Magnetic Circuit Lecture 2 PDF

Magnetic Circuit Lecture 2 UNIVAC magnetic core RAM  Magnet Magnet is the ability of attracting objects. The poles of a magnet will align with the earth’s magnetic field. Magnetic field lines exist from N-pole to S-pole of the magnet. Every m...

Magnetic Circuit Lecture 2 UNIVAC magnetic core RAM  Magnet Magnet is the ability of attracting objects. The poles of a magnet will align with the earth’s magnetic field. Magnetic field lines exist from N-pole to S-pole of the magnet. Every magnetic molecule has it’s own North Pole and South Pole. Every magnetic material consists of magnetic domains and magnetic molecules.  Magnetic Circuits Basic Principles: – A current carrying wire produces magnetic field around it Transformer Action [i→B→E→i] Motor Action [i+B→F] Generator Action [F+B→i] i – Current, B – Magnetic, E – EMF, F - Force In what application we can find magnetic circuits in our real life?  Simple magnetic circuit Three basic components to build a magnetic circuit! Current Wire Coils Ferromagnetic material Current (I) carrying coils are winded around a ferromagnetic material Magnetic field (H) is produced based on the Ampere’s law. Magnetic flux (φ) moves around the core along the mean path length (lc) which is bounded within a cross-sectional area (Ac)  Maxwell’s Equation Cited: http://dumlphyshead.wordpress.com/2009/07/21/maxwells-equations/ Gauss’s law for Electricity Gauss’s law for Magnetism Faraday’s law Ampere’s law  Ampere’s law (1) Assume the “magneto-quasi-static” form (the magnetic field has no change with respect to time)  H  dl =  J  d a c s i = H l N c c Definition: The magnitude of magnetic flux in a closed path equals to the total current passing through an unit surface.  Ampere’s law (2) Assume uniform area Ac and high permeability magnetic materials [μc =μ0 μr ; μ0 = 4π × 10-7] – With a given closed loop magnetic path, the total current running through the N turns coils produces the magneto-motive force, MMF. f =  Ni =  Hl  Faraday’s Law (1) When magnetic field varies in time, an electric field is produced in space. Definition: Line integral of electric field intensity (E) around a closed loop is equal to the rate of magnetic flux linking with the surface defined by the loop.  Faraday’s Law (2) Since the winding (and hence the contour C) links the core flux N times, then d d e(t ) = N = dt dt The induced voltage represents the voltage due to a time-varying flux linkage. It is also known as electromotive force (EMF).  B vs H Air-space Magnetic field intensity (H) – Unit: A.turn/m – Representing the effort exerted by the current to establish a magnetic field Magnetic flux density (B) – Unit: tesla (T) or Weber/m2 or Nm-1A-1 – Representing the flux per unit area in the magnetic core Magnetic material  B-H Curve and Relationship The relationship between flux density and magnetic force (H) is given: B = H =  0  r H  Magnetic permeability (µ) Unit: Weber/ A-turn-m; Henry/m; NA-2 Represents the relative ease of establishing a magnetic field in a given material µ0 → air gap/ free space permeability (µ0 = 4π × 10-7) µc → core material permeability µr = µc / µ0 → relative permeability The highest value of permeability, the easier magnetic flux moving in the material Non-linearity due to B-H characteristics c = r 0  Types of Magnetic material Ferro/ferritemagnetic – Iron, nickel, cobalt, and most of the alloys – Retain magnetization and become magnets Paramagnetic – Platinum, aluminum and oxygen liquid – Do not retain any magnetization in the absence of an externally magnetic field Diamagnetic – Carbon, copper, water and plastic – Have the opposition of magnetic field to the externally applied magnetic field  Ferromagnetic Basic mechanism to form permanent magnets or attracted to the magnets Can be magnetized by an external magnetic field and remain magnetized after the external field is removed. Two types of ferromagnetic: (1) Soft Ferromagnetic for which a linearization of the B-I variation in a region is possible, (2) Hard Ferromagnetic for which difficult to define the term permeability and it may refer to permanent magnets.  Magnetic domain Magnetic domain of ferromagnetic materials under the influence of external magnetic-field strength H  Total Magnetic flux (φ) Magnetic flux φ (crossing a surface S) is the surface integral of normal component of B  =  B  da s Assuming uniform B across Ac  = Bc Ac  Reluctance  Bc Ac f = Ni = H c lc = lc = lc c c lc f =  c Ac lc 1 Reluctance/Resistance R= = c Ac  Permeance /Conductance Unit: A.turns/Wb  Air Gap What if the magnetic circuit has air gaps? Fringing effect (the cross-sectional area (Ag) in the air gap is larger than the core (Ac)) Ag  Ac  Air Gap Assuming no fringing effect (Ac = Ag = A) f = Ni = H c lc + H g l g  Bc = Bg = Note: (Researchers seek for perfect material) A In real life, we will choose high permeability Bc Bg ferromagnetic material. Therefore, f = lc + lg c 0 If µc>>µ0, Rc = 0, Φ=f/Rg lc lg f = ( + ) f =  ( Rc + R g ) c A 0 A Magnetic circuit has a similar operation as the electrical circuit  Analogy between Electric and Magnetic Circuit Electric circuit Magnetic circuit Formula Unit Formula Unit J A/m2 B Wb/m2 V A F I= = Wb ( R1 + R2 ) ( Rc + Rg ) l Ω l A.turns/Wb R= R= A A V = IR V f = R A.turns RT = R1 + R2 Ω RT = Rc + Rg A.turns/Wb  Example 2-1 (Fringing Effect) Neglect leakage flux (φ11 and φ22) Compensate for fringing effect g1 and g2 ( Add gap length to each side of gap area) Compute φT , φ1 and φ2 1cm  Solution (1) Since µ →∞, therefore no flux resistance in the core material, Rc = 0. Hint: Always convert to magnetic equivalent circuit  2  1  Solution (2) Steps to calculate flux F = Ni F  = R l l Mean path R= A Which cross sectional area does it refer? A Core or Airgap area? Fringing Effect in the airgap?  Airgap: μ=μ0 Core: μc= μrμ0  Solution (3) WARNING: Only accept three Fringing Effect (g1) significant points in the exam and assignments. Ag1 = (0.5 + 0.1)  (1 + 0.1)  10 −4 = 0.66  10 −4 m 2 l g1 = 0.1 10 −2 m WARNING: Use SI unit for all calculation! l g1 0.110 −2 R g1 = = = 1.206 107 At / Wb  0 Ag1 4 10 −7  0.66 10 − 4 Ni 900  0.2 −5  g1 = = = 1.49  10 Wb Rg1 1.206 10 7  Solution (4) Fringing Effect (g2) Ag 2 = (1 + 0.25)  (1 + 0.25)  10 −4 = 1.56  10 −4 m 2 l g 2 = 0.25  10 −2 m lg 2 0.25 10−2 Rg 2 = = = 1.273  10 7 At / Wb 0 Ag 2 4 10 1.56 10 −7 −4 Ni 900  0.2 g 2 = = = 1.413  10 −5 Wb Rg 2 1.273 10 7  Solution (5) Total Flux T =  g1 +  g 2 = (1.49 + 1.413) 10 −5Wb = 2.913 10 −5Wb  Behavior of Ferromagnetic Materials f = Ni = Hl Non-linear  = BA i H B φ linear linear B = H The strength of the magnetic flux produced in the core depends on its material permeability (µ), current (i), and core dimension (L, A) The non-linearity of magnetic circuit is caused by B-H characteristic.  B-H curve Ferro-material maximum permeability r knee The advantage of using ferromagnetic material for cores is that: we can get more flux for a given magneto-motive force (f) in iron than in air. Most non-linear systems (motors, generators and transformers) are designed to operate near to the knee of the magnetizing curve.  Magnetizing curve for commonly used material  Energy losses (1) Hysteresis losses in the core – Magnetic material does not return to its original value after the decrease of external field – The internal field 'lags' behind the external field. This behaviour results in a loss of energy, called the hysteresis loss, when a sample is repeatedly magnetized and demagnetized. (2) Eddy current losses – Induced voltages cause circulate of current flow within the core on the conductor surface  Hysteresis Loss Residual Flux – The magnetic flux density that remains in a material when magnetizing force is zero Coercive force – The amount of reverse magnetic field which must be applied to a magnetic material to make the magnetic flux return to zero.  Soft Vs Hard ferromagnetic The difference of soft and hard ferro-magnetic material is: The coercive force in the soft magnetic is smaller than hard magnetic. Types of magnetic material Hysteresis loop as RAM Memory  Eddy Current loss Faraday’s law explains the eddy current losses of iron cores Induced voltage caused swirls of current flow within the core (like the eddies seen at the edges of a river) Lamination  Hysteresis and Eddy current losses calculation Consider a core of volume (V), maximum flux density Bmax and frequency of supply (f) Hysteresis Losses (Ph) Ph = k hVfBmax n watt – n is a steinmetz constant depends on material and it is between 1.5 to 2.5 – kh is a constant Eddy current losses (Pe) Pe = ke (d Bmax f ) 2 watt – d is the material thickness – ke is a constant  Inductance Assuming linear B-H characteristics We can define the λ-I relationship by inductance (L)  L= i μ= constant Name Symbol Unit Total flux linkage λ=Nφ Wb.turn Inductance L H Current i A  Self inductance Definition: – The ratio of the electromotive force produced in a circuit by self-induction to the rate of change of current producing it – Self-Inductance caused by magnetic flux created by its current flowing in the coils  N L= = i i N 2 0  r A  f = Ni = Hl L= = i l B = 0 r H  Self Inductance Proof Number of turn B =  o nI n =N/l Core length Faraday’s Law (Induced EMF) d B d ( o nIA )  = −N = −N dI = − N o nA = − NBA dI dt dt dt I dt N B dI =− = −L dI I dt dt N B Define: Self Inductance L= I  Example 2-2 Compute inductance, L (linear property) No Fringing  Solution  N = Ni L= = Rc + R g i i Ni N NiA = =  lc lg i lc lg + +  c Ac  0 Ag c 0 NiA = 0 N 2 A lc + lg No Fringing Effect, Ac = Ag=A = c 0  0 lc + lg c Note: if core material has infinity permeability, inductance L L  0 N 2 A lg  Induced EMF vs Inductance d d ( Li ) Faraday’s law e= = dt dt For static magnetic circuit (L= constant, constant flux) d (i) e=L dt For rotating magnetic circuit (L=L(t), rotating flux) di dL e=L +i dt dt  Power & Energy (Work done) d dW P(t ) = ei = i = dt dt t2 2 W =  P(t ) dt =  i d t1 1  Assuming L = constant: L = i μ= constant 1 2 1 2 Wtotal =  = Li 2L 2  Magnetic Circuits with two windings Self inductance: Self inductance is a property of a single coil, due to the fact that the magnetic field set up by the coil current links the coil itself. Mutual Inductance: It is the phenomenon in which a change of current in one coil causes an induced Emf in another coil placed to the first coil.  Mutual Inductance Definition: – The mutual inductance M can be defined as the proportionality between the EMF generated in coil 2 to the change in current in coil 1 which produced it. 1 =  + 11 2 =  + 22  = 12 = 21 = mutual flux 11 = primary leakage flux 22 = secondary leakage flux  Inductance Relationship (1) N11 L11 = When i2=0 i1 Self Inductance N 22 L22 = When i1=0 i2 N112 N1 ( K 22 ) N 2 K 2 N1 L12 = =  = L22 i2 i2 N2 N2 Mutual Inductance N N (K  ) N KN L21 = 2 21 = 2 1 1  1 = 1 2 L11 i1 i1 N1 N1 Φ1 = Flux produced by i1, which links only the N1 turns Φ21 = Flux produced by i1, which links with N2 as well as N1  Inductance Relationship (2) K 2 N1 L12 = L22 N2 Multiplication L12 L21 = K1 K 2 L11 L22 K1 N 2 L21 = L11 N1 M = L12 = L21 = K L11L22 Mutual Inductance K = K1 K 2 Coupling Coefficient  Derivation of flux linkage (1) Assuming no leakage flux 11 = 22 = 0 and high core permeability f = Ni = N1i1 + N 2 i2 f  0 Ac Mutual flux: = = ( N1i1 + N 2i2 ) Rg g It is broken into terms attributed by i1 and i2: 0 Ac 0 Ac 1 = N1 = N 1 2 i1 + N1 N 2 i2 g g = L11i1 + L12i2  Derivation of flux linkage (2) 0 Ac 0 Ac 1 = N1 = N 1 2 i1 + N1 N 2 i2 g g = L11i1 + L12i2 L11 = self-inductance of coil 1 L12 = Mutual inductance between coil 1 and 2 0 Ac 0 Ac 2 = N 2 = N1 N 2 i1 + N 2 2 i2 g g = L21i1 + L22i2 L22 = self-inductance of coil 2 L21 = Mutual inductance between coil 1 and 2  Inductance under high core permeability 0 Ac N12 L11 = N12 = g Rg 0 Ac N 22 L22 = N 22 = g Rg 0 Ac N1 N 2 L12 = L21 = N1 N 2 = =M g Rg  Coupled circuits In a single core, φP=φS All flux links in the two coils Coupling factor 0≤K≤1 and M= K√ (LpsLsp) d S d p Lsp = M = N S and L ps = M = N p di p dis  Series connection & Mutually-coupled coils (1)  di1 di2 e1 = L1 dt + M 12 dt  e = L di2 + M di1  2 2 dt 21 dt di di1 di2 di2 di1 e = LT = L1 + M 12 + L2 + M 21 dt dt dt dt dt di1 di2 = ( L1 + M 21 ) + ( M 12 + L2 ) M 12 = M 21 dt dt  Series connection & Mutually-coupled coils (2) Scenario 1 Scenario 2  Parallel Connection Mutually-coupled coils Mutually coupled coils in Mutually coupled coils in parallel –aiding connection parallel –opposing connection L1 L2 − M 2 L1 L2 − M 2 Le = Le = L1 + L2 − 2M L1 + L2 + 2M  Dot Convention A dot is placed on one terminal of each winding; when current enters the dotted terminal of a coil ➔polarity of induced voltage (into other coil) is positive at its dotted terminal Use to decide the Positive/Negative sign of mutual inductance  Possibility of Dot Convention at the transformer Input current at Input current at the primary side the Secondary side  Dot Convention Example (1) Loop 1 Loop 2 di1 di2 Loop 1: v g = i1 R1 + L1 −M dt dt di2 di1 Loop 2: 0 = i2 R2 + L2 −M dt dt  Dot convention Example (2) coil 1 : L1 + M 12 − M 13 = 5 + 2 − 1 = 6 H  coil 2 : L2 + M 12 − M 23 = 10 + 2 − 3 = 9 H coil 3 : L − M − M = 15 − 3 − 1 = 11H  3 23 13 LT = 6 + 9 + 11 = 26 H  Dot convention Example (3) Mutual inductance between 2 4H and 9H inductors 1 3 1 2 3  Example 2-6 30mm r *Stacking Factor (SF) = Actual Magnetic Area / Gross Cross Area  Solution (1) Magnetic equivalent circuit  Solution (2) Limb ab lab = Aab = 2.5x2.5×10-4 8  10 −2 (0.5+6.0+1.5)×10-2 = 6.25 ×10-4m2 Rab = 1200  4  10 −7  6.25  10 − 4 = 8×10-2m = 84.9kAt / Wb Limb bc lbc = Abc = 3x2.5×10-4 Rbc = 66.3 kAt/Wb (1.25+5.0+1.25)×10-2 = 7.5×10-4m2 = 7.5×10-2m Limb cd lcd = 8×10-2m Acd = 6.25×10-4m2 Rcd = 84.9 kAt/Wb Limb de lde = 0.5+4+1.0 Ade = 6.25×10-4m2 Rde = 58.4 kAt/Wb = 5.5×10-2m Limb ef lef = 7.5×10-2m Aef = 2.0x2.5 ×10-4 Ref = 99.5 kAt/Wb = 5×10-4m2 Limb fa lfa = 5.5×10-2m Afa = 6.25×10-4m2 Rfa = 58.4 kAt/Wb Limb ad lad = 7.5×10-2m Aad =1.0x2.5×10-4 Rad = 198.9kAt/Wb = 2.5×10-4m2  Solution (3) L11 is occurred due to φ by i1 (& i2=0) R11 = Rad || ( Rab + Rbc + Rcd ) + Ref + R fa + Rde = (1.989 105 || 2.361105 ) + 0.995 105 + 0.584 105  2 = 3.2425 105 At / Wb N12 10002 L11 = = = 3.084H R11 3.2425 10 5  Solution (4) L22 is occurred due to φ by i2 (=>i1=0) R22 = Rad || ( Ref + R fa + Rde ) + Rab + Rbc + Rcd = (1.989 105 || 2.163 105 ) + 2.361105 = 3.397 105 At / Wb N 22 5002 L22 = = = 0.736H R22 3.397 10 5  Solution (5) L12 and L21 are occurred due to φ by the opposite current φ in coil 2 is losing N 221 N 2 1 L21 = i = i (0.4571 ) Rad 21 = Rad + ( Rab + Rbc + Rcd ) 1 1 1.989 N2 N1i1 = 1 = (0.457)  1.989 + 2.361 i1 R11 = 0.4571 1000  500 = 0.457  = K11 3.2425 105 = 0.705H  Solution (6) L12 and L21 are occurred due to φ by the opposite current φ in coil 1 is losing N1 12 N1 Rad L12 = = (0.4792 ) 12 = 2 i2 i2 Rad + ( Ref + R fa + Rde ) N1 N 2i2 1.989 = (0.479)  = 2 i2 R22 1.989 + 2.163 = 0.4792 1000  500 = 0.479  = K 22 3.397  10 5 = 0.705H  L21  Solution (7) Find the mutual constant M = K L11 L22 M 0.705 K= = = 0.468 L11L22 3.084  0.736 OR K = K1 K 2 = 0.457  0.479 = 0.468 To be continue…