-25) Tad = 1581.54C

Home Work Example : The following data refers to a heavy oil fired in a boiler C=85.4 % N=0.10 % H=11.4 % MC=0.10 % S=2.8 % Ash=0.10 % O=0.1 % GCV=42.900 kJ/kg Cpfg = 1.36 kJ/kg-K It is operated at 25% excess air levels. Determine :- (i) Amount of air required per kg of fuel (ii) Flue gas analysis on dry basis (iii) Density of wet flue gas (iv) Adiabatic flame temperature

ESTIMATION OF BOILER EFFICIENCY DIRECT METHOD : 1.2 Efficiency of Boiler :-

NUMERICAL EXAMPLE : A two hour boiler trial was conducted on a coal fired, smoke-tube boiler in a process house & the following data was collected. Rating : Equivalent Evaporation = 5 T/h Max. Pressure = 10.5 kg/cm2 Av. Steam Pressure during trial = 7.5 kg/cm2 (g) Barometric Pressure = 1.0132 bar Av. Steam temperature = 179C O2 in flue gas (dry basis) = 6.2 % Ambient temperature = 35C Unburnt in Ash = 7.2 % Size of feed tank = 2.5m  x 3.0 m Depression of water level during trial = 1.87m Temperature of feed water = 70C GCV of fuel = 27000 kJ/kg Coal consumption during trial = 1025 kg Determine :- (i) Boiler output in kW (ii) Equivalent evaporation in T/h (iii) Boiler efficiency, in %

SOLUTION :- Psg = 7.5 kg/cm2 (g) = 7.3575 bar Psabs = Psg + Pb = 7.3575 +1.0132 bar = 8.3707 bar Corresponding to Psabs = 8.3707 bar from steam tables :- At P = 8.2 bar tsat = 171.5   C hg=2770.2   kJ/kg At p = 8.4 bar tsat = 172.5   C hg=2771.2   kJ/kg  At p = 8.3707 bar tsat = 172.35 C hg=2771.05 kJ/kg Now as ts =179C > tsat = 172.35C steam is superheated. From steam tables. At p = 8.0 bar t = 200C hg = 2839.3 kJ/kg P = 9.0 bar t = 200C hg = 2833.6 kJ/kg At p = 8.3707 bar t = 200C hg = 2837.19 kJ/kg At p = 8.3707 bar t = 172.35C hg = 2771.05 kJ/kg p = 8.3707bar t = 179C hg = 2786.96 kJ/kg  At

(iii) hffw = ? From steam table at tfw=70C, hfw=293 kJ/kg

(vi) Equivalent Evaporation : (vii) Steam to Fuel Ratio (viii) Specific Equivalent Evaporation :

:II: INDIRECT OR LOSS METHOD : Estimation of Boiler Efficiency By Indirect Method [BS-2885] BOILER

BASIS OF THE METHOD : Applying Energy Balance on Boiler :-

This is the Basis of Loss or Indirect Method

DETERMINATION OF VARIOUS LOSSES 1. Dry Gas Loss : [Stack Loss] 2. Losses in Ash (b) Sensible heat loss in Ash :

BOILER EFFICIENCY =100-SUM OF VARIOUS LOSSES (3) Loss due to hydrogen in fuel : (4) Loss due to moisture in fuel : Unaccounted Losses :- (i) Radiation loss = 0.3 to 1.0 % (ii) Blow down loss=0.1 to 0.5 % (iii) Loss due to unburnt gas  0.1 to 1.0 % Total unaccounted losses =1.5 % THUS BOILER EFFICIENCY =100-SUM OF VARIOUS LOSSES

NUMERICAL EXAMPLE : The following data refers to a boiler trial conducted on a 5 T/h, 10kg/cm2, coal fired smoke tube type boiler :- Duration of trial : 1 hour Coal consumption : 510 kg/h Stack Temperature : 210C O2 in flue gas : 6.2 % [By orsat apparatus ] Ambient Temperature : 35 C Unburnt in Ash : 7.2 % Steam pressure : 7.5kg/cm2 (g) Steam temperature : 179 C Feed water Temperature : 70 C Fuel Analysis : C=66.0 %, H=4.1 %, S=1.7 %, O=7.2 %, N=1.3%, MC=12.0%, Ash=7.7% GCVf : 27000 kJ/kg Determine :- (i) Various Losses, (ii) Boiler Efficiency (iii) Boiler Output (iv) Steam generation in T/h (v) Equivalent Evaporation in T/h

SOLUTION : Stoichiometry of Reactions : Combustion of Carbon : Combustion of Hydrogen : Combustion of Sulphur :

FLUE GAS ANALYSIS CO2 (MC+ H2O) SO2 O2]ex N2]f+N2]th N2]ex N2]act Constituent Mass Kg/kg of fuel Mol Mass Moles (wet) Moles (dry) % by Vol. on Dry basis CO2 2.42 44 0.055 (MC+ H2O) 0.489 18 0.02717 0.00 SO2 0.034 64 0.000531 O2]ex 32 X 6.2 N2]f+N2]th [0.013+6.806] = 6.819 28 0.243536 - N2]ex 3.762 X* N2]act Air]th 8.839 Air]act %Excess Xwet Xdry 0.451525 0.299067+4.762X 100.00 * =79/21 by volume Xwet=Co2+MC+H2o+SO2+O2ex+N2act

X=0.02631 moles of excess O2 3.762 x =0.098978 moles of excess N2 Total moles of dry flue gas = 0.424355 moles Total mass of dry flue gas = 12.8863 kg/ kg of fuel

FLUE GAS ANALYSIS CO2 (MC+ H2O) SO2 O2]ex N2]f+N2]th N2]ex N2]act Constituent Mass Kg/kg of fuel Mol Mass Moles (wet) Moles (dry) % by Vol. on Dry basis CO2 2.42 44 0.055 12.961 (MC+ H2O) 0.489 18 0.02717 0.00 SO2 0.034 64 0.000531 0.1250 O2]ex 0.84192 32 0.02631 X=0.02631 6.2 N2]f+N2]th [0.013+6.806] = 6.819 28 0.243536 - N2]ex 2.771384 3.762 X*=0.098978 N2]act 9.59038 0.342514 80.714 Air]th 8.839 Air]act 12.4653 %Excess 41.02% Xwet Xdry 13.3753 12.8863 0.451525 0.424355 100.00 * =79/21 by volume Xwet=Co2+MC+H2o+SO2+O2ex+N2act

FLUE GAS ANALYSIS CO2 (MC+ H2O) SO2 O2]ex N2]f+N2]th N2]ex N2]act Constituent Mass Kg/kg of fuel Mol Mass Moles (wet) Moles (dry) % by Vol. on Dry basis CO2 2.42 44 0.055 12.961 (MC+ H2O) 0.489 18 0.02717 0.00 SO2 0.034 64 0.000531 0.1250 O2]ex 0.84192 32 0.02631 X=0.02631 6.2 N2]f+N2]th [0.013+6.806] = 6.819 28 0.243536 - N2]ex 2.77138 3.762 X= 0.098978 N2]act 9.59038 0.342514 80.714 Air]th 8.839 Air]act 12.4523 %Excess 40.88% Xwet Xdry 13.3753 12.8869 0.451525 0.424355 100.00 Xwet=Co2+MC+H2o+SO2+O2ex+N2act

X=0.02631 moles of excess O2 3.762 x =0.098978 moles of excess N2 Total moles of dry flue gas = 0.424355 moles Total mass of dry flue gas = 12.8863 kg/ kg of fuel Calculation of Losses : (i) Dry Gas Loss :[Stack Loss] (ii) Losses in Ash (a) Loss Due to Combustibles in Ash

(b) Sensible heat loss in Ash : (iii) Loss due to H2 in Fuel :

(iv) Loss due to moisture in fuel : (v) Unaccounted losses : (a) Radiation loss : 1.0% (b)Blow down loss : 0.5 (c) Loss due to unburnt gas : 0.2% ------------------- Qun = 1.5% From Steam table, at 70C Water Temperature hfw=293 kJ/kg

For hs=? Ps =7.5 kg/cm2 (g) +1.033 kg/cm2 (atm) Pabs = 7.3575 x 105 N/m2 +1.0132 x 105 = 8.3707 x 105 N/m2 = 8.3707 bar At this pressure Tsat= ? hg= ? At p=8.2 bar tsat= 171.5 C hg=2770.2 kJ/kg At p=8.4 bar tsat= 172.5 C hg=2771.2 kJ/kg At p=8.3707bar tsat=172.35 C hg=2771.05 kJ/kg Now tsteam = 179 C > Tsat  steam is super heated from superheated steam table : At p=8.0 t=200 C hg=2839.3 kJ/kg At p=9.0 t=200 C hg=2833.6 kJ/kg At p=8.3707 t=200 C hg=2837.19 kJ/kg At p=8.3707 & tsat=172.35, hg=2771.05 kJ/kg At p=8.3707 & tsup = 179 C, hg=2786.96 kJ/kg

EFFICIENCY CALCULATION OF UKAI THERMAL POWER STATION 210 MW UNIT. INPUT DATA: Steam Generation = 625 T/h Steam Pressure = 137 bar (abs.) Steam Temperature = 520oC Carbon % 39.93 Flue gas Temp. o C 152.00 Hydrogen % 4.55 Unburnt in Fly Ash % 0.75 Nitrogen % 0.12 Unburnt in Bottom Ash % 10.66 O 2 (FUEL) % 7.24 % of Fly Ash 80.00 Sulphur % 0.44 % of Bottom Ash 20.00 Ash % 42.12 GCV kCal/kg 3812.71 Moisture % 5.60 GCV p=constt. kJ/kg 16075.25 O 2 (FLUE GAS) % 4.20 Ambient Temp. o C 35.00

STOICHIOMETRY OF REACTION Combustion of Carbon : C + O2 CO2 12 kg 32 kg 44 kg 0.3993 kg 1.0648 kg 1.4641kg (2) Combustion of Hydrogen : H + 1/2 O2 H2O 2 kg 16 kg 18 kg 0.0455 kg 0.364 kg 0.4095 kg (3) Combustion of Sulphur : S + O2 SO2 32 kg 32 kg 64 kg 0.0044 kg 0.0044kg 0.0088 kg

CALCULATION O2 Theoretical = 1.3607734 kg/kg of fuel. = 2.666 x % C + 8.0 x % H + % S - % O2 Fuel 100 100 100 100 (2) Air Theoretical = 5.9164060 kg/kg of fuel = O2 Theoretical + N2 Theoretical = O2 Theoretical + 77 x O2 Theoretical 23 X = 0.0116560 = %O2 in Flue Gas x No. of MolesWBof (CO2 + SO2 + H2O + N2fuel + N2Theoritical ) 100 – (%O2 in Flue Gas x 4.762)

FLUE GAS ANALYSIS COMPO NENT Mol. Wt CO2 SO2 H20+MC O2-Excess N2 fuel MASS (WB) MASS (DB) COMPOSITION ON MASS BASIS NO. OF MOLES (WB) COMPOSITION ON VOL/MOLE BASIS  KG/KG %WET BASIS %DRY BASIS CO2 44.0 1.46410 18.084 19.187 0.0332750 0.033275 11.989979 13.222083 SO2 64.0 0.00880 0.109 0.115 0.0001375 0.000137 0.049545 0.054637 H20+MC 18.0 0.46550 0.00000 5.750 0.000 0.0258611 0.000000 9.318533 O2-Excess 32.0 0.37299 4.607 4.888 0.0116560 0.011656 4.200000 4.631597 N2 fuel 28.0 0.0012 0.00120   0.0000429 0.000042 N2 Theoritical 4.55563 0.1627012 0.162701 N2 Excess 1.22779 0.0438498 0.043849 N2actual 5.78463 71.45027 75.80910 0.2065938 0.206593 74.441943 82.091684 Airtheoritical 28.84 5.91641 0.20515 Airactual 7.51719 % Excess Air 27.0567 TOTAL 8.0960 7.6305 100.00 0.27752 0.25166

(1) Theoretical Nitrogen: (Kg/ Kg of fuel) = 4.5556326 Calculated Results: (1) Theoretical Nitrogen: (Kg/ Kg of fuel) = 4.5556326 = 77 x Theoretical Oxygen 23 Theoretical Air: (Kg/ Kg of fuel) = 5.9164060 = Theoretical Oxygen + Theoretical Nitrogen Actual Air: (Kg/ Kg of fuel) = 7.5171922 % Excess Air = 27.05673

= [Total of Mass (DB) x 1.03x(Flue Gas Temp.–Ambient Temp.)] x 100 CALCULATION OF LOSSES (5) % Dry Gas Losses = 5.7203 = [Total of Mass (DB) x 1.03x(Flue Gas Temp.–Ambient Temp.)] x 100 G.C.V. of fuel, kJ/Kg (6) % Loss due to combustibles in fly ash = 0.536 = % Unburnt in fly ash x % Ash in fuel x % Fly ash x 33820 100 100 100 100 - % Unburnt in fly ash GCV of fuel, kJ/Kg

(7) % Sensible Heat Loss in Fly Ash = 0.2076 = % Unburnt in fly ash x % Fly ash x % Ash in Fuel + 100 100 (100 - % Unburnt in fly ash) % Fly ash x % Ash in fuel x 0.84x(Flue gas Temp. – Ambient Temp.) x100 ( GCV of Fuel, kJ/Kg )

(8) % Loss due to combustibles in Bottom Ash = 2.1147 = % Unburnt in Bottom ash x % Bottom ash x % Ash in fuel x 33820 100 100 x 100 (100 - % Unburnt in Bottom Ash) G.C.V. of fuel, kJ/Kg (9) % Sensible Heat Loss in Bottom Ash = 0.29202 = % Unburnt in Bottom ash x % Bottom ash x % Ash in fuel + 100 100 100 - % Unburnt in Bottom ash % Bottom ash x % Ash in fuel x [0.84 (627.68 – Ambient Temp.)] x 100 100 100 GCV of fuel, kJ/Kg

(10) % Loss due to Hydrogen in fuel = 6.72273 = 9 x % Hydrogen in fuel x 4.2 x (25 – Ambient Temp.) + 2442.3 100 + 1.88 x (Flue gas Temp. – 25) x 100 G.C.V. of Fuel, kJ/Kg (11) % Loss due to Moisture in fuel = 0.91970 = % Moisture in fuel x 4.2 x (25 – Ambient Temp.) + 2442.30 + 100 1.88 x (Flue gas Temp. - 25) x 100 G.C.V. of Fuel, kJ/Kg

(12) % Unaccounted loss = 1.5 (13)Total Losses = (5)+(6)+(7)+………..+(12) = 18.0127 % (14) Efficiency = 100 - Total Losses = 100.00 – 18.0127 = 81.987 %

EFFICIENCY CALCULATION OF GIPCL, SLPP, MANGROL 2*125 MW UNIT, CFBC BOILERS WITH LIMESTONE ADDITION] INPUT DATA: Steam Generation = 390 T/h Steam Pressure = 130 kg/cm2 (g) Steam Temperature = 540 oC Carbon % 33.50 Flue gas Temp. o C 140.00 Hydrogen % 2.50 Unburnt in Fly Ash % 1.50 Nitrogen % 0.50 Unburnt in Bottom Ash % 0.05 O 2 (FUEL) % 9.00 % of Fly Ash 70.00 Sulphur % 0.60 % of Bottom Ash 30.00 Ash % 12.50 GCV kCal/kg 3322.98 Moisture % 41.40 GCV p=c kJ/kg 13999.60 O 2 (FLUE GAS) % 4.0 Ambient Temp. o C 35.00

INPUT DATA: MODIFIED CONSIDERING LIME STONE ADDITTION = 8.00 % & FUEL = 92.00 % Carbon % 30.82 Flue gas Temp. o C 140.00 Hydrogen % 2.30 Unburnt in Fly Ash % 1.50 Nitrogen % 0.46 Unburnt in Bottom Ash % 0.05 O 2 (FUEL) % 8.28 % of Fly Ash 80.00 Sulphur % 0.552 % of Bottom Ash 20.00 Ash % + CaO (from limestone) =11.50+4.48 =15.98 GCV kCal/kg 3057.14 Moisture % 38.09 GCV p=constt. kJ/kg 12891.28 O 2 (FLUE GAS) % 4.0 Ambient Temp. o C 35.00 CO2 % (from lime stone ) 3.52

STOICHIOMETRY OF REACTION Combustion of Carbon : C + O2 CO2 12 kg 32 kg 44 kg 0.3082 kg 0.82187 kg 1.13007 kg (2) Combustion of Hydrogen : H + 1/2 O2 H2O 2 kg 16 kg 18 kg 0.0230 kg 0.1840 kg 0.2070 kg (3) Combustion of Sulphur : S + O2 SO2 32 kg 32 kg 64 kg 0.00552 kg 0.00552 kg 0.01104 kg

CALCULATION O2 Theoretical = 0.9285661kg/kg of fuel = 2.666 x % C + 8.0 x % H + % S - % O2 Fuel 100 100 100 100 (2) Air Theoretical = 4.0372440 kg/kg of fuel = O2 Theoretical + N2 Theoretical = O2 Theoretical + 77 x O2 Theoretical 23 X = 0.0084250 = %O2 in Flue Gas x No. of MolesWBof (CO2 + SO2 + H2O + N2fuel + N2Theoritical ) 100 – (%O2 in Flue Gas x 4.762)

COMPOSITION ON MASS BASIS FLUE GAS ANALYSIS COMPO NENT Mol. Wt MASS (WB) MASS (DB) COMPOSITION ON MASS BASIS NO. OF MOLES(WB) MOLES COMPOSITION ON VOL/MOLE BASIS KG/KG %WET BASIS %DRY BASIS CO2 44.0 1.16527 19.310 21.394 0.0264833 12.573 14.881 SO2 64.0 0.01104 0.183 0.203 0.0001725 0.0819 0.0969 H20 + MC 18.0 0.58788 0.00000 9.742 0.000 0.0326600 0.0000000 15.506 0.0000 O2-Excess 32.0 0.26960 4.468 4.950 0.0084250 4.0000 4.7340 N2 fuel 28.0 0.0046 0.00460   0.0001643 N2 Theoritical 3.10868 0.1110242 N2 Excess 0.88745 0.0316947 N2actual 4.00073 66.2974 73.45323 0.1428832 67.838 80.287 Airtheoritical 28.84 4.03724 0.13999 Airactual 5.19429 % Excess Air 28.65941 28.6594 TOTAL 6.03451 5.44663 100.00 0.21062 0.17796

(1) Theoretical Nitrogen: (Kg/ Kg of fuel) = 3.1086779 Calculated Results: (1) Theoretical Nitrogen: (Kg/ Kg of fuel) = 3.1086779 = 77 x Theoretical Oxygen 23 Theoretical Air: (Kg/ Kg of fuel) = 4.0372440 = Theoretical Oxygen + Theoretical Nitrogen Actual Air: (Kg/ Kg of fuel) = 5.1942942 % Excess Air = 28.65941

= [Total of Mass (DB) x 1.03x(Flue Gas Temp.–Ambient Temp.)] x 100 CALCULATION OF LOSSES (5) % Dry Gas Losses = 4.56940 = [Total of Mass (DB) x 1.03x(Flue Gas Temp.–Ambient Temp.)] x 100 G.C.V. of fuel, kJ/Kg (6) % Loss due to combustibles in fly ash = 0.511 = % Unburnt in fly ash x % Ash in fuel x % Fly ash x 33820 100 100 100 100 - % Unburnt in fly ash GCV of fuel, kJ/Kg

(7) % Sensible Heat Loss in Fly Ash = 0.0888 = % Unburnt in fly ash x % Fly ash x % Ash in Fuel + 100 100 (100 - % Unburnt in fly ash) % Fly ash x % Ash in fuel x 0.84x(Flue gas Temp. – Ambient Temp.) x 100 ( GCV of Fuel, kJ/Kg )

(8) % Loss due to combustibles in Bottom Ash = 0.0042 = % Unburnt in Bottom ash x % Bottom ash x % Ash in fuel x33820 x 100 100 100 (100 - % Unburnt in Bottom Ash) G.C.V. of fuel, kJ/Kg (9) % Sensible Heat Loss in Bottom Ash = 0.12349 = % Unburnt in Bottom ash x % Bottom ash x % Ash in fuel + 100 100 100 - % Unburnt in Bottom ash % Bottom ash x % Ash in fuel x [0.84 (627.68 – Ambient Temp.)] x 100 100 100 GCV of fuel, kJ/Kg

(10) % Loss due to Hydrogen in fuel = 4.20141 = 9 x % Hydrogen in fuel x 4.2 x (25 – Ambient Temp.) + 2442.3 100 + 1.88 x (Flue gas Temp. – 25) x 100 G.C.V. of Fuel, kJ/Kg (11) % Loss due to Moisture in fuel = 7.73355 = % Moisture in fuel x 4.2 x (25 – Ambient Temp.) + 2442.3 + 100 1.88 x (Flue gas Temp. - 25) x 100 G.C.V. of Fuel, kJ/Kg

(12)% Unaccounted loss = 1.5 (13) Total Losses = (5)+(6)+(7)+………..+(12) = 18.7316 % (14)Efficiency = 100 - Total Losses = 100.00 – 18.7316 = 81.268 %

Conclusion Maximize Useful Heat Energy Minimize Losses -Improve combustion Efficiency -Improve Heat Transfer Efficiency

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