1. Stoichiometry of Combustion and Boiler Efficiency CalculationsBy
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. 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 = 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 = kJ/kg mole- K
T = Temp. in K=298 K

5. STOICHIOMETRY OF COMBUSTION
DATA :
Coal :
Ultimate analysis : Proximate : Analysis
C = % FC = %
H = % VM= %
S = % Ash = %
O =7.2 % MC = %
N= %
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 : kg/kg of fuel
(b) Actual Air [40% excess] :-
O2  N2  Air
kg  kg  kg
Air/Fuel = :1

7. 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
0.118
0.126
O2]ex
0.8132*
6.115
6.349 32
5.661
6.026
{N2}fuel
{N2}air
0.013
9.5284
N2]fg
9.5414
71.753
74.492 28
75.915
80.806
 mwet
kg/kg
100.00
wet
 mdry
kg/kg
dry *

8. (d) Flue Gas Density :-

9. 1 kg.-mole of any gas  22.4 m3 at 0 & 1 atm.
Check : We know that
1 kg.-mole of any gas  m3 at 0 & 1 atm.
  m3
Adiabatic Flame Temperature :-
GCV = LCV+mH2O. Levap
= LCV x 2305
LCV = kJ/kg
= x 1.25 x (Tad-25)
Tad = C

10. Home Work Example :
The following data refers to a heavy oil fired in a boiler
C=85.4 % N= %
H=11.4 % MC=0.10 %
S=2.8 % Ash=0.10 %
O=0.1 % GCV= 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 = 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 = 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) = bar
Psabs = Psg + Pb = bar
= bar
Corresponding to Psabs = bar from steam tables :-
At P = 8.2 bar tsat = 171.5   C hg=2770.2   kJ/kg
At p = 8.4 bar tsat =   C hg=   kJ/kg
 At p = bar tsat =  C hg= kJ/kg
Now as ts =179C > tsat = C steam is superheated. From steam tables.
At p = 8.0 bar t = 200C hg = kJ/kg
P = 9.0 bar t = 200C hg = kJ/kg
At p = bar t = 200C hg = kJ/kg
At p = bar t = C hg = kJ/kg
p = bar t = 179C hg = 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. :II: INDIRECT OR LOSS METHOD :
Estimation of Boiler Efficiency By Indirect Method [BS-2885]
BOILER

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. 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

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 : 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 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
0.00
SO2
0.034 64
O2]ex 32 X
6.2
N2]f+N2]th
[ ] = 6.819 28 -
N2]ex
3.762 X*
N2]act
Air]th
8.839
Air]act
%Excess
Xwet
Xdry X
100.00
* =79/21 by volume
Xwet=Co2+MC+H2o+SO2+O2ex+N2act

24. X= moles of excess O2
3.762 x = moles of excess N2
Total moles of dry flue gas = moles
Total mass of dry flue gas = kg/ kg of fuel

25. 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
0.00
SO2
0.034 64
0.1250
O2]ex 32 X=
6.2
N2]f+N2]th
[ ] = 6.819 28 -
N2]ex
3.762 X*=
N2]act
80.714
Air]th
8.839
Air]act
%Excess
41.02%
Xwet
Xdry
100.00
* =79/21 by volume
Xwet=Co2+MC+H2o+SO2+O2ex+N2act

26. 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
0.00
SO2
0.034 64
0.1250
O2]ex 32 X=
6.2
N2]f+N2]th
[ ] = 6.819 28 -
N2]ex
3.762 X=
N2]act
80.714
Air]th
8.839
Air]act
%Excess
40.88%
Xwet
Xdry
100.00
Xwet=Co2+MC+H2o+SO2+O2ex+N2act

27. X= moles of excess O2
3.762 x = moles of excess N2
Total moles of dry flue gas = moles
Total mass of dry flue gas = 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) kg/cm2 (atm)
Pabs = x 105 N/m x 105
= x 105 N/m2 = bar
At this pressure
Tsat= ? hg= ?
At p=8.2 bar tsat= C hg= kJ/kg
At p=8.4 bar tsat= C hg= kJ/kg
At p=8.3707bar tsat= C hg= kJ/kg
Now tsteam = 179 C > Tsat  steam is super heated
from superheated steam table :
At p=8.0 t=200 C hg= kJ/kg
At p=9.0 t=200 C hg= kJ/kg
At p= t=200 C hg= kJ/kg
At p= & tsat=172.35, hg= kJ/kg
At p= & tsup = 179 C, hg= 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
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
Moisture %
5.60
GCV p=constt. kJ/kg
O 2 (FLUE GAS) %
4.20
Ambient Temp. o C
35.00

32. STOICHIOMETRY OF REACTION
Combustion of Carbon :
C O2 CO2
12 kg kg kg
kg kg kg
(2) Combustion of Hydrogen :
H /2 O2 H2O
2 kg kg kg
kg kg kg
(3) Combustion of Sulphur :
S O2 SO2
32 kg kg kg
kg kg kg

33. CALCULATION
O2 Theoretical = kg/kg of fuel.
= x % C x % H + % S % O2 Fuel
(2) Air Theoretical = kg/kg of fuel
= O2 Theoretical + N2 Theoretical
= O2 Theoretical x O2 Theoretical 23
X =
= %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 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
18.084
19.187
SO2
64.0
0.109
0.115
H20+MC
18.0
5.750
0.000
O2-Excess
32.0
4.607
4.888
N2 fuel
28.0
0.0012
N2 Theoritical
N2 Excess
N2actual
Airtheoritical
28.84
Airactual
% Excess Air
TOTAL
8.0960
7.6305
100.00

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

36. = [Total of Mass (DB) x 1.03x(Flue Gas Temp.–Ambient Temp.)] x 100
CALCULATION OF LOSSES
(5) % Dry Gas Losses =
= [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 =
= % Unburnt in fly ash x % Ash in fuel x % Fly ash x
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 - % Unburnt in fly ash)
% Fly ash x % Ash in fuel x 0.84x(Flue gas Temp. – Ambient Temp.) x100
( 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
x 100
(100 - % Unburnt in Bottom Ash)
G.C.V. of fuel, kJ/Kg
(9) % Sensible Heat Loss in Bottom Ash =
= % Unburnt in Bottom ash x % Bottom ash x % Ash in fuel
100 - % Unburnt in Bottom ash
% Bottom ash x % Ash in fuel x [0.84 ( – Ambient Temp.)] x 100
GCV of fuel, kJ/Kg

39. (10) % Loss due to Hydrogen in fuel = 6.72273
= 9 x % Hydrogen in fuel x x (25 – Ambient Temp.)
x (Flue gas Temp. – 25) x 100
G.C.V. of Fuel, kJ/Kg
(11) % Loss due to Moisture in fuel =
= % Moisture in fuel x x (25 – Ambient Temp.)
100
1.88 x (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)
= %
(14) Efficiency = Total Losses
= –
= %

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
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
Moisture %
41.40
GCV p=c kJ/kg
O 2 (FLUE GAS) %
4.0
Ambient Temp. o C
35.00

42. INPUT DATA: MODIFIED CONSIDERING
LIME STONE ADDITTION = 8.00 % & FUEL = %
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) =
=15.98
GCV kCal/kg
Moisture %
38.09
GCV p=constt. kJ/kg
O 2 (FLUE GAS) %
4.0
Ambient Temp. o C
35.00
CO2 % (from lime stone )
3.52

43. STOICHIOMETRY OF REACTION
Combustion of Carbon :
C O2 CO2
12 kg kg kg
kg kg kg
(2) Combustion of Hydrogen :
H /2 O2 H2O
2 kg kg kg
kg kg kg
(3) Combustion of Sulphur :
S O2 SO2
32 kg kg kg
kg kg kg

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

45. 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
19.310
21.394
12.573
14.881
SO2
64.0
0.183
0.203
0.0819
0.0969
H20 + MC
18.0
9.742
0.000
15.506
0.0000
O2-Excess
32.0
4.468
4.950
4.0000
4.7340
N2 fuel
28.0
0.0046
N2 Theoritical
N2 Excess
N2actual
67.838
80.287
Airtheoritical
28.84
Airactual
% Excess Air
TOTAL
100.00

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

47. = [Total of Mass (DB) x 1.03x(Flue Gas Temp.–Ambient Temp.)] x 100
CALCULATION OF LOSSES
(5) % Dry Gas Losses =
= [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
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 - % 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 )

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

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

51. (12)% Unaccounted loss = 1.5
(13) Total Losses = (5)+(6)+(7)+………..+(12)
= %
(14)Efficiency = Total Losses
= –
= %

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

53. Thank You