ME352 Project 4. Internal Combustion Engine Analysis. Jeremy Bourque Lab Division 9 April 24 th, 2018

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1 ME352 Project 4 Internal Combustion Engine Analysis Jeremy Bourque Lab Division 9 April 24 th, 2018 Summary of Report This project aims to model an internal combustion engine from a lawnmower as a slidercrank mechanism. The forces on the piston will be determined by using pressure forces derived from a given p-ν diagram. Using the power equation, the equation of motion of the system will be found. Using the ode45 function in MATLAB, the equation of motion will be used to find and plot the dynamic response of the system. A flywheel will then be designed to minimize the speed fluctuations in the system.

2 Analysis Procedure The first step in modeling this engine is completing preliminary kinematics on the slider-crank mechanism. This is completed by creating a vector loop with three vectors, one across each link. The unknowns of these vectors are the angle of link three and the length of link four. By splitting the vector loop into x and y coordinates and deriving, the kinematic coefficients can be found. The resulting equations from this kinematic analysis are provided in Appendix A. After kinematic analysis is completed, the provided p-ν diagram is used to determine the cylinder pressure throughout a four-stroke cycle. By using the givens pressures and the given p-ν relationship, the pressure can be found throughout the entire 720-degree cycle. The pressure is constant for intake and exhaust, but for compression and expansion the pressure follows the relationship of pν 1.37 being constant. Pressure calculations were completed and the equations used are shown in Appendix B. The resulting pressure was then plotted (Figure 1) for the four strokes to ensure it matches the shape of the provided p-ν diagram (Figure 7). Cylinder Pressure (psi) Figure 1: Pressure Diagram, plotted vs. the length of vector R4 Using the cylinder pressure as the driving force of the system, the force on the piston can be found for a complete four-stroke cycle. This is simply done by summing the forces in the x- direction, where the forces are found by multiplying pressure by the cylinder diameter. These calculations and the final equation used is provided in Appendix C. The resulting forces for an entire 720-degree cycle of the input link are shown in Figure 2 below. The tabulated output for the piston force is also provided below the plot in Table 1.

3 Pressure Force (lb) Figure 2: Piston force for a complete four-stroke cycle Table 1: Tabulated Piston Force for a Complete Cycle θ 2 F P (lb) Now that the piston force has been found, the power equation can be used to derive the equation of motion of the mechanism. This has been completed and the full derivation of the equation and

4 its components are provided in Appendix D. Rearranging the equation of motion into the state space representation allows for use of the ode45 function in MATLAB. This function can be used to determine the dynamic response of the system. This has been completed and the complete code for all parts of this project has been provided in Appendix E. Results and Discussion The first thing completed was the dynamic response of the system using the ode45 MATLAB function. The angle and the angular velocity of the input link are plotted below versus time in the first two graphs of Figure 3. The last plot in Figure 3 shows a zoomed in version of the steady state response of the angular velocity. As seen, the angular velocity of link two has speed fluctuations ranging from around 206 to 216 rad/s. 2 (rad/s) 2 (rad/s) 2 (deg) Figure 3: Dynamic response of internal combustion engine To adjust for some of these large fluctuations in angular velocity, a flywheel can be added at the input link to increase the moment of inertia. By increasing the moment of inertia of link 2 from 1.5 to 2.5 lb-s 2 in, the dynamic response becomes the one shown in Figure 4. The fluctuations now range from around 208 to 215. This leads to an approximate 30% decrease in steady state speed fluctuations which will lead to smoother motion. Although, it seems to take a longer amount of time to reach steady state. This might not be desired in some situations where a higher acceleration is desired.

5 300 Four-Stroke Engine Response idling at 401 RPM (With Flywheel) 2 (rad/s) (rad/s) t (sec) Figure 4: Dynamic response of internal combustion engine with flywheel To closer compare the original configuration to the one with the flywheel, the steady state angular velocities are plotted below on the same graph (Figure 5). Again, it s around a 30-35% reduction in speed fluctuation once the flywheel is added. Figure 5: Comparison of steady state fluctuations with and without the flywheel In conclusion, adding a flywheel can be a very effective method to reduce steady state speed fluctuation of a piston-cylinder assembly. In this case, increasing the moment of inertia from 1.5 to 2.5 lb-s 2 in led to around 30% less fluctuation in the angular speed of input link two. Although, as seen in the complete dynamic responses, the larger moment of inertia led to a longer settling time. This might not be desired in cases where a larger acceleration is desired. As a recommendation, the addition of the flywheel is beneficial in this example. The settling time is still very low at around four seconds, but less fluctuation will feel much smoother. In this situation, less fluctuation seems worth the tradeoff of having slower acceleration.

6 Appendix Appendix A: Preliminary Kinematics Figure 6: Piston-Cylinder system modelled as a Slider-Crank Vector Loop: R + R R = 0 R = R cos θ + (R ) (R sin θ ) θ = tan R sin θ R R cos θ h = R cos θ R cos θ h = R sin θ R cos θ + tan θ h f = R sin θ + R cos θ tan θ f = R cos θ R cos θ h R sin θ h

7 Appendix B: Pressure Equation Figure 7: Provided p-v diagram Cylinder Pressure Calculations, given: P = 14.7 psi, Pν. = constant P = 14.7 psi, P = 13 psi, P = 13 psi - Intake (0 to 180 ) P = P = P = 13 psi - Compression (180 to 360 ) P = P L. 4 psi L R - Expansion (360 to 540 ) - Exhaust (540 to 720 ) P = P L. 8 psi L R P = P = P = 14.7 psi Where L0, the piston depth, is found by using the compression ratio: L 4 L 8 = 8, L = 60 7

8 Appendix C: Force Equation Slider pressure diagram (link 4): Figure 8: Force diagram of the piston Pressure Force, F = F F F = P Area P Area F = π 4 D (P P )

9 Appendix D: Power Equation Power Equation F v T w = A θ θ F f θ T θ = A θ θ + B θ + m gf θ + B θ + m gf θ Cancel a θ Solve for θ Where, F f 0.005θ = (A + A ) θ + (B + B )θ + W f θ = (B + B )θ + F f W f A + A f = sin θ f = cos θ f = cos θ f = sin θ A = m f + f + I B = m (f f + f f ) A = m f B = m f f

10 Appendix E: Complete MATLAB Code Main Code clear; clc; %Constants R2 = 2; R3 = 6; D = 3.5; area = (pi*d^2)/4; W2 = 8; M2 = W2/(32.17*12); Ig2 = 1.5; W4 = 4.5; M4 = W4/(32.17*12); P1 = 13; P2 = P1; P7 = 18; Patm = 14.7; P4 = 600; w2 = 42; %input rad/s 401rpm = 42 rad/s w2p = 0; CR = 8; %Force Calculations i=1; for T2 = 0:pi/36:4*pi R4(i) = R2*cos(T2)+sqrt(R3^2-(R2*sin(T2))^2); T3 = atan2(-r2*sin(t2),(r4(i)-r2*cos(t2))); h3 = (-R2*cos(T2))/(R3*cos(T3)); h3p = (R2*sin(T2))/(R3*cos(T3))+tan(T3)*h3^2; f4 = -R2*sin(T2)+R2*cos(T2)*tan(T3); f4p = -R2*cos(T2)-R3*cos(T3)*h3^2-R3*sin(T3)*h3p; L0 = 60/7; %total length of piston if T2 >= 0 && T2 <=pi %Intake P(i) = P1; elseif T2 > pi && T2 <= 2*pi %Compression P(i) = P2*((L0-4)/(L0-R4(i)))^1.37; elseif T2 > 2*pi && T2 <= 3*pi %Expansion P(i) = P4*((L0-8)/(L0-R4(i)))^1.37; elseif T2 > 3*pi && T2 <= 4*pi %Exhaust P(i) = P7; end Fp(i) = - P(i)*area + Patm*area; i = i + 1; end %Force and Pressure Plots fprintf('%10s%10s\n','input','forcep'); fprintf('%8d%10.3f\n',[0:5:720;fp(1:end)]); figure(1) plot(-r4,p) %Pressure curve vs slider position xlabel('length of R4 (in)') ylabel('cylinder Pressure (psi)') xlim([-9-3]) %Similar to pv diagram title('pressure Diagram (vs. Slider Position)')

11 figure(2) plot(0:5:720,fp) xlabel('input Angle theta2 (deg)') ylabel('pressure Force (lb)') title('piston Force for Complete Cycle') %ODE45 Function Call tstart = 0; tstop = 10; th2initial = 0; th2dotinitial = 42; options = odeset('reltol',1e-6, 'AbsTol', 1e-6); [t,x]=ode45('odefunction',[tstart,tstop],... [th2initial,th2dotinitial],options); [t2,x2]=ode45('odefunctionfly',[tstart,tstop],... [th2initial,th2dotinitial],options); figure(3) subplot(311) plot(t,x(:,1)*180/pi,'k','linewidth',2); ylabel('\theta_{2} (deg)','fontsize',16); title('four-stroke Engine Response idling at 401 RPM','Fontsize',18) subplot(312) plot(t,x(:,2),'b','linewidth',2); ylabel('\omega_{2} (rad/s)','fontsize',16); subplot(313) plot(t,x(:,2),'b','linewidth',2); hold on plot(t2,x2(:,2),'r','linewidth',2); xlim([8.6 9]) ylim([ ]) xlabel('t (sec)','fontsize',16); ylabel('\omega_{2} (rad/s)','fontsize',16); title('steady State Fluctuations','Fontsize',16) legend('standard','with Flywheel') ODE Function with flywheel function th2d = odefunctionfly(t,theta2) %% Initialize all your variables R2, R3,... R2 = 2; R3 = 6; D = 3.5; area = (pi*d^2)/4; W2 = 8; M2 = W2/(32.17*12); Ig2 = 2.5; W4 = 4.5; M4 = W4/(32.17*12); P1 = 13; P2 = P1; P7 = 18; Patm = 14.7; P4 = 600; CR = 8; if theta2(1) > 0 theta2(1) = rem(theta2(1),4*pi); else

12 theta2(1) = rem(theta2(1),-4*pi); end %% Include your preliminary kinematics code here. This function takes % the time t and theta2 as the input. theta2 is a vector which has both % the position theta2 as well as the velocity theta2dot. R4 = R2*cos(theta2(1))+sqrt(R3^2-(R2*sin(theta2(1)))^2); T3 = atan2(-r2*sin(theta2(1)),(r4-r2*cos(theta2(1)))); h3 = (-R2*cos(theta2(1)))/(R3*cos(T3)); h3p = (R2*sin(theta2(1)))/(R3*cos(T3))+tan(T3)*h3^2; f4 = -R2*sin(theta2(1))+R2*cos(theta2(1))*tan(T3); f4p = -R2*cos(theta2(1))-R3*cos(T3)*h3^2-R3*sin(T3)*h3p; L0 = 60/7; %total length of piston if theta2(1) >= 0 && theta2(1) <=pi %Intake P = P1; elseif theta2(1) > pi && theta2(1) <= 2*pi %Compression P = P2*((L0-4)/(L0-R4))^1.37; elseif theta2(1) > 2*pi && theta2(1) <= 3*pi %Expansion P = P4*((L0-8)/(L0-R4))^1.37; elseif theta2(1) > 3*pi && theta2(1) <= 4*pi %Exhaust P = P7; end Fp = - P*area + Patm*area; %Point Path fxg2 = -sin(theta2(1)); fyg2 = cos(theta2(1)); fxg2p = -cos(theta2(1)); fyg2p = -sin(theta2(1)); A2 = M2*(fxg2^2 + fyg2^2)+ig2; B2 = M2*(fxg2*fxg2p + fyg2*fyg2p); A4 = M4*(f4^2); B4 = M4*(f4*f4p); %% Equation of motion expressed as two first order ODEs th2d=zeros(2,1); th2d(1) = theta2(2); % integrate to get position th2d(2) = ((-(B2+B )*theta2(2)^2-W2*fyg2+Fp*f4))/(A2+A4);