818 E lectrical M achinery__________________________________________________________ [

(c) Two 3-ph ase induction m otors A , B a re identical in all re sp e cts e xce p t t h a t m o to r A h a s a la rg e r air
gap th an m otor B . E xp lain which of th e two m otors will have
( i ) m ore no-load cu rren t
( ii ) poorer no-load power factor and
(H i) b e tte r full-load power factor. lA ns. ( c ) : ( i) and ( ii ) Motor A , (Hi) Motor B]

6 .3 . ( a ) Describe th e principle of operation of a 3-p h ase induction m otor. E x p la in w hy th e ro to r is forced

to ro ta te in th e direction of ro ta tin g m agnetic field.
(6) D iscuss the differences betw een 3-ph ase induction m otors an d tra n sfo rm e rs.
6 .4 . A 3-ph ase induction m otor is som etim es called a generalized tra n s fo rm e r in so far as voltage and
frequency tran sfo rm atio n s a re concerned. As p er this sta te m e n ts , discu ss how a 3-p h a se induction m otor
operates under the following conditions : 1
(a) rotor frequency f 2 = sta to r frequency f\

(b )f2 < f l

(O f2> h
( d ) rotor g enerated voltage and rotor cu rre n t a re m axim u m
(e) rotor em f E 2 and ro to r cu rre n t a re zero
(/) both E 2 and / 2 are minim um
(g ) both E 2 an d 72 a re negative.

[A ns. (a ) A t stan d still ( b ) U n d er norm al runn ing conditions (c) R otor is d riven a g a in s t th e direction of
ro ta tin g m agnetic field (d ) A t stan d still (e) A t synchronous speed (f) A t no-load ( g ) W h en o p e ra tin g as a 3-phase
induction generator]
6 .5 . D iscuss the production of s ta rtin g torque, throu g h th e con cep t of in te ra ctio n of flux an d m m f waves
in a 3-ph ase slip-ring induction motor.
H ence show th a t th e ro to r is forced to ro ta te in the direction of ro ta tin g flux w ave.
6 .6 . E xp lain the production of torque in a 3-p h ase slip -rin g induction m o to r w hen th e ro to r is ru n n in g with
a slip s. H ence introduce th e concept of load angle.
D iscuss th e conditions und er which optim um torqu e is developed in a 3 -p h a s e in d u ction m otor.
6 .7 . D escribe th e developm ent of electrom agn etic torque in a sq u irre l-ca g e in d u ction m o to r through the
in teraction of flux and m m f w aves, w hen the ro to r is ru n n in g a t a speed less th a n sy n ch ro n o u s speed.

.. c^ en.C* Sh. ™ th a t th e slip' ria g and sq u irrel-cage induction m o to rs a re id e n tica l in so fa r a s th e ir rotor-reac-

tions on the s ta to r a re concerned.

6 .8 . In a 3-p h ase induction m otor, electrom agn etic torq u e is given by

Tt = ^ P2 <t>F 2 cos 02 ,

T he above torq u e exp ression can also be exp ressed a s T = K i n cos On I ,* ,., •
Kl a u* • n. i , e 2 2 E x p lain h °w th is torque expression
K I 2 cos 62 can be used to obtain the torque-slip ch a ra cte ristics of a 3-p h ase in d u ction m otor.

- 6 .9 . (a) T h e speed of rotor field, w ith resp ect to sta to r, is alw avs eau al tn 5vnr(im T,n„o j * ui
speeds o f th e induction motor. E xplain . * q syn ch ron ou s speed a t all possible

( « E xp lain why th e rotor of a polyphase induction m otor can n ev er a tta in syn ch ron ou s speed.
(c) I he rotor o f a sh p -n n g induction m otor is connected to an n r « « , , , « „ v i 4 . ,. •
short-circuited. I f rotatin g m agnetic field produced by rotor w in d in g 1 ! , ’ , W a S 1 5 s ta to r wmding IS
which the rotor m ust revolve. g ro ta tes clock w ise, ex p lain th e direction in
_ n . . i [Ans. (c) Anti-clockwise]

4-Mes but « . ^ is

to p l d ’uce fh™ tam e3 n u m b e r * o f W i l l I n d u X n ’S X t a r t k l t o r a n ’ E x Z m " ' h ° WeVer' d“ iB" ed

M
A3-phase, 50 Hr induction motor has a full-load speed of 960 rpm. Calculate
(/) number of the poles
(ii) slip freq u en cy

Sc a nn e d by C a m Sc a nn er
 — ob‘ _____________________________________ Polyphase In

(m) speed o f rotor field w ith resp ect to rotor stru ctu re ; w ith resp ect to s ta to r str u c tu r e and w ith resp ect
to sta to r field. (Ans (fl) Nq (ft) Yeg (c) g 2 Hz 4() ^ 1000 rpm flnd zero

6 .1 1 . (a) Show t h a t th e voltage gen erated in th e rotor circu it of a 3-p h ase in d u ction m otor a t a n y slip s is
equ al to s tu n es th e voltage gen erated a t stan d still.
(b ) W ith th e help of rotor eq u iv alen t circu it o f an induction m otor, show th a t th e pow er tra n sfe rre d

m ag n etically from sta to r to rotor is given by l\ — p er phase.

s
(c) E x p la in th e term s air-gap power P in tern a l m ech an ical power developed Pm and sh a ft pow er P ,h. How
a re th e se term s rela te d w ith each oth er ? H ence show th a t
Pg : rotor ohm ic loss : Plu = 1 ; s : (1 - s)
6 -1 2 . (a) Ju s tify the following sta tem en ts for a 3-ph ase induction m otor :
( ’) R otor leak ag e im pedance a t sta rtin g is different from its value a t norm al ru n n in g conditions.
( « ) R e la tiv e sp eed betw een s t a t o r field and ro to r field is zero.
(h i) S ta to r cu rre n t rises as the sh a ft load is increased.

m otor ^ v a n a b ^e *®sses m a 3-phase induction m otor ? G ive the power-flow d iagram for th is
m otor and d iscu ss th e various losses involved in it.

te rm in a ls ^ L ' a u e n c f n f M U *** ’ 4 ^ m otor is connected to 5 0 H z supply. A t th e rotor

L X ra tio o f Shn r Yu Fm d th e P° 8sible 8peeds a t w hich th e rotor m u st be ^ v e n W h at
is th e ra tio o f sh p -n n g em fs a t th e se speeds a t no load ? | Ans. 600 rpm, 2400 rpm. unityl

d ir. 6 ' 1 4 ' Y 4 ‘P° le ’ 3 *p h as®- 5 0 Hz synchronous m achine h as its rotor directly coupled to th a t o f a 3-p h ase
! i 3 Y r T 0 S ta to rs o f both m achines are connected to th e sam e 3-p h ase, 5 0 H z supply I t is
D eterm in e th e n u m b « 150 Hz across th e rotor term in als o f th e in d u ction m otor.
D eterm in e th e n u m b er of poles for w hich th e induction m achine should be wound. G ive a ll p o ssib ilities.
(7.A.S., 1987) (Ans. 8 poles or 16 poles)
a t t h t ’ n n p r a t i ^ Y 8' del^ ' C0Tm®cted - 4 'P ole<5 0 Hz induction motor h a s a s ta to r re sista n ce o f 0 .4 f i p er p h ase
? tem p eratu re. F o r a lu ie cu rren t of 2 0 A, th e total sta to r inpu t is 4 0 0 0 w atts. F o r n egligible
s ta to r core lo sses, find out th e in te rn a l torque. (Ang 2 4 4 5 Nml

6 ' i ? ‘ A. 3 ' P.haSe; 4 3 0 V ’ 5 0 Hz induction motor ta k es a power input of 3 5 kW a t its full-load speed of 9 8 0
r.p.m . T h e to tal s ta to r lo sses are 1 kW and the friction and windage lo sses are 1.5 kW . C a lcu la te (a ) slip ( b )
) rotor ohm ic lo sses (c) sh a ft power (d) sh a ft torque and (e) efficiency.

(A ns. (a) 0 .0 2 (6) 6 8 0 W (c) 3 1 .8 2 kW (d) 3 1 0 .0 6 Nm (e) 90.91% )

u 6 1 7 * ^ 4 0 0 V ’ 3 Ph a s e > 6 P °le * 5 0 H z induction m otor d raw s a pow er of 2 k W a t no load an d a t ra te d

V 1 /Y x Y x t Y Uen2Cy a fulM oad shp ° f 3 % ’ the P ° w e r in Pu t t0 m o to r is 5 0 k W an d th e s t a t o r o hm ic loss
,S i k ^ , NCg eCt/ R ! T a t n 0 ,l0ad - I f the sta to r core loss and m ech an ical losses a re assu m ed eq u al th e n at
a slip o f 3% ca lc u la te (a) rotor ohm ic loss (6) sh a ft (or output) pow er (c) sh a ft torque (d) in te rn a l torq u e and
(e) efficiency. (Ans. (a) 1.425 kW (6) 45.075 kW (c) 443.75 Nm (d) 453.60 Nm (e) 9 ? l S i j
6 .1 8 . A 2 0 kW , 6 pole, 4 0 0 V , 50 Hz 3-ph ase induction m otor h as a full-load slip o f 0 0 2 I f th e torou e lost
in m ech an ical (fn c tio n and w indage) losses is 20 N m, find th e rotor ohm ic loss, m otor in p u t and e f L e n c v
S ta to r lo sses to ta l 9 0 0 w atts. F wm -iency.
^ 20,000 x 60......... .................
IH“ ‘ - T" = 2 . » 1 0 0 0 x 0 . 9 8 * 194 88 ^ N" ’-
M ech an ical torq u e developed = 1 94 .81 + 2 0 = 2 1 4 .8 8 Nm.
. D 2n x 1000 (0.98) ............ . . I4 „
” m~ go (214.88) = ...J [Ans. 450.04 w atts, 23402.1 w atts. 85 .462%1

6 .1 9 . A 10 kW , 3 -p h a se, 50 Hz, 4 pole induction m otor h as a full-load slip of 0 .0 3 . M ech an ical and stra y
load lo sses a t full-load a re 3.5% o f output power. Com pute
(a ) pow er d elivered by s ta to r to rotor,
( b ) electro m a g n etic (in te rn a l) torque a t full load, and 1
(c) rotor ohm ic lo sses at full load. | Ans. (a) 10.67 kW (6) 67.93 Nm (c) 320.10 W|

J II IVy
 (Prob . 6
820 Electrical M a ch in e ry _________________________________________ ____ ________________________________________

6 .2 0 . A 3-ph ase induction m otor has its s ta to r copper loss e q u a l to th e su m o f th e [® "d

losses. Its rotor copper loss is equal to one-third of s ta to r copper loss. I f its e icien y <, ^ 0347)
Take m echanical loss equal to iron loss.
6 .2 1 . The power supplied to a 3-phase induction m otor is 4 0 kW an d th e co rresp on d in g s t a t o r losses a re
1.5 kW. C alculate the n et m echanical power developed and th e ro to r I R loss w |jen ® lf c m r h r n n l i l
W hat will be the net power developed if the speed of the above m otor is reduce o c ^
speed by m eans of extern al rotor resistors, assum ing th e torque and s ta to r losses to re m a in u n a e re ^ ^ jggQ \
and windage losses may be assum ed to be 0 .8 kW. ^ ^ kw ^ k — kW]

6 .2 2 . (a) Discuss why the speed of a 3-phase induction m otor falls as its load to rq u e is in cre a se d .
(6) Two w attm eters a re connected to m easure the power input to a 3 -p h a se induction m o to r ru n n in g a t no
load. One of the two w attm eters gives negative reading. W hy ? E xp lain .
(c) Explain why a 3-phase induction motor, in gen eral sim ilar to a tra n sfo rm e r, ta k e s m ore m a g n etizin g
cu rren t as com pared to a transform er.
6 .2 3 . (a) Explain why slip in a 3-phase induction m otor is directly p roportion al to to rq u e w hen o p e ra tin g
n ear synchronous speed.
(ft) A 4-pole, 2 0 kW, 5 0 Hz, 4 0 0 V SCIM has a sta rtin g torque of 1 6 0 N m and a full-load to rq u e o f 1 2 0 Nm .
C alculate
(i) sta rtin g torque for a s ta to r voltage of 3 0 0 V,
(ii ) voltage so th a t m otor operates satisfactorily a t full load from a 60 -H z so u rce,
(lii) voltage applied to sta to r so th a t full-load torque is developed a t sta rtin g .
V'
[H int. (ft) Keep j constant) [A ns. (ft) 9 0 Nm, 4 8 0 V, 3 4 6 .1 4 V]

6 .2 4 . ( a ) A 3-phase, 5 0 Hz source feeds the sta to rs of both 3 -p h a se SR IM h a v in g 8 poles an d a synchronous

m otor having 2 poles. The synchronous m otor, coupled m ech an ically w ith ind u ction m o to r, ru n s clockwise
w hereas the ro ta tin g field in 3-<J> induction m otor ro ta te s counterclockw ise.
C alcu late frequency of the voltages taken from th e slip rin gs of SR IM .
(6) Find the num ber of synchronous-m otor poles and the conditions for o b tain in g a freq u en cy of 150 Hz
from the ro to r of SRIM of p a rt (a).

(c) R epeat p a rt (ft) for obtaining a slip-ring frequency of 16 - Hz.

3
[A ns. ( a ) 250 H z (ft) 4 poles, synchronous m otor ro ta tio n a g a in st th e d irectio n o f ro ta tin g field in SRIM (c)
6 poles, synchronous m otor rotation in the direction of ro ta tin g field in SRIM.
6 .2 5 . A 4-pole, 3-ph ase SRIM is coupled m echan ically w ith a sv n rh rrm m ,c u , r™
synchronous m otor and sta to r of the induction m otor a re fed from n 7 ,n 8 P
frequency o f th e em fs a t the ro to r te rm in a ls if the sy n ch ro n ou s moTor k d r i v e n ^ S° UrCe' * * ‘ he
(a) in a direction opposite to the induction m otor s ta to r ro ta tin g field,
(ft) in the direction of ro ta tin g field in SRIM .

If the frequency of th e ro to r voltage is req u ired to be 3 0 0 H z, th e n calculate

(c) th e n u m b er of poles th a t th e SR IM m u st h av e ,a , v, ,
c . IAns - (n ) 150 Hz (ft) 5 0 Hz (c) 10 poles)

sta to rs of both t h e m achines a re given a S O -H ^ su p p lT w h a t f r e a ^ ^ 4 P° le ’ 3 ‘ p h a s e sy n ch ro n o u s m otor. If

(6) A 3-p h ase, 5 0 H z indue Uon m otor has a ^ 2 * ^ ^ ^ °f ^
m axim um torque w hich is 2 .5 tim es fuil-Ioad torq u e N e e l e d W S t 15 Um eS fulM oad tor(lue and a
assu m in g co n sta n t ro to r re s is ta n c e , find : “ s t a t o r r e s is ta n c e a n d ro ta tio n a l losses and

( i) the slip a t full load ( ii ) th e slip a t m a x im u m to rq u e and

(H
i)th e ro to r c u rr e n t a t s ta rtin g in p er u n it of full-lond ro to r c u r r e n t. , , . E .S .. W ?S |

n , , lA n s .(n ) 2 5 Hz or 125 Hz (ft) 0 . 0 5 6 , 0 . 2 6 8 . 4.7221

diagrum of aCtransformer ?80r ^ * * " P“'y,,1,“SC induCtil>'' » — •How does i, differ from the phasor

^s S tD
„Th?tprall ? : i ; t i : lt c “/eu irlyphasc induction ra°i°r-

Sc a nn e d by C a m Sc a nn er
 P ro b . 6]
Polyphase In d u ction M o to rs 821

circ u it p a ra m e te rs ? ' nt*u c t' on n iotor eq u iv alen t circu it, w h a t should be k ep t in m ind re g a rd in g th e e q u iv a le n t

W T h ^ s h u n t b r 1110 C m u ii ° f 0 0 mot°r- W h at do th e various p a ra m e te rs rep resen t ?

d u rin g th e a n a ly s is nf im *s t ' nS ° ^ c a n d X ,,, in p a ra lle l, is e ith e r o m itted or m oved to th e p rim a ry te rm in a ls

ind u ction m o to r e q u iv a le n t^ r c u T ^ E x p y a in '6111 C' rCU*t- T ^ is’ how ever, is n ° t p erm issib le in th e a n a ly sis o f

W> b" W,!en th e tr a n s f “r m c r “ d ta d “ «<>" " » t » r e q u iv a le n t c i r c u i t ,

I H i n t .( d ^ B e c a u s e o f t h ' 13" ' 63 * " “ * • ' * ™ l ™ ,h a " in ‘ " - “ fo rm ers. E x p la in ,
le a k a g e re a c ta n c e .) ^ ° Ce ° ^ a *r ®a P> induction m otor h a s m ore le ak ag e flux an d , th e re fo re , m ore

to rq u e p e r p h a se is g iven by Auction m o to r is n eglected , show from its e q u iv a le n t circu it t h a t m a x im u m

T - 1 ^
em 2 n n ,2 T 2
and hence show th a t T'
T..„
em ssmT ^ S
S smT
F o r sm a ll v a lu e o f slip o ccu rrin g in th e sta b le o p eratin g region prove th a t

r1' = -2 F„„ S
smT

m o d ified ' S k e ‘ Ch ‘ h e lyPiCa' * a r a c t e r i s , i c o f an induction m otor. How is th is c h a ra c te ris tic

(i) if its ro to r-circu it re s is ta n c e is in cre a se d ,

(ii) if its ro to r circ u it re a c ta n c e is in cre a se d

a t f„Sh,„“ „e e t rrg e " ? " " “ * d' » ° “ 'd * •

m a r i m u m t t u ? a U U r i t a g ° f “ ‘ e m a ' reS iS ,a" Ce ‘ h a l be i,1Serted “ *h ' » • « * « » in o rd e r to obtain

^ [H in t. (W U ) I f s t a t o r re s is ta n c e w ere con sid ered then th e slip a t w hich m a x im u m to rq u e o ccu rs, is given

s,nT=^ J 7 F
S in ce s mT i s re d u ce d w ith th e co n sid e ra tio n of s t a t o r re s is ta n c e , th e full-load slip w ould be sm a lle r.
r2 0 fl4
( « ) y = s , uT = 0 .2 o r X = = 0 .2 « etc.)

[A ns. (a) (,) See Fig. 6 .1 7 ( « ) All the three T „ „ Ttm and s mT a re reduced. (6) « ) 0 .0 2 5 4 , sm aller, (ii) 0 .1 6 O).

6 .3 2 . (a ) S k e tc h th e torq u e-slip c h a r a c te r is tic o f an induction m o to r w ork in g a t ra te d v o lta re and

E x p la in a n d d ra w th e s e c h a r a c te r is tic s , w ith re s p e c t to th e n orm al o- e, if th e follow ing ch a n g e s a re m a d e ^
(i) A pplied s t a t o r v o lta g e is red u ced to h a lf a t ra te d frequ ency.

(ti) B o th th e ap p lied v o lta g e an d freq u en cy a r e red u ced to half.

(6 ) F o r a 3 -p h a s e in d u ction m o to r, th e ro to r o hm ic loss a t m a x im u m to rq u e is 16 tim e s t h a t a t full lnnd

torq u e. T h e slip a t fu ll-load to rq u e is 0 .0 3 . I f s t a t o r re s is ta n c e an d ro ta tio n a l losses a re n eg le cte d , th e n c a lc u la te
(i) th e slip a t m a x im u m to rq u e ,

( ii ) th e m a x im u m to rq u e in te rm s o f full-load to rq u e an d

( iii ) th e s t a r t i n g to rq u e in te rm s o f full load torq u e.

u u a i ii i c u vjy u a i i ix j^ai 11 i c i
r
[Prob. 6
822 Electrical M achinery _______ "

T'f, 2
IHint- (6)
Sp s mT

I 21
T.n sfl S,nT- — SmT
Now -J-
7 ? LL = - —
r . = j75------------
2 7 ~ 16
7Sn 16 S fl
i , 2 2 ‘ 2m T 2 '* '
. / r , . T --- ----------
/2'"r *,nr
1 s-nr 2 .e|c_
" . 16 Sf l SmT + Sfl
Sfl s mT
[Ans. (a) (i) Ttt l and Ttm are reduced to ^, but n , and s mT remain unchanged.
w Z- increase, T „ re ra -in , same but a . is reduced te half.
(b ) (i) 0.167 (ii) 2.874 T , n (Hi) 0.934 Tt fl.\
6 .3 3 . (a ) Show th a t th e m axim um in tern al torque developed by a polyphase induction m
depend on th e rotor circuit resistan ce.
(6) A 3-ph ase squirrel-cage induction m otor has a rotor sta rtin g cu rre n t o f 6 tim es its full load value. The
m otor h as a full load slip of 5% . D eterm ine
( i ) th e sta rtin g torque in te rm s of full-load torque ;
( ii ) th e slip a t which m axim um torque occurs ; and
(Hi) m axim um torque in term s of full-load torque.
(H in t, (i) U se Eq. (6 .5 1 a ). ( ii ) U se Eq. 6 .3 7 . lAns. ( b ) ( i ) 1.8 Tr „ (ii ) 0 .3 1 (Hi) 3 .18 Tt f l ]

6 .3 4 . ( a ) W ith s ta to r resistan ce neglected, the torque-slip ch a ra c te ris tic can be obtained from th e expression

Tt
T e ni s mT S
s s Tm

D erive this expression and show th a t

s = s m7 > ± VA:2 - 1]

where ^ = -sr-
1e
(U se n egative sign if s mT > s , e . g ., s mT > and use positive sign in ca se s mT < s , e . g . a t s ta rtin g ).

(b ) The m axim u m torque of a 3-p h ase sq u irrel-cage induction m otor is 4 tim e s th e full-load torq u e and the
sta rtin g torque is 1.6 tim es th e full-load torque. N eglect s ta to r re s is ta n c e . C a lc u la te

(i) th e slip a t th e m axim u m torque

( ii ) full-load slip and
(iii) th e ro to r cu rre n t a t sta rtin g in te rm s of full-load ro to r cu rre n t. (A ns. (i) 0 .2 1 , (it) 0 .0 2 6 7 , (Hi) 7.7591

6 .3 5 . (a ) S k etch th e torque-speed curve of a conventional in d uction m o to r an d in d ica te how th is will change

w hen
( i ) th e ro to r re s is ta n c e is doubled, keeping s ta to r v oltag e an d freq u en cy u n ch a n g e d ;

( ii ) th e applied voltag e is halv ed , th e freq u en cy and ro to r re s is ta n c e re m a in u n ch a n g e d ;

( iii ) both th e applied v oltag e and freq u en cy a re h alv ed , ro to r re s is ta n c e re m a in s u n ch an g ed .

(I.A .S .. 1987)

( b ) A 3-p h a se induction m o to r a t ra te d voltage an d freq u en cy h a s a s t a r t in g to rq u e of 1 5 0 % an d a m axim um

torque of 2 0 0 p e rce n t of full-load torq u e. N eglectin g s ta to r re s is ta n c e a n d ro ta tio n a l lo sse s, ca lcu la te the slip
a t full load and slip a t m a x im u m torq u e. (/.A .S ..1 9 5 4 )
[Ans. (6) 0.452, 0.1211

Sc a nn e d by C a m Sc a nn er
 Prob. 6]
Polyphase Induction Motors 823

N m a t a sp eed o f 1 2 o \ ) r n n ! d e l t a - c o n n e c t e d in d u ction m o to r develops a m a x im u m to rq u e o f 2 4 0

is 0 .2 fl p e r p h a se C a lc u la te t h J • a i^ e & r o ta tio n a l losses a r e a ssu m e d n egligib le. T h e ro to r r e s is ta n c e
(a ) 7 W r ^ alCUlate th e ad d itio n al ro to r re s is ta n c e re q u ire d to give a s t a r t i n g to rq u e o f :
( ) 7 5 % o f m a x im u m to rq u e a t ra te d v o lta g e and

( ) 7 5 % o f m a x im u m to rq u e w hen th e su p ply v o lta g e is red u ced to 3 6 0 v olts.

,Hint‘ s mT = 0.2 ; x2 “ = 1 II.

0. s'"^
S in ce r t a n d x , a r e n egligib le,

T - i K
«». 2x2 ~ 2
or 240 = -
2
K = 480.

fa ) I f R is th e su m o f r o to r re s is ta n c e a n d e x te rn a l re s is ta n c e , th e n a t s t a t i n g

re sl ~- J - xX_ 3V? -
[«2 + *^l
or 0 . 7 5 x 2 4 0 = —^ — R etc
R 2 + 1 ,e tc -

<6 > 0 .7 5 x 2 4 0 = f M f _ i 8 0 _ R
1,400 J fl2 + i ’ ‘ IA n s. (a) 0 .2 5 1 4 Q, (6) 0 .4 7 2 fl]

t h a t to in c re a s e th e s t a r t i n g t o r q u ^ e x t S S ^ M t t e r o t o ! * ^ m 0 t° r Ke“ “ P r° V<i

P h a s e r ™ ts ^ V o P2hr c S . t X ^ l de ^ 'r 3 "17 * ^ l0">” at * • *«* °f W » ^ ™ «P-

e a ch r o , h fa p ro d u ce a ~ r £ e ^ ^ ^

[H in t, (a ) C on d ition for m a x im u m to rq u e is

r2
...(6 .2 5 )
T h e e x te r n a l r e s is ta n c e in ro to r c irc u it a t s ta r tin g is
(V ^ + J^ -r,).
(8) *- ^
2 smr, 1
*f
1 s mT1

an H _ r 2 * added resistance
2 sistance
s m T, =
. [A ns. (6) 0 .3 3 6 fl)
6 .3 8 . ( a ) E x p la in th e d ifferen ces b etw een th e c h a ra c te ris tic s nf «lirx ,
in d u ction m o to rs. S k e tc h a ty p ica l c h a r a c te r is tic for e a ch . sq u irre l-c a g e polyphasi

(6 ) A 4-pole, 3-phase, 5 0 H z induction motor has a full-load sliD o f 3% »nH » mnar- . a

full-load to rq u e . C le a rly sp ecifyin g a n y a p p ro x im a tio n s you m ay n eed to m ak e calcuTaTe^he s t a T ' th<
th e m o to r a s a p e rc e n ta g e of full-load to rq u e ca lc u la te th e s ta r tin g to rq u e o
( I .E .S .. 1975
co o t o u . , . [A ns. (6) 44.24%
b .d 9 . In a 3 -p h a s e in d u ctio n m oto r, th e s t a t o r re a c ta n c e eq u als th e ro to r r e a c ta n c e a t et a „ j c n i u . i
r e s is ta n c e is o n e -fo u rth o f th is v a lu e . I f th e m o to r develops 2 2 0 N m a t 3% slip, w h a t w ill b T i J f
(a ) s t a r t i n g an d

< W p u ll-o u t to rq u e s. Ig n o re no-load c u rr e n t. IA n .. 1 2 0 .6 7 N m . ( » 4 5 2 737 Nm

mot°r ha> a ro,M °f 0 2 0 P" ph“ « “ 0 * —

(a ) th e to rq u e for a fu ll-lo ad slip o f 4 % an d

Sc a nn e d by C a m Sc a nn er
 [Prob. 6
824 E lectrical M achinery _----------- !— :---------- ------------- - " 7 ..
------------ " . • , u* n of full-load torque a t starting.
(fe ) t h e resistance to b e added to the rotor circu it o o a | Ans. 9 2.89 Nm, 0.192 Q]

R otatio n al losses and s ta to r im pedance are neglected. ^ ^ ^ ^ ^ ^

6 .4 1 . A 4-pole, 3-p h ase, star-co n n ected slip-ring induction mo F o r negligib]e s ta to r impedance
m ains. Its rotor h as a stan d still leakage im pedance of 0 .4 + y2 ohm s p f

and rotation al losses, com pute :

( ) m axim um torque in new ton-m etres, g ^ m axim u m torq u e a t s ta rt.
( ) the resistan ce to be included in the rotor circu it to ae v • ^ ^ ^ ^ m (fc) Q6 Qj

j tor h as s ta to r im p ed an ce of 0 .0 7 + j 0 .3 0 O and
6 .4 2 . A 4 20-V , 6-pole, 5 0 H z s ta r ^ m agn etizin g c u rr e n t is n eglected . D eterm ine
stan d still rotor im pedance referred to s ta to r is 0 .0 8 + 7 0 .3 7 U. i n g
(a) the m axim um internal power developed and the corresponding slip and

(b) the m axim um internal torque and the slip at ^ RW Q 1Q44 . (6) 1088.1 76 Nm, 0.1187]

. v d in a t normal voltage. Find th e slip o fth e induction motor

6 .4 3 . (a ) A 3-phase inductionp^ t ° r ^ s^ a sli,p ^ ^ ^ th e n o rm ai voltage.
w hen developm g the sam e torque bu g tnrmlP a n d a t n orm al voltage. T h e ro to r resistan ce
^ .(b ) A 3-ph ase induction m otor h as 5% slip a t full-loa q W h a t should be th e percentage
- 3 ,4 th of n o ™ , full-load speed.

N eglect s ta to r im pedance.
[Hint, (a) U se E q . (6 .3 1 )
V\ 0.10
(*>) T* P = T T T Iv x 0.05
r0.10] + ( 0 .6)2
0.05 J
(*V?) 0.10 [Ans. (a) 4.938% , (6) 20.356%]
X 0.2875 CtC‘
6 4 4 A 440-V 3-phase, 4-pole, 5 0 Hz slip-ring star-co n n ected induction m otor h a s a voltag e of 8 0 V between
slin Hnirs when full-voltage is applied to the sta to r and the slip-rings a re o p e n -c rc u ite d w ith the rotor
s a t i o n a i The sta to r cu rren t a t no-load is 2A a t a p f of 0 .2 lagging. T h e ro to r is sta r-co n n e cte d w ith a per
phase stan d still leakage im pedance of 0 .0 5 + j 0 .2 5 fi referred to rotor. F o r th e m o to r ru n n in g w ith th e slip-nngs
sh ort-circuited and a t a slip of 5% , calculate
(a) the torque developed in Nm,
{ b ) the mechanical power developed,
(c) the rotor ohmic loss and •
(id ) the stator current and power factor. -9
Neglect stator leakage impedance and rotational l’g sses.
[H in t, (d) R otor cu rre n t a t 0 .0 5 slip, / 2 = 4 4 .8 1 A.

Rotor pf at 0.0 5 slip, cos 02 = 0.970

Rotor current in phasor form, 72 = / 2 (cos 02 - j sin 02) = 43.4 7 - j 10.89.
Stator current required to balance the rotor current is given by
" _1_T
5.5

T otal s ta to r cu rre n t, I ] = ( / 1 ' + I q)

= - b h + ( 0 . 4 - 7 1.96) = (8 .3 - j 3 .9 4 ) A e tc.].
5 .5 J
[Ans. (a) 3 8 .3 5 Nm (6) 5 7 2 2 .8 w atts (c) 301.2 w atts (d) 9 .1 8 8 A at 0.903 pf lagging !

6 .4 5 . (a ) A 3 -p h a se induction m otor h a s o p eratin g pf of 0 .8 5 a t full-load sp eed of 9 6 0 r.p .m . and a t 400 \

supply voltage. In ca se th e supply voltage falls to 3 5 0 V, find th e o p e ra tin g p f a t th e sa m e full-load torque.
( b ) A 3 -p h a se induction m o to r h as a ro to r re s ista n ce of 0 .5 D p er p h ase an d ro to r sta n d still leak ag e roa' ta " 2
of 1 5 O p er p h ase. I f th c ra tio of m axim u m s ta rtin g torq u e to full-load to rq u e is 2 , find th e ratio o
s ta rtin g torque to full-load torque for d ire ct sta rtin g . N eglect s ta to r im p ed an ce a n d ro ta tio n a l losses.

Sc a nn e d by C a m Sc a nn er
Prob. 6] Polyphase Induction Motors 825

[H in t, (a) 0 .8 5 = t= s ^ = y •
Vr22 + ( 0 .0 4 * 2)2
This gives r 2 = 0 .0 6 4 5 4 jc2.
Slip a t reduced voltage = 0 .0 5 2 2 etc.
(6) M axim um startin g torque can, at the most, be equal to maximum torque Te m . H ere s mT = 1 /3 .

Te-it 2
etc. (A ns. (a) 0 .7 7 7 5 lag (6 ) 0.6 1 4 3 )
T<m ' 1 /3 1
1 1 /3
.4 6 . A 3 -p h a s e , 5 0 H z, 4 0 0 -V w ound-rotor induction m otor ru n s a t 9 6 0 r.p .m . a t fu ll-load . T h e ro to r
resis a n ce an d sta n d still re a c ta n c e p er p h ase a re 0 .2 11 an d 1 £1 resp ectiv ely . I f a re s is ta n c e of 1 .8 £2 is ad ded
o e a c h p h a se o f th e ro to r a t sta n d still, w h a t would be th e ra tio o f s ta rtin g to rq u e w ith full v o lta g e an d th e
added re s is ta n c e to th e full-load torq u e u n d er n o rm al r u n n i n g conditions ? S ta te a ssu m p tio n s m a d e in y o u r
ca cu la tio n s. C a n th e sa m e s ta r tin g torq u e be obtained w ith a n o th e r v alu e of th e a d d ition al r e s is ta n c e ? E x p la in .
It th e a n sw e r is y e s, find its v alu e. ( I E S 1979 )

H in t . Te.,t = — • | v a ; r e/r = ^ - . - 5 - V 2 e tc
e s l (os 5 ’ ef l (og 2 6

[A n s. A ssu m p tio n s, ( i ) s ta to r im p edance and ro ta tio n a l losses ignored an d ( ii ) in d u ction m o to r p a ra m e te rs

re m a in co n sta n t. 2 .0 8 . Y e s , b u t w ith an additional re sista n ce of 0 .3 £1).

6 .4 7 . A 4 0 k W , 3 -p h a s e slip rin g induction m otor of negligible s ta to r im pedan ce ru n s a t a sp eed o f 0 .9 6

tim e s syn ch ron ou s speed a t ra te d torque. The slip a t m axim u m torq ue is 4 tim e s th e full load v alu e. I f th e
ro to r re s is ta n c e of th e m o to r is in cre a se d by 5 tim e s, determ in e

(a ) th e sp eed, pow er o u tp u t and ro to r ohm ic loss a t ra te d torque

( b ) th e sp eed co rresp on d in g to m axim u m torque.
N eglect m e ch a n ica l losses. ( G A T E 1993 )
[Ans. (a) 0.5 2 Ns, 2 1 .67 kW , 20 kW ( b ) 0 .0 4 Ns]
6 .4 8 . A 5 k W , 4 0 0 V , 5 0 H z, 4 poles d elta conn ected 3-p h ase induction m o to r is su p p lied by a cab le of
negligible in d u cta n ce . On s ta rtin g th e m otor using a s ta r-d e lta s ta r te r , it is found t h a t th e s ta r tin g to rq u e is
th e sa m e on s t a r a s w ell a s d elta connection, due to th e v oltag e drop in th e feed er re s is ta n c e . T h e eq u iv alen t
circu it p a ra m e te rs of th e m o to r a re a s follows :

ri - 1 £1, * i = 4 .5 £1, r 2' = 1.4 Cl and x 2' = 4.5 Cl

D eterm ine th e feeder resistance. (G A T E 1991 )
[Ans. 5 .3 7 8 £1]
6 .4 9 . A 4 0 0 V, 5 0 H z, 3 -p h a se , d e lta connected, 6-pole induction m o to r h a s th e follow ing sta n d still le a k a g e
im p ed an ces p e r p h a se re fe rre d to s ta to r:
z 1 = 1.2 +>2.1 £1, z2 = 1.32 + >2.2 £1

N eg lectin g m a g n e tiz in g im p ed an ce, d eterm in e

(a ) g ro s s o u tp u t p ow er a t a slip of 0 .0 4 ,

( b ) th e ch a n g e in line c u rr e n t w hen th e s ta to r te rm in a ls are sw itch ed from s t a r to d e lta a t a slip of 0 4

(c) th e re s is ta n c e to be added in e a ch ro to r p h ase to obtain a s ta rtin g torq u e of 3 5 0 N m

[Ans. ( a ) 127 98.6 W (b ) 3 7 .1 0 5 A, 111 .309 A (c) 1.08 £1 or 6.9 8 Cl]

6 .5 0 . A 1 0 k W , 6-pole, 5 0 H z, 3 -p h a se induction m otor h a s lin e a r torq u e-slip c h a ra c te ris tic s b etw een zero
torq u e an d m a x im u m to rq u e. Th e slip a t w hich m a xim u m torq u e of 5 2 0 N m o ccu rs is 0 .2 . F o r m e ch a n ica l losses
of 6 0 0 W , fm d th e sp eed a t w hich th e m o to r would ru n w hen d eliverin g ra te d sh a ft pow er. [Ans. 9 5 9 .5 rpm)

6 .5 1 . T h e n o rm a l full-load slip an d s h a ft torqu e o f a 3 7 3 kW , 5 0 H z, 3 -p h a s e ind u ction m o to r a re resp ectiv ely

1.9% a n d 2 4 0 0 N m . T h e ro to r w in ding h a s a re s is ta n c e of 0 .2 5 £1 an d a sta n d still le a k a g e re a c ta n c e of 1 .5 £1
p er p h ase. E s tim a te th e slip an d th e pow er o u tp u t for th e sa m e full-load c u rre n t w hen e x te rn a l re s is to rs of
2 £1 p e r p h a se a r e in s e rte d in th e ro to r circu it. N eg lect no-load cu rre n t. [/.A.S., 79971

[H in t. F o r so lu tion , re fe r to C h a p te r on P oly p h ase Induction m a ch in e s in th e book “G en eralised T h eory

of E le c tric a l M a ch in e s” by th e sa m e a u th o r.) [Ans. 17.1.% , 3 1 5 .2 0 6 kW)
 826 Electrical M a c h i n e r y _____________________________________________________________________ [Prob . 6

6 .5 2 . A 3 -p h a se indu ction m otor h a s 4-pole sta r-co n n e cted s ta to r w in d in g an d ru n s on 4 0 0 V , 5 0 Hz supply.

T h e rotor re sista n c e is 0 .1 Q and re a cta n c e 0 .9 O. T h e s ta to r to ro to r tu r n s ra tio is 1 .7 5 an d th e fu ll-lo ad slip
is 5%. C a lcu la te th e full-load pow er output. D ete rm in e also th e m a x im u m to rq u e an d th e co rresp o n d in g speed.
[7.A.S., 1998 \
[A ns. 1 2 1 .0 8 0 6 kW , 184.791 Nm, 1333.5 rpm]
6 .5 3 . A 3-p h a se, 6 -pole, 5 0 H z, 4 0 0 V S C IM h a s p e r-p h a se ro to r le a k a g e im p e d a n ce = 0.8 + j 2 .4 Q a t
sta n d still. R oto r copp er b a rs o f th is m otor a re rep laced by a lu m in iu m b a r s . D e te r m in e how th e follow ing
perfo rm an ce p a ra m e te rs ch an g e w ith re sp ect to th e ir previou s v a lu es :
S lip a t m ax im u m to rq u e, s ta r tin g torque, m axim u m to rq u e, fu ll-lo ad slip .
T a k e r e sis tiv ity = 2.1 p fl cm fo r copper an d 3 .4 p fl cm for a lu m in iu m . [A ns. 1 .1 6 2 5 , 1.2 267, 1.0, 1.1625]
6 .5 4 . A 3-p h a se, 5 kW , 4 0 0 V , 5 0 Hz S C IM h a s th e follow ing d a ta :
F lu x p er pole = 4 0 m W b, m ean conductor le n g th = 18 cm
C opper d en sity = 8 .9 4 gm / cm 3, cu rre n t d en sity = 4 A / m m 2.
F u ll-lo ad efficien cy = 0 .8 4 and pow er fa cto r = 0 .8 5 .
C om p are th e am o u n t o f copper requ ired for s ta to r w ind ing i f th e in d u ctio n m o to r is d esig n e d to o p erate
w ith s ta to r w in d ing in (a) d elta an d ( b ) s ta r .
[A ns. ( a ) D elta : T o ta l tu rn s = 144, cond u ctor w eig h t = 0 .6 7 6 2 kg.
(b ) S t a r : T o ta l tu rn s = 8 4 , cond u ctor w eig h t = 0 .6 8 3 1 kg.]
6 .5 5 . A 3 -p h a se in d u ction m otor is design ed to o p era te a t ra te d v o lta g e V a n d r a te d fre q u e n c y f . In ca se
both sou rce voltage an d freq u en cy a r e re sp ectiv ely ch a n g ed to (a ) ~ , f / 2 (6 ) V , f / 2 (c) — , f
£ 2
(d ) V , 2f ; find th e m axim u m an d s ta r tin g to rq u e s in te r m s o f th e ir r a te d v a lu e s . N e g le c t a ll s t a to r losses.

A n s . (a) 1, 2 ( 6 ) 4, 8 (c) i i (d ) J , |

6 .5 6 . (a ) W h a t a re th e fa cto rs th a t govern th e o p e r a tin g c h a r a c te r is tic s o f p o ly p h a s e in d u c tio n m otors

(W In ca s e o f p o ly p h ase in d u ctio n m otors, ex p la in w hy th e ro to r sp e ed f a lls a s th e lo a d to r q u e is in creased .
(c ) R ow er fa c to r o f a p o ly p h ase in d u ctio n m o to r is low a t n o -lo ad b u t it im nrn v p s
is in crea sed . E x p la in . ’ im p ro v es a s th e lo a d on th e m otor

one offfie? Z £ £ Z Z S S tZ S S X ; *“ “ “ * —**

(a ) M o to r A h a s open s ta to r s lo ts b u t m o to r B h a s sem i-clo se d s t a t o r s lo ts .
Cb) M o to r A h a s sem i-clo sed ro to r slo ts b u t m o to r B h a s clo sed r o to r slo ts .
(c) M o to r A h a s lo n g er a ir-g a p th a n m o to r B.
F o r e a c h o f th e th r e e d iffe re n c e s lis te d a b o v e, d isc u ss w h ich o f t h e tw o m o to r s h a v e •
(i) b e tte r s t a r tin g to rq u e
( i i ) h ig h e r b rea k d o w n to rq u e

( i i i ) b e tte r fu ll-lo a d pow er fa c to r and

( i v ) h ig h e r fu ll-lo a d speed . ..

6.58. fa) In case of induction motors exnlain „h„ ,h • , ^

possible. ’ P hy the air-gap length is kept as small as is mechanically
(b) Show that polyphase induction motors possess shunt .
W A 3-phase squirrel cape induction motor I d o lled “ o h * 7 “ '
aluminium is used for rotor bars, exnlain what hnrm.n. t •* [ copper bars. If, instead of using copper,
conditions. ’ ^ What haM>ens >«• speed, efficiency etc. under normal running

therete% ffic^ yda ^ l c ^ r main8*— •*— °*'mic 1— increase, output and,'

from no-load and blocked-rotor tests^Md* pw p L ^ ^ t O T w r n d ^ B ^ ^ i i t ^ c r 0" m°l°r be daten"ined

CMA 400 V. 50 Hz, 3-phase star-connected squirrel-cage induction motor^ave the following test results :

Sc a nn e d by C a m Sc a nn er
 P rob. 6] Polyphase Induction M o to rs 827

No-lond or o p en -circu it te s t (lin e v a lu es) : 4 0 0 V , 9 A, 5 6 0 W . B lo c k e d -ro to r o r s h o r t c ir c u it t e s t (lin e

v a l u e s ) : 2 1 0 V , 3 6 A, 4 8 2 0 W.
T h e effectiv e s ta to r re sista n ce is 0 .7 2 f l per p h ase. C a lcu la te th e eq u iv a le n t c irc u it p a ra m e te rs .
IA n s. r , = 0.72 fi, r 2 = 0.59 fl, x , = x 2 = 1.567 Q , X , n = 24 D|

6 .6 0 . (a) W h a t is re p resen ted by th e circle d iag ram o f an induction m otor? W h a t in fo rm a tio n ca n b e o b ta in e d

from it ?

D erive th e cu rren t-lo cu s for th e ro to r-circu it o f a polyphase ind u ction m otor.

(6 ) Show th a t th e d ia m e te r o f current-locbs^circle o f a polyphase in d u ction m o to r is V j / X j + x 2 , w h e re Xj
is th e per p h ase s ta to r le a k a g e re a cta n c e , x 2 is th e 's ta n d s till per p h a se ro to r le a k a g e r e a c ta n c e re fe rre d to
s ta to r and V , is th e per p h ase s ta to r voltage.

6 .6 1 . (a) E x p la in how th e circle diagram for a polyphase induction m otor ca n be d raw n from its t e s t d a ta .
( 6 ) A 4 0 0 V , 3 -p h a se, 8 pole, 5 0 Hz sta r-co n n ected induction m otor gave th e follow ing t e s t r e s u lts :
No-load test (line v a lu e s ): 400 V, 10 A, cos 0O= 0.2.
Blocked-rotor test (line v a lu e s ): 160 V, 30 A, cos 0,f = 0 . 35.

If, a t full load and rated voltage, th e power fa cto r is a t its m axim u m , th e n c a lc u la te fu ll-lo ad c u r r e n t,
pow er factor, torq u e in n ew ton -m etres, speed, power output and efficiency. S ta to r and ro to r o h m ic lo sse s a re
equaL lA ns. 28.75 A, 0.806, 174.224 Nm, 698.7 r.p.m ., 12.644 kW, 78.834%1

r n A 4 klV’ 400 Y’ 50 Hz‘ 3 ‘p h a se >4 -Pole d elta connected slip rin g ind u ction m otor h a s s ta to r r e s is ta n c e
o f 0 .3 6 D p er p h ase, rotor re sista n ce o f 0 .0 6 D per phase and per p h ase s ta to r to ro to r tu r n s r a tio o f 2 T h e
following d a ta p e rta in s to th e line v alu es d uring lig h t load te sts :
No load ; 400 V, 3.3 A, cos 0(1= 0.174
Locked rotor : 210 V, 16 A, cos 0 = 0.45

D raw th e circle d iag ram and com pute :

(a) lin e c u rre n t, pow er factor, slip, torque and efficiency a t full load,,

torque^ ^ ° p e ra tin g pow er fa cto r. m axim um power output, m axim u m to rq u e in N m and slip a t m a x im u m

startin g 1116 eX te m a l re s ista n c e th a t m u st be in se rted in series w ith ro to r circ u it to o b ta in m a x im u m to rq u e a t

(Ans. (a) 8.66 A, 0.8434, 0.057, 26.738 Nm. 77.65% (6 ) 0.852. 6.42 kW, 50.04 Nm, 0 .2 1 (c) 0 .2 2 6 4 f l referred to
1ULUI *J *

6 .6 3 . (a ) I f no-load pow er fa cto r a n gle 0O for a polyphase induction m otor is ta k e n to b e e q u a l to 9 0 °, show

w ith th e help o f circ le diagram th a t th e b est possible op erating pow er fa cto r is given by V*
V , + 2 / 0 (x , + x 2) ‘
(b) Tw o in d u ction m otors A an d B a re id en tical in a ll resp ects exceDt th a t m ntnr 4 c i

r : M t T p o t ^ r r win possess hie,,er v a J °f £ « r ^ x pow:sr “ dr

(H in t. (6 ) In d u ctio n m otor le a k a g e re a cta n ce is d irectly proportional to th e n u m b er o f poles.)
lA n s. ( b ) M otor B . C on sequ en tly slow -speed ind u ction m otors (m otor A h e re ) o p era te a t poor pow er fa cto rs I
6 .6 4 . (a ) W h a t a re th e cau ses o f an induction m otor o p eratin g alw ays a t lag g in g pow er fa cto rs ?
E x p la in how th e pow er fa cto r o f a n induction m otor is controlled by s ta tic ca p a cito rs Show th a t for n
co n sta n t ca p a cita n ce , th e d egree o f pow er fa cto r co rrectio n is n ot th e sa m e a t d iffere n t load s.
(6 ) A d elta-co n n ected ind u ction m otor o p eratin g from a b alanced 3 -p h a se, 4 0 0 V 5 0 Hz suddIv tn k e« n
cu n-ent o f 5 0 A a t 0 .7 6 p.f. lag. C a lcu la te per p h ase v alu e o f th e ca p a cita n ce and to tal’ kV A ra tin g o f th e 3 D h o s o
d elta-con nected ca p a cito r b a n k w hich w hen connected to th e m otor te rm in a ls would im prove th e lin e pow er
factor to 0 .9 lag. | w 6 4 „ mF „ 768PkVA]

6 .6 5 . (a) D iscu ss why th e pow er fa cto r o f a 3 -p h a se ind u ction m otor is low a t (a ) no-load and ( b ) also u n d er
overloads. UCI

(6 ) A th re e -p h a se ind u ction m otor ra ted a t 4 0 0 V, 5 0 Hz, coupled to a pum p is ru n n in g a t a low pow

la cto r o f 0 .6 . T h e in p u t is 4 .5 kVA. It is proposed to im prove th e pow er fa cto r to 0 .8 by co n n ectin g a
d elta-conn ected ca p a cito r b a n k . F in d th e v alu e o f th e ca p a cito r p er p h ase. (G A T E , 1 9 9 -i) (A ns 18 104 pFl

iKJNNriiwnnroi wi
Sc a nn e d by C a m Sc a nn er
828 Electrical Machinery ________________________ [Prob. 6

6 .6 6 . (a ) W h a t is th e o b je ctio n to th e re d u ce d -v o lta g e s t a r tin g o f p o ly p h a se in d u c tio n m o to rs ?

( b ) U n d e r w h a t co n d itio n s is th e d ire ct-o n -lin e s t a r tin g o f p o ly p h a se ca g e in d u c tio n m o to rs p re fe rre d ?

(c) T h e s ta r tin g period from zero to n o rm a l sp eed is m o re s e v e r e in c a s e o f d .c. m o to rs th a n in in d u ctio n

m o to rs. E x p la in .
(d) A 3 -p h a s e in d u ctio n m o to r a t n o rm a l sp eed is c a r r y in g a b o u t h a l f th e r a t e d lo a d . I f o n e o f th e su p p ly -lin e
fu se s blow s off, w ill th e m o to r sto p or co n tin u e ru n n in g ? E x p la in .
lA n s . (c) L a rg e s ta r tin g c u r r e n ts in c a s e o f d.c. m o to rs g iv e r is e to c o m m u ta tio n d iffic u ltie s etc.
(d) M o to r w ill co n tin u e ru n n in g a s a sin g le -p h a s e in d u ctio n m o to r a t a s lig h tly re d u ce d speed .]
6 .6 7 . A 3 -p h a s e in d u ctio n m otor h a s ro to r im p ed a n ce o f 0 .0 3 + . / 0 .1 8 f i p e r p h a s e re fe r r e d to sta to r.
F u ll-lo a d slip a t ra te d v o lta g e is 3% . E s tim a te th e p e rc e n ta g e re d u ctio n in s t a t o r v o lta g e to d evelop full-load
to rq u e a t h a lf o f th e fu ll-lo ad speed . A lso, d e te rm in e th e pow er fa cto r. N e g le c t s t a t o r im p e d a n c e an d m ech a n ica l
lo sse s. (A n s. 22.85% , 0.308)

6 . 68 . A 3 -p h a se , 4 -pole, 5 0 Hz in d u ctio n m o to r, d u rin g th e s h o r t-c ir c u it t e s t , to o k 1 0 0 A a n d 3 0 kW . In

c a s e s ta to r r e s is ta n c e is eq u a l to e q u iv a le n t s ta n d s till ro to r r e s is ta n c e , co m p u te t h e s t a r t i n g to rq u e .
(A n s. 95.493 Nm]

6 .6 9 . A 3 -p h a se sta r-c o n n e cte d , 4 4 0 V , 5 0 H z, 4-p o le in d u ctio n m o to r h a s th e fo llo w in g c o n s ta n ts in ohm s

p e r p h a se re fe rre d to s ta to r sid e :
r , = 0.2 9 4 , x , = 0 .5 0 3, r 2 = 0.1 4 4 , x 2 = 0 .2 0 9 , X m = 13.25

T h e s ta to r core lo sses a r e n eg lig ib le.

T o ta l frictio n and o th e r lo sses (a ssu m ed co n s ta n t) = 1 4 0 0 W
F in d th e pow er o u tp u t in k W an d th e ra te d o u tp u t in N m i f th e m o to r is b e in g o p e r a te d a t ra te d voltage
and freq u en cy w ith 2 p e rce n t slip . ( I .A .S ., 1 9 7 9 ] | Ans. 2 1 .0 9 3 kW , 137.02 Nm]

6 .7 0 . A 2 0 kW , 4 0 0 V, 3 -p h a se in d u ctio n m o to r h a s fu ll-lo a d pow er fa c to r o f 0 .8 6 a n d fu ll-lo a d efficiency

o f 0 .8 8 . W ith s ta to r w in d ing in d e lta , s h o r t-c ir c u it lin e c u r r e n t a t 2 0 0 V is 7 0 A. I f th is m o to r is fitte d w ith a
s ta r -d e lta s t a r te r , find
(a ) th e r a tio o f s ta r tin g to fu ll-lo ad lin e c u r r e n t and
( 6 ) th e s ta r tin g to rq u e in te rm s o f fu ll-lo a d to rq u e , for a fu ll-lo a d s lip o f 5 % .
(H in t. ( b ) U s e E q . (6.63)1. [A n s. ( a ) 1.223 ( 6 ) 0.2244]
6 .7 1 . (a ) W h y is i t n e c e s s a r y to em ploy sp e c ia l s t a r tin g a r r a n g e m e n ts fo r in d u c tio n m o to rs ?
( b ) C a lc u la te th e re la tiv e v a lu es o f s ta r tin g c u r r e n ts an d s t a r tin g to r q u e s o f a 3 -p h a s e squ irrel-cage
in d u ctio n m o to r, w h en i t is s ta r te d by
(i) d ire ct-o n -lin e s t a r t e r , •
(ii) s ta r -d e lta s t a r t e r an d
(tii) a u to -tr a n s fo r m e r s t a r t e r w ith 6 0 % ta p p in g . . ,
(A n s. (o) 1 : - : 0.361

I- A 7.2 ’ ^ 3 'P ^ a s e s q u ir r e l ca g e in d u ctio n m o to r h a s a s h o r t-c ir c u it c u r r e n t o f 5 tim e s th e fu ll-lo a d current.

It s fu ll-lo a d slip is 5% . C a lc u la te th e s t a r tin g to rq u e as a p e r c e n ta g e o f fu ll-lo a d to r q u e i f th e m otor is started

( ) d ire c t-o n -lin e s t a r t e r ,

( ) s t a r -d e lta s t a r t e r ,

find f r t " v lim it: r 6 th e " ? 0 , ° r « ™ n t to tw ic e t h e m o to r fu ll-lo a d cu rren t. Also

supp y’ m t o r m s o f m o t o r f u u -,o a d “ wha* - ihc

th e e u !o at ^ " ; se0 S" P Ply H" e S la r tin B fu ll-lo a d Fiad

(Hint. (6) and (c). Read auto-transformer starter between lines.]

(A n s. (a) 1.25 ( 6 ) 0 .4 1 7 (c) 0 .2 0 , 0 .8 Ip . 40% (d ) 0.5, 63.2%l

an d d r a t « f f r r tet t V° lta g ? ’ 8 p ° ly p h a s e in d u ctio ? m o to r d ev e lo p s a s t a r t i n g to r q u e o f tw ic e th e full-load torque

and d ra w s fro m th e su p p ly a s t a r t in g c u r r e n t o f 6 tim e s th e fu ll-lo a d c u r r e n t.

D e t e ^ ™ t h T ™ . “ S.U w ed " $ i" B ‘ M m Pc n s a t o r ta u to -tr n n s fo r m e r ) w h ic h s t e p s dow n th e v o ltage to 7 0 * .

e te r m in e th e m o to r s t a r t in g c u r r e n t, s t a r t in g c u r r e n t d ra w n fro m th e s u p p ly a n d th e s t a r t in g torque.
 —------------------------------------------------------------------------------------------Polyphase Induction Motors 829

a u ^° tra n sfo rm e r s ta r te r lim its the s ta rtin g cu rre n t from th e supply to tw ice th e fu ll-load c u rre n t,
d eterm in e th e s ta r tin g torque and au to -tran sfo rm er tapping.

H in t. From Eq. (6.51). s p = (Ans. (a) 4.2 Ip , 2.94 I p , 0.98 T , p (6 ) 0.667 T , p . 57 . 74 % tappingl

tim e * thn n . n T ' * ! 0/ Cage ‘n d uct‘on m °to r when sta rte d by sta r-d e lta s ta r te r , develops a s ta r tin g to rq u e o f 0 .4
th e su n nlv lin orque and l a ^es from th e supply a s ta rtin g cu rren t o f tw ice th e full-load c u rre n t. C a lc u la te
SUpply h n e cu rre n t and s ta r tin g torque if th is induction m otor is sta rte d by
( ) a cro ss-th e -lin e sta r te r ,
( ) a u to -tra n sfo rm e r s ta r te r w ith 80%. tapping.

H in t. Here sn = — ,. . . _.
. n 30 | Ans. (a ) 6 I p , \ .2 T t p ( b ) 3.84 l p , 0 .7 6 8 T e p \

powe6r factor oPfO S ’ w S Y ’ J 0? V Squi.r r e l‘ca e e induction m otor h a s a full-load efficien cy o f 0 .8 7 an d a fu ll-lo ad
ra ted voltage Its full lu d e lta * th e m ° t0 r ta k es a s ta r tin S c u rre n t o f 8 0 a m p e re s a t
th a t s ta r tin g torq u e eo u al to h a lf th ' f* u i ; m m im um s ta r lin E cu rre n t to be ta k e n from th e su pply in o rd er
6 t0rqUe eq u al t0 h a lf th e full-load torque is developed w hen sta rte d by
(a ) a n a u to -tra n sfo rm e r sta r te r ,
( b ) s ta to r r e sisto r s ta r te r .

C a lc u la te th e p e rce n ta g e tap p in g on a u to -tra n sfo rm er also. | Ana. (a) 47.432 A, 77% U pping <(,) 51 6 6 4 Al

r o t o n r la c U m e 'o f 2 ^ Sft'p e r T h e r a t io IT o to r ,h a s • ,">*” e fO .3 n p e r p h a se an d a s ta n d s till

T
[H in t. W ith d ire ct sw itch in g , esl ^
3T eP ~ 0.12 + 1
2 0.12
W ith s ta r -d e lta s ta r tin g torq u e is reduced to o n e-th ird o f th a t d urin g d ire ct sw itch in g ]. [A n s. 0.2366]

6 .7 7 . A 5 0 kVA 4 4 0 V , 3 -p h a se, 5 0 Hz in d u ction m otor is provided w ith a 3 -p h a s e step -d ow n a u to tr-xn*

form er s t a r t e r w h ich ste p s down th e v o ltage to 50% o f th e input. G iven th a t th e s ta r tin g c u r r e n t n f t l i
on ra te d v o ltag e ,s 6 tim e s th e ra te d full-load c u rre n t, e s tim a te th e c u r r e n t d raw n hv r / a! u „ t h e fm o to r
from th e m a in s a t s ta r tin g . W h a t is th e s ta r tin g kV A draw n by th e a u to -tra n sfo r m e r ? u T s w Y )

IA n s . 9 8 .4 1 5 A, 75 kVA]
6 .7 8 . T h e s t a r tin g c u r r e n t o f a d elta -co n n ected 3-p h a se ind u ction m o to r a t ra te d v o lt * ™ ic * *• ‘
fu ll-load c u r r e n t an d th e slip a t fu ll load is 5% . T h e no-load c u rre n t is n eg lig ib le . S
( ) I f a n a u to -tr a n s fo r m e r s t a r t e r is u sed to lim it th e s ta r tin g c u r r e n t from m a in s to 9 r n i
c u r r e n t, e s tim a te th e s ta r tin g to rq u e th e n o b ta in ed a s a p e rce n ta g e o f th e fu ll-lo ad to rq u e ° °3
<M N e g le c tin g s t a to r im p ed a n ce, d e te rm in e th e slip a t w h ich m a x im u m to rq u e o ccu rs in th e m otor.

( I .A .S ., 1 9 9 1 ) [A n s. (a) 5 0 per ce n t (6 ) 0.253]

6.79. A 3-phase, 4 0 0 V , 5 0 A, 4-pole, 1 4 4 0 r.p.m. induction motor ta k e s a hlnrkod rn in * ro •
it s fu ll-lo a d c u r r e n t a t 0 .4 p.f. la g a t ra te d v o lta g e an d develops a to rq u e o f 1 8 tim e s it s fu ll lo a d toroY
in d u ctio n m o to r is s ta r te d by a n a u to -tr a n s fo r m e r w ith 6 0 % ta p p in g th e n a t th e tim e o f t i i
m o to r p o w er fa c to r ( 6 ) m o to r c u r r e n t (c) lin e c u r r e n t « ) p o w e r 't a p * to m o to r “' f
o f fu ll-lo a d to rq u e . IA n, (a )P „ , 4 ,a g ^ m A

6 .8 0 . A SCIM has a starting current of six times the full-load current at a slip of 0.04 Calculate the lin e
current and starting torque in p.u. of full-load values for the following methods of starting :
ja ) Direct switching

( ) Auto-transformer starting with motor current limited to 2.0 p.u.

(c) Star-delta starting.
.a , a. U.E.S.,20021
I la ) 6 Pu - 144 Pu (fc) 0.6 7 p.u., 0 .1 6 pu (c) 2 pu, 0 .4 8 pul

caCe'nduclio)^otors ?°l °dViSab'C^ S‘art W°“nd'r0l0r induction molors b* lhc employed for stnrtinC

" .o to ? m h e " ^ r « n B T CXtemal rCSiStnnC° lhC ° f " « « » * * — induction

3 830 Electrical M achinery____________________________________________________________ [P ro b ^

(c) How is th e induction m otor operation affected i f th e e x te rn a l r e s is ta n c e in th e ro to r circ u it is n ot fully

cu t o ff ? ,
(d) T h e re sista n ce m easu red a t th e s ta to r te rm in a ls o f a p olyphase in d u ctio n ^ ° to r 9 , or a g’ ven
lin e cu rre n t show th a t th e sta to r ohm ic losses would be sa m e w h eth er it is s a r or
6 .8 2 . C a lcu la te th e value o f th e re sista n ce elem en ts o f a 5 -step (o r 6 -stu d ) s t a r t e r for a 3 p h a se , 400l V
w ound-rotor ind u ction m otor. T h e full-load slip is 2.5 % and th e m ax im u m s t a r in g c u rr
load value. R otor re sista n ce per phase is 0 .0 2 12.
D erive an y form ula used sta tin g th e variou s assu m p tio n s m ade.
|Ana. 0 .4 1 8 12, 0 .2 0 G, 0 .0 9 6 12. 0 .046 G, 0.022 G|

6 .8 3 . D esign a 4-step s ta r te r for a 3 -ph ase wound rotor ind u ction m otor. T h e fu ll-lo a d slip is 2 .5 % and th e
m axim um sta r tin g cu rre n t is lim ited to 1.6 tim es its full load value. R oto r r e s is ta n c e p e r p h a se is 0 .0 2 12.
D erive th e form ula used for ca lcu la tin g th e re sista n ce sectio n s and s t a te th e v a rio u s a ssu m p tio n s m ade.
[A ns. 0 .2 7 7 12, 0 .1 2 4 12, 0 .0 5 5 12, 0.025 111

6 .8 4 . (a) E x p la in an y two m ethods for perform ing th e p o la rity t e s t on a 3 -p h a s e in d u ctio n m otor.

(b ) A p o larity te s t is perform ed on a 3-p h a se, 4 0 0 V w ound-rotor in d u ctio n m o to r. P h a s e A is g iven 100 V
and th e v o ltm eter connected across th e two te rm in a ls o f p h a se s A an d B is s e e n to reco rd a v o lta g e o f 141 volts,
w ith th e ro to r open circuited. W h a t a re th e p o larity m a rk in g s o f p h a se B , r e la tiv e to p h a se A ?
[A n s. (6 ) T h e two te rm in a ls o f p h ases A and B , w hich a re co n n ected to g e th e r , h a v e th e s a m e p o larity
m arkin gs.)
6 .8 5 . ( a ) W h a t is a n ind u ction g en era to r ? D escrib e, w ith th e h elp o f c irc le d ia g ra m , th e o p e ra tio n o f an
ex tern a lly -e x cited 3-p h ase induction g en erator.
(6 ) D iscu ss th e follow ing for a 3-p h ase in d u ction g e n e r a to r :
(i) I t o p era tes alw ays a t a lead in g pow er factor.
(ii) C ontrol o f load on ind u ction g en era to r and its lim it.
(iii) A p p lication o f S C IM in th e g en e ra tin g mode.
(iv) It re q u ires rea ctiv e pow er for its op eration from a n e x is tin g su p p ly n etw o rk .
(o) C om p arison o f th e use o f induction g e n e ra to r w ith a sy n ch ro n o u s g e n e r a to r.
6 . 86 . (a) E x p la in th e p rin cip le o f op eration o f a self-e x cited 3 -p h a se in d u ctio n g e n e r a to r .
Give th e condition u nd er w hich th is g e n e ra to r m ay fa il to b u ild up.
(6 ) G ive th e ap p licatio n s o f both e x te rn a lly -e x c ite d an d se lf-e x c ite d 3 -p h a s e in d u c tio n g e n e r a to r s .
6 .8 7 . ( a ) C om p are th e re la tiv e ad v a n ta g es and d isa d v a n ta g e s o f c a g e -ro to r a n d w o u n d -ro to r in d u ction
m otors o f th e sa m e pow er ra tin g .

motor*? H°Wd° y°UCOmpare the operation of a P°lypbase induction motor with that of a polyphase synchronous

S canned B y X a m b c a n n e r