AAE 439
5. COMBUSTION AND THERMOCHEMISTRY
Ch5 –1
AAE 439
Overview
Definition & mathematical determination of chemical equilibrium,
Definition/determination of adiabatic flame temperature,
Prediction of composition and temperature of combusted gases as a function
of initial temperature,
Prediction of amounts of fuel & oxidizer,
Thermochemical changes during expansion process in nozzle.
Performance Parameters:
γ +1 ⎡ γ −1
⎤
2γ ⎛ 2 ⎞
2 γ −1
⎢ ⎛ pe ⎞ γ
⎥ pe − pa
CF = 1−
⎢ ⎜⎝ p ⎟⎠ ⎥ + ⋅ε
γ −1 ⎜⎝ γ +1 ⎟⎠ p
⎢⎣ 0
⎥⎦ 0
γ +1
RT0 ⎡ γ +1 ⎤ γ −1
Performance depends on:
c* =
γ ⎢⎣ 2 ⎥⎦ T, MW, p0, pe, pa, γ
Ch5 –2
AAE 439
Overview
Important Concepts & Elements of Analysis
Conversion of Chemical Energy to Heat
Simple Treatment of Properties of Gases
Balancing Chemical Reactions - Stoichiometry
Adiabatic Flame Temperature
Chemical Equilibrium and Gibbs Free Energy
Nozzle Expansion Effects
Thermochemical Calculations
Ch5 –3
AAE 439
5.1 THERMODYNAMICS OF GAS MIXTURES
Ch5 –4
AAE 439
Perfect Gas
Perfect Gas Law relates pressure, temperature and density for a perfect gas/
mixture of gases :
p V = n ℜT = mR T ⇔ pv = RT
J
Universal Gas Constant: ℜ = 8.314
mol ⋅K
ℜ
Gas Constant: R=
M
Calorically Perfect Gas:
Internal Energy du = c v dT u2 − u1 = c v (T2 − T1 )
Enthalpy dh = c p dT h 2 − h1 = c p (T2 − T1 )
cp
Specific Heat Relationships: cp − cv = R γ =
cv
Definition of “Mole”:
A mole represents the amount of gas, which contains Avogadro’s number of gas
molecules: 6.02·1023 molecules/mol.
Ch5 –5
AAE 439
Gibbs-Dalton Law
Properties of a mixture is determined by the properties of
constituents according to Gibbs–Dalton Law:
VContainer
The pressure of a mixture of gases is equal to the sum of the TContainer
pressure of each constituent when each occupies alone the pContainer
volume of the mixture at the temperature of the mixture.
The internal energy and the entropy of a mixture are equal,
respectively, to the sums of the internal energies and the
entropies of its constituents when each occupies alone the
volume of the mixture at the temperature of the mixture.
Temperature Tmix = T1 = T2 = … = TN
N
Pressure pmix = p1 + p2 + p3 …+ pN = ∑ pi
i=1
Volume Vmix = m mix v mix = m1v1 = m2 v2 = … = m N v N
N “Bar” denotes Property
E mix = m mix emix = m1e1 + m2 e2 +…+ m N eN = ∑ m i ei
with respect to
Energy Molar Quantity
i=1
Entropy Smix = m mix smix = m1 s1 + m2 s2 + … + m N sN smix = Smix n mix
Enthalpy H mix = m mix h mix = m1h1 + m2 h 2 + … + m N h N h mix = H mix n mix
Ch5 –6
AAE 439
Mixture of Gases
Composition of a gas mixture is expressed by either the constituent mass
fractions or mole fractions.
m m N
Definition of Mass Fraction: yi = i = N i
m mix
⇒ ∑y i
=1
∑ mi i=1
1
VContainer
1 m mix
Equivalent Molecular Weight: Mmix = N
= TContainer
n mix
equiv
∑(y i
Mi ) pContainer
i=1
Perfect Gas Law p i V = m i R i T = n i ℜT
N
Pressure (Gibbs-Dalton Law) p = ∑ pi
i=1
Enthalpy h mix = ∑ yi h i
i
Entropy smix (T, p) = ∑ yi si (T, p i )
i
pi
where species entropy is si (T, pi ) = si (T, p ref ) − R ln
p ref
Ch5 –7
AAE 439
Mixture of Gases
n n N
Definition of Mole Fraction: xi = i = N i
n mix
⇒ ∑x i
=1
∑ ni i=1
1
N
m mix
Equivalent Molecular Weight: Mmix = ∑ x i Mi =
equiv i=1 n mix
VContainer
Perfect Gas Law p i V = m i R i T = n i ℜT TContainer
N
pContainer
Pressure (Gibbs-Dalton Law) p = ∑ pi
i=1
Partial Pressure: pi = xi p
Enthalpy h mix = ∑ x i h i
i
Entropy smix (T, p) = ∑ x i si (T, p i )
i
pi
where species entropy is si (T, pi ) = si (T, p ref ) − ℜ ln
p ref
Ch5 –8
AAE 439
Mixture of Gases
Relationship between Mass and Mole Fractions:
Mmix
x i = yi
Mi
Other Relationships for a Gas Mixture:
N
Specific Heat: c p,mix = ∑ c p,i yi
i=1
c p,mix c p,mix
Ratio of Specific Heat: γ mix = =
c v,mix c p,mix − R mix
Ch5 –9
AAE 439
5.2 1st LAW OF THERMODYNAMICS
Ch5 –10
AAE 439
1st LTD - Fixed Mass
First law of thermodynamics embodies the fundamental principle of
conservation of energy.
Q and W are path functions and occur only at the system boundary.
E is a state variable (property), ∆E is path independent.
System Boundary enclosing Fixed Mass
m, E
W
Q − W = ΔE1→2
Heat added to Work done by system Change in total system
system in going on surrounding in energy in going
from state 12 going from state 12 from state 12
Q − W = dE dt
q − w = de dt
⎛ 1 2 ⎞
E = m ⎜u + v + g z⎟
⎝ 2 ⎠
Ch5 –11
AAE 439
1st LTD - Control Volume
Conservation of energy for a steady-state, steady-flow system.
Control Surface (CS) enclosing Control Volume (CV)
(
m e + p v ) inlet dmCV dE CV (
m e + p v )
outlet
=0 =0
dt dt
QCV WCV
Q CV −
W CV
= m eoutlet − m einlet (
+ m po vo − p i v i )
Rate of heat Rate of all work Rate of energy Rate of energy Net rate of work
transferred across done by CV, flowing out flowing into associated with pressure
the CS, from the including shaft work of CV. CV. forces where fluid
surrounding to the CV. but excluding flow work. crosses CS, flow work.
= m ⎡ h − h + 1 v2 − v2 + g z − z ⎤
Q CV − W ( ) ( ) ( )
CV ⎢ o i
2 o i o i ⎥
⎣ ⎦
Assumptions:
Control Volume is fixed relative to the coordinate system.
Eliminates any work interactions associated with a moving boundary,
Eliminates consideration of changes in kinetic and potential energies of CV itself.
Properties of fluid at each point within CV, or on CS, do not vary with time.
Fluid properties are uniform over inlet and outlet flow areas.
There is only one inlet and one exit stream. Ch5 –12
AAE 439
TD PROCESSES in CHEM. SYSTEMS
Chemical systems (chemical reactions) are treated as either constant-volume or
constant-pressure processes.
Energy Equation (1st Law of TD)
E = U + E potential + E kinetic = Q − Wshaft − Wflow
Inside a rocket combustion chamber, fluid velocity (Ekin) is small and height changes
of the fluid mass (Epot) is negligible. Energy contained in the fluid is governed by
the internal energy of the hot combustion gas.
E =U ⇔ dE = dU = (δ Q − δ Wshaft − δ Wflow )
Work contribution in a rocket combustion chamber results from changes in specific
volume of pressure. The fluid doesn’t perform any mechanical work (Wshaft=0).
V2
W = − ∫ p(ext ) dV ⇔ δ Wflow = p dV
V1
Constant–Volume (Isochoric) Process: dU = Q
Constant–Pressure (Isobaric) Process: dU = Q − p dV ⎫
⎬ dH = Q
H = U + pV ⎭
Ch5 –13
AAE 439
5.3 REACTANT AND PRODUCT MIXTURES
Ch5 –14
AAE 439
STOICHIOMETRY
The stoichiometric quantity of oxidizer (substance A) is just that amount
needed to completely burn a quantity of fuel (substance B):
An oxidizer-fuel mixture is LEAN, when there is more than a stoichiometric
quantity of oxidizer in the mixture.
An oxidizer-fuel mixture is RICH, when there is less than a stoichiometric quantity
of oxidizer in the mixture.
Stoichiometric Chemical Reaction:
Examples:
CH 4 + 2O2 → CO2 + 2H 2O
One mole of methane and 2 moles of oxygen form one mole of carbon dioxide and 2
mole of water.
H 2 + 12 O2 → H 2O
One mole of H2 and a half mole of O2 form one mole of H2O.
Ch5 –15
AAE 439
STOICHIOMETRY
Stoichiometric Oxidizer-Fuel Ratio:
⎛O ⎞ ⎛m ⎞ ⎛ A⎞ ⎛ m air ⎞ n air Mair 4.76 ⋅ a Mair
⎜ ⎟ = ⎜⎜ ⎟⎟
oxidizer
⎜⎝ F ⎟⎠ = ⎜ ⎟ = ⋅ = ⋅
⎝ F ⎠stoic ⎝ m fuel ⎠stoic stoic ⎝ m ⎠
fuel stoic
n fuel
M fuel
1 M fuel
Equivalence Ratio Φ :
Φ=
( O F)
=
F O ⎛ O ⎞ n oxygen Moxygen
where ⎜ ⎟ =
stoic
O F ( )
F O stoic ⎝ F ⎠ n fuel M fuel
This ratio is a quantitative indicator whether a fuel-oxidizer mixture is
Lean: !<1
Rich: !>1
Stoichiometric: ! = 1
Other Parameters:
100%
Percent Stoichiometric Oxidizer: % stoichiometric oxidizer =
Φ
(1− Φ)
Percent Excess Oxidizer: % excess oxidizer = 100%
Φ
Ch5 –16
AAE 439
AIR (O2)/FUEL COMBUSTION
Stoichiometric Combustion of Air and Fuel (Hydrocarbon)
( ) y
C xH y + a ⋅ O2 + 3.76 N 2 → x ⋅CO2 + ⋅H 2O + 3.76a ⋅ N 2
2
x & y define the
y
hydrocarbon fuel!
⇒ a= x+
4
Lean Combustion of Air and Fuel
(
C xH y + a ⋅ O2 + 3.76 N 2 ) → b ⋅CO2 + c ⋅H 2O + d ⋅O2 + 3.76 a ⋅ N 2
Balancing Chemical Reaction: C: x=b b=x
H : y = 2c c = 12 y
O : 2a = 2b + c + 2d a = x + 14 y + d
Rich Combustion of Air and Fuel
(
C xH y + a ⋅ O2 + 3.76 N 2 ) → b ⋅CO2 + c ⋅H 2O + d ⋅C xH y + 3.76 a ⋅ N 2
Balancing Chemical Reaction: C : x = b + xd b = x (1 − d)
H : y = 2c + y d c = 12 y (1 − d)
O : 2a = 2b + c a = (x + 14 y) (1 − d)
Ch5 –17
AAE 439
Examples
Example #1:
A small, low-emission, stationary gas-turbine engine operates at full load (3,950
kW) at an equivalence ratio of 0.286 with an air flowrate of 15.9 kg/s. The
equivalent composition of the fuel (natural gas) is C1.16H4.32. Determine the fuel
mass flow rate and the operating air-fuel ratio for the engine.
Ch5 –18
AAE 439
Examples
Example #2:
A natural-gas-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. Treat the natural gas as methane.
Ch5 –19
