Combustion
►
Combustion is a chemical reaction accompanied by the evolution of light and heat.
►
Combustion in the furnace of a boiler is the rapid chemical union of the combustible elements in a fuel and the oxygen in the air, with the resultant liberation of heat.

Combustion
►
Fuel oil is burned in a boiler by pumping oil into a furnace in a fine atomized spray.
►
The air for combustion is admitted at high velocity through openings around the burner and mixes thoroughly with the oxidation of oil and the liberation of heat.
►
The finer the oil particles the more rapid and efficient the combustion

Combustion
►
The oil must be under pressure and it must be reduced to a viscosity by preheating before it is sent to the burners.
►
The complete and efficient combustion of the fuel oil depends of the following factors:
 Proper mixing of the air and fuel to be burned.
 Air in proper quantity.
 Temperature that is high enough to cause combustion reactions.
 Time to complete the combustion reaction.

Combustion
►
More air is actually admitted than that amount which is theoretically needed for complete combustion.
►
The amount of air which is added above the theoretical amount needed is called “Excess Air”.
►
The excess air is necessary in order that all the atomized fuel oil particles will come in contact with the oxygen in the air to ensure complete combustion

Combustion
►
If only the theoretical amount of air is used
Incomplete Combustion will occur.
►
If to much excess air is added at too high velocity, some of the atomized fuel oil might be swept out of the furnace and up into the stack before it has time to undergo combustion in the furnace.

Combustion
►
The three main combustible elements in fuel oil that produce heat during combustion are carbon, hydrogen, and sulfur.
►
Air is the medium which provides the oxygen for the combustion of these elements.
►
The percentage of oxygen in the air by weight is nearly 23 %.

Combustion
Possible combustion reactions
►
►
C + O
2
= CO
2
14,600 BTU/lb C
2C + O
2
=2CO 4,500 BTU/lbm C
►
►
►
2CO + O
2H
2
+ O
S + O
2
2
2
=2CO
2
= SO
2
10,100 BTU/lbm C
=2H
2
O 61,950 BTU/lbm H
2
4,000 BTU/lbm S

Combustion
►
If CO
2 is formed, the combustion is said to be complete.
►
If CO is formed, the combustion is said to be incomplete.
►
If CO is burned to CO
2 before it leaves the furnace the energy would be recovered.

Combustion
►
In the total process of combustion of released 4,500 + 10,100 BTU of heat or the same amount heat as when the carbon is burned to carbon dioxide directly.
►
When carbon monoxide is formed, only
4,500 BTU are liberated per pound of carbon versus 14,600 BTU per pound of carbon when carbon dioxide is formed.

Fuel Oil Composition
Specific Gravity
Carbon
Hydrogen
Sulfur
Nitrogen
Oxygen
0.86
84.0 %
13.0 %
0.3 %
0.2 %
1.0 %
0.90
85.0 %
12.0 %
0.5 %
0.2 %
1.0 %
0.94
86.0 %
11.0 %
0.8 %
0.2 %
1.0 %
0.99
87.0 %
9.5 %
1.1 %
0.2 %
1.0 %

Higher Heating Value
►
The Higher Heating Value (HHV) of a fuel can be calculate using Dulong’s Formula.
HHV

14100 BTU lbm C
*

61100 BTU lbm H
2


    
2
8
O
2 



4000 BTU lbm S

Lower Heating Value
►
The Lower Heating Value (LHV in BTU/lbm fuel), which does not include the latent heat of the water vapor in the exhaust gas formed from hydrogen in the fuel
LHV

HHV

1040 BTU lbm H
2
O


9 lbm H
2
O



* H
2
 lbm fuel

Combustion
Calculate the Higher Heating Value (HHV) and
Lower Heating Value (LHV) of a Heavy Fuel
Oil with the below composition.
►
Carbon – 85%
►
Hydrogen – 13%
►
Sulfur – 1%
►
Oxygen – 0.5%
►
Nitrogen – 0.5%

Higher Heating Value
►
The Higher Heating Value (HHV) of a fuel can be calculate using Dulong’s Formula.
HHV

14100 BTU lbm C
*

61100 BTU lbm H
2


    
2
8
O
2 



4000 BTU lbm S

Higher Heating Value
HHV
HHV




14100
Btu lb
C



0 .
85 lb
C



11985
 
7904 .
8
  
Btu
61100 lb
H
2


0 .
13 lb
H
2

0 .
005
8
HHV

19930
Btu lb
Fuel


 4000
Btu lb
S



0 .
01


Lower Heating Value
LHV
LHV


HHV
19930




1040
Btu





1040 lb
W ater
 
0 .
13

9 lb
W ater lb
Fuel
LHV

18713
Btu lb
Fuel
2



Adiabatic Flame Temperature
AFT

LHV
C
Exh aust

T

Adiabatic Flame Temperature
►
AFT = Adiabatic Flame Temperature
►
LHV = Lower Heating Value
►
► ṁ
Exhaust
C
P
= Fuel + Air
= Specific Heat of the Exhaust Gas
►
T in
= Temperature of the air entering the boiler.

Adiabatic Flame Temperature
►
Calculate the AFT of the heavy fuel if it is used in a boiler which requires 15% excess air for complete combustion. The inlet air to the furnace is 220 °F. Assume the specific heat of the exhaust gas is 0.31 Btu/lbm-°F.
►
Theoretically each lbm of fuel requires
13.75 lbm of air for complete combustion.

Adiabatic Flame Temperature
►
AFT with theoretical air requirement
►
►
AFT = LHV/(ṁ
Exhaust
X C
P Exhaust
) + T in
AFT = (18713/((1 + 13.75) X .31)) + 220
►
AFT = 18713/4.57 + 220
►
AFT = 4315 °F

Adiabatic Flame Temperature
►
AFT with 15% excess air
►
►
AFT = LHV/(ṁ
Exhaust
X C
P Exhaust
) + T in
AFT = (18713/((1+(13.75 X 1.15)) X .31) +
220
►
AFT = 18713/((1 + 15.81) X .31) + 220
►
AFT = 18713/5.21 + 220
►
AFT = 3812 °F