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.
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 báo này, tối ưu hóa giai đoạn đốt cháy phân tích thời gian đánh lửa động SI kiểm sốt để tối đa hóa hiệu nhiên liệu, thảo luận điểm đặt pha tối ưu giai đoạn đốt chiến lược kiểm sốt trung bình dịch chuyển để thu thơng tin hiệu suất nhiên liệu từ trình cháy Mục đích nghiên cứu tối ưu hóa góc đánh lửa tỷ lệ nhiên liệu khơng khí động xe máy để có mơmen xoắn cực đại Chúng sử dụng Matlab để giải vấn đề góc đánh lửa dựa mơ hình động đốt Thuật tốn tối ưu hóa sử dụng cho hai biến đầu vào tỷ lệ nhiên liệu khơng khí góc đánh lửa Đối với mơ hình động đốt xây dựng phần mềm Matlab, cấu trúc tham số động thay đổi linh hoạt Key words - góc đánh lửa; tối ưu hóa; động đốt trong; hiệu nhiên liệu; mơmen xoắn cực đại Từ khóa - ignition advance angle; optimization; combustion engine; fuel efficiency; maximum brake torque 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 degree crank angle because the maximum pressure will be too late ✓ Air fuel ratio is not higher 1than 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 max pressure =15o Step 𝜕𝐹 𝜕𝐹 𝜕𝐹 𝜕𝐹 𝐹 = (𝐹(𝑥 ) − 𝑥 − 𝑥 )+ 𝑥 − 𝑥 𝜕𝑥1 𝜕𝑥2 𝜕𝑥1 𝜕𝑥2 Where: x1: ignition advance angle; x2: Air fuel ratio 𝜕𝐹 𝐹(𝑥1 + ∆𝑥1 , 𝑥2 ) − 𝐹(𝑥1 − ∆𝑥1 , 𝑥2 ) = 𝜕𝑥1 2∆𝑥1 𝜕𝐹 𝐹(𝑥1 , 𝑥2 + ∆𝑥12 ) − 𝐹(𝑥1 , 𝑥2− ∆𝑥12 ) = 𝜕𝑥2 2∆𝑥2 Step Design variable: x1, x2 Step Objective function: ✓ The position of max pressure after Top Dead Center is >= 15 (CA) Approach and Methods Five steps to define OPT problem [1]: Step • • • • 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 Max 𝐹 = (𝐹(𝑥 ) − 𝜕𝐹 𝜕𝑥1 𝑥10 − 𝜕𝐹 𝜕𝑥2 𝑥20 ) + 𝜕𝐹 𝜕𝑥1 𝑥1 − 𝜕𝐹 𝜕𝑥2 Step Constraints: • g1(x1) = x1 – 50 ≤ • g2(x1) = x1 – 10 ≥ – x1 + 10 ≤ • g3(x2) = x2 – 18 ≤ • g4(x2) = x2 – 10 ≥ – x2 + 10 ≤ • g5(x1, x2) = 𝜕𝑃 𝜕𝑃 𝜕𝑃 𝜕𝑃 (𝑃(𝑥 ) − 𝑥 − 𝑥 )+ 𝑥 − 𝑥 ≤0 𝜕𝑥1 𝜕𝑥2 𝜕𝑥1 𝜕𝑥2 • g6(x1, x2) = 𝜕𝐴 𝜕𝐴 𝜕𝐴 𝜕𝐴 (𝐴(𝑥 ) − 𝑥 − 𝑥 )+ 𝑥 − 𝑥 ≤0 𝜕𝑥1 𝜕𝑥2 𝜕𝑥1 𝜕𝑥2 𝑥2 ISSN 1859-1531 - THE UNIVERSITY OF DANANG, JOURNAL OF SCIENCE AND TECHNOLOGY, NO 11(120).2017, VOL where: 𝜕𝐹 𝐹(𝑥1 + ∆𝑥1 , 𝑥2 ) − 𝐹(𝑥1 − ∆𝑥1 , 𝑥2 ) = 𝜕𝑥1 2∆𝑥1 𝜕𝐹 𝐹(𝑥1 , 𝑥2 + ∆𝑥12 ) − 𝐹(𝑥1 , 𝑥2− ∆𝑥12 ) = 𝜕𝑥12 2∆𝑥2 ) 𝜕𝑃 𝑃(𝑥1 + ∆𝑥1 , 𝑥2 − 𝑃(𝑥1 − ∆𝑥1 , 𝑥2 ) = 𝜕𝑥1 2∆𝑥1 𝜕𝑃 𝑃(𝑥1 , 𝑥2 + ∆𝑥12 ) − 𝑃(𝑥1 , 𝑥2− ∆𝑥12 ) = 𝜕𝑥12 2∆𝑥2 𝜕𝐴 𝐴(𝑥1 + ∆𝑥1 , 𝑥2 ) − 𝐴(𝑥1 − ∆𝑥1 , 𝑥2 ) = 𝜕𝑥1 2∆𝑥1 𝜕𝐴 𝐴(𝑥1 , 𝑥2 + ∆𝑥12 ) − 𝐴(𝑥1 , 𝑥2− ∆𝑥12 ) = 𝜕𝑥12 2∆𝑥2 The problem will become: Design variable: x1, x2 Object to: Min: −𝐹 = − ((𝐹(𝑥 ) − 65 Example: Motorcycle optimized Building motorcycle engine model Table Engine specification Vd = 125e-6; b = 0.0515 s = 0.06 l = 0.1 Vm = 127e-6 CR = 10.0 Cd=0.85 D=0.024 d=0.0069 0= 0.926 (deg) Engine displacement volume [m^3] Bore [m] Stroke [m] Connecting rod [m] Intake manifold volume [m^3] Compression ratio Discharge coefficient of throttle Throttle bore diameter (m) Throttle shaft diameter (m) Zero flow angle Figure 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) 𝜕𝐹 𝜕𝐹 𝜕𝐹 𝜕𝐹 𝑥 − 𝑥 )+ 𝑥 − 𝑥 ) 𝜕𝑥1 𝜕𝑥2 𝜕𝑥1 𝜕𝑥2 Subject to: • g1(x1) = x1 – 50 ≤ • g2(x1) = – x1 + 10 ≤ • g3(x2) = x2 – 18 ≤ • g4(x2) = – x2 + 10 ≤ • g5(x1, x2) = 𝜕𝑃 𝜕𝑃 𝜕𝑃 𝜕𝑃 (𝑃(𝑥 ) − 𝑥1 − 𝑥2 ) + 𝑥1 − 𝑥 ≤ 50 𝜕𝑥1 𝜕𝑥2 𝜕𝑥1 𝜕𝑥2 • g6(x1, x2) = 𝜕𝐴 𝜕𝐴 𝜕𝐴 𝜕𝐴 (𝐴(𝑥 ) − 𝑥1 − 𝑥2 ) + 𝑥1 − 𝑥 ≤ 15 𝜕𝑥1 𝜕𝑥2 𝜕𝑥1 𝜕𝑥2 Algorithm/ Programming Process: This process is used to calculate the optimum value of this project Figure 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 shows the limit of position and peak pressure curve Figure The limitation of pressure curve For example at 4000 rpm and full throttle, we have: Figure Process to solve OPT problem x10 25 Crank shaft degree 66 Hoang Thang, Nguyen Thi Hai Van, Tran Quoc An x1 = 25 x 13 We get the equation from calculation tool and constraint for maximum pressure curve [5]: A/F g x1 , x2 65.66 1.461x1 3.175x2 50 x2 = or Table Engine specification Engine speed = 4000rpm Full throttle g x 01 , x 02 g(25, 13.6) 59 197 g(50, 13.6) 100.6 182 g(0, 13.6) 27.55 220 g(25, 14.6) 56.55 197 g(25, 12.6) 62.9 197 gx x , x gx , x x gx , x x g x 01 x1 , x 02 0 1 2 0 Max pressure Peak pressure (bar) after TDC (0CA) (1) In this constraint, maximum pressure curve is normally lower than 50 bar for motorcycle engine Constraint for peak pressure after TDC g x1 , x2 216 0.76x1 180 15 or Apply Calculation Tool to find constraint equation in Figure g x1 , x2 15.66 1.461x1 3.175x2 g x1 0.76x1 21 (2) The position of peak pressure after TDCis always higher than 150CA 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 x10 = 10 Crank shaft degree x1 = x2 = 13.6 A/F x2 = 0.5 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: Figure Calculation tool Table Results with another engine speed Engine speed 1000 2000 3000 4000 5000 6000 7000 8000 x1 x2 10 14.26 10 15 10 15 10 15 17 10 15 20 20 25 25 26.32 25 30 31.7 13.6 14.1 13.6 13.6 13.6 13.1 13.6 13.1 12.9 13.6 13.1 12.99 13.1 13.51 13.6 13.1 13.6 13.1 12.6 F(x1,x2) F(x1+∆x1,x2 9.7 9.8 10 11 10 10 9.6 10 10 9.1 9.7 10 9.6 9.9 9.3 9.4 8.5 8.9 9.911 1.912 10.44 10.59 10.39 10.58 10.1 10.33 10.33 9.645 10.13 10.26 9.941 10 9.561 9.535 8.872 9.084 9.096 Full throttle F(x1F(X1,x2+∆x2 ∆x1,X2) 9.276 9.604 9.738 9.829 9.64 10.04 10.38 10.48 9.463 9.907 10.31 10.43 9.071 9.533 10 10.14 10.2 10.33 8.537 9.049 9.219 9.654 9.996 10.12 9.14 9.519 9.756 9.856 8.853 9.203 9.277 9.382 8.084 8.474 8.611 8.872 8.987 9.024 F(X1,X2∆x2) 9.722 9.863 10.17 10.52 10.06 10.48 9.715 10.2 10.38 9.219 9.806 10.17 9.675 9.9 9.323 9.428 8.611 9.004 9.061 x1* X2* f F(new) %