Induction machines

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Induction machines

  1. 1. Induction Motors
  2. 2. 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
  3. 3. 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
  4. 4. 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.
  5. 5. Construction Squirrel cage rotor Wound rotor Notice the slip rings
  6. 6. Construction Slip rings Cutaway in a typical wound- rotor IM. Notice the brushes and the slip rings Brushes
  7. 7. 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 120 f e nsync = rpm PWhere fe is the supply frequency andP is the no. of poles and nsync is called the synchronous speed in rpm (revolutions per minute)
  8. 8. 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
  9. 9. Rotating Magnetic Field
  10. 10. Rotating Magnetic Field
  11. 11. Rotating Magnetic FieldBnet (t ) = Ba (t ) + Bb (t ) + Bc (t ) = BM sin(ωt )∠0° + BM sin(ωt − 120°)∠120° + BM sin(ωt − 240)∠240° = BM sin(ωt )x ˆ 3 −[0.5 BM sin(ωt − 120°)]x − [ ˆ BM sin(ωt − 120°)]y ˆ 2 3 −[0.5 BM sin(ωt − 240°)]x + [ ˆ BM sin(ωt − 240°)]yˆ 2
  12. 12. Rotating Magnetic Field 1 3 1 3Bnet (t ) = [ BM sin(ωt ) + BM sin(ωt ) + BM cos(ωt ) + BM sin(ωt ) − BM cos(ωt )]x ˆ 4 4 4 4 3 3 3 3 +[ − BM sin(ωt ) − BM cos(ωt ) + BM sin(ωt ) − BM cos(ωt )]y ˆ 4 4 4 4 = [1.5 BM sin(ωt )]x − [1.5BM cos(ωt )]y ˆ ˆ
  13. 13. Rotating Magnetic Field
  14. 14. 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 τ ind = kBR × BsWhere τind is the induced torque and BR and BS are the magnetic flux densities of the rotor and the stator respectively
  15. 15. 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
  16. 16. 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 nslip = nsync − nm Where nslip= slip speed nsync= speed of the magnetic field nm = mechanical shaft speed of the motor
  17. 17. The Slip nsync − nm s= nsyncWhere s is the slipNotice that : if the rotor runs at synchronous speed s=0 if the rotor is stationary s=1Slip may be expressed as a percentage by multiplying the aboveeq. by 100, notice that the slip is a ratio and doesn’t have units
  18. 18. 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
  19. 19. Frequency The frequency of the voltage induced in the rotor is given by P×n 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
  20. 20. 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 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
  21. 21. 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 Pout 2π nm τ load = N .m and ωm = rad / s ωm 60
  22. 22. 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 hp = 746 watts
  23. 23. ExampleA 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?
  24. 24. Solution 120 f e 120(60) 1. nsync = = = 1800 rpm P 4 2. nm = (1 − s )ns = (1 − 0.05) ×1800 = 1710 rpm 3. f r = sf e = 0.05 × 60 = 3Hz Pout Pout 4. τ load = = ωm 2π nm 60 10 hp × 746 watt / hp = = 41.7 N .m 1710 × 2π × (1/ 60)
  25. 25. 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
  26. 26. 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? ER = sER 0 Where ER0 is the largest value of the rotor’s induced voltage obtained at s = 1(loacked rotor)
  27. 27. Equivalent Circuit The same is true for the frequency, i.e. fr = s fe It is known that X = ω L = 2π f L So, as the frequency of the induced voltage in the rotor changes, the reactance of the rotor circuit also changes X = ω L = 2π f L r r r r rWhere Xr0 is the rotor reactance = 2π sf e Lr at the supply frequency(at blocked rotor) = sX r 0
  28. 28. 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
  29. 29. 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 ER 0 IR = RR ( + jX R 0 ) s Where ER0 is the induced voltage and XR0 is the rotor reactance at blocked rotor condition (s = 1)
  30. 30. Equivalent Circuit Now we can have the rotor equivalent circuit
  31. 31. 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 X 2 = aeff X R 0 2 R2 = aeff RR 2 IR I2 = aeff E1 = aeff ER 0 NS aeff = NR
  32. 32. 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?
  33. 33. Power flow in induction motor
  34. 34. Power relations Pin = 3 VL I L cos θ = 3 V ph I ph cos θ PSCL = 3 I12 R1 PAG = Pin − ( PSCL + Pcore ) PRCL = 3I 2 R2 2 Pconv = PAG − PRCL Pconv Pout = Pconv − ( Pf + w + Pstray ) τ ind = ωm
  35. 35. Equivalent Circuit We can rearrange the equivalent circuit as follows Resistance Actual rotor equivalent to resistance mechanical load
  36. 36. Power relations Pin = 3 VL I L cos θ = 3 V ph I ph cos θ PSCL = 3 I12 R1 R2 PRCL PAG = Pin − ( PSCL + Pcore ) = Pconv + PRCL = 3I 2 2 = s s PRCL = 3I 2 R2 2 PRCL (1 − s ) Pconv = PAG − PRCL = 3I 2 R2 (1 − s ) 2 = s s Pconv = (1 − s ) PAG Pconv (1 − s ) PAG Pout = Pconv − ( Pf + w + Pstray ) τ ind = = ωm (1 − s )ωs
  37. 37. Power relations PAG Pconv 1 1-s PRCL s PAG : PRCL : Pconv 1 : s : 1-s
  38. 38. ExampleA 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.
  39. 39. Solution 1. Pin = 3VL I L cos θ = 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
  40. 40. Solution 37.3 Pout = = 50 hp 0.746 Pout 4. η= × 100% Pin 37.3 = × 100 = 88% 42.4
  41. 41. ExampleA 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= 0.641Ω R2= 0.332ΩX1= 1.106 Ω X2= 0.464 Ω XM= 26.3 ΩThe 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 4. P and P conv out 1. Speed 2. Stator current 5. τind and τload 3. Power factor 6. Efficiency
  42. 42. Solution 120 f e 120 × 60 1. nsync = = = 1800 rpm P 4 nm = (1 − s )nsync = (1 − 0.022) ×1800 = 1760 rpm R2 0.332 2. Z 2 = + jX 2 = + j 0.464 s 0.022 = 15.09 + j 0.464 = 15.1∠1.76° Ω 1 1 Zf = = 1/ jX M + 1/ Z 2 − j 0.038 + 0.0662∠ − 1.76° 1 = = 12.94∠31.1° Ω 0.0773∠ − 31.1°
  43. 43. Solution Z tot = Z stat + Z f = 0.641 + j1.106 + 12.94∠31.1° Ω = 11.72 + j 7.79 = 14.07∠33.6° Ω 460∠0° Vφ 3 I1 = = = 18.88∠ − 33.6° A Z tot 14.07∠33.6° 3. PF = cos 33.6° = 0.833 lagging 4. Pin = 3VL I L cos θ = 3 × 460 ×18.88 × 0.833 = 12530 W PSCL = 3I12 R1 = 3(18.88) 2 × 0.641 = 685 W PAG = Pin − PSCL = 12530 − 685 = 11845 W
  44. 44. Solution Pconv = (1 − s ) PAG = (1 − 0.022)(11845) = 11585 W Pout = Pconv − PF &W = 11585 − 1100 = 10485 W 10485 = = 14.1 hp 746 PAG 11845 5. τ ind = = = 62.8 N.m ωsync 2π ×1800 60 Pout 10485 τ load = = = 56.9 N.m ωm 2π ×1760 60 Pout 10485 6. η = ×100% = ×100 = 83.7% Pin 12530
  45. 45. Torque, power and Thevenin’sTheorem 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
  46. 46. Torque, power and Thevenin’sTheorem jX M XM VTH = Vφ | VTH |=| Vφ | R1 + j ( X 1 + X M ) R12 + ( X 1 + X M )2 RTH + jX TH = ( R1 + jX 1 ) // jX M
  47. 47. Torque, power and Thevenin’sTheorem Since XM>>X1 and XM>>R1 XM VTH ≈ Vφ X1 + X M Because XM>>X1 and XM+X1>>R1 2  XM  RTH ≈ R1  ÷  X1 + X M  X TH ≈ X 1
  48. 48. Torque, power and Thevenin’sTheorem VTH VTHI2 = = ZT  R2  2  RTH + ÷ + ( X TH + X 2 ) 2  s Then the power converted to mechanical (Pconv) R2 (1 − s )Pconv = 3I2 2 sAnd the internal mechanical torque (Tconv) 2 R2 Pconv Pconv 3I s = PAG 2 τ ind = = = ωm (1 − s )ωs ωs ωs
  49. 49. Torque, power and Thevenin’sTheorem 2    ÷ 3  VTH ÷  R2 τ ind =  ÷ s ÷ ωs  ÷  2  R + R2  + ( X + X ) 2   TH s ÷ TH 2 ÷     2  R2  3V  ÷ TH 1  s  τ ind = ωs  R2  2  RTH + ÷ + ( X TH + X 2 ) 2  s 
  50. 50. Torque-speed characteristics Typical torque-speed characteristics of induction motor
  51. 51. Comments1. The induced torque is zero at synchronous speed. Discussed earlier.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.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.
  52. 52. Comments4. 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.5. The torque of the motor for a given slip varies as the square of the applied voltage.6. If the rotor is driven faster than synchronous speed it will run as a generator, converting mechanical power to electric power.
  53. 53. Complete Speed-torque c/c
  54. 54. 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) R2 = RTH + ( X TH + X 2 ) 2 2 sTmax R2 sTmax = RTH + ( X TH + X 2 ) 2 2
  55. 55. Maximum torque The corresponding maximum torque of an induction motor equals 1  2 3VTH  τ max =  ÷ 2ωs  R + R 2 + ( X + X )2 ÷  TH TH TH 2  The slip at maximum torque is directly proportional to the rotor resistance R2 The maximum torque is independent of R2
  56. 56. 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.
  57. 57. Maximum torque Effect of rotor resistance on torque-speed characteristic
  58. 58. 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?
  59. 59. Solution 120 f e 120 × 50 1. nsync = = = 3000 rpm P 2 nsync − nm 3000 − 2950 s= = = 0.0167 or 1.67% nsync 3000 2. Q no Pf +W given ∴ assume Pconv = Pload and τ ind = τ load Pconv 15 ×103 τ ind = = = 48.6 N.m ωm 2950 × 2π 60
  60. 60. 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 nm = (1 − s )nsync = (1 − 0.0333) × 3000 = 2900 rpm 4. Pconv = τ ind ωm 2π = (2 × 48.6) × (2900 × ) = 29.5 kW 60
  61. 61. 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 circuitR1= 0.641Ω R2= 0.332ΩX1= 1.106 Ω X2= 0.464 Ω XM= 26.3 Ω 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.
  62. 62. Solution XM VTH = Vφ R12 + ( X 1 + X M )2 460 × 26.3 = 3 = 255.2 V (0.641) + (1.106 + 26.3) 2 2 2  XM  RTH ≈ R1  ÷  X1 + X M  2  26.3  ≈ (0.641)  ÷ = 0.590Ω  1.106 + 26.3  X TH ≈ X 1 = 1.106Ω
  63. 63. Solution R2 1. sTmax = RTH + ( X TH + X 2 ) 2 2 0.332 = = 0.198 (0.590) + (1.106 + 0.464) 2 2 The corresponding speed is nm = (1 − s )nsync = (1 − 0.198) ×1800 = 1444 rpm
  64. 64. Solution The torque at this speed is 1  2 3VTH  τ max =  ÷ 2ωs  R + R 2 + ( X + X )2 ÷  TH TH TH 2  3 × (255.2) 2 = 2π 2 × (1800 × )[0.590 + (0.590) 2 + (1.106 + 0.464) 2 ] 60 = 229 N.m
  65. 65. Solution 2. The starting torque can be found from the torque eqn. by substituting s = 1 2  R2  3VTH  ÷ 1  s  τ start = τ ind s =1 = ωs  R2  2  RTH + ÷ + ( X TH + X 2 ) 2  s  s =1 2 3VTH R2 = ωs [( RTH + R2 ) + ( X TH + X 2 ) 2 ] 2 3 × (255.2) 2 × (0.332) = 2π 1800 × × [(0.590 + 0.332) 2 + (1.106 + 0.464) 2 ] 60 = 104 N.m
  66. 66. Solution 3. If the rotor resistance is doubled, then the slip at maximum torque doubles too R2 sTmax = = 0.396 RTH + ( X TH + X 2 ) 2 2 The corresponding speed is nm = (1 − s )nsync = (1 − 0.396) × 1800 = 1087 rpm The maximum torque is still τmax = 229 N.m
  67. 67. Solution The starting torque is now 3 × (255.2) 2 × (0.664) τ start = 2π 1800 × × [(0.590 + 0.664) 2 + (1.106 + 0.464) 2 ] 60 = 170 N.m
  68. 68. 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).
  69. 69. 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.
  70. 70. DC test - then VDC RDC = I DC - If the stator is Y-connected, the per phase stator resistance is RDC R1 = 2 - If the stator is delta-connected, the per phase stator resistance is 3 R1 = RDC 2
  71. 71. 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
  72. 72. No-load test 4. At this small slip R2 (1 − s ) R 2 (1 − s ) ? R2 & ? X2 s s The equivalent circuit reduces to…
  73. 73. No-load test 5. Combining Rc & RF+W we get……
  74. 74. 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 Pin = PSCL + Pcore + PF &W = 3I12 R1 + Prot Where Prot = Pcore + PF &W
  75. 75. No-load test 9. The equivalent input impedance is thus approximately Vφ Z eq = ≈ X1 + X M I1, nl If X1 can be found, in some other fashion, the magnetizing impedance XM will be known
  76. 76. 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.
  77. 77. 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 Pin PF = cos θ = 3Vl I l The magnitude of the total impedance Vφ Z LR = I
  78. 78. Blocked-rotor test Z LR = RLR + jX LR = Z LR cos θ + j Z LR sin θ RLR = R1 + R2 X LR = X 1 + X 2 Where X’1 and X’2 are the stator and rotor reactances at the test frequency respectively R2 = RLR − R1 f rated X LR = X LR = X 1 + X 2 f test
  79. 79. Blocked-rotor test X1 and X2 as function of XLRRotor Design X1 X2Wound rotor 0.5 XLR 0.5 XLRDesign A 0.5 XLR 0.5 XLRDesign B 0.4 XLR 0.6 XLRDesign C 0.3 XLR 0.7 XLRDesign D 0.5 XLR 0.5 XLR
  80. 80. Midterm Exam No.2
  81. 81. 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 V IDC = 28.0 A No-load Test: Vl = 208 V f = 60 Hz I = 8.17 A Pin = 420 W Locked-rotor Test: Vl = 25 V f = 15 Hz I = 27.9 A Pin = 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.

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