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:

• 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

2

1

2

1

2

1

2

1

2

1

2

1

)(t

t

t

t

B

B

B

B

B

B

core HdBVHdBlAAdBN

HlNNididt

dt

dNeidtPdt

nKBmax

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:

• Using a wattmeter core loss

can be measured• However It is not easy to know

what portion is eddy & hysteresis

pdynamicloo

corec HdBfVP

Eddy Current Core Loss Sl & St

• Effect of lamination thickness (at 60 Hz)

Eddy Current Core LossSl & 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 CurrentUsing ac Excitation

• Current which establish the flux in the core

• The term : Iφ 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, 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 Currenta- 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 Currentb- 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 InductorIdeal 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 Wbλ = 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