Bài giảng Electromechanical energy conversion - Chapter V: Polyphase Induction Machines - Nguyễn Công Phương

Nguyễn Công Phương ELECTROMECHANICAL ENERGY CONVERSION Polyphase Induction Machines Contents I. Magnetic Circuits and Magnetic Materials II. Electromechanical Energy Conversion Principles III. Introduction to Rotating Machines IV. Synchronous Machines V. Polyphase Induction Machines VI. DC Machines VII.Variable – Reluctance Machines and Stepping Motors VIII.Single and Two – Phase Motors IX. Speed and Torque Control sites.google.com/site/ncpdhbkhn 2 Polyphase Induction Machines 1. Introduction to Polyphase Induction Machines 2. Currents and Fluxes in Polyphase Induction Machines 3. Induction – Motor Equivalent Circuit 4. Analysis of the Equivalent Circuit 5. Torque and Power by Use of Thevenin’s Theorem 6. Parameter Determination from No – Load and Blocked – Rotor Tests 7. Effects of Rotor Resistance sites.google.com/site/ncpdhbkhn 3 Introduction to Polyphase Induction Machines (1) • Induction motor: alternating current is supplied to the stator directly & to the rotor by induction or transformer action from the stator. • Two kinds of rotor: – Wound rotor (relatively uncommon) – Squirrel-cage rotor (the most commonly used) sites.google.com/site/ncpdhbkhn 4 Introduction to Polyphase Induction Machines (2) • The rotor speed: n (r/min). • The synchronous speed of the stator field: ns (r/min). • The rotor slip: ns – n (r/min) n n • The fractional slip: s  s (%) ns •  the rotor speed: n = (1 – s)ns • The mechanical angular velocity: ωm = (1 – s)ωs • The slip frequency: fr = sfe sites.google.com/site/ncpdhbkhn 5 Introduction to Polyphase Induction Machines (3) • The rotor terminals of an induction motor are short circuited. • The rotating air – gap flux induces slip – frequency voltages in the rotor windings. • The operating speed can never equal the synchronous speed. • The rotor currents produce a rotating flux wave which rotate at sns (r/min) with respect to the rotor. • The rotor speed: n (r/min) • The speed of the rotor’s flux wave: sns + n = sns + ns(1 – s) = ns sites.google.com/site/ncpdhbkhn 6 Introduction to Polyphase Induction Machines (4) • The speed of the stator field: ns (r/min). • The speed of the rotor field: ns (r/min). •  the rotor currents produce a field which rotates at synchronous speed and hence in synchronism with that produced by the stator currents. •  the stator & rotor fields are stationary with respect to each other, & produce a steady torque, called asynchronous torque. sites.google.com/site/ncpdhbkhn 7 Introduction to Polyphase Induction Machines (5) sites.google.com/site/ncpdhbkhn 8 Polyphase Induction Machines 1. Introduction to Polyphase Induction Machines 2. Currents and Fluxes in Polyphase Induction Machines 3. Induction – Motor Equivalent Circuit 4. Analysis of the Equivalent Circuit 5. Torque and Power by Use of Thevenin’s Theorem 6. Parameter Determination from No – Load and Blocked – Rotor Tests 7. Effects of Rotor Resistance sites.google.com/site/ncpdhbkhn 9 Currents and Fluxes in Polyphase Induction Machines (1) Resultant flux – density Rotor – mmf wave wave a c b a c b a   90o Rotation Resultant flux – density Rotor – mmf wave wave a c b a c b a 90o   Rotation sites.google.com/site/ncpdhbkhn 10 Currents and Fluxes in Polyphase Induction Machines (2) Flux-density wave Instantaneous bar – voltage magnitudes s 1 2 3 4 5 6 7 8 9 11 12 13 14 15 16 1 (1 s )s Instantaneous bar – current s magnitudes 1 2 3 4 5 6 7 8 9 11 12 13 14 15 16 1  (1 s )s sites.google.com/site/ncpdhbkhn 11 Currents and Fluxes in Polyphase Induction Machines (3) Rotor – mmf wave Flux-density wave  s 1 2 3 4 5 6 7 8 9 11 12 13 14 15 16 1 (1 s )s Fundamental component of rotor – mmf wave sites.google.com/site/ncpdhbkhn 12 Polyphase Induction Machines 1. Introduction to Polyphase Induction Machines 2. Currents and Fluxes in Polyphase Induction Machines 3. Induction – Motor Equivalent Circuit 4. Analysis of the Equivalent Circuit 5. Torque and Power by Use of Thevenin’s Theorem 6. Parameter Determination from No – Load and Blocked – Rotor Tests 7. Effects of Rotor Resistance sites.google.com/site/ncpdhbkhn 13 Induction – Motor Equivalent Circuit a X1 Iˆ X 2 R1 ˆ  Iˆ  I1 2  R2 ˆ Iˆ Iˆ ˆ s V1 c m E2   Rc X m b sites.google.com/site/ncpdhbkhn 14 Polyphase Induction Machines 1. Introduction to Polyphase Induction Machines 2. Currents and Fluxes in Polyphase Induction Machines 3. Induction – Motor Equivalent Circuit 4. Analysis of the Equivalent Circuit 5. Torque and Power by Use of Thevenin’s Theorem 6. Parameter Determination from No – Load and Blocked – Rotor Tests 7. Effects of Rotor Resistance sites.google.com/site/ncpdhbkhn 15 Analysis of the Equivalent Circuit (1) a X1 ˆ X 2 2 R2 R I ˆ 1 Iˆ I2 Pgap n ph I2  1  R s 2 ˆ Iˆ Iˆ ˆ s 2 V1 c m E2 Protor n ph I2 R 2   Rc X m PPPmech gap  rotor b R n I22  n I 2 R ph2s ph 2 2 a X1 ˆ X 2 R I ˆ 1 Iˆ I2 2 1 s  1  R2  nph I2 R 2 s ˆ Iˆ Iˆ ˆ V1 c m E2 Pmech(1  s ) P gap   R X  1 s  c m R Protor sP gap s 2 b sites.google.com/site/ncpdhbkhn 16 Analysis Ex. 1 of the Equivalent Circuit (2) A three – phase, two – pole, 60-Hz induction motor is operating at 3502 r/min with an input power of 15.7 kW and a terminal current of 22.6 A, the stator – winding resistance is 0.20 Ω/phase. Calculate the power dissipated in rotor? 2 2 PIRstator 31 1  3  22.6  0.2  306W PPPgap input  stator 15.7  0.3  15.4 kW 120 120 n f 60  3600r/min spoles e 2 n n 3600  3502 s s   0.0272 ns 3600 Protor sP gap 0.0272  15.4  419 W sites.google.com/site/ncpdhbkhn 17 Analysis of the Equivalent Circuit (3) PTmech  m mech m(1  s )  s Pmech (1  s ) s T mech Pmech(1  s ) P gap Pgap Tmech  s R P n I 2 2 gap ph 2 s 2 nph I2(/) R 2 s Tmech  s PPPshaft mech  rot TTTshaft mech  rot sites.google.com/site/ncpdhbkhn 18 Analysis Ex. 2 of the Equivalent Circuit (4) Given a three – phase, six – pole, Y – connected, 220-V (line-to-line) 60-Hz 7.5-kW induction motor, R1 = 0.249Ω/phase, R2 = 0.144, X1 = 0.503, X2 = 0.209, Xm = 13.25, Prot = 403W (independent of load), s = 2%. Compute the speed, output torque and power, stator current, power factor, and efficiency? a R  2  jX jX X X 2  m R 1 ˆ 2 s  1 Iˆ I2 Z   1  ab R 2 jX  jX 2 m ˆ Iˆ ˆ s V1 m E2 R2 0.144   X  s  j0.209  j 13.25 m 0.02    0.144 b j0.209  j 13.25 0.02 5.43 j 3.11  220 Vˆ ˆ 1 3 o I1   15.93 j 10.14  18.88   32.5 A R1 jX 1  Zab 0.249j 0.503  5.43  j 3.11 sites.google.com/site/ncpdhbkhn 19 Analysis Ex. 2 of the Equivalent Circuit (5) Given a three – phase, six – pole, Y – connected, 220-V (line-to-line) 60-Hz 7.5-kW induction motor, R1 = 0.249Ω/phase, R2 = 0.144, X1 = 0.503, X2 = 0.209, Xm = 13.25, Prot = 403W (independent of load), s = 2%. Compute the speed, output torque and power, stator current, power factor, and efficiency? ˆ o I1 15.93  j 10.14  18.88   32.5 A pf cos(  32.5o )  0.844 lagging 120 120 n f 60  1200 r/min spoles e 6 n(1  s ) ns  (1  0.02)1200  1176 r/min 2 2 2   2 f  2   60  125.7 rad/sec spoles e poles e 6 m(1 s )  s  (1  0.02)125.7  123.2 rad/sec sites.google.com/site/ncpdhbkhn 20 Analysis Ex. 2 of the Equivalent Circuit (6) Given a three – phase, six – pole, Y – connected, 220-V (line-to-line) 60-Hz 7.5-kW induction motor, R1 = 0.249Ω/phase, R2 = 0.144, X1 = 0.503, X2 = 0.209, Xm = 13.25, Prot = 403W (independent of load), s = 2%. Compute the speed, output torque and power, stator current, power factor, and efficiency? ˆ o a I1 18.88   32.5 A X1 X 2 R1 ˆ Iˆ Zab 5.43  j 3.11   I1 2  2 ˆ ˆ Im ˆ PIRgap 3 1 ab V1 E2 R2 2   s 3  18.8  5.41  5740W X m PPPshaft mech  rot b (1 s ) Pgap  P rot (1  0.02)5740  403  5220W Pshaft 5220 Tshaft    42.4Nm m 123.2 sites.google.com/site/ncpdhbkhn 21 Analysis Ex. 2 of the Equivalent Circuit (7) Given a three – phase, six – pole, Y – connected, 220-V (line-to-line) 60-Hz 7.5-kW induction motor, R1 = 0.249Ω/phase, R2 = 0.144, X1 = 0.503, X2 = 0.209, Xm = 13.25, Prot = 403W (independent of load), s = 2%. Compute the speed, output torque and power, stator current, power factor, and efficiency? Pshaft  5220W ˆ ˆ* PVIin  3Re{1 1 } 3Re{127(18.88   32.5o )}  6060W P 5220  shaft   86.1% Pin 6060 sites.google.com/site/ncpdhbkhn 22 Polyphase Induction Machines 1. Introduction to Polyphase Induction Machines 2. Currents and Fluxes in Polyphase Induction Machines 3. Induction – Motor Equivalent Circuit 4. Analysis of the Equivalent Circuit 5. Torque and Power by Use of Thevenin’s Theorem 6. Parameter Determination from No – Load and Blocked – Rotor Tests 7. Effects of Rotor Resistance sites.google.com/site/ncpdhbkhn 23 Torque and Power by Use of Thevenin’s Theorem (1) a jX m VVˆ ˆ X1 X 2 1,eq 1 R1 ˆ Iˆ  I1 2  R1 j() X 1  X m ˆ Vˆ Im Eˆ jXm () R1 jX 1 1 2 R Z1,eq  2 R1 j() X 1  X m   s X m R  jX 1,eq 1, eq b a ˆ V1,eq X X Iˆ  R 1,eq ˆ 2 2  1,eq I2  Z1,eq  jX 2  R 2 / s Vˆ Eˆ n I2 (/) R s 1,eq 2 R T  ph 2 2 2 mech   s s 2 b 1 nph V1, eq (/) R 2 s  2 2 s(/)()R1, eq R 2 s  X 1, eq  X 2 sites.google.com/site/ncpdhbkhn 24 Torque and Power by Use of Thevenin’s Theorem (2) 2 1 nph V1, eq (/) R 2 s ns  n Tmech 2 2 , s  s(/)()R1, eq R 2 s  X 1, eq  X 2 n s 600 400 Motor 200 0 Braking region Motor region Generator region -200 Torque -400 -600 -800 Generator -1000 -1200 2 1.5 1 0.5 0 -0.5 -1 -1.5 Fractional slip s sites.google.com/site/ncpdhbkhn 25 Torque and Power by Use of Thevenin’s Theorem (3) 2 1 nph V1, eq (/) R 2 s Tmech  2 2 s(/)()R1, eq R 2 s  X 1, eq  X 2 2 2 2 2 dTmech nph V1, eq R 1, eq()(/) X 1, eq  X 2  R 2 s  2 2 2  0 d( R2 / s )s [( R 1, eq R 2 / s )  ( X 1, eq  X 2 ) ] R 2 RXX2 ()  2 s 1,eq 1, eq 2 2  1 0.5nph V1, eq Tmax   2 2  s RRXX1,eq 1, eq () 1, eq  2    R2 smaxT   RXX2()  2  1,eq 1, eq 2 sites.google.com/site/ncpdhbkhn 26 Torque and Power by Use of Thevenin’s Theorem (4) 600 I 1 P T 500 400 300 200 100 0 -100 -2 -1.8 -1.6 -1.4 -1.2 -1 -0.8 -0.6 -0.4 -0.2 0 s sites.google.com/site/ncpdhbkhn 27 Torque and Power by Use of Ex. Thevenin’s Theorem (5) Given a three – phase, six – pole, Y – connected, 220-V (line-to-line) 60-Hz 7.5-kW induction motor, R1 = 0.249Ω/phase, R2 = 0.144, X1 = 0.503, X2 = 0.209, Xm = 13.25, s = 3%. Compute (a) I2, Tmech, Pmech; (b) the maximum electromechanical torque & the corresponding speed; (c) the electromechanical starting torque Tstart & the corresponding stator load current I2,start? a jX m VVˆ ˆ X1 X 2 1,eq 1 R1 ˆ Iˆ  I1 2  R1 j() X 1  X m ˆ Vˆ Im Eˆ 220j 13.25 1 2 R  2   s 3 0.249j (0.503  13.25) X m 122.4  1o V b jX() R jX a m 1 1 X X Z1,eq  R 1, eq  jX 1, eq R 1,eq ˆ 2  1,eq I2  R1 j() X 1  X m Vˆ Eˆ j13.25(0.249 j 0.503) 1,eq 2 R  2 0.249j (0.503  13.25)   s 0.231 j 0.449  b sites.google.com/site/ncpdhbkhn 28 Torque and Power by Use of Ex. Thevenin’s Theorem (6) Given a three – phase, six – pole, Y – connected, 220-V (line-to-line) 60-Hz 7.5-kW induction motor, R1 = 0.249Ω/phase, R2 = 0.144, X1 = 0.503, X2 = 0.209, Xm = 13.25, s = 3%. Compute (a) I2, Tmech, Pmech; (b) the maximum electromechanical torque & the corresponding speed; (c) the electromechanical starting torque Tstart & the corresponding stator load current I2,start? a o ˆ X X V1,eq122.4  1 V; Z 1, eq  0.231  j 0.449  R 1,eq ˆ 2  1,eq I2  Vˆ ˆ RX1,eq 0.231  ; 1, eq  0.449  1,eq E2 R2   s Vˆ ˆ 1,eq I2  b Z1,eq  jX 2  R 2 / s 122.4 1o  0.231j 0.449  j 0.209  0.144 / 0.03 23.916 j 2.877  24.089   6.9o A sites.google.com/site/ncpdhbkhn 29 Torque and Power by Use of Ex. Thevenin’s Theorem (7) Given a three – phase, six – pole, Y – connected, 220-V (line-to-line) 60-Hz 7.5-kW induction motor, R1 = 0.249Ω/phase, R2 = 0.144, X1 = 0.503, X2 = 0.209, Xm = 13.25, s = 3%. Compute (a) I2, Tmech, Pmech; (b) the maximum electromechanical torque & the corresponding speed; (c) the electromechanical starting torque Tstart & the corresponding stator load current I2,start? ˆ o I2 24.089   6.9 A R 0.144 P n I2 2 (1  s ) 3  24.0892 (1  0.03)  8105W mech ph 2 s 0.03 2 2 2   2 f  2   60  125.7 rad/sec spoles e poles e 6 m(1 s )  s  (1  0.03)125.7  121.9 rad/sec Pmech 8105 Tmech    66.5Nm m 121.9 sites.google.com/site/ncpdhbkhn 30 Torque and Power by Use of Ex. Thevenin’s Theorem (8) Given a three – phase, six – pole, Y – connected, 220-V (line-to-line) 60-Hz 7.5-kW induction motor, R1 = 0.249Ω/phase, R2 = 0.144, X1 = 0.503, X2 = 0.209, Xm = 13.25, s = 3%. Compute (a) I2, Tmech, Pmech; (b) the maximum electromechanical torque & the corresponding speed; (c) the electromechanical starting torque Tstart & the corresponding stator load current I2,start? ˆ o V1,eq122.4  1 V; Z 1, eq  0.231  j 0.449  ; s  125.7 rad/sec 2 2 1 0.5 3  122.4  1 0.5nph V1, eq  Tmax  2 2  2 2 125.7 0.231 0.231  (0.449  0.209)  s RRXX1,eq 1, eq () 1, eq  2  0.144 R   2 2 2 smaxT  0.231 (0.449  0.209)  RXX2()  2  1,eq 1, eq 2   Tmax  192.6 Nm   s  0.2065  maxT sites.google.com/site/ncpdhbkhn 31 Torque and Power by Use of Ex. Thevenin’s Theorem (9) Given a three – phase, six – pole, Y – connected, 220-V (line-to-line) 60-Hz 7.5-kW induction motor, R1 = 0.249Ω/phase, R2 = 0.144, X1 = 0.503, X2 = 0.209, Xm = 13.25, s = 3%. Compute (a) I2, Tmech, Pmech; (b) the maximum electromechanical torque & the corresponding speed; (c) the electromechanical starting torque Tstart & the corresponding stator load current I2,start? Vˆ ˆ 1,eq s1  I2,start  Z1,eq  jX 2  R 2 /1 122.4 1o  154.5   60.7o A 0.231j 0.449  j 0.209  0.144 2 2 nph I2 R 2 3 154.5  0.144 Tstart    82.04 Nm s 125.7 sites.google.com/site/ncpdhbkhn 32 Torque and Power by Use of Thevenin’s Theorem (10) 100 90 80 70 60 50 Torque 40 30 R = 0.1 2 R = 0.2 20 2 R = 0.5 2 R = 1.0 10 2 R = 1.5 2 0 0 200 400 600 800 1000 1200 1400 1600 1800 rpm sites.google.com/site/ncpdhbkhn 33 Polyphase Induction Machines 1. Introduction to Polyphase Induction Machines 2. Currents and Fluxes in Polyphase Induction Machines 3. Induction – Motor Equivalent Circuit 4. Analysis of the Equivalent Circuit 5. Torque and Power by Use of Thevenin’s Theorem 6. Parameter Determination from No – Load and Blocked – Rotor Tests a) No – Load Test b) Blocked – Rotor Test 7. Effects of Rotor Resistance sites.google.com/site/ncpdhbkhn 34 No – Load Test 2 Prot P nl  n ph I1, nl R 1 This test is ordinarily performed d P m rot at rated frequency & with balanced JT rot   dt m polyphase voltages applied to the stator terminals dm PJrot() m    m dt a X X 2 R 1 ˆ 2 1 Iˆ I2 Pcore P nl  P rot  n ph I1, nl R 1  1  ˆ Vˆ Im Eˆ 2 1 2 R nph V1, nl 2 R    s c X m Pcore b XXXXnl11  1  m 2 2 Qnl S nl  P nl, S nl  n ph V1, nlnl I 1, Qnl V1,nl X nl 2  nph I1, nl I 1, nl sites.google.com/site/ncpdhbkhn 35 Blocked – Rotor Test (1) The rotor is blocked so that it can not rotate (hence the slip is equal to unity), and balaced polyphase voltages are applied to the stator terminals 2 2 a Qbl S bl  P bl, S bl  n ph V1, bl I 1, bl X X R 1 ˆ 2 1 Iˆ I2 f Q  1  X  r bl bl 2 ˆ Iˆ ˆ V1 m E2 fbl n ph I1, bl R2  X  s P m R  bl bl 2 b nph I1, bl Zbl R1  jX 1 [( R 2  jX 2 )in parallel with jX m ] 2 2 XXRXXXm m[2 2 ( m  2 )]  R1  R 22 2  j X 1  2 2  RXXRXX2()()m  2 2  m  2  RX2  m 2 XXXm2 m  Zbl  R1  R 22  j X 1   ()XXXXm2 2  m  sites.google.com/site/ncpdhbkhn 36 Blocked – Rotor Test (2) 2 XXXm2 m  Zbl  R1  R 22  j X 1   Rbl  jX bl ()XXXXm2 2  m  2  X m RRRbl 1  2 2 2  RXX2()m  2   XRXXX[2  (  )] XX  m2 2 m 2  bl 1 2 2  RXX2()m  2  2 XX2  m  RRR2()bl  1    X m    X XXX()  m  2bl 1  XXXm1  bl XXXXnl11  1  m XXXm  nl  1 XXnl  1 XXX2 ()bl  1 XXnl bl sites.google.com/site/ncpdhbkhn 37 Polyphase Induction Machines 1. Introduction to Polyphase Induction Machines 2. Currents and Fluxes in Polyphase Induction Machines 3. Induction – Motor Equivalent Circuit 4. Analysis of the Equivalent Circuit 5. Torque and Power by Use of Thevenin’s Theorem 6. Parameter Determination from No – Load and Blocked – Rotor Tests 7. Effects of Rotor Resistance sites.google.com/site/ncpdhbkhn 38 Effects of Rotor Resistance 100 90 80 70 60 50 Torque 40 30 R = 0.1 2 R = 0.2 20 2 R = 0.5 2 R = 1.0 10 2 R = 1.5 2 0 0 200 400 600 800 1000 1200 1400 1600 1800 rpm sites.google.com/site/ncpdhbkhn 39