Chapter 3-Principle of Electromechanical Energy Conversion...Electromechanical Energy Conversion...

Electric Machines I

2017 Shiraz University of Technology Dr. A. Rahideh

In The Name of God The Most Compassionate, The Most Merciful

2

Table of Contents1. Introduction to Electric Machines

2. Electromagnetic Circuits

3. Principle of Electromechanical Energy Conversion

4. Principle of Direct Current (DC) Machines

5. DC Generators

6. DC Motors


3

Chapter 3Principle of Electromechanical Energy Conversion

3.1. Electromechanical Relations

3.2. Slow versus Fast Movements

3.3. Mechanical Force Calculations

3.4. Developed Torque in Doubly Excited Systems


4

Electromechanical RelationsMotoring Case

Electrical System

Coupling Field

Mechanical System

Electrical losses

elecP mechP

Magnetic losses Mechanical losses

Ohmic (copper) losses

Hysteresis losses

Eddy current losses

Bearing losses

Ventilation

Windage

mechfieldelec WWW mechfieldelec dWdWdW


5

Magnetic System with Single Excitation

irve

i

e

r

v

2irviei

dtirvidteidt 2

mechfieldelec dWdWdW 0mechdW

eidtdWdW fieldelec dtde

N

iddWfield


i

N

+ v _

+ e _

r

g

Fixed part

Mov

able

par

t

Spring

6

Magnetic Relay with Single ExcitationAssumption 1: The movable part cannot move or is not allowed to move

0mechdW iddWdW fieldelec


7

Magnetic Relay with Single ExcitationAssumption 1: The movable part cannot move

0mechdW

iddWdW fieldelec

0

idWfield

N

ggcc lHlHNi

B ggcc

field NAdBN

lHlHW

0

0BHg

B

gccfield AdBlBlHW0

0

i

N

+ v _

+ e _

r

g

Fixed part

Mov

able

par

t

Spring


8


where is the core volumeis the air‐gap volume

B

gccfield AdBlBlHW0

0

B

gccfieldBVdBHVW

00

2

2

cVgV

i

N

+ v _

+ e _

r

g

Fixed part

Mov

able

par

t

Spring

Stored magnetic energy in the core

Stored magnetic energy in the air‐gap


9


In the case of linear systems:

0

2

0

2

22 BVBVW g

rcfield

i

N

+ v _

+ e _

r

g

Fixed part

Mov

able

par

t

Spring

Stored magnetic energy in the core

Stored magnetic energy in the air‐gap

constantr

rc

BH0


10

Magnetic Relay with Single ExcitationExample 1: In the following system if the air‐gap flux density is 1 T and air‐gap length is constant and fringing effect is neglected, calculate:a) The DC source voltage;b) Stored magnetic energy.

1.0

(At/m)H

(T)B

200

0.8

400

1.2

600

0.6

1000800

cm5

i 250N

+ v _

+ e _

5r

cm10

Fixed part

Mov

able

par

t

Spring

mmg 5cm10 Not in scale cm5 cm5

cm10

cm5


11

Magnetic Relay with Single ExcitationSolution 1: From the curve

1.0

(At/m)H

(T)B

200

0.8

400

1.2

600

0.6

1000800

T1 cg BBAt/m670cH

At/m10795104

1 37

0

g

g

BH

cm5

i 250N

+ v _

+ e _

5r

cm10

Fixed part

Mov

able

par

t

Spring

mmg 5cm10 Not in scale cm5 cm5

cm10

cm5

ggcc lHlHNi 2

m6.0clmm5gl

A4.3321 ggcc lHlH

Ni

V167 rivdc


12

Magnetic Relay with Single ExcitationSolution 1:

1.0

(At/m)H

(T)B

200

0.8

400

1.2

600

0.6

1000800

T1 cg BB

1

00

2

2BVdBHVW gccfield

33 m10305.01.06.0 cV

35 m10505.01.0005.02 gV

7

253

10421105335103

fieldW

J9.20895.19005.1 fieldW


13

Energy and CoenergyEnergy

Coenergy

0

idWfield

)A(i

(Wb) Stored Energy

Coenergy

i

field diW0

iWW fieldfield


14

-i curve in system with changing air-gap

The ‐i curve varies with the air‐gap length

)A(i

(Wb.t)

Increasing air-gap length

i

N

+ v _

+ e _

r

g

Fixed part

Mov

able

par

t

Spring


i

N

+ v _

+ e _

r

g

Fixed part

Mov

able

par

t

Spring

15

Magnetic Relay with Single ExcitationAssumption 2: The movable part can move but slowly

In this case the current remains constant during the movement.


i

N

+ v _

+ e _

r

g

Fixed part

Mov

able

par

t

Spring

16


From o to a: No movement yet, therefore

o )A(i

(Wb.t) Closed

Open a

b

1i

2

1

c

d

oadfieldelec AiddWdW

0mechdW

mechfieldelec dWdWdW


i

N

+ v _

+ e _

r

g

Fixed part

Mov

able

par

t

Spring

17


From a to b:

o )A(i

(Wb.t) Closed

Open a

b

1i

2

1

c

d

abcdelec AiiddW 121

oabmech AdW


oadobcafieldbfieldfield AAWWdW )()(


i

N

+ v _

+ e _

r

g

Fixed part

Mov

able

par

t

Spring

18

Magnetic Relay with Single ExcitationAssumption 3: The movable part can move but very fast

In this case the flux linkage remains constant during the movement.


i

N

+ v _

+ e _

r

g

Fixed part

Mov

able

par

t

Spring

19




0mechdW


b

o )A(i

(Wb.t) Closed

Open a

1i

2

1

e

d c

2i


i

N

+ v _

+ e _

r

g

Fixed part

Mov

able

par

t

Spring

20


From a to c:


b

o )A(i

(Wb.t) Closed

Open a

1i

2

1

e

d c

2i

0 iddWelec

oacmech AdW

oadocdafieldcfieldfield AAWWdW )()(


i

N

+ v _

+ e _

r

g

Fixed part

Mov

able

par

t

Spring

21


From c to b: No movement, therefore

cbedfieldelec AiddWdW

0mechdW


b

o )A(i

(Wb.t) Closed

Open a

1i

2

1

e

d c

2i


1‐Why does the current remain constant during the slow movement?

2‐Why does the flux linkage remain constant during the fast movement?

3‐Why should the currents at the open and closed stages be the same?

i

N

+ v _

+ e _

r

g

Fixed part

Mov

able

par

t

Spring

22

Some Questions (H.W)


i

N

+ v _

+ e _

r

g

Fixed part

Mov

able

par

t

Spring

23

Magnetic Relay with Single ExcitationAssumption 4: The movable part moves with normal speed


i

N

+ v _

+ e _

r

g

Fixed part

Mov

able

par

t

Spring

24




0mechdW


b

o )A(i

(Wb.t) Closed

Open a

1i

3

1

f

d

c

2i

e 2


i

N

+ v _

+ e _

r

g

Fixed part

Mov

able

par

t

Spring

25


From a to c:


acedelec AiddW

oacmech AdW

oadoceafieldcfieldfield AAWWdW )()(

b

o )A(i

(Wb.t) Closed

Open a

1i

3

1

f

d

c

2i

e 2


i

N

+ v _

+ e _

r

g

Fixed part

Mov

able

par

t

Spring

26


From c to b: No Movement, therefore


b

o )A(i

(Wb.t) Closed

Open a

1i

3

1

f

d

c

2i

e 2

cbfefieldelec AiddWdW

0mechdW


27

Electromechanical Energy ConversionExample 2: The energy conversion cycles of two machines are OABO and OABCO curves shown below. If the energy conversion efficiency is defined as follows, calculate R1 and R2

EnergyElectricalInputEnergyConverted

R

B

O )A(i

(Wb.t)

A

4

3

1

(1)

B

O )A(i

(Wb.t)

A

4

3

1

(2)

C

2


28

Electromechanical Energy ConversionSolution 2: Part (1)

4.0104

1

11

elec

mech

WW

R

B

O )A(i

(Wb.t)

A

4

3

1

(1)

10443

1

1

01

ddididW B

A

A

Oelec

41 OABOmech AW


29

Electromechanical Energy ConversionSolution 2: Part (2)

7.0107

2

22

elec

mech

WW

R

10443

3

3

1

1

02

idddidididW C

B

B

A

A

Oelec

72 OABCOmech AW B

O )A(i

(Wb.t)

A

4

3

1

(2)

C

2


30

Mechanical ForceThe average force is calculated as the mechanical work divided by the displacement

ntDisplacemeWorkMechanical

aveF

g

dx

i

N

+ v _

r

Fixed part

Mov

able

par

t

Spring

x

)A(i

(Wb.t)

a

0x

gx

x dxx

Fully open (no movement yet)

Fully closed

o

c h b

k

d

f


31

Mechanical Force

• If the flux linkage is constant (fast movement),

)A(i

(Wb.t)

a

0x

gx

x dxx

Fully open (no movement yet)

Fully closed

o

c h b

k

d

f

oabmech AdW

oahmech AdW

0elecdW


fieldmech dWdW

fielddWFdx

cte

field

xxW

F

),(


32

Mechanical Force

• If the current is constant (slow movement),

)A(i

(Wb.t)

a

0x

gx

x dxx

Fully open(no movement yet)

Fully closed

o

c h b

k

d

f

oabmech AdW

oacmech AdW

okafokcdacdfelec AAAdW

mechfieldelec dWdWdW fieldmech WddW

ctei

field

xxiW

F

),(

fieldfieldafieldafieldcfieldcfieldelec WddWWWWWdW )()()()(

fieldWdFdx


Translational movementForce

Rotational movement Torque

33

Force and Torque Calculation

where is the angular position.

ctei

field

xxiW

F

),(

cte

field

xxW

F

),(

ctei

field iWT

),(

cte

fieldWT

),(


34

Electromechanical Energy ConversionExample 3: In an electromechanical system for

and . Calculate the force exerted on the movable part if the current is 3 A and air‐gap length is 5 cm.

Solution: 1st method (Coenergy) 2nd method (Energy)

209.0xi 40 i cm103 x

i

field diW0

ctei

field

xxiW

F

),(

cte

field

xxW

F

),(

0

idWfield


35

Electromechanical Energy Conversion

Solution 3: 1st method (Coenergy)

For and

209.0xi 40 i cm103 x

2/3

00 3209.009.0 i

xdi

xidiW

ii

field

2/32 3

209.0),(i

xxxiW

Fctei

field

A3i cm5x ?F

xi09.0

A3i cm5x N1243209.0 2/3

2 ix

F


36

Electromechanical Energy Conversion

Solution 3: 2nd method (Energy)

For and

209.0xi 40 i cm103 xA3i cm5x ?F

A3i cm5x

309.009.0

3

2

2

0

2

0

xdxidWfield

3

2

3

2

09.031

09.02

309.02),(

x

ixxx

xWF

cte

field

N7.12409.031

09.02

3

2

xixF


37

Electromechanical Energy ConversionExample 4: In the following system assume flux is constant during the movement, calculate the average force.

g

Fixed part A

Movable part N

xxg


38

Electromechanical Energy ConversionSolution 4: Since the permeability of the core goes to infinity, the system is linear.

g

Fixed part A

Movable part N

x xg

Li

2NL

221

2

21

212

21

LiLiWfield

N

2

021

AxgWfield

AxxW

Fcte

field

0

221 1),(


39

Electromechanical Energy ConversionExample 5: In an electromechanical system the characteristics is defined as for , calculate the average force when .

i 22/3 15.2 xi m10 xm6.0x

0

223

0)1(5.2 dxidWfield

)1(245),( 2

xx

xWF

cte

field

224525

52 )1( xWfield

26.0 xF


40

Developed Torque in Doubly Excited Systems

Consider the following electric motor with double excitation.

wherestator currentrotor currentstator flux linkagerotor flux linkagestator self‐inductancerotor self‐inductancemutual inductance

rrsselec dididW

is

vs

2sN

2sN

Stator

vr

rN

ir

rrssrr

rsrsss

iLiMiMiL

sr

risi

sLrLsrM


41


The inductances are defined as follows

wherestator number of turnsrotor number of turnsreluctance seen by stator fluxreluctance seen by rotor fluxreluctance seen by resultant flux

s

ss

NL

2

is

vs

2sN

2sN

Stator

vr

rN

ir

s

r

rNsN

sr

r

rr

NL

2

sr

rssr

NNM


42


Assumption 1: The rotor cannot rotate

is

vs

2sN

2sN

Stator

vr

rN

ir

rrsselec dididW

rrssrr

rsrsss

iLiMiMiL

rrssrrrsrssselec iLiMdiiMiLdidW

rrrsrsrrssrssselec diiLdiiMdiiMdiiLdW


(1)

rrrrssrsssfield diiLiidMdiiLdW

rrss i

rrr

ii

rssr

i

sssfield diiLiidMdiiLW000

0

2212

21

rrrssrssfield iLiiMiLW


43


Assumption 2: The rotor can rotate

Taking differentiation from the relationyields

is

vs

2sN

2sN

Stator

vr

rN

ir

rrsselec dididW

rrssrr

rsrsss

iLiMiMiL

rrssrrrsrssselec iLiMdiiMiLdidW

rrrrrsrrssrsr

srsrrssrssssselec

dLidiiLdMiidiiM

dMiidiiMdLidiiLdW2

2

rrrrrsrsr

srsrrssrsssssfield

dLidiiLdiiM

dMiidiiMdLidiiLdW2

21

221

2212

21

rrrssrssfield iLiiMiLW


44



is

vs

2sN

2sN

Stator

vr

rN

ir

rrsselec dididW

rrssrr

rsrsss

iLiMiMiL

rrrrrsrrssrsr

srsrrssrssssselec

dLidiiLdMiidiiM

dMiidiiMdLidiiLdW2

2

rrrrrsrsr

srsrrssrsssssfield

dLidiiLdiiM

dMiidiiMdLidiiLdW2

21

221


rrsrrsssmech dLidMiidLidW 2212

21


45



Since the torque is defined as

it yields

is

vs

2sN

2sN

Stator

vr

rN

ir

rrsrrsssmech dLidMiidLidW 2212

21

ddWT mech

ddL

id

dMiiddL

iT rr

srrs

ss

2212

21

Electromagnetic torque

Reluctance torque Reluctance torque


46


Case 1: Both rotor and stator have salient structures:, and are function of .

Reluctance torque is independent of current direction.

is

vs

2sN

2sN

Stator

vr

rN

ir

ddL

id

dMiiddL

iT rr

srrs

ss

2212

21


Reluctance torque due to rotor saliency

Reluctance torque due to stator

saliency

sL rL srM


47


Case 2: Rotor has salient structures but stator is cylindrical:and are function of .

ddL

id

dMiiddL

iT rr

srrs

ss

2212

21



sL srM

0

ddMii

ddL

iT srrs

ss 2

21

is

vs

Stator

vr

rN

ir


48


Case 3: Stator has salient structures but rotor is cylindrical:and are function of .

ddL

id

dMiiddL

iT rr

srrs

ss

2212

21



rL srM

0

ddL

id

dMiiT rr

srrs

221

ir

is

vs

2sN

2sN

Stator

vr

rN


49


Case 4: Both rotor and stator are cylindrical (non‐salient):only is a function of .

ddL

id

dMiiddL

iT rr

srrs

ss

2212

21


srM

0

ddMiiT sr

rs

0

is

ir

Stator

vr

rN vs


50


Example 6: In the following figure the rotor has no winding and the self‐inductance is assumed to be where is the rotor position. If stator current is .a) Calculate the torque exerted on the rotor.b) Find the conditions in which the average torque is not zero if

where is the angular velocity of the rotor and is the initial position of the rotor.

ddL

id

dMiiddL

iT rr

srrs

ss

2212

21

2cos10 LLLs tIi ms sin

tm m

si

N

Stator

Rotor

m


51


Solution 6:Part a)

Since rotor has no winding: d

dLi

ddMii

ddL

iT rr

srrs

ss

2212

21


si

N

Stator

Rotor

m

0ri

ddL

iT ss2

21

2cossin 102

21 LL

ddtIT m

tLIT m 21

2 sin2sin


52


Solution 6:Part b)

To have non‐zero average torque the coefficient of t in one of the above sin terms should be zero:

or or


si

N

Stator

Rotor

m

0ritLIT m 2

12 sin2sin

tm

2

2cos12sin12 ttLIT mm

tttLIT mmmm 2sin2sin2sin 21

21

12

21

0m m m


53


Solution 6:Part b)

1) If

2) If

si

N

Stator

Rotor

m

tttLIT mmmm 2sin2sin2sin 21

21

12

21

0m

m

2sin12

21 LIT mave

2sin12

41 LIT mave


54

Electromechanical Energy ConversionExample 7: In the following system assume the movable part is fixed and the air‐gap length is .1) Calculate the current and voltage if flux density in air‐gap is 0.5 T.2) Calculate the stored energy.

mm1d

3) Obtain the force exerted on the movable part.

4) Compute the inductance of the winding.

1.0

(At/m)H

(T)B

200

0.8

400

1.2

600

0.6

1000800

0.4 0.2

4r

V

Movable

cm5

i

500N

cm5 Spring

cm5

cm10

Not in scale cm5

d

cm10

cm5


55

Electromechanical Energy ConversionSolution 7:Part 1)

From the curve

mm1 gld

T5.0 cg BB At/m350cH

1.0

(At/m)H

(T)B

200

0.8

400

1.2

600

0.6

1000800

0.4 0.2

NllH

NlHlH

i gB

ccggccg

0

m6.0cl 70 104

A215.1i

V86.4 riV

?i ?V

4r

V

Movable

cm5

i

500N

cm5 Spring

cm5

cm10

Not in scale cm5

d

cm10

cm5


56


mm1 gldT5.0 cg BBB At/m350cH

1.0

(At/m)H

(T)B

200

0.8

400

1.2

600

0.6

1000800

0.4 0.2

m6.0cl 70 104

B

gccfieldBVdBHVW

00

2

2

33 m105.105.005.06.0 cV36 m105.205.005.0001.0 gV

J38.0fieldW

?fieldW

4r

V

Movable

cm5

i

500N

cm5 Spring

cm5

cm10

Not in scale cm5

d

cm10

cm5


57


mm1 gld

T5.0 cg BBB

m6.0cl 70 104

B

ggcccfieldBAldBHAlW

00

2

2

xlg

0

2

2

BAx

WF g

cte

field

23 m105.205.005.0 gA

N7.248F

?F

4r

V

Movable

cm5

i

500N

cm5 Spring

cm5

cm10

Not in scale cm5

d

cm10

cm5


58


mm1 gld

T5.0 cg BBB

m6.0cl 70 104

iNAB

iN

iL

23 m105.205.005.0 gA

H514.0L

?L

215.15.0105.2500 3

L

4r

V

Movable

cm5

i

500N

cm5 Spring

cm5

cm10

Not in scale cm5

d

cm10

cm5


59

Electromechanical Energy ConversionExample 8: In the following system assume the current is 1.215 A and the air‐gap is fully closed.1) Calculate the flux density.2) Calculate the force exerted on spring.3) Calculate the stored energy

1.0

(At/m)H

(T)B

200

0.8

400

1.2

600

0.6

1000800

0.4 0.2

0d

4r

V

Movable

cm5

i

500N

cm5 Spring

cm5

cm10

Not in scale cm5

d

cm10

cm5


60


From the curve

0 gld

T21.1cB

1.0

(At/m)H

(T)B

200

0.8

400

1.2

600

0.6

1000800

0.4 0.2

ggcc lHlHNi

m6.0cl 70 104

?B

cclHNi At/m1012c

c lNiH

A215.1i

4r

V

Movable

cm5

i

500N

cm5 Spring

cm5

cm10

Not in scale cm5

d

cm10

cm5


61


0 gldT21.1cB At/m1012cH

1.0

(At/m)H

(T)B

200

0.8

400

1.2

600

0.6

1000800

0.4 0.2

m6.0cl 70 104

B

gccfieldBVdBHVW

00

2

2

33 m105.105.005.06.0 cV

0gV

J92.0fieldW

?fieldW

B

ccfield dBHVW0

4r

V

Movable

cm5

i

500N

cm5 Spring

cm5

cm10

Not in scale cm5

d

cm10

cm5


62


0 gld

T21.1 cg BBB

m6.0cl 70 104

B

ggcccfieldBAldBHAlW

00

2

2

xlg

0

2

2

BAx

WF g

cte

field

23 m105.205.005.0 gA

N4.1456F

?F

4r

V

Movable

cm5

i

500N

cm5 Spring

cm5

cm10

Not in scale cm5

d

cm10

cm5


63

Developed TorqueExample 9: In the following figure the rotor has no winding and the self‐inductance is assumed to be where is the rotor position. If stator current is 10 Aac with frequency of 60 Hz: .a) In which rotational velocity the average torque is not zero if

where is the angular velocity of the rotor and is the initial position of the rotor.

b) In the velocity obtained above calculate the maximum torque and mechanical power.

c) Obtain maximum torque at zero velocity.

ddL

id

dMiiddL

iT rr

srrs

ss

2212

21

4cos2.02cos3.01.0 sL

tm m

si

N

Stator

Rotor

m


64

Developed TorqueSolution 9:Part a)

ddL

iT ss2

21

4cos2.02cos3.01.0 sL tm

ti ss cos10

si

N

Stator

Rotor

m

4cos2.02cos3.01.0cos10 221

ddtT s

120602 s

4sin8.02sin6.0cos50 2 tT s

)(4sin8.0)(2sin6.02

2cos150

tttT mm

s


65

Developed TorqueSolution 9:Part a)

or or


ti ss cos10

si

N

Stator

Rotor

m

120602 s

)(4sin8.0)(2sin6.02

2cos150

tttT mm

s

422sin4.0422sin4.022sin3.022sin3.0

)(4sin8.0)(2sin6.025

tttt

ttT

smsm

smsm

mm

)sin()sin(cossin 21 bababa

0m 120 sm 6021 sm


66

Developed TorqueSolution 9:Part b)

at

at


si

N

Stator

Rotor

m

120602 s

120 sm

6021 sm

2sin5.7aveT

Nm5.7max aveT 4 k

4sin10aveT

W28271205.7max maveTP

Nm10max aveT82 k

W18856010max maveTP


67

Developed TorqueSolution 9:Part c)


si

N

Stator

Rotor

m

120602 s

0m 4sin202sin15 aveT

Nm31max aveT

04cos802cos300 aveT

012cos282cos3 2

082cos32cos16 2 62.02cos

81.02cos o8.25o9.71 Nm10max aveT


Chapter 3-Principle of Electromechanical Energy Conversion...Electromechanical Energy Conversion...

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Transcript of Chapter 3-Principle of Electromechanical Energy Conversion...Electromechanical Energy Conversion...