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ENERGY CONVERSION ONE (Course 25741) Chapter one Electromagnetic Circuits …continued.

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Presentation on theme: "ENERGY CONVERSION ONE (Course 25741) Chapter one Electromagnetic Circuits …continued."— Presentation transcript:

1 ENERGY CONVERSION ONE (Course 25741) Chapter one Electromagnetic Circuits …continued

2 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

3 Hysteresis Losses Energy transfer during t 1 to t 2 is: V core =A l, volume of core Power loss due to hysteresis in core: P h =V core W h 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 = B max is the max flux density

4 Hysteresis Losses n varies from 1.5 to 2.5, K is a constant Therefore the hysteresis power loss: P h =K h ( B max)^n f K h a constant depends on - ferromagnetic material and - core volume

5 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 i e passes, resistance of core cause: P e =i e ^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

6 EDDY CURRENT LOSS Application of Laminated Core Eddy current loss: P e =K e B max ^2 f^2 K e : constant depends on material & lamination thickness which varies from 0.01 to 0.5 mm

7 CORE LOSS P c =P h +P e 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

8 Eddy Current Core Loss Sl & St Effect of lamination thickness (at 60 Hz)

9 Eddy Current Core Loss Sl & St Effect of Source Frequency

10 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 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

11 Sinusoidal Excitation a - e = N dφ/dt => N.φ=E.t E constant => 500(2φ max )=Ex1/120 Φ max =100/(1000x120)Wb=0.833x10^-3 Wb b - B max =1.2 T (to find maximum value of E) Φ max =B max x A=1.2 x 0.001=1.2 x10^-3 Wb N(2φ max )=E x 1/120 E max =120x500x2x1.2x10^-3=144 V

12 Exciting Current Using ac Excitation Current which establish the flux in the core The term : I φ if B-H nonlinear, non- sinusoid 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, i φ1 lags V 90° no loss (since Hysteresis neglected)

13 Exciting Current Realizing Hysteresis, Exciting Current recalculated Now i φ determined from multi-valued φ-I curve Exciting current nonsinusoid & nonsymmetric It can split to 2 components: i c in phase with e (represents loss), i m in ph. With φ & symmetric

14 Simulation of an RL Cct with Constant Parameters Source sinusoidal i=I m. sin ωt V = L di/dt + R i v dt = L.di + Ri.dt λ=L di +R i. dt = = L I m sinωt + R/ω I m cosωt Now drawing λ versus i: However with magnetic core L is nonlinear and saturate Note: Current sinusoidal

15 Wave Shape of Exciting Current a- ignoring hysteresis From sinusoidal flux wave & φ-i curve for mag. System with ferromagnetic core, i φ determined i φ as expected nonsinusoidal & in phase with φ and symmetric w.r.t. to e Fundamental component i φ1 of exciting current lags voltage e by 90 (no loss) Φ-i saturation characteristic & exciting current

16 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

17 Wave Shape of Exciting Current It can be presented by summation of a series of harmonics of fundamental power frequency i e = i e1 + i e3 + i e5 + … A It can be shown that main components are the fundamental & the third harmonic

18 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

19 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

20 Equivalent Circuit of an Inductor Ideal Inductor v = e = dλ / dt Volts λ = L i e Wb v = L d i e /dt Volts realizing winding resistance in practice v = L d i e /dt + R w i e Volts

21 Equivalent Circuit of an Inductor Realizing the core losses and simulating it by a constant parallel resistance R c with L

22 Equivalent Circuit of an Inductor In practice Inductors employ magnetic cores with air gap to linearize the characteristic φ = φ m + φ l N φ m = L m i e Wb N φ l = L l i e Wb λ = N φ = L m i e + L l i e e = dλ/dt = =L m di e /dt + L l di e /dt

23 Equivalent Circuit of an Inductor Example: A inductor with air gap in its magnetic core has N=2000, and resistance of R w =17.5 Ω. When i e passes the inductor a measurement search coil in air gap measures a flux of 4.8 mWb, while a search coil close to inductors winding measures a flux of 5.4 mWb Ignoring the core losses determine the equivalent circuit parameters

24 Equivalent Circuit of an Inductor φ = 5.4 mWb, φ m = 4.8 mWb φ l = φ – φ m =0.6 mWb L m =N φ m / i e =2000x4.8/0.7=13.7 H L l =N φ l / i e = 2000 x 0.6 / 0.7=1.71 H

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