# Electromechanical Energy Conversion

Topics: Magnetic field, Force, Electric motor Pages: 42 (3503 words) Published: April 18, 2013
Introduction
Chapter 3

Electromechanical Energy
Conversion
Topics to cover:
1. Introduction
3. Force and Torque
5. Friction

2. Electro-Motive Force (EMF)
4. Doubly-Excited Actuators
6. Mechanical Components

Introduction (Cont.)

For energy conversion between electrical and mechanical forms, electromechanical devices are developed. In general, electromechanical energy conversion devices can be divided into three categories:

– Transducers (for measurement and control), which transform signals of different forms. Examples are microphones, pickups, and speakers
– Force producing devices (linear motion devices), which
produce forces mostly for linear motion drives, such as relays, solenoids (linear actuators), and electromagnets.
– Continuous energy conversion equipment, which operate in rotating mode. A device would be known as a generator if it
convert mechanical energy into electrical energy, or as a motor if it does the other way around (from electrical to mechanical).

Lorentz Force & EMF
Lorentz force is the force on a point charge due to electromagnetic fields. It is given by the following equation in terms of the electric and magnetic fields

F  q(E vB)

The induced emf in a conductor of length l moving with a speed v in a uniform magnetic field of flux density B can be determined by a

e  vB  dl

b

In a coil of N turns, the induced emf can be calculated by e  

Concept map of electromechanical system modeling

d
dt

where  is the flux linkage of the coil and the minus sign indicates that the induced current opposes the variation of the field. It makes no difference whether the variation of the flux linkage is a result of the field variation or coil movement.

EMF

EMF

- Example: EMF in a Linear Actuator

- Example Solution

Sketch L(x) and calculate the induced emf in the excitation coil for a linear actuator shown below.

Assuming infinite permeability for the magnetic core and ignore the fringing effect, we can express the self inductance of the coil as

L x  
where

Rg  x 

N2
o N 2 l
d  x

Rg  x 
2g

L(x)
L(0)

2g
o  d  x l
O

is the air gap reluctance.

 e

 N 2l
d d  Li 
di dL dx
di

 L i
=L x   i o
v
2g
dt
dt
dt
dx dt
dt

EMF

- A Single Conductor in a Uniform Field

e   I dc

If i=Imsint ,

e

Force and Torque

- Example Solution (Cont.)
If i=Idc ,

o N 2 l
2g

 Im
 Im

o N 2l
2g

For a single conductor in a uniform
magnetic field, we have

v

 d  x I m cos t  vI m sin t

o N l
2

2g

o N 2 l
2g

d

Fm  Il  B

o N 2 l

In a rotating system, the torque about an
axis can be calculated by

2g

 d  x  cost  v sin t 

T r Fm
v



  d  x   

 d  x  2  2  v 2 cost  arctan

where r is the radius vector from the
axis towards the conductor.

B

Fm

l
I

X

Force and Torque
- A Singly Excited Actuator
Consider a singly excited linear actuator. After a time interval dt, we notice that the plunger has moved for a distance dx under the action of the force F. The mechanical work done by the force acting on the plunger during this time interval is thus

dWm  Fdx

Force and Torque
- A Singly Excited Actuator
The amount of electrical energy that has been transferred into the magnetic field and converted into the mechanical work during dt is

dWe  dWf  dWm ; dWe  eidt  vidt Ri2dt

e  d dt  v  Ri

Because

dWf  dW  dW  eidt  Fdx  id  Fdx
e
m

we obtain

From the total differential
dW f   , x  



 W f  , x 

i

Therefore,

 W f  , x 



d 

and

 W f  , x 
x

F

dx

W f   , x 
x

Force and Torque

Force and Torque

- A Singly Excited Actuator (Cont.)

- A Singly Excited Actuator (Cont.)

From the knowledge of electromagnetics, the energy stored in a...