Week 10 Part 3 PE 6282Mecchanical Liquid and Electrical

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

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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/Viscous


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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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Week 10 Part 3 PE 6282Mecchanical Liquid and Electrical

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