Energy Converters CAD and System Dynamics

Transcription

1 Energy Converters CAD and System Dynamics 1. Basic design rules for electrical machines 3. Heat transfer and cooling of electrical machines 4. Dynamics of electrical machines 5. Dynamics of DC machines 6. Space vector theory 7. Dynamics of induction machines 8. Dynamics of synchronous machines Source: SPEED program 2/1

2 Energy Converters - CAD and System Dynamics Source: Breuer Motors, Germany 2/2

3 Energy Converters CAD and System Dynamics 2. Design of induction machines 2.1 Main dimensions and basic electromagnetic quantities of induction machines 2.2 Scaling effect in electric machines 2.3 Stator winding low and high voltage technology 2.4 Stator winding design 2.5 Rotor cage design 2.6 Wound rotor design 2.7 Design of main flux path of magnetic circuit 2.8 Stray flux and inductance 2.9 Influence of saturation on inductance 2.10 Masses and losses 2/3

4 Demands for design Torque Stator current 0 0 Output characteristic of induction motors vs. speed Speed 2/4

5 Typical demands acc. to International Standard IEC : Rated operation: Tolerances for measured values: Measured rated slip s N ±20% of calculated rated slip s N,calc Measured power factor cosϕ N -(1-cosϕ N,calc )/6 Measured efficiency η N (1-η N,calc )for P N 50 kw (1-η N,calc )for P N > 50 kw Overload capability: Measured breakdown torque M b : Demand: for 15 s: > 1.6M N -10% of calculated breakdown torque M b,calc Starting parameters (at s = 1): Measured starting torque M 1 Measured starting current I 1-15% % of calculated starting torque M 1,calc +20% of calculated starting current I 1,calc Measured minimum torque M min ("saddle" torque): Demands: M min > 0.5M 1 2/5

6 Example: Efficiency tolerance Calculated efficiency η N = 0.9 Efficiency tolerance: Tol= (1-η N,calc )for P N > 50 kw Tol = 0.1 (1 0.9) = 0.01 Minimum admissible measured efficiency: η meas, min = η N + Tol = = /6

7 Main dimensions of induction machines yoke air gap tooth slot STATOR These dimensionshave to be defined during design process to get wanted machine output characteristic! yoke cooling duct ROTOR rotor shaft 2/7

8 1: Statorhousing 2: End shield 3: Oilcooler for sleeve bearings 4: Statoriron stack 5: Overhangof two layer form wound stator winding 6: Rotoriron stack with axial ducts in rotor iron back 7: Shaft end Source: ABB, Birr, Switzerland 8: Overhangof rotor two layer winding 9: Three-phase rotor slip rings with brush contacts and hand wheel to lift brushes and to short circuit the rotor 10: Shaft mounted radial fan for generating cooling air flow, which passes through openings in stator housing 2/8

9 Design calculation must yield equivalent circuit parameters! a) Equivalent circuit per phasefor fundamentalair gap flux density distribution, sinusoidal currents and voltages (here: iron losses P Fe,s neglected) b) Corresponding phasor diagram for motor operation U h a) b) 2/9

10 Energy Converters CAD and System Dynamics Summary: Main dimensions and basic electromagnetic quantities of induction machines - Design requirements by standards (IEC 60034) and by customers - Electromechanical design implies magnetic, thermal and mechanical issues - Design must be translated into equivalent circuit for performance prediction 2/10

11 Energy Converters CAD and System Dynamics 2. Design of induction machines 2.1 Main dimensions and basic electromagnetic quantities of induction machines 2.2 Scaling effect in electric machines 2.3 Stator winding low and high voltage technology 2.4 Stator winding design 2.5 Rotor cage design 2.6 Wound rotor design 2.7 Design of main flux path of magnetic circuit 2.8 Stray flux and inductance 2.9 Influence of saturation on inductance 2.10 Masses and losses 2/11

12 Scaling effect of power - Velocity of rotor at synchronous speed at stator bore vsyn = 2f s τ p is proportional to velocity of air flow in air gap of open ventilated machines with fans mounted on shaft. - Better cooling at bigger pole pitch, so higher current loading possible: As ~ τ p. 2 - Flux Φ ˆ h = τ p l Fe Bδ, 1 is excited by coils. For minimum copper losses at given π flux, surface τ p lfe needs minimum length per turn of 2 ( τ p + lfe), yielding τ p = l Fe. - Result: As ~ τ p, lfe ~ τ p and dsiπ = 2 pτ p dsi ~ τ p p - Rated apparent power S N with Esson s equation: SN = 3U snisn, S = 3U h Is S S N N S ~ P δ N ~ 2 π = 2 p τ k 4 p w1 ~ A s p A Bˆ 4 s δ1 ~ d 2 si p l l 4 Fe Fe fs p = (2 2 k w1 f s Bˆ δ1 ) p A Increase of pole pitch, iron length and current loading with 4 τ p ~ 4 P /, ~ 4 /, ~ 4 N p lfe PN p As PN / p s τ 2 p l Fe δ = K P N / p! p τ 4 p, 2/12

13 Main dimensions may be chosen from curves, where parameters of already built machines are summarized. Behind these curves ESSON s utilization number is hidden for - a given induction machine type - an open ventilation cooling system - a winding temperature limit 125 C τ p ~ 4 P N / p Increase of pole pitchof induction machines for pole numbers 2, 4, 6, 8, 12 y= lg( ) ~lg( 4 P / p) τ p y= lg( τ p ) ~lg( P )/ 4 lg( p)/ 4 y~ x/ 4 lg( p)/ 4 Diagram is a straight line in double logarithmic scaling N N 2/13

14 l b Iron stack length vs. pole pitch Pole pitch τ p shrinkswith increasing pole count 2p τ = p d si π /( 2p) l b l e l e l e W τ p 2p = 4 l e τ p At low pole count 2p= 4the condition l e τ p for minimum copper losses at given pole area l e. τ p is used W τ p 2p > 4 l e τ p At higher pole count 2p> 4for the condition l e τ p the induction machines become axially very short, with a dominating ratio l b /l e > 1 Therefore: W τ p 2p > 4 l e > τ p At higher pole count 2p> 4the iron length is increased l e >τ p. 2/14

15 Increase of (equivalent) iron stack length l e of induction machines for pole count 2p= 2, 4, 6, 8, 12 l e : equivalent stack length (due to radial ventilation ducts) 2p= 4: l e ~ τ ~ 4 p PN 2p> 4: l e ~ k( p) 4 PN k( p) > 1 2/15

16 Stator current loading A s & average air gap flux density B δ,av for pole count 2p= 2, 4, 6, 8, 12; open ventilated air cooled machines, Thermal Class B (temperature rise 80 K over 40 C ambient); A ~ τ s Same pole pitch, but higher pole count, means bigger machine (= ESSON-number C ), so higher A s is possible. p B 2 ˆ av = π B δ, δ1 Saturation limits tooth flux density to 2 T, so: peak air gap B δ to 1 T, thus: average B δ,av to 0.66 T. 2/16

17 Comparison of 2- and 4-pole machine cross section For the same bore diameter d si the pole pitch of the 2-pole machine is twice of the 4-pole machine. Hence for the same air gap flux density the flux Φper pole is twice. Therefore for the same yoke flux density B y we need twice big yoke height h ys for 2-pole machines, yielding a huge stator iron mass. 2 τ p = d si π /( 2p) Φ = τ ˆ p l e Bδ1 Φ / 2=hysleBy π half B-field line In order to reducethe iron mass of the 2-pole machine, the air gap flux density and the yoke height are reduced, yielding the same yoke flux density = reduced ESSON number! ˆ δ 1 B = 1.0T 0. 7T h ys =100% 70% C ~ A ˆ 1= 100% 70% sb δ B y h ys Φ/2 B δ Φ d si 2-pole machine l e B y h ys Φ/2 B δ Φ d si 4-pole machine 2/17

18 Decrease of current density Jof 6 kv induction machines for pole count 2p= 2, 4 and 6; open ventilated air cooled machines, Thermal Class B (temperature rise 80 K over 40 C ambient) P J lgj lgj ~ L ~1/ 4, J 8 P ~lgp ~1/ 1/8 L ~ lgp /18

19 2/19 Pole count scaling effect of current densityjvs. powerp p L A L d p d A s si si p s / ~ ) /(2 ~ = π τ / 1 1 / ˆ 2 ˆ 2 K p P L J K p P L P p L K p l d A f B k p f l d B A k S S N N N Fe si s s w s Fe si s w N = = = = = = δ δ δ π π l Fe L 4 ) lg( 8 ) lg( 8 ) lg( ) lg( p P K J y N = = 4 ) lg( 8 ) lg( 8 ) lg( p K x J y + = = y =lg(j) x = lg(p N ) p 1 p 2 > p 1 0

20 Increase of air gap width δwith increasing induction machine size (pole pitch τ p ) for pole count 2p= 2, 4, 6, 8 and 12 For the same pole pitch τ p the higher pole count 2pgives a bigger diameter d si : need for bigger air gap δto avoid touching of rotor, caused by shaft bending δ d si δ ~ const. τ d si ~ τ p p p = dsi π 2p /20

21 Increase of shaft diameter d sh (= inner rotor stack diameter d ri ) with increasing power Pand size for pole count 2p= 2, 4, 6, 8 and 12 Bigger power Pat higher pole count 2p means -lower speed n, so higher torquem, so -bigger shaft diameter d ri is required lg( dri ) ~lg( PN )/ 3+ lg( p)/3 2/21

22 Torsion stress requires a certain shaft diameter d ri Torsion stress σ t at torque M: σ t,max = M ( dri/ 2) I p I p π d 2 2 = ri 4 d ri M = PN PN = 2π n 2π ( f / p) ~ s P N p σ t,max shaft σ P p ~ P p N 3 t, max d ~ 3 ri dri N lg( dri ) ~lg( PN )/ 3+ lg( p)/3 2/22

23 Increase of motor efficiency η with increasing induction machine power P at pole count 2p= 2, 4, 6, 8 and 12 for cage and wound rotor, 6 kv, 50 Hz, Thermal Class B, open ventilated machines Power rise with lengths: 3 3 P C L ~ A Bˆ s δ1l At constant loss density p d = P d /Vlosses rise with volume: P d = p d V ~ L So efficiency increases with bigger machines (P= P out ). η = ~ ~ L P~ L 3 4 P 1 1 = = P+ Pd 1+ Pd / P 1+ k/ L k: constant 4 2/23

24 Efficiency increases with speed at given torque At a given torque M efficiency η increases at dominant I 2 R-losses with speed n. Torque M ~ A. B δ. d 2. l Fe ~ p. Φ.. I = "Flux x Current Efficiency η = P P out in = P Pout + P out d 2π n M = 2π n M + 3 R I 2 ~ n η ~ I = I sn =const. n+ k I n n p Φ I p Φ I + k R I 2 2/24

25 Increase of power factor cosϕ with increasing induction machine power P at pole count 2p= 2, 4, 6, 8 and 12 for cage and wound rotors, 6 kv, 50 Hz, Thermal Class B, open ventilated machines a) In bigger machinesthe ratio stray flux vs. main flux is smaller due to higher slot numbers Q, yielding higher power factor cosϕ. b) High pole countmachines have (1) low slot number qper pole & phase, so higher harmonic leakage, (2) smaller ratio τ p /δ, so lower magnetizing reactance, yielding lower power factor cosϕ. 2/25

26 Induction machines with high pole count have poor cosϕ Example: m= 3, q= 1 e Magnetizing current I m ~ U s /(2πL s ) ~ δ/τ p Inductance per phase L s = L sσ + L h L h ~ l e. τ p /δ (a) High pole count 2p pole pitch τ p = d si π/(2p)small (b) Mechanical lower limit for air gap δ min τ p /δ min gets too small for big 2p! Example: Machine for very low speed 28/min: I: High pole count induction machine (not feasible!) for low speed, because cosϕ is too low II:4-pole induction machine frame size 400mm with gear i= 50 is feasible P / kw n /min -1 f / Hz 2p d si / m τ p /mm δ/mm δ/d si /% τ p /δ l / m cosϕ I m / I N I II /26

27 Design example: Motor, P N = 500 kw, 6 kv, 50 Hz, 4 poles 2/27

28 2p= 4, 500 kw, cage machine:estimation of motor efficiency η 94.4% 2/28

29 500 kw, induction machine: Estimation of power factor cosϕ /29

30 2p= 4,500 kw, induction machine:estimation of pole pitch τ p 360 mm 2/30

31 2p= 4, 500 kw, cage machine:estimation of (equivalent) stack length l e 380 mm 2p τ = 4: = 360mm p l e = 380mm 2/31

32 Pole pitch τ p = 360 mm, 2p= 4, 500 kw:estimation of air gap δ 1.4 mm 36 2/32

33 2p= 4, 500 kw:estimation of rotor shaft diameter d ri 200 mm 2/33

34 Pole pitchτ p = 360 mm, 2p= 4:Estimation of current loading & air gap flux density 500 A/cm (2/π)= = 0.59 T 36 2/34

35 500 kw, 2p= 4:Estimation of current density J 5.5 A/mm 2 2/35

36 Motor, P N = 500 kw, 6 kv, 50 Hz, 4 poles: Estimated values: η N = 94.4 %, cosϕ N = Motor output power: 500 kw, apparent power: SsN 610 Motor current: IsN = = = 59 A, 3 U 3 6 sn Design example PN cosϕ 2/36 = 610 kva s synchronous speed: nsyn = fs / p=1500 /min Xh Pole pitch 360 mm, stack length: 380 mm, air gap: 1.4 mm, shaft diameter: 200 mm, Current loading: 500 A/cm, current density: 5.5 A/mm 2, A (A/cm)(A/mm 2 s Js =2750 ) Stator bore diameter: d si = 2 pτ p / π = 458 mm. Internal apparent power: (σ s = 0.08/2 = 0.04): S S/( 1+ σ ) = 610/1.04= 587kVA Electromagnetic utilization: C = Sδ /( dsi le nsyn) = 4.9 kva. min/m 3 2 π Flux density (k w1 = 0.91 estimated): C k A Bˆ Bˆ = T. 2 2 S N = η N δ = s = w1 s δ1 δ1 N U U s h = 1+ X X h sσ + X X h sσ = 1+ σ =

37 Energy Converters CAD and System Dynamics Summary: Scaling effect in electric machines - Typical length Lused as scale -Scaling laws for power P, air gap flux density B δ, current loading A, current density J - Assumptions for given cooling system & type of construction must be kept in mind 2 -Pole count influence on active iron mass ~ dsa l e, efficiency η, power factor cosϕ - Scaling laws verified with results of built machines 2/37

38 Energy Converters CAD and System Dynamics 2. Design of induction machines 2.1 Main dimensions and basic electromagnetic quantities of induction machines 2.2 Scaling effect in electric machines 2.3 Stator winding low and high voltage technology 2.4 Stator winding design 2.5 Rotor cage design 2.6 Wound rotor design 2.7 Design of main flux path of magnetic circuit 2.8 Stray flux and inductance 2.9 Influence of saturation on inductance 2.10 Masses and losses 2/38

39 Low voltage stator winding manufacturing for a wind generator Source: Winergy Germany 2/39

40 Coil groups per pole and phase Example:q= 3 per group L: Total axial iron stack length including radial ventilation ducts τ p Example: = τ p concentric coils for single layer winding coils with identical span for double layer winding 2/40

41 Single layer winding Two different sizes of concentric coil groups possible for 2p= 4, 8, K Example:q= 2 coils per pole and phase, Coil arrangement (without coil connectors) -Cross section of winding overhang 2/41

42 Identical sizes for all coils Two layer winding Example:q= 2 coils per pole and phase Coil arrangement with full pitched coils: W= τ p W winding overhang 2/42

43 Pitched two layer lapwound winding W < τ p Arrangement of pitched coils per phase and connection of coil groups for N-and S-pole Example:q= 3 coils per pole and phase, pitch W/τ p = 8/9 Cross section of coil arrangement in slots, showing all three phases U, V, W in upper and lower layer, being symbolized by phase belts Phase belt without depicting single slots 2/43

44 s Upper layer Lower layer Screen-shot of a MMF V s (x) (magneto-motive force) distribution of a two-layer winding s Is ˆ = I 3/ 2 s Example: q= 2 slots per pole and phase s Coil pitch 5/6 Slot pitch 2/44

45 Winding factor k wsν = k dsν k psν k ds ν = ν π sin 2 m ν π q sin 2 q m Distribution factor k psν π W = sin( ν 2 τ p ) Pitching factor Reduction of harmonic induced voltage e.g. for q = 3, m= 3 (distribution factor) -for ν = 5: k ds, 5 = 0.218, ν = 7: kds, 7 = Also fundamental flux linkage is reduced a little bit: ν = : kds, = The flux linkage with stator winding of harmonic flux density waves with pole pairs is reduced for W/τ p = 8/9 e.g. (pitching factor) For ν = 5: k ps, 5 = 0.64, ν = 7: kps, 7 = Note that also fundamental flux linkage is reduced a little bit: ν = : kps, = ν 2/45

46 Winding overhang High voltage form wound stator coil with several turns Nc for two-layer winding coil side, inserted in slot coil terminals Source: Andritz Hydro, Austria 2/46

47 Ventilation duct Tooth Slot Massive iron clamping finger Pressing plate upper layer lower layer winding overhang Source: Andritz Hydro, Austria Inserting a two-layer winding 2/47

48 Two-layer winding: Slot and coil arrangement, winding insulation Wedge top lining main insulation inter-turn insulation Insulation between upper and lower layer coil conductors Oval semi-closed slot for round wire low voltage Rectangular slot for form wound high voltage coil arrangement with N c = 8 turns per coil 2/48

49 bar Wedge top lining main insulation strand iron stack press plate press finger insulation Z between upper and lower layer slot lining Roebel bar Cross section in slot with 24 strands per bar N c = 1 Winding overhang with clearing Eto end shields (on earth potential) Clearing ZFof end of main insulation from pressing construction of iron stack end 2/49

50 High voltage insulation for stator winding 3.3 k V kv Rated voltage (line-to-line, r.m.s.) kv Slot main insulation, thickness d mm Inter-layer insulation, thickness Z mm Top lining thickness mm Winding overhang main insulation, thickness d s mm Clearing of slot main insulation ZF mm Clearingbetween coils in winding overhang λ mm Clearingfrom winding overhang to earth potential E mm Total conductor insulation thickness (both sides), glass fibre, inter-turn voltage < 80 V mm Slot lining thicknessin vertical direction and slot play mm /50

51 Energy Converters CAD and System Dynamics Example: High voltage insulation for stator winding atu N = 16.5 kv (Dimensions in mm) Drawings NOT to scale! /51

52 Energy Converters CAD and System Dynamics Summary: Stator winding low and high voltage technology -Low voltage winding up to U N 1000 V, usually single-layer winding - Round wire winding and enamel insulation for low voltage -High voltage: U N >1000 V - Special high voltage insulation system, open slots, rectangular conductors - High voltage: Usually double-layer winding - Large machines: Prefabricated, fully insulated coils -Smaller machines: Complete impregnation of stator with insertedcoils 2/52

53 Energy Converters CAD and System Dynamics 2. Design of induction machines 2.1 Main dimensions and basic electromagnetic quantities of induction machines 2.2 Scaling effect in electric machines 2.3 Stator winding low and high voltage technology 2.4 Stator winding design 2.5 Rotor cage design 2.6 Wound rotor design 2.7 Design of main flux path of magnetic circuit 2.8 Stray flux and inductance 2.9 Influence of saturation on inductance 2.10 Masses and losses 2/53

54 Cleaning of the stator winding of surplus resin after stator impregnation Source: Winergy Germany 2/54

55 Stator two-layer winding design -Chosen number of slots: q s = 5, Qs = 2 p ms qs = 4 3 5= 60, leads to 15 slots per pole, d si = 458 mm - Slot pitch: τqs = dsiπ / Qs =24.0 mm - Coil pitching is possible in steps of one slot pitch: W/τ p = 14/15, 13/15 etc., chosen pitch W/τ p = 12/15 leads to k ps, 1= The influence of 5 th space harmonic is completely eliminated: k ps,5 = 0. - Distribution factor: kds, 1= 0.957, stator winding factor: k ws, 1= = With chosen air gap flux density 0.9 T main flux per pole of fundamental ν= 1 is 2 2 Φ h = τ p l Fe Bˆ δ, 1= = 78. 4mWb π π -Choice of number of turns per phase N s : U Estimated induced voltage per phase: N / / 3 Uh = = = 3330V 1+ σs 1.04 U 2πf N k 1 Φ N = h = s s ws h s Nc = Ns as/( 2pqs) = /(2 2 5) = 10.5 a s = 1 - Final values: Nc =10 N = 200 U h = 3330V Bˆ δ, 1 = T s 2/55

56 Low voltage winding with round copper wire Low voltagewinding is usually manufactured of round wire: slot fill factor k f for calculating slot is used. ACu k f = A single layer winding double layer winding slot fill factor k f Round wire diameter d Cu < 1.1 mm for easy bending! Example: A Q = 194 mm 2, single layer winding with round copper, N c = 13, a s = 1. Each turn per winding consists of a i = 8 parallel wires with diameter d Cu = 1.0 mm: AL = ai ( dcu π / 4) = 8 (1 π / 4) = 6.28mm -Slot fill factor: N A = = 81.7mm 2, =81.7/194= 0. A 421 Cu = c L k f Q 2/56

57 Breadth (mm) b L (mm) Conductor height h L b L h L Selection of available profile copper wire: -Dimensions (without enamel coating) and cross section -Edges of wire rounded by 0.5 mm mm radius Note:12.42 mm 2 < 1.8 x 7.1 = mm 2 due to rounded edges -h L < b L to minimize eddy currents, induced by the AC slot stray flux 2/57

58 Profile copper winding: -Number of adjacent turns: n ne -Number of strands per turn: a i -Cross section area of strand A TL Example:6.6 kv, a i = 1, N c = 8 Height: 4.0 mm 0.4 mm 2.2 mm 0.3 mm 4.0 mm A h L =1. 8 TL n ne = 2 2/58

59 Example: -Current per phase 59 A, voltage per phase 3460 V, -Current density limit 5.5 A/mm 2, N s = 200, N c = 10, τ Qs = 24.0mm -Parallel paths: a s = 1, a i = 1. -Conductor cross section: A I /( J a a ) = 59/( ) = 10.73mm 2 b = s s s i -Chosen slot breadth: mm ( = ) Qs TL = τ - Chosen conductor dimensions(table 2.4-2): b =7.1 mm, h L =1. 8mm U s / N s = 3460/ 200= Inter-turn voltage: V< 80 V: -Table 2.3-3: Conductor insulation thickness d ic = 0.4 mm (both sides) -Additional inter-turn insulation d i = 0.3 mm Nc ( hl + dic) + ( Nc 1) di = 10 ( ) = 24. 7mm n ( b + d ) = 1 ( ) = 7. mm ne L ic 5 L Qs 2/59

60 Slot height design: mm Number of insulated turns per coil one above the other = Main insulation 4.4 Insulated coil side = Two coils per slot 58.2 Inter-layer insulation Z = Slot lining (thickness 0.15 mm) 0.45 Wedge h wedge = 4.0, h 4 = 0.5 mm 4.5 Top and bottom lining 0.8 Vertical play 1.05 Slot height h Qs + h Slot width design: mm Number of adjacent insulated turns n ne = Main insulation 4.4 Slot lining (thickness 0.15 mm) 0.3 Play 0.3 Slot width b Q /60 h 4 = 0.5 mm s Q = 12.5 mm h 2 = = = 6.6 mm h 1= 24.7 mm h = = = 8.4 mm h 1= 24.7 mm b Q = 12.5 mm

61 Slot fill factor: k f k = f = A A Cu Q = 0.3 Low slot fill factor due to high voltage winding! k f 0. 6 k f 0.95 a) b) c) a) Two coils (two-layer winding) in one slot for 6 kv, 500 kw induction machine b) Wound rotor induction machine: rotor slot design for low voltage winding < 1kV c) Cross section of deep rotor bar in slot: very low voltage, no insulation! 2/61

62 Check of winding design: -Limit of height of conductor h L (or strand) to avoid eddy current losses (current displacement): reduced conductor height ξ < 0.3 K Conductor cross section concerning current density J s -Coil main insulation thicknesswith respect to rated voltage U N - Conductor insulation with respect to inter-turn voltage - Sufficient tooth width to avoid increased iron saturation: < 2.2 T -Resulting thermal utilization A s. J s b Q Checking of eddy currents at 20 C: ξ= h L / d E ξ = conductor height / penetration depth of eddy currents d E d E = = ξ = 2 bq µ 0 ω κ n s 4π 10 7 ne b / 0.125= 0.144< L π = 12.5mm h L = µ 0, κ n ne. bl Example:n ne = 1 2/62

63 - Checking of current density and thermal utilization: J I /( a a A ) = 59/( ) = 4.75A/mm 2 < 5.5 A/mm 2 A s s = s i TL = 2msNsI 2pτ p s = 4 36 = 492A/cm A = = 2337(A/cm)(A/mm 2 ): s Js Result fits to limits for open ventilated machine with 80 K temperature rise. - Electromagnetic utilization: C = 2 2 π ˆ π kws As B δ 1= = VAs/m 3 = 4.93 kva. min/m /63

64 Energy Converters CAD and System Dynamics Summary: Stator winding design -Increased insulation thickness dwith increased rated high voltage U N -Choice of winding type and number of slots per pole and phase q -Choice of current density Jand current loading A - Detailed slot design for HV winding -Slot fill factor k f for LV winding, depending on manufacturing abilities 2/64

65 Energy Converters CAD and System Dynamics 2. Design of induction machines 2.1 Main dimensions and basic electromagnetic quantities of induction machines 2.2 Scaling effect in electric machines 2.3 Stator winding low and high voltage technology 2.4 Stator winding design 2.5 Rotor cage design 2.6 Wound rotor design 2.7 Design of main flux path of magnetic circuit 2.8 Stray flux and inductance 2.9 Influence of saturation on inductance 2.10 Masses and losses 2/65

66 Die-cast aluminium cage rotor with skewed rotor bars and small fan blades at the rings Source: H. Kleinrath, Studientext 2/66

67 Induced phase shifted bar voltages Example:Q r /p= 14 bars per pole pair Example:Q r /p= 10 bars per pole pair s δ δ Example: 2p = 4, τ s= 1: U p i, bar = 36cm, = s v syn f s l Bˆ / 2 = = 25.5 V; s δ U ˆ i, bar = s vsyn l B δ1 Use of non-insulated bars due to low bar voltage! = 50Hz, l = 1m, Bˆ = 1T: v = 2f τ = 36m/s δ1 δ1 syn s N p = 1%: U i, bar = 0.26V / 2 2/67

68 Squirrel cage: Bar and ring currents a) Bar currents b) Ring currents α =2πp/ Qr Q r i + 23 = i12 i2 R bar Ring r R Ring 2 2 * 2 2 Cu, r = QrRbarIr + 2Qr RRingIRing = Qr( Rbar + RRing) Ir QrRrIr P = * 1 RRing = RRing Equivalent bar series resistance 2 2sin ( πp/ Q ) r I IRing = I Ir /( 2 sin( pπ / Qr)) 2/68

69 Copper bars: Shape of rotor bars in rotor slots Increasing current displacement effect! Source: T. Bohn, Energietechnik, TÜV Rheinland Bronze bars Current displacement : Resulting rotor bar current (inclusive eddy currents) flows mainly in upper half of bar! So it is using only part of rotor bar cross section, which leads to increase of effective rotor bar resistance (AC resistance). Copper bars Double cage for big starting torque M e,1 P 2 Cu, r = QrRrIr = s Pδ = Pδ = ( ωs/ p) Me,1 s= 1 2/69

70 Choice of rotor slot numbers 1. No cogging s r 2. For skewed cage: s Minimum inter-bar currents 3. Minimization of acoustic noise: 3a) Q Q 0.8Q Q < 2r = 4 Qs Qr 0,1, 2,... r*, 2p, 2p± 1, 2p± 2,...,2p± r* 3b) Avoid (Q s, Q r ) pairs with high number of common dividers! 4. No pulsating rotor bending force:choose Q r evento avoid radial force waves with 2r= 2 nodes! r 5. Minimize flux pulsation in teeth = reduce additional losses: Q r /Q s not below0.8 and not above1.2, otherwise slot frequent flux pulsation is too big Q s r*: maximum avoided order of deformation 2/70

71 Cogging torque & bending vibrations (from lecture: Motor development for electrical drive systems ) B r F n e n e r x If r F x B t r e << B x A n : 2 n B 2µ 0 r ( e n r e x ) da F F= 0 Q s = Q r Big cogging force: FORBIDDEN! : 2r= 2 F res F res > 0 n A Even rotor slot number: Yields with even stator slot number Q s = 2p. m s q s a symmetrical upper an lower flux density with symmetrical upper and lower radial force. Odd rotor slot number: Single-sided magnetic rotating pull F res excites bending shaft vibrations with frequency f = n F 2/71

72 Slot harmonic field wave Losses due to inter-bar currents (from lecture: Motor development for electrical drive systems ) bar a) b) c) a) Unskewedcage:Nointer-bar current, because R Ring << R q b) Skewed cage:slot numbers equal Q r =Q s : nointer-bar current c) Q r =Q s /1.5:BIGharmonic inter-bar current flows, as harmonic voltages add up. 2/72

73 Electromagnetic acoustic noise (from lecture: Motor development for electrical drive systems ) Magnetic pull as force wave Radial deflection Vibration Acoustic sound -The stator iron may be regarded as a steel ring, whereas the rotor is a steel cylinder. -Therefore the stator is less stiff than the rotor and isbentby the radial force waves. -As the iron surface is shaken with this frequency, the surrounding air is compressed and de-compressed with frequency f Ton. -So acoustic sound wavesare generated with that tonal frequencyf Ton to be heard by e.g. human beings. Source: Seinsch, H.-O., Teubner- Verlag, Stuttgart Vibration 2/73

74 Deformation of the stator yoke Acoustic noise (from lecture: Motor development for electrical drive systems ) 2r = 0 : Stator surface oscillates in phase along the stator circumference, so a far reachingsound pressure wave p air is generated. 2r > 0 : With increased node number rthe sound pressure p air is equalized strongly along the circumference = nofar reaching sound. Source: Jordan, H.; Der geräuscharme Elektromotor, Verlag Girardet, Essen, 1950 Stator is approximated for the far sound pressure field as a vibrating sphere. 2/74

75 Example: Choice of rotor slot numbers: 2p= 4, Q s = 60, unskewedrotor: Rule 1:Q r 60 Rule 5: Q r = ( )Q s = Rule 4: Take only even numbers! (44) (46) (74) (76) Rule 3:Choose e.g. r* = 4: Qs Qr 0,1, 2,... r*, 2p, 2p± 1, 2p± 2,...,2p± r* = 0,1, 2, 3, Rule 2:For skewedrotor only slot numbers Q r < Q s remain: 4, 4, 3, 5, 2, 6,1, 7, 0, 8 e.g. Q r = 50 is chosen. 2/75

76 Energy Converters CAD and System Dynamics Typical slot numbers of 3-phase cage induction motors Typical stator slot numbers Q s : q s = integer ( integer slot winding ) 2p / q s Typical stator/rotor slot numbers for skewed line-start cage motors Q s < Q r : (The rules 1 K5 are not always obeyed strictly!) 2p Q s / Q r 24/22 36/28 48/36 48/40 60/48 36/28 48/36 48/44 60/44 60/50 36/30 54/48 72/54 48/44 2/76

77 Estimation of rotor bar current Current transfer ratio: ü I Rotor bar current: I r k = k = I wr r ws m m s r / ü I cosϕ I N N s r 2k = s m Q ws r s s N s Ir = Ir/ üi Is cosϕ s = = 51.33A ü I = 2k ws m Q r s N s = = Ir = üi Ir = = 1121A k wr = r r r 1, m = Q, N = 1/ 2 2/77

78 Choice for rotor slot shape Deep bar rotor to increase starting torque: -ratio h Cu /b Cu 8 -Choice: h Cu = 40 mm b Cu = 5 mm cross section: A Cur = 200 mm 2 hqr h Cu bq b Cu Semi-closed rotor slot to hold the bar against the centrifugal force -Choice: s Cu = 2.5 mm h 1 = h Cu = 40 mm, b 1 = b Cu = 5 mm (+ 0.1 mm play) h 4 = 3.4 mm, s Qr = 2.5 mm h Qr = = 43.5mm, b Qr = 5.1mm 2/78

79 Choice of deep bar ratio h Cu /d E = ξ R AC = k R R ξ >2 R DC ξ DC h Cu = Example: 50 Hz, copper bar: d E = ca. 10 mm, h Cu = 40 mm ξ= h Cu /d E = 4: expected bar resistance increase: factor 4 at stand still 2/79

80 Cage design Deep bar rotor to increase starting torque: ratio h Cu /b Cu 8 Choice: h Cu = 40 mm, b Cu = 5 mm, cross section: A Cur = 200 mm 2 Rotor bar current density: J I / A = 1121/ 200= 5.6A/mm 2 r = r Cur Rotor ring current I Ring = I /( 2 sin( p / Q )) = 1121/(2 sin(2π / 50)) = r π r 4472A Necessary ring cross section: A I / J = 4472/ 5.6= 800mm 2 Ring = Ring Ring 2/80

81 Energy Converters CAD and System Dynamics Summary: Rotor cage design -Choice of number of slots per pole -Rotor slot count must differ from stator slot count Q r Q s - Bar shapes for increased starting torque via current displacement -Skewing of rotor cage - Die-cast cage (aluminum, copper) for small machines - Brazed copper cage for large machines - Ring cross section much bigger than bar cross section 2/81

82 Energy Converters CAD and System Dynamics 2. Design of induction machines 2.1 Main dimensions and basic electromagnetic quantities of induction machines 2.2 Scaling effect in electric machines 2.3 Stator winding low and high voltage technology 2.4 Stator winding design 2.5 Rotor cage design 2.6 Wound rotor design 2.7 Design of main flux path of magnetic circuit 2.8 Stray flux and inductance 2.9 Influence of saturation on inductance 2.10 Masses and losses 2/82

83 Wound three-phase rotor winding manufacturing Especially for big doubly-fed wind generators with rotor slip rings Source: Calculation example: see text book Winergy Germany 2/83

84 Wound rotor doubly-fed induction wind generator Doubly-fed induction generator for wind power generation 4 poles 2000 kw, 1800/min, 50 Hz, slip range: -/+ 20% Source: Winergy, Germany Details: See lecture: Large Generators & High power drives 2/84

85 Wound rotor winding of slip-ring induction machines - Wound rotor winding is usually a three-phase two-layer distributed winding. -Due to the slip ring system usually a low voltage winding (U r,ll < 1000 V) is chosen by a big voltage transformer ratio ü U, so a low N r. - Choice: N cr = 1, Q s Q r (reduced cogging torque, unskewedmachine) ü = U N N s r k k ws wr N / r Q Q =2p q =2p q m m = 2p qr Ncr ar r r r - Advantage of low voltage rotor winding: s s s Usually: m s = m r = 3, so: q s q r -Thin slot insulation allowed= high copper fill factor = low rotor winding resistance = low rotor copper losses -N cr = 1: One turn per coil = Rotor winding is manufactured as wave winding= 50% reduction in winding overhang conductors = further reduction of rotor winding resistance 2/85

86 Rotor wave winding brazing l Fe τ p τ p Upper layer bar Lower layer bar N cr = 1 Upper layer Lower layer Bars are inserted into the semi-closed or closed rotor slots and are afterwards brazed together to form the phase winding! One turn= two bars, which are brazed together in the winding overhang 2/86

87 Q s = 60 Q r = 72 b 12.8 mm N Cur cs h = 10 Cur 2 = mm Total copper cross section area: mm = mm J s 2 = 4.8 A/mm 2 2 = 2 a) b) 81mm a) Two-layer high-voltage stator winding,6 kv, N cs = 10, lap winding, open slot, low fill factor k f 0.3, bigslot cross section necessary for given A s. J s b) Two-layer low voltage rotor winding< 1 kv, N cr = 1, wave winding, semi-closed slot, high fill factor k f 0.6, smallslot cross section necessary for given A s. J s b N Cur cr f h = 1 k = Cur 2 A A Cu Q = mm Total copper cross section area: mm = = mm J r 2 = 5.2 A/mm 2 2 2/87

88 Energy Converters CAD and System Dynamics Summary: Wound rotor design - Design of three-phase rotor winding similar to stator winding -Usually two-layer winding for low voltage < 1 kv -Wave winding with 50% reduction of winding overhang possible, if one turn per coil N cr = 1 -Rotor slot count must differ from stator slot count Q r Q s 2/88