# What are the advantages of distributed winding

## 8. Distributed winding - tooth coil winding

Transcript

1 8. Distributed Winding - Toothed Coil Winding

2 A) Distributed winding

3 Single layer winding Per slot only one coil side is placed. Coils manufactured as: a) Coils with identical coil span: W = p b) Concentric coils Example: Three-phase, 12-pole machine with q = 3 coils per pole and phase: Total slot number: Q = m. 2p. q = = 108 North and south poles are generated by ONE coil group per phase. Problem with single layer windings: Crossing of coils in winding overhang part, as all coils are lying in the same plane. Thus some coils must be bent upward in winding overhang region (2nd plane).

4 Example: Single layer winding with short and long coils Unrolled winding system gives winding scheme: here: four-pole machine: 2p = 4, m = 3, q = 2, Q = 24 Winding manufactured with concentric coils. Long coils: Winding overhang part of coils is longer; so these coils may be bent upwards! Each phase has one pole pair with short and one pole pair with long coils! So resistance per phase is equal, but minimum of 4 poles required!

5 Two-layer winding coils with equal span Two-layer winding: Per slot TWO coil sides are placed one above the other. North and south poles are generated by two coil groups. Direction of current flow in N- and S-pole coils opposite! Changing of current flow direction by reversal connector. Bigger machine ratings typically above 500 kw: Profiled coil conductors (rectangular cross section), round wire with smaller machines! Example: For 4-pole machine we need four coil groups per phase!

6 Winding overhang of two-layer winding a) Two form wound coils before being put into the stator slots: Due to S-shape in winding overhang part of coils there are NO crossing points of the coils. b) Form wound coil with profiled conductor, placed in stator slot, with left coil side in lower and right coil side in upper layer. Manufacturing much more expensive than with round wire single-layer winding, therefore used usually only in bigger machines: e.g. high voltage machines up to 30 kv (High voltage: U> 1000 V (rms)!). a) b)

7 Series and parallel connection of coil groups Series and parallel connection of coil groups to get one winding phase Example: Eight-pole machine: Two-layer winding: 8 coil groups, which may be connected as follows: a = 1: Series connection of all 8 coil groups a = 2: 4 coil groups in series, then paralleling the two series sections a = 4: 2 coil groups in series, then paralleling the four series sections a = 8: All 8 coil groups are connected in parallel Single- layer winding: 4 coil groups, which may be connected as follows: a = 1: Series connection of all 4 coil groups a = 2: 2 coil groups in series, then paralleling the two series sections a = 4: All 4 coil groups are connected in parallel Resulting number of turns per phase N: pqnc 2 pqnc N single-layer winding N two-layer winding aa Example: 2p = 4, q = 2, eleven turns per coil (N c = 11), series connection of all coil groups: a = 1: number of turns per phase: N = / 1 = 88

8 Pitching (chording) of coils W

9 Example: Pitched Two-layer winding Four pole machine, m = 3, Q = 24, q = 2: Pitching W / p = 5/6.

10 Distributed winding: Full-hole winding - broken-hole winding Full-hole winding: The number of coils per pole and strand q is an integer Example: Four-pole machine2p = 4, number of strands m = 3, number of slots Q = 24, q = Q / (2p. M) = 2 broken-hole winding : The number of coils per pole and strand q is a real fraction Example: Four-pole machine2p = 4, number of strands m = 3, number of slots Q = 18, q = Q / (2p. M) = 1.5 That means: Each pole has each Strand alternately 1 or 2 coils in series, therefore an average of 1.5 coils per pole and strand.

11 Comparison of full-hole with broken-hole winding a) b) UVW Χ Χ -W -U -VV (x) -distribution fundamental wave U -WV -UW -VV (x) -distribution xx fundamental wave xxa) broken-hole winding q = 1/2 b) all-hole winding q = 1 current flow for a) and b) for the moment when i U = 0, i V = - i W = I. The magnetic field of the full-hole winding is identical for N and S poles, so the field has ordinal numbers 1, -5, 7, -11, 13, ... The magnetic field of the broken hole winding is different for the N and S pole. It therefore has increased harmonics and sometimes also lower waves: 1, -2, 4, -5, 7, -8, 10, -11, 13, ...

12 B) Toothed coil winding

13 Hole winding as tooth coil winding a) b) UVW Χ Χ -W -U -VV (x) distribution Fundamental wave U -WV -UW -V xxxa) Hole winding q = 1/2 b) Full hole winding q = 1 V (x) distribution Fundamental wave x broken-hole windings with q <1 have only one tooth with a coil width. Therefore, the individual strands can be separated. Each coil becomes a tooth coil winding with a short winding head.

14 Toothed coil winding q = 1/2 Advantages of toothed coil windings: - Simple electrical insulation of the phases in the slots and winding heads - Multi-pole design possible - Simple coil production - Ev. Plug-on spools with a high slot fill factor - short winding heads - low current heat losses BUT: increased harmonic scattering due to the increased number of harmonics: cos phi decreases at q = 1/2: no lower waves, only harmonics: 1, -2, 4, -5, 7, -8, 10, -11.13, ... BUT: Coil width W = 2/3 of the pole pitch, hence the stretching factor of the fundamental wave: Flux loss of 13%! Idle field, 24-pole machine, q = 1/2, 36 stator slots

15 Toothed coil winding q = 1/4 With q = 1/4: The spacing of the slot openings of a coil is set via the intermediate tooth so that the coil width = pole pitch. Therefore: Longing factor of the fundamental wave: 1. No loss of flow! BUT: If q = 1/2: Also sub-waves, which the wave spectrum: 1, -2, 4, -5, 7, -8, 10, -11,13, ... Refers to the longest wavelength in each case (original scheme the winding). This wavelength includes four rotor poles. Therefore the rotor poles ONLY form a constant torque with the shaft of atomic number -2. Ordinal number 1: Lower wave Ordinal numbers 4, -5, 7, -8, 10, ... Harmonics Due to the high magnetic energy of the lower wave, the lower / upper wave scatter increases, the cos phi drops significantly! Idle field, 24-pole machine, q = 1/4, 36 stator slots by dividing the 18 slots by intermediate teeth.

16 Modular synchronous machine - principle arrangement Example: Three-phase U, V, W, on 4 rotor poles there are zs = 3 modules, results in an original scheme Each module: 1 strand coil: results in 24 poles: 6 original schemes with 36 slots Lowest sub-wave: Wavelength corresponds to 2 = 1 Original diagram of the working wave: Wavelength corresponds to 2/2 (2nd harmonic ") Source: TU Darmstadt, ICEM publication, Bruges, 2002

17 Modular synchronous machine as torque motor Example: three-phase U, V, W, tooth coil technology, each tooth is a module, coil sequence U, V, W, 2p = 28 rotor poles, zs = 21 modules, permanent magnet rotor, 4 rotor magnet poles each 3 modules = 1 Original scheme deepest lower wave: wave length corresponds to 2 working wave: wave length corresponds to 2/2 (2nd harmonic ") water jacket-cooled stators of PM torque motors: left: coil arrangement, right: executed stators; source: Siemens AG

18 Toothed shaft winding in the modular synchronous machine Example: Three-phase U, V, W, with 2p = 10 rotor poles and zs = 3 modules results in an original scheme Per module two slots plus limit slot: results in 3rd (2 + 1) = 9 slots with 10 poles, so very GROBNUTIG deepest lower wave: wave length corresponds to 2 working wave: wave length corresponds to 2/5 ("5th harmonic")

19 Multi-pole machines: "light" active masses Comparison of the number of poles 2p and DOUBLE the number of poles 2p = 2. (2p) with - the same cross-sectional dimensions per pole, - thus also the same end connector lengths lb, (lb / l Fe = 0.2 with 2p) - the same power P and M, n, U, I and the same B, A, J. Pole number 2p 2p = 2.2p 2p Bore diameter d = 2p p 200% 100% active iron length l Fe ~ M / (d 2 AB) 25% 100% electrical Stator frequency f = p. n 200% 100% active iron mass stator m Fe ~ d. l 50% 100% active copper mass m Cu ~ (l Fe + l b). 2p 75% 100% magnetic mass m M ~ h. M d. l Fe> 50% *) 100% OHM losses RI 2 ~ m Cu 75% 100% centrifugal tension F / A ~ m. R d. n 2 / (d. l Fe) 200% 100% stator iron losses P Fe ~ f 1.8. B 2 350% 100% *) Because of increased pole spread approx. 65% for the same air gap flow High number of poles Lightweight construction with the same utilization C Esson BUT: - low-loss sheets, - "medium frequency" converter

20 C) Example: Comparison of two tooth coil windings

21 Metro direct drive with PM synchronous machine Requirements: Rated power P N: Rated rotational speed n N: Maximum rotational speed n max: Rated torque M N: DC link voltage U dc: Main dimensions: 250 kw 398 / min 772 / min 6000 Nm 750 V

22 Winding Parameters for Motor 1 and Motor 2 Motor 1 (q (q = 1/4) 1/4) Motor 2 (q (q = 1/2) 1/2) Number of poles 2p: Number of slots per pole and phase q: Number of turns per coil N c: Number of parallel paths a: Number of turns per phase N s: Copper space factor per slot k ff: Winding factor kw, = 1: Resistance (at 150 C) per phase R s : m 91.9 m

23 Winding Factor for Motor 1 and Motor 2 (1) Motor 1 (q (q = 1/4) 1/4) Motor 2 (q (q = 1/2) 1/2) kw, ν kd, ν kp, ν Winding factor = distribution factor pitch factor z Number of slots per pole and phase: qz, n: integer numbers nn: even 2 ordinal numbers: (1 mg) g 0, 1, 2, 3 nm 3 ordinal numbers (q = 1 / 4): Subharmonic Fundamental wave Harmonics ... Ordinal numbers (q = 1/2):

24 Winding Factor for Motor 1 and Motor 2 (2) kw, ν kd, ν kp, ν kd, ν 1 kp, ν W sin p 2 Motor 1 (q (q = 1/4) 1/4) Motor 2 ( q (q = 1/2) 1/2) kd, = 1 Distribution factor kd, kd, = mm changeable narrow tooth / wide tooth Pole pitch p Coil width W mm fix and unchangeable equal tooth width = 0.5: kw, = 0.5 = 0.68 = 1: kw, = = 1: kw, = 0.866

25 Motor 1 Motor 2 Flux patterns of Motor 1 and Motor 2 at no-load operation B r (T) Circumference coordinate x (m) Motor 1: B r, = 1 = 0.88 T Motor 2: B r, = 1 = 0.88 T

26 Fourier Spectrum of the flux density at rated speed Fundamental wave wave Motor 1 Motor 2 B (T) Subharmonic Ordinal number

27 Phasor diagram of Motor 1 and Motor 2 Motor 1 (q (q = 1/4) 1/4) Motor 2 (q (q = 1/2) 1/2) q - axis q - axis j X q I sj X q I s R s I s R s I s U p U h U p U h U s U s I s I sd - axis d - axis

28 Calculated losses at sinusoidal supply at rated speed P = 250 kw; n = 398 / min; M = 6000 Nm; f = 93 Hz, 6 Motor 1 Motor 2 22.5 28.0 P kw, Ohmic losses 2.1 2.1 Iron losses (wide teeth) 0.5 0.0 0.7 0.3 Iron losses (narrow teeth Iron losses (stator yoke) Total losses

29 Calculated losses at sinusoidal supply at maximum speed P = 250 kW; n = 772 / min; M = 3094 Nm; f = 180 Hz 30 motor 1 motor, 6 23.6 P kw, 0 10 9.6 5 0 Ohmic losses 4.7 1.8 Iron losses (wide teeth) 1.2 Iron losses (narrow teeth) 0.0 0 .5 0.2 Iron losses (stator yoke) Total losses

30 Comparison of the results at rated / maximum speed At rated speed and power (q = 1/4) (q = 1/2) Rated phase voltage U 1, N / V:% Stator phase current IS = I sq / A: 349 (-)% Power factor cos: 0.74 (-)% Thermal load A * J / A / cm A / mm 2: 7952 (++)% Efficiency: 91.7 (+)% At maximum speed and rated power (q = 1/4) (q = 1/2) Rated phase voltage U 1, N / V:% Stator phase current IS = I s / A: 247 (+)% Power factor cos: 0.99 (+)% Thermal load A * J / A / cm A / mm 2: 3982 (++)% Efficiency: 93.9 (+)%

31 Benefits and Drawbacks of Motor 1 compared to Motor 2 Motor 1 Motor 2 Bigger total leakage inductance Bigger inverter rating: (-) (+) Lower flux-weakening I sd: (+) (-) Bigger winding factor Increased motor utilization: ( +) (-) Subharmonic air gap wave Additonal oscillating torque: (-) (+) Additional radial oscillating forces: (-) (+) Reduced phase resistance Increased motor efficiency: (+) (-) Reduced thermal load: (+) (-)

32 Comparison of tooth coil winding q = 1/4 and q = 1/2 Motor 1 (q = 1/4) Motor 2 (q = 1/2) Both machines meet the requirements given by space and power rating Cost reduction is possible by round wire winding (tooth coils) Motor 1 has an increased calculated efficiency of about 2-3% Motor 1 has a considerably lower thermal load: - 26% at rated speed - 56% at maximum speed Motor 1 requires a higher inverter current rating of 15% than motor 2

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