CALCULATION OF STOICHIOMETRIC AIR OF COAL AND COMBUSTION

Calculation of Stoichiometric Air of Coal and Combustion Calculation

The specifications of furnace oil from lab analysis is given below:

Constituents	% By weight
Carbon	85.9
Hydrogen	12
Oxygen	0.7
Nitrogen	0.5
Sulphur	0.5
H2O	0.35
Ash	0.05

GCV of fuel: 10880 kCal/kg

Calculation for Requirement of Theoretical Amount of Air

Considering a sample of 100 kg of furnace oil. The chemical reactions are:

Element	Molecular Weight Kg / Kg mole
C	12
O2	32
H2	2
S	32
N2	28
CO2	44
SO2	64
H2O	18

C   +    O₂      →  CO₂
H₂   + 1/2O₂  →    H₂O
S    +   O₂      →     SO₂

Constituents of fuel

C      +     O₂   →         CO₂
12     +     32  →          44

12 kg of carbon requires 32 kg of oxygen to form 44 kg of carbon dioxide therefore 1 kg of carbon requires 32/12 kg i.e 2.67 kg of oxygen

(85.9) C  +  (85.9 × 2.67) O₂  →    315.25 CO₂

2H₂      +    O₂   →     2H₂O
4          +    32   →     36

4 kg of hydrogen requires 32 kg of oxygen to form 36 kg of water, therefore 1 kg of hydrogen requires 32/4 kg i.e 8 kg of oxygen

(12) H₂  +  (12 × 8) O₂    →       (12 x 9 ) H₂O
S    +    O₂   →    SO₂
32   +    32  →    64

32 kg of sulphur requires 32 kg of oxygen to form 64 kg of sulphur dioxide, therefore 1 kg of sulphur requires 32/32 kg i.e 1 kg of oxygen

(0.5) S + (0.5 × 1) O₂   →     1.0 SO₂

Total Oxygen Required (229.07 + 96 + 0.5)	= 325.57 kg
Oxygen already present in [100 kg fuel (given)]	= 0.7 kg
Additional Oxygen Required	= 325.57 - 0.7
	= 324.87 kg
Therefore quantity of dry air required.	= (324.87) / 0.23
(air contains 23% oxygen by wt.)	= 1412.45 kg of air
Theoretical Air required	= (1412.45)/100
	= 14.12 kg of air / kg of fuel

Calculation of theoretical CO₂ content in flue gases

Nitrogen in flue gas	= 1412.45 - 324.87
	= 1087.58 kg

Theoretical CO₂% in dry flue gas by volume is calculated as below :

Mole of CO2 in flue gas	= (314.97) / 44	= 7.16
Mole of N2 in flue gas	= (1087.58) / 28	= 38.84
Mole of SO2 in flue gas	= 1/64	= 0.016

Theoretical CO2 % by Volume =	(Moles of CO2) x 100
	Total moles (dry)
=	7.16
	7.16 + 38.84 + 0.016
	= 15.5%

Calculation of constituents of flue gas with excess air:

% Excess air   =  [(15.5/10)-1] x 100
                        = 55%

Theoretical air required for 100 kg  of fuel burnt  =  1412.45 kg 
Total quantity of air supply required with 55% excess 
                                                                         air  = 1412.45 X 1.55 
                                                                              = 2189.30 kg
Excess Air quantity = 2189.30 - 1412.45 
                                = 776.85 kg  

                             O₂ = 776.85 x 0.23 = 178.68  kg
                            N₂ = 776.85 - 178.68 = 598.17 kg

The final constitution of flue gas with 55% excess air for every 100 kg fuel.

CO₂ = 314.97 kg
H₂O = 108.00 kg
SO₂ = 1 kg
O₂ = 178.68 kg
N₂ = 1087.58 + 598.17 = 1685.75 kg

Calculation of Theoretical CO2% in Dry Flue Gas By Volume

Moles of CO₂ in flue gas = 314.97/44 = 7.16
Moles of SO₂ in flue gas = 1/64 = 0.016
Moles of O₂ in flue gas = 178.68 / 32 = 5.58
Moles of N₂ in flue gas = 1685.75 / 28 = 60.20



     =  (7.16 * 100) / (7.16 + 0.016 + 5.58 + 60.20)

   =  7.16 *100/ 72.956
  = 10%

Theoretical O₂ % by volume = 5.58 *100 *100/72.956
                                              = 7.5%

Optimizing Excess Air and Combustion:

For complete combustion of every one kg of fuel oil 14.1 kg of air is needed. In practice, mixing is never perfect, a certain amount of excess air is needed to complete combustion and ensure that release of the entire heat contained in fuel oil. If too much air than what is required for completing combustion were allowed to enter, additional heat would be lost in heating the surplus air to the chimney temperature. This would result in increased stack losses. Less air would lead to the incomplete combustion and smoke. Hence, there is an optimum excess air level for each type of fuel.