Chapter 5/1-19 (Includes Solutions to Examples)

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:
CF =
c* =
2γ ⎛ 2 ⎞
γ −1 ⎜⎝ γ +1 ⎟⎠
2
RT0 ⎡ γ +1 ⎤
γ ⎢⎣ 2 ⎥⎦
γ +1
γ −1
γ +1
γ −1
γ −1
⎡
⎤
γ
⎛
⎞
pe
pe − pa
⎢
⎥
1−
+
⋅ε
⎢ ⎜⎝ p ⎟⎠ ⎥
p
0
0
⎢⎣
⎥⎦
Performance depends on:
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
⇔
 Universal Gas Constant:
ℜ = 8.314
 Gas Constant:
R=
pv = RT
J
mol ⋅K
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 )
 Specific Heat Relationships:
cp − cv = R
γ =
cp
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
pressure of each constituent when each occupies alone the
volume of the mixture at the temperature of the mixture.
TContainer
pContainer
 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
 Energy
E mix = m mix emix = m1e1 + m2 e2 +…+ m N eN = ∑ m i ei
“Bar” denotes Property
with respect to
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.
 Definition of Mass Fraction:
m
m
yi = i = N i
m mix
∑ mi
⇒
1
Mmix =
N
∑(y
equiv
i=1
 Perfect Gas Law
 Pressure (Gibbs-Dalton Law)
i
Mi
∑y
i=1
1
 Equivalent Molecular Weight:
N
)
m mix
=
n mix
i
=1
VContainer
TContainer
pContainer
p i V = m i R i T = n i ℜT
N
p = ∑ pi
i=1
 Enthalpy
h mix = ∑ yi h i
i
 Entropy
smix (T, p) = ∑ yi si (T, p i )
i
 where species entropy is
si (T, pi ) = si (T, p ref ) − R ln
pi
p ref
Ch5 –7
AAE 439
 Definition of Mole Fraction:
Mixture of Gases
n
n
xi = i = N i
n mix
∑ ni
⇒
N
∑x
i=1
i
=1
1
N
 Equivalent Molecular Weight:
Mmix = ∑ x i Mi =
equiv
i=1
m mix
n mix
VContainer
 Perfect Gas Law
TContainer
p i V = m i R i T = n i ℜT
pContainer
N
 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
 where species entropy is
si (T, pi ) = si (T, p ref ) − ℜ ln
pi
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
 Ratio of Specific Heat:
γ mix =
c p,mix
c v,mix
=
c p,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
Q
m, E
Q
−
Heat added to
system in going
from state 12
W
W
=
Work done by system
on surrounding in
going from state 12
ΔE1→2
Change in total system
energy in going
from state 12
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
)
dmCV
inlet
dt
Q CV
Rate of heat
transferred across
the CS, from the
surrounding to the CV.

W
CV
−
QCV
Rate of all work
done by CV,
including shaft work
but excluding flow work.
=
=0
dE CV
dt
(
m e + p v
=0
m eoutlet
Rate of energy
flowing out
of CV.
WCV
−
)
(
+ m po vo − p i v i
m einlet
Rate of energy
flowing into
CV.
 = m ⎡ h − h + 1 v2 − v2 + g z − z ⎤
Q CV − W
⎢ o
CV
i
o
i ⎥
2 o i
⎣
⎦
(
) (
) (
outlet
)
)
Net rate of work
associated with pressure
forces where fluid
crosses CS, flow work.
 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
V1
⇔
δ Wflow = p dV
 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:
⎛m
⎞
⎛O ⎞
oxidizer
⎟⎟
⎜ ⎟ = ⎜⎜
⎝ F ⎠stoic ⎝ m fuel ⎠stoic
⎛ m air ⎞
n air Mair 4.76 ⋅ a Mair
⎛ A⎞
=
=
⋅
=
⋅
⎜
⎟
⎜⎝ F ⎟⎠
m
n
M
1
M fuel
⎝
⎠
fuel stoic
fuel
fuel
stoic
 Equivalence Ratio Φ :
O F)
(
Φ=
stoic
O F
F O
=
F O stoic
(
)
⎛ O ⎞ n oxygen Moxygen
where ⎜ ⎟ =
⎝ F ⎠ n fuel M fuel
 This ratio is a quantitative indicator whether a fuel-oxidizer mixture is
!<1
 Lean:
 Rich:
!>1
 Stoichiometric: ! = 1
 Other Parameters:
 Percent Stoichiometric Oxidizer:
% stoichiometric oxidizer =
 Percent Excess Oxidizer:
% excess oxidizer =
100%
Φ
(1− Φ)
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
H : y = 2c
b=x
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
H : y = 2c + y d
b = x (1 − 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

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