EEC Principles

8/11/2019 EEC Principles

1/49

Nguyn Cng Phng

ELECTROMECHANICAL ENERGY

CONVERSION

Electromechanical Energy

Conversion Principles


2/49

Contents

I. Magnetic Circuits and Magnetic MaterialsII. Electromechanical Energy Conversion

Principles

III. Introduction to Rotating Machines

IV. Synchronous MachinesV. Polyphase Induction Machines

VI. DC Machines

VII.VariableReluctance Machines and Stepping

MotorsVIII.Single and TwoPhase Motors

IX. Speed and Torque Control

sites.google.com/site/ncpdhbkhn 2


3/49

ElectromechanicalEnergy


1. Forces and Torques in Magnetic Field Systems2. Energy Balance

3. Energy in SinglyExcited Magnetic FieldSystems

4. Determination of Magnetic Force and Torquefrom Energy and Coenergy

5. MultiplyExcited Magnetic Field Systems

6. Forces and Torques in Systems with Permanent

Magnets7. Dynamic Equations

8. Analytical Techniques



4/49


Forces and Torques in Magnetic

Field Systems (1)

( )q F E v B

qF Ev

B

F

( )q F v B

( )v F v B

J v

v F J B


5/49



Field Systems (2)Ex.A nonmagnetic rotor contains a single

turn coil, it

is in a uniform magnetic field. The rotor is of radius

R and of lengthl. Find thedirected torque as a

function of?

0 B y

x

I

I

v F J B

1( )S F J B

I B

0

sinin

F IB l

0 sinoutF IB l

0(2 ) 2 sin (Nm)T R F RIB l


6/49



Field Systems (3)

Lossless magnetic

energy storage system

i

,e

fldf

x

Electrical

terminal

Mechanical

terminal

Lossless winding

i1

e

Magnetic core

v

Winding

resistance x fld

f

Movable

magnetic

plunger


7/49



Field Systems (4)

Lossless magnetic


i

,e

fldf

x

Electrical

terminal

Mechanical

terminal

fld

fld

dW dxei f

dt dt

d

e dt

fld flddW id f dx


8/49












9/49


Energy Balance

Energy is neither created or destroyed, it is merely changed in form

electrical mechanical field dW eidt dW dW


10/49












11/49


12/49


Energy in SinglyExcited

Magnetic Field Systems (2)

21 ( )

2

fldW L x i

2201 (1 / ) 1 J

2 2fld

N ld x d xW i K

g d

LosslessNturns coil

+

i

h

g

x g

d

l

x

d

d x g

gMagneticflux

Ex.

A relay is made of infinitely

permeable magnetic

material,h >>g. Compute the magnetic energy as

a function of plunger position?

02( )2

gS

L x Ng

( ) 1g

xS l d x ld

d


13/49












14/49


Determination of Magnetic Force and

Torque from Energy and Coenergy (1)

( , )fld flddW x id f dx

const

const

( , )

( , )

fld

x

fld

fld

W x

i

W xf

x

const const

( , ) ( , )( , )

fld fld

fld

x

W x W xdW x d dx

x

21( , )

2 ( )fldW x

L x

2 2

2

const

1 ( )

2 ( ) 2 ( )fld

dL xf

x L x L x dx

( )L x i

2 ( )

2fld

i dL xf

dx


15/49



Torque from Energy and Coenergy (2)Ex. 1

Plot the force as a function of position for a current of 1A.

x (cm) 0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0

L (mH) 2.8 2.26 1.78 1.52 1.34 1.26 1.20 1.16 1.13 1.11 1.10


16/49



Torque from Energy and Coenergy (3)Ex. 2

Plot the force as a function of position for a flux of 2mWb.

x (cm) 0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0

L (mH) 2.8 2.26 1.78 1.52 1.34 1.26 1.20 1.16 1.13 1.11 1.10


17/49





const

( , )fld

fld

WT

( , )fld flddW id T d

const const

( , ) fld fld

fld

W WdW d d

21

( , )2 ( )

fldW xL x

2 2

2

const

1 ( )

2 ( ) 2 ( )fld

dLT

L x L d

( )L i

2 ( )

2fld

i dLT

d


18/49




Rotor

Air gap

i

0 2( ) cos(2 )L L L

2 ( )

2fld

i dLT

d

2

2( ) [ 2 sin(2 )]2

fld

iT L


19/49




( , ) ( , )fld fldW i x i W x

( , ) ( ) ( , )fld fld

dW i x d i dW x

( )d i id di


( , )fld fld

dW i x di f dx

const const

( , ) fld fld

fld

x i

W WdW i x di dx

i x

const

const

( , )

( , )

fld

x

fld

fld

i

W i x

i

W i xf

x


20/49




LosslessNturns coil

+

i

h

g

x g

d

l

Ex. 3

A relay is made of infinitely

permeable magneticmaterial,h >>g. Compute the magnetic energy as

a function of plunger position, ifi(x) =I0x/d(A)?

2

0 (1 / )

( )2

N ld x dL x

g

02( )2

gSL x N

g

( ) 1g xS l d x ld d

2 ( )

2fld

i dL x

f dx

2 2

0( )2 2

i N lL x

g

0( ) x

i x Id

22 2

0 0( )4

I N l xL x

g d


21/49


22/49




2ag

gH Ni

g r

i

Ex. 4

Find the torque acting on the rotor as a function ofthe dimensions and the magnetic field in the two

air gap, suppose that the reluctance of the steel is

negligible ( ). The axial length is h.

1Energy density :2

BH

21Energy densityof the core : 0

2 2

steelsteel steel

BB H

2

0Energy densityof the air-gap :

2 2

ag ag agB H H

2

0[2 ( 0.5 ) ]

2

ag

ag

HW gh r g

2

0( ) ( 0.5 )

4

Ni h r g

g

const

( , )agfld

i

W iT

2

0( ) ( 0.5 )

4fld

Ni h r gT

g


23/49




02( )2

gSL N

g

g r

i

Ex. 5

Find the inductance as a function of, then extractthe expression for the torque acting on the rotor as

a function ofi &. The axial length ish.

( 0.5 )gS h r g

2

0 ( 0.5 )

( )

2

N h r gL

g

2 ( )

2fld

i dLT

d

2

0( ) ( 0.5 )

4fld

Ni h r gT

g


24/49












25/49


MultiplyExcited Magnetic

Field Systems (1)

Lossless magnetic


1i

1

fldT

Electrical

terminals

Mechanical

terminal

2i

2

1 2 1 1 2 2( , ) ( , , )fld fld fld flddW id T d dW i d i d T d

2 22 1

1 2 1 2 1 2

1 2

1 2 , are const, are const , are const

( , , ) ( , , ) ( , , ); ;

fld fld fld

fld

W W Wi i T

1 22 1

1 2 1 21 2 , are const, are const , are const

( , , )

fld fld fld

fld

W W W

dW d d d


26/49



Field Systems (2)

Lossless magnetic


1i

1

fldT

Electrical

terminals

Mechanical

terminal

2i

2

1 11 1 12 2

2 21 1 22 2

L i L i

L i L i

22 1 12 2 22 1 12 21

11 22 12 21

21 1 11 2 21 1 11 22

11 22 12 21

L L L Li

L L L L D

L L L Li

L L L L D

2 2 12 010 20 0 11 0 20 22 0 10 10 20

0 0 0

1 1 ( )( , , ) ( ) ( )

2 ( ) 2 ( ) ( )fld

LW L L

D D D

1 1,e


27/49



Field Systems (3)1 2 1 1 2 2( , , )fld fldW i i i i W

1 2 1 1 2 2( , , )fld flddW i i di di T d

2

1

1 2

1 2

1

1 , are const

1 2

2

1 , are const

1 2

, are const

( , , )

( , , )

( , , )

fld

i

fld

i

fld

fld

i i

W i i

i

W i i

i

W i iT

2 2

1 2 11 1 22 2 12 1 2

1 1( , , ) ( ) ( ) ( )

2 2fld

W i i L i L i L i i

1 2

2 21 2 1 11 2 22 12

1 2

, are const

( , , ) ( ) ( ) ( )

2 2

fld

fld

i i

W i dL i dL dLT i i

d d d


28/49



Field Systems (4)

+

+

1 1,e 2 2,e

1i

2i

mechT

fldT

Ex.

L11 =a + cos2,L12 =bcos,L22 =c +ecos2. FindTfld()?

2 2

1 11 2 22 121 2

( ) ( ) ( )

2 2fld

i dL i dL dLT i id d d

2 2

1 2 1 2sin 2 sin 2 sini ei bi i


29/49



Field Systems (5)2 2

1 2 1 2sin 2 sin 2 sinfldT i ei bi i

0 1 2 3 4 5 6-4

-3

-2

-1

0

1

2

3

4x 10

-3

(rad)

Torque(Nm)

Reluctance torque

Mutual - interaction torque

Total torque


30/49



Field Systems (6)

1 2

1 2

1 2

, are const

2 21 2 1 11 2 22 12

1 2

, are const

2 2

1 2 11 1 22 2 12 1 2

( , , )

( , , ) ( ) ( ) ( )

2 2

1 1( , , ) ( ) ( ) ( )2 2

fld

fld

fld

fld

i i

fld

WT

W i i i dL i dL dLT i i

d d d

W i i L i L i L i i

1 2

1 2

1 2

, are const

2 21 2 1 11 2 22 12

1 2

, are const

2 2

1 2 11 1 22 2 12 1 2

( , , )

( , , ) ( ) ( ) ( )2 2

1 1( , , ) ( ) ( ) ( )

2 2

fld

fld

fld

fld

i i

fld

W xf

x

W i i x i dL x i dL x dL xf i ix dx dx dx

W i i x L x i L x i L x i i


31/49







6. Forces and Torques in Systems with

Permanent Magnets7. Dynamic Equations




32/49


Forces and Torques in Systems

with Permanent Magnets (1)

x

Fictitious windingNf turns

x

fi

f

( , )fld flddW i x di f dx

( , )fld f f f flddW i x di f dx

const

( 0, )

f

fld f

fld

i

W i xf

x

0

0

( 0, ) ( , )f

fld f f f fI

W i x i x di


33/49




Fictitious

winding

x

f

i

fN

gW

0W

d

mW

0g

DepthDEx. 1

Find an expression for:a) the coenergy of the system as a function of plunger

positionx?

b) the force on the plunger as a function ofx?

0 0f f m gN i H d H x H g

0 0 0m g m m g gB S B S B S

0 0m m g gB W D B W D B W D

( )m R m cB H H

0

0 0

( )R f f cm

Rm

g

N i H dBx g

d WW W

f f m f m m f m mN N B S N B W D

0

0 0

( )f m R f f cf

Rm

g

N W D N i H d

x gd W

W W


34/49



with Permanent Magnets (3)Ex. 1Find an expression for:a) the coenergy of the system as a function of plunger

positionx?

b) the force on the plunger as a function ofx?

Fictitious

winding

x

f

i

fN

gW

0W

d

mW

0g

DepthD

0

0 0

( )f m R f f cf

R

mg

N W D N i H d

x gd W

W W

00 c

f f f

f

H di I

N

0

0

( )f

fld f fI

W x di 2

0

0 0

( )

2

m R c

Rm

g

W D H d

x gd WW W

2 2

const 0

0 0

( 0, ) ( )

2f

fld f m R cfld

i Rg m

g

W i x W D H d f

x x gW d W

W W


35/49




External

magnetic

circuit

1

d

eF

S

mH

( )m R m cB H H

External

magnetic

circuit

2

d

eF

S

m R mB H

( )equivNi

0m eH d F

1 ( )

m R m cB S S H H

eR c

FS H

d

eNi F H

2 RBS HS

2

( )equiv eR

Ni FS

d d

If ( )equiv cNi H d

1 2


36/49



with Permanent Magnets (5)Ex. 2a) Find thex

directed force on the plunger when the

current in the excitation winding is zero and x = 3 mm?

b) Find the current in the excitation winding required to

reduce the plunger force to zero? x

1i

1N

gW

W

d

W

0g

DepthD

( )equiv cNi H d

+

+

( )equiv

Ni

1 1N i

0gR

mR

xR

00

0 0

; ;m x

R R g

d d x gR R R

S WD W D WD

2

1

1

2fldW Li

2

1

total

NL

R

2

0

( )1

2

equiv

fld

m x

NiW

R R R


37/49



with Permanent Magnets (6)Ex. 2a) Find thex

directed force on the plunger when the

current in the excitation winding is zero?

b) Find the current in the excitation winding required to

reduce the plunger force to zero? x

1i

1N

gW

W

d

W

0g

DepthD

+

+

( )equiv

Ni

1 1N i

0gR

mR

xR

2

0

( )1

2

equiv

fld

m x

NiW

R R R

2

2

0const

( )

( )equiv

fld equiv xfld

m xi

W Ni dRf

x R R R dx

2

2

0 0

( )

( )

equiv

g m x

Ni

W D R R R

1 1( ) 0equivNi N i 1

1

( )equivNi

iN


38/49












39/49


Dynamic Equations (1)

+

Electromechanical

energyconversion

system

x

,e

0v

fldf

i

R

K

B

M

0f

0

dv Ri

dt

( )L x i0

( )( )

di dL x dxv Ri L x idt dx dt


40/49



+

Electromechanical

energyconversion

system

x

,e

0v

fldf

i

R

K

B

M

0f0( )Kf K x x

D

dxf B

dt

2

2M

d xf M

dt

0 0

fld K D Mf f f f f

2

0 02( ) 0flddx d x

f K x x B M fdt dt


41/49



+

Electromechanical

energyconversion

system

x

,e

0v

fldf

i

R

K

B

M

0f

0

2

0 0 2

( )( ) ( )

( ) ( ) ( , )fld

di dL x dxv t Ri L x i

dt dx dt dx d x

f t K x x B M f x idt dt


42/49


Dynamic Equations (4)Ex.

Extract the dynamic equations of motion of theelectromechanical system?

Length of flux path in the direction of field

(area of flux path perpendicular to field)R

1

0

g

gR

dx

2

0

g

gR

da

1 2

0 0

1 1g g

g g a xR R R

d x a da x

2 2

0

2

0

( ) ,N daN x x

L x LR g a x a x

daNL

g

0l

1l

x a

a

h

d

g

CoilSpring,K

Cylindrical

steel plunger,

MApplied force,ft


43/49


Dynamic Equations (5)Ex.


2 2

2

const

( , )

2 2 ( )

fld

fld

i

W i x i dL i aLf

x dx a x

( )d Li di dL di dL dxe L i L i

dt dt dt dt dx dt

2( )

x di ai dxL L

a x dt a x dt

2

0( ) ,x daN

L x L La x g

0l

1l

x a

a

h

d

g

CoilSpring,K

Cylindrical

steel plunger,

MApplied force,ft


44/49



0l

1l

x a

a

h

d

g

CoilSpring,K

Cylindrical

steel plunger,

MApplied force,ft

Ex.


2 2

2

2

2 2 ( )

( )

fld

i dL i aLf

dx a x

x di ai dxe L L

a x dt a x dt

0

2

0 0 2

( )( ) ( )

( ) ( ) ( , )fld

di dL x dxv t Ri L x i

dt dx dt

dx d xf t K x x B M f x i

dt dt

0 2

2 2

0 0 2 2

( )( )

( ) ( )2 ( )

x di a dxv t Ri L L i

a x dt a x dt

dx d x i aLf t K x l B M

dt dt a x


45/49












46/49


Analytical Techniques (1)

2

2 2

0 2 2

( )( )

( ) ( )2 ( )


a x dt a x dt


dt dt a x

2

(0)If

,( )

v V

x di a dxRi L L i

a x dt a x dt

Vi

R

0If

0

f

M

2

02

1( ) ( )

2 ( )

dx a V B L K x l f x

dt a x R

0 ( )

X Bt dx

f x


47/49



2

2 2

0 2 2

( )( )

( ) ( )2 ( )


a x dt a x dt


dt dt a x

2

(0)If

,( )

v Vx di a dx

Ri L L ia x dt a x dt

Vi

R

0If

0

f

B

22

02 2

1( ) ( )

2 ( )

d x a V M L K x l f x

dt a x R

0

2( )

( )

xdx Bv x dx

dt M f x


48/49


49/49



0 00 0 2

0 0

2 2

00 0 0

2 20

( ) ( )( )

( )

1 ( )( )

2 ( ) t

L X x di L a I i dxV v R I i

a X x dt a X x dt

L a I i dx d xK X x l B M f f

a X x dt dt

0 0

2

0 0

2 2

0 0

2 2 3

0 0

( )

( ) ( )

L X di L aI dxv Ri

a X dt a X dt

L aI dx d x L aIi B M K x f

a X dt dt a X

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