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Stoichiometry of Combustion and Boiler Efficiency Calculations

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.

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Stoichiometry of Combustion and Boiler Efficiency Calculations

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  1. 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

  2. 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

  3. 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]

  4. 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

  5. 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 :

  6. 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

  7. (c) Flue Gas Composition (With 40% Excess Air) : *2.8462-2.033

  8. (d) Flue Gas Density :-

  9. 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

  10. 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

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

  12. 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 %

  13. 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

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

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

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

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

  18. This is the Basis of Loss or Indirect Method

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

  20. (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

  21. 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

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

  23. FLUE GAS ANALYSIS Xwet=Co2+MC+H2o+SO2+O2ex+N2act * =79/21 by volume

  24. 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

  25. FLUE GAS ANALYSIS * =79/21 by volume Xwet=Co2+MC+H2o+SO2+O2ex+N2act

  26. FLUE GAS ANALYSIS Xwet=Co2+MC+H2o+SO2+O2ex+N2act

  27. 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

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

  29. (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

  30. 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

  31. 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

  32. 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

  33. 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)

  34. FLUE GAS ANALYSIS

  35. 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

  36. 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

  37. (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 100 100 ( GCV of Fuel, kJ/Kg )

  38. (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

  39. (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 1.88x (Flue gas Temp. - 25) x 100 G.C.V. of Fuel, kJ/Kg

  40. (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 %

  41. 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

  42. INPUT DATA: MODIFIED CONSIDERING LIME STONE ADDITTION = 8.00 % & FUEL = 92.00 %

  43. 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

  44. 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)

  45. FLUE GAS ANALYSIS

  46. 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

  47. 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

  48. (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 100 100 100 ( GCV of Fuel, kJ/Kg )

  49. (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

  50. (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 1.88x (Flue gas Temp. - 25) x 100 G.C.V. of Fuel, kJ/Kg

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