Transcript Document 7466434

6.11s Notes for Lecture 1
Elements of Energy Flows in Electromechanics
June 12, 2006
J.L. Kirtley Jr.
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Take this as a prototypical machine form
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
2R


T

2

R



R

2

R

2
 
P  T  2R   2R u
u  R
2
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K zR  ReK R e j t p  ReK R e j tkx 
p
x  R
k
R
F J B
  K  B
  ReK R e j t p ReB re j t p 
1
1
*
   ReK R B  K R Br cos
2
2
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
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Field Description of Forces: Maxwell Stress Tensor
If we include forces due to changing permeability,
We find force is the divergence of something we call a Tensor
1 2
F  J  B  H 
2
F   T

Fk  
Tik
x i
i
1
2
Tik  H i H k  ik H
2

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Force on an object is the integral of force density:
f 
 Fdv     Tdv   T  nda
vol
fi 


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vol
surface
  T n da
ik
k
surface k
Tr  K z Br  0 H Hr  
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Forces due to fields normal to a surface
Assume these are highly permeable poles
1 2
1 2
Tzz  0 H  0 Hz  0 Hz
2
2
2
z
Force is UP on lower pole, DOWN on upper pole

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Poynting’s Energy Flow is:
S  E H
Power flow INTO a volume is:
P 


surface
S  nda  
   Sdv
volume
  S    E  H  H   E  E   H

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This means we must invoke our old
friends, Farady’s Law and Ampere’s Law
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Using the two induction equations, we find
B
H J
t
B
  S  H 
 E J
t

B 
P   E  J  H  dv
t 
volume
E 
Of course J is flowing in something, and if
 that is stationary:
J  E
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
so
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E  J  E 2 
J2

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If the magnetic material is linear and lossless, the
magnetic term is clearly rate of change of energy
stored. More on this later
B  H
B  1
2
H

H
t t 2
If the material is moving,

E  E  v  B
E  J  E  J  v  B  J
v  B  J  v  B  J  v  J  B
And this last is clearly energy conversion
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(velocity times force density)
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Now look at a prototypical situation: like an air-gap. We are
looking in the axial direction and there is no variation tha way.
Excitation is z-directed and like a traveling wave
K z  ReK ze j tkx 
E  iz ReE ze j tkx 
H  ix ReH x e j tkx  iy ReH y e j tkx 
Since Faraday’s Law is:
ix

E 

x
Ex
The interesting part
of this is:

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iy

x
Ey
H y
E z

 0
x
t
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i H 
x x

  
 0 ix H y 
x
t  
Ez
ix H z 
iz
or E z  0

k
Hy
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Now suppose the lower surface is moving
x  x  ut
t  kx    kut  kx   st  kx  st  kx 
Now energy flows are related: In the stationary frame:
Sy 


k
0 H x H y 

k
Txy
And in the moving frame:
Sy 

s
k
0 H x H y 
s
k
Txy
The difference between these must be energy converted
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Sy Sy 
  s
k
0 H x H y 
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
k
Txy (1 s)
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Now consider about any type of system
with voltage and current defined at a
pair of terminals:
Energy into those terminals over a time
interval is:
And over a periodic cycle, energy input

per cycle is:

p  vi  i
t
w

dt   id
time t
i
w cycle 
 id
cycle

Here we would have net energy IN to the
system (possibly a motor?)

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If there is no motion so there
are no mechanical terminals,
we have an analysis of
hysteretic material loss
There is more to this than fits
on a page -- see the notes
 Hd
     B  nda
N
w
   H  dBdv
Ni 
cycle
vol cycle

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K zs  ReK se j tkx 
Here is a simple ‘cartoon’
model of a linear
induction motor. The
analysis follows steps we
have already outlined
H y
 K zs  K zr
x

K r   s s 0 H y
g
k

 
 jkg  s s 0 H y  K s

k 
j
1
Hy  Ks
kg 1 j 0 s s
k 2g
Txy  0K zsH y
 Txy 
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
0 K s
 Txy 
2
2 kg
0
2
ReK s H

y
0 s s
k 2g
0 s s 2
1  2 
 k g 
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
Surface Impedance is important: For energy flow,
Sy  E z Hx 
s
k
0Hx Hy  ZsHx
For a layer of magnetically linear conductive stuff excited
by a traveling wave:
Z s  j


cothh 

if h  , Z s 
  j  k 2
1 j


2

In iron, if we can idealize the saturation curve,

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Zs 
8 2 j
3 
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
2Hx
Bs
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Iron Loss Consists of
Eddy Current (linear)
Hysteresis (nonlinear and hard to characterize)
Semi-Empirical ‘Curve Fit’
 B  b  f  f
P  P0    
B0   f 0 
 B  1
 B  2
Q  Q01   Q02 
B0 
B0 

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