MOTOR BAKAR ( 3 SKS)
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MOTOR BAKAR
( 3 SKS)
Combustion
Particle diagram equationCombustion
C2 + 2O2 2CO2
CC O
O
OO+
OO C
OO C
Hydrocarbon Oxidationmethane (CH4), the primary constituent of liquefied or compressed natural gas
propane (C3H8), the primary constituent of liquid petroleum gas
isooctane (C8H18), typical of the molecules found in gasoline
n-hexadecane (C16H34), typical of diesel fuel
If sufficient oxygen is available, a hydrocarbon fuel can be completely oxidized, the carbon is converted to carbon dioxide (CO2) and the hydrogen is converted to water (H2O).
The overall chemical equation for the complete combustion of one mole of propane (C3H8) with oxygen is:
OcHbCOaOHC 22283
Elements cannot be created or destroyed, soC balance: 3 = b b= 3H balance: 8 = 2c c= 4O balance: 2a = 2b + c a= 5
Thus the above reaction is:OHCOOHC 22283 435
# of moles species
Hydrocarbon Oxidation
Hydrocarbon OxidationAir Composition
Oxygen : 21%Nitrogen : 79%
Oxygen - Nitrogen Ratio in Air 1 : 3.76
2222283 8.184376.35 NOHCONOHC
Combustion StoichiometryAir contains molecular nitrogen N2, when the products are low temperature the nitrogen is not significantly affected by the reaction, it is considered inert.
The complete reaction of a general hydrocarbon CxHy with air is:
22222 )76.3( dNOcHbCONOaHC yx
The above equation defines the stoichiometric proportions of fuel and air.
22222 476.32)76.3(4 NyxOHyxCONOyxHC yx
C balance: x = b b = xH balance: y = 2c c = y/2O balance: 2a = 2b + c a = b + c/2 a = x + y/4N balance: 2(3.76)a = 2d d = 3.76a/2 d = 3.76(x + y/4)
Combustion StoichiometryThe stoichiometric quantity of oxidizer is just that amount needed to completely burn a quality of fuel. If more than a stoichiometric quantity of oxidizer is supplied, the mixture is said to be fuel lean
While supplying less than the stoichiometric oxidizer result in fuel rich
1*12
)28*76.332(41
)/(1/
xy
xy
AFFA
ss
Combustion Stoichiometry
Substituting the respective molecular weights and dividing top and bottom by x one gets the following expression that only depends on the ratio of thenumber of hydrogen atoms to hydrogen atoms (y/x) in the fuel.
Example: For Octane (C8H18), y/x = 2.25 (A/F)s = 15.1 Benzene (C8H16), y/x = 2.0 (A/F)s = 14.7
The stoichiometric mass based air/fuel ratio for CxHy fuel is:
HC
NO
fuelii
airii
fuel
airs MyMx
MyxMyx
MnMn
mmFA
22 476.34/
MO2 : 32
MN2 : 28
MC : 12MH : 1
Fuel Lean Mixture• Fuel-air mixtures with more than stoichiometric air (excess air) can burn
• With excess air you have fuel lean combustion
• At low combustion temperatures, the extra air appears in the products in unchanged form:
for a fuel lean mixture have excess air, so > 1
222222 2)76.3)(4( eOdNOHyxCONOyxHC yx
• Fuel-air mixtures with less than stoichiometric air (excess fuel) can burn.
• With less than stoichiometric air you have fuel rich combustion, there is insufficient oxygen to oxidize all the C and H in the fuel to CO2 and H2O.
• Get incomplete combustion where carbon monoxide (CO) and molecular hydrogen (H2) also appear in the products.
222222 2)76.3)(4( fHeCOdNOHyxCONOyxHC yx
where for fuel rich mixture have insufficient air < 1
Fuel Rich Mixture
The equivalence ratio, , is commonly used to indicate if a mixture is stoichiometric, fuel lean, or fuel rich.
s
mixture
mixture
s
AFAF
FAFA
//
//
Off-Stoichiometric Mixtures
stoichiometric = 1 fuel lean < 1 fuel rich > 1
Products)76.3(4 22
NOyxHC yx
Stoichiometric mixture:
Off-stoichiometric mixture:
Products)76.3(41
22
NOyxHC yx
Example: Consider a reaction of octane with 10% excess air, what is ?
Off-Stoichiometric Conditions
22222188 4798)76.3(5.12 NOHCONOHC
10% excess air is:222222188 98)76.3)(5.12(1.1 bNaOOHCONOHC
91.01/)76.4)(5.12(1.1
1/)76.4(5.12//
mixture
s
FAFA
16 + 9 + 2a = 1.1(12.5)(2) a = 1.25, b = 1.1(12.5)(3.76) = 51.7
Other terminology used to describe how much air is used in combustion:110% stoichiometric air = 110% theoretical air = 10% excess air 1.1 )76.3)(4
83( 2283 NOHC mixture is fuel lean
Stoichiometric :
Example
9.58286.085.16//
s
mixtureFAFA
85.161*16.132.412
)28*76.332(416.132.41
1*12
)28*76.332(41
)/(1/
xy
xy
AFFA
ss
A small stationary gas turbine engine operates at full load (3950 kW) at an equivalence ratio of 0.286 with an air flow rate of 15.9 kg/s. The equivalent composition of the fuel is C1.16H4.32. Determine the fuel mass flow rate and operating air fuel ratio for the engineSolution
skgFAmm
mixture
airfuel /27.09.58
9.15/
Example
aa
abb
NN
mix
OO 76.41
276.321
22
A natural gas (methane / CH4) – fired industrial boiler operates with an oxygen concentration of 3 mole percent in the flue gases. Determine the operating air-fuel ratio and the equivalence ratio.Solution3% of O2 in flue gases Fuel lean mixtureIf all fuel C is found in CO2 and all fuel H is found in H2O
CH4 + a(O2 + 3.76N2) 1CO2 + 2H2O + bO2 + 3.76a N2
O balance ; 2a = 2 + 2 + 2b b = a - 2Mole fraction of O2
aa76.41203.0
a = 2.368
3.201684.28
176.4)/(
)/(
aFA
MWMW
NNFA
mixture
fuel
air
fuel
airmixture
1.171*1412
)28*76.332(4141
1*12
)28*76.332(41
)/(1/
xy
xy
AFFA
ss
MWair = 21% x 32 + 79% x 28 = 28.84MWfuel = 12 + 4 x 1 = 16
Air Fuel Ratio Stoichiometry
Air Fuel Ratio
Equivalence ratio
84.03.20
1.17//
mixture
s
FAFA
Example
1. O2 Concentration in flue gases is 8 mole percent on oxidation of 200 kg/h diesel fuel
Determine the operating air-fuel ratio, the equivalence ratio and the air flow rate, if:
Solution 1
aa
abb
NN
mix
OO 76.47
2376.31416
22
O2 Concentration in flue gases is 8 mole percent on oxidation of 200 kg/h diesel fuel
8% of O2 in flue gases Fuel lean mixtureIf all fuel C is found in CO2 and all fuel H is found in H2O
C16H28 + a(O2 + 3.76N2) 16CO2 + 14H2O + bO2 + 3.76a N2
O balance ; 2a = 32 + 14 + 2b b = a - 23Mole fraction of O2
aa
76.472308.0
a = 30.05
74.2322084.28
176.4)/(
)/(
aFA
MWMW
NNFA
mixture
fuel
air
fuel
airmixture
35.141*162812
)28*76.332(416281
1*12
)28*76.332(41
)/(1/
xy
xy
AFFA
ss
MWair = 21% x 32 + 79% x 28 = 28.84MWfuel = (12 x 16) + (28 x 1) = 220
Air Fuel Ratio Stoichiometry
Air Fuel Ratio
Equivalence ratio
604.074.23
35.14//
mixture
s
FAFA
Air flow rate
mair = mfuel x (A/F)mixture
mair = 200 x 23.74 = 4748 Kg/h
The maximum amount of energy is released from a fuel when reacted with astoichiometric amount of air and all the hydrogen and carbon contained in thefuel is converted to CO2 and H2O
This maximum energy is referred to as the heat of combustion or the heating value and it is typically given per mass of fuel
Heat of Combustion
22222 476.32)76.3(4 NyxOHyxCONOyxHC yx
HR(298K)
alcohols
Fuel Energydensit
y(MJ/L)
Air-fuelratio
Specificenergy(MJ/kg air)
Heat ofvaporiza
tion
Gasoline and biogasoline
32 14.7 2.9 0.36 MJ/kg
Butanol fuel 29.2 11.2 3.2 0.43 MJ/kg
Ethanol fuel 19.6 9.0 3.0 0.92 MJ/kg
Methanol 16 6.5 3.1 1.2 MJ/kg
Heat of Formation
Heat of Formation for 1 Bar and 298.15 K
Heat Transfer in a Chemically Reacting Flow
Example
Example
Combustion Flame Temperature
Example
Heat Combustion and Heating Value
Example
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