Upload
hafiz-azman
View
223
Download
0
Embed Size (px)
Citation preview
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
8/14/2019 part 5 emachines EEP 3243
2/36
2
RECAP/ADDITION
8/14/2019 part 5 emachines EEP 3243
3/36
8/14/2019 part 5 emachines EEP 3243
4/36
8/14/2019 part 5 emachines EEP 3243
5/36
8/14/2019 part 5 emachines EEP 3243
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.
8/14/2019 part 5 emachines EEP 3243
7/36
7
8/14/2019 part 5 emachines EEP 3243
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.
8/14/2019 part 5 emachines EEP 3243
9/36
9
8/14/2019 part 5 emachines EEP 3243
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.pptx8/14/2019 part 5 emachines EEP 3243
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
8/14/2019 part 5 emachines EEP 3243
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.
8/14/2019 part 5 emachines EEP 3243
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.
8/14/2019 part 5 emachines EEP 3243
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.
8/14/2019 part 5 emachines EEP 3243
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.
8/14/2019 part 5 emachines EEP 3243
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
8/14/2019 part 5 emachines EEP 3243
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
8/14/2019 part 5 emachines EEP 3243
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
8/14/2019 part 5 emachines EEP 3243
19/36
8/14/2019 part 5 emachines EEP 3243
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
8/14/2019 part 5 emachines EEP 3243
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.
8/14/2019 part 5 emachines EEP 3243
22/36
22
Cont.Transformer possesses 2 leakage fluxes and amutual flux
Voltage Esand Epcomposed of:two partsEf2induced by leakage flux f2
E2induced by mutual flux m
Ef2and E2 .are not in phase
8/14/2019 part 5 emachines EEP 3243
23/36
23
Cont.Transformer possesses 2 leakage fluxes and amutual flux
Ef1induced by leakage fluxf1
E1 induced by mutual flux m
Induced Ep = applied voltageEg.
8/14/2019 part 5 emachines EEP 3243
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
8/14/2019 part 5 emachines EEP 3243
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
8/14/2019 part 5 emachines EEP 3243
26/36
26
PROBLEM 2
Equivalent Circuit of
http://smb//tmp/svlh6.tmp/EEP%203243%20part%205%20Exmp.pptxhttp://smb//tmp/svlh6.tmp/EEP%203243%20part%205%20Exmp.pptx8/14/2019 part 5 emachines EEP 3243
27/36
27
Equivalent Circuit ofa Practical Transformer
Complete equivalent circuit of a.practical transformer The shaded.box T is an ideal transformer
8/14/2019 part 5 emachines EEP 3243
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.
8/14/2019 part 5 emachines EEP 3243
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
8/14/2019 part 5 emachines EEP 3243
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
8/14/2019 part 5 emachines EEP 3243
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
8/14/2019 part 5 emachines EEP 3243
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
8/14/2019 part 5 emachines EEP 3243
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
8/14/2019 part 5 emachines EEP 3243
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
8/14/2019 part 5 emachines EEP 3243
35/36
35
PROBLEM 3
http://smb//tmp/svlh6.tmp/EEP%203243%20part%205%20Exmp.pptxhttp://smb//tmp/svlh6.tmp/EEP%203243%20part%205%20Exmp.pptx8/14/2019 part 5 emachines EEP 3243
36/36
36
END OF PART 5