Transformator Ptt

8/10/2019 Transformator Ptt

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TRANSFORMATOR


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1.TRANSFORMATOR DAYA2.TRANSFORMATOR INSTRUMEN :

(A).TRANSFORMATOR POTENSIAL (PT)(B).TRANSFORMATOR ARUS (CT)

3.AUTO TRAFO

.


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1.TRANSFORMATOR DAYA


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TRANSFORMATOR INSTRUMEN

TRANSFORMATOR UNTUK PENGUKURAN


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A.TRANSFORMATOR POTENSIAL

DIGUNAKAN UNTUK MENGUKUR TEGANGAN


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.


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.


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B.TRANSFORMATOR ARUS

UNTUK MENGUKUR ARUS YANG


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PENGUKURAN DAYA


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3.AUTO TRAFO

BELITAN PRIMER DAN SEKUNDER MEMILIKI INTI BERSAMATEGANGAN SEKUNDER DAPAT BERHARGA TETAP


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UNTUK MENDAPATKAN TITIK NETRAL


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TEORI SIGKAT TRANSFORMATOR DAYA SATU FASA


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fluksi, yang ditimbulkan oleh arus listrik yang . ,

besi dibuat dari lempengan lempengan besi ,

(sebagai rugi rugi besi) yang ditimbulkan oleh


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membentuk suatu kumparan. Kumparan ,

maupun terhadap kumparan lain disebelahnya, , .


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luar melalui sebuah bushing , yaitu sebuah,

sekaligus berfungsi sebagai penyekat antara.


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PENDINGINTAP CHANGER

ALAT PERNAPASANPENGAMAN


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transformasi untuk mendapatkan tegangan

(diinginkan) dari tegangan jaringan/sisi primer .

dilakukan baik dalam keadaan berbeban (on

load) tergantung pada jenisnya.


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Rele Differensial pengaman trafo darigangguan u ung s ng at a am tra o


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Load% Over load

10% 20% 30% 40% 50%

Jam jam jam menit menit

0.5 3 1.5 1 30 15

0.75 2 1 0.5 15 8

0.9 1 0.5 0.25 8 4


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Keadaaan Transformator Tan a beban

Transformator tanpa beban

Vektor transformator tanpa beban


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dihubungkan dengan sumber tegangan V 1 yangsinusoid, akan mengalirlah arus primer Io yang jugasinusoid dan dengan mengannggap belitan N 1 reaktif murni, Io akan tertinggal 90 o dari V

1 (lihat gambar ).

rus pr mer o men m u an u s yang se asadan juga berbentuk sinusoid.

=maks


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Keadaan Tanpa Beban

Fluks yang sinusoid ini akan menghasilkan tegangan1 .

ubah memotong suatu kumparan maka padakumparan tersebut akan di induksikan suatutegangan listrik :

dt N e 11 =

wt N dt N e maksmaks

cos111 == (tertinggal 90 o dari )

maksmaks f N E 1

11 44,4

2==Harga efektifnya

d b


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Keadaan Tanpa Beban Pada rangkaian sekunder, fluks () bersama tadi

menimbulkan

dt d N e 22 = wt w N e m cos22 = maks f N E 22 44,4=

2

1

2

1

N N

E E =

Dengan mengabaikan rugi tahanan dan adanya fluks bocor,

a N V E

===2

1

2

1

2

1a = perbandingan transformasi

1

berlawanan arah dengan tegangan sumber V1.


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Keadaaan Transformator Berbeban

Apabila kumparan sekunder dihubungkan dengan beban ZL, I2 mengalir pada kumparan sekunder, di = .

d f b b


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Keadaaan Transformator Berbeban Arus beban I 2 ini akan menimbulkan gaya gerak

magnet (ggm) N 2 I2 yang cenderung menentang

pemagnetan I M. Agar fluks bersama itu tidakberubah nilainya, pada kumparan primer harusmenga r arus 2, yang menen ang u s yangdibangkitkan oleh arus beban I 2, hingga

keseluruhan arus an men alir ada rimermenjadi :

'

'21o

2o1

III =

K d T f B b b


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Keadaaan Transformator Berbeban

Bila rugi besi diabaikan (IC diabaikan) maka Io = IM= +

Untuk menjaga agar fluks tetap tidak berubah sebesar ggm yang dihasilkan oleh arus pemagnetan IM saja, berlaku hubungan : N1IM = N1I1 N2I2

1 M 1 M 2 2 2 Sehingga

=

Karena nilai IM dianggap kecil maka I2 = I1N1I1 = N2I2 atau I 1/I 2 = N2/N 1

A GA A A A


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RANGAIAKAN EKIVALEN TRANSFORMATOR


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DIPINDAHKAN KE SISIS PRIMER


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RANGAKAIAN EKIVALEN PENDEKATAN


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The no load current ran es from 1% to 3% of the full load current.

a (b)

Therefore, the circuit can be simplified to circuit (b).


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''eqeq


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e percen regu a on

The transformer efficiency


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''' s ncos 2221 eqeq +=


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The efficienc of the transformer is the ratio of out ut secondar ower to the input (primary) power. Formally the efficiency is :

1

2PP=

,

P1 : The input power (Primary) = V1I1 cos1

LPPP += 21

P2 : The output power (Secondary) = V2I2 cos2

Where,

PL is the power loss in the transformer = Pcopper + Piron

''

ironeq P R I I V ++

= 2'22

'

2

'

2

222

cos

E l


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Example

A 100 kVA, 400/2000 V, single phase transformer has the following parameters

R1 = 0.01 R2 = 0.25 ohms

= . = . The transformer supplies a load of 90 kVA at 2000 V and 0.8 PF lagging.

Calculate the primary voltage and current using the simplest equivalent circuit.

Find also the V.R. and efficiency for the transformer

S l ti


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Solution


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sincos '2

'

2

'

21 eqeq X I R I V V VR +=

6.006.02258.002.0225 +VR

7.11 V VR

%84.27.11

% =VR.


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Determining the Values of Components in the Transformer Model


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Determining the Values of Components in the Transformer Model

approximate the equivalent circuit. An adequate approximation ofthese values can be obtained with only two tests.

open-circuit test

short-circuit test

Circuit Parameters: Open-Circuit Test


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Transformer's secondary winding is open-circuited

- .input current must be flowing through the excitation branch of thetransformer.

p p C

X M to cause a significant voltage drop, so essentially all the inputvoltage is dropped across the excitation branch.

, ,measured.


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Circuit Parameters: Short-Circuit Test


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Transformer's secondary winding is short-circuited

Primary winding is connected to a fairly low-voltage source.

The input voltage is adjusted until the current in the short-circuitedwindings is equal to its rated value.

In ut volta e, in ut current, and in ut ower to the transformer are

measured. Excitation current is negligible, since the input voltage is very low.

Thus the volta e dro in the excitation branch can be i nored. All thevoltage drop can be attributed to the series elements in the circuit.

Circuit Parameters: Short-Circuit Test


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sc

scSE I

V Z =

e magn tu e o t e ser es mpe ance:

= sc sc P P 1

The short-circuit power factor and power factor angle:

= sc sc sc sc I V I V

Therefore the series impedance is:

PF cos I V

X a X j R a R

X R Z

sc

sc s p s p

eqeqSE

122

It is possible to determine the total series impedance, but there is no easyway to split the series impedance into the primary and secondarycom onents. These tests were erformed on the rimar side so the circuit impedances are referred to the primary side.

Example 2 (Example 2-2, page 92 of your text)


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p ( p , p g y )

The equivalent circuit impedances of a 20-kVA, 8000/240-V, 60-Hztransformer are to be determined. The open-circuit test and the short-

,the following data were taken:

Open-circuit test Short-circuit teston pr mary

V oc = 8000 V

on pr mary

V sc = 489 V

I oc = 0.214 A

P oc = 400 W

I sc = 2.5 A

P sc = 240 W

Find the impedances of the approximate equivalent circuit referred tothe primary side, and sketch the circuit.

A E l 2


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Answer to Exam le 2

Transformer Voltage Regulation


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g g

Because a real transformer has series impedance within it, the output voltageof a transformer varies with the load even if the input voltage remains

[ ] [ ]

.magnitude of the secondary terminal voltage from no-load to full-load.

[ ] [ ][ ]

[ ] [ ]

100 =

load full V oa fuV oa noV

gulation ReVoltage% s

s s

[ ] [ ][ ]

100

load full V p

p p

Referred to the primary side

Transformer Efficiency


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= Input Power Output Power

=

=

Losses Input Power

Losses Input Power

++

+=

cos

1sslosscorelosscopper

losscorelosscopper

I V PP

PP

Input Power

Usually the efficiency for a power transformer is between 0.9 to 0.99., .


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A single-phase, 100-kVA, 1000:100-V, 60-Hz transformer has the

Open-circuit test (HV side open): 100 V, 6 A, 400 WShort-circuit test LV side shorted : 50 V, 100 A, 1800 W

Draw the equivalent circuit of the transformer referred to the high-voltage side. Label impedances numerically in ohms and in per unit.

Determine the voltage regulation at rated secondary current with 0.6 power factor lagging. Assume the primary is supplied with ratedvoltage

Determine the efficiency of the transformer when the secondary currentis 75% of its rated value and the power factor at the load is 0.8 laggingwith a secondary voltage of 98 V across the load

PU System


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er un sys em, a sys em o mens on ess parame ers, s use orcomputational convenience and for readily comparing the performanceof a set of transformers or a set of electrical machines.

Quantity BaseQuantity Actual

Value PU =

[ ]base

VA=

, , , .[VA]base and [ V ]base are chosen first.

[ ] [ ] [ ][ ] [ ] [ ]basebasebase

basebasebasebasebasebase

basease

V V V Z X R

I V VAS QP

V

=====

====22

[ ] [ ]

[ ]

VAVA sec base pri base =

[ ][ ]base

basebase

basebasebase

V I Y

VAS I

=

[ ]

[ ] ratio turnsV

V

sec base

pri base

=

base

ohmPU Z

Z =

Example 4 (Problem No. 2-2, page 144 of your text)


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A 20-kVA, 8000:480-V distribution transformer has the followingresistances and reactances:

=P

X P

= 45 ohm

RC = 250,000 ohm

S .

X S = 0.06 ohm

X M = 30,000 ohm

The excitation branch impedances are referred to the high-voltage side.

a) Find the equivalent circuit of the transformer referred to the high-vo tage s e.

b)

b) Find the per unit equivalent circuit of this transformer.

power factor lagging. What is this transformers input voltage? What isits voltage regulation?

d What is this transformers efficienc under the conditions of art c ?

Pengaturan Tengan Pada


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Transformator

PENGATURAN TENGAN GAN DENGAN LTC ( LOAD TAPE CHANGER)

ADA 2 JENIS(A). NO LOAD LTC

.


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PERUBAHAN JUMLAH

LILITAN DILAKUKAN PADA BEBAN NOL

BIASANYA PADA SISI PRIMER

CONTOH TEKNIS PERUBAHAN SAKLAR


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UNTUK MENGUBAH TAPTAP

CONTOH SPESIFIKASI TEKNIS TAPTAP


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SUATU TRANSFORMATOR


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BIASANYA PERUNAHAN JUMLAH LILITAN DALAM KEADAAN BERBEBAN

BIASANYA DILAKUKAN PADA

SAKLAR SEBELAH KIRI DIOPERASIKAN SECARA

BERANTIAN DENGAN SAKLAR SEBELAH KANAN PADA GAMBAR DI SAMPING


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SISTEM PENDINGIN TRANSFORMATOR


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FUNGSI PENDINGIN TRAFO1.MEMBUANG PANAS YANG TIMBUL PADA INTI

2.MINYAK TRAFO SELAIN UNTUK PENDINGIN INTI


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Pada inti besi dan kumparan kumparan akan,

tembaga. Bila panas tersebut mengakibatkankenaikan suhu yang berlebihan, akan merusakisolasi (di dalam trafo), maka untuk mengurangikenaikan suhu yang berlebihan tersebut trafo

untuk menyalurkan panas keluar trafo. Mediayang dipakai pada system pendingin dapat

, , .pengalirannya (sirkulasi) dapat dengan caraalamiah (natural) atau tekanan/paksaan.

PENDINGIN


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MEDIA

No MACAM SISTEM*

Dalam Trafo Luar Trafo

r u asAlamiah

r u asPaksa

r u asAlamiah

r u asPaksa

1 AN - - Udara -

2 AF - - - Udara

3 ONAN Minyak - Udara -

4 ONAF Minyak - - Udara

5 OFAN - Minyak Udara -

6 OFAF - Minyak - Udara

7 OFWF - Minyak - Air

8 ONAN/ONAF Kombinasi3 dan 4

9 ONAN/OFAN Kombinasi

3 dan 510 ONAN/OFAF Kombinasi

3 dan 6

11 ONAN/OFWF Kombinasi3 dan 7

SINGKATAN SINGKATAN SISTEM


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PENDINGIN TRAFO

1.ONAF Cooling. 3.OFWF Cooling 4.ODAF Cooling 5.ODWF Cooling


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KONSTRUKSI TRAFO DAN PENDINGIN


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DIAGRAM SISTEM KONTROL PENDINGIN

TRAFODANALATPROTEKSI


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TRAFO DAN ALAT PROTEKSI


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KONSTRUKSI TRAFO TEGANGAN

TINGGI


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TINGGI

Transformator Ptt

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