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Sanaa University Faculty of Engineering Electrical Engineering Department Do.Mo amme! "l-ya!

omi
Final E#am in Electrical Mac ine II

Name:------------------ID#:---------------------

$uestion %
a- Drive an e#pression for t e open circuit r.m.s p ase voltage in a sync ronous mac ine in terms of fre&uency' flu# per pole an! num(er of turns per p ase. Discuss t e assumptions t at are ma!e. Define t e term )in!ing factor an! s o) o) it may (e calculate!. (- " *-poles' +- sync ronous generator fee!s a +- (alance! , connecte! loa! aving a p ase impe!ance of -./+ . 0 e rotor is !riven at %122 r.p.m an! t e resultant air gap flu# is sinusoi!ally !istri(ute!. 0 e armature )in!ings of t e generator ave 133 con!uctors an! are !istri(ute! in 41 slots aroun! t e stator. 0 e coils span %2 slots. 0 e flu# per pole is 2.21- 5(. Fin! t e )in!ing factor 6) ' ence calculate t e mac ine internal voltage. Fin! t e loa! current' voltage an! t e consume! electrical po)er

$uestion 1
" t ree-p ase ,-connecte! sync ronous generator is rate! %12 M7"' %+.1 67' 2.3 8F lagging' an! *2 9:. Its sync ronous reactance is 2.4 ' an! its resistance may (e ignore!. a- 5 at is its voltage regulation; (- If t e fiel! current is constant' ) at is t e ma#imum po)er possi(le out of t is generator; 9o) muc reserve po)er or tor&ue !oes t is generator ave at full loa!; c- 5 at )oul! t e voltage an! apparent po)er rating of t is generator (e if it )ere operate! at <2 9: )it t e same armature an! fiel! losses as it a! at *2 9:; !- 5 at )oul! t e voltage regulation of t e generator (e at <2 9:;

$uestion +
a- =an Sync ronous motor (e starte! (y itself )it out any ai!; If your ans)er is no' e#plain t e !ifferent met o!s of starting sync ronous motor. b- For a *-pole' *2 9:' < p sync ronous motor' )it parameters are: >a ?2.2< an! @s = 0. 11 . Determine t e no-loa! spee!' t e full-loa! spee!' an! t e alf-loa! spee! of t e motor. c- It is re&uire! to supply a t ree p ase loa! of +2265' *2 9: from t e national gri!. 0 e gri! fre&uency is <2 9:. It is !eci!e! to generate t e re&uire! po)er (y means of a motor-generator set consisting of a sync ronous motor !riving a sync ronous generator. 9o) many poles s oul! eac of t e t)o mac ines ave in or!er to convert <2 9: po)er to *2 9: po)er. 1

$uestion " -327' +4< 67"' 2.3 lagging po)er factor' , connecte! sync ronous generator as a sync ronous reactance of 2.- an! a negligi(le armature resistance. 0 e generator supplies a -327 32 65' o.3 lea!ing po)er factor , connecte! sync ronous motor )it a sync ronous reactance of %.% a negligi(le armature resistance. 0 e generator is a!/uste! to ave a terminal voltage of -32 7 ) en t e motor is !ra)ing t e rate! po)er at unity po)er factor. a- =alculate t e magnitu!es an! angles of E" for (ot mac ines (- If t e flu# of t e motor is increase! (y %2 percent' ) at appens to t e terminal voltage of t e po)er system' ) at is t e ne) value; c- 5 at is t e ne) po)er factor of t e motor;

$uestion <
%Descri(e one of t e follo)ing mac ines un!er t e operation' types' an! application: %- Stepper motor 1- Servo motor +- 8ulls transformer -- Microp one <- Aou! spea6er *- Electromec anical relay 4- Matc ing transformer 3- Isolation transformer B- Croa! (an! transformer %2- "u!io transformer %%- Measurement transformer D=.0. or 8.0.E ea!ings: Mac ine construction' principle of

Solution:

$uestion %
c- Drive an e#pression for t e open circuit r.m.s p ase voltage in a sync ronous mac ine in terms of fre&uency' flu# per pole an! num(er of turns per p ase. Discuss t e assumptions t at are ma!e. Define t e term )in!ing factor an! s o) o) it may (e calculate!. !- " *-poles' +- sync ronous generator fee!s a +- (alance! , connecte! loa! aving a p ase impe!ance of -./+ . 0 e rotor is !riven at %122 r.p.m an! t e resultant air gap flu# is sinusoi!ally !istri(ute!. 0 e armature )in!ings of t e generator ave 133 con!uctors an! are !istri(ute! in 41 slots aroun! t e stator. 0 e coils span %2 slots. 0 e flu# per pole is 2.21- 5(. Fin! t e )in!ing factor 6) ' ence calculate t e mac ine internal voltage. Fin! t e loa! current' voltage an! t e consume! electrical po)er

Solution
a- The E.M.F Equation e.m.f/cond = e = Blv emax/cond= Bmax l v , v= Dn s 60

D = oto diamete , ns = oto s!eed in .!.m 2# max = fund / !o"e = l ! p = D/p


= fund / !o"e = 2# max l D 2 l D = # max ! !

# max =

! 2l D

Dn s ! l 2l D 60 conside tota" no of conducto s in se ies !e !hase equa" $!h Dn s ! E max/ !hase = l $ !h 2l D 60 120 f $ !h n ns = E max/ !hase = ! s ! 2 60 E max/ cond =
E max/ !hase = $ !h 2 2 = f % 2 ! 120 f 60 !
E
. m .s / !hase

E max/ !hase = 2 f %
= &.&& f %

. m .s / !hase

E max 2

.m .s / !hase

b- Fu"" '!o"e( !itch and f actiona" 'coi"( !itch

Fu"" !itch) is the distance bet*een t*o successi+e !o"es and it is 1,0
.

e"ect ica" de- ee

F actiona" !itch 'coi" s!an(. /s the distance bet*een the t*o coi" sides of the same coi". /t is measu ed in
s"ots, in e"ect ica" o mechanica" de- ee. 0 coi" s!an is 1e!t as nea 1,0o 'e"ect( as !ossib"e to ensu e +o"ta-e summin- u!. c- Define the *indin- facto and sho* ho* it ma2 be ca"cu"ated. 3hen the coi" s!an in a matu e *indin- is made "ess than a !o"e !itch it is said that the *indin- is cho ded. The mino disad+anta-e of usin- f actiona"-!itch coi" 'cho ded *indin-( , is that the e *i"" be a s"i-ht eduction in -ene ated e.m.f as com!a ed to fu"" !itch coi"s. 4o*e+e , the man2 ad+anta-es -ained, such as a unifo m coi" sha!e on the *indin- and a !u e"2 -ene ated sinusoida" +o"ta-e, fa out*ei-h this disad+anta-e. /t is customa 2 to ex! ess the eduction in the -ene ated +o"ta-e b2 a !itch facto 'cho din- facto ( 1! *hich can be ex! essed in te ms of the co din- an-"e.
Et

e1

/2 e1

Et = e 1 cos 2 2
cho din- facto 1! =

E t = 2 e1 cos

Et actua" E = cos = idea" E 2e 1 2

1 ! = cos 2 1 ! = sin 2

&

Dist ibution facto


The *indin- fo each !hase a e usua""2 dist ibuted in a numbe of s"ots. 0ddin- these +o"ta-es shou"d be achie+ed +ecto ia""2. The induced +o"ta-e in each conducto occu!2in- a ce tain s"ots is shifted in !hase f om the +o"ta-e induced in a conducto occu!2in- the fo""o*in- s"ot b2 an an-"e *he e is -i+en b2)
= ! 5

= s"ot an-"e 'e"ect(,

!= no of !o"es ,
e1

5= no of a matu e s"ots the s"ot an-"e = /'5/!(


e1 e1

%o. of s"ots !e !o"e = 5/! ,

the !o"e !itch = e"ect,


e1

e1

Et

Fo an ac *indin- *ith q= no s"ots/!o"e/!hase, if the +o"ta-e induced in the conducto s occu!2in- one s"ot 6e6 then ne-"ectin- the !hase dis!"acement bet*een s"ots *indin-s +o"ta-es the +a"ue of !hase +o"ta-e is 4o*e+e , The actua" +a"ue esu"tin- f om +ecto sum is)
E q = a sin 2 2 e = a sin 2 2

qe.

q sin E 2 = e sin 2

actua" +o"ta-e E = = dist ibution facto 1 d = idea" +o"ta-e qe

q 2 q sin 2 sin

*indin- facto = 7*=1d1!

E ms=&.&& f % 1*

.-!hase 59, $=2,,, 5=:2 n=1200 !m ;=6 %o of s"ts/!o"e =:2/6=12 s"ots ='!(/5= '1,06(/:2=18o e"ect %o of s"ots/!o"e/ !hase= q = 12/. =& s"ots q & 18 sin sin o 2 = 2 = sin .0 1d = = 0.<8, o 18 & sin : . 8 q sin & sin 2 2 The !o"e !itch=:2/6=12 s"ots
coi" !itch

=0.02& *b

The coi" s!an in e"ect ica" de- ee = !o"e !itch 1,0 = 12 1,0 = 180 =1,0-180=.0o
1 ! = cos = cos18 = 0.<66 2

10

The *indin- facto = 1d 1!=0.<660.<8,=0.<28& The emf -ene ated !e !hase is -i+en b2)

E ms=&.&& f % 1*
%o of conducto s !e !hase=2,,/.=<6 conducto s %o of tu ns !e !hase %=<6/2=&, f='!ns(/ 120=61200/120=60 4= ?oad +o"ta-e >?=E=2,& 0o ?oad cu ent
/? = >? 2,&0 o 2,&0 o = = = 86.:, .6.,: o o $? & + @. 8.6.,:
E=284 0
0

tu ns

E ms=&.&& 0.02&60&,0.<28&=2,& >


IL R L =3

+?
X L =3

?oad !o*e ;?=.'/?(2A?=.'86.,(2&=.,:1&.,, 3

$uestion 1
" t ree-p ase ,-connecte! sync ronous generator is rate! %12 M7"' %+.1 67' 2.3 8F lagging' an! *2 9:. Its sync ronous reactance is 2.4 ' an! its resistance may (e ignore!. e- 5 at is its voltage regulation; f- If t e fiel! current is constant' ) at is t e ma#imum po)er possi(le out of t is generator; 9o) muc reserve po)er or tor&ue !oes t is generator ave at full loa!; g- 5 at )oul! t e voltage an! apparent po)er rating of t is generator (e if it )ere operate! at <2 9: )it t e same armature an! fiel! losses as it a! at *2 9:; - 5 at )oul! t e voltage regulation of t e generator (e at <2 9:;

a)

EA

j.7

1..2 0 . 120 MVA .

F
V = 1..2 = :.621kV .

pf=.8 lagging

120 106 IA = .6.,: o = 82&< .6.,: o A . . '1..2 10 (

E A = V + jX S I A = :6210 + j '.:('82&< .6.,: o ( E A = 10,28616.: o V

VR =

10,286 :621 100B = .&.6B :621

c) Ignoring RA, the ma im!m o!tp!t po"er of the generator i# gi$en %&
Pmax = .V E A sin XS
= <0 o

.':<6:('1&260( 1,

Pmax = 1,.< MW IA = E A V = <02.8.8o A RA + jX S

E A = 1&260<0o V

V = :<6:0o V

%) 'ince the $oltage pro(!ce( %& a generator i# (irectl& proportional to #pee(, the $oltage )an( hence the apparent po"er) rating of the generator "ill %e re(!ce( %& a factor of *+,
Vrated = 80 '1..2kV ( = 11.0kV 60 80 '120 MVA( = 100MVA 60

S rated =

c) At *- ./ rate( con(ition#,

V =

11 = 6..81kV .

I A = 82&< A

E A = 6.810 + j '.8,.('82&< .6.,: o ( = ,8&816.6o V

0ote that the #&nchrono!# reactance i# al#o re(!ce( %& a factor of *+,, #ince reactance i# a f!nction of fre1!enc&. 2he $oltage reg!lation i#
VR = ,8&8 6.81 100B = .&.6B 6.81

$uestion +
d- =an Sync ronous motor (e starte! (y itself )it out any ai!; If your ans)er is no' e#plain t e !ifferent met o!s of starting sync ronous motor. e- For a *-pole' *2 9:' < p sync ronous motor' )it parameters are: >a ?2.2< an! @s = 0. 11 . Determine t e no-loa! spee!' t e full-loa! spee!' an! t e alf-loa! spee! of t e motor. f- It is re&uire! to supply a t ree p ase loa! of +2265' *2 9: from t e national gri!. 0 e gri! fre&uency is <2 9:. It is !eci!e! to generate t e re&uire! po)er (y means of a motor-generator set consisting of a sync ronous motor !riving a sync ronous generator. 9o) many poles s oul! eac of t e t)o mac ines ave in or!er to convert <2 9: po)er to *2 9: po)er.

<

$uestion " -327' +4< 67"' 2.3 lagging po)er factor' , connecte! sync ronous generator as a sync ronous reactance of 2.- an! a negligi(le armature resistance. 0 e generator supplies a -327 32 65' o.3 lea!ing po)er factor , connecte! sync ronous motor )it a sync ronous reactance of %.% a negligi(le armature resistance. 0 e generator is a!/uste! to ave a terminal voltage of -32 7 ) en t e motor is !ra)ing t e rate! po)er at unity po)er factor. !- =alculate t e magnitu!es an! angles of E" for (ot mac ines e- If t e flu# of t e motor is increase! (y %2 percent' ) at appens to t e terminal voltage of t e po)er system' ) at is t e ne) value; f- 5 at is t e ne) po)er factor of t e motor;

10

11

12

1.

1&

18

16

1:

$uestion <
1Descri(e one of t e follo)ing mac ines un!er t e ea!ings: Mac ine construction' principle of operation' types' an! application: %1- Stepper motor %+- Servo motor %-- 8ulls transformer %<- Microp one %*- Aou! spea6er %4- Electromec anical relay %3- Matc ing transformer %B- Isolation transformer 12- Croa! (an! transformer 1%- "u!io transformer 11- Measurement transformer D=.0. or 8.0.E

1,

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