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Tiêu đề Optimization of Ignition Advance Angle and Air Fuel Ratio of Combustion Engines
Tác giả Hoang Thang, Nguyen Thi Hai Van, Tran Quoc An
Trường học College of Technology - The University of Danang
Chuyên ngành Mechanical Engineering
Thể loại Graduation project
Thành phố Da Nang
Định dạng
Số trang 4
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Nội dung

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.

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64 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

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ISSN 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 3

66 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

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ISSN 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 )

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