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Unformatted text preview: Introduction to Tansportation Engineering Fundamentals of Ground Vehicle Performance Dr. Antonio A. Trani Professor of Civil and Environmental Engineering Virginia Polytechnic Institute and State University Blacksburg, Virginia Fall and Spring 2011 Virginia Tech 1 Introductory Remarks • Simplified analysis of various modes of transportation allow us to integrate the elements of technology needed to predict cost and performance measures of effectiveness associated with a mode of transportation (i.e, speed, travel time, station spacing. level of service, etc.). • Need to understand the basic forces that make all transportation systems move and behave the way they do • Knowledge of the operational limitations of existing and future transportation technology can influence our decisions with respect to the transportation mode to be used Virginia Tech 2 Ground Vehicle Performance Basics Virginia Tech 3 Basic Forces Acting on a Ground Vehicle The following diagram applies to cars, buses, trains, etc. Lift Thrust Drag φ Friction Force mg Assumed all forces act through a point mass system (center of gravity) Thrust is also called Tractive Effort (TE) Virginia Tech 4 Analysis of Individual Forces 12 L = -- ρ V SC l 2 (1) 12 D = -- ρ V SC D 2 (2) T = f ( V, ρ ) (3) F f = ( mg cos φ – L ) f roll (4) G t = mg sin φ (5) Virginia Tech 5 Nomenclature T is the net engine thrust force or sometimes called tractive force or tractive effort (Newtons), L is the lifting force (Newtons), V is the vehicle speed (m/s), D is the drag force (Newtons), Ff is the friction force (Newtons), G is the component of the gravity force (Newtons), ρ is the air density (kg/cu.-m.), Virginia Tech 6 S is the reference area (m2), Cl lift coefficient (nondimensional), CD is the drag coefficient (nondimensional), froll is the rolling friction coefficient (nondimensional) m is the vehicle mass g = gravitational acceleration (9.81 m/s2 or 32.2 ft/s2) Virginia Tech 7 Application of Newton’s Second Law m ax = T – D – Ff – Gf 12 ma x = f ( V, ρ ) – -- ρ V SC D – ( mg cos φ – L ) f roll 2 (6) (7) – m g sin φ • The gravity term is important and thus has been added • We could add another term to account for curvature resistance (good for rail transportation) Virginia Tech 8 Some Remarks • Acceleration capability of the vehicle decreases as speed is gained during the acceleration phase • If expression (7) is integrated twice between an initial speed, V0 and the desired cruising speed, Vcr the distance covered during the acceleration phase can be found (this might require a numerical integration procedure) Virginia Tech 9 Types of Forces • There are two types of forces acting on the vehicle: Tractive Resistance Tractive Force Resistance Forces • The following pages explain these forces in more detail Virginia Tech 10 Variations of Tractive Effort and froll • T decreases with speed for all transportation modes • froll is a function of the contact material between vehicle and guideway or road and the vehicle speed T (N) f roll Bias-Ply Tire Sea Level Radial Tire High Elevation V (m/sec) Virginia Tech Metal Rim Surface V (m/sec) 11 Rolling Friction Coefficient ( f roll ) • The value of f roll varies with speed, tire construction, tire pressure, etc. • f roll is usually determined experimentally • f roll is associated with a tire rolling free on a road. • f roll doubles with speed increments from 0 to 200 km/hr. Virginia Tech 12 froll Value (dimensionless) Variation of froll with Speed Source: W. Hucho (1998) Data good for radial car tires Speed (km/hr) Virginia Tech 13 Example Problem # 1 Find the operating speed schedule of a maglev train operating from Orlando International Airport to the Main Plaza at Epcot Center in Disneyworld. The MBB (Krause Maffei) maglev has a linear thrustvelocity diagram as shown in Table 1 (if necessary linear extrapolation beyond 89.4 m/s is possible). Using a minimum constant speed segment time of 60 seconds and a final deceleration of 0.1 g's to estimate the velocity-time profile of the vehicle. Assume that grades are zero throughout the route. Florida is pretty flat like a tortilla. Virginia Tech 14 Maglev Train Parameters The following parameters apply to the Maglev train. S=25 m2, froll=.025 (magnetic rolling friction),ρ=1.225 kg/cu-m., CD=.55,Cl=.20,g=9.81 m/s2 and m=350,000 kg MAGLEV Thrust (or tractive effort) Parameters for the Hypothetical Orlando Line. Speed (m/s) Thrust Force (N) 0.00 300,000 89.40 150,000 Virginia Tech 15 Analysis of the Maglev Train Simplify the basic equation of motion by eliminating the angle between the track and the horizontal, 1 ma x = f ( V, ρ ) – -- ρ V 2 SC D – ( mg – L ) f roll 2 Substitute the numerical values for the train in equation (6) to obtain, dV --- ( 214, 163 – 1, 678 V – 8.345 V 2 ) ----- = a t = 1 m dt Virginia Tech (8) 16 Integration of Equation of Motion This can be done both numerically or analytically Analytical: the equation is a perfect quadratic so there is a solution to the definite integral (consult CRC tables for example) t V dV -------------------------------------------------------------------------------- = ∫ dt ∫ 1 2 --- 214, 163 – 1, 678 V – 8.345 V 0 Vo m (9) The denominator is of the quadratic form: A + BV + CV 2 Virginia Tech 17 Numerical Integration A general procedure to solve any complex differential equation encountered in practice. Start with, 1 dV --- ( 214, 163 – 1, 678 V – 8.345 V 2 ) ----- = a t = m dt then take small increments of time into the future and dV integrate ----dt Breaking down the path into small intervals is the key to the numerical integration process Virginia Tech 18 How it Works Virginia Tech (A.A. Trani) 19 Graphical Interpretation • The numerical integration procedure is illustrated graphically in the diagram • Note: the Euler integration scheme assumes a constant value for dV ⁄ dt in the interval ( t – ∆ t , t ) • To achieve good accuracy make the step size ( ∆ t ) small watching that the round-off errors in the computation do not become excessive Virginia Tech 20 Finding the Distance Traveled The distance traveled is the second integral of the acceleration function or the integral of the vehicle velocity function d S S t = S t – ∆t + ----∆ t = S t – ∆t + V t ∆ t dt ( t – ∆t, t ) Note that we are trying to find the distance to accelerate (named S1 hereon) and reach some desired cruise speed ( V cruise ) Virginia Tech 21 More Analysis for Maglev Train A sketch of the velocity profile is shown in the figure illustrating three possible regimes of motion where the vehicle accelerates, cruises and then decelerates to come to a full stop at the station Area under the speed vs time curve is the distance traveled (S) Speed S2 S1 T1 Virginia Tech S3 T2 T3 Time 22 Cruise and Deceleration Segments • For second interval (i.e., constant speed segment) we estimate distances as a function of speed allowing 60 seconds of cruise time. S 2 = t cruise V cruise = 60 V cruise (10) For the deceleration segment assume a uniform deceleration (assume 0.1g or 1.0 m/s2), V – VS 3 = --------------2a 2 2 2 0 (11) Virginia Tech 23 Analysis • To find Vcruise and unknown distances S1, S2 and S3 we use a numerical integration procedure (i.e., Euler, modified Euler or Runge-Kutta of fourth order) • The pseudocode to find distance, velocity and acceleration of the vehicle is shown in flowchart corresponding flowchart illustrating the main steps of the numerical integration procedure • Note that a maximum speed of 72.25 m/s is reached in the 16 km trajectory followed by the required 60 seconds cruising time. Note that this problem has a unique solution for Vcruise and thus the numerical integration technique advocated here is suitable for any type of acceleration function Virginia Tech 24 More Analysis • The travel time is estimated directly from the integration procedure or from the solution of basic travel time kinematic equations once each segment of the trajectory is known. • In this particular case the total travel time is known to be 342 seconds from the numerical integration procedure. The acceleration time is 204 seconds whereas the deceleration time is 78 seconds • As a matter of curiosity you can check that the maglev's maximum speed turns out to be around 90 m/s in this configuration (i.e., weight and aerodynamic characteristics) Virginia Tech 25 Numerical Results 1 : V elocit y 1: 2: 2: Dista nce 1 72.25 16090.03 1 1: 2: 2 2 35.75 8045.01 1 2 1 1: 2: -0.75 0.00 2 0.00 100.00 200.00 Time 300.00 400.00 Velocity units are m/s Distance units are m Virginia Tech 26 Force Diagram The force diagram shows all forces acting on the vehicle Data #2 350000 Te (N) 300000 Treq (N) Force (N) 250000 200000 150000 Maximum Speed 100000 50000 0 20 40 60 80 100 Speed (m/s) Virginia Tech 27 Energy Consumption The model shown described so far can be modified to estimate the energy consumed in the trajectory between two stations Knowing that the power required to move the vehicle at speed V is the product of the tractive force (or thrust) needed to move the vehicle at V and the speed itself. Mathematically this becomes, V -------P = TR η (12) where P is the power (Watts), TR is the tractive force required (in Newtons) to overcome the resistance forces Virginia Tech 28 opposing the motion of the vehicle,η is an efficiency factor (typically between 0.75 to 0.90 ground vehicles) and V is the speed (m/s). Also known is the fact that power is the time rate of doing work which has units of energy consumption (Joules), t E= ∫ Pdt (13) 0 where E is the energy consumed in Joules and P is the power required to overcome the motion of the vehicle. You can convert energy consumption into KWh by diving the result for E by 3.6 106. Virginia Tech 29 Numerical Integration Using Excel Excel or Matlab can be used to numerically integrate the expressions presented and do the analysis. 1 dV --- ( 214, 163 – 1, 678 V – 8.345 V 2 ) ----- = a t = m dt Set several columns to estimate values of time, velocity, rate of change of velocity, the desired step size interval (∆t), and the product of the step size and the rate of change of velocity (in that order) Virginia Tech 30 Excel Representation (Maglev Problem) Sample Formulas B3 = B2+E2 C3 = 1/350000*(214163-1678*B3-8.345*B3^2) D3 = A4-A3 E3 = D3 * C3 Virginia Tech 31 Observations • The Maglev train accelerates modestly reaching 5.96 m/s in about 10 seconds according to the sheet presented • Note that in order to have a better accuracy in the predictions we would want to use smaller step sizes (typically one second or less) • The acceleration capability of the train diminishes with speed Virginia Tech 32 General Models for Ground Vehicles • The characteristics of rail vehicles are very important in determining travel time, energy consumed and other measures of effectiveness in intercity travel • From our previous derivation of the basic equations of motion for transportation vehicles it is evident that two types of forces arise while trying to derive the performance of rail vehicles: Traction and Resistance • General models attempt to describe the performance using these two types of forces in the model Virginia Tech 33 Highway Vehicle Performance Estimation • Highway vehicles obtain their power from internal combustion engines. Two types of engines developed over the last century have dominated: 1) Four-cycle Internal Combustion (IC) and 2) Diesel engines • Define a simple procedure to extract performance from engine torque and engine speed (in revolutions per minute or RPM) diagrams. The discussion that follows is adopted primarily from Vuchic (1982) and Hucho (1998) • Performance curves for rubber-tired vehicles are usually expressed by manufacturers in terms of torque, power developed (e.g., shaft horsepower), and fuel consumption curves versus engine speed (N) • Virginia Tech 34 Trends • The general trend is that at higher N values the torque decreases and power output of the engine increases. • The fuel consumption has a minima at intermediate vehicle speeds (e.g., 45 m.p.h for the average car). • Since Thrust vs. vehicle speed (not engine speed) is not given directly it is necessary to derive a diagram that will approximate the general tractive effort (thrust) vs. speed function hypothesized in this course Virginia Tech 35 Procedure to Derive Force - Velocity Diagrams • The procedure requires power information from the engine or vehicle manufacturer. The speed of the vehicle (V), just like in rail technology, is related to the powerplant engine speed (N) by, V = (N)(D)(π) -------------------------Ju i (14) where: J is the differential reduction gearing ratio (usually varies from 3:1 to 6:1 for typical IC vehicle applications), u is the transmission gearing ratio for the ith gear (the value of u is high for low gears and low for high gears and varies from 5:1 to 1:1), D is the diameter of the tractive wheels, and N is the engine speed. i i Virginia Tech 36 Procedure - Force vs. Speed Diagrams Since N is usually expressed in Revolutions per Minute (RPM) a conversion will be necessary as D and V will usually take on units such as meters, and km/hr, respectively. Apply a conversion factor to change m/minute to km/hr in Equation (14) to get: 60 ( N ) ( D ) ( π ) V = ------------------------------1000 Ju i (15) where the units of each variable are: N in revolutions per minute, D in meters, J and u are dimensionless, and V is expressed in km/hr. i Virginia Tech 37 Force vs. Speed Diagrams Using the expression to convert power output to tractive effort we have, T = ηP -----V (16) where: P is the power output, η is the efficiency of the engine ( varies from 0.7 to 0.85 for most IC type applications), V is the vehicle speed and T is the thrust or tractive effort produced by the engine. Since P is usually expressed in horsepower (hp), V in km/ hr and T in Newtons, it is necessary to convert to proper units before applying this equation to car, truck and bus vehicles. Virginia Tech 38 Froce vs. Speed Diagrams Once the conversion factors are applied we have: T= ηP V 2650 ------ (17) where: T is expressed in Newtons P in horsepower V in km/hr Virginia Tech 39 Force vs. Speed Diagram The procedure to estimate force vs. speed performance is: • Compute V for several values of N and iterate for all values of u in Equation 15. i • Substitute the corresponding velocity (V) values into Equation 17 to estimate T. Since each vehicles has several feasible transmission gearing ratios, u , it is necessary to compute several values of T for each value N in Equation 15. i The resultant performance of a typical IC engine vehicle is shown in the following page. Virginia Tech 40 Power (hp) Given Estimate T (N) First Gear Second Gear Engine Speed - N - (RPM) Third Gear Fourth Gear Given: Vehicle by Manufacturer Vehicle Speed (m/s) 60 ( N ) ( D ) ( π ) Use V = ------------------------------- and T = 2650 η P -----1000 Ju V i Virginia Tech 41 Estimation of Vehicle Resistance Many manufacturers find useful to distinguish between the resistance and the propulsive forces required to do performance analysis This section presents typical equations found in texts or vehicle performance tables to familiarize yourself with some of the basic computations of ground vehicle performance The analysis that follows is similar to that explained for the Maglev vehicle except that resistance forces are treated independently in the computation. Sometimes this makes the analysis more clear. Virginia Tech 42 Resistance Formulas for Highway Vehicles According to the Society of Automotive Engineers (SAE), the basic resistance of a highway vehicle (in Newtons) can be represented by, R = [ c 1 + c 2 V ] W + c a AV 2 b (18) where: c 1, c 2 are coefficients known to have values of 7.6, and 0.056, respectively for buses and trucks (per SAE data); W is the total weight of the vehicle (kN), A is the frontal vehicle area (m2) and V is the vehicle speed in km/ hr., and R is the basic resistance in Newtons. b Virginia Tech 43 Vehicle Resistance A typical value for ca is .018 for buses and 0.020 for streamlined trucks (0.025 for non-streamlined trucks). Ground vehicles are also subjected to gradient resistance. Since equation 18 expresses the weight of the vehicle in kN the gradient resistance becomes, R G = 10 Wi (19) where: W is the weight in kN and i is the grade expressed as a percent. For a 100 kN bus this would imply 1000 Newtons of resistance for each one percent of gradient. Virginia Tech 44 Example # 2 - Urban Bus Analysis To illustrate the use of equations presented let us look at an urban bus vehicle traveling in hilly terrain (say Blacksburg). The vehicle considered is a Flxible bus. Flxible Bus (Blacksburg Transit) Virginia Tech 45 Bus Characteristics The technical characteristics of the bus are given below; Table 1: Flxible Bus Power Characteristics. N (RPM) 800 1200 1600 1800 2000 2200 2400 P (hp) 75 130 165 170 190 203 210 Transmission gearing ratios are: 4.5:1, 2.6:1, 1.5:1, and 1:1 for the four speed bus gear system. The differential ratio is constant at 4.6:1. The bus weighs 26,400 lbs (120 kN), the wheel diameter is 37 inches (.94 m.) and the frontal area of the bus is 6.97 m2 (75 ft2) . The resistance function for the bus is given by, Virginia Tech 46 R = [ c 1 + c 2 V ] W + c a AV 2 (20) The values for each constant are: ca = 0.020, c1 = 7.6, and c2 = 0.056; A surveying team estimates the most demanding grade of the Windsor Hills Route is 7%. The maximum speeds for each gear are as follows: V max for first gear = 20 km/hr V max for second gear = 35 km/hr V max for third gear = 55 km/hr V max for fourth gear = 85 km/hr Virginia Tech 47 With this information we construct a Force vs Speed diagram using the process explained in previous pages. Virginia Tech 48 Construction of TE vs. V Diagram a) Start with a low value of N and compute the power output from the engine (look at the power vs. N table or data) b) Estimate V for the value of N selected for every gear (note that ui changes for every gear) from equation (15) c) Find the value of T from equation (17) for a given value of P and V. Do this for every gear. d) Repeat the process (a-c) for other values of N. I suggest you repeat the process for at least 5-10 values of N A sample T vs. V diagram is shown below. Virginia Tech 49 Force vs. Speed Diagram for Flxible Bus x 104 3 First gear T (Newtons) Tractive Effort (N) 2.5 2 Second gear 1.5 Third gear TextEnd 1 Fourth gear 0.5 Total resistance (0% grade) 0 0 10 20 30 40 50 Speed (km/hr) 60 Virginia Tech 70 80 90 100 50 Notes • The first gear provides the highest value of T as expected • Higher gears provide modest T values • The resistance equation for zero grade is shown for illustration • Other resistance curves (say for other grade conditions) can be easily superimposed in the T vs. V diagram to show the maximum vehicle speed for any gradient condition • For example, if the grade is 5% the maximum speed the bus can achieve is only 43 km/hr as shown in the following diagram Virginia Tech 51 T vs V Diagram (5% grade resistance forces) 3 x 104 First gear TTractive Effort (N) (Newtons) 2.5 5 100 2 Second gear 1.5 TextEnd Third gear Total resistance (5% grade) Fourth gear 1 0.5 0 10 20 30 40 50 Speed (km/hr) Virginia Tech 60 70 80 90 100 52 Approximation of T vs. V Diagram x 104 3 First gear Approximation Profile 2.5 Tractive Effort (N) 2 Second gear 1.5 Third gear TextEnd 1 Fourth gear 0.5 Total resistance (0% grade) 0 0 10 20 30 40 50 Speed (km/hr) 60 70 80 90 100 Use linear segments to approximate the T vs. V curve. Virginia Tech 53 Finding Acceleration T vs. V Diagram x 104 3 2.5 Excess Force to Accelerate Tractive Effort (N) 2 1.5 TextEnd 1 0.5 Total resistance (0% grade) 0 0 10 20 30 40 50 Speed (km/hr) 60 Virginia Tech 70 80 90 100 54 Sample Kinematic Models Two models are useful in the analysis of ground transportation systems: a) Constant acceleration/deceleration) b) Variable acceleration/deceleration The use of the model depends on the characteristics of the system studied and thelevel of accuracy needed. Example 1: in the analysis of deceleration profiles for cars and trains we can use a constant deceleration model Example 2: in the analysis to determine the acceleration lane distance of a highway a variable acceleration model is recommended Virginia Tech 55 Kinematic Models dV/dt Constant acceleration/deceleration k V dV/dt k1 k2 k1 k1/k2 Virginia Tech V 56 Constant Acceleration Model Well known equations. the acceleration of the vehicle is equal to a constant ( k ), V d ---- = k dt Integrate once to get speed (V) and twice to get distance traveled (S) V t = V 0 + kt where: V t is the speed at time t, V 0 is the initial speed of the vehicle and t is the time from start of motion Virginia Tech 57 Constant Acceleration Model The distance traveled (S) is, 1 S t = V 0 t + -- kt 2 + S 0 2 where: S 0 is the intial position of the vehicle at time 0 Since k is just the acceleration of the vehicle this equation is equivalent to the well known relationship, 12 S t = V 0 t + -- at + S 0 2 where a is the constant acceleration of the vehicle Virginia Tech 58 Variable Acceleration Model Start with, V d ---- = k 1 – k 2 V dt Integrate once to get speed (V) –k t –k t k V t = ---1 ( 1 – e ) + V 0 e k2 2 2 where: V t is the speed at time t, V 0 is the initial speed of the vehicle, k 1 and k 2 are constants and t is the time from start of motion Virginia Tech 59 Variable Acceleration Model The distance traveled (S) is, –k t –k t k V - kS t = ---1 t – ---1 ( 1 – e ) + ----0 ( 1 – e ) k2 k2 k2 2 2 Virginia Tech 2 60 Problem 1 The Blacksburg Middle school board hires you as a transportation engineer to ease complaints from parents driving vehicles and making a left turn to the school entrance during the peak hour in the morning (see Figure below). The road is divided and has a left turn queueing island allowing cars to stop before making the turn. Measurements at the road by the town engineer indicate that traffic flows in this section at 35.7 km/hr (restricted by the speed limit). Virginia Tech 61 School Entrance To Blacksburg R = 12 m. Spacing = S Car Traffic Counters Car To Radford FIGURE 1. Blacksburg Middle School Traffic Situation. The typical acceleration model for a car is known to be: a = 4.0 – 0.1 V where: a is the acceleration of the car (in m/s2) and V is the vehicle speed in m/s. Virginia Tech 62 a) Find the typical spacing (S) and the average headway (h) between vehicles traveling from Radford to Blacksburg during the peak morning period. b) Find if the average headway (h) allows a typical driver to make a left turn if the driver has a perception/reaction time of 0.5 seconds. The radius of the curve to make a left turn is 12 meters. According to AASHTO standards, the critical vehicle length is 5.8 meters. Virginia Tech 63 Solution to Part (a) Find the Spacing (S) between vehicles. Since the density of the traffic flow is known to be 20 veh/km-la we compute the spacing as the reciprocal of the density 1S = 1 = ----- = 0.05 kilometers -k 20 S = 50 meters To find the headway we need to figure out how fast the cars are traveling on the road. We use Greenshield’s model to estimate the speed when k = 20 veh/km-la. Virginia Tech 64 Solution to Part (a) Traveling at 35.71 km/hr (9.92 m/s) the headway (h) is, 50 h = --------- = 5.04 seconds 9.92 Virginia Tech 65 Solution to Part (b) To check if the turning vehicle can make a safe maneuver, check the time to turn against the headway (h) calculated in part (a). Account for the reaction time of the turning vehicle. The time available to execute a safe turn is (h) - 0.5 seconds to account for reaction time, t available = 5.04 – 0.5 = 4.54 seconds Technically we should use the gap between two successive vehicles to estimate the time to turn left. In this case we have to subtract the time traveled by the oncoming vehicle to cover its car length at 9.92 m/s Virginia Tech 66 t gap 5.8- = 3.96 seconds = 5.04 – 0.5 – --------9.92 The distance traveled by a vehicle with a non-uniform acceleration is, –k t –k t k1 V0 k 1 t ---S = ----- – - ( 1 – e ) + ---- ( 1 – e ) 2 k2 k2 k2 2 2 Note that t is either 3.96 or 4.54 seconds (depending on your assumption on when the stopped vehicle starts the left turn). Using values of k 1 , k 2 of 4 and 0.1, respectively, the left turning vehicle travels 35.6 meters in 4.54 seconds and 28 meters in 3.96 seconds. Virginia Tech 67 A plot of distance traveled vs. time is shown in the following diagram. The total distance to be traveled in the left turn maneuver to reach a safe point is, 2 π R + L = 2 π ( 12 ) + 5.8 = 24.65 d = ------------------------meters 4 4 The vehicle can execute the turn safely. Virginia Tech 68 Distance vs. Time Profile (Turning Car) The car reaches 24.65 m in 3.73 seconds The car travels 35.65 m in 4.54 seconds Virginia Tech 69 Problem 2 A new Light Rail Transit (LRT) system to be installed in San Diego, California has a fully loaded weight (W) of 300 kilo-Newtons (kN). The LRT vehicle manufacturer states that the basic resistance (in Newtons) can be represented by a quadratic formula, R B = a + bV + cV 2 where: a , b , and c have been determined to be: 750 (Newtons), 18 (Newtons * hr/km) and 0.14 (Newtons * hr2/ km2), recpectively. V is the speed of the LRT train in km/hr. Virginia Tech 70 As usual, the vehicle grade resistance, RG (in Newtons) is expressed as, R G = 10 Wi where: W is the weight in kN and i is the grade expressed as percent (%). The tractive effort (TE) curve vs. speed for the same vehicle can be approximated by a line with negative slope, TE = 15, 000 – 70 V where: TE is in Newtons and V in km/hr. The tractive effort and total resistance curves as function of speed are shown in Figure 2 for flat terrain. Virginia Tech 71 Tractive Effort (N) Total Resistance (N) at zero grade FIGURE 2. Tractive Effort and Total Resistance Diagram for LRT Vehicle. Virginia Tech 72 Questions for Problem 2 a) Find the maximum acceleration the vehicle can generate while traveling at 40 km/hr in a +2% grade in San Diego. The basic resistance (RB) at 40 km/hr is, R B = 750 + 18 ( 40 ) + 0.14 ( 40 ) 2 = 1, 694 Newtons The grade resistance at +2% is estimated to be, R G = 10 ( 300 ) ( 2 ) = 6, 000 Newtons The tractive force at 40 km/hr is, Virginia Tech 73 TE = 15, 000 – 70 ( 40 ) = 12, 200 Newtons The acceleration is just the sum of forces divided by the mass of the vehicle, m 12, 200 N – ( 7, 694 N ) a = dV = ------------------------------------------------------ = 0.15 ---2 -----dt s m 9.81 --- ( 300, 000 N ) ⁄ s2 b) Find the maximum speed the LRT can develop in another application in Florida where the terrain is flat. By inspection of Figure 2 the maximum speed is 133 km/ hr. Virginia Tech 74 Plot of TE and Total Resistance vs. Speed Tractive Force Resistance at 2% grade Virginia Tech 75 ...
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