fundamemtal of i. c engine 01
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Lecture –1: Basics of Combustion EngineeringCombustion Stoichiometry
1.1 Introduction
1.2 Definitions
1.3 Combustion stoichiometry for gaseous fuels
1.4 Combustion stoichiometry for liquid and solid fuels
1.5 Combustibles burnout for solid fuels
1.2 Definitions
Chemical Reactions, Atoms and Molecules in Combustion
H12
O H O2 2 2+ →
The atoms are conserved (neither created nor destroyed)
Molecules are not conserved
Water is the productHydrogen and oxygen arereactants
Atoms relevant in combustion are: C,H,O,N,S,Cl
1.1 Definitions
Compounds of carbon and hydrogen are called hydrocarbons.
HYDROCARBONS
Aliphatic
Alkanes CnH2n+2Alkenes CnH2nAlkynes CnH2n-2
Alicyclic
CH2
CH2 CH2
Aromatic
BenzeneBenzene derivativesNaphtalene
The first ten members of the unbranched-chain alkanes series are:
CH4 methane C6H14 hexaneC2H6 ethane C7H16 heptaneC3H8 propane C8H18 octaneC4H10 butane C9H20 nonaneC5H12 pentane C10H22 decane
Table 1.1 Names of aliphatic hydrocarbons
CnH2n+1-CnH2n-2CnH2nCnH2n+2n
C5H11- PentylC5H8 PentyneC5H10 PenteneC5H12 Pentane5
C4H9-ButylC4H6 ButyneC4H8 ButeneC4H10 Butane4
C3H7-PropylC3H6 ButyneC3H10PropeneC3H8 Propane3
C2H5- EthylC2H2 EthyneC2H4 EtheneC2H6 Ethane2
CH3- MethylCH4 Methane1
Alkylgroup
AlkyneAlkeneAlkaneNo.of Catoms
Other molecules relevant in combustion are:
Haloalkanes R-X CH3Cl (chloromethane)
Alcohols R-OH C2H5OH (ethanol)
Amines R-NH2 CH3NH2 (methylamine)
Aldehyde R-COH CH3COH (ethanal)
Ketons R-CO-R CH3COCH3 (propanone)
Carboxylic Acid R-COOH CH3COOH (ethanoic acid)
Amount of substances, mole and mass fractions
1 mole contains 6.023 x 1023 particles (atoms, molecules)
For a mixture of species:
n (total number of moles) =∑ni
xnnii= M x Mmean i i=∑
Mole (volume) fractions and mass fractions:
wi = =num ber of kg of species " i"
total num ber of kg in the system
= =∑ ∑
n Mn M
x Mx M
i i
k kk
i i
k kk
xi = =number of moles of species "i" in 1kg of mixture
total number of moles in 1kg of mixture
= =∑
w MM
w Mw M
i i
m e a n
i i
k kk
//
//1
Equation of state for gases and gas mixtures
F p T c( , , ) = 0or
F p T( , , )ρ = 0
The perfect gas equation:
pV nR T= or cp
R T=
ρ = =∑
pMRT
p
RTwM
mean
i
i
The perfect gas law:
pV nR T=Under constant pressure and temperature one mole (kmol) of any ideal gas occupies the same volume.
At normal conditions:p = 760 Tr (1 Tr=133,322 N/m2)T = 273.15 K (0 C)
1 kmol of gas = 22.418 mn3
1 mol of gas = 22.418 dmn3
R = 8,314 J/kmol/K
1.2 Combustion Stoichiometry for Gaseous Fuels
Stoichiometric Combustion
Combustion is said to be stoichiometric if fuel and oxidizerconsume each other completely forming only CO2 and H2O
CH 2O 2H O CO4 2 2 2+ → + stoichiometric
CH 3O 2H O CO O4 2 2 2 2+ → + + lean
CH O H O 0.5CO 0.5CH4 2 2 2 4+ → + + rich
1.3 Combustion Stoichiometry for Gaseous Fuels
Mole fraction of fuel in a stoichiometric mixture
1 kmol fuel O products (CO H O)2 2+ → +ν 2
xfuel stoich in oxygen, _ _ =+
=number of moles of fuel
total number of moles (fuel oxygen)
=+1
1 ν
x fuel stoich in air, _ _ / . .=
+=
+1
1 0 211
1 4 762ν ν
For example:
CO 0.5O CO2 2+ →xfuel,stoich_in_oxygen = 1/1.5=2/3xfuel,stoich_in_air = 1/(1+0.5/0.21)=0.2958
C H 5O 3CO 4H O3 8 2 2 2+ → +
Xfuel,stoich_in_oxygen = 1/(1+5)=1/6
xfuel,stoich_in_air = 1/(1+5/0.21)=0.0403
Excess air ratio (air equivalence ratio)
λ = =( / )
( / )( / )
( / )x xx x
w ww w
air fuel
air fuel stoich
air fuel
air fuel stoich
Fuel equivalence ratio Φ =1λ
Rich combustion Φ > 1λ < 1Stoichiometric combustion λ = 1Lean combustion λ > 1 Φ < 1
Minimum oxygen requirement for a mixture of gaseous fuels
Fuel composition
Molar fractions of species are known
xH2, xCO, xCH4,
xO2, xCnHm
Chemical reactionsH 0.5O H O2 2 2+ →CO 0.5O CO2 2+ →
CH O CO H O4 2 2 2+ → +2 2Om/2HnCOm/4)O(nHC 222mn +→++
Minimum oxygen requirement is then uniquely determined
lO2,min = 0.5xH2+ 0 5. xCO+2xCH4 mnHCm/4)x(n++ − xO2
lO2,min is in kmol O2/kmol fuel
Minimum air requirement for a mixture of gaseous fuels
ll0.21
4.7619 lair,minO ,min
O ,min2
2= = ×
λ = =amount of dry air supplied per kmol of fuel
minimum dry air requirement per kmol of fuell
lair
air,min
The excess air ratio
Amount of combustion products
Fuel composition Chemical reactions
Composition and amount (per a unit of fuel) of combustion products are uniquely determined for a given excess air ratio
Fuel component
Combustion productsCO2 H2O N2 O2
H2 - 1 - -CO 1 - - -CH4 1 2 - -CnHm n m/2 - -
O2 - - - -N2 - - 1 -
Vwet = xH2+ xCO+ 3xCH4
+ +(n m / 2)xC Hn m+ xN2 +0.79lair,min
+ −( )λ 1 lair,min
Vdry = + xCH4 mnHCnx+ + xN2+0.79lair,minCOx min,)1( airl⋅−+ λ
Composition of combustion products
Fuel composition Chemical reactions
Composition and amount (per a unit of fuel) of combustion products are uniquely determined for a given excess air ratio
Fuel component
Combustion productsCO2 H2O N2 O2
H2 - 1 - -CO 1 - - -CH4 1 2 - -CnHm n m/2 - -
O2 - - - -N2 - - 1 -
Vwet = xH2+ xCO+ 3xCH4
+ +(n m / 2)xC Hn m+ xN2 +0.79lair,min
+ −( )λ 1 lair,min
Vdry = + xCH4+ nxC Hn m
+ xN2+0.79lair,min min,)1( airl⋅−+ λCOx
1.3 Combustion stoichiometry for liquid and solid fuels
For solid and liquid fuels the fuel composition is usually expressesin mass fraction (percentage)
The following applies:
c+h+s+o+n+moisture+ash =1
cc
1 moisturedryas_received=
−
cc
1 moisture ashdry_ash_freeas_received=
− −
1%3%1%
4%
82%
4%
4%
Coal Fettnuss mvb
onh
ash
H2O
Carbon
Coal Analysis – “as fired” (“as received”)
Minimum oxygen and air requirements for a solid fuel
Fuel composition
Mass fractions of species are known: c,h,s,o,n,
moisture,ash
Chemical reactionsC O CO2 2+ →H 0.5O H O2 2 2+ →S O SO2 2+ →
Minimum oxygen and air requirements are then uniquely determined
lO ,min2= c
12 + ×12
h2 +
s32
−o
32 ll0 .2 1air, m inO , m in2=
in kmol O2/kg of fuel in kmol air/kg of fuel
Amount of combustion products
Fuel composition
c,h,s,o,n,
moisture,ash
Chemical reactions
Composition and amount (per unit of fuel) of combustion products are uniquely determined for a given excess air ratio
C O CO2 2+ →H 0.5O H O2 2 2+ →S O SO2 2+ →
Vwet = +h2
+moisture
18 +s
3 2+ 0.79 lair,minλ + −( )λ 1 lO ,min2
Vdry = +s
3 2+ 0.79 lair,minλ + −( )λ 1 lO ,min2
12c
12c
28n+
28n+
1.0 1.1 1.2 1.3 1.40
2
4
6
coal Fettnuss
Dutch Natural Gas
C2H6
CH4
O2 %
vol
, dry
Excess air ratio λλλλ
Oxygen (%,vol,dry) content in combustion products as a functionof excess air ratio
1.0 1.1 1.2 1.3 1.40
2
4
6
8
10
12
14
16
18
20
Dutch Natural Gas
C2H6
CH4
coal Fettnuss
CO
2 % v
ol, d
ry
Excess air ratio λλλλ
Carbon dioxide concentration (%,vol,dry) as a function ofexcess air ratio
1.4 Combustibles burnout for solid fuels
Total combustibles = 1 - ash
FURNACESolid fuel
Combustion air
Co+Ash0=1
mo (kg/s)
Combustion products
Unburned solids
C1+Ash1=1
m1 (kg/s)
Question: What fraction of combustibles has been burned ?
FURNACESolid fuel
Combustion air
Combustion products
Unburned solids
Co+Ash0=1 mo (kg/s) C1+Ash1=1 m1 (kg/s)
Mass balance of combustibles:
moCo = m1C1 + bmoCo
Mass balance of ash (ash is assumed to be an inert):
m0Ash0 = m1Ash1
bC A s hC A s h
A s hA s hA s h
= −××
=−
−1
1
11 0
0 1
0
1
0