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Transcript
ENERGY CONVERSION ONE (Course 25741) Chapter one Electromagnetic Circuits …continued Hysteresis Losses • As I of coil slowly varying in a coil energy flows to coil-core from source • However, Energy flowing in > Energy returns • The net energy flow from source to coil is the heat in core (assuming coil resistance negligible) • The loss due to hysteresis called : Hysteresis Loss • hysteresis loss ~ Size of hysteresis loop • Voltage e across the coil: e=N dφ/dt Hysteresis Losses • Energy transfer during t1 to t2 is: 2 2 2 2 d Hl t Pdt t eidt N dt idt Nid B N N AdB lA B HdB (Vcore ) B HdB 1 1 1 1 1 1 t2 t2 B B B • Vcore=A l, volume of core • Power loss due to hysteresis in core: Ph=Vcore Wh f • f freq. of variation of i • Steinmetz of G.E. through large no. of experiment for machine magnetic materials proposed a relation: n Area of B-H loop = KBmax • Bmax is the max flux density Hysteresis Losses • n varies from 1.5 to 2.5, • K is a constant • Therefore the hysteresis power loss: Ph=Kh (Bmax)^n f • Kh a constant depends on - ferromagnetic material and - core volume EDDY CURRENT LOSS • Another power loss of mag. Core is due to rapid variation of B (using ac source) • In core cross section, voltage induced and ie passes, resistance of core cause: Pe =ie^2 R (Eddy Current loss) • this loss can be reduced as follows when: a- using high resistive core material, few % Si b- using a laminated core EDDY CURRENT LOSS • Application of Laminated Core Eddy current loss: Pe=KeBmax^2 f^2 Ke: constant depends on material & lamination thickness which varies from 0.01 to 0.5 mm CORE LOSS • Pc=Ph+Pe • If current I varies slowly eddy loss negligible • Total core loss determined from dynamic B-H loop: Pc Vcore f HdB dynamicloop • Using a wattmeter core loss can be measured • However It is not easy to know what portion is eddy & hysteresis Eddy Current Core Loss Sl & St • Effect of lamination thickness (at 60 Hz) Eddy Current Core Loss Sl & St • Effect of Source Frequency Sinusoidal Excitation • Example: • A square wave voltage E=100 V & f=60 Hz applied coil on a closed iron core, N=500 • Cross section area 0.001 mm^2, assume coil has no resistance • a- max value of flux & sketch V & φ vs time • b- max value of E if B<1.2 Tesla Sinusoidal Excitation • a - e = N dφ/dt => • • • • • • N.∆φ=E.∆t E constant => 500(2φmax)=Ex1/120 Φmax=100/(1000x120)Wb=0.833x10^-3 Wb b - Bmax=1.2 T (to find maximum value of E) Φmax=Bmax x A=1.2 x 0.001=1.2 x10^-3 Wb N(2φmax)=E x 1/120 Emax =120x500x2x1.2x10^-3=144 V Exciting Current Using ac Excitation • Current which establish the flux in the core • The term : Iφ if B-H nonlinear, nonsinusoid • a - ignoring Hysteresis: • B-H curve Ξ φ-i curve (or the rescaled one) • Knowing sine shape flux, exciting current waveform by help of φ-i curve obtained • The current non-sinusoidal, iφ1 lags V 90° no loss (since Hysteresis neglected) Exciting Current • Realizing Hysteresis, Exciting Current recalculated • Now iφ determined from multi-valued φ-I curve • Exciting current nonsinusoid & nonsymmetric • It can split to 2 components: ic in phase with e (represents loss), im in ph. With φ & symmetric Simulation of an RL Cct with Constant Parameters • Source sinusoidal i=Im . sin ωt • V = L di/dt + R i • ∫ v dt = ∫L.di + ∫ Ri.dt λ=L ∫di +R ∫i . dt = = L Im sinωt + R/ω Im cosωt • Now drawing λ versus i: • However with magnetic core L is nonlinear and saturate Note: Current sinusoidal Wave Shape of Exciting Current a- ignoring hysteresis • From sinusoidal flux wave & φ-i curve for mag. System with ferromagnetic core, iφ determined • iφ as expected nonsinusoidal & in phase with φ and symmetric w.r.t. to e • Fundamental component iφ1 of exciting current lags voltage e by 90◦ (no loss) • Φ-i saturation characteristic & exciting current Wave Shape of Exciting Current b- Realizing hysteresis • Hysteresis loop of magnetic system with ferromagnetic core considered • Waveform of exciting current obtained from sinusoidal flux waveform &multivalued φ-i curve • Exciting current nonsinusoidal & nonsymmetric Wave Shape of Exciting Current • It can be presented by summation of a series of harmonics of fundamental power frequency • ie = ie1 + ie3 + ie5 +… A • It can be shown that main components are the fundamental & the third harmonic Equivalent Circuit of an Inductor • Inductor: is a winding around a closed magnetic core of any shape without air gap or with air gap • To build a mathematical model we need realistic assumptions to simplify the model as required, and follow the next steps: • Build a System Physical Image • Writing Mathematical Equations Equivalent Circuit of an Inductor • Assumptions for modeling an Ideal Inductor: 1- Electrical Fields produced by winding can be ignored 2- Winding resistance can be ignored 3- Magnetic Flux confined to magnetic core 4- Relative magnetic permeability of core material is constant 5- Core losses are negligible Equivalent Circuit of an Inductor Ideal Inductor • • • • • v = e = dλ / dt Volts λ = L ie Wb v = L d ie /dt Volts realizing winding resistance in practice v = L d ie /dt + Rw ie Volts Equivalent Circuit of an Inductor • Realizing the core losses and simulating it by a constant parallel resistance Rc with L Equivalent Circuit of an Inductor • In practice Inductors employ magnetic cores with air gap to linearize the characteristic φ = φm + φl N φm = Lm ie Wb N φl = Ll ie Wb λ = N φ = Lm ie + Ll ie e = dλ/dt = =Lm die/dt + Ll die/dt Equivalent Circuit of an Inductor • Example: A inductor with air gap in its magnetic core has N=2000, and resistance of Rw=17.5 Ω. When ie passes the inductor a measurement search coil in air gap measures a flux of 4.8 mWb, while a search coil close to inductor’s winding measures a flux of 5.4 mWb • Ignoring the core losses determine the equivalent circuit parameters Equivalent Circuit of an Inductor • • • • φ = 5.4 mWb, φm= 4.8 mWb φl= φ – φm =0.6 mWb Lm=N φm/ ie =2000x4.8/0.7=13.7 H Ll=N φl/ ie = 2000 x 0.6 / 0.7=1.71 H