konsep dasar.pdf
TRANSCRIPT
BAB 1. KONSEP DASAR
1.1 Daya Listrik pada Rangkaian 1
Fasa
1.2 Rangkaian Tiga Fasa
1.3 Daya Listrik pada Rangkaian 3
Fasa
BAB 1. 1
Daya Listrik pada Rangkaian 1 Fasa
Real (Active) and Reactive Power
Real (Active) and Reactive Loads
Power Triangle
Real (Active) and Reactive Power Flow
Sine Wave Basics (Review) RMS – a method for computing the effective value of a time-varying e-m
wave, equivalent to the energy under the area of the voltage waveform.
Real, Reactive and Apparent
Power in AC Circuits in DC circuits: P=VI but…= in AC circuits: average
power supplied to the load will be affected by the phase angle between the voltage and the current.
If load is inductive the phase angle (also called impedance angle) is positive; (i.e, phase angle of current will lag the phase angle of the voltage) and the load will consume both real and positive reactive power
If the load is capacitive the impedance angle will be negative (the phase angle of the current will lead the phase angle of the voltage) and the load will consume real power and supply reactive power.
Resistive and Reactive Loads
Impedance Angle, Current Angle
& Power
Inductive loads positive impedance angle current angle lags voltage angle
Capacitive loads negative impedance angle current angle leads voltage angle
Both types of loads consume real power
One (inductive) consumes reactive as well while the other (capacitive) supplies reactive power
Tegangan, Arus dan Daya
First term is average or Real power (P)
Second term is power transferred back and forth
between source and load (Reactive power Q)
More equations
Real term averages to P = VI cos (+)
Reactive term Q = VI sin (+ for inductive load,
- for capacitive load)
Reactive power is the power that is first stored and then released
in the magnetic field of an inductor or in the electric field of a capacitor
Apparent Power (S) is just = VI
Tegangan, Arus dan Daya sbg Fungsi Waktu
Tegangan, Arus (sefasa) dan Daya sbg Fungsi
Waktu
Tegangan, Arus (lag 90°) dan Daya sbg Fungsi
Waktu
Loads with Constant Impedance
V = IZ
Substituting…
P = I2Z cos
Q = I2Z sin
S= I2Z
Since… Z = R + jX = Z cos + jZ sin
P = I2R and Q = I2X
Review V, I, Z
If load is inductive then the Phase Angle
(Impedance Angle Z) is positive, If phase
angle is positive, the phase angle of the current
flowing through the load will lag the voltage
phase angle across the load by the impedance
angle Z.
Complex Power
S = P + jQ
S = VI cos + j VI sin
S = VI (cos + j sin)
S = VI ej
S = VI
I = I- dan V = V0o)
S = VI* Since S=√(P2 + Q2)
Rangk. Induktif
I-
V0°
S
PCos
Complex Power and Key
Relationship of Phase Angle to V&I
S = P + jQ
S = VI* (complex conjugate operator)
If V = V30o and I = I15o
THEN….. COMPLEX POWER SUPPLIED TO
LOAD = S = (V30o)(I-15o) = VI (30o-15o )
= VI cos(15o ) + jVI sin(15o )
NOTE: Since Phase Angle = v - i
S = VI cos() + jVI sin() = P + jQ
The Power Triangle
Aliran Daya Aktif
0VV
BILA SEPHASE DENGAN , BERARTI
DAYA LISTRIK DIBANGKITKAN (SUMBER
ADALAH GENERATOR) DAN MENGALIR MENUJU
SISTEM (ARUS KELUAR DARI TERMINAL
POSITIP)
P = Re (VI*) MEMPUNYAI TANDA POSITIP.
II
cosI
V
I
cosI V
Aliran Daya Aktif II
BILA MEMPUNYAI BEDA PHASE 180°
TERHADAP , BERARTI DAYA LISTRIK DISERAP
(SUMBER ADALAH MOTOR), DAN ARUS MENUJU
TERMINAL POSITIP DARI SUMBER.
P = Re (VI*) MEMPUNYAI TANDA NEGATIP.
0VV
cosIV
I
cosI
V
Aliran Daya Reaktif
0VV
DAYA REAKTIF SEBESAR I2 XL (DENGAN
TANDA POSITIP) DIBERIKAN PADA INDUKTANSI
ATAU INDUKTANSI MENYERAP DAYA REAKTIF.
ARUS TERBELAKANG (LAGGING) 90°
TERHADAP
Q = Im (VI*) MEMPUNYAI TANDA POSITIF
90II
V
I90
LX
VI
Aliran Daya Reaktif
II
DAYA REAKTIF SEBESAR I2 XC (DENGAN TANDA
NEGATIF) DIBERIKAN PADA KAPASITOR ATAU
SUMBER MENERIMA DAYA REAKTIF DARI
KAPASITOR.
ARUS MENDAHULUI (LEADING) 90° TERHADAP
Q = Im (VI*) MEMPUNYAI TANDA NEGATIF.
0VV V
I90
I V
Contoh soal 1
V = 1200o V
Z = 20-30o
Calculate current I, Power Factor (is it leading or
lagging), real, reactive, apparent and complex
power supplied to the load
BAB 1.2
Rangkaian Tiga Fasa (3-)
What are they?
Benefits of 3- Systems
Wye (Y) and delta () connections
One line diagram (of a balanced 3 phase
system)
What does Three-Phase mean?
A 3- circuit is a 3- AC-generation system
serving a 3- AC load
3 - 1- AC generators with equal voltage but
phase angle differing from the others by 120o
Balanced 3 phase systems
SISTEM TEGANGAN TIGA FASA YANG
SEIMBANG TERDIRI DARI TEGANGAN SATU
FASA YANG MEMPUNYAI MAGNITUDE DAN
FREKWENSI YANG SAMA TETAPI ANTARA
SATU DENGAN LAINNYA MEMPUNYAI BEDA
FASA SEBESAR 120°.
Tegangan & Arus 3 Fasa
Balanced
Same amplitude
120° phase diff.
Phase shift
ia lags ua angle j
Phase sequence
abc
Fasor Tegangan/Arus
Urutan Fasa abc
Seimbang: Ia+ Ib+ Ic=0
No return current
Losses reduced
No return conductor
a
b
c
Benefits of 3- circuits
GENERATION SIDE:
More power out
Constant power out (vs. pulsating sinusoidal)
………
LOAD SIDE:
Induction Motors (no starters required)
Common Neutral
A 3- circuit can have the negative ends of the
3- generators connected to the negative ends
of the 3- AC loads and one common neutral
wire can complete the system
If the three loads are equal (or balanced) what
will the return current be in the common neutral?
If loads are equal….
the return current can be calculated to be…
ZERO!
Neutral is actually unnecessary in a balanced
three-phase system (but is provided since
circumstances may change)
Wye (Y) and delta () connection
Delta ()
Hubungan Y
n : TITIK NETRAL
Vab=Vbc=Vca = VL : TEGANGAN ANTAR
FASA
Van=Vbn=Vcn = Vp : TEGANGAN FASA
Hubungan Arus dan Tegangan
Bila IL adalah Arus Saluran dan Ip adalah Arus
Fasa, maka berlaku :
IL = Ip
VL = √3 Vp
Dimana VL, Vp, IL , Ip adalah harga efektif dari
tegangan dan arus
Diagram Fasor (Hub. Y)
Sumber = Beban
Vab
30o
opcn 120VV
opan 0VV
o
pbn VV 120
Hubungan ∆
TITIK NETRAL tidak ada
Iab=Ibc=Ica = Ip : ARUS FASA
Ia=Ib=Ic = IL : ARUS SALURAN
Hubungan Arus dan Tegangan
Bila VL adalah Tegangan Antar Fasa dan Vp
adalah Tegangan Fasa, maka berlaku :
VL = Vp
IL = √3 Ip
Dimana VL, Vp, IL , Ip adalah harga efektif
dari tegangan dan arus
Diagram Fasor (Hub. ∆)
o
pbc II 120
o
pca II 120
Sumber ≠ Beban
o
pab II 0
bI
o30
o
pbc II 120
o
pca II 120
o
pab II 0
aI
o30
One-Line Diagram (of a BALANCED 3 PHASE SYSTEM)
since all phases are the same (except for phase
angle) and loads are typically balanced only one
of the phases is usually shown on an electrical
diagram… it is called a one-line diagram
Typically include all major components of the
system (generators, transformers, transmission
lines, loads, other [regulators, swithes])
Daya pada Rangkaian 3 Fasa
=uiR
=uiL
Daya 3 Fasa
Daya 3 Fasa
ptotal(t)= pa(t)+ pb(t)+ pc(t)
Daya 3 fasa = Jumlah Daya tiap-tiap Fasa
ptotal(t)=constant
If voltages and currents balanced
cosj need not be zero
Constant ptotal(t) => constant torque
ppp
ppp
SinIVQ
CosIVP
j
j
j
j
3
3
3
3
LpL
p IIV
V ; 3
3 ; L
pLp
IIVV
Untuk Sistem 3 fasa seimbang
φp adalah sudut antara
Arus Fasa (Lagging) dan
Tegangan Fasa
Hubungan Y
Hubungan ∆
Rumus Daya 3 Fasa
pLL
pLL
SinIVQ
CosIVP
j
j
j
j
3
3
3
3
LLIV
QPS
3
22
Watt
Var
VA
TUGAS
PELAJARI CONTOH SOAL (HAND OUT)