BIẾN ĐỔI NĂNG LƯỢNG ĐIỆN CƠ

Chương 4.1/ a) Mạch từ tương đương: a/ b/ v (t )  L L di dt  1  N 2W 0 (  ) i g 2x c/ i ( , x )   1 N 2W i 0 (  ) g 2x CuuDuongThanCong.com https://fb.com/tailieudientucntt  Wm   i( , x) d   fe  Wm ( , x )  x f e (i, x)  2 1 2.N 2W 2i 0 (  ) g 2x i   x2  2.N W i o   x  g  2 d) W 'm    (i, x)di  o2 ( 1 i2  ).N 2x g  ' Wm  2 Ni  o x 4x 4.2/ a) Mạch từ tương đương: Ni Ni Theo đề: L()  L1  L2cos(2) =>  L(  0)  L1  L2   L(  90)  L1  L2 N 2 o Ag o N 21Ag1  N N Ni  N N Ni L(  0)   , L(  90)       i i i 2g o 2g o i i i 2g1 2g1  o Ag o o Ag1   N 2o Ag o N 2o Ago Ag1 L1 L2 L1 ( )       2g o go g1     2  L1  L2  N 1Ag1  L2  N o ( Ago  Ag1 )   2g1 go g1 CuuDuongThanCong.com https://fb.com/tailieudientucntt b)   L()i  L1i  L 2cos(2)i i W 'm    (i, x)di  L1 Te  i2 i2  L cos(2) 2 W 'm  i sin(2)L x 4.3/ a) Rg  g  o Wd , Rx  x  o Wd b) (N  N )i  1   Rg 1Rg  N 2i  N1i    (1   )2Rx=N 2i   N 2i(Rg  2Rx)  N1i2Rx  2RxRg 4.4/ Mạch từ tương đương: CuuDuongThanCong.com https://fb.com/tailieudientucntt Rx    x o w Ni Ni o w  Rx x N i o w x i N 2i  o w ' W m    (i, x)di  x W 'm N 2i  o w  f (i, x)   x x2 e i x Wm  i  W 'm  , i= 2 N o w  x Wm   i( , x)d  2N 2 o w 4.5/ ias i bs ir 0 W 'm   (L0  L1cos2)i 'as di 'as   (L0  L1cos2)i 'bs di 'bs   (L2cos4)i 'r di 'r   N  b) Ni  bD  o aD  Ni( o  ) (2Rx / /2Ry) 2y 2x  1 di N 2i o bD 1 dy N 2i o aD 1 dx v(t)  N (  )   t 2Rx 2Ry dt y dt x dt i N 2i 1 ' c) W m    (i, x)di  (  ) 2Rx 2Ry W 'm N 2i o aD  f (i, x)   x 2x e 4.7/ CuuDuongThanCong.com https://fb.com/tailieudientucntt i1 i2 i3 0 a) W 'm   1 (i '1 ,0,0, )di '1    (i1 ,i '2 ,0, )di'2    (i1 ,i ,i '3 , )di '3 i1 i2 i3 0   (L11i '1 )di '1   (L 22i '2 )di '2   (Mcosi1  M sini  L33i '3 )di '3  (L11i12  L 22i 2  L33i32 )  Mcosi1i3  Mcosi 2i3 b) T e (i1 ,i ,i3 , )  W 'm   Msin i1i3  Mcosi 2i3 x 4.8 1  (5  cos2)103 i1  0.1cosi   0.1cosi1  (50  10cos2)i i1 i2 W 'm   (5  cos2)103 i '1 di '1   (0.1cosi1  (50  10cos2))i '2 di '2 0 3 i i 22  (5  cos2)10  0.1cosi1i  (50  10cos2) 2 T e ()  W 'm  103 sin 2i12  0.1sin i1i  10i 2 sin 2 x CuuDuongThanCong.com https://fb.com/tailieudientucntt 4.9/ i as ibs 0 W 'm   (L  L1cos2)i 'as di 'as   (L  L1cos2)i 'bs di 'bs ir   ((Mcos)ias  (M sin )ias  L cos4))i 'r di 'r i as i bs ir  (L  L1cos2)  (L  L1cos2)  Mcosi asi r  M sin i bsi r  L cos4 2 4.10/ a/ Ta có: Wm '  Wm ' 3i i3   4x 4x x b/ f e (i, x)  Wm' i3  x 4x c/ Wm  i  Wm'  i  i3 3i3 i i3    4x x 4x x 4.11/ s  ( L cos 2 )i s  ( M sin  )ir r  ( M sin  )is  ( L cos 2 )ir a/ vs (t )  d s d  Mir cos  MI cos dt dt b/ is ir W 'm   s (is ' ,0, )dis '   r (is , ir ' ,  )dir '  0 Lis Li cos 2  Mis ir sin   r cos 2 2 c/ T e (is , ir ,  )  Wm '   Lis sin 2  Misir cos  Lir sin 2  CuuDuongThanCong.com https://fb.com/tailieudientucntt 4.12 / a/ d vs  s  M I .cos  dt b/ is ir W ' m  i1 , i2 , x    s  i 's ,0,  di 's   r  is , i 'r ,   di 'r 0 is i2 cos 2  M is ir sin   L r cos 2 2  ' W m  M is ir cos   (is2  ir2 ).L.sin 2 fe (is , ir ,  )    L 4.14 / a/  dx  dv  dt  v  dt  0,1.v.i  x  x    d x  dv  di   (2.v.i  i  5) x  dt dt  dt b/ ve  x    x  x  i       x  1 i     i  CuuDuongThanCong.com https://fb.com/tailieudientucntt 4.15 / a/ d x dv  , dt dt 1 W ' m   L0 I fe  x  x a 1    a (2) v e   f e   M g (1)  L I  0   a Tu (1), (2)  x    2.M g a    b/  L I  0   a  x0  2.M g a     I0  2.M g a L0 4.17 / a/ i W 'm (i, x)    (i ', x).di '  fe  C.i 3.x dW 'm (i, x) C.i3  dx 3.x b/ Phương trình động lực học CuuDuongThanCong.com https://fb.com/tailieudientucntt M g  f e  M a  M d 2x dv  M dt dt dv fe  g dt M d I (t )  i R dt d  di i  R di dt 2.C.i di i  R x dt di x.R.( I (t )  i )   dt 2.c Wm (i, x)  i.  W 'm (i , x )   i C.i C.i 2.C.i   x 3.x 3.x 4.18 / a/ t  0.1s x1 (0,1)  x1 (0)  t.x2 (0)  1,05 x2 (0,1)  x2 (0)  t  0,1.x2 (0).x3 (0)  x13 (0)  x1 (0)   0, 45 x3 (0,1)  x3 (0)  t  2.x2 (0).x3 (0)  x3 (0)    8,5 x1 (0, 2)  x1 (0,1)  t.x2 (0,1)  1, 095 x2 (0, 2)  x2 (0,1)  t  0,1.x2 (0,1).x3 (0,1)  x13 (0,1)  x1 (0,1)   0, x3 (0, 2)  x3 (0,1)  t  2.x2 (0,1).x3 (0,1)  x3 (0,1)    7,39 b/ x1  x2  x3   x2     x1  1, 1,0 x   CuuDuongThanCong.com https://fb.com/tailieudientucntt 4.21/   i ( x  a) i  W 'm (i, x)    i3 ( x  a) di   i 4.( x  a )  Wm ( , x)  i.  W 'm (i, x)  3. i 4.( x  a ) 4.22 / a/ i1 i2 W 'm   1 (i1 ,0, ).di1   2 (i1 , i2 ,  ) 0   L0  M cos 2  Te  i i2  M i1.i2 sin 2   L0  M cos 2  2 W 'm  2.M i1.i2 cos 2  M i12 sin 2  M i2 sin 2  b/ Te  W 'm  M I sin  2 s t  2.   4.23/ f e ( x, i)  9.96.10 6 x2 f e ( x, i )  K ( x  l )  Bv  v=0 9.96.106 f ( x, i)  K ( x  l )   K ( x  4.922.103 ) x e Vậy: x1e  2.64.103 x1e  3.85 (mm) 4.24/ x1e (0,1)  x1 (0)  t * x2 (0)  x2e (0,1)  x2 (0)  t *( 5sin x1 (0)  x2 (0)) = 0+0.1(-5sin(1)-0)=-8.73* 103 CuuDuongThanCong.com https://fb.com/tailieudientucntt x1 (0, 2)  x1 (0,1) t ( x2 (0,1))   0,1( 8, 72.103 )  0,999 x2 (0,2)  x2 (0,1) t.(5sin1  8,73.103 )  16,58.103 x2 (0,3)  x1 (0, 2) t ( x2 (0, 2))  0,999  0,1.16,58.103  4.26/  x  x    x32 x   40( x1  0,1)  80 x2 )  2 x   xx  x3   1000 x2  1000 x1  x3 x1  x1 (0, 001)  x1 (0) t x2 (0)  0,1  x (0)  x2 (0,001)  x2 (0) t   32  40( x1 (0)  0,01)  80 x2 (0)    x1 (0)   x (0).x3 (0)   1000 x1 (0)  1000 x1 (0).x3 (0)   0,1 x3 (0,001)  x3 (0) t   x1 (0)  x1 (0,002)  x1 (0,001) t.x2 (0,001)  0,1  x (0,001)   40( x1 (0,001)  0,1)  80 x2 (0,001)   0,001 x2 (0,002)  x2 (0,001) t   32  x1 (0,001)   x (0,001) x3 (0,001)  x3 (0,002)  x3 (0,001) t   1000 x1 (0,001)  1000 x1 (0,001) x3 (0,001)  x1 (0,001)    0,1  0,001  1000.0,1  1000.0,1.0,1  0,19 4.27/ a/ CuuDuongThanCong.com https://fb.com/tailieudientucntt Mơ hình khơng gian trạng thái x  x2 x2  0,1.x2 i  x3  x i 2 x2 i i 5 x x x x(0,1)  x(0) t.x2 (0)   0,1.0,5  1,05 x2 (0,1)  x2 (0) t  0,1.x2 (0).i (0)  x (0)  x(0)   0,5  0,1  0,1.0,5.0  13  1  0,5  2 x2 (0) i (0)  i (0,1)  i (0) t  i x(0)   x (0) 5  0,5   0,1  2     0, 1  x(0, 2)  x(0,1) t.x2 (0,1)  1,1 x2 (0, 2)  x2 (0,1) t  0,1x2 (0,1)i (0,1)  x (0,1)  x(0,1)   0,5  0,1( 0,1.0,5.0,  1, 053  1, 05)  0, 4872  2 x2 (0,1) i (0,1)  i (0, 2)  i (0,1) t  i (0,1)   x(0,1) x(0,1)   x (0,1) CuuDuongThanCong.com https://fb.com/tailieudientucntt