part 5 emachines EEP 3243

8/14/2019 part 5 emachines EEP 3243

1/36

1

ELECTRICAL MACHINE

EEP 3243

Lt C d r O n g K h ye Lia t R M N

27 Jan 2010


2/36

2

RECAP/ADDITION


3/36


4/36


5/36


6/36

6

Shifting Impedances fromSecondary to Primary

While shifting from secondary to primary,these rules applied:

Shifted impedance values are multipliedby a2.

Shifted voltage E values becomes aE.

Shifted current I values becomes I/a. The secondary of the transformer is on

open-circuit and both current are zero,

therefore can remove the transformer.


7/36

7


8/36

8

Shifting Impedances fromPrimary to Secondary

While shifting from primary to secondary, theserules applied:

Shifted impedance values are divided by a2

. Shifted voltage E values becomes E/a.

Shifted current I values becomes aI.

The primary of the transformer is on open-circuit

and both current are zero, therefore can removethe ideal transformer completely.


9/36

9


10/36

10

PROBLEM 1
http://smb//tmp/svlh6.tmp/EEP%203243%20part%205%20Exmp.pptxhttp://smb//tmp/svlh6.tmp/EEP%203243%20part%205%20Exmp.pptx


11/36

11

Practical Transformer

In th e re a l w o u ld th e w in d in g s o ftra n sfo rm e r h a v e re sista n ce a n d th e.co re s a re n o t in fin ite ly p e rm e a b le S o

th e flu x p ro d u ce d b y th e p rim a ry is n o t.co m p le te ly ca p tu re d b y th e se co n d a ry

-A n d th e iro n co re s p ro d u ce e d d y cu rre n t,a n d h y ste re sis lo sse s w h ich co n trib u te

.to th e tra n sfo rm e r te m p e ra tu re rise


12/36

Ideal Transformer withAn Imperfect Core

, -C o re h a v in g h y ste re sis lo ss e d d y cu rre n t.lo ss a n d lo w p e rm e a b ility

:IG 1 An imperfect core represented by a reactance Xmand aresistance Rm.


13/36

Cont.

The Rm represents the

iron loss andresulting heat and asmall current Ifis

drawn from the line.

Magnetizing reactance

Xm is a measure ofthe permeability ofthe transformercore.


14/36

14

Cont.

Thus if permeabilityis low Xm is

relatively low. The

current Im flowingthoughXmrepresents the

magnetizing

current needed tocreate the flux m

in the core.


15/36

15

Cont.

The total current,Io(exciting current)

needed to producethe flux m in an

imperfect core isequal to the phasor

sum of Ifand Im.


16/36

16

Cont

Rm and Xm can be

foundexperimentally

by connectingthe transformerto an ac sourceunder no-load

condition andmeasuring theactive power andreactive power it

absorbs.

:IG 2 Instruments used to measure E,I, P, and Q .in a circuit


17/36

17

Cont.

The followingequations apply:

WhereRm = resistance

representing the ironlosses,

Xm = magnetizing

reactance of theprimary winding,

E1 = primary voltage, VPm = iron losses, W

Qm = reactive power

needed to set up the

Id l T f ith


18/36

18

Ideal Transformer withLoose Coupling

A ssu m e a tra n sfo rm e r h a v in g p e rfe ctco re b u t lo o se co u p lin g b e tw e e n its

p rim a ry a n d se co n d a ry w in d in g s

.w h ich h a v e n e g lig ib le re sista n ce S e q u e n ce o f sim p le o p e ra tio n se ts o ff

:a tra in o f e v e n ts


19/36


20/36

20

Cont.Mutual fluxes and leakage fluxes produced by atransformer under load. The leakage fluxes are

due to the imperfect coupling between the coils.I 1 and I2immediately flow in

.windingI2produces an mmf N2I2 while I1produces mmf N1I1.

mmf N

2I

2produces ac flux

2.Portion of 2 (m2) links withprimary winding while another portion

f2 .does not Flux f2is called.secondary leakage flux

mmf N1I1produces ac flux 1.

Portion of 1 (m1) links withprimary winding while another portion

f1 .does not Flux f1is called.primary leakage flux


21/36

21

Cont.Transformer possesses 2 leakage fluxes and amutual flux

Total flux produced by I1composed of new mutual flux m1 and primary leakage fluxf1 . Total flux produced by I2

composed of new mutual flux m2 and primary leakage fluxf2 . Combine m1 and m2 into a

single mutual flux

m. flux f1 is in phase with I1and flux f2 is in phase withI2.


22/36

22


Voltage Esand Epcomposed of:two partsEf2induced by leakage flux f2

E2induced by mutual flux m

Ef2and E2 .are not in phase


23/36

23


Ef1induced by leakage fluxf1

E1 induced by mutual flux m

Induced Ep = applied voltageEg.


24/36

24

Cont. Ef2 is really a voltage

drop across thesecondary leakagereactance Xf2.

Ef1 is simply a voltage

drop across primaryleakage reactance Xf1.

IG 3Separating the various inducedvoltages due to the mutual flux and theleakage fluxes


25/36

25

Cont. A lso a d d e d th e p rim a ry a n d se con d a ry

w in d in g re sista n ce R 1and R 2in series.w ith th e re sp e ctiv e w in d in g s


26/36

26

PROBLEM 2

Equivalent Circuit of


27/36

27

Equivalent Circuit ofa Practical Transformer

Complete equivalent circuit of a.practical transformer The shaded.box T is an ideal transformer


28/36

28

Cont. Typical value of transformer

parameters ranging from

1kVA to 400MVA.

Io always much

smaller than therated primarycurrent Inp .

Enp In = Ens Ins = Sn,where Sn is

transformer ratedpower.


29/36

29

Simplifying the Equivalent Circuit.

T h e co m p le te e q u iv a le n t circu it o fth e tra n sfo rm e r g iv e s fa r m o re

d e ta il th a n is n e e d e d in m o st

.p ra ctica l p ro b le m s


30/36

30

Cont. A t n o lo a d

I1and I2 = .0 Ioflows in R1and Xf1 . R1and Xf1are so small compare to Xmand Rmand I2 .is zero

So R1, R2, Xf1and Xf2 .can be neglected


31/36

31

Cont. A t fu ll lo a d

Ipisat least 20 times larger than Io , Io a n d. (m a g n e tizin g b ra n ch ca n b e n e g le cte d a lso

%a p p ly w h e n th e lo a d is o n ly 1 0 o f ra te d)ca p a city


32/36

32

Cont.

Fu rth e r sim p lify b y sh iftin g th e im p e d a n ce s to.p rim a ry sid e


33/36

33

Cont.Rp = R1 + a

2R2X

p= X

f1+ a2X

f2

WhereRp = total transformer

resistance referred to theprimary side

Xp = total transformerleakage reactance referredto the primary side

Zp = total impedance referredto primary side whichproduces an internal

voltage drop whentransformer is loaded then


34/36

34

Cont.

Transformer above 500kVA possess a leakage

reactance Xpthat is at least 5 times greaterthan Rp. So we can neglect Rp (out of

)standpoint of temperature rise and efficiency


35/36

35

PROBLEM 3


36/36

36

END OF PART 5

Documents

part 5 emachines EEP 3243