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advertisement ```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
• https://www.unr.edu/nevada-today/news/2014/college-of-engineering-distinguishedlecture-series
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
• Steady, dependent on location
• 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
𝑊
• Constituents
• reactants and products, 𝑖 = 1, 2, … 𝑀 (book uses 𝑁)
• P and m constant
• Find as a function of time, t
• Temperature 𝑇
• To find use conservation of energy
• Molar concentration 𝑖 =
𝑁𝑖
𝑉
𝑄
(book calls this 𝑋𝑖 )
• use species generation/consumption rates from chemical kinetics
• 𝑉=
𝑚
, 𝑛𝑒𝑒𝑑
𝜌
𝜌
• state, mixture
• Highly coupled
• Assume we know “production rates” per unit volume
•
1 𝑑𝑁𝑖
𝑉 𝑑𝑡
= 𝜔𝑖 = 𝑘(𝑇)
𝑀
𝑖=1
𝑖
𝑛𝑖
= 𝑓𝑛 1 , 2 , … 𝑀
• Rate depends on current molar concentration (per volume) of each constituent, and temperature
• From chemical Kinetics
First Law (Energy
Conservation)
• 𝑄−𝑊 =
=𝑚
𝑑𝑢
𝑑𝑡
𝑑𝑉
• Only boundary work: 𝑊 = 𝑃 𝑑𝑡
𝑑𝑉
𝑑𝑈
=
𝑑𝑡
𝑑𝑡
𝑑𝑈
𝑑𝑉
+𝑃
𝑑𝑡
𝑑𝑡
• 𝑄−𝑃
• 𝑄=
𝑊
𝑑𝑈
𝑑𝑡
=
𝑑 𝑈+𝑃𝑉
𝑑𝑡
=
𝑑𝐻
𝑑𝑡
• Where enthalpy 𝐻 = 𝑈 + 𝑃𝑉, For a mixture H =
• 𝑄=
𝑑𝐻
𝑑𝑡
=
𝑑𝑁𝑖
ℎ
𝑑𝑡 𝑖
𝑁𝑖 ℎ𝑖 𝑇
𝑑 ℎ𝑖
𝑑𝑡
+ 𝑁𝑖
1 𝑑𝑁𝑖
;
𝑑𝑡
𝑑ℎ
; 𝑑𝑡𝑖 = 𝑐𝑝,𝑖
• Production rate 𝜔𝑖 = 𝑉
•
• 𝑄=
•
𝑄
•
𝑄
𝑉
=
𝑑𝑁𝑖
𝑑𝑡
= 𝑉 𝜔𝑖 ; 𝑁𝑖 = V 𝑖
𝑉𝜔𝑖 ℎ𝑖 + V 𝑖 𝑐𝑝,𝑖
𝜔𝑖 ℎ𝑖 + 𝑖 𝑐𝑝,𝑖
𝑑𝑇
𝑑𝑡
𝑑𝑇
𝑑𝑡
=
𝑇
𝑑𝑇
𝑑𝑡
; Divide by 𝑉
𝜔𝑖 ℎ𝑖 +
𝑑𝑇
𝑑𝑡
𝑖 𝑐𝑝,𝑖 ; Solve for
𝑑𝑇
𝑑𝑡
𝑄
−
𝑉
𝜔 𝑖 ℎ𝑖 𝑇
𝑑𝑇
=
𝑑𝑡
𝑖 𝑐𝑝,𝑖 𝑇
• First order differential equation, Initial conditions (IC): 𝑇 = 𝑇0 𝑎𝑡 𝑡 = 0
• At each time step, to find the change in 𝑇
• Need 𝑄, 𝑉, 𝜔𝑖 (which depends on 𝑖 and T), 𝑖 ,𝑇, ℎ𝑖 𝑇 , and 𝑐𝑝,𝑖 (𝑇)
Change in Molar Concentrations
•
𝑑𝑖
𝑑𝑡
=
𝑑
𝑁𝑖
𝑉
=
𝑑𝑡
1 𝑑𝑁𝑖
𝑉 𝑑𝑡
−
𝑁𝑖 𝑑𝑉
𝑉 2 𝑑𝑡
= 𝜔𝑖 −
𝑖 𝑑𝑉
𝑉 𝑑𝑡
*
• Species production and volume change affect molar concentration
• Find the volume V from ideal gas equation of state
• 𝑃𝑉 = 𝑁𝑅𝑢 𝑇 =
𝑁𝑖 𝑅𝑢 𝑇
• Take time derivative
• 𝑃
𝑑𝑉
𝑑𝑡
= 𝑅𝑢
𝑑 𝑇 𝑁𝑖
𝑑𝑡
𝑑
to see how volume changes with time
𝑑𝑡
= 𝑅𝑢 𝑇
• Divide both sides by 𝑃𝑉 =
•
1 𝑑𝑉
𝑉 𝑑𝑡
=
1
𝑁𝑖
𝑑𝑁𝑖
𝑑𝑡
1 𝑑𝑇
+
𝑇 𝑑𝑡
=
𝑑
𝑁𝑖
𝑑𝑡
+
𝑑𝑇
𝑑𝑡
𝑁𝑖 = 𝑅𝑢 𝑇
𝑑𝑖
𝑑𝑡
= 𝜔𝑖 − 𝑖
𝜔𝑖
𝑖
+
+
𝑁𝑖 𝑅𝑢 𝑇
1
𝑁𝑖
𝑉
1 𝑑𝑁𝑖
𝑉 𝑑𝑡
1 𝑑𝑇
+
𝑇 𝑑𝑡
=
𝜔𝑖
𝑖
• Plug into *
•
𝑑𝑁𝑖
𝑑𝑡
1 𝑑𝑇
𝑇 𝑑𝑡
• Initial Conditions: at t = 0, 𝑖 = 𝑖 0 , 𝑖 = 1, 2, … 𝑀
1 𝑑𝑇
+
𝑇 𝑑𝑡
𝑑𝑇
𝑑𝑡
𝑁𝑖
𝑀 + 1 coupled System of 1st order differential equations
• Initial Conditions, at t = 0
• 𝑖 = 𝑖 0 , 𝑖 = 1, 2, … 𝑀, and
• 𝑇 = 𝑇0
• Assume we also know 𝜔𝑖 = 𝐴𝑒𝑥𝑝(−
𝐸𝐴
)
𝑅𝑢 𝑇
𝑀
𝑖=1
𝑖
𝑛𝑖
• Use the first order differentials to find 𝑖 and 𝑇 at time t + Δ𝑡
•
•
𝑑𝑖
𝑑𝑡
𝑑𝑇
𝑑𝑡
= 𝜔𝑖 − 𝑖
=
𝑄
−
𝑉
𝜔𝑖 ℎ𝑖 𝑇
𝑖 𝑐𝑝,𝑖 𝑇
𝜔𝑖
𝑖
≈
+
1 𝑑𝑇
𝑇 𝑑𝑡
≈
Δ𝑖
Δ𝑡
;
𝑖
Δ𝑇
;
Δ𝑡
𝑡+Δ𝑡
= 𝑖 𝑡+
𝑑𝑖
𝑑𝑡
𝑑𝑇
𝑑𝑡
Δ𝑡
𝑇𝑡+Δ𝑡 = 𝑇𝑡 +
Δ𝑡
• System Volume
• 𝑉 𝑡 =
t
T
𝑚
𝜌 𝑇
=
 
𝑚
;
𝑖 𝑀𝑊𝑖
𝜌=
𝑚
𝑉
=
… [M] w1 w2
0 T0 0 0 … [M]0
Dt
2Dt
𝑚𝑖
𝑉
=
…
𝑁𝑖 𝑀𝑊𝑖
𝑉
wM
V
=
𝑖 𝑀𝑊𝑖
Q d/dt d/dt
…
d[M]/dt dT/dt
Constant-Volume Fixed-Mass Reactor
• Constant V and m
• Find T, P, 𝑖 (𝑓𝑜𝑟 𝑖 = 1,2, … , 𝑀) versus time 𝑡
• 1st Law
• 𝑄−𝑊 =
• 𝑈=
𝑑𝑁𝑖
𝑢𝑖
𝑑𝑡
• 𝜔𝑖 =
• 𝑁𝑖 =
• 𝑄=
•
•
=
𝑑𝑇
𝑑𝑡
=
=
𝑑
𝑁𝑖 𝑢𝑖 𝑇 ;
• 𝑄=
𝑄
𝑉
𝑑𝑈
𝑑𝑡
𝑁𝑖 𝑢𝑖 𝑇
𝑑𝑡
𝑑 𝑢𝑖
+ 𝑁𝑖
𝑑𝑡
1 𝑑𝑁𝑖 𝑑𝑁𝑖
;
= 𝑉 𝜔𝑖
𝑉 𝑑𝑡 𝑑𝑡
𝑑𝑢
𝑑𝑇
V 𝑖 , 𝑖 = 𝑐𝑣,𝑖 𝑇
𝑑𝑡
𝑑𝑡
𝑢𝑖 𝑉 𝜔𝑖 + V 𝑖 𝑐𝑣,𝑖
𝑢𝑖 𝜔𝑖 + 𝑖
𝑄
−
𝑉
𝑢 𝑖 𝜔𝑖
𝑖 𝑐𝑣,𝑖
𝑑𝑇
𝑐𝑣,𝑖
𝑑𝑡
𝑑𝑇
𝑑𝑡
=
, divide by V
𝑢𝑖 𝜔𝑖 +
𝑑𝑇
𝑑𝑡
𝑖 𝑐𝑣,𝑖 , solve for
𝑑𝑇
𝑑𝑡
Tabulated Data
• Need to evaluate
𝑑𝑇
𝑑𝑡
=
𝑄
−
𝑉
• However, tables only have
𝑢𝑖 𝜔𝑖
𝑖 𝑐𝑣,𝑖
(true but not useful)
• 𝑐𝑝,𝑖 (= 𝑐𝑣,𝑖 + 𝑅𝑢 ), so use
• 𝑐𝑣,𝑖 = 𝑐𝑝,𝑖 − 𝑅𝑢
• ℎ𝑖 (= 𝑢𝑖 + 𝑃𝑣 = 𝑢𝑖 + 𝑅𝑢 𝑇) , so use
• 𝑢𝑖 = ℎ𝑖 − 𝑅𝑢 𝑇
•
𝑑𝑇
𝑑𝑡
=
𝑄
−
𝑉
ℎ𝑖 𝜔𝑖 +𝑅𝑢 𝑇
𝑖 𝑐𝑝,𝑖 −𝑅𝑢
𝜔𝑖
(true and useful)
• Initial Condition: 𝑇 = 𝑇0 at 𝑡 = 0
• Species Production
•
𝑑𝑖
𝑑𝑡
𝑁
=
𝑑 𝑉𝑖
𝑑𝑡
=
1 𝑑𝑁𝑖
𝑉 𝑑𝑡
= 𝜔𝑖 = 𝑘(𝑇)
𝑀
𝑖=1
𝑖
𝑛𝑖
Reactor Pressure
• Ideal Gas Law
• 𝑃𝑉 =
𝑁𝑖 𝑅𝑢 𝑇
• Divide by 𝑉 (constant)
• 𝑃=
𝑁𝑖
𝑉
𝑅𝑢 𝑇 =
𝑖 𝑅𝑢 𝑇
• Pressure Rate of change (affects detonation)
•
𝑑𝑃
𝑑𝑡
= 𝑅𝑢 𝑇
𝑑
𝑖
𝑑𝑡
+
𝑖
𝑑𝑇
𝑑𝑡
= 𝑅𝑢 𝑇
𝑑
𝑁𝑖
𝑉
𝑑𝑡
+
𝑑𝑇
𝑑𝑡
𝑖
= 𝑅𝑢 𝑇
𝜔𝑖 +
𝑑𝑇
𝑑𝑡
𝑖
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
• 𝐶𝑥 𝐻𝑦 + 𝑥 +
𝑦
4
𝑂2
𝑘𝐺
𝑦
2
𝑥𝐶𝑂2 + 𝐻2 𝑂
• stoichiometric mixture with 𝑂2 , not air
•
𝑑 𝐶𝑥 𝐻𝑦
𝑑𝑡
= −𝐴𝑒𝑥𝑝
𝐸𝑎 𝑅𝑢
𝑇
𝐶𝑥 𝐻𝑦
𝑚
𝑂2
𝑛
=
𝑔𝑚𝑜𝑙𝑒
𝑐𝑚3 𝑠
• Page 157, Table 5.1: 𝐴, 𝐸𝑎 𝑅𝑢 , 𝑚 𝑎𝑛𝑑 𝑛 for different fuels
• These values are based on flame speed data fit (Ch 8)
• In Table 5.1 units for 𝐴 =
• However, we often want 𝐴
•
1 𝑔𝑚𝑜𝑙𝑒 1−𝑚−𝑛
𝑠
𝑐𝑚3
• 𝐴
1 𝑘𝑚𝑜𝑙𝑒 1−𝑚−𝑛
𝑠
𝑚3
𝑔𝑚𝑜𝑙𝑒 1−𝑚−𝑛
−
𝑚+𝑛
𝑔𝑚𝑜𝑙𝑒
𝑔𝑚𝑜𝑙𝑒
𝑐𝑚3
=
𝑐𝑚3 𝑠
𝑐𝑚3
𝑠
1−𝑚−𝑛
1 𝑘𝑚𝑜𝑙𝑒
in units of
𝑠
𝑚3
𝑘𝑚𝑜𝑙𝑒
1000 𝑔𝑚𝑜𝑙𝑒
=𝐴
1−𝑚−𝑛
100 𝑐𝑚 3
𝑚
1 𝑔𝑚𝑜𝑙𝑒 1−𝑚−𝑛
𝑠
𝑐𝑚3
=
1 𝑔𝑚𝑜𝑙𝑒 1−𝑚−𝑛
𝑠
𝑐𝑚3
Sometimes Want These Units
1000
1−𝑚−𝑛
=
1 𝑘𝑚𝑜𝑙𝑒 1−𝑚−𝑛
𝑠
𝑚3
10001−𝑚−𝑛 = 𝐴 𝑇𝑒𝑥𝑡𝑏𝑜𝑜𝑘 10001−𝑚−𝑛
Given in Table 5.1, p. 157
•
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