In this paper, the combustion phase optimization is analyzed and ignition timing of SI engines is controlled to maximize the fuel efficiency. Particularly, the optimum set point of the combustion phase and moving average based control strategy are discussed to extract information about the fuel efficiency from the combustion.
Trang 164 Hoang Thang, Nguyen Thi Hai Van, Tran Quoc An
OPTIMIZATION OF IGNITION ADVANCE ANGLE AND AIR FUEL RATIO
OF COMBUSTION ENGINES
TỐI ƯU HÓA GÓC ĐÁNH LỬA VÀ TỈ LỆ NHIÊN LIỆU TRONG
ĐỘNG CƠ ĐỐT TRONG
Hoang Thang, Nguyen Thi Hai Van, Tran Quoc An
College of Technology - The University of Danang; waveletthang@gmail.com
Abstract - In this paper, the combustion phase optimization is
analyzed and ignition timing of SI engines is controlled to maximize the
fuel efficiency Particularly, the optimum set point of the combustion
phase and moving average based control strategy are discussed to
extract information about the fuel efficiency from the combustion The
purpose of this study is to optimize the ignition advance angle and air
fuel ratio of the motorcycle engine to get maximum brake torque We
use Matlab to solve the problem of the ignition angle based on the
model of combustion engine The optimization algorithm used for the
two input variables is Air fuel ratio and ignition advance angle For the
internal combustion engine model built on Matlab software, it is
possible to change the engine parameter structure
Tóm tắt - Trong bài báo này, tối ưu hóa giai đoạn đốt cháy được
phân tích và thời gian đánh lửa của động cơ SI được kiểm soát để tối đa hóa hiệu quả nhiên liệu, trong đó thảo luận về điểm đặt pha tối
ưu của giai đoạn đốt và chiến lược kiểm soát trung bình dịch chuyển
để thu được thông tin về hiệu suất nhiên liệu từ quá trình cháy Mục đích của nghiên cứu này là tối ưu hóa góc đánh lửa và tỷ lệ nhiên liệu không khí của động cơ xe máy để có được mômen xoắn cực đại Chúng tôi sử dụng Matlab để giải quyết vấn đề về góc đánh lửa dựa trên mô hình động cơ đốt trong Thuật toán tối ưu hóa được sử dụng cho hai biến đầu vào là tỷ lệ nhiên liệu không khí và góc đánh lửa Đối với mô hình động cơ đốt trong được xây dựng trên phần mềm Matlab, cấu trúc tham số động cơ có thể thay đổi linh hoạt
Key words - góc đánh lửa; tối ưu hóa; động cơ đốt trong; hiệu quả
nhiên liệu; mômen xoắn cực đại
Từ khóa - ignition advance angle; optimization; combustion
engine; fuel efficiency; maximum brake torque
1 Introduction
As we have known, the target developments for the
future vehicle are how to reduce fuel consumption,
pollutant emission while maintaining high level of the
engine performance Much research on engine has attracted
great attention from scientists for a long period of time
However, it is too difficult and urgent to decrease
consumption, pollutant emission and increase the engine
performance simultaneously For this reason, our team
would like to choose this topic
The project has to find the spark advance angle and air
fuel ratio optimization values at every operation engine
speed (1000 to 8000 rpm) and full throttle condition
And we have some constraint conditions:
✓ Ignition timing is not higher than 35 degree crank
angle, because of knocking and not lower than 5 degree
crank angle because the maximum pressure will be too late
✓ Air fuel ratio is not higher 1than 8 because of miss
fire and not lower than 10 because at this point, torque
decreases and unburned fuel increases very quickly
✓ The max pressure in the combustion chamber has to
be less than 50 (Bar)
✓ The position of max pressure after Top Dead Center
is >= 150 (CA)
2 Approach and Methods
Five steps to define OPT problem [1]:
Step 1
• Maximum brake torque
• Ignition timing ≤ 50o crank shaft degree (Before TDC)
• Ignition timing ≥ 10o crank shaft degree (Before TDC)
• Air fuel ratio A/F ≤ 18 and A/F ≥ 10
• The max pressure <=50 bar
• The max pressure position >=15o
Step 2
𝐹 = (𝐹(𝑥0) −𝜕𝐹
𝜕𝑥1
𝑥1 − 𝜕𝐹
𝜕𝑥2
𝑥2) +𝜕𝐹
𝜕𝑥1
𝑥1− 𝜕𝐹
𝜕𝑥2
𝑥2
Where: x1: ignition advance angle;
x2: Air fuel ratio
𝜕𝐹
𝜕𝑥1
=𝐹(𝑥1+ ∆𝑥1, 𝑥2) − 𝐹(𝑥1− ∆𝑥1, 𝑥2)
2∆𝑥1
𝜕𝐹
𝜕𝑥2=
𝐹(𝑥1, 𝑥2+ ∆𝑥12) − 𝐹(𝑥1, 𝑥2−∆𝑥12)
2∆𝑥2
Step 3
Design variable: x1, x2
Step 4
Objective function:
Max 𝐹 = (𝐹(𝑥0) − 𝜕𝐹
𝜕𝑥1𝑥1 − 𝜕𝐹
𝜕𝑥2𝑥2) + 𝜕𝐹
𝜕𝑥1𝑥1− 𝜕𝐹
𝜕𝑥2𝑥2
Step 5
Constraints:
• g1(x1) = x1 – 50 ≤ 0
• g2(x1) = x1 – 10 ≥ 0 – x1 + 10 ≤ 0
• g3(x2) = x2 – 18 ≤ 0
• g4(x2) = x2 – 10 ≥ 0 – x2 + 10 ≤ 0
• g5(x1, x2) = (𝑃(𝑥0) −𝜕𝑃
𝜕𝑥1𝑥1−
𝜕𝑃
𝜕𝑥2𝑥1) +
𝜕𝑃
𝜕𝑥1𝑥1−
𝜕𝑃
𝜕𝑥2𝑥2≤ 0
• g6(x1, x2) = (𝐴(𝑥0) −𝜕𝐴
𝜕𝑥1𝑥1−
𝜕𝐴
𝜕𝑥2𝑥1) +
𝜕𝐴
𝜕𝑥1𝑥1−
𝜕𝐴
𝜕𝑥2𝑥2≤ 0
Trang 2ISSN 1859-1531 - THE UNIVERSITY OF DANANG, JOURNAL OF SCIENCE AND TECHNOLOGY, NO 11(120).2017, VOL 4 65 where:
𝜕𝐹
𝜕𝑥1=
𝐹(𝑥1+ ∆𝑥1, 𝑥2) − 𝐹(𝑥1− ∆𝑥1, 𝑥2)
2∆𝑥1
𝜕𝐹
𝜕𝑥12
=𝐹(𝑥1, 𝑥2+ ∆𝑥12) − 𝐹(𝑥1, 𝑥2−∆𝑥12)
2∆𝑥2
𝜕𝑃
𝜕𝑥1=
𝑃(𝑥1+ ∆𝑥1, 𝑥2) − 𝑃(𝑥1− ∆𝑥1, 𝑥2)
2∆𝑥1
𝜕𝑃
𝜕𝑥12 =
𝑃(𝑥1, 𝑥2+ ∆𝑥12) − 𝑃(𝑥1, 𝑥2−∆𝑥12)
2∆𝑥2
𝜕𝐴
𝜕𝑥1=
𝐴(𝑥1+ ∆𝑥1, 𝑥2) − 𝐴(𝑥1− ∆𝑥1, 𝑥2)
2∆𝑥1
𝜕𝐴
𝜕𝑥12 =
𝐴(𝑥1, 𝑥2+ ∆𝑥12) − 𝐴(𝑥1, 𝑥2−∆𝑥12)
2∆𝑥2
The problem will become:
Design variable: x1, x2
Object to: Min:
−𝐹 = − ((𝐹(𝑥0) −𝜕𝐹
𝜕𝑥1𝑥1−
𝜕𝐹
𝜕𝑥2𝑥2) +
𝜕𝐹
𝜕𝑥1𝑥1−
𝜕𝐹
𝜕𝑥2𝑥2)
Subject to:
• g1(x1) = x1 – 50 ≤ 0
• g2(x1) = – x1 + 10 ≤ 0
• g3(x2) = x2 – 18 ≤ 0
• g4(x2) = – x2 + 10 ≤ 0
• g5(x1, x2) =
(𝑃(𝑥0) −𝜕𝑃
𝜕𝑥1𝑥1 −
𝜕𝑃
𝜕𝑥2𝑥2) +
𝜕𝑃
𝜕𝑥1𝑥1−
𝜕𝑃
𝜕𝑥2𝑥2≤ 50
• g6(x1, x2) =
(𝐴(𝑥0) −𝜕𝐴
𝜕𝑥1𝑥1 −
𝜕𝐴
𝜕𝑥2𝑥2) +
𝜕𝐴
𝜕𝑥1𝑥1−
𝜕𝐴
𝜕𝑥2𝑥2≤ 15
3 Algorithm/ Programming
Process:
This process is used to calculate the optimum value of
this project
Figure 1 Process to solve OPT problem
Example: Motorcycle optimized Building motorcycle engine model
Table 1 Engine specification
Vd = 125e-6; Engine displacement volume [m^3]
b = 0.0515 Bore [m]
s = 0.06 Stroke [m]
l = 0.1 Connecting rod [m]
Vm = 127e-6 Intake manifold volume [m^3]
CR = 10.0 Compression ratio Cd=0.85 Discharge coefficient of throttle D=0.024 Throttle bore diameter (m) d=0.0069 Throttle shaft diameter (m)
0= 0.926 (deg) Zero flow angle
Figure 2 is motorcycle model [2]; data input is throttle position (%), engine speed (rpm), air fuel ratio and ignition timing (crank angle) Data output is fuel consumption (g/kWh), torque (Nm) and the pressure curve (bar)
Figure 2 Motorcycle engine mode
Calculating constraints:
Constraints for two variables are calculated from the model by fix throttle position and engine speed, and change spark advance angle and air fuel ratio (A/F) [3], [4] Because the value and position of peak pressure curve are dependents as first order with ignition time and A/F, so at each engine speed we find one constraint equation for value and position of the pressure curve Figure 3 shows the limit of position and peak pressure curve
Figure 3 The limitation of pressure curve
For example at 4000 rpm and full throttle, we have:
25
0
1
Trang 366 Hoang Thang, Nguyen Thi Hai Van, Tran Quoc An
x1 = 25
6 13
0
2
x2 = 1
Table 2 Engine specification
Engine speed = 4000rpm
Full throttle
Max pressure (bar)
Peak pressure after TDC (0CA)
2
1,
x
g x g(25, 13.6) 59 197
1
0
2
x
g x x g(50, 13.6) 100.6 182
1
0
2
x
g x x g(0, 13.6) 27.55 220
0
0
2
1,
x
g x x g(25, 14.6) 56.55 197
0
0
2
1,
x
g x x g(25, 12.6) 62.9 197
Apply Calculation Tool to find constraint equation in
Figure 4
Figure 4 Calculation tool
We get the equation from calculation tool and constraint for maximum pressure curve [5]:
x1, x2 65 66 1 461 x1 3 175 x2 50
g
or
x1, x2 15 66 1 461 x1 3 175 x2 0
In this constraint, maximum pressure curve is normally lower than 50 bar for motorcycle engine
Constraint for peak pressure after TDC
x1, x2 216 0 76 x1 180 15
g
or
x1 0 76 x1 21 0
The position of peak pressure after TDCis always higher than 15 0 CA
Use equation (1) and (2) to calculate for every optimum point (OPT) at 4000 rpm and full throttle
To find OPT point, first, we start at 100 CA of ignition timing and 13.6 A/F
x1 = 10 Crank shaft degree x1 = 5
Because the engine runs at full throttle, the engine needs maximum torque and does not care about fuel consumption, the cost function at full throttle is maximum torque
Apply the same method with another engine speed,we have the results as below:
Table 3 Results with another engine speed
Engine
speed x1 x2
F(x1,x2) F(x1+∆x1,x2
F(x1-∆x1,X2) F(X1,x2+∆x2
F(X1,X2-∆x2) x1* X2* f F(new) %<0.01
1000 10 13.6 9.7 9.911 9.276 9.604 9.722 14 14 10 9.9 0.0219 14.26 14.1 9.8 1.912 9.738 9.829 9.863 15 13 10 9.9 0.0003
2000 10 13.6 10 10.44 9.64 10.04 10.17 15 13 10 11 0.0459
15 13.6 11 10.59 10.38 10.48 10.52 15 13 11 11 0.0001
3000 10 13.6 10 10.39 9.463 9.907 10.06 15 13 10 11 0.0541
15 13.1 10 10.58 10.31 10.43 10.48 16 13 10 10 0.0012
4000
10 13.6 9.6 10.1 9.071 9.533 9.715 15 13 10 10 0.0618
15 13.1 10 10.33 10 10.14 10.2 17 13 10 10 0.0192
17 12.9 10 10.33 10.2 10.33 10.38 17 13 10 11 0.002
5000
10 13.6 9.1 9.645 8.537 9.049 9.219 15 13 9 9.8 0.07
15 13.1 9.7 10.13 9.219 9.654 9.806 20 13 10 10 0.0486
20 12.99 10 10.26 9.996 10.12 10.17 20 13 10 10 0.0006
6000 20 13.1 9.6 9.941 9.14 9.519 9.675 25 14 10 9.9 0.0351
25 13.51 9.9 10 9.756 9.856 9.9 25 14 10 9.9 0.0007
7000 25 13.6 9.3 9.561 8.853 9.203 9.323 26 13 9 9.4 0.0165 26.32 13.1 9.4 9.535 9.277 9.382 9.428 26 13 9 9.4 0.0024
8000
25 13.6 8.5 8.872 8.084 8.474 8.611 30 13 9 9 0.0542
30 13.1 8.9 9.084 8.611 8.872 9.004 32 13 9 9.1 0.0167 31.7 12.6 9 9.096 8.987 9.024 9.061 32 12 9 9.1 0.0016
Trang 4ISSN 1859-1531 - THE UNIVERSITY OF DANANG, JOURNAL OF SCIENCE AND TECHNOLOGY, NO 11(120).2017, VOL 4 67
4 Result and discussion
From the result we can see that, there are differences
between the characteristics of the engine using a carburetor
and OPT Because with the engine using a carburetor, the
ignition time and A/F ratio cannot be controlled, ignition
time curve is first order curve, and A/F ratio is
approximately 13.5 So the torques curves are always lower
than OPT curve
At 3000 rpm the OPT point is coincident with
carburetor It means the engine using carburetor is adjusted
to optimize at 3000 rpm which is peak torque curve
From the torque curve, it is clear that, by using OPT
solution, we can find the best torque curve, and with this
torque curve, engine accelerator time will be shortened
than with a carburetor
Figure 5 Compare Ignition time
Figure 6 Compare air fuel ratio
Figure 7 Compare torque curve
5 Conclusion
From the result we see that:
✓ The OPT point always lies on the boundary At each OPT step, at least one constraint is active
✓ The degree of accuracy of OPT point depends on how we choose initial data at each step and step size Step size is smaller and data is more accurate
✓ The number of steps depends on knowledge of author
✓ With multivariable function problem, using OPT solution will decrease the experiment time and decrease the charge
REFERENCES
[1] Jasbir S Arora, "Introduction to Optimum Design", The University
of Iowa, Technology & Engineering, 2004
[2] B Hnatiuc, M Hnatiuc, C Petrescu and D Astanei, "Ignition modelling of a double sparking plug for internal combustion
engines", 2014 International Conference and Exposition on
Electrical and Power Engineering (EPE), Iasi, 2014, pp 226-230
[3] J Gao, Y Wu and T Shen, "Real-time optimization and control of
combustion phase of SI engines using statistical analysis”, 2016 35th
Chinese Control Conference (CCC), Chengdu, 2016, pp
8986-8992
[4] H Cui, "The Fuel Control System and Performance Optimization of
a Spark-Ignition LPG Engine”, 2009 International Conference on
Measuring Technology and Mechatronics Automation, Zhangjiajie,
Hunan, 2009, pp 901-904 [5] W T Reiersen, J A Schmidt and D B Montgomery, "Physics
optimization of the Compact Ignition Tokamak (CIT)”, IEEE
Thirteenth Symposium on Fusion Engineering, Knoxville, TN, 1989,
pp 1222-1225 vol.2
(The Board of Editors received the paper on 06/09/2017, its review was completed on 19/10/2017)
0
5
10
15
20
25
30
35
Engine speed (rpm)
Ignition time OPT Ignition time carburetor
12
12.5
13
13.5
14
14.5
0 1000 2000 3000 4000 5000 6000 7000 8000 9000
Engine speed (rpm)
A/F OPT A/F carburetor
8.8 9 9.2 9.4 9.6 9.8 10 10.2 10.4 10.6
0 1000 2000 3000 4000 5000 6000 7000 8000 9000
Engine speed (rpm )