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Build equations relating the power ?? and torque of the motor ?? to the number of revolutions of the motor ?? .... The relationship between the number of revolutions of the engine  e

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

1 Build equations relating the power (𝑷𝒆) and torque of the motor (𝑻𝒆) to the

number of revolutions of the motor (𝝎𝒆) 4

2 The working range when shifting gears: (𝝎𝟏,𝝎𝟐) in the power region reaching

90% PM 5

3 Design a gearbox with 5 gears according to the multiplier scheme, with a straight gear at hand 4 (n 4 = 1) 6

4 The relationship between the number of revolutions of the engine ( e ) - the speed

of the car (v) in each gear 8

5 Characterization of the relationship between traction torque (T w ) - vehicle speed (v) in each gear 8

6 The characteristic of traction relationship (F x ) - vehicle speed (v) in each gear 9

7 Determine the acceleration time of the car from v min = 0 to v = 100 km/h 11

Theory of automobile

3

Theory of automobile Midterm exam

11.5 in ≈ 292,1𝑐𝑚 73 kW 649 rad/s

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 5 gears according to the multiplier scheme, with a straight gear at hand 4 (n4 = 1), so that the acceleration time of the car from 𝑣𝑚𝑖𝑛 to v = 100 km/h is the

Theory of automobile

smallest (The car's speed reaches 100 km/h in straight gear (4th gear) at the maximum number of revolutions 2 of the engine)

4) Graph the relationship between the number of revolutions of the engine (𝜔𝑒) - the speed of the car (v) in each gear

5) Characterization of the relationship between traction torque (𝑇𝑊) - vehicle speed (v) in each gear

6) 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

7) 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= 0 to v = 100 km/h

• Tire parameters: 185/55R15

- Tire height of a tire:

ℎ𝑇 = 185 𝑥 55% = 101,75 𝑚𝑚 ≈ 4 𝑖𝑛

- The radius of a tire 185/55R15:

𝑅𝑤 = 2ℎ𝑇+

2𝑥4 + 15

2 = 11,5 𝑖𝑛 ≈ 292,1 𝑚𝑚

1 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:

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5

𝑃𝑒 = ∑ 𝑃𝑖 𝜔𝑒𝑖 = 𝑃1 𝜔𝑒 + 𝑃2 𝜔𝑒2

3

𝑖=1

+ 𝑃3 𝜔𝑒3 (3.1)

𝑃1 = 𝑃𝑀

𝜔𝑀 𝑃2 =

𝑃𝑀

𝜔𝑀2 𝑃3 = −

𝑃𝑀

𝜔𝑀3 (3.2) Substitude into (3.2):

𝑃1 = 73000

649 = 112,48 (𝑊𝑠) 𝑃2 =

73000

649 2 = 0,17331 (𝑊𝑠 2 )

𝑃3 = −73000

649 3 = −2,67𝑥10−4(𝑊𝑠 3 )

𝑃𝑒 = 112,48𝜔𝑒+ 0,17331𝜔𝑒2 − 2,67𝑥 10−4 𝜔𝑒3(W)

- The torque of the motor is determined by the following formula:

𝑇𝑒 = 𝑃𝑒

𝜔𝑒 = 𝑃1+ 𝑃2 𝜔𝑒 + 𝑃3 𝜔𝑒

2 (3.5)

Substitute into (3.5), the torque performance equation:

𝑇𝑒 = 112,48 + 0,17331𝜔𝑒 − 2,67𝑥10 −4 𝜔𝑒2

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+ 0,17331𝜔𝑒2 − 2,67𝑥 10−4 𝜔𝑒3= 65700 W  1 = 494 rad/s;  2 = 787 rad/s

Substitute e = 0 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 1

Theory of automobile

Figure 1: Engine’s external characteristics

3 Design a gearbox with 5 gears according to the multiplier scheme, with a straight gear

at hand 4 (n 4 = 1)

Determine the gear ratio of the main transmission n d :

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= 𝑅𝑤𝜔2

𝑛𝑑 ⇒ 𝑛𝑑 =

𝑅𝑤𝜔2

𝑣4𝑚𝑎𝑥 =

0,292 ∗ 787 27,78 = 8,272

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 =𝑛𝑖𝑛𝑑

𝑅𝑤 𝑣𝑥 ⇒ 𝑣𝑥 =

𝑅𝑤𝜔𝑒

𝑛𝑖𝑛𝑑 ; 𝑛𝑖 =

𝑅𝑤𝜔𝑒

𝑣𝑥𝑛𝑑

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7

• At vx = 27.78 m/s, e = 2 = 787 rad/s, the gearbox operates in number 4: n4 = 1

At the 4th gear:

𝑣𝑥,4 = 𝑅𝑤𝜔𝑒

𝑛4𝑛𝑑 = 𝑅𝑤𝜔𝑒

𝑛𝑑 (4*)

At  e =  2 = 787 rad/s, we have: 𝑣𝑥,4 = 𝑅𝑤 𝜔𝑒

𝑛4𝑛𝑑 = 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

We have: 𝑛3=𝑅𝑤 𝜔2

𝑣𝑥𝑛𝑑 = 0.292∗787

17.438∗8.272 = 1,5931

• At the 3rd gear:

𝑣𝑥,3 =𝑅𝑤 𝜔𝑒

𝑛3𝑛𝑑 = 𝑅𝑤 𝜔𝑒

1,5931∗𝑛𝑑 (3*)

At  e =  1 = 494 rad/s, at the 3rd gear => v x,3 =𝑅𝑤𝜔𝑒

𝑛 3 𝑛𝑑 = 𝑅𝑤 𝜔𝑒

1,5929∗𝑛𝑑= 0,292∗494

1,5931∗8,272 = 10,946 𝑚/𝑠

The gearbox shifts to 2nd gear, the engine jumps to the high revolution e = 2 = 787 rad/s

We have: 𝑛2 =𝑅𝑤 𝜔2

𝑣 𝑥 𝑛𝑑 = 0,292∗787

10,946∗8.272 = 2.538

• At the 2nd gear:

𝑣𝑥,2= 𝑅𝑤 𝜔 𝑒

𝑛2𝑛𝑑 = 𝑅𝑤 𝜔 𝑒

2,538∗𝑛𝑑 (2*)

At e = 1 = 494 rad/s, ở tay số 2 => 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:

𝑛 1 =𝑅𝑤𝜔2

𝑣𝑥𝑛𝑑 =

0,292 ∗ 787 6,87 ∗ 8,272= 4.044

• At the 1st gear:

𝑣𝑥,1 =𝑅𝑤 𝜔𝑒

𝑛 1 𝑛𝑑 = 𝑅𝑤 𝜔𝑒

4,044𝑛𝑑 (1*)

At e = 1 = 494 rad/s, => vx,1 = 4,312 m/s

The 5th gear at e = 1 = 494 rad/s

𝑛5=𝑅𝑤𝜔𝑒

𝑣𝑥𝑛𝑑 =

0,292 ∗ 494 27,78 ∗ 8,272= 0.628

Theory of automobile

At  e =  1 = 494 rad/s: 𝑣𝑥,5 =𝑅𝑤 𝜔 𝑒

𝑛5𝑛𝑑 = 𝑅𝑤 𝜔 𝑒

0,628∗𝑛𝑑= 27,78 (m/s) (5*)

So, the gear ratio of the gearbox:

n1 = 4,044; n2 = 2,538; n3 = 1,5931; n4 = 1; n5 = 0,628

4 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 = 0 to 900 rad/s)

5 Characterization of the relationship between traction torque (T w ) - 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, 2 ,5):

𝑇w,𝑖 = 𝜂𝑃1𝑛𝑑𝑛𝑖 + 𝜂𝑃2

𝑅𝑤𝑛𝑑2𝑛𝑖2𝑣𝑥+ 𝜂𝑃3

𝑅𝑤2 𝑛𝑑3𝑛𝑖3𝑣𝑥2 (3.59)

• At n1 = 4,044; vx,1 = 4,312 m/s

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9

𝑇w,1 = 𝜂𝑃1𝑛𝑑𝑛1+ 𝜂𝑃2

𝑅𝑤𝑛𝑑2𝑛12𝑣𝑥,1+ 𝜂𝑃3

𝑅𝑤2 𝑛𝑑3𝑛13𝑣𝑥,12 = 4002 (Nm)

• At n2 = 2,538; vx,2 = 6,87 m/s

𝑇w,2= 𝜂𝑃1𝑛𝑑𝑛2 + 𝜂𝑃2

𝑅 𝑤 𝑛𝑑2𝑛22𝑣𝑥,2+ 𝜂 𝑃3

𝑅𝑤2 𝑛𝑑3𝑛23𝑣𝑥,22 = 2512 (Nm)

• At n3 = 1,5931; vx,3 = 10,946 m/s

𝑇w,3= 𝜂𝑃1𝑛𝑑𝑛3 + 𝜂𝑃2

𝑅𝑤𝑛𝑑2𝑛32 𝑣𝑥,3+ 𝜂 𝑃3

𝑅𝑤2 𝑛𝑑3𝑛33𝑣𝑥,32 = 1576 (Nm)

• At n4 = 1; vx,4 = 17,438 m/s

𝑇w,4 = 𝜂𝑃1𝑛𝑑𝑛4+ 𝜂 𝑃2

𝑅𝑤𝑛𝑑2𝑛42𝑣𝑥,4+ 𝜂 𝑃3

𝑅𝑤2 𝑛𝑑3𝑛43𝑣𝑥,42 = 990 (Nm)

• At n5 = 0,628; vx,5 = 27,78 m/s

𝑇w,5 = 𝜂𝑃1𝑛𝑑𝑛5+ 𝜂 𝑃2

𝑅 𝑤 𝑛𝑑2𝑛52𝑣𝑥,5+ 𝜂 𝑃3

𝑅𝑤2 𝑛𝑑3𝑛53𝑣𝑥,52 = 621 (Nm)

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

Figure 3:The relationship between traction torque - vehicle speed

6 The characteristic of traction relationship (F x ) - vehicle speed (v) in each gear

Traction at the active wheel in each gear is determined by the formula:

w

i i

T F R

= (3.51)

Combining formula (3.59), we have:

0 1000

2000

3000

4000

5000

6000

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)

The relationship between traction tourue - speed at each gear

Theory of automobile

𝐹x,𝑖 = 𝑇w,𝑖

𝑅 𝑤 = 𝜂𝑃1 𝑛𝑑𝑛𝑖

𝑅 𝑤 + 𝜂 𝑃2

𝑅𝑤2 𝑛𝑑2𝑛𝑖2𝑣𝑥+ 𝜂 𝑃3

𝑅𝑤3 𝑛𝑑3𝑛𝑖3𝑣𝑥2 (6*)

Substitude into (6*) we have:

𝐹x,1 = 𝜂𝑃1𝑛𝑑𝑛1

𝑅𝑤 + 𝜂

𝑃2

𝑅𝑤2 𝑛𝑑2𝑛12𝑣𝑥+ 𝜂 𝑃3

𝑅𝑤3 𝑛𝑑3𝑛13𝑣𝑥2 = 11597,3 + 2047𝑣𝑥− 361,3𝑣𝑥2

𝐹x,2 = 𝜂𝑃1 𝑛𝑑𝑛2

𝑅 𝑤 + 𝜂 𝑃2

𝑅𝑤2 𝑛𝑑2𝑛22𝑣𝑥+ 𝜂 𝑃3

𝑅𝑤3 𝑛𝑑3𝑛23𝑣𝑥2 = 7278,4 + 806,3𝑣𝑥− 89,3𝑣𝑥2

𝐹x,3 =𝜂𝑃1 𝑛𝑑𝑛3

𝑅 𝑤 + 𝜂 𝑃2

𝑅𝑤2 𝑛𝑑2𝑛32𝑣𝑥+ 𝜂𝑃3

𝑅𝑤3 𝑛𝑑3𝑛33𝑣𝑥2 = 4568,6 + 317,7𝑣𝑥− 22,1𝑣𝑥2

𝐹x,4 =𝜂𝑃1𝑛𝑑𝑛4

𝑅𝑤 + 𝜂

𝑃2

𝑅𝑤2 𝑛𝑑2𝑛42𝑣𝑥+ 𝜂 𝑃3

𝑅𝑤3 𝑛𝑑3𝑛43𝑣𝑥2 = 2867,8 + 125,2𝑣𝑥− 5,5𝑣𝑥2

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

Figure 4: The relationship between traction force - vehicle speed

0.0 2000.0

4000.0

6000.0

8000.0

10000.0

12000.0

14000.0

16000.0

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)

The relationship between traction force - speed at each gear

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11

7 Determine the acceleration time of the car from v min = 0 to v = 100 km/h

At vxmax, ni = n5:

𝐹𝑅 = 𝐹𝑥= 𝜂𝑛𝑖 𝑛𝑑

𝑅 𝑤

𝑃 𝑒

𝜔 𝑒 = 0.9 ∗0,628∗8,272

0.292 ∗65,7

787 = 1,336(𝑘𝑁) = 1336(𝑁)

• Acceleration time from v = 0 to v = 6,87 m/s (tay số 1): m=1379 kg

𝐹𝑥,1− 𝐹𝑅𝑑𝑣𝑥

6,87

0

= 0,83𝑠

• Acceleration time from v = 6,87 to v = 10,946 m/s (tay số 2):

𝑡2 = 𝑚 ∫ 𝐹 1

𝑥,2 −𝐹 𝑅 𝑑𝑣𝑥

10,946

• Acceleration time from v = 10,946 to v = 17,438 m/s (tay số 3):

𝐹𝑥,3− 𝐹𝑅𝑑𝑣𝑥

17,438

10,946

= 2,89𝑠

• Acceleration time from v = 17,438 to v = 27,78 m/s (100 km/h) (tay số 4):

𝐹𝑥,4− 𝐹𝑅𝑑𝑣𝑥

27,78

17,438

= 10,21𝑠

• Total acceleration time from 0 đến 100 km/h:

𝑡 = 𝑡1 + 𝑡2+ 𝑡3+ 𝑡4 = 14,906(𝑠)