Lecture 5 - The Pennsylvania State University Department of...

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Unformatted text preview: The Pennsylvania State University Department of Civil and Environmental Engineering CE 321: Highway Engineering Lecture 5: Tractive Effort and Vehicle Acceleration Fall 2007 Overview Available tractive effort How much can the engine generate? How much can the tire/road deliver? Vehicle acceleration Maximum Tractive Effort At some point, the engine can develop additional power beyond which can be supported by the pavement / tire interface. At this point, the additional power will merely spin the wheels. No vehicle acceleration. Available Tractive Effort 1 2 Available Tractive Effort To Calculate Wr mah f r Wr = Ra h + Wl f cos g + mah Wh sin g L Available Tractive Effort To Calculate Wf mah f r Wf = Ra h - Wlr cos g + mah Wh sin g L Looking Back and Substituting Wr = Ra h + Wl f cos g + mah Wh sin g L YIELDS h Wr = W + ( Ra + ma + W sin g ) L L lf Looking Back and Substituting RECALL F = ma + Ra + Rrl + Rg YIELDS F - Rrl = ma + Ra + Rg Wr = lf L W+ Eq. 2.2 h ( Ra + ma + W sin g ) L h Wr = W + ( F - Rrl ) L L lf For a Rear Wheel Drive Car h Wr = W + ( F - Rrl ) L L lf Fmax = Wr FMAX W (l f - f rl h) L = h 1- L For a Front Wheel Drive Car W (lr + f rl h) L = h 1+ L FMAX Example Problem 1 A 2,500 lb car is at rest. It has a 120-inch wheel base. The center of gravity is 40-inches above the pavement and 40-inches behind the front axle. What is the maximum tractive effort if: a) b) the car is front wheel drive, the car is rear wheel drive. (Assume the coefficient of road adhesion is 0.6) Solution to Example Problem 1 (con't.) For the front-wheel-drive case: Fmax W(l r + f rl h) / L = 1+ h / L V 147 Eq. 2.5 Find frl : frl = 0.01 1 + V = 0 ft /s Therefore: fr = 0.01 Solution to Example Problem 1 (con't.) For the front-wheel-drive case: Fmax = {0.6 x 2,500 x [80 + 0.01(40)]} / 120 1 + (0.6 x 40) / 120 Fmax = 837.50 lbs Solution to Example Problem 1 (con't.) For the rear-wheel-drive case: Fmax W(lf - f rl h) / L = 1- h / L Fmax = {0.6 x 2,500 x [40 - 0.01(40)]} / 120 1 - (0.6 x 40) / 120 Fmax = 618.75 lbs Engine Generated Tractive Effort Common measures: Torque (work output) ft-lb Power = torque / time (rate of work) ft-lb / sec or horsepower (hp) Engine Generated Tractive Effort Problems with gasoline engines The tractive effort needed for acceptable vehicle performance is greatest at low speed Inside the engine, max torque is developed at high speeds Hence gear reductions (see Figure 2.5) Engine Generated Tractive Effort (Figure 2.5) Engine Generated Tractive Effort There are two factors that will determine the amount of tractive effort reaching the driving wheels: d = efficiency of driveline (Typically 0.75 to 0.95) 0 = gear reduction ratio (The relationship between the number of turns of the engine and the number of road wheel turns) i.e. 4:1 = 4 turns of the engine per every one turn of the wheel Engine Generated Tractive Effort Equation for Fe = engine generated tractive effort reaching drive wheels M e 0 d Fe = r Me = Torque Output (ft-lb) r = radius of the drive wheels (ft) d = mechanical efficiency 0 = Gear reduction ratio Engine Generated Tractive Effort Equation for relationship between vehicle speed and engine speed 2 r ne V = (1 - i) 0 V = vehicle speed (ft/sec) ne = engine speed (crankshaft rev/sec) i = engine slippage (typically 2 to 5 percent) Engine Generated Tractive Effort (cont) Tractive effort is the lesser of: Fe = engine tractive effort or FMAX = maximum tractive effort at pavement / tire interface Example Problem 2 A 2,200 lb car has an engine that produces 250 ft-lb of torque. The gear reduction ratio is 4.5 to 1.0. Gear efficiency is 80% and the wheel radius is 14 in. What is the tractive effort generated by the engine? Example Problem 2 Solution: M e 0 d Fe = r Fe = (250)(4.5)(0.8) (14/12) Fe = 771.4 lbs Vehicle Acceleration Vehicle Acceleration Tractive Effort can be used to determine acceleration and top speed. In acceleration, the force available to accelerate is: FNET = F - summation of resistances = ma Vehicle Acceleration (cont) Include a mass factor, gamma F - R = mm a ( m accounts for inertia of the vehicle's rotating parts) Where: m = 1.04 + 0.0025 o2 o = gear ratio Example Problem 3 A vehicle is traveling at 10 mi / hr on a snow covered surface (f = 0.20). The car has a 14 in wheel radius and a wheel base of 120 in. It has a center of gravity 40 in above the roadway surface and 50 in behind the front axle. ( CD = 0.3, Af = 20 ft2, W = 3,000 lbs and = 0.002045 slugs/ft3 ) What is the acceleration rate if: a) It is a front wheel drive? b) It is a rear wheel drive? Example Problem 3 (con't) First compute: the resistances tractive effort generated by the engine the mass factor These will be the same for both front and rear wheel drive vehicles Example Problem 3 (cont) The air resistance is (from Eq. 2.3): 2 Ra = Cd A f V 2 0.002045 (0.30)(20)(10 x 1.47)2 Ra = 2 Ra = 1.33 lbs Example Problem 3 (con't) The rolling resistance is (from Eq. 2.6): Rrl = f rlW 10 x 1.47 Rr = 0.01 1 + 147 x 3,000 Rr = 33.0 lbs Example Problem 3 (con't) The engine-generated tractive effort is (from Eq. 2.17): M e 0 d Fe = r Assume 0.75 to 0.95 Fe = (110)(4.5)(0.8) (14/12) Fe = 339.4 lbs Example Problem 3 (con't) The mass factor is (from Eq. 2.20): m = 1.04 + 0.0025 2 o 2 m = 1.04 + 0.0025 (4.5) m = 1.091 Example Problem 3 (con't) The available tractive effort, F is equal to the lesser of Fe or Fmax. For the front-wheel-drive car: Fmax W(l r + f rl h) / L = 1+ h / L Fmax = {(0.2)(3,000)[70 + 0.011(40)]} / 120 1 + (0.2)(40) / 120 Fmax = 330.1 lbs Example Problem 3 (con't) For the front-wheel-drive car, F = 330.1 lbs (the lesser of 339.4 lbs and 330.1 lbs) F - R = m ma 330.1 34.33 F-R a = mm = 1.091 x 3,000/32.2 = 2.91 ft/s2 Example Problem 3 (con't) For the rear-wheel-drive car, determine Fmax Fmax W(lf - f r h) / L = 1- h / L Fmax = {(0.2)(3,000)[50 - 0.011(40)]} / 120 1 - (0.2)(40) / 120 Fmax = 265.5 lbs Example Problem 3 (con't) For the rear-wheel-drive car, F = 265.5 lbs (the lesser of 339.4 lbs and 265.5 lbs) 265.5 34.33 F-R a = mm = 1.091 x 3,000/32.2 = 2.27 ft/s2 ...
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This note was uploaded on 04/11/2008 for the course C E 321 taught by Professor Pietrucha,martinkeller,michaelwi during the Spring '07 term at Pennsylvania State University, University Park.

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