This timely new edition offers up-to-date theory and guidelines for the design of electrical machines, taking into account recent advances in permanent magnet machines as well as synchronous reluctance machines. Manual arts senior high. Design of Rotating Electrical Machines. Design of rotating electrical machines by Juha Pyrhonen, Tapani Jokinen, and Valeria Hrabovcova. The objective of this book is to provide students in electrical engineering with an adequate basic knowledge of rotating electric machines, for an understanding of the operating principles of these machines as well as developing elementary skills in machine. OF SYNCHRONOUS MACHINES The synchronous electrical generator (also called alternator) belongs to the. Free movie budget software mac. Standing of the principles governing the machine’s design and operation, but. Widely used in several types of electric rotating machines, including synchronous machines. However, due to mechanical, as well as operational reasons, perma.
In one complete volume, this essential reference presents an in-depth overview of the theoretical principles and techniques of electrical machine design. https://posterskyey316.weebly.com/virtual-dj-705-download.html. This timely new edition offers up-to-date theory and guidelines for the design of electrical machines, taking into account recent advances in permanent magnet machines as well as synchronous reluctance machines.
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Brand new material on the ecological impact of the motors, covering the eco-design principles of rotating electrical machines
An expanded section on the design of permanent magnet synchronous machines, now reporting on the design of tooth-coil, high-torque permanent magnet machines and their properties
Large updates and new material on synchronous reluctance machines, air-gap inductance, losses in and resistivity of permanent magnets (PM), operating point of loaded PM circuit, PM machine design, and minimizing the losses in electrical machines>
End-of-chapter exercises and new direct design examples with methods and solutions to real design problems>
A supplementary website hosts two machine design examples created with MATHCAD: rotor surface magnet permanent magnet machine and squirrel cage induction machine calculations. Also a MATLAB code for optimizing the design of an induction motor is provided
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1.65 T 1.5 T 2 0.278 kg = 2.85 W. 13. The factor α i and the saturation factor ksat were defined in point 7. Recall that the factor αi was given as a function of the factor ksat in Figure 7.2. Now, we have to check the saturation factor and determine a new α i . If α i does not correspond with sufficient accuracy to the factor selected in the initial phase of the calculation, the peak value Bˆ δ of the air-gap flux density has to be recalculated according to Equation (7.7) because N is now fixed. Simultaneously, the flux density values of the stator and rotor teeth have to be corrected, and new magnetic voltages have to be calculated for the teeth and the air gap. The factor α i has to be iterated gradually to a correct value. For PMSMs with rotor surface magnets of uniform thickness, this step is not valid. 14. The flux density maxima Bˆ ys and Bˆ yr of the stator and rotor yokes are selected according to Table 6.1. With the peak value of the flux of the machine, together with the flux density peaks Bˆ ys and Bˆ yr , we are able to determine the heights h ys and h yr of the rotor and stator yokes that realize the selected flux density maxima. 15. When the air-gap diameter Ds , the heights h ds and h dr of the teeth, and the heights h ys and h yr of the stator and rotor yokes are known, we obtain the outer diameter Dse of the stator and the inner diameter Dri of the machine; cf. Figures 3.1 and 3.2. 16. Now all the main dimensions of the machine have been determined, next we have to check the magnetic voltages required by different parts of the machine. The sum of the magnetic voltages has to be covered by the current linkage Θ produced by some (or multiples) of the windings or by permanent magnets. Magnetizing is accomplished by different methods in different machines. A DC machine is magnetized with a separate l B, T H, A/m PFe/V , W/m3 Figure 7.5 Principle of the flux density distribution B of a nonuniform stator tooth, the absolute value of the field strength H and the iron loss density PFe /V on the integration path in the middle of the tooth in Figure 7.4. The curves are derived according to the stator tooth flux, the local tooth width and the BH curve of the tooth material P1: OTA/XYZ P2: ABC c07 JWBK312-Pyrhonen November 25, 2008 16:47 312 Printer Name: Yet to Come Design of Rotating Electrical Machines field winding or permanent magnets, whereas an induction machine is magnetized with the magnetizing current of the stator winding, or a synchronous machine with the current of the rotor field winding or by permanent magnets, and so on. In this book, systematically, half of the magnetic circuit is calculated (e.g. half of the stator yoke, one stator tooth, one air gap, one rotor tooth, half of the rotor yoke). The magnetic voltage sum of half of the magnetic circuit has to be covered by the current linkage amplitude produced by the rotating-field winding or by the current linkage of one pole winding of an excitation winding or by one permanent magnet. At this point, the height h PM of the permanent magnets is calculated. The magnetic voltage sum of the magnetic circuit is Uˆ m,ys Uˆ m,yr Hc + BPM h PM + Uˆ m,tot = Hc h PM = Uˆ m,δe + Uˆ m,ds + Br 2 2 (7.15b) from which h PM Uˆ m,ys Uˆ m,yr + Uˆ m,δe + Uˆ m,ds + 2 2 = Hc Hc − BPM Br (7.15c) ˆ A stator √ rotating field produces the fundamental current linkage of amplitude Θs1 = mkw1 Ns 2Is /(π p) according to Equation (2.15) modified for the fundamental component. The required magnetizing current can be calculated by keeping the sum Um,tot of the magnetic ˆ s1 . Now we obtain the voltages of half the magnetic circuit equal to the amplitude Uˆ m,tot = Θ RMS value of the stator magnetizing current Is, mag = Uˆ m, tot π p √ . mkw1 Ns 2 For machines magnetized with pole windings, the calculation of the no-load magnetizing current is a simple task, since the current linkage of a single pole winding is Nf If . This, again, has to equal the sum of the magnetic voltages of half a magnetic circuit. When dimensioning a synchronous machine excitation winding, for instance, it must be remembered that the pole winding must also capable of compensating for the armature reaction and, hence, the amount of copper must be large enough to carry a much larger current than at no load. 17. Since the dimensions have been defined and the winding has been selected, the resistances and inductances of the machine are now calculated. The magnetizing inductance was discussed in Chapter 3, the leakage inductances in Chapter 4 and the resistances in Chapter 5. With them, the equivalent circuit parameters of the machine per phase is obtained. Now the losses, efficiency, temperature rise and torques of the machine can be determined. In Figure 7.5, the iron losses in the teeth are solved by manual calculations. The frequency of the flux density alternations corresponds in principle to the rated frequency of the machine. The specific power loss (W/kg) corresponding to this frequency and the peak flux density can be checked from the manufacturer’s catalogue, and then a dissipation power density curve can be constructed corresponding to the flux density curve of Figure 7.5 for the determination of losses in a single tooth, when the weight of the tooth is known. Using the same principle, P1: OTA/XYZ P2: ABC c07 JWBK312-Pyrhonen November 25, 2008 16:47 Printer Name: Yet to Come Design Process and Properties of Rotating Electrical Machines 313 all the iron parts of the machine can be analysed. Since stresses and burrs occur in punched plates, and since the shaped parts on both sides of the air gap cause flux pulsations, there are notably higher losses in the teeth and also in other iron parts of the machine than the power loss calculated with the base frequency would suggest. Further, the power losses given by the manufacturer are presented for AC magnetizing, not for rotating magnetizing, which is the dominant form of motion of the field in the stator yoke. These facts have led to the utilization of the empirical factors in Table 3.2. These factors in Table 3.2 are used to correct the iron loss calculations in the most significant parts of the magnetic circuit of a machine. In machine design, special attention must be paid to the fact that the frequencies in all parts of the machine are not equal. In the stators of rotating-field machines, the base frequency is the input frequency f s of the machine. However, in the teeth of the stator and the rotor, high-frequency flux components occur, which are based on the motion of the teeth with respect to each other. For instance, in the rotor of a synchronous machine, the base frequency is zero, but other pulsation losses occur on the rotor surface because of stator slotting. Resistive losses are defined with the methods discussed in Chapter 9 by determining the resistances of the windings. Following the definition of the iron and resistive losses, the windage and friction losses and the additional losses of the machine can be defined according to the guidelines given in Chapter 9. Now the efficiency of the machine has been resolved. The thermal rise, on the other hand, finally determines the resistances of the machine, and thus also the resistive losses, and therefore the heat transmission calculations should still be carried out before the analysis of final losses of the machine. The previously discussed calculation procedure can be illustrated in compact form by the flow chart in Figure 7.6. Next, some special characteristics of the calculation of the most common machine types are discussed. 7.1 Asynchronous Motor An induction motor is the most common motor type applied in industry. Usually, the rated output powers follow a geometric series, in which the ratio of rated output powers is approxi√ mately n 10. The ordinal n of the root defines a power series that can be denoted ‘a series n’. Typically, the series 5, series 7 and series 9 are employed. Different series are employed at different power ranges. The recommended power ranges are given in Table 7.2. The output powers of induction motors do not strictly follow the recommendations. Usually, the rated powers of the motors below 1000 V are the following: 0.18, 0.25, 0.37, 0.55, 0.75, 1.1, 1.5, 2.2, 3.0, 4.0, 5.5, 7.5, 11, 15, 18.5, 22, 30, 37, 45, 55, 75, 90, 110, 132, 160, 200, 250, 315, 400, 450, 500, 560, 630 and 710 kW. We see that also at the lower end of this power series, the series 7 is approximately followed. Machines above 710 kW are often constructed for high voltages, however, with an inverter drive; for instance, even a power of 5 MW can be reached with a 690 V voltage. Each rated voltage has an optimum power range, the lower and upper limits of which should not be exceeded if a favourable construction is to be produced. If the voltage is excessively high with respect to the power, the conductors become rather thin, and the number of turns increases drastically. In the opposite case, the winding has to be made of preformed copper, which increases the winding costs. In the above series, 630 kW is, typically, the limit for P1: OTA/XYZ P2: ABC c07 JWBK312-Pyrhonen November 25, 2008 16:47 Printer Name: Yet to Come 314 Design of Rotating Electrical Machines 1. set the starting values of the machine 2. determine a suitable tangential stress σtan or machine constant Cmec 3. select a suitable length ratio χ, D and l' 4. define a suitable physical air gap δ 5. select a winding, define Qs, Qr, and kw 6. define an air gap flux density Bˆ 7. select αi and the number of coil turns N 8. seek a coil zQ for a slot and then a new N 9. define a new Bˆ 10. define the teeth widths bd 11. define the teeth heights hd 12. define the magnetic voltages of the teeth 13. define a new αi. If the accuracy is not sufficient, return to 9. 14. define the flux densities in the yokes 15. define rest of the diameters of the machine 16. define the total magnetic voltage Um,tot of the magnetic circuit 17. define all required machine characteristics of the dimensions solved above Figure 7.6 Design process of a rotating electrical machine in brief. This chart was originally intended for induction motor design but may also be applied to other rotating-field machine types. The factor α i behaves in a different way, especially in surface permanent magnet machines. The relative magnet width may be used as an initial value for α i in PMSMs with rotor surface magnets of uniform thickness Table 7.2 Power series of induction motors √ Series (n) Power ratio ( n 10) Power range/kW 5 7 9 <1.1 1.1–40 >40 1.58 1.39 1.29 P1: OTA/XYZ P2: ABC c07 JWBK312-Pyrhonen November 25, 2008 16:47 Printer Name: Yet to Come Design Process and Properties of Rotating Electrical Machines 315 Ω = ω/3 0 − 6π π Ω =ω 0 − 2π Θˆ Ds 2π, Θˆ 3π (a) 4π α Ds 5π (b) π Figure 7.7 (a) Six-pole fundamental current linkage propagating at a physical angular speed Ω = ω/3, and (b) a two-pole fundamental, propagating at a speed Ω = ω. If the angular frequency ω of the input current is equal in both windings, the current linkage propagates in a six-pole winding locally at one-third speed when compared with the flux of the two-pole winding. Both distributions propagate one wavelength during one supply frequency period, which explains the propagation speed difference. The influence of the number of pole pairs on the rotation speed of a machine is based on this fact a 400 V machine. The rated current of such a machine is about 1080 A, which is a value that can be considered a practical maximum for the rated current of a direct-on-line (DOL) induction motor. The starting current of such a machine may reach even 10 kA. According to the IEC 60034-1 standard, the minimum rated output is 100 kW for a rated voltage UN = 1–3 kV, 150 kW for UN = 3–6 kV, and 800 kW for UN = 6–11 kV. 7.1.1 Current Linkage and Torque Production of an Asynchronous Machine The poly-phase stator winding of a rotating-field machine creates a magnetic flux wave propagating in the air gap, when supplied with a symmetric poly-phase current. Figure 7.7a illustrates a fundamental wave ν = 1 of a six-pole machine. Figure 7.7b illustrates the fundamental of a two-pole machine. The amplitude of the νth harmonic of a current linkage created by an m-phase winding is written analogously to Equation (2.15) ˆ ˆ ν = m kwν Ns i, Θ π νp 2π i U = iˆ cos ωt, i V = iˆ cos ωt − , 3 4π i W = iˆ cos ωt − . 3 (7.16) When a three-phase winding is fed with symmetrical three-phase currents i U , i V , i W , we obtain, according to Figure 7.7, the fundamental current linkage propagating in the air gap, depending on the structure of the winding. The current linkage harmonic can thus be expressed as a function of time and position angle α (Figure 7.8) as ˆ ν ωt − να = Θ ˆ ν ej(ωt−να) . Θν = Θ (7.17) P1: OTA/XYZ P2: ABC c07 JWBK312-Pyrhonen November 25, 2008 16:47 Printer Name: Yet to Come 316 Design of Rotating Electrical Machines H1 + Aˆ1 two-pole linear current density ∂H 1 ∂α dα c H1 d A1 (α ) A (α ) b dα Θ (α ) a A1 δe α 0 r − Aˆ1 (a) (b) Figure 7.8 (a) Linear current density A(α), its fundamental A1 (α) and its integral, the current linkage Θ(α). (b) The linear current density A1 created by the poly-phase currents of a two-pole rotating-field stator. Only the fundamental harmonic of the linear current density is illustrated. The linear current density creates field strength in the air gap and a corresponding flux density. In this figure, the linear current density is illustrated as poles in (a), the height of which is the slot current divided by the slot opening width. A = z Q I /b1 This current linkage is exerted on half of the main magnetic circuit, that is a single effective air gap δ ef including the influence of the iron (cf. Chapter 3, Equation (3.56)). The flux density created by the stator current linkage is inversely proportional to the effective air gap. The νth flux density harmonic is written as µ0 ˆ Θν = Bˆ δν ωt − να = Bˆ δν e j(ωt−να) , δef µ0 m kwsν Ns ˆ i. = δef π νp Bδν = (7.18) Bˆ δν (7.19) This harmonic represents the flux penetrating the air gap, the peak value of which is 0001 0001 ˆ ˆ mν = Dsl Bˆ δν = µ0 m Dsl kwsν Ns i. Φ νp π p 2 δef ν 2 (7.20) The peak value of the total air-gap flux is the sum ˆm = Φ ±∞ ±∞ 0001 kwsν µ0 m Dsl 0001 kw1 ˆ ˆ mν = µ0 m Dsl Ns iˆ ≈ Ns i. Φ 2 2 π p δef ν π p 2 δef ν=1 ν=1 (7.21) P1: OTA/XYZ P2: ABC c07 JWBK312-Pyrhonen November 25, 2008 16:47 Printer Name: Yet to Come Design Process and Properties of Rotating Electrical Machines 317 The series converges so rapidly that it normally suffices to take only the fundamental into account in the calculation. The composition of the main flux of the fundamental and harmonics is illustrated by Equation (7.21). It is worth remembering that the harmonics analogously induce a voltage of base frequency in the windings of the machine. A current is supplied to the windings mounted in the slots of the machine. The slot currents can be replaced with sufficient accuracy by a slot’s local linear current density values Au of the width of the slot opening b1 : Au = zQ I , at the slot opening, elsewhere A = 0. b1 (7.22) Note that in some cases in this book (e.g. in Tables 6.2 and 6.3), for convenience, the linear current density is defined as an RMS value of the fundamental component of the linear current density. In some cases, a step function Au = z Q I /τs is also used. The amplitude of the fundamental component of the current linkage is nevertheless the same. We may also assume that the slot opening width is infinitesimal. In such a case, by integrating the linear current density values we obtain a stepped current linkage function Θ(α) varying as a function of the periphery electrical angle α. The current linkage waveform has an electrical phase shift of π/2 compared with the fundamental linear current density, Figure 7.8a. The linear current density distribution can be developed into a Fourier series. Each term ν in the series produces a corresponding flux density harmonic as a function of position. Figure 7.8b illustrates a linear current density A1 of this type on the inner surface of the stator. An axial current Aν r dα flows in an element of the equivalent rotor length l 0001 of the cylinder dIν = Aν r Ds dα = Aν dα. p 2p (7.23) Here dα/p is used, since the angle α obtains values from 0 to p2π as the rotor periphery is traversed. When the effective air gap δ e is notably smaller than the radius of the rotor (δef 0007 r ), the air-gap field strength Hν can be assumed constant in the direction of the radius r . According to the penetration law, when travelling along the path a–b–c–d–a of the length l, the rotor being currentless, we obtain Ds ∂ Hν dα δef = Aν dα. Hν dl = Hν δef − Hν + ∂α 2p (7.24) Here, we obtain the linear current density Aν = − 2 pδef ∂ Hν 2 pδef ∂ Bν 2 p ∂Θν =− =− . Ds ∂α µ0 Ds ∂α Ds ∂α (7.25) Ds is the stator bore diameter and p is the number of pole pairs. By substituting the differential of the current linkage (Equation 7.17) into this equation, we obtain a basic equation for the P1: OTA/XYZ P2: ABC c07 JWBK312-Pyrhonen November 25, 2008 16:47 Printer Name: Yet to Come 318 Design of Rotating Electrical Machines linear current density, which is phase shifted by π/2 with the current linkage ˆ ν ωt − να + π/2 , Aν = A (7.26) where ˆ ˆ ν = 2 pm kwsν Ns i. A π Ds ν p (7.27) The local slot currents are, hence, replaced with an imaginary, infinitely thin, linear current density. The linear current density creates a current linkage, which is phase shifted by 90 electrical degrees to the linear current density. The flux density distribution (7.18) created by the current linkage is, however, in the same phase as the current linkage distribution. A current flows in both the stator and rotor windings of an asynchronous machine operating under load. Both currents create a current linkage of their own, and therefore the current linkage exerted on half of a magnetic circuit is the sum of these two linkages Θm (α) = Θr (α) + Θs (α) . (7.28) This sum creates the real magnetic flux density in the air gap. When the machine is linearized for the calculations, also the local flux densities can be superposed. Now we obtain for the air-gap flux density Bδ (α) = Bs0001 (α) + Br0001 (α) . (7.29) The imaginary flux densities Bs0001 (α) and Br0001 (α) presented here cannot be measured, yet they can be calculated as follows: Br0001 (α) = µ0 Θr (α) ; δef Bs0001 (α) = µ0 Θs (α) . δef (7.30) Figure 7.9 illustrates a linear current density on the boundary of the rotor and air gap. At an arbitrary position α of the air gap, the differential current is written as dI (α) = A (α) r dα. p (7.31) The prevailing linear current density at the position α can be defined according to Equation (7.26). The air-gap flux density vector field Bδ and the rotor current I dl (where l is the unit vector in the rotor axis direction) are perpendicular to each other and, according to Lorentz force equation, they cause a peripheral force element dF parallel to the tangent to the cylinder dF (α) = l 0001 Bδ (α) dI (α) = Ds l 0001 A (α) Bδ (α) dα. 2p For simplicity, Ds ≈ Dr ≈ D ≈ 2r is used in the following. (7.32) P1: OTA/XYZ P2: ABC c07 JWBK312-Pyrhonen November 25, 2008 16:47 Printer Name: Yet to Come Design Process and Properties of Rotating Electrical Machines 319 Bδ dF Idl A dα l' α 0 r Figure 7.9 Definition of the torque T acting on the rotor. The torque is defined with a peripheral force dF exerted on the linear current density element I dl Since the peripheral forces are tangential everywhere, their vector sum around the rotor is zero, yet they can be employed in the calculation of the torque. The peripheral force of a machine is solved by line integrating dF around the surface of the rotor over an angle 2πp. Simultaneously, we obtain the electromagnetic torque of the machine by multiplying the obtained force by the diameter of the rotor radius (r ≈ D/2) Tem D 2l 0001 = 4p 00032π p A (α)Bδ (α) dα. (7.33) 0 This result is in line with the results found in Section 1.5 where the tangential stress was first studied based on the Lorentz force. The fundamental v = 1 of the linear current density of the rotor is solved by substituting the rotor current into Equations (7.26) and (7.27), and by taking into account the angular difference ζ r of the linear current density of the rotor with respect to the stator current linkage ˆ r ωt − α − ζr + π/2 , Ar = A (7.34) where ˆ r = 2 pm r kwr Nr iˆr = 2 m r kwr Nr iˆr . A π D p π D (7.35) P1: OTA/XYZ P2: ABC c07 JWBK312-Pyrhonen November 25, 2008 16:47 320 Printer Name: Yet to Come Design of Rotating Electrical Machines ζ r represents the phase angle of the rotor caused by the rotor impedance Rr + sjωs L rσ = Z r ζr . The real air-gap flux density follows Equation (7.18). When the elements are substituted in Equation (7.33), we may write the electromagnetic torque of the induction machine (see also Equation 6.2) as Tem = πDlσtan r = πDl ˆ Bˆ δ cos ζr π A ˆ Bˆ δ cos ζr . r = D 2l A 2 4 (7.36) ˆ and Bˆ δ are used. That is why In (7.36), the peak values of the fundamental distributions of A the result must be divided by two. Note also that cos ζ r corresponds to the rotor power factor. In Chapter 6, Equation (6.1), we have cos ϕ, which corresponds to the stator power factor. In Equation (1.115) instead, there is no angle involved, since instantaneous local values are used. In principle, Equation (7.36) is the same equation we discussed previously in the definition of tangential stress in Section 1.5. This is a general torque equation based on the Lorentz force and derived from the flux and current distributions of the machine. Formally, the equation is valid for all machine types. The torque acts upon both the rotor and the stator with equal magnitude but in opposite directions. When Equation (7.36) is repeated for stator quantities, it may be applied to any rotating-field machine type. For example, if a synchronous machine is operating with an overlapping linear current density and flux density distributions – which are found close to stator unity power factor – the electromagnetic torque will be ˆ s Bˆ δ /4. Tem = πDs2l A This equation shows that the torque of a machine is proportional to the volume of the rotor ˆ Bˆ δ . The maximum value that a machine can produce in continuous operaand the product A tion is determined by the temperature rise of the machine. The maximum of the fundamental component of the air-gap flux density is usually limited to the order of 1 T or slightly above. If the air-gap flux density curve is rectangular and the maximum flux density is 1 T, the fundamental peak value reaches 4/π. On the other hand, the variation in the linear current density A is large and depends on the cooling of the machine. 7.1.2 Impedance and Current Linkage of a Cage Winding Figure 7.10 illustrates a simplified cage winding (a) and phasors (b) of the bar and ring currents for the cage winding. The bars are numbered from 1 to Q r . The bar and ring currents are denoted respectively, the arrows indicating the positive directions. If the resistance of a single bar is Rbar (the DC value of the resistance of the bar is obtained by the area S of the slot of the rotor, the length l of the rotor core, the skew α skew and the specific conductivity σ of the conductor material Rbar = l/(cos αskew · σ S)) and the inductance is L bar , and finally Rring and L ring are the respective values of the ring section between two bars, then the resultant impedance of the complete winding can be calculated as follows. A current phasor diagram for the rotor bars is constructed for the harmonic ν in such a way that the angular phase shift of the bar currents is ναu p. As a squirrel cage rotor can operate with different stator pole pair numbers p, we use α u here as a mechanical angle, so then the corresponding electrical angle is pαu (cf. Figure 7.10). Next, a polygon is constructed for P1: OTA/XYZ P2: ABC c07 JWBK312-Pyrhonen November 25, 2008 16:47 Printer Name: Yet to Come Design Process and Properties of Rotating Electrical Machines 4 3 Iring2 αu Iring1 Iring, x-1 x ναup 2 Ibar2 Ibar1 321 Ibar2 Ibar1 IbarQr Qr Iring1 IringQr ναup Qr - 1 Ibarx bar currents 1 Iring2 Ibar3 Iring3 Iring, x Ibar4 ring currents (a) (b) Figure 7.10 (a) Diagram for a cage winding; (b) a sector of a polygon of the phasors of the bar currents and a section of the current phasor diagram. Here α u is given in mechanical degrees, and the corresponding electrical degrees are given as pα u . Note that in reality the currents in the opposite sides of the cage have different signs the bar currents. A single sector of this polygon is illustrated by the phasors I bar1 − I bar4 in Figure 7.10. The phasors from I ringQr to I ring3 drawn from the centre of the polygon to the angle points are the phasors of the ring currents. They follow Kirchhoff’s first law at each connection point of the bar and the ring – see Figure 7.10a: I ring,x = I bar,x + I ring,x−1 . (7.37) The mutual phase angle between the ring currents is also ναu p. Since the bar alone comprises the phase winding of the rotor (Nr = 1/2), the bar current is the phase current of the rotor. Now, the RMS value of the bar current induced by the νth flux density harmonic is generally denoted by Ibar ν . Correspondingly, the RMS value of the ring current is now denoted by Iring ν . Thus, based on Figure 7.10b, we obtain Iringν = Ibarν αuν ; 2 sin 2 The currents create a resistive loss in the rotor 000e αuν = ν 2π p . Qr 2 2 2 = Q r Ibarν PCuν = Q r Rbarν Ibarν + 2Rring Iringν Rbar + (7.38) Rring αuν , 2 sin2 2 (7.39) where Rring is the resistance of that part of the ring belonging to one bar. The resistance of a single rotor phase is higher than the mere bar resistance Rbar ν by an amount equal to the term of the element in the large brackets. The phase inductance is calculated by the same principle. When we take into account the electric angular frequency supplying the machine, we obtain the equation for the phase impedance of the rotor of an induction machine for the air-gap flux harmonic ν for a locked rotor, s = 1 (7.40) Z rν = Rrν + jωs L rν = Z rν ζrν , P1: OTA/XYZ P2: ABC c07 JWBK312-Pyrhonen November 25, 2008 16:47 Printer Name: Yet to Come 322 Design of Rotating Electrical Machines Note that the angular frequency ωs of the stator is employed in the calculation of the impedance. The effect of slip will be taken into account later. Now we can write for the resistance and leakage inductance of the rotor Rrν = Rbar + Rring νπ p ; 2 sin2 Qr L rν = L bar + L ring νπ p . 2 sin2 Qr (7.41) In Equations (7.40) and (7.41), L bar and L ring are the leakage inductances of the bar and ring sections. The leakage inductance of the end ring can be solved simply with Equation (4.67), which replaces the latter part of Equation (7.41). When analysing the equations, we note that the value of the phase impedance of the rotor is a function of the ordinal ν of the inducing flux density harmonic. A cage rotor reacts only to such flux density harmonics created by the stator, the ordinals ν of which meet the condition νp = cQ r , where c = 0, ±1 ± 2, ±3, . . . . (7.42) This explains the fact that for the harmonics, the ordinal of which is ν0001 = cQ r , p (7.43) the impedance of the rotor phase is infinite. The pitch factor kpν of these harmonics is zero. The wavelength of the harmonic ν 0001 is equal to the slot pitch of the rotor, or an integral part of it. In that case, each bar always has a flux density of equal magnitude. This creates an equal induced current linkage in each bar, and the emfs of a closed electric circuit compensate each other, and thus the voltages induced by the harmonic ν 0001 do not generate currents. According to Equations (7.40) and (7.41), a cage winding can be replaced by such an equivalent cage, the impedance of the short-circuit rings of which is zero and the impedance of the bars is Z¯ rν . The impedance of the cage winding or the resistance and inductance are usually transferred to the stator side. The case is analysed next. The currents in the bars of a single pole pitch of a cage winding are all of different phases. In a symmetrical m-phase system, the angle between the phases is 360◦ /m. Thus, there are as many phases in the rotor as the number of rotor bars. If the number of bars is Q r in the rotor, the phase number of the rotor is accordingly mr = Qr. (7.44) Generally, there has to be at least two conductors in a coil turn at a distance of about 180◦ from each other. We may thus consider that a single rotor bar comprises half a turn, and we can write Nr = 1/2. The number of effective coil turns in a stator is m s kw1s Ns and in the rotor m r kW1r Nr . Here kW1r = 1 and Nr = 1/2. If the slots of the stator and rotor are skewed with respect to each other, we also have to take the skewing factor ksq into account. When considering the current linkages of a harmonic ν, the rotor current Iν0001 r referred to the stator and flowing in the stator winding has to produce an equal current linkage to the original rotor P1: OTA/XYZ P2: ABC c07 JWBK312-Pyrhonen November 25, 2008 16:47 Printer Name: Yet to Come Design Process and Properties of Rotating Electrical Machines 323 current when flowing in the rotor. Thus we may write m s Ns kwνs Iν0001 r = m r Nr ksqνr kwνr Iν r . (7.45) The quantities with the prime symbol are referred to the stator from the rotor. The transformation ratio for the harmonic ν from the rotor to the stator therefore becomes Iν r m s kwνs Ns = . 0001 Iν r m r kwνr ksqνr Nr K rs,ν = (7.46) By applying the above to the cage winding and the fundamental component, we obtain K rs,1 = m s kw1s Ns m s kw1s Ns 2m s kw1s Ns = = . m r kw1r ksq1r Nr Q r · 1 · ksq1r · 1/2 Q r ksq1r (7.47) If Rr is the resistance of the rotor bar added to the proportion of the short-circuit rings, and Ir is the RMS value of the current of the rotor bar, we obtain, by writing the resistive loss of the rotor equal in both the stator and the rotor, 0001 m s Ir 2 Rr0001 = Q r Ir2 Rr . (7.48) The phase resistance of the rotor referred to the stator can now be written as Q r Ir2 Rr . m s Ir0001 2 Rr0001 = (7.49) Since Ir = K rs , Ir0001 (7.50) the resistance of the rotor referred to the stator now becomes Rr0001 Qr = ms 0015 Ir Ir0001 00162 Qr 2 Qr Rr = K rs,1 Rr = ms ms 0015 2m s kw1s Ns Q r ksq1r 00162 Rr = 4m s (kw1s Ns )2 Rr . 2 Q r ksq1r (7.51) When it is necessary to refer the rotor resistance to the stator, in general, it has to be multiplied by the term ms ρν = mr
Ns kwνs Nr ksqνr kwνr 2 . (7.52) In a squirrel cage induction motor, this is written in the form ρν = 4m s Qr
Ns kwνs ksqνr 2 . (7.53) P1: OTA/XYZ P2: ABC c07 JWBK312-Pyrhonen November 25, 2008 16:47 Printer Name: Yet to Come 324 Design of Rotating Electrical Machines If there is no skewing, we write further ρν = 4m s (Ns kwνs )2 . Qr (7.54) The same referring factor is valid also in the referring of the inductances. It cannot, however, be obtained based on equal stator and rotor losses (I 2 R), but on equal energy stored in the inductance (1/2 L I 2 ). Hence we get Rν0001 r = ρν Rν r ; L 0001ν r = ρν L νr . (7.55) Here it is worth noting that in a rotating electrical machine, the referring deviates from the referring in a transformer by the fact that the impedance quantities are not referred directly with the square of the transformation ratio of the current, but we also have to take into account the ratio of the numbers of phases. With low values of slip, the resistance Rr can be given as a DC value, but for instance at start-up, the rotor frequency is so high that the skin effect in the squirrel cage has to be taken into account. Also, in rapid transients, the rotor resistance deviates notably from its DC value. In a cage winding, there are no coils, and there can be an odd number of bars in the winding. Therefore, the definition of the current linkage is not quite as straightforward as was the case with the coil windings discussed previously. First, a current linkage of a single bar is defined, and next the current linkages of all the bars are summed. The analysis can be carried out in a reference frame attached to the rotor, since we are now analysing phenomena that occur only between the resulting air-gap flux density harmonic ν and the rotor cage. The cage winding itself does not form poles, but the number of pole pairs always settles to be the same as the stator harmonic influencing it. In the analysis of a cage winding, geometrical angles ϑ are employed as here. Now the electric angles are at the fundamental pϑ and at harmonics pνϑ. Figure 7.10 is investigated at the instant t = 0, when the peak Bˆ δν of the air-gap flux density of the harmonic ν occurs at the first bar. In a bar that is at an arbitrary position angle ϑx = x pϑ, an emf is induced with a slip sν of a certain harmonic eνx (t) = eˆ νx cos (sν ωs t − xν pϑ) . (7.56) This harmonic emf is found by calculating the flux and its time derivative and is written with the absolute value and the phase angle as a complex number eνx (t) = sν ωs πDrl ˆ Bδν sν ωs t − xν pϑ . ν 2p (7.57) The bar current i νx (t) is determined by dividing the emf by the equivalent bar impedance Z rν (s). When the case is investigated in the rotor coordinate system, the slip becomes important with respect to the impedance. The imaginary part of the impedance changes as a function of the slip angular frequency, the phase angle of the rotor impedance for the νth harmonic being ζ rν (sν ). Correspondingly, the angle of the current depends on the slip as sν ωs t i νx (t) = sν ωs πDl ˆ Bδν sν ωs t − xν pϑ − ζrν (sν ) . 2 pν Z rν (7.58) P1: OTA/XYZ P2: ABC c07 JWBK312-Pyrhonen November 25, 2008 16:47 Printer Name: Yet to Come Design Process and Properties of Rotating Electrical Machines Θbar1 ir ν 0 −π 325 Θr1 ϑx 2 (a) +π i rν 0 +π Θbar2 −π ϑx 0 (b) ir ν 0 Θbar1 + Θbar2 −π 0 +π ϑx (c) Figure 7.11 (a), (b) Current linkage created by a single rotor bar at ϑ x = 0 and ϑ x = π. (c) The current linkage created by two bars together The amplitude of this current corresponds to the rotor bar current needed in (7.60). When considering a current linkage of a single bar, we have to find a suitable graph for the current linkage created by the current of the bar. With intuitive inference, we may construct a sawtooth wave as illustrated in Figure 7.11a. Normally, a current loop is always required to produce a current linkage, as discussed with Figure 2.19. In that figure, a single current penetrates the system at two points, thus creating a closed loop. This kind of a loop is not unambiguously created for a cage winding, since in a cage winding the number of bars is not necessarily even. The current of a certain bar under investigation may be divided between several bars on the other side of the pole pitch. Therefore, the current linkage created by a single bar has to be observed in the analysis. The intuitive deduction can be complemented by investigating Figures 7.11a–c. Now a second, separate bar is mounted at a point ϑ x = π (i.e. at a distance of a pole pitch from the first bar). The current of this bar is opposite to the current of the bar in the position ϑ x = 0. As a result, the graph in Figure 7.11b that is opposite to the current linkage graph of 7.11a is created, and half of the graph is shifted a full pole pitch left in the case of two poles. By combining the graphs in Figures 7.11a and b, we obtain a familiar graph for the current linkage of a single loop that corresponds to Figure 2.19. Therefore, with intuitive deduction, we may draw the graph of a single bar that corresponds to Figure 7.11a. The figure also illustrates the fundamental Θ 1r of the current linkage of one bar Θ bar (ϑ x ). We may now write the current linkage for the bar at x = 0. Its current linkage is thus Θbar, 0 = iˆν0 (π − ϑx ) , 2 π ϑx [0, 2π] . (7.59) P1: OTA/XYZ P2: ABC c07 JWBK312-Pyrhonen November 25, 2008 16:47 Printer Name: Yet to Come 326 Design of Rotating Electrical Machines The current linkage of an arbitrary bar at x is obtained by substituting the current of the original bar i ν0 with the current i νx , Equation (7.58), iˆrν (π − ϑx + νx pϑ) , 2 π Θbar, x = ϑx [0, 2π] . (7.60) The change in sign results from the fact that the temporal phase shift −νx pϑ corresponds to the local position angle +νx pϑ. The function is continuous only in the range 0–2π. When the function Θ bar,x of the current linkage of the rotor is developed into Fourier series, its term ν r becomes iˆrν iˆrν j(sν ωs t−νr ϑ+βrν −(νp−νr )xϑ) e , sν ωs t − νr ϑ + βrν − (νp − νr ) xϑ = 2πνr 2πνr where βrν = − π2 − ζrν (sν ) . Θνxνr = (7.61) Only the last term of the exponent of the constant e (Napier’s constant, also known as Euler’s number) depends on the ordinal number x. The sum of the Fourier series taking all the bars into account is Qr x=0 e−j(νp−νr )xϑ = 1 − e−j(νp−νr )Q r ϑ . 1 − e−j(νp−νr )ϑ (7.62) Since (νp − ν r ) is an integer and Qr ϑ = 2π, the numerator is always zero. The sum obtains values other than zero only when νp − νr = cQ r , (7.63) where c = 0, ±1, ±2, ±3, . . . The harmonic ν of the resultant flux density of the air gap can create rotor harmonics ν r that meet the condition (7.63). Equation (7.62) then takes the limit value Q r . The ordinals ν and νr can be either positive or negative. The cage winding thus creates current linkages ˆ νr sν ωs t − νr ϑ + βrν , Θνr = Θ (7.64) where ˆ νr = Q r iˆrν = sν ωs Dl Bˆ δν . Θ 2πνr 4πνpνr Z rν (7.65) The corresponding linear current density of the rotor is ˆ νr sν ωs t − νr ϑ − ζrν , Aνr = A (7.66) P1: OTA/XYZ P2: ABC c07 JWBK312-Pyrhonen November 25, 2008 16:47 Printer Name: Yet to Come Design Process and Properties of Rotating Electrical Machines 327 where ˆ νr = Q r iˆrν . A πD (7.67) ζ rν is the phase angle of the rotor impedance for the harmonic ν. 7.1.3 Characteristics of an Induction Machine The characteristics of an induction motor greatly depend on the fulfilment of Equations (7.42) and (7.63). When the number of bars Q r is finite, c = 0 always meets the condition (7.42). Each flux-density-producing air-gap harmonic ν that meets the condition (7.42) can induce a vast number of rotor current linkage harmonics ν r (7.61). Now, the equivalent circuit of an asynchronous motor per phase, the quantities of which are calculated in the machine design, is worth recollecting. Figure 7.12 illustrates a single-phase equivalent circuit of an ordinary induction motor per phase, a simplified equivalent circuit and a phasor diagram. In Figure 7.12, the stator is supplied with the voltage Us . The stator resistance Rs is the resistance of the stator winding at the operating frequency and temperature, Us0001 is the stator Us Is Rs Ps Lsσ Rs Us PsCu Ψs U 's RFe PFe Lr'σ Ψm Us RFe I Fe Rs Um R'r Um PrCu Em Lm Ψ ' r I Fe Is I′r Pδ Is Us ' (a) I′r Lm L' Ψs R'r(1-s)/s Im −I r 'Rr ' R'r R'r(1-s)/s Is −Ir' Pmec, em Im L jωLsσIs (1−s) Ψr ' −Ir' s −Ir ' Rr' 1 Er '−I r ' Rr ' s Ψm θr Er Em Lsσ I s Lrσ' Ir ' Es =−Us ' −jωLs′σ I′r Im (b) (c) Figure 7.12 (a) Steady-state equivalent circuit of an asynchronous machine per phase. (b) A simplified equivalent circuit of the machine, the parameters of which are calculated in the machine design. (c) A phasor diagram of an asynchronous machine. The input stator power is Ps . In the stator resistance, a resistive loss PsCu takes place. The iron loss is the loss PFe created in the magnetic circuit. The air-gap power Pδ flows across the air gap to the rotor. In the rotor, some resistive losses PrCu take place. The power Pmec,em is the electromechanical power of the machine. When friction and windage losses are subtracted, the output power P of the machine can be found P1: OTA/XYZ P2: ABC c07 JWBK312-Pyrhonen November 25, 2008 16:47 Printer Name: Yet to Come 328 Design of Rotating Electrical Machines voltage after the resistive voltage drop is subtracted, L sσ is the leakage inductance of the stator, L m is the magnetizing inductance of the machine at the rated point, RFe is the resistance describing the losses of the iron circuit of the machine, L 0001rσ is the rotor leakage inductance of the machine referred to the stator, and Rr0001 is the rotor resistance referred to the stator. s is the slip of the rotor. The term Rr0001 (1 − s)/s describes the electromechanical power produced by the machine. A part of the mechanical power is consumed in the friction and windage losses of the machine. Ψ s is the stator flux linkage, which includes the air-gap flux linkage Ψ m and the stator leakage flux Ψ sσ . Correspondingly, Ψ r is the flux linkage of the rotor, which includes the air-gap flux linkage Ψ m and the rotor leakage flux Ψ rσ . The stator voltage Us0001 creates the stator flux linkage, from which, in turn, a back emf E s is derived. A voltage E m is induced over L m by the air-gap flux linkage Ψ m . This voltage is consumed completely in the rotor apparent resistance Rr0001 (1/s) and in the rotor leakage reactance. The iron loss current IFe is small and is therefore not illustrated in the phasor diagram. Stator power Ps is fed to the motor. Some power is consumed in the stator resistance and in the iron loss resistance. An air-gap power Pδ crosses the air gap. In the rotor, a part of the air-gap power is lost in the rotor resistance Rr0001 and a part is converted into mechanical power in Rr0001 (1 − s)/s. The fundamental flux density created by the stator and rotor currents causes a varying flux, which in turn induces a voltage and a current acting against the flux variation. At no load, the rotor rotates approximately with a synchronous speed, and the rotor frequency and the currents approach zero. If the rotor is loaded, its speed decreases and the relative speed with respect to the fundamental flux propagating in the air gap increases. Now the emf induced in the rotor increases. Also the inductive reactance of the rotor increases as the rotor frequency increases. The peripheral force created by the fundamental flux density of the air gap and the rotor torque reaches its maximum at a certain slip. If required, the rotor of an asynchronous machine can be driven by an above synchronous speed. In that case, the rotor currents create a torque opposing the accelerating torque and the machine acts as a generator, the slip being negative. Let us assume that the fundamental ν = 1 of the resultant air-gap flux density follows Equation (7.18). As the rotor rotates with a slip s with respect to the fundamental air-gap wave, an emf is induced in the phase winding of the rotor. The peak value of the emf induced in the rotor bars depends on the slip eˆ r (s) = s eˆ rk sωs t − π/2 . (7.68) In this equation, there is a peak value of the emf induced in the rotor at slip s = 1 eˆ rk = ωs Ψˆ r = ωs 2 πD ˆ Bδlkwr Nr . π 2p (7.69) The impedance of the rotor circuit depends on the slip angular frequency Z r (s) = Rr + jsωs L rσ = Z r (s) ζr (s) . (7.70) P1: OTA/XYZ P2: ABC c07 JWBK312-Pyrhonen November 25, 2008 16:47 Printer Name: Yet to Come Design Process and Properties of Rotating Electrical Machines 329 The peak value for the rotor current phasor now becomes s eˆ rk (s) ˆ iˆr (s) = = i r (s) sωs t − ζr (s) − π/2 . Z r (s) (7.71) s eˆ rk s eˆ rk iˆr (s) = =0017 . 2 Z r (s) Rr + s 2 (ωs L rσ )2 (7.72) The amplitude is In the case of a slip-ring machine, additional impedances can be attached to the rotor circuit. Now the total impedance of the circuit has to be substituted into the equation. The current iˆr (s) represents the linear current density of the rotor surface, the amplitude of which can be calculated from the induced voltage and the rotor impedance Nr 2 p ˆ r (s) = 2 p sωs 0017 Bˆ δ . A π Rr2 + s 2 (ωs L rσ )2 2 m rlkwr (7.73) When the above is substituted in the general torque Equation (7.36), and noting that cos ζr (s) = 0017 Rr Rr2 + s 2 (ωs L rσ )2 , (7.74) which corresponds to the power factor of the rotor impedance, we obtain the electromagnetic torque Tem (s) = s Rr pm r E2 . ωs Rr2 + s 2 (ωs L rσ )2 rk (7.75) Here E rk is the RMS value √ of the emf of the phase winding of the rotor when the machine is held at stall, E rk = eˆ rk / 2. Also, the power distribution leads to the same equation. Thus, the RMS value of the current in the rotor circuit of the equivalent circuit with slip s is Ir (s) = 0018 s E rk Rr2 + (sωs L rσ )2 = 0019 E rk . 2 Rr + (ωs L rσ )2 s (7.76a) Transferring Equation (7.76a) to the stator side we get Ir0001 (s) = 0019 0001 E rk . Rr0001 2 + (ωs L 0001rσ )2 s (7.76b) The rotor circuit of the equivalent circuit in Figure 7.12 follows this equation. In the figure, the rotor resistance has been divided into two parts, the sum of which is Rr /s. Correspondingly, P1: OTA/XYZ P2: ABC c07 JWBK312-Pyrhonen November 25, 2008 16:47 Printer Name: Yet to Come 330 Design of Rotating Electrical Machines the active power Pδ crossing the air gap is divided into the resistive loss PrCu of the rotor and the mechanical power Pmec,em Pδ = Rr0001 I 0001 r + 2 1 − s 0001 00012 R0001 0001 Rr I r = r Ir 2 = PrCu + Pmec, em , s s PrCu s = . Pmec, em 1−s (7.77) (7.78) In motor drive, the air-gap power Pδ is the power transmitted from the stator via the air gap to the rotor. Of this power, the proportion PrCu is consumed in the resistive losses of the rotor, and the rest is electromechanical power Pmec,em . The mechanical power Pmec is obtained from the shaft when friction and windage losses are subtracted from the electromechanical power. We can write Equations (7.77) and (7.78) in the form PrCu = s Pδ ; Pmec, em = (1 − s) Pδ p p p Pmec, em = Tem (s) = Pmec, em = PrCu = Pδ . Ω sωs ωs (1 − s) ωs (7.79) The torque can thus be solved with the resistive loss power of the rotor. The torque is always (including when the machine is at stall) proportional to the air-gap power Pδ : Tem = 2 2 pm r E rk m r E rk Rr /s s Rr m r E rk Ir cos ϕr Pδ = , = = 2 ωs / p ωs / p ωs / p Rr ωs Rr2 + (sωs L rσ )2 + (ωs L rσ )2 s (7.80a) Tem ≈ m s Us2 ωs / p Rr0001 /s . Rr0001 2 2 Rs + + (ωs L k ) s (7.80b) Equation (7.80b) above is obtained from the simplified equivalent circuit of Figure 7.12b by substituting Ir = s E rk /Z r and cos ϕr = Rr /Z r and, based on the simplified equivalent circuit, by assuming that the air-gap voltage is equal to the terminal voltage of the machine. Furthermore, we employ the short-circuit inductance L k ≈ L sσ + L 0001rσ . Neglecting the effects of Rs , the highest value of Tem , the pull-out torque Tb can be found from the slip sb = ± Rr0001 Rr0001 = , 0001 ωs L sσ + ωs L r σ ωs L k (7.81) Tb = ± 3 p Us2 . 2ωs2 L k (7.82) Here we remember that L k ≈ L sσ + L 0001rσ ≈ 2L 0001rσ . We see that the peak torque is inversely proportional to the short-circuit inductance of the machine. If the per unit value of L k is 0.2 for instance, the maximum torque is about 5Tn . Contrary to the slip value of the peak torque, P1: OTA/XYZ P2: ABC c07 JWBK312-Pyrhonen November 25, 2008 16:47 Printer Name: Yet to Come Design Process and Properties of Rotating Electrical Machines 331 the maximum torque is independent of the rotor resistance. By substituting the peak value of the slip in Equation (7.74) we obtain cos ζr (s) = 0019 s sb 1 2 . (7.83) +1 √ The term takes the value cos ζr (s) = 1/ 2 at peak torque slip. When the peak value of the resultant flux density Bˆ δ is kept constant, according to Equations (7.73), (7.74), (7.79), (7.80), (7.81) and (7.82), 2 T ˆ r (s) cos ζr (s) = C A s sb = C Tb , + sb s (7.84) where C is the machine constant. The calculation of torque introduced above can thus be simplified with the equivalent circuit to the definition of the power of the resistance. With the single-phase equivalent circuit, the load variations can chiefly be observed in the changes in the slip. Somewhat more accurate results can be obtained for an asynchronous machine by again employing a simplified equivalent circuit, but applying a reduced voltage in the calculation of the rotor current of the machine L sσ Us 1 − Lm Ir0001 = 00180001 00022 . 00022 0001 Rs + Rr0001 /s + ωs L sσ + ωs L 0001rσ (7.85a) Now, the generated electromechanical torque is Tem = ωs p 0016 0015 L sσ 2 Rr0001 3 Us 1 − Lm s 001a0001 00022 001b . 00022 0001 0001 Rs + Rr /s + ωs L sσ + ωs L 0001rσ (7.85b) The starting torque produced by the fundamental is obtained by setting s = 1 in Equation (7.80b). However, the rotor resistance is higher at large slips because of the skin effect, and the resistance must be defined at every slip value before calculating the torque. The pull-out torque is solved by derivation with respect to Rr0001 /s, after which we can write for the slip of the maximum torque, if Rs is taken into account, Rr0001 sb = ± 0018 00022 , 0001 2 0001 (Rs ) + ωs L sσ + ωs L rσ (7.85c) P1: OTA/XYZ P2: ABC c07 JWBK312-Pyrhonen November 25, 2008 16:47 Printer Name: Yet to Come 332 Design of Rotating Electrical Machines where the plus sign is for motoring and the minus sign for generating operation. The corresponding torque for motoring is 0016 0015 L sσ 2 3 Us 1 − Lm 0016, 0015 Tb = 0018 00022 0001 ωs Rs + Rs2 + ωs L sσ + ωs L 0001rσ 2 p (7.85d) 0015 0016 L sσ 2 3 Us 1 + Lm 0015 0016. Tb = − 0018 00022 0001 ωs Rs − Rs2 + ωs L sσ + ωs L 0001rσ 2 p (7.85e) and for generating 7.1.4 Equivalent Circuit Taking Asynchronous Torques and Harmonics into Account The parameters of a cage winding are investigated next. When the number of bars Q r of the rotor is finite and the factor c = 0, the condition (7.63) is always met. Each resulting air-gap flux density harmonic ν that meets the condition (7.42) can induce a large number of rotor harmonics ν r (7.61). In the case where c = 0, we investigate the harmonic ν r , the number of pole pairs of which is the same as the inducing air-gap harmonic ν. The torque equation for the pair of harmonics ν and ν r is derived in the same way as for the torque of the fundamental (7.75). First we obtain two equations Tν (sν ) = Tν (sν ) = D 2l 2 Rrν sν ωs Qr 2 Bˆ 2 , 8νp Rrν + sν2 ωs2 L 2rσν mν (7.86a) sν Rrν νp Qr 2 E2 . ωs Rrν + sν2 ωs2 L 2rσν rkν (7.86b) Here E rkν is the RMS value of the emf induced by the harmonic ν at the slip sν = 1. We may now infer that each harmonic meeting the conditions behaves accordingly. Torque is a continuous function of slip that becomes zero at the slip sν = 0, hence the term ‘synchronous torque’. When the machine is running at a fundamental slip s1 , the slip of the rotor with respect to the νth stator harmonic is written as sν = 1 − ν (1 − s1 ). (7.87) The angular frequency of the νth harmonic in the rotor is thus ωνr = ωs (1 − ν (1 − s1 )) . (7.88) P1: OTA/XYZ P2: ABC c07 JWBK312-Pyrhonen November 25, 2008 16:47 Printer Name: Yet to Come Design Pr