# pdfslide.net me-475675-introduction-to-combustion-lecture-21

```ME 475/675 Introduction to
Combustion
Lecture 21
Announcements
• HW 8, Numerical Solution to Example 6.1
• Due Friday, Oct. 17, 2014 (?)
• College Distinguished Lecture
• The future of drone technology
• Saturday, October 18, 2014,
• 5 pm posters;
• 6 pm Lecture
Chapter 6 Coupling Chemical and Thermal Analysis of
Reacting systems
• Four simple reactor systems, p 184
1. Constant pressure and fixed Mass
• Time dependent, well mixed
2. Constant-volume fixed-mass
• Time dependent, well mixed
3. Well-stirred reactor
• Steady, different inlet and exit conditions
4. Plug-Flow
• Coupled Energy, species production,
and state constraints
• For plug flow also need momentum
since speeds and pressure vary with
location
Constant pressure and fixed Mass Reactor
π
• Constituents
• reactants and products, π = 1, 2, … π (book uses π)
• P and m constant
• Find as a function of time, t
• Temperature π
• To find use conservation of energy
• Molar concentration π =
ππ
π
π
(book calls this ππ )
• use species generation/consumption rates from chemical kinetics
• π=
π
, ππππ
π
π
• state, mixture
• Highly coupled
• Assume we know “production rates” per unit volume
•
1 πππ
π ππ‘
= ππ = π(π)
π
π=1
π
ππ
= ππ 1 , 2 , … π
• Rate depends on current molar concentration (per volume) of each constituent, and temperature
• From chemical Kinetics
First Law (Energy
Conservation)
• π−π =
=π
ππ’
ππ‘
ππ
• Only boundary work: π = π ππ‘
ππ
ππ
=
ππ‘
ππ‘
ππ
ππ
+π
ππ‘
ππ‘
• π−π
• π=
π
ππ
ππ‘
=
π π+ππ
ππ‘
=
ππ»
ππ‘
• Where enthalpy π» = π + ππ, For a mixture H =
• π=
ππ»
ππ‘
=
πππ
β
ππ‘ π
ππ βπ π
π βπ
ππ‘
+ ππ
1 πππ
;
ππ‘
πβ
; ππ‘π = ππ,π
• Production rate ππ = π
•
• π=
•
π
•
π
π
=
πππ
ππ‘
= π ππ ; ππ = V π
πππ βπ + V π ππ,π
ππ βπ + π ππ,π
ππ
ππ‘
ππ
ππ‘
=
π
ππ
ππ‘
; Divide by π
ππ βπ +
ππ
ππ‘
π ππ,π ; Solve for
ππ
ππ‘
π
−
π
π π βπ π
ππ
=
ππ‘
π ππ,π π
• First order differential equation, Initial conditions (IC): π = π0 ππ‘ π‘ = 0
• At each time step, to find the change in π
• Need π, π, ππ (which depends on π and T), π ,π, βπ π , and ππ,π (π)
Change in Molar Concentrations
•
ππ
ππ‘
=
π
ππ
π
=
ππ‘
1 πππ
π ππ‘
−
ππ ππ
π 2 ππ‘
= ππ −
π ππ
π ππ‘
*
• Species production and volume change affect molar concentration
• Find the volume V from ideal gas equation of state
• ππ = πππ’ π =
ππ ππ’ π
• Take time derivative
• π
ππ
ππ‘
= ππ’
π π ππ
ππ‘
π
to see how volume changes with time
ππ‘
= ππ’ π
• Divide both sides by ππ =
•
1 ππ
π ππ‘
=
1
ππ
πππ
ππ‘
1 ππ
+
π ππ‘
=
π
ππ
ππ‘
+
ππ
ππ‘
ππ = ππ’ π
ππ
ππ‘
= ππ − π
ππ
π
+
+
ππ ππ’ π
1
ππ
π
1 πππ
π ππ‘
1 ππ
+
π ππ‘
=
ππ
π
• Plug into *
•
πππ
ππ‘
1 ππ
π ππ‘
• Initial Conditions: at t = 0, π = π 0 , π = 1, 2, … π
1 ππ
+
π ππ‘
ππ
ππ‘
ππ
π + 1 coupled System of 1st order differential equations
• Initial Conditions, at t = 0
• π = π 0 , π = 1, 2, … π, and
• π = π0
• Assume we also know ππ = π΄ππ₯π(−
πΈπ΄
)
ππ’ π
π
π=1
π
ππ
• Use the first order differentials to find π and π at time t + Δπ‘
•
•
ππ
ππ‘
ππ
ππ‘
= ππ − π
=
π
−
π
ππ βπ π
π ππ,π π
ππ
π
≈
+
1 ππ
π ππ‘
≈
Δπ
Δπ‘
;
π
Δπ
;
Δπ‘
π‘+Δπ‘
= π π‘+
ππ
ππ‘
ππ
ππ‘
Δπ‘
ππ‘+Δπ‘ = ππ‘ +
Δπ‘
• System Volume
• π π‘ =
t
T
π
π π
=
[1] [2]
π
;
π πππ
π=
π
π
=
… [M] w1 w2
0 T0 [1]0 [2]0 … [M]0
Dt
2Dt
ππ
π
=
…
ππ πππ
π
wM
V
=
π πππ
Q d[1]/dt d[2]/dt
…
d[M]/dt dT/dt
Constant-Volume Fixed-Mass Reactor
• Constant V and m
• Find T, P, π (πππ π = 1,2, … , π) versus time π‘
• 1st Law
• π−π =
• π=
πππ
π’π
ππ‘
• ππ =
• ππ =
• π=
•
•
=
ππ
ππ‘
=
=
π
ππ π’π π ;
• π=
π
π
ππ
ππ‘
ππ π’π π
ππ‘
π π’π
+ ππ
ππ‘
1 πππ πππ
;
= π ππ
π ππ‘ ππ‘
ππ’
ππ
V π , π = ππ£,π π
ππ‘
ππ‘
π’π π ππ + V π ππ£,π
π’π ππ + π
π
−
π
π’ π ππ
π ππ£,π
ππ
ππ£,π
ππ‘
ππ
ππ‘
=
, divide by V
π’π ππ +
ππ
ππ‘
π ππ£,π , solve for
ππ
ππ‘
Tabulated Data
• Need to evaluate
ππ
ππ‘
=
π
−
π
• However, tables only have
π’π ππ
π ππ£,π
(true but not useful)
• ππ,π (= ππ£,π + ππ’ ), so use
• ππ£,π = ππ,π − ππ’
• βπ (= π’π + ππ£ = π’π + ππ’ π) , so use
• π’π = βπ − ππ’ π
•
ππ
ππ‘
=
π
−
π
βπ ππ +ππ’ π
π ππ,π −ππ’
ππ
(true and useful)
• Initial Condition: π = π0 at π‘ = 0
• Species Production
•
ππ
ππ‘
π
=
π ππ
ππ‘
=
1 πππ
π ππ‘
= ππ = π(π)
π
π=1
π
ππ
Reactor Pressure
• Ideal Gas Law
• ππ =
ππ ππ’ π
• Divide by π (constant)
• π=
ππ
π
ππ’ π =
π ππ’ π
• Pressure Rate of change (affects detonation)
•
ππ
ππ‘
= ππ’ π
π
π
ππ‘
+
π
ππ
ππ‘
= ππ’ π
π
ππ
π
ππ‘
+
ππ
ππ‘
π
= ππ’ π
ππ +
ππ
ππ‘
π
Example 6.1 (p. 189) This will be HW
• In spark-ignition engines, knock occurs when the unburned fuel-air mixture
ahead of the flame reacts homogeneously, i.e., it auto-ignites. The rate-ofpressure rise is a key parameter in determining knock intensity and propensity for
mechanical damage to the piston-crank assembly. Pressure-versus-time traces
for normal and knocking combustion in a spark-ignition engine are illustrated in
Fig. 6.2. Note the rapid pressure rise in the case of heavy knock. Figure 6.3
shows schleiren (index-of-refraction gradient) photographs of flame propagation
for normal and knocking combustion
•
Example 6.1
• Create a simple constant-volume model of the autoignition process and determine the
temperature and the fuel and product concentration histories. Also determine the
dP/dt as a function of time. Assume initial conditions corresponding to compression of
a fuel-air mixture from 300 K and 1 atm to top-dead-center for a compression ratio of
10:1. The initial volume before compression is 3.68*10-4 m3, which corresponds to an
engine with both a bore and a stroke of 75 mm. Use ethane as fuel. Assume:
• One-step global kinetics using the rate parameters for ethane C2H6 (Table 5.1)
• Fuel, air, and products all have equal molecular weights: MWF= MWOx= MWP= 29
• The specific heats of the fuel, air and products are constants and equal:
• cp,F= cp,Ox= cp,Pr= 1200 J/kgK
• The enthalpy of formation of the air and products are zero, and that of the fuel is
• 4*107j/kg
• The stoichiometric air-fuel ratio is 16.0 and restrict combustion to stoichiometric or
lean conditions.
Global and Quasi-global mechanisms
• Empirical
• πΆπ₯ π»π¦ + π₯ +
π¦
4
π2
ππΊ
π¦
2
π₯πΆπ2 + π»2 π
• stoichiometric mixture with π2 , not air
•
π πΆπ₯ π»π¦
ππ‘
= −π΄ππ₯π
πΈπ ππ’
π
πΆπ₯ π»π¦
π
π2
π
=
πππππ
ππ3 π
• Page 157, Table 5.1: π΄, πΈπ ππ’ , π πππ π for different fuels
• These values are based on flame speed data fit (Ch 8)
• In Table 5.1 units for π΄ =
• However, we often want π΄
•
1 πππππ 1−π−π
π
ππ3
• π΄
1 πππππ 1−π−π
π
π3
πππππ 1−π−π
−
π+π
πππππ
πππππ
ππ3
=
ππ3 π
ππ3
π
1−π−π
1 πππππ
in units of
π
π3
πππππ
1000 πππππ
=π΄
1−π−π
100 ππ 3
π
1 πππππ 1−π−π
π
ππ3
=
1 πππππ 1−π−π
π
ππ3
Sometimes Want These Units
1000
1−π−π
=
1 πππππ 1−π−π
π
π3
10001−π−π = π΄ πππ₯π‘ππππ 10001−π−π
Given in Table 5.1, p. 157
•
```