topic 4 principlesmachines
TRANSCRIPT
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Jan-May 2010 Electrical Machines & Drives - ECNG 3010 2 -
Basic Principles of ElectromechanicalEnergy Conversion
Electric Drive
Electrical
System
Electrical
Machine
Mechanical
System
Motoring Mode
Braking (Generating)Mode
Pelec Pmech
Pelec Pmech
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Jan-May 2010 Electrical Machines & Drives - ECNG 3010 2 -
Magnetic Field
! Magnetic field produced by a current carrying conductor
! Ampere’s Law
!" H dl =Closed path
i
dl
2
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Flux Density
BsatBm
!0
!0
!m
HmHm
Bm
" Units : Weber/meter2 (Wb/m2)) or Tesla (T)
" In air B = !oH , !o = 4" x 107 Wb/A/m
" Ferro-magnetic materials (higher permeability)
# Linear approximation Bm = !mHm# Bsat~ 1.6 – 1.8 Tesla
# In saturation !m approaches !0
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Jan-May 2010 Electrical Machines & Drives - ECNG 3010 2 -
Flux, Flux Linkage & MMF
! Flux !m (Wb)
(assuming uniform flux density)
!m = Bm A m
Bm= !mHm and Hm = Ni / l m
!m
= Ni
l m"mAm
= F
R
Am
!m
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Jan-May 2010 Electrical Machines & Drives - ECNG 3010 2 -
Flux, Flux Linkage, and MMF
! Reluctance
! FluxLinkage
! MMF
" m = N !m
F = N i
m
=l m
"mAmR
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Jan-May 2010 Electrical Machines & Drives - ECNG 3010 2 -
Analogy between Electric and MagneticCircuits
Electric Circuit Unit Magnetic Circuit Unit
Driving Force emf (V) V mmf (F) At
Response current (I) A flux (!) Wb
Impedance resistance (R) " reluctance ( R ) 1/H
Equivalentcircuit
Field intensityrelationship
# E . dl = V V # H . dl = I A
Potentialdifference
V = IR V At
V R +
_ I
V = IR
F+
_ !
F = !
F = !
R
R
R
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Jan-May 2010 Electrical Machines & Drives - ECNG 3010 2 -
Inductance
! For linear magnetic conditions inductance depends only on magnetic circuit
Energy stored in
magnetic circuits
Energy density
"m = Lmi
i A m
!mN
!m
Lm=i
= N
l m
"mAm N# m
=
N2
l m "mAm
= N2
R
N2
l m
"mAm
Lm=
i Hm Bm !m ! m N l rx x (!m) x (A m) x (N)
volume
w = W = 1 B2m
volume 2"m
W= (1/2) L i2 = 1 B2 A ml m
2"m
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Rotating Machine
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Rotating Machine
Stator
Rotor
Air Gap
Two sets of windings: stator and rotor… why?
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Rotating Machine
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Jan-May 2010 Electrical Machines & Drives - ECNG 3010 2 -
Basic Principles of Operation
! Force on a current carrying conductor subjected toan externally-established magnetic field (extensionof Lorentz law – motor equation)
$ f em = il xB
! emf induced in a conductor moving in a magnetic
field ( extension of Faraday law)$ e = B l u
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Jan-May 2010 Electrical Machines & Drives - ECNG 3010 2 -
Electromagnetic Force
! Force Direction – Higher concentration field to lower.
External B field
f em f em
subtract add
f em
resultant
B
il
f em = B l i
[Nm] [Wb/m2] [m] [A]
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Jan-May 2010 Electrical Machines & Drives - ECNG 3010 2 -
Electromagnetic Force
Torque produced by forcescaused by interaction ofcurrent-carrying conductors
and magnetic fields
ferromagnetic pieces
Principle of Alignment
Magnetic flux
lines
F
F
Principle of interaction
F F
F
F
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! Force on positive charges
f q = q ( u x B )
In this example a net positive chargeaccumulates at the top and a net
negative charge accumulates at thebottom (Lorentz force)
! Magnitude of induced emf
e = B l u
[V] [Wb/m2] [m] [m/s]
Polarity of induced emf is given by f q
and is independent of current flowingthrough the conductor.
Induced EMF
f q-
f q+
+
_
B
(into paper)
f q+
u
B
(into paper)
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Jan-May 2010 Electrical Machines & Drives - ECNG 3010 2 -
Rotating Electrical Machine:Two windings
! One winding: producesmain magnetic field (couldbe replaced by magnets).
! Next winding: produces thetorque due to the “motorequation”. Carries thecurrent associated withtorque.
! Additional issues: inducedvoltages, more than onemagnetic field.
F
F
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Jan-May 2010 Electrical Machines & Drives - ECNG 3010 2 -
Production of Magnetic Field
! Radial field (H,F,B) in the air gap
H positive if away from center
Rotorsurface
Nsis#02l g
-$/2 0 $ 2$
Statorsurface B s(t)
+ Nsis#
02l g
!
Magnetic
axis
!
" = $
r
is(t)
is(t) N s
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Jan-May 2010 Electrical Machines & Drives - ECNG 3010 2 -
Application of Basic Principles
Assumptions
Uniform B s , radial in direction
- Rotor current of constant magnitudebut polarity changes with position
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Counter- clockwise torque ispositive
Force acting on conductor
f em = B s (N r I) l
and torque on coil
T em = 2 f emr = 2 B s (N r I) l r
Emf induced in coil
er =2 econd = 2 N r B s l r"m r
i r
I
- I
er E
- E
0
0
0
1800
3600
T em
%
%
%
Magnetic
axisElectricalsystem
!
&m
Tem
+ _ er ir
stator
MotoringMode
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