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Lecture Electromechanical energy conversion - Magnetic circuits and magnetic materials presents the following content: Introduction to magnetic circuits; flux linkage, inductance, and energy; properties of magnetic materials; AC excitation; permanent magnets; application of permanent magnet materials.
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 Magnetic Circuits and Magnetic Materials Introduction to Magnetic Circuits Flux Linkage, Inductance, and Energy Properties of Magnetic Materials AC Excitation Permanent Magnets Application of Permanent Magnet Materials sites.google.com/site/ncpdhbkhn Introduction to Magnetic Circuits (1) ∫ H.dL = ∫ S ∫ S J.dS B.dS = B = àH H: J: B: à: magnetic field intensity, A/m current density, A/m2 magnetic flux density, Wb/m2 (T) permeability, H/m sites.google.com/site/ncpdhbkhn Introduction to Magnetic Circuits (2) -2 -4 -6 1 0.5 0.5 0 ∫ H.dL = ∫ S -0.5 J.dS -0.5 -1 -1 sites.google.com/site/ncpdhbkhn Introduction to Magnetic Circuits (3) ∫ H.dL = ∫ S J.dS sites.google.com/site/ncpdhbkhn Introduction to Magnetic Circuits (4) ∫ S B.dS = sites.google.com/site/ncpdhbkhn Introduction to Magnetic Circuits Magnetic flux Φ (5) Mean core length, l c i + Cross-sectional area, Sc – Air gap, length g, permeability µ0, area Sg Winding, N turns 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 Introduction to Magnetic Circuits Magnetic flux Φ (6) Mean core length, l c i + Cross-sectional area, Sc – Air gap, length g, permeability µ0, area Sg Winding, N turns Magnetic core permeability µ Φ = ∫ B.dS F = Ni = S B c ≈ const ∫ H.dL H c ≈ const → Φ c = Bc S c sites.google.com/site/ncpdhbkhn → F = H c lc Introduction to Magnetic Circuits Magnetic flux Φ (7) Mean core length, l c i + Cross-sectional area, Sc – Air gap, length g, permeability µ0, area Sg Winding, N turns Magnetic core permeability µ F = Ni = ∫ H.dL = H l c c + H glg →F= Bc lc + Bg g µ µ0 Bc = µ H c ; Bg = µ0 H g Φ c = Bc Sc ; Φ g = Bg S g ; Φ c = Φ g = Φ lc g → F = Φ + µ Sc µ0 S g sites.google.com/site/ncpdhbkhn 10 AC Excitation (2) Erms = 2π fNBmax Sc ; E rms Iϕ ,rms = 2π fNSc Bmax Pa = Erms Iϕ ,rms mass = Iϕ ,rms = lc H c , rms N lc H c , rms = 2π fBmax H c , rms ( Sc lc ) N Erms Iϕ , rms Sc lc ρ c = 2π f ρc sites.google.com/site/ncpdhbkhn Bmax H c ,rms 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 (RI2) – 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) Ex The winding is excited with a 60-Hz voltage to produce B = 1.6sinωt T, the steel occupies 0.94 of the core cross-sectional area, ρc = 7.65 g/cm3 Find: a) The applied voltage? b) The peak current? c) The rms exciting current? d) The core loss? B = 1.6sin(2π × 60t ) = 1.6sin 377t T dϕ dS B dB e= N = N c = NSc dt dt dt = 300 × (25 × 10 −4 ) × 377 cos377 t i 20 cm cm + e – 15 cm cm 300 turns = 283cos 377t V Bmax = 1.6T → H max = 70 A.turns/m lH I max = c max N (15 + 10 + 15 + 10) × 10 −2 × 70 = 300 = 0.12A sites.google.com/site/ncpdhbkhn 38 AC Excitation (6) Ex The winding is excited with a 60-Hz voltage to produce B = 1.6sinωt T, the steel occupies 0.94 of the core cross-sectional area, ρc = 7.65 g/cm3 Find: a) The applied voltage? b) The peak current? c) The rms exciting current? d) The core loss? Bmax = 1.6T → Pa = VA/kg i 20 cm cm + e – 15 cm cm 300 turns Vc = (5 × 5) × (15 + 10 + 15 + 10) = 1250 cm Wc = ρ cVc = 7.65 × 1250 = 9562.5g = 9.56 kg Pa = → Iϕ ,rms = E rms Iϕ ,rms Wc PaW × 9.56 = E rms 283/ = 95.57A sites.google.com/site/ncpdhbkhn 39 AC Excitation (7) Ex The winding is excited with a 60-Hz voltage to produce B = 1.6sinωt T, the steel occupies 0.94 of the core cross-sectional area, ρc = 7.65 g/cm3 Find: a) The applied voltage? b) The peak current? c) The rms exciting current? d) The core loss? Bmax = 1.6T → Pc = 1.4 W/kg i 20 cm cm + e – 15 cm cm 300 turns Pc ,total = PW c c = 1.4 × 9.56 = 13.39 W sites.google.com/site/ncpdhbkhn 40 Magnetic Circuits and Magnetic Materials Introduction to Magnetic Circuits Flux Linkage, Inductance, and Energy Properties of Magnetic Materials AC Excitation Permanent Magnets Application of Permanent Magnet Materials sites.google.com/site/ncpdhbkhn 41 Permanent Magnets (1) sites.google.com/site/ncpdhbkhn 42 Permanent Magnets (2) Ex cm2, g = 0.2cm, lm = 2cm, Sm = Sg = Bg = ? l F = = H g g + H m lm → H g = − m H m g S φ = Bg S g = Bm S m → Bg = m Bm Sg Sm lm µ→∞ Magnetic Air gap, g material µ , S g µ→∞ Bg = µ0 H g S l → Bm = − µ0 g m H m Sm g = −4π × 10−7 × × Hm 0.2 = −126 × 10 −7 H m H m = −40kA/m → Bm = 0.50T Bg = Bm = 0.60T sites.google.com/site/ncpdhbkhn 43 Permanent Magnets (3) Sm S Bg = m Bm Sg lm H mlm = −1 Hgg µ→∞ Magnetic Air gap, g material µ , S g µ→∞ Bg = µ0 H g Point of maximum energy product → Bg2 = µ0 = µ0 → Volmag = Sm lm ( − H m Bm ) Sg g Volmag Volair gap ( − H m Bm ) Volair gap Bg2 µ0 ( − H m Bm ) sites.google.com/site/ncpdhbkhn 44 Magnetic Circuits and Magnetic Materials Introduction to Magnetic Circuits Flux Linkage, Inductance, and Energy Properties of Magnetic Materials AC Excitation Permanent Magnets 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→∞ lm sites.google.com/site/ncpdhbkhn Permanent magnetic material i N turns 47 Application of Permanent Magnet Materials (3) Ex g = 0.2cm, Sm = 2cm2, 2cm2 ≤ Sg ≤ 4cm2 a) Find the magnet length such that the system will operate on a recoil line which intersects the maximum B – H product point on the magnetization curve b) Calculate the flux density in the air gap as the plunger moves back and forth Bg = Sm Bm Sg H m lm = −1 Hg g S m Bm → lm = g S g − µ0 H m Sm µ →∞ Permanent magnetic material lm Sm g/2 x µ→∞ g/2 Point of maximum energy product H m 2cm = −40kA/m; Bm 2cm = 1T 2 → lm = 0.2 × 10 = 3.98cm 4π × 10 −7 × × 104 −2 sites.google.com/site/ncpdhbkhn 48 Ex Application of Permanent Magnet Materials (4) g = 0.2cm, Sm = 2cm2, 2cm2 ≤ Sg ≤ 4cm2 a) Find the magnet length such that the system will operate on a recoil line which intersects the maximum B – H product point on the magnetization curve b) Calculate the flux density in the air gap as the plunger moves back and forth S g lm Bm = − µ0 H m Sm g → Bm 4cm2 Sm µ →∞ Permanent magnetic material lm g/2 x µ→∞ g/2 Point of maximum energy product 3.98 = −4π × 10 H m 4cm2 0.2 −7 = −5 × 10−5 H m 4cm2 2cm ≤ S g ≤ 4cm ֏ 1.0T ≤ B ≤ 1.08T sites.google.com/site/ncpdhbkhn 49 ... sites.google.com/site/ncpdhbkhn 26 Magnetic Circuits and Magnetic Materials Introduction to Magnetic Circuits Flux Linkage, Inductance, and Energy Properties of Magnetic Materials AC Excitation Permanent... Machines and Stepping Motors VIII.Single and Two – Phase Motors IX Speed and Torque Control sites.google.com/site/ncpdhbkhn Magnetic Circuits and Magnetic Materials Introduction to Magnetic Circuits. .. ∫ H.dL = ∫ S -0 .5 J.dS -0 .5 -1 -1 sites.google.com/site/ncpdhbkhn Introduction to Magnetic Circuits (3) ∫ H.dL = ∫ S J.dS sites.google.com/site/ncpdhbkhn Introduction to Magnetic Circuits (4)