Embed Size (px)
Citation preview
1
EE 554, HW#3, Spring 2020, Dr. McCalley, Due Tuesday 2/25.
SOLUTIONS
1. We argued in class that, for a 2-pole machine, the mutual inductances for rotor windings in quadrature (e.g., LFG,
LFQ, LGD, LDQ) are zero because the flux from one winding does not link with the coils of the other winding. Explain,
using a physical structure of a 4-pole machine together with a drawing of d-axis flux lines, why the d-axis flux does
not link with the q-axis windings.
Solution:
Consider a four-pole configuration, as shown below.
Consider if we could cut the rotor along the blue dashed line in the center of the above figure, then we could lay the
four different poles out side-by-side as shown below (here, there is no significance to the colors of the colored vertical
bars on either side of each pole).
In studying the lines of flux in the above figure, it is easy to see that most lines of flux are perpendicular to the q-axis
winding. The only exception to this are the two lines shown in the very center of the concentric flux-circles. However,
these two would cancel each other since they will be equal but opposite direction. The implication of this is the D-axis
windings do not link with Q-axis windings.
D1
d
N
S D1
d
D2
D2
Q1
Q1 Q2
Q2
STATOR
ROTOR
2
2. Problem 4.1
Parks transformation is an orthogonal one because P-`=PT. This question
requests that we observe what happens to the voltage equations when we
use a transformation that is not orthogonal. In fact, the original trans-
formation used by Park in his 1929 paper was not orthogonal. Let’s try it.
0
0 0
abc abc abc abc n
FGDQFGDQ FGDQ FGDQ
v iR v
v R i
Applying the Q-transformation, we get
4 4 4 4
0 0 0 00
0 00 0 0 0
term 1 term 2 term 4term 3
abc abc abc abc n
FGDQFGDQ FGDQ FGDQ
Q Q Q Qv iR v
v R iU U U U
Term 1:
0
4
0
0
dqabc abc
FGDQ FGDQFGDQ
Qv vQ v
v vvU
Term 2:
4
0 0
00
abc abc
FGDQ FGDQ
Q iR
R iU
But
10 0
4 4
0 00 0
dq dqabc abc
FGDQ FGDQFGDQ FGDQ
i ii iQ QU Ui ii i
Substitution yields
3
10
44
0 0 0000
dqabc
FGDQ FGDQ
iQ R QUR iU
110 0
4
0 00000
abc dq dqabc
FGDQFGDQ FGDQFGDQ
QR i iQR QQRU i iR
Before, when using transformation P, at this point we had 1
abcPR P
, and
because P is orthogonal and Rabc is diagonal, it is true that PRabcP-1=Rabc,
assuming Rabc also has equal diagonal elements (i.e., so that ra=rb=rc=r). Here,
however, Q is not orthogonal therefore we must do the math. For this we
need Q-1. You can get this using the method of cofactors. Then…
1
1 1 11 cos sin0 02 2 22
cos cos 120 cos 120 0 0 1 cos 120 sin 1203 0 0sin sin 120 sin 120 1 cos 120 sin 120
2 2 22cos cos 120 cos 120
3 sin sin 120 sin 120
abcQR Q
rr
r
r r r
r r rr r r
1 cos sin1 cos 120 sin 1201 cos 120 sin 120
3 0 02 0 02 30 0 0 0
23 0 030 02
abc
rr
r r Rrr
This is a nice surprise! Why did it happen? Does it mean that QAQ-1=A works
for any diagonal matrix A independent of whether Q is orthogonal or not?
No! Careful analysis of the math above will show that, if ra≠rb≠rc, we would
have obtained something quite different. Our conclusion is that QAQ-1=A if A
4
is diagonal and with equal elements on the diagonal. Does this work for any
matrix Q or just this one?
Answer: Let A=kU where k is the element along the diagonal (in our case,
k=r). Then QAQ-1=QkUQ-1=kQUQ-1=kQQ-1=kU=A. We should have noticed this
at the beginning to save ourselves some work!
Therefore:
110 0
4
0
0 00000
0 0 0 0 0 00 0 0 0 0 00 0 0 0 0 00 0 0 0 0 00 0 0 0 0 00 0 0 0 0 00 0 0 0 0 0
abc dq dqabc
FGDQFGDQ FGDQFGDQ
d
q
F F
G G
D D
Q Q
QR i iQR QQRU i iR
iririr
r ir i
r ir i
Now let’s go to term 3.
4
0
0
abc abc
FGDQFGDQ
Q Q
U
Using the same analysis we used in class for P, we find:
1
0 0abc dq dqQ QQ
And so term 3 is
5
-10 0dq
4
Term 3bTerm 3a
0 QQ λ
00
abc abc dq
FGDQFGDQFGDQ
Q Q
U
Let’s work on Term 3b. Given that
Re Re
1 1 12 2 22
cos cos 120 cos 1203 sin sin 120 sin 120
/ 2
Q
t
we can compute the derivative of Q:
0 0 02sin sin 120 sin 120
3 cos cos 120 cos 120Q
2
10 0 0 1 cos sin2
sin sin 120 sin 120 1 cos 120 sin 1203 cos cos 120 cos 120 1 cos 120 sin 120
0 0 0 0 0 04 3 2
0 0 0 09 2 3
3 20 0 0 0
2 3
And then 1
0dqQQ becomes
6
01
0
0 0 0 0
2 20 0
3 32 2
0 03 3
dq d q
q
d
Then Term 3 becomes
0
-10 0dq
Term 3bTerm 3a
02
32
QQ λ30
0000
qd
qabc dq d
FFGDQFGDQ
G
D
Q
Q
Now let’s get term 3a. From the lecture notes (see pp. 24-25 of “macheqts”),
0
4 4
110 0
14 4
0 00 0
0 00 0
dqabc abcaa aR
FGDQ FDQGRa RRFGDQ
aa aRdq dqaa aR
Ra RR FGDQ FDQGRa RR
Q Q iL LiL LU U
QL Q QLi iQ L L QL L i iU U L Q L
Again, from lecture notes (p. 2 of “macheqts,” eq. L-ex), we have cos 2 [ cos 2( 30 )] [ cos 2( 150 )] cos
[ cos 2( 30 )] cos 2( 120 ) [ cos 2( 90 )] cos( 120 )[ cos 2( 150 )] [ cos 2( 90 )] co
a S m S m S m F
b S m S m S m F
c S m S m S m
F
G
D
Q
L L M L M L MM L L L M L MM L M L L L
s 2( 240 ) cos( 240 )cos cos( 120 ) cos( 240 )sin sin( 120 ) sin( 240 ) 0cos cos( 120 ) cos( 240 )sin sin( 120 ) sin( 240 ) 0
F
F F F F
G G G
D D D R
Q Q Q
MM M M LM M MM M M MM M M
sin cos sinsin( 120 ) cos( 120 ) sin( 120 )sin( 240 ) cos( 240 ) sin( 240 )
0 00
0 00
G D Q a
G D Q b
G D Q b
FR
GG Y
DD
QY Q
M M M iM M M iM M M i
iMiL MiLiM L
where the submatrices of the inductance matrix are defined as
aa aR
Ra RR
L LL L
7
Now we have to evaluate each of the terms in the 2x2 matrix 1
00
1
dq aa aR dq
FDQGRa RRFGDQ
QL Q QL iiL Q L
This is tedious, but we must do it. So here we go.
Submatrix (1,1):
1
1 1 1cos [ cos(2( 30))] [ cos(2( 150))]2 2 22
cos cos 120 cos 120 [ cos(2( 30))] cos(2( 120)) [ cos(2( 90))]3 [ cos(2( 150))] [sin sin 120 sin 120
aa
S m S m S m
S m S m S m
S m
QL Q
L L M L M LM L L L M LM L
1 cos sin1 cos 120 sin 120
cos(2( 90))] cos(2( 240)) 1 cos 120 sin 120S m S mM L L L
1 1 1[ 2 ] [ 2 ] [ 2 ]
2 2 2 1 cos sin2 3 3 3[ ]cos( ) [ ]cos( 120) [ ]cos( 120) 1 cos 120 sin 1
3 2 2 23 3 3
[ ]sin( ) [ ]sin( 120) [ ]sin( 120)2 2 2
S S S S S S
S m S S m S S m S
S m S S m S S m S
L M L M L M
L L M L L M L L M
L L M L L M L L M
201 cos 120 sin 120
3[ 2 ] 0 0
2 0 022 3 3 3
0 [ ] 0 0 03 2 2 2
3 3 30 0 [ ] 0 0
2 2 2
S SS S
S m S S m S
S m S S m S
L ML M
L L M L L M
L L M L L M
Now we define:
0
3 32 ; ;
2 2S S d S m S q S m SL L M L L L M L L L M
Then
01
0
0 00 00 0
aa dqd
q
LQL Q L L
L
Submatrix (1,2):
1 1 1cos sin cos sin2 2 22
cos cos 120 cos 120 cos( 120) sin( 120) cos( 120) sin( 120)3 cos( 240) sin( 240) cos( 240) sin( 240)sin sin 120 sin 120
F G D Q
ar F G D Q
F G D Q
M M M MQL M M M M
M M M M
And this is the same thing we
obtained using P (see p. 27 of
“macheqts” notes).
8
0 0 0 00 0 0 02 3 3
0 0 0 03 2 2 0 0
3 30 0
2 2
F D F D
G Q
G Q
M M M MM M
M M
Submatrix (2,1):
1
cos cos( 120) cos( 240)1 cos sinsin sin( 120) sin( 240)1 cos 120 sin 120cos cos( 120) cos( 240)1 cos 120 sin 120sin sin( 120) sin( 240)
F F F
G G GRa
D D D
Q Q Q
M M MM M M
L Q M M MM M M
30 0
23
0 02
30 0
23
0 02
F
G
D
Q
M
M
M
M
Submatrix (2,2): This submatrix does not change. It is given by
0 0
0 0
0 0
0 0
F R
G Yrr
R D
Y Q
L M
L ML
M L
M L
So the term 3a is found by differentiating the below expression: 0
0
1
0 0
1
0 0 0 0 0 0
0 0 0 0
0 0 0 0
30 0 0 0
23
0 0 0 02
30 0 0 0
23
0 0 0 02
d F D
q G Q
d
F F R qdq aa aR dq
FFDQGFGDQ G G YRa RR
D
G
D R DQ
Q Y Q
L
L M MiL M Mi
M L M iQL Q QL i
ii M L ML Q L i
iM M L
i
M M L
We can then differentiate both sides to obtain:
9
0
0
0
0 0 0 0 0 0
0 0 0 0
0 0 0 0
30 0 0 0
23
0 0 0 02
30 0 0 0
23
0 0 0 02
d F D
q G Q
d
F F R qdq
F
G G YFGDQG
DD R D
Q
Q Y Q
L
L M MiL M Mi
M L M i
iM L M
i
iM M L
i
M M L
It is important to observe at this point that QLar≠[LarQ-1], i.e., the lower
left 4x3 submatrix is not the transpose of the upper right 3x4
submatrix. Using the P transformation does result in PLar=[LarP-1]. The
significance of this is that the elements in the submatrices represent
mutual inductances; for a physically realizable circuit, it must be the
case that Mij=Mji. Thus, the P transformation results in a system that
can be physically realized, but the Q transformation does not. You can
obtain the correct answers either way, but with the P transformation,
you can actually draw a circuit that corresponds to the equations and
thus better conceptualized the various relationships.
Finally, we express term 4:
4
0
000
1 1 12 2 22
cos cos 120 cos 1203 sin sin 120 sin 120
n n
n
n n
n
Q v Qv
U
vQv v
v
10
The product of the second row and vn, or of the third row and vn, will
include a summation of symmetrical components, which will be zero!
So the only non-zero element in Qvn will be the product of the first row
of Q and vn, i.e., the first element of the term 4 vector, which is
1 1 1
3 3 3
n
n n
n
vv vv
But recall from our circuit the voltage equation indicates that:
)()( cbanncbannnnn iiiLriiiiLriv (**)
Also, recall that from the Park’s transformation iodq=Qiabc that
1 1 12 2 22
cos cos 120 cos 1203 sin sin 120 sin 120
a
odq abc b
c
ii Qi i
i
And therefore the i0 current is:
0 0
2 1 1 1 1( ) ( ) ( ) 3
3 2 2 2 3a b c a b c a b ci i i i i i i i i i i ; 0( ) 3a b ci i i i
And so equation (**) above becomes
0 03 3n n nv i r L i
The problem asks us to express the voltage equations similar to
equation (4.39) in VMAF. Recall our originally transformed voltage
equations, below.
11
44 4 4
0 0 00 0
0 0 00 0 0
term 1 term 2 term 4term 3
abc abc abc abc n
FGDQFGDQ FGDQ FGDQ
Q Q Qv iR vP
v R i UU U U
We have done all the work necessary to express each term. Making those
substitutions, we obtain:
0
0
0
Term 1
Term 2
0 0 0 0 0 00 0 0 00 0 0 0
0 0 0 0 0 030 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 230 0 0 0 0 0
0 0 00 0 0 0 0 0 20 0 0 0 0 00 0 0 0 0 0
d F D
q G Q
dF F R
qdq
F FG
FGDQ G G
D D
Q Q
LL M M
L M Mirir M L Mirv
r iM Lv r i
r ir i
00 0
Term 3b
Term 3a
02 3 3
032 0
00 3 0003 00 0 0 0 002
030 0 0 0
2
n nqd
q
dF
G YG
DD R D
Q
Q Y Q
i i r L iiiiMii
M M Li
M M L
Term 4
On page 8 of this solution, we wrote:
0
0
1
0 0
1
0 0 0 0 0 0
0 0 0 0
0 0 0 0
30 0 0 0
23
0 0 0 02
30 0 0 0
23
0 0 0 02
d F D
q G Q
d
F F R qdq aa aR dq
FFDQGFGDQ G G YRa RR
D
G
D R DQ
Q Y Q
L
L M MiL M Mi
M L M iQL Q QL i
ii M L ML Q L i
iM M L
i
M M L
From this we may extract the following two equations:
d d d F F D D
q q q G G Q Q
L i M i M i
L i M i M i
Substitution of the previous equations into Term 3b of the voltage equations
results in
12
0
0
0
Term 1
Term 2
0 0 0 0 0 00 0 0 00 0 0 0
0 0 0 0 0 030 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 230 0 0 0 0 0
0 0 00 0 0 0 0 0 20 0 0 0 0 00 0 0 0 0 0
d F D
q G Q
dF F R
qdq
F FG
FGDQ G G
D D
Q Q
LL M M
L M Mirir M L Mirv
r iM Lv r i
r ir i
0
Term 3a
Term 3b
0
30 0 0 0
23
0 0 0 02
02
( )32
( )3
0000
d
q
FG Y
G
DD R D
Q
Q Y Q
q q G G Q Q
d d F F D D
iiiiMii
M M Li
M M L
L i M i M i
L i M i M i
0 0
Term 4
3 3000000
n ni r L i
Compare to the answer we got when using P:
0 0
0
Term 1 Term 2
0 0 0 0 0 0
0 0 0 0 0 00 0 0 0 0 00 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
a
d db
d F Dq qc
F F F
G G
D D
Q Q
v ir
Lv irL kM kM
v ir
v r i
r i
r i
r i
0
Term 3a
0 0 0 00 0 0 00 0 0 00 0 0 00 0 0 0
03 3
2 23 3
2 20000
d
qq G Q
F F R F
G G YG
D R D
Q Y Q D
Q
q q G G Q Q
d d F F D D
i
i
iL kM kM
kM L M ikM L M
ikM M LkM M L i
i
L i M i M i
L i M i M i
0 0
Term 3bTerm 4
3 3
0
0
0
0
0
0
n nr i L i
Terms 1 and 2 are the same; term 3a differs (and is not symmetric!);
term 3b is also a bit different; term 4 is the same.
13
3. Problem 4.2
4. Problem 4.6
This problem is asking us to take a step which is the same as a step taken in
obtaining term 3a in our voltage equation when using the “P” transformation.
See eq. (tve4) on pg. 23 of “macheqts” which has
14
4 term3b term3a term2 term1 term
000
000
1000
dqdq
FDQG
dq
FDQG
dq
FDQG
abc
FDQG
dq nPP
i
i
R
R
v
v
We started this work by looking at term 3a, but without the derivatives (indeed, this is exactly (4.20) in your text for which this problem asks us to verify). This will be
0
4
00
dqabc
FGDQFGDQ
PU
and then replacing the flux linkage quantities on the right with their equivalent inductance*current expressions results in
0
4 4
0 00 0
dqabc abcaa aR
FGDQ FDQGRa RRFGDQ
iL LP PiU U L L
And finally, replacing the abc currents on the right using 1
0
4
00
dqabc
FGDQ FGDQ
ii Pi iU
we obtain
100
4 4
0 00 0
dqdqaa aR
Ra RR FGDQFGDQ
iL LP PU L L iU
which results in
FDQG
dq
RRRa
aRaa
FDQG
dq
i
i
LPL
LPPLP 0
1
10
Again, from lecture notes (p. 2 of “macheqts,” eq. L-ex), we have
15
cos 2 [ cos 2( 30 )] [ cos 2( 150 )] cos[ cos 2( 30 )] cos 2( 120 ) [ cos 2( 90 )] cos( 120 )[ cos 2( 150 )] [ cos 2( 90 )] co
a S m S m S m F
b S m S m S m F
c S m S m S m
F
G
D
Q
L L M L M L MM L L L M L MM L M L L L
s 2( 240 ) cos( 240 )cos cos( 120 ) cos( 240 )sin sin( 120 ) sin( 240 ) 0cos cos( 120 ) cos( 240 )sin sin( 120 ) sin( 240 ) 0
F
F F F F
G G G
D D D R
Q Q Q
MM M M LM M MM M M MM M M
sin cos sinsin( 120 ) cos( 120 ) sin( 120 )sin( 240 ) cos( 240 ) sin( 240 )
0 00
0 00
G D Q a
G D Q b
G D Q b
FR
GG Y
DD
QY Q
M M M iM M M iM M M i
iMiL MiLiM L
where the submatrices of the inductance matrix are defined as
aa aR
Ra RR
L LL L
Submatrix (1,1) 1
1 1 1
cos 2 [ cos(2( 30))] [ cos(2( 150))]2 2 22cos cos( 120) cos( 120) [ cos(2( 30))] cos(2( 120)) [ cos(2( 90))]
3 sin sin( 120) sin( 120) [ cos(2(
aa
S m S m S m
S m S m S m
S m
PL P
L L M L M LM L L L M LM L
1cos sin
22 1
cos( 120) sin( 120)3 2150))] [ cos(2( 90))] cos(2( 240))
1cos( 120) sin( 120)
2
S m S mM L L L
2 0 03
0 02
30 0
2
S S
S m S
S m S
L M
L L M
L L M
Now we define:
0
3 32 ; ;
2 2S S d S m S q S m SL L M L L L M L L L M
Then
01
0
0 00 00 0
aa dqd
q
LPL P L L
L
Submatrix (1,2) 1 1 1
cos sin cos sin2 2 22cos cos( 120) cos( 120) cos( 120) sin( 120) cos( 120) sin( 120)
3 sin sin( 120) sin( 120) cos( 240) sin( 240) cos( 240) sin( 240)
F G D Q
aR F G D Q
F G D Q
M M M M
PL M M M M
M M M M
0 0 0 0
3 30 0
2 2
3 30 0
2 2
mF D
G Q
M M L
M M
16
Submatrix (2,1)
1
1cos sin
cos cos( 120) cos( 240) 22 1sin sin( 120) sin( 240)
cos( 120) sin( 120)cos cos( 120) cos( 240) 3 2sin sin( 120) sin( 240) 1
cos( 120) sin( 120)2
F F F
G G GRa
D D D
Q Q Q
M M MM M M
L PM M MM M M
30 0
2
30 0
2
30 0
2
30 0
2
F
GT
m
D
Q
M
M
L
M
M
So the desired matrix relation of (4.20) is given by 00 0 0 0 0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
dd d F D
qq q G Q
F F F R F
G G G Y G
D D R D D
Q Q Y Q Q
iL
iL kM kM
iL kM kM
kM L M i
kM L M i
kM M L i
kM M L i
5. Problem 4.11
17
6. Problem 4.12
Equation (4.20) is
00 0 0 0 0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
dd d F D
qq q G Q
F F F R F
G G G Y G
D D R D D
Q Q Y Q Q
iL
iL kM kM
iL kM kM
kM L M i
kM L M i
kM M L i
kM M L i
from which we obtain: 3 3
2 2
3 3
2 2
d d d F F D D
q q q G G Q Q
L i M i M i
L i M i M i
7. Problem 4.18
18
19
