Determination of Equivalent Circuit parameters and performance characteristics Circle Diagram

Includes the testing of induction motor to draw the circle diagram of induction motor with step wise procedure and calculation for the same. Also explains the working and application of Induction generator

1. EEE 2003 ELECTROMECHANICAL ENERGY CONVERSION Dr. P. Vijayapriya Associate Professor, SELECT Module: 6 Testing of Induction Machine Determination of Equivalent Circuit parameters – performance characteristics Circle Diagram –Speed Control –Induction Generator Applications

2. Equivalent Circuit The induction motor is similar to the transformer with the exception that its secondary windings are free to rotate When the rotor is locked (or blocked), i.e. s =1, the largest voltage is induced in the rotor at highest frequency (supply frequency f) 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

3. Equivalent Circuit So the voltage in rotor under running condition is given by E 2r = sE 2 Where E 2 is the rotor’s induced voltage obtained at s = 1(locked rotor) The same is true for the frequency, i.e. f` = sf The same is true for the rotor reactance as frequency changes X 2r = sX 2

4. Equivalent Circuit Then, we can draw the rotor equivalent circuit as follows Where E 2r is the induced voltage in the rotor and R 2 is the rotor resistance jX 2r = jsX 2 E 2r = sE 2 R2 I 2

5. 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 2 is the induced voltage and X 2 is the rotor reactance at blocked rotor condition ( s = 1)

6. Equivalent Circuit Now we can have the rotor equivalent circuit E 2 jX 2 R2 S I 2

7. Equivalent Circuit Splitting the resistances in to two we get the modified equivalent circuit as E 2 jX 2 R2 S I 2 Actual rotor resistance Resistance equivalent to mechanical load

8. Equivalent Circuit We can rearrange the equivalent circuit as follows Actual rotor resistance Resistance equivalent to mechanical load R X I I` 2

9. 9 Problem-1 Dr. J. Belwin Edward, Asso . Prof., SELECT, VIT, Vellore The maximum torque of a 3-phase induction motor occurs at a slip of 12%. The motor has an equivalent secondary resistance of 0.08  /phase. Calculate the equivalent load resistance R L , the equivalent load voltage V L and the current at this slip if the gross power output is 9,000 watts.

10. 10 Problem - 2 Dr. J. Belwin Edward, Asso . Prof., SELECT, VIT, Vellore A 220-V, 3-  , 4-pole, 50-Hz, Y-connected induction motor is rated 3.73 kW. The equivalent circuit parameters are: R 1 = 0.45  , X 1 = 0.8  ; R 2 ’ = 0.4  , X 2 ’ = 0.8  , B O = – 1/30 mho The stator core loss is 50 W and rotational loss is 150 W. For a slip of 0.04, find : input current p.f . air-gap power mechanical power electro-magnetic torque output power and efficiency.

11. 11 Problem - 2 Dr. J. Belwin Edward, Asso . Prof., SELECT, VIT, Vellore

12. 12 Problem - 3 Dr. J. Belwin Edward, Asso . Prof., SELECT, VIT, Vellore A 440-V, 3- φ 50-Hz, 37.3 kW, Y-connected induction motor has the following parameters: R 1 = 0.1  , X 1 = 0.4  , R 2 ’ = 0.15 Ω , X 2 ′ = 0.44 Ω Motor has stator core loss of 1250 W and rotational loss of 1000 W. It draws a no-load line current of 20 A at a p.f . of 0.09 (lag). When motor operates at a slip of 3%, calculate : input line current and p.f . electromagnetic torque developed in N-m output and efficiency of the motor.

13. 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).

14. 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.

15. 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

16. No-load test The motor is allowed to spin freely The only load on the motor is the friction and windage losses, so all P conv is consumed by mechanical losses The slip is very small

17. No-load test At this small slip The equivalent circuit reduces to… >> >>

18. No-load test Combining R c & R F+W we get……

19. No-load test At the no-load conditions, the input power measured by meters must equal the losses in the motor. The P RCL is negligible because I 2 is extremely small because R 2 (1- s )/ s is very large. The input power equals Where

20. 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.

21. 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

22. The core is very small as only small voltage is required to circulate the rated current under blocked rotor condition and also there is no load delivered by the motor Hence the entire power consumed by the motor is only to meet the total copper loss (stator +rotor)

23. The Heyland  diagram  is an approximate representation of circle diagram applied to induction motors, which assumes that stator input voltage, rotor resistance and rotor reactance are constant. First conceived by A. Heyland in 1894 and B.A. Behrend in 1895, the circle diagram is the graphical representation of the performance of the electrical machine drawn in terms of the locus of the machine's input voltage and current. CIRCLE DIAGRAM

24. Draw the voltage axis (Y axis) and X axis (Input line), mark intersection point as O From the no load test and blocked rotor test calculate the following Fix the current scale such that Isn does not exceed 25 cm At an angle φ o from Y axis, draw Io according to scale and mark O` Draw O`B parallel to X axis which is the constant loss line At an angle φ sc from Y axis, draw Isn according to scale and mark the point A Join O`A (which is the output line ) and draw perpendicular bisectors to O`A. Extend it to meet the constant loss line at C. With C as centre and CO` as radius draw a semicircle. The circle will pass through A. Step wise Procedure

25. With C as centre and CO` as radius draw a semicircle. The circle will pass through A. From A drop a perpendicular to X axis to intersect constant loss line at F. AF represents the total cu loss ie power input at blocked rotor condition at rated voltage (ie Psn) Fix the power scale as follows Location of point E: Check for data of R1 (stator resistance) If given calculate stator cu loss = 3 I 2 snR1 . Calculate Or calculate EF = stator cu loss / power scale and mark E from the point F If no data given, assume stator cu loss equals rotor copper loss and divide as AE = EF Join EO` which is the torque line

26. 15. Join EO` which is the torque line. For full load condition (if the power rating of the machine is given in horse power convert to watts ( * 746 w (mks unit)). Find how many cm represents output power and extend it from A to A` From A` draw a line parallel to output line. It will intersect circle at maximum two different places. Take the point near to Y axis as operating point H. From H drop a perpendicular to X axis to meet at N. Mark K, L, M as intersection with output line, torque line and constant loss line Calculate the required parameters Out put = HK x power scale Input = HN x power scale rotor input = HL x power scale rotor cu loss = KL x power scale rotor efficiency = HK / HL slip = KL / HL total efficiency = HK / HN power factor = HN / OH To find Maximum quantities: draw a line parallel to the corresponding line and tangent to the circle. Drop a perpendicular from the circle intersection point to the corresponding line. When multiplied by power scale we get the maximum value of that particular quantity

27. O V Draw the X Axis and Y Axis

28. ɸ O’ O V Draw the Vector OO’ = I at ɸ

29. ɸ O’ O V Draw the Horizontal line from O’ parallel to X axis

30. ɸ O’ O ɸ sc V A Draw the Vector OA = I SN at ɸ SC

31. ɸ O’ O ɸ sc V A Join O’A this is output line

32. ɸ O’ O ɸ sc V A Draw the Perpendicular Bi-Sector of AO’, to meet the horizontal line from O’ at C. This is the centre of the circle A Radius Centre

33. O’ Output Line Draw a Semi Circle to meet the horizontal line from O’ at B with C as a centre and CO’ as radius

34. O’ Output Line D Draw perpendicular from A on X-axis, meeting at point D

35. O’ Output Line D E F Draw perpendicular from A on X-axis, meeting at point D AD = AE + EF + FD

36. O’ Output Line Torque Line D F E AF = AE + EF AF = W SN = Y 1 cm Power Scale = W SN /AF Calculate stator cu loss = 3 I 2 SN R 1 Where R1 = stator resistance / phase I SN = stator current / phase under short circuit with normal voltage EF = Stator cu loss : Measure EF from F Power scale Join O`E which is the torque line AD = Total Losses AE = Rotor Cu Loss EF = Stator Cu Loss FD = Fixed Loss

37. O’ Output Line Parallel Output Line P Q S T R Torque Line AF = W SN = Y 1 cm Power Scale = W SN /AF D F E Base Line X-Axis AF = AE + EF A’ Draw the line from A’ parallel to the output line , it intersects the circle at P AA’ = Output Power / Power Scale For Rated output :

38. O’ Output Line Parallel Output Line P Q S T R Torque Line AF = W SN = Y 1 cm Power Scale = W SN /AF D F E Total motor input = PT x Power scale Fixed loss = ST x Power scale Stator copper loss = SR x Power scale Rotor copper loss = QR x Power scale Total loss = QT x Power scale Rotor output = PQ x Power scale Rotor input = PQ + QR = PR x Power scale Draw the Vertical Line from P to intersect output line at Q intersect torque line at R intersect base line at S intersect X-axis at T Base Line X-Axis A’

39. O’ Output Line Parallel Output Line P Q S T R Torque Line AF = W SN = Y 1 cm Power Scale = W SN /AF D F E Total motor input = PT x Power scale Fixed loss = ST x Power scale Slip s = QR/PR = rotor cu loss / rotor input Stator copper loss = SR x Power scale Rotor copper loss = QR x Power scale Total loss = QT x Power scale Rotor output = PQ x Power scale (PQ = AA`) Rotor input = PQ + QR = PR x Power scale Power factor cos = PT/OP Motor efficiency = Output / Input = PQ/PT ɸ A’ 0P = Full Load Line Current Rated Current = OP x Current scale

40. O’ Output Line Parallel Torque Line P Q S T R Torque Line AF = W SN = Y 1 cm Power Scale = W SN /AF D F E Total motor input = PT x Power scale Fixed loss = ST x Power scale Slip s = QR/PR Stator copper loss = SR x Power scale Rotor copper loss = QR x Power scale Total loss = QT x Power scale Rotor output = PQ x Power scale Rotor input = PQ + QR = PR x Power scale Power factor cos = PT/OP Motor efficiency = Output / Input = PQ/PT ɸ J K Max Torque = JK * power scale A’

41. Construction of circle diagram Radius Centre C O O’ Io I 2 ’ I SN A F E B V φ o φ sc Output Line Torque Line Input Line I 1 Stator Cu loss Rotor Cu loss Core loss H K L M N P’ P S’ S T T’ Out put = HK x power scale Input = HN x power scale rotor input = HL x power scale rotor cu loss = KL x power scale slip = KL / HL total efficiency = HK / HN power factor = HN / OH

42. Draw the voltage axis (Y axis) and X axis (Input line), mark intersection point as O From the no load test and blocked rotor test calculate the following Fix the current scale such that Isn does not exceed 25 cm At an angle φ o from Y axis, draw Io according to scale and mark O` Draw O`B parallel to X axis which is the constant loss line At an angle φ sc from Y axis, draw Isn according to scale and mark the point A Join O`A (which is the output line ) and draw perpendicular bisectors to O`A. Extend it to meet the constant loss line at C. With C as centre and CO` as radius draw a semicircle. The circle will pass through A. Step wise Procedure

43. With C as centre and CO` as radius draw a semicircle. The circle will pass through A. From A drop a perpendicular to X axis to intersect constant loss line at F. AF represents the total cu loss ie power input at blocked rotor condition at rated voltage (ie Psn) Fix the power scale as follows Location of point E: Check for data of R1 (stator resistance) If given calculate stator cu loss as 3 I 2 snR1. Calculate EF = 3I 2 SN R1 Power scale If not assume stator cu loss equals rotor copper loss and divide as AE = EF Join EO` which is the torque line

44. 15. Join EO` which is the torque line. For full load condition (if the power rating of the machine is given in horse power convert to watts ( * 746 w (mks unit)). Find how many cm represents output power and extend it from A to A` From A` draw a line parallel to output line. It will intersect circle at maximum two different places. Take the point near to Y axis as operating point H. From H drop a perpendicular to X axis to meet at N. Mark K, L, M as intersection with output line, torque line and constant loss line Calculate the required parameters Out put = HK x power scale Input = HN x power scale rotor input = HL x power scale rotor cu loss = KL x power scale rotor efficiency = HK / HL slip = KL / HL total efficiency = HK / HN power factor = HN / OH To find Maximum quantities: draw a line parallel to the corresponding line and tangent to the circle. Drop a perpendicular from the circle intersection point to the corresponding line. When multiplied by power scale we get the maximum value of that particular quantity

45. Problem 1 A 3 phase, 400V, star connected induction motor gave the following test reading. No load : 400V 9A 1250W Blocked : 150V 38A 4kW Draw the circle diagram. IF the normal rating is 20.27 hp find from the circle diagram, the full load values of current, power factor slip, efficiency and rotor efficiency. Also find all the maximum values. Assume rotor copper loss equals stator copper loss.

46. Problem 2

47. I SN = Short circuit @ Normal Voltage W SN = Short circuit power / blocked rotor input with normal voltage

48. Problem 3

49. N S – Synchronous Speed f – Supply frequency in Hz P – No. of Stator Poles   Speed Control of 3-Phase IM D.C. shunt motors can be made to run at any speed within wide limits, with good efficiency and speed regulation, merely by manipulating a simple field rheostat, the same is not possible with induction motors. In their case, speed reduction is accompanied by a corresponding loss of efficiency and good speed regulation . That is why it is much easier to build a good adjustable-speed d.c. shunt motor than an adjustable speed induction motor.

50. Speed Control Methods

51. Speed Control There are 2 types of speed control of 3 phase induction machines From stator side Variation in supply voltage Variation in supply frequency Variation in number of poles From Rotor side i rotor rheostat control Ii cascaded operation Iii by injecting emf in rotor circuit

52. Maximum torque changes The speed which at max torque occurs is constant (at max torque, X R =R R /s Relatively simple method – uses power electronics circuit for voltage controller Suitable for fan type load Disadvantages : Large speed regulation since ~ n s T n s ~n NL T n r1 n r2 n r3 n n r1 > n r2 > n r3 V 1 V 2 V 3 V 1 > V 2 > V 3 V decreasing Varying supply Voltage

53. The best method since supply voltage and supply frequency is varied to keep V / f constant Maintain speed regulation uses power electronics circuit for frequency and voltage controller Constant maximum torque T n NL1 T n r1 n r2 n r3 n f decreasing n NL2 n NL3 Varying supply voltage and supply frequency

54. Rotor Rheostatic Control One serious disadvantage of this method is that with increase in rotor resistance , I 2 R losses also increase which decrease the operating efficiency of the motor. In fact, the loss is directly proportional to the reduction in the speed. The second disadvantage is the double dependence of speed , not only on R 2 but on load as well. Because of the wastefulness of this method, it is used where speed changes are needed for short periods only. T n s ~n NL T n r1 n r2 n r3 n n r1 < n r2 < n r3 R 1 R 2 R 3 R 1 < R 2 < R 3

55. Cascade or Concatenation or Tandem Operation In this method, two motors are used and are ordinarily mounted on the same shaft, so that both run at the same speed (or else they may be geared together).

56. Injecting an e.m.f. in the Rotor Circuit - Kramer In this method, the speed of an induction motor is controlled by injecting a voltage in the rotor circuit, it being of course, necessary for the injected voltage to have the same frequency as the slip frequency. There is, however, no restriction as to the phase of the injected e.m.f. One big advantage of this method is that any speed, within the working range , can be obtained instead of only two or three, as with other methods of speed control. Yet another advantage is that if the rotary converter is over-excited, it will take a leading current which compensates for the lagging current drawn by main motor M and hence improves the power factor of the system.

57. Injecting an e.m.f. in the Rotor Circuit – Scherbius System The slip energy is not converted into d.c. and then fed to a d.c. motor, rather it is fed directly to a special 3-phase (or 6-phase) a.c . commutator motor-called a, Scherbius machine. The polyphase winding of machine C is supplied with the low-frequency output of machine M through a regulating transformer RT . The commutator motor C is a variable-speed motor and its speed (and hence that of M ) is controlled by either varying the tappings on RT or by adjusting the position of brushes on C .

58. 63 Same basic construction as squirrel-cage induction motors Drive at a speed greater than the synchronous speed Not started as a motor Operated by wind turbines, steam turbines, etc. Induction Generator

59. 64 Motor – to – Generator Transition

60. 65 1 - Motor shaft coupled to a steam turbine Initially, the turbine valve is closed 2 - Motor started at full voltage by closing the breaker 3 –Motor drives the turbine at less than synchronous speed Typical setup for induction-generator operation

61. 66 Operation as an Induction-Generator continued Gradually open the turbine valve, causing a buildup of turbine torque, adding to the motor torque, resulting in an increase in rotor speed.

62. 67 When the speed approaches synchronous speed, the slip = 0, R s /s becomes infinite, rotor current I r = 0, and no motor torque is developed. (The motor is neither a motor or a generator – it is “floating” on the bus. The only stator current is the exciting current to supply the rotating magnetic field and the iron losses.

63. 68 The speed of the rotating flux is independent of the rotor speed – only a function of the number of poles and the frequency of the applied voltage. Increasing the rotor speed above the synchronous speed causes the slip [(n s – n r )/n s ] to become negative! The gap power, P gap = P rcl /s becomes negative, now supplying power to the system!

64. Applications of Induction Generator The use of induction generator started in the early twentieth century. In 1960’s and 1970’s its usage has become very less but later it usage started again. They are used with the alternative energy sources, such as windmills (WEGS), Hydro Electric Power Plants , Diesel Generators (DGs). They are also used to supply additional power to a load in a remote area that is being supplied by a weak transmission line. Energy recovery systems in the industrial processes. Externally excited generators are widely used for regenerative breaking of hoists driven by the three phase induction motors .

65. 70 Applications of Induction Generator

66. 71 Applications of Induction Generator

67. 72 Applications of Induction Generator

68. 73 Applications of Induction Generator