Transcript System Models - University of Ottawa

System Models
Mathematical Models
Mechanical System Building Blocks
Electrical System Building Blocks
Fluid System Building Blocks
Thermal Systems Building Blocks
Mathematical Models
• Think how systems behave with time when subject to
some disturbances.
• In order to understand the behaviour of systems,
mathematical models are required.
• Mathematical models are equations which describe the
relationship between the input and output of a system.
• The basis for any mathematical model is provided by the
fundamental physical laws that govern the behaviour of
the system.
Building Blocks
• Systems can be made up from a range of building
blocks.
• Each building block is considered to have a single
property or function.
• Example: an electric circuit system which is made up
from blocks which represent the behaviour of resistance,
capacitance, and inductor, respectively.
• By combining these building blocks a variety of electrical
circuit systems can be built up and the overall inputoutput relationship can be obtained.
• A system built in this way is called a lumped parameter
system.
Mechanical System Building Blocks
• Basic building block: spring, dashpots, and masses.
• Springs represent the stiffness of a system
• Dashpots represent the forces opposing motion, for
example frictional or damping effects.
• Masses represent the inertia or resistance to
acceleration.
• Mechanical systems does not have to be really made up
of springs, dashpots, and masses but have the
properties of stiffness, damping, and inertia.
• All these building blocks may be considered to have a
force as an input and displacement as an output.
Rotational Systems
• The mass, spring, and dashpot are the basic building blocks for
mechanical systems where forces and straight line displacements are
involved without any rotation.
• If rotation is involved, then the equivalent three building blocks are a
torsional spring, a rotary damper and the moment of inertia (i.e. the
inertia of a rotating mass).
• With a torsional spring the angle  rotated is proportional to the
torque: T = k.
• With a rotary damper a disc is rotated in a fluid and the resistive
torque T is proportional to the angular velocity .
• The moment of inertia block exhibit the property that the greater the
moment of inertia J the greater the torque needed to produce an
angular acceleration
d
T  c  c
; T  Ja
dt
Stiffness of a Spring
• Stiffness of a spring is described as the relationship
between the force F used to extend or compress a
spring and the resulting extension or compression x.
• In the case of spring where the extension or
compression is proportional to the force (linear spring): F
= kx, where k is a constant, the bigger the value of k the
greater the forces have to be to stretch or compress the
spring and so the greater the stiffness.
F
x
Spring
Translational Spring, k (N)
Appiedforce Fa (t ) in Newton
Linear velocityv(t ) (m/sec)
Linearpositionx (t ) (m)
Fa (t )  k s x (t )
1
x (t ) 
Fa (t )
ks
dx(t ) 1 dFa (t )
v(t ) 

dt
k s dt
t
Fa (t )  k s  v (t )dt
t0
x(t)
Fa(t)
Rotational Spring, ks (N-m-sec/rad)
AppiedtorqueTa (t ) (N - m)
Angular velocity (t ) (rad/sec)
Angulardisplacement  (t ) (rad)
Ta (t )  Bm (t )
1
 (t )  Ta (t )
ks
 (t ) 
d (t ) 1 dTa (t )

dt
k s dt
t
Ta (t )  k s   (t )dt
t0
 (t)  (t)
Fa(t)
ks
Dashpot
• The dashpot block represents the types of forces
experienced when pushing an object through a fluid or
move an object against frictional forces. The faster the
object is pushed the greater becomes the opposing
forces.
• The dashpot which represents these damping forces that
slow down moving objects consists of a piston moving in
a closed cylinder.
• Movement of the piston requires the fluid on one side of
the piston to flow through or past the piston. This flow
produces a resistive force. The damping or resistive
force is proportional to the velocity v of the piston: F = cv
or F = c dv/dt.
Translational Damper, Bv (N-sec)
Appiedforce Fa (t ) in Newton
Linear velocityv(t ) (m/sec)
Linearpositionx (t ) (m)
Fa (t )  Bm v (t )
1
v(t ) 
Fa (t )
Bm
dx(t )
Fa (t )  Bm v (t )  Bm
dt
1 t
x (t ) 
Fa (t ) dt

Bv t0
x(t)
Fa(t)
Bm
Rotational Damper, Bm (N-m-sec/rad)
AppiedtorqueTa (t ) (N - m)
Angular velocity (t ) (rad/sec)
Angulardisplacement  (t ) (rad)
Ta (t )  Bm (t )
1
 (t ) 
Ta (t )
Bm
d (t )
Ta (t )  Bm (t )  Bm
dt
1 t
 (t ) 
Ta (t )dt

Bm t 0
 (t)  (t)
Fa(t)
Bm
Mass
• The mass exhibits the property that the bigger the mass the
greater the force required to give it a specific acceleration.
• The relationship between the force F and acceleration a is
Newton’s second law as shown below.
• Energy is needed to stretch the spring, accelerate the mass and
move the piston in the dashpot. In the case of spring and mass we
can get the energy back but with the dashpot we cannot.
Force
Acceleration
Mass
dv
d 2x
F  ma  m
m
2
dt
dt
Mechanical Building Blocks
Building Block
Spring
Dashpot
Mass
Spring
Damper
Moment of inertia
Equation
Translational
F = kx
F = c dx/dt
F = m d2x/dt2
Rotational
T = k
T = c d/dt
T = J d2/dt2
Energy representation
E = 0.5 F2/k
P = cv2
E = 0.5 mv2
E = 0.5 T2/k
P = c2
P = 0.5 J2
Building Mechanical Blocks
Output, displacement
• Mathematical model of a
machine mounted on the
ground
Mass
d 2x
dx
m
 c  kx  F
2
dt
dt
Ground
Input, force
Building Mechanical Blocks
Moment of inertia
Torque
Torsional resistance
• Mathematical model
rotating a mass
d 2
of
a
d
J
c
 k  T
2
dt
dt
Block model
Shaft
Physical situation
Electromechanical Analogies
•
From Newton’s law or using Lagrange equations of motions, the secondorder differential equations of translational-dynamics and torsionaldynamics are found as
m
d 2x
dt 2
d 2
 Bv
dx
 k s x  Fa (t ) (Translationaldynamics)
dt
d
j 2  Bm
 k s  Ta (t ) (Torsionaldynamics)
dt
dt
Electrical System Building Blocks
•
The basic building blocks of electrical systems are resistance, inductance and
capacitance.
2
Resistor: v  iR; P  i R
1
1 2
Inductor: i   vdt; E  Li
L
2
dv
1 2
Capacitor: i  C ; E  Cv
dt
2
Resistance, R (ohm)
Appiedvoltagev(t )
i(t)
Currenti(t )
v(t )  Ri(t )
1
i(t )  v(t )
R
v(t)
R
Inductance, L (H)
Appiedvoltagev(t )
Currenti (t )
di(t )
v(t )  L
dt
t
1
i (t )   v(t )dt
L t0
i(t)
v(t)
L
Capacitance, C (F)
Appiedvoltagev(t )
Currenti (t )
1 t
v(t )   i (t )dt
C t0
dv(t )
i (t )  C
dt
i(t)
v(t)
C
For a series RLC circuit, find the characteristic equation
and define the analytical relationships between the
characteristic roots and circuitry parameters.
d 2i
R di 1
1 dva


i
2
L dt LC
L dt
dt
R
1
2
s  s
0
L
LC
The characteristicrootsare
2
s1  
R
1
 R 
 


2L
LC
 2L 
2
s2  
R
1
 R 
 


2L
LC
 2L 
Fluid System Building Blocks
•
The basic building blocks of fluid systems are the volumetric rate of
flow q and the pressure difference.
Input
Output
Pressure difference
Volumetric rate of flow
Fluid system can be divided into two types: hydraulic and pneumatic.
Hydraulic resistance is the resistance to flow of liquid as the liquid flow
through valves or changes in pipe diameter takes place.
p1  p2  Rq
p1 - p2 is pressure difference
R is the hydraulic resistance
q is the volumetric rate of flow
•
Hydraulic capacitance is the term used to describe energy storage with a
liquid where it is stored in the form of potential energy. A height of liquid in a
container is one form of such a storage. For such capacitance, the rate of
change of volume V in the container (dV / dt) is equal to the difference
between the volumetric rate at which liquid enters the container q1 and the
rate at which it leaves q2.
dV
q1  q2 
; V  Ah
dt
dh
q1  q2  A
dt
A dp
q1  q2 
pg dt
( p is liquid density;g is theacceleration due to gravity)
A
dp
C
; q1  q2  C
pg
dt
•
Hydraulic inertance is the equivalent of inductance in electrical systems or a
spring in mechanical systems. To accelerate a fluid and so increase its
velocity a force is required.
F1=p1A
Mass m
F1  F2  p1 A  p2 A  ( p1  p2 ) A
( p1  p2 ) A  m a
( p1  p2 ) A  m
dv
dv
dq
 ALp  Lp
dt
dt
dt
dq
Lg
; I is thehydraulicinertance
dt
A
L is thelengthof theblock and g is thedensity
p1  p2  I
F2=p2A
L
•
With pneumatic systems the three basic buildings blocks are as with
hydraulic systems, resistance, capacitance, and inertance. However,
gasses differ from liquids in being compressible.
dm p1  p2
Resistance

dt
R
dm
d ( p1  p2 )
Capacitance
C
dt
dt
dm 1
Inertance   ( p1  p2 )dt
dt L
A fluid system
p is theliquid density
g is theacceleration due to gravity
q is the volumetric rateof flow
q1
dp
q1  q2  C
(Capacitor)
dt
p1  p2  Rq2 (Resistance)
h
q2
hpg
p1-p2  hpg; q2 
R
hpg
d (hpg)
dh pgh
q1 
C
A 
R
dt
dt
R
Thermal System Building Blocks
• There are only two basic building blocks for thermal systems:
resistance and capacitance.
• There is a net flow of heat between two points if there is a
temperature difference between them.
• The value of the resistance depends on the mode of heat transfer.
T T
T T
q  2 1  Ak 2 1
R
L
A : Crosssectionalarea of the material through which theheat is beingconducted
L is the lengthof material between the pointsat which he
t temperature are T1 and T2 .
k is the thermal conductivity
Thermal System
q is thenet rateof heat flow
C is thecapacitance
R is the thermalresistance
T
TL
q
TL  T
q
R
dT
dT
q1  q2  C
;q  C
dt
dt
dT TL  T
C

dt
R
dT
RC
 T  TL
dt