Induction MotorsIntroduction 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 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 winding Stator of IM 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. Construction Squirrel cage rotor Wound rotor Notice the slip rings Construction Cutaway in a typical wound- rotor IM. Notice the brushes and the slip rings Brushes Slip rings 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 f e is the supply frequency and P is the no. of poles and n sync is called the synchronous speed in rpm (revolutions per minute) 120 e sync f n rpm P = Synchronous speed P 50 Hz 60 Hz 2 3000 3600 4 1500 1800 6 1000 1200 8 750 900 10 600 720 12 500 600 Rotating Magnetic Field Rotating Magnetic Field Rotating Magnetic Field ( ) ( ) ( ) ( ) net a b c B t B t B t B t = + + sin( ) 0 sin( 120 ) 120 sin( 240) 240 M M M B t B t B t e e e = Z ° + ÷ ° Z ° + ÷ Z ° ˆ sin( ) 3 ˆ ˆ [0.5 sin( 120 )] [ sin( 120 )] 2 3 ˆ ˆ [0.5 sin( 240 )] [ sin( 240 )] 2 M M M M M B t B t B t B t B t e e e e e = ÷ ÷ ° ÷ ÷ ° ÷ ÷ ° + ÷ ° x x y x y Rotating Magnetic Field 1 3 1 3 ˆ ( ) [ sin( ) sin( ) cos( ) sin( ) cos( )] 4 4 4 4 3 3 3 3 ˆ [ sin( ) cos( ) sin( ) cos( )] 4 4 4 4 net M M M M M M M M M B t B t B t B t B t B t B t B t B t B t e e e e e e e e e = + + + ÷ + ÷ ÷ + ÷ x y ˆ ˆ [1.5 sin( )] [1.5 cos( )] M M B t B t e e = ÷ x y Rotating Magnetic Field 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 t ind is the induced torque and B R and B S are the magnetic flux densities of the rotor and the stator respectively ind R s kB B t = × 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 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 n slip = slip speed n sync = speed of the magnetic field n m = mechanical shaft speed of the motor slip sync m n n n = ÷ The Slip sync m sync n n s n ÷ = 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 doesn’t have units 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 Frequency The frequency of the voltage induced in the rotor is given by Where f r = the rotor frequency (Hz) P = number of stator poles n = slip speed (rpm) 120 r P n f × = ( ) 120 120 s m r s e P n n f P sn sf × ÷ = × = = Frequency What would be the frequency of the rotor’s induced voltage at any speed n m ? 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 e f s f = 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 . out load m P N m t e = 2 / 60 m m n rad s t e = 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 = 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? Solution 1. 2. 3. 4. 120 120(60) 1800 4 e sync f n rpm P = = = (1 ) (1 0.05) 1800 1710 m s n s n rpm = ÷ = ÷ × = 0.05 60 3 r e f sf Hz = = × = 2 60 10 746 / 41.7 . 1710 2 (1/ 60) out out load m m P P n hp watt hp N m t e t t = = × = = × × 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 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 E R0 is the largest value of the rotor’s induced voltage obtained at s = 1(loacked rotor) 0 R R E sE = 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 X r0 is the rotor reactance at the supply frequency (at blocked rotor) r e f s f = 2 X L f L e t = = 0 2 2 r r r r r e r r X L f L sf L sX e t t = = = = Equivalent Circuit Then, we can draw the rotor equivalent circuit as follows Where E R is the induced voltage in the rotor and R R is the rotor resistance Equivalent Circuit Now we can calculate the rotor current as Dividing both the numerator and denominator by s so nothing changes we get Where E R0 is the induced voltage and X R0 is the rotor reactance at blocked rotor condition (s = 1) 0 0 ( ) ( ) R R R R R R R E I R jX sE R jsX = + = + 0 0 ( ) R R R R E I R jX s = + Equivalent Circuit Now we can have the rotor equivalent circuit 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 2 2 0 2 2 2 1 0 eff R eff R R eff eff R S eff R X a X R a R I I a E a E N a N = = = = = Power losses in Induction machines Copper losses - Copper loss in the stator (P SCL ) = I 1 2 R 1 - Copper loss in the rotor (P RCL ) = I 2 2 R 2 Core loss (P core ) Mechanical power loss due to friction and windage How this power flow in the motor? Power flow in induction motor Power relations 3 cos 3 cos in L L ph ph P V I V I u u = = 2 1 1 3 SCL P I R = ( ) AG in SCL core P P P P = ÷ + 2 2 2 3 RCL P I R = conv AG RCL P P P = ÷ ( ) out conv f w stray P P P P + = ÷ + conv ind m P t e = Equivalent Circuit We can rearrange the equivalent circuit as follows Actual rotor resistance Resistance equivalent to mechanical load Power relations 3 cos 3 cos in L L ph ph P V I V I u u = = 2 1 1 3 SCL P I R = ( ) AG in SCL core P P P P = ÷ + 2 2 2 3 RCL P I R = conv AG RCL P P P = ÷ ( ) out conv f w stray P P P P + = ÷ + conv RCL P P = + 2 2 2 3 R I s = 2 2 2 (1 ) 3 R s I s ÷ = RCL P s = (1 ) RCL P s s ÷ = (1 ) conv AG P s P = ÷ conv ind m P t e = (1 ) (1 ) AG s s P s e ÷ = ÷ Power relations AG P RCL P conv P 1 s 1-s : : 1 : : 1- AG RCL conv P P P s s 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 P AG. 2. The power converted P conv . 3. The output power P out . 4. The efficiency of the motor. Solution 1. 2. 3. 3 cos 3 480 60 0.85 42.4 kW in L L P V I u = = × × × = 42.4 2 1.8 38.6 kW AG in SCL core P P P P = ÷ ÷ = ÷ ÷ = 700 38.6 37.9 kW 1000 conv AG RCL P P P = ÷ = ÷ = & 600 37.9 37.3 kW 1000 out conv F W P P P = ÷ = ÷ = Solution 4. 37.3 50 hp 0.746 out P = = 100% 37.3 100 88% 42.4 out in P P q = × = × = 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: R 1 = 0.641O R 2 = 0.332O X 1 = 1.106 O X 2 = 0.464 O X M = 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 motor’s 1. Speed 2. Stator current 3. Power factor 4. P conv and P out 5. t ind and t load 6. Efficiency Solution 1. 2. 120 120 60 1800 rpm 4 e sync f n P × = = = (1 ) (1 0.022) 1800 1760 rpm m sync n s n = ÷ = ÷ × = 2 2 2 0.332 0.464 0.022 15.09 0.464 15.1 1.76 R Z jX j s j = + = + = + = Z ° O 2 1 1 1/ 1/ 0.038 0.0662 1.76 1 12.94 31.1 0.0773 31.1 f M Z jX Z j = = + ÷ + Z÷ ° = = Z ° O Z÷ ° Solution 3. 4. 0.641 1.106 12.94 31.1 11.72 7.79 14.07 33.6 tot stat f Z Z Z j j = + = + + Z ° O = + = Z ° O 1 460 0 3 18.88 33.6 A 14.07 33.6 tot V I Z | Z ° = = = Z÷ ° Z ° cos33.6 0.833 lagging PF = ° = 3 cos 3 460 18.88 0.833 12530 W in L L P V I u = = × × × = 2 2 1 1 3 3(18.88) 0.641 685 W SCL P I R = = × = 12530 685 11845 W AG in SCL P P P = ÷ = ÷ = Solution 5. 6. (1 ) (1 0.022)(11845) 11585 W conv AG P s P = ÷ = ÷ = & 11585 1100 10485 W 10485 = 14.1hp 746 out conv F W P P P = ÷ = ÷ = = 11845 62.8 N.m 1800 2 60 AG ind sync P t e t = = = × 10485 56.9 N.m 1760 2 60 out load m P t e t = = = × 10485 100% 100 83.7% 12530 out in P P q = × = × = Torque, power and Thevenin’s Theorem Thevenin’s theorem can be used to transform the network to the left of points ‘a’ and ‘b’ into an equivalent voltage source V TH in series with equivalent impedance R TH +jX TH Torque, power and Thevenin’s Theorem 1 1 ( ) M TH M jX V V R j X X | = + + 1 1 ( ) // TH TH M R jX R jX jX + = + 2 2 1 1 | | | | ( ) M TH M X V V R X X | = + + Torque, power and Thevenin’s Theorem Since X M >>X 1 and X M >>R 1 Because X M >>X 1 and X M +X 1 >>R 1 1 M TH M X V V X X | ~ + 2 1 1 1 M TH M TH X R R X X X X | | ~ | + \ . ~ For the calculation Purpose Torque, power and Thevenin’s Theorem Then the power in the air gap (P 2 ) 2 2 2 2 2 ( ) TH TH T TH TH V V I Z R R X X s = = | | + + + | \ . And the internal mechanical torque (T) {Putting Value of I 2 } Torque, power and Thevenin’s Theorem Gives, Torque-speed characteristics Typical torque-speed characteristics of induction motor Comments 1. The induced torque is zero at synchronous speed. Discussed earlier. (i.e. N=N s and s=0) 2. The curve is nearly linear between no-load and full load. In this range, the rotor resistance is much greater than the reactance, so the rotor current, torque increase linearly with the slip. (T o s) 3. There is a maximum possible torque that can’t be exceeded. This torque is called pullout torque and is 2 to 3 times the rated full-load torque.(it occurs on s Tmax for which sX 2 ~R 2 ) Comments 4. The starting torque of the motor is slightly higher than its full-load torque, so the motor will start carrying any load it can supply at full load.(i.e. s=1 at N=0) 5. The torque of the motor for a given slip varies as the square of the applied voltage.( To V 2 ) 6. If the rotor is driven faster than synchronous speed it will run as a generator, converting mechanical power to electric power. Complete Speed-torque c/c Complete Torque-slip c/c Maximum torque Maximum torque occurs when the power transferred to R 2 /s is maximum. This condition occurs when R 2 /s equals the magnitude of the impedance R TH + j (X TH + X 2 ) max 2 2 2 2 ( ) TH TH T R R X X s = + + max 2 2 2 2 ( ) T TH TH R s R X X = + + [i.e. only resistive] Maximum torque The corresponding maximum torque of an induction motor equals The slip at maximum torque is directly proportional to the rotor resistance R 2 The maximum torque is independent of R 2 Maximum torque Rotor resistance can be increased by inserting external resistance in the rotor of a wound-rotor induction motor. The value of the maximum torque remains unaffected but the speed at which it occurs can be controlled. Maximum torque Effect of rotor resistance on torque-speed characteristic Example A two-pole, 50-Hz induction motor supplies 15kW to a load at a speed of 2950 rpm. 1. What is the motor’s 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? Solution 1. 2. 120 120 50 3000 rpm 2 3000 2950 0.0167 or 1.67% 3000 e sync sync m sync f n P n n s n × = = = ÷ ÷ = = = 3 no given assume and 15 10 48.6 N.m 2 2950 60 f W conv load ind load conv ind m P P P P t t t t e + = = × = = = × 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 4. (1 ) (1 0.0333) 3000 2900 rpm m sync n s n = ÷ = ÷ × = 2 (2 48.6) (2900 ) 29.5 kW 60 conv ind m P t e t = = × × × = 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 R 1 = 0.641O R 2 = 0.332O X 1 = 1.106 O X 2 = 0.464 O X M = 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. Solution 2 2 1 1 2 2 ( ) 460 26.3 3 255.2 V (0.641) (1.106 26.3) M TH M X V V R X X | = + + × = = + + 2 1 1 2 26.3 (0.641) 0.590 1.106 26.3 M TH M X R R X X | | ~ | + \ . | | ~ = O | + \ . 1 1.106 TH X X ~ = O Solution 1. The corresponding speed is max 2 2 2 2 2 2 ( ) 0.332 0.198 (0.590) (1.106 0.464) T TH TH R s R X X = + + = = + + (1 ) (1 0.198) 1800 1444 rpm m sync n s n = ÷ = ÷ × = Solution The torque at this speed is 2 max 2 2 2 2 2 2 3 1 2 ( ) 3 (255.2) 2 2 (1800 )[0.590 (0.590) (1.106 0.464) ] 60 229 N.m TH s TH TH TH V R R X X t e t | | | = | + + + \ . × = × × + + + = Solution 2. The starting torque can be found from the torque eqn. by substituting s = 1 ( ) 2 2 2 1 2 2 2 1 2 2 2 2 2 2 2 2 2 3 1 ( ) 3 [ ( ) ] 3 (255.2) (0.332) 2 1800 [(0.590 0.332) (1.106 0.464) ] 60 104 N.m TH start ind s s TH TH s TH s TH TH R V s R R X X s V R R R X X t t e e t = = | | | \ . = = | | + + + | \ . = + + + × × = × × + + + = 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 t max = 229 N.m max 2 2 2 2 0.396 ( ) T TH TH R s R X X = = + + (1 ) (1 0.396) 1800 1087 rpm m sync n s n = ÷ = ÷ × = Solution The starting torque is now 2 2 2 3 (255.2) (0.664) 2 1800 [(0.590 0.664) (1.106 0.464) ] 60 170 N.m start t t × × = × × + + + = Induction Generator •At speed N operating point is ‘a’. •At N=Ns operating point is ‘B’. •At N>Ns operating point is ‘c’. When Rotor speed made more than Ns its slip becomes Negative and m/c operates as induction generator Induction Generator Contd… As Rotating Magnetic field set up by Stator current i.e. via supply mains hence induction generator will continue to take power from the line. In other words its not self excited. Can not work in isolation, i.e it must work in parallel to the bus bar. Capacitor Bank is used to ease the extra burden over weak transmission line which already operates on a poor (lagging) power factor. Isolated Induction Generator Does not Required an existing supply system for obtaining magnetizing current or reactive power. A three phase capacitor bank is used to supply reactive power for both load as well as generator. Hence reactive power supplied by capacitor bank = Reactive power needed by 3-phase Induction generator + Reactive power needed by the load. Isolated(Self Excited) Induction Gen. Load line b c d e V T I m V 1 I m1 OA is the voltage due to residual flux. No residual Voltage, hence no voltage build-up. Ob capacitor current due to residual voltage OA. Note: Vc = I.Xc, and Xc = 1/ eC Voltage build- up depends upon the value of capacitor i.e for higher C voltage will be high provided capacitor load line cuts the magnetization curve. Application of Induction Generator Externally Excited: Runs in parallel to bus bar used in Hydro Power plants for interconnection purpose. However they suffers from poor efficiency and considerable amount of reactive power is always required. Self Excited: Wind Mills (Non Conventional Energy Resources), like wind turbine where wind energy is converted into electrical energy. Provide a clean and environment friendly form of energy conversion. Summary Q & A 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 R 1 - 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). DC test - The purpose of the DC test is to determine R 1 . 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. 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 DC DC DC V R I = 1 2 DC R R = 1 3 2 DC R R = 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 P conv is consumed by mechanical losses 3. The slip is very small No-load test 4. At this small slip The equivalent circuit reduces to… 2 2 2 2 (1 ) R (1 ) & R s s R X s s ÷ ÷ No-load test 5. Combining R c & R F+W we get…… No-load test 6. At the no-load conditions, the input power measured by meters must equal the losses in the motor. 7. The P RCL is negligible because I 2 is extremely small because R 2 (1-s)/s is very large. 8. The input power equals Where & 2 1 1 3 in SCL core F W rot P P P P I R P = + + = + & rot core F W P P P = + No-load test 9. The equivalent input impedance is thus approximately If X 1 can be found, in some other fashion, the magnetizing impedance X M will be known 1 1, eq M nl V Z X X I | = ~ + 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. 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 impedance cos 3 in l l P PF V I u = = LR V Z I | = Blocked-rotor test Where X’ 1 and X’ 2 are the stator and rotor reactances at the test frequency respectively ' cos sin LR LR LR LR LR Z R jX Z j Z u u = + = + 1 2 ' ' ' 1 2 LR LR R R R X X X = + = + 2 1 LR R R R = ÷ ' 1 2 rated LR LR test f X X X X f = = + Blocked-rotor test X 1 and X 2 as function of X LR Rotor Design X 1 X 2 Wound rotor 0.5 X LR 0.5 X LR Design A 0.5 X LR 0.5 X LR Design B 0.4 X LR 0.6 X LR Design C 0.3 X LR 0.7 X LR Design D 0.5 X LR 0.5 X LR Midterm Exam No.2 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: V DC = 13.6 V I DC = 28.0 A No-load Test: V l = 208 V f = 60 Hz I = 8.17 A P in = 420 W Locked-rotor Test: V l = 25 V f = 15 Hz I = 27.9 A P in = 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. 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 - 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 winding Stator of IM 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. - Construction Squirrel cage rotor Wound rotor Notice the slip rings . Notice the brushes and the slip rings Brushes .Construction Slip rings Cutaway in a typical woundrotor IM. i. rotating with a speed nsync 120 f e P rpm Where fe is the supply frequency and P is the no. mechanically displaced 120 degrees form each other.e.Rotating Magnetic Field Balanced three phase windings. fed by balanced three phase source A rotating magnetic field with constant magnitude is produced. of poles and nsync is called the synchronous speed in rpm (revolutions per minute) . Synchronous speed P 2 4 50 Hz 3000 1500 60 Hz 3600 1800 6 8 10 12 1000 750 600 500 1200 900 720 600 . Rotating Magnetic Field . Rotating Magnetic Field . Rotating Magnetic Field Bnet (t ) Ba (t ) Bb (t ) Bc (t ) BM sin(t )0 BM sin(t 120)120 BM sin(t 240)240 ˆ BM sin(t )x 3 ˆ BM sin(t 120)]y 2 3 ˆ ˆ [0.5BM sin(t 120)]x [ .5BM sin(t 240)]x [ BM sin(t 240)]y 2 ˆ [0. 5BM cos(t )]y .5BM sin(t )]x [1.Rotating Magnetic Field 1 3 1 3 ˆ Bnet (t ) [ BM sin(t ) BM sin(t ) BM cos(t ) BM sin(t ) BM cos(t )]x 4 4 4 4 3 3 3 3 ˆ [ BM sin(t ) BM cos(t ) BM sin(t ) BM cos(t )]y 4 4 4 4 ˆ ˆ [1. Rotating Magnetic Field . 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 ind kBR Bs Where ind is the induced torque and BR and BS are the magnetic flux densities of the rotor and the stator respectively .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. Induction motor speed At what speed will the IM run? Can the IM run at the synchronous speed. which is the same speed of the rotating magnetic field. 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. then the rotor will appear stationary to the rotating magnetic field and the rotating magnetic field will not cut the rotor. why? If rotor runs at the synchronous speed. the rotating magnetic field will cut the rotor windings and a torque is produced - . 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 nslip nsync nm Where nslip= slip speed nsync= speed of the magnetic field nm = mechanical shaft speed of the motor .Induction motor speed So. by 100. notice that the slip is a ratio and doesn’t have units nsync nm nsync .The Slip s 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. the secondary windings can move Due to the rotation of the rotor (the secondary winding of the IM).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. the induced voltage in it does not have the same frequency of the stator (the primary) voltage . in the case of the induction motor. Frequency The frequency of the voltage induced in the rotor is given by Pn fr 120 Where fr = the rotor frequency (Hz) P = number of stator poles n = slip speed (rpm) P (ns nm ) fr 120 P sns sf e 120 . if the rotor runs at synchronous speed (s = 0).Frequency What would be the frequency of the rotor’s induced voltage at any speed nm? fr s fe When the rotor is blocked (s=1) . the frequency will be zero . the frequency of the induced voltage is equal to the supply frequency On the other hand. m and 2 nm m 60 rad / s . This torque is related to the motor output power and the rotor speed load Pout m N . 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.Torque While the input to the induction motor is electrical power. as an electrical engineers.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. deal with watts as a unit to measure electrical power. there is a relation between horse power and watts hp 746 watts . What is the synchronous speed of this motor? What is the rotor speed of this motor at rated load? What is the rotor frequency of this motor at rated load? What is the shaft torque of this motor at rated load? .Example A 208-V. 4. four pole. 2. 10hp. Y-connected induction motor has a full-load slip of 5 percent 1. 60 Hz. 3. 05) 1800 1710 rpm 3. load Pout Pout 2 nm 60 10 hp 746 watt / hp 41.05 60 3Hz m 4.m 1710 2 (1/ 60) .7 N . f r sf e 0. nm (1 s)ns (1 0. nsync 120 f e 120(60) 1800 rpm P 4 2.Solution 1. 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 . e. s = 0. Why? On the other side. the induced voltage and frequency in the rotor will be equal to zero. i.Equivalent Circuit When the rotor is locked (or blocked). s =1.e. the largest voltage and rotor frequency are induced in the rotor. Why? ER sER 0 Where ER0 is the largest value of the rotor’s induced voltage obtained at s = 1(loacked rotor) . i. if the rotor rotates at synchronous speed. the reactance of the rotor circuit also changes X L 2 f L Where Xr0 is the rotor reactance at the supply frequency (at blocked rotor) r r r r r 2 sf e Lr sX r 0 . fr s fe It is known that X L 2 f L So. i. as the frequency of the induced voltage in the rotor changes.Equivalent Circuit The same is true for the frequency.e. we can draw the rotor equivalent circuit as follows Where ER is the induced voltage in the rotor and RR is the rotor resistance .Equivalent Circuit Then. Equivalent Circuit Now we can calculate the rotor current as ER IR ( RR jX R ) sER 0 ( RR jsX R 0 ) Dividing both the numerator and denominator by s so nothing changes we get IR ( ER 0 RR jX R 0 ) s Where ER0 is the induced voltage and XR0 is the rotor reactance at blocked rotor condition (s = 1) . Equivalent Circuit Now we can have the rotor equivalent circuit . we can combine the stator and rotor circuits in one equivalent circuit Where 2 X 2 aeff X R 0 2 R2 aeff RR I2 IR aeff NS NR E1 aeff ER 0 aeff .Equivalent Circuit Now as we managed to solve the induced voltage and different frequency problems. 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? . Power flow in induction motor . Power relations P 3 VL I L cos 3 Vph I ph cos in PSCL 3 I12 R1 PAG Pin ( PSCL Pcore ) 2 PRCL 3I 2 R2 Pconv PAG PRCL Pout Pconv ( Pf w Pstray ) ind Pconv m . Equivalent Circuit We can rearrange the equivalent circuit as follows Actual rotor resistance Resistance equivalent to mechanical load . Power relations P 3 VL I L cos 3 Vph I ph cos in PSCL 3 I12 R1 PAG Pin ( PSCL Pcore ) Pconv PRCL 2 PRCL 3I 2 R2 R2 3I s 2 2 PRCL s 2 Pconv PAG PRCL 3I 2 R2 (1 s) s Pconv (1 s) PAG PRCL (1 s) s Pconv Pout Pconv ( Pf w Pstray ) ind m (1 s ) PAG (1 s )s . Power relations PAG 1 Pconv 1-s PRCL s PAG : PRCL : Pconv 1 : s : 1-s 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. 2. 3. 4. The air-gap power PAG. The power converted Pconv. The output power Pout. The efficiency of the motor. Solution 1. Pin 3VL I L cos 3 480 60 0.85 42.4 kW PAG Pin PSCL Pcore 42.4 2 1.8 38.6 kW 2. Pconv PAG PRCL 700 38.6 37.9 kW 1000 3. Pout Pconv PF &W 600 37.9 37.3 kW 1000 Solution 37.3 Pout 50 hp 0.746 4. Pout 100% Pin 37.3 100 88% 42.4 464 XM= 26. 2. four-pole. The core loss is lumped in with the rotational losses. 3.2 percent at the rated voltage and rated frequency. ind and load 6. find the motor’s 1. Speed Stator current Power factor 4. Efficiency . Pconv and Pout 5. 60 Hz.3 The total rotational losses are 1100 W and are assumed to be constant.641 R2= 0.106 X2= 0.332 X1= 1.Example A 460-V. Y-connected induction motor has the following impedances in ohms per phase referred to the stator circuit: R1= 0. For a rotor slip of 2. 25-hp. 464 2.0773 31.332 jX 2 j 0.09 j 0.038 0.76 1 1 Zf 1/ jX M 1/ Z 2 j 0.0662 1.9431.11.Solution 120 f e 120 60 1800 rpm 1.1 .022) 1800 1760 rpm R2 0.464 15. Z 2 s 0.76 1 12. nsync P 4 nm (1 s )nsync (1 0.022 15.1 0. 6 A Z tot 14.0733.72 j 7.9431.88 33.88) 2 0.641 j1.79 14.833 12530 W in PSCL 3I12 R1 3(18. PF cos33.6 0.641 685 W PAG Pin PSCL 12530 685 11845 W . P 3VL I L cos 3 460 18.88 0.0733.6 4600 V 3 I1 18.833 lagging 4.106 12.1 11.6 3.Solution Ztot Z stat Z f 0. m load 60 Pout 10485 100% 100 83.8 N. ind 62.1 hp 746 PAG 11845 sync 2 1800 5. Pin 12530 10485 m 2 1760 Pout .Solution Pconv (1 s) PAG (1 0.9 N.m 60 56.022)(11845) 11585 W Pout Pconv PF &W 11585 1100 10485 W 10485 = 14.7% 6. power and Thevenin’s Theorem Thevenin’s 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 .Torque. power and Thevenin’s Theorem VTH jX M V R1 j ( X 1 X M ) | VTH || V | XM R12 ( X 1 X M )2 RTH jX TH ( R1 jX 1 ) // jX M .Torque. Torque. power and Thevenin’s Theorem Since XM>>X1 and XM>>R1 VTH XM V X1 X M Because XM>>X1 and XM+X1>>R1 RTH XM R1 X1 X M X TH X 1 2 For the calculation Purpose . Torque. power and Thevenin’s Theorem VTH VTH I2 2 ZT R2 RTH ( X TH X 2 )2 s Then the power in the air gap (P2) And the internal mechanical torque (T) {Putting Value of I2 } . Torque. . power and Thevenin’s Theorem Gives. Torque-speed characteristics Typical torque-speed characteristics of induction motor . torque increase linearly with the slip. (T s) 3. The induced torque is zero at synchronous speed.Comments 1. the rotor resistance is much greater than the reactance. Discussed earlier. This torque is called pullout torque and is 2 to 3 times the rated full-load torque. There is a maximum possible torque that can’t be exceeded.e. so the rotor current. N=Ns and s=0) 2. (i. The curve is nearly linear between no-load and full load.(it occurs on sTmax for which sX2 R2) . In this range. s=1 at N=0) 5. .Comments 4.( T V2 ) 6. If the rotor is driven faster than synchronous speed it will run as a generator.(i.e. The torque of the motor for a given slip varies as the square of the applied voltage. converting mechanical power to electric power. so the motor will start carrying any load it can supply at full load. The starting torque of the motor is slightly higher than its full-load torque. Complete Speed-torque c/c . Complete Torque-slip c/c . only resistive] . This condition occurs when R2/s equals the magnitude of the impedance RTH + j (XTH + X2) R2 2 RTH ( X TH X 2 )2 sTmax sTmax R2 2 RTH ( X TH X 2 )2 [i.Maximum torque Maximum torque occurs when the power transferred to R2/s is maximum.e. 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 . Maximum torque Rotor resistance can be increased by inserting external resistance in the rotor of a wound-rotor induction motor. . The value of the maximum torque remains unaffected but the speed at which it occurs can be controlled. Maximum torque Effect of rotor resistance on torque-speed characteristic . What is the induced torque in the motor in N. What will be the operating speed of the motor if its torque is doubled? 4. What is the motor’s slip? 2. 1.Example A two-pole. 50-Hz induction motor supplies 15kW to a load at a speed of 2950 rpm.m under these conditions? 3. How much power will be supplied by the motor when the torque is doubled? . nsync 3000 rpm P 2 nsync nm 3000 2950 s 0.Solution 120 f e 120 50 1.6 N.m 2 2950 60 .67% nsync 3000 2.0167 or 1. no Pf W given assume Pconv Pload and ind load ind Pconv m 15 103 48. Solution 3. Pconv ind m 2 (2 48. So. if the torque is doubled the new slip will be 3.6) (2900 ) 29.5 kW 60 . the torque-speed curve is linear and the induced torque is direct proportional to slip. In the low-slip region.33% and the motor speed will be nm (1 s )nsync (1 0.0333) 3000 2900 rpm 4. Calculate and plot the T-s c/c for both cases.464 XM= 26.3 1. 25-hp. 60-Hz.332 X1= 1. what is the speed at which the maximum torque now occur? What is the new starting torque of the motor? 4.106 X2= 0. .Example A 460-V. What is the starting torque of this motor? 3. Y-connected woundrotor induction motor has the following impedances in ohms per phase referred to the stator circuit R1= 0. What is the maximum torque of this motor? At what speed and slip does it occur? 2.641 R2= 0. If the rotor resistance is doubled. four-pole. 641) (1.3 3 255.590 1.3) RTH XM R1 X1 X M 2 26.106 2 .106 26.3 X TH X 1 1.3 (0.2 V 2 2 (0.106 26.641) 0.Solution VTH V XM R12 ( X 1 X M ) 2 460 26. 198) 1800 1444 rpm . sTmax R2 2 RTH ( X TH X 2 )2 0.590) (1.464) 2 2 0.332 (0.106 0.198 The corresponding speed is nm (1 s )nsync (1 0.Solution 1. m 1 2s .590 (0.106 0.2) 2 2 2 (1800 )[0.590) 2 (1.464) 2 ] 60 229 N.Solution The torque at this speed is max 2 3VTH R R 2 ( X X )2 TH TH 2 TH 3 (255. The starting torque can be found from the torque eqn.332) .2) 2 (0.590 0.106 0.Solution 2. by substituting s = 1 2 R2 3VTH 1 s start ind s 1 2 s R2 RTH ( X TH X 2 ) 2 s s 1 s [ RTH R2 ( X TH X 2 ) 2 ] 2 2 3VTH R2 2 1800 [(0.m 3 (255.332) 2 (1.464) 2 ] 60 104 N. If the rotor resistance is doubled. then the slip at maximum torque doubles too R2 sTmax 0.396) 1800 1087 rpm The maximum torque is still max = 229 N.Solution 3.m .396 2 RTH ( X TH X 2 )2 The corresponding speed is nm (1 s )nsync (1 0. 664) .664) 2 (1.Solution The starting torque is now start 2 1800 [(0.464) 2 ] 60 170 N.106 0.2) 2 (0.590 0.m 3 (255. When Rotor speed made more than Ns its slip becomes Negative and m/c operates as induction generator . •At N=Ns operating point is ‘B’.Induction Generator •At speed N operating point is ‘a’. •At N>Ns operating point is ‘c’. Capacitor Bank is used to ease the extra burden over weak transmission line which already operates on a poor (lagging) power factor. In other words its not self excited. i.e it must work in parallel to the bus bar.Induction Generator Contd… As Rotating Magnetic field set up by Stator current i. via supply mains hence induction generator will continue to take power from the line. Can not work in isolation. .e. A three phase capacitor bank is used to supply reactive power for both load as well as generator. . Hence reactive power supplied by capacitor bank = Reactive power needed by 3-phase Induction generator + Reactive power needed by the load.Isolated Induction Generator Does not Required an existing supply system for obtaining magnetizing current or reactive power. Note: Vc = I. No residual Voltage.Isolated(Self Excited) Induction Gen. b d Im Im1 OA is the voltage due to residual flux. and Xc = 1/ C .e for higher C voltage will be high provided capacitor load line cuts the magnetization curve. Ob capacitor current due to residual voltage OA. hence no voltage build-up.Xc. VT e Load line V1 c Voltage buildup depends upon the value of capacitor i. Application of Induction Generator Externally Excited: Runs in parallel to bus bar used in Hydro Power plants for interconnection purpose. Provide a clean and environment friendly form of energy conversion. like wind turbine where wind energy is converted into electrical energy. However they suffers from poor efficiency and considerable amount of reactive power is always required. . Self Excited: Wind Mills (Non Conventional Energy Resources). Summary Q&A . same tests are used to determine the values of the motor parameters.Determination of motor parameters Due to the similarity between the induction motor equivalent circuit and the transformer equivalent circuit. Locked-rotor test: determine the rotor and stator impedances (similar to short-circuit test in Transformers). - . DC test: determine the stator resistance R1 No-load test: determine the rotational losses and magnetization current (similar to no-load test in Transformers). 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. - .DC test The purpose of the DC test is to determine R1. A variable DC voltage source is connected between two stator terminals. the per phase stator resistance is RDC R1 2 If the stator is delta-connected.DC test then RDC VDC I DC - If the stator is Y-connected. the per phase stator resistance is 3 R1 RDC 2 - . The slip is very small . The only load on the motor is the friction and windage losses. so all Pconv is consumed by mechanical losses 3.No-load test 1. The motor is allowed to spin freely 2. At this small slip R2 (1 s) s R2 & R 2 (1 s) s X2 The equivalent circuit reduces to… .No-load test 4. Combining Rc & RF+W we get…… .No-load test 5. No-load test 6. The input power equals Pin PSCL Pcore PF &W 3I12 R1 Prot Where Prot Pcore PF &W . 7. 8. At the no-load conditions. The PRCL is negligible because I2 is extremely small because R2(1-s)/s is very large. the input power measured by meters must equal the losses in the motor. nl X1 X M If X1 can be found.No-load test 9. The equivalent input impedance is thus approximately Z eq V I1. the magnetizing impedance XM will be known . in some other fashion. and the resulting voltage. the rotor is locked or blocked so that it cannot move. a voltage is applied to the motor. . current and power are measured.Blocked-rotor test In this test. The locked-rotor power factor can be found as Pin PF cos 3Vl I l The magnitude of the total impedance Z LR V I .Blocked-rotor test The AC voltage applied to the stator is adjusted so that the current flow is approximately full-load value. Blocked-rotor test ' Z LR RLR jX LR Z LR cos j Z LR sin RLR R1 R2 ' ' X LR X1' X 2 Where X’1 and X’2 are the stator and rotor reactances at the test frequency respectively R2 RLR R1 X LR f rated ' X LR X 1 X 2 ftest . 5 XLR 0.5 XLR 0.7 XLR Design D 0.4 XLR 0.5 XLR 0.5 XLR .5 XLR 0.Blocked-rotor test X1 and X2 as function of XLR Rotor Design Wound rotor Design A Design B Design C X1 0.5 XLR 0.6 XLR 0.3 XLR X2 0. Midterm Exam No.2 . 6 V No-load Test: Vl = 208 V I = 8. 60Hz. Y-connected IM having a rated current of 28 A. and find the value of the pull-out torque. .0 A f = 60 Hz Pin = 420 W f = 15 Hz Pin = 920 W (a) Sketch the per-phase equivalent circuit of this motor. four-pole.5-hp.9 A IDC = 28. design A. DC Test: VDC = 13.17 A Locked-rotor Test: Vl = 25 V I = 27. 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