50400lecture_2

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ME 50400 / ECE 59500Automotive Control

Lecture Notes: 2

InstructorSohel Anwar, Ph.D., P.E.

Dept. of Mechanical EngineeringIUPUI

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Recap from Lecture 1

Background on Automotive Control

Importance of Automotive Control

Examples of Automotive Control Systems

Drive-By-Wire System Overview

Benefits of Automotive Control Systems

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THERMODYNAMICS BASICS

• To understand how an engine converts the thermal energy into mechanical energy in a quantitative way, it is necessary to review the fundamentals of thermodynamics. • First law of thermodynamics essentially states the law of conservation of energy:

dq = du + dw

dq = change in thermal energy in a control volume, positive when thermal energy is added to itdu = change in the internal energy, positive when the internal energy increase inside the control volumedw = change in mechanical work on the control volume, negative when it is worked on

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Thermodynamics Basics

• Consider an engine cylinder, the air is compressed by the piston, fuel is injected, a spark (SI) initiates combustion, pressure rises pressing the piston outward creating mechanical energy. All of these events can be described by thermodynamic equations.

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• The work done by the piston can be expressed by one of the following:

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• Review the definitions of Enthalpy and Specific Heat Constant (pages 6-9 of the textbook).

• Note that:h = u + PV where h is enthalpyR = cp – cv where R is universal gas constantk = cp / cv where k is adiabatic exponent

• The ideal gas law states:PV = mR

Wherem = mass of the gas, kgR = 287.4 m2/(s2K)P = pressure in N/m2

V = volume in m3

= gas temperature in K

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State Change of Ideal Gas

1. Isothermal change – Gas temperature remains constant

2. Isobaric change – Gas pressure remains constant3. Isochoric change – Gas volume remains constant4. Isentropic or Adiabatic change – Gas thermal energy

remains constant• Isothermal change:

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Ideal Gas State Change

• Isobaric change: Pressure P = const.

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• Isochoric change: Volume V = const.

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• Isentropic Change: q = constant

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Thermodynamic Cycles

• The operation of engines can be described by an appropriate thermodynamic cycle for the purpose of analysis.

• Thermodynamic cycles also give the basis for ideal efficiency of an engine, i.e. what is upper limit of energy conversion?

• While most real processes are not reversible from a thermodynamic sense, these cycles assume that the processes or the state changes are reversible.

• A reversible adiabatic process is known as insentropic process where entropy remains constant. Entropy is defined as: dS = dq/

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The Carnot Cycle

• An ideal engine will run on a Carnot cycle:

in

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Carnot Cycle

• Since heat energy due to combustion enters the cylinder near point 1, the only heat input in the cycle is assumed to be during process 1-2.• Also the work input during isentropic compression (4-1) and isentropic expansion (2-3) are identical and cancel each other.

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Engine Basics: Thermodynamic Cycle

• Two fundamental types of IC engines: 1) Four Stroke Cycle, and 2) Two Stroke Cycle. Four stroke cycle engines are most common. Two stroke cycle engines are principally used in marine applications.

• Four stroke engine runs on a thermodynamic cycle that has the following piston strokes: air-fuel intake, compression, combustion/expansion, and exhaust. Two stroke engines combine intake and expansion strokes into one and compression and exhaust strokes into one.

• Based on ignition, engines are classified as: 1) Spark Ignited (SI) or 2) compression ignited (CI). SI engines operate on low compression ratio whereas CI engines operate on higher compression ratio which cause self-ignition.

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First introduced by Nikolaus Otto in 1862.

The P-V diagram has four stages: compression, combustion (constant volume), expansion, and exhaust (constant volume).

Compression ratio is given by:

The expansion and compression processes are considered isentropic in an Otto cycle.

Spark-ignited (SI) Engines

2

1

VV

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p-V Diagram for SI Engines

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The thermal efficiency of an engine is defined as the ratio of all kinetic energies to the total thermal energy (q2,3) of combustion in one cycle. To derive the expression of thermal efficiency th in terms of pressure, volume, temperature, we need to consider all four stages of an Otto cycle. Work done in process 1-2: Isentropic compression

Thermal Efficiency of an SI Engine

)(

0____0__

1221

2

1

vv

v

mcdcmw

dmcdudwdoneworkenergynternalinchangedwdudq

changeheatnetdq

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• The combustion of the gas inside the cylinder causes the pressure to rise at constant volume (process 2-3).

Thermal Efficiency of Otto Cycle

)(

0

00

0__

2332

32

3

2

3

2

vv

v

V

V

mcdcmq

dmcdudwdudq

pdVw

pdVdwdq

changevolumenetdV

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• Following combustion and pressure rise, the piston moves resulting in expansion of the gas (process 3-4). Assuming isentropic expansion (dq=0), we obtain:

)(

00

3443

4

3

vv

v

mcdcmw

dmcdudwdwdudq

dq

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• The last stage of the cycle is assumed to be a constant volume process when heat is lost via exhaust gas exiting the cylinder and new air-fuel mixture entering the cylinder (process 4-1):

)(

0

00

0__

4114

14

1

4

1

4

vv

v

V

V

mcdcmq

dmcdudwdudq

pdVw

pdVdwdq

changevolumenetdV

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• Now the thermal efficiency of the Otto cycle is given by:

)()(1

)()()(

)(0)(0)(

23

14

23

1423

23

3412

32

14433221

v

vv

v

vv

th

mcmcmc

mcmcmc

qwwww

22


1

1

2

23

14

1223

1114

122

111

123

114

23

14

)()(

)()(

)()(1

k

kk

kk

kk

th

VV

VVTherefore

VV

VVNow

23

2

11

1

4

3

3

4 1

k

k

VV

• Utilizing the ideal gas law for an isentropic process, we can obtain the following relationship:

1

11 kth


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Observations

• Higher value of the compression ratio will result in higher thermal efficiency.

• For a compression ratio of 11 and adiabatic constant k = 1.4, the ideal thermal efficiency of an Otto cycle is 0.617.

617.011

1111 14.11 kth

•Is there a limit on the compression ratio?

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Diesel Engines

• Rudolf Diesel developed the first compression ignition (CI) engine in 1893-1897.

• The CI engine (widely known as Diesel Engine) also has two fundamental types: Four Stroke Cycle and Two Stroke Cycle.

• It operates on higher compression ratio than SI engines. Combustion takes place due to high temperature developed in the compression stroke, which cause gas expansion at nearly constant pressure. Constant pressure is maintained by controlling the fuel injection in the chamber.

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p-V Diagram for a CI Engine

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Diesel Engines

• The constant pressure combustion process becomes longer if more fuel is injected into the combustion chamber.

• Injection ratio or load is defined as the ratio of the volume at the end of combustion to the volume at the beginning of combustion.

2

3

2

3

VV

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Thermal Efficiency of a Diesel Engine

• The four stages of a diesel cycle are: isentropic compression, isobaric combustions, isentropic expansion, and isochoric heat loss.

• Process 1-2: Isentropic compression dq=0:

)(

0____0__

1221

2

1

vv

v

mcdcmw

dmcdudwdoneworkenergynternalinchangedwdudq

changeheatnetdq

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• The combustion of the gas inside the cylinder causes the gas to expand at constant pressure (process 2-3).

)(

)(

0

2332

2332

3

2

3

2

mRdmRw

mRdpdVdw

mcdcmq

dmcdqdp

pp

p

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• Following combustion, the piston moves resulting in expansion of the gas (process 3-4). Assuming isentropic expansion (dq=0), we obtain:

)(

00

3443

4

3

vv

v

mcdcmw

dmcdudwdwdudq

dq

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• The last stage of the cycle is assumed to be a constant volume process when heat is lost via exhaust gas exiting the cylinder and new air-fuel mixture entering the cylinder (process 4-1):

)(

0

000__

4114

14

1

4

1

4

vv

v

V

V

mcdcmq

dmcdudwdudq

pdVw

pdVdwdqchangevolumenetdV

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• Now the thermal efficiency of the Diesel cycle is given by:

][)];([

)1(

)1(11

)(0)())(()(

2

3

1

4

2

1

23

342312

32

14433221

v

pvp

p

vvpv

th

cc

kccR

k

mcmcccmmc

qwwww


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• After simplification using the law of gases, the following efficiency equation is obtained:

kk

kk

kk

kk

kk

kk

kkk

VV

VV

CpVVV

VV

pp

pp

pp

3

2

1

4

1

3

21

1

41

1

21

3

4

1

4

11

1

2

2

11

1

3

4

4

3

2

1

4

3

1

2

3

4

1

4

1

4

;

][

11111 1

k

kth k


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Example Problem

• Find the ideal thermodynamic (thermal) efficiency of a diesel engine, if the compression ratio for the engine is 20 and the injection ratio is 2. Assume the adiabetic gas constant k = 1.4.

• Solution:

647.01212

4.11

2011

11111

4.1

14.1

1

k

kth k

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Observations

• Efficiency of a diesel engine is not only dependent of compression ratio but also, on the injection ratio or load.

• The diesel efficiency decreases with increase in load .

• At high load, a diesel engine has a lower efficiency than an Otto engine at the same compression ratio.

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Comparison of Different Engines

• SI engines have lower compression ratio to keep the maximum pressure below allowable values since isochoric combustion can generate high pressures.

• CI engines are on the other hand can work at high compression ratios since the combustion is isobaric.

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Comparison of Different Engines

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Supplemental Material

• Please review example on engine efficiency posted on OnCourse under Supplemental Materials.

50400lecture_2

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