Dynamic Analysis of the Internal Combustion Engine

Review

1. The induction (intake) stroke: The inlet valve opens, and the piston travels from TDC (state 0) to BDC(state 1). As the piston moves, low pressure forms in the cylinder, and an air-fuel mixture at theambient temperature and pressure is sucked into the cylinder.

2. The compression stroke: The inlet valve closes, and the piston travels from the BDC (state 1) to theTDC (state 2). In this process, both the air-fuel mixture’s pressure and temperature increase. Duringthe compression stroke, the piston does work on the gas in the cylinder. At some point during thecompression process, the spark plug fires, ignition occurs, and the fuel combusts, raising both thetemperature and the pressure of the gas even further (state 2 to state 3).

3. The expansion (power) stroke: The piston moves towards BDC (state 4) while the combustion processcontinues. The gases push the piston. Towards the end of the power stroke, the exhaust valve opensand the combustion products start escaping.

4. The exhaust stroke: As the exhaust valve opens, the pressure drops (state 4 to state 1). The pistonmoves from the BDC to the TDC expelling the combustion products (state 1 to state 0). During theexpulsion process, the combustion products are at a temperature and pressure above ambientconditions. At the end of the expulsion process, the exhaust valve closes, and we are back where westarted.

1

2

3

4

0

V

P

Vc Vt

patm

Figure 1 The P-V diagram for the ideal air cycle for the four stroke internal combustion engine.

Process 1-2: The piston moves from BDC (point 1) to TDC, compressing the gas in the cylinder. No heatexchange occurs between the gas and its surroundings during the compression process. The process isadiabatic. The ratio of the gas’ volume at the beginning and the end of the compression process is knownas the compression ratio,

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r = V1

V2

= V4

V3

.

Process 2-3: The cylinder remains at TDC without moving. Heat is added to the gas. The heat addition isassumed to occur instantaneously. During the heat addition process, the gas’ volume remains fixed whileits temperature and pressure increase. In other words, this is a constant volume process.Process 3-4: The piston moves from TDC to BDC. The gas expands while pushing the piston. Theexpansion process is assumed to be reversible and adiabatic.Process 4-1: The piston remains at BDC. Heat is removed from the gas instantaneously before thecylinder has an opportunity to move. This is a constant volume process.

Some terminology is appropriate here. Vc is called the clearance volume and is denoted by Vc. Similarly,the volume of the cylinder at the bottom dead center is denoted by Vt. The difference between the two iscalled the swept volume, or the displacement, Vs. The compression ratio, r, is given by the ratio of thevolume at the bottom dead center position to the clearance volume:

c

scV

VVr

+= .

Dynamics: An overview

Here is a brief overview of what is going on in the internal combustion engine.1. The pressure of the gases (air + fuel, or by products of combustion) exerts a force on the piston. Think

of this force as being an “input force”, although during some parts of this cycle, we know that the gasesin the cylinder are not doing positive work.

2. All the parts of the internal combustion engine have a finite mass (inertia). Thus a fraction of the inputforces are “spent” on accelerating or decelerating the masses.

3. Some of the input is used to overcome the friction on the piston walls and the friction at the bearings.(Gravitational forces are insignificant for internal combustion engines).

4. The crank is coupled to the crankshaft which in turn is coupled via a power train to the wheels of theautomobile. The crankshaft may also power the water pump, camshaft, power steering pump, the airconditioning compressor and other accessories. A significant part of the input force is used to drive theautomobile. Since the crank has a rotary motion, the fraction of the input force used to drive the crankis effectively a moment, and is called the turning moment.

We want a formal description of these ideas and we want to develop an equation that will relate thepressure in the cylinder to the turning moment.

Basic definitions

DisplacementThe swept volume of the engine. = n (Vt – Vc)where n = number of cylinders.

Brake horse powerThe net engine output power at the crankshaft, 1 horsepower = approx. 745 Watts.

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Maximum torqueThe maximum average turning moment at the crankshaft.

Engine speed (RPM)The number of revolutions per minute for the crank shaft.

Curb weightThe mass of the vehicle when ready to use (excluding the weight of the driver).

Table 1 Specifications for some commercially available automotive engines

Audi A4,1.8

BMW740i

Escort1.8

Mercedes S 600

Porsche911

Porsche911

Turbo

Rolls-Royce

HondaCivic

ChryslerVoyager

Displacement (liters) 1.781 3.982 1.753 5.987 3.6 3.6 6.75 1.493 2.972Brake horse power (BHP) 125 210 77 394 272 408 245 90 147Brake horse power (kW) 93 157 57 294 203 304 183 67 110Max Torque (Nm) 173 400 156 570 330 540 500 119 225Engine speed for BHP peak(RPM)

3950 4500 4000 3800 6000 5750 4000 6000 5100

Curb Weight (kg) 1225 1790 1065 2180 1370 1500 2430 935 1585Weight to power ratio(kg/kW)

13.14 11.43 18.55 7.42 6.75 4.93 13.30 13.93 14.46

Compression Ratio 10.3 10 10 10 11.3 8 8 9.2 8.9

Particle dynamics

Consider a single particle of mass m in the plane and consider a force F acting on the particle. Let itsposition vector be given by r. If i and j represent unit vectors along the x and y axes, the force andposition vectors can be written in terms of components in the following form:

F = Fx i + Fy j (1)

r = x i + y j (2)As the position of the particle changes, after an infinitesimal time dt, the new position vector is r+dr givenby:

dr = dx i + dy j (3)

m

F

x

y

r

r+dr

dr

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Figure 2 A particle of mass m acted on by a force F as its position vector changes from r to r+dr.

The work done on the particle by the force F is given by: dW = F . dr (4)

where the product between F and dr is the scalar product between the two vectors. The power associatedwith the force is given by

vFr

F ⋅=⋅==dtd

dtdW

P (5)

From Newton’s second law we know that the force is related to the acceleration (a) of the particle by theequation:

F = m a (6)If the velocity of the particle is denoted by v,

dtd

mv

F = (7)

Combining (5) and (7), we get:

vv ⋅=

dtd

mP (8)

We will now try to relate the right hand side the above equation to the rate of change of kinetic energy ofthe particle. The kinetic energy K is half the mass times the square of the velocity of the particle:

vv ⋅= mK 21 (9)

The time derivative the kinetic energy is given by:

⋅+⋅=

dtd

dtd

mdtdK v

vvv

21 (10)

Recognizing that the scalar product is commutative, we see that the right hand sides of (8) and (10) are thesame. Therefore, we get the simple result:

dtdK

P = (11)

The power associated with the force acting on a particle is equal to the rate of change ofkinetic energy of the particle.

There are couple of obvious extensions to this. First, if there is more than one force acting on the particle,we can let F be the resultant force and the net power associated with the forces acting on the particle willequal the rate of change of kinetic energy. Second, (11) is also true for a system of n particles. If there aren particles and we compute the power associated with the net external force acting on each particle, we canapply (11) to each particle. We will find that the net power is equal to the rate of change of the total kineticenergy of the system. Therefore we can write:

The total power associated with external forces acting on a system of n particles is equal tothe rate of change of the total kinetic energy of the system.

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Rigid body dynamics

It is possible to characterize the dynamics of any rigid body by considering an equivalent system of a finitenumber of particles. In particular, any link (see Figure 3) can be approximated by a system of twoparticles. If the center of mass of the system is at C, as shown in the figure, the equivalent system of asystem of two particles of mass mA and mB is given by:

la

mmlb

mm BA == , (12)

where m is the total mass of the original link.

A BC

lba

l

BA

Figure 3 Any given link with the center of mass at C (see left) can be approximated by a massless rod withparticles of mass mA and mB at A and B respectively.

Dynamics of the slider crank mechanism

Consider the special case of the slider crank mechanism shown in Figure 4. Define the following variables:θ = θ2, φ = 2π− θ3, and x=r1.

X

Y

rQ

θ 2

θ 3

r1

r2 r3

OP

φ

Figure 4 The slider crank mechanism in an internal combustion engine

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The position closure equations are:x = r2 cos θ2 + r3 cos φ

r2 sin θ2 - r3 sin φ = 0 (13)From these equations, substituting for θ, we can see that:

θ=φ − sinsin

3

21rr

(14)

Differentiating equation (12) and solving we get:

θφθ=φ &&

coscos

3

2rr

( )θφ

φ+θ−= &&cos

sin2rx (15)

We are now in a position to calculate the kinetic energy of the mechanism. Let the masses of the crank, thepiston, and the connecting rod be denoted by mcrank, mpiston, and mconn respectively. Let us assume that allthe rigid bodies are symmetric. In other words, the center of mass for each rigid body is at its geometriccenter. Following the procedure outlined above in Equation (12), we obtain the approximate model shownin Figure 5,

pistonconnP

conncrankQ

crankO

mmm

mmm

mm

+=

+=

=

21

21

2121

(16)

X

Y

θ x

r2r3

φ

mQ

mO

mP

Figure 5 The approximate dynamic model for the slider crank mechanism

The total kinetic energy of the mechanism is given by:

( ) ( ) ( ) 222

22

222

2 21

cossin

21

21 θ=θ

φφ+θ+=

+θ= &&&& JrmrmxmrmK PQPQ (17)

The effective inertia of the mechanism as “seen” at the crankshaft is:

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( ) ( )

φφ+θ+=

2

22

2 cossin

rmrmJ PQ

If the slider crank mechanism is viewed as a system, there are two primary sources of energy interactionthat will change the kinetic energy of the mechanism. First, there is the work done by the gases in thecombustion chamber. Second, the crankshaft of the mechanism is coupled to a load, and the momentrequired to turn the load (or alternatively, the resistance offered by the load), called the turning momentdoes work on the system.

X

Y

θO

Q

P

x

Mt

pApatm A

Figure 6 The external forces and moment acting on the slider crank mechanism.

Denote the pressure in the combustion chamber by p, the turning moment acting on the crankshaft by Mt,and the atmospheric pressure by patm. The power associated with external forces acting on the mechanismis given by:

( ) xAppMP atmt && −+θ=where A is the cross-sectional area of the piston given by

4

2DA

π= ,

D being the bore (diameter) of the piston.

Since the kinetic energy is given by (17), we can apply Equation (11) to get:

( ) ( ) xAppMJdtdJ

atmt &&&&&& −+θ=θθ+θ 2212

21 (18)

Let us assume that the engine is running at constant speed. This is at best an approximation since theengine speed is never constant but instead fluctuates within a narrow band. If this assumption is valid, theterm associated with the angular acceleration drops out and we get the following equation:

( ) xAppMdtdJ

atmt &&& −+θ=θ221

and an expression for the turning moment that looks like:

( ) ( ) ( ) 22

22

221

cossin θ

φφ+θ++−=θ &&& rmrm

dtd

xAppM PQatmt

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Substituting from (15) and dividing by the crank speed, we get an expression for the turning moment:

( ) ( ) ( ) ( ) ( ) 2

3

22

32

22 coscoscos

cos

cos

sincos

sin θ

φ+θφ+

φθ

φφ+θ+

φφ+θ−= &

rr

rmArppM Patmt (19)

If the slider crank mechanism were massless, the turning moment would be equal to be equal to the momentdue to the pressure in the cylinder:

( ) ( )φφ+θ−=

cossin

2ArppM atmp (20)

The moments due to the inertia of the moving parts in the mechanism is given by the inertial moment, Mi,

( ) ( ) ( ) 2

3

22

32

2 coscoscos

cos

cos

sin θ

φ+θφ+

φθ

φφ+θ= &

rr

rmM Pi (21)

Typical plots for the three moments are shown in Figure 7.

0 100 200 300 400 500 600 700-200

-100

0

100

200

300

400

500

600

700

Crank Angle (degrees)

Mom

ents

(New

ton-

met

ers)

Figure 7 The moment due to the cylinder pressure (Mp, dotted), the inertial moment (Mi, dashed), and theturning moment (Mt, solid) for a single cylinder of the sample engine shown in

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Table 2.

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Table 2 Specifications for a sample four cylinder engine

Symbol Definition Valuer compression ratio 10.3k expansion coefficient 1.4

pmax maximum pressure (after combustion), p3 100 atmospherespatm atmospheric pressure 14.7 psi or 1.01325 barVc clearance volume 0.05 litersL stroke (2 times crank radius) 86.4 mmD bore 81 mmα ratio of crank to conn. rod length 0.35N Engine speed 3000 RPM

mpiston mass of the piston 1 lbsmcrank effective mass of the crank shaft for one cylinder 15 lbs/4mconn mass of the connecting rod 1.75 lbs

0 100 200 300 400 500 600 7000

10

20

30

40

50

60

70

80

90

100

110

Crank Angle (degrees)

Cyl

inde

r Pre

ssur

e (a

tmos

pher

es)

Figure 8 The cylinder pressure in atmospheres for a single cylinder of the sample engine shown in the tablebelow (

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Table 2). An ideal air cycle is assumed except for two changes. The intake pressure is assumed to be 0.9atmospheres and the exhaust pressure is assumed be 1.0 atmospheres.

0 100 200 300 400 500 600 700-200

-100

0

100

200

300

400

500

600

700

Crank Angle (degrees)

Mom

ents

(New

ton-

met

ers)

Figure 9 The turning moment for a balanced four cylinder engine. The dotted lines show the turningmoments from each cylinder while the solid line shows the total turning moment. The dashed line shows the

average torque at the crankshaft. The data for the plot comes from

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Table 2.

Exercise: Computer aided analysis of a four-stroke internal combustion engine

The goal of this exercise is to develop the mathematical model and a software package that will allow youto analyze a four-stroke internal combustion engine. Specifically we want to be able to do the followingtasks.

1. Plot the cylinder pressure against the crank angle for 720 degrees (a complete cycle). Note that Figure1 shows the pressure against the cylinder volume – we want pressure against the crank angle.

2. Derive an expression for the piston speed as a function of crank angle and crank speed and plot thepiston speed against the crank angle for a constant crank speed. & . &x vs θ (see Figure 4).

3. Plot the effective inertia against the crank angle.

4. Plot the inertial moment against the crank angle

5. Plot the turning moment against the crank angle.

6. Calculate the average turning moment and estimate the engine torque.

7. Plot the turning moment for the engine for a 720 degree rotation by overlaying the plots for a singlecylinder at the appropriate intervals.

8. Find the crank angles at which the engine speed is the lowest and the angle at which the engine speed isthe highest.

9. Suggest a suitable remedy for reducing the engine fluctuations.

NoteYou may choose any engine to analyze. The data in Table 2 is a good starting point. Data for more enginesis available on the course web site. However, there are four variables that you will need to fix for yourself.• You will need to increase the masses mcrank, mpiston, and mconn, if the engine is larger. A good rule of

thumb is to assume that the masses scale with the product of the square of the bore and the stroke.• Another variable is the maximum pressure in the engine. The maximum pressure pmax will be greater in

high performance engines.

Report

1. Provide a table with specifications (analogous to

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Table 2) and representative plots for the engine of your choice for each of the 9 items listed above.Each plot must be clearly labeled.

2. Attach a copy of the matlab script file that was used to generate the data for the plots.3. Discuss each plot briefly. What does the plot tell you about engine performance? What do you find

interesting about the plot?

Table 4 Miscellaneous units and useful conversions

Pressure1 bar 105 Newtons/meter2

1 atmosphere 14.7 lbs/inch2 (psi)1 psi 6895 Newtons/meter2

Volume1 liter 1000 cubic centimeters (cc)1 cc 106 mm3

1 cc 10-3 meter3

Mass1 kg (weight) 2.2 pounds

Angular velocity1 rotation per minute (RPM) 2π/60 radians per second

Power1 Watt 1 Newton meter/second1 Horsepower 745.6 Watts