ECE330 Fall 16 Lecture4 PDF
ECE330 Fall 16 Lecture4 PDF
ECE330 Fall 16 Lecture4 PDF
http://www4.hcmut.edu.vn/~nqnam/lecture.php
Lecture 4 1
In general, both the currents in the coils and the force or torque are
time-varying.
becomes
Contour C
Hl = Ni
Faradays law
d
B da v=
d
( N ) = d
dt S
E dl = becomes
C dt dt
Application of Gausss law depends on geometry and is required for
system with different H. Conservation of charge leads to KCL.
Lecture 4 3
d di dx
v= = +
dt i dt x dt
transformer voltage speed voltage
Lecture 4 4
Electrically linear system
= L( x )i
Hence,
dL( x ) dx
v = L(x )
di
+i
dt dx dt
For system with no moving parts
di
= Li and v=L
dt
For multi-port system
d k N k di j M k dx j
vk = = j =1 + j =1 k = 1,2,..., N
dt i j dt x j dt
Force and flux linkage can be functions of all variables.
Lecture 4 5
Example 4.1
Find H1, H2, , and v, with the following assumptions: 1) = for the
core, 2) g >> w, x >> 2w and 3) no leakage flux.
Ex. 4.2: Fig. 4.7. Find s, r as a function of is, ir, and , and find vs and
vr of the cylindrical rotor. Assuming = , and g << R and l.
N s is N r ir N s is + N r ir
H r1 = = H r3 Hr2 = = H r 4
2g 2g
s = N s s = N s 0 H r1 Rl + N s 0 H r 2 R( )l
Simplified to
2
s = N s2 L0 i s + N s N r L0 1 ir 0 < <
Similarly,
2
r = N s N r L0 1 i s + N r L0 i r
2
0 < <
Practical machines,
dis di d
v s (t ) = Ls + M cos( ) r ir M sin ( )
dt dt dt
Lecture 4 7
Example 4.4
0W 2
1 = N 1 1 =
3x
(2 N 2
1 1 i N1 N 2 i2 )
0W 2
2 = N 2 2 =
3x
( N N i
1 2 1 + 2 N 22 i2 )
Can we identify self and mutual inductances?
Lecture 4 8
Forces of Electric Origin Using Energy
Lecture 4 9
Path B
b
Wm (b , xb ) Wm ( a , x a ) = i ( , x a )d f (b , x )dx
xb
e
a xa
Wm ( , x ) = i ( , x )d
This can then be generalized as
0
Lecture 4 10
Forces energy relations
Recall that
dWm = id f e dx
Wm ( , x ) Wm ( , x )
dWm = d + dx
x
Comparing the two equations, yielding
Wm ( , x )
i=
Wm ( , x )
fe =
x
Lecture 4 11
Example 4.5
2 wd 0 N 2i 2 wd 0 N 2 i i
= N = = = L0
g+x g 1+ x g 1+ x g
Solve for i
i= (1 + x g )
L0
2
Wm = i ( , x )d = (1 + x g )d = (1 + x g )
0 0 L0 2 L0
Calculate fe
Wm 2
f =
e
( , x ) =
x 2 L0 g
L20 i 2 1 L0 i 2
f (i, x ) =
e
=
2 L0 g (1 + x g ) 2 (1 + x g )2
2
Lecture 4 12
Force of Electric Origin Using Co-Energy
To compute Wm(, x), i = i(, x) is required. This could be complicated. It would
be more convenient to compute fe directly from = (i, x).
dWm = id f e dx d (i ) = id + di id = d (i ) di
dWm = d (i ) di f e dx d (i Wm ) = di + f e dx
Define co-energy as
fe
Lecture 4 13
Example 4.8
Find fe for the system in Fig. 4.22.
Ni
Riron
l 2x
Riron = c R gap = Rgap
A 0 A
Ni Ni Ni
= = =
Riron + R gap A + 0 A
lc 2x R(x )
Flux linkage and co-energy
N 2i N 2i 2
= N = W = (i, x )di =
i
'
R( x ) m
0 2 R(x )
Force of electric origin
Wm' N 2i 2 d 1 N 2i 2
f =
e
= =
x 2 dx R( x ) 0 A lcA + 20xA( )2
Lecture 4 14
Graphical interpretation of energy and co-energy
In electrically linear systems, both energy and co-energy are numerically
equal. In Fig. 4.24,
Wm = i( , x )d = Area A Wm' = (i, x )di = Area B
i
0 0
If (i, x) is a nonlinear function as illustrated in Fig. 4.25, then the two areas
are not numerically equal. However, fe derived using either energy or co-energy
will be the same.
Wm Wm'
f = lim
e
f = lim
e
x 0 x x 0 x
Lecture 4 15
dWm dx d d dx
= v1i1 + v 2 i2 f e = i1 1 + i2 2 f e
dt dt dt dt dt
or dWm = i1 d1 + i 2 d 2 f e dx
Consider
i1 d1 + i2 d2 = d (1i1 + 2 i2 ) 1 di1 2 di2
Hence,
d (1i1 + 2i2 Wm ) = 1di1 + 2 di2 + f e dx
Lecture 4 16
Force of electric origin in general multi-port system
Consider a system with N electrical and M mechanical ports, the flux linkages
are 1(i1, ..., iN, x1, ..., xM), ..., N(i1, ..., iN, x1, ..., xM).
Wm'
Wm'
i = i = 1,..., N
ii
Wm'
fi =
e
i = 1,..., M
xi
Lecture 4 17
Computation of Wm
To compute Wm, integration is done along the xi axes first, then along each of
the ii axes. While integrating along xi, Wm = 0 since fe is zero. Then,
( i2
+ 2 i1 , i2' ,...,0, x1 , x2 ,...xM di2' + ...
0
)
+ (i , i ,..., i )
iN
N 1 2 N 1 , iN' , x1 , x2 ,...xM diN'
0
Note the dummy variable of integrations. For special case of two-electrical and
two-mechanical port system,
i1
( ) i2
Wm' = 1 i1' ,0, x1 , x 2 di1' + 2 i1 , i2' , x1 , x 2 di2'
0 0
( )
And,
Wm' Wm'
f = 1
e
f =
2
e
dx1 dx 2
Lecture 4 18
Example 4.10
Compute Wm and torques of electric origin of a three-electrical and one-
mechanical port system.
1 (i1' ,0,0, , )di1' + 2 (i1 , i2' ,0, , )di2' + 3 (i1 , i2 , i3' , , )di3'
i1 i2 i3
Wm' =
0 0 0