Transcript Slides

ME 475/675 Introduction to
Combustion
Lecture 23
Example 6.1 Excel programing, Effect of equivalence ratio
Announcements
• HW 8, Numerical Solution to Example 6.1
• Due Monday, Oct. 19, 2015
• Course Project
• Construct a homemade camp stove fueled with absolute alcohol
• See, for example:
• http://www.backpacking.net/makegear/stove/
• https://www.google.com/search?q=homemade+backpacking+stove&biw=1461&bih=708&tb
m=isch&tbo=u&source=univ&sa=X&sqi=2&ved=0CBwQsARqFQoTCN_ctrCmx8gCFck0iAodLok
G7g
• To bring 0.5 liters of room-temperature water to a boil, determine
• How much time is required
• How much fuel is consumed
• Is soot deposited on the water container?
• Work in groups of two with our teaching assistant
• Hasibul Alam [email protected]
• Due date and group assignments soon
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?
Last Lecture found Numerical values
• Species Production
• Assuming 𝑂2 = 0.21 𝐴𝑖𝑟 = 0.21 𝑂𝑥 (only true initially since 𝑁2 is not consumed)
•
•
•
𝑑 𝐹𝑢
𝑑𝑡
𝑑 𝑂𝑥
𝑑𝑡
𝑑 𝑃𝑟
𝑑𝑡
= 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)
•
𝑃
𝑇
=
𝑁𝑅𝑢
𝑉
=
𝑃0 𝑑𝑃
;
𝑇0 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
Numerical Solution (Excel)
dt [sec] phi
1.00E-07
t [sec]
[Fuel]
1
0 0.023888
1.00E-07 2.39E-02
2.00E-07 2.39E-02
[Oxidizer] [Products] T
P [kPa]
d[F]/dt
0.382206
0 753.5659 2544.541 -0.13195
3.82E-01 2.24E-07 7.54E+02 2.54E+03 -0.13196
3.82E-01 4.49E-07 7.54E+02 2.54E+03 -0.13196
• Starting point
• http://wolfweb.unr.edu/homepage/greiner/teaching/MECH.475.675.Combustion/Ex6.1.start.xlsx
• Make sure [Fu] and [Ox] are ≥ 0 (use MAX(0,value) function in Excel)
• Results, Φ = 1, Δ𝑡 = 10−7 𝑠𝑒𝑐,
• Low temperature, slow start
• Temperature and pressure spike and rapid reaction at t = 0.00307 sec = 3.07 ms
• Extinction when both fuel and oxidizer are consumed
• Realistic results, but not realistic reaction model, which should be multi-stepped
d[Ox]/dt
-2.11122
-2.11129
-2.11137
d[Pr]/dt
2.243168
2.243251
2.243333
dT/dt [K/s] dP/dT
14231.31 48054.4
14231.84 48056.17
14232.36 48057.94
Dependence on Φ, Δ𝑡 =
−7
10
𝑠𝑒𝑐
Φ = 0.5
Φ = 0.1
• As before the reaction is “delayed”
• Initial [Fuel] is decreases
• Due to excess air, the mixture takes longer to heat up, and reaches lower temperatures and pressures
• Lower temperatures delay reaction to 3.12 and 3.53 ms (3.07 ms for Φ = 1)
•