Bài report được đánh giá 8.7đ . Thiết kế trên lý thuyết hoặc tính toán lại 1 loại động cơ trên xe oto. Thông qua project này có thể giúp các sinh viên ngành công nghệ ký thuật oto biết được những bước cơ bản để thiết kế ra một động cơ mới. Ngoài ra, sinh viên còn biết sử dụng các ứng dụng hỗ trợ học tập như Matlab,...
HO CHI MINH CITY UNIVERSITY OF TECHNOLOGY AND EDUCATION FACULTY FOR HIGH QUALITY TRAINING FINAL REPORT TOPIC: ENGINE CALCULATION BASED ON TOYOTA 1NZ-FE Student name: Nguyen Tran Binh 20145402 Vo Ngoc Khoi Nguyen 20145015 Du Thanh Vinh 20145020 Major: Internal Combustion Engine Calculation Adviser: Ly Vinh Dat Ho Chi Minh City, December 2022 HO CHI MINH CITY UNIVERSITY OF TECHNOLOGY AND EDUCATION FACULTY FOR HIGH QUALITY TRAINING FINAL REPORT TOPIC: ENGINE CALCULATION BASED ON TOYOTA 1NZ-FE Student name: Nguyen Tran Binh 20145402 Vo Ngoc Khoi Nguyen 20145015 Du Thanh Vinh 20145020 Major: Internal Combustion Engine Calculation Adviser: Ly Vinh Dat Ho Chi Minh City, December 2022 THE SOCIALIST REPUBLIC OF VIETNAM Independence – Freedom– Happiness -Ho Chi Minh City, December 4th, 2022 PROJECT ASSIGNMENT Student name: Student ID: Student name: Student ID: Student name: Student name: Student name: Student ID: Student ID: Student ID: Major: Internal Combustion Engine Calculation Class: 20145CLA Advisor: Ly Vinh Dat Phone number: Date of assignment: Date of submission: Project title: Engine calculation based on 1NZ-FE Initial materials provided by the advisor: Content of the project: Final product: CHAIR OF THE PROGRAM ADVISOR (Sign with full name) (Sign with full name) THE SOCIALIST REPUBLIC OF VIETNAM Independence – Freedom– Happiness Ho Chi Minh City, December 4th, 2022 ADVISOR’S EVALUATION SHEET Student name: Student ID: Student name: Student ID: Student name: Student ID: Student name: Student ID: Student name: Student ID: Major: Project title: Advisor: EVALUATION Content of the project: Strengths: Weaknesses: Approval for oral defense? (Approved or denied) Overall evaluation: (Excellent, Good, Fair, Poor) Mark: ……………… (in words: ) Ho Chi Minh City, December 4th, 2022 ADVISOR (Sign with full name) THE SOCIALIST REPUBLIC OF VIETNAM Independence – Freedom– Happiness Ho Chi Minh City, December 4th, 2022 CRITICAL TEACHER’S EVALUATION SHEET Student name: Student ID: Student name: Student ID: Student name: Student ID: Student name: Student ID: Student name: Student ID: Major: Project title: Critical Teacher: EVALUATION Content of the project: Strengths: Weaknesses: 10 Approval for oral defense? (Approved or denied) 11 Overall evaluation: (Excellent, Good, Fair, Poor) …… 12 Mark: ……………… (in words: …………………………………………………….) Ho Chi Minh City, December 4th , 2022 FHQ-HCMUTE (Sign with full name) CONTENTS CHAPTER 1: ENGINE PARAMETER CHAPTER 2: THERMAL CALCULATE AND BUILD INDICATED WORK DIAGRAM IN ENGINE AND KINETIC AND DYNAMIC OF SLIDER CRANK MECHANISM I Choosing parameter II Thermal Calculation 10 CHAPTER 3: DYNAMICS CALCULATION OF CRANKSHAFTCONNECTING ROD STRUCTURE CHAPTER 1: ENGINE PARAMETER Engine parameters Engine: 1,5L 1NZ-FE Max Brake Power, Pmax (kW): 80 Stroke, 𝜏: RPM: 6000 Compression ratio, ɛ: 11 Air equivalence ratio, 𝜆 (𝛼): Cooling system: Air No of cylinder, i: in-line Report contents 2.1 Thermal calculate and build indicated work diagram in engine 2.2 Kinetic and dynamic of slider crank mechanism Curves 3.1 P-V figure 3.2 P − φ, P𝐽 , P1 figures 3.3 Kinetic and dynamic figures 0.9 CHAPTER 2: THERMAL CALCULATE AND BUILD INDICATED WORK DIAGRAM IN ENGINE AND KINETIC AND DYNAMIC OF SLIDER CRANK MECHANISM I Choosing parameter Intake air pressure: po = 0.1013 MPa Intake air temperature To Choose T0 = 29 ˚C due to the average temperature in the South of Vietnam T0 = Tenvironment = 29 ˚C= 302 K Intake air pressure at the intake valve pk: We decide to choose a 4-stroke engine without turbo, so: pk = p0= 0.1013 MPa Intake air temperature at the intake valve Tk: The 4-stroke engine without turbo: Tk= T0= 302 K Pressure at the end of intake process: pa Base on the experimental calculation, pa = (0.8 ÷ 0.95) p0 Choose pa = 0.95 p0 = 0.0962 MPa Residual gas pressure pr: For the gasoline engine pr = (0.11 ÷ 0.12) MPa Choose pr = 0.12 MPa Residual gas temperature Tr: For the gasoline engine Tr = 900 K ÷ 1000 K Choose Tr = 1000 K Temperature rises on new charge coefficient ΔT: For gasoline engine: ΔT = ÷ 20 ˚C Choose ΔT = 15˚C Heat added coefficient λ1: Limit coefficient of heat added λ1 = 1.02 ÷ 1.07 Choose λ1 = 1.05 10 Combustion chamber sweep coefficient λ2: With 4-stroke engine without turbo λ2 = 0.8 ÷ 0.9 Choose λ2 = 0.85 11 Temperature revise coefficient λt : For gasoline engine, choose air residue coefficient α = 0.9 So λt = 1.15 12 The heat ultilization coefficient at point z ξz For gasoline ξz = 0.75 ÷ 0.92 Choose ξz = 0.85 13 The heat ultilization coefficient at point b ξb For gasoline engine, ξb = 0.85 ÷ 0.95 Choose ξb = 0.9 14 Excess air value α For gasoline engine, α = 0.85 ÷ 0.95 Choose α = 0.9 15 The coefficient of diagram rounding-off φr For gasoline engine, φr = 0.93 ÷ 0.97 Choose φr = 0.95 II Thermal Calculation Turbocharge ratio Gasoline engine: λ = 3.00 ÷ 4.00 Choose λ = Heat calculation with m= 1.45 1 𝑇𝑘 𝑝𝑎 𝑝𝑟 𝑚 𝜂𝑣 = × × × [𝜀𝜆1 − 𝜆𝑡 𝜆2 ( ) ] 𝜀 − 𝑇𝑘 + ΔT 𝑝𝑘 𝑝𝑎 302 0.0962 = × × 11 − 302 + 15 0.1013 × [11 × 1.05 − 1.15 × 0.85 × ( 0.12 ) 0.0962 1.45 ] = 0.942 The coefficient of residual gases γr : γ𝑟 = 𝜆2 𝑝𝑟 𝑇𝑘 0.85 0.0962 302 = × × = 0.0259 (𝜀 − 1)𝜂𝑣 𝑝𝑘 𝑇𝑟 (11 − 1) × 0.942 0.1013 1000 Temperature at the end of induction: Ta 𝑚−1 𝑚 𝑝 (𝑇𝑘 + ΔT) + 𝜆𝑡 𝛾𝑟 𝑇𝑟 ( 𝑎 ) 𝑝𝑟 𝑇𝑎 = + γ𝑟 1.45−1 1.45 0.0962 (302 + 15) + 1.15 × 0.0259 × 1000 × ( 0.12 ) = + 0.0259 = 336 𝐾 The mean molar specific heat of: 3.1 Fresh mixture 𝑏 0.00419 𝑚𝑐𝑣 = 𝑎𝑣 + 𝑇 = 19.806 + ̅̅̅̅̅ 𝑇 2 3.2 Residual gases Excess air value: 0.7< α < 1 −5 ′′ ̅̅̅̅̅̅̅ 𝑚𝑐 𝑣 = (17.997 + 3.504𝛼) + × (360.34 + 252.4𝛼) × 10 𝑇 CHAPTER 3: DYNAMICS CALCULATION OF CRANKSHAFT- CONNECTING ROD STRUCTURE 3.1 Kinematic Analysis of Crankshaft-Connecting Rod Structure In an internal combustion engine: piston, connecting rod, crankshaft are the moving parts and they work on the following principles: - The reciprocating piston group transmits air force to the connecting rod - The connecting rod group is an intermediate moving part with complex motion to convert the reciprocating motion of the piston into the rotation of the crankshaft - The crankshaft is the most important part that rotates and transmits the power from the engine out to drive the machines There are three main types of crankshaft connecting rod construction: a) The crankshaft cuts the cylinder center line b) The crankshaft deviates from the cylinder centerline by a small distance a < 0.1S, (S - piston stroke) c) Structure for V-engine - this type has two connecting rods mounted on the same crank pin (called double connecting rod) - In all type above, the crankshaft cuts the cylinder centerline structure is the most common one nowadays 3.2 Piston Kinematics With the hypothesis that the crankshaft rotates with angular velocity 𝜔 = const, then the crankshaft angle 𝛼 is variable proportional to time, and all kinematic quantities are functions dependent on the variable 𝛼 However, this assumption for modern high-speed engines gives an error negligible because of the variation of angular velocity (𝜔) due to the inhomogeneity uniformity of the motor torque caused when the motor is operating at steady state is very small 3.2.1 Piston displacement When the crankshaft rotates at an angle α, the piston moves a distance Sp from its initial position The displacement of the piston in the cylinder is calculated by the formula: 𝜆 Sp = R[1- cos𝛼 + (1- cos2𝛼)] = Sp1 + Sp2 With: Sp1 = R(1- cos𝛼) 𝜆 Sp2 = R* (1- cos2𝛼) λ= 𝑅 𝐿 : Engine structure parameter 3.2.2 Piston speed Speed of the piston is a function depend on the crankshaft angle and λ: 𝜆 Vp = R𝜔(sin𝛼 + sin2𝛼) = Vp1 + Vp2 With Vp1 = R𝜔sin𝛼 𝜆 Vp2 = R𝜔 sin2𝛼 𝜔= 2𝜋𝑛 60 : angular velocity of connecting rod Average speed of piston, to classify the different type of engine: Vtb = 𝑆𝑛 30 = *R𝜔 𝜋 At 𝛼 = 73 ÷ 78°, velocity of piston has maximum value: Vpmax = (1,6 ÷ 1,66)Vtb 3.2.3 Acceleration of the piston Taking the derivative V over time, we have the piston acceleration formula: J = R.𝜔2 (cosα + λ.cos2α) = Jp1 + Jp2 The acceleration of the piston is the sum of two harmonic functions of class I and class II with: Jp1 = R.𝜔2 cosα Jp2 = R.𝜔2 λcos2α 3.3 STRUCTURE DYNAMICS OF CRANKSHAFT – CONNECTING ROD 3.3.1 Gas pressure forces Pgas Gas pressure force is a quantity that varies with the angle of rotation of the crankshaft, defined as obtained from gas pressure P at the thermal calculation of the engine Pgas = (pgas )*Fp Where: pgas : gas pressure in the cylinder p0 = 0,1 MN/m2 – atmospheric pressure Fp = πD2 : cross sectional area of piston (m2) - Intake process: Pgas = Pa - Compression process: Pgas = Pa.in1 , with i from (1800 ) to ε (3600 – θs) Pz - Expansion process: Pgas = n ɛ - Expansion process: Pgas = Pr 3.3.2 Inertial forces of moving details Translational inertial force: Pj = −mj.J = −mj Rω2(cosα + λcos2α) Centrifugal inertial force: PK = −mr Rω2 = const With: mj = mpg + mA mr = mK + mB mA = (0,275 ÷ 0,35)mtt mB = (0,725 ÷ 0,65)mtt Where: Pj – translational inertial force PK – centrifugal inertial force mj – the mass of translational motion details mr – the mass of rotational movement details mcr – the mass of connecting rod mpg – the mass of piston group mC – the mass of the crankshaft mA – the mass of connecting rod small end mB – the mass of connecting rod big end 3.3.3 The force system acting on the crankshaft – connecting mechanism Total force acting on the piston pin is the synergy of the gas pressure force Pkt and the translational inertial force Pj, valid equal to the algebraic sum of these two forces: P1 = Pgas + Pj Vertical force acting on the connecting rod Pcr = P∑ cosβ with β = sin−1(λ sinα) Thrust force N N = Pcr.tgβ Tangential force T sin(α + β) T = Pcr*sin(α + β) = P∑ cosβ Normal force Z: cos(α + β) Z = Pcr*cos(α + β) = P∑ cosβ 3.3.4 Rotation moment M of the engine Calculation of the working deflection angle of the engine: δK = 180o Working order of the engine: – – – Determination of the working phase of each cylinder: + Cylinder 1: α + Cylinder 2: α + 180 + Cylinder 3: α + 540 + Cylinder 4: α + 360 Total moment ∑ Mi defined by the relation: ∑Mi = R.∑𝑖𝑖=1 Ti Where: ∑ Mi: total moment ∑ Ti – total tangential force 3.3.5 Forces acting on crankpins At the crankpin there is the following force: tangential force T, normal force Z, ⃗ defined centrifugal force PKO The force acting on the crankpin is the force vector 𝑄 by force balance equation: ⃗ =𝑇 ⃗ + 𝑍 + ⃗⃗⃗⃗⃗⃗⃗ 𝑄 𝑃𝐾𝑂 with PKO = −mB R ω2 % Calculated parameters R = 0.4285; %(dm) B = 0.758; %(dm) S = 0.857; %(dm) l = 1.575; %(dm) T = 11; %(Compression ratio) Vd = 0.387; %(lit) Sp = pi*B^2/4; %Area of piston lamda = R/l; Pa = 0.0962; %MPa Pr = 0.12; Pz = 10.47; Pc = 2.57; ro = 1; Vc = Vd/(11-1);%dm^3 Vz = ro*Vc; %Va = Vd+Vc n1 = 1.37; n2 = 1.25; w = 628.3185307 ; %rad/s, n=6000 rpm % Calculation grid on Xa = R*(1-cosd(180) + (lamda/4).*(1-cosd(2.*180))); Va = Xa*Sp + Vc; % (lit) % Exhaust and Intake Adjustment ahc1 = [0 20 40]; phc1 = [Pr (Pr+Pa)/2 Pa]; a1 = linspace(0,40,1000); X1 = R*(1-cosd(a1) + (lamda/4).*(1-cosd(2.*a1))); V1 = X1*Sp + Vc; P1 = interp1(ahc1,phc1,a1, 'sline'); J1 = R*w^2.*(cosd(a1)+lamda.*cosd(2.*a1)); Vt1 = R*w*(sind(a1) + (lamda/2).*sind(2*a1)); % Intake process a2 = linspace(40,180,1000); X2 = R*(1-cosd(a2)+(lamda/4)*(1-cosd(2*a2))); V2 = X2*Sp+Vc; P2 = linspace(Pa,Pa,1000);%MN/m2 J2 = R*(w^2).*(cosd(a2)+lamda.*cosd(2.*a2)); Vt2 = R*w*(sind(a2) + (lamda/2).*sind(2*a2)); % Compression process a3 = linspace(180,350,1000); X3 = R*(1-cosd(a3)+(lamda/4)*(1-cosd(2*a3))); V3 = X3*Sp+Vc; P3 = Pa*((Va./V3).^n1); %MN/m2 J3 = R*(w^2).*(cosd(a3)+lamda.*cosd(2.*a3)); Vt3 = R *w*(sind(a3) + (lamda/2).*sind(2*a3)); % Compression and Combustion Adjustment Pcc =(Pz-Pc)/3 + Pc; ahc4 = [350 360 370]; phc4 = [max(P3) (max(P3)+Pcc)/2 Pcc]; a4 = linspace(350,370,1000); X4 = R*(1-cosd(a4)+(lamda/4)*(1-cosd(2*a4))); V4=X4*Sp+Vc; P4 = interp1(ahc4,phc4,a4, 'sline'); J4 = R*(w^2).*(cosd(a4)+lamda.*cosd(2.*a4)); Vt4 = R*w*(sind(a4) + (lamda/2).*sind(2*a4)); % z” point X41 = R*(1-cosd(370) + (lamda/4).*(1-cosd(2.*370))); V41 = X41*Sp + Vc; % Expansion process a6 = linspace(380,500,1000); X6 = R*(1-cosd(a6) + (lamda/4).*(1-cosd(2.*a6))); V6 = X6.*Sp + Vc; P6 = Pz.*(Vz./V6).^n2; J6 = R*(w^2).*(cosd(a6)+lamda.*cosd(2.*a6)); Vt6 = R*w*(sind(a6) + (lamda/2).*sind(2*a6)); % Combustion an Expansion Adjustment ahc5 =[370 372 380]; phc5 =[Pcc 0.49*(Pcc+max(P6)) max(P6)]; a5 = linspace(370,380,16000); X5 = R*(1-cosd(a5) + (lamda/4).*(1-cosd(2.*a5))); V5 = X5.*Sp + Vc; P5 = interp1(ahc5,phc5,a5,'sline'); J5 = R*(w^2).*(cosd(a5)+lamda.*cosd(2.*a5)); Vt5 = R*w*(sind(a5) + (lamda/2).*sind(2*a5)); % Expansion and Exhaust Adjustment % b’ point Pb = min(P6); % b” point Vbb = Va; Pbb =(Pb-Pr)/2+Pr ahc7 = [500 540 580]; phc7 = [Pb (Pb+Pr)/2 Pr]; a7 = linspace(500,580,1000); X7 = R*(1-cosd(a7) + (lamda/4).*(1-cosd(2.*a7))); V7 = X7.*Sp + Vc; P7 = interp1(ahc7,phc7,a7,'sline'); J7 = R*w^2.*(cosd(a7)+lamda.*cosd(2.*a7)); Vt7 = R*w*(sind(a7) + (lamda/2).*sind(2*a7)); % Exhaust process a8 = linspace(580,720,1000); X8 = R*(1-cosd(a8) + (lamda/4).*(1-cosd(2.*a8))); V8 = X8*Sp + Vc; P8 = linspace(Pr,Pr,1000); J8 = R*w^2.*(cosd(a8)+lamda.*cosd(2.*a8)); Vt8 = R*w*(sind(a8) + (lamda/2).*sind(2*a8)); % -V = [V1 V2 V3 V4 V5 V6 V7 V8]; P = [P1 P2 P3 P4 P5 P6 P7 P8]; a0 = [a1 a2 a3 a4]; ap= [a1 a2 a3 a4 a5 a6 a7 a8]; X = [X1 X2 X3 X4 X5 X6 X7 X8]; Vt = [Vt1 Vt2 Vt3 Vt4 Vt5 Vt6 Vt7 Vt8]; J = [J1 J2 J3 J4 J5 J6 J7 J8]; Pj = (-0.0002052*R/10*(w^2)).*(cosd(ap)+lamda.*cosd(2.*ap)); Pkt = (P-0.1); Phl = Pkt+Pj; figure(1) % P-V Diagram plot(V,P); title('P-V Diagram'); xlabel('Volume (dm^3)'); ylabel('Pressure P (MN/ dm^2)'); grid on hold on figure(2) % P-phi diagram plot(ap,Pkt,'r'); hold on plot(ap,Pj,'g'); hold on plot(ap,Phl,'b'); hold on title('P-phi Diagram'); xlabel('Crank shaft angle (Degree)'); ylabel('Pkt, Pj, P1 (MN/m2)'); legend('Pkt','Pj','P1'); grid on %xuat du lieu excel A = [V' P' ap' Pj' Pkt' Phl']; xlswrite('filesolieu.xlsx' ,A); a1 = linspace(0,180,13) ; p1 = 0.32 ; D=0.075 ; p0=0.1 ; pz=10.5 ; pr=0.11 ; Pkt1=(p1-p0)*10^-2*pi/(4*D^2) ; P1=linspace(Pkt1,Pkt1,13) ; a2=linspace(195,345,11) ; i1=linspace(1,9.7,11) ; p2=p1*(i1.^1.37) ; Pkt2=(p2-p0)*10^-2*pi/(4*D^2) ; a5=linspace(350,375,6) ; i3=linspace(9.7,9.7,6) ; p5=p1*(i3.^1.37) ; Pkt5=(p5-p0)*10^-2*pi/(4*D^2) ; a3=linspace(375,540,12) ; i2=linspace(1.869,11.43,12) ; p3=pz*(i2.^1.27).^-1 ; Pkt3 =(p3-p0)*10^-2*pi/(4*D^2) ; a4=linspace(555,720,12) ; p4=pr ; Pkt4=(p4-p0)*10^-2*pi/(4*D^2) ; P4=linspace(Pkt4,Pkt4,12) ; a=[a1 a2 a5 a3 a4] ; Pkt=[P1 Pkt2 Pkt5 Pkt3 P4] ; Pj=-0.016845*77.86*(cosd(a)+0.263*cosd(2*a)) ; Psum=Pkt+Pj ; N=Psum.*(sind(a)/sqrt(1-0.0692*(sind(a)).^2)) ; T=Psum.*(sind(a)+0.263.*cosd(a).*sind(a)/sqrt(1 0.0692*(sind(a)).^2)); Z=Psum.*(cosd(a)+0.263.*sind(a).*sind(a)/sqrt(1 0.0692*(sind(a)).^2)); figure(1) plot(a,Pkt,a,Pj,a,Psum) title('Graph of gas force Pkt, inertial force Pj, combined force Psum') xlabel('Angular crankshaft (degrees)') ylabel('Pressure (MN/m^2)') figure(2) plot(a,N) title('Horizontal force N') xlabel('Angular crankshaft (degrees)') ylabel('Pressure (MN/m^2)') figure(3) plot(a,T) title('Tangent force T') xlabel('Angular crankshaft (degrees)') ylabel('Pressure (MN/m^2)') figure(4) plot(a,Z) title('Normal force Z') xlabel('Angular crankshaft (degrees)') ylabel('Pressure (MN/m^2)')