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