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

24
ENERGY CONVERSION ONE (Course 25741) Chapter one Electromagnetic Circuits …continued

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

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

ENERGY CONVERSION ONE (Course 25741)

Chapter one

Electromagnetic Circuits

…continued

Page 2: 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

Page 3: ENERGY CONVERSION ONE (Course 25741) Chapter one Electromagnetic Circuits …continued.

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

Page 4: ENERGY CONVERSION ONE (Course 25741) Chapter one Electromagnetic Circuits …continued.

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

Page 5: ENERGY CONVERSION ONE (Course 25741) Chapter one Electromagnetic Circuits …continued.

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

Page 6: ENERGY CONVERSION ONE (Course 25741) Chapter one Electromagnetic Circuits …continued.

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

Page 7: ENERGY CONVERSION ONE (Course 25741) Chapter one Electromagnetic Circuits …continued.

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

Page 8: ENERGY CONVERSION ONE (Course 25741) Chapter one Electromagnetic Circuits …continued.

Eddy Current Core Loss Sl & St

• Effect of lamination thickness (at 60 Hz)

Page 9: ENERGY CONVERSION ONE (Course 25741) Chapter one Electromagnetic Circuits …continued.

Eddy Current Core LossSl & St

• Effect of Source Frequency

Page 10: ENERGY CONVERSION ONE (Course 25741) Chapter one Electromagnetic Circuits …continued.

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

Page 11: ENERGY CONVERSION ONE (Course 25741) Chapter one Electromagnetic Circuits …continued.

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

Page 12: ENERGY CONVERSION ONE (Course 25741) Chapter one Electromagnetic Circuits …continued.

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)

Page 13: ENERGY CONVERSION ONE (Course 25741) Chapter one Electromagnetic Circuits …continued.

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

Page 14: ENERGY CONVERSION ONE (Course 25741) Chapter one Electromagnetic Circuits …continued.

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

Page 15: ENERGY CONVERSION ONE (Course 25741) Chapter one Electromagnetic Circuits …continued.

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

Page 16: ENERGY CONVERSION ONE (Course 25741) Chapter one Electromagnetic Circuits …continued.

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

Page 17: ENERGY CONVERSION ONE (Course 25741) Chapter one Electromagnetic Circuits …continued.

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

Page 18: ENERGY CONVERSION ONE (Course 25741) Chapter one Electromagnetic Circuits …continued.

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

Page 19: ENERGY CONVERSION ONE (Course 25741) Chapter one Electromagnetic Circuits …continued.

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

Page 20: ENERGY CONVERSION ONE (Course 25741) Chapter one Electromagnetic Circuits …continued.

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

Page 21: ENERGY CONVERSION ONE (Course 25741) Chapter one Electromagnetic Circuits …continued.

Equivalent Circuit of an Inductor

• Realizing the core losses and simulating it by a constant parallel resistance Rc with L

Page 22: ENERGY CONVERSION ONE (Course 25741) Chapter one Electromagnetic Circuits …continued.

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

Page 23: ENERGY CONVERSION ONE (Course 25741) Chapter one Electromagnetic Circuits …continued.

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

Page 24: ENERGY CONVERSION ONE (Course 25741) Chapter one Electromagnetic Circuits …continued.

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