Bài giảng Electromechanical energy conversion - Chapter I: Magnetic circuits and magnetic materials - Nguyễn Công Phương

Nguyễn Công Phương ELECTROMECHANICAL ENERGY CONVERSION Magnetic Circuits and Magnetic Materials 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 Magnetic Circuits and Magnetic Materials 1. Introduction to Magnetic Circuits 2. Flux Linkage, Inductance, and Energy 3. Properties of Magnetic Materials 4. AC Excitation 5. Permanent Magnets 6. Application of Permanent Magnet Materials sites.google.com/site/ncpdhbkhn 3 Introduction to Magnetic Circuits (1) HLJS.d= . d ∫ ∫ S BS.d = 0 ∫S BH= µ • H: magnetic field intensity, A/m • J: current density, A/m2 • B: magnetic flux density, Wb/m2 (T) • µ: permeability, H/m sites.google.com/site/ncpdhbkhn 4 Introduction to Magnetic Circuits (2) 6 4 2 0 -2 -4 -6 1 0.5 1 0.5 0 0 -0.5 = -0.5 HLJS.d . d -1 -1 ∫ ∫ S sites.google.com/site/ncpdhbkhn 5 Introduction to Magnetic Circuits (3) HLJS.d= . d ∫ ∫ S sites.google.com/site/ncpdhbkhn 6 Introduction to Magnetic Circuits (4) BS.d = 0 ∫S sites.google.com/site/ncpdhbkhn 7 Introduction to Magnetic Circuits Magnetic flux Φ (5) Mean core length, lc i Cross-sectional area, S + c Air gap, length g, – permeability µ0, area S Winding, N turns g Magnetic core permeability µ µ >> µ0 : the magnetic flux is confined almost entirely to the core F = Ni : magnetomotive force (mmf) sites.google.com/site/ncpdhbkhn 8 Introduction to Magnetic Circuits Magnetic flux Φ (6) Mean core length, lc i Cross-sectional area, S + c Air gap, length g, – permeability µ0, area S Winding, N turns g Magnetic core permeability µ Φ = BS.d F= Ni = HL. d ∫S ∫ ≈ ≈ Bc const Hc const → Φ = → = cB c S c F Hc l c sites.google.com/site/ncpdhbkhn 9 Introduction to Magnetic Circuits Magnetic flux Φ (7) Mean core length, lc i Cross-sectional area, S + c Air gap, length g, – permeability µ0, area S Winding, N turns g Magnetic core permeability µ = = = + FNi HL. d Hl Hl B Bg ∫ cc gg →F =c l + g =µ = µ µc µ BHBHc c; g0 g 0 Φ = Φ = Φ = Φ = Φ cBS cc; g BS gg ; c g l g  →F = Φc +  µ µ  Sc0 S g sites.google.com/site/ncpdhbkhn 10  Introduction to Magnetic Circuits (8) l g  F = Φc +  µ µ  Sc0 S g  → = Φ + FRR(c g ) l g c =RR; = µc µ g Sc0 S g FFF → Φ = = = = PF + l g total RRRc gc + total µ µ Sc0 S g sites.google.com/site/ncpdhbkhn 11 Introduction to Magnetic Circuits Φ lc Ex. 1 (9) Given Bc = 1T, lc = 40cm, g = 0.05cm, i Sc 2 N = 1000 turns, Sc = Sg = 16cm , + µ = 65000 µ . Find: g, 0 µ , a) The reluctances R & R ? 0 c g – S b) The flux Φ? g c) The current i? N turns µ l 40× 10−2 A.turns R =c = = 3061 c µ π ×−7 × − 4 Sc 65000(4 10 )(16 10 ) Wb g 0.0510× −2 A.turns R == =×2.49 10 5 g µ π ×−7 × − 4 0Sg (4 10)(1610) Wb Φ= =××−4 =× − 4 Bc S c 1 16 10 16 10 Wb F Φ(R + R ) 16× 10−4 (3061 + 2.49 × 10 5 ) i ===c g = 0.41A NN 1000 sites.google.com/site/ncpdhbkhn 12 Introduction to Magnetic Circuits Φ lc Ex. 2 (10) 2 Given Φ = 0.141 mWb, Sc = Sg = 4cm , i Sc lc = 44cm, g = 0.02cm, N = 400 turns. + The core is made of nickel – iron alloy. g, µ , Find the current i? 0 – S − g Φ0.141 × 10 3 B = = = 0.35T N turns c −4 µ Sc 4× 10 → = Hc 850 A/m B (T) sites.google.com/site/ncpdhbkhn H (A/m) 13 Introduction to Magnetic Circuits Φ lc Ex. 2 (11) 2 Given Φ = 0.141 mWb, Sc = Sg = 4cm , i Sc lc = 44cm, g = 0.2cm, N = 400 turns. + The core is made of nickel – iron alloy. g, µ , Find the current i? 0 – S − g Φ0.141 × 10 3 B = = = 0.35T N turns c −4 µ Sc 4× 10 → = Hc 850 A/m − Φ0.141 × 103 B 0.35 B== =→==0.35T H g =× 2.7910A/m5 g −4 g µ − 7 Sg 410×0 410π × = + =× + ×××5− 3 = F Hlcc Hl gg 850 0.44 2.79 10 2 10 931A.turns F 931 →=i = = 2.33 A N 400 sites.google.com/site/ncpdhbkhn 14 Introduction to Magnetic Circuits Ex. 3 (12) g 2 Given i = 10 A, Sg = 1500 cm , g = 2 cm, N = 500 turns. Rotor and stator are made i of iron (infinite permeability). Find the flux of the air gap? Φ =F = Ni N turns + l g RRc g c + µ µ Sc0 S g µ → ∞ Niµ S 500× 10(4π × 10−7 )1500 × 10 − 4 →Φ==0 g = 0.024 Wb 2g 2210× × −2 sites.google.com/site/ncpdhbkhn 15 Introduction to Magnetic Circuits Ex. 4 (13) l1 l3 i + Ф1 l Ф Ф2 2 3 – N turns = = + F Ni Hl11 Hl 22  = Hl22 Hl 33 Φ = Φ + Φ  1 2 3 sites.google.com/site/ncpdhbkhn 16 Magnetic Circuits and Magnetic Materials 1. Introduction to Magnetic Circuits 2. Flux Linkage, Inductance, and Energy 3. Properties of Magnetic Materials 4. AC Excitation 5. Permanent Magnets 6. Application of Permanent Magnet Materials sites.google.com/site/ncpdhbkhn 17 Flux Linkage, Inductance, and Energy (1) d ∫EL.d= − ∫ BS . d Cdt S dϕ d λ e= N =, λ = N ϕ dt dt λ L = i ϕ = F Rtotal F= Ni N 2 →L = Rtotal sites.google.com/site/ncpdhbkhn 18 Flux Linkage, Inductance, and Ex. 1 Energy (2) i Area µ → ∞ Find the inductance of the winding and S1 the flux density of gap 2? + Area S1 g g Gap1 g g R=1, R = 2 λ 1 2 1µ 2 µ 01S 02 S − Gap 2 RR gg N turns R =12 = 12 total +µ + R1 R 2 012( gS gS 21 ) λ Nφ φ L = = i i φ φ + 1 2 Ni φ = R R R Ni 1 2 total − S S  → =µ 2 1 + 2 LN0   g1 g 2  sites.google.com/site/ncpdhbkhn 19 Flux Linkage, Inductance, and Ex. 1 Energy (3) i Area µ → ∞ Find the inductance of the winding and S1 the flux density of gap 2? + Area S1 λ Gap1 g1 g2 − φ Gap 2 = 1 N turns B1 S1 µ φ =Ni = 0 S 1 Ni φ 1 R g φ φ 1 1 + 1 2 Ni R1 R2 µ Ni − → = 0 B1 g1 sites.google.com/site/ncpdhbkhn 20 Flux Linkage, Inductance, and Φ lc Ex. 2 Energy (4) 2 i Sc Given lc = 40cm, Sc = Sg = 16cm , Bc = 1T, N = 1000 turns, g = 0.05cm, find L for + g, a) µ = 65000 µ0, and b) µ = 3000 µ0, µ0, lc – R = Sg c, a µ N turns Sc 0.4 A.turns µ = =3.07 × 10 3 65000(4π × 10−7 )(16 × 10 − 4 ) Wb 0.4 A.turns R = =6.61 × 10 4 c, b 3000(4π × 10−7 )(16 × 10 − 4 ) Wb g 0.0510× −2 A.turns R == =×2.49 10 5 g µ π ×−7 × − 4 0Sg (4 10)(1610) Wb A.turns A.turns RRR=+=×2.52105 , RRR =+=× 3.1510 5 total,, a c a g Wbtotalb, cb , g Wb N21000 2 N 2 1000 2 L== =3.97 H , L == = 3.17 H a ×5 b × 5 Rtotal, a 2.52 10 R total, b 3.15 10 sites.google.com/site/ncpdhbkhn 21 Flux Linkage, Inductance, and Energy (5) 4.5 4 3.5 3 2.5 2 Inductance (H) Inductance 1.5 1 0.5 0 0 1 2 3 4 5 6 7 8 9 10 Core relative permeability 4 x 10 sites.google.com/site/ncpdhbkhn 22 Flux Linkage, Inductance, and Energy (6) Air gap i = + i1 2 F Ni11 Ni 22 + g + φ =F ≈ F R+ R R λ λ g c g 1 2 µ S − − →φ = + 0 c (Ni11 Ni 22 ) g N1 turns µ, lc , Sc N2 turns µS µ S λ=N φ = N2 0c iNN + 0 c i 11 1g 112 g 2 µS µ S λ=N φ = NN0c iN + 2 0 c i 22 12g 12 g 2 λ =Li + Li →  1 111 122 µS µ S λ = + LN=20c, LN = 2 0 c (self-inductance)  2Li 211 Li 222 11 1g 22 2 g µ S L= NN0 c = L (mutual inductance) 12 12g 21 sites.google.com/site/ncpdhbkhn 23 Flux Linkage, Inductance, and Energy (7) dλ e = dt di e= L dt di dL e= L + i dt dt sites.google.com/site/ncpdhbkhn 24 Flux Linkage, Inductance, and Energy (8) dλ p= ie = i dt t λ ∆W =2 pdt = 2 id λ ∫ ∫ λ t1 1 λ λ λ2− λ 2 =2 dλ = 2 1 ∫λ 1 LL2 Li 2 λ =0 →W = 1 2 sites.google.com/site/ncpdhbkhn 25 Flux Linkage, Inductance, and Φ lc Ex. 3 Energy (9) 2 i Sc Given lc = 40cm, Sc = Sg = 16cm , Bc = 1T, N = 1000 turns, g = 0.05cm, µ = 65000 µ . + 0 g, a) Find the magnetic stored energy? µ0, b) Bc = sin(314 t) (T), find e? – Sg l A.turns N turns R =c =3.07 × 10 3 c µ Sc Wb µ g A.turns R = =2.49 × 10 5 g µ 0Sg Wb N 21000 2 L = = = 3.97 H + ×+×3 5 Rc R g 3.07 10 2.49 10 Li 23.97(0.41) 2 W = = = 0.67J 2 2 dλ dN ϕ dB − e = = = NS c =1000 ×× (16 104 ) × 314cos(314t ) = 502cos(314t ) dt dt c dt sites.google.com/site/ncpdhbkhn 26 Magnetic Circuits and Magnetic Materials 1. Introduction to Magnetic Circuits 2. Flux Linkage, Inductance, and Energy 3. Properties of Magnetic Materials 4. AC Excitation 5. Permanent Magnets 6. Application of Permanent Magnet Materials sites.google.com/site/ncpdhbkhn 27 Properties of Magnetic Materials (1) Magnetic materials can be used: • To constrain and direct magnetic fields in well – defined paths • In transformer: – To maximize the coupling between the windings, and – To lower the excitation current required for transformer operation • In electric machinery: – To shape the fields to obtain desired torque – production and electrical terminal characteristics • The most common: ferromagnetic materials sites.google.com/site/ncpdhbkhn 28 Properties of Magnetic Materials (2) sites.google.com/site/ncpdhbkhn 29 Properties of Magnetic Materials (2) sites.google.com/site/ncpdhbkhn 30 Properties of Magnetic Materials Φ µ lc Ex. 1 (3) Given Bc = 0.9T, lc = 40cm, g = 0.05cm, i Sc 2 N = 1000 turns, Sc = Sg = 16cm . Find + the current i? g, µ0, = – H c 10 A.turns/m Sg = N turns Fc H c l c =10 × 0.4 = 4 A.turns B F= Hg = c g g g µ 0 × × −4 = 1 (16 10 ) 4π × 10 −7 = 1273 A.turns F+ F 4+ 1273 i =c g = N 1000 = 1.28A sites.google.com/site/ncpdhbkhn 31 Properties of Magnetic Materials Φ µ lc Ex. 2 (4) Given Bc = 2T, lc = 40cm, g = 0.05cm, i Sc 2 N = 1000 turns, Sc = Sg = 16cm . Find + the current i? g, µ0, = – H c 10000 A.turns/m Sg = = × N turns Fc H c l c 10000 0.4 = 4000 A.turns B F= Hg = c g g g µ 0 × × −4 = 1 (16 10 ) 4π × 10 −7 = 1273 A.turns F+ F 4000+ 1273 i =c g = N 1000 = 5.27A sites.google.com/site/ncpdhbkhn 32 Magnetic Circuits and Magnetic Materials 1. Introduction to Magnetic Circuits 2. Flux Linkage, Inductance, and Energy 3. Properties of Magnetic Materials 4. AC Excitation 5. Permanent Magnets 6. Application of Permanent Magnet Materials sites.google.com/site/ncpdhbkhn 33 AC Excitation (1) ϕ iϕ ( t ) i1 0 i2 t t t 0 i iϕ 1 2 i2 1 ϕ(t ) e( t ) ϕφω= = ω (t )max sin tBS max c sin t d et()= [ Ntϕωφ ()] = N cos()2 ωπ t = fNBS cos() ω t dt max max c π 2 fNBmax S c lc H c, rms E= = 2π fNB S ; Iϕ = rms max c ,rms N 2 sites.google.com/site/ncpdhbkhn 34 AC Excitation (2) lc H c, rms E=2π fNB S ; I ϕ = rmsmax c , rms N lc H c, rms EIϕ =2π fNSB = 2 π fBH () Sl rms, rms c maxN maxc , rms c c EIϕ EI ϕ 2π f P=rms, rms = rms , rms = B H a ρ ρ maxc , rms mass Sccc l c sites.google.com/site/ncpdhbkhn 35 AC Excitation (3) • The exciting current i supplies the mmf needed to produce: – the core flux, and – the power input associated with the energy in the magnetic • This energy: – Part is dissipated as losses & results in heating of the core ( RI 2) – The rest appears as reactive power, it is not dissipated in the core, but cyclically supplied & absorbed by the excitation source sites.google.com/site/ncpdhbkhn 36 AC Excitation (4) sites.google.com/site/ncpdhbkhn 37 AC Excitation (5) 20 cm Ex. The winding is excited with a 60-Hz voltage to 5 cm produce B = 1.6sin ωt T, the steel occupies 0.94 of the i 3 core cross-sectional area, ρc = 7.65 g/cm . Find: + a) The applied voltage? e 15 cm 5 cm b) The peak current? c) The rms exciting current? – d) The core loss? 300 turns B=1.6sin(2π × 60 t ) = 1.6sin 377 t T dϕ dSB dB eN= = Nc = NS dt dtc dt =300 ×× (25 10−4 ) × 377cos377 t = 283cos377t V = → = Bmax1.6 T H max 70 A.turns/m l H I = c max max N +++ ×−2 × = (15 10 15 10) 10 70 300 = 0.12A sites.google.com/site/ncpdhbkhn 38 AC Excitation (6) 20 cm Ex. The winding is excited with a 60-Hz voltage to 5 cm produce B = 1.6sin ωt T, the steel occupies 0.94 of the i 3 core cross-sectional area, ρc = 7.65 g/cm . Find: + a) The applied voltage? e 15 cm 5 cm b) The peak current? c) The rms exciting current? – d) The core loss? 300 turns = → = Bmax 1.6 T P a 2VA/kg =×× + + + Vc (5 5) (15 10 15 10) = 1250 cm 3 =ρ = × Wc c V c 7.65 1250 =9562.5g = 9.56kg E I = rmsϕ , rms Pa Wc × → =PWa = 2 9.56 Iϕ ,rms Erms 283/ 2 = 95.57A sites.google.com/site/ncpdhbkhn 39 AC Excitation (7) 20 cm Ex. The winding is excited with a 60-Hz voltage to 5 cm produce B = 1.6sin ωt T, the steel occupies 0.94 of the i 3 core cross-sectional area, ρc = 7.65 g/cm . Find: + a) The applied voltage? e 15 cm 5 cm b) The peak current? c) The rms exciting current? – d) The core loss? 300 turns = → = Bmax 1.6 T P c 1.4W/kg = Pctotal, PW c c =1.4 × 9.56 = 13.39 W sites.google.com/site/ncpdhbkhn 40 Magnetic Circuits and Magnetic Materials 1. Introduction to Magnetic Circuits 2. Flux Linkage, Inductance, and Energy 3. Properties of Magnetic Materials 4. AC Excitation 5. Permanent Magnets 6. Application of Permanent Magnet Materials sites.google.com/site/ncpdhbkhn 41 Permanent Magnets (1) sites.google.com/site/ncpdhbkhn 42 Permanent Magnets (2) Ex. µ → ∞ 2 g = 0.2cm, lm = 2cm, Sm = Sg = 5 cm , Bg = ? Sm lm F=0 = HgHl + →H = − H l Magnetic Air gap, g g m m gg m m material µ0, Sg S φ = = → = m BSgg BS mm Bg B m µ → ∞ Sg = µ Bg0 H g S  l  → = − µ g m Bm0     H m Sm   g  − 5 2 =−4π × 10 7 ×× H 5 0.2 m = − × −7 126 10 H m =− → = Hm40kA/m B m 0.50T = = Bg B m 0.60T sites.google.com/site/ncpdhbkhn 43 Permanent Magnets (3) µ → ∞ S Sm = m Bg B m Sg Magnetic lm Air gap, g material µ , S H l 0 g m m = − 1 µ → ∞ Hg g = µ Bg0 H g Point of maximum energy product S l →2 =µ m m − Bg0 ( H m B m ) Sg g Vol =µ mag − 0 (Hm B m ) Vol air gap Vol B 2 →Vol = airgap g mag µ − 0 (Hm B m ) sites.google.com/site/ncpdhbkhn 44 Magnetic Circuits and Magnetic Materials 1. Introduction to Magnetic Circuits 2. Flux Linkage, Inductance, and Energy 3. Properties of Magnetic Materials 4. AC Excitation 5. Permanent Magnets 6. Application of Permanent Magnet Materials sites.google.com/site/ncpdhbkhn 45 Application of Permanent Magnet Materials (1) sites.google.com/site/ncpdhbkhn 46 Application of Permanent Magnet Materials (2) µ → ∞ i Permanent lm magnetic N material turns sites.google.com/site/ncpdhbkhn 47 Application of Permanent Magnet Ex. Materials (3) Sm 2 2 2 µ → ∞ g = 0.2cm, Sm = 2cm , 2cm ≤ Sg ≤ 4cm . g / 2 a) Find the magnet length such that the system will Sm operate on a recoil line which intersects the Permanent x µ → ∞ maximum B – H product point on the magnetization lm magnetic curve. material b) Calculate the flux density in the air gap as the plunger moves back and forth. g / 2 S B= m B Point of gS m maximum energy product g   → = Sm B m lm g   H l S−µ H  m m = − 1 g0 m Hg g = − = H2 40kA/m; B 2 1T m2cm m 2cm →=×−2 2 1  = l 0.2 10−  3.98cm m 24π × 107 × 410 × 4  sites.google.com/site/ncpdhbkhn 48 Application of Permanent Magnet Ex. Materials (4) 2 2 2 µ → ∞ g = 0.2cm, Sm = 2cm , 2cm ≤ Sg ≤ 4cm . g / 2 a) Find the magnet length such that the system will Sm operate on a recoil line which intersects the Permanent x µ → ∞ maximum B – H product point on the magnetization lm magnetic curve. material b) Calculate the flux density in the air gap as the plunger moves back and forth. g / 2 S  l  Point of = − µ g m Bm0     H m maximum energy product Sm   g  → =−×π −7 4 3.98  B2 4 10   H 2 m 4cm 2 0.2  m 4cm = − × −5 5 10 H 2 m 4cm 2≤≤ 2 ֏ ≤≤ 2cmSg 4cm 1.0T B 1.08T sites.google.com/site/ncpdhbkhn 49