Transcript Slides

ME 475/675 Introduction to
Combustion
Lecture 22
Example 6.1 solution, Constant volume and mass reactor
Announcements
• HW 7 Due Now
• Course Projects (instructions soon)
• HW 8, Numerical Solution to Example 6.1
• Due Monday, Oct. 19, 2015
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 Schlieren (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*107 J/kg
• The stoichiometric air-fuel ratio is 16.0 and restrict combustion to stoichiometric or
lean conditions, Φ ≤ 1.
• Constant Pressure, or constant Volume?
Constant-Volume Fixed-Mass Reactor
• Constant V and m
• Find T, P, 𝑖 (𝑓𝑜𝑟 𝑖 = 1,2, … , 𝑀) versus time 𝑡
• 1st Law
•
𝑑𝑇
𝑑𝑡
=
𝑄
−
𝑉
ℎ𝑗 𝜔𝑗 +𝑅𝑢 𝑇
𝜔𝑗
𝑗 𝑐𝑝,𝑗 −𝑅𝑢
(true and useful)
• Species Production
•
𝑑𝑖
𝑑𝑡
𝑁
=
𝑑 𝑉𝑖
𝑑𝑡
=
1 𝑑𝑁𝑖
𝑉 𝑑𝑡
= 𝜔𝑖 = 𝑘(𝑇)
𝑀
𝑗=1
𝑗
𝑛𝑗
• Initial Conditions:
• At t = 0, 𝑇 = 𝑇0 , and 𝑖 = 𝑖 0 , 𝑖 = 1, 2, … 𝑀
• State Equation
• 𝑃=
𝑁𝑗
𝑉
𝑅𝑢 𝑇 =
𝑗 𝑅𝑢 𝑇
• Pressure Rate of change (affects detonation)
•
𝑑𝑃
𝑑𝑡
= 𝑅𝑢 𝑇
𝜔𝑗 +
𝑑𝑇
𝑑𝑡
𝑗
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
Numerical values
• Species Production
• Assuming 𝑂2 = 0.21 𝐴𝑖𝑟 = 0.21 𝑂𝑥 (only true initially since 𝑁2 is not consumed)
• 𝜔𝐹𝑢𝑒𝑙 = −6.186 ∗ 109 𝑒𝑥𝑝
−15098𝐾
𝑇
𝐹
0.1
0.21 𝑂𝑥
1.65
• 𝜔𝑂𝑥 = 𝐴 𝐹 16𝜔𝐹𝑢𝑒𝑙 = 16𝜔𝐹𝑢𝑒𝑙
• 𝜔𝑃𝑟𝑜𝑑 = − 𝐴 𝐹 + 1 𝜔𝐹𝑢𝑒𝑙 = −17𝜔𝐹𝑢𝑒𝑙
• 1st Law
•
𝑑𝑇
𝑑𝑡
=
𝑄
−
𝑉
ℎ𝑗 𝜔𝑗 +𝑅𝑢 𝑇
𝑗 𝑐𝑝,𝑗 −𝑅𝑢
𝜔𝑗
=
ℎ𝐹𝑢𝑒𝑙 𝜔𝐹𝑢𝑒𝑙
𝑃
𝑐𝑝 −𝑅𝑢 𝑅 𝑇
𝑢
=
𝑘𝑃𝑚3 𝜔𝐹𝑢𝑒𝑙 𝑇
−364,184
𝑘𝑚𝑜𝑙
𝑃
• Pressure Rate of change (affects detonation)
•
𝑑𝑃
𝑑𝑡
=
𝑃0 𝑑𝑇
𝑇0 𝑑𝑡
• Initial Conditions at t = 0
• 𝑇0 =
• 𝑃0 =
𝑉𝐵𝐷𝐶 𝛾−1
𝑇𝐵𝐷𝐶
= 300𝐾 10 0.4 = 754 𝐾
𝑉𝑇𝐷𝐶
𝑉𝐵𝐷𝐶 𝛾
𝑃𝐵𝐷𝐶
= 101.3𝑘𝑃 10 1.4 = 2545 𝑘𝑃
𝑉𝑇𝐷𝐶
• 𝑃𝑟
• 𝑂𝑥
0
0
=0
=
1
𝑃
𝜙
+1 𝑅𝑢 𝑇
𝐴 𝐹
• 𝐹𝑢
0
=
1
𝑃
𝐴 𝐹
1+ 𝜙 𝑅𝑢 𝑇
•
Results, phi = 1, dt = 10-7 sec
1
Temperature
0.1
0.01
Fuel
0.001
Oxidizer
dP/dt
3500
1E+10
3000
1E+09
2500
100000000
2000
10000000
1500
1000000
100000
1000
Products
10000
500
1000
0
0.0001
0
0.001
dt
Phi
1.00E-07
t
1
0.002
Fuel
0 0.023890721
0.0000001 0.023890707
0.0000002 0.023890694
0.003
0
0.001
Oxidizer Products Temperature
0.382252
0
753.5659295
0.382251 2.25E-07
753.5673536
0.382251 4.49E-07
753.5687778
0.002
0
0.003
Press
d[F]/dt
2544.541 -0.1320642
2544.546 -0.132069
2544.551 -0.1320739
d[Ox]/dt
-2.1130267
-2.1131046
-2.1131826
• Starting point
• http://wolfweb.unr.edu/homepage/greiner/teaching/MECH.475.675.Combustion/Ex6.1.start.xlsx
0.001
d[Pr]/dt
2.245091
2.245174
2.245257
0.002
dT/dt
14241.26
14241.79
14242.31
0.003
dP/dt
48087.99
48089.77
48091.54