15 trans vr

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1

1-phase Transformer

Vita Lystianingrum

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2

References

1. Theraja, B. L., ‘Electrical Technology’, S.Chand Com!any Ltd., 1"#$.

2. Schmidt%&alter, '., (ories, )., ‘ElectricalEngineering* + oc-et )eerence’, +rtech

'ouse /nc., 200#.. (arady, ., ‘Lecture 3otes on Single

hase Transormers’,444.eas.asu.edu56-arady5708!!.html.

9. Cha!man, S. :., ‘Electric ;achinery<undamentals’, ;cra4%'ill, 200=.

=. >uhal, ‘?asar Tenaga Listri-’, ener@it/TB, 1""1.

http://www.eas.asu.edu/~karady/360_pp.html

http://www.eas.asu.edu/~karady/360_pp.html

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Transformers Outlines

1* .Transormer.EAuialent circuit o a transormer 1D

19* .EAuialent circuit o a transormer 2D.reerred to !rimary or sec side

.Test or transormer’s !arameters

1=* .Transormer’s hasor

.Loaded Transormer

.Voltage )egulation

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9

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=

Equivalent circuit (4)

V1/a

jXe

I2

Re

Rc/a2 jXm/a2

ZLV2

aI1

)eerred to the secondary side

V1

jXe

I’2= I2 /a

Re

Rc jXm

a2

ZLaV2I1

)eerred to the !rimary side

a

N

N

E

E

I

I ===

2

1

2

1

1

2 E’2 = aE2 = E1.

V’2 = aV2

I’2 = I2 /a

R’2 = a2R2

X’2 = a2X2

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7

Transformer Test

Equivalent circuit parameters

.These !ur main parameters can

"e #etermine# "$ t%! tests&

.'pencircuit test

.h!rtcircuit test

V1

jXe Re

Rc jXm

a2ZLaV2

)eerred to the !rimary side

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#

Transformer Test

Open-circuit or !o-loa" Test #1$

A

V

W

Vopencircuit

Rate"%olta&e

HV LV

'oIo

Vo

Re

Rc jXm

jXe

HV LV

Io

Vo

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$

Transformer Test

(hort-circuit or Impe"ance Test #1$

R e " u c e "

% o l t a & e V

A

W

L VH V

A

s h o r t

c i r c u i t

I s c

V s c

' s c

Re jXe

Isc secIsc

Vsc

HV LV

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"

p. *4+ ,r. - (c!nt’#)

The result of a )** V/2** V transformer test is the follo+in&

,Open-circuit test #LV si"e$ V = 2** V I = *,. ' = 0 ,

,(hort-circuit test #3V si"e$ V = 1 V I = 2* ' = 14* ,

5in" the transformer’s e6ui%alent circuit referre" to lo+-%olta&e si"e7

!luti!n

89amples #Re%ie+$

,Open-circuit test -: Rc an" Xm

R c = V*2 / '* = )22 ;,

V/R$<4,1*2# 2

*2

** =−= P S QΩ== 4<<

*

2

*

Q

V X m

,(hort-circuit test -: Xe an" Re -: this test +as "one at )**-V si"e

Ze = Vsc / Isc = *,. ;,

Re = 'sc / Isc2 = *,42 ;,

Ω=−= =.,*22eee R Z X Then calculate the parameters to LV-

si"e referre" #Re/a2 an" Xe/a

2$

ZLV1/a

jXe/a2

I2

Re/a2

Rc jXm

V2

aI1

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10

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11

Transformer Volta&e Re&ulation #1$

+ real transormer has series im!edance4ithin it.

Thereore, the out!ut oltage aries 4iththe load, een i the in!ut oltage remainsconstant.

To coneniently com!are transormers inthis res!ect, it’s customary to deine aAuantity called voltage regulation.

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12

Transformer Volta&e Re&ulation #2$

<ull%oltage regulation*com!ares out!ut oltage atno%load 4ith out!ut oltage atull%load

+t no%load, V2 V1 5a

<or an ideal transormer,

V) 0F

%100,2

,2,2⋅

−=

fl

fl nl

V

V V VR

%100,2

,21

⋅−

= fl

fl a

V

V

V VR

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1

The Transformer 'hasor >ia&ram #1$

To determine V), 4e should understand theoltage dro! 4ithin it %G 4e deal 4ithcurrent !hase angle %G !hasor diagram.

V1/a

jXe

I2

Re

Rc/a2 jXm/a2

V2

aI1

hasor oltage V2 is the reerence.

hasor oltage V2 is assumed at an angle 0H.

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19


+!!lying (ircho’s oltage la4*

2221 V I jX I R eea

V ++=

V1/a

jXe

I2

Re

Rc/a2 jXm/a2

V2

aI1

I2

V2

V1/a

Re I2

jXeI2

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1=


I2

V2

V1/a

ReI2

jXeI2

I2

I2

V2

V2

ReI2

ReI2

jXeI2

jXeI2

V1/a

V1/a

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Ruirui #an eisiensi &Ruirui #an eisiensi &

(um?er @umparan

'rimer

5luAs

Bersama

@umparan

(eAun"er

Ru&i 5luAs Bocor

Ru&i Besi

3isterisis "an

8""C current

Out put

Ru&i Tem?a&a

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Rui Tem"aa (0Rui Tem"aa (0uu) &) &

R I P Cu

2=

Rui esi (0Rui esi (0ii) &) &

watt fB K P makshh

6.1=

watt B f K P maksee

22=

Ru&i 3isterisi

Ru&i e""C current

ehi P P P +=

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Eisensi &Eisensi &

rugirugikeluar dayakeluar daya

masuk dayakeluar daya

−∑+==η

masuk daya

rugirugi −∑−=1

iCu P P rugirugi +=−∑

2222

2

I

P R I CosV

CosV

i

ek

++

=

φ

φ η

0eru"ahan eisiensi terha#ap "e"an &0eru"ahan eisiensi terha#ap "e"an &

0eru"ahan eisiensi terha#ap a3t!r 3era c!s0eru"ahan eisiensi terha#ap a3t!r 3era c!s

"e"an &"e"an &

φ

φ

φ η

cos/1

cos/1

cos1

X

X

X

X

+−=

+−=

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1"

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20

89amples 2- #Dhapman$

1-AVE 24**/24*-V transformer is teste" #?oth tests

are "one in the primarC si"e77$

,OD Test ,(D Test

VOD = 24** V V(D = ). V

IOD = *,21 I(D =

'OD = * '(D = 1*

a, 5in" the e6ui%alent circuit referre" to hi&h-%olta&e si"e7

?, 5in" the e6ui%alent circuit referre" to lo+-%olta&e si"e7

c, Dalculate full-loa" VR at *,< la&&in& po+er factor7", hat is the efficiencC of the transformer at full loa"

+ith a po+er factor of *,< la&&in&F

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21

89amples 2- #Dhapman$ G (olution #1$

a, OD test

R c = V*2 / '* = 1*<** ;,

Ω== 11*11*

2*

Q

V X m

SC test

Re = 'sc / Isc2 = ),)) ;,

Xe = ,) ;

Because ?oth tests are "one in the primarC si"eE these parametersare also referre" to the primarC si"e,

V1

jXe

I’2= I2 /a

Re

Rc jXm

aV2

I1

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22


?, To o?tain the e6ui%alent circuit referre" to sec si"eE +e "i%i"e the

impe"ance ?C a2,

Rc = 1*< ;, Re = *,*)) ;,

Xm = 11* ; Xe = *,* ;

V1/a

jXe

/a2

I2

Re

/a2

Rc/a2 jXm/a2

V2

aI1

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2


c, 5ull-loa" VR

AV

S I

rated

rated

rated

2,=E2

E2 ==

%100,2

,21

⋅

−

= fl

fl a

V

V

V

VR

5ull-loa" current in the secon"arC

t pf = *,< la&&in&E current I2 = ,2H-4,0 ,

Then V1/a = 24),<)H*,)* V

2221 V X jI R I eea

V ++=

J1,2J1**24*

24*<,24)=⋅

−=VR

I2

V2

V1/a

Re I2

jXeI2

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29

89amples 2- #Dhapman$ G (olution #)$

", To fin" efficiencCE first fin" its losses,

The copper losses are

'cu = #I2$2 9 Re = #,2$2 9 *,*) = 1<0

The core losses are

( ) ( ) W R

P C

a

V

core 5.521050

85.234 22

1

===

The output po+er of the transformer at this po+er factor is

'out = V2I2cosK

= #24*V$#,2$cos 4,0 = 12***,

The efficiencC

%03.98%100%100 =⋅++

=⋅=

out corecopper

out

in

out

P P P

P

P

P η

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!nt!h 1 &!nt!h 1 &

'en&uAuran hu?un&an sin&Aat transformator fasa tun&&al 1 AV Can&

mempunCai pera?an"in&an te&an&an 2)** %olt / 2)* %oltE f = * 3E

men&hasilAan "ata pen&uAuran se?a&ai ?eriAut

rus hu?un& sin&Aat I hs

= ,2

Te&an&an C& "ipasan& V hs

= 141 V

>aCa masuA 'hs = 21)

3itun& prosentase pen&aturan untuA ?e?an "en&an cos φ = *,< la&&in&

R e " u c e "% o l t a & e

V

A

W

L VH V

A

s h o r t

c i r c u i t

I s c

V s c

' s c

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0en$elesaian c!nt!h 1 &0en$elesaian c!nt!h 1 &

'0540261.025.6131

214==

×=

×=

hshs

hs

I V

P Cosφ # Tertin&&al $

'0

'0

0

54496.2054425.6

0131∠=

−∠

∠==hs

hsek I

V Z

Ω=×=

Ω=×=

9.1954490.20

49.554490.20'0

'0

in X

Cos R

ek

ek

!olt

j j j

j jV

2.2502

8.15643.19548.1243.3414401920

!9.1949.5"25.6!6.08.0"24001

=

+=+++=

+++=

%26.4%1002400

24002.2502% =

−= " pengaturan

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2#

-8n"e-

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2$

Transformer

5uesti!ns t! p!n#er.

M 6h$ the #isc!ver$ ! the trans!rmer accelerate# the #evel!pment

! the use ! electricit$ 7

M 6h$ the ir!n !r maneti8ati!n l!sses are m!re imp!rtant than the

l!sses cause# "$ the %in#in resistance 7

M 6h$ the trans!rmer has t! "e c!!le# 7 9!% it is #!ne 7

M 6hat is the c!nnecti!n ! the trans!rmer that supplies $!ur

h!use an# %here is it 7

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2"

Transformer

: 2* 3V:+ 24;; V / 24; V sinle phase trans!rmer %as teste#.

In the sh!rtcircuit tests+ the 9V si#e %as sh!rte# an# the

parameters %ere measure# at the <V si#e.

In the !pencircuit test+ the <V si#e %as !pene# an# theparameters %ere measure# at the 9V si#e.

The results ! the tests are &

Mh!rtcircuit test (9V sh!rte#) Vs = -; v!lt+ Is = 4;; amp+

0 = -;; %att

M'pencircuit test (<V !pen) V! = 24;; v!lt+ I! = 2 amp+

0! = ;; %att

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0

Transformer

a) alculate the trans!rmer parameters.

") >ra% the simpliie# equivalent circuit

c) alculate an# pl!t the trans!rmer v!ltae reulati!n vs. l!a#

an# #etermine the l!a# %hen the reulati!n is *?#) alculate an# pl!t the eicienc$ vs. l!a#. >etermine the l!a#

%hen the eicienc$ is ma@imum an# the ma@imum eicienc$

:ssume& p!%er act!r ;.- lain an# l!a# v!ltae = rate#

v!ltae

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1

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2

The use o iron core and magneticcore.

/ron core

15 trans vr

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