Lecture Set 5 Distributed Windings and Rotating MMF

Transcription

1 Lecture Set 5 Distributed Windings and Rotating MMF S.D. Sudhoff Spring 2017

2 Distributed Windings and Rotating MMF Objective In this chapter, we will set the stage to study ac machinery including permanent magnet synchronous machines as well as induction machines Reading Power Magnetic Devices: A Multi-Objective Design Approach, , (Available through IEEE Xplore at no cost) 2

3 Parts of an AC Machine 3

4 Parts of an AC Machine Stator 4

5 Operation of AC Machines N-Pole: Flux leaves material S-Pole: Flux enters material 5

6 P-Pole Machines For given ac frequency having more than 2 poles reduces machine speed by P/2 6

7 Position Measurement 7

8 Electrical Positions 8

9 Distributed Windings Thus far, we have mostly considered lumped windings. We must now consider distributed winding. This is a more difficult concept. Two Representations Discrete Winding Representation Continuous Winding Representation 9

10 Spring 2010 ECE321 10

11 Discrete Winding Representation Let N x,i denote number of conductors of a x- phase winding in the i th slot i designates the slot number x denotes winding or phase and is typically as, bs, or cs y is s for stator or r for rotor A positive value indicates the conductor directed out of the page 11

12 Slot and Tooth Locations φ = π (2i 2) / S + φ ys, i y ys,1 φ = π (2i 3) / S + φ yt, i y ys,1 12

13 Discrete Winding Representation S y i= 1 N x, i = 0 S y = N N u( N ) x x, i x, i i= 1 13

14 Developed Diagram 14

15 Continuous Winding Description Conductor density (conductors/rad): n x (φ ym ) x denotes winding/phase as, bs, or cs y is location ( s for stator, r for rotor) positive value indicates conductors out of the page For example, the a-phase conductor density may be given by n ( φ ) = N sin( Pφ / 2) N sin(3 Pφ / 2) as sm s1 sm s3 sm Total conductors per phase N 2π = n ( φ) u(n ( φ)) dφ xy xy xy 0 15

16 Symmetry Conditions N x, i+ 2 S / P = Nx, i y N + = N x, i S / P x, i y n ( φ + 4 π / P) = n ( φ ) x m x m n ( φ + 2 π / P) = n ( φ ) x m x m 16

17 Symmetry Conditions 17

18 Winding Arrangements N as [ ] = N

19 Winding Arrangements 19

20 The Winding Function The winding function is a measure of how many times a winding links flux at a given place The winding function will be used to Find self-inductance of a distributed winding Find mutual-inductance between distributed windings Find the mmf associated with a distributed winding It can be expressed in terms of both discrete and continuous winding descriptions 20

21 The Discrete Winding Function W x,i is the discrete winding function of phase x of the stator or rotor It is the number of times the winding links flux through the i th tooth It is the number of turns around tooth i 21

22 The Discrete Winding Function Consider the following N x, i +1 Tooth i+1 N x, i N x, i 1 Tooth i β turns α turns 22

23 The Discrete Winding Function N x, i +1 Tooth i+1 N x, i N x, i 1 Tooth i β turns α turns 23

24 The Discrete Winding Function N x, i +1 Tooth i+1 N x, i N x, i 1 Tooth i β turns α turns 24

25 The Discrete Winding Function 25

26 The Discrete Winding Description Summarizing: W S y / P 1 = N x,1 x, i 2 i= 1 W = + W N x, i 1 x, i x, i 26

27 Example Suppose we have [ ] T Nx = 10 27

28 Example Finally we obtain [ ] T Wx = 10 28

29 The Continuous Winding Function Notation φ = φ rm for rotor windings φ = φ sm for stator windings w x (φ) = winding function Number of times a winding links flux at position φ 29

30 The Continuous Winding Function 30

31 The Continuous Winding Function 31

32 The Continuous Winding Function 32

33 The Continuous Winding Function Thus we have 2 π / P m 1 w x ( φm ) = n x ( φm ) dφm n x ( φm ) dφm 2 φ

34 Example Consider the following geometry We might have as = 100sin( φsm ) Find the winding function n 34

35 Example 35

36 Example 36

37 Comparison of Winding Functions N as = N [ ] W as = N [ ] n = N 7.221sin( 2φ ) sin(6 φ ) ( ) as sm sm was = N cos( 2φsm ) cos(6 φsm )

38 Comparison of Winding Functions 38

39 Distributed MMF In a distributed machine, the MMF source is associated with a winding is a function of position as well as the current We will focus on the continuous winding distribution We can show F S ( φm ) = w x ( φm ) ix F ( φ ) = w ( φ ) i x X S R m x m x x X R F ( φ ) = F ( φ ) + F ( φ ) g m S m R m 39

40 Distributed MMF

41 Airgap MMF Before we start, define airgap MMF in a rotating machine: F stator ( φ ) = H ( φ ) d l g m m rotor Since the integration is in radial direction stator Fg ( φm ) = H r( φm ) dl rotor 41

42 Air Gap MMF Now consider the following 42

43 Air Gap MMF 43

44 Air Gap MMF 44

45 Air Gap MMF 45

46 Air Gap MMF 46

47 Air Gap MMF 47

48 Air Gap MMF 48

49 Distributed MMF Once we have the MMF what do we do with it??? 49

50 Air Gap Flux Density Thus we have B g ( φ ) m = µ 0 Fg ( φm ) g ( φm ) 50

51 Distributed MMF First, it gives the total field = F( φ) F ( φ) B = x X ( φ) x µ 0F( φ) g 51

52 Distributed MMF Second, it gives us the fields associated with a winding B x µ F = 0 g x ( φ) ( φ) 52

53 Example Suppose w as =100cos(4φ sm ); i as =5 A w bs =50sin(4φ sm ); i bs = 10 A g = 1 mm What is the peak B-field? At what position is the peak B-field? 53

54 Example 54

55 Example 55

56 Rotating MMF Another concept from the MMF is that the fields rotate 56

57 Rotating MMF in Two-Phase Systems Suppose n n as bs = N s = N And that i as = ( Pφ 2) sin sm / s ( Pφ 2) cos sm / cos( ω t + φ ) 2Is e i w w as bs = = 2N P 2N P s s cos sin ( Pφ / 2) sm ( Pφ / 2) sm i bs = sin( ω t + φ ) 2Is e i What is the MMF? 57

58 Rotating MMF in Two-Phase Systems 58

59 Rotating MMF in Two-Phase Systems 59

60 Rotating MMF in Two-Phase Systems Result F 2 2N si s = cos Pφ / 2 ω t ϕ P ( ) s sm e i 60

61 Affect of Number of Poles F 2 2N si s = cos Pφ / 2 ω t ϕ P ( ) s sm e i 61

62 Unbalanced Two-Phase Systems Let s consider another case. Again starting with 2N s F = cos ( Pφ / 2) i + sin ( Pφ / 2) i P ( ) s sm as sm bs Suppose the b-phase current is zero; the a- phase current given by i as = cos( ω t + φ ) 2Is e i 62

63 Unbalanced Two-Phase Systems 63

64 Unbalanced Two-Phase Systems 64

65 Unbalanced Two-Phase Systems 65

66 Rotating MMF in Three-Phase Systems In this case suppose n ( φ ) = N sin( Pφ / 2) N sin(3 Pφ / 2) as sm s1 sm s3 sm n ( φ ) = N sin( Pφ / 2 2 π / 3) N sin(3 Pφ / 2) bs sm s1 sm s3 sm n ( φ ) = N sin( Pφ / π / 3) N sin(3 Pφ / 2) cs sm s1 sm s3 sm Question: why would I do this? 66

67 Rotating MMF in Three-Phase Systems Proceeding, we have s1 s3 w as ( sm ) = cos( P sm / 2) cos(3 P sm / 2) Now φ 2N 2N P φ 3P φ φ 2N 2N P φ π 3P φ s1 s3 w bs ( sm ) = cos( P sm / 2 2 / 3) cos(3 P sm / 2) φ 2N 2N P φ π 3P φ s1 s3 w cs ( sm ) = cos( P sm / / 3) cos(3 P sm / 2) F ( φ ) = w ( φ ) i + w ( φ ) i + w ( φ ) i S sm as sm as bs sm bs cs sm cs 67

68 Rotating MMF in Three-Phase Systems Now let us assume a three-phase balanced set of currents i i bs cs i as = cos( ω t + φ ) 2Is e i 2I cos( ω t + φ 2π / 3) = s e i 2I cos( ω t + φ + 2π / 3) = s e i 68

69 Rotating MMF in Three-Phase Systems Combining yields F 3 2N I = cos / 2 P ( Pφ ω t φ ) s1 s S sm e i 69

70 Flux Linkage and Magnetizing Inductance Flux linkage: Inductance λx = λxl + λxm L xy = λ x due to iy i y 70

71 Flux Linkage and Magnetizing Inductance Partitioning Inductance Inductance Terms Lxy = Lxyl + Lxym L xyl = λ xl due to i y i y L xym = λ xm due to iy i y 71

72 Flux Linkage and Magnetizing Inductance We can show w ( ) w ( ) L µ rl dφ 2π λxym x φm y φm = xym = 0 iy 0 g( φm ) m 72

73 Flux Linkage and Magnetizing Inductance Stator dφ r φ 73

74 Flux Linkage and Magnetizing Inductance Stator dφ r φ 74

75 Flux Linkage and Magnetizing Inductance Stator dφ r φ 75

76 Example 1 Suppose n as n bs = N s sin( Pφsm / 2) = N sin( Pφsm / 2 2π/3) s and that the air gap is uniform. Find the magnetizing inductance between the two phases w w as bs 2N = s cos( Pφsm P 2N = s cos( Pφsm P / 2) / 2 2π / 3) 76

77 Example We obtain L 2πµ rln = 0 P g asbs 2 2 s 77

78 Another Example (IM) n ( φ ) = N sin( Pφ / 2) as sm s1 sm n ( φ ) = N sin( Pφ / 2) ar rm r1 rm 78

79 Another Example (Cont.) 79

80 Another Example (Cont.) 80

81 Another Example (Cont.) 81