Chapter 8 VOLTAGE SOURCE CONNECTED SYNCHRONOUS MACHINES 8.1 Introduction The synchronous machine has traditionally been used for power generation purposes. For motor applications (when mains connected) its use is ideal when the operating speed must remain constant, i.e. independent of load changes. Starting up however needs special measures. Modern drivesoftenuseaconverterwhichgives us more flexibilityintermsof controlling the machine and enable the machine to self-start. The synchronous machine when used with a converter has become an important player in the field of drives. In this chapter we will look at the basicoperationofthesynchronousmachine, where we will apply the theory of the previous chapter in general and section 7.3 on page 178 in particular. 8.2 Machine configuration The machine has a non-rotating component known as the stator which is shown in figure 8.1. The stator consists of a ‘frame’ within which a laminated stator core stack is positioned. This core has a series of slots which house the three-phase windings of the machine. Of these windings the so-called active sides (named thus because these are responsible for the energy conversion) are distributed appropriately in the core slots. The stator-coil end-winding- parts are at each end of the core stack as shown in figure 8.1. The three- phase windings will, when connected to a three-phase supply source, produce a rotating magnetic field which is, as was discussed in the previous chapter, an essential requirement for producing constant torque. More details on ma- chines and rotatingfieldsare given in B ¨ odefeld [B ¨ odefeld and Sequenz,1962] or Hughes [Hughes, 1994]. Note that the same stator is also used when discussing 194 FUNDAMENTALS OF ELECTRICAL DRIVES Figure 8.1. Stator of three-phase synchronous machine the asynchronous/induction machine. The rotor configuration for the synchro- nous machines may take on several forms. A very simple configuration as shown in figure 8.2 conveys the basic structure. A more extensive discussion on the machine structure is given by Hughes [Hughes, 1994]. Shown in fig- ure 8.2 is a rotor in the form of a single coil (referred to as the field winding) with sides A and B which are connected to sliprings 1 and 2 respectively. These copper sliprings are linked to a set of brushes which in turn are connected to a stationary DC power supply. The use of the slipring/brush combination allows Figure 8.2. Simple rotor for synchronous machine Voltage Source connected Synchronous Machines 195 us to excite a rotating rotor coil with a DC ‘field’ current via a stationary source. Note that this rotor can be replaced by a permanent magnet which means that the slipring/brush combination is then avoided. However, the excitation can under these circumstances not be varied. Such machines belong to the class of so-called ‘Brushless DC machines’. 8.3 Operating principles The synchronous machine with a slipring/brush combination is, as was dis- cussed in the previous section, fedon therotor side with a DC field current. This implies that the current  i xy (as introduced in section 7.3 on page 178) is given by equation (7.15) with ω r =0, hence  i xy = ˆ ie jρ r . A choice remains with re- spect to the angle ρ r ; its value can be taken to be zero or π rad. The latter choice amounts to reversing the polarity of the DC current source shown in figure 7.9. This option is chosen here for reasons which will become apparent shortly, hence  i xy = − ˆ i. The magnitude of the current ˆ i is in context of synchronous machines more commonly known as the field current i F , i.e.  i xy = −i F .On the basis of the speed condition (see equation (7.21), with ω r =0) constant torque operation is only possible when the shaft speed is equal to the rotational speed ω s of the rotating flux vector produced by the three-phase stator winding. The term ‘synchronous machine’ reflects this type of operation, i.e. rotor speed synchronized to the rotating field. The torque produced by this machine can be calculated using equation (7.22) on page 179 with,  i xy = −i F , ρ r = π which yields T e = − ˆ ψ M i F sin (ρ m ) (8.1) If a mechanical load is applied to the machine the load angle ρ m will be non- zero. This explains why this angle is referred to as the ‘load angle’. When a load is applied, the rotor will lag behind the magnetic field (ρ m < 0), which leads to a positive torque value that matches the applied load. The load angle can for a given load torque be modified by varying the amplitude of the field or rotor current. The maximum torque ˆ T e that can be delivered by this machine is reached when the load angle reaches π 2 rad. Speed changes are implemented by changing the stator frequency ω s , which nowadays require the use of a power electronic converter with the machine. The general space vector diagram according to figure 7.10 reduces to the form shown in figure 8.3 given the present choice of rotor excitation. Several interesting observations can be made with respect to figure 8.3. Firstly, the current vector  i xy is tied to the negative real axis of the rotor given that the rotor is fed with a DC current i F . Secondly, the vectors  ψ m ,  i are stationary with respect to each other. The angle between the two can and will vary depending on the load torque. For example, an increase in the load torque will see the rotor slip back momentarily (in the clockwise direction) so that the machine adjusts 196 FUNDAMENTALS OF ELECTRICAL DRIVES Figure 8.3. Space vector diagram forsynchronousmachine, motoring operation shown ρ m < 0 its torque T e to match the load torque T l . As indicated above the highest torque (known as the ‘pull out’ torque) that can be delivered by the machine is equal to ˆ T e = ˆ ψ m i F . The machine will stall (or rotate uncontrolled in the opposite direction) when a load torque above this value is applied. In this situation, the two vectors will no longer be stationary with respect to each other, i.e. the machine will deliver a pulsating torque with zero average value. Also shown in figure 8.3 is the voltage vector e (which is equal to the supply voltage vector u) which leads the flux vector by π/2. The projection of the current onto this vector shown as i torque in figure 8.3 is proportional to the torque. If the option ρ r =0would have been selected then the field current would be aligned with the positive real axis. Under the circumstances shown in figure 8.3 the torque would be negative, hence the rotor would rotate until the current vector would be diametrically opposite its present position. 8.4 Symbolic model The model to be discussed is based on the simplified generalized model as given in figure 7.15 which in this case is configured to operate as a synchronous machine. The symbolic model of the machine is given in figure 8.4. The rotor resistance is purposely omitted in figure 8.4 given that the rotor winding is connected to a current source i F with the new polarity, as discussed above. The stator resistance and leakage are also not included in this model. The equation set which corresponds to figure 8.4 is found using Kirchhoff’s law which gives u s = d  ψ M dt (8.2a)  ψ M = L M   i s + i F e jθ  (8.2b) Voltage Source connected Synchronous Machines 197 Figure 8.4. Synchronous machine, zero resistance and leakage inductance T e − T l = J dω m dt (8.2c) ω m = dθ dt (8.2d) Note that in this equation set the flux vector  ψ M (see equation (8.2b)) is given in its stator based form. This expression is in fact found by using Kirchhoff’s current law on the rotor side which gives  ψ xy M = L M   i xy s −  i xy R  ,where  i xy R = −i F . A rotating magnetizing flux vector  ψ M = ˆ ψ M e jω s t is established as a result of the machine being connected to a three-phase grid with angular frequency ω s . 8.4.1 Generic model A generic representation of the(two-pole) synchronous machine inits present form is given in figure 8.5. The model follows directly from equation (8.2). Central to this model is the IRTF sub-module which is represented by fig- ure 7.5(a). Theload torqueis provided by a sub-module T l (ω m ), which has been configured to deliver a quadratic torque speed characteristic T l = k l  ω m ω R m  2 where ω R m is set to ω s (rad/sec). The value of the constant k l is set by the user. The initial speed of the machine is set to ω m = ω s (rad/sec), which corresponds to the rotational speed of the flux vector  ψ M . 8.5 Generalized symbolic model The model discussed in section 8.4 was given in its simplified form, i.e. without stator resistance and leakage inductance. The rotor resistance was also not included but this is of less importance, given that we assume a current source connected to the rotor. The complete model (without rotor resistance) shown in figure 8.6, is applicable for so-called ‘non-salient’ machines, which generally do not carry any damper windings (short-circuited windings on the rotor). Salient machines show different L M values for the x and y direction. The equation set which corresponds to this machine is given by equation (8.3). 198 FUNDAMENTALS OF ELECTRICAL DRIVES Figure 8.5. Generic representation of a voltage source connected synchronous machine, cor- responding to figure 8.4 Figure 8.6. Synchronous machine u s =  i s R s + d  ψ s dt (8.3a)  ψ s =  i s L σ +  ψ M (8.3b) e xy M = d  ψ xy M dt (8.3c)  ψ xy M = L M   i xy s + i F  (8.3d) T e − T l = J dω m dt (8.3e) ω m = dθ dt (8.3f) The symbolic model according to figure 8.6 can also be redrawn in the form given by figure 8.7. The difference between the two models lies in the posi- tioning of the magnetizing inductance L M which is now on the stator side. The Voltage Source connected Synchronous Machines 199 Figure 8.7. Revised synchronous machine model revised model allows use to move towards a generic model where we are able to group the two inductances L σ and L M . To realize this it is helpful to introduce the flux ψ F = L M i F . This allows us to rewrite equation (8.3d) (in its stator coordinate based form) in the form given by equation (8.4).  ψ M = L M  i s + ψ F e jθ (8.4) Substitution of equation (8.4) into equation (8.3b) leads to equation (8.5), which for completeness contains the complete set needed to derive a generic model of this machine. u s =  i s R s + d  ψ s dt (8.5a)  ψ s =  i s (L σ + L M )    L s +ψ F e jθ (8.5b) e R = d  ψ F e jθ  dt (8.5c) T e − T l = J dω m dt (8.5d) ω m = dθ dt (8.5e) In equation (8.5b)the sum ofthe two inductances known asthe stator inductance L s appears as intended. Furthermore, the voltage vector e R is introduced, given that the vectore M is no longer directly available as a result of using the parameter L s . 8.5.1 Generic model The generic model of the non-salient synchronous machine without damper- winding is directly found using equation (8.5). An example of implementation as given in figure 8.8, shows that the ‘IRTF-flux’ model is replaced with an ‘IRTF-current’ model (see figure 7.5(b)). Observation of figure 8.8 learns that 200 FUNDAMENTALS OF ELECTRICAL DRIVES Figure 8.8. Full generic synchronous machine model, corresponding to model in figure 8.7 the IRTF module has as ‘inputs’ the flux vector  ψ xy F =(ψ F , 0) and stator current vector  i s =(i sα ,i sβ ). The mechanism through which a stator current will occur is readily shown using this generic model. Note that the reasoning presented here differs from that given in section 8.3, given the way in which the IRTF module is now used. Both approaches to describing machine operation must of course give the same result. For simplicity we will ignore the stator resistance in this discussion. If we assume that the stator is connected to a three-phase sinusoidal supply then this leads to a rotating stator flux vector  ψ s . The rotor winding carries a current i F which corresponds to a flux vector ψ F . If we assume that the rotor rotates at the same speed as the stator flux field then the rotor and stator flux vectors will move at the same speed. If we initially assume that the magnitude of both vectors is equal and aligned then no stator current will be present. Under these circumstances the vector into the gain module 1/L s (see figure 8.8) is zero, hence the current  i s will be zero. If, for example, we increase the rotor current then the rotor flux will increase and a stator current component will occur which will lead the voltage vector by π 2 radians. No torque will be realized under these circumstances. If we return to our initial conditions (flux vectors equal magnitude and aligned) then the application of a mechanical load will momentarily cause the rotor to slow down, until an angle between the two flux vectors occurs. The difference vector between the two vectors is proportional to the stator current. Hence, a current vector will occur which means that the machine will produce a torque to counteract the new load torque (after transient effects have died down following the load torque change). Voltage Source connected Synchronous Machines 201 8.6 Steady-state characteristics In this section we look to the steady-state performance of the synchronous machine in case we connect the stator windings to a three-phase sinusoidal source. This implies that the stator phase voltage equals the grid voltage with its fixed amplitude and frequency. We also assume the shaft speed to run at synchronous speed, although different shaft angles are possible with respect to the rotating field in the stator. Consequently the torque and field current are the only independent variables left at this stage. Steady-state analysis provides insight with regard to the trajectory of the stator current vector and load angle when either of these independent variables is varied. The approach taken is to consider the simplified model first and develop the so-called ‘Blondel diagram’ and torque angle curves on the basis of the phasor equation set applicable to this model. This model is then extended to a full machine model. Synchronous operation is assumed which means that the condition ω s = ω m is met. The stator is connected to a three-phase sinusoidal voltage supply which is represented by the space vector u s . Uptonowwe have chosen the flux vector of the form  ψ M = ˆ ψ M e jω s t which corresponds to a supply vector u s = jω s ˆ ψ M e jω s t . The corresponding phasor representations are according to equation (4.59) of the form ψ M = ˆ ψ M and u s = jω s ˆ ψ M . The phasor diagrams which are linked to AC machines, will be discussed in line with the general convention where the supply voltage phasor is chosen along the real axis, i.e u s =ˆu s . With our present choice of flux vector the voltage phasor is along the imaginary axis. If we redefine (for the purpose of examining the steady-state performance only) the relationship between space vectors and phasors as, x = x e j ( ω s t+ π 2 ) (8.6) then the flux and voltage phasor will be of the form ψ M = ˆ ψ M e −j π 2 and u s = ω s ˆ ψ M respectively, i.e. rotated clockwise so that the voltage phasor is real, as preferred for steady-state analysis. 8.6.1 Steady-state characteristics, simplified model The steady-state characteristics of the non-salient synchronous machine are studied with the aid of figure 8.5. The magnetizing flux vector  ψ M will also rotate at the same speed but will lag the voltage vector u s by π 2 radians. This vector is derived from the voltage vector using u s = d  ψ M dt ,oru s = jω s ψ M in phasor form. The basic characteristics of the machine relate to the stator current of the machine in phasor form and the torque load angle curve. The flux phasor ψ M according to equation (8.4) can, with the aid of equations (8.6) and (7.17) (with 202 FUNDAMENTALS OF ELECTRICAL DRIVES ω m = ω s ) be written as ψ M = L M  i s + i F e j ( ρ m − π 2 )  (8.7) The current i s can with the aid of equation (8.7) and expression ψ M = ˆu s jω s ,be written as i s = ˆu s jω s L M − i F e ( jρ m − π 2 ) (8.8) Equation (8.8) can also be represented in terms of an equivalent circuit as given by figure 8.9. Equation (8.8) may also be rewritten in the following form Figure 8.9. Simplified synchronous machine, phasor based model i s = ˆu s jω s L M    i s1 − ˆu s k F e jρ m jω s L M    i s2 (8.9) in which the factor k F is defined as k F = ω s L M i F ˆu s (8.10) which can also be written in the form given by equation (8.11). k F = L M i F ˆ ψ M (8.11) If the value of k F is greater than 1 the machine is said to be operating under ‘over-excited’ conditions. For k F value less than 1 a so-called ‘under-excited’ machine operating condition is present. The current phasor i s can be plotted as a function of the load angle ρ m with i F as parameter. This type of diagram as given in figure 8.10, shows the two current components according to equation (8.9) together with lines of constant output power p out . These output power curves are found by making use of the energy balance equation, which for the current machine with zero stator resistance is of the form {u s i ∗ s } = p out (8.12) [...]... machine-simplified model 215 Voltage Source connected Synchronous Machines Torque (Nm) 1.5 Sim MATLAB 1 0.5 0 -9 0 -8 0 -7 0 -6 0 1 -5 0 -4 0 (a) ρm (deg) -3 0 -2 0 -1 0 0 Is (A) Sim MATLAB 0.5 0 -9 0 -8 0 -7 0 -6 0 400 -5 0 -4 0 (b) ρm (deg) -3 0 -2 0 -1 0 0 Ps (W) Sim MATLAB 200 0 -9 0 -8 0 -7 0 -6 0 Qs (VA) 600 -5 0 -4 0 (c) ρm (deg) -3 0 -2 0 -1 0 0 Sim MATLAB 400 200 0 -9 0 -8 0 Figure 8. 17 -7 0 -6 0 -5 0 -4 0 (d) ρm(deg) -3 0 -2 0 -1 0... 373.06 241.73 130.96 25 .81 -7 6 .84 -1 78. 16 -2 78. 66 -3 78. 62 -5 3.13 -3 2.23 -2 3. 58 -1 8. 66 -1 5.47 -1 3.21 -1 1.54 -1 0.24 Build an m-file which will display the data from your simulation in the form of three sub-plots: Is (iF ) , Qs (iF ) , ρm (iF ) In addition, plot the stator current phasor is in the form of a ‘Blondel’ diagram Add to these plots the results as calculated via a steady-state (phasor) analysis... machine: no-load→load Te (Nm) 0.0 0.125 0.250 0.375 0.50 0.625 0.750 0 .87 5 1.00 1.125 1.250 ρm (◦ ) Ps (W) Qs (VA) Is (A) 0 -5 .74 -1 1.54 -1 7.46 -2 3. 58 -3 0.00 -3 6 .87 -4 4.43 -5 3.13 -6 4.16 -8 9.29 0 39.27 78. 54 117 .81 157. 08 196.35 235.62 274 .89 314.16 353.43 392.67 98. 14 100.14 106.11 116.26 130.96 150.79 176.71 210.43 255.25 319.43 485 . 98 0.14 0.16 0.19 0.24 0.30 0.36 0.43 0.51 0.60 0.70 0.92 Build an m-file... %measured, from display, load angle (deg) rhoM=[0 -5 .74 -1 1.54 -1 7.46 -2 3. 58 -3 0.00 -3 6 .87 -4 4.43 -5 3.13 -6 4.16 -8 9.29]; %measured, from display, real power (W) PM=[0 39.27 78. 54 117 .81 157. 08 196.35 235.62 274 .89 314.16 353.43 392.67]; %measured, from display, reactive power (VA) QM=[ 98. 17 100.14 106.11 116.26 130.96 150.79 176.71 210.43 255.25 319.43 485 . 98] ; %measured, from display, RMS stator current... iFM=[0.5:0.25:2.25]; % selected field current values %measured, from display, load angle (deg) rhoM= [-5 3.13 -3 2.23 -2 3. 58 -1 8. 66 -1 5.47 -1 3.21 -1 1.54 -1 0.24]; %measured, from display, real power (W) PM=157. 08; %measured, from display, reactive power (VA) QM=[373.06 241.73 130.96 25 .81 -7 6 .84 -1 78. 16 -2 78. 66 -3 78. 62]; %measured, from display, RMS stator current (A) IM=[0.60 0.42 0.30 0.23 0.26 0.35 0.47 0.60];... appear from this m-file is given in figures 8. 17, and 8. 18 Also shown (not to scale) in figure 8. 18 by way of reference, is the stator voltage phasor us Clearly noticeable from figure 8. 18 is that the locus of the stator current phasor is is part of a circle which has its center at 0, -1 .25A An example of an m-file implementation is given below m-file Tutorial 3, Chapter 8 %Tutorial 3, Chapter 8 %Tutorial synchronous... equation (8. 16) us = is Rs + jωs ψ s (8. 16a) ψ s = Ls is + ψF ej (ρm − 2 ) π (8. 16b) Elimination of the flux phasor ψ s from equation (8. 16), leads to the following current phasor expression is = us 1 − kF ejρm ˆ Rs + jωs Ls (8. 17) 206 FUNDAMENTALS OF ELECTRICAL DRIVES Expression (8. 17) may also be presented in terms of a circuit representation as given in figure 8. 12 A normalization of equation (8. 17) is... example of an m-file implementation for this tutorial is given below: m-file Tutorial 4, Chapter 8 %Tutorial 4, Chapter 8 %Tutorial synchronous machine-simplified model 219 Voltage Source connected Synchronous Machines 1 Sim MATLAB s I (A) 0 .8 0.6 0.4 0.2 0.4 0.6 0 .8 1 1.2 1.4 (a) i (A) 1.6 1 .8 2 2.2 400 Sim MATLAB 200 Qs (VA) 2.4 F 0 −200 −400 0.4 0.6 0 .8 1 1.2 ρm (deg) 0 1.4 (b) iF (A) 1.6 1 .8 2 2.2... speed and torque are to be plotted as a function of time An example of the results which should appear for these variables is given in figures 8. 26, 8. 27 and 8. 28 An observation of these simulation results reveal some interesting details namely 226 FUNDAMENTALS OF ELECTRICAL DRIVES 250 40Hz max 60Hz max shaft speed (rad/s) 200 150 100 50 0 0 0.05 0.1 Figure 8. 26 0.15 0.2 0.25 time (s) 0.3 0.35 0.4 0.45... given by equation (8. 25) pn = − out 4kF 1+ ωs Ls Rs where γ = arctan form ωs Ls Rs 2 sin ρm + γ − π 2 − kF sin γ − π 2 (8. 25) The normalization used in equation (8. 25) is of the pn = out pout pmax out (8. 26) pmax = out u2 ˆ 4Rs (8. 27) where Voltage Source connected Synchronous Machines 209 A graphical representation of equation (8. 25) as function of the load angle ρm s is shown in figure 8. 14 with kF = . figure 8. 8, shows that the ‘IRTF-flux’ model is replaced with an ‘IRTF-current’ model (see figure 7.5(b)). Observation of figure 8. 8 learns that 200 FUNDAMENTALS OF ELECTRICAL DRIVES Figure 8. 8. Full. by equation (8. 3). 1 98 FUNDAMENTALS OF ELECTRICAL DRIVES Figure 8. 5. Generic representation of a voltage source connected synchronous machine, cor- responding to figure 8. 4 Figure 8. 6. Synchronous. i F e ( jρ m − π 2 ) (8. 8) Equation (8. 8) can also be represented in terms of an equivalent circuit as given by figure 8. 9. Equation (8. 8) may also be rewritten in the following form Figure 8. 9. Simplified synchronous