ELKA DAYA Pertemuan 3

2/11/2010

1

Review of Basic Concepts

Pekik Argo Dahono

School of Electrical Engineering and Informatics

Institute of Technology Bandung

Average and RMS Concepts

• Periodic signals

• Average value

• RMS value

)()( Ttxtx +=

∫+

=Tt

t

o

o

dttxT

x )(1

∫+

=Tt

t

o

o

dttxT

X )(1 2

2Pekik A. Dahono : Elektronika Daya

2/11/2010

2

Single-Phase Power Concept

• Sinusoidal voltage and current

• Instantaneous power

( )tVv ωcos2=

( )φω −= tIi cos2

( )[ ] ( )44 344 21444 3444 21

part reactivepart resistive

2sinsin2cos1cos tVItVIvip ωφωφ ++==



• Active or Average Power

• Reactive Power

• Apparent Power

φcos1

VIdtpT

PTt

t

o

o∫

+

=⋅=

φsinVIQ =

VIS =4Pekik A. Dahono : Elektronika Daya

2/11/2010

3


• Power triangle

• Power factor

222QPS +=

φcos==S

PPF



LoadE V

jQP +

R jX

I

E

IVφ

δ

V∆

Vδ

( )

( )

( )

( ) ( )[ ]22

22

22

222

//

/

sincos

VQVPR

VSRRI

V

XQ

V

XQRP

φXIφRIE-V∆V

VVV

VVVE

+=

==

≈+

=

+=≈

∆+<<

+∆+=

Losses

Thus

δ

δ


2/11/2010

4

Balanced Three-Phase Power

Sinusoidal voltage and current:

( )

( )( )

32

32

cos2

cos2

cos2

π

π

ω

ω

ω

+=

−=

=

tVv

tVv

tVv

c

b

a

( )

( )( )φω

φω

φω

π

π

−+=

−−=

−=

32

32

cos2

cos2

cos2

tIi

tIi

tIi

c

b

a




• Instantaneous power is a constant that is

equal to average power

ccbbaa ivivivp ++=

φcos3VIp =


2/11/2010

5


• Reactive power is defined as

• Apparent power is defined as

φsin3VIQ =

VIS 3=


Three-Phase Power System

sR

sR

sR

sR

I

I

I

oR oR oR

23RI Losses =

0.1=PF

sR

sR

sR

sR

I3

3/oR

2RI18 Losses =

?=PF


2/11/2010

6

Three-Phase Power System

sR

sR

sR

sR

?=PF

11j1j−


Three-Phase Four-Wire Systems

( )

( )

e

eee

cabcabcnbnane

nocoboaoe

sh

e

sh

nocoboao

ncbae

es

ncbas

SP

IVS

VVVVVVV

VVVVV

R

VP

R

VVVV

IIIII

IrP

IIIIr

/

3

12/

3

3

3

3

(

222222

2222

2

2222

2222

2

2222

Factor Power

Power Apparent

or

Thus,

(balanced)

d)(unbalance P

:ionConsiderat LossesShunt

Thus,

(balanced)

)unbalancedP

:ionConsiderat Losses Series

=

=

+++++=

+++=

=∆

+++=∆

+++=

=∆

+++=∆ ( )( )( ) 9/

3/

3/

222

222

222

cabcab

cnbnane

cbae

VVV

VVVV

IIII

++=

++=

++=

System Wire-Three Phase-Three


2/11/2010

7

Fourier Series Representation

• Fourier series

• Average value

• RMS Value

( )∑∞

=

++=1

sin2)(

n

nno tnCctx θω

ocx =

∑∞

=

+=1

22

n

no CcX


Fourier Series

Pekik A. Dahono : Elektronika Daya 14

2/11/2010

8

Total Harmonic Distortion (THD)

• Voltage signal

• Total Harmonic Distortion

( )∑∞

=

+=1

sin2

n

nn tnVv θω

1

2/1

2

2

V

V

THDn

n

=∑

∞

=


The same definition applicable to current

Power concept under nonsinusoidal waveforms

• Voltage

• Current


( )∑∞

=

++=1

cos2

n

nno tnVVv αω

( )∑∞

=

++=1

cos2

n

nno tnIIi βω

vip =


2/11/2010

9

Power concept under nonsinusoidal waveforms

• Average power

• Apparent power

• Power factor

( )∑∞

=

−+=1

cos

n

nnnnoo IVIVP βα

rmsrms IVS =

( )

rmsrms

n

nnnnoo

IV

IVIV

S

PPF

∑∞

=

−+

== 1

cos βα


Example

• Voltage

• Current

• Average power

• RMS Voltage and Current

• Power factor

( ) ( )ttv ππ 300cos22100cos1021 ++=

( ) ( )ttio ππ 300cos260100cos52 −−=

( ) ( ) 25cos260cos50 =+= πoP

1052101 222 =++=V 261522 =+=I

478,026105

25===

S

PPF


2/11/2010

10

Sinusoidal voltage case

Average power :

Power factor :

where :

( )1111 cos βα −= IVP

( ) 12/1

2

221

111

1 coscos φβα

+

=−==

∑∞

=n

n

rms

II

I

I

I

S

PPF

111 βαφ −=

2

1

1

cos

THD

PF

+=

φ

Relationship between power factor and THD:


Transformer and inductor

• Inductor is used to store temporary energy and also used to smoothing the current

• Transformer is used for voltage conversion, galvanic isolation, and also used to store temporary energy.

• Transformer and inductor are the heaviest component in power electronics system.

• In power electronics applications, transformer sometimes has to operate with both dc and ac voltages or currents.


2/11/2010

11

Electromagnetism

c

c

c

c

c

c

l

AN

IL

Il

NIl

ABA

l

NIHB

l

NIH

NIHl

2

2

2

ANN

Wb

Wb/m

At/m

At NIH.dl

:Ampere Hukum

µλ

µλ

µ

µµ

==

=Φ=

==Φ

==

=

=

=ℑ=∫

IΦ

A

lilit N


Airgap Influence

IΦ

lilit N g

gAN

A

g

A

l

NL

A

g

A

l

INN

A

g

A

l

NI

AAA

A

g

A

lg

Bl

BNI

gHlHNI

o

oor

c

oor

c

oor

c

or

gc

goc

c

o

g

cc

gcc

/22

2

µ

µµµ

µµµ

λ

µµµ

µµµ

µµµµ

≈

+

=

+

=Φ=

+

=Φ

=

=≈

+Φ=+=

+==ℑ


2/11/2010

12

Transformer

1v

1i

1N 2N

2i

2v


Ideal Transformer

• •

1N 2N1v 2v

1i 2i

energy storesnor dissipates

neither er transformIdeal

1

2

2

1

2

1

i

i

N

N

v

v−==


2/11/2010

13

Practical Transformer

• •

1N 2N1v2v

1i 2i

mL1lL

2lL


Symmetrical Components

Any unbalanced three-phase quantities can be

composed into three symmetrical components:

- Positive sequence components

- Negative sequence components

- Zero sequence components

1aI

1bI

1cI

2aI

2cI

2bI

coboao III ==


2/11/2010

14


=

c

b

a

a

a

ao

V

V

V

aa

aa

V

V

V

2

2

2

1

1

1

111

3

1

=

ao

a

a

c

b

a

V

V

V

aa

aa

V

V

V

2

1

2

2

1

1

111

3

2πj

ea =

1aI

1bI

1cI

2aI

aoI

2bI

boI

2cI

coI

cI

bI

aI



• In three-phase three-wire systems we have

no neutral current and, therefore, we have

no zero sequence current.

• The neutral current is three times the zero

sequence current.


2/11/2010

15

Balanced nonsinusoidal quantities

Let assume:

For n=1:

For n=2:

For n=3:

( )[ ]∑∞

=

=1

cos

n

na tnVv ω ( )[ ]∑∞

=

−=1

32cos

n

nb tnVv πω ( )[ ]∑∞

=

+=1

32cos

n

nc tnVv πω

( )tVva ωcos11 = ( )3

211 cos πω −= tVvb

( )3

211 cos πω += tVvc

( )tVva ω2cos22 = ( )3

222 2cos πω += tVvb

( )3

222 2cos πω −= tVvc

( )tVva ω3cos33 = ( )tVvb ω3cos33 = ( )tVvc ω3cos33 =


Balanced nonsinusoidal quantities

For n=3k-2, The harmonics are similar to

positive sequence quantities.

For n=3k-1, the harmonics are similar to

negative sequence quantities.

For n=3k, the harmonics are similar to zero

quantities.


Inilah alasan mengapa pada sistem tiga-fasa tiga-kawat yang

seimbang, kita tidak menemui harmonisa kelipatan tiga.

2/11/2010

16

Symmetrical components

• Symmetrical components theory can be

applied to both steady-state phasor

quantities and instantaneous quantities.

• Symmetrical components theory can be

derived also by using linear algebra and

treated as variable transformation


Voltage and Current across the inductor

Steady-state:

Thus

Average voltage across the inductor under steady-state condition is zero.

dt

diLv L

L = )(1

oL

tt

t LL tidtvL

i o

o+= ∫

+

)()( Ttiti LL += )()( Ttvtv LL +=

0=∫+Tt

t Ldtv


2/11/2010

17

Voltage and current across the capacitor

Steady-state:

Thus

Average current through the capacitor under steady-state conditions is zero.

dt

dvCi C

C = )(1

oC

tt

t CC tvdtiC

v o

o+= ∫

+

)()( Ttiti CC += )()( Ttvtv CC +=

0=∫+Tt

t Cdti


Batasan Topologi

• Sumber tegangan hanya boleh diparalel jika sama

besar

• Sumber arus hanya boleh diseri jika sama besar

• Sumber tegangan tidak boleh dihubungsingkat

• Sumber arus tidak boleh dibuka

• Sumber tegangan bisa berupa kapasitor

• Sumber arus bisa berupa induktor


2/11/2010

18

Hubungan Berikut Harus Dihindari

+

+


Dualitas

Sumber tegangan Sumber arus

Hubungan paralel Hubungan seri

Induktor Kapasitor

Resistor Konduktor

Reverse conducting switch Reverse blocking switch

Variabel tegangan Variabel arus


2/11/2010

19

Contoh Dualitas

+

R L C

sV si LCpvpI G

)0(1

0 C

t

ss

ss vdtiCdt

diLRiV +++= ∫ )0(

1

0 L

t

pp

pp idtvLdt

dvCGvI +++= ∫


Contoh Dualitas

+ +


2/11/2010

20

Penggunaan Komputer

• PSIM, MATLAB, EMTP, PSPICE, etc.

• Switching Concept

• Averaging Concept


Switching Concept

dE

1S

2S

ii

ov

oi

R

L

Sw=1 IF S1 ON and S2 OFF

Sw=0 IF S1 OFF and S2 ON

vo=SwEd

Vo(ω)=Sw(ω)Ed

Io(ω)=Vo(ω)/Z(ω)

Ii(ω)=Sw(ω)Io(ω)


2/11/2010

21

Switching Concept

( )

owi

odwo

wwref

iSi

LRiESdt

di

SSv

=

−=

==>

/

0 ELSE 1 THENcar IFdE

1S

2S

ii

ov

oi

R

L

+

−

refv

car


Averaging Concept

( ) ( )

o

s

ONd

oo

s

ONdOFFONo

OFFONo

oo

sON

do

o

vT

TE

dt

idLiR

T

TETTdt

diLTTRi

dt

diLRi

TtT

Edt

diLRi

==+

=+++

=+

<<

=+

<<

in resultsby Divided

Averaging

0

Tt0

ON

dE

1S

2S

ii

ov

oi

R

L

+

−

refv

car


2/11/2010

22

D-Q Transform

( ) [ ]( ) [ ]

( ) ( )( ) ( )

( ) o

t

cbaT

odqT

d

fff

fff

θςςωθ

θθθ

θθθππ

ππ

+=

+

+

=

=

=

=

∫0

21

21

21

32

32

32

32

abc

qdo

abcqdo

sin-sinsin

cos-coscos

3

2K

f

f

Kff

( ) ( )( ) ( )

++

−−=

=

1sincos

1sincos

1sincos

K

fKf

32

32

32

321-

qdo-1

abc

ππ

ππ

θθ

θθ

θθ


f bisa dipakai untuk tegangan, arus, maupun fluksi.

Bentuk gelombang f bebas, tidak harus sinusoidal.

ω menyatakan kecepatan putar kerangka referensi.

D-Q Transform

af

dfcf

bf

qf

θ


2/11/2010

23

Example

( )( )

( ) ( ) ( ) ( )[ ]

( )

( ) ( ) ( ) ( )[ ]

( )0

sin2

sinsinsinsinsinsin3

22

cos2

coscoscoscoscoscos3

22

cos2

cos2

cos2

32

32

32

32

32

32

32

32

32

32

=

−=

−−++++=

−=

−−++++=

−=

+=

=

o

sd

sssd

sq

sssq

sc

sb

sa

i

tIi

ttttttI

i

tIi

ttttttI

i

tIi

tIi

tIi

ωω

ωωωωωω

ωω

ωωωωωω

ω

ω

ω

ππππ

ππππ

π

π


D-Q Transform

( )

( )qddq

ooddqqqdoabc

ccbbaaabc

ivivq

ivivivPP

ivivivP

−=

++==

++=

2

3

:Power Reactive ousInstantane

22

3

InvariancePower


2/11/2010

24

DQ Transform of Stationary Elements

rKrK

iKrKv

riv

1-

qdo1-

qdo

abcabc

=

=

=

[ ] [ ]

[ ] ( ) ( )( ) ( )

[ ]

−=

++−

−−−

−

=

+==

=

000

001

010

KK

Thus

0cossin

0cossin

0cossin

K

KKKKKKv

v

1-

32

32

32

321-

qdo1-

qdo1-

abc1-

qdo

abcabc

ω

θθ

θθ

θθ

ω

λλλ

λ

ππ

ππ

p

p

ppp

p



( ) [ ]

oo

dqd

qdq

qdT

dq

qdodq

pv

pv

pv

p

λ

λωλ

λωλ

λλλ

λωλ

=

+−=

+=

−=

+=qdov


2/11/2010

25


a

b

c

n

ai

bi

ci

R L qv

qi R L

+

dLiω

dv

di R L

+

qLiω

ov

oi R L


DQ Transform

• If the speed of reference frame ω is equal to the supply frequency, it is called synchronous reference frame.

• If the speed of reference frame is zero, it is called stationary reference frame.

• If the speed of reference frame is not equal to the supply frequency, it is called asynchronous reference frame.


2/11/2010

26

Space Vector

( )( ) ( )

( )( ) ( )

current. and voltageapply to sdefinition same The

current. of vector space called is

3

2

8

3

8

3

44

22cos

2

22cos

2

22cos

2

32

32

32

32

32

32

32

32

32

32

*

32

32

++=

+=

+++

++=

++=

+=+=

+=−=

+==

−

−

−−−

+−+

−−−

−

ππ

ππππ

ππ

ππ

θθ

θθ

θθπ

θθπ

θθ

θ

θ

θ

j

c

j

ba

jj

j

c

j

bajj

c

j

baj

ccbbaaa

jj

c

jj

b

jj

a

eieiii

ieN

ieN

eieiieN

eieiieN

inininF

eeNNn

eeNNn

eeNNn

r

rr


Example of Space Vector

[ ]

( ) ( ) ( )

( ) ( ) ( )

[ ] ( ) ( ) ( ) ( )

tj

jtjtjjtjtjtjtj

tjtj

sc

tjtj

sb

tjtjsa

s

ssssss

ss

ss

ss

Iei

eeeeeeeeIi

eeItIi

eeItIi

eeItIi

ω

ωωωωωω

ωωπ

ωωπ

ωω

ππππππ

ππ

ππ

ω

ω

ω

2

3

2

2

2cos2

2

2cos2

2

2cos2

32

32

32

32

32

32

32

32

32

32

32

32

=

++

+++=

+=−=

+=+=

+==

−−−−+−+−

−−−

+−+

−

r

r


Besaran sinusoidal seimbang akan nampak sebagai suatu vektor

yang bergerak melingkar pada kecepatan tetap.

2/11/2010

27

Example of Space Vector


( )

( ) ( )

dt

idLiR

iiidt

dLiiiR

vvvv

dt

diLRiv

dt

diLRiv

dt

diLRiv

cbacba

cba

ccc

bbb

aaa

rr

rrrr

rrr

+=

+++++=

++=

+=+=+=

aa3

2aa

3

2

aa3

2

:form vector spaceIn

:elements Stationary

22

2

Transformation of Space Vector

( )

[ ] [ ] [ ]32

32

ReReRe

:noted be shouldIt

2

example Previous

ππ

ωωω

ωω

j

c

j

ba

tj

tj

exxexxxx

Iei

exx

s

−

−

−

===

=

=

rrr

r

rr


2/11/2010

28

Numerical Methods To Solve Differential Equations

( ) ( )

( )

( )

( ) ( ) ( )4321

34

23

12

1

'

226

,

2,

2

2,

2

,

:Method Kutta-RungeOrder -Fourth

),('

:MethodEuler

;)( x);,(

Let

kkkkh

txhtx

htkxfk

ht

kxfk

ht

kxfk

txfk

txhftxhtx

hxttxf

oo

oo

oo

oo

oo

oooo

oo

++++=+

++=

++=

++=

=

+=+

=


Example

( ) ( )( )( ) ( )( ) ( ) 875.175.15101.075.14.0

75.15.15101.05.13.0

5.1)1510(1.012.0

105101.001.0

1.0

0

510

0

=×−×+=

=×−×+=

=×−×+=

=×−×+=

=

=

−=

i

i

i

i

h

i

idt

di


2/11/2010

29

Numerical Integration

++=

=

−=

∑∫

∫ ∑−

=

−

=

1

1

0

1

0

22

rule)simplest (the

Let

: to from interval in the function for the data N have We

N

n

iN

b

a

b

a

N

n

i

yyyh

ydt

yhydt

N

abh

bay


Example

T 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9

i 1 2 3 1 2 3 1 2 3 1

∑∫

∑∫−

−

==

==

=

=

1

0

29.0

0

2

1

0

9.0

0

3.4

9.1

10

1.0

N

N

ihdti

ihidt

N

h


2/11/2010

30

The End


Documents

ELKA DAYA Pertemuan 3