Tính toán thiết kế động cơ đốt trong

Bài report được đánh giá 8.7đ . Thiết kế trên lý thuyết hoặc tính toán lại 1 loại động cơ trên xe oto. Thông qua project này có thể giúp các sinh viên ngành công nghệ ký thuật oto biết được những bước cơ bản để thiết kế ra một động cơ mới. Ngoài ra, sinh viên còn biết sử dụng các ứng dụng hỗ trợ học tập như Matlab,...

HO CHI MINH CITY UNIVERSITY OF TECHNOLOGY AND

EDUCATION FACULTY FOR HIGH QUALITY TRAINING

FINAL REPORT TOPIC: ENGINE CALCULATION BASED ON TOYOTA

1NZ-FE

Major: Internal Combustion Engine Calculation

Adviser: Ly Vinh Dat

Ho Chi Minh City, December 2022

HO CHI MINH CITY UNIVERSITY OF TECHNOLOGY AND

EDUCATION FACULTY FOR HIGH QUALITY TRAINING

FINAL REPORT TOPIC: ENGINE CALCULATION BASED ON

TOYOTA 1NZ-FE

Major: Internal Combustion Engine Calculation

Adviser: Ly Vinh Dat

Ho Chi Minh City, December 2022

THE SOCIALIST REPUBLIC OF VIETNAM Independence – Freedom– Happiness

-

Ho Chi Minh City, December 4th, 2022

PROJECT ASSIGNMENT

Student name: Student ID:

Student name: Student ID:

Major: Internal Combustion Engine Calculation Class: 20145CLA

Advisor: Ly Vinh Dat Phone number:

Date of assignment: Date of submission:

1 Project title: Engine calculation based on 1NZ-FE

2 Initial materials provided by the advisor:

3 Content of the project:

4 Final product:

CHAIR OF THE PROGRAM

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ADVISOR

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THE SOCIALIST REPUBLIC OF VIETNAM

Independence – Freedom– Happiness

Ho Chi Minh City, December 4th, 2022

ADVISOR’S EVALUATION SHEET

Student name: Student ID:

Student name: Student ID:

Student name: Student ID:

Student name: Student ID:

Student name: Student ID:

Major:

Project title:

Advisor:

EVALUATION 1 Content of the project:

2 Strengths:

3 Weaknesses:

4 Approval for oral defense? (Approved or denied)

5 Overall evaluation: (Excellent, Good, Fair, Poor)

6 Mark: ……… (in words: )

Ho Chi Minh City, December 4th, 2022

ADVISOR

(Sign with full name)

THE SOCIALIST REPUBLIC OF VIETNAM

Independence – Freedom– Happiness

Ho Chi Minh City, December 4th, 2022

CRITICAL TEACHER’S EVALUATION SHEET

Student name: Student ID:

Student name: Student ID:

Student name: Student ID:

Student name: Student ID:

Student name: Student ID:

Major:

Project title:

Critical Teacher:

EVALUATION 7 Content of the project:

8 Strengths:

9 Weaknesses:

10 Approval for oral defense? (Approved or denied)

11 Overall evaluation: (Excellent, Good, Fair, Poor) ……

12 Mark: ……… (in words: ……….)

Ho Chi Minh City, December 4 th , 2022

FHQ-HCMUTE

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CONTENTS

CHAPTER 1: ENGINE PARAMETER

CHAPTER 2: THERMAL CALCULATE AND BUILD INDICATED WORK

DIAGRAM IN ENGINE AND KINETIC AND DYNAMIC OF SLIDER CRANK MECHANISM

I Choosing parameter 8

II Thermal Calculation 10

CHAPTER 3: DYNAMICS CALCULATION OF CRANKSHAFT-

CONNECTING ROD STRUCTURE

CHAPTER 1: ENGINE PARAMETER

No of cylinder, i: 4 in-line

2 Report contents

2.1 Thermal calculate and build indicated work diagram in engine

2.2 Kinetic and dynamic of slider crank mechanism

CHAPTER 2: THERMAL CALCULATE AND BUILD INDICATED WORK DIAGRAM IN ENGINE AND KINETIC AND DYNAMIC OF SLIDER CRANK

MECHANISM

I Choosing parameter

1 Intake air pressure: po = 0.1013 MPa

2 Intake air temperature To

Choose T0 = 29 ˚C due to the average temperature in the South of Vietnam

T0 = Tenvironment = 29 ˚C= 302 K

3 Intake air pressure at the intake valve pk:

We decide to choose a 4-stroke engine without turbo, so:

pk = p0= 0.1013 MPa

4 Intake air temperature at the intake valve Tk:

The 4-stroke engine without turbo:

Tk= T0= 302 K

5 Pressure at the end of intake process: pa

Base on the experimental calculation, pa = (0.8 ÷ 0.95) p0

Choose pa = 0.95 p0 = 0.0962 MPa

6 Residual gas pressure pr:

For the gasoline engine pr = (0.11 ÷ 0.12) MPa

Choose pr = 0.12 MPa

7 Residual gas temperature Tr:

For the gasoline engine Tr = 900 K ÷ 1000 K

Choose Tr = 1000 K

8 Temperature rises on new charge coefficient ΔT:

For gasoline engine: ΔT = 0 ÷ 20 ˚C

Choose ΔT = 15˚C

9 Heat added coefficient λ1:

Limit coefficient of heat added λ1 = 1.02 ÷ 1.07

Choose λ1 = 1.05

10 Combustion chamber sweep coefficient λ2:

With 4-stroke engine without turbo λ 2 = 0.8 ÷ 0.9

Choose λ2 = 0.85

11 Temperature revise coefficient λt :

For gasoline engine, choose air residue coefficient α = 0.9

So λt = 1.15

12 The heat ultilization coefficient at point z ξz

For gasoline ξz = 0.75 ÷ 0.92

Choose ξz = 0.85

13 The heat ultilization coefficient at point b ξb

For gasoline engine, ξb = 0.85 ÷ 0.95

Choose ξb = 0.9

14 Excess air value α

For gasoline engine, α = 0.85 ÷ 0.95

Choose α = 0.9

15 The coefficient of diagram rounding-off φr

For gasoline engine, φr = 0.93 ÷ 0.97

Choose φr = 0.95

II Thermal Calculation

1 1.45

] = 0.942 The coefficient of residual gases γr :

0.09620.1013×

1 + γ𝑟

=(302 + 15) + 1.15 × 0.0259 × 1000 × (0.09620.12 )

1.45−1 1.45

Excess air value: 0.7< α < 1

𝑚𝑐𝑣′′

̅̅̅̅̅̅̅ = (17.997 + 3.504𝛼) +1

2× (360.34 + 252.4𝛼) × 10

−5𝑇

𝑝𝑐 = 𝑝𝑎𝜀𝑛1 = 0.0962 × 111.37 = 2.57 𝑀𝑃𝑎 3.6 Temperature at the end of compression

0.208× (𝐶

12+𝐻

4 − 𝑂

32) Then, we have M0 = 0.516 (kmole of air/ kg of fuel)

4.1.2 The amount of combustible mixture M1

𝑀1 = 𝛼𝑀0 + 1

𝑚𝑓𝑀1 = 𝛼 𝑀𝑜 + 1

𝑚𝑓 = 0,9 0,516 +

1114

= 0,473 (𝑘𝑚𝑜𝑙𝑒 𝑜𝑓 𝑐𝑜𝑚 𝑚𝑖𝑥/𝑘𝑔 𝑜𝑓 𝑓𝑢𝑒𝑙) Where: mf is the molecular mass of gasoline;

mf = 110 ÷ 114 kg/kmole

4.1.3 The total amount of combustion products M2

𝑘𝑔 𝑛𝑙4.1.5 Gas molecular change coefficient (practicality): β

𝛽 = 1 + 𝛽0− 1

1 + 𝛾𝑟 = 1 +

1.078 − 1

1 + 0.0324 = 1.076 4.1.6 Gas molecular change coefficient at point z: βz

The specific heat of residual gases α<1 (if α>1 then, ΔQH = 0), because of the lack of oxygen, it leads to the residual gases, so that heat losses

∆𝑄𝐻 = 120 103 (1 − α) 𝑀0 = 120 103 (1 − 0,9) 0,516 = 6192 (kJ

kg) 4.1.8 The mean molar specific heat of solvent at point z

̅̅̅̅̅ = 19,806 +0,00419

2 𝑇 4.1.9 Temperature at the end of combustion process : Tz

𝜉𝑏(𝑄𝐻 − ∆𝑄𝐻)𝑀1(1 + 𝛾𝑟) + 𝑚𝑐̅̅̅̅̅̅ × 𝑇𝑣′ 𝐶 = 𝛽𝑍 × 𝑚𝑐̅̅̅̅̅̅̅̅ × 𝑇𝑣𝑧" 𝑍

4.1.10 Pressure at the end of combustion process: P z

𝑃𝑧 = 𝛽𝑧𝑇𝑧

𝑇𝑐𝑃𝑐 = 1,071 ×3105

816 × 2,57 = 10,47 𝑀𝑃𝑎

4.2 The expansion process

4.2.1 The pre-expansion ratio: ρ = 1

4.2.2 The after-expansion ratio: δ = ε/ρ = 11

4.2.3 The mean expansion adiabatic and polytropic indices:

(𝜉𝑏 − 𝜉𝑧) × 𝑄𝐻

𝑀1(1 + 𝛾𝑟) = 𝛽 × 𝑚𝑐̅̅̅̅̅̅̅ × 𝑇𝑣𝑏" 𝑏 − 𝛽𝑧 × 𝑚𝑐̅̅̅̅̅̅̅ × 𝑇𝑣𝑧" 𝑧+

8,314

𝑛2− 1 (𝛽𝑧𝑇𝑧 − 𝛽𝑇𝑏)

At temperature from 1200 ÷ 2600 K, the difference of the mean specific heat is

no need to consider, so that we can accept: a’vb = a’vz ; bb = bz ; β=βz

We have:

(𝜉𝑏 − 𝜉𝑧) × 𝑄𝐻𝑀1(1 + 𝛾𝑟) × 𝛽 × (𝑇𝑧 − 𝑇𝑏) + 𝑎" 𝑣𝑧 +𝑏"2𝑧 × (𝑇𝑧 + 𝑇𝑏) With 𝑇𝑏 = 𝑇𝑧

𝜀 (𝑛2−1) ; 𝑎′′𝑣𝑧 = 19.842; 𝑏′′𝑧 = 0.0022

= 1672 × (0.12

0.522 )

1.45−1 1.45

= 1080 𝐾 4.2.7 Tolerable

∆𝑇𝑟

𝑇𝑟 =|1080−1000|

1000 = 8% < 10%

5 The indicated parameters of working cycle

5.1 The theoretical mean indicated pressure

5.4 The mean effective pressure: pe

η𝑖 = 8.314 × 𝑀1𝑝𝑖𝑇𝑘

𝑄ℎη𝑣𝑝𝑘 = 8.314 ×

0.473 × 1.38 × 302

43960 × 0.942 × 0.1013 = 0.39 5.7 The effective efficiency: ηe

η𝑒 = 8.314 ×𝑀1𝑝𝑒𝑇𝑘

𝑄ℎη𝑣𝑝𝑘 = 8.314 ×

0.473 × 1.052 × 302

43960 × 0.942 × 0.1013 = 0.3 5.8 Calculate fuel consumption rate (gi indicated)

6 The basic parameters and indices of the engine:

6.1 The cylinder displacement: Vh

𝑉ℎ = 30𝜏𝑁𝑒

𝑝𝑒𝑛𝑒𝑖 =

30 × 4 × 761.052 × 5600 × 4= 0.387 𝑙 Where:

τ: number of engine stroke

i: number of cylinders of the engine

ne: number of revolutions at design performance

Ne: performance of the designed engine, kW

pe: the mean of effective pressure, 𝑀𝑁

The value of Stroke:

𝑆 = 𝐷 ×𝑆

𝐷 = 75.8 × 1.13 = 85.7 𝑚𝑚

35 gi Kg/Kw.h 0.21

P-V AND P-Φ DIAGRAM

CHAPTER 3: DYNAMICS CALCULATION OF CRANKSHAFT- CONNECTING ROD STRUCTURE

3.1 Kinematic Analysis of Crankshaft-Connecting Rod Structure

In an internal combustion engine: piston, connecting rod, crankshaft are the moving parts and they work on the following principles:

- The reciprocating piston group transmits air force to the connecting rod

- The connecting rod group is an intermediate moving part with complex motion to convert the reciprocating motion of the piston into the rotation of the crankshaft

- The crankshaft is the most important part that rotates and transmits the power from the engine out to drive the machines There are three main types of crankshaft - connecting rod construction:

a) The crankshaft cuts the cylinder center line

b) The crankshaft deviates from the cylinder centerline by a small distance a < 0.1S, (S - piston stroke)

c) Structure for V-engine - this type has two connecting rods mounted on the same crank pin (called double connecting rod)

- In all type above, the crankshaft cuts the cylinder centerline structure is the most common one nowadays

3.2 Piston Kinematics

With the hypothesis that the crankshaft rotates with angular velocity 𝜔 = const, then the crankshaft angle 𝛼 is variable proportional to time, and all kinematic quantities are functions dependent on the variable 𝛼

However, this assumption for modern high-speed engines gives an error negligible because of the variation of angular velocity (𝜔) due to the inhomogeneity uniformity

of the motor torque caused when the motor is operating at steady state is very small

3.2.1 Piston displacement

When the crankshaft rotates at an angle α, the piston moves a distance Sp from its initial position The displacement of the piston in the cylinder is calculated by the formula:

V p1 = R𝜔sin𝛼

V p2 = R𝜔𝜆

2sin2𝛼

𝜔 = 2𝜋𝑛

60 : angular velocity of connecting rod

Average speed of piston, to classify the different type of engine:

3.2.3 Acceleration of the piston

Taking the derivative V over time, we have the piston acceleration formula:

3.3 STRUCTURE DYNAMICS OF CRANKSHAFT – CONNECTING ROD

3.3.1 Gas pressure forces Pgas

Gas pressure force is a quantity that varies with the angle of rotation of the crankshaft, defined as obtained from gas pressure P at the thermal calculation of the engine

- Intake process: P gas = P a

- Compression process: P gas = P a in1 , with i from 1 (1800) to ε (360 0 – θs)

- Expansion process: P gas = Pz

ɛn2

- Expansion process: P gas = P r

3.3.2 Inertial forces of moving details

Translational inertial force:

Pj = −m j J = −m j Rω 2 (cosα + λcos2α) Centrifugal inertial force:

P K = −m r Rω 2 = const With: m j = m pg + m A

m r = m K + m B

m A = (0,275 ÷ 0,35)m tt

m B = (0,725 ÷ 0,65)m tt

Where:

Pj – translational inertial force

P K – centrifugal inertial force

m j – the mass of translational motion details

m r – the mass of rotational movement details

m cr – the mass of connecting rod

m pg – the mass of piston group

m C – the mass of the crankshaft

m A – the mass of connecting rod small end

m B – the mass of connecting rod big end

3.3.3 The force system acting on the crankshaft – connecting mechanism

Total force acting on the piston pin is the synergy of the gas pressure force Pkt

and the translational inertial force Pj, valid equal to the algebraic sum of these two forces:

P1 = Pgas + Pj Vertical force acting on the connecting rod

Pcr = P∑

cosβ with β = sin−1(λ sinα) Thrust force N

N = Pcr.tgβ

Tangential force T

T = Pcr*sin(α + β) = P∑ sin(α + β)

cosβNormal force Z:

Z = Pcr*cos(α + β) = P∑ cos(α + β)

cosβ

3.3.4 Rotation moment M of the engine

Calculation of the working deflection angle of the engine: δK = 180o Working order of the engine: 1 – 3 – 4 – 2

Determination of the working phase of each cylinder:

+ Cylinder 1: α + Cylinder 2: α + 180 + Cylinder 3: α + 540 + Cylinder 4: α + 360 Total moment ∑ Mi defined by the relation:

3.3.5 Forces acting on crankpins

At the crankpin there is the following force: tangential force T, normal force Z, centrifugal force PKO The force acting on the crankpin is the force vector 𝑄⃗ defined

by force balance equation:

𝑄⃗ = 𝑇⃗ + 𝑍 + 𝑃⃗⃗⃗⃗⃗⃗⃗ with P𝐾𝑂 KO = −mB R ω2

% Expansion process

a6 = linspace(380,500,1000);

X6 = R*(1-cosd(a6) + (lamda/4).*(1-cosd(2.*a6))); V6 = X6.*Sp + Vc;

title( 'P-phi Diagram' );

xlabel( 'Crank shaft angle (Degree)' );

title( 'Horizontal force N' )

xlabel( 'Angular crankshaft (degrees)' )

ylabel( 'Pressure (MN/m^2)' )

figure(3)

plot(a,T)

title( 'Tangent force T' )

xlabel( 'Angular crankshaft (degrees)' )

ylabel( 'Pressure (MN/m^2)' )

figure(4)

plot(a,Z)

title( 'Normal force Z' )

xlabel( 'Angular crankshaft (degrees)' )

ylabel( 'Pressure (MN/m^2)' )