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7/30/2019 Week 10 Part 3 PE 6282Mecchanical Liquid and Electrical
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Week 5 Topic:
MechanicalModeling of Systems
Control System EngineeringPE-3032Prof. CHARLTON S. INAODefense Engineering College,Debre Zeit , Ethiopia
1
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Instructional Objectives
In this lesson students will:
1) Review the mechanical, electrical, hydraulic ,pneumatic, and
fluid fundamentals
2) Learn how to find and construct mathematical model forlinear time invariant mechanical, electrical, pneumatic ,
hydraulic, and fluid systems.
3) Review of Laplace transform as applied to transfer function
4) Solve practical samples and application
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System Modeling Definition
Systems modeling or system modeling is theinterdisciplinary study of the use of models to
conceptualize and construct systems in
engineering.(mechanical, hydraulic, fluid,liquid level,
electrical , electromechanical and thermal). System analysis, acquiring information on various
aspects of system performance. system analysis was
carried out using the physical system subjected to
test input signals, observing its corresponding
response.
System model, a simplified representation of
the physical system under analysis .
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Dynamic systems
To be able to describe how the output of a system
depends on its input and how the output changes withtime when the input changes, we need a mathematical
equation relating the input and output. The following
describes how we can arrive at the input-output
relationships for systems by considering them to becomposed of just a few simple basic elements.
Thus, if we want to develop a model for a car suspension
we need to consider how easy it is to extend or compress
it, i.e. its stiffness, the forces damping out any motion of
the suspension and the mass of the system and so its
resistance of the system to acceleration, i.e. its inertia.
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So we think of the model as having the
separate elements of stiffness, damping
and inertia which we can represent by a
spring, a dashpot and a mass (Figure ) and
then write down equations for thebehaviour of each element using the
fundamental physical laws governing the
behaviour of each element. This way of
modelling a system is known as lumped-
parameter modelling.
Car Suspension System
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Mechanical systems
Mechanical systems, however complex, have stiffness(or springiness),damping and inertia and can be
considered to be composed of basic elements which
can be represented by springs, dashpots and masses.
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1 Spring
The 'springiness' or 'stiffness' of a system can be represented
by an ideal spring (ideal because it has only springiness and
no other properties). For a linear spring (Figure a), the
extension y is proportional to the applied extending force F
and we have:
F=ky
where kis a constant termed the stiffness.
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2 Dashpot
The 'damping' of a mechanical system can be represented by a dashpot.
This is a piston moving in a viscous medium, e.g. oil, in a cylinder (Figureb). Movement of the piston inwards requires the trapped fluid to flow out
past edges of the piston; movement outwards requires fluid to flow past
the piston and into the enclosed space. For such a system, the resistive
force F which has to be overcome is proportional to the velocity of the
piston and hence the rate of change of displacement y with time, i.e.
dy/dt. Thus we can write:
where c is a constant. ; i. e., c is the viscous damping coefficient, given in
units ofnewton seconds per meter (N s/m)
Dashpot --this
device uses the
viscous drag of a
fluid, such as oil,
to provide a
resistance that is
related linearly to
velocity.
http://en.wikipedia.org/wiki/Viscoushttp://en.wikipedia.org/wiki/Newton_secondhttp://en.wikipedia.org/wiki/Newton_secondhttp://en.wikipedia.org/wiki/Viscous7/30/2019 Week 10 Part 3 PE 6282Mecchanical Liquid and Electrical
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3 Mass
The 'inertia' of a system, i.e. how much it resists being accelerated
can be represented by mass. For a mass m (Figure c), the
relationship between the applied force F and its acceleration a is
given by Newton's second law as F = ma. But acceleration is the
rate of change of velocity v with time /, i.e. a = dy/dt, and velocity
is the rate of change of displacement y with time, i.e. v = dy/dt.Thus a = d(dy/dt)/dt and so we can write
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Example
Derive a model for themechanical system
represented by the
system of mass,
spring and dashpotgiven in Figure a. The
input to the system
is the force Fand the
output is thedisplacement y.
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To obtain the system model
we draw free-body
diagrams, these being
diagrams of masses showing
just the external forces
acting on each mass. For the
system in Figure a ,we have
just one mass and so justone free-body diagram and
that is shown in Figure b. As
the free-body diagram
indicates, the net forceacting on the mass is the
applied force minus the
forces exerted by the spring
and by the dashpot:
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Then applying Newton's second law, this force must be equal to
ma, where a is the acceleration, and so:
The term second-order is used because the equation
includes as its highest derivative d2y/dt2.
The Net force is the force
applied to the mass to cause it
to accelerate.
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Application Example: Mechanical
Spring-dashpot-mass modelProblem 1. Derive the differential equation describing the
relationship between the input force F and the output of the
displacement x for the system shown below.
Solution:
Netforce=F- k1x-k2x; but Netforce= md2x/dt2;
md2x/dt2;therefore =F- k1x-k2x md2x/dt2;+ x(k1
-k2
) = FAns..
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Problem No.2.Derive the differential equation describing
the motion of the mass m1 in the figurewhen a force F is applied.
Solution:
Using Hookes Law
Consider first just m1 and the force acting
on it. ; thus the force on the lower spring
is k(x2-x1);
then the force exerted by the upper
spring is k2(x3-x2).
Net force=k1(x2-x1) k2(x3-x2)
The net force will cause the mass to havean acceleration
md2x/dt2; =k1(x2-x1) k2(x3-x2)
But F=k1(x2-x1), the force
causing the extension of the
lower spring.
Hence, the final equation is
md2x/dt2 + K2(x3-x2)=F;
F
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Problem No. 3
Derive a differential equation relating the input and output for
each of the systems shown in figure a.
Answer.
Figure a
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Rotational systems
In control systems we are often concernedwith rotational systems, e.g. we might want amodel for the behavior of a motor drive shaft
(Figure) and how the driven load rotation willbe related to the rotational twisting input tothe drive shaft.
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For rotational
systems the basicbuilding blocks
are a torsion
spring, a rotary
damper and the
moment of
inertia (Figure a,
b, c).
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1 Torsional spring
The 'springiness' or 'stiffness' of a rotational spring is
represented by a torsional spring. For a torsional spring, the
angle rotated is proportional to the torque T:
where k is a measure of the stiffness of the spring.
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2 Rotational dashpot
The damping inherent in rotational motion is represented bya rotational dashpot. For a rotational dashpot, i.e. effectively
a disk rotating in a fluid, the resistive torque T is proportional
to the angular velocity and thus:
where c is the damping constant.
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3 .Inertia
The inertia of a rotational system is represented by the moment
of inertia of a mass.
A torque T applied to a mass with a moment ofinertia Iresults
in an angular acceleration a and thus, since angularacceleration
is the rate of change of angular velocity with time, i.e. d/dt,and angular velocity is the rate of change of angle with time,
i.e. d/dt, then the angular acceleration is d(d/dt)/dt and so:
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Example
Develop a model for thesystem shown inFigure a of the
rotation of a disk as aresult of twisting ashaft. Figure (b) showsthe free-body diagram
for the system.
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The torques acting on the disk are the applied torque T,
the spring torque kand the damping torque cw. Hence:
We thus have the second-order differential equation
relating the input of the torque to the output of the angle of
twist:
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Application Example:
Rotational system
a) Rotating mass on the end of the shaft
b) The building block model
A motor is used to rotate a load.
Devise a model and obtain a
differential equation for it.
Answer:
Id2/dt2 + c d/dt + k=T
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Problem 2
Figure
Answer.
Derive a differential equation relating the input and output for
each of the systems shown in the figure .
From T- cd/dt - k
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Analogous Quantities
(Force-Voltage Analogy)
Mechanical SystemElectrical System
Translatory Rotational
Force (f) Torque (T) Voltage (e)
Mass (M) Moment of Inertia (J) Inductance (L)Viscous friction
Coefficient (C)
Viscous friction
Coefficient (C)
Resistance (R)
Spring Stiffness (K) Torsional Spring
Stiffness (K)
Reciprocal of
Capacitance (1/C)Displacement (x) Angular Displacement () Charge (q)
Velocity(x) Angular Velocity() Current (i)
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Analogous Quantities
(Force-Current Analogy)Mechanical System
Electrical SystemTranslatory Rotational
Force (f) Torque (T) Current (i)
Displacement (x) Angular
Displacement ( ()
Flux linkages (F)
Velocity(x) Angular Velocity () Voltage (e)
Mass (M) Moment of Inertia (J) Capacitance (C)
Viscous friction
Coefficient (B)
Viscous friction
Coefficient (f)
Reciprocal of
Resistance (1/R)
Spring (K) Torsional Spring
Constant (K)
Reciprocal of
Inductance (1/L)
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Electrical systems
The basic elements ofelectrical systems are the
pure components of
resistor, inductor and
capacitor (Figure), the termpure is used to indicate
that the resistor only
possesses the property of
resistance, the inductor
only inductance and the
capacitor only capacitance.
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1 Resistor
For a resistor, resistance R, the potential
difference v across it when there is a current i
through it is given by:
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2 Inductor
For an inductor, inductance L, the potential
difference v across it atany instant depends
on the rate of change of current iand is:
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dt
diLRivv a
aba
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Electrical Application problems
Problem 1.
Derive the transfer function shown below:
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RCTwheresT
sRCsVi
sVoFunctionTransfer
sRCsVosVithensVoRsVior
ssCVosIorsIsC
sVo
sVosRIsVi
getwetransformLaplacetheTaking
1
1)1(
1
)(
)(
)1)(()()()(
)()()(1
)(
)()()(
,
CVo ( s )
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Example 1
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H d li & P ti
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Hydraulic & Pneumatic
Fundamentals
Pneumatic Hydraulic
Compressed Air Industrial Oil
Light loads,6-8 bars Heavy loads, unlimited, no OL
Fast, erratic Slow, stable
Compressor Pump
Compressible Incompressible
Air Receiver/Air Reservoir Tank
Exhaust to Atmosphere Liquid back to Tank
PU tubes Hi pressure Wire braided hose
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Hydraulic System
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Pneumatic System
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Component parts of Pneumatic
System
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Pneumatic Service Units
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Comparison Between Pneumatic
Systems and Hydraulic Systems
The fluid generally found in pneumatic systems isair; in hydraulic systems it is oil. And it is primarily
the different properties of the fluids involved that
characterize the differences between the twosystems.
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These differences can be listed as follows:
1. Air and gases are compressible, whereas oil is
incompressible (except at high pressure).
2. Air lacks lubricating property and always
contains water vapor. Oil functions as a hydraulic
fluid as well as a lubricator.
3. The normal operating pressure of pneumatic
systems is very much lower than that of hydraulicsystems.
4 O f i
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4. Output powers of pneumatic systems are
considerably less than those of hydraulic systems.
5. Accuracy of pneumatic actuators is poor at lowvelocities, whereas accuracy of hydraulic actuators
may be made satisfactory at all velocities.
6. In pneumatic systems, external leakage ispermissible to a certain extent, but internal leakage
must be avoided because the effective pressure
difference is rather small. In hydraulic systems
internal leakage is permissible to a certain extent,
but external leakage must be avoided.
7 No return pipes are required in pneumatic systems
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7. No return pipes are required in pneumatic systemswhen air is used, whereas they are always neededin hydraulic systems.
8. Normal operating temperature for pneumaticsystems is 5 to 60C (41 to 140F). The pneumaticsystem, however, can be operated in the 0 to 200C(32 to 392F) range. Pneumatic systems are
insensitive to temperature changes, in contrast tohydraulic systems, in which fluid friction due toviscosity depends greatly on temperature. Normaloperating temperature for hydraulic systems is 20to 70C (68 to 158F).
9. Pneumatic systems are fire- and explosion-proof,whereas hydraulic systems are not, unlessnonflammable liquid is used.
Advantages and Disadvantages of
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Advantages and Disadvantages of
Hydraulic Systems.
There are certain advantages and
disadvantages in using hydraulic systemsrather than other systems.
S f th d t th f ll i
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Some of the advantages are the following:
1. Hydraulic fluid acts as a lubricant, in additionto carrying away heat generated in the
system to a convenient heat exchanger.
2. Comparatively small-sized hydraulic actuatorscan develop large forces or torques.(Pascals
Law)
3. Hydraulic actuators have a higher speed of
response with fast starts, stops, and speed
reversals.
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4. Hydraulic actuators can be operated under
continuous, intermittent, reversing, and
stalled conditions without damage.
5. Availability of both linear and rotary actuators
gives flexibility in design.
6. Because of low leakages in hydraulic
actuators, speed drop when loads are applied
is small.
O th th h d l di d t t d
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On the other hand, several disadvantages tend
to limit their use.
1. Hydraulic power is not readily available
compared to electric power.
2. Cost of a hydraulic system may be higher than
that of a comparable electrical system
performing a similar function.
3. Fire and explosion hazards exist unless fire-resistant fluids are used.
4 Because it is difficult to maintain a hydraulic
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4. Because it is difficult to maintain a hydraulicsystem that is free from leaks, the system tendsto be messy.
5. Contaminated oil may cause failure in the properfunctioning of a hydraulic system.
6. As a result of the nonlinear and other complex
characteristics involved, the design ofsophisticated hydraulic systems is quite involved.
7. Hydraulic circuits have generally poor damping
characteristics. If a hydraulic circuit is notdesigned properly, some unstable phenomenamay occur or disappear, depending on theoperating condition.
Fluid and Liquid Systems
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Fluid and Liquid SystemsA common fluid control system involves liquid flowing into a container and out
of it through a valve, the requirement being to control the level of the liquid in the
container. For such a system we need a model which indicates how the height of liquid inthe container is related to the rates of inflow and outflow.
For a fluid system the three building blocks are resistance, capacitance and
inertance; these are the equivalents of electrical resistance, capacitance and inductance.
The equivalent of electrical current is the volumetric rate of flow and of potential
difference is pressure difference.
Hydraulic Resistance
Hydraulic Capacitance
Hydraulic Inertance
ALI gACqpR 2
Figure shows the basic form of building blocks for hydraulic
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Figure shows the basic form of building blocks for hydraulic
systems.
1 Hydraulic resistance
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1 Hydraulic resistance
Hydraulic resistance R is the resistance to
flow which occurs when a liquid flows fromone diameter pipe to another (Figure a) and
is defined as being given by the hydraulic
equivalent of Ohm's law:
R=p1-p2/q
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2 Hydraulic capacitance
Hydraulic capacitance C is the term used to
describe energy storage where the hydraulicliquid is stored in the form of potentialenergy (Figure b). The rate of change of
volume V of liquid stored is equal to thedifference between the volumetric rate atwhich liquid enters the container q1 and therate at which it leaves q2, i.e.
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;h=p/g
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3 Hydraulic inertance
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3 Hydraulic inertance
Hydraulic inertance is the equivalent of
inductance in electrical systems. To acceleratea fluid a net force is required and this is
provided by the pressure difference (Figure
c). Thus:
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E l
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Example
Develop a model for the hydraulic systemshown in Figure where there is a liquid
entering a container at one rate q1 andleaving
through a valve at another rate q2.
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Review Questions:
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Review Questions:
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