CHAPTER 3INDUCTION MACHINE3.1 INTRODUCTION Induction motor is the common type of AC motor. Induction motor was invented by Nicola Tesla (185!1"#$% in 1888. Also &nown as asynchronous motor. It has a stator and a rotor mounted on bearin's and separated from the stator by an air 'ap. It re(uires no electrical connection to the rotating member. )uch motor are classified induction machines because the rotor volta'e (which produce the rotor current and the rotor ma'netic field% is induced in the rotor windin'rather than bein' physically connected by wires. The transfer of ener'y from the stationary member to the rotatin' member is by means of electroma'netic induction. This motor is widely used by the industries because*! +u''ed. ! )imple construction. ! +obust. ! +eliable.! ,i'h efficiency. ! -ood power factor. ! +e(uire less maintenance ! .asy to start.! +otates itself without e/ternal assistant.! 0ess e/pensive than direct current motor of e(ual power and speed. The wea&nesses of this machine are*! 0ow startin' tor(ue if compared to dc shunt motor.! )peed will be reduced when load increased.! )peed can1t be chan'ed without reducin' efficiency. )mall sin'le phase induction motors (in fractional horsepower ratin'% are used in many household appliances such as*! 2lenders ! 0awn mowers! 3uice mi/ers! 4ashin' machines! +efri'erators 49 Two phase induction motors are used primarily as servomotor in control system. 0ar'e three phase induction motors (in ten or hundreds of horsepower% are used in*! 5umps! 6ans ! Compressors! 5aper mills! Te/tile mills7 and so forth.3.2 INDUCTION MOTOR CONTRUCTION 8nli&e dc machine7 induction machine have a uniform air 'ap. Composed by two main parts*! )tator! +otor 6i'ure #.1 and #.9 show the inside of induction machine.!ig"re 3.150!ig"re 3.2tator Con#tr"ctionThe stator and the rotor are electrical circuits that perform as electroma'nets. The stator isthe stationary electrical part of the motor. The stator core of a N.:A motor is made up ofseveral hundred thin laminations.!ig"re 3.3$tator coretator %in&ing#)tator laminations are stac&ed to'ether formin' a hollow cylinder. Coils of insulated wireare inserted into slots of the stator core.51!ig"re 3.'$tator (in&ing.ach 'roupin' of coils7 to'ether with the steel core it surrounds7 form an electroma'net. .lectroma'netism is the principle behind motor operation. The stator windin's are connected directly to the power source.Rotor Con#tr"ction The rotor also consists of laminated ferroma'netic material7 with slot cuts on the outer surface. The rotor are of two basic types *! )(uirrel ca'e! 4ound rotor)(uirrel ca'e rotor It consist of a series of a conductin' bars laid into slots carved in the face of the rotor and shorted at either ends by lar'e shortin' rin'. This desi'n is referred to as s(uirrel ca'e rotor because the conductors would loo&li&e one of the e/ercise wheels that s(uirrel or hamsters run on. )mall s(uirrel ca'e rotors use a slotted core of laminated steel into which moltenaluminums cast to form the conductor7 end rin's and fan blades. 0ar'er s(uirrel ca'e rotors use brass bars and brass end rin's that are bra;ed to'ether to form the s(uirrel ca'e.52 )&ewin' the rotor slots help to*! Avoid crawlin' (loc&in' in at sub!synchronous speeds%! +educe vibration )(uirrel ca'e rotor is better than wound rotor because it is*! )impler ! :ore ru''ed! :ore economical! +e(uire less maintenance!ig"re 3.)$*"irrel cage Rotor!ig"re 3.+ $ Rotor core53!ig"re 3.,4ound rotor ,as a complete set of three phase insulated windin's that are mirror ima'es of the windin' on stator. Its three phase windin' are usually wye connected and ends of three rotor wires are tied to a slip rin's on the rotor shaft. The rotor windin' are shorted throu'h carbon brushes ridin' on the slip rin's. The e/istence of rheostat enable user to modify the tor(ue speed characteristic of the motor. It is used to ad no. of poles +otatin' ma'netic field will cause the rotor to rotate the same direction as the stator flu/. Tor(ue direction is always the same as the flu/ rotation. At the time of startin' the motor7 rotor speed is =. The rotatin' ma'netic field will cause the rotor to rotate from = speed to a speed that is lower than the synchronous speed.If the rotor speed is e(ual to the synchronous speed7 there will be no cuttin' of flu/ and rotor current e(uals ;ero. Therefore7 it is not possible for the rotor to rotate at ns.3.' 0IP AND ROTOR PEED)lip is defined as *ssn n ns=wherens > synchronous speed in rpmn > rotor speed in rpm)lip can also represented in percent.The fre(uency of the rotor7 fr is*sf fr=where s > slipf > supply fre(uencyE1am2le 1Calculate the synchronous speed of a $!phase induction motor havin' 9= poles when it is connected to a 5= ,; source.ol"tionrpmp fns$==9=% 5= ( 19= 19== = =E1am2le 256A =.5 hp7 !pole induction motor is e/cited by a $ ?phase7 = ,; source. If the full!load speed is 11#= rpm7 calculate the slip.ol"tion@ 5 =5 . =19==11#= 19==19==

% = ( 19= 19== ==== = =sssn n nsrpmp fnE1am2le 3The !pole7wound!rotor induction motor is e/cited by a $!phase7 = ,; source. Calculate the fre(uency of the rotor current under the followin' conditions*(i% at standstill(ii% motor turnin' at 5== rpm in the same direction as the revolvin' field(iii% motor turnin' at 5== rpm in the opposite direction to the revolvin' field(iv% motor turnin' at 9=== rpm in the same direction as the revolvin' fieldol"tionns > 19=f A p > 19=(=A% > 19== rpm(i% n>=ssn n ns=> 119=== 19===fr> sf > 1 / = > =,;(ii% n > B5==ssn n ns=>58$ . =19==5== 19===fr> sf > =.58$ / = > $5 ,;(iii% n > !5==ssn n ns=> #1C . 119==% 5== ( 19=== (sD1 motor is operatin' as a bra&e%fr> sf > 1.#1C / = > 85 ,;(iv%n > B9===57ssn n ns=>C . =19==9=== 19== =fr> sf > !=.C / = > !#=,; (!ve means that the phase se(uence of the volta'es inducedin the rotor windin' is reversed%E1am2le 'A $!phase7 # pair of poles7 #==&47#==E7=,;induction motor is C8= rpm full!load speed. Fetermine the fre(uency of the rotor current under full load condition.ol"tionf rotor > sfn s > rpmp f"==8% = ( 19= 19== =ssn n ns=>1$$ . ="==C8= "===Hz sf fr8 % = ( 1$$ . = = = =3.) PER3PHAEE4UI5A0ENTCIRCUITO!THREE3PHAEINDUCTIONMOTORThe per!phase e(uivalent circuit of a three!phase induction motor is I07 EL>$L V5 in >$ELILcos> cos $ L LI VI0>

kWVPinL5" . 15#%. = %( #15 ( $C . cos $= =IL > I0>15#.5" = 1$ . 5$ 5" . 15#. = cos1A(iii%n s > rpmp f==1=% 5= ( 19= 19== =

ssn n ns= n > n s (1!s% > == (1!=.=5% > 5C= rpm(iv%fr > sf > =.=5(5=% > 9.5,;(v% T > !mnkWnPwPm moutmm9 . 1==5= A 9== A 9= = = Mr ws>s radfp fA 8$ . 9599 A9= = wm > ws(1!s% > 9.8$(1!=.=5% > 5"." radAs!ms radkWT 9 . 1==5A " . 5"== =E1am2le +61A $!phase7 delta connection7 # pole7 ##=E7 = ,;induction motor havin' a rotor speed 19==rpm and 5=&4 input power at =.8 power factor la''in'. The copper losses and iron losses in the stator amount to 9&4 and the winda'e and friction lossesare $&4. Fetermine*(i% Net output power(ii% .fficiency(iii% Input currentol"tion(i% ns > 19=fAp > 19=(=%A# > 18== rpm ssn n ns= > $$ . =18==19== 18=== 5 net output > 9"&4(ii% K > @ 585=9"= =kWkWPPinout(iii%N connection I

kWV PII V PLpinpp p in89 $ $5 . #C8C . $ $5 . #C 8 . = cos $5 . #C% 8 . = %( ##= ( $5=cos $cos $1= = = = = ==E1am2le ,625 input(to stator% >5=&45 input rotor A5a'>5=&4!9&4 >#8&45 stator losses >9&45 rotor losses>s(5 input rotor%> =.$$(#8&4%>1&45mech>5out rotor >(#8!1%&4 >$9&45 wind and fric losses >$&45 out net >($9!$%&4>9"&4I1 1.#O+91 > +otor resistance> =.OG91 > +otor lea&a'e inductance> 9O+m > no!load loses resistance > 15=OGm > ma'neti;in' reactance > 9=OCalculate*(i% Input power(ii%)peed of the rotor(iii% :echanical power(iv% Feveloped tor(ue(v% .fficiency ol"tion 1.0k1.0m1.0m 1.0k 1.0m(i%5 in >$EIcosHE > ##=EP total > + =+ +++ + "" "" "" #5 . 11 #5 . "9 9= 9=% 9 9= ( 9=# . 1 # . = "" #VItotal = =+= = # . 5= # . 9" 8 . 99 8C . 18#5 . 11 #5 . "##=15in > $(##=%(9".#%cos(!5=.#=% > 9#"=C.54(ii%rpmp fns C9=1=% = ( 19= 19== = = ssn n ns=7 n > ns(1!s% 7 n > C9=(1!=.=$% > "8.# rpm(iii% 5m>$(I99+9As ? I99+9%E9>##= ? I1P1> ##= ? (18.8C!99.8 = =+= 1" . 8 "# . 1" 8# . 9 C# . 1"9 9=9C . 1C #5 . #==99"""#VA5m > $ >1"."#9=$ . = . = ! 1"."#9(=.%? > 9$1#=.54(iv% T dev > mmsagwPwP=5a'>$I99+9As > $(1"."#%9(=.%A(=.=$% > 9$85.94ws >s radp fA # . C55% = ( 999= = T>!m # . $1# . C59 . 85 . 9$=(v% @ 815 . 9#"=1$ 5 . 9$1#=== =WkW WPPinoutputE1am2le -A $!phase induction motor7 wye connection7 = ,; is connected to a 99=E source.The slip is 5@ and rotor speed is 855 rpm.The e(uivalent circuit perphase is*! +1> )tator resistance > =.#OG1 > )tator lea&a'e inductance > 1O+91 > +otor resistance> =.8OG91 > +otor lea&a'e inductance> $.5O+m > no!load loses resistance > 15=OGm > ma'neti;in' reactance > 1=OCalculate*(i% Number of poles(ii% Input power(iii% :echanical power(iv% Feveloped tor(ue(v% .fficiencyol"tion64=.#O 19=fApssn n ns=7sns > ns ? n 7n > ns ? sns > ns(1!s%ns > sn 1 > rpm "===5 . = 1855= ns > 19=fApp > polerpm n fs8"==% = ( 19= 19== =(ii% 5in > cos $L LI VP total >" . C =5 . #1= 5 . $ 1% 1= %( 5 . $ 1 (1 # . = "" "" "" + =+ +++ +O ""I = =+= 8# . 9 $ . 1# C$ . 19 5$ . " . C =5 . #$ 99=15in > W $ . 9#8C % 8# . 9 %(cos $ . 1# %( 99= ( $ = (iii% 5m>$(I99+9As ? I99+9% E9> $99= ? I1P1>$99=? (.5$! = =+= 1$ 8 .5# . 1 # . 5 . $ 1#$8 . 1 8 . 11199"""#VA5m > $ >.89=5 . =8 . = ! .89(=.8%? > 91=8.5#4(iv%T dev > mmsagwPwP=5a'>$I99+9As65+9As>=.8A=.=5>1O=,;799=E7J)ws >s radp fA 95 . "##% = ( 999= = T> ($I99+9As% Aws >Q$(.8%9(=.8%A(=.=5%R A "#.95 > 9$.55Nm(v%@ 85 85 . =$ . 9#8C5# . 91=8= = = =inoutputPP3., TOR4UE PEED CHARACTERITIC!ig"re 3.12There are $ re'ions involve in a $!phase induction motor*!(i%2ra&in'A5lu''in' 2ra&in' process occurs at sD=(positive slip%. In this case the motor acts as a bra&ewhere it rotates in opposite direction respect to the rotor.(9SslipS1%.(ii% :otorin':otorin' is the re'ion where induction motor acts as a motor. )lip is reducin'from 1 into =. )lip e(uals to = at synchronous speed7ns.(1SslipS=%.(iii% -eneratin'66-eneratin' re'ion is a re'ion where motor acts as a 'enerator. Furin' this timethe slip is ne'ative. At this time7 the motor acts as a 'enerator.(slipS=%3.- DETERMINATION O! CIRCUIT MODE0 PARAMETERThe parameter of the e(uivalent circuit can be determined from the results of a*! No load test! 2loc&ed rotor test!.FC testThe bloc&ed rotor test*! To determine G1 and G9! 4hen combines with FC test7 it also determines +9! Test is performed by bloc&in' the rotor so that it cannot turn and measurin' the line volta'e7 line current and three phase power input to the stator! Connection for the test is shown in 6i'ure #.1$ !ig"re 3.13 @a#ic circ"it 6or blocAe& rotor te#t an& no loa& te#tThe no load test*! To determine ma'neti;in' reactance7 Gm7 and the combined core7 friction and winda'e losses (these losses are essentially constant for all load condition%! The connection for the no load test are identical to those shown in 6i'ure 19! ,owever7 the rotor is unbloc&ed and allowed to run unloaded at rated volta'e and rated fre(uencyFC test*! To determine +1! Accomplished by connectin' any two stator leads to a variable volta'e FC source as67shown in 6i'ure #.1#. ! The FC source is ad 98 AIc > 9C. A)&etch the per!phase e(uivalent circuit for this motor.68ol"tion!romthe DC te#tB = = = 9#$ . =% = . 98 ( 9 . 1$91

VIVR$%$%!rom the no3loa& te#tB9 1 C . 1#1C . 819=19=$9=81C . 8$18 . 8 9 . 8 19 . 8X X

V#V V

IL+ = = == ==+ +=5scl > $I19+1 > $(8.1CA%9(=.9#$O% > #8.C45a' >5in ?5scl > #9=4! #8.C4 > $C1.$4!rom the locAe&3rotor te#t7

IL " . 9C$ . 9C 98 1 . 98=+ +=cos $51C . =" . 9C$ 95L L in I V P IV#= = = =H > cos!1 = = # . #=% " . 9C %( 95 ( $"9=$V WI VPL Lin+1 B +9 > =.51C(cos #=.#T% > =.$"#OFC test7 R1 : 7.2'3CBR2 : 7.3.'C D 7.2'3C : 7.1)1CAt 15 ,;7 G > =.51C(sin #=.#T%>=.$$5OAt = ,;7 G > = $# . 1 % $$5 . = (15=HzHz> G1 B G96or class A induction motor7 this reactance is assumed to be divided e(ually between the rotor and stator769 E1 : E2 : 7.+, CEm : 1'., 37.+, : 1'.73C3..TARTIN/ O! INDUCTION MOTOR(i% Firect Mn 0ine )tarter(FM0%! A widely!used startin' method of electric motors. ! The simplest motor starter.! A FM0 starter connects the motor terminals directly to the power supply. ! ,ence7 the motor is sub =.#O G9 > 9 OG1 > 9 O +m > 15=O+9 > 9 O Gm > 9=OCalculate*(i% Input power(ii% Air!'ap power(iii% :echanical power(iv% Feveloped tor(ueAtor(ue induced(v% .fficiency). A ##=E7 5=,;7 1= pole7 deltaAJ connected induction motor is rated at 1==&4. The e(uivalent parameter for the motor are*phase Xphase Xphase RmssA A $ . =A =8 . = = = = = = =cRRRphase Xphase RA 9 . =A 1 . =VVAt full load condition 7 the friction and winda'e losses are #==47 the miscellaneous losses are 1==4 and the core losses are 1===4. The slip of the motor is =.=#.(i% Calculate the input power(ii% Calculate the stator copper loss(iii% Calculate the air 'ap power(iv%Calculate theconverted power(v% Calculate the tor(ue induced by the motor(vi%Calculate the load tor(ue(vii% Calculate the startin' tor(ue(viii% Calculate the ma/imum tor(ue and slip(i/% Calculate the efficiency of the motor+. )(uirrel ca'e and wound rotors are the two common types of rotor used in induction machines. -ive four(#% advanta'es of s(uirrel ca'e rotor. ,. A # pole induction machine is supplied by 5= ,; source and havin' #@ of full load slip. 6ind the rotor fre(uency durin'*(i% )tartin'(ii% 6ull load-. A $!phase7 J!connected75= ,;7 # pair of poles7 induction motor havin' C9= rpm full load speed. The motor is connected to a #15 E supply. The machine has the followin' 72impedances in ohms per phase referred to the stator circuit*+1>=.9 O G1 >9.= O+9 >=." O G9 >#.= OGm > = OIf the total friction and winda'e losses are 9== 47 (i% 6ind the slip7 s.(ii% 6ind the input power7 5in.(iii% 6ind the air 'ap power7 5a'.(iv% 6ind the mechanical power7 5m.(v% 6ind the tor(ue induced by the motor7 W ind.(vi% 6ind the efficiency of the motor... Induction machine is a common type of AC machine. )tate three wea&nesses of the induction machine.17. A $!phase7 delta!connected75= ,;7 9 pair of poles7 induction motor havin' 1#55 rpm full load speed. The motor is connected to a #15 E supply. The machine has the followin' impedances in ohms per phase referred to the stator circuit*+1>=.9 O G1 >=. O+9 >=." G9 >=.# OGm > 9= OIf the total friction and winda'e losses are 1=== 47 calculate*(i% )lip(ii% Input power7 5in(iii%Air 'ap power7 5a'(iv% :echanical power7 5conv (v% Tor(ue induced by the motor7 W ind(vi%.fficiency of the motor11. A $!phase7 J!connected7poles7 #15 E7 5= ,; induction motor havin' a rotor speed"5= rpm. The input power is 1== &4 at =.85 power factor la''in'. Thecopperand ironlosses in the stator are # &4 and the winda'e and friction losses are # &4.Fetermine theoutput power of the motor.73