Sample Combustion Calculations by weight method
Combustion calculations are starting
point for Boiler or any Thermal equipment design. Here is a sample calcualtion
done for Coal with a calorific value of approximately 7200 Kcal/kg (12960
BTU/Lb).
Coal composition in % is as follows :
Carbon , C -
76.0 ; Hydrogen, H2 - 4.1 ; Nitrogen , N2
- 1.0 ; Oxygen, O2 - 7.6
; Suphur, S - 1.3 ; Moisture, H2O
- 3.0 ; Ash - 7.0 ;
Excess Air : 30 %
Molecular Wt of C - 12 , O2 -32
, H2 - 2, S - 32
The Chemicl equations are as follows:
C + O2 --> CO2 ; S + O2 --> SO2 ; H2 + O --> H2O
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Comp Wt. O2 reqd CO2 H2O N2 SO2
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C 0.76 2.0267 2.7867 - - -
H2 0.041 0.328 - 0.369 - -
O2 0.076 (-0.076)
N2 0.01 - - - 0.01 -
S 0.013 0.013 - - - 0.026
H2O 0.03 - - 0.03 - -
ASH 0.07
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100.0 2.2917
2.7867
0.399 0.01 0.026
From the above table, Theoritical O2
required = 2.2917 kg/kg of coal
There fore, Theoritical dry Air
required = 2.2917/0.23 = 9.964 kg of dry air/ kg of fuel
considering 30% Excess Air , Th air
reqd = 9.964 x 1.3 = 12.9532 kg/kg
Wet air required (assuming 60%
RH) = 12.9532 x 1.013 = 13.12 kg/kg of fuel
Gas composition :
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Gas
Weight %Wt %WT/MW %Volume
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CO2 2.7867 19.832 0.4507 13.381
H2O 0.567 4.035 0.2242 6.655
SO2 0.026 0.185 0.0029 0.086
N2 9.9837 71.055 2.5376 75.339
O2 0.6875
4.893 0.1529 4.539
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14.0509 100.0 3.3654 100
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