Handout Rangkaian Listrik II

RANGKAIAN LISTRIK II

Dosen : Rahmad Hidayat, S.T., M.T.

Bagian - 2

Rangkaian Tiga-Phase

Dosen : Rahmad Hidayat ST MT Dosen : Rahmad Hidayat,ST,MT

Analisa

Analisa Rangkaian

Rangkaian

Rangkaian

Rangkaian Poliphase

Poliphase

2

2.1

.1 Sistem

Sistem Poliphase

Poliphase

2

2.2

.2 Notasi

Notasi

2

2.2

.2 Notasi

Notasi

2

2.3

.3 Sistem

Sistem Tiga

Tiga Kawat

Kawat Phase Tunggal

Phase Tunggal

2

2 4

4 Hubungan

Hubungan

Y

Y

tiga

tiga phase

phase

2

2.4

.4 Hubungan

Hubungan

tiga

tiga phase

phase

2

2.5

.5 Hubungan

Hubungan Delta ( )

Delta ( )

Y

Y

−

Δ

2

2.1

2.1 Sistem

Sistem Poliphase

Poliphase

Sistem poliphase : sistem dengan sumber poliphase p p g p p Sumber tunggal (Vs)

Perhatikan tegangan sesaat dapat berharga nol Æ Daya sesaat akan berharga nol

V

T

2

T

3

T

V

t

o

o

o

V

1 2 3

,

,

Vs

Vs

Vs

Multi sumber ( )

Berbeda phase120o _{satu sama lain}

Æ Daya sesaat tidak akan pernah nol.

V

t

T

3

T

2

V

V

V

V

+

+

2.1

2.1 Sistem

Sistem Poliphase

Poliphase

3 2

1

Vs

Vs

Vs

V

=

+

+

• Terhindar dari daya sesaat yang berharga nol.

• Daya sumber dapat diberikan lebih stabil.

• Dapat memberikan banyak level tegangan output

2

2.2

.2 Notasi

Notasi

A 8

a b

A 4

c d e f

A 5 A 2 = de I ? = cd I ef I A 3 I

A

3

A

8

A

5

=

+

I

_d

I

_d

=

−

Titik c :

g _h i j

A 5 ij I A 6

− Ifj =3A

A

3

,

A

8

A

5

=

+

I

_cd

I

_cd

=

Titik f :

A

7

,

A

3

A

4

+

=

=

_ef

ef

I

I

k l

A

2 10 A

,

_ef

ef

Titik j :

A

7

,

A

10

A

4

A

3

=

=

+

_ij

ij

I

2

2.2

.2 Notasi

Notasi

V

V

=

100

∠

0

0

o

c

V

V

an

100

∠

0

V

V

_bn

=

100

∠

−

120

0

V

V

=

100

∠

240

0

o

a

n

+

+

− −

V

V

_cn

=

100

∠

−

240

V

V

V

_ab

=

_an

+

_nb

o

a

+

+

−

V

V

V

V

_{an bn}

0 0

120

100

0

100

∠

−

∠

−

=

−

=

o

b

Tegangan titik a

terhadap b a +; b -;

V

0

30

2

.

173

∠

=

T d li fik ? ( k di h ) Serupa, I_ab menunjukkan arus dari a ke b.

2

2.3

.3 Sistem

Sistem Tiga

Tiga kawat

kawat Phase

Phase--tunggal

tunggal

2

2.3

.3 Sistem

Sistem Tiga

Tiga kawat

kawat Phase

Phase tunggal

tunggal

Fungsi: memungkinkan perangkat elektronik rumah tangga beroperasi d d j i ilih l l t

pada dua jenis pilihan level tegangan

a

Karakteristik tegangan

nb

an

V

V

=

nb an

ab

V

V

V

=

2

=

2

Sumber

1-phase

3-kawat

a

n

b

3 kawat

b

a

Perangkat elektronik rumah tangga dapat beroperasi

1

V

a

n

V

110

220

V

V

V

=

∠

∠

∠V = −∠Vb

atau

Karakteristik Phase

2

V

b

nb an

V

V

=

∠

∠

∠Van = ∠Vbn

0

=

∠

+

2

2.3

.3 Sistem

Sistem Tiga

Tiga kawat

kawat Phase

Phase--tunggal

tunggal

Karakteristik arus

A

2

2.3

.3 Sistem

Sistem Tiga

Tiga kawat

kawat Phase

Phase tunggal

tunggal

aA bB

Nn

I

I

I

=

+

V

V

1

V

A

a

p

Z

p bB

Z

V

I

=

1

p Aa

Z

V

I

=

1

0

=

I

1

V

N

n

p

Z

0

=

Nn

I

Tidak ada arus pada kawat netral.

B

b

S

ISTEM

P

HASE

- T

UNGGAL

z

Single phase is used primarily only in low voltage,

l

i

h

id

i l d

low power settings, such as residential and some

commercial.

z

Single phase transmission used for electric trains in

z

Single phase transmission used for electric trains in

2.3

2.3 Sistem

Sistem Tiga

Tiga kawat

kawat Phase

Phase--tunggal

tunggal

Contoh 9.1 (P242)

Tentukan daya yg diberikan ke beban

Ω

1

,

50Ω

100

Ω

₂₀ + _j10Ω

y yg dan the

Tentukan daya yang hilang di ketiga

Ω

3

Ω

50

20

Ω

V

0 115∠ 0

1

I

rms

Ω

1 3Ω

1

Ω

saluran yg dilewati , dan

Tentukan efisiensi transmisinya ?

Ω

3

Ω

100

Ω

10

j

10

Ω

V 0 115∠ 0

2

I

3

I

Ω

10

j

η = total daya yang diserap beban total daya yang dihasilkan sumber rms

Gunakan KVL :

Gunakan KVL :

(

)

3

(

)

0

50

1

V

0

115

∠

0

+

Ω

⋅

₁

+

Ω

⋅

₁

−

₂

+

Ω

₁

−

₃

=

−

I

I

I

I

I

(

20

+

j

10

)

Ω

⋅

I

₂

+

100

Ω

⋅

(

I

₂

−

I

₃

)

+

50

Ω

(

I

₂

−

I

₁

)

=

0

(

j

)

₂

(

_{2 3}

)

(

_{2 1}

)

(

)

100

(

)

1

0

3

V

0

115

∠

0

+

Ω

⋅

₃

−

₁

+

Ω

₃

−

₂

+

Ω

⋅

₃

=

−

I

I

I

I

I

Susun ke dalam matriks :

⎥

⎤

⎢

⎡

∠

⎥

⎤

⎢

⎡

⎥

⎤

⎢

⎡

+

−

−

0

1

115

0

Sehingga dapat dihitung :

Sehingga dapat dihitung :

A

47

.

24

389

.

9

A

83

.

19

24

.

11

0 2 0 1

−

∠

=

−

∠

=

I

I

I

₁

I

₂

=

2

.

02

2

.

27

o

A

A

12

.

2

08

.

1

2 3 o

I

I

=

rms rms rms rms

A

80

.

21

37

.

10

0

3

=

∠

−

I

Daya rata-rata yang diberikan ke tiap beban adalah :

A

3

.

2

947

.

0

1 3 o

I

I

=

rms rms

⎪⎪

⎫

=

⋅

−

=

_{1 2} 2

50

206

W

50

I

I

P

y y g p

⎪

⎪

⎭

⎪⎪

⎬

=

⋅

=

=

⋅

−

=

+

20

1763

W

W

117

100

2 2 10 20 2 2 3 100

I

P

I

I

P

j W 2086 =

Daya yang hilang di ketiga kawat :

Daya yang hilang di ketiga kawat :

⎪ ⎪⎪ ⎬ ⎫ = ⋅ = = ⋅ = W 108 1 W 126 1 2 2 2 3 2 1 I P I P bB aA

W

237

=

Total daya hilang

⎪ ⎪ ⎭ = ⋅ − = ⋅

= I 2 3 I₃ I₁ 2 3 3W

P_{nN nN}

%

100

×

=

generated

power

total

load

the

to

delivered

Power

Efisiensi Transmisi , η

generated

power

total

(

11

24

)

cos

19

83

115

(

10

37

)

cos

21

80

115

0

+

0

=

P

Total daya yang dihasilkan kedua sumber (power generated) :

(

)

(

)

W

2323

W

1107

W

1216

80

.

21

cos

37

.

10

115

83

.

19

cos

24

.

11

115

=

+

=

+

=

sources

P

% 8 89 % 100 W 2086

Efisiensi Transmisi _{100% 89.8%}

W

2323 × =

=

S

ISTEM

POL

I

PHASE

|

Two Phase System

:

|

Two Phase System

:

y

A generator consists of two coils placed perpendicular to each

other

y

The voltage generated by one lags the other by 90

°

.

|

Three Phase System

:

y

A generator consists of three coils placed 120

°

apart

y

A generator consists of three coils placed 120 apart.

y

The voltage generated are equal in magnitude but, out of

phase by 120

°

.

3 PHASE 4 K

AWAT

D

D

EFINISI

|

4 kawat

y

3 phase “aktif” : A, B, C

y

1 “ground”, atau “neutral”

K d

|

Kode warna

y

Phase A

Merah

y

Phase B Hitam

Phase B Hitam

y

Phase C

Biru

y

Neutral Putih atau Abu-abu

ECE ECE

G

TIGA PHASE

G

ENERATOR

TIGA PHASE

|

2-pole (North-South)

rotor turned by a

“prime mover”

|

Sinusoidal voltages

are induced in each

stator winding

stator winding

ECE 4411

T

HREE

P

HASE

C

IRCUITS

T

HREE

-P

HASE

C

IRCUITS

|

In three phase circuits the 3 voltages

22

|

In three-phase circuits the 3 voltages

sources are 120° apart

|

Polyphase generation and transmission of

|

Polyphase generation and transmission of

electricity is more advantageous and

economical

(1) three-phase instantaneous power is

constant over time

Lect

(2) single-phase line losses are 50% greater

than three-phase losses (for the same load

Lect ure 9

B

ALANCED

S

YSTEM

B

ALANCED

S

YSTEM

23

|

A

balanced

system is one in which the 3

sinusoidal voltages have the same

magnitude and frequency and each is 120°

magnitude and frequency, and each is 120°

out-of-phase with the other two

( )

cos

)

(

t

V

t

v

( )

ω

(

)

(

−

°

)

(

)

=

=

120

cos

)

(

cos

)

(

t

V

t

v

t

V

t

v

M bn

M an

ω

ω

Lect

(

−

°

)

=

(

+

°

)

=

cos

240

cos

120

)

(

t

V

t

V

t

v

_{cn M}

ω

_M

ω

T

HREE

P

HASE

V

OLTAGES

T

HREE

-P

HASE

V

OLTAGES

a

24

V_an +

–

+ V_b

b

– V_bn

+ – V_cn

c

n

Lect

Balanced If:

V

_an

=V

_rms

∠

0°

V

_bn

=V

_rms

∠

-120°

V

_cn

=V

_rms

∠

-240°

25

A

DVANTAGES OF

3

φ

P

OWER

z

Can transmit more power for same amount of

p

wire (twice as much as single phase).

z

Total torque produced by 3

φ

machines is

t

t

l

ib

ti

27

constant, so less vibration.

z

Three phase machines use less material for same

power rating.

power rating.

z

Three phase machines start more easily than

A

DVANTAGES OF

3

φ

P

OWER

IMPORTANCE OF THREE PHASE SYSTEM

IMPORTANCE OF THREE PHASE SYSTEM

|

All electric power is generated and distributed in

three phase.

y

One phase, two phase, or more than three phase input can

be taken from three phase system rather than generated

independently.

IMPORTANCE OF THREE PHASE SYSTEM

|

Uniform power transmission and less vibration of three

phase machines

phase machines.

y

The instantaneous power in a 3

φ

system can be constant (not

pulsating).

Hi h

t

f

t

d

t

i ll

IMPORTANCE OF THREE PHASE SYSTEM

|

Three phase system is more economical than the single

phase

phase.

y

The amount of wire required for a three phase system is less

than required for an equivalent single phase system.

3

PHASE POWER

|

A single phase generator is an alternator with a

g

p

g

single set armature coil producing a single

voltage waveform.

A th

h

lt

t

h

th

t f il

|

A three –phase alternator has three sets of coils

spaced at 120

o

apart and generates three sets of

W

YE

|

Neutral conductor is

|

Allows each phase to

connected between all

3-phase conductors

p

D

ELTA

|

Neutral conductor is

|

High leg serves only

3-centered between

two-phase conductors

g

g

y

phase loads and

cannot be used with

the neutral

T

RANSFORMERS

|

Wye is typical for office

|

Wye is typical for office

buildings and shopping centers

l

i

ll

d i

|

Delta is usually used in

2

2.4

.4 Hubungan

Hubungan

Y

−

Y

tiga

tiga--phase

phase

Karakteristik arus : Karakteristik arus :

aA

I

A

B

bB

I

b

o

p

Z

A

+

+

a

o

b

−

−

Z

p

Z

p

N

n

−

Z

C

cC

I

+

Z

P

2

2.4

.4 Hubungan

Hubungan

Y

−

Y

tiga

tiga--phase

phase

Z

Pertimbangan ketiga impedansi terhubung antar tiap kawat menuju

p

Z

Pertimbangan ketiga impedansi terhubung antar tiap kawat menuju kawat netral.

V

0 0

120

120

∠

−

∠

b

V

V

p an aA

Z

V

I

=

=

=

∠

120

=

_aA

∠

−

120

0

p an p bn bB

I

Z

V

Z

V

I

0 0

240

240

∠

−

∠

p cn

I

V

V

I

=

=

=

_aA

∠

−

240

0 p p p cn cC

I

Z

Z

I

0

=

+

+

_{bB C}

A

I

I

I

Maka

+

+

₀

cC bB

aA

I

I

I

2

2.4

.4 Hubungan

Hubungan

Y

−

Y

tiga

tiga--phase

phase

A

B

a

o

o

o

B

bn

V

+

+

a

b

o

o

−

−

o

Karakteristik tegangan

Sumber 3-phase seimbang

N

an

V

bn

n

o

=

=

_{bn cn}

an

V

V

V

Su be 3 p ase se ba g (tegangan phasor )

cn

V

+

− an

+

bn

+

cn

=

0

cn bn

an

V

V

V

C

2

2.4

.4 Hubungan

Hubungan

Y

−

Y

tiga

tiga--phase

phase

Urutan phase positip (abc)p p p ( ) (rotasi searah jarum jam)

0 0 120 0 − ∠ = ∠ = p bn p an V V V

V

V

cn

p

V

0

240

( j j )

0 240 − ∠ = _p cn p bn V

V

V

an

V

p 0

240

−

0

120

−

Urutan phase negatip (cba)

0

0

∠

V

V

bn

V

0

120

bn

V

(rotasi berlawanan arah jarum jam)

2.4

2.4 Hubungan

Hubungan

Y

−

Y

tiga

tiga--phase

phase

0 0

Tegangan line-to-line (urutan abc sebagai contoh)

0 0 0 30 3 2 3 2 3 60 sin 60 cos 60 0 ∠ = + = ° + ° + = ∠ + ∠ = + = p p p p p p p p nb an ab V V j V jV V V V V V V V V _V V V 0 0 0 90 3 2 3 2 1 2 3 2 1 60 120 − ∠ = − + − − = − ∠ + − ∠ = + = p p p p p p p nc bn bc V V j V V j V V V V V

V Vcn

V nb V ab V ca

V V_na

nb V 2

2 2

2 p j p p j p p

0 0

3 1

180 240 + ∠

− ∠ =

+

= _{cn na p p}

ca V V V V

V an V bn V na V nc V 0 210 3 0 2 3 2 1 − ∠ = + − + −

= V_p j V_p V_p V_p

0

+

+

V

V

V

S hi

bn

V V_nc

bc

V

nc V

0

=

+

+

_{bc ca}

ab

V

V

V

2

2.4

.4 Hubungan

Hubungan

Y

−

Y

tiga

tiga--phase

phase

0

30

3

∠

=

_p

ab

V

V

0

90 3 ∠−

= _p

bc V

V V_ca = 3V_p∠ −2100

Jenis tegangan

h (

V

)

magnitude Beda Phase

V

0

120

Tegangan phase ( )

L

V

p

V

Tegangan line-to-line ( )

p

V

p

V

a

a

_ +

V∠0

Wye Connected

So rce

_ _

+ +

n

b

V∠-120 V∠-240

Source

b

c

b

Sumber Delta

a

a

+

+ _

_

Delta

Source

V

ab

= |

V

ab

|

∠

0

V

V

∠

120

b

c

b

+

_

V

_bc

= V

_ab

∠

-120

V

_ca

= V

_ab

∠

-240

c

+

S

is

te

m

Wy

e

–

Wy

e

Z

aA

Zl

ZL

ZL

ZL

n

N

Zl

L

L

b

cB

C

Sistem Wye – Delta

a a

A

_ +

V∠0

Z _Z

I_{aA I}

CA

I_AB

_ _

+ +

n

b _b V∠-120

V∠-240

B C

Z _Z

I_BC c

c

Sistem

Delta –

D

elta

Zl

aA

ZL

Z

L

+

+

_

Z

b

cB

C

Z

+

+

_

_

Zl Zl

Sistem

Delta –

W

ye

Zl

aA

+

+

_

ZL

Z

b

cB

C

+

+

_

_

ZL ZL

Rangkaian

Y

-Y

4

kawat

4

d

a b c

V

V

V

I

I

I

Four - wire

Rangkaian Y-Y ; 4 kawat seimbang

,

, and

a b c

aA bB cC

A B C

=

=

=

I

I

I

Z

Z

Z

I

_nN

=

I

_aA

+

I

_bB

+

I

_cC

I

I

I

I

Daya rata rata yang diberikan sumber 3 phase ke beban 3 phase : Daya rata-rata yang diberikan sumber 3-phase ke beban 3-phase :

A B C

P

=

P

_A

+

P

_B

+

P

_C

P

P

+

P

+

P

Saat Z_A = Z_B = Z_C , beban dikatakan seimbang (balanced)

0

120

120

,

, and

a b c

aA bB cC

A B C

V

V

V

Z

θ

Z

θ

Z

θ

∠ °

∠ −

°

∠

°

=

=

=

=

=

=

∠

∠

∠

V

V

V

I

I

I

Z

Z

Z

,

(

120 ), and

(

120 )

A B C

aA bB cC

V

V

V

Z

θ

Z

θ

Z

θ

= ∠ − °

= ∠ − −

°

= ∠ − +

°

Rangkaian Y-Y ; 4 kawat seimbang

Tidak ada arus di kawat netral sumber ke kawat netral beban :

0

nN

=

aA

+

bB

+

cC

=

I

I

I

I

Daya rata-rata yang diberikan ke beban adalah :

A B C

P

=

P

+

P

+

P

cos(

)

cos(

)

cos(

)

V

V

V

V

V

V

Z

θ

Z

θ

Z

θ

=

− +

− +

−

2

3

V

cos( )

Z

θ

=

Contoh 1 Ditanyakan S = ?

rms

110 0

V

a

=

∠ °

V

Z

_A

=

50

+

j

80

Ω

Untuk 4-wire Seimbang (balance), dg data sbb :

rms rms

110

120

V

110 120

V

b c

=

∠ −

°

=

∠

°

V

V

50

80

50

80

B C

j

j

=

+

Ω

=

+

Ω

Z

Z

rms

110 0

1.16

58

A

50

80

a aA A

j

∠ °

=

=

=

∠ − °

+

V

I

Z

_A

j

*

68

109

VA

A

=

aA a

=

+

j

S

I V

Total daya komplex yang diberikan ke beban 3 phase adalah : Total daya komplex yang diberikan ke beban 3 phase adalah :

3

204

j

326

VA

S

=

3

S

_A

=

204

+

j

326

VA

S

S

Juga

rms rms

1.16

177

A

,

1.16 62

A

bB

=

∠ −

°

cC

=

∠ °

I

_{bB rms}

,

I

_{cC rms}

68

109

VA

B

=

+

j

=

C

Rangkaian

Y

-Y

(lanjutan)

3 - wire

Rangkaian Y-Y (lanjutan)

0

=

V

a

−

V

Nn

+

V

b

−

V

Nn

+

V

c

−

V

Nn

0

0

120

120

A B C

Nn Nn Nn

V

V

V

=

+

+

∠ ° −

∠ −

° −

∠

° −

+

+

Z

Z

Z

V

_Nn

V

_Nn

V

_Nn

A B C

=

+

+

Z

Z

Z

Solusi untuk V_Nn

(

120 )

_{A C}

120

_{A B}

0

_{B C}

Nn

V

∠ −

°

+ ∠

V

°

+ ∠ °

V

=

+

+

Z Z

Z Z

Z Z

V

Z Z

_{A C}

+

Z Z

_{A B}

+

Z Z

_{B C}

Z Z

Z Z

Z Z

d

a

−

Nn b

−

Nn c

−

Nn

V

V

V

V

V

V

I

_aA a Nn

,

I

_bB b Nn

, and

I

_cC c Nn

A B C

=

=

=

I

I

I

Saat rangkaian seimbang (balanced) yaitu saat Z = Z = Z

Rangkaian Y-Y (lanjutan)

Saat rangkaian seimbang (balanced) yaitu saat Z_A = Z_B = Z_C

(

V

∠ −

120 )

°

+ ∠

V

120

°

+ ∠ °

V

0

=

ZZ

ZZ

ZZ

V

0

Nn

=

+

+

=

V

ZZ

ZZ

ZZ

Daya rata-rata yang diberikan ke beban :

2

A B C

P

P

P

P

V

=

+

+

3

V

cos( )

Z

θ

Solusi (contoh 2) :

Tegangan phase :

Tegangan saluran : Tegangan saluran :

Contoh 3 : Sumber 3-phase seimbang dengan koneksi Y memberikan

daya ke beban Y seimbang. Magnitude tegangan saluran 150 V. Jik i d i b b ti h d l h 36+j12 h

Jika impedansi beban setiap phase adalah 36+j12 ohm, tentukan arus saluran jika diketahui

Transmission lines

Rangkaian Y-Y (lanjutan)

s ss o es

Contoh 4 :

Sistem 3-phase Y-Y seimbang urutan abc memiliki masing-masingp g g g impedansi saluran 0,6+j1 ohm dan impedansi beban 18+j14 ohm. Jika tegangan beban di phase a sebesar

Contoh 5 Ditanyakan S = ? 3-wire Seimbang (balance) dengan data sbb :

rms

110 0

V

a

=

∠ °

V

Z

_A

=

50

+

j

80

Ω

rms rms

110

120

V

110 120

V

b

c

=

∠ −

°

=

∠

°

V

V

50

80

50

80

B

C

j

j

=

+

Ω

=

+

Ω

Z

Z

rms

110 0

1.16

58

A

50

80

a aA

A

j

∠ °

=

=

=

∠ − °

+

V

I

Z

_A

j

*

68

109

VA

A

=

aA a

=

+

j

S

I V

Total daya kompleks yang diberikan ke beban 3 phase adalah : Total daya kompleks yang diberikan ke beban 3-phase adalah :

3

204

j

326

VA

S

=

3

S

_A

=

204

+

j

326

VA

Contoh 6 Ditanyakan P_Load= ? P_Line= ? P_Source= ? untuk data sbb :

Rangkaian ekivalen

3-kawat seimbang

Rangkaian ekivalen per-phase

100 0

( )

1.894

18.7

A

50

(377)(0.045)

a

aA

A

j

ω

=

=

∠ °

=

∠ −

°

+

V

I

Z

_A

j

(

)(

)

Tegangan phase di beban adalah :

( )

(40

(377)(0.04))

( )

81 2

V

AN

ω

=

+

j

aA

ω

= ∠ °

Contoh 6 (lanjutan)

Daya yang diberikan oleh sumber adalah :

cos(

)

2

m m

a V I

V I

P

=

θ

−

θ

(100)(1.894)

cos(18.7 )

89.7

W

2

=

° =

di i b b d l h Daya yang diterima beban adalah :

2 2

(1.894)

Re(

)

40

71 7

W

m

I

P

=

Re(

Z

)

=

40

=

71.7

W

2

2

A A

P

=

Z

=

=

Kehilangan daya di saluran adalah :

2 2

(1.894)

Re(

)

10

17.9

W

2

2

m

aA Line

I

P

=

Z

=

=

2

2

Contoh 7 (p247) Contoh 7 (p247)

Suatu

Suatu hubunganhubungan YY--Y Y tigatiga--phase : phase :

Vrms

240

200

,

Vrms

120

200

,

Vrms

0

200

∠

0

=

∠

−

0

=

∠

−

0

=

_{bn cn}

an

V

V

V

Tegangan phase :

Tegangan line-to-line :

0 A li Vrms 210 3 200 , Vrms 90 3 200 , Vrms 30 3

200 ∠ 0 = ∠ − 0 = ∠ − 0

= _{bc ca}

ab V V

V Arms 60 2 60 100 0 200 ₀ 0 0 − ∠ = ∠ ∠ = = p an aA Z V I Arms 180

2∠− 0

=

bB

I

Arus line :

Arms

300

2

∠

0

=

I

Arms 180 2∠ = bB I

Daya yang diserap ketiga beban :

Arms

300

2

∠

−

=

cC

I

W

600

60

2

200

3

o

P

=

3

×

200

×

2

×

cos

60

o

=

600

W

Contoh 7 (lanjutan)

Bagaimana dengan daya sesaat-nya ? Bagaimana dengan daya sesaat-nya ?

Note: V = 200V rms

( )

( )

( )

2

2

cos

(

60

)

A

V

cos

2

200

0

−

=

=

t

t

i

t

t

v

_an

ω

ω

Note: V_an = 200V rms

( )

(

)

( )

( )

( )

( )

(

)

(

2

60

)

W

cos

400

200

A

60

cos

2

2

V

cos

2

200

A

60

cos

2

2

0 0

−

+

=

−

×

=

×

=

−

=

t

t

t

t

i

t

v

t

P

t

t

i

aA an aA aA

ω

ω

ω

ω

Serupa , total daya sesaat yang diserap beban :

( )

P

( )

P

( )

P

( )

P

(

2

60

)

W

cos

400

200

+

ω

t

( )

( )

( )

( )

(

)

(

)

(

)

W

600

W

180

2

cos

400

300

2

cos

400

60

2

cos

400

600

=

°

−

+

°

+

+

°

−

+

=

+

+

=

t

t

t

t

P

t

P

t

P

t

P

_{A B C}

ω

ω

ω

Total daya sesaat TIDAK PERNAH NOL

Contoh 8 (p249)

Sistem 3-phase seimbang dengan tegangan kawat (line) 300Vrms diberikan pada

b b h b b d f ( ) l di i l h

beban hubungan Y sebesar 1200W pada power factor (PF) 0.8 leading. Hitunglah arus kawat (line) I_L dan impedansi beban Z_p untuk setiap phase.

T h V 300/ V

I

Tegangan phase : V_p = 300/ V_rms. Daya per-phase : 1200W/3 = 400W.

Sehingga , and I_L = 2.89A_rms

3 8 0 ) ( 300 400 p

Z

V 3 300 V

−

+

~

L

I

Impedansi phase :

8 . 0 × ) ( 3 =

400 I_L

Ω 60 = 89 . 2 3 300 = = | | L P P I V Z Vrms 3 300 = p V

PF sebesar 0.8 leading berakibat arus mendahului tegangan, dan sudut impedansi :

-arccos(0.8) = -36.9o

dan Z = 60 -36.9o Ω

L

dan Z_p 60 36.9 Ω

Catatan : Daya nyata hubungan Y-Y pada beban adalah P = V_an × I_AN

2.5

2.5 Hubungan

Hubungan Delta ( )

Delta ( )

Δ

Tidak terdapat saluran kawat netral. Impedansi seimbang terhubung antara tiap pasangan kawat (line)

A

B

b

Z

o

p

Z

A

B

+

+

a

o

b

−

−

o

P

Z

+

n

−

p

Z

C

6.5

6.5 Hubungan

Hubungan Delta ( )

Delta ( )

Δ

cn bn

an

p

V

V

V

V

=

=

=

Karakteristik tegangan Tegangan phase

0

30

3

3

V

V

V

∠

V

p

Tegangan line

V

_L

=

V

_ab

=

V

_bc

=

V

_ca

﹠

0

30

3

3

=

∠

=

_{p ab p}

L

V

V

V

V

﹠

Karakteristik arus

CA BC

AB

p

I

I

I

I

=

=

=

Arus phase

2

2.5

.5 Hubungan

Hubungan Delta ( )

Delta ( )

Δ

hubungan

Tegangan phase

Y

hubungan

Δ

p

V √ V_p

g g p Tegangan line

I

Arus phase

p p

p

L V

V = 3 V_L = 3V_p

I

√ √

p

I

Arus phase

Arus line

I

_L

=

3

I

_p

p

I

p L

I

I

=

√

2

2.5

.5 Hubungan

Hubungan Delta ( )

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 400 0 8 3 300 0 8 3 d 1 66 A

Sehingga, 400W = V_L · I_P · 0.8 = 300V · I_P · 0.8 , and I_P = 1.667Arms Arus saluran (line), I_L = I_P = 1.667A = 2.89Arms

PF lagging berarti bahwa tegangan mendahului arus sebesar arccos(0.8) = 36.9o

3

3 3

3

Impedansinya :

Catatan : Daya nyata pada beban (hubunganΔ) P = V _b × I

Ω =

∠ =

= o o

P P P

I V

Z 36.9 180 36.9 667

. 1

300

& &

Catatan : Daya nyata pada beban (hubunganΔ) , P V_ab × I_AB

The

Δ

-Y and Y-

Δ T

ransformation

1 3

Z Z

Z

A

Z Z_B

1 3

1 2 3

2 3

A

=

+

+

Z

Z

Z

Z Z

Z

Z

A

Z Z_B

C

Z

2 3

1 2 3

1 2

B

=

+

+

=

Z

Z

Z

Z Z

Z

Z

1 2 3

C

=

+

+

Z

Z

Z

Z

+

+

Z Z

Z Z

Z Z

3

Z 1

A B B C A C

B

+

+

+

+

=

Z Z

Z Z

Z Z

Z

Z Z

Z Z

Z Z

Z

1

Z Z₂

2

A B B C A C

A

+

+

+

=

+

Z Z

Z Z

Z Z

Z

Z

Z Z

Z Z

Z Z

3

A B B C A C

C

+

=

+

Z

Z Z

Z Z

Z Z

Rangkain Y-

Δ

I

I

I

3

AB AB

=

V

I

Z

aA AB CA

bB BC AB

=

−

=

−

I

I

I

I

I

I

dengan : 1

BC BC

=

V

I

Z

cC

=

CA

−

BC

I

I

I

2

CA CA

=

V

I

Rangkain Y-

Δ

(lanjutan)

I

I

I

cos

sin

cos(

120 )

sin(

120 )

aA AB CA

I

φ

j

φ

I

φ

j

φ

=

−

=

+

−

+

° −

+

°

I

I

I

3

I

(

φ

30 )

=

∠ − °

atau

3

3

aA

=

I

⇒

I

L

=

I

p

Contoh 10 IP = ? IL = ?

rms

220

30

V

3

a

=

∠ − °

V

rms

3

220

150

V

3

b

=

∠ −

°

V

rms

220

90

V

3

c

=

∠ °

V

Beban yg terhubung Δ seimbang :

10 50

Δ

= ∠ °

Z

220 0

V

=

=

∠ °

V

V

V

22 50

A

rms

AB

AB

=

=

∠ °

V

I

Z

rms

rms

220 0

V

220

120

V

220

240

V

AB a b

BC b c

CA

=

−

=

∠

=

−

=

∠ −

°

=

−

=

∠ −

°

V

V

V

V

V

V

V

V

V

22

70

A

rms

BC BC Δ Δ

=

=

∠ − °

Z

V

I

Z

⇒

rms

220

240

V

CA c a

∠

V

V

V

rms

22

190

A

CA CA Δ Δ

=

=

∠ −

°

Z

V

I

Z

Arus saluran :

22 3 20 ,

22 3

100 ,

22 3

220

aA

=

AB

−

CA

=

∠ °

bB

=

∠ −

°

cC

=

∠ −

°

The Balanced Three-Phase

Circuits

Y-to-

Δ

circuit

equivalent Y-to-Y circuit

Δ

Z

Z

3

Y

Δ

=

Z

Contoh 11 IP = ?

rms

110 0

V

a

=

∠ °

V

rms rms

110

120

V

110 120

V

b

c

=

∠ −

°

=

∠

°

V

V

10

5

75

225

L

j

j

Δ

=

+

Ω

=

+

Ω

Z

Z

25

j

75

Δ

+

Ω

Z

Z

25

75

⇓

3

Y

j

Δ

=

=

+

Ω

Z

⇓

⇐

⇐

1.26

66 A

a

A

=

=

∠ − °

V

I

_aA

1.26

66 A

_rms

L Y

∠

+

I

Contoh 11 (lanjutan)

1 26

186 A

and

1 26

54 A

=

∠ −

°

=

∠ − °

I

_bB

=

1.26

∠

186 A

_rms

and

I

_cC

=

1.26

∠

54 A

_rms

I

I

99 6 5

V

=

=

∠ °

V

I Z

Tegangan di rangkaian ekivalen per-phase adalah :

rms rms

rms

99.6 5

V

99.6

115

V

99.6 125

V

AN aA Y

BN

CN

=

=

∠

=

∠ −

°

=

∠

°

V

I Z

V

V

_CN

99.6

5

V

_rms

V

Tegangan line-to-line adalah :

rms

0.727

36

A

AB

AB

=

=

∠ − °

V

I

Z

rms

rms

172 35

V

172

85

V

172 155

V

AB AN BN

BC BN CN

=

−

=

∠ °

=

−

=

∠ − °

∠

°

V

V

V

V

V

V

V

V

V

rms

0.727

156

A

BC BC Δ Δ

=

=

∠ −

°

Z

V

I

Z

⇒

rms

172 155

V

CA

=

CN

−

AN

=

∠

°

V

V

V

rms

0.727 84

A

CA CA

Δ

Δ

=

V

=

∠ °

I

Contoh 12 P = ?

rms

110 0

V

a

=

∠ °

V

rms rms

110

120

V

110 120

V

b

c

=

∠ −

°

=

∠

°

V

V

10

5

75

225

L

j

j

Δ

=

+

Ω

=

+

Ω

Z

Z

rms

1.26

66 A

a aA

L Y

=

=

∠ − °

+

V

I

Z

_L

Z

_Y

rms

99.6 5

V

AN

=

aA Y

=

∠ °

V

I Z

3(99.6)(1.26) cos(5

( 66 ))

122.6

W

Pengukuran

Pengukuran Daya

Daya Rangkaian

Rangkaian Poliphase

Poliphase

V

I

P

=

⋅

Wattmeter diukur dg coil arus diukur dg coil potential

I

coil arus

V

+

+

coil potential /tegangan Pasif Network

E

I

11

18

∠

153

4

°

A

/tegangan

E.g.

(

ang

ang

)

cos

Vrms

0

100

Arms

4

.

153

18

.

11

−

⋅

=

°

∠

=

°

∠

=

I

V

I

V

P

V

I

(

)

P

OWER

M

EASUREMENT

Wh

i

AC

i d t

i

d

t

l b th

|

When using AC,

power

is determined not only by the

r.m.s. values of the

voltage

and

current

, but also by

the

phase angle

p

g

(which determines the

(

power factor

p

)

)

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

|

In

three-phase systems

we need to sum the

P

OWER

M

EASUREMENT

p

y

power taken from the various phases

y

in three-wire arrangements we can deduce the total

power from measurements using 2 wattmeter

power from measurements using 2 wattmeter

y

in a four-wire system it may be necessary to use 3

wattmeter

(

Electr

o

dynamic W

a

Digital Power

Meter

V

AR Meter

pf

Two-Wattmeter Power Measurement

cc = current coil cc = current coil vc = voltage coil

W1 d W1 read

1 AB A

cos

1

P

=

V I

θ

W2 read

2 CB C

cos

2

P

₂

=

V I

_{CB C}

cos

θ

₂

P

V I

θ

For balanced load with abc phase sequence

30

and

30

θ θ

_{1 a}

+ °

30

and

θ

₂

θ

_a

30

°

θ θ

=

+ °

θ

=

θ

− °

is the angle between phase current and phase voltage of phase

a

a

Two-Wattmeter Power Measurement(cont.)

P

= +

P

₁

P

₂

2

_{L L}

cos cos 30

P

P

P

V I

θ

= +

=

°

3

V I

_{L L}

cos

θ

=

To determine the power factor angle

1 2 L L

2 cos cos 30

P

+

P

=

V I

θ

°

1 2 L L

( 2 sin sin 30 )

P

₁

−

P

₂

=

V I

_{L L}

(

−

θ

°

)

1 2

2 cos cos 30

3

( 2 sin sin 30 )

tan

L L

P

P

V I

P

P

V I

θ

θ

θ

+

=

°

=

−

−

−

°

1 2 L L

( 2 sin sin 30 )

tan

P

−

P

V I

−

θ

θ

1

1 2 1 2

tan

θ

3

P

−

P

or

θ

tan

−

⎛

3

P

−

P

⎞

∴

=

=

⎜

⎟

1 2 1 2

Contoh 14 P = ?

10 45

∠ °

Z

= ∠ °

10 45

Z

Tegangan line-to-line = 220Vrms

Tegangan phase

220

220

30

3

A

=

∠ − °

V

Arus saluran Arus saluran

220

30

12.7

75

10 3 45

A A

∠ − °

=

=

=

∠ − °

∠ °

V

I

Z

10 3 45

∠

dan

I

B

=

12.7

∠ −

195

°

Z

1

cos

1

2698

W

723

W

AC A

P

V I

P

V I

θ

θ

=

=

1 2

3421 W

P

= +

P

P

=

⇒

2 BC B

cos

2

723

W

2

2..6

6 Pengukuran

Pengukuran Daya

Daya

A

B

θ

∠

I&bB

B

+

a

+

θ

∠

P

Z

o

aA

I& I&_AB

BC

I&

b

+

θ

∠

P

Z

Z

P

∠

θ

1

o

C

c

o

cC

I&

CA

I&

+

+

2

2

2..6

6 Pengukuran

Pengukuran Daya

Daya

(

−

)

=

(

−

( )

−θ

)

⋅

= 0

1 VAB IaA cos angVAB angIaA VLIL cos 30 P

(

)

(

)

(

(

)

)

(

θ

)

θ θ − ° − = − ⋅ = + = 0 0 2 0 120 90 cos ang ang cos 30 cos L L cC CB cC CB L L I V I V I V P I V

(

−θ

)

= 0 30 cos L LI V

(

)

(

)

θ θ θ θ θ θ θ θ θ θ tg tg tg P P + − = − = + − = − + = 3 3 1 3 2 1 2 3 sin 30 sin cos 30 cos sin 30 sin cos 30 cos 30 cos 30 cos 0 0 0 0 0 0 2 1

(

)

_θ g

tg + 2 2 2 1 2 1 2 3 P P P P tg + − = θ 1 2 1 2 3 P P P P arctg + − = θ

reaktif (PF=0) kapasitif / induktif (0<PF<1) resistif (PF=1) ±∞

= ±

= π θ θ , tg

2 − < < θ < +∞ θ > −∞ θ =0 , tgθ = 0

π θ π

tg tg , , 2 2 iti 0 < < > P π θ

P inductive , 2 0 , 2 1 π θ < < < P P 2 1 P

P = − P1 = P2

capacitive , 0 2 , 2

1 > P − <θ <

2

2..6

6 Pengukuran

Pengukuran Daya

Daya

.

o

| Contoh 15 (p256)

.

_{4Ω j15Ω}

A +

a

+ 1

o

Vrms 0

230∠ ° =

ab

V& _{dg urutan phase positip}

(1) Berapa pembacaan tiap wattmeter. ₊

b

+ 2

o

.

B N

.

(2) Total daya diserap beban. Dg urutan phase positip :

Vrms 0

230∠ ° =

ab V&

c

2

o

C

.

ab

Vrms 120

230∠+ ° =

ca V&

Vrms 120

230∠− ° =

bc V&

Wattmeter 1 membaca dan :I&_aA V&_ac

Vrms 60

230∠− ° =

− = _ca

ac V V& &

30 230 _⎟⎞_{∠ °} ⎜ ⎛ A 1 . 105 554 . 8 15 4 30 3 230 15

4 + = ∠− °

2.6

2.6 Pengukuran

Pengukuran Daya

Daya

.

o

| Contoh 15 (p256)

.

_{4Ω j15Ω}

A +

a

+ 1

o

(

ang ang

)

cos

1 = Vac IaA Vac − IaA P & & & & Wattmeter 1 membaca :

+

b

+ 2

o

.

B N

.

(

60 105.1

)

1389W cos

554 . 8

230× × − °+ ° = =

150 230 _⎟⎞_{∠− °} ⎜

⎛

&

Wattmeter 2 baca dan :I&_bB V&_bc

c

2

o

C

.

A 9 . 134 554 . 8 15 4 150 3 15

4 + = ∠ °

∠ ⎟ ⎠ ⎜ ⎝ = + = j j V I bn aB & &

(

& &

)

&

&

(

)

(

120 134.9

)

512.5W cos 554 . 8 230 ang ang cos 2 − = ° − ° − × × = −

= V_bc I_bB V_bc I_bB P

S hi P = P₁ + P₂ =876.5W Sehingga

,

TERIMA

TERIMA

KASIH

KASIH

TERIMA

TERIMA

KASIH

KASIH

http://risetwpt.wordpress.com email : rhidayat4000@gmail com email : rhidayat4000@gmail.com

6