Fundamentals of Electrical Drivess - Chapter 7 pps

Chapter 7 INTRODUCTION TO ELECTRICAL MACHINES 7.1 Introduction This chapter considers the basic working principles of the so-called ‘clas- sical’ set of machines. This set of machines represents the asynchronous (in- duction), synchronous, DC machines, and variable reluctance machines. The latter will be discussed in the book ‘Advanced Electrical Drives’ currently un- der development by the authors of this book. Of these classical machines, the asynchronous machine is most widely used in a large range of applications. Note that the term ‘machine’ is used here, which means that the unit is able to operate as a motor (converting electrical power into mechanical power) or as a generator (converting mechanical power into electrical power). The machine can be fed via a power electronic converter or connected directly to an AC or DC supply. Central to this chapter is the development of an ‘ideal rotating transformer’, which is in fact a logical extension of the two-phase ITF module discussed in chapter 6. We will then look to the conditions required for producing constant torque in an electrical machine. This in turn will allow us to derive the principle of operation for the three classical machine types. A general model concept will be introduced at the end of this chapter which forms the backbone of the machine models discussed in this book. 7.2 Ideal Rotating Transformer (IRTF) concept The fundamental building block for rotating machines used in this book is the IRTFmodule which isdirectly basedon thework by A.Veltman[Veltman, 1994] which is directly derived from the two-phase space vector ITF concept given in figure 6.1. The new IRTF module, shown in figure 7.1, differs in two points. Firstly, the inner (secondary) part of the transformer is assumed to be able to 170 FUNDAMENTALS OF ELECTRICAL DRIVES rotate freely with respect to the outer (primary) side. The airgap between the two components of this model remains infinitely small. Secondly, the number of ‘effective’ turns on the primary and secondary winding are assumed to be equal, i.e. n 1 = n 2 = n. Furthermore, the windings are (like the ITF) taken to be sinusoidally distributed (see appendix A). This implies that the winding representation as shown in figure 7.1 is only symbolical as it shows where the majority of conductors for each phase are located. In the future the primary and secondary will be referred to as the stator and rotor respectively. A second Figure 7.1. Two-phase IRTF model complex plane (in addition to the stator based complex plane) with axis  xy ,  xy is introduced in figure 7.1 which is tied to the rotor. Note the use of the superscript xy which indicates that a vector is represented in rotor coordinates. For a stationary coordinate system, as used on the stator side, we sometimes use the superscript αβ. However, in most cases this superscript is omitted to simplify the mathematical expressions. Hence, no superscript implies a stationary coordinate based vector. The angle between the stationary and rotating complex plane is given as θ and this is in fact the machine shaft angle of rotation (relative to the stationary part of the motor). If the angle of rotation θ is set to zero then the IRTF module is reduced to the two-phase ITF concept (with n 1 = n 2 ) as given by figure 6.1. The symbolic representation of the IRTF module as given in figure 7.2 shows similarity with the ITF module (see figure 6.2(a)). The IRTF is a three-port unit (stator circuit, rotor circuit and machine shaft). In figure 7.2 a symbolic shaft (shown in ‘red’) is introduced which is physically Introduction to Electrical Machines 171 Figure 7.2. Symbolic IRTF representation connected to the rotor circuit and appears via the stator circuit on the outside of the machine. The flux linked with the stator and rotor is equal to  ψ m = n  φ m and can be expressed in terms of the components seen by each winding namely  ψ m = ψ mα + jψ mβ (7.1a)  ψ xy m = ψ mx + jψ my (7.1b) An illustration of the flux-linkage seen by the rotor and stator winding is given in figure 7.3(a). The relationship between the stator and rotor oriented flux- (a) Flux-linkage space vector (b) Current space vector Figure 7.3. Flux-linkage and current space vector diagrams linkage space vectors is given as  ψ xy m =  ψ m e −jθ (7.2) The relationship between the rotor and stator oriented current space vectors as shown in figure 7.3(b) can be written as  i =  i xy e jθ (7.3) Figure 7.3 emphasizes the fact that there is only one single flux-linkage and one current space vector present in the IRTF. The components of these vectors can be projected onto a rotating or stationary complex reference frame. The rela- tionship between for example the variables i α , i β and i x , i y , θ of equation (7.3) 172 FUNDAMENTALS OF ELECTRICAL DRIVES can also be written as  i α i β  =  cos θ −sin θ sin θ cos θ  i x i y  (7.4a) The symbolic diagram of figure 7.2 can also be shown in terms of its space vector components as was done for the ITF case (see figure 6.3). The result, given in figure 7.4, shows that the rotor side of the IRTF now rotates with the rotor shaft given that it is physically attached to it. The energy balance for the Figure 7.4. Three-dimen- sional IRTF representation IRTF module is found by making use of the power expressions (5.30), (5.33) which are directly linked with the incremental energy dW = pdt. The input (stator) power p in for the IRTF module is of the form p in =   u   i  ∗  (7.5) Note that u and  i are taken to be time dependent complex numbers as indicated by equation (5.8) on page 123. Equation (7.5) can with the aid of u = d  ψ m dt be expressed in terms of the stator (input) incremental energy dW in =   d  ψ m   i  ∗  (7.6) Similar to the ITF, the IRTF is defined with a positive electrical and positive (rotor) electrical ‘power out’ convention (see figure 3.4). Unlike the ITF, the IRTF has a second output power component formed by the product of the shaft torque T e (Nm) and shaft speed ω m (rad/sec). If this product is positive the machine is said to operate as a motor. The total (mechanical plus electrical) output power is now of the form p out =   u xy   i xy  ∗  + T e ω m (7.7) The incremental mechanical and electrical output energy linked with the rotor side of the IRTF can with the aid of u xy = d  ψ xy m dt be written as dW out =   d  ψ xy m   i xy  ∗  + T e dθ (7.8) Introduction to Electrical Machines 173 The overall IRTF incremental energy balance can with the aid of equations (7.6) and (7.8) and use of the energy conservation law be written as   d  ψ m   i  ∗  −  d  ψ xy m   i xy  ∗  = T e dθ (7.9) The left hand side of equation (7.9) shows the electrical incremental energy IRTF components. In equation (7.9), the term d  ψ xy m may be developed further using equation (7.2) and the differential ‘chain rule’, which leads to d  ψ xy m = e −jθ d  ψ m − j  ψ m e −jθ dθ (7.10) Multiplication of equation (7.10) by the vector   i xy  ∗ =  i ∗ e jθ (see equa- tion (7.3)) gives d  ψ xy m   i xy  ∗ = d  ψ m  i ∗ − j  ψ m  i ∗ dθ (7.11) Substitution of equation (7.11) into equation (7.9) leads to the following ex- pression for the electromechanical torque on the rotor. T e =   j  ψ m  i ∗  (7.12) Equation (7.12) can with the aid of expression   ja  b ∗  =   a ∗  b  be rewrit- ten as T e =    ψ ∗ m  i  (7.13) Hence, the torque acting on the rotor is at its maximum value in case the two vectors  ψ m ,  i asshown in figure 7.3, are perpendicularwith respect toeach other. Under these circumstances the torque is directly related to the product of the rotor radius and the Lorentz force. The latter is proportional to the magnitudes of the flux and current vectors. The generic diagram of the IRTF module that corresponds to the symbolic representation shown in figure 7.2 is based on the use of equations (7.2), (7.3) and (7.13). The IRTF generic module as given in figure 7.5(a) is shown with a stator to rotor coordinate flux conversion module and rotor to stator current conversion module. The two coordinate conversion modules can also be reversed as shown in figure 7.5(b). The IRTF version used is application dependent as will become apparent at a later stage. The torque computation is not affected by the version used. Nor for that matter is the torque affected by the choice of coordinate system. The rotor angle θ required for the IRTF module must be derived from the mechanical equation-set of the machine which is of the form T e − T l = J dω m dt (7.14a) ω m = dθ dt (7.14b) 174 FUNDAMENTALS OF ELECTRICAL DRIVES (a) IRTF-flux (b) IRTF-current Figure 7.5. Generic representations of IRTF module with T l and J representing the load torque and inertia of the rotor/load combi- nation respectively (as discussed in section 1.4.2). Finally, it is noted that the IRTF has a unity winding ratio, which implies that an inductance component (unlike a resistive component) may be moved from one side to the other without having to change its value. 7.2.1 IRTF example The IRTF module forms the backbone to the electrical machine concepts presented in this book. Consequently, it is particularly important to fully un- derstand this concept. In the example given here we will discuss how stator currents and torque can be produced in the event that stator windings are con- nected to a voltage source and the rotor windings to a current source as shown in figure 7.6). In the discussion to come we will make use of figure 7.1 in a stylized form for didactic reasons. Furthermore, we will assume that we can hold the motor shaft at any desired position. The voltage source shown in fig- Figure 7.6. IRTF module connected to voltage and current source Introduction to Electrical Machines 175 Figure 7.7. Voltage excita- tion and flux-linkage for the α winding ure 7.6 delivers a pulse to the α winding at t = t o as shown in figure 7.7. The flux-linkage ψ mα versus time waveform which corresponds with the applied pulse is also shown in figure 7.7. The β stator winding is short circuited with condition ψ mβ =0. We will consider events after t = t o + T in which case a flux distribution will be present in the IRTF where the majority of the flux is concentrated along the  α,β axis as shown in figure 7.8(a). Furthermore, the flux-linkage value will be equal to ψ mα = ˆ ψ m . The corresponding space vector representation will be of the form  ψ m = ˆ ψ m . We wouldlike torealize arotor excitationof theform  i xy = j ˆ i, which implies that they rotor winding must carry a current i y = ˆ i. We have omittedfor didactic reasons the x winding from figure 7.8 because this winding is not in use (open circuited). We will now examine the IRTF model and corresponding space vector diagrams for the excitation conditions indicated above and three rotor positions. For each case (rotor postion) we will assume that the rotor is initially set to the required rotor position after which the stator and rotor excitation as discussed above is applied. The aim is to provide some understanding with respect to the currents which will occur on the stator side and the nature of torque production based on first principles. To assist us with this discussion two ‘contours’ namely x and y are introduced in figure 7.8 which are linked to the rotating complex plane. These contours are helpful in determining the currents which mustappear onthe statorside. If weassume thatsuch acontour represents a flux tubethen therewouldneed tobe a correspondingresultant MMFwithin the contour in case the latter would contain some form of magnetic reluctance. The magnetic reluctance of the IRTF model is zero (infinite permeability material and infinitely small airgap), hence the MMF ‘seen’ inside either contour must always be zero. 176 FUNDAMENTALS OF ELECTRICAL DRIVES (a) θ =0, i α =0, i β = ˆ i, T e = ˆ ψ m ˆ i (b) θ = π 4 , i α = − ˆ i √ 2 , i β = ˆ i √ 2 , T e = ˆ ψ m ˆ i √ 2 (c) θ = π 2 , i α = − ˆ i, i β =0, T e =0 Figure 7.8. IRTF Symbolic model and space vector diagrams, for: θ =0, θ = π 4 and θ = π 2 Introduction to Electrical Machines 177 Rotor position θ =0: if we consider the x contour in figure 7.8(a), then it coincides with the flux distribution that exists in the model. The MMF seen by this contour is equal to ni α . The excited rotor winding cannot contribute (given both halves are in the contour) to this contour. Hence, the current i α must be zero. By observing the y contour and the MMF ‘seen’, we note the presence of the imposed rotor current i y = ˆ i. The number of winding turns on rotor and stator are equal, hence a stator current i β = ˆ i (with the direction shown) must appear in the short-circuited β coil to ensure that the zero MMF condition with this contour is satisfied. The space vector representation of the current and flux as given in figure 7.8(a), shows that they are π 2 radian apart. Note (again) that there is only one set of space vectors and their components may be projected onto either the rotor or stator complex plane. The torqueaccording to equation(7.13) will underthese circumstances (with the current vector leading the flux vector) equal to T e = ˆ ψ m ˆ i. From first principles (see figure 1.9) we note that forces will be exerted on the rotor winding in case the latter carries a current and is exposed to a magnetic field. In this case the flux and flux density distributions are at their highest level along the α axis (see appendix A). The y winding carries a current in the direction shown and this will cause a force on the individual conductors of the y rotor winding and thus a corresponding torque in the anti-clockwise (positive) direction. Rotor position θ = π 4 : if we consider the x contour in figure 7.8(b), then we note that a MMF due to the α winding would be less than ni α . The reason for this is that the contour also encloses part of α winding which gives a negative MMF contribution to the total MMF. The resultant MMF is found by integrating equation (A.9) over the angle range π/4 → π, −π →−3π/4 and comparing the outcome of this integral with the integral over the range 0 → π. This analysis will show that the MMF is reduced by a factor 1/ √ 2 (when compared to the previous case) hence the resultant MMF of the α winding is equal to ni α / √ 2. The same x contour also encloses part of the β winding and its MMF contribution is equal to ni β / √ 2. There are no other contributions, hence the sum of these two MMF’s is of the form ni α / √ 2+ni β / √ 2=0because the MMF in the contour must be zero. From this analysis it follows that, for the given angle, the currents must be in opposition, i.e. i α = −i β . If we now consider the MMF enclosed by the y contour we note that the α will contribute a MMF component −ni α / √ 2 (now negative because the other side of this winding), while the β will contribute a component ni β / √ 2. Furthermore, the y winding will add a component n ˆ i. The resultant MMF (the sum of these three contributions), gives together with the condition i α = −i β , the required stator currents i α = − ˆ i/ √ 2, i β = ˆ i/ √ 2 respectively. The space vector representation shown in 178 FUNDAMENTALS OF ELECTRICAL DRIVES figure 7.8(b) confirms the presence of the two current components. Note also that the angle between the current and flux vectors is equal to 3π 4 radian. The torque according to equation (7.13) will under these circumstances be equal to T e = ˆ ψ m ˆ i/ √ 2. The torque must be less (than the previous case) because a number of the conductors of the y winding now ‘see’ a flux density value which is in opposition to that seen by the majority of the conductors. Rotor position θ = π 2 : if we consider the x contour in figure 7.8(c), then we can conclude that the current i β must be zero (the other winding cannot contribute, given both halves are in the contour). The reason for this is that the MMF seen by this contour must be zero. By observing the y contour and the MMF ‘seen’ by this contour we note the presence of the rotor current i y = ˆ i. The number of winding turns on rotor and stator are equal, hence a stator current i α = − ˆ i must appear to ensure that the zero MMF condition with this contour is satisfied. The space vector representation of the current and flux as given in figure 7.8(c) shows that they are π radian apart. The torque according to equation (7.13) will under these circumstances be equal to T e =0. From first principles we note that half the conductors of the y winding will experience a force in opposition to the other half, hence the net torque will be zero. 7.3 Conditions required to realize constant torque The ability of a machine to produce a torque which has a non-zero average component is of fundamental importance. In this section we will consider the conditions under which electrical machines are able to produce a constant (time independent) torque. For this analysis it is sufficient to re-consider the model according to figure 7.6. The rotor is again connected to a current source which in this case is assumed to be of the form ˆ ie j(ω r t+ρ r ) ,whereω r represents the angular velocity of this vector relative to the rotor of the IRTF. The ITF current  i xy is therefore of the form  i xy = ˆ ie j(ω r t+ρ r ) (7.15) The stator is connected to a three-phase sinusoidal voltage source such that the voltage vector is taken to rotate at a constant angular velocity ω s .The corresponding flux vector  ψ m can then be found using u = d  ψ m dt which will in general terms result in a rotating flux vector of the form  ψ m = ˆ ψ m e jω s t (7.16) The rotor angle θ is a function of rotor speed ω m and load angle ρ m and is (for constant speed operation) defined as θ = ω m t + ρ m (7.17) [...]... introduction of the two-phase IRTF module as discussed in the previous section Figure 7. 13 Generalized two-phase machine model The resultant two-inductance model (with zero coil resistance at this stage) of the generalized machine is given in figure 7. 14 The parameters shown in figure 7. 14 are found using equation (7. 23) kr = κ LM Ls Lr = κ2 Ls (7. 23a) (7. 23b) 184 FUNDAMENTALS OF ELECTRICAL DRIVES Figure 7. 14... aid of vectors ψM , is which gives ∗ ψM is Te = (7. 29) This expression is found upon substitution of equation (7. 28c) into (7. 28g) The complete set of equations in (7. 28) can for the purpose of numerical simulation be rewritten in the following form: ψs = xy ψM = ψM us − is Rs dt (7. 30a) xy xy uR + iR RR dt (7. 30b) xy = ψM ejθ ψs − ψM Lσ ψs = is − LM −jθ = iR e (7. 30c) is = (7. 30d) iR (7. 30e) xy iR (7. 30f)... equation set for the model according to figure 7. 15 is of the form dψ s dt + is Lσ us = is Rs + (7. 28a) ψs = ψ M (7. 28b) ψM LM = is − iR xy xy uR = −iR RR + xy iR = iR ejθ xy ψM = ψM e−jθ Te = ∗ ψM iR (7. 28c) xy dψ M dt (7. 28d) (7. 28e) (7. 28f) (7. 28g) 186 FUNDAMENTALS OF ELECTRICAL DRIVES It is noted that the magnetizing inductance LM can be placed to either side of the IRTF If LM is placed on the rotor... given that a quasi-steady-state operadt ˆ tion is assumed, i.e we assume that the flux amplitude ψm is constant On the basis of this assumption equation (7. 18) reduces to xy xy em = j (ωs − ωm ) ψm (7. 19) The torque produced by this machine is found using equation (7. 13) Further mathematical handling of this torque equation and use of equations (7. 16) (in rotor coordinates), (7. 2) and (7. 17) leads to Te... rated torque of the machine, given that the latter is constrained by the rotor-volume, as was discussed in section 1 .7 on Figure 7. 12 Use of IRTF module for multi-pole pair models 183 Introduction to Electrical Machines page 22 Figure 7. 12 shows the two additional gain modules which must be added to the IRTF module (figure 7. 5(a)), in order to use the IRTF concept for multi-pole models 7. 4 General machine... 179 Introduction to Electrical Machines The importance of the angle variable ρm will be discussed at a later stage The Figure 7. 9 Simplified twophase, machine model xy xy induced voltage on the rotor side in figure 7. 9 is in this case em = dψm , which dt with the aid of equations (7. 16), (7. 2) and (7. 17) may also be written as xy em = xy ˆ dψ m dψm xy = ej((ωs −ωm )t−ρm ) + j (ωs − ωm ) ψm dt dt (7. 18)... form of an array named ‘dat’ will be generated when the simulation is run Input to the ‘dat’ module are the stator currents (both components), the time, rotor angle and of course the torque A set of MATLAB commands needs to be given to display the data Below an example of such a file m-file Tutorial 1, chapter 7 %Tutorial 1, chapter 7 close all; plot(dat(:,3),dat(:,2),’r’); % i_alpha as function of rotor... process of modelling the behaviour of a multi-pole machine with a two-pole IRTF model can thus be initiated by introducing (as a first step) a ‘gain’ module with gain p to the rotor angle input side of the IRTF model The torque per amp`re produced by the multi-pole machine will increase e when a larger number of pole pairs is used A qualitative explanation of this statement is as follows: in the two-pole... have more than one pole-pair given that a more efficient winding configuration can be realized in for example a four-pole (p = 2) machine The flux distribution which will occur in a four-pole machine is shown (symbolically) in figure 7. 11(b) The corresponding four-pole (sinusoidally distributed) phase (a) Two-pole flux distribution Figure 7. 11 (b) Four-pole flux distribution Two and four-pole flux distributions... (7. 30c) is = (7. 30d) iR (7. 30e) xy iR (7. 30f) 7. 5 Tutorials for Chapter 7 7.5.1 Tutorial 1 This tutorial is aimed at providing a better understanding of the IRTF module As a first step a Simulink model of the IRTF module must be built The Simulink implementation follows the generic version given in figure 7. 5(a) An example of possible implementation is given in figure 7. 16, where the IRTF submodule ‘x/y→ alpha/beta’ . Chapter 7 INTRODUCTION TO ELECTRICAL MACHINES 7. 1 Introduction This chapter considers the basic working principles of the so-called ‘clas- sical’ set of machines. This set of machines. J dω m dt (7. 14a) ω m = dθ dt (7. 14b) 174 FUNDAMENTALS OF ELECTRICAL DRIVES (a) IRTF-flux (b) IRTF-current Figure 7. 5. Generic representations of IRTF module with T l and J representing the load torque and inertia of the. i x , i y , θ of equation (7. 3) 172 FUNDAMENTALS OF ELECTRICAL DRIVES can also be written as  i α i β  =  cos θ −sin θ sin θ cos θ  i x i y  (7. 4a) The symbolic diagram of figure 7. 2 can also