elka daya pertemuan 3
DESCRIPTION
ELKA DAYA Pertemuan 3TRANSCRIPT
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 ++==
3Pekik A. Dahono : Elektronika Daya
Single-Phase Power Concept
• 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
Single-Phase Power Concept
• Power triangle
• Power factor
222QPS +=
φcos==S
PPF
5Pekik A. Dahono : Elektronika Daya
Single-Phase Power Concept
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
6Pekik A. Dahono : Elektronika Daya
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
7Pekik A. Dahono : Elektronika Daya
Balanced Three-Phase Power
• Instantaneous power
• Instantaneous power is a constant that is
equal to average power
ccbbaa ivivivp ++=
φcos3VIp =
8Pekik A. Dahono : Elektronika Daya
2/11/2010
5
Balanced Three-Phase Power
• Reactive power is defined as
• Apparent power is defined as
φsin3VIQ =
VIS 3=
9Pekik A. Dahono : Elektronika Daya
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
10Pekik A. Dahono : Elektronika Daya
2/11/2010
6
Three-Phase Power System
sR
sR
sR
sR
?=PF
11j1j−
11Pekik A. Dahono : Elektronika Daya
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
12Pekik A. Dahono : Elektronika Daya
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
13Pekik A. Dahono : Elektronika Daya
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
=∑
∞
=
15Pekik A. Dahono : Elektronika Daya
The same definition applicable to current
Power concept under nonsinusoidal waveforms
• Voltage
• Current
• Instantaneous power
( )∑∞
=
++=1
cos2
n
nno tnVVv αω
( )∑∞
=
++=1
cos2
n
nno tnIIi βω
vip =
16Pekik A. Dahono : Elektronika Daya
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
17Pekik A. Dahono : Elektronika Daya
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
18Pekik A. Dahono : Elektronika Daya
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:
19Pekik A. Dahono : Elektronika Daya
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.
20Pekik A. Dahono : Elektronika Daya
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
21Pekik A. Dahono : Elektronika Daya
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
22Pekik A. Dahono : Elektronika Daya
2/11/2010
12
Transformer
1v
1i
1N 2N
2i
2v
23Pekik A. Dahono : Elektronika Daya
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−==
24Pekik A. Dahono : Elektronika Daya
2/11/2010
13
Practical Transformer
• •
1N 2N1v2v
1i 2i
mL1lL
2lL
25Pekik A. Dahono : Elektronika Daya
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 ==
26Pekik A. Dahono : Elektronika Daya
2/11/2010
14
Symmetrical Components
=
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
27Pekik A. Dahono : Elektronika Daya
Symmetrical Components
• 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.
Pekik A. Dahono : Elektronika Daya 28
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 =
29Pekik A. Dahono : Elektronika Daya
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.
30Pekik A. Dahono : Elektronika Daya
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
31Pekik A. Dahono : Elektronika Daya
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
32Pekik A. Dahono : Elektronika Daya
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
33Pekik A. Dahono : Elektronika Daya
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
34Pekik A. Dahono : Elektronika Daya
2/11/2010
18
Hubungan Berikut Harus Dihindari
+
+
35Pekik A. Dahono : Elektronika Daya
Dualitas
Sumber tegangan Sumber arus
Hubungan paralel Hubungan seri
Induktor Kapasitor
Resistor Konduktor
Reverse conducting switch Reverse blocking switch
Variabel tegangan Variabel arus
36Pekik A. Dahono : Elektronika Daya
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 +++= ∫
37Pekik A. Dahono : Elektronika Daya
Contoh Dualitas
+ +
38Pekik A. Dahono : Elektronika Daya
2/11/2010
20
Penggunaan Komputer
• PSIM, MATLAB, EMTP, PSPICE, etc.
• Switching Concept
• Averaging Concept
39Pekik A. Dahono : Elektronika Daya
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)
40Pekik A. Dahono : Elektronika Daya
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
41Pekik A. Dahono : Elektronika Daya
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
42Pekik A. Dahono : Elektronika Daya
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
43Pekik A. Dahono : Elektronika Daya
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
44Pekik A. Dahono : Elektronika Daya
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
45Pekik A. Dahono : Elektronika Daya
D-Q Transform
( )
( )qddq
ooddqqqdoabc
ccbbaaabc
ivivq
ivivivPP
ivivivP
−=
++==
++=
2
3
:Power Reactive ousInstantane
22
3
InvariancePower
46Pekik A. Dahono : Elektronika Daya
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
47Pekik A. Dahono : Elektronika Daya
DQ Transform of Stationary Elements
( ) [ ]
oo
dqd
qdq
qdT
dq
qdodq
pv
pv
pv
p
λ
λωλ
λωλ
λλλ
λωλ
=
+−=
+=
−=
+=qdov
48Pekik A. Dahono : Elektronika Daya
2/11/2010
25
DQ Transform of Stationary Elements
a
b
c
n
ai
bi
ci
R L qv
qi R L
+
dLiω
dv
di R L
+
qLiω
ov
oi R L
49Pekik A. Dahono : Elektronika Daya
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.
50Pekik A. Dahono : Elektronika Daya
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
51Pekik A. Dahono : Elektronika Daya
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
52Pekik A. Dahono : Elektronika Daya
Besaran sinusoidal seimbang akan nampak sebagai suatu vektor
yang bergerak melingkar pada kecepatan tetap.
2/11/2010
27
Example of Space Vector
Pekik A. Dahono : Elektronika Daya 53
( )
( ) ( )
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
54Pekik A. Dahono : Elektronika Daya
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
++++=+
++=
++=
++=
=
+=+
=
55Pekik A. Dahono : Elektronika Daya
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
56Pekik A. Dahono : Elektronika Daya
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
57Pekik A. Dahono : Elektronika Daya
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
58Pekik A. Dahono : Elektronika Daya
2/11/2010
30
The End
Pekik A. Dahono : Elektronika Daya 59