mechanical , Liquid and Electrical Modeling

1. Control System Engineering
PE-3032
Prof. CHARLTON S. INAO
Defense Engineering College,
Debre Zeit , Ethiopia
Week 5 – Topic:
Mechanical
Modeling of Systems
1

2. 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 for
linear 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

3. 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 .

4. Dynamic systems
To 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.

5. 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

6. 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.

7. 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.

9. 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 (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:
Dashpot --this
device uses the
viscous drag of a
fluid, such as oil,
to provide a
resistance that is
related linearly to
velocity.
where c is a constant. ; i. e., c is the viscous damping coefficient, given in
units of newton seconds per meter (N s/m)

12. 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

15. 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.

16. 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:

17. Then applying Newton's second law, this force must be equal to
ma, where a is the acceleration, and so:
The Net force is the force
applied to the mass to cause it
to accelerate.
The term second-order is used because the equation
includes as its highest derivative d2y/dt2.

18. Application Example: Mechanical
Spring-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;
therefore md2x/dt2; =F- k1x-k2x md2x/dt2; + x(k1-k2) = F
Ans..

19. 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 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 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

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

21. 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.

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

23. 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.

24. 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.

25. 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 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:

28. 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.

29. 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:

30. 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

31. Problem 2
Derive a differential equation relating the input and output for
each of the systems shown in the figure .
Figure
Answer.
From T- cdθ/dt - k θ

33. Analogous Quantities
(Force-Voltage Analogy)
Mechanical System
Electrical System
Translatory Rotational
Force (f) Torque (T) Voltage (e)
Mass (M) Moment of Inertia (J) Inductance (L)
Viscous friction
Viscous friction
Resistance (R)
Coefficient (C)
Coefficient (C)
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)

34. Analogous Quantities
(Force-Current Analogy)
Mechanical System
Electrical System
Translatory Rotational
Force (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
Viscous friction
Coefficient (B)
Coefficient (f)
Reciprocal of
Resistance (1/R)
Spring (K) Torsional Spring
Constant (K)
Reciprocal of
Inductance (1/L)

35. 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.

36. 1 Resistor
For a resistor, resistance R, the potential
difference v across it when there is a current i
through it is given by:

37. 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:

45. v v Ri L dia
a - b = a +
dt

46. Electrical Application problems
Problem 1.
Derive the transfer function shown below:

47. Taking the Laplace transform we get
Vi s RI s Vo s
( ) = 1 I ( s ) or I ( s ) =
sCVo ( s
)
or Vi s = R +
Vo s
( ) ( )
then Vi s = Vo s +
sRC
( ) ( )(1 )
Transfer Function Vo s
( )
where T RC
sT
1
Vi s sRC
sC
Vo s
=
+
=
+
= =
= +
1
1
(1 )
( )
( ) ( ) ( )
,
s CVo(s)

48. Example 1

50. 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

51. Hydraulic System

52. Pneumatic System

53. Component parts of Pneumatic
System

54. Pneumatic Service Units

56. 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 primarily
the different properties of the fluids involved that
characterize the differences between the two
systems.

57. 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.

58. 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.

59. 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.

60. Advantages and Disadvantages of
Hydraulic Systems.
• There are certain advantages and
disadvantages in using hydraulic systems
rather than other systems.

61. 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.

62. 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.

63. 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.

64. 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.

65. Fluid and Liquid Systems
A 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 Resistance
Hydraulic Capacitance
Hydraulic Inertance
R = p1 -
p2
C A
=
ρg
I Lρ
A
q
=

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

67. 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

68. 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.

69. ;h=p/ρg

71. 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:

73. 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.

75. Review Questions: