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Internal Combustion Engines – ME422 Yeditepe Üniversitesi
THERMODYNAMICS of COMBUSTION
Prof.Dr. Cem Soruşbay
Internal Combustion Engines
Thermodynamics of Combustion
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Introduction
Properties of mixtures
Combustion stoichiometry
Combustion in IC-engines
Chemical energy
Heat of reaction
Heat of formation
Chemical equilibrium
Adiabatic flame temperature
Dissociation reactions
1
Introduction
Thermodynamics deals with equilibrium states and how chemical
composition can be calculated for a system of known
atomic/molecular composition if two independent thermodynamic
properties are known.
Systems undergoing chemical reactions are not in general, in chemical
equilibrium - reactions are rate-controlled
Chemical composition at a given instant in time is controlled by the
thermodynamic properties, chemical reaction rates and fluid
dynamics of the system.
Introduction
First Law of Thermodynamics
for a thermodynamic system, the time rate of change of energy of
the system is equal to the rate at which work is done on the system
+ the rate at which heat is transferred to the system.
Neglecting PE and KE within system, energy consists of internal energy
due to,
- thermal energy due to translation, rotation and vibration of
molecules - so called sensible energy,
- chemical energy due to chemical bonds between atoms in the
molecules,
For system of mass m, work done W and heat transfer rate to the
system q, first law can be written as,
d (mu ) &
dt
=W + q
2
Introduction
mechanical work -
power is,
dV
dx
W& = − pA = − p
dt
dt
dV
d (mu )
= −p
+q
dt
dt
first law becomes,
thermodynamic properties can be evaluated by assuming that the whole
system is uniform or by dividing the system into subsystems that are
assumed uniform,
j number of cells with volumes Vj
J
mu = ∑ ρ j V j u j
j =1
Introduction
For a closed system energy balance can be obtained integrating first law
equation with-respect-to time,
m(u2 − u1 ) = −W12 + Q12
where
t2
t2
dV
W12 = ∫ p
dt
dt
t1
for constant pressure,
integrating,
Q12 = ∫ q dt
t1
d (mu + pV )
=q
dt
d (mh)
=q
dt
m(h2 − h1 ) = Q12
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Introduction
If the chemical composition is constant, chemical energy does not
change - for ideal gases,
T2
u2 − u1 = ∫ cv dT
T1
T2
h2 − h1 = ∫ c p dT
T1
c p = cv + R
Properties of Mixtures
For mixtures of gases, system mass is obtained from the sum of masses
of separate species,
m = ∑ mi
i
density of system is the sum of species densities,
ρ = ∑ ρi
i
mass fraction is mass of species i, divided by total mass,
yi =
and by definition
mi ρ i
=
m ρ
∑y
i
=1
i
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Properties of Mixtures
Similarly mole fraction is mole of species i, divided by total number of
moles, or ratio of molar concentration of species i to total molar
concentration,
xi =
∑x
and
i
N i ni
=
N
n
=1
i
M =∑
Molecular weight is,
i
mi
=∑ xi M i
N i
relation between mole fraction and mass fraction,
xi =
N i mi / M i M yi
=
=
N
m/ M
Mi
Properties of Mixtures
mixture internal energy and enthalpy per unit mass,
u = ∑ yi u i
i
h = ∑ yi hi
i
internal energy and enthalpy per mole of mixture,
uˆ = ∑ xi uˆi
hˆ = ∑ xi hˆi
i
i
pressure of mixture – for an ideal gas, is equal to sum of partial p of
component gases if each existed alone in the mixture volume at
mixture temperature,
∑ p =∑x
i
i
i
p= p
i
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Properties of Mixtures
volume of mixture – for an ideal gas, is equal to sum of partial volumes
which the component gases would occupy if each existed alone at the
pressure and temperature of the mixture,
∑V = ∑ x V = V
i
i
i
i
Combustion Stoichiometry
When molecules undergo chemical reaction, the reactant atoms are
rearranged to form new combinations.
For example, hydrogen and oxygen react to form water :
1
H 2 + O2 → H 2O
2
two atoms of hydrogen and one atom of oxygen form one molecule of
water - number of atoms of H and O must be the same on both sides
Such reaction equation represents initial and final states and does not
indicate actual path of reaction, which may involve many
intermediate steps and intermediate species.
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Combustion Stoichiometry
Relative masses of molecules are obtained by multiplying number of
moles of each species by the molecular weights (kg/kg-mol)
For hydrogen-oxygen reaction,
1
H 2 + O2 → H 2O
2
⎛
⎞ ⎛1
2 kg
⎞⎛ 32 kg ⎞
⎟⎟ + ⎜ mol O 2 ⎟⎜⎜
⎟⎟
kg
mol
H
2
kg
mol
O
−
⎝
⎠
2 ⎠
2 ⎠
⎝
⎝
(1 mol H 2 )⎜⎜
⎞
⎛
18 kg
⎟⎟
= (1 mol H 2 O )⎜⎜
⎝ kg − mol H 2 O ⎠
mass of reactants equals mass of products, although mols of reactants
do not equal moles of products - fixed p and T, ideal gas,
1 volume H 2 + 1 / 2 volume O 2 = 1 volume H 2 O
Combustion Stoichiometry
Stoichiometric calculations are done by performing atom balance for
each of the elements in mixture.
The theoretical amount of air required to burn a fuel completely to
products with no dissociation is defined as stoichiometric air.
In most combustion calculations dry air is assumed as a mixture of
79% (vol) N2 and 21% (vol) O2 or 3.764 moles of N2 per mole O2
Molecular weight of pure air is 28.96, as it also contains small amounts
of argon, carbon dioxide, hydrogen etc.
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Dry Air
N2
O2
Ar
CO2
% (vol)
78.09
20.95
0.93
0.03
% (mass)
75.02
23.15
1.28
0.05
Ne , neon
He , helium
Kr , kripton
Xe , xenon
H2 , hydrogen
IC Engine Combustion
In IC engines, complete combustion of the fuel with air, under ideal
conditions - stoichiometric mixture with no dissociation, gives,
m⎞
⎛
Cn H m + ⎜ n +
⎟ (O2 + 3.76 N 2 ) →
4⎠
⎝
m
m⎞
⎛
n CO2 + H 2O + 3.76 ⎜ n +
⎟ N2
2
4⎠
⎝
Under real engine conditions, CO2 , CO , H2 , O2, H , O , NO are also
produced as a result of the dissociation reactions (parçalanma
reaksiyonları).
Sulphur in the fuel produces SO2 etc.
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Fuel / Air Ratio
Mols of stoichiometric air per mol of fuel,
nair − stoich
m⎞
⎛
= 4.76 ⎜ n + ⎟
n fuel
4⎠
⎝
Stoichiometric Fuel / Air ratio by weight,
f stoich =
m fuel
mair
The percent excess air is,
% excess air =
=
=
M fuel n fuel
M air nair − stoich
=
M fuel
29.0 (n + m / 4) 4.76
(mair actual air used, mair-stoich stoichiometric air)
100 (mair − mair − stoich )
mair − stoich
(
100 (nair − nair − stoich ) 100 nO2 − nO2 − stoich
=
nair − stoich
nO2 − stoich
)
Equivalence Ratio
The fuel/air equivalence ratio is defined as the actual fuel / air mass
ratio, divided by stoichiometric fuel / air mass ratio,
φ=
f
f stoich
% excess air is defined as,
air excess ratio is defined as,
(hava fazlalık katsayısı)
% excess air =
λ=
100 (1 − φ )
φ
1
φ
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Equivalence Ratio
Stoichiometric mixture,
φ = λ = 1.0
Lean mixture, (fakir karışım)
φ <1
λ >1
φ >1
λ <1
Rich mixture, (zengin karışım)
Stoichiometric Combustion
Fuel composition, for 1 kg of fuel,
c
h
o
n
s
w
a
kg
kg
kg
kg
kg
kg
kg
of
of
of
of
of
of
of
carbon
hydrogen
oxygen
nitrogen
sulphur
water
ash, etc
c + h + o + n + s + w + a = 1 kg
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Stoichiometric Combustion
For complete combustion of the fuel the following reactions take place
carbon :
C + O2 → CO2
1 mol C + 1 mol O2
12 kg C + 32 kg O2
1 kg C + (32/12) kg O2
1 mol CO2
44 kg CO2
(44/12) kg CO2
hydrogen :
H2 +
1
O2 → H 2O
2
2 kg H2 + 16 kg O2
1 kg H2 + 8 kg O2
18 kg H2O
9 kg H2O
Stoichiometric Combustion
sulphur :
S + O2 → SO2
32 kg C + 32 kg O2
1 kg C + 1 kg O2
64 kg CO2
1 kg CO2
oxygen required will be,
for c kg carbon
for h kg hydrogen
for s kg sulphur
8c/3 kg
8h kg
s kg
minimum amount of O2 needed for complete combustion of 1 kg of fuel
mO − min =
8c
+ 8h + s − o
3
[kg-O2 / kg-fuel]
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Stoichiometric Combustion
mO −min =
8cσ
3
[kg-O2 / kg-fuel]
where
3 ⎛ 8h + s − o ⎞
⎟
8⎝
c
⎠
σ = 1+ ⎜
The amount of oxygen in air is 23.3 % by mass,
mair −min =
mo − min
0.233
[kg-air / kg-fuel]
[kg-air / kg-fuel]
mair −min = 11.44cσ
Stoichiometric Combustion
For IC engine fuels
mair −min = 14 − 15 [kg-air / kg-fuel]
the ratio of air included in the combustion process, to the
theoretically mimimum air required gives the air excess ratio,
λ=
λ
mair
mair − min
equivalence ratio
φ=
1
λ
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Chemical Energy
To describe the chemical energy released when fuel reacts with air to
form products, chemical species in the reactants and products and
their states are specified.
Heat of vaporization of liquid fuels and heat of pyrolysis of solid fuels is
small compared with the chemical energy relased by combustion.
But effect of water condensation can be important.
For lean HC-air mixtures with low T, products may be assumed to be
complete (usually CO2, H2O, O2 and N2). But with high product T and
rich mixtures it is generally necessaary to include other species and
assume chemical equilibrium to determine species mole fractions. If
products are not in chemical equilibrium, then chemical kinetic
analysis (or measurements) is required to determine end state.
Heat of Reaction
Consider F and A mixture of mass m, constant-volume combustion with
heat transfer, Qv ,from state 1 to state 2
m[(u2 − u1 )sensible + (u2 − u1 )chemical ] = Qv
for n species, u at state 1 can be obtained by,
n
(u1 ) sensible = (u (T1 )) sensible = ∑ yi (ui (T1 )) sensible
i =1
n
T1
T1
i =1
To
To
= ∑ yi ∫ (cvi ) r dT = ∫ (cv ) r dT
To is ref temperature, (cv)r is specific heat of reactant mixture
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Heat of Reaction
Similarly,
T2
(u2 ) sensible = ∫ (cv ) p dT
To
If heat heat transfer is just large enough to bring products’ temperature
back to the reactants temperature, and if this T is taken as the ref T,
To for sensible energy, then
(u2 − u1 ) sensible = 0
and Qv is the chemical energy released by the reaction
(sabit hacimdeki reaksiyon ısısı)
The quantity [(1+f)/f](-Qv/m) is the lower heating value (LHV) of the
fuel for constant-volume combustion.
If the water in the products is condensed, that quantity becomes the
higher heating value (HHV) of the fuel for constant-V combustion
Heat of Reaction
If reaction takes place at constant pressure,
m[(h2 − h1 )sensible + (h2 − h1 )chemical ] = Q p
again if T1 = T2 = To , then Qp is the chemical energy released.
For constant-pressure case, if moles of gaseous products Np are larger
than moles of gaseous reactants Nr , then some of the chemical is
expended to push aside the ambient pressure. Thus,
Q p − Qv = ∆( pV ) = ( N p − N r ) Rˆ To = ∆NRˆ To
Here
“overbar” shows quantity per mole
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Heat of Reaction
m⎞
⎛
Cn H m + ⎜ n +
⎟ (O2 + 3.76 N 2 ) →
4⎠
⎝
m
m⎞
⎛
n CO2 + H 2O + 3.76 ⎜ n +
⎟ N2
2
4⎠
⎝
For the reaction,
assume fuel is in gas phase and gaseous water,
∆N =
m
−1
4
if m > 4, ∆N>0 , since both Qv and Qp are negative, |Qv| > |Qp|
for most cases of interest, the difference is negligable.
Heat of Reaction
Heat of reaction can be obtained for reactions at T other than To
To
T2
T1
To
Q p = m ∫ (c p ) r dT + Q p (To ) + m ∫ (c p ) p dT
Q p − Q p (To ) = m(hs 2 − hs1 )
hs is the sensible enthalpy
Qp(To) is negative for an exothermic reaction
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Heat of Reaction
Enthalphy of reactants and products
Heat of Formation
Heat of reaction of fuels combusting in air or oxygen with starting and
ending points at 25 OC and 1 atm gives fuel heating value. These
values are tabulated for common fuels.
This cannot be applied to all reactions - needs enormous amounts of
tabulated data. Instead selected reactions and their heat of reactions
can be added to obtain any given reaction and its heat of reaction.
Heat of formation of a particular species is defined as the heat of
reaction per mole of product formed isothermally from elements in
their standard states. Standard state is chosen as the most stable
form of the element at 1 atm and 25 OC
Heat of Formation of elements, ∆ho in their standard state is assigned
a value of zero. These are given in JANAF tables.
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Heat of Formation
Absolute enthalpy of a substance can be calculated from, sensible
enthalpy relative to reference T plus the heat of formation at ref T,
T
hˆ = ∫ cˆ p dT + ∆hˆ o
To
where ∆ĥ
Here
o
is the heat of formation
overbar, “ ^ ”
shows quantity per mole
Adiabatic Flame Temperature
at constant-pressure combustion
( H R )TR = ( H P )TP
H R (TR ) − H P (TP ) + ∆H o = 0
at constant-volume combustion
U R (TR ) − U P (TP ) + ∆U o = 0
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Adiabatic Flame Temperature
Products
Chemical Equilibrium
Chemical equilibrium is achieved for constant T and p systems when rate
change of concentrations goes to zero for all species - very fast
reaction rates or very small change in concentrations.
For a system of J species in chemical equilibrium, p and T do not
change, which may be specified by stating that Gibbs free energy of
the system (G = H – TS) does not change :
(dG )T , p = 0
J
G = ∑ N j gˆ j
j =1
gˆ j = hˆ j − Tsˆ j
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Chemical Equilibrium
Considering the following reaction,
aA + bB = cC + dD
for chemical equilibrium, the equilibrium constants,
Kp =
pCc pDd xCc xDd c + d − a −b
=
p
p Aa pBb x Aa xBb
Dissociation Reactions
For example,
CO + H 2O ⇔ H 2 + CO2
k f = A f exp − ( E f / RT )
kb = Ab exp − ( Eb / RT )
here kf and kb are the forward and backward reaction rate constants,
A is the pre-exponential factor, E is the activation energy.
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Dissociation Reactions
λ =1
Combustion with dissociation,
m⎞
⎛
Cn H m + ⎜ n +
⎟ (O2 + 3.76 N 2 ) →
4⎠
⎝
β1 CO2 + β2CO + β3 H 2O + β4 H 2 + β5O2 + β6 N 2
K p1 =
CO + H 2O ⇔ H 2 + CO2
1
CO + O2 ⇔ CO2
2
K p2 2 =
β1 β 4
β 2 β3
β12
β22 β5
NR
pR
Dissociation Reactions
Carbon balance,
n = β1 + β 2
Hydrogen balance,
m / 2 = β3 + β4
Oxygen balance,
2(n + m / 4) = 2 β1 + 2 β 2 + β 3 + 2 β 5
Equation of state,
pV = NRT
main reaction equation + 6 equations to find the values of 6 unknowns
βi
i=1,2, … 6
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Dissociation Effects on Flame Temperature
Flame temperature reduces as a result of dissociation reactions.
H
With dissociation
R
P
Tb
T
Temperature
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