ME 475/675 Introduction to
Combustion
Lecture 21
Coupled chemical/thermal analysis, Constant pressure, Constant
Volume, Start Ex 6.1
Announcements
• HW 8, Numerical Solution to Example 6.1
• Due Monday, Oct. 19, 2015
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 speed 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
𝑗=1
𝑗
𝑛𝑗
= 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, and for a mixture H =
• 𝑄=
𝑑𝐻
𝑑𝑡
=
𝑑𝑁𝑗
𝑑𝑡
ℎ𝑗 + 𝑁𝑗
𝑁𝑗 ℎ𝑗 𝑇
𝑑 ℎ𝑗
𝑑𝑡
1 𝑑𝑁𝑗
;
𝑑𝑡
𝑑ℎ𝑗
• Production rate 𝜔𝑗 = 𝑉
•
𝑑𝑁𝑗
𝑑𝑡
• 𝑄=
•
𝑄
𝑄
𝑉
=
= 𝑉𝜔𝑗 ; 𝑁𝑗 = V 𝑗 ;
𝑑𝑡
𝑉𝜔𝑗 ℎ𝑗 + V 𝑖 𝑐𝑝,𝑗
𝑑𝑇
𝑑𝑡
𝜔𝑗 ℎ𝑗 + 𝑗 𝑐𝑝,𝑗
𝑑𝑇
𝑑𝑡
=
= 𝑐𝑝,𝑗 𝑇
𝑑𝑇
𝑑𝑡
; Divide by 𝑉
𝜔𝑗 ℎ𝑗 +
𝑑𝑇
𝑑𝑡
𝑗 𝑐𝑝,𝑗 ; Solve for
𝑑𝑇
𝑑𝑡
𝑄
•
− 𝜔𝑗 ℎ𝑗 𝑇
𝑑𝑇
𝑉
=
𝑑𝑡
𝑗 𝑐𝑝,𝑗 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, 𝜔𝑗 (which depends on 𝑗 and T), 𝑗 ,𝑇, ℎ𝑗 𝑇 , and 𝑐𝑝,𝑗 (𝑇)
Change in Molar Concentrations
•
𝑑𝑖
𝑑𝑡
𝑁
=
𝑑 𝑉𝑖
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
𝑑
• Take time derivative
•
𝑑𝑉
𝑃
𝑑𝑡
= 𝑅𝑢
𝑑 𝑇
𝑁𝑗
𝑑𝑡
to see how volume changes with time
= 𝑅𝑢 𝑇
𝑑𝑡
• Divide both sides by 𝑃𝑉 =
•
1 𝑑𝑉
𝑉 𝑑𝑡
=
1
𝑁𝑗
𝑑𝑁𝑗
𝑑𝑡
+
1 𝑑𝑇
𝑇 𝑑𝑡
=
𝑑
𝑁𝑗
𝑑𝑡
𝑑𝑇
+
𝑑𝑡
𝑁𝑗 = 𝑅𝑢 𝑇
•
= 𝜔𝑖 − 𝑖
𝜔𝑗
𝑗
+
𝑑𝑡
+
𝑁𝑗 𝑅𝑢 𝑇
1
𝑁𝑗
𝑉
1 𝑑𝑁𝑖
𝑉 𝑑𝑡
+
1 𝑑𝑇
𝑇 𝑑𝑡
=
𝜔𝑗
𝑗
• Plug into *
𝑑𝑖
𝑑𝑡
𝑑𝑁𝑗
1 𝑑𝑇
𝑇 𝑑𝑡
• Initial Conditions: at t = 0, 𝑖 = 𝑖 0 , 𝑖 = 1, 2, … 𝑀
+
1 𝑑𝑇
𝑇 𝑑𝑡
𝑑𝑇
𝑑𝑡
𝑁𝑗
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,𝑖
𝑅𝑢 𝑇
)
𝑀
𝑗=1
𝑖
𝑛𝑗
• Use the first order differentials to find 𝑖 and 𝑇 at time t + Δ𝑡
•
𝑑𝑖
𝑑𝑡
•
𝑑𝑇
𝑑𝑡
= 𝜔𝑖 − 𝑖
=
𝑄
−
𝑉
𝜔𝑗 ℎ𝑗 𝑇
𝑗 𝑐𝑝,𝑗 𝑇
𝜔𝑗
𝑗
≈
+
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
=
[1] [2]
𝑚
;
𝑗 𝑀𝑊𝑗
𝜌=
𝑚
𝑉
=
… [M] w1 w2
0 T0 [1]0 [2]0 … [M]0
Dt
2Dt
𝑚𝑗
𝑉
=
…
𝑁𝑗 𝑀𝑊𝑗
𝑉
wM
V
=
𝑗 𝑀𝑊𝑗
(algebraic not differential eqn.)
Q d[1]/dt d[2]/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=
• 𝑄=
•
•
•
•
=
𝑑𝑇
𝑑𝑡
=
=
𝑑
𝑁𝑗 𝑢𝑗 𝑇
𝑁𝑗 𝑢𝑗 𝑇 ;
𝑢𝑗
𝑑𝑁𝑗
𝑑𝑡
+ 𝑁𝑗
𝑑𝑡
𝑑 𝑢𝑗
𝑑𝑡
𝑑𝑁𝑗
1 𝑑𝑁𝑗
𝜔𝑗 =
; so
= 𝑉𝜔𝑗
𝑉 𝑑𝑡
𝑑𝑡
𝑑 𝑢𝑗
𝑑𝑇
𝑁𝑗 = V 𝑗 ,
= 𝑐𝑣,𝑗 𝑇
𝑑𝑡
𝑑𝑡
• 𝑄=
𝑄
𝑉
𝑑𝑈
𝑑𝑡
𝑢𝑗 𝑉 𝜔𝑗 + V 𝑗 𝑐𝑣,𝑗
𝑢𝑗 𝜔𝑗 + 𝑗
𝑄
−
𝑉
𝑢 𝑗 𝜔𝑗
𝑗 𝑐𝑣,𝑗
;
𝑑𝑇
𝑐𝑣,𝑗
𝑑𝑡
𝑑𝑇
𝑑𝑡
=
, divide by V
𝑢𝑗 𝜔𝑗 +
𝑑𝑇
𝑑𝑡
𝑗 𝑐𝑣,𝑗 , solve
in contrast for constant pressure
𝑑𝑇
𝑑𝑡
𝑑𝑇
for
𝑑𝑡
𝑄
− 𝜔 𝑗 ℎ𝑗 𝑇
𝑉
=
𝑗 𝑐𝑝,𝑗 𝑇
Tabulated Data
• Need to evaluate
𝑑𝑇
𝑑𝑡
=
𝑄
−
𝑉
𝑢𝑗 𝜔𝑗
𝑗 𝑐𝑣,𝑗
(true, but not useful)
• However, tables only have
• 𝑐𝑝,𝑗 (= 𝑐𝑣,𝑗 + 𝑅𝑢 ), so use
• 𝑐𝑣,𝑗 = 𝑐𝑝,𝑗 − 𝑅𝑢
• ℎ𝑗 (= 𝑢𝑗 + 𝑃𝑣 = 𝑢𝑗 + 𝑅𝑢 𝑇) , so use
• 𝑢𝑗 = ℎ𝑗 − 𝑅𝑢 𝑇
•
𝑑𝑇
𝑑𝑡
=
𝑄
−
𝑉
ℎ𝑗 𝜔𝑗 +𝑅𝑢 𝑇
𝑗 𝑐𝑝,𝑗 −𝑅𝑢
𝜔𝑗
(true and useful)
• Initial Condition: 𝑇 = 𝑇0 at 𝑡 = 0
• Species Production (no volume change)
•
𝑑𝑖
𝑑𝑡
=
𝑑
𝑁𝑖
𝑉
𝑑𝑡
=
1 𝑑𝑁𝑖
𝑉 𝑑𝑡
= 𝜔𝑖 = 𝑘(𝑇)
𝑀
𝑗=1
𝑗
𝑛𝑗
Reactor Pressure
• Ideal Gas Law
• 𝑃𝑉 =
𝑁𝑗 𝑅𝑢 𝑇
• Divide by 𝑉 (constant)
• 𝑃=
𝑁𝑗
𝑉
𝑅𝑢 𝑇 =
𝑗 𝑅𝑢 𝑇
• Pressure Rate of change (affects detonation)
•
𝑑𝑃
𝑑𝑡
= 𝑅𝑢 𝑇
𝑑
𝑗
𝑑𝑡
+
𝑗
𝑑𝑇
𝑑𝑡
= 𝑅𝑢 𝑇
𝑑
𝑁𝑗
𝑉
𝑑𝑡
+
𝑑𝑇
𝑑𝑡
𝑗
= 𝑅𝑢 𝑇
𝜔𝑗 +
𝑑𝑇
𝑑𝑡
𝑗
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/kg K
• The enthalpy of formation of the air and products are zero, and that of the fuel is
• 4*107 J/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
•