Combustion

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Combustion
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 + O2 = CO2
14,600 BTU/lb C
► 2C + O2 =2CO
4,500 BTU/lbm C
► 2CO + O2 =2CO2
10,100 BTU/lbm C
► 2H2 + O2 =2H2O
61,950 BTU/lbm H2
► S + O2 = SO2
4,000 BTU/lbm S
Combustion
► If
CO2 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 CO2 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
0.86
0.90
0.94
0.99
Carbon
84.0 %
85.0 %
86.0 %
87.0 %
Hydrogen
13.0 %
12.0 %
11.0 %
9.5 %
Sulfur
0.3 %
0.5 %
0.8 %
1.1 %
Nitrogen
0.2 %
0.2 %
0.2 %
0.2 %
Oxygen
1.0 %
1.0 %
1.0 %
1.0 %
Higher Heating Value
► The
Higher Heating Value (HHV) of a fuel can be
calculate using Dulong’s Formula.
14100BTU
61100BTU 
1
 4000BTU
* C  
* S 
HHV 
 * H 2   * O2  
lbm C
lbm H 2 
8
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
1040BTU  9 lbm H 2O 

* H 2 
LHV  HHV 
lbm H 2O  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.
14100BTU
61100BTU 
1
 4000BTU
* C  
* S 
HHV 
 * H 2   * O2  
lbm C
lbm H 2 
8
lbm S

Higher Heating Value


Btu 
Btu 
0.005 
Btu 




0.01


HHV  14100
0.85lbC  61100
 0.13lbH 2 
   4000



lbC 
lbH 2 
8  
lbS 


HHV  11985  7904.8  40
Btu
HHV  19930
lbFuel
Lower Heating Value


Btu  lbWater
 9
H 2 
LHV  HHV  1040
lbWater  lbFuel


LHV  19930 104090.13
Btu
LHV  18713
lbFuel
Adiabatic Flame Temperature
LHV
AFT 
 TIn
m Exhaust C PExhaust
Adiabatic Flame Temperature
► AFT
= Adiabatic Flame Temperature
► LHV = Lower Heating Value
► ṁExhaust = Fuel + Air
► CP = Specific Heat of the Exhaust Gas
► Tin = 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
► AFT
► AFT
► AFT
► AFT
with theoretical air requirement
= LHV/(ṁExhaust X CP Exhaust) + Tin
= (18713/((1 + 13.75) X .31)) + 220
= 18713/4.57 + 220
= 4315 °F
Adiabatic Flame Temperature
► AFT
with 15% excess air
► AFT = LHV/(ṁExhaust X CP Exhaust) + Tin
► 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
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