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.

10/15/2017


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

2 10/15/2017



• 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



• 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


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



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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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



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



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


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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ece 330 power circuits and electromechanics lecture … 17.pdf · sauer’s ece 330 lecture notes....

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