Transcript Slide 1

7. Modeling of Electromechanical Systems
Electromechanical systems consist of an electrical subsystem and a mechanical
subsystem with mass and possibly elasticity and damping.
In some devices, such as motors and speakers, the mass is driven by a force
generated by the electrical subsystem.
In other devices, such as microphones, the motion of the mass generates a
voltage or current in the electrical subsystem.
DC MOTORS
There are many types of electric motors, but the two main categories are direct
current (dc) motors and alternating current (ac) motors.
Within the dc motor category there are the armature-controlled motor and the
field-controlled motor.
We aim to control the speed or motion of dc motors.
There are many different types of servo-drivers. Most are designed to
control the speed of dc motors, which improves the efficiency of
operating servomotors.
Here we shall discuss only armature control of a dc motor and obtain its
mathematical model.
Elements of DC Motor
The basic elements of a motor, as shown in the
Figure are the stator, the rotor, the armature, and
the commutator.
The stator is stationary and
magnetic field.
provides
the
The rotor is an iron core that is supported by
bearings and is free to rotate.
The coils are attached to the rotor, and the
combined unit is called the armature.
Basic Principle of DC Motor
The majority of electromechanical devices utilize a magnetic field.
The basic principle of Dc motor is based on a wire carrying a current
within a magnetic field: a force is exerted on the conductor by the field.
Right Hand Rule for Magnetic Field
The direction of the force (F) due
to a magnetic field (B) is
perpendicular to the direction of
motion.
We will use right hand rule to find the
direction of the force of a magnetic field
q a
B
F
r
Tm  rF
Tm: Motor torque (moment)
The product of the magnetic force (F) and the
radius (r) will generate the motor moment.
Example 7.1
System with DC Motor
1  m  2  m (Rigid shaft)
4  L
Jm , Bm
Vk +
-
z1
1
2
By
Motor
in shaft 2:
4
3
JL
Ra , La
2
Ki , Kb
Ra : Motor’s resistance
La : Motor’s inductance
K2: Rotational spring constant of
shaft numbered 2
z2
K2
Jm : Motor’s mass moment of inertia
Bm : Motor’s rotational damping coefficient
Ki : Motor’s torque constant
Kb : Motor’s back emf constant
Vk : Motor’s supply voltage
q a
: Motor’s current
JL : Load’s mass moment of inertia
By : Rotational damping coefficient in bearings
N
z1
z2
1
E 2  K 2 ( L   3 )2
2
Nm    3
Tm  K iq a
The torque Tm developed by the
motor is proportional to the product of
Motor’s torque constant and the
current . When the sign of the current
is reversed, the sign of the torque will
be reversed.
Vb  K b  m
When the armature is rotating, the
voltage (back emf) Vb is directly
proportional to the angular velocity of
the motor.
Jm , B m
Vk +
-
m  1   2 (Rigid shaft)
z1
1
2
By
DC Motor
4
3
Ra , La
K i , Kb
2
z2
1
E2  K 2 (L   3 )2
2
z
N 1
Nm    3
z2
Tm  K iq a
Vb  K b m
In shaft 2 :
JL
K2
Energy equations for Lagrange equation:
1
1
1 2 1 2
2
E1  Laq a  J mm  J LL
E2  K 2 (L  Nm )2
2
2
2
2
W  Vk qa  R aq aqa  Bm mm  K iq a m K b mqa
 By  mm  By  mm By (  N m )(  Nm )  By  LL
W  ( Vk  Raq a  K b m )qa  ( Bm m  K iq a  2By  m  N 2By  m )m
 By  LL
Input : Vk; Generalized variables : qa, θm, θL
Lagrange Equation→
Homework 07-Problem 1
Example 7.2 Movable plate capacitor
x(t)
k/2
q
b/2
R
-
The plate of the capacitor at left hand side
is fixed. The other plate is movable. The
moving plate is fixed to the body with the
elements of the spring k and damper c.
fa(t)
+
Vk
Movable, m
C
Vk is the power supply. Vk is
connected to the lines with the
resistor R and the capacitor C in
serial.
k/2
b/2
Fixed
The force fa is applied to movable plate. The
displacement of movable plate is x(t). The value
of the capacitor depends on the changing of the
distance between the plates changes.
d0
C( x)  C0
d0  x
C 0 , d 0 are the constants.
Inputs: Vk(t) ve fa(t)
Generalized variables: q(t) ve x(t)
x(t)
Inputs: Vk(t) ve fa(t)
k/2
b/2
q
Generalized variables: q(t) ve x(t)
Movable, m
R
C( x)  C0
fa(t)
Vk
C
+
k/2
b/2
Sabit
d0
d0  x
For the electromechanical system, We can
write the energy and virtual work equation as
follows.
1
E1  mx 2
2
1 k 2 1 (d 0  x ) 2
E2  2
x 
q
22
2 C0d0
b
W  Vk q  Rq q  fax  2 x x
2
x(t)
Inputs: Vk(t) ve fa(t)
Generalized variables: q(t) ve x(t)
k/2
b/2
q
Movable, m
R
fa(t)
Vk
C
+
k/2
b/2
Sabit
1 k 2 1 (d 0  x ) 2
E2  2
x 
q
22
2 C0d0
b
W  Vk q  Rq q  fax  2 x x
2
1
E1  mx 2
2
The equations of motion of the system are
obtained by applying the Lagrange equation to the
general variables.
1
mx  kx 
q 2  fa  bx
2C0d0
(d 0  x )
q  Vk  Rq
C0d0
Set of non-linear differential equations
Runge-Kutta method
Linearization
Homework 07- Problem 2: Movable core inductance