BITS Pilani, Pilani Campus Induction Motors Induction Motor BITS Pilani, Pilani Campus Introduction Three-phase induction motors are the most common and frequently encountered machines in industry simple design, rugged, low-price, easy maintenance wide range of power ratings: fractional horsepower to 10 MW run essentially as constant speed from no-load to full load Its speed depends on the frequency of the power source not easy to have variable speed control requires a variable-frequency power-electronic drive for optimal speed control BITS Pilani, Pilani Campus Construction An induction motor has two main parts a stationary stator consisting of a steel frame that supports a hollow, cylindrical core core, constructed from stacked laminations (why?), having a number of evenly spaced slots, providing the space for the stator windingStator of IM BITS Pilani, Pilani Campus Construction a revolving rotor composed of punched laminations, stacked to create a series of rotor slots, providing space for the rotor winding one of two types of rotor windings conventional 3-phase windings made of insulated wire (wound-rotor)similar to the winding on the stator aluminum bus bars shorted together at the ends by two aluminum rings, forming a squirrel-cage shaped circuit (squirrel-cage) Two basic design types depending on the rotor design squirrel-cage: conducting bars laid into slots and shorted at both ends by shorting rings. wound-rotor: complete set of three-phase windings exactly as the stator. Usually Y-connected, the ends of the three rotor wires are connected to 3 slip rings on the rotor shaft. In this way, the rotor circuit is accessible. BITS Pilani, Pilani Campus Construction Squirrel cage rotor Wound rotor Notice the slip rings BITS Pilani, Pilani Campus Construction Cutaway in a typical wound-rotor IM. Notice the brushes and the slip rings

Brushes Slip rings BITS Pilani, Pilani Campus Rotating Magnetic Field Balanced three phase windings, i.e. mechanically displaced 120 degrees form each other, fed by balanced three phase source A rotating magnetic field with constant magnitude is produced, rotating with a speed Where fe is the supply frequency and P is the no. of poles and nsync is called the synchronous speed in rpm (revolutions per minute) 120esyncfn rpmP= BITS Pilani, Pilani Campus Synchronous speed P50 Hz60 Hz 230003600 415001800 610001200 8750900 10600720 12500600 BITS Pilani, Pilani Campus Rotating Magnetic Field BITS Pilani, Pilani Campus Rotating Magnetic Field BITS Pilani, Pilani Campus Rotating Magnetic Field ( ) ( ) ( ) ( )net a b cB t B t B t Bt = + +sin( ) 0 sin( 120 ) 120 sin( 240) 240M M MB t B t B t e e e = Z+ Z + Z sin( )3 [0.5 sin( 120 )] [ sin( 120 )]23 [0.5 sin( 240 )] [ sin( 240 )]2MM MM MB tB t B tB t B tee ee e= + xx yx yBITS Pilani, Pilani Campus Rotating Magnetic Field 1 3 1 3 ( ) [ sin( ) sin( ) cos( ) sin( ) cos( )]4 4 4 43 3 3 3 [ sin( ) cos( ) sin( ) cos( )]4 4 4 4net M M M M MM M M MB t B t B t B t B t B tB t B t B t B te e e e ee e e e= + + + + + xy [1.5 sin( )] [1.5 cos( )]M MB t B t e e = x yBITS Pilani, Pilani Campus Rotating Magnetic Field BITS Pilani, Pilani Campus Principle of operation This rotating magnetic field cuts the rotor windings and produces an induced voltage in the rotor windings Due to the fact that the rotor windings are short circuited, for both squirrel cage and wound-rotor, and induced current flows in the rotor windings The rotor current produces another magnetic field A torque is produced as a result of the interaction of those two magnetic fields Where tind is the induced torque and BR and BS are the magnetic flux densities of the rotor and the stator respectively ind R skB B t = BITS Pilani, Pilani Campus Induction motor speed At what speed will the IM run? Can the IM run at the synchronous speed, why? If rotor runs at the synchronous speed, which is the same speed of the rotating magnetic field, then the rotor will appear stationary to the rotating magnetic field and the rotating magnetic field will not cut the rotor. So, no induced current will flow in the rotor and no rotor magnetic flux will be produced so no torque is generated and the rotor speed will fall below the synchronous speed When the speed falls, the rotating magnetic field will cut the rotor windings and a torque is produced BITS Pilani, Pilani Campus Induction motor speed So, the IM will always run at a speed lower than the synchronous speed The difference between the motor speed and the synchronous speed is called the Slip Where nslip= slip speed nsync= speed of the magnetic field nm = mechanical shaft speed of the motor slip sync mn n n = BITS Pilani, Pilani Campus The Slip

sync msyncn nsn =Where s is the slip Notice that : if the rotor runs at synchronous speed s = 0 if the rotor is stationary s = 1 Slip may be expressed as a percentage by multiplying the above eq. by 100, notice that the slip is a ratio and doesnt have units BITS Pilani, Pilani Campus Induction Motors and Transformers Both IM and transformer works on the principle of induced voltage Transformer: voltage applied to the primary windings produce an induced voltage in the secondary windings Induction motor: voltage applied to the stator windings produce an induced voltage in the rotor windings The difference is that, in the case of the induction motor, the secondary windings can move Due to the rotation of the rotor (the secondary winding of the IM), the induced voltage in it does not have the same frequency of the stator (the primary) voltage BITS Pilani, Pilani Campus Frequency The frequency of the voltage induced in the rotor is given by Where fr = the rotor frequency (Hz) P = number of stator poles n = slip speed (rpm) 120rP nf=( )120120s mrseP n nfP snsf == =BITS Pilani, Pilani Campus Frequency What would be the frequency of the rotors induced voltage at any speed nm? When the rotor is blocked (s=1) , the frequency of the induced voltage is equal to the supply frequency On the other hand, if the rotor runs at synchronous speed (s = 0), the frequency will be zero r ef sf =BITS Pilani, Pilani Campus Torque While the input to the induction motor is electrical power, its output is mechanical power and for that we should know some terms and quantities related to mechanical power Any mechanical load applied to the motor shaft will introduce a Torque on the motor shaft. This torque is related to the motor output power and the rotor speed and .outloadmPN m te=2/60mmnrad ste =BITS Pilani, Pilani Campus Horse power Another unit used to measure mechanical power is the horse power It is used to refer to the mechanical output power of the motor Since we, as an electrical engineers, deal with watts as a unit to measure electrical power, there is a relation between horse power and watts 746 hp watts =BITS Pilani, Pilani Campus Example A 208-V, 10hp, four pole, 60 Hz, Y-connected induction motor has a full-load slip of 5 percent 1. What is the synchronous speed of this motor? 2. What is the rotor speed of this motor at rated load? 3. What is the rotor frequency of this motor at rated load? 4. What is the shaft torque of this motor at rated load? BITS Pilani, Pilani Campus Solution 1. 2. 3. 4.

120 120(60)18004esyncfn rpmP= = =(1 )(1 0.05) 1800 1710m sn s nrpm= = =0.05 60 3r ef sf Hz = = =26010 746 /41.7 .1710 2 (1/ 60)out outloadmmP Pnhp watt hpN mtett= == = BITS Pilani, Pilani Campus Equivalent Circuit The induction motor is similar to the transformer with the exception that its secondary windings are free to rotate As we noticed in the transformer, it is easier if we can combine these two circuits in one circuit but there are some difficulties BITS Pilani, Pilani Campus BITS Pilani, Pilani Campus BITS Pilani, Pilani Campus Equivalent Circuit When the rotor is locked (or blocked), i.e. s =1, the largest voltage and rotor frequency are induced in the rotor, Why? On the other side, if the rotor rotates at synchronous speed, i.e. s = 0, the induced voltage and frequency in the rotor will be equal to zero, Why? Where ER0 is the largest value of the rotors induced voltage obtained at s = 1(loacked rotor) 0 R RE sE =BITS Pilani, Pilani Campus Equivalent Circuit The same is true for the frequency, i.e. It is known that So, as the frequency of the induced voltage in the rotor changes, the reactance of the rotor circuit also changes Where Xr0 is the rotor reactance at the supply frequency(at blocked rotor) r ef sf =2 X L f L e t = =022r r r r re rrX L f Lsf LsXe tt= ===BITS Pilani, Pilani Campus Equivalent Circuit Then, we can draw the rotor equivalent circuit as follows Where ER is the induced voltage in the rotor and RR is the rotor resistance BITS Pilani, Pilani Campus Equivalent Circuit Now we can calculate the rotor current as Dividing both the numerator and denominator by s so nothing changes we get Where ER0 is the induced voltage and XR0 is the rotor reactance at blocked rotor condition (s = 1) 00( )( )RRR RRR REIR jXsER jsX=+=+00( )RRRREIRjXs=+BITS Pilani, Pilani Campus Equivalent Circuit Now we can have the rotor equivalent circuit BITS Pilani, Pilani Campus Equivalent Circuit Now as we managed to solve the induced voltage and different frequency problems, we can combine the stator and rotor circuits in one equivalent circuit Where 22 02221 0eff Reff RReffeff RSeffRX a XR a RIIaE a ENaN=====BITS Pilani, Pilani Campus Power losses in Induction machines Copper losses Copper loss in the stator (PSCL) = I12R1 Copper loss in the rotor (PRCL) = I22R2 Core loss (Pcore) Mechanical power loss due to friction and windage How this power flow in the motor? BITS Pilani, Pilani Campus Power flow in induction motor BITS Pilani, Pilani Campus Power relations 3 cos 3 cosin L L ph phP V I V I u u = =21 13SCLP I R =( )AG in SCL coreP P P P = +22 23RCLP I R =conv AG RCLP P P = ( )out conv f w strayP P P P+= +convindmPte=BITS Pilani, Pilani Campus Equivalent Circuit We can rearrange the equivalent circuit as follows Actual rotor resistance Resistance equivalent to mechanical load BITS Pilani, Pilani Campus BITS Pilani, Pilani Campus BITS Pilani, Pilani Campus BITS Pilani, Pilani Campus BITS Pilani, Pilani Campus BITS Pilani, Pilani Campus Power relations 3 cos 3 cosin L L ph phP V I V I u u = =21 13SCLP I R =( )AG in SCL coreP P P P = +22 23RCLP I R =conv AG RCLP P P = ( )out conv f w strayP P P P+= +conv RCLP P = +2223RIs=222(1 )3R sIs=RCLPs=(1 )RCLP ss =(1 )conv AGP s P = convindmPte=(1 )(1 )AGssPs e=BITS Pilani, Pilani Campus Power relations AGPRCLP convP1 s 1-s : :1::1-AG RCL convP P Ps sBITS Pilani, Pilani Campus Example A 480-V, 60 Hz, 50-hp, three phase induction motor is drawing 60A at 0.85 PF lagging. The stator copper losses are 2 kW, and the rotor copper losses are 700 W. The friction and windage losses are 600 W, the core losses are 1800 W, and the stray losses are negligible. Find the following quantities: 1. The air-gap power PAG. 2. The power converted Pconv. 3. The output power Pout. 4. The efficiency of the motor. BITS Pilani, Pilani Campus Solution 1. 2. 3. 3 cos3 480 60 0.85 42.4 kWin L LP V I u == =42.4 2 1.8 38.6 kWAG in SCL coreP P P P = = =70038.6 37.9 kW1000conv AG RCLP P P = = =&60037.9 37.3 kW1000out conv F WP P P = = =BITS Pilani, Pilani Campus Solution 4. 37.350 hp0.746outP = =100%37.3100 88%42.4outinPPq = = =BITS Pilani, Pilani Campus Example A 460-V, 25-hp, 60 Hz, four-pole, Y-connected induction motor has the following impedances in ohms per phase referred to the stator circuit: R1= 0.641O R2= 0.332O X1= 1.106 OX2= 0.464 OXM= 26.3 O The total rotational losses are 1100 W and are assumed to be constant. The core loss is lumped in with the rotational losses. For a rotor slip of 2.2 percent at the rated voltage and rated frequency, find the motors 1. Speed 2. Stator current 3. Power factor 4.Pconv and Pout 5. tind and tload 6. Efficiency BITS Pilani, Pilani Campus Solution 1. 2.

120 120 601800 rpm4esyncfnP= = =(1 ) (1 0.022) 1800 1760 rpmm syncn s n = = =22 20.3320.4640.02215.09 0.464 15.1 1.76RZ jX jsj= + = += + = Z O21 11/ 1/ 0.038 0.0662 1.76112.94 31.10.0773 31.1fMZjX Z j= =+ + Z = = Z OZ BITS Pilani, Pilani Campus Solution 3.4. 0.641 1.106 12.94 31.111.72 7.79 14.07 33.6tot stat fZ Z Zjj= += + + Z O= + = Z O1460 0318.88 33.6 A14.07 33.6totVIZ|Z= = = Z Z cos33.6 0.833 lagging PF = =3 cos 3 460 18.88 0.833 12530 Win L LP V I u = = =2 21 13 3(18.88) 0.641 685 WSCLP I R = = =12530 685 11845 WAG in SCLP P P = = =BITS Pilani, Pilani Campus Solution 5. 6. (1 ) (1 0.022)(11845) 11585 Wconv AGP s P = = =&11585 1100 10485 W10485= 14.1hp746out conv F WP P P = = ==1184562.8 N.m1800260AGindsyncPtet= = =1048556.9 N.m1760260outloadmPtet= = =10485100% 100 83.7%12530outinPPq = = =BITS Pilani, Pilani Campus Torque, power and Thevenins Theorem Thevenins theorem can be used to transform the network to the left of points a and b into an equivalent voltage source VTH in series with equivalent impedance RTH+jXTH BITS Pilani, Pilani Campus Torque, power and Thevenins Theorem 1 1( )MTHMjXV VR j X X|=+ +1 1( ) //TH TH MR jX R jX jX + = +2 21 1| | | |( )MTHMXV VR X X|=+ +BITS Pilani, Pilani Campus Thevenins Equivalent Circuit BITS Pilani, Pilani Campus Torque, power and Thevenins Theorem Since XM>>X1 and XM>>R1

Because XM>>X1 and XM+X1>>R1

1MTHMXV VX X|~+2111MTHMTHXR RX XX X| |~ |+\ .~BITS Pilani, Pilani Campus Torque, power and Thevenins Theorem Then the power converted to mechanical (Pconv) 22222( )TH THTTH THV VIZRR X Xs= =| |+ + + |\ .222(1 )3convR sP Is=And the internal mechanical torque (Tconv) convindmPte=(1 )convsPs e=2223AGs sRIPse e= =BITS Pilani, Pilani Campus Torque, power and Thevenins Theorem 2222223( )THindsTH THV RsRR X Xste| | | | | |= | |\ .| | |+ + + | |\ .\ .22222231( )THindsTH THRVsRR X Xste| | |\ .=| |+ + + |\ .BITS Pilani, Pilani Campus Torque-speed characteristics Typical torque-speed characteristics of induction motor BITS Pilani, Pilani Campus Comments 1. Theinducedtorqueiszeroatsynchronous speed. Discussed earlier. 2. Thecurveisnearlylinearbetweenno-loadand fullload.Inthisrange,therotorresistanceis muchgreaterthanthereactance,sotherotor current, torque increase linearly with the slip.3. Thereisamaximumpossibletorquethatcant be exceeded. This torque is called pullout torque and is 2 to 3 times the rated full-load torque. BITS Pilani, Pilani Campus Comments 4. Thestartingtorqueofthemotorisslightly higher than its full-load torque, so the motor willstartcarryinganyloaditcansupplyat full load. 5. Thetorqueofthemotorforagivenslip varies as the square of the applied voltage. 6. If the rotor is driven faster than synchronous speeditwillrunasagenerator,converting mechanical power to electric power. BITS Pilani, Pilani Campus Complete Speed-torque c/c BITS Pilani, Pilani Campus Maximum torque Maximum torque occurs when the power transferred to R2/s is maximum. This condition occurs when R2/s equals the magnitude of the impedance RTH + j (XTH + X2) max2 222( )TH THTRR X Xs= + +max22 22( )TTH THRsR X X=+ +BITS Pilani, Pilani Campus Maximum torque The corresponding maximum torque of an induction motor equals The slip at maximum torque is directly proportional to the rotor resistance R2 The maximum torque is independent of R2 2max2 223 12( )THsTH TH THVR R X Xte| | |= |+ + +\ . BITS Pilani, Pilani Campus Maximum torque Rotor resistance can be increased by inserting external resistance in the rotor of a wound-rotor induction motor. Thevalue of the maximum torque remains unaffectedbutthe speed at which it occurs can be controlled. BITS Pilani, Pilani Campus Maximum torque Effect of rotor resistance on torque-speed characteristic BITS Pilani, Pilani Campus Example A two-pole, 50-Hz induction motor supplies 15kW to a load at a speed of 2950 rpm. 1. What is the motors slip? 2. What is the induced torque in the motor in N.m under these conditions? 3. What will be the operating speed of the motor if its torque is doubled? 4. How much power will be supplied by the motor when the torque is doubled? BITS Pilani, Pilani Campus Solution 1. 2. 120 120 503000 rpm23000 29500.0167 or 1.67%3000esyncsync msyncfnPn nsn= = == = =3 no given assume and 15 1048.6 N.m2295060f Wconv load ind loadconvindmPP PPt ttte+ = == = =BITS Pilani, Pilani Campus Solution 3. In the low-slip region, the torque-speed curve is linear and the induced torque is direct proportional to slip. So, if the torque is doubled the new slip will be 3.33% and the motor speed will be

(1 ) (1 0.0333) 3000 2900 rpmm syncn s n = = =2(2 48.6) (2900 ) 29.5 kW60conv ind mP t et== =BITS Pilani, Pilani Campus Example A 460-V, 25-hp, 60-Hz, four-pole, Y-connected wound-rotor induction motor has the following impedances in ohms per phase referred to the stator circuit R1= 0.641O R2= 0.332O X1= 1.106 OX2= 0.464 OXM= 26.3 O 1. What is the maximum torque of this motor? At what speed and slip does it occur? 2. What is the starting torque of this motor? 3. If the rotor resistance is doubled, what is the speed at which the maximum torque now occur? What is the new starting torque of the motor? 4. Calculate and plot the T-s c/c for both cases. BITS Pilani, Pilani Campus Solution

2 21 12 2( )46026.33255.2 V(0.641) (1.106 26.3)MTHMXV VR X X|=+ += =+ +211226.3(0.641) 0.5901.106 26.3MTHMXR RX X| |~ |+\ .| |~ = O |+\ .11.106THX X ~ = OBITS Pilani, Pilani Campus Solution 1. The corresponding speed is max22 222 2( )0.3320.198(0.590) (1.106 0.464)TTH THRsR X X=+ += =+ +(1 ) (1 0.198) 1800 1444 rpmm syncn s n = = =BITS Pilani, Pilani Campus Solution The torque at this speed is 2max2 2222 23 12( )3 (255.2)22 (1800 )[0.590 (0.590) (1.106 0.464) ]60229 N.mTHsTH TH THVR R X Xtet| | |= |+ + +\ .= + + +=BITS Pilani, Pilani Campus Solution 2. The starting torque can be found from the torque eqn. by substituting s = 1 ( )2221222122222 222 231( )3[ ( ) ]3 (255.2) (0.332)21800 [(0.590 0.332) (1.106 0.464) ]60104 N.mTHstart indssTH THsTHs TH THRVsRR X XsV RR R X Xt teet==| | |\ .= =| |+ + + |\ .=+ + + = + + +=BITS Pilani, Pilani Campus Solution 3. If the rotor resistance is doubled, then the slip at maximum torque doubles too The corresponding speed is The maximum torque is still tmax = 229 N.m max22 220.396( )TTH THRsR X X= =+ +(1 ) (1 0.396) 1800 1087 rpmm syncn s n = = =BITS Pilani, Pilani Campus Solution The starting torque is now 22 23 (255.2) (0.664)21800 [(0.590 0.664) (1.106 0.464) ]60170 N.mstarttt = + + +=BITS Pilani, Pilani Campus Determination of motor parameters Due to the similarity between the induction motor equivalent circuit and the transformer equivalent circuit, same tests are used to determine the values of the motor parameters. DC test: determine the stator resistance R1

No-load test: determine the rotational losses and magnetization current (similar to no-load test in Transformers). Locked-rotor test: determine the rotor and stator impedances (similar to short-circuit test in Transformers). BITS Pilani, Pilani Campus DC test The purpose of the DC test is to determine R1. A variable DC voltage source is connected between two stator terminals. The DC source is adjusted to provide approximately rated stator current, and the resistance between the two stator leads is determined from the voltmeter and ammeter readings. BITS Pilani, Pilani Campus DC test then If the stator is Y-connected, the per phase stator resistance is If the stator is delta-connected, the per phase stator resistance is DCDCDCVRI=12DCRR=132DCR R =BITS Pilani, Pilani Campus No-load test 1. The motor is allowed to spin freely 2. The only load on the motor is the friction and windage losses, so all Pconv is consumed by mechanical losses 3. The slip is very small BITS Pilani, Pilani Campus No-load test 4. At this small slip The equivalent circuit reduces to 2 22 2(1 ) R (1 )&R s sR Xs s BITS Pilani, Pilani Campus No-load test 5. Combining Rc & RF+W we get BITS Pilani, Pilani Campus No-load test 6. At the no-load conditions, the input power measured by meters must equal the losses in the motor. 7. The PRCL is negligible because I2 is extremely small because R2(1-s)/s is very large. 8. The input power equals Where &21 13in SCL core F WrotP P P PI R P= + += +& rot core F WP P P = +BITS Pilani, Pilani Campus No-load test 9. The equivalent input impedance is thus approximately If X1 can be found, in some other fashion, the magnetizing impedance XM will be known 11,eq MnlVZ X XI|= ~ +BITS Pilani, Pilani Campus Blocked-rotor test In this test, the rotor is locked or blocked so that it cannot move, a voltage is applied to the motor, and the resulting voltage, current and power are measured. BITS Pilani, Pilani Campus Blocked-rotor test The AC voltage applied to the stator is adjusted so that the current flow is approximately full-load value. The locked-rotor power factor can be found as The magnitude of the total impedancecos3inl lPPFV Iu = =LRVZI|=BITS Pilani, Pilani Campus Blocked-rotor test

Where X1 and X2 are the stator and rotor reactances at the test frequency respectively 'cos sinLR LR LRLR LRZ R jXZ j Z u u= += +1 2' ' '1 2LRLRR R RX X X= += +2 1 LRR R R = '1 2ratedLR LRtestfX X X Xf= = +BITS Pilani, Pilani Campus Blocked-rotor test X1 and X2 as function of XLR Rotor DesignX1X2 Wound rotor0.5 XLR 0.5 XLR Design A0.5 XLR 0.5 XLR Design B0.4 XLR 0.6 XLR Design C0.3 XLR 0.7 XLR Design D0.5 XLR 0.5 XLR BITS Pilani, Pilani Campus Example The following test data were taken on a 7.5-hp, four-pole, 208-V, 60-Hz, design A, Y-connected IM having a rated current of 28 A. DC Test: VDC = 13.6 VIDC = 28.0 A No-load Test: Vl = 208 Vf = 60 Hz I = 8.17 APin = 420 W Locked-rotor Test: Vl = 25 V f = 15 Hz I = 27.9 APin = 920 W (a) Sketch the per-phase equivalent circuit of this motor. (b) Find the slip at pull-out torque, and find the value of the pull-out torque.