Analisa Sistem Tenaga

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)