Author: Ion Boldea, S.A.Nasar………… ……… Chapter 6 LEAKAGE INDUCTANCES AND RESISTANCES 6.1. LEAKAGE FIELDS Any magnetic field (H i , B i ) zone within the IM is characterized by its stored magnetic energy (or coenergy) W m . () 2 I LdVHB 2 1 W 2 i i V mi =⋅= ∫∫∫ (6.1) Equation (6.1) is valid when, in that region, the magnetic field is produced by a single current source, so an inductance “translates” the field effects into circuit elements. Besides the magnetic energy related to the magnetization field (investigated in Chapter 5), there are flux lines that encircle only the stator or only the rotor coils (Figure 6.1). They are characterized by some equivalent inductances called leakage inductances L sl , L rl . end turn (connection) leakage field lines A C’ A A A A magnetisation flux lines zig - zag leakage flux lines airgap leakage rotor slot leakage flield lines stator slot leaka g e flield lines ma g netisation flux lines x Figure 6.1 Leakage flux lines and components There are leakage flux lines which cross the stator and, respectively, the rotor slots, end-turn flux lines, zig-zag flux lines, and airgap flux lines (Figure 6.1). In many cases, the differential leakage is included in the zig-zag leakage. Finally, the airgap flux space harmonics produce a stator emf as shown in Chapter 5, at power source frequency, so it should also be considered in the leakage category. Its torques occur at low speeds (high slips) and thus are not there at no–load operation. © 2002 by CRC Press LLC Author: Ion Boldea, S.A.Nasar………… ……… 6.2. DIFFERENTIAL LEAKAGE INDUCTANCES As both the stator and rotor currents may produce space flux density harmonics in the airgap (only step mmf harmonics are considered here), there will be both a stator and a rotor differential inductance. For the stator, it is sufficient to add all L 1m ν harmonics, but the fundamental (5.122), to get L ds . 1Km ; K K gKp WL6 L 1 1 s 2 2 w c1 2 2 1e0 ds ±=ν νπ τµ = ∑ ≠ν ν ν (6.2) The ratio σ d of L ds to the magnetization inductance L 1m is ν ≠ν ν ⋅         ν ==σ ∑ s s 1 2 1ws 2 2 ws m1 ds 0dS K K K K L L (6.3) where K Ws ν , K Ws1 are the stator winding factors for the harmonic ν and for the fundamental, respectively. K s and K s ν are the saturation factors for the fundamental and for harmonics, respectively. As the pole pitch of the harmonics is τ/ν, their fields do not reach the back cores and thus their saturation factor K s ν is smaller then K s . The higher ν, the closer K s ν is to unity. In a first approximation, ssts KKK <≈ ν (6.4) That is, the harmonics field is retained within the slot zones so the teeth saturation factor K st may be used (K s and K st have been calculated in Chapter 5). A similar formula for the differential leakage factor can be defined for the rotor winding. µ ≠µ µ ⋅         µ =σ ∑ s s 1 2 1wr 2 2 wr 0dr K K K K (6.5) As for the stator, the order µ of rotor harmonics is 1mK 22 ±=µ (6.6) m 2 –number of rotor phases; for a cage rotor m 2 = Z 2 /p 1 , also K wr1 = K wr µ = 1. The infinite sums in (6.3) and (6.5) are not easy to handle. To avoid this, the airgap magnetic energy for these harmonics fields can be calculated. Using (6.1) ν ν ν µ = sc m0 g KgK F B (6.7) We consider the step-wise distribution of mmf for maximum phase A current, (Figure 6.2), and thus © 2002 by CRC Press LLC Author: Ion Boldea, S.A.Nasar………… ……… () () 2 m1 N 1 2 j s 2 m1 2 0 2 0ds F F N 1 2F dF s ∑ ∫ θ = π θθ =σ π (6.8) A A A C’ C’ C’ C’ B B B B A’ A’ A’ π/ p F 1m θ F( ) θ 1 0.5 0.5 1 Figure 6.2 Step-wise mmf waveform (q = 2, y/τ = 5/6) The final result for the case in Figure 6.2 is σ dso = 0.0285. This method may be used for any kind of winding once we know the number of turns per coil and its current in every slot. For full-pitch coil three-phase windings [1], 1 Km 2 q12 1q5 2 1w 2 1 2 2 2 ds − π ⋅ + ≈σ (6.9) Also, for standard two-layer windings with chorded coils with chording length ε , σ ds is [1] ( ) 1 q12 1 y 1q9 y 1 4 3 1q5 Km 2 q3/y1y 2 2 22 2 1w 2 1 2 ds −         +               τ −       τ −−+ ⋅ π =σ τ−=−τ=ε (6.10) In a similar way, for the cage rotor with skewed slots, 1 'K 1 2 1r 2 skew 0r − η =σ (6.11) with () r 1 er er er r1 r er r er skew N p 2 ; 2/ 2/sin ; c c sin 'K π=α α α =η τ ⋅α         τ ⋅α = (6.12) © 2002 by CRC Press LLC Author: Ion Boldea, S.A.Nasar………… ……… The above expressions are valid for three-phase windings. For a single- phase winding, there are two distinct situations. At standstill, the a.c. field produced by one phase is decomposed into two equal traveling waves. They both produce a differential inductance and, thus, the total differential leakage inductance (L ds1 ) s=1 = 2L ds . On the other hand, at S = 0 (synchronism) basically the inverse (backward) field wave is almost zero and thus (L ds1 ) S=0 ≈ L ds . The values of differential leakage factor σ ds (for three- and two-phase machines) and σ dr , as calculated from (6.9) and (6.10) are shown on Figures 6.3 and 6.4. [1] A few remarks are in order. • For q = 1, the differential leakage coefficient σ ds0 is about 10%, which means it is too large to be practical. • The minimum value of σ ds0 is obtained for chorded coils with y/ τ ≈ 0.8 for all q s (slots/pole/phase). • For same q, the differential leakage coefficient for two-phase windings is larger than for three-phase windings. • Increasing the number of rotor slots is beneficial as it reduces σ dr0 (Figure 6.4). Figures 6.3 do not contain the influence of magnetic saturation. In heavily saturated teeth IMs as evident in (6.3), K s /K st > 1, the value of σ ds increases further. Figure 6.3 Stator differential leakage coefficient σ ds for three phases a.) and two-phases b.) for various q s The stator differential leakage flux (inductance) is attenuated by the reaction of the rotor cage currents. Coefficient ∆ d for the stator differential leakage is [1] d0dsds 1mq2 1 2 1w w rskew 0ds d 1 ; K K 'K 1 1 ∆σ=σ         ν η σ −≈∆ ∑ + ≠ν ν νν (6.13) © 2002 by CRC Press LLC Author: Ion Boldea, S.A.Nasar………… ……… Figure 6.4 Rotor cage differential leakage coefficient σ dr for various q s and straight and single slot pitch skewing rotor slots (c/τ r = 0,1,…) Figure 6.5. Differential leakage attenuation coefficient ∆ d for cage rotors with straight (c/ τ r = 0) and skewed slots (c/ τ s = 1) As the stator winding induced harmonic currents do not attenuate the rotor differential leakage: σ dr = σ dr0 . A rather complete study of various factors influencing the differential leakage may be found in [Reference 2]. Example 6.1. For the IM in Example 5.1, with q = 3, N s = 36, 2p 1 = 4, y/ τ = 8/9, K w1 = 0.965, K s = 2.6, K st = 1.8, N r = 30, stack length L e = 0.12m, L 1m = 0.1711H, W 1 = 300 turns/phase, let us calculate the stator differential leakage inductance L ds including the saturation and the attenuation coefficient ∆ d of rotor cage currents. © 2002 by CRC Press LLC Author: Ion Boldea, S.A.Nasar………… ……… Solution We will find first from Figure 6.3 (for q = 3, y/ τ = 0.88) that σ ds0 = 1.16 ⋅ 10 -2 . Also from Figure 6.5 for c = 1 τ s (skewing), Z 2 /p 1 = 30/2 = 15, ∆ d = 0.92. Accounting for both saturation and attenuation coefficient ∆ d , the differential leakage stator coefficient K ds is 22 d st s 0dsds 105415.192.0 8.1 6.2 1016.1 K K KK −− ⋅=⋅⋅⋅=∆⋅= (6.14) Now the differential leakage inductance L ds is H102637.01711.0105415.1LKL 32 m1dsds −− ⋅=⋅⋅== (6.15) As seen from (6.13), due to the rather large q, the value of K ds is rather small, but, as the number of rotor slots/pole pair is small, the attenuation factor ∆ d is large. Values of q = 1,2 lead to large differential leakage inductances. The rotor cage differential leakage inductance (as reduced to the stator) L dr is m1 2 m1 2 m1 ts s 0drdr L1004.4L 8.1 6.2 108.2L K K L −− ⋅=⋅⋅=σ= (6.16) σ dr0 is taken from Figure 6.4 for Z 2 /p 1 = 15, c/ τ r = 1, : σ dr0 = 2.8 ⋅ 10 -2 . It is now evident that the rotor (reduced to stator) differential leakage inductance is, for this case, notable and greater than that of the stator. 6.3. RECTANDULAR SLOT LEAKAGE INDUCTANCE/SINGLE LAYER The slot leakage flux distribution depends notably on slot geometry and less on teeth and back core saturation. It also depends on the current density distribution in the slot which may become nonuniform due to eddy currents (skin effect) induced in the conductors in slot by their a.c. leakage flux. Let us consider the case of a rectangular stator slot where both saturation and skin effect are neglected (Figure 6.6). b s b os h os h s H(x) n I s n I/b s os n I/b c s slot mmf n I. x h s s b os b s Figure 6.6 Rectangular slot leakage © 2002 by CRC Press LLC Author: Ion Boldea, S.A.Nasar………… ……… Ampere’s law on the contours in Figure 6.6 yields () () 0sssss s s s s hhxh ; inbxH hx0 ; h xin bxH +≤≤⋅= ≤≤ ⋅⋅ = (6.17) The leakage inductance per slot, L sls , is obtained from the magnetic energy formula per slot volume. ()       +µ=⋅µ⋅== ∫ + os os s s e 2 s0se hh 0 2 0 2 ms 2 sls b h b3 h LnbLdxxH 2 1 i 2 W i 2 L 0ss (6.18) The term in square parenthesis is called the geometrical specific slot permeance. () m1031h ;5.25.0 b h b3 h 3 os os os s s s − ÷=÷≈+=λ (6.19) It depends solely on the aspect of the slot. In general, the ratio h s /b s < (5 − 6) to limit the slot leakage inductance to reasonable values. The machine has N s stator slots and N s/ m 1 of them belong to one phase. So the slot leakage inductance per phase L sl is qp LW2L m qmp2 L m N L 1 s e 2 10sls 1 11 sls 1 s sl λ µ=== (6.20) The wedge location has been replaced by a rectangular equivalent area on Figure 6.6. A more exact approach is also possible. The ratio of slot leakage inductance L sl to magnetizing inductance L 1m is (same number of turns/phase), () s 2 1W sc 2 m1 sl q K KgK 3L L λ τπ = (6.21) Suppose we keep a constant stator bore diameter D i and increase two times the number of poles. The pole pitch is thus reduced two times as τ = π D/2p 1 . If we keep the number of slots constant q will be reduced twice and, if the airgap and the winding factor are the same, the saturation stays low for the low number of poles. Consequently, L sl /L 1m increases two times (as λ s is doubled for same slot height). Increasing q (and the number of slots/pole) is bound to reduce the slot leakage inductance (6.20) to the extent that λ s does not increase by the same ratio. Our case here refers to a single-layer winding and rectangular slot. Two-layer windings with chorded coils may be investigated the same way. © 2002 by CRC Press LLC Author: Ion Boldea, S.A.Nasar………… ……… 6.4. RECTANGULAR SLOT LEAKAGE INDUCTANCE/TWO LAYERS We consider the coils are chorded (Figure 6.7). Let us consider that both layers contribute a field in the slot and add the effects. The total magnetic energy in the slot volume is used to calculate the leakage inductance L sls . () () [] () xbdxxHxH i L2 L st h 0 2 210 2 e sls ⋅+µ= ∫ (6.22) b s h h b sl h i h su h o h w h os os st self mutual 1 2 H (x) 1 x mmfs H (x) 2 fields (H(x)) x b s b(x) b w b os mutual field zone Figure 6.7 Two-layer rectangular semiclosed slots: leakage field () () () oswi suisl i kcu i cl suislisl su isl s kcu s cl islsl s cl sl sls cl 21 bor bb with hhhfor x ; b cosIn b In hhhxhhfor ; h hhx b cosIn b In hhxhfor b In hx0for ; h x b In xHxH =                  ++> γ + ++<<+ −−γ + +<< <<⋅ =+ (6.23) The phase shift between currents in lower and upper layer coils of slot K is γ K and n cl , n cu are the number of turns of the two coils. Adding up the effect of all slots per phase (1/3 of total number of slots), the average slot leakage inductance per phase L sl is obtained. © 2002 by CRC Press LLC Author: Ion Boldea, S.A.Nasar………… ……… While (6.23) is valid for general windings with different number of turn/coil and different phases in same slots, we may obtain simplified solutions for identical coils in slots n cl = n cu = n c . () () ()             ++γ+++     + γ ++ γ+ = µ =λ os os w w s o 2 s i s ksu s su s k 2 susl e 2 c0 sl sk b h b h b h cos1 b h b cosh b h b3 coshh 4 1 Ln2 L (6.24) Although (6.24) is quite general–for two-layer windings with equal coils in slots–the eventual different number of turns per coil can be lumped into cos γ as Kcos γ with K = n cu /n cl . In this latter case the factor 4 will be replaced by (1 + K) 2 . In integer and fractionary slot windings with random coil throws, (6.24) should prove expeditious. All phase slots contributions are added up. Other realistic rectangular slot shapes for large power IMs (Figure 6.8) may also be handled via (6.24) with minor adaptations. For full pitch coils (cos γ K = 1.0) symmetric winding (h su = h sl = h s ′ ) (6.24) becomes () os os w w s 0 s i s s 'hhh 0 sk b h b h b h b4 h b3 'h2 sslsu k ++++=λ == =γ (6.25) Further on with h i = h o = h w = 0 and 2h s ′ = h s , we reobtain (6.19), as expected. h sl h i h su h os b os n I c n Icos γ c k H (t)=H (x)+H (x) t 1 2 n I/b c s n I(1+cos b γ )/ c ks h sl h i h su b ss b b os h o h w h os a.) b.) Figure 6.8 Typical high power IM stator slots © 2002 by CRC Press LLC Author: Ion Boldea, S.A.Nasar………… ……… 6.5. ROUNDED SHAPE SLOT LEAKAGE INDUCTANCE/TWO LAYERS Although the integral in (6.2) does not have exact analytical solutions for slots with rounded corners, or purely circular slots (Figure 6.9), so typical to low-power IMs, some approximate solutions have become standard for design purposes: • For slots a.) in Figure 6.9, () 2 1 r,os 1 o r,os r,os 21 1r,s r,s K785.0 b2 b b h b h bb3 Kh2         +−++ + ≈λ (6.26) with () 21 y y 2 y y 2 y y 2 K 4 3 4 1 K 21for ; 4 123 K 3 2 3 1 for ; 4 16 K 1 y 3 2 for ; 4 31 K +≈ ≤β≤ +β− ≈ ≤β≤ −β ≈ ≤ τ =β≤ β+ ≈ (6.27) • For slots b.), () 2 r,os1 w 1 o r,os r,os 21 1r,s r,s K b2b h3 b h b h bb3 Kh2         + +++ + ≈λ (6.28) • For slots c.), or or or or 1 or r b h 66.0 b h b2 b 785.0 +≈+−=λ (6.29) • For slots d.), or or 1 or 2 b 2 1 1 r r b h b2 h 66.0 A8 b 1 b3 h +−+         π −=λ (6.30) where A b is the bar cross section. If the slots in Figure 6.9c, d are closed (h o = 0) (Figure 6.9e) the terms h or /b or in Equations (6.29, 6.30) may be replaced by a term dependent on the bar current which saturates the iron bridge. [m]in b ;10b5I ; I 10 h12.13.0 b h 1 3 1b 2 b 3 or or or >+≈→ (6.31) © 2002 by CRC Press LLC [...]... dependent only on the airgap/slot opening, in (6.34) and (6.35) the airgap enters directly the denominator of L1m (magnetization inductance) and, in general, (6.34) and (6.35) includes the number of slots of stator and rotor, Ns and Nr As the term in parenthesis is a very small number an error here will notably “contaminate” the results On the other hand, iron saturation will influence the zig-zag flux... (6.63)), the equivalent bar resistance Rbe is R be = 2.7 ⋅ 10 − 5 + 0.9 ⋅ 10 −6 = 3.804 ⋅ 10 − 5 Ω 2π 2 sin 30 (6.70) As we can see from (6.70), the contribution of the end ring to the equivalent bar resistance is around 30% This is more than a rather typical proportion So far we did consider that the distribution of the currents in the bars and end rings are uniform However, there are a.c currents in the. .. already calculated as “seen from the stator side,” in terms of frequency because only in this case Φ1 is the same in both stator and rotor phases In reality the flux in the rotor varies at Sf1 frequency, while in the stator at f1 frequency The amplitude is the same and, with respect to each other, they are at stall Now we may proceed as for transformers by dividing the two equations to obtain WK E1... cross the airgap to define the magnetization inductance L1m, there are leakage fields that encircle either the stator or the rotor conductors The leakage fields are divided into differential leakage, zig-zag leakage, slot leakage, end-turn leakage, and skewing leakage Their corresponding inductances are calculated from their stored magnetic energy Step mmf harmonic fields through airgap induce emfs in the. .. rotor bars and end rings Consequently, the distribution of the current in the bar is not uniform and depends essentially on the rotor frequency f2 = sf1 Globally, the skin effect translates into resistance and slot leakage correction coefficients KR > 1, Kx < 1 (see Chapter 9) In general, the skin effect in the end rings is neglected The bar resistance Rb in (6.69) and the slot permeance λb in (6.53) are... saturation, in the sense of reducing them For the former, we already introduced the partial teeth saturation factor Kst = Ksd to account for it (6.14) For the latter, in (6.48), we have assumed that the level of saturation is implicitly accounted for L1m (that is, it is produced by the magnetization current in the machine) In reality, for large rotor currents, the skewing rotor mmf field, dependent... effect, for rotor frequencies f2 = Sf1 > (4 − 5)Hz; KR > 1 The skin effect also reduces the slot geometrical permeance (Kx < 1) and, finally, also the leakage inductance of the rotor The rotor cage (or winding) has to be reduced to the stator to prepare the rotor resistance and leakage inductance for utilization in the equivalent circuit of IMs The equivalent circuit is widely used for IM performance... density jcob = 6 A/mm2 in the bar and jcoi = 5 A/mm2 in the end ring, the average ring diameter Dir = 0.15 m, lb = 0.14 m, let us calculate the bar, ring cross section, the end ring current, and bar and equivalent bar resistance Solution The bar cross-section Ab is Ab = © 2002 by CRC Press LLC Ib = 1000 / 6 = 166mm 2 jcob (6.64) Author: Ion Boldea, S.A.Nasar………… ……… The current in the end ring Ii (from... (Figure 6.10a) while the zig-zag flux “snakes” out through the teeth around slot openings In general, they may be treated together either by conformal transformation or by FEM From conformal transformations, the following approximation is given for the geometric permeance λzs,r [3] λ zs, r ≈ 5gK c / b os, r 5 + 4gK c / b os, r ⋅ 3β y + 1 4 < 1.0; β y = 1 for cage rotors (6.32) The airgap zig-zag leakage... than the magnetization flux as the airgap is crossed many times (Figure 6.10b) Finally, the influence of chorded coils is not included in (6.34) to (6.35) We suggest the use of an average of the two expressions (6.33) and (6.34 or 6.35) In Chapter 7 we revisit this subject for heavy currents (at standstill) including the actual saturation in the tooth tops Example 6.2 Zig-zag leakage inductance For the . As the pole pitch of the harmonics is τ/ν, their fields do not reach the back cores and thus their saturation factor K s ν is smaller then K s . The. from (6.13), due to the rather large q, the value of K ds is rather small, but, as the number of rotor slots/pole pair is small, the attenuation factor