ANALISIS SISTEM TENAGA LISTRIK (KONSEP DASAR)
[email protected] Lab. Simulasi Sistem Tenaga Listrik
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. 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… 2 P = I Z 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 V0°
S =
P + jQ
S =
VI cos + j VI sin
I-
S =
VI (cos + j sin) j S = VI e S = VI I =
Rangk. Induktif
I- dan V = V0o)
S = VI*
Since S=√(P
2 +
Q2)
Cos
P S
Complex Power and Key Relationship of Phase Angle to V&I = P + jQ
S
= VI* (complex conjugate operator)
S
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 I I
V V0
I cos
I BILA I cos SEPHASE DENGAN V , BERARTI DAYA LISTRIK DIBANGKITKAN (SUMBER ADALAH GENERATOR) DAN MENGALIR MENUJU SISTEM (ARUS KELUAR DARI TERMINAL POSITIP) P = Re (VI *) MEMPUNYAI TANDA POSITIP.
V
Aliran Daya Aktif I I
I cos
V
V V0
I
BILA I cos MEMPUNYAI BEDA PHASE 180° TERHADAP V , BERARTI DAYA LISTRIK DISERAP (SUMBER ADALAH MOTOR), DAN ARUS MENUJU TERMINAL POSITIP DARI SUMBER. P = Re (VI *) MEMPUNYAI TANDA NEGATIP.
Aliran Daya Reaktif I I 90
V V0
V
X L
I
90
DAYA REAKTIF SEBESAR I2 XL (DENGAN TANDA POSITIP) DIBERIKAN PADA INDUKTANSI ATAU INDUKTANSI MENYERAP DAYA REAKTIF. ARUS I TERBELAKANG (LAGGING) 90° TERHADAP V
Q = Im (VI *) MEMPUNYAI TANDA POSITIF
Aliran Daya Reaktif I I I 90
V V0
DAYA REAKTIF SEBESAR I2 XC (DENGAN TANDA NEGATIF) DIBERIKAN PADA KAPASITOR ATAU SUMBER MENERIMA DAYA REAKTIF DARI KAPASITOR. ARUS I MENDAHULUI (LEADING) 90° TERHADAPV
Q = Im (VI *) MEMPUNYAI TANDA NEGATIF.
V
Contoh soal 1 = 1200o V
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 120 o
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 a c
Urutan Fasa abc
Seimbang: Ia+ Ib+ Ic=0
No return current Losses reduced No return conductor
b
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) Vcn
V p 120o V ab
30o
Van
V bn
V p 120
o
Sumber = Beban
V p 0
o
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 I = √3 Ip
L
Dimana VL, Vp, IL , Ip adalah harga efektif dari tegangan dan arus
Diagram Fasor (Hub. ∆) I ca
I p 120
I ca
o
I ab
o
I ab
I p 120o
I b 30
I bc
o
I p 120
Sumber
I p 0
30
o
I bc
≠
I p 120o
Beban
I p 0o
o
I a
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
Untuk Sistem 3 fasa seimbang
P 3j
3V p I p Cosj p
Q3j 3V p I p Sinj p
V p
V p
V L
;
3
V L
;
I p
I p
I L
I L 3
φ p adalah sudut antara Arus Fasa (Lagging) dan Tegangan Fasa
Hubungan Y
Hubungan ∆
Rumus Daya 3 Fasa
P 3j
3V L I L Cosj p
Watt
Q3j
3V L I L Sin j p
Var
2
S P
Q
3 V L I L
2 VA
