ENERGY CONVERSION ONE Course 25741 Chapter one Electromagnetic

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 t 1 to t 2 is: • 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: Area of B-H loop = • 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=Ke. Bmax^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: • 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)=Ex 1/120 Φmax=100/(1000 x 120)Wb=0. 833 x 10^-3 Wb b - Bmax=1. 2 T (to find maximum value of E) Φmax=Bmax x A=1. 2 x 0. 001=1. 2 x 10^-3 Wb N(2φmax)=E x 1/120 Emax =120 x 500 x 2 x 10^-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 = ie 1 + ie 3 + ie 5 +… 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 = L m 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 m. Wb, while a search coil close to inductor’s winding measures a flux of 5. 4 m. Wb • Ignoring the core losses determine the equivalent circuit parameters

Equivalent Circuit of an Inductor • • φ = 5. 4 m. Wb, φm= 4. 8 m. Wb φl= φ – φm =0. 6 m. Wb Lm=N φm/ ie =2000 x 4. 8/0. 7=13. 7 H Ll=N φl/ ie = 2000 x 0. 6 / 0. 7=1. 71 H