Bài giảng Electromechanical energy conversion - Chapter II: Electromechanical Energy Conversion Principles - Nguyễn Công Phương

Nguyễn Công Phương ELECTROMECHANICAL ENERGY CONVERSION Electromechanical – Energy – Conversion Principles Contents I. Magnetic Circuits and Magnetic Materials II. Electromechanical Energy Conversion Principles III. Introduction to Rotating Machines IV. Synchronous Machines V. Polyphase Induction Machines VI. DC Machines VII.Variable – Reluctance Machines and Stepping Motors VIII.Single and Two – Phase Motors IX. Speed and Torque Control sites.google.com/site/ncpdhbkhn 2 Electromechanical – Energy – Conversion Principles 1. Forces and Torques in Magnetic Field Systems 2. Energy Balance 3. Energy in Singly – Excited Magnetic Field Systems 4. Determination of Magnetic Force and Torque from Energy and Coenergy 5. Multiply – Excited Magnetic Field Systems 6. Forces and Torques in Systems with Permanent Magnets 7. Dynamic Equations 8. Analytical Techniques sites.google.com/site/ncpdhbkhn 3 Forces and Torques in Magnetic Field Systems (1) Fq() E  v  B v FE q B Fq() v  B F Fv () v  B J  v FJBv   sites.google.com/site/ncpdhbkhn 4 Forces and Torques in Magnetic Ex. Field Systems (2) B0 yˆ ˆ A nonmagnetic rotor contains a single – turn coil, it is in a uniform magnetic field. The rotor is of radius R and of length l. Find the θ – directed torque as a function of α? I FJBv    xˆ FJB S 1(  ) I IB  Fin  IB0 l sin Fout  IB0 l sin T(2 R )  F  2 RIB0 l sin (Nm) sites.google.com/site/ncpdhbkhn 5 Forces and Torques in Magnetic Field Systems (3) i f fld  Lossless magnetic  ,e x  energy storage system  Electrical Mechanical terminal terminal i 1 Magnetic core  Winding  f resistance x fld v e Movable   magnetic Lossless winding plunger sites.google.com/site/ncpdhbkhn 6 Forces and Torques in Magnetic Field Systems (4) i f fld  Lossless magnetic  ,e x  energy storage system  Electrical Mechanical terminal terminal dW dx eifld  f dtfld dt d e  dt dWfld  id  f fld dx sites.google.com/site/ncpdhbkhn 7 Electromechanical – Energy – Conversion Principles 1. Forces and Torques in Magnetic Field Systems 2. Energy Balance 3. Energy in Singly – Excited Magnetic Field Systems 4. Determination of Magnetic Force and Torque from Energy and Coenergy 5. Multiply – Excited Magnetic Field Systems 6. Forces and Torques in Systems with Permanent Magnets 7. Dynamic Equations 8. Analytical Techniques sites.google.com/site/ncpdhbkhn 8 Energy Balance Energy is neither created or destroyed, it is merely changed in form dWelectrical eidt  dW mechanical  dW field sites.google.com/site/ncpdhbkhn 9 Electromechanical – Energy – Conversion Principles 1. Forces and Torques in Magnetic Field Systems 2. Energy Balance 3. Energy in Singly – Excited Magnetic Field Systems 4. Determination of Magnetic Force and Torque from Energy and Coenergy 5. Multiply – Excited Magnetic Field Systems 6. Forces and Torques in Systems with Permanent Magnets 7. Dynamic Equations 8. Analytical Techniques sites.google.com/site/ncpdhbkhn 10 Energy in Singly – Excited Magnetic Field Systems (1) x Mechanical i source   L() x i  f  fld + 1 – 2 ,e Massless Wfld  Li 2 magnetic  armature 1  2 Lossless W fld  coil 2L ( x ) Magnetic core sites.google.com/site/ncpdhbkhn 11 Energy in Singly – Excited Ex. Magnetic Field Systems (2) A relay is made of infinitely – permeable magnetic material, h >> g. Compute the magnetic energy as i a function of plunger position? + x g  h l 1 2 – g Wfld  L() x i d 2 Lossless – N turns coil  S L() x N 2 0 g 2g d Magnetic g flux x  x Sg  l( d  x )  ld  1   d  d x g 2 1N0 ld (1 x / d ) 2  x  Wfld  i  K 1   J 2 2g d  sites.google.com/site/ncpdhbkhn 12 Electromechanical – Energy – Conversion Principles 1. Forces and Torques in Magnetic Field Systems 2. Energy Balance 3. Energy in Singly – Excited Magnetic Field Systems 4. Determination of Magnetic Force and Torque from Energy and Coenergy 5. Multiply – Excited Magnetic Field Systems 6. Forces and Torques in Systems with Permanent Magnets 7. Dynamic Equations 8. Analytical Techniques sites.google.com/site/ncpdhbkhn 13 Determination of Magnetic Force and Torque from Energy and Coenergy (1) dWfld(,) x id   f fld dx Wfld(,)(,) x    W fld  x  dWfld (,) x  d     dx  x xconst    const   Wfld (,) x  i    xconst   W(,) x  f   fld  fld x  const 1  2 W(,) x  fld 2L ( x )  12   2 dL ( x ) 2 f fld     2 i dL() x x2 L ( x )  2 L ( x ) dx f  const fld 2 dx   L() x i sites.google.com/site/ncpdhbkhn 14 Determination of Magnetic Force and Torque from Energy and Coenergy (2) Ex. 1 x (cm) 0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0 L (mH) 2.8 2.26 1.78 1.52 1.34 1.26 1.20 1.16 1.13 1.11 1.10 Plot the force as a function of position for a current of 1A. sites.google.com/site/ncpdhbkhn 15 Determination of Magnetic Force and Torque from Energy and Coenergy (3) Ex. 2 x (cm) 0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0 L (mH) 2.8 2.26 1.78 1.52 1.34 1.26 1.20 1.16 1.13 1.11 1.10 Plot the force as a function of position for a flux of 2mWb. sites.google.com/site/ncpdhbkhn 16 Determination of Magnetic Force and Torque from Energy and Coenergy (4) dWfld(,) x id   f fld dx dWfld(,)  id   T fld d  WWfld    fld  dW(,)   d     d  fld    const     const  W fld (,)  Tfld    const 1  2 W(,) x  fld 2L ( x )  12   2 dL (  ) Tfld     2 2 2L ( x ) 2 L (  ) d  i dL()   const T  fld 2 d  L()  i sites.google.com/site/ncpdhbkhn 17 Determination of Magnetic Force and Torque from Energy and Coenergy (5) i   LLL( )  cos(2  ) 0 2  Rotor i2 dL() T  fld 2 d  Air gap i2 TL( )  [  2 sin(2  )] fld 2 2 sites.google.com/site/ncpdhbkhn 18 Determination of Magnetic Force and Torque from Energy and Coenergy (6) Wfld (,)(,) i x i  W fld  x dWfld (,)()(,) i x  d i  dW fld  x d() i id    di dWfld(,) x id   f fld dx dWfld (,) i x  di  f fld dx WWfld    fld   dW (,) i x  di    dx fld i  x xconst   i  const   Wfld (,) i x     i xconst   W (,) i x  f  fld  fld x  iconst sites.google.com/site/ncpdhbkhn 19 Determination of Magnetic Force and Torque from Energy and Coenergy (7) Ex. 3 A relay is made of infinitely – permeable magnetic material, h >> g. Compute the magnetic energy as i g a function of plunger position, if i(x) = I0x/d (A)? + x  h 2 0Sg L() x N l 2g – g d x  Lossless – Sg  l( d  x )  ld  1   d  N turns coil 2 N0 ld(1 x / d ) L() x  i2 N 2 l 2g L() x   0 i2 dL() x 2 2g f fld  x 2 dx i() x I0 d 2 2 2 I0 0 N l x  L() x     4g d  sites.google.com/site/ncpdhbkhn 20 Determination of Magnetic Force and Torque from Energy and Coenergy (8)  ()i  0 Energy W fld Coenergy W fld 0 i0 i Wfld W fld   i 1 1 2 1 Ifki : W  W   i   Li2 fld fld 2 2L 2 sites.google.com/site/ncpdhbkhn 21 Determination of Magnetic Force and Torque from Energy and Coenergy (9) Ex. 4  Find the torque acting on the rotor as a function of g the dimensions and the magnetic field in the two r air gap, suppose that the reluctance of the steel is negligible (μ → ∞). The axial length is h. 2gHag  Ni 1 Energy density : BH 2 i 1 B2 Energy densityof the core :BH steel  0 2steel steel 2 BHH 2 Energy densityof the air-gap : ag ag 0 ag 2 2  H 2 (Ni )2 h ( r 0.5 g )  W 0 ag [2 gh ( r  0.5 g ) ]  0 ag 2 4g 2 0 (Ni ) h ( r 0.5 g ) W (,) i  Tfld  ag 4g Tfld   iconst sites.google.com/site/ncpdhbkhn 22 Determination of Magnetic Force and Torque from Energy and Coenergy (10) Ex. 5  Find the inductance as a function of θ, then extract g the expression for the torque acting on the rotor as r a function of i & θ. The axial length is h.  S LN()  2 0 g 2g i Sg  h( r  0.5 g ) N2 h( r 0.5 g )  L()  0 2g 2 0 (Ni ) h ( r 0.5 g ) Tfld  2 4g i dL() T  fld 2 d sites.google.com/site/ncpdhbkhn 23 Electromechanical – Energy – Conversion Principles 1. Forces and Torques in Magnetic Field Systems 2. Energy Balance 3. Energy in Singly – Excited Magnetic Field Systems 4. Determination of Magnetic Force and Torque from Energy and Coenergy 5. Multiply – Excited Magnetic Field Systems 6. Forces and Torques in Systems with Permanent Magnets 7. Dynamic Equations 8. Analytical Techniques sites.google.com/site/ncpdhbkhn 24 Multiply – Excited Magnetic Field Systems (1) i1 Tfld    1  i Lossless magnetic  2 energy storage system   2   Electrical Mechanical terminals terminal dWfld(,)(,,)  id   T fld d   dW fld 1  2   i 1 d  1  i 2 d  2  T fld d        WWWfld  fld  fld dWfld (,,)1  2   d  1    d  2    d  1     2    ,  are const 2,  are const    1 ,  are const  1 2    WWWfld(,,)(,,)(,,)1  2   fld  1  2   fld  1  2  i1 ;; i 2  Tfld   1   2   , are const  2,  are const  1 ,  are const 2  2 sites.google.com/site/ncpdhbkhn 25 Multiply – Excited Magnetic Field Systems (2) i1 Tfld    1  i Lossless magnetic  2 energy storage system   2   Electrical  ,e Mechanical terminals 1 1 terminal 1L 11 i 1  L 12 i 2  2L 21 i 1  L 22 i 2  LLLL     i 22 1 12 2  22 1 12 2  1  LLLLD11 22 12 21    LLLL21 1  11  2  21  1  11  2 i2    LLLLD11 22 12 21 12 1 2 L12 ( 0 ) WLLfld (,,)()()10  20  0 11  0  20  22  0  10   10  20 2DDD (0 ) 2 (  0 ) (  0 ) sites.google.com/site/ncpdhbkhn 26 Multiply – Excited Magnetic Field Systems (3) Wfld (,,) i1 i 2  1 i 1   2 i 2  W fld dWfld (,,) i1 i 2  1 di 1   2 di 2  T fld d   Wfld (,,) i1 i 2  1  i1  i2 , are const   Wfld (,,) i1 i 2  2  i  1 i1, are const  W (,,) i i  T  fld 1 2  fld   i1, i 2 are const 1 1 W (,,)()()() i i L  i2  L  i 2  L  i i fld 1 22 11 1 2 22 2 12 1 2 W  (,,)   i2 dL()()() i 2 dL  dL  Tfld 1 2 1 11  2 22  i i 12 fld 2d  2 d 1 2 d  i1, i 2 are const sites.google.com/site/ncpdhbkhn 27 Multiply – Excited Magnetic Ex. Field Systems (4)  L11 = a + cos2θ, L12 = bcosθ, L22 = c + ecos2θ. Find Tfld(θ)? i1  + Tfld  – –  ,e + 1 1 2,e 2   i Tmech 2 i2 dL()()() i 2 dL  dL  T1 11  2 22  i i 12 fld 2d 2 d 1 2 d  2 2  i1sin 2  ei 2 sin 2   bi 1 i 2 sin  sites.google.com/site/ncpdhbkhn 28 Multiply – Excited Magnetic Field Systems (5) 2 2 Tfld   i1sin 2  ei 2 sin 2   bi 1 i 2 sin  -3 x 10 4 3 2 1 0 Torque (Nm)Torque -1 -2 Reluctance torque -3 Mutual - interaction torque Total torque -4 0 1 2 3 4 5 6  (rad) sites.google.com/site/ncpdhbkhn 29 Multiply – Excited Magnetic Field Systems (6)  W fld (,,)1  2   Tfld     1,  2 are const   2 2  Wfld (,,) i1 i 2  i1 dL 11()()() i 2 dL 22  dL 12   Tfld     i1 i 2 2d  2 d  d   i1, i 2 are const  12 1 2 Wfld (,,)()()() i12 i L 11  i 1  L 22  i 2  L 12  i 12 i  2 2  Wfld (,,)1  2 x  f fld   x  1,  2 are const   2 2  Wfld (,,) i1 i 2 x idLx1 11()()() idLx 2 22 dLx 12  ffld     i1 i 2 x2 dx 2 dx dx  i1, i 2 are const  12 1 2 Wiixfld (,,)()()()12 Lxi 11 1  Lxi 22 2  Lxii 12 12  2 2 sites.google.com/site/ncpdhbkhn 30 Electromechanical – Energy – Conversion Principles 1. Forces and Torques in Magnetic Field Systems 2. Energy Balance 3. Energy in Singly – Excited Magnetic Field Systems 4. Determination of Magnetic Force and Torque from Energy and Coenergy 5. Multiply – Excited Magnetic Field Systems 6. Forces and Torques in Systems with Permanent Magnets 7. Dynamic Equations 8. Analytical Techniques sites.google.com/site/ncpdhbkhn 31 Forces and Torques in Systems with Permanent Magnets (1)    dWfld (,) i x di  f fld dx x    dWfld (,) i f x f di f  f fld dx Wfld ( i f  0, x ) f     fld x i f const i f x 0     Wfld( i f 0, x )   f ( i  f , x ) di  f I f 0  f  Fictitious winding Nf turns sites.google.com/site/ncpdhbkhn 32 Forces and Torques in Systems Ex. 1 with Permanent Magnets (2) Depth D Find an expression for:    Wm W a) the coenergy of the system as a function of plunger g position x? i f d x b) the force on the plunger as a function of x?  BHHm R() m  c N f Nf i f H m d  H g x  H0 g 0  g0 Fictitious W0    m  g  0 BSBSBS m m  g g  0 0 winding BWDBWDBWDm m  g g  0 0 R()N f i f H c d Bm     R x g0 Nf W m D R() N f i f H c d d Wm       WW  f 0 g 0   x g  d W R  0 m    0WWg 0 fNNBSNBWD f  m  f m m  f m m    sites.google.com/site/ncpdhbkhn 33 Forces and Torques in Systems Ex. 1 with Permanent Magnets (3) Depth D Find an expression for:    Wm W a) the coenergy of the system as a function of plunger g position x? i f d x b) the force on the plunger as a function of x?  N W D () N i H d   f m R f f c f N  x g  f d W R  0 m    0 WWg 0   g0 Fictitious W0     Hc d winding  f0 i f  I f 0  N f 2 0 Wm D R() H c d Wfld() x  f di  f  I f 0  x g   2 d W R   0   m  WW  0 g 0   W ( i  0, x ) W2 D () H d 2 f fld f   m R c fld x    i f const R  x g0 2Wg d W m       WW  0 g 0   sites.google.com/site/ncpdhbkhn 34 Forces and Torques in Systems with Permanent Magnets (4)  S 1 H d F  0  m e External  d F 1 BSSHHm  R() m  c H e magnetic m circuit  Fe    RSH  c   d BHHm R() m  c   S 2 Ni F  H  e External 2 BS  R HS d Fe magnetic ()Ni equiv  1   2 circuit ()Ni F   equiv e  2 RS    d d  BHm  R m If (Ni )equiv  H c d sites.google.com/site/ncpdhbkhn 35 Forces and Torques in Systems Ex. 2 with Permanent Magnets (5) Depth D a) Find the x – directed force on the plunger when the    W W current in the excitation winding is zero and x = 3 mm? g b) Find the current in the excitation winding required to i1 d reduce the plunger force to zero? x ()Ni  H d equiv c N1 g0 d d x g W    RRR ;;   0 mS  WD x  W D0  WD R R0 g 0  ()Ni equiv 1 2 + W  Li – R fld 2 1 m 1 ()Ni 2 W   equiv fld 2 RRR  + 2 m x 0 – N i Rx N 1 1 L  1 R Rtotal g0 sites.google.com/site/ncpdhbkhn 36 Forces and Torques in Systems Ex. 2 with Permanent Magnets (6) Depth D a) Find the x – directed force on the plunger when the    W W current in the excitation winding is zero? g b) Find the current in the excitation winding required to i1 d reduce the plunger force to zero? x 2 1 ()Ni equiv N W   1 fld 2 RRR  m x 0 g W 0    W () Ni 2 dR f fld   equiv x fld x() R  R  R2 dx iequiv const m x 0  2 ()Ni equiv + ()Ni equiv   2 – Rm 0WDRRRg() m x  0 + – R ()Ni equiv N1 i 1 x (Ni )equiv  N1 i 1  0 i1  N1 Rg0 sites.google.com/site/ncpdhbkhn 37 Electromechanical – Energy – Conversion Principles 1. Forces and Torques in Magnetic Field Systems 2. Energy Balance 3. Energy in Singly – Excited Magnetic Field Systems 4. Determination of Magnetic Force and Torque from Energy and Coenergy 5. Multiply – Excited Magnetic Field Systems 6. Forces and Torques in Systems with Permanent Magnets 7. Dynamic Equations 8. Analytical Techniques sites.google.com/site/ncpdhbkhn 38 Dynamic Equations (1) x K i B v0  f fld + R Electromechanical – – ,e energy – conversion system M  f0 d v Ri  di dL() x dx 0 dt v  Ri  L() x  i 0 dt dx dt   L() x i sites.google.com/site/ncpdhbkhn 39 Dynamic Equations (2) x K i B v0  f fld + R Electromechanical – – ,e energy – conversion system M  f fK   K() x  x0 0 dx fD   B dt dx d2 x ffld  K( x  x0 )  B  M2  f 0  0 d2 x dt dt f  M M dt2 ffld f K  f D  f M  f0  0 sites.google.com/site/ncpdhbkhn 40 Dynamic Equations (3) x K i B v0  f fld + R Electromechanical – – ,e energy – conversion system M  f0  di dL() x dx v()() t Ri  L x  i  0 dt dx dt  dx d2 x  f()()(,) t  K x  x  B  M  f x i  0 0 dt dt 2 fld sites.google.com/site/ncpdhbkhn 41 Dynamic Equations (4) l0 Ex. Extract the dynamic equations of motion of the Spring, K electromechanical system? Coil l1 Length of flux path in the direction of field R  (area of flux path perpendicular to field) a x g R  g1     0 dx     g h Rg 2        da 0     g1 1  g a x RRRg1  g 2      d x a    da x 0 0 g a 2 2 N0  daN x x L(), x   L Cylindrical R g a x a  x d steel plunger, 2 M 0 daN L  Applied force, ft g sites.google.com/site/ncpdhbkhn 42 Dynamic Equations (5) l0 Ex. Extract the dynamic equations of motion of the Spring, K electromechanical system? Coil l1 x  daN 2 L(), x L L   0 a x g a x     W (,) i x i2 dL i 2 aL f fld       fld x2 dx 2 ( a  x )2 h iconst          dLi() di dL di dLdx e  L  i  L  i dt dt dt dt dxdt g a Cylindrical x di ai dx d steel plunger, LL   a x dt() a  x2 dt M Applied force, ft sites.google.com/site/ncpdhbkhn 43 Dynamic Equations (6) l0 Ex. Extract the dynamic equations of motion of the Spring, K electromechanical system? Coil l1 i2 dL i 2 aL f   fld 2dx 2 ( a x )2 a x x di ai dx e L  L  a x dt() a  x2 dt          di dL() x dx h v()() t Ri  L x  i       0 dt dx dt      dx d2 x  f()()(,) t  K x  x  B  M  f x i  0 0 dt dt 2 fld g a  x di a dx v() t Ri  L  L  i  0 2 Cylindrical  a x dt() a  x dt d  steel plunger, dx d2 x i 2 aL M  Applied force, f f0()() t  K x  l 0  B  M 2  2 t  dt dt2 ( a x ) sites.google.com/site/ncpdhbkhn 44 Electromechanical – Energy – Conversion Principles 1. Forces and Torques in Magnetic Field Systems 2. Energy Balance 3. Energy in Singly – Excited Magnetic Field Systems 4. Determination of Magnetic Force and Torque from Energy and Coenergy 5. Multiply – Excited Magnetic Field Systems 6. Forces and Torques in Systems with Permanent Magnets 7. Dynamic Equations 8. Analytical Techniques sites.google.com/site/ncpdhbkhn 45 Analytical Techniques (1) v(0)  V  If  x di a dx Ri L, L  i   2 V  a x dt() a  x dt i  R  x di a dx v() t Ri  L  L  i  2  a x dt() a  x dt  2 2  dx d x i aL f()() t  K x  l0  B  M 2  2  dt dt2 ( a x )  f  0 If  M  0 dx1 a V 2    B  L2    K()() x  l0  f x dt2 ( a x )  R  X B t  dx 0 f() x sites.google.com/site/ncpdhbkhn 46 Analytical Techniques (2) v(0)  V  If  x di a dx Ri L, L  i   2 V  a x dt() a  x dt i  R  x di a dx v() t Ri  L  L  i  2  a x dt() a  x dt  2 2  dx d x i aL f()() t  K x  l0  B  M 2  2  dt dt2 ( a x )  f  0 If  B  0 d2 x1 a V 2    M2  L 2    K()() x  l0  f x dt2 ( a x )  R  dx2 x B v() x   dx dt M0 f() x sites.google.com/site/ncpdhbkhn 47 Analytical Techniques (3) di dx If 0,  0 dt dt V0 RI 0   x di a dx  1 L aI 2 v() t Ri  L  L  i 0 K() X  l  f  2  2 0 0t 0  a x dt() a  x dt 2 (a l0 )  2 2  dx d x i aL f()() t  K x  l0  B  M 2  2  dt dt2 ( a x ) IfiIiff0  ,t  t 0  fvVvxXx  , t  0   ,  0    LX()() xdi   LaI   i  dx  V v  R() I  i  0  0  0 0 2  a X0  x dt() a  X 0  x  dt   1L a ( I i  )2 dx  d 2 x   0 K() X  x  l  B  M  f  f   20 0 2 t 0 2 (a X0  x ) dt dt sites.google.com/site/ncpdhbkhn 48 Analytical Techniques (4)  LX()() xdi   LaI   i  dx  V v  R() I  i  0  0  0 0 2  a X0  x dt() a  X 0  x  dt  1L a ( I i  )2 dx  d 2 x   0 K() X  x  l  B  M  f  f   20 0 2 t 0 2 (a X0  x ) dt dt  L X di  L  aI dx  v Ri  0  0  2  a X0 dt() a  X 0 dt   L aI dx  d2 x  L  aI 2   0i B  M  K  0 x   f   2 2 3  ()()a X0 dt dt a  X 0  sites.google.com/site/ncpdhbkhn 49