Sistem Listrik AC Tiga Fasa - Spada UNS

Sistem Listrik AC Tiga Fasa

• Pertemuan 4 Teknik Tenaga Listrik

MENGAPA LISTRIK AC ?

• Transmisi listrik harus

menggunakan tegangan yang sangat tinggi agar rugi-rugi

rendah

• Untuk distribusi dan pemakaian tegangan diturunkan kembali

menggunakan trafo. Trafo

bekerja untuk tegangan AC tidak

bisa DC

MENGAPA TIGA FASA ?

• Daya sesaat yang dikirimkan ke beban akan

“melonjak tinggi”pada sistem 1 fasa. Pada Sistem tiga fasa daya yang dikirimkan lebih

“stabil/ steady” (ingat mesin mobil dengan

multi silinder)

MENGAPA TIGA FASA ?

• Untuk mengirimkan daya yang sama, ukuran konduktor/ kabel

dan komponen lainnya lebih kecil dibanding dengan

menggunakan 1 fasa.

3 Ph ase

cv 1

Ph as e

MENGAPA TIGA FASA ?

• Daya listrik yang dibangkitkan pada pembangkit adalah fasa banyak dengan frekuensi 50 Hz atau 60

Hz

LISTRIK DI INDONESIA

• Tegangan rms fasa = 220 V →Vp

• Tegangan fasa ke fasa (Line Voltage) = 380 V

→ VL.

• frekuensi 50 Hz

LISTRIK DI AMERIKA

• Tegangan rms fasa = 115 V

• frekuensi 60 Hz

Listrik 1 fase

• Sistem listrik yang masuk ke rumah kita, jika menggunakan sistem listrik 1 fase biasanya terdiri atas 3 kabel (Ideal):

– Kabel fase, yang merupakan sumber listrik bolak-balik → kabel yang membawa tegangan dari pembangkit tenaga listrik (PLN misalnya); kabel ini biasanya dinamakan kabel panas (hot), dapat dibandingkan seperti kutub positif pada sistem listrik arus searah.

– Kabel Netral. Kabel ini pada dasarnya adalah kabel acuan tegangan nol, yang biasanya

disambungkan ke tanah di pembangkit tenaga listrik dapat dibandingkan seperti kutub negatif pada sistem listrik arus searah; jadi jika listrik ingin dialirkan ke lampu misalnya, maka satu kaki lampu harus dihubungkan ke kabel fase dan kaki lampu yang lain dihubungkan ke kabel netral

– Kabel Tanah/Ground. Kabel ini adalah acuan nol di lokasi pemakai, yang biasanya disambungkan ke tanah di rumah pemakai; kabel ini benar-benar berasal dari logam yang ditanam di tanah dekat rumah kita; kabel ini merupakan kabel pengamanan yang biasanya disambungkan ke badan

(chassis) alat2 listrik di rumah untuk memastikan bahwa pemakai alat tersebut tidak akan mengalami kejutan listrik.

• Listrik 3 Phase adalah jaringan listrik yang menggunakan tiga kawat Phase (R,S,T) dan satu kawat neutral (N) atau sering dibilang kawat ground.

Menurut istilah Listrik 3 Phase terdiri dari 3 kabel bertegangan listrik dan 1 kabel neutral. Umumnya listrik 3 Phase bertegangan 380 volt yang banyak digunakan Industri atau pabrik.

• Listrik 3 fasa adalah listrik AC (Alternating Current) yang menggunakan 3 kawat penghantar yang mempunyai tegangan pada masing-masing

Phasenya sama, tetapi berbeda dalam sudut curvenya sebesar 120 derajat.

• Ada 2 macam tegangan listrik yang dikenal dalam sistem 3 phase ini, yaitu : - Tegangan antar phase (Vpp : voltage phase to phase atau ada juga yang

menggunakan istilah Voltage line to line atau VL)

- Tegangan phase ke neutral (Vpn : Voltage phase to neutral atau Voltage line

to neutral atau Vp).

V_bn

V_cn V_an

a b

n

c n

V_an V_bn

V_cn

- Vab/VL : Tegangan antar phase (Vpp : voltage phase to

phase atau ada juga yang menggunakan istilah Voltage

line to line) 380 V

- Van/Vp : Tegangan phase ke neutral (Vpn : Voltage phase to

neutral atau Voltage line to neutral). 220V

220V AC dll

Rangkaian Tiga Fase Seimbang

• Sistem Tenaga Listrik pada umumnya disalurkan menggunakan sistem 3 fase,

karena penyaluran daya dapat lebih efektif.

• Pada sistem tenaga listrik 3 fase, idealnya

daya listrik yang dibangkitkan, disalurkan dan diserap oleh beban semuanya seimbang,

P pembangkitan = P pemakain,

Kelebihan 3 fase dibanding 1 fase

• Kapasitas penyaluran daya lebih besar dibandingkan 1 fase.

• Dengan nilai daya yang sama, harga peralatan 3 fase lebih ekonomis

• Listrik 3 fase lebih stabil daripada 1 fase.

Terhubung Bintang (Wye, Y, Star, Bintang)

• Pada hubungan bintang (Y, wye), ujung-ujung tiap fase dihubungkan menjadi satu dan menjadi titik netral atau titik bintang.

• Dengan adanya saluran / titik netral maka besaran tegangan fase dihitung terhadap saluran / titik netralnya, juga membentuk sistem tegangan 3 fase yang seimbang dengan magnitudenya (akar 3 dikali magnitude dari tegangan fase).

• Sedangkan untuk arus yang mengalir pada semua fase mempunyai nilai yang sama,

ILine = Ifase

Ia = Ib = Ic

Terhubung Bintang (Wye, Y)

Relationship between line and phase value in the Y- Connection

• The voltages and currents in a given phase are called phase quantities.

• The voltages between lines and currents in the lines connected to the generators are called line quantities.

• Line voltage,

• Phase voltage,

V_ab ; V_bc ; V_ca V_an ; V_bn ; V_cn

• The phase voltage in this generator are given by:

• The relationship between the magnitudes of the line-to-line(line) voltage and line-to-neutral (phase) voltageis:

− V_BN V_AB = V_AN

= V_AN0 −V_BN −120

=V_AN0 −V_AN −120

= V_AN(10 −1 −120)

= V_AN((1+ j0) −(−0.5 − j0.866))

= V_AN (1.5 +j0.866)

= V_AN(1.73230)

V_L leads V_φ by 30°

= 3V_AN30

V_AB

V_BC = V_BN −V_CN

= 3V_BN −90

V

_CA

= V

_CN

− V

_AN

= 3V

_CN

150

• Since the load connected to this generator is assumed to be resistive, the current in each phase of the generator will be at the same angle.

• The current in each phase:

• The current in any line is the same as the current in the corresponding phase which is:

•Figures show the line-to-lineand phase voltages forthe

Y-connection.

(abc phase sequence)

a) Phasor diagram of the line and phase voltages of a three-phase generator; (b) demonstrating that the vector sum of the line voltages of a three-phase system is zero.

Phase sequence (Y-connected generator)

ABC (Positive) Phase sequence

Phase voltage (in phasor):

= E

_AN

0(reference)

= E

_BN

−120

E

_AN

E

_BN

E

_CN

= E

_CN

120

Line voltage (in phasor):

E

_CA

= E

_CA

120

= E

_AB

0(reference)

= E

_BC

 −120

E

_AB

E

_BC

16

= E

_BN

120

E

_BN

E

_CN

= E

_CN

 −120

= E

_AN

0(reference) E

_AN

Line voltage (in phasor):

= E

_BC

120

E

_BC

E

_CA

= E

_CA

 −120

= E

_AB

0(reference) E

_AB

17

Phase sequence (Y-connected generator)

ACB (Negative) Phase sequence

Phase voltage (in phasor):

Y-Connected Generator with a Y-Connected Load

• Loads connected with three-phase supplies are of two types: the Y and the ∆.

• If a Y-connected load is connected to a Y-connected generator, the system is symbolically represented by Y-Y

• If the load is balanced, the neutral connection can be removed without affecting the circuit in any manner; that is, if Z1 =Z2 =Z3, then I^Nwill be zero

The currents flowing in the three phases are:

I_A I_B I_C

Contoh 1

Calculate the line currents in the three-wire Y-Y system as shown below

SOLUTION

• Due to the three-phase circuit isbalanced; we may replace it with its single-phase equivalent circuit

• Phase “a”equivalent circuit:

= = 6.81 − 21.8

16.15521.8

I I Z

Aa

T

= V

_AN

Aa

;

Z

_T

= (5 − j2) + (10 + j8) = 16.15521.8

1100

• Since the source voltage are in positive (ABC) phase sequence, the line currents are also in positive sequence:

I

_Bb

= I

_Aa

−120

= 6.81 −141.8A I

_Cc

= I

_Aa

− 240

= 6.81 − 261.8 = 6.81 98.2 A

Contoh 2 → Latihan

(a) Hitung Arus Line.

(b) Hitung tegangan line-line dan tegangan fasa.

Balanced Y-Δ Connection

• There is no neutral connection for the Y-∆system as shown below.

• Any variation in the impedance of a phase that produces an unbalanced system will simply vary the line and phase currents

of the system.

• For a balanced load, Z₁ = Z₂ = Z₃ =Z_Δ

• The voltage across each phase of the load is equal to the line voltage of the generator for a balanced or an unbalanced

load: Vɸ =E_L.

• Assuming the positivesequence:

– the phase voltageare:

– The line voltages are:

– The phase currentis

V_ab = 3V_p30 =V_AB V_bc= 3V_BN −90 =V_BC ^{Vca= 3VCN150 =VCA}

Contoh 3

• A balanced positive sequence Y-connected source with V_AN

= 10010V is connected to a -connected balanced load (8+j4)  per phase. Calculate the phase and line currents.

SOLUTION

• Balanced WYE source, V_AN = 10010V

• Balanced DELTA load, Z_= 8 + j4

• Phasecurrents:

V

_AB

= 3 V

_AN

30 = V

_ab

= 3 (10010)30

V

_ab

= 173.210+ 30 = 173.240V

ab

Δ

I = V Z

_ab

 I

ab

= 173.240  = 19.3613.43 A

8 + j4

I

_bc

= I

_ab

 −120 = I

_ab

13.43 −120

I

_bc

= 19.36 −106.57 A

I

_ca

= I

_ab

120 = 19.3613.43+120

I

_ca

= 19.36133.43 A

Linecurrents:

I

_Aa

= 3 I

_ab

 − 30

= 3 (19.36) 13.43 − 30

I

_Aa

= 33.53  −16.57 A

 I

_Bb

= I

_Aa

 −120 = 33.53  −136.57 A

 I

_Cc

= I

_Aa

 +120 = 33.53 103.43 A

Contoh 4

(a) Hitung Arus Line.

(b) Hitung tegangan line-line dan tegangan fasa.

Terhubung Segitiga (delta, Δ, D)

• Pada hubungan segitiga (delta, Δ, D) ketiga fase saling dihubungkan sehingga membentuk hubungan segitiga 3 fase

• Dengan tidak adanya titik netral, maka besarnya

tegangan saluran dihitung antar fase, karena tegangan saluran dan tegangan fasa mempunyai besar magnitude yang sama, maka:

Vline = Vfase

Terhubung Segitiga (delta, Δ, D)

I fase

I saluran

Relationship between line and phase value in the ∆ - Connection

• Phase voltage, V_ab ; V_bc ; V_ca

• The phase voltage in this generator are given by:

• The line-to-line voltage between any two lines will be the same as the voltage in the corresponding phase. So :

▪Unlike the line current for the Y-connected generator, the line currentfor the ∆-connected system is not equal to the phase current. The relationship between the two can be found by applying Kirchhoff’s current law at one of the nodes and solving for the line current in terms of the phase current; that is, at node A,

I_BA = I_Aa +I_AC or

I_Aa = I_BA - I_AC = I_BA +I_CA

35

In 3-phase system, for ∆-connected, the current that flow from one phase to another is called a phase current.

I_BA– phase A current I_CB– phase B current I_AC –phase C current

36

Definition of Phase Current

In 3-phase system, for ∆-connected, the current that flow through the line is called a line current.

I_Aa– line A current I_Bb – line B current I_Cc –line C current

37

Definition of Line Current

Line current:

I_Aa ; I_Bb ;I_Cc

Phase current:

for generator:

I_BA ; I_AC ;I_CB

38

∆ -Connected system(Generator)

Line current:

I_Aa ; I_Bb ;I_Cc Phase current:

for load:

I_ab ; I_bc ;I_ca

39

∆ -Connected system(Load)

Current in ∆ -Connected system (Generator Side)

For 3-phase ∆-connected system (generator), if the phase current I_BA is taken as the reference, so

I BA = I BA 0

I CB = I CB −120

I AC = I AC 120

40

I Aa = 3I BA  − 30 A

Current in ∆ -Connected system (Generator Side)

By applying Kirchhoff’s Current Law, the line current can be written as

I

_Aa

= I

_BA

−I

_AC

= I

_BA

0  − I

_BA

120

= I

_BA

(10 −1120)

= I

_BA

(1+ j0 − (−0.5 + j0.866))

= I

_BA

(1.5 − j0.866)

= I

_BA

(1.732 − 30)

41

Current in ∆ -Connected system (Generator Side)

With the same method,

I Bb = I CB − I BA

= 3I CB  −150

and

I Cc = I AC − I CB

= 3I AC 90

42

Current in ∆ -Connected system (Load Side)

For 3-phase ∆-connected system (load), if the phase current I_ab is taken as the reference, so

I ab = I ab  0

I bc = I bc −120 

I ca = I ca 120

43

I Aa = 3I ab  − 30 A

Current in ∆ -Connected system (Load Side)

By applying Kirchhoff’s Current Law, the line current can be written as

I

_Aa

=I

_ab

−I

_ca

= I

_ab

0 − I

_ab

120

= I

_ab

(10−1120)

= I

_ab

(1+ j0 − (−0.5 + j0.866))

= I

_ab

(1.5 − j0.866)

= I

_ab

(1.732 − 30)

44

I Bb = I bc − I ab

= 3I bc  −150

I Cc = I ca − I bc

= 3I ca 90

45

Current in ∆ -Connected system (Load Side) With the same method,

and

∆ -Connected system

The relationship between the line current and the phase current can be represented as

Where; I_L : linecurrent I_φ = I_p : phase current

I L = 3I φ  − 30

I

_L

lags I

_φ

by 30°

46

▪

The phasor diagram is shown below for a balanced load.

▪ In general, line current is:

= 3I

_BA

 −30  I

_Aa

I

_Bb

= 3I

_CB

− 150 

= 3I

_AC

 90  I

_Cc

47

Phase current Line current

48

Phase sequence (Δ -connected generator)

▪

Even though the line and phase voltages of a ∆ -connected system are the same, it is standard practice to describe the phase sequence in terms of theline voltages

▪In drawing such a diagram, one must take care to have the sequence of the first and second subscripts the same

▪ In phasor notation,

E_AB = E_AB 0^o = E_p 0^o E_BC = E_BC −120^o = E_p −120^o E_CA = E_CA 120^o = E_p 120^o

ABC (positive) phase sequence

▪

In phasor notation, E_AB = E_AB 0^o E_BC = E_BC 120^o E_CA = E_CA -120^o

ACB (negative) Phase sequence

49

Phase sequence (Δ -connected generator)

• Because the load is resistive, the phase current are given by:

• The relationship between the magnitudes of the line and phase currentsis:

• Figures show theline and phase currents forthe

∆-connection.

(abc phase sequence)

Contoh5 → (∆ -Connected Generator with a ∆ - Connected Load)

• A balanced delta connected load having an impedance 20 - j15  is connected to a delta connected, positive sequence

generator having V_AB = 3300V. Calculate the phase currentsof the load and the line currents.

 Z

_Δ

= 20 − j15  = 25 − 36.87

 V

_AB

= 3300 V

SOLUTION

= I

_ab

 +120 + 36.87 = 13.2156.87A I

_ca

I

_bc

= I

_ab

 −120 + 36.87 = 13.2 - 83.13A

= 13.236.87A 3300

I

Z

Δ

25 − 38.87

ab

= V

_ab

=

= ( 13.2  36.87 ) ( 3 − 30  )

= 22.866.87 A

I

_Bb

= I

_Aa

 −120 + 6.87 = 22.86 -113.13 A I

_Cc

= I

_Aa

 +120 + 6.87 = 22.86126.87 A

3 − 30

I

_Aa

= I

_ab

• Phase currents of the load

• Line currents.

Contoh6 → (∆ -Connected Generator with a Y- Connected Load) → Latihan

Impedan fasa = 40 + j25  dengan sumber-benan setimbang, dengan urutan positif V line = 210 V. Hitung Arus Line dan Arus Fasa.

Summary of relationships in Y and ∆ -Connections

Y-Connections ∆ -Connections

Voltage Magnitude

C u rren t Magnitude abc phase sequence

acb phase sequence

Perbandingan hub Delta-Y

Delta Wye Terdapat 1 jenis tegangan, yaitu

tegangan fase ke fase

Terdapat dua nilai tegangan, fase ke fase dan fase netral

Tidak perlu titik ground Harus memasang ground pada netralnya

Power in three-phase circuit (Phase Quantities)

• Apply to each phase of a Y- or ∆- connected three-phase load.

1) Real, P

P = 3 V

_

I

_

cos 

2) Reactive, Q

Q = 3 V

_

I

_

sin 

3) Apparent, S

S = 3 V

_

I

_

2

Q = 3I Z sin 

S = 3I

²

Z

The angle θ is again the angle between the voltage and current in any phase of the load ( it is the same in all phases), and the power factor of the load is the

cosine of the impedance angle θ

2

P = 3I Z cos 

Power in three-phase circuit (Line Quantities)

3) Apparent, S

S = 3V

_LL

I

_L

• Y-Connected load:

1) Real, P

The power consumed by a load is given by

P = 3V_I_cos

 I

_L

= I

_

V

_LL

= 3 V

_

3

LL

) I

L

cos 

P = 3( V



P = 3V

_LL

I

_L

cos 

2) Reactive, Q

Q = 3V

_LL

I

_L

sin 

• ∆-Connected load:

1) Real, P

P = 3V_I_cos



I_L = 3I_ V_LL =V_

LL 3

P = 3V ( I_L

) cos



 _{P = 3V I cos}



LL L

2)Reactive, Q

Q = 3V

_LL

I

_L

sin 

3)Apparent, S

S = 3 V

_LL

I

_L

Contoh 7

A 208-V three-phase power system is shown in above figure. It consists of an ideal 208-V Y-connected three-phase generator connected through a three-

phase transmission line to a Y-connected load. The transmission line has an impedance of 0.06+ j0.12Ωper phase, and the load has an impedance of 12

+ j9Ω per phase. Find

(a) the real, reactive and apparent powers consumed by the load (b) the power factor of the load

SOLUTION

(a) The real power consumed by the load is

(b) The load power factor is

=3V_L I_L cos

P_load

=3(119.1V)(7.94A) cos36.9

= 2270W

The reactive power consumed by the load is

= 3V_LI_L sin



Q_load

= 3(119.1V)(7.94A) sin 36.9

=1702 var

The apparent power consumed by the load is

= 3V_LI_L S_load

= 3(119.1V)(7.94A)

= 2839VA

PF _load= cos

= cos 36.9

= 0.8lagging

Contoh 8 → Latihan

Hitung

(a) S, Q, dan P (b) PF

Impedans, Admitans, Resistans, Reaktans

Impedans dan Admitans

• Impedansi adalah perbandingan fasor tegangan V dan fasor arus I pada suatu elemen kutub dua dengan adanya sinyal masukan gelombang sinusoidal dalam keadaan setimbang atau mantap atau tunak (steady state).

• Admitansi merupakan kebalikan dari Impedansi.

• Impedansi dan admitansi bukan merupakan fasor.

• Impedansi dapat dihubungkan seri atau paralel seperti halnya pada

Resistansi.

Impedans dan Admitans

• Impedansi Z = V / I [Ohm]

Z = R ± jX

R: resistansi;

X : induktans : resultan antara kapasitans dan induktans

• Admitansi Y = I / V [Mho]

Y = 1/ Z Y = G ± jB

G:konduktansi = 1/R; B:suseptansi = 1/X

Pengertian Impedansi Dan Admitansi

Pengertian Impedansi Dan Admitansi

◼

Elemen L

Pengertian Impedansi Dan Admitansi

◼

Elemen C

• Impedansi Z = V / I [Ohm]

Z = R ± jX

TERIMAKASIH