r
se
/'2 1: --u,+ 1 iL / :u{
t-r.rn
r\s Z
a"4+
lnduction
Motor
1267
Example 34,14 (a). A 3-phase,lQQlffiW Y-Y connected wound-rotor induction motor has
ffiand0,3dlstandstillreactanceperphase,Findtheadditionalresistance
required in the rotor ci.rcuit to make the starting torqae equal to the maximum torque oJ'the-motor.
(Electrical Technology, Bombay Univ. f990)
Solution.
a2+ | = 2q
e2 -La+ , fi.r?
(a -t)Cq-,).=- -t
r=
l=
#;
Since 7,,,=T^u
lf;
or
@Now,
external resista nce per phase
0,06 +
r
r = 0.3
03
- 0.06 = 0.24 S)
Example 344&L3-phase, 50-Hz, 8-pole, inductionmotor hasfull-load slip of 2Vo. The rotorstill rotor-reactance per phase are 0.001 ohm and 0.005 ohm respectively.
Find the ratio of the maximum to full-load torque and the speed at which the maximum torque
(Amravati University, 1999)
occurs.
resiitffiifstand
Solution. Synchronous speed.
Slip at maximunr
Thlrs,
torque.
.en,t.
letO\CoVS.)
Corresponding speed
t -'.l Maximum torque
x 50/8 = 750 rpm
=
=
r.r/,\.,
=
(l-0.2)x750=600rpm
r, _0.001 _,,,
= .r. 0.(X)5 - "'-
rqtl-!gg9-!stgl'
T-
120
N.,
?)yt'.t- ".
--o
,1,,, +
,'rr
L+,--?]!4eg
Lo*
o ,1gg
o.zo2 + 0.022
Example 34.L4 (c). A I2-pole,3-phase,600-V 50-H2., star-connected, induction motor has
rotor-resistance and stand-still reactance of 0,03 and 0.5 ohm per phase respectively. Calculate:
(a) Speed of maximum torque. (b) ratio of full-load torque to maximum torque, if rhe.full-load
(Nagpur University, April 1999)
speed is 495 rpm.
Solution. For a l2-pole, 50 Hz motor,
Synchronous speed = l20x 50 / I 2 = 500 rpm
For r = 0.03 and .r = 0.5 ohm. the slip fur maximum torque is related as
S,rr= u = rl.r= 0,03/0,-5 = 0.06
(a) Corresponding speed
(b) Full-load speed
Full-load
M^^*rr
torque
=
=
500 ( | --
.r,,,7.)
495 rpnr, slip
:
= 470 rTrnt
.s
= 0,0
| , at
tull load.
2 a.s _ 2 x 0,06 x 0.01 t\ ,4
!*q* = i * r, = o.ootig.gli = u'524 Zz= ftz +j l+
a
15. A 746-kw 3-phase, 50-Hz, I6-pole induction motor has a rc!2!@gof
0.02 + j 0.15)lfrIat standstill. Full-loadtorque is obtainedat 360 rpm. Calculate (i) the ratio of
maximum to full-load torque (ii) the speed of mqximum torque and (iii) the rotor resistance to be
(Elect, Machines, Nagpur Univ. 1993)
added to get maximum starting torque.
Solution. Let us first find out the value of full-load slip.r,
N, = 120x50/16=375rprn,; F.L,Speed=360rpm,
..
sr
=
(315 -360)1375= 0.04;
a=
RJXr=0.0210,15 =2115
1,,2.68
ElectricalTechnology
rt
(r)
]:y_ =?t_e!)2l.991
r.o*
o' + tj
Qtts)'+
=0.55 *
(0.04)2
+= #= r.8r8
TJ
(rr) Atmaximumtorque, a = s, = R/Xr=0,02/0.15 =2115
N = N,(l -s) =375 (l -2115) =325 r.p.m.
(ttt) For maximum starting torque, Rz== Xz. Hence, total rotor resistance per phase = 0.15O
external resistance required/phase = 0.l5 - 0.02 = 0.f 3 A
-2.,
16. The
of a 4-pole, 5
3-phase
r are
simpLifiing
0.
at maxt
(ii)
I
Starting. D
rotor
give three-.
t of a single-
s=Rr/Xr=0
i) At maximum t
N.,
(tr) It is gi
r phase requ
res'i
the equivalent c
=
7,,
t2oxjn6=l5oo
.=
rpm
ll88 rpm
7n,o^ Nory
.75
C',^
81
3a2-8a+3=
t
36
= 0.45
tional rotor
et,
a=
Rr+ r
R2
0.45 =
induction rnotor has EL.:sli;p of 4Vo. The rotor,reststqnce/phdse
hase = 0. I ohm. Find the ratio of maximum to full-load torque
matcimum torque occurs,
Solution.
Now,
rl
4;
lT:tr
a=
Tt =
h;;
N"
=
RzlXz=Q.Q1/0,
2
x.0: l
o.l?
l20x
r 0 .qj
l=0. l,{/=0.04
..-g..q,q.q-
=
+0.042 o'oll6
= 0.6e
50/8 = 750 rpm, s,,=
T
|
I =1.45
=0,69
0,I
N = (l-0,1)x750=675rpm
e3:phaa?
:.''.;''
Cuo
iduction mdb.r; thi
m
is 2-S
(Elect Mrcldn$, A,M,[E,
Sec-
tines
E
1
8,,1992)
Solution. Given,Zn.o*= 2.5fr 4,= 1 .5Tl.;Tr/T^o = 1.512.5=315.
2-( 1o
3 S,.2-{oSn+3=6
Srr
- ( -'l o )
=
S,",
+ ,--lo
-
lo
f
2x3
V
-)
'-Yx3 x3
2al
t /
to
1-
toL
e ,o
a
bR
T.
2a
?
T;.Jt
Now,
lnduction
\
_
-
-
8
Motor l 269
n2
or )a
-lja+3=0 or a=ll3
l+u"
--=
F*@
,33
FrC)
S rra
(Elect. Machines-I, Punjab Univ. l99f )
Solution. lt should be noted that stalling speed corresponds to maximum torque (also called
stalling torque) and to maximum slip under running conditions.
N, = 120 x 50/8 = 750 r,p.m,t stalling speed is = 650 r,p.m,
sb
sh
Now.
= (750 - 650)/750 =2115 = 0, | 333 or l3.33Vo
= R.,lXz X: = 0.08 x l5l2 = 0.6 Q
T'' - 2o-..
T,=7.I'nu*
r"=
t* -= ;;i ' Since
29
l-
|+
, orct=l
o' -"
Let r be the external resistance per phase added to the rotor circuit. Then
R"+r or
.7-
N, = ;,N1x5014
rng to rna
r500-
1365
=
'=
1500
0.08+r:
--tU
r.p.mf=
.'.
r=0.52Slperphase.
1365 r.p.m.
um torque ls
= 0.09
But
R,lx.,
).09
oE.')
=2.22
...Art34.20
K
2
' be the external
x
2.22
stance introduced
Starting torque
Solving the quadratic
ation for
'r
circuit, then
K (0,2 + r)
r
get r = 0.4
3/'t
o
127:0
ElectricalTechnology
34.21. A 4-pole, 50-Hz 7.46.kW motor has, at rated voltage andfrequency, a starting
torque of 160 per cent and a maximum torque of 200 per cent of full-load torque. Determine (i) fuU(Electrical Technology-I, Osmania Univ. 1990)
load speed (ii) speed at maximum torque,
T.
Solution.
Now,
or
1,6 and
'r')^
a i.\,
Inr*
a=
or
-
.!
Tr,
=l^6=0.g
"'4r'n* 2
2o
- =o,g
l+a2
2as.,
Also,
...
Tl
lra = 2a
^ T,no* l+a2
0.8a'-2a+08 =Q a=0.04
T
(i)
L-,*=2
Tl
RzlXz= 0.04
2 x 0.04
s,
0,001 6 + ^r,
--ut
full-load speed occurs at a slip of 0.01 or' I per ccnt.
Now.
N, = '20x5014 = l-5(X) r,p,m.;N=
or
j
=
1500
orRr= 0.04X2
I
sl= 0.01
- l5 = 1485 r.p,m.
(ii) Maximum torque occurs at a slip given Otl;OrrXr, As seen from above slip corresponding
totllttum torqueis0'04'
N = r500- r500x0,04= 1440r.p.m.
Y.22, A 3-phase induction motor having a 6-pole, star-connected stator winding
rans on 240-U SO--Hz supply. The rotor resistance and standstill reactance ere 0.12 ohm and
0,85 ohm per phase. The ratio of stator to rotor turns is /.8. Full load slip is 4Vo.
Calculate the developed torque at full load, maximum torque and speed at maximum torque.
Solution. Here,
tr
u2
Nr=
+
.
0,04 x 17- x0.12
2ntstitlj
g
t2i;
to.l+xo,ssl,
= 52.4 N-nr
.t = R,,X"=0,12l0,85=0.14
For maximum torque,
/ rt&t
0.14x77?x0.l2
3
= li fi,ot3) O. tzt r i+ *
o sjii
tg.
= 99.9 N-m
Alternatively, as seen from Art 34.20.
T
'max
4,* =
Speed corresponding
'pecti
3E:
2nN
,'
2 X.,
a--
,
ee.
e N-m
:i,. ,s =
to maximum torque, N = 1000( | - 0, l4) = 860 rpm
zn rior 3)
2
rotorl4sistance and standstill reactance2f a 3-phase luduction motor are
0.01Wand 0.0/Q per phas{ At no,r6al volrage,lrt.fult-toad sltf, is 3vo. /srimate t
tion iny'ator voltgg4 ro dey'lop full-loqd4orque ar lilfy yutt-lofispeed. .,{lso,
(Adv.
f=
100 r.p,
I\'I
aclifes, A,N[,W, I 989)
f
Example 34.31. The power input to the rotor
o,f a
400
\
nduction
Motor
1287
50.H2, 6-pole, 3-phase induction motor
3vo,Calcutate(i)thefrequencyofrotorcurrents(ii)rotorspeed(iii)rotor'
/
copper losses and (iv) rotor resistance per phase if rotor current is 60 A.
(Blect. Engg.Pur\iab Univ. 1991)
Solution. (i) Frequency of rotor current = ,s./= 0.03 x
-50 = I .5 Hz.
(rt) N, = 120 x 50/6 = 1000 rpm; N= 1000(l - 0.03) = 700 rpm
(iii) rotor Cu loss = s x rotor input = 0.03 x 20 =0.6kW = 600 W
(iv) rotor Cu loss/phase = 200 W; .'. 602Rr= 200; Rz = 0.055 ft
tor devel
.73 kW at 960 rpm.
raj Univ
powe,
in rotor =
3O/rotor
l0!0/r.p.rn,
3
96ry1(
i
t + stator
I
s=3885+2
Example34.33. Thepowerinputtotherotorofa400U50-H2,6-pole,3-QinductionmotorisT5
kW. The rotor electromotive force is observed to make 100 complete alteration per minute. Calculate:
(i) slip (ii) rotor speed (iii) rotor copper losses per phase (iv) mechanical power developed.
(Elect. Engg.I, Nagpur Univ. 1993)
Solution. Frequency of rotor emf ,.t''' = I 00/60 = 513 Hz.
(t) Now,,f'' -,f or 5/3 = s x 50; s = l/30 = 0.033 or 3.33Vo
(tt) N,= 120x5016= 1000rpm;N=N. (l -s)= 1000 (l - l/30)=966.7
(iii) P2=15 kW; total rotor Cu loss = sP2 = (1/30) x75 = 2.5 kW
rpm
rotor Cu loss/phase = 2.513 = 0.833 kW
(iv)
P,,,=
(l -
s)Pz
- (l -
l/30) x75 =72.5 kW
Example 34.34, The power input to a 500 V 50-H2., 6-pole, 3-phase induction motor running at
975 rpm is 40 kW. The stator losses are I kW and the friction and windage losses total 2 kW.
Calculate : (i) the slip (ii) the rotor copper loss (iii) shaft power and (iv) the efficiencv.
(Elect. Engg.- II, Pune Univ. 1989)
Solution. (i) N"
-
20 x 50/6
= I 000 rpm ;,s = ( | 00 - 91 5)l I 000 = 0,025 or 2.5Vo
; stator loss = I kW; rotor input P.,= 40 - I = 39 kW
rotor Cu loss = .r x rotor input = 0,025 x 39 = 0.975 kW
|
(tt) Motor input Pr=40 kW
(iii) P,,- Pz- rotor Cu loss = 39 -0.975
Pour= P,,,
-
= 38.025 kW
friction and windage loss = 38.025 -2=36.025 kW
(iv) rt = Pu,/P t=36.025140 = 0,9 or90Vo
Example 34.35. A 100-kW (output), 3300-V 50-H2,, 3-phase, star-connected induction motor
np.m. The futl-load slip is L\Vo and EL. power factor 0.85. Stator
copper loss = 2440 W, Iron loss = 3500 W. Rotational losses = I200 W. Calculate (i) the rotor copper
(Elect. Machines, Nagpur Univ. 1993)
loss (ii) the line curcent (iii) the full-load efficiency.
hqs a synchronous speed of 500
1288
ElectricalTechnology
Solution. P,=output+rotationalloss=
(t)
rotorCuloss
(tt)
rotor input,
Stator
( p,; =fi Vuf L C-fl
,A-lr
\
The entire power
100
+ 1,2=
101.2 kW
= *"p,,=J'ffi'xl0l'2=1.855kW
P, =
P
n,
+ rotor Cu loss = l0l .2 + 1.855 = 103.055 kW
Cu and iron losses
input = P, + stator
= t03,0ss + 2,44 + 3,5 = r'8.ees kw
109,995 =u5x3300x/,x0,85; lt=22.4A
flow in the motor
is given below.
Rotor
Stator
Motor output
Mechanical
Rotational
lron
losses
Cu &
losses
Cu loss
1855W
1200w
5940W
(iii)
F,L, efficiency
(v) speed of stator field
=
100,000/108,995 =0,91'7 or9l.7
Vo
power input to the rotor of a 440 V, 50-Hz, 6-pole, 3-phase induction motor
romotive force is observed to make 120 cycles per minute, Calculate (i) the
(iii) mechanical power developed (iv) the rotor copper loss per phase and
(Elect. Engg.AMIETE Sec. A June 1991)
with respect to roton
Solution. (i)f '= q/or (120160) = s x 50; s = 0.0{r
qclQ
f,
2l-F*
l2Ox5Ol6=1000rpm;N=1000(l-0.09JbrV lv = l?-o
=
l- s)Pr= (l- 0.0O x 100 =96kW
6OS€c
P.. L
(iv) total rotorCuloss=.sP,=0,0t[.x 100 'tftW;Culoss/phase=tll3 * l'33jk\rl
(ii)
lV,=
(iii) P,7(
(v)
N.= 1000 rpm; N = )(Orpm. Hence, speed of stator field with respect to rotor
= f 000 -gfp=Q4 rpm.
S=
is
+=+$o
=
Exg_!rnp!g_3437. An induction motor h.a.s an efficiency of 0.9 w,hen delivering on output of
kW. At this load, the stator Cu loss and rotor Cu loss each equals the stator iron loss. The
mechanical losses are one-third of the no-load loss. Calculate the slip.
(Adv. Elect. Machines, A.M.I.E. Sec. B Winter f 993)
Solution.Motorrnlu!=37,000/0,9=4l,lllW z
€
totalloss =4l,ll I -37,000=4,11lw
-t
r; ,L
t_eeac
- rl
t-r ' - l-ar'[
laSS€S =
This includes (i) stator Cu and iron losses (ii) rotor Cu loss (its iron loss being negligibly small)
and (fii) rotor mechanical losses.
Now, no-load loss of an induction notor consists of (i) stator iron loss and (ii) mechanical losses
provided we neglect the small amount of stator Cu loss under no-load condition. Moreover, these two
losses are independent of the load on the motor.
no-load loss
Wn
=
W,+W,,,=3Wn,
NotoJ
= Wilz
l.w -= P+,
3
where W, is the stator iron loss andW,,, is the rotor mechanical losses.
Let, stator iron loss =.r; then stator Cu loss =.x, rotor Cu loss = x; nrechanical loss =.r/2
= ?r'^, l ?^D =x
P *;+?6*r + 3Pr"{V'";
t ?g''r )
f;Jr.= 2
^gt
YCvl=
'-'/
t\r"
=
o .ot't
Motor
Induction
Now, rotor input
-
gross output + mechanical losses + rotor Cu loss
?lZ
-
shaft
to
rotor
+ l 175 = 38,752 W
37,000 + (||1512)
=
.
PD7:
1289
inPut 38,152
0 V 50-Hz.
and runs
resistance
rotor Cu I
n motor consunxes 45 kW
is 1200 :W, windage and
12 {2, calculate (i) power
Pou, Qv)
fficiency and (v)
rque developed.
Solution. Cos (r =
-:lLt{L-
J3x400x75
= 0 u66 lag
vJ i.e. 43.3 arnp
A Iine current bt'75 amp nreans a phasc-currcnt of' 75
Next, winding resistance has to be worked
Refer to Fig,
out
N.. ...
34,40.
.
..
a
r and}rin
parallel have an equivalent resistance measured at
and b in delta connected motor as r x 2rl3r = 2rl3
A
c
I
ohms
o -42 JL
From the data given
+rb = 0.12, r = 0,18
|
t"
Total statorcopperloss = 3 x43.32 x0.lti= l0l2
Watts
Total input to stator
Stator copper loss
= 45,000
=
lx
r
'
iri
Fig. 34.40
Watts
1012 Watts, stator core loss
=
1200
Watts A
.
(?o *","111;::H = ffilfi5ffiJ,'j'iT i,r?=",,;;:( ?*)
Shafi
output
Efficicncy
Shatt output torque,
-
=
= 4l'504 Watts
Mech, oLttput ot'rotor - Mech losses
=
41504 -'900 = 40604 Watts
Rotor mech, output power= 42,'788
1284
x 100% =92.23o/o
= +g
4s000
r = q91::o
2nx970
= 400 Nw'm
34.39(a),.A 3-phase induction motor has a 4-pole, star-connected stator winding and
runs on a 220-V 50-Hz supply. The rotor resistance per phase is 0.1 d2 and reactance 0.9 Q. The
ratio of stator to rotor turns is 1.75. The full-load slip is 5Vo. Calculate for this Load:
(a) the load torque I h o1l< cL mech 4nicrl [.AS€S '
(b) speed at mqximum torque
(Electrical Machines-I, South GujaratUniv. 1985)
(c) rotor e.m.f. at maximum torque.
Solution.
(c)
rotor turns/stator turns = 111.15
statorvoltage/phase,
Et = ZZOIJlV
standstill rotor e.m.f,/phase, E,
tE
+
I
)
(rX;)
=
4"+
J:
t
'7s
+ (0.0-s x 0,9)
= 72.6V
1
'
=0.11()
F-V
r-l-
t
)
(?,
Ue
P.
'
fs
(L
\-P*
rr9s5 ,rlr;
N = {, (l -,r) = 1500 (t _ 0,05) = 1425 rpm
TE 9,55x621311425 = 4t,6N-m
(b) For maximum torque, sr, = Rz lXr= 0. l/0,g | = I/9
)
Calculate(i)sltpandrotorspeed,(x+end(iii)shaft-torque,
[Nagpur University, April 1998]
Solution.
Pr=P,'*-
= .f
Input to motor
x a00
x l0 x 0,86
= 5958 Watrs
=
= star.r outpLrt = -595u - 2gg = 5660 watts
RotorCopper-loss = 4 % of -5660 = 226.4 Warts <?a = S P_t'
Mechanical losses = 3o/a t'tt' 5660 = 169.g Watts
Shaftoutput = 5660 - 226,4 - 169.g = 5264 Wans
(,)
slip, s = Rotor-copper-loss / rotor-inpu t= 4 vo,as grven
Synchronousspeed, N = l20xflp=1500 rpm
Rotor speed,
N = N,(l-s)=1500 (l-0.04)=1440rpm
Plg
(-
Total stator losses
5%,
of 5958 = 29g Watts
Rotor input
(;t
olr,Angttlar
speea
Aec
U)
oss
_
(.
(iii)
Z-crl,.
=. 54?i
f"Crl,
= rotor i"nttt;n
Shaft-torque
Exam
=
=
se,44
res$tance
')
6,
glving
I,:-
=<1r3,8
/
r50,?2 _ 3b,05 I.{v,
output at
ip of
Shafr ourrput in Watts /co
5264
V,
I
t50,l2 =
34,93 N
nA
50-Hz, 4Q-pctle, Y-conneg*ed indut.rton
s'tcttor tct
5Vo
orque m
tion.
Phase
ng slip,
age
rns
andstill
e.
,
per rotor p
L)
OO
of
gro.
n_
is E.,
- KE,
t;
I
35
2.6 Y
tor has rotor
= t$o.a24
5rc
L
r=
rit
?
U)
N
Induction Motor
=
3.63 V
=0.1096
1l)2
330 x 0.95
gross outp
output
(b) For
mum
R.,i N',X,
R1 lX.,
=
Q.
.9
=
I/9;
+ (0.919.
P"rJt
1291
E,
x o.l
7,250 W
ll9) x72,6 = 8.
414=57,1 A
I rotor Q6 loss
rotor Cu loss
t
,9
&"4
p
s ,/
/
978
ll9
Example 34.41. An 18165-kW 4-pole, 50-H2., 3-phase induction motor ha.s.friction and windage
(Elect" Machines-Il, lndore Univ 1987)
Solution. Motor output 'p ttltl
r8,650 W
Friction and windage loss 'p !t'
Rotor gross output
P=
-n|
(c)
Rotor
input
T,n=9.55 Pu,,,l N
g- PXn
t5>
P2
19. I
l6 W
0'91
. rotor Cu loss = ,.
,
-L
l -.r
(l - 0.04)
=
I
rotor Cu loss
.s
=
7?S
0.u*+
x l9,l
16 =
796.6 W
= 19,912.5 \\,
19,913 W)
= 120x5014= 1500 r.p,m,
N - (l-0,04)x 1500= 1440r.p,m
N.,
\
. >.J)
T,n :='9'55x
18,650/ll440=
l o,oJU/ t ++U = 123.7N-m
l4J.l N-t
'T
T"
=
\i&&Wl':c..ll =-f-ou* 't Pgul
?otd
+ 466 =
(orrotorinput= l9,ll6 +j96,6=
(
xample 34.42.
of | 8,650 = 466 W
18,650
rotor Cu loss
rotor gross output
(a)
(b\
2.5%,
n <cn
^
t^r
r\F.
9.55 P,,,lN
=9.55 x
(or
= Pzl2 fiN., =
l*
f
l9,l 16ll440= 126.8 N-m
l9,9l3l2nx25 = l2l N-m)
An
3-phase, 50 Hz, induction motor is running at a speed of 7 t 0 rpm
Losses at this operating condirion are known to be 1200 W
y:t,t:,:!, yt-ationlllolse,s are 600 W. Find (i).the roror ('opper los.s, (ii) the gross rorque developed,
(iii) the Sross mechanical power oyed, (iv) the net torque and (v) rhe
mechanical power orrrrr,.
(Elect. Engg.AMrE'r'E Sec, A l99l ET.ajiv--hi
ith an input power of 35-8-pole,
kW. The stator
Solution.
750 = 0.0533
(i)Pz=35- 1.2=33.8kW;N, =120x50/g=750rpm;N=7l0rpm;,r=(750 _il0)l
Rotor Cu loss
(iii)
(iv)
=
sP" = 1.8 kW
P,, = Pz- rotor Cu loss = 33,8 - L8 = 32 kW
Tr = 9,5-5 P,,,lN = 9,55 x 3200011l0 = 430.42 N_m
t;;l
..
Pu,,,
= P,,__ rotational l<,rsses = 32000 _ 600 = 31400 W
1
292,ElectricalTechnology
(v)
Trn
=
9,5-5 P,u/
/N = 9,55 x3l400ll l0 = 422,35 N-m
3-phase,
rim
1,620 W. Calcul
-load with 4Vo slip
50-H2.,
at
its
pulley
are 200 W and the
Cu los.c and(c) the
fficiency atfull-load.
Solution.
Now,
(Elect. Technology, Mysore Univ. 1989)
N. = 120 x 50/6 = 1,000 r,p,m; N= (l - 0.04) x 1,000 = 960 r.p.m.
Out put poWer = T,nx2 IrN = 2 n x (960160) x I 49,3 = l5 kW
output = 15,000 w
Friction and windage
losses =
Pr,
P2
(b)
rotor Cu
:
\
stator
N
=
|
= 15,200 W
rotor input Pz= 15,200
fi
loss =
(
I rotor Cu loss is given by
(c)
200 W ; Rotor gross output
5,833
-
x
l ,000/960= 15,833
W
I5.200 = 633 W
IgALq!1g[ - .\ )
rotor oLrtpLrt r-i j
output = rotor input =
l-5.833 W
w
+ 1,620 = 17,453 W
I 5,000 x 100/17,453 = 86Vo
StatorCu and iron losses = r,620
tlD
.'. Stator input
f
n=
overall efficiency,
5,833
An I 8.65-kW, 6-pole, 50-Hz, 3-Q slip-ring induction motor runs al 960 r.p.m.
on full-load with a rotor current per phase of 35 A. Allowing I kW
for mechanical losses, find the
resistance per phase of 3-phase rotor winding,
(Elect Engg-I, Nagpur Univ. lgg?)
Solution.
Motor output
=
18,65
.'. Mechanical power developed by rotor,
N, =
Now,
rotor Cu loss =
P
n,= 18.65 + I = | 9.65 kW
120x5016= 1000r,p.m.;
*"t,
31.,2R., = 819
,/
kw; Mechanicar losses = I kw
=
s
-
( I 000
-
960y I 000 = 0.04
x 19'65=0.819kw=819w
ffi
or 3x35'xR,-819
R: = 0.023 {-Vphase
@+4??\4-pold,3-phase,50-Hz,inductionmotorhasarotorresistanceand
reactance per phase
0.01
of
dl and 0.1 A respectiveb). Determine (a) marimum torque in N-m and
the corresponding slip (b) the.futl-load slip and power output in watt5, if maximum torque is nvice
the full-load torque. The rario of stator to rotor turns is 4.
Sofution, Applied voltage/phase E,
= 400frE
= 23 I V
Standstille,m.f. induced in rotor, Ez= KEt=23114=51.15y
(a) Slip fbr maximum torque, s,, = R2 lX.,=1'1.9110, | = 0. I or l\Vo
T
' t1ta.r N, =
T
I n,,,.r' =
(b)
"
3_*
Ei
.,,
ZnN,,, 2 X.,
Art. 34.20
120x5014=1500 r.p.m. = 25 r.p,s.
3 .,57.15- _",
,""r5" ffi = 320 N-m
+t = 2=
l;
n-*.s- /.
N,rw ,
(t=
R.,
-
tx"-0.01/0,1 = 0,1
2:€ o .1
Sr. ca Sf = o'3?3 f svrr,4
\re s e\ecp sp = o. o2],
T1'
' 72olz :
\'y\
JA4 opr.^hor
rqi.: ,i*^.;J
>to(e(l-o'oz))= l9sg.s
tso Vrn -+P""t = 2TNT+/go
n
rpv,a
=2_+;;;;@,