Transcript 25471_energy_conversion_3

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:
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2
2
2
2
d
Hl
t Pdt  t eidt   N dt  idt   Nid  B N  N AdB  lA B HdB  (Vcore ) B HdB
1
1
1
1
1
1
t2
t2
B
B
B
• 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:
n
Area of B-H loop = KBmax
• Bmax is the max flux density
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:
Pc  Vcore f
 HdB
dynamicloop
• Using a wattmeter core loss
can be measured
• However It is not easy to know
what portion is eddy & hysteresis
Eddy Current Core Loss
Sl & St
• Effect of lamination thickness (at 60 Hz)
Eddy Current Core Loss
Sl & 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 =>
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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 Current
Using ac Excitation
• Current which establish the flux in the
core
• The term : Iφ if B-H nonlinear, nonsinusoid
• 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 Current
a- 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 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
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 Inductor
Ideal Inductor
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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 Wb
λ = 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
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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