Download presentation

Presentation is loading. Please wait.

Published byMohammad Pebley Modified over 4 years ago

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 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

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

Similar presentations

Presentation is loading. Please wait....

OK

ENERGY CONVERSION ONE (Course 25741)

ENERGY CONVERSION ONE (Course 25741)

© 2018 SlidePlayer.com Inc.

All rights reserved.

To ensure the functioning of the site, we use **cookies**. We share information about your activities on the site with our partners and Google partners: social networks and companies engaged in advertising and web analytics. For more information, see the Privacy Policy and Google Privacy & Terms.
Your consent to our cookies if you continue to use this website.

Ads by Google

Ppt on content addressable memory cell Ppt on areas of parallelograms and triangles worksheets Ppt on motivation and leadership Ppt on precautions of tsunami wave Ppt on our country india class 6 Ppt on revolution and rotation of the earth Download ppt on indus valley civilization images Ppt on australian continent highest Ppt on political parties and electoral process lesson Ppt on computer graphics applications