Handout Rangkaian Listrik II
RANGKAIAN LISTRIK II
Dosen : Rahmad Hidayat, S.T., M.T.
STT Mandala
Bandung
Bagian - 2
Rangkaian Tiga-Phase
Dosen : Rahmad Hidayat,ST,MT
Hidayat ST MT
STT Mandala - 2016
Analisa Rangkaian
Rangkaian Poliphase
2.1 Sistem Poliphase
2.2 Notasi
2.3 Sistem Tiga Kawat Phase Tunggal
2.44 Hubungan Y − Y tiga phase
2.5 Hubungan Delta ( Δ )
2.6 Pengukuran Daya
2.1 Sistem Poliphase
Sistem p
poliphase
p
: sistem dengan
g sumber poliphase
p p
Sumber tunggal (Vs)
Perhatikan tegangan sesaat dapat berharga nol
Æ Daya sesaat akan berharga nol
V
o
T
V
o
o
3T t
2T
Berbeda phase120o satu sama lain
Æ Daya sesaat tidak akan pernah nol.
Multi sumber (Vs1 , Vs2 , Vs3 )
T
2T
3T
t
2.1 Sistem Poliphase
V = Vs
V 1 + Vs
V 2 + Vs
V3
• Terhindar dari daya sesaat yang berharga nol.
• Daya sumber dapat diberikan lebih stabil.
• Dapat memberikan banyak level tegangan output
2.2 Notasi
a
b
8A
Titik c :
5A = 8A + I cdd , I cdd = −3A
Titik f :
I ef = 4A + 3A , I ef = 7 A
Titik j :
I ij + 3A = 4A10
0A , I ij = 7 A
4A
e
d
c
Icd = ?
I de = 2A
5A
Ief
− 6A
g
h
f
Iij
j
i
2A
10 A
k
I fj = 3A
l
2.2 Notasi
co
Van = 100∠00V
+
−
n
−
Vbn = 100∠ − 1200V
−
+ oa
Vab = Van + Vnb
+
b
o
Vcn = 100
100∠
∠ − 240 0 V
Tegangan titik a
terhadap b
a +; b -;
= Van − Vbn
= 100∠00V − 100∠ − 1200V
= 173.2∠300V
Serupa, Iab menunjukkan arus dari a ke b.
T dg
Tes
d analisa
li grafik
fik ? (gunakan
(
k diagram
di
phasor)
h
)
2.3 Sistem Tiga kawat Phase
Phase--tunggal
Fungsi: memungkinkan perangkat elektronik rumah tangga beroperasi
pada
d dua
d jenis
j i pilihan
ilih level
l l tegangan
t
a
n
b
Sumber
1-phase
33-kawat
kawat
a
V1
V2
n
b
Karakteristik tegangan
Van = Vnb
Vab = 2Van = 2Vnb
Perangkat elektronik rumah tangga dapat beroperasi
110V
atau 220V
Karakteristik Phase
∠Van = ∠Vnb
∠Van + ∠Vbn = 0
∠Van = −∠
∠Vbn
b
2.3 Sistem Tiga kawat Phase
Phase--tunggal
a
A
Zp
V1
n
N
Zp
V1
B
b
Karakteristik arus
I Nn = I bB + I aA
V1
V
I bB =
I Aa = 1
Zp
Zp
I Nn = 0
Tidak ada arus pada kawat netral.
Bgm jika kedua Z p tidak sama, dan semua kawat mempunyai impedansi ?
SISTEM PHASE - TUNGGAL
z
z
Single phase is used primarily only in low voltage,
l
low
power settings,
i
such
h as residential
id
i l and
d some
commercial.
Single phase transmission used for electric trains in
Europe.
2.3 Sistem Tiga kawat PhasePhase-tunggal
Contoh 9.1 (P242)
Tentukan daya
y yg diberikan ke beban
1Ω
115∠0 V
rms
0
115∠0 0 V
rms
I1
3Ω
I3
10Ω
50Ω
20Ω
I2
100Ω
j10Ω
50Ω, 100Ω dan the 20 + j10Ω
Tentukan daya yang hilang di ketiga
saluran yg dilewati 1Ω , 3Ω dan 1Ω
Tentukan efisiensi transmisinya ?
total daya yang diserap beban
η=
total daya yang dihasilkan sumber
Gunakan KVL :
− 115∠0 0 V + 1Ω ⋅ I1 + 50Ω ⋅ (I1 − I 2 ) + 3Ω(I1 − I 3 ) = 0
(20 +
j10 )Ω ⋅ I 2 + 100Ω ⋅ (I 2 − I 3 ) + 50Ω(I 2 − I1 ) = 0
− 115∠0 0 V + 3Ω ⋅ (I 3 − I1 ) + 100Ω(I 3 − I 2 ) + 1Ω ⋅ I 3 = 0
Susun ke dalam matriks :
− 50
− 3 ⎤ ⎡ I1 ⎤ ⎡115∠00 ⎤
⎡+ 54
⎢ − 50 170 + j10 − 100⎥ ⎢ I ⎥ = ⎢ 0 ⎥
⎥
⎥⎢ 2 ⎥ ⎢
⎢
⎢⎣ − 3
104 ⎥⎦ ⎢⎣ I 3 ⎥⎦ ⎢⎣115∠00 ⎥⎦
− 100
Sehingga dapat dihitung :
I1 = 11.24∠ − 19.830 A rms
I 2 = 9.389∠ − 24.47 0 A rms
I 3 = 10.37∠ − 21.80 0 A rms
I1 I 2 = 2.02∠2.27 o A rms
I 3 I 2 = 1.08∠2.12o A rms
I 3 I1 = 0.947∠2.3o A rms
y rata-rata y
yangg diberikan ke tiap
p beban adalah :
Daya
2
P50 = I1 − I 2 ⋅ 50 = 206 W ⎫
⎪⎪
2
P100 = I 3 − I 2 ⋅100 = 117 W ⎬ Total daya di beban = 2086W
2
P20+ j10 = I 2 ⋅ 20 = 1763W ⎪⎪
⎭
Daya yang hilang di ketiga kawat :
⎫
⎪⎪
2
PbB = I 3 ⋅1 = 108W
⎬Total daya hilang = 237W
2
2
PnN = I nN ⋅ 3 = I 3 − I1 ⋅ 3 = 3W ⎪⎪
⎭
2
PaA = I1 ⋅1 = 126 W
Efisiensi Transmisi , η
=
Power delivered to the load
× 100%
total power generated
Total daya yang dihasilkan kedua sumber (power generated) :
Psources = 115(11.24 ) cos19.830 + 115(10.37 ) cos 21.800
= 1216 W + 1107 W = 2323W
Efisiensi Transmisi =
2086 W
× 100% = 89.8%
2323W
SISTEM POLIPHASE
|
Two Phase System:
A generator consists of two coils placed perpendicular to each
other
y The voltage generated by one lags the other by 90°.
y
|
Three Phase System:
A generator consists of three coils placed 120
120° apart.
apart
y The voltage generated are equal in magnitude but, out of
phase by 120°.
y Three phase is the most economical polyphase system
y
SISTEM DUA PHASE - TIGA KAWAT
3 PHASE -4
4 KAWAT
DEFINISI
|
4 kawat
3 phase “aktif” : A, B, C
y 1 “ground”, atau “neutral”
y
|
K d warna
Kode
Phase A
y Phase B
y Phase C
y Neutral
y
Merah
Hitam
Biru
Putih atau Abu-abu
ECE
441
GENERATOR TIGA PHASE
| 2-pole
(North-South)
rotor turned by a
“prime mover”
| Sinusoidal voltages
are induced in each
stator winding
19
20
ECE 4411
DAYA 3 PHASE
THREE-P
PHASE CIRCUITS
| In
22
three-phase
three
phase circuits the 3 voltages
sources are 120° apart
| Polyphase generation and transmission of
electricity is more advantageous and
economical
(1) three-phase instantaneous power is
constant over time
(2) single-phase line losses are 50% greater
than three-phase losses (for the same load
Lect
power, voltage, pf), i.e., PSingle=3/2×PThree
ure 9
BALANCED SYSTEM
|A
23
balanced system is one in which the 3
sinusoidal voltages have the same
magnitude and frequency
frequency, and each is 120°
out-of-phase with the other two
v an (t ) = VM cos(ω t )
vbn (t ) = VM cos(ω t − 120°)
vcn (t ) = VM cos(ω t − 240°) = VM cos(ω t + 120°)
Lect
ure 9
THREE-P
PHASE VOLTAGES
a
Van
+
–
b
24
Vbn
b
+
–
c
Vcn
+
–
n
Balanced If:
Van=Vrms ∠0°
Vbn=Vrms ∠-120°
Vcn=Vrms ∠-240°
Lect
ure 9
Lect
ure 9
25
ADVANTAGES OF 3φ POWER
z
z
z
27
z
Can transmit more p
power for same amount of
wire (twice as much as single phase).
Total torque produced by 3φ machines is
constant,
t t so lless vibration.
ib ti
Three phase machines use less material for same
power rating.
Three phase machines start more easily than
single phase machines.
ADVANTAGES OF 3φ POWER
28
IMPORTANCE OF THREE PHASE SYSTEM
|
All electric power is generated and distributed in
three phase.
One phase, two phase, or more than three phase input can
be taken from three phase system rather than generated
independently.
y Melting purposes need 48 phases supply.
y
IMPORTANCE OF THREE PHASE SYSTEM
|
Uniform power transmission and less vibration of three
phase machines.
machines
The instantaneous power in a 3φ system can be constant (not
pulsating).
y High
Hi h power motors
t
prefer
f a steady
t d torque
t
especially
i ll one
created by a rotating magnetic field.
y
IMPORTANCE OF THREE PHASE SYSTEM
|
Three phase system is more economical than the single
phase.
phase
The amount of wire required for a three phase system is less
than required for an equivalent single phase system.
y Conductor: Copper, Aluminum, etc
y
3 PHASE POWER
A single
g p
phase g
generator is an alternator with a
single set armature coil producing a single
voltage waveform.
| A three
th
–phase
h
alternator
lt
t has
h three
th
sets
t off coils
il
spaced at 120o apart and generates three sets of
voltage waveforms.
|
WYE
|
Neutral conductor is
connected between all
3-phase conductors
|
Allows each p
phase to
be used for single
phase loads
DELTA
|
Neutral conductor is
centered between twophase conductors
|
High
g leg
g serves only
y 3phase loads and
cannot be used with
the neutral
TRANSFORMERS
|Wye
is typical for office
buildings and shopping centers
|Delta
l iis usually
ll used
d in
i
industrial applications
2.4 Hubungan
tiga-phase
Y − Y tiga-
Karakteristik arus :
I aA
I bB
Zp
a o
+
+
b
−
−
oB
A
Zp
Zp
n
N
−
+
c
I cC
ZP
C
2.4 Hubungan
tiga-phase
Y − Y tiga-
Pertimbangan ketiga impedansi Z p terhubung antar tiap kawat menuju
kawat netral.
V
I aA= an
Zp
0
Vbn
V
∠
−
120
I bB= b = an
= I aA∠ − 1200
Zp
Zp
Vcn V p ∠ − 240
I cC=
=
= I aA∠ − 2400
Zp
Zp
0
Maka
I aAA + I bB + I cCC = 0
Dengan kondisi seimbang , maka tidak ada arus pada kawat netral.
2.4 Hubungan
tiga-phase
Y − Y tiga-
ao
+
Van
o
+
−
−
n
b
oA
oB
Vbn
Su be 3-phase
Sumber
3 p ase seimbang
se ba g
(tegangan phasor )
oN
Van = Vbn = Vcn
Van + Vbn + Vcn = 0
−
+
Karakteristik tegangan
Vcn
o C
2.4 Hubungan
tiga-phase
Y − Y tiga-
Urutan pphase positip
p
p (abc)
( )
((rotasi searah jjarum jam)
j )
Vcn
Van = V p ∠00
Vbn = V p ∠ − 1200
Vcn = V p ∠ − 240
0
− 240
0
Vp
− 120
0
Van
Vbn
Urutan phase negatip (cba)
Van = V p ∠0
Vbn
0
Vbn = V p ∠120
(rotasi berlawanan arah jarum jam)
120 0
0
Vcn = V p ∠2400
240 0
Vcn
Van
2.4 Hubungan
tiga-phase
Y − Y tiga-
Tegangan line-to-line (urutan abc sebagai contoh)
Vab = Van + Vnb = V p ∠00 + V p ∠60 0 = V p + V p cos 60° + jV p sin 60°
3
3
= Vp + j
V p = 3V p ∠30 0
2
2
Vbc = Vbn + Vnc = V p ∠ − 120 + V p ∠ − 60
0
Vca
Vna
Vcn
0
1
3
1
3
= − Vp − j
Vp + Vp − j
V p = 3V p ∠ − 90 0
2
2
2
2
Vnb
Vnb
Vna
Van
Vca = Vcn + Vna = V p ∠ − 240 0 + V p ∠180 0
1
3
= − Vp + j
V p − V p + 0 = 3V p ∠ − 210 0
2
2
S hi
Sehingga
Vab + Vbc + Vca = 0
Vab
Vbn
Vnc
Vnc
Vbc
2.4 Hubungan
tiga-phase
Y − Y tiga-
Vab = 3V p ∠30 0
Vbc = 3V p ∠ − 90 0
1200
Jenis tegangan
magnitude
Tegangan phase
h
( Vp )
Vp
Tegangan line-to-line ( VL )
3V p
Vca = 3V p ∠ − 210 0
Beda Phase
120 0
a
a
+
_
V∠0
Wye Connected
So rce
Source
n
_
V∠-240
+
c
_
V∠-120
+
b
b
c
Sumber Delta
a
a
_
+
+
Delta
Source
_
Vbc = Vab ∠ -120
120
b
c
_
Vab = | Vab | ∠ 0
b
+
c
Vca = Vab ∠ -240
c
a
n
b
Zl
Zl
Zl
Sistem Wye – Wye
B
ZL
ZL
A
N
ZL
C
Sistem Wye – Delta
a
a
A
I aA
I AB
+
_
I CA
V∠0
Z
Z
n
_
V∠-240
+
_
I BC
V∠-120
+
b
b
c
C
B
Z
c
c
+
_
_
a
+
+
_
b
Zl
Zl
Zl
Sistem Delta – Delta
B
ZL
A
ZL
ZL
C
c
+
_
_
a
+
+
_
b
Zl
Zl
Zl
Sistem Delta – Wye
B
ZL
ZL
A
ZL
C
4 kawat
Rangkaian Y-Y
Rangkaian Y-Y ; 4 kawat seimbang
Four - wire
Va
Vb
Vc
I aA =
, I bB =
, and
d I cC =
ZA
ZB
ZC
I nN = I aA + I bB + I cC
Daya rata-rata
rata rata yang diberikan sumber 3-phase
3 phase ke beban 3-phase
3 phase :
P = PA + PB + PC
Saat ZA = ZB = ZC , beban dikatakan seimbang (balanced)
Va V ∠0°
Vb V ∠ − 120°
Vc V ∠120°
=
, I bB =
=
, and I cC =
=
I aA =
Z ∠θ
Z ∠θ
Z A Z ∠θ
ZB
ZC
V
V
V
I aA = ∠ − θ °, I bB = ∠( −θ − 120°), and I cC = ∠( −θ + 120°)
Z
Z
Z
Rangkaian Y-Y ; 4 kawat seimbang
Tidak ada arus di kawat netral sumber ke kawat netral beban :
I nN = I aA + I bB + I cC = 0
Daya rata-rata yang diberikan ke beban adalah :
P = PA + PB + PC
V
V
V
= V cos( −θ ) + V cos( −θ ) + V cos( −θ )
Z
Z
Z
2
V
= 3 cos(θ )
Z
Contoh 1 Ditanyakan S = ? Untuk 4-wire Seimbang (balance), dg data sbb :
Va = 110∠0° Vrms
Z A = 50 + j80 Ω
Vb = 110∠ − 120° Vrms
Z B = 50 + j80 Ω
Vc = 110∠120° Vrms
ZC = 50 + j80 Ω
I aA =
Va 110∠0°
=
Z A 50 + j80
= 1.16∠ − 58° A rms
S A = I*aA Va = 68 + j109 VA
Total daya komplex yang diberikan ke beban 3 phase adalah :
S = 3S A = 204 + j 326 VA
Juga
I bB = 1.16∠ − 177° A rms , I cC = 1.16∠62° A rms
S B = 68 + j109 VA = SC
3 kawat
Rangkaian Y-Y (lanjutan)
Rangkaian Y-Y (lanjutan)
3 - wire
Va − VNn Vb − VNn Vc − VNn
0=
+
+
ZA
ZB
ZC
V ∠0° − VNn V ∠ − 120° − VNn V ∠120° − VNn
+
+
=
ZA
ZB
ZC
Solusi untuk VNn
(V ∠ − 120°) Z AZC + V ∠120°Z AZ B + V ∠0°Z B ZC
VNn =
Z AZC + Z AZ B + Z B ZC
Va − VNn
Vb − VNn
Vc − VNn
I aA =
, IbB =
, and
d I cC =
ZA
ZB
ZC
Rangkaian Y-Y (lanjutan)
Saat rangkaian seimbang (balanced) yaitu saat ZA = ZB = ZC
(V ∠ − 120°)ZZ + V ∠120°ZZ + V ∠0°ZZ
VNn =
ZZ + ZZ + ZZ
=0
Daya rata-rata yang diberikan ke beban :
P = PA + PB + PC
V2
= 3 cos(θ )
Z
Contoh 2 : Skets diagram phasor sistem 3-phase seimbang untuk tegangan
phase dan tegangan saluran, dengan urutan abc.
Solusi (contoh 2) :
Tegangan phase :
Tegangan saluran :
Magnitude =
x magnitude tegangan phase
Sudut = sudut tegangan phase +
Solusi :
Contoh 3 : Sumber 3-phase seimbang dengan koneksi Y memberikan
daya ke beban Y seimbang. Magnitude tegangan saluran 150 V.
Jik impedansi
Jika
i
d i beban
b b setiap
ti phase
h
adalah
d l h 36+j12 ohm,
h
tentukan arus saluran jika diketahui
Solusi (contoh 3) :
Rangkaian Y-Y (lanjutan)
Transmission
s ss o lines
es
3 wire Y
3-wire
Y-Y
Y dengan impedansi saluran
Contoh 4 :
Sistem 3-phase
p
Y-Y seimbang
g urutan abc memiliki masing-masing
g
g
impedansi saluran 0,6+j1 ohm dan impedansi beban 18+j14 ohm.
Jika tegangan beban di phase a sebesar
Hitunglah tegangan di input saluran !
Solusi (contoh 4) :
Contoh 5 Ditanyakan S = ?
3-wire Seimbang (balance) dengan data sbb :
Va = 110∠0° Vrms
Z A = 50 + j80 Ω
Vb = 110∠ − 120° Vrms
Z B = 50 + j80 Ω
Vc = 110∠120° Vrms
ZC = 50 + j80 Ω
I aA =
Va 110∠0°
=
Z A 50 + j80
= 1.16∠ − 58° A rms
S A = I*aA Va = 68 + j109 VA
Total daya kompleks yang diberikan ke beban 3-phase
3 phase adalah :
S = 3S A = 204 + j 326 VA
Contoh 6 Ditanyakan PLoad = ? PLine = ? PSource = ? untuk data sbb :
Rangkaian ekivalen
per-phase
3-kawat seimbang
Va
100∠0°
I aA (ω ) =
=
Z A 50 + j ((377)(0.045)
)(
)
= 1.894∠ − 18.7° A
Tegangan phase di beban adalah :
VAN (ω ) = (40 + j(377)(0.04))IaA (ω ) = 81∠2° V
Contoh 6 (lanjutan)
Daya yang diberikan oleh sumber adalah :
Vm I m
Pa =
cos(θV − θ I )
2
(100)(1.894)
=
cos(18.7°) = 89.7 W
2
Daya yang diterima
di i beban
b b adalah
d l h:
I m2
(1.894) 2
PA =
Re( Z A ) =
40 = 71.7
71 7 W
2
2
Kehilangan daya di saluran adalah :
(1.894) 2
I m2
Re( Z Line ) =
10 = 17.9 W
PaA =
2
2
Line loss ≈ 20 %
Contoh 7 (p247)
Suatu hubungan Y-Y tiga
tiga--phase :
Tegangan phase :
Van = 200 ∠00 Vrms , Vbn = 200 ∠ − 120 0 Vrms , Vcn = 200 ∠ − 240 0 Vrms
Tegangan line-to-line :
Vab = 200 3∠30 0 Vrms , Vbc = 200 3∠ − 90 0 Vrms , Vca = 200 3∠ − 210 0 Vrms
A li
Arus
line :
Van 200 ∠00
0
I aA=
=
=
2
∠
−
60
Arms
Z p 100 ∠60 0
I bB= 2∠
2∠ − 1800 Arms
I cC= 2∠ − 3000 Arms
Daya yang diserap ketiga beban :
P = 3 × 200 × 2 × cos 60 o = 600 W
Contoh 7 (lanjutan)
Bagaimana dengan daya sesaat-nya ?
van (t ) = 200 2 cos (ωt )V
(
)
iaA (t ) = 2 2 cos ωt − 60 0 A
Note: Van = 200V rms
(
)
PaA (t ) = van (t )× iaA (t ) = 200 2 cos (ωt )V × 2 2 cos ωt − 60 0 A
(
)
= 200 + 400 cos 2ωt − 60 0 W
Serupa , total daya sesaat yang diserap beban :
P (t ) = PA (t ) + PB (t ) + PC (t )
= 600 + 400 cos (2ωt − 60 ° ) + 400 cos (2ωt + 300 ° ) + 400 cos (2ωt − 180 ° )W
= 600 W
Total daya sesaat TIDAK PERNAH NOL
Contoh 8 (p249)
Sistem 3-phase seimbang dengan tegangan kawat (line) 300Vrms diberikan pada
b b hubungan
beban
h b
Y sebesar
b
1200W pada
d power factor
f
((PF)) 0.8 lleading.
di
Hitunglah
i
l h
arus kawat (line) IL dan impedansi beban Zp untuk setiap phase.
T
Tegangan
phase
h
: Vp = 300/
3
Vrms.
Daya per-phase : 1200W/3 = 400W.
Sehingga
400 =
300
3
, and IL = 2.89Arms
( I L ) × 0.8
+
~−
V p = 300
I
L
Z
3 Vrms
V
Impedansi phase :
VP 300 3
=
= 60Ω
IL
2.89
PF sebesar 0.8 leading berakibat arus mendahului tegangan, dan sudut impedansi :
| Z P |=
-arccos(0.8) = -36.9o
dan Zp = 60 ∠-36.9
36.9o Ω
Catatan : Daya nyata hubungan Y-Y pada beban adalah P = Van × IAN
(tegangan phase × arus line)
p
2.5 Hubungan Delta ( Δ )
Tidak terdapat saluran kawat netral. Impedansi seimbang terhubung
antara tiap pasangan kawat (line)
Zp
a
o
+
+
b
A
ZP
−
−
Zp
n
+
−
c
C
oB
6.5 Hubungan Delta ( Δ )
Karakteristik tegangan
Tegangan phase V p = Van = Vbn = Vcn
Tegangan line
VL = 3V p
VL = Vab = Vbc = Vca
﹠
Vab = 3V p ∠30 0
Karakteristik arus
Arus phase
I p = I AB = I BC = I CA
Arus line
I L = I aA = I bB = I cC = 3 I p
2.5 Hubungan Delta ( Δ )
hubungan Y
Tegangan
g g pphase
Tegangan line
Arus phase
Arus line
Vp
√
VL = 3V p
Ip
IL = I p √
hubungan
Δ
Vp
VL = 3V p √
Ip
I L = 3I p
√
2.5 Hubungan Delta ( Δ )
|
Contoh 9 (p251)
Hitunglah amplitudo arus kawat (line) sistem 3-phase dengan tegangan kawat (line)
300Vrms yang memberikan 1200W pada beban dg hubungan Δ padaPF 0.8
lagging !
Daya rata-rata per-phase : 1200W/3 = 400W
S hi
Sehingga,
400 = VL · IP · 0.8
400W
0 8 3 = 300
300V · IP · 0.8
0 8 3 , and
d IP = 1.667Arms
1 66 A
Arus saluran (line), IL = 3 IP = 3 1.667A = 2.89Arms
PF lagging berarti bahwa tegangan mendahului arus sebesar arccos(0.8) = 36.9o
Impedansinya :
300
V&P
ZP =
=
∠36.9o = 180∠36.9o Ω
I&P 1.667
Catatan : Daya nyata pada beban (hubungan Δ) , P = Vabb × IAB
(tegangan line × arus phase)
The Δ-Y and Y- Δ Transformation
Z1Z3
ZA =
Z1 + Z 2 + Z3
ZA
ZB
ZC
Z1Z 2
ZC =
Z1 + Z 2 + Z3
Z AZ B + Z B ZC + Z AZC
Z1 =
ZB
Z3
Z1
Z2 Z3
ZB =
Z1 + Z 2 + Z3
Z2
Z AZ B + Z B ZC + Z AZC
Z2 =
ZA
Z AZ B + Z B ZC + Z AZC
Z3 =
ZC
Contoh
Rangkain Y- Δ
I AB
I aA = I AB − ICA
I bB = I BC − I AB
I cC = ICA − I BC
dengan :
VAB
=
Z3
I BC =
VBC
Z1
ICA =
VCA
Z2
Rangkain Y- Δ
(lanjutan)
I aA = I AB − ICA
= I cos φ + j sin φ − I cos(φ + 120°) − j sin(φ + 120°)
= 3I ∠(φ − 30°)
atau
I aA = 3 I
⇒
I L = 3I p
Contoh 10 IP = ? IL = ?
220
∠ − 30° Vrms
3
220
Vb =
∠ − 150° Vrms
3
220
Vc =
∠90° Vrms
3
Va =
Beban yg terhubung Δ seimbang :
VAB = Va − Vb = 220∠0° Vrms
VBC = Vb − Vc = 220∠ − 120° Vrms
VCA = Vc − Va = 220∠ − 240° Vrms
Arus saluran :
Z Δ = 10∠50°
I AB
⇒I
BC
VAB
=
= 22∠50° A rms
ZΔ
VBC
=
= 22∠ − 70° A rms
ZΔ
ICA =
VCA
= 22∠ − 190° A rms
ZΔ
I aA = I AB − ICA = 22 3∠20°, I bB = 22 3∠ − 100°, I cC = 22 3∠ − 220°
The Balanced Three-Phase
Y-to-Δ circuit
Circuits
equivalent Y-to-Y circuit
ZΔ
ZY =
3
per-phase equivalent circuit
Contoh 11 IP = ?
Va = 110∠0° Vrms
Vb = 110∠ − 120° Vrms
Vc = 110∠120° Vrms
Z L = 10 + j5 Ω
Z Δ = 75 + j 225 Ω
ZY =
ZΔ
= 25 + j 75 Ω
3
⇐
Va
I aAA =
= 1.26∠ − 66° A rms
Z L + ZY
⇓
Contoh 11 (lanjutan)
IbB = 1.26
1 26∠ − 186° A rms
and I cC = 1.26
1 26∠ − 54° A rms
Tegangan di rangkaian ekivalen per-phase adalah :
VAN = I aAZY = 99.6
99 6∠5° Vrms
VBN = 99.6∠ − 115° Vrms
VCN = 99.6∠125
5° Vrms
Tegangan line-to-line adalah :
I AB =
VAB = VAN − VBN = 172∠35° Vrms
VBC = VBN − VCN = 172∠ − 85° Vrms
⇒I
BC
VCA = VCN − VAN = 172∠155° Vrms
ICA
VAB
= 0.727∠ − 36° A rms
ZΔ
VBC
=
= 0.727∠ − 156° A rms
ZΔ
VCA
=
= 0.727∠84° A rms
ZΔ
Contoh 12 P = ?
Va = 110∠0° Vrms
Vb = 110∠ − 120° Vrms
Vc = 110∠120° Vrms
Z L = 10 + j5 Ω
Z Δ = 75 + j 225 Ω
Va
I aA =
= 1.26∠ − 66° A rms
Z L + ZY
VAN = I aAZY = 99.6∠5° Vrms
P = 3(99.6)(1.26) cos(5° − ( −66°)) = 122.6 W
Contoh 13 Hitunglah arus saluran masing-masing dan magnitude
tegangan beban masing-masing ! (tugas mandiri )
Pengukuran Daya Rangkaian Poliphase
P = I ⋅V
Wattmeter
diukur dg
coil arus
I
coil arus
+
coil potential
/tegangan
+
E
E.g.
V
diukur dg
coil potential
Pasif
Network
I = 11 .18∠153 .4°Arms
A
V = 100 ∠0°Vrms
P = V ⋅ I cos (ang V − ang I )
= 100 ⋅11 .18 ⋅ cos (0° − 153 .4° ) = −1000 W
POWER MEASUREMENT
|
When using
Wh
i AC,
AC power is
i determined
d t
i d nott only
l by
b the
th
r.m.s. values of the voltage and current, but also by
the p
phase angle
g ((which determines the p
power factor))
y
|
consequently, you cannot determine the power from
independent measurements of current and voltage
In single-phase systems power is normally measured
using an electrodynamic wattmeter
y
measures power directly using a single meter which
effectively multiplies instantaneous current and voltage
POWER MEASUREMENT
|
In three-phase
p
systems
y
we need to sum the
power taken from the various phases
in three-wire arrangements we can deduce the total
power from measurements using 2 wattmeter
y in a four-wire system it may be necessary to use 3
wattmeter
y in balanced systems (systems
(
that take equal power
from each phase) a single wattmeter can be used, its
reading being multiplied by 3 to get the total power
y
Electrodynamic Wattmeter
Digital Power Meter
pf Meter
VAR Meter
Two-Wattmeter Power Measurement
cc = current coil
vc = voltage coil
W1 read
d
P1 = VAB I A cosθ1
W2 read
P2 = VCB I C cos θ 2
For balanced load with abc phase sequence
θ1 = θ a + 30° and θ 2 = θ a − 30°
θ a is the angle between phase current and phase voltage of phase a
Two-Wattmeter Power Measurement(cont.)
P = P1 + P2
= 2VL I L cosθ cos 30°
= 3VL I L cosθ
To determine the power factor angle
P1 + P2 = VL I L 2 cos θ cos 30°
P1 − P2 = VL I L ( −2sin θ sin 30°)
P1 + P2
VL I L 2 cos θ cos 30°
− 3
=
=
P1 − P2 VL I L ( −2sin θ sin 30°) tan θ
P1 − P2
∴ tan θ = 3
P1 + P2
⎛
P1 − P2 ⎞
or θ = tan ⎜ 3
⎟
P
+
P
1
2 ⎠
⎝
−1
Contoh 14 P = ?
Z = 10∠45°
Tegangan line-to-line = 220Vrms
Tegangan phase
220
VA =
∠ − 30°
3
Arus saluran
VA 220∠ − 30°
=
= 12.7∠ − 75° dan
IA =
Z
10 3∠45°
P1 = VAC I A cosθ1 = 2698 W
P2 = VBC I B cosθ 2 = 723
⇒
W
I B = 12.7∠ − 195°
P = P1 + P2 = 3421 W
2.6 Pengukuran Daya
A
o
a
I&aA
+
+
Z P ∠θ
I&AB
B
I&BC
Z P ∠θ
Z P ∠θ
1
I&bB
b o
I&CA
c o
I&cC
+
2
C
+
Periksalah pembacaan daya yang diserap
/dibawa oleh ketiga
g impedansi.
p
2.6 Pengukuran Daya
(
)
P1 = V AB ⋅ I aA cos (ang V AB − ang I aA ) = V L I L cos 30 0 − (− θ )
(
)
(
)
= V L I L cos 30 0 + θ
(
)
P2 = VCB ⋅ I cC cos (ang VCB − ang I cC ) = V L I L cos 90 0 − (120 ° − θ )
= V L I L cos 30 0 − θ
3 1
− tgθ
3 − tgθ
cos 30 cos θ − sin 30 sin θ
P1 cos 30 + θ
2 2
=
=
=
=
cos 300 cos θ + sin 30 0 sin θ
P2 cos 300 − θ
3 1
3 + tg
gθ
+ tgθ
P2 − P1
P2 − P1 2 2
tgθ = 3
θ = arctg 3
P2 + P1
P2 + P1
(
(
0
)
)
reaktif (PF=0)
π
θ = ± , tgθ = ±∞
2
P1 = − P2
0
0
kapasitif / induktif (0 P2 , − < θ < 0, capacitive
iti
2 π
P1 < P2 , 0 < θ < , inductive
2
resistif (PF=1)
θ = 0 , tgθ = 0
P1 = P2
2.6 Pengukuran Daya
a
.
o
A 4Ω j15Ω
+
|
Contoh 15 (p256)
+
1
V&ab = 230∠0°Vrms dg urutan phase positip
bo
(1) Berapa pembacaan tiap wattmeter.
B
+
2
(2) Total daya diserap beban.
+
co
Dg urutan phase positip :
V&ab = 230∠0°Vrms
V&bc = 230∠ − 120°Vrms
V&ca = 230∠ + 120°Vrms
Wattmeter 1 membaca I&aA dan V&ac :
V& = −V& = 230∠ − 60°Vrms
ac
.
ca
⎛ 230 ⎞∠ − 30°
⎜
⎟
&
V
3
⎝
⎠
an
I&aA =
=
= 8.554∠ − 105.1°A
4 + j15
4 + j15
.
C
.
N
2.6 Pengukuran Daya
a
.
o
A 4Ω j15Ω
+
|
Contoh 15 (p256)
+
1
Wattmeter 1 membaca :
P1 = V&ac I&aA cos(angV&ac − angI&aA )
bo
.
B
+
+
= 230 × 8.554 × cos(− 60° + 105.1°) = 1389 W
co
Wattmeter 2 baca I&bB dan V&bc :
⎛ 230 ⎞∠ − 150°
⎜
⎟
&
V
3
⎠
bn
I&aB =
=⎝
= 8.554∠134.9°A
4 + j15
4 + j15
2
.
C
P2 = V&bc I&bB cos (ang V&bc − ang I&bB )
= 230 × 8.554 × cos (− 120 ° − 134 .9° ) = −512 .5W
Sehingga
S
hi
P = P1 + P2 = 876.5W
, Cobalah buktikan kedua wattmeter membaca daya pada ketiga impedansi
Q:
.
N
TERIMA KASIH
http://risetwpt.wordpress.com
email : rhidayat4000@gmail.com
rhidayat4000@gmail com
STT MANDALA
6
Rahmad Hidayat, ST, MT
Dosen : Rahmad Hidayat, S.T., M.T.
STT Mandala
Bandung
Bagian - 2
Rangkaian Tiga-Phase
Dosen : Rahmad Hidayat,ST,MT
Hidayat ST MT
STT Mandala - 2016
Analisa Rangkaian
Rangkaian Poliphase
2.1 Sistem Poliphase
2.2 Notasi
2.3 Sistem Tiga Kawat Phase Tunggal
2.44 Hubungan Y − Y tiga phase
2.5 Hubungan Delta ( Δ )
2.6 Pengukuran Daya
2.1 Sistem Poliphase
Sistem p
poliphase
p
: sistem dengan
g sumber poliphase
p p
Sumber tunggal (Vs)
Perhatikan tegangan sesaat dapat berharga nol
Æ Daya sesaat akan berharga nol
V
o
T
V
o
o
3T t
2T
Berbeda phase120o satu sama lain
Æ Daya sesaat tidak akan pernah nol.
Multi sumber (Vs1 , Vs2 , Vs3 )
T
2T
3T
t
2.1 Sistem Poliphase
V = Vs
V 1 + Vs
V 2 + Vs
V3
• Terhindar dari daya sesaat yang berharga nol.
• Daya sumber dapat diberikan lebih stabil.
• Dapat memberikan banyak level tegangan output
2.2 Notasi
a
b
8A
Titik c :
5A = 8A + I cdd , I cdd = −3A
Titik f :
I ef = 4A + 3A , I ef = 7 A
Titik j :
I ij + 3A = 4A10
0A , I ij = 7 A
4A
e
d
c
Icd = ?
I de = 2A
5A
Ief
− 6A
g
h
f
Iij
j
i
2A
10 A
k
I fj = 3A
l
2.2 Notasi
co
Van = 100∠00V
+
−
n
−
Vbn = 100∠ − 1200V
−
+ oa
Vab = Van + Vnb
+
b
o
Vcn = 100
100∠
∠ − 240 0 V
Tegangan titik a
terhadap b
a +; b -;
= Van − Vbn
= 100∠00V − 100∠ − 1200V
= 173.2∠300V
Serupa, Iab menunjukkan arus dari a ke b.
T dg
Tes
d analisa
li grafik
fik ? (gunakan
(
k diagram
di
phasor)
h
)
2.3 Sistem Tiga kawat Phase
Phase--tunggal
Fungsi: memungkinkan perangkat elektronik rumah tangga beroperasi
pada
d dua
d jenis
j i pilihan
ilih level
l l tegangan
t
a
n
b
Sumber
1-phase
33-kawat
kawat
a
V1
V2
n
b
Karakteristik tegangan
Van = Vnb
Vab = 2Van = 2Vnb
Perangkat elektronik rumah tangga dapat beroperasi
110V
atau 220V
Karakteristik Phase
∠Van = ∠Vnb
∠Van + ∠Vbn = 0
∠Van = −∠
∠Vbn
b
2.3 Sistem Tiga kawat Phase
Phase--tunggal
a
A
Zp
V1
n
N
Zp
V1
B
b
Karakteristik arus
I Nn = I bB + I aA
V1
V
I bB =
I Aa = 1
Zp
Zp
I Nn = 0
Tidak ada arus pada kawat netral.
Bgm jika kedua Z p tidak sama, dan semua kawat mempunyai impedansi ?
SISTEM PHASE - TUNGGAL
z
z
Single phase is used primarily only in low voltage,
l
low
power settings,
i
such
h as residential
id
i l and
d some
commercial.
Single phase transmission used for electric trains in
Europe.
2.3 Sistem Tiga kawat PhasePhase-tunggal
Contoh 9.1 (P242)
Tentukan daya
y yg diberikan ke beban
1Ω
115∠0 V
rms
0
115∠0 0 V
rms
I1
3Ω
I3
10Ω
50Ω
20Ω
I2
100Ω
j10Ω
50Ω, 100Ω dan the 20 + j10Ω
Tentukan daya yang hilang di ketiga
saluran yg dilewati 1Ω , 3Ω dan 1Ω
Tentukan efisiensi transmisinya ?
total daya yang diserap beban
η=
total daya yang dihasilkan sumber
Gunakan KVL :
− 115∠0 0 V + 1Ω ⋅ I1 + 50Ω ⋅ (I1 − I 2 ) + 3Ω(I1 − I 3 ) = 0
(20 +
j10 )Ω ⋅ I 2 + 100Ω ⋅ (I 2 − I 3 ) + 50Ω(I 2 − I1 ) = 0
− 115∠0 0 V + 3Ω ⋅ (I 3 − I1 ) + 100Ω(I 3 − I 2 ) + 1Ω ⋅ I 3 = 0
Susun ke dalam matriks :
− 50
− 3 ⎤ ⎡ I1 ⎤ ⎡115∠00 ⎤
⎡+ 54
⎢ − 50 170 + j10 − 100⎥ ⎢ I ⎥ = ⎢ 0 ⎥
⎥
⎥⎢ 2 ⎥ ⎢
⎢
⎢⎣ − 3
104 ⎥⎦ ⎢⎣ I 3 ⎥⎦ ⎢⎣115∠00 ⎥⎦
− 100
Sehingga dapat dihitung :
I1 = 11.24∠ − 19.830 A rms
I 2 = 9.389∠ − 24.47 0 A rms
I 3 = 10.37∠ − 21.80 0 A rms
I1 I 2 = 2.02∠2.27 o A rms
I 3 I 2 = 1.08∠2.12o A rms
I 3 I1 = 0.947∠2.3o A rms
y rata-rata y
yangg diberikan ke tiap
p beban adalah :
Daya
2
P50 = I1 − I 2 ⋅ 50 = 206 W ⎫
⎪⎪
2
P100 = I 3 − I 2 ⋅100 = 117 W ⎬ Total daya di beban = 2086W
2
P20+ j10 = I 2 ⋅ 20 = 1763W ⎪⎪
⎭
Daya yang hilang di ketiga kawat :
⎫
⎪⎪
2
PbB = I 3 ⋅1 = 108W
⎬Total daya hilang = 237W
2
2
PnN = I nN ⋅ 3 = I 3 − I1 ⋅ 3 = 3W ⎪⎪
⎭
2
PaA = I1 ⋅1 = 126 W
Efisiensi Transmisi , η
=
Power delivered to the load
× 100%
total power generated
Total daya yang dihasilkan kedua sumber (power generated) :
Psources = 115(11.24 ) cos19.830 + 115(10.37 ) cos 21.800
= 1216 W + 1107 W = 2323W
Efisiensi Transmisi =
2086 W
× 100% = 89.8%
2323W
SISTEM POLIPHASE
|
Two Phase System:
A generator consists of two coils placed perpendicular to each
other
y The voltage generated by one lags the other by 90°.
y
|
Three Phase System:
A generator consists of three coils placed 120
120° apart.
apart
y The voltage generated are equal in magnitude but, out of
phase by 120°.
y Three phase is the most economical polyphase system
y
SISTEM DUA PHASE - TIGA KAWAT
3 PHASE -4
4 KAWAT
DEFINISI
|
4 kawat
3 phase “aktif” : A, B, C
y 1 “ground”, atau “neutral”
y
|
K d warna
Kode
Phase A
y Phase B
y Phase C
y Neutral
y
Merah
Hitam
Biru
Putih atau Abu-abu
ECE
441
GENERATOR TIGA PHASE
| 2-pole
(North-South)
rotor turned by a
“prime mover”
| Sinusoidal voltages
are induced in each
stator winding
19
20
ECE 4411
DAYA 3 PHASE
THREE-P
PHASE CIRCUITS
| In
22
three-phase
three
phase circuits the 3 voltages
sources are 120° apart
| Polyphase generation and transmission of
electricity is more advantageous and
economical
(1) three-phase instantaneous power is
constant over time
(2) single-phase line losses are 50% greater
than three-phase losses (for the same load
Lect
power, voltage, pf), i.e., PSingle=3/2×PThree
ure 9
BALANCED SYSTEM
|A
23
balanced system is one in which the 3
sinusoidal voltages have the same
magnitude and frequency
frequency, and each is 120°
out-of-phase with the other two
v an (t ) = VM cos(ω t )
vbn (t ) = VM cos(ω t − 120°)
vcn (t ) = VM cos(ω t − 240°) = VM cos(ω t + 120°)
Lect
ure 9
THREE-P
PHASE VOLTAGES
a
Van
+
–
b
24
Vbn
b
+
–
c
Vcn
+
–
n
Balanced If:
Van=Vrms ∠0°
Vbn=Vrms ∠-120°
Vcn=Vrms ∠-240°
Lect
ure 9
Lect
ure 9
25
ADVANTAGES OF 3φ POWER
z
z
z
27
z
Can transmit more p
power for same amount of
wire (twice as much as single phase).
Total torque produced by 3φ machines is
constant,
t t so lless vibration.
ib ti
Three phase machines use less material for same
power rating.
Three phase machines start more easily than
single phase machines.
ADVANTAGES OF 3φ POWER
28
IMPORTANCE OF THREE PHASE SYSTEM
|
All electric power is generated and distributed in
three phase.
One phase, two phase, or more than three phase input can
be taken from three phase system rather than generated
independently.
y Melting purposes need 48 phases supply.
y
IMPORTANCE OF THREE PHASE SYSTEM
|
Uniform power transmission and less vibration of three
phase machines.
machines
The instantaneous power in a 3φ system can be constant (not
pulsating).
y High
Hi h power motors
t
prefer
f a steady
t d torque
t
especially
i ll one
created by a rotating magnetic field.
y
IMPORTANCE OF THREE PHASE SYSTEM
|
Three phase system is more economical than the single
phase.
phase
The amount of wire required for a three phase system is less
than required for an equivalent single phase system.
y Conductor: Copper, Aluminum, etc
y
3 PHASE POWER
A single
g p
phase g
generator is an alternator with a
single set armature coil producing a single
voltage waveform.
| A three
th
–phase
h
alternator
lt
t has
h three
th
sets
t off coils
il
spaced at 120o apart and generates three sets of
voltage waveforms.
|
WYE
|
Neutral conductor is
connected between all
3-phase conductors
|
Allows each p
phase to
be used for single
phase loads
DELTA
|
Neutral conductor is
centered between twophase conductors
|
High
g leg
g serves only
y 3phase loads and
cannot be used with
the neutral
TRANSFORMERS
|Wye
is typical for office
buildings and shopping centers
|Delta
l iis usually
ll used
d in
i
industrial applications
2.4 Hubungan
tiga-phase
Y − Y tiga-
Karakteristik arus :
I aA
I bB
Zp
a o
+
+
b
−
−
oB
A
Zp
Zp
n
N
−
+
c
I cC
ZP
C
2.4 Hubungan
tiga-phase
Y − Y tiga-
Pertimbangan ketiga impedansi Z p terhubung antar tiap kawat menuju
kawat netral.
V
I aA= an
Zp
0
Vbn
V
∠
−
120
I bB= b = an
= I aA∠ − 1200
Zp
Zp
Vcn V p ∠ − 240
I cC=
=
= I aA∠ − 2400
Zp
Zp
0
Maka
I aAA + I bB + I cCC = 0
Dengan kondisi seimbang , maka tidak ada arus pada kawat netral.
2.4 Hubungan
tiga-phase
Y − Y tiga-
ao
+
Van
o
+
−
−
n
b
oA
oB
Vbn
Su be 3-phase
Sumber
3 p ase seimbang
se ba g
(tegangan phasor )
oN
Van = Vbn = Vcn
Van + Vbn + Vcn = 0
−
+
Karakteristik tegangan
Vcn
o C
2.4 Hubungan
tiga-phase
Y − Y tiga-
Urutan pphase positip
p
p (abc)
( )
((rotasi searah jjarum jam)
j )
Vcn
Van = V p ∠00
Vbn = V p ∠ − 1200
Vcn = V p ∠ − 240
0
− 240
0
Vp
− 120
0
Van
Vbn
Urutan phase negatip (cba)
Van = V p ∠0
Vbn
0
Vbn = V p ∠120
(rotasi berlawanan arah jarum jam)
120 0
0
Vcn = V p ∠2400
240 0
Vcn
Van
2.4 Hubungan
tiga-phase
Y − Y tiga-
Tegangan line-to-line (urutan abc sebagai contoh)
Vab = Van + Vnb = V p ∠00 + V p ∠60 0 = V p + V p cos 60° + jV p sin 60°
3
3
= Vp + j
V p = 3V p ∠30 0
2
2
Vbc = Vbn + Vnc = V p ∠ − 120 + V p ∠ − 60
0
Vca
Vna
Vcn
0
1
3
1
3
= − Vp − j
Vp + Vp − j
V p = 3V p ∠ − 90 0
2
2
2
2
Vnb
Vnb
Vna
Van
Vca = Vcn + Vna = V p ∠ − 240 0 + V p ∠180 0
1
3
= − Vp + j
V p − V p + 0 = 3V p ∠ − 210 0
2
2
S hi
Sehingga
Vab + Vbc + Vca = 0
Vab
Vbn
Vnc
Vnc
Vbc
2.4 Hubungan
tiga-phase
Y − Y tiga-
Vab = 3V p ∠30 0
Vbc = 3V p ∠ − 90 0
1200
Jenis tegangan
magnitude
Tegangan phase
h
( Vp )
Vp
Tegangan line-to-line ( VL )
3V p
Vca = 3V p ∠ − 210 0
Beda Phase
120 0
a
a
+
_
V∠0
Wye Connected
So rce
Source
n
_
V∠-240
+
c
_
V∠-120
+
b
b
c
Sumber Delta
a
a
_
+
+
Delta
Source
_
Vbc = Vab ∠ -120
120
b
c
_
Vab = | Vab | ∠ 0
b
+
c
Vca = Vab ∠ -240
c
a
n
b
Zl
Zl
Zl
Sistem Wye – Wye
B
ZL
ZL
A
N
ZL
C
Sistem Wye – Delta
a
a
A
I aA
I AB
+
_
I CA
V∠0
Z
Z
n
_
V∠-240
+
_
I BC
V∠-120
+
b
b
c
C
B
Z
c
c
+
_
_
a
+
+
_
b
Zl
Zl
Zl
Sistem Delta – Delta
B
ZL
A
ZL
ZL
C
c
+
_
_
a
+
+
_
b
Zl
Zl
Zl
Sistem Delta – Wye
B
ZL
ZL
A
ZL
C
4 kawat
Rangkaian Y-Y
Rangkaian Y-Y ; 4 kawat seimbang
Four - wire
Va
Vb
Vc
I aA =
, I bB =
, and
d I cC =
ZA
ZB
ZC
I nN = I aA + I bB + I cC
Daya rata-rata
rata rata yang diberikan sumber 3-phase
3 phase ke beban 3-phase
3 phase :
P = PA + PB + PC
Saat ZA = ZB = ZC , beban dikatakan seimbang (balanced)
Va V ∠0°
Vb V ∠ − 120°
Vc V ∠120°
=
, I bB =
=
, and I cC =
=
I aA =
Z ∠θ
Z ∠θ
Z A Z ∠θ
ZB
ZC
V
V
V
I aA = ∠ − θ °, I bB = ∠( −θ − 120°), and I cC = ∠( −θ + 120°)
Z
Z
Z
Rangkaian Y-Y ; 4 kawat seimbang
Tidak ada arus di kawat netral sumber ke kawat netral beban :
I nN = I aA + I bB + I cC = 0
Daya rata-rata yang diberikan ke beban adalah :
P = PA + PB + PC
V
V
V
= V cos( −θ ) + V cos( −θ ) + V cos( −θ )
Z
Z
Z
2
V
= 3 cos(θ )
Z
Contoh 1 Ditanyakan S = ? Untuk 4-wire Seimbang (balance), dg data sbb :
Va = 110∠0° Vrms
Z A = 50 + j80 Ω
Vb = 110∠ − 120° Vrms
Z B = 50 + j80 Ω
Vc = 110∠120° Vrms
ZC = 50 + j80 Ω
I aA =
Va 110∠0°
=
Z A 50 + j80
= 1.16∠ − 58° A rms
S A = I*aA Va = 68 + j109 VA
Total daya komplex yang diberikan ke beban 3 phase adalah :
S = 3S A = 204 + j 326 VA
Juga
I bB = 1.16∠ − 177° A rms , I cC = 1.16∠62° A rms
S B = 68 + j109 VA = SC
3 kawat
Rangkaian Y-Y (lanjutan)
Rangkaian Y-Y (lanjutan)
3 - wire
Va − VNn Vb − VNn Vc − VNn
0=
+
+
ZA
ZB
ZC
V ∠0° − VNn V ∠ − 120° − VNn V ∠120° − VNn
+
+
=
ZA
ZB
ZC
Solusi untuk VNn
(V ∠ − 120°) Z AZC + V ∠120°Z AZ B + V ∠0°Z B ZC
VNn =
Z AZC + Z AZ B + Z B ZC
Va − VNn
Vb − VNn
Vc − VNn
I aA =
, IbB =
, and
d I cC =
ZA
ZB
ZC
Rangkaian Y-Y (lanjutan)
Saat rangkaian seimbang (balanced) yaitu saat ZA = ZB = ZC
(V ∠ − 120°)ZZ + V ∠120°ZZ + V ∠0°ZZ
VNn =
ZZ + ZZ + ZZ
=0
Daya rata-rata yang diberikan ke beban :
P = PA + PB + PC
V2
= 3 cos(θ )
Z
Contoh 2 : Skets diagram phasor sistem 3-phase seimbang untuk tegangan
phase dan tegangan saluran, dengan urutan abc.
Solusi (contoh 2) :
Tegangan phase :
Tegangan saluran :
Magnitude =
x magnitude tegangan phase
Sudut = sudut tegangan phase +
Solusi :
Contoh 3 : Sumber 3-phase seimbang dengan koneksi Y memberikan
daya ke beban Y seimbang. Magnitude tegangan saluran 150 V.
Jik impedansi
Jika
i
d i beban
b b setiap
ti phase
h
adalah
d l h 36+j12 ohm,
h
tentukan arus saluran jika diketahui
Solusi (contoh 3) :
Rangkaian Y-Y (lanjutan)
Transmission
s ss o lines
es
3 wire Y
3-wire
Y-Y
Y dengan impedansi saluran
Contoh 4 :
Sistem 3-phase
p
Y-Y seimbang
g urutan abc memiliki masing-masing
g
g
impedansi saluran 0,6+j1 ohm dan impedansi beban 18+j14 ohm.
Jika tegangan beban di phase a sebesar
Hitunglah tegangan di input saluran !
Solusi (contoh 4) :
Contoh 5 Ditanyakan S = ?
3-wire Seimbang (balance) dengan data sbb :
Va = 110∠0° Vrms
Z A = 50 + j80 Ω
Vb = 110∠ − 120° Vrms
Z B = 50 + j80 Ω
Vc = 110∠120° Vrms
ZC = 50 + j80 Ω
I aA =
Va 110∠0°
=
Z A 50 + j80
= 1.16∠ − 58° A rms
S A = I*aA Va = 68 + j109 VA
Total daya kompleks yang diberikan ke beban 3-phase
3 phase adalah :
S = 3S A = 204 + j 326 VA
Contoh 6 Ditanyakan PLoad = ? PLine = ? PSource = ? untuk data sbb :
Rangkaian ekivalen
per-phase
3-kawat seimbang
Va
100∠0°
I aA (ω ) =
=
Z A 50 + j ((377)(0.045)
)(
)
= 1.894∠ − 18.7° A
Tegangan phase di beban adalah :
VAN (ω ) = (40 + j(377)(0.04))IaA (ω ) = 81∠2° V
Contoh 6 (lanjutan)
Daya yang diberikan oleh sumber adalah :
Vm I m
Pa =
cos(θV − θ I )
2
(100)(1.894)
=
cos(18.7°) = 89.7 W
2
Daya yang diterima
di i beban
b b adalah
d l h:
I m2
(1.894) 2
PA =
Re( Z A ) =
40 = 71.7
71 7 W
2
2
Kehilangan daya di saluran adalah :
(1.894) 2
I m2
Re( Z Line ) =
10 = 17.9 W
PaA =
2
2
Line loss ≈ 20 %
Contoh 7 (p247)
Suatu hubungan Y-Y tiga
tiga--phase :
Tegangan phase :
Van = 200 ∠00 Vrms , Vbn = 200 ∠ − 120 0 Vrms , Vcn = 200 ∠ − 240 0 Vrms
Tegangan line-to-line :
Vab = 200 3∠30 0 Vrms , Vbc = 200 3∠ − 90 0 Vrms , Vca = 200 3∠ − 210 0 Vrms
A li
Arus
line :
Van 200 ∠00
0
I aA=
=
=
2
∠
−
60
Arms
Z p 100 ∠60 0
I bB= 2∠
2∠ − 1800 Arms
I cC= 2∠ − 3000 Arms
Daya yang diserap ketiga beban :
P = 3 × 200 × 2 × cos 60 o = 600 W
Contoh 7 (lanjutan)
Bagaimana dengan daya sesaat-nya ?
van (t ) = 200 2 cos (ωt )V
(
)
iaA (t ) = 2 2 cos ωt − 60 0 A
Note: Van = 200V rms
(
)
PaA (t ) = van (t )× iaA (t ) = 200 2 cos (ωt )V × 2 2 cos ωt − 60 0 A
(
)
= 200 + 400 cos 2ωt − 60 0 W
Serupa , total daya sesaat yang diserap beban :
P (t ) = PA (t ) + PB (t ) + PC (t )
= 600 + 400 cos (2ωt − 60 ° ) + 400 cos (2ωt + 300 ° ) + 400 cos (2ωt − 180 ° )W
= 600 W
Total daya sesaat TIDAK PERNAH NOL
Contoh 8 (p249)
Sistem 3-phase seimbang dengan tegangan kawat (line) 300Vrms diberikan pada
b b hubungan
beban
h b
Y sebesar
b
1200W pada
d power factor
f
((PF)) 0.8 lleading.
di
Hitunglah
i
l h
arus kawat (line) IL dan impedansi beban Zp untuk setiap phase.
T
Tegangan
phase
h
: Vp = 300/
3
Vrms.
Daya per-phase : 1200W/3 = 400W.
Sehingga
400 =
300
3
, and IL = 2.89Arms
( I L ) × 0.8
+
~−
V p = 300
I
L
Z
3 Vrms
V
Impedansi phase :
VP 300 3
=
= 60Ω
IL
2.89
PF sebesar 0.8 leading berakibat arus mendahului tegangan, dan sudut impedansi :
| Z P |=
-arccos(0.8) = -36.9o
dan Zp = 60 ∠-36.9
36.9o Ω
Catatan : Daya nyata hubungan Y-Y pada beban adalah P = Van × IAN
(tegangan phase × arus line)
p
2.5 Hubungan Delta ( Δ )
Tidak terdapat saluran kawat netral. Impedansi seimbang terhubung
antara tiap pasangan kawat (line)
Zp
a
o
+
+
b
A
ZP
−
−
Zp
n
+
−
c
C
oB
6.5 Hubungan Delta ( Δ )
Karakteristik tegangan
Tegangan phase V p = Van = Vbn = Vcn
Tegangan line
VL = 3V p
VL = Vab = Vbc = Vca
﹠
Vab = 3V p ∠30 0
Karakteristik arus
Arus phase
I p = I AB = I BC = I CA
Arus line
I L = I aA = I bB = I cC = 3 I p
2.5 Hubungan Delta ( Δ )
hubungan Y
Tegangan
g g pphase
Tegangan line
Arus phase
Arus line
Vp
√
VL = 3V p
Ip
IL = I p √
hubungan
Δ
Vp
VL = 3V p √
Ip
I L = 3I p
√
2.5 Hubungan Delta ( Δ )
|
Contoh 9 (p251)
Hitunglah amplitudo arus kawat (line) sistem 3-phase dengan tegangan kawat (line)
300Vrms yang memberikan 1200W pada beban dg hubungan Δ padaPF 0.8
lagging !
Daya rata-rata per-phase : 1200W/3 = 400W
S hi
Sehingga,
400 = VL · IP · 0.8
400W
0 8 3 = 300
300V · IP · 0.8
0 8 3 , and
d IP = 1.667Arms
1 66 A
Arus saluran (line), IL = 3 IP = 3 1.667A = 2.89Arms
PF lagging berarti bahwa tegangan mendahului arus sebesar arccos(0.8) = 36.9o
Impedansinya :
300
V&P
ZP =
=
∠36.9o = 180∠36.9o Ω
I&P 1.667
Catatan : Daya nyata pada beban (hubungan Δ) , P = Vabb × IAB
(tegangan line × arus phase)
The Δ-Y and Y- Δ Transformation
Z1Z3
ZA =
Z1 + Z 2 + Z3
ZA
ZB
ZC
Z1Z 2
ZC =
Z1 + Z 2 + Z3
Z AZ B + Z B ZC + Z AZC
Z1 =
ZB
Z3
Z1
Z2 Z3
ZB =
Z1 + Z 2 + Z3
Z2
Z AZ B + Z B ZC + Z AZC
Z2 =
ZA
Z AZ B + Z B ZC + Z AZC
Z3 =
ZC
Contoh
Rangkain Y- Δ
I AB
I aA = I AB − ICA
I bB = I BC − I AB
I cC = ICA − I BC
dengan :
VAB
=
Z3
I BC =
VBC
Z1
ICA =
VCA
Z2
Rangkain Y- Δ
(lanjutan)
I aA = I AB − ICA
= I cos φ + j sin φ − I cos(φ + 120°) − j sin(φ + 120°)
= 3I ∠(φ − 30°)
atau
I aA = 3 I
⇒
I L = 3I p
Contoh 10 IP = ? IL = ?
220
∠ − 30° Vrms
3
220
Vb =
∠ − 150° Vrms
3
220
Vc =
∠90° Vrms
3
Va =
Beban yg terhubung Δ seimbang :
VAB = Va − Vb = 220∠0° Vrms
VBC = Vb − Vc = 220∠ − 120° Vrms
VCA = Vc − Va = 220∠ − 240° Vrms
Arus saluran :
Z Δ = 10∠50°
I AB
⇒I
BC
VAB
=
= 22∠50° A rms
ZΔ
VBC
=
= 22∠ − 70° A rms
ZΔ
ICA =
VCA
= 22∠ − 190° A rms
ZΔ
I aA = I AB − ICA = 22 3∠20°, I bB = 22 3∠ − 100°, I cC = 22 3∠ − 220°
The Balanced Three-Phase
Y-to-Δ circuit
Circuits
equivalent Y-to-Y circuit
ZΔ
ZY =
3
per-phase equivalent circuit
Contoh 11 IP = ?
Va = 110∠0° Vrms
Vb = 110∠ − 120° Vrms
Vc = 110∠120° Vrms
Z L = 10 + j5 Ω
Z Δ = 75 + j 225 Ω
ZY =
ZΔ
= 25 + j 75 Ω
3
⇐
Va
I aAA =
= 1.26∠ − 66° A rms
Z L + ZY
⇓
Contoh 11 (lanjutan)
IbB = 1.26
1 26∠ − 186° A rms
and I cC = 1.26
1 26∠ − 54° A rms
Tegangan di rangkaian ekivalen per-phase adalah :
VAN = I aAZY = 99.6
99 6∠5° Vrms
VBN = 99.6∠ − 115° Vrms
VCN = 99.6∠125
5° Vrms
Tegangan line-to-line adalah :
I AB =
VAB = VAN − VBN = 172∠35° Vrms
VBC = VBN − VCN = 172∠ − 85° Vrms
⇒I
BC
VCA = VCN − VAN = 172∠155° Vrms
ICA
VAB
= 0.727∠ − 36° A rms
ZΔ
VBC
=
= 0.727∠ − 156° A rms
ZΔ
VCA
=
= 0.727∠84° A rms
ZΔ
Contoh 12 P = ?
Va = 110∠0° Vrms
Vb = 110∠ − 120° Vrms
Vc = 110∠120° Vrms
Z L = 10 + j5 Ω
Z Δ = 75 + j 225 Ω
Va
I aA =
= 1.26∠ − 66° A rms
Z L + ZY
VAN = I aAZY = 99.6∠5° Vrms
P = 3(99.6)(1.26) cos(5° − ( −66°)) = 122.6 W
Contoh 13 Hitunglah arus saluran masing-masing dan magnitude
tegangan beban masing-masing ! (tugas mandiri )
Pengukuran Daya Rangkaian Poliphase
P = I ⋅V
Wattmeter
diukur dg
coil arus
I
coil arus
+
coil potential
/tegangan
+
E
E.g.
V
diukur dg
coil potential
Pasif
Network
I = 11 .18∠153 .4°Arms
A
V = 100 ∠0°Vrms
P = V ⋅ I cos (ang V − ang I )
= 100 ⋅11 .18 ⋅ cos (0° − 153 .4° ) = −1000 W
POWER MEASUREMENT
|
When using
Wh
i AC,
AC power is
i determined
d t
i d nott only
l by
b the
th
r.m.s. values of the voltage and current, but also by
the p
phase angle
g ((which determines the p
power factor))
y
|
consequently, you cannot determine the power from
independent measurements of current and voltage
In single-phase systems power is normally measured
using an electrodynamic wattmeter
y
measures power directly using a single meter which
effectively multiplies instantaneous current and voltage
POWER MEASUREMENT
|
In three-phase
p
systems
y
we need to sum the
power taken from the various phases
in three-wire arrangements we can deduce the total
power from measurements using 2 wattmeter
y in a four-wire system it may be necessary to use 3
wattmeter
y in balanced systems (systems
(
that take equal power
from each phase) a single wattmeter can be used, its
reading being multiplied by 3 to get the total power
y
Electrodynamic Wattmeter
Digital Power Meter
pf Meter
VAR Meter
Two-Wattmeter Power Measurement
cc = current coil
vc = voltage coil
W1 read
d
P1 = VAB I A cosθ1
W2 read
P2 = VCB I C cos θ 2
For balanced load with abc phase sequence
θ1 = θ a + 30° and θ 2 = θ a − 30°
θ a is the angle between phase current and phase voltage of phase a
Two-Wattmeter Power Measurement(cont.)
P = P1 + P2
= 2VL I L cosθ cos 30°
= 3VL I L cosθ
To determine the power factor angle
P1 + P2 = VL I L 2 cos θ cos 30°
P1 − P2 = VL I L ( −2sin θ sin 30°)
P1 + P2
VL I L 2 cos θ cos 30°
− 3
=
=
P1 − P2 VL I L ( −2sin θ sin 30°) tan θ
P1 − P2
∴ tan θ = 3
P1 + P2
⎛
P1 − P2 ⎞
or θ = tan ⎜ 3
⎟
P
+
P
1
2 ⎠
⎝
−1
Contoh 14 P = ?
Z = 10∠45°
Tegangan line-to-line = 220Vrms
Tegangan phase
220
VA =
∠ − 30°
3
Arus saluran
VA 220∠ − 30°
=
= 12.7∠ − 75° dan
IA =
Z
10 3∠45°
P1 = VAC I A cosθ1 = 2698 W
P2 = VBC I B cosθ 2 = 723
⇒
W
I B = 12.7∠ − 195°
P = P1 + P2 = 3421 W
2.6 Pengukuran Daya
A
o
a
I&aA
+
+
Z P ∠θ
I&AB
B
I&BC
Z P ∠θ
Z P ∠θ
1
I&bB
b o
I&CA
c o
I&cC
+
2
C
+
Periksalah pembacaan daya yang diserap
/dibawa oleh ketiga
g impedansi.
p
2.6 Pengukuran Daya
(
)
P1 = V AB ⋅ I aA cos (ang V AB − ang I aA ) = V L I L cos 30 0 − (− θ )
(
)
(
)
= V L I L cos 30 0 + θ
(
)
P2 = VCB ⋅ I cC cos (ang VCB − ang I cC ) = V L I L cos 90 0 − (120 ° − θ )
= V L I L cos 30 0 − θ
3 1
− tgθ
3 − tgθ
cos 30 cos θ − sin 30 sin θ
P1 cos 30 + θ
2 2
=
=
=
=
cos 300 cos θ + sin 30 0 sin θ
P2 cos 300 − θ
3 1
3 + tg
gθ
+ tgθ
P2 − P1
P2 − P1 2 2
tgθ = 3
θ = arctg 3
P2 + P1
P2 + P1
(
(
0
)
)
reaktif (PF=0)
π
θ = ± , tgθ = ±∞
2
P1 = − P2
0
0
kapasitif / induktif (0 P2 , − < θ < 0, capacitive
iti
2 π
P1 < P2 , 0 < θ < , inductive
2
resistif (PF=1)
θ = 0 , tgθ = 0
P1 = P2
2.6 Pengukuran Daya
a
.
o
A 4Ω j15Ω
+
|
Contoh 15 (p256)
+
1
V&ab = 230∠0°Vrms dg urutan phase positip
bo
(1) Berapa pembacaan tiap wattmeter.
B
+
2
(2) Total daya diserap beban.
+
co
Dg urutan phase positip :
V&ab = 230∠0°Vrms
V&bc = 230∠ − 120°Vrms
V&ca = 230∠ + 120°Vrms
Wattmeter 1 membaca I&aA dan V&ac :
V& = −V& = 230∠ − 60°Vrms
ac
.
ca
⎛ 230 ⎞∠ − 30°
⎜
⎟
&
V
3
⎝
⎠
an
I&aA =
=
= 8.554∠ − 105.1°A
4 + j15
4 + j15
.
C
.
N
2.6 Pengukuran Daya
a
.
o
A 4Ω j15Ω
+
|
Contoh 15 (p256)
+
1
Wattmeter 1 membaca :
P1 = V&ac I&aA cos(angV&ac − angI&aA )
bo
.
B
+
+
= 230 × 8.554 × cos(− 60° + 105.1°) = 1389 W
co
Wattmeter 2 baca I&bB dan V&bc :
⎛ 230 ⎞∠ − 150°
⎜
⎟
&
V
3
⎠
bn
I&aB =
=⎝
= 8.554∠134.9°A
4 + j15
4 + j15
2
.
C
P2 = V&bc I&bB cos (ang V&bc − ang I&bB )
= 230 × 8.554 × cos (− 120 ° − 134 .9° ) = −512 .5W
Sehingga
S
hi
P = P1 + P2 = 876.5W
, Cobalah buktikan kedua wattmeter membaca daya pada ketiga impedansi
Q:
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N
TERIMA KASIH
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Rahmad Hidayat, ST, MT