ECE 330 POWER CIRCUITS AND ELECTROMECHANICS LECTURE … 17.pdf · Sauer’s ECE 330 lecture notes....
Transcript of ECE 330 POWER CIRCUITS AND ELECTROMECHANICS LECTURE … 17.pdf · Sauer’s ECE 330 lecture notes....
Copyright © 2017 Hassan Sowidan
ECE 330
POWER CIRCUITS AND ELECTROMECHANICS
LECTURE 17
FORCES OF ELECTRIC ORIGIN – ENERGY
APPROACH
Acknowledgment-These handouts and lecture notes given in class are based on material from Prof. Peter
Sauer’s ECE 330 lecture notes. Some slides are taken from Ali Bazi’s presentations
Disclaimer- These handouts only provide highlights and should not be used to replace the course textbook.
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MECHANICAL EQUATIONS
• Newton’s law states that acceleration force in the
positive x direction is equal to the algebraic sum of
all the forces acting on the mass in the positive x direction
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MECHANICAL EQUATIONS
• In a linear or translational system with displacement
x, velocity , mass M , and different forces (along x):
• In a rotational system with displacement θ,rotational
speed ω, moment of inertia J, and different torques:
3
, e
spring damper ext
d dJ T T T T
dt dt
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e
spring damper ext
dx
dt
dM f f f f
dt
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MECHANICAL EQUATIONS
• Forces and torques with superscript e are of
electrical origin.
• In a translational system, power is
• In a rotational system, power is
dxP f f
dt
dP T T
dt
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ENERGY CONVERSION
In an electro-mechanical energy conversion system,
energy is transferred from the electrical side to the
mechanical side as follows: (mechanical to electrical
transfer is backwards):
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ENERGY CONVERSION
• By differentiating the energy with respect to time,
power transfer can be shown to be:
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ENERGY CONVERSION
Neglecting the field losses, we get the simple relation
for the coupled system (shown dotted)
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ENERGY CONVERSION
The force of electrical origin can be derived using the
energy function Wm :
Choosing λ and x as independent variables:
,
= emdW dxvi f
dt dt
= emdW d dxi f
dt dt dt
, , ,=
m m mdW x W x W xd dx
dt dt x dt
,( , ) =
mW xi x
,( , ) =
meW x
f xx
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ENERGY CONVERSION
In a rotational system
Choosing λ and θ as independent variables:
,
= emdW d di T
dt dt dt
, , ,=
m m mdW W Wd d
dt dt dt
,( , ) =
mWi
,( , ) =
meW
f xx
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FORCES OF ELECTRICAL ORIGIN
When the system moves from one operating point to
another
Change variable
= ( ( , ) ( , ) )t tb b em
t ta a
dW d dxdt i x f x dt
dt dt dt
= ( , ) ( , )m x
b b b e
mm x
a a a
dW i x d f x dx
= ( , ) ( , )x
b b e
mb max
a a
W W i x d f x dx
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FORCES OF ELECTRICAL ORIGIN
Integrate x keeping λ constant at λa , then integrate λ
keeping x at xb
If λa = 0 and Wma = 0 (f e = 0), then:
Letting λb any λ and xb any x, then:
0= ( , )
b
mbW i x d
0( , ) = ( , )mW x i x d
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= ( , ) ( , )x
b b e
mb ma b ax
a a
W W i x d f x dx
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ENERGY AND CO-ENERGY
To compute , we need to express
from the flux linkage equation. This would require
solving for from . Particularly in multi-port
systems, this could be quite complicated.
We can avoid this problem by defining a quantity
called co-energy directly computable from
and then using it to compute .
Choosing λ and x as independent variables, define:
Using i and x as independent variables
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( , )mW x = ( , )i i x
i = ( , )i x
= ( , )i x ef
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ENERGY AND CO-ENERGY
Choosing λ and x as independent variables, define:
Using i and x as independent variables
m mW i W
( , )=m mdW i x dWdi d
idt dt dt dt
( , ) ( , ) ( , )=m m mdW i x W i x W i xdi dx
dt i dt x dt
( , )= ( )emdW i x di d d dx
i i fdt dt dt dt dt
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ENERGY AND CO-ENERGY
The left sides of these equations are the same,
comparing
terms:
Computation of
Method # 1
( , ) = mWi x
i
( , ) =e mWf i x
x
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mW
( , ) ( , ) ( ( , ))m mW i x i x W i x
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ENERGY AND CO-ENERGY
Method # 2
Change of variables using i and x as independent variables
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= emdW di dxf
dt dt dt
= ( )t tb b em
t ta a
dW d i dxdt f dt
dt dt dt
= ( , ) ( , )i xb b e
mb mai xa a
W W i x di f i x dx
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ENERGY AND CO-ENERGY
Choose a path
Integrate x keeping i constant at ia , then integrate λ
keeping x at xb
Use ia = 0 ( so f e = 0 ) and assume = 0
Let ib be any i and xb be any x
= ( , ) ( , )i xb b e
mb ma b ai xa a
W W i x di f i x dx
maW
0( , ) = ( , ) ( .)
i
mW i x i x di x const 10/15/2017 16
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EXAMPLE
: The mean length of the magnetic path
A : Cross-sectional area, the magnetic circuit has finite μ.
Find the force of electric origin f e.
c
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EXAMPLE
Ampere’s circuital law gives:
Applying Gauss’s law around the moving part:
Applying Gauss’s law to the closed surface around the upper
fixed surface
1 2 =c cH H x H x Ni
0 1 0 2=H A H A
1 2=H H
0 1=cH A H A
01=cH H
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EXAMPLE
Solve for H1:
The flux linkage of the coil:
= N
0 1= N H A
2
0
0
0
= =2
2 cc
N Ni N iA
xx
A A
12 =c cH H x Ni
1
0
=
2c
NiH
x
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EXAMPLE
The co-energy is
The force of electric origin:
Note that is the reluctance of the iron path, and
is the reluctance of the two air gaps in series.
2 2
0
0
= ( , ) =2
2
i
m
c
N iW i x di
x
A A
=e mWf
x
2 2 2 2
2 2
0 0
0 0
2= =
2 22
e
c c
N i N if
A Ax x
A A A A
mW
c A0
2x
A
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