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Assignment 3: Vehicle Forces and Kinematics Date Due: September 17, 2009 Instructor: Trani

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Assignment 3: Vehicle Forces and Kinematics Date Due: September 17, 2009 Instructor: Trani CEE 3604: Introduction to Transportation Engineering Fall 2009 Assignment 3: Vehicle Forces and Kinematics Date Due: September 17, 2009 Instructor: Trani Problem 1 Wind tunnel testing of a new, low-profile high-speed train has the following equivalent Davis equation coefficients: A = 8.57 ; % units are kN B = 0.0983; % units are kN s/m C = 0.0129; % units are kN s-s/m-m The vehicle tested is expected to have a mass of 256,000 kg for a 6-car train unit (includes the locomotive unit to pull the passenger cars) and maximum axle load of 16 metric tons per axle. The front locomotive has nearly constant power out (P) rated at 15,000 HP with 0.8 efficiency. To avoid slippage of the rail wheels at low speeds, the Tractive Force (T) of the engine is limited to a constant 400 kN for all speeds below 20 m/s by an electromechanical system. Solution % Wind tunnel testing of a new, low-profile high-speed train has the % following equivalent Davis equation coefficients: % A = 8.57 ; % units are kN % B = 0.0983; % units are kN s/m % C = 0.0129; % units are kN s-s/m-m % The vehicle tested is expected to have a mass of 256,000 kg for a % 6-car train unit (includes the locomotive unit to pull the passenger cars) % and maximum axle load of 16 metric tons per axle. The front locomotive % has nearly constant power out (P) rated at 15,000 HP with 0.8 efficiency. % To avoid slippage of the rail wheels at low speeds, the Tractive Force (T) % of the engine is limited to a constant 400 kN for all speeds below 20 m/s % by an electromechanical system. % Programmer: A. Trani clear clc % Coefficients of Davis equation applied to Japanese Shinkansen system % Series 200 (new system) A = 8.57; % units are kN CEE 3604 A3 Trani Page 1 of 15 B = 0.0983; % units are kN s/m C = 0.0129; % units are kN s-s/m-m % Grade resistance equation M = 256; % metric tons g = 9.81; % gravity force (m/s-s) X = 1e12; % grade (1 in X) Rg = M * g / X; % or use the standard analysis with sin(phi) on the equations of motion w=M*1000*g; % weight in Newtons phi = 0; % angle of railway (radians) RgMod = w * sin(phi); % in Newtons % Create a speed vector V = 0:1:100; % speed in meters/second % Create a speed vector V = 0:1:100; % speed in meters/second % Calculate Resistance (in KiloNewtons) according to % modified Davis equation % Added the grade resistance R = A + B * V + C * V.^2 + RgMod/1000; % Make a plot of total resistance vs speed plot(V,R,'o--') xlabel(' Speed (m/s)') ylabel('Resistance (kN) or T (kN)') title('Reisistance of High-Speed Rail System') grid hold on % Calculate the Tractive Effort (T) profile CEE 3604 A3 Trani Page 2 of 15 P = 15000; % horsepower (hp) Vkmhr = V*3.6; % velocity in km/hr (needed in the T equation) nu = 0.8; % efficiency % cap the value of T - tractive force - to 400 kN npoints = length(Vkmhr); for p = 1:npoints T(p) = 2650 * nu * P ./ Vkmhr(p) / 1000; % in kN if T(p) > 400 T(p) = 400; % caps the tractive effort to 400 kN end end plot(Vkmhr/3.6,T,'^-r'); % plot the tractive force vs speed (km/hr) a) Plot the total resistance and tractive force diagram for this vehicle. In your diagram include the limitation of tractive force. Plot of T and R vs. Speed. Flat Terrain. b) Find the maximum speed (in meters/second and in miles per hour) for the new high-speed rail in level ground. By inspection the maximum speed is 83.25 m/s (299.7 km/hr). CEE 3604 A3 Trani Page 3 of 15 Point of maximum speed for HS rail. 83.25 m/s (299.7 km/hr) is the maximum speed. c) If the train negotiates a positive (uphill) 1 in 50 slope (one foot vertical rise for every 50 feet horizontally) estimate the gravity force (in kN) acting against the vehicle motion. A 1 to 50 slope is equivalent to (1/50) radians (phi). This is also equal to X = 50 in the handout equation. The new plot is shown below. Plot of T and R vs. Speed. Slope 1:50. d) Find the maximum speed for the train for condition (c). Make a plot of the new resistance curve and tractive force to find the maximum speed. CEE 3604 A3 Trani Page 4 of 15 Point of maximum speed for HS rail. Slope 1:50 (uphill). Maximum speed is 69.3 m/s (240 km/hr). e) Repeat (d) when the train returns and now goes downhill Plot of T and R vs. Speed. Slope -1:50 (downhill). Balancing Sped is 97.5 m/s (351 km/hr). Point of maximum speed for HS rail. Slope -1:50 (downhill). Maximum speed is 97.5 m/s (351 km/hr). f) Comment on the values of maximum speed calculated in parts (d) and (e) and contrast with (b). What is the impact of topography (or terrain) in travel time for a rail system? CEE 3604 A3 Trani Page 5 of 15 g) Estimate the travel time between two stations located 40 km apart if the driver uses 90% of the maximum speed as the cruise speed for the train. Consider the acceleration and deceleration phases in your analysis. Assume flat terrain so the target cruise speed is 74.9 m/s (262 km/hr). Acceleration Phase This requires that we set up the acceleration function of the HS train. Recall: ma = ∑Fx = T − R x 1 a = (T − R) m Both T and R are known functions of velocity (V). So we can setup a simple numerical procedure (see pages 18-34 in handout - Transportation_technology.pdf) to calculate speed and distance profile in the acceleration phase of the HS profile. A simple Matlab script is shown below. This process can be done in Excel following the analysis shown on page 30 of the same handout. % Calculation of speed and distance traveled in the acceleration phase of % the HS rail motion between stations deltaT = 1; % step size for numerical integration - seconds Vx(1) = 0; % initial speed (m/s) Sx(1) = 0; % intial distance traveled (meters) t(1) = 0; % time clock starts for i=1:1:400 % Try 400 steps to integrate numerically the ODE if Vx(i) < 20; % limits the tractive force to 400 kN TractiveForce(i) = 400; end TractiveForce(i) = interp1(V,T,Vx(i)); % calculates T at any speed Vx - use linear interpolation (interp1) Resistance(i) = interp1(V,R,Vx(i)); % calsculates R at Vx - use linear interpolation t(i+1)=t(i)+deltaT; % updates the time during the acceleration phase Ax(i) = 1/(M*1000) * ((TractiveForce(i) - Resistance(i)) * 1000); % acceleration m/s-s Vx(i+1) = Vx(i) + Ax(i) * deltaT; % m/s Sx(i+1) = Sx(i) + Vx(i) * deltaT; % meters end % Make plots of speed, distance and acceleration figure plot(t,Vx) CEE 3604 A3 Trani Page 6 of 15 grid xlabel('Time (seconds)') ylabel('Speed (m/s)') figure plot(t,Sx) grid xlabel('Time (seconds)') ylabel('Distance Traveled (meters)') figure plot(t(1:400),Ax) grid xlabel('Time (seconds)') ylabel('Acceleration (m/s-s)') Plot of Speed Profile of the HS Train in the Acceleration Phase. CEE 3604 A3 Trani Page 7 of 15 Looking at the figure (use the Matlab zoom to determine the approximate time to stop the acceleration phase), the acceleration phase lasts 144.25 seconds as the train reaches the desired 74.9 m/s cruise speed. This process consumes 7,700 meters (using the numerical integration). Plot of Distance vs. Time in the Acceleration Phase. Deceleration Phase Assume -1 m/s-s as a constant deceleration rate to provide a nice and gentle deceleration for passengers on board the HS train. 1 S = S +V t + at 2 t o 0 2 V 2 −V 2 S = f 0 t 2a The distance traveled in the deceleration phase is: 2,805 meters. The time for deceleration is then 64.6 seconds Cruise Phase The distance remaining for the cruise phase is 30,215 meters. At 74.9 m/s this phase lasts 403.4 seconds. Total Travel Time The total travel time between stations is then 612.25 seconds. if we had not used the numerical integration procedure and should we had assumed a constant cruise speed of 74.9 seconds, the travel time would have been 534.1 seconds. This represents a 12.5% difference. _____________________________________________ Problem 2 Blacksburg Transit wants to re-equip their fleet of 24 Flxible Buses with a new more powerful engine that delivers 15% more power than the baseline engine shown on page 46 of the handout (transportation_technology.pdf). Refer to table on page 46 to see the current baseline engine parameters. All other characteristics of the vehicle will remain the same. CEE 3604 A3 Trani Page 8 of 15 a) Using the Matlab script provided construct a Force vs. Speed diagram. Plot the resistance and tractive force in the same diagram. Solution % Script to estimate Force vs. Speed Diagram for a Bus % Calculates and resistances for the bus (SAE formulas) % Uses a linear and quadratic approximation for basic and % Aero resistances % Programmer: Toni Trani % % Version 1 (01/20/09) % Define model parameters (bus model) Ca = 0.020; % Coefficient for trucks C1 = 7.6; % C2 = 0.056; % W = 120; % Bus weight in kN A = 6.97; % Frontal area i = 0; % Grade in percent (%) % Define for Tractive Effort estimation D = 0.94; % Wheel diameter (m) u(1) = 4.5; u(2) = 2.6; u(3) = 1.5; u(4) = 1.1; % gear ratios in explicit vector form J = 4.6; % Differential reduction ratio con1 = .06; % conversion factor (to convert km/hr to m/s) % conversion to obtain V in km/hr con2 = 2650; % conversion factor (to convert hp to N) % conversion from Hp to Newtons nu = 0.85; P = 1.15 * [75 130 165 170 190 203 210]; % power output vector RPM = [800 1200 1600 1800 2000 2200 2400]; % RPM vector % ______________________________________________ % Compute resistances for a=1:1:130 % Starts loop V(a) = a - 1; % Speed vector (km/hr) R_basic(a) = (C1 + C2 * V(a)) * W; % Basic resistance (N) R_aero(a) = Ca * A * V(a) ^ 2; % Aerodynamic resistance (N) R_grade = 10 * W * i ; % Grade resistance (N) R_total(a) = R_basic(a) + R_aero(a) + R_grade; % Total resistance (N) CEE 3604 A3 Trani Page 9 of 15 end % Ends the comp.
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