Solution for Internal test-1
2009-2010
M.Kaliamoorthy
Solution Key for Test-1
PART-A
1. Principles of Electromechanical Energy Conversion
a) To establish an expression for electromagnetic torque in terms of machine
variables.
b) The derivation of Equivalent circuit representation of magnetically coupled
circuits.
c) The concept of sinusoidally distributed winding.
d) The concept of rotating air gap mmf.
e) The derivation of winding inductances.
2. Transducer converts energy from one form to another. Types of Transducers are
(1) Transducers (for measurement and control)
These devices transform the signals of different forms. Examples are microphones,
pickups, and speakers.
(2) Force producing devices (linear motion devices)
These type of devices produce forces mostly for linear motion drives, such as
relays, Solenoids (linear actuators), and electromagnets.
(3) Continuous energy conversion equipment
These devices operate in rotating mode. A device would be known as a generator if
it converts mechanical energy into electrical energy, or as a motor if it does the
other way around (from electrical to mechanical).
3. Torque produced in a machine because of self inductance of the machine is called
reluctance torque. Generally they are expressed as
1 L  
T  i 2 11
Where L11 is the self inductance of the coil
2

 Energy input from
4. 
 Electrical Sources
 Increase in energy
  Mechanical Energy  
  
   Stored in magnetic
  Output
  Field


  Energy Converted in 

  
to
Heat




Solution for Internal test-1
2009-2010
M.Kaliamoorthy
5.
6. L12 
N2
N
Lm1  1 Lm 2  L21
N1
N2
7. Leakage Flux: The part of the Total magnetic flux that has its path wholly with in the
magnetic circuit is called the useful magnetic flux. The magnetic paths having paths
partly in the magnetic circuit and partly in the air is called the leakage flux.
Fringing Flux: When the flux line crosses the air gaps they tend to bulge out across the
edges of the air gap. This effect is called Fringing. The effect of fringing makes the
effective gap area larger than that of the original air gap.

W f   i  , x d - - - Energy
0
8.
i
W f    i, x di - - - Co - Energy
'
0
Solution for Internal test-1
2009-2010
M.Kaliamoorthy
1
Wf i, x  1 2 dL( x)
'
 i
9. W f i, x   i 2 L( x) and F 
2
x
2
dx
'
1
Wf  ,  1 2 dL( )
 i
10. W f  ,    i 2 L( ) and F 
2

2
d
PART-B
11. Reluctance at the air gap =  
2l
2 x 5 x10 -3

 2.2104 x 106
0 A 4 x 6 x 6 x 10-7 x 10-4
Therefore Inductance of the coil =
N2
3002

 0.0407 H
 2.2104 x 106
(a) Current in the coil due to 120 V DC source =
Therefore field energy stored = W fld 

Lifting Force =
V 120

 20 A
R
6
1 2 1
Li  x 0.0407 x 20 2  8.143 Joules
2
2
W fld
x
1  N 2 x  0 A  2 1  300 2 x 4 x 10-7 x 36 x 10-4  2
 
i  
i
2
2x
2
2x



1  300 2 x 4 x 10-7 x 36 x 10-4  2 0.0407

 20 
4
x
x

F  
  0.0407  0.0407
0.0407


 1628.6 Nm
2


x  x 
x
25 x 10-6
(b) For AC excitation the impedance of the coil is
Z  R  jL  6  j 2 x 50  x 0.0407  6  j15.34
I 
120
6  15.34 2
2
 7.29 A
 Field Energy Stored 
1
x0.0407 x7.29 2  1.0814 Joules
2
Solution for Internal test-1

2009-2010
M.Kaliamoorthy
W fld
x
1  N 2 x  0 A  2 1  300 2 x 4 x 10-7 x 36 x 10-4  2
 
i  
i
2
2x
2
2x


Lifting Force =
2
-7
-4
1  300 x 4 x 10 x 36 x 10 
0.00540941
2
 
 7.29 
4
x
x

F  
  0.00540941 0.00540941 0.00540941

 216.3Nm
 
x 
x
x2
25 x 10-6
This is utmost one-eighth of the lifting force obtained with a dc supply voltage. Hence
Lifting magnets are normally operated from dc sources.
11. (b) (i)
The electromechanical energy conversion devices operate with electrical system on one
side and mechanical system on the other side. The behavior of the entire
electromechanical system must be satisfactorily under steady state as well as under the
electromechanically transients. In view of this, the operation of the entire system,
comprising of electrical system conversion device and mechanical system, need to be
investigated in detail.
The voltage equation for the electrical system shown in the figure above is given by
Vt  iR 
d
       (1)
dt
If the flux linkage  can be expressed as   L( x).i, Therefore the above equation
becomes,
Vt  iR  L( x)
di
dL( x) dx
i
.        (2)
dt
dx dt
Solution for Internal test-1
The term L
2009-2010
M.Kaliamoorthy
di
of the above expression is the transformer voltage (of self inductance)
dt
term, because it involves the time derivative of current. The third term of equation (2) is
the speed or rotational voltage term, because of the presence of speed
dx
in it and it is
dt
this term which is responsible for the energy transfer between the external electric system
and the energy conversion means.
The mechanical portion of the above figure shows symbols for a spring (spring
constant K), a damper (damper constant B), a mass M and an external mechanical
excitation force Fi. The force and displacement x can be related as follows
Spring : FK  K x  x0 
Damper: FD  B
Mass: FM  M
dx
dt
d 2x
dt 2
Where X0 is the value of x with the spring in unstreched position. Force equilibrium thus
requires
Fm  FK  FD  FM  Fi
 K x - x 0   B
dx
d 2x
 M 2  Fi      (3)
dt
dt
Thus equations (2) and (3) describe the total behavior of a linear electromechanical
system.
11 (b) (ii)
Wfld  Total Energy Stored
1
1
2
2
L1i1  L2i2  Mi1i2
2
2
1
1
 11  3 cos 2 0.7 2  7  2 cos 2 0.82  11cos  0.7 x 0.8
2
2
 0.24511  3 cos 2   0.327  2 cos 2   0.5611 cos  
 2.695  0.735cos2  2.24  0.64cos2  6.16cos

W fld  4.935  1.375 cos 2  6.16 cos 
Solution for Internal test-1
Torque  
2009-2010
M.Kaliamoorthy
W fld

1
 1

 - 0.7 2 x 6sin2  - 0.82 x 4sin2  - 0.56 x 11sin  
2
 2

1
1
 0.7 2 x 6sin2   0.82 x 4sin2   0.56 x 11sin 
2
2
 1.47sin2   1.28sin2   6.16sin 

 

T  2.75 sin 2  6.16 sin 
Given that   50 0
T  2.75 x sin 2 x - 50  6.16 x sin(-50)  7.426 Nm
12 (a)
The general principle for force and torque calculation discussed above is equally
applicable to multi-excited systems. Consider a doubly excited rotating actuator shown
Schematically in the diagram below as an example. The differential energy and co energy
functions can be derived as following:
A doubly excited actuator
Solution for Internal test-1
2009-2010
M.Kaliamoorthy
For magnetically linear systems, currents and flux linkages can be related by constant
inductances as following
Where
The magnetic energy and co energy can then be expressed as
Solution for Internal test-1
2009-2010
M.Kaliamoorthy
Respectively, and it can be shown that they are equal.
Therefore, the torque acting on the rotor can be calculated as
Because of the salient (not round) structure of the rotor, the self inductance of the stator is
a function of the rotor position and the first term on the right hand side of the above
torque expression is nonzero for that
dL11
 0 . Similarly, the second term on the right
d
hand side of the above torque express is nonzero because of the salient structure of the
stator. Therefore, these two terms are known as the reluctance torque component. The last
term in the torque expression, however, is only related to the relative position of the
stator and rotor and is independent of the shape of the stator and rotor poles.
12 (b)
Consider a singly excited linear actuator as shown below. The winding resistance is R. At
a certain time instant t, we record that the terminal voltage applied to the excitation
winding is v, the excitation winding current i, the position of the movable plunger x, and
the force acting on the plunger F with the reference direction chosen in the positive
direction of the x axis, as shown in the diagram. 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
done by the force acting on the plunger during this time interval is thus
Solution for Internal test-1
2009-2010
M.Kaliamoorthy
A singly excited linear actuator
The amount of electrical energy that has been transferred into the magnetic field and
converted into the mechanical work during this time interval can be calculated by
subtracting the power loss dissipated in the winding resistance from the total power fed
into the excitation winding as
From the above equation, we know that the energy stored in the magnetic field is a
function of the flux linkage of the excitation winding and the position of the plunger.
Mathematically, we can also write
Therefore, by comparing the above two equations, we conclude
Solution for Internal test-1
2009-2010
M.Kaliamoorthy
From the knowledge of electromagnetic, the energy stored in a magnetic field can be
expressed as
For a magnetically linear (with a constant permeability or a straight line magnetization
curve such that the inductance of the coil is independent of the excitation current) system,
the above expression becomes
and the force acting on the plunger is then
In the diagram below, it is shown that the magnetic energy is equivalent to the area above
the magnetization or  -i curve. Mathematically, if we define the area underneath the
magnetization curve as the co energy (which does not exist physically), i.e.
we can obtain
Therefore
Solution for Internal test-1
2009-2010
M.Kaliamoorthy
From the above diagram, the co energy or the area underneath the magnetization curve
can be calculated by
For a magnetically linear system, the above expression becomes
and the force acting on the plunger is then
12 (b) (ii)
Since it is the case of current excitations, the expression of co energy will be used
W f i1 , i2 , x  
1
1
2
2
L11i1  L12i1i2  L22i2
2
2
1 
1
1 


  2   x 100  (200)  1   x 50
2x 
2x

 2x 
25
 450 
x
'
(a)
Ff 
W f
x
'

25
x2
1
1
0.5
0.5
 Wm   F f dx   
25
dx  25 Joules
x2
Solution for Internal test-1
(b)
We1 
2  x 1
 i d
1
1  x  0.5 
1
2009-2010
 i1 1  x  1  2  x  0.5
1  L11i1  L12i2
1 
1
5

  2   x 20  (10)  40 
2x 
2x
x

1 x  0.5  50, 2 x  1  45
 We1  2045  50   100 Joules
similarly We 2  i2 1  x  1  2  x  0.5
2  L12i1  L22i2
1
1 
5

(20)  1  (10)  10 
2x
x
 2x 
 2  x  0.5  0, 2  x  1  5

We 2  (10)( 5)  50 joules
Net electrical Energy input  We  We1  We 2
 100  50  50 Joules
M.Kaliamoorthy