Energy Converters – Computer-Aided Design (CAD) and System ... · Prof. Dr. A. Binder, Energy...

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TECHNISCHE UNIVERSITÄT DARMSTADT INSTITUT FÜR ELEKTRISCHE ENERGIEWANDLUNG Prof. Dr. A. Binder, Energy Converters - CAD and System Dynamics 1/1 Energy Converters – Computer-Aided Design (CAD) and System Dynamics A. Binder Institut für Elektrische Energiewandlung Technische Universität Darmstadt

Transcript of Energy Converters – Computer-Aided Design (CAD) and System ... · Prof. Dr. A. Binder, Energy...

Page 1: Energy Converters – Computer-Aided Design (CAD) and System ... · Prof. Dr. A. Binder, Energy Converters - CAD and System Dynamics 1/3 Learning outcomes Understanding of design

TECHNISCHE UNIVERSITÄT

DARMSTADTINSTITUT FÜRELEKTRISCHE ENERGIEWANDLUNG

Prof. Dr. A. Binder, Energy Converters - CAD and System Dynamics

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Energy Converters –

Computer-Aided Design (CAD)

and

System Dynamics

A. Binder

Institut für Elektrische Energiewandlung

Technische Universität Darmstadt

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Lecturer

M. Sc. Sascha NeusüsInstitut für Elektrische Energiewandlung

TU Darmstadt 64283, Landgraf-Georg-Strasse 4, Darmstadt

tel.: +49-6151-16-24192fax.:+49-6151-16-24183

e-mail: [email protected]

Tutorial

Prof. Dr.-Ing. habil. Dr. h.c. Andreas BinderInstitut für Elektrische Energiewandlung

TU Darmstadt 64283, Landgraf-Georg-Strasse 4, Darmstadt

tel.: +49-6151-16-24181 o. 24182fax.:+49-6151-16-24183

e-mail: [email protected]

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Learning outcomes

Understanding of design rules for electrical machines

- scaling laws (typical for power engineering)

i.e. AC and DC motors and generators

Knowledge of design of AC machinery, here: cage induction machine

- Magnetic circuit and winding topology

- Calculation of resistances and inductances

- Losses and efficiency

Knowledge of basics in cooling systems and temperature calculation

- applications in electrical motors and generators

Understanding of dynamics of DC and AC machinery

- system dynamics of variable speed drives & space vector theory

- transients in generator systems & power stability

Calculation examples for self-training

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Contents

1. Basic design rules for electrical machines

2. Design of Induction Machines

3. Heat transfer and cooling of electrical machines

4. Dynamics of electrical machines

5. Dynamics of DC machines

6. Space vector theory

7. Dynamics of induction machines

8. Dynamics of synchronous machines

Source:

SPEED program

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Organization

- Moodle: Down-load of slides, text book & collected exercises

- CityCopies, Holzstrasse 5: Paper copies of slides, text book & collected exercises

- Demo videos:Via Moodle platform link

- Introduction of two computer programs:

- SPEED program: Induction machine design

- Matlab/SIMULINK for dynamic calculations

- Excursion offered

Source:

SPEED program

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Examination

- Examination: In written form

Details see - Moodle or

- “Blue-board” (Institute)

a) Three calculation examples

b) Theoretical questions

- Homework:Dynamics examples – add-on points for examination

Source:

SPEED program

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Type of examination

Written examination

Winter term: 2 partial exams

Summer term: 1 exam

Calculation examples & List of theory questions: see “Collection of Exercises”

Examination

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Energy Converters – CAD and System Dynamics

1. Basic design rules for electrical machines

2. Design of Induction Machines

3. Heat transfer and cooling of electrical machines

4. Dynamics of electrical machines

5. Dynamics of DC machines

6. Space vector theory

7. Dynamics of induction machines

8. Dynamics of synchronous machines

Source:

SPEED program

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1. Basic design rules for electrical machines

Source:

Winergy, Germany

Energy Converters - CAD and System Dynamics

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Energy Converters – CAD and System Dynamics

1. Basic design rules for rotating machines

1.1 Torque generation and internal power

1.2 Electromagnetic utilization

1.3 Thermal utilization

1.4 Overload capability of AC machines

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Air-gap between

stator and rotor

Stator

tooth

Rotor

tooth

1. Basic design rules for electrical machines

Cross section of a cage induction machine

Stator

yoke

Rotor

yoke

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1. Basic design rules for electrical machines

Air gap field, excited by slot conductors

sQsQ

ΘQ

A = ΘQ/ sQ A = 0

Air gap δ

Assumption: µFe → ∞ : HFe = 0

Ampere´s law: ∫∫∫∫ ⋅=⋅+⋅=⋅=δδ

δδΘCCC

Fe

C

Q sdHsdHsdHsdH

Fe

rrrrrrrr

Case 1: Tangential air gap field along path 1:

Case 2: Radial air gap field along path 2:

QQQ

C

Q sHsHsdH /11

1

ΘΘ δδδ

δ

=⇒⋅=⋅= ∫rr

)2/(2 22

2

δΘδΘ δδδ

δ

Q

C

Q HHsdH =⇒⋅=⋅= ∫rr

Current loading A introduced to replace

the slot geometry

The same air gap field is obtained at µFe → ∞a) for the real slot geometry and

b) for the current loading model.

Stator

RotorSource:

H. Kleinrath, Studientext 1975,

Wiesbaden

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1. Basic design rules for electrical machines

Resulting air gap field at the slot

ΘQ ΘQ=0Stator

Rotor

- Only stator self field

of slot conductor

- No external rotor field

δ

Θµδ

2

0 QsB =

0=rBδ

δsBδ− sBδ+

- No stator self field

of slot conductor

- Only external rotor field

0=sBδ

0>rBδ

rBδ+ rBδ+

Source:

H. Kleinrath, Studientext 1975,

Wiesbaden

Superposition of:- Stator self field

of slot conductor

and external rotor field

δ

Θµδ

2

0 QsB =

0>rBδ

sr BB δδ +sr BB δδ −

Note: For simple superposition we have to

assume µFe(x,y) = const.

x

y

ΘQ

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1. Basic design rules for electrical machines

Magnetic force Fe on a slot conductor

ΘQ=0

δ

External rotor field is at the slot

conductor location very small:

rrc BB δ<<,

rBδ+ rBδ+

Source:

H. Kleinrath, Studientext 1975,

Wiesbaden

sr BB δδ +sr BB δδ −

LORENTZ-force Fc on the

slot conductor is very small:

rcB ,

rcQc BlF ,⋅⋅= Θ

Magnetic pull force FM due

to MAXWELL stress fn on

magnetized iron dominates!

ΘQ

)( yBleftx

y

)( yBright

)2/( 02 µnnFe

A

nM BfdAfF

Fe

≈⋅= ∫ leftMrightMM FFF ,, −=

∫ ⋅−⋅=y

leftrightM dyBBl

F )(2

22

Fc

FMResulting magnetic force Fe:

Mce FFF +=

ca. 10 % ca. 90 %

rQe BlF ,δΘ ⋅⋅=

Mathematical proof:

Williams-Mamak: IEE Trans. C, 1961,

Monograph no. 456U

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1. Basic design rules for electrical machines

Simplified model for magnetic force Fe on a conductor

δ

Source:

H. Kleinrath, Studientext 1975,

Wiesbaden

sr BB δδ +sr BB δδ −

LORENTZ-force Fe on a

SURFACE conductor yields

the same force as the exact

model of a slot conductor:

ΘQ

)( yBleft )( yBright

Fc

FM

Mce FFF += rQe BlF ,δΘ ⋅⋅=

rBδ

ΘQ

rQe BlF ,δΘ ⋅⋅=

The same force is obtained

a) for the real slot geometry and

b) for the surface conductor model, leading again to a current loading model A(x).

Slotting replaced by “discrete” current loading

e.g.: m = 3, q = 2, W/τp = 1

Slotting

Current loading

sQ

sQ → 0

⊗Fe

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1. Basic design rules for electrical machines

m.m.f. V(x) and current loading A(x) in AC machines

)/sin(ˆp1 txV ωτπ −⋅

Distributed three-phase winding excites a distributed stator field:

δγµγδ /)()( 0 VB ⋅=

Example:

q = 2, m = 3,

full-pitched coils at time t = 0

Fundamental of m.m.f. distribution V(x) is a

sine wave:

Current loading is derivative of m.m.f.:

))/sin(ˆ(),( p11 txVdx

dtxA ωτπ −⋅=

)/cos(ˆ

),( p

p

11 tx

VtxA ωτπ

τπ

−⋅=

pw1p11 /2

/ˆˆ τππ

τπ ⋅

⋅⋅⋅⋅=⋅= IkN

p

mVA

Fundamental current loading A1(x) is a continuous function,

not any longer “discrete” like single conductors!

V1(x)

0 πp/τπγ x=

δδ ⋅= )()( xHxV

∫ ⋅= dxxAxV )()(

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1. Basic design rules for electrical machines

Fundamental m.m.f. V1(x), air-gap field Bδ1(x) & current loading A1(x)

)/sin(ˆ),( p11 txVtxV ωτπ −⋅=

)/sin(ˆ/),(),( 10 txBtxVtxB p ωτπδµ δδ −=⋅=

Stator winding:

Fundamental m.m.f. distribution V(x) as sine wave:

Current loading is derivative of m.m.f.:

)/cos(ˆ/),(),( p111 txAdttxdVtxA ωτπ −⋅==

V1(x)

-τp 0 τp x

Bδ1(x)

-τp 0 τp x

t = 0:

Stator fundamental air-gap field distribution Bδ(x) as sine

wave at µFe →∞:

A1(x)

-τp -τp/2 0 τp/2 τp xFundamental current loading A1(x) is a continuous function,

not any longer “discrete” like single conductors!

1V̂

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1. Basic design rules for electrical machines

• LORENTZ force: Stator field and rotor current

OR

rotor field and stator current: e.g.:

Number of conductors per element dx:

With current loading As:

ers ltxBtxidztxdF ⋅⋅⋅= ),(),(),( ,δ

pp

dxzdz

τ2⋅=

z: total number of conductors

Total tangential force:(acting on stator)

p

sss

p

txiztxA

τ2

),(),(

⋅=

dxtxBtxAltF

pp

rse ⋅⋅⋅= ∫τ

δ

2

0

, ),(),()(

dγ ~ dx

r

s: stator

r: rotor

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1. Basic design rules for electrical machines

Self-field does not produce resulting tangential force

0),(),()(

2

0

, =⋅⋅⋅= ∫ dxtxBtxAltF

pp

sse

τ

δ

( ) dxtxBtxBtxAldxtxBtxAltF

pp p

rsse

p

rse ⋅+⋅⋅=⋅⋅⋅= ∫∫τ

δδ

τ

δ

2

0

,,

2

0

, ),(),(),(),(),()(

dxtxBtxAltF

pp

se ⋅⋅⋅= ∫τ

δ

2

0

),(),()(Tangential force on stator:

Tangential force on rotor: “Actio est reactio” (Newton´s 3rd law):

dxtxBtxAltF

pp

se ⋅⋅⋅−= ∫τ

δ

2

0

),(),()(

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1. Basic design rules for electrical machines

Fundamental sine wave magnetic air gap travelling field in AC machines

1ˆ2δτ

πΦ Bleph ⋅⋅=

Magnetic flux per pole:

1

Resulting air gap field

fundamental:

)/cos(ˆ),( 1, txBtxB p ωτπδδ −⋅=

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1. Basic design rules for electrical machines

Torque generation in AC machines by fundamental fields

dxtxBtxAltF

pp

se ⋅⋅⋅= ∫τ

δ

2

0

),(),()( 2/)()( sie dtFtM ⋅=

dxtxBtxAlF

pp

se ⋅⋅⋅= ∫τ

δ

2

0

1,1,1 ),(),( 2/1, siACe dFM ⋅=

Generally the tangential force depends on time due to slotting and phase

bands +U, -W, +V, -U, +W, -V. Therefore we have a torque ripple ∆Me(t):

Me(t) = Me,av + ∆Me(t)

Considering only fundamental fields we get a CONSTANT torque:

Me(t) = Me,av ∆Me(t) = 0

2

sisi

dr =

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1. Basic design rules for electrical machines

Torque generation in AC machines by fundamental fields

πϕτ δδ /cosˆˆ)( 1,1

2

, ⋅⋅⋅⋅= BAplM speACe

γπ

τωγϕωγ

π

δ dpltBtAM

p

s ⋅

⋅⋅⋅−⋅−−= ∫

2

0

2

p

eδ,11ACe, )cos(ˆ)cos(ˆ

px τπγ /⋅=

dxtxBtxAldM

pp

sesiACe ⋅⋅⋅⋅= ∫τ

δ

2

0

1,1,, ),(),()2/(Torque on stator:

psi pd τπ 2=

απ

ταϕα

πω

ωδ dplBAM

pt

t

s ⋅

⋅⋅⋅⋅−= ∫

+−

2 2

p

eδ,11ACe, )cos(ˆ)cos(ˆ

αωγ =− t

[ ] ααϕαϕαπ

τ πω

ωδδ dplBAM

pt

t

s ⋅+⋅

⋅⋅⋅= ∫

+−

22

p

eδ,11ACe, cossinsincoscosˆˆ2/))2cos(1(cos2 αα +=

2/)2sin(sincos ααα =

Torque on rotor: “Actio est reactio” (Newton´s 3rd law):

πϕτ δδ /cosˆˆ)( 1,1

2

, ⋅⋅⋅⋅−= BAplM speACe

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1. Basic design rules for electrical machines

Torque generation in AC machines by fundamental fields

πτ δ /315cosˆˆ)( 1,1

2

, °⋅⋅⋅⋅= BAplM speACe

Example:

Phase shift between stator current loading and air-gap field: ϕδ = 315°

ϕδπ

τ δ

1

2

1ˆˆ)( 1,1

2

, ⋅⋅⋅⋅⋅= BAplM speACe

Positive torque on the stator = in counter-clockwise direction

)cos(ˆ))2(cos(ˆ)cos(ˆ111 δδδ ϕωγπϕωγϕωγ −−=−−−=−− tAtAtA sss

πϕϕ δδ 2−=

Note: Due to

we also get: °−= 45δϕ πτ δ /)45cos(ˆˆ)( 1,1

2

, °−⋅⋅⋅⋅= BAplM speACe

πτ δ

1

2

1ˆˆ)( 1,1

2

, ⋅⋅⋅⋅⋅= BAplM speACe

δϕ

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Internal phase angle between internal

voltage and stator current:

1. Basic design rules for electrical machines

Example: Torque generation in an induction machine

Bδ,1

As(γ)

ϕδ

Main flux Φh ~ Bδ,1 ~ Im and internal voltage Uh ~ j.Φh:

Bδ,1 Φh Im

Uh

Bδ,1,s ~ Is

Is

As

ϕδ

ϕi

πϕϕ δ −=i

Torque on rotor: πϕτπϕτ δδδ /cosˆˆ)(/cosˆˆ)( 1,1

2

1,1

2

, ispespeACe BAplBAplM ⋅⋅⋅⋅=⋅⋅⋅⋅−=

)coscos( iϕϕδ =−

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1. Basic design rules for electrical machines

πϕτππ δδ /cosˆˆ)(22 11

2

, ispes

ACesyn BAplp

fMnP ⋅⋅⋅⋅⋅==

Phase shift ϕi between Is and Uh!

Internal phase shift ϕi in an induction machine

Internal power:

2/2 1 hswssh NkfU Φπ ⋅=1

ˆ2δτ

πΦ Bleph ⋅⋅=

ishs IUmP ϕδ cos⋅=

Example: 0 < ϕi < π/2: Me,AC > 0 on rotor = motor operation

Phase shift ϕs between Is and Us!

Uh

sss

s IkNp

mA ⋅⋅⋅⋅= ws1

p1

2ˆτ

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As(γ)ϕδ

1. Basic design rules for electrical machines

Internal phase shift ϕi of a synchronous machineBδ1

Main flux Φh ~ Bδ1

Bδ1 ~ Im , Uh ~ j.Φh

Bδ1 ~ Im

⊗Bδ,1,s ~ Is

As(γ)ϕδ

πϕϕ δ −=i

Bδ1 ~ Im

Is

Uh

Example: Cylindrical rotor, generator operation, over-excited

Source:

Siemens AG

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1. Basic design rules for electrical machines

Phasor diagram of a synchronous machine

Rotor pole axis

ϕi

Field axis

Phase shift ϕi between Is and Uh

Phase shift ϕs between Is and Us

Example: -π/2 > ϕi > -π: Me,AC < 0 on rotor = generator operation

πϕτππ δδ /cosˆˆ)(22 11

2

, ispes

ACesyn BAplp

fMnP ⋅⋅⋅⋅⋅==

Internal power:

2/2 1 hswssh NkfU Φπ ⋅=1

ˆ2δτ

πΦ Bleph ⋅⋅=

ishs IUmP ϕδ cos⋅=

Uh

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1. Basic design rules for electrical machines

Torque generation in DC machines

dxtxBtxAlF

pp

sre ⋅⋅⋅= ∫τ

δ

2

0

, ),(),( 2/sie dFM ⋅=

πατ δ /ˆ)(2 ,2

, serpeDCe BAplM ⋅⋅⋅=Φτ

⋅⋅⋅=⋅⋅⋅⋅⋅= ∫ rr AdpdxxBAld

pM si

0

sδ,esi

e

p

)(2

2

.)(si

constd

IzAxA c

rsr =⋅

⋅==

πsδ,pee

0

sδ,eˆ)(:

p

BldxxBlxx s ⋅⋅⋅=⋅== ∫ ταΦτ

Bδ,s

s

r

s s

s

Tangential force on rotor:

s

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1. Basic design rules for electrical machines

Internal power in DC machines

sδ,pee B̂l ⋅⋅⋅= ταΦ

πατππ δδ /ˆ)(222 ,2

, serpeDCe BAplnMnP ⋅⋅⋅⋅=⋅=

π⋅⋅

=sid

IzA c

rΦ⋅⋅⋅

= na

pzUi ca IaI ⋅= 2

ai IUP ⋅=δ

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1. Basic design rules for electrical machines

• Torque generation

dxtxBtxAlF

pp

⋅⋅⋅= ∫τ

δ

2

0

),(),(

πατ δ /ˆ)(2 ,2

, serpeDCe BAplM ⋅⋅⋅=

DC machines:

Current load and air gap flux density constant along αe~0.7 of pole pitch:

2/sie dFM ⋅=

AC machines: Current loading As and air gap flux density Bδ are sinusoidal distributed, phase shift ϕi

between Is and Uh:

πϕτ δ /cosˆˆ)( 112

, ispeACe BAplM ⋅⋅⋅⋅=Rotor torque:

Rotor torque:

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1. Basic design rules for electrical machines

)/( ldF sie ⋅⋅= πτ

Specific air gap thrust

Specific air gap thrust: Force per surface:

2/cosˆˆ11 isAC BA ϕτ δ ⋅⋅=)2/(cosˆˆ

)( 2211

2

lpBApl

pisp

AC τπϕπ

ττ δ ⋅⋅⋅⋅

⋅=

Surface: lpldArea psi τπ 2=⋅⋅=

)/()2//( pesiee pMdMF τπ== )2/( 22 lpM pe τπτ =

AC machines:

DC machines:

)2/(ˆ)(2 22

,

2

lpBApl

pserp

DC τπαπ

ττ δ ⋅⋅⋅

⋅= serDC BA ,

ˆδατ ⋅=

ell =

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1. Basic design rules for electrical machines

Effective current loading

)/cos(ˆ

),( pp

1s1 πϕωτπ

τπ

−−−⋅= is tx

VtxA

swsss

ss AkIkNp

mVA ⋅⋅=⋅⋅⋅⋅⋅=⋅= 1pws1p11 2/

2/ˆˆ τπ

πτπ

p2

2

τ⋅⋅⋅⋅

=p

INmA sss

s

AC machines:

AC: „Fictive“ effective current loading is constant along the air gap circumference!

DC machines:

.2

)2/()(

si

constp

aIz

d

IzAxA

p

acrsr =

⋅⋅

=⋅

⋅==

τπ

DC: Current loading is constant along the air gap circumference!

Ia: Armature current

Ic = Ia/(2a): Coil current

2/cosˆˆ11 isAC BA ϕτ δ ⋅⋅=

serDC BA ,ˆδατ ⋅=

AsAr

2/cosˆ11 iwsAC BkA ϕτ δ ⋅⋅⋅=

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Electromagnetic torque Me is determined by air gap flux density Bδ and

current loading A ( = ampere-turns per unit length) and corresponds with

internal power Pδ (air gap power).

- Air gap flux density peak value

- Typical maximum current loading for air cooling with open

ventilation:

A = 700 A/cm (DC-machines), corresponding amplitude for

AC-machines

a) DC-machines: αe = 0.7:

N/m2 ≅ 0.5 bar

b) AC-machines: kws1 ≈ 0.95, = 940 A/cm, maximum thrust at :

N/m2 ≅ 0.5 bar

In reality:

cosϕi ~ 0.9, so thrust for AC lower than for DC machines.

sws AkA ⋅⋅= 11 2ˆ

490000.17.070000ˆ, =⋅⋅=⋅= serDC BA δατ

470002/11940002/cosˆˆ1,1 =⋅⋅=⋅⋅= isAC BA ϕτ δ

1. Basic design rules for electrical machines

sA

Specific air gap thrust

T,0.1ˆ:DCT,0.1ˆ:AC ,1, == sBB δδ

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Summary: Torque generation and internal power

- Radial and tangential magnetic forces on magnetized iron and on conductors

- Tangential forces lead to torque

- Equivalent current loading represents slot conductor arrangement

- Fundamental waves for torque in AC machines

- Internal phase angle between resulting field wave and current loading

- DC machines have bigger torque at same peak current loading and field

- Specific air gap thrust as force per area in the range of 0.5 N 1 bar

Energy Converters – CAD and System Dynamics

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Energy Converters – CAD and System Dynamics

1. Basic design rules for rotating machines

1.1 Torque generation and internal power

1.2 Electromagnetic utilization

1.3 Thermal utilization

1.4 Overload capability of AC machines

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1. Basic design rules for electrical machines

Air gap torque Me, air gap power Pδ, internal apparent power Sδ

Air gap torque: Me

Air gap power (internal power):

AC machines:

DC machines:

)/(22 pfMnMP sesyne ⋅⋅=⋅= ππδ

nMP e ⋅⋅= πδ 2n: Rotor speed !

Internal apparent power:

AC machines:

DC machines:

DC:

si IUS ⋅= 3δ

ai IUS ⋅=δ Ui, Ia: Induced armature voltage

& armature current

Ui (or Uh), Is: Internal stator phase voltage

& stator phase current (r.m.s. values!)

δδ PS =

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DC machines:

Induced armature voltage:

With and current loading

we get:

Internal apparent power Sδ : Induced voltage x current:

AC machines: Induced phase voltage r.m.s:

With , r.m.s. current loading and

we get:p

ssss

p

INmA

τ2

2 ⋅⋅⋅=

1. Basic design rules for electrical machines

hwsssi kNfU Φπ ⋅⋅⋅⋅= 12

sis IUmS ⋅⋅=δ

1ˆ2δτ

πΦ Blph ⋅⋅=

sepeh Bl ,ˆδταΦ ⋅⋅=

hi na

pzU Φ⋅⋅

⋅=

p

ar

p

aIzA

τ2

)2/(⋅=

sreesi BAnldPS ,22 ˆ

δδδ απ ⋅⋅⋅⋅⋅⋅==

ai IUPS ⋅== δδ

Internal apparent power Sδ

swss

esishwsssssis ABkp

fldIkNfmIUmS 11

22

22 δδ

πΦπ ⋅⋅=⋅⋅⋅⋅⋅==

πτ /2 psi pd =

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Electromagnetic utilization (Esson´s number) C:

AC machines:

11

2

δ πBAk

nld

SC sws

synesi

AC ⋅⋅⋅=⋅⋅

=

1. Basic design rules for electrical machines

sre

esi

DC BAnld

PC ,

2

2ˆδ

δ απ ⋅⋅⋅=⋅⋅

=

Electromagnetic utilization C = Sδ per volume & speed

ACACC τπ ⋅= 2

DC machines:

at cosϕi = 1

DCDCC τπ ⋅= 2

Rotor volume: esiFesir ldldV ⋅≈⋅⋅≈ 22 4/π

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The internal apparent power Sδ per stator bore volume (usually

neglecting π/4) and per speed nsyn is called electromagnetic utilization or Esson´s number C. It increases with current loading A and air gap flux density Bδ.

ldsi ⋅)4/(2 π

11

2

δ πBAk

nld

SC sws

synesi

AC ⋅⋅⋅=⋅⋅

= sre

esi

DC BAnld

PC ,

2

2ˆδ

δ απ ⋅⋅⋅=⋅⋅

=

p

ssss

p

INmA

τ2

2 ⋅⋅⋅=

p

ar

p

aIzA

τ2

)2/(⋅=

32 ~/~ LldCnSM siN ⋅⋅=δ

Torque determines size of machine, NOT power !

r.m.s. current loading

1. Basic design rules for electrical machines

Electromagnetic utilization (Esson´s number) C

Machine volume (roughly): 3LV ≈

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Mounting of air-air heat exchanger on slip ring

induction wind generatorSource:

Winergy

Germany

Doubly-fed induction wind generator

1500 kW at 1800/min

Air inlet fan

Air-air heat

exchanger

Generator terminal

box

1. Basic design rules for electrical machines

- Internal “open”

ventilation

- Closed air-circuit to

reduce air acoustic noise L

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Example:

AC induction machines:

four poles, winding temperature rise 80 K,

air-cooled, open ventilated, for larger rated power

rated apparent power SN kVA 100 1000 10000

current loading As A/cm 300 550 1000

air gap flux density Bδ,1 T 1.0 1.05 1.1

Esson´s number C kVA.min/m3 3.3 6.4 12.2

Facit:

At given rotational speed: Bigger power = bigger torque = bigger machine size

= bigger rotor diameter = higher rotor surface speed = better air cooling!

With a better cooling higher current loadings are possible,

so electromagnetic utilization increases with rated power.

1. Basic design rules for electrical machines

11

πBAkC swsAC ⋅⋅⋅=

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Example: AC induction machine:four poles, winding temperature rise 80 K, air-cooled, open ventilated

rated apparent power SN 10 000 kVA

current loading As 1000 A/cm

air gap flux density Bδ,1 1.1 T

Esson´s number C 12.2 kVA.min/m3

1. Basic design rules for electrical machines

23 N/m732000VAs/m732000 ==C

ACC τπ 22N/m732000 ==

bar67.09.074.0:9.0cosAt

1cosatbar74.0N/m74176/ 22

=⋅==

====

ACi

iAC C

τϕ

ϕπτ

3352

11

2

min/mkVA2.12VAs/m7320001.11095.02

ˆ2

⋅=≈⋅⋅⋅≅⋅⋅⋅=ππ

δBAkC sws

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Summary: Electromagnetic utilization

- Internal apparent power = apparent air gap power Sδ

- Apparent air gap power per rotor volume and speed = electromagnetic utilization

- ESSON´s utilization C = „figure of merit“

- Utilization C ≈ 10 x Specific air gap thrust τ- Utilization C increases with current loading A and air-gap flux density Bδ

- Better cooling = higher current loading = higher utilization

Energy Converters – CAD and System Dynamics

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Energy Converters – CAD and System Dynamics

1. Basic design rules for rotating machines

1.1 Torque generation and internal power

1.2 Electromagnetic utilization

1.3 Thermal utilization

1.4 Overload capability of AC machines

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AC machines: ms phase winding: 22 )(2s

ca

bFesssssCu I

Aa

llNmIRmP ⋅

⋅⋅+⋅

⋅=⋅=κ

1. Basic design rules for electrical machines

Armature copper losses

sssibFe

si

sss

c

assibFeCu AJ

dll

d

INm

A

aIdllP ⋅⋅

⋅+=

⋅⋅⋅⋅

⋅+=

κπ

πκπ )(2/)(

Current density per conductor: cc AIJ /= )/( asc aII =

DC machines: z armature conductors:

222

2

⋅=⋅==a

IRzIRzIRP a

cccaaCu

AJdll

d

Iz

A

IdllI

A

llzP sibFe

si

c

c

csibFec

c

bFeCu ⋅⋅

⋅+=

⋅⋅⋅

⋅+=

⋅+

⋅=κ

ππκ

πκ

)()(2

AJdll

P sibFeCu ⋅⋅

⋅+=

κπ)(

(c: “conductor”)

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Heat transfer coefficient αk, W/(m2K): skkCu AP ϑ∆α ⋅⋅=

sskbFesik

Cus AJ

lld

P⋅⋅

⋅=

+⋅⋅=

καπαϑ∆

1

)(

Facit:

Temperature rise in armature winding for given

machine torque is determined by product of current

density J and current loading A.

It may be reduced by superior cooling (increased αk)

or decreased losses (increased κ).

sss AJ ⋅~ϑ∆

1. Basic design rules for electrical machines

Thermal utilization

Stator resistive losses

Cooling surface Ak

dsi

Steady state temperature rise:

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1. Basic design rules for electrical machines

Thermal scaling effectIncreasing motor size: Increasing surface of conductors

Example:

Surface cc ld ⋅π for round conductor wire (wire diameter dc & length lc),

Heat transfer coefficient at conductor surface: αc

Losses per conductor:

222, J

lAI

A

lIRP

c

ccc

cc

ccccCu ⋅=⋅==

κκ ⇒ 23

2

~ JLPJ

lA

P

V

PCu

ccc

Cu

c

Cu ⋅⇒=⋅

Temperature rise in conductor:

222

,

4

)4/(J

dJ

ld

ld

ld

P

cc

c

cccc

cc

ccc

cCu ⋅⋅⋅

=⋅⋅⋅⋅

⋅=

⋅⋅=

καπακπ

παϑ∆ ⇒ 2~ JL ⋅ϑ∆

Facit:

Admissible current density for SAME temperature rise ∆ϑ and cooling lower for bigger machines:

LJ /1~

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Source:

Winergy

Germany

Totally enclosed doubly-fed induction wind generator

Surface air-cooled with iron-cast cooling fin housing

600 kW at 1155/min

Fan hood Cooling fins Feet Power terminal box Slip ring

Shaft mounted fan inside terminal box

1. Basic design rules for electrical machines

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Lower admissible winding temperature rise means lower thermal utilization !

a) winding temperature rise 105 K:

rated power PN kW 5 650

current loading As A/cm 280 430

current density Js A/mm2 7.6 5.0

thermal utilization As.Js A/cm.A/mm2 2100 2150

b) winding temperature rise 80 K:

rated power PN kW 4 570

current loading As A/cm 240 380

current density Js A/mm2 6.6 4.4

thermal utilization As.Js A/cm.A/mm2 1580 1670

1. Basic design rules for electrical machines

Example: AC machines thermal utilization

totally enclosed, surface cooled, for smaller rated power

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Lower admissible winding temperature rise means lower utilization !

1. Basic design rules for electrical machines

Example:

Thermal utilization at Class F (105K) and Class B (80K)

totally enclosed, surface cooled

76.0105

80==

∆∆

F

B

ϑϑ

76.02100

1580

)(

)(==

⋅⋅

F

B

JA

JA

8.05

4

,

, ==FN

BN

P

P86.0

280

240

)(

)(

,

, ==≈=F

B

F

B

FN

BN

A

A

AB

AB

P

P

δ

δ

87.076.0~ 2 ==⇒=⋅F

B

A

AAAJA

Rough estimate of power scaling with lower admissible winding temperature:

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Air-cooled, four-pole induction machines, internal air flow

temperature rise ∆ϑ = 80 K (atϑamb = 40 °C), rated speed of about 1450/min.

Rated power / kW 100 1000 10000

Bδ,1 / T 1.0 1.0 1.0

As / A/cm 300 550 1000

C / kVA.min/m3 3.3 6.1 11.1

machine volume ~ L3 / p.u. 1.0 5.4 29.7

machine size L / p.u. 1.0 1.75 3.1

1.0 0.75 0.55

/A/mm2 6.8 5.1 3.7

/ (A/cm).(A/mm2) 2040 2850 3700

L/1LJ s /1~

ss JA ⋅

1. Basic design rules for electrical machines

Example: AC machines thermal utilizationopen ventilated

Facit:

Thermal utilization A.J rises with increased power.

Increased current density J for big machines needs improved cooling.

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Summary: Thermal utilization

- I2.R losses per cooling surface ~ thermal utilization A.J

- Reduced current density J at bigger machine size L for same cooling system

- Thermal scaling laws

- Increased thermal utilization A.J leads to increased ESSON´s number C ~ A.B

- Thermal Class (B, F, H, N) of insulation system determines thermal utilization

Energy Converters – CAD and System Dynamics

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DARMSTADTINSTITUT FÜRELEKTRISCHE ENERGIEWANDLUNG

Prof. Dr. A. Binder, Energy Converters - CAD and System Dynamics

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Energy Converters – CAD and System Dynamics

1. Basic design rules for rotating machines

1.1 Torque generation and internal power

1.2 Electromagnetic utilization

1.3 Thermal utilization

1.4 Overload capability of AC machines

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DARMSTADTINSTITUT FÜRELEKTRISCHE ENERGIEWANDLUNG

Prof. Dr. A. Binder, Energy Converters - CAD and System Dynamics

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Rated stator impedance x of AC machines

pshd A

B

xx τδδ ⋅≈ 1

ˆ~

11

1. Basic design rules for electrical machines

N

dd

sN

sNN

N

ss

Z

Xx

I

UZ

Z

Xx === or

hshs xxxx ≈+= σ

δ

τ

πµπ

l

p

mkNfX

pswssh 2

210

2)(2 ⋅=

11

10

ˆ~

ˆ

2

2

2

δδ

δ

τδ

τ

τ

πµ

B

A

B

p

INm

kx

ps

p

p

sNss

wsh

⋅⋅≈

1

10

11

2

210

ˆ

2

2

2

ˆ22

2)(2

δδ

δ

τ

τ

πµ

τπ

π

δ

τ

πµπ

B

p

INm

k

BlkNf

Il

p

mkNf

U

IX

U

IXx

p

p

sNss

ws

pwss

sNps

wss

h

sNh

sN

sNhh ⋅=

=≈=

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DARMSTADTINSTITUT FÜRELEKTRISCHE ENERGIEWANDLUNG

Prof. Dr. A. Binder, Energy Converters - CAD and System Dynamics

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Per-unit pull-out power

1. Basic design rules for electrical machines

Synchronous machines:

d

p

d

sNsN

sN

p

sNsN

dsNp

N

pqds

x

u

X

IU

U

U

IU

XUU

S

PXXR =⋅=

°===

/

3

/)90sin(3:,0

0

Induction machines:

sssN

sN

sNsN

ssN

sN

sN

sNsN

bsN

N

b

N

bs

xXI

U

IU

XU

p

p

IU

Mp

S

P

S

PR

⋅−

=−

⋅=

−⋅

=⋅

=≈=σ

σσ

σσσ

ωωω

δ

2

1

2

1

3

2

13

3:0

2

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DARMSTADTINSTITUT FÜRELEKTRISCHE ENERGIEWANDLUNG

Prof. Dr. A. Binder, Energy Converters - CAD and System Dynamics

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Overload capability of AC machines:

psdd

p

N

p

A

B

xx

u

S

P

τδδ ⋅≈= 10

ˆ~

1

psssN

b

A

B

xxS

P

τδ

σσσ δ ⋅⋅⋅

−≈ 1

ˆ~

11~

2

1

1. Basic design rules for electrical machines

Synchronous machines: pull out torque Induction machines: Breakdown torque

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DARMSTADTINSTITUT FÜRELEKTRISCHE ENERGIEWANDLUNG

Prof. Dr. A. Binder, Energy Converters - CAD and System Dynamics

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Summary: Overload capability of AC machines

- Overload capability = Maximum vs. rated torque Mmax/MN

- DC machine :

Maximum torque simply limited by maximum armature current Ia,max

- AC machines:

Maximum torque given by breakdown torque Mb or pull-out torque Mp0

- Small stray reactance Xσ and small synchronous reactance Xd

lead to increased maximum torque

Energy Converters – CAD and System Dynamics