ECE 320 Energy Conversion and Power Electronics Dr. Tim Hogan. Chapter 7: Synchronous Machines and Drives (Textbook Chapter 5)

Transcription

1 ECE 30 Energy Converion and Power Electronic Dr. Tim Hogan Chapter 7: ynchronou Machine and Drive (Textbook Chapter 5) Chapter Objective For induction machine, a the rotor approache ynchronou peed, the frequency of the current in the rotor decreae, a doe the amplitude of thee current. The reaon an induction motor produce no torque at ynchronou peed i not that the current are DC, but that their amplitude i zero. t i poible to operate a three-phae machine at ynchronou peed if DC i externally applied to the rotor and the rotor i rotating at ynchronou peed. n thi cae torque will be developed only at thi peed, i.e. if the rotor i rotated at peed other than ynchronou, the average torque will be zero. Machine operating on thi principle are called ynchronou machine, and cover a great variety. A generator they can be quite large, rated a few hundred MVA, and almot all power generation i done uing ynchronou machine. Large ynchronou motor are not very common, but can be an attractive alternative to induction machine for ome application. mall ynchronou motor with permanent magnet in the rotor, rather than coil with DC, are rapidly replacing induction motor in automotive, indutrial and reidential application ince they lighter and more efficient. 7.1 Deign and Principle of Operation The tator of a ynchronou machine i of the type that we have already dicued with three winding carrying a three-phae ytem of current. A we aw in Chapter 5, thi reult in a tator magnetic field that patially rotate around the tator at a contant angular peed, ω. Unlike induction machine, ynchronou machine have zero lip, and the rotor maintain the ame angular peed, ω, a the tator generated field. n a ynchronou machine, the rotor winding carry DC current, or are compoed of permanent magnet a dicued next Wound Rotor Carrying DC n thi cae the rotor teel tructure can be either cylindrical, that that in Figure 1(a), or alient like the one in Figure 1(b). n either cae, the rotor winding carrie DC, delivered through lip ring, or through a rectified voltage of an inide-out ynchronou generator mounted on the ame haft. n thi handout, dicuion i limited to cylindrical rotor Permanent Magnet Rotor ntead of upplying DC to the rotor, the rotor contain permanent magnet in configuration uch a thoe hown in Fig.. The effect of permanent magnet rotor include: The rotor flux can no longer be controlled externally. t i defined by the magnet and the geometry The machine become impler to contruct, at leat for mall ize. 1-1

2 inuoidally ditributed winding Concentrated winding (a) (b) Figure 1. Wound rotor configuration of ynchronou machine. (a) urface PM (PM) ynchronou machine (b) urface inet PM (PM) ynchronou machine (c) nterior PM (PM) ynchronou machine (d) nterior PM with circumferential orientation ynchronou machine Figure. Poible magnet placement in PMAC motor. 1-

3 7. Equivalent Circuit The flux in the air gap can be conidered to be due to two ource: the tator current, and the rotor current or permanent magnet. Recall the tator current produce a flux that rotate at ynchronou peed, and that thi flux could alo be produced by one equivalent winding that i rotating at ynchronou peed and carrying a current equal to the magnitude of the tator-current pace vector. The rotor i itelf uch a winding, a real one, inuoidally ditributed, carrying DC and rotating at ynchronou peed. t produce an airgap flux, which could alo be produced by an additional et of three phae tator current, giving a pace vector i F. The amplitude of thi pace vector would be: i F = i f (7.1) where i the number of the tator turn of the one equivalent winding (when the three tator winding are repreented by a ingle equivalent rotating winding a dicued above) and R i the number of the turn in the rotor winding. t angle φ R would be the ame a the angle of the rotor poition: The tator current pace vector ha amplitude: R φ = + (7.) R ωt φr0 3 i = (7.3) where i the rm current of one phae. The tator current pace vector will have an intantaneou angle, φ i, of φ = + (7.4) i ωt φi0 The airgap flux i then produced by both thee current pace vector (rotor and tator). Thi flux induce in the tator winding a voltage, e. n quai teady-tate everything i inuoidal and the voltage pace vector correpond to three phae voltage E 1, E, E 3. n thi cae we can create an equivalent circuit for the tator a hown in Figure 3. Here i the tator AC current, that if it were to flow in the tator winding, it would have the ame effect a the rotor current, i f. n our analyi we can ue a reference either the tator voltage, V, or the tator current, a hown in Figure 4. From the relationhip of the pace vector in Figure 4, the angle θ i the power factor angle (the angle between and V ). We call β, the angle between V and, and the power angle, δ, i between and. + V E jx = β M Figure 3. tator equivalent circuit for a ynchronou machine. 1-3

4 δ β θ V α Figure 4. pace vector diagram of a ynchronou machine. From Figure 3 and Figure 4 we can note the following two relationhip: The pace vector of the voltage induced in the tator, e, i 90º ahead of the magnetizing current pace vector,. Thi can be undertood by the fact that i what caue all the airgap flux that link the tator and induce e. For a given frequency, the amplitude of thi voltage, e, i proportional to the current. A permanent magnet machine can be conidered equivalent to that with a winding, carrying a direct current, i f, that i contant (and cannot be controlled). 7.3 Operation of the Machine Connected to a Bu of Contant Voltage and Frequency Thi i uually the cae for large ynchronou generator or motor. We can conider any bu a one of contant voltage, by making a few modification to the equivalent circuit a hown in Figure

5 jx + V jx M = β jx jx M + V + V F = jx M β + V jx j(x M+ X ) = M F β j(x + X ) M Figure 5. Equivalent circuit that account for a non-zero bu impedance in the ytem. ynchronou machine are very efficient, and mot of the time we can neglect the tator reitance. All power then i converted to mechanical power and: P = 3V coθ = Tω (7.5) p P = 3V F co β (7.6) + F = (7.7) V jx M = (7.8) n thi operation, V and ω (and therefore peed) remain contant. The only input variable are the torque, T, which affect output power, Pout = Tω, and the field current, i f, which i p proportional to ; the magnetizing current i contant, ince it i tied to the voltage V. Auming the machine i operated o the power to it varie while the frequency and field current remain contant. ince thi i a ynchronou machine, the peed will not vary with the load. From equation (7.6) we can ee the power, and therefore the torque, varie inuoidally with the angle β. Remember that β i the angle between the axi of the rotor winding, and the tator voltage pace vector. ince thi voltage pace i 90º ahead of the pace vector of the magnetizing current (β 90º) 1-5

6 i the angle between the rotor axi and the magnetizing current pace vector (ame a the airgap flux). When there i no torque thi angle i 0, i.e. the rotor rotate aligned with the flux, when external torque i applied to the rotor in the direction of rotation the rotor will accelerate. A it accelerate (with the flux rotating at contant peed) the flux fall behind the rotor, and negative torque i developed, making the rotor low down and rotate again at ynchronou peed, but now ahead of the flux. imilarly, when load torque i applied to the rotor, the rotor decelerate; a it doe o, the angle β decreae beyond -90º, i.e. the rotor fall behind the flux. Poitive torque i developed that bring the rotor back to ynchronou peed, but now rotating behind the tator flux. n both cae when the load torque on a motor or the torque on a motor or the torque of the prime mover in a generator increae beyond a maximum, correponding to co β = ±1, the machine cannot develop adequate torque and it loe ynchronization. Let u now dicu the effect of varying the field current while keeping the power contant. From equation (7.6), when power and voltage are kept contant, the product co β remain contant a well, but thi product i the projection of on the horizontal axi. Thi mean that a the field current change while power tay the ame, the tip of travel on a vertical line, a hown in Figure 7(a). imilarly equation (7.5) mean that at the ame time the tip of travel on another vertical line, alo hown in Figure 7(a). t i clear from Figure 7(a) that once the field current ha exceeded a value pecific to the power level, the power factor become leading and the machine produce reactive power. Thi i different from the operation of an induction machine, which alway aborb reactive power. When the machine operate a a generator, the input power i negative. Figure 7(b) how thi operation for both leading and lagging load power factor. Here the angle between tator voltage and tator current defined in the direction hown in the equivalent circuit (Figure 5) i outide the range -90º < θ < 90º. Motor Generator Torque π π/ 0 π/ β Figure 6. Torque and angle β in a ynchronou motor. 1-6

7 1 3 V V (a) Figure 7. Equivalent circuit that account for a non-zero bu impedance in the ytem. (b) Example A 3-phae Y-connected ynchronou machine i fed from a 300(V), 60(Hz) ource. The ratio of the AC tator equivalent current to the rotor DC i /i f = 1.8. The magnetizing inductance of the machine i 00(mH). The machine i operated a a motor and i aborbing 110 (kw) at 0.89 pf leading. Calculate the required field current and the load angle. Draw the correponding phaor diagram. Uing Figure 8: 7.1º V -131º Figure 8. ynchronou machine a a motor. We find: X M = π = 75.4 (Ω) 1-7

8 V 300 = = 138 (V) = / 7.1º = 31/ 7.1º (A) uing the tator voltage we can calculate and from it. 138 = = 17.6 / -90º (A) 75.4 = M = 4 / -131º (A) i f 1.8 = F = 3.4 (A) Repeating thi proce for operation a a generator at 110 (kw), 0.8 pf leading give Figure 9 G 35º 3.56 V -145º Figure 9. ynchronou machine a a generator. G = 33.7 / 35º (A) = / -145º (A) = M = 7.66 / 3.56º (A) i f = (A) 1.8 = F What i the maximum power the above machine can produce (or aborb when operating a a generator and at the field current jut calculated? 1-8

9 We know that aborbed power i: P = 3V F co β for i f = (A) we have = 7.66 (A), and P generated i maximum for β = 0, thu: P = = 110. (kw) f the terminal voltage remain at 300 (V), 60 (Hz), what i the minimal field current required to maintain operation a a motor with load 70 (kw)? or P 3 V co β = = (W) = F F = (A) Operation from a ource of Variable Frequency and Voltage Thi operation require that our ynchronou machine i upplied by an inverter. The operation now i entirely different than before. We no longer have an infinite bu, but rather whatever tator voltage or current and frequency we deire. Moreover, with a pace-vector controlled inverter, the phae of thi voltage or current can be arbitrarily et at any intant, i.e. we can define the tator current or voltage pace vector, and obtain it at will. The conideration for the motor operation are alo different: There i no concern for aborbing or upplying reactive power. ntead, there i a limit on the total tator current, determined by thermal conideration. There i a limit to the maximum voltage the ource can upply, which lead to modification of the machine mode of operation at high peed. Operation from a ource of variable frequency and voltage i mot common for permanent magnet machine, where the value of i contant. n imple term, when the machine i tarting a a motor the frequency applied hould be zero, but the voltage pace vector hould be of uch angle with repect to the rotor that torque i developed a dicued in the previou ection. A torque develop, the machine accelerate, and the applied tator current have to create a rotating pace vector leading the rotor flux. Voltage and frequency have to be increaed, o that thi torque i maintained. t i important therefore to monitor the poition of the rotor in order to determine the location of the tator current or voltage pace vector. Two poible control technique are implemented: either voltage control, where the tator voltage pace vector i determined and applied, or current control, where the tator current pace vector i applied. For a fixed tator voltage and power (and torque) level, the tator loe are minimal when the tator voltage and current are in phae. Figure 10 how thi condition. 1-9

10 θ V β Figure 10. Operation of a ynchronou PM drive at contant voltage and frequency. otice that a the power change with the voltage contant two thing happen: 1. The voltage pace vector varie in amplitude and the magnetizing current change with it.. The amplitude of tay contant, but it angle with repect to the voltage change. From the developed torque and peed we can calculate the frequency, the value of and, and the angle between the tator voltage pace vector and the rotor, ince: 3 3 p p T = V = LM M (7.9) ω F M = + (7.10) where i a contant in permanent magnet machine. More common i the cae when the tator voltage i not contant. Here we monitor the poition of the rotor and ince the rotor flux and rotor pace current are attached to it, we are actually monitoring the poition of. To make matter imple we ue thi current rather than the tator voltage a reference, a hown in Figure 11. j X M j X M V θ γ Figure 11. Operation of a ynchronou PM drive below bae peed. Although the previou equation for power and torque are till true, they are not a ueful. ew formulae can be created that have the tator current,, and magnetizing current,, a variable. We 1-10

11 alo make ue of the angle γ, between and, ince we can control it. tarting from what we already know: P * * [ V ] = Re[ jx ( + ) ] * * [ jx M ] Re[ jx M ] m[ ] X inγ g = Re M (7.11) = Re + (7.1) = X M * F = M F (7.13) For a given torque, minimum loe require minimum value of the tator current. To minimize the value of with contant power and we chooe γ = 90º to obtain: P = X = g [ * F ] X M F p m[ * F ] 3 LM F M m (7.14) p Pg p T = 3 = 3 LM = ω (7.15) which mean that for contant power, the projection of the tator current on an axi perpendicular to i contant. A the rotor peed increae, even if tay contant, the tator voltage V = ω L M increae. At ome bae peed, ω B, the required voltage exceed the maximum that the power ource can provide. To increae the peed beyond the bae peed, we no longer keep γ = 90º. On the other hand, at that peed we know that the voltage ha reached it upper limit, V = V,max, therefore the value of = V,max /X M i known. n thi cae equation (7.9) and (7.10) become: 3p 3p T = V coθ = LM M coθ (7.16) ω F M = + + M inθ (7.17) Figure 1 how uch an operation with the variable having the ubcript 1. ote that we calculate the torque from power. P = 3X M F inγ (7.18) p [ ] 3 L inγ P p p T = = 3 LM m * F = M F (7.19) ω 1-11

12 j 1 X M j X M V 1 j X M V θ θ 1 1 γ γ 1 1 Figure 1. Field weakening of a PM AC motor. The two diagram are at the ame frequency, but the econd one ha γ > 90º and a lower V. Example A permanent magnet, Y connected, three-phae, -pole motor ha = 40 (A), and X M = 0.9 (Ω) at 100 (Hz). 1. f it i aborbing P = 1.5 (kw) at 100 (Hz) with minimum tator current, calculate thi current, the angle between and, the peed, the tator voltage (line-neutral) and the power factor. The minimum current will exit when the angle between and, γ, i 90º. Under thi condition: 1500 = or = = (A) = /90º = 4.34/19.15º (A) P 3X M F = F V = jω LM M = 38.1 /109.15º (V) where ω = π 100 = 68 (rad/) or 6000 (rpm) The power factor i then calculated a: ( º 90º ) pf = co = lagging. t i deired to increae the motor peed to 6900 (rpm) while keeping the power the ame, P = 1500 (W), but the upplied voltage ha reached it upper limit of V = 38.1 (V). ow the motor aborb the ame power at the voltage calculated in the previou quetion, but at frequency of 115 (Hz). Thi can be accomplihed by having the tator current no longer at a minimum value and g no longer at 90º. Calculate again the angle between and, the peed, and the power factor. 1-1

13 1500 P 3 F β = º and V = /109.14º (V) V º = / = = /19.15º (A) jx M 115 j = M F = /113.18º (A) = V co( β ) or co( β ) = = 0. 3 ω = π 115 = 7.57 (rad/) or 6900 (rpm) The power factor i then calculated a: ( º º ) pf = co = leading j X M j X M V θ γ M Cae #1 j X M j X M V θ γ Cae # 1-13

14 Example 7.4. A permanent magnet machine ha 4-pole, = 40 (A), and the value of the magnetizing reactance i X M =.9 (Ω) at 100 (Hz). The maximum tator current i,max = 50 (A) and the maximum voltage the inverter can provide i V ln,max = 00 (V). Firt calculate the maximum peed the machine can operate with minimal tator Joule loe. = X L m m = 4.6 (mh) π 100 For minimal loe in(γ) = 1, o: T p = 3 L m F in 4 [ ] ( ) 3 ( γ ) = ( H) [ 50 A ][ 40( A) ]() 1 = 55 ( m) Then calculate the maximum peed the machine can reach for operation a above: m ˆ + ˆ = = F 64( A) ω V 00( V) 3 [ 64( A) ] ( H) [ ], max1 = = = Lmm and the maximum mechanical peed under thee condition will be: ω p ω mech, max1 =,max1 = (rad/) 679 (rad/) ext aume operation above thi peed, ω mech,max1, for example at 400 (rad/). Obviouly thi cannot be done with minimal tator Joule loe, o the field ha to be weakened by increaing γ above 90º. Calculate the maximum torque that can be developed. We now have the value of V and of, ince they correpond to the limit, V = 00 (V), = 50 (A). From the voltage and peed we can find V p [ 400( rad/) ] L = 00( V) = m m m 00 = = 54.35( A) From the triangle of, m, (cae in the lat example): 50 F m F m co = + co (/ m ) = (/ m ) (/ m ) = 61.8º b = = 151.8º 1-14

15 T p = 3 m co β ( ) = 5.8( m) ote that the demagnetizing component of i which might be too high. ( ) = ( A) = in β 3.6 d 7.5 Controller for Permanent Magnet AC Machine Figure 13 how a typical controller for an AC Machine. t require a DC power upply, uually a rectifier fed from an AC ource, an inverter and a controller. Figure 14 how the controller in lightly higher detail. 3-phae AC Rectifier DC Link nverter Variable voltage and frequency Motor Voltage or Current Command peed or Torque Command Controller Figure 13. Generic Controller for a PMAC Machine. ω + _ P Controller T * Calculate Calculate * * * a, ib, ic i nverter PMAC Motor Optical encoder ω ο γ Poition of Rotor poition Rotor peed Figure 14. Field Oriented controller for a PMAC Machine. The calculation for are baed on equation (7.19) and the calculation of i from equation (5.3). * * * a, ib, ic are calculated from the pace vector 1-15

16 7.6 Bruhle DC Machine While it would be difficult to find the difference between a PM AC machine decribed above and a bruhle DC machine by jut looking at them, the concept of operation i quite different a i the analyi. The winding in the tator in a bruhle DC machine are not inuoidally ditributed but intead they are concentrated, each occupying one third of the pole pitch. The flux denity on the magnet urface and in the airgap i alo not inuoidally ditributed over the magnet but almot uniform in the air gap. A the tator current interact with the flux coming from the magnet torque i developed. t hould be clear that for the ame direction of flux, current in oppoite direction reult in oppoite force, and therefore in reduction of the total torque. Thi in turn make it neceary that all the current in the tator above the rotor i in the ame direction. To accomplih thi, the following are needed: enor on the tator that ene the direction of the flux coming from the rotor, A fat power upply that will provide current to the appropriate tator winding a determined by the flux direction A way to control thee current, e.g. through Pule Width Modulation A controller with input of the deired peed, the flux direction in the tator, and the tator current; and output of the deired current in the tator Fig how the rotor poition, the tator current and the witche of the upply inverter for two rotor poition. b' a b c c a' orth outh a c' Figure 15. Energizing the winding in a bruhle DC motor. b b' a b c c orth outh a c' a' Figure 16. Energizing the winding in a bruhle DC motor, a little later. b 1-16

17 The operation of the ytem can be decribed imply ince the developed torque i proportional to the tator current a: T = k (7.0) At the ame time, the rotating flux induce a voltage in the energized winding: E = k ω (7.1) Finally the terminal voltage differ from the induced voltage by a reitive voltage drop: V = E R (7.) term + The equation are imilar to thoe for a DC motor (4.6), (4.7), (4.8). Thi i the reaon that although thi machine i entirely different from a DC motor, it i commonly referred to a a bruhle DC motor. 1-17