Week 5 – Topic:

Mechanical Modeling of

Systems

Control System EngineeringPE-3032Prof. CHARLTON S. INAODefense Engineering College, Debre Zeit , Ethiopia

1

Instructional Objectives

In this lesson students will:1) Review the mechanical, electrical, hydraulic ,pneumatic, and

fluid fundamentals2) Learn how to find and construct mathematical model for

linear time invariant mechanical, electrical, pneumatic , hydraulic, and fluid systems.

3) Review of Laplace transform as applied to transfer function4) Solve practical samples and application

System Modeling Definition

Systems modeling or system modeling is the interdisciplinary 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 .

Dynamic systemsTo be able to describe how the output of a system

depends on its input and how the output changes with time 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 be composed 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.

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 the behaviour 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

Mechanical systemsMechanical 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.

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 k is a constant termed the stiffness.

2 DashpotThe '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 (Figure b). 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 of newton 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.

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

Example

Derive a model for the mechanical system represented by the system of mass, spring and dashpot given in Figure a. The input to the system is the force F and the output is the displacement y.

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 just one free-body diagram and that is shown in Figure b. As the free-body diagram indicates, the net force acting on the mass is the applied force minus the forces exerted by the spring and by the dashpot:

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.

Application Example: MechanicalSpring-dashpot-mass model

Problem 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..

Problem No.2. Derive the differential equation describing the motion of the mass m1 in the figure when a force F is applied.

Solution: Using Hooke’s LawConsider 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 have an 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

Problem No. 3Derive a differential equation relating the input and output for each of the systems shown in figure a.

Answer.

Figure a

Rotational systems

• In control systems we are often concerned with rotational systems, e.g. we might want a model for the behavior of a motor drive shaft (Figure) and how the driven load rotation will be related to the rotational twisting input to the drive shaft.

• For rotational systems the basic building blocks are a torsion spring, a rotary damper and the moment of inertia (Figure a, b, c).

1 Torsional spring

The 'springiness' or 'stiffness' of a rotational spring is represented by a torsional spring. For a torsional spring, the angle θ rotated is proportional to the torque T:

where k is a measure of the stiffness of the spring.

2 Rotational dashpot

The damping inherent in rotational motion is represented by a 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 θ and thus:

where c is the damping constant.

3 .InertiaThe 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 of inertia I results in an angular acceleration a and thus, since angular acceleration is the rate of change of angular velocity ω with time, i.e. dω/dt, and angular velocity ω is the rate of change of angle θ with time, i.e. dθ/dt, then the angular acceleration is d(dθ /dt)/dt and so:

Example

Develop a model for the system shown in Figure a of the rotation of a disk as a result of twisting a shaft. Figure (b) shows the free-body diagram for the system.

The torques acting on the disk are the applied torque T, the spring torque kθ and 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:

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:

Id2θ/dt2 + c dθ/dt + kθ=T

Problem 2

Figure

Answer.

Derive a differential equation relating the input and output for each of the systems shown in the figure .

From T- cdθ/dt - k θ

Analogous Quantities (Force-Voltage Analogy)

Mechanical SystemElectrical System

Translatory RotationalForce (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 (θ)

Charge (q)

Velocity(x) Angular Velocity(¨θ) Current (i)

Analogous Quantities (Force-Current Analogy)

Mechanical SystemElectrical System

Translatory RotationalForce (f) Torque (T) Current (i)Displacement (x) Angular

Displacement ( (θ)Flux linkages (F)

Velocity(x) Angular Velocity (θ) 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)

Electrical systems

The basic elements of electrical systems are the pure components of resistor, inductor and capacitor (Figure), the term pure is used to indicate that the resistor only possesses the property of resistance, the inductor only inductance and the capacitor only capacitance.

1 Resistor For a resistor, resistance R, the potential

difference v across it when there is a current i through it is given by:

2 Inductor• For an inductor, inductance L, the potential

difference v across it at any instant depends on the rate of change of current i and is:

dt

diLRivv a

aba

Electrical Application problems

Problem 1.Derive the transfer function shown below:

RCTwheresT

sRCsVi

sVoFunctionTransfer

sRCsVosVithen

sVoRsVior

ssCVosIorsIsC

sVo

sVosRIsVi

getwetransformLaplacetheTaking

1

1

)1(

1

)(

)(

)1)(()(

)()(

)()()(1

)(

)()()(

,

CVo(s)s

Example 1

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

Hydraulic System

Pneumatic System

Component parts of Pneumatic System

Pneumatic Service Units

Comparison Between Pneumatic Systems and Hydraulic Systems

The fluid generally found in pneumatic systems is air; in hydraulic systems it is oil. And it is primarilythe different properties of the fluids involved that characterize the differences between the two systems.

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 hydraulic systems.

4. Output powers of pneumatic systems are considerably less than those of hydraulic systems.

5. Accuracy of pneumatic actuators is poor at low velocities, whereas accuracy of hydraulic actuators may be made satisfactory at all velocities.

6. In pneumatic systems, external leakage is permissible 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 when air is used, whereas they are always needed in hydraulic systems.

8. Normal operating temperature for pneumatic systems is 5° to 60°C (41° to 140°F). The pneumatic system, however, can be operated in the 0° to 200°C (32° to 392°F) range. Pneumatic systems are insensitive to temperature changes, in contrast to hydraulic systems, in which fluid friction due to viscosity depends greatly on temperature. Normal operating temperature for hydraulic systems is 20° to 70°C (68° to 158°F).

9. Pneumatic systems are fire- and explosion-proof, whereas hydraulic systems are not, unless nonflammable liquid is used.

Advantages and Disadvantages of Hydraulic Systems.

• There are certain advantages and disadvantages in using hydraulic systems rather than other systems.

Some of the advantages are the following:

1. Hydraulic fluid acts as a lubricant, in addition to carrying away heat generated in the system to a convenient heat exchanger.

2. Comparatively small-sized hydraulic actuators can develop large forces or torques.(Pascal’s Law)

3. Hydraulic actuators have a higher speed of response with fast starts, stops, and speed reversals.

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.

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 system that is free from leaks, the system tends to be messy.

5. Contaminated oil may cause failure in the proper functioning of a hydraulic system.

6. As a result of the nonlinear and other complex characteristics involved, the design of sophisticated hydraulic systems is quite involved.

7. Hydraulic circuits have generally poor damping characteristics. If a hydraulic circuit is not designed properly, some unstable phenomena may occur or disappear, depending on the operating condition.

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 in the 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 ResistanceHydraulic CapacitanceHydraulic Inertance

ALρ

I

ρgA

C

qpp

R 21

Figure shows the basic form of building blocks for hydraulic systems.

1 Hydraulic resistance• Hydraulic resistance R is the resistance to

flow which occurs when a liquid flows from one diameter pipe to another (Figure a) and is defined as being given by the hydraulic equivalent of Ohm's law:

R=p1-p2/q

2 Hydraulic capacitance• Hydraulic capacitance C is the term used to

describe energy storage where the hydraulic liquid is stored in the form of potential energy (Figure b). The rate of change of volume V of liquid stored is equal to the difference between the volumetric rate at which liquid enters the container q1 and the rate at which it leaves q2, i.e.

;h=p/ρg

3 Hydraulic inertance• Hydraulic inertance is the equivalent of

inductance in electrical systems. To accelerate a fluid a net force is required and this is provided by the pressure difference (Figure c). Thus:

Example• Develop a model for the hydraulic system

shown in Figure where there is a liquid entering a container at one rate q1 and leaving through a valve at another rate q2.

Review Questions: