Chapter 15 IM DESIGN BELOW 100 KW AND CONSTANT V AND f 15.1. INTRODUCTION The power of 100 kW is traditionally considered the border between a small and medium power induction machine. In general, sub 100 kW motors use a single stator and rotor stack (no radial cooling channels) and a finned frame washed by air from a ventilator externally mounted at the shaft end (Figure 15.1). It has an aluminum cast cage rotor and, in general, random wound stator coils made of round magnetic wire with 1 to 6 elementary conductors (diameter ≤ 2.5mm) in parallel and 1 to 3 current paths in parallel, depending on the number of pole pairs. The number of pole pairs 2p 1 = 1, 2, 3, … 6. Figure 15.1 Low power 3 phase IM with cage rotor Induction motors with power below 100 kW constitute a sizable portion of the world electric motor markets. Their design for standard or high efficiency is a nature mixture of art and science, at least in the preoptimization stage. Design optimization will be dealt with separately in a dedicated chapter. For the most part, IM design methodologies are proprietory. Here we present what may constitute a sample of such methodologies. For further information, see also [1]. © 2002 by CRC Press LLC Author: Ion Boldea, S.A.Nasar………… ……… 15.2. DESIGN SPECIFICATIONS BY EXAMPLE Standard design specifications are • Rated power: P n [W] = 5.5kW • Synchronous speed: n 1 [rpm] = 1800 • Line supply voltage: V 1 [V] = 460V • Supply frequency: f 1 [Hz] = 60 • Number of phases m = 3 • Phase connections: star • Targeted power factor: cos ϕ n = 0.83 • Targeted efficiency: η n = 0.895 (high efficiency motor) • p.u. locked rotor torque: t LR = 1.75 • p.u. locked rotor current: i LR = 6 • p.u. breakdown torque: t bK = 2.5 • Insulation class: F; temperature rise: class B • Protection degree: IP55 – IC411 • Service factor load: 1.0 • Environment conditions: standard (no derating) • Configuration (vertical or horizontal shaft etc.): horizontal shaft 15.3. THE ALGORITHM The main steps in IM design are shown in Figure 15.2. The design process may start with (1) design specs and assigned values of flux densities and current densities and (2) calculate in the stator bore diameter D is , stack length, stator slots, and stator outer diameter D out , after stator and rotor currents are found. The rotor slots, back iron height, and cage sizing follows. All dimensions are adjusted in (3) to standardized values (stator outer diameter, stator winding wire gauge, etc.). Then in (4), the actual magnetic and electric loadings (current and flux densities) are verified. If the results on magnetic saturation coefficient (1 + K st ) of stator and rotor tooth are not equal to assigned values, the design restarts (1) with adjusted values of tooth flux densities until sufficient convergence is obtained in 1 + K st . Once this loop is surpassed, stages (5) to (8) are traveled by computing the magnetization current I 0 (5); equivalent circuit parameters are calculated in (6), losses, rated slip S n , and efficiency are determined in (7) and then power factor, locked rotor current and torque, breakdown torque, and temperature rise are assessed in (8). In (9) all this performance is checked and if found unsatisfactory, the whole process is restarted in (1) with new values of flux densities and/or current densities and stack aspect ratio λ = L/τ (τ – pole pitch). The decision in (9) may be made based on an optimization method which might result in going back to (1) or directly to (3) when the chosen construction and geometrical data are altered according to an optimization method (deterministic or evolutionary) as shown in Chapter 18. © 2002 by CRC Press LLC Author: Ion Boldea, S.A.Nasar………… ……… All construction and geometrical data are known and slightly adjusted here Sizing the electrical & magnetic circuits A =I/J A = /B Φ Co Co tooth tooth t Verification of electric and magnetic loadings: J =I/A B = /A Φ tooth toothf t Cof Cof Design specs electric & magnetic loadings: J , J , B B , B , λ Cos Cor g tc start seeking convergence in teeth saturation coefficient 1 + K st 1 2 4 3 Computation of magnetisation current I 0 5 Computation of equivalent circuit electric parameters R , X , R' , X' ,X 6 Co ssl r rl m Computation of loss, S (slip), efficiency 7 n Computation of power factor, starting current and torque, breakdown torque, temperature rise 8 is performance satisfactory? 9 NO YES END Figure 15.2 The design algorithm So, IM design is basically an iterative procedure whose output–the resultant machine to be built–depends on the objective function(s) to be minimized and on the corroborating constraints related to temperature rise, starting current (torque), breakdown torque, etc. The objective function may be active materials or costs or (efficiency) -1 or global costs or a weighted combination of them. Before treating the optimization stage in Chapter 18, let us perform here a practical design. © 2002 by CRC Press LLC Author: Ion Boldea, S.A.Nasar………… ……… 15.4. MAIN DIMENSIONS OF STATOR CORE Here we are going to use the widely accepted D is 2 L output constant concept detailed in the previous chapter. For completely new designs, the rotor tangential stress concept may be used. Based on this, the stator bore diameter D is (14.15) is 97.0p005.098.0K ; C S f pp2 D 1E 3 0 gap 1 1 is =−= πλ = (15.1) with τ = π =λ ϕη = L D 2p L ; cos PK S is 1 n1n nE gap (15.2) From past experience, λ is given in Table 15.1. Table 15.1. Stack aspect ratio λ 2p 1 2 4 6 8 λ 0.6 – 1.0 1.2 – 1.8 1.6 – 2.2 2 -3 From (15.2), the apparent airgap power S gap is VA8.7181 83.0895.0 105.597.0 S 3 gap = ⋅ ⋅⋅ = C o is extracted from Figure 14.14 for S gap = 7181.8VA, C 0 = 147⋅10 3 J/m 3 and λ = 1.5, f 1 = 60Hz, p 1 = 2. So D is from (15.1) is m1116.0 10147 8.7181 6015 222 D 3 3 is = ⋅ ⋅ ⋅⋅π ⋅⋅ = The stack length L (from 15.2) is m1315.0 22 1116.05.1 L = ⋅ ⋅π = The pole pitch m0876.0 22 1116.0 = ⋅ ⋅π =τ The number of stator slots per pole 3q may be 3⋅2 = 6 or 3⋅3 = 9. For q = 3, the slot pitch τ s will be around m10734.9 33 0876.0 q3 3 s − ⋅= ⋅ = τ =τ (15.3) In general the larger q gives better performance (space field harmonics and losses are smaller). © 2002 by CRC Press LLC Author: Ion Boldea, S.A.Nasar………… ……… The slot width at airgap is to be around 5 to 5.3 mm with a tooth of 4.7 to 4.4 mm which is mechanically feasible. From past experience (or from optimal lamination concept, developed later in this chapter), the ratio of the internal to external stator diameter D is /D out , bellow 100 kW for standard motors is given in Table 15.2. Table 15.2. Inner/outer stator diameter ratio 2p 1 2 4 6 8 out is D D 0.54 – 0.58 0.61 – 0.63 0.68 – 0.71 0.72 – 0.74 With 2p 1 = 4, we choose out is D D = K D = 0.62 and thus m18.0 62.0 1116.0 K D D D is out === (15.4) Let us suppose that this value is normalized. The airgap value has also been introduced in Chapter 14 as ( ) () 22pfor m10P012.01.0g 22pfor m10P02.01.0g 1 3 3 n 1 3 3 n ≥⋅⋅+= =⋅⋅+= − − (15.5) In our case, ( ) m1035.0103111.0105500012.01.0g 333 3 −−− ⋅≈⋅=⋅⋅+= As known, too small airgap would produces large space airgap field harmonics and additional losses while a too large one would reduce the power factor and efficiency. 15.5. THE STATOR WINDING Induction motor windings have been presented in Chapter 4. Based on such knowledge, we choose the number of stator slots N s . 363322qmp2N 1s =⋅⋅⋅== (15.6) A two layer winding with chorded coils: y/τ = 7/9 is chosen as 7/9 = 0.777 is close to 0.8, which would reduce the first (5 th order) stator mmf space harmonic. The electrical angle between emfs in neighboring slots α ec is 936 22 N p2 s 1 ec π = π = π =α (15.7) © 2002 by CRC Press LLC Author: Ion Boldea, S.A.Nasar………… ……… The largest common divisor of N s and p 1 (36, 2) is t = p 1 = 2 and thus the number of distinct stator slot emfs N s /t = 36/2 = 18. The star of emf phasors has 18 arrows (Figure 15.3a) and the distribution of phases in slots of Figure 15.3b. 1,19 18,36 17,35 16,34 15,33 14,32 13,31 12,30 11,29 10,28 9,27 8,26 7,25 6,24 5,23 4,22 3,21 2,20 A B’ C A’ B C’ A A A C’ C’ C’B B B A’ A’ A’ C C C B’B’ B’ A A A 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 C’ C’C’ B B B A’ A’ A’ C C C B’ B’B’ A C’ C’C’B B B A’ A’ A’ C C C B’B’ B’A A A C’C’C’ B B B A’ A’ A’ C C C B’ B’B’ A A Figure 15.3. A 36 slots, 2p 1 = 4 poles, 2 layer, chorded coils (y/ τ = 7/9) three phase winding The zone factor K q1 is 9598.0 18 sin3 5.0 q6 sinq 6 sin K 1q = π = π π = (15.8) The chording factor K y1 is 9397.0 9 7 2 sin y 2 sinK 1y = π = τ π = (15.9) So, the stator winding factor K w1 becomes 9019.09397.09598.0KKK 1y1q1w =⋅== The number of turns per phase is based on the pole flux φ, gi LBτα=φ (15.10) The airgap flux density is recommended in the intervals © 2002 by CRC Press LLC Author: Ion Boldea, S.A.Nasar………… ……… ( ) () () () 82pfor T85.075.0B 62pfor T82.07.0B 42pfor T78.065.0B 22pfor T75.05.0B 1g 1g 1g 1g =−= =−= =−= =−= (15.11) The pole spanning coefficient α i (Chapter 14, Figure 14.3) depends on the tooth saturation factor 1 + K st . Let us consider 1 + K st = 1.4 with α i = 0.729, K f = 1.085. Now from (15.10) with B g = 0.7T: Wb10878.57.01315.00876.0729.0 3− ⋅=⋅⋅⋅=φ The number of turns per phase W 1 (from Chapter 14, (14.9)) is: phase/turns8.186 10878.560902.0085.14 3 460 97.0 fKK4 VK W 3 11wf ph1E 1 = ⋅⋅⋅⋅⋅ ⋅ = φ = − (15.12) The number of conductors per slot n s is qp Wa n 1 11 s = (15.13) where a 1 is the number of current paths in parallel. In our case, a 1 = 1 and 33.31 32 8.1861 n s = ⋅ ⋅ = (15.14) It should be an even number as there are two distinct coils per slot in a double layer winding, n s = 30. Consequently, W 1 = p 1 qn s = 2⋅3⋅30 = 180. Going back to (15.12), we have to recalculate the actual airgap flux density B g . T726.0 180 8.186 7.0B g =⋅= (15.15) The rated current I 1n is A303.9 46073.183.0895.0 5500 V3cos P I 1nn n n1 = ⋅⋅⋅ = ϕη = (15.16) As high efficiency is required and, in general, at this power level and speed, winding losses are predominant from the recommended current densities. ( ) () 8,62pfor mm/A85J 2,42pfor mm/A74J 1 2 cos 1 2 cos == == K K (15.17) © 2002 by CRC Press LLC Author: Ion Boldea, S.A.Nasar………… ……… we choose J cos = 4.5A/mm 2 . The magnetic wire cross section A Co is 2 1cos n1 Co mm06733.2 15.4 303.9 aJ I A = ⋅ == (15.18) Table 15.3. Standardized magnetic wire diameter Rated diameter [mm] Insulated diameter [mm] 0.3 0.327 0.32 0.348 0.33 0.359 0.35 0.3795 0.38 0.4105 0.40 0.4315 0.42 0.4625 0.45 0.4835 0.48 0.515 0.50 0.536 0.53 0.567 0.55 0.5875 0.58 0.6185 0.60 0.639 0.63 0.6705 0.65 0.691 0.67 0.7145 0.70 0.742 0.71 0.7525 0.75 0.749 0.80 0.8455 0.85 0.897 0.90 0.948 0.95 1.0 1.0 1.051 1.05 1.102 1.10 1.153 1.12 1.173 1.15 1.2035 1.18 1.2345 1.20 1.305 1.25 1.305 1.30 1.356 1.32 1.3765 1.35 1.407 1.40 1.4575 1.45 1.508 1.5 1.559 With the wire gauge diameter d Co mm622.1 06733.24A4 d Co Co = π ⋅ = π = (15.19) © 2002 by CRC Press LLC Author: Ion Boldea, S.A.Nasar………… ……… In general, if d Co > 1.3 mm in low power IMs, we may use a few conductors in parallel a p . mm15.1 2 06733.24 a A4 'd p Co Co = ⋅π ⋅ = π = (15.20) Now we have to choose a standardized bare wire diameter from Table 15.3. The value of 1.15 mm is standardized, so each coil is made of 15 turns and each turn contains 2 elementary conductors in parallel (diameter d Co ’ = 1.15 mm). If the number of conductors in parallel a p > 4, the number of current paths in parallel has to be increased. If, even in this case, a solution is not found, use is made of rectangular cross section magnetic wire. 15.6. STATOR SLOT SIZING As we know by now, the number of turns per slot n s and the number of conductors in parallel a p with the wire diameter d Co ’, we may calculate the useful slot area A su provided we adopt a slot fill factor K fill . For round wire, K fill ≈ 0.35 to 0.4 below 10 kW and 0.4 to 0.44 above 10 kW. 2 2 fill sp 2 Co su mm7.155 40.04 30215.1 K4 na'd A = ⋅ ⋅⋅⋅π = π = (15.21) For the case in point, trapezoidal or rounded semiclosed shape is recommended (Figure 15.4). Figure 15.4 Recommended stator slot shapes For such slot shapes, the stator tooth is rectangular (Figure 15.5). The variables b os , h os , h w are assigned values from past experience: b os = 2 to 3 mm ≤ 8g, h os = (0.5 to 1.0) mm, wedge height h w = 1 to 4 mm. The stator slot pitch τ s (from 15.3) is τ s = 9.734 mm. Assuming that all the airgap flux passes through the stator teeth: Fetstssg LKbBLB ≈τ (15.22) K Fe ≈ 0.96 for 0.5 mm thick lamination constitutes the influence of lamination insulation thickness. © 2002 by CRC Press LLC Author: Ion Boldea, S.A.Nasar………… ……… h h b b s2 os s os h w b s1 h cs b ts D out D is Figure 15.5 Stator slot geometry With B ts = 1.5 – 1.65 T, (B ts = 1.55 T), from (15.22) the tooth width b ts may be determined. m1075.4 96.055.1 10734.9726.0 b 3 3 ts − − ⋅= ⋅ ⋅⋅ = From technological limitations, the tooth width should not be under 3.5⋅10 - 3 m. With b os = 2.2⋅10 -3 m, h os = 1⋅10 -3 m, h w = 1.5⋅10 -3 m, the slot lower width b s1 is () () m1042.51075.4 36 105.12126.111 b N h2h2D b 33 3 ts s wosis 1s −− − ⋅=⋅− ⋅⋅+⋅+π = =− ++π = (15.23) The useful area of slot A su may be expressed as: () 2 bb hA 2s1s ssu + = (15.24) Also, s s1s2s N tanh2bb π +≈ (15.25) From these two equations, the unknowns b s2 and h s may be found. s su 2 1s 2 2s N tanA4bb π =− (15.26) © 2002 by CRC Press LLC [...]... 15.9 The T equivalent circuit (core losses not evident) The stator phase resistance, © 2002 by CRC Press LLC Author: Ion Boldea, S.A.Nasar………… ……… R s = ρCo lc W1 A co a1 (15.63) The coil length ls includes the active part 2L and the end connection part 2lend lc = 2(L + lend ) (15.64) The end connection length depends on the coil span y, number of poles, shape of coils, and number of layers in the winding... 609.6W The efficiency ηn (from 15.87) becomes ηn = 5500 = 0.9002! 5500 + 609.6 The targeted efficiency was 0.895, so the design holds If the efficiency had been smaller than the target value, the design should have returned to square one © 2002 by CRC Press LLC Author: Ion Boldea, S.A.Nasar………… ……… by adopting a larger stator bore diameter Dis, then smaller current densities, etc A larger size machine. .. density Increasing the slot height is not practical as already d2 = 1.2⋅10-3m This bar current density increase could reduce the efficiency below the target value We may alternatively increase 1 + Kst, and redo the design from (15.10) When the power factor constraint is not too tight, this is a good solution To maintain same efficiency, the stator bore diameter has to be increased So the design should... choose Ns ≠ Nr = 36/28 As the starting current is rather large–high efficiency is targeted the skin effect is not very pronounced Also, as the locked rotor torque is large, the leakage inductance will not be large Consequently, from the four typical slot shapes of Figure 15.6, that of Figure 15.6c is adopted a.) b.) c.) d.) e.) f.) Figure 15.6 Typical rotor cage slots First, we need the value of rated rotor... iron iron loss The fundamental core losses occur only in the teeth and back iron (pt1, py1) of the stator as the rotor (slip) frequency is low (f2 < (3 – 4)Hz) The stator teeth fundamental losses (see Chapter 11) are: h 1.3 f 1.7 p t1 ≈ K t p10 1 Bts G t1 50 (15.98) where p10 is the specific losses in W/Kg at 1.0 Tesla and 50 Hz (p10 = (2 – 3)W/Kg; it is a catalog data for the lamination... ⋅10− 6 m 2 J b 3.42 ⋅106 (15.36) The end ring current Ier is Ier = Ib 279.6 = 628.255A = πp1 2π 2 sin 2 sin Nr 28 (15.37) The current density in the end ring Jer = (0.75 – 0.8)Jb The higher values correspond to end rings attached to the rotor stack as part of the heat is transferred directly to rotor core With Jer = 0.75⋅Jb = 0.75⋅342⋅106 = 2.55⋅106A/m2, the end ring cross section, Aer,... KI = 1, the rotor and stator mmf would have equal magnitudes In reality, the stator mmf is slightly larger K I ≈ 0.8 ⋅ cos ϕ1n + 0.2 = 0.8 ⋅ 0.83 + 0.2 = 0.864 From (15.34), the bar current Ib is © 2002 by CRC Press LLC (15.35) Author: Ion Boldea, S.A.Nasar………… ……… Ib = 0.864 ⋅ 2 ⋅ 3 ⋅180 ⋅ 0.9019 ⋅ 9.303 = 279.6A 28 For high efficiency, the current density in the rotor bar jb = 3.42 A/mm2 The rotor... verify the rotor teeth mmf Fmtr for Btr = 1.6T, Htr = 2460A/m (Table 15.4) (d + d ) (1.2 + 5.70) ⋅10− 3 Fmtr = H tr h r + h or + 1 2 = 2460 20 + 0.5 + 2 2 = 60.134Aturns (15.45) This is rather close to the value of Vmtr = 55.11 Aturns of (15.39) The design is acceptable so far If Vmtr had been too large, we might have reduced the flux density, thus increasing tooth width btr and the. .. general, it is not easy to make a few changes to get the desired operation characteristics Here the optimization design methods come into play 15.12 TEMPERATURE RISE Any electromagnetic design has to be thermally valid (Chapter 12) Only a coarse verification of temperature rise is given here First the temperature differential between the conductors in slots and the slot wall ∆θco is calculated ∆θco ≈ © 2002... 0.25 λ ins = = 833W / m 2 K h ins 0.3 ⋅10− 3 (15.124) λins is the insulation thermal conductivity in (W/m0K) and hins total insulation thickness from the slot middle to teeth wall The stator slot lateral area Als is A ls ≈ (2h s + b s 2 ) ⋅ L ⋅ N s = (2 ⋅ 21.36 + 9.16 ) ⋅10 −3 ⋅ 0.1315 ⋅ 36 = 0.2456m 2 (15.125) The frame area Sframe (including the finns area) is A frame = πD out (L + τ) ⋅ K fin = π ⋅190(0.1315 . As the starting current is rather large–high efficiency is targeted the skin effect is not very pronounced. Also, as the locked rotor torque is large, the. (15.37) The current density in the end ring J er = (0.75 – 0.8)J b . The higher values correspond to end rings attached to the rotor stack as part of the