Dynamic Analysis of the Internal Combustion Engine Review 1. The induction (intake) stroke: The inlet valve opens, and the piston travels from TDC (sta te 0) to BDC (state 1). As the piston moves, low pressure forms in the cylinder, and an air-fuel mixture at the ambient 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 the TDC (state (stat e 2). In this process, both the air-fuel air-fu el mixture’s pressure and temperatur temperatur e increase. increase. During the compression stroke, the piston does work on the gas in the cylinder. At some point during the compression process, the spark plug fires, ignition occurs, and the fuel combusts, raising both the temperature 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 process continues. The gases push the piston. Towards the end end of the power power stroke, the exhaust valve opens opens and the combustion products start escaping. 4. The exhaust stroke: As the exhaust valve opens, the pressure drops (state 4 to state 1). The piston moves moves from the BDC to the TDC expelling the combustion products (state 1 to state stat e 0). During the expulsion process, the combustion products are at a temperature and pressure above ambient conditions. conditions. At the end end of the expulsion process, the exhaust valve closes, and we are back bac k where where we started.
3
P
2
patm
0 V c
4
1 V t t
V
Figure 1 The P-V diagram for the ideal ideal air cycle for the four stroke internal combustion engine. engine.
Process 1-2: The T he piston moves moves from BDC (point (point 1) to TDC, compressing the gas in the cylinder. cylinder. No heat exchange occurs between the gas and its surroundings during the compression process. The process is adiabatic. The ratio of the gas’ volume at the beginning and the end of the compression process is known as the compression ratio,
r =
V 1 V 2
=
V 4 V 3
.
Process 2-3: The cylinder remains at TDC without moving. Heat is added to the gas. The heat addition is assumed to occur instantaneously. During the heat addition process, the gas’ volume remains fixed while its 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. The expansion process is assumed to be reversible and adiabatic. Process 4-1: The piston remains at BDC. Heat is removed from the gas instantaneously before the cylinder has an opportunity to move. This is a constant volume process.
Some terminology is appropriate here. V c is called the clearance volume and is denoted by V c. Similarly, the volume of the cylinder at the bottom dead center is denoted by V t. The difference between the two is called the swept volume, or the displacement, V s. The compression ratio, r , is given by the ratio of the volume at the bottom dead center position to the clearance volume: V + V s . r = c V c
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 gases in 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 input forces 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 the automobile. The crankshaft may also power the water pump, camshaft, power steering pump, the air conditioning compressor and other accessories. A significant part of the input force is used to drive the automobile. Since the crank has a rotary motion, the fraction of the input force used to drive the crank is 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 the pressure in the cylinder to the turning moment.
Basic definitions Displacement The swept volume of the engine. = n (V t – V c) where n = number of cylinders. Brake horse power The net engine output power at the crankshaft, 1 horsepower = approx. 745 Watts.
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Maximum torque The maximum average turning moment at the crankshaft. Engine speed (RPM) The number of revolutions per minute for the crank shaft. Curb weight The 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
BMW 740i 3.982
Escort Mercede Porsche Porsche Rolls1.8 s S 600 911 911 Royce Turbo 1.753 5.987 3.6 3.6 6.75
Displacement (liters)
1.781
Brake horse power (BHP) Brake horse power (kW) Max Torque (Nm) Engine speed for BHP peak (RPM) Curb Weight (kg) Weight to power ratio (kg/kW) Compression Ratio
Honda Chrysler Civic Voyager 1.493
2.972
125 93 173 3950
210 157 400 4500
77 57 156 4000
394 294 570 3800
272 203 330 6000
408 304 540 5750
245 183 500 4000
90 67 119 6000
147 110 225 5100
1225 13.14
1790 11.43
1065 18.55
2180 7.42
1370 6.75
1500 4.93
2430 13.30
935 13.93
1585 14.46
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 its position vector be given by r. If i and j represent unit vectors along the x and y axes, the force and position vectors can be written in terms of components in the following form: F = F x i + F y j
(1)
r = x i + y j
(2)
As the position of the particle changes, after an infinitesimal time d t , the new position vector is r+dr given by: dr = d x i + d y j
(3)
F
dr
m
r r+dr x
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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 associated with the force is given by P=
dW dt
=F⋅
d r dt
=F⋅v
(5)
From Newton’s second law we know that the force is related to the acceleration ( a) of the particle by the equation: F=ma
(6)
If the velocity of the particle is denoted by v, F=m
d v dt
(7)
Combining (5) and (7), we get: P=m
d v dt
⋅v
(8)
We will now try to relate the right hand side the above equation to the rate of change of kinetic energy of the particle. The kinetic energy K is half the mass times the square of the velocity of the particle: K = 1 m v ⋅ v 2
(9)
The time derivative the kinetic energy is given by: d v d v = 12 m ⋅ v + v ⋅ dt dt dt
dK
(10)
Recognizing that the scalar product is commutative, we see that the right hand sides of (8) and (10) are the same. Therefore, we get the simple result: P=
dK dt
(11)
The power associated with the force acting on a particle is equal to the rate of change of kinetic 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 will equal the rate of change of kinetic energy. Second, (11) is also true for a system of n particles. If there are n particles and we compute the power associated with the net external force acting on each particle, we can apply (11) to each particle. We will find that the net power is equal to the rate of change of the total kinetic energy of the system. Therefore we can write: The total power associated with external forces acting on a system of n particles is equal to the 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 finite number of particles. In particular, any link (see Figure 3) can be approximated by a system of two particles. If the center of mass of the system is at C , as shown in the figure, the equivalent system of a system of two particles of mass m A and m B is given by:
m A
=m
b l
,
=m
m B
a
(12)
l
where m is the total mass of the original link.
A
B
C
A
B
b
a l
l
Figure 3 Any given link with the center of mass at C (see left) can be approximated by a massless rod with particles of mass m A and m B 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=r 1.
Y
θ3 Q r
r r
2
O
3
θ2
φ
r1
P
X
Figure 4 The slider crank mechanism in an internal combustion engine
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The position closure equations are: x = r 2 cos θ2 + r 3 cos φ r 2 sin θ2 - r 3 sin φ = 0
(13)
From these equations, substituting for θ, we can see that:
r φ = sin −1 2 sin θ r 3
(14)
Differentiating equation (12) and solving we get:
φ& =
r 2 cos θ & θ r 3 cos φ
r sin (θ + φ) & x& = − 2 θ cos φ
(15)
We are now in a position to calculate the kinetic energy of the mechanism. Let the masses of the crank, the piston, and the connecting rod be denoted by mcrank , m piston , and mconn respectively. Let us assume that all the rigid bodies are symmetric. In other words, the center of mass for each rigid body is at its geometric center. Following the procedure outlined above in Equation (12), we obtain the approximate model shown in Figure 5, mO
= 12 mcrank
mQ
= 12 mcrank + 12 mconn
mP
= 12 mconn + m piston
(16)
Y
m Q
r
3
r m O
2
θ
φ
x
m P
X
Figure 5 The approximate dynamic model for the slider crank mechanism The total kinetic energy of the mechanism is given by: 1 2 K = mQ (r 2 θ& ) 2
2 1 sin (θ + φ) 2 θ& 2 = 1 J θ& 2 + m P x& = mQ (r 2 ) + mP r 2 2 cos φ 2 2
The effective inertia of the mechanism as “seen” at the crankshaft is:
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(17)
sin(θ + φ) 2 2 J = mQ (r 2 ) + m P r 2 cos φ If the slider crank mechanism is viewed as a system, there are two primary sources of energy interaction that will change the kinetic energy of the mechanism. First, there is the work done by the gases in the combustion chamber. Second, the crankshaft of the mechanism is coupled to a load, and the moment required to turn the load (or alternatively, the resistance offered by the load), called the turning moment does work on the system. Y
M t
Q
θ
O
patm A
P
pA
X
x
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 M t, and the atmospheric pressure by patm. The power associated with external forces acting on the mechanism is given by: P = M t θ& + ( p atm
− p ) A x&
where A is the cross-sectional area of the piston given by A =
π D 2 4
,
D being the bore (diameter) of the piston.
Since the kinetic energy is given by (17), we can apply Equation (11) to get: 1 dJ & 2 2 dt
θ + 12 J (2 θ& &θ&) = M t θ& + ( patm − p ) A x&
(18)
Let us assume that the engine is running at constant speed. This is at best an approximation since the engine speed is never constant but instead fluctuates within a narrow band. If this assumption is valid, the term associated with the angular acceleration drops out and we get the following equation: 1 dJ & 2 2 dt
θ = M t θ& + ( p atm − p ) A x&
and an expression for the turning moment that looks like:
M θ& = ( p − p t
atm ) A x&
+
1 d m (r )2 Q 2 2 dt
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+ m P r 2
sin(θ + φ) & 2 θ cos φ 2
Substituting from (15) and dividing by the crank speed, we get an expression for the turning moment: M t = ( p − p atm ) Ar 2
sin (θ + φ) cos φ
2
+ m P (r 2 )
2 sin (θ + φ ) r 2 cos θ
cos 3 φ
+ cos φ cos(θ + φ) θ& 2 r 3 cos φ
(19)
If the slider crank mechanism were massless, the turning moment would be equal to be equal to the moment due to the pressure in the cylinder: M p
= ( p − p atm ) Ar 2
sin (θ + φ)
(20)
cos φ
The moments due to the inertia of the moving parts in the mechanism is given by the inertial moment, M i, M i
= m P (r 2 )
2
sin (θ + φ ) r 2 cos
cos 3 φ
θ + cos φ cos(θ + φ ) θ& 2 r 3 cos φ 2
(21)
Typical plots for the three moments are shown in Figure 7. 700 600 500 ) s r e 400 t e m n o 300 t w e N ( 200 s t n e m 100 o M
0 -100 -200
0
100
200
300 400 500 Crank Angle (degrees)
600
700
Figure 7 The moment due to the cylinder pressure ( M p, dotted), the inertial moment ( M i, dashed), and the turning moment ( M t, 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 r k pmax patm V c L D α
N m piston mcrank mconn
Definition compression ratio expansion coefficient maximum pressure (after combustion), p3 atmospheric pressure clearance volume stroke (2 times crank radius) bore ratio of crank to conn. rod length Engine speed mass of the piston effective mass of the crank shaft for one cylinder mass of the connecting rod
Value 10.3 1.4 100 atmospheres 14.7 psi or 1.01325 bar 0.05 liters 86.4 mm 81 mm 0.35 3000 RPM 1 lbs 15 lbs/4 1.75 lbs
110 100 90 ) s e r e h p s o m t a ( e r u s s e r P r e d n i l y C
80 70 60 50 40 30 20 10 0
0
100
200
300 400 500 Crank Angle (degrees)
600
700
Figure 8 The cylinder pressure in atmospheres for a single cylinder of the sample engine shown in the table below (
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Table 2). An ideal air cycle is assumed except for two changes. The intake pressure is assumed to be 0.9 atmospheres and the exhaust pressure is assumed be 1.0 atmospheres. 700 600 500 ) s r e 400 t e m n o 300 t w e N ( 200 s t n e m 100 o M
0 -100 -200
0
100
200
300 400 500 Crank Angle (degrees)
600
700
Figure 9 The turning moment for a balanced four cylinder engine. The dotted lines show the turning moments from each cylinder while the solid line shows the tota l 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 you to analyze a four-str oke internal combustion engine. Specifically we want to be able to do the following tasks. 1. Plot the cylinder pressure against the crank angle for 720 degrees (a complete cycle). Note that Figure 1 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 the piston 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 single cylinder 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 is the highest. 9. Suggest a suitable remedy for reducing the engine fluctuations.
Note
You may choose any engine to analyze. The data in Table 2 is a good starting point. Data for more engines is 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 , m piston, 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
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specifications
(analogous
to
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 Pressure 1 bar 1 atmosphere 1 psi
105 Newtons/meter2 14.7 lbs/inch2 (psi) 2 6895 Newtons/meter
1 liter 1 cc 1 cc
1000 cubic centimeters (cc) 6 3 10 mm 10-3 meter3
1 kg (weight)
2.2 pounds
1 rotation per minute (RPM)
2π /60 radians per second
1 Watt 1 Horsepower
1 Newton meter/second 745.6 Watts
Volume
Mass Angular velocity Power
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