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HANOI UNIVERSITY OF SCIENCE AND TECHNOLOGY SCHOOL OF TRANSPORTATION ENGINEERING ***** THEORY OF AUTOMOBILE MID-TERM PROJECT Instructor: Minh Hoang Trinh Class’s ID: 128355 Student: Vi Van Nguyen Student’s ID: 20186132 Hanoi, February -2022 Theory of automobile Table of Contents Build equations relating the power (𝑷𝒆) and torque of the motor (𝑻𝒆) to the number of revolutions of the motor (𝝎𝒆) The working range when shifting gears: (𝝎𝟏,𝝎𝟐) in the power region reaching 90% PM Design a gearbox with gears according to the multiplier scheme, with a straight gear at hand (n4 = 1) The relationship between the number of revolutions of the engine (e) - the speed of the car (v) in each gear Characterization of the relationship between traction torque (Tw) - vehicle speed (v) in each gear The characteristic of traction relationship (Fx) - vehicle speed (v) in each gear Determine the acceleration time of the car from vmin= to v = 100 km/h 11 Nguyen Van Vi 20186132 Theory of automobile Theory of automobile Midterm exam TT Vehicle 36 Vinfast Fadil m (kg) Tire 1379 185/55R15 11.5 in ≈ 292,1𝑐𝑚 𝑃𝑀 (hp) 98 73 kW 𝜔𝑀 (rpm) 6200 649 rad/s 𝜂 0.9 Requirement: 1) Build equations relating the power (𝑃𝑒 ) and torque of the motor (𝑇𝑒 ) to the number of revolutions of the motor (𝜔𝑒 ) 2) 2) Determine the working range when shifting gears: (𝜔1 , 𝜔2 ) in the power region reaching 90% PM; Plot the power characteristic (𝑃𝑒 ), torque (𝑇𝑒 ) against the number of revolutions of the motor (𝜔𝑒 ), mark the main working area 3) Design a gearbox with gears according to the multiplier scheme, with a straight gear at hand (n4 = 1), so that the acceleration time of the car from 𝑣𝑚𝑖𝑛 to v = 100 km/h is the Nguyen Van Vi 20186132 Theory of automobile 4) 5) 6) 7) smallest (The car's speed reaches 100 km/h in straight gear (4 th gear) at the maximum number of revolutions of the engine) Graph the relationship between the number of revolutions of the engine (𝜔𝑒 ) - the speed of the car (v) in each gear Characterization of the relationship between traction torque (𝑇𝑊 ) - vehicle speed (v) in each gear Assuming constant rolling resistance (the maximum traction force balances the total rolling resistance force when the vehicle is running at the maximum speed), Building the characteristic of traction relationship (𝐹𝑥 ) - vehicle speed (v) in each gear Assuming that the time to first gear is ignored and the time of gear shifting operation is ignored, determine the acceleration time of the car from 𝑣𝑚𝑖𝑛 = to v = 100 km/h • Tire parameters: 185/55R15 - Tire height of a tire: ℎ𝑇 = 185 𝑥 55% = 101,75 𝑚𝑚 ≈ 𝑖𝑛 - The radius of a tire 185/55R15: 𝑅𝑤 = 2ℎ𝑇 + 2𝑥4 + 15 = = 11,5 𝑖𝑛 ≈ 292,1 𝑚𝑚 2 Build equations relating the power (𝑷𝒆) and torque of the motor (𝑻𝒆) to the number of revolutions of the motor (𝝎𝒆) The power of the motor is determined by the following formula: Nguyen Van Vi 20186132 Theory of automobile 𝑃𝑒 = ∑ 𝑃𝑖 𝜔𝑒𝑖 = 𝑃1 𝜔𝑒 + 𝑃2 𝜔𝑒2 + 𝑃3 𝜔𝑒3 (3.1) 𝑖=1 𝑃1 = 𝑃𝑀 𝜔𝑀 𝑃2 = 𝑃𝑀 𝜔𝑀 𝑃3 = − 𝑃𝑀 𝜔𝑀 (3.2) Substitude into (3.2): 73000 73000 = 112,48 (𝑊𝑠) 𝑃2 = = 0,17331 (𝑊𝑠2 ) 649 6492 73000 𝑃3 = − = −2,67𝑥10−4 (𝑊𝑠3 ) 6493 𝑃1 = 𝑃𝑒 = 112,48𝜔𝑒 + 0,17331𝜔𝑒 − 2,67𝑥 10−4 𝜔𝑒 (W) - The torque of the motor is determined by the following formula: 𝑇𝑒 = 𝑃𝑒 = 𝑃1 + 𝑃2 𝜔𝑒 + 𝑃3 𝜔𝑒2 𝜔𝑒 (3.5) Substitute into (3.5), the torque performance equation: 𝑇𝑒 = 112,48 + 0,17331𝜔𝑒 − 2,67𝑥10−4 𝜔𝑒 2 The working range when shifting gears: (𝝎𝟏,𝝎𝟐) in the power region reaching 90% PM From equation (3.1), with P = 90%.PM = 73 (kW), the range of revolution allowed to reach this power is from 494 rad/s (≈ 4717 rpm) to 787 rad/s (≈ 7515 rpm) 𝑃𝑒 = 112,48𝜔𝑒 + 0,17331𝜔𝑒 − 2,67𝑥 10−4 𝜔𝑒 = 65700 W 1 = 494 rad/s; 2 = 787 rad/s Substitute e = to 900 rad/s, we can draw power characteristic graphs (Pe), torque (Te) against the number of revolutions of the motor (we), mark the main working of the engine in the figure Nguyen Van Vi 20186132 Theory of automobile Figure 1: Engine’s external characteristics Design a gearbox with gears according to the multiplier scheme, with a straight gear at hand (n4 = 1) Determine the gear ratio of the main transmission nd: According to the problem, the car speed reaches 100km/h (27.78 m/s) at the straight gear (number 4) at the maximum number of revolutions we of the engine, we have: 𝑣4𝑚𝑎𝑥 = 𝑅𝑤 𝜔2 𝑅𝑤 𝜔2 0,292 ∗ 787 ⇒ 𝑛𝑑 = = = 8,272 𝑛𝑑 𝑣4𝑚𝑎𝑥 27,78 Determine the gear ratio in the gears: The relationship between the number of engine revolutions and the speed in the ith gear according to the formula: 𝜔𝑒 = 𝑛𝑖 𝑛𝑑 𝑅𝑤 𝜔𝑒 𝑅𝑤 𝜔𝑒 𝑣𝑥 ⇒ 𝑣𝑥 = ; 𝑛𝑖 = 𝑅𝑤 𝑛𝑖 𝑛𝑑 𝑣𝑥 𝑛𝑑 Nguyen Van Vi 20186132 Theory of automobile • At vx = 27.78 m/s, e = 2 = 787 rad/s, the gearbox operates in number 4: n4 = At the 4th gear: 𝑣𝑥,4 = 𝑅𝑤 𝜔𝑒 𝑛4 𝑛𝑑 At e = 2 = 787 rad/s, we have: 𝑣𝑥,4 = = 𝑅𝑤 𝜔𝑒 𝑛4 𝑛 𝑑 𝑅𝑤 𝜔𝑒 𝑛𝑑 = (4*) 𝑅𝑤 𝜔𝑒 𝑛𝑑 = 0.292∗494 8,272 = 17,438 𝑚/𝑠 The gearbox shifts to 3rd gear, the engine jumps to the high revolution e = 2 = 787 rad/s 𝑛3 = We have: 𝑅𝑤 𝜔2 𝑣𝑥 𝑛𝑑 0.292∗787 = 17.438∗8.272 = 1,5931 • At the 3rd gear: 𝑣𝑥,3 = 𝑅𝑤 𝜔𝑒 𝑛 𝑛𝑑 = 𝑅𝑤 𝜔𝑒 1,5931∗𝑛𝑑 (3*) 𝑅 𝜔 𝑅 𝜔 0,292∗494 𝑤 𝑒 At e = 1 = 494 rad/s, at the 3rd gear => vx,3 = 𝑛𝑤𝑛 𝑒 = 1,5929∗𝑛 = 1,5931∗8,272 = 10,946 𝑚/𝑠 𝑑 𝑑 The gearbox shifts to 2nd gear, the engine jumps to the high revolution e = 2 = 787 rad/s We have: 𝑛2 = 𝑅𝑤 𝜔2 𝑣𝑥 𝑛𝑑 0,292∗787 = 10,946∗8.272 = 2.538 • At the 2nd gear: 𝑣𝑥,2 = 𝑅𝑤 𝜔𝑒 𝑛2 𝑛𝑑 𝑅 𝜔 𝑤 𝑒 = 2,538∗𝑛 (2*) 𝑑 At e = 1 = 494 rad/s, tay số => vx,2 = 6.87 m/s The gearbox shifts to 1st gear, the engine jumps to the high revolution: e = 2 = 787 rad/s We have: 𝑛1 = • 𝑅𝑤 𝜔2 0,292 ∗ 787 = = 4.044 𝑣𝑥 𝑛𝑑 6,87 ∗ 8,272 At the 1st gear: 𝑣𝑥,1 = 𝑅𝑤 𝜔𝑒 𝑛1 𝑛𝑑 𝑅 𝜔 𝑤 𝑒 = 4,044𝑛 𝑑 (1*) At e = 1 = 494 rad/s, => vx,1 = 4,312 m/s The 5th gear at e = 1 = 494 rad/s 𝑛5 = Nguyen Van Vi 20186132 𝑅𝑤 𝜔𝑒 0,292 ∗ 494 = = 0.628 𝑣𝑥 𝑛𝑑 27,78 ∗ 8,272 Theory of automobile At e = 1 = 494 rad/s: 𝑣𝑥,5 = 𝑅𝑤 𝜔𝑒 𝑛5 𝑛𝑑 𝑅 𝜔 𝑤 𝑒 = 0,628∗𝑛 = 27,78 (m/s) (5*) 𝑑 So, the gear ratio of the gearbox: n1 = 4,044; n2 = 2,538; n3 = 1,5931; n4 = 1; n5 = 0,628 The relationship between the number of revolutions of the engine (e) - the speed of the car (v) in each gear Figure 2: The relationship between the number of revolutions and the speed From equations (1*), (2*), (3*), (4*) and (5*), we can draw graphs of the relationship between the number of revolutions of the engine (e) and the speed of the car in each gear (Substitute with e = to 900 rad/s) Characterization of the relationship between traction torque (Tw) - vehicle speed (v) in each gear Traction torque (Tw) in each gear is calculated according to the formula (3.59) with the velocity relationship vx in each gear (i = 1, ,5): 𝑃 𝑃 𝑇w,𝑖 = 𝜂𝑃1 𝑛𝑑 𝑛𝑖 + 𝜂 𝑅2 𝑛𝑑2 𝑛𝑖2 𝑣𝑥 + 𝜂 𝑅23 𝑛𝑑3 𝑛𝑖3 𝑣𝑥2 (3.59) 𝑤 𝑤 • At n1 = 4,044; vx,1 = 4,312 m/s Nguyen Van Vi 20186132 Theory of automobile 𝑃 𝑃 𝑇w,1 = 𝜂𝑃1 𝑛𝑑 𝑛1 + 𝜂 𝑅 𝑛𝑑2 𝑛12 𝑣𝑥,1 + 𝜂 𝑅32 𝑛𝑑3 𝑛13 𝑣𝑥,1 = 4002 (Nm) 𝑤 𝑤 • At n2 = 2,538; vx,2 = 6,87 m/s 𝑃 𝑃 𝑇w,2 = 𝜂𝑃1 𝑛𝑑 𝑛2 + 𝜂 𝑅2 𝑛𝑑2 𝑛22 𝑣𝑥,2 + 𝜂 𝑅23 𝑛𝑑3 𝑛23 𝑣𝑥,2 = 2512 (Nm) 𝑤 𝑤 • At n3 = 1,5931; vx,3 = 10,946 m/s 𝑃 𝑃 𝑇w,3 = 𝜂𝑃1 𝑛𝑑 𝑛3 + 𝜂 𝑅2 𝑛𝑑2 𝑛32 𝑣𝑥,3 + 𝜂 𝑅32 𝑛𝑑3 𝑛33 𝑣𝑥,3 = 1576 (Nm) 𝑤 𝑤 𝑃 𝑃 • At n4 = 1; vx,4 = 17,438 m/s 𝑇w,4 = 𝜂𝑃1 𝑛𝑑 𝑛4 + 𝜂 𝑅2 𝑛𝑑2 𝑛42 𝑣𝑥,4 + 𝜂 𝑅23 𝑛𝑑3 𝑛43 𝑣𝑥,4 = 990 (Nm) 𝑤 𝑤 • At n5 = 0,628; vx,5 = 27,78 m/s 𝑃 𝑃 𝑇w,5 = 𝜂𝑃1 𝑛𝑑 𝑛5 + 𝜂 𝑅2 𝑛𝑑2 𝑛52 𝑣𝑥,5 + 𝜂 𝑅23 𝑛𝑑3 𝑛53 𝑣𝑥,5 = 621 (Nm) 𝑤 𝑤 Combined with the values vx,i (the formulas (1*), (2*) to (5*) above), we can draw the characteristic lines of the drag torque (Tw) - car speed (v) in each gear The relationship between traction tourue - speed at each gear 6000 Tw (Nm) 5000 4000 3000 2000 1000 0.00 5.00 10.00 15.00 20.00 25.00 30.00 35.00 40.00 45.00 50.00 vx (m/s) Tw1 Tw2 Tw3 Tw4 Tw5 Figure 3:The relationship between traction torque - vehicle speed The characteristic of traction relationship (Fx) - vehicle speed (v) in each gear Traction at the active wheel in each gear is determined by the formula: Fx,i = Tw,i Rw (3.51) Combining formula (3.59), we have: Nguyen Van Vi 20186132 Theory of automobile 𝐹x,𝑖 = 𝑇w,𝑖 = 𝑅𝑤 𝜂𝑃1 𝑛𝑑 𝑛𝑖 𝑅𝑤 𝑃 𝑃 + 𝜂 𝑅22 𝑛𝑑2 𝑛𝑖2 𝑣𝑥 + 𝜂 𝑅33 𝑛𝑑3 𝑛𝑖3 𝑣𝑥2 (6*) 𝑤 𝑤 Substitude into (6*) we have: 𝐹x,1 = 𝜂𝑃1 𝑛𝑑 𝑛1 𝑃2 𝑃3 + 𝜂 𝑛𝑑2 𝑛12 𝑣𝑥 + 𝜂 𝑛𝑑3 𝑛13 𝑣𝑥2 = 11597,3 + 2047𝑣𝑥 − 361,3𝑣𝑥2 𝑅𝑤 𝑅𝑤 𝑅𝑤 𝜂𝑃1 𝑛𝑑 𝑛2 𝐹x,2 = 𝑅𝑤 𝐹x,3 = 𝐹x,4 = 𝑃 𝑤 𝜂𝑃1 𝑛𝑑 𝑛3 𝑅𝑤 𝑃 + 𝜂 𝑅22 𝑛𝑑2 𝑛22 𝑣𝑥 + 𝜂 𝑅33 𝑛𝑑3 𝑛23 𝑣𝑥2 = 7278,4 + 806,3𝑣𝑥 − 89,3𝑣𝑥2 𝑤 𝑃 𝑃 + 𝜂 𝑅22 𝑛𝑑2 𝑛32 𝑣𝑥 + 𝜂 𝑅33 𝑛𝑑3 𝑛33 𝑣𝑥2 = 4568,6 + 317,7𝑣𝑥 − 22,1𝑣𝑥2 𝑤 𝑤 𝜂𝑃1 𝑛𝑑 𝑛4 𝑃2 𝑃3 + 𝜂 𝑛𝑑2 𝑛42 𝑣𝑥 + 𝜂 𝑛𝑑3 𝑛43 𝑣𝑥2 = 2867,8 + 125,2𝑣𝑥 − 5,5𝑣𝑥2 𝑅𝑤 𝑅𝑤 𝑅𝑤 Combined with the values vx,i (formulas (1*), (2*) to (5*) above), it is possible to draw characteristic lines of tractionv force (Fx) - vehicle speed (v) in each gear The relationship between traction force - speed at each gear 16000.0 14000.0 Fx (N) 12000.0 10000.0 8000.0 6000.0 4000.0 2000.0 0.0 0.00 5.00 10.00 15.00 20.00 25.00 30.00 35.00 40.00 vx (m/s) Fx1 Fx2 Fx3 Fx4 Fx5 Figure 4: The relationship between traction force - vehicle speed 10 Nguyen Van Vi 20186132 45.00 50.00 Theory of automobile Determine the acceleration time of the car from vmin= to v = 100 km/h At vxmax, ni = n5: 𝐹𝑅 = 𝐹𝑥 = 𝜂 𝑛𝑖 𝑛𝑑 𝑃𝑒 𝑅𝑤 𝜔𝑒 = 0.9 ∗ 0,628∗8,272 0.292 65,7 ∗ 787 = 1,336(𝑘𝑁) = 1336(𝑁) • Acceleration time from v = to v = 6,87 m/s (tay số 1): m=1379 kg 6,87 𝑡1 = 𝑚 ∫ 𝑑𝑣 = 0,83𝑠 𝐹𝑥,1 − 𝐹𝑅 𝑥 • Acceleration time from v = 6,87 to v = 10,946 m/s (tay số 2): 10,946 𝑡2 = 𝑚 ∫6,87 𝐹𝑥,2 −𝐹𝑅 𝑑𝑣𝑥 = 0,976𝑠 • Acceleration time from v = 10,946 to v = 17,438 m/s (tay số 3): 17,438 𝑡3 = 𝑚 ∫ 10,946 𝑑𝑣 = 2,89𝑠 𝐹𝑥,3 − 𝐹𝑅 𝑥 • Acceleration time from v = 17,438 to v = 27,78 m/s (100 km/h) (tay số 4): 27,78 𝑑𝑣𝑥 = 10,21𝑠 17,438 𝐹𝑥,4 − 𝐹𝑅 𝑡4 = 𝑚 ∫ • Total acceleration time from đến 100 km/h: 𝑡 = 𝑡1 + 𝑡2 + 𝑡3 + 𝑡4 = 14,906(𝑠) 11 Nguyen Van Vi 20186132