III. Electromechanical Energy
Conversion
Schematic representation of an
electromechanical energy conversion device
copper losses
core losses
(field losses)
mechanical losses
Differential energy input from electrical source:
dWelec  Vt i dt  i 2 Rdt  ei dt
dWmech  net mechanical energy output  mechanical losses
Differrential energy balance eqn.:
dWelec  dWmech  dW fld  field losses
For a lossless magnetic energy storage system:
dWelec  dWmech  dW fld
1
1. Energy in Magnetic Field
dWelec  ei dt
d
dWelec  N
idt
dt
dWelec  Nid
dWelec  F d
Welec  1 F d
0
 W fld
 01 F d  00 F d
Coenergy
with 0 = 0 and F0 = 0
  0F1  dF
W fld
W fld  W fld  F1 1
2
For a linear magnetic circuit:
1
  F1 1
W fld  W fld
2
1
 R 12
2
1
 PF1 2
2
where R and P are the reluctance and permeance of the magnetic circuit respectively
2. Energy & Force in Singly-Excited
E.M.C device
Electromechanical relay
• Assume mobile armature is initially at open position
• When the coil is excited by i(t), (t) is produced in M.C. and
an electromagnetic force ffld is exerted on mobile armature tending
to align it with the densest part of M.F.
• When the armature moves  
3
Armature Movement
• Slow motion case
• Fast motion case
( is gradually increased during motion)
( is constant during motion)
• Normal motion case
(i) Slow motion
Welec  W fld  W mech
Welec  1 F d  A1  A3
0
W fld  0 F d  00 F d   A1  A2    A2  A4 
1
F is constant during motion
W mech  Welec  W fld  A3  A4
Shaded area gives Wmech
4
(ii) Fast motion
Welec  W fld  W mech
Welec  1 F d  A1
0
W fld   F d   F d   A1  A2    A2  A3  A4 
 is constant during motion
1
0
0
0
W mech  Welec  W fld  A3  A4
Shaded area gives Wmech
(iii) Normal motion
Welec  W fld  W mech
Shaded area gives Wmech
5
dWmech  f fld dx
• if dWmech > 0 then electical system does work on the armature
• if dWmech  0 then mechanical system does work on the armature
(a) Mechanical force in fast motion
dWelec  dW fld  dWmech
F d  dW fld  f fld dx
dW fld  F d  f fld dx
Note that
dW fld  , x  
So
F 
W fld  , x 

d 
W fld  , x 
x
dx
W fld  , x 
and

 is independent of x
f fld  
W fld  , x 
x
6
(a) Mechanical force in slow motion
W fld F, x   F   W fld F, x 
dW fld F, x   d F    dW fld F, x 
 F, x    dF  F d  F d  f fld dx 
dW fld
 F, x    dF  f fld dx
dW fld
Note that dW fld
 F , x  
So

 F , x 
W fld
F
dF 
 F , x 
W fld
x
dx
W fld F, x 
and
F
F is independent of x
f fld 
W fld F, x 
x
For a linear magnetic circuit
1
1
1
2
W fld  W fld  F   R   PF
2
2
2
2
where R and P are the reluctance and permeance of
the magnetic circuit respectively
1.
When  is constant during motion
1 dR
f fld    2
2
dx
2.
When F is constant during motion
1
dP 1 2 dL
f fld  F 2
 i
2
dx 2 dx
7
Ex-1: Derive an expression for the electromagnetic force ffld
produced by the E.M.D. shown in the figure and then plot ffld
against x.
Note: Initial gap length is c and i is constant during motion
Ex-2: For the electromagnet shown below:
(a) For V = 120 Vdc, determine the energy stored in magnetic field
and the lifting force produced
(b) For V = 120 Vac at 60 Hz, determine the average lifting force
produced
Note: Initial gap length is 5 mm, internal resistance of the winding r = 6,
N = 300, Ac = Ag = 6 x 6 = 36 cm2.
8
3. Rotating Machines
Linear
Motion
Linear
Displacement
x, m
Linear
Speed
v, m/s
Rotational
Motion
Angular
Displacement
, rad
Angular
Speed
, rad/s
fast motion
slow motion
f fld  
f fld 
W fld  , x 
x
W fld F, x 
x
Mass
Force
m, kg
f, N
Inertia
Torque
J, kg m2
Te, Nm
For flux independent of  during motion:

Te  
W fld  ,  

For mmf independent of  during motion:

Te 
W fld F,  

(a) Singly-excited device
  r t  0
For a linear M.C.
Te 
1 2 dL
i
2 d
Pmech  Te r
Pmech 
r 
d
dt
1 2 dL
i
2 dt
9
Ex: The self inductance L  in the figure as a function of rotor position is given
by
L  = L0 + L2 cos( 2)
where the stator current is i(t) = Imsin(t)
(i) Obtain an expression for electromagnetic torque produced
(instantaneous torque)
(ii) Let  = rt + 0, find necessary condition for non-zero
average torque
(b) Doubly-excited device
1  L1i1  Mi2
2  L2 i2  Mi1
1  L1  i1 t   M  i2 t 
2  L2  i2 t   M  i1 t 
e1 
e1  L1
di1
dL d
di
dM d
 i1 1
 M 2  i2
dt
d dt
dt
d dt
e2  L2
di2
dL d
di
dM d
 i2 2
 M 1  i1
dt
d dt
dt
d dt
d1
dt
v1  e1  r1i1
e2 
d2
dt
v 2  e2  r2 i2
r 
Terminal voltage equation (in matrix form):
 v1   r1
v    0
 2 
0   i1   L1

r2   i 2   M
M  d  i1  d  L1

L2  dt  i 2  d  M
d
dt
M   i1 
ω
L2   i 2  r
10
 v1   r1
 
v 2   0
v1i1  v 2 i2  i1
Total power
0   i1   L1

r2   i 2   M
M   i1 
ω
L2   i 2  r
 v1 
i2  
v2 
v1 i1  v 2 i 2  I T V  I T RI  I T L
Pelec
M  d  i1  d  L1

L2  dt  i 2  d  M
dI T dL
I
I r
dt
d
(in matrix form)
Pcu
Total electrical power input:
Pelec  i12 r1  i22 r2  L1i1
Copper loss
di1
di
di
di
dL
dM
dL
 Mi1 2  L2 i2 2  Mi2 1  i12 1  r  2i1i2
 r  i22 2  r
d
dt
dt
dt
d
d
dt
Rate of increase in stored energy in MF + Electr.power converted to mech. en.
Pelec  i12 r1  i22 r2  L1i1
dL
dM
dL
di
di
di
di1
 Mi1 2  L2 i2 2  Mi 2 1  i12 1  r  2i1i2
 r  i22 2  r
d
d
d
dt
dt
dt
dt
Magnetic stored energy
W fld   i 1d1   i 2 d2

P fld 
P fld  L1i1
1
1
L1i12  L2 i22  Mi1i2
2
2
dW fld
dt
di1
di
di
di 1 dL
1 dL
dM
r
 Mi1 2  L2 i2 2  Mi2 1  i12 1  r  i22 2  r  i1i2
dt
dt
dt
dt 2 d
2 d
d
Pmech  Pelec  Pcu  P fld
Pmech 
1 2 dL1
1 dL
dM
 r  i22 2  r  i1i2
r
i1
2 d
2 d
d
Te 
1 2 dL1 1 2 dL2
dM
i1
 i2
 i1i2
2 d 2 d
d
Pmech  Te r
In matrix form: Te 
1 T dL
I
I
2 d
11
Ex1: Te ?
Ex2: Te ?
12