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Lecture Electromechanical energy conversion: Polyphase Induction Machines - Nguyễn Công Phương

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This chapter include contents: Introduction to Polyphase Induction Machines, currents and fluxes in Polyphase Induction Machines, induction – motor equivalent circuit, analysis of the equivalent circuit, torque and power by use of Thevenin’s theorem, parameter determination from no – load and blocked – rotor tests, effects of rotor resistance.

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 Polyphase Induction Machines Introduction to Polyphase Induction Machines Currents and Fluxes in Polyphase Induction Machines Induction – Motor Equivalent Circuit Analysis of the Equivalent Circuit Torque and Power by Use of Thevenin’s Theorem Parameter Determination from No – Load and Blocked – Rotor Tests Effects of Rotor Resistance sites.google.com/site/ncpdhbkhn 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 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) ns  n (%) • The fractional slip: 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 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 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 Introduction to Polyphase Induction Machines (5) sites.google.com/site/ncpdhbkhn Polyphase Induction Machines Introduction to Polyphase Induction Machines Currents and Fluxes in Polyphase Induction Machines Induction – Motor Equivalent Circuit Analysis of the Equivalent Circuit Torque and Power by Use of Thevenin’s Theorem Parameter Determination from No – Load and Blocked – Rotor Tests Effects of Rotor Resistance sites.google.com/site/ncpdhbkhn Currents and Fluxes in Polyphase Induction Machines (1) Resultant flux – density wave Rotor – mmf wave a c b a   90o c b Rotation Resultant flux – density wave Rotor – mmf wave a c a b a   90o   c b a Rotation sites.google.com/site/ncpdhbkhn 10 Torque and Power by Use of Thevenin’s Theorem (2) Tmech n phV1,2eq ( R2 / s )  , s ( R1,eq  R2 / s )  ( X 1,eq  X ) s ns  n ns Motor 600 400 200 Braking region Motor region Generator region Torque -200 Generator -400 -600 -800 -1000 -1200 1.5 0.5 -0.5 -1 -1.5 Fractional slip s sites.google.com/site/ncpdhbkhn 25 Torque and Power by Use of Thevenin’s Theorem (3) Tmech n phV1,2eq ( R2 / s )  s ( R1,eq  R2 / s )  ( X 1,eq  X ) n phV1,2eq R1,2eq  ( X 1,eq  X )  ( R2 / s ) dTmech   0 2 d ( R2 / s ) s [( R1,eq  R2 / s )  ( X 1,eq  X ) ]  R2  R1,2eq  ( X 1,eq  X ) s  0.5n phV1,2eq Tmax  2  R  R  ( X  X ) s  1,eq 1,eq 1,eq  R2 s  2  max T R  ( X  X ) 1, eq 1, eq  sites.google.com/site/ncpdhbkhn 26 Torque and Power by Use of Thevenin’s Theorem (4) 600 I1 P T 500 400 300 200 100 -100 -2 -1.8 -1.6 -1.4 -1.2 -1 s -0.8 sites.google.com/site/ncpdhbkhn -0.6 -0.4 -0.2 27 Ex Torque and Power by Use of 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? Vˆ1,eq Z1,eq jX m  Vˆ1 R1  j ( X  X m ) 220 j13.25  0.249  j (0.503  13.25)  122.41o V jX m ( R1  jX )   R1,eq  jX 1,eq R1  j ( X  X m ) j13.25(0.249  j 0.503)  0.249  j (0.503  13.25)  0.231  j 0.449  a R1  Iˆ1 X1 Iˆ2 Iˆm Vˆ1   X2 Eˆ R2 s  Xm b a  Vˆ1,eq R1,eq X 1,eq Iˆ2  Eˆ R2 s   sites.google.com/site/ncpdhbkhn X2 b 28 Ex Torque and Power by Use of 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? Vˆ1,eq  122.41o V; Z1,eq  0.231  j 0.449   R1,eq  0.231; X 1,eq  0.449 Iˆ2  a  Vˆ1,eq R1,eq X 1,eq Iˆ2  Eˆ  Vˆ1,eq X2 R2 s  b Z1,eq  jX  R2 / 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 Thevenin’s Theorem (7) Ex 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? Iˆ2  24.089  6.9o A Pmech R2 0.144 n I (1  s )   24.089 (1  0.03)  8105W s 0.03 ph 2 2 s  e  2 f e  2  60  125.7 rad/sec poles poles m  (1  s )s  (1  0.03)125.7  121.9 rad/sec Tmech  Pmech 8105   66.5 Nm m 121.9 sites.google.com/site/ncpdhbkhn 30 Ex Torque and Power by Use of 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? Vˆ1,eq  122.41o V; Z1,eq  0.231  j 0.449 ; s  125.7 rad/sec 2 0.5   122.4  0.5n phV1,eq  T   max 125.7 0.231  0.2312  (0.449  0.209)2 2  R1,eq  ( X 1,eq  X ) s R1,eq    0.144 R  s max T  0.2312  (0.449  0.209)2 2  R1,eq  ( X 1,eq  X )   Tmax  192.6 Nm   smax T  0.2065 sites.google.com/site/ncpdhbkhn 31 Ex Torque and Power by Use of 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? s   Iˆ2,start  Vˆ1,eq Z1,eq  jX  R2 /1 122.41o  154.5  60.7o A  0.231  j 0.449  j 0.209  0.144 Tstart  n ph I 22 R2 s  154.52  0.144   82.04 Nm 125.7 sites.google.com/site/ncpdhbkhn 32 Torque and Power by Use of Thevenin’s Theorem (10) 100 90 80 70 Torque 60 50 40 30 R2 = 0.1 R2 = 0.2 20 R2 = 0.5 R2 = 1.0 10 R2 = 1.5 0 200 400 600 800 1000 1200 1400 1600 1800 rpm sites.google.com/site/ncpdhbkhn 33 Polyphase Induction Machines Introduction to Polyphase Induction Machines Currents and Fluxes in Polyphase Induction Machines Induction – Motor Equivalent Circuit Analysis of the Equivalent Circuit Torque and Power by Use of Thevenin’s Theorem Parameter Determination from No – Load and Blocked – Rotor Tests a) No – Load Test b) Blocked – Rotor Test Effects of Rotor Resistance sites.google.com/site/ncpdhbkhn 34 No – Load Test Prot  Pnl  n ph I1,2nl R1 d m Prot J  Trot   dt m  Prot (m )  m J d m dt a Pcore  Pnl  Prot  n I R1 ph 1,nl Rc  This test is ordinarily performed at rated frequency & with balanced polyphase voltages applied to the stator terminals ph 1,nl n V Pcore R1  Vˆ1  X nl  X 11  X  X m Iˆ1 X1 Iˆ2 Iˆm  X2 Eˆ R2 s  Xm b Qnl  Snl2  Pnl2 , Snl  n phV1,nl I1,nl X nl  V1,nl Qnl  n ph I1,2nl I1,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 Qbl  Sbl2  Pbl2 , Sbl  n phV1,bl I1,bl f r Qbl X bl  f bl n ph I1,2bl Pbl Rbl  n ph I1,2bl a R1  Vˆ1  Iˆ1 X1 Iˆ2 Iˆm  X2 Eˆ R2 s  Xm b Z bl  R1  jX  [( R2  jX ) in parallel with jX m ]  X m2 X m [ R22  X ( X m  X )]   R1  R2  j  X1   2 R2  ( X m  X ) R2  ( X m  X )   R2  X m  X m2 X2Xm   Z bl  R1  R2  j X    ( X m  X )2 X  X m   sites.google.com/site/ncpdhbkhn 36 Blocked – Rotor Test (2)  X m2 X2Xm  Z bl  R1  R2  j X     Rbl  jX bl (Xm  X2) X2  Xm    X m2  Rbl  R1  R2 R  ( X  X )  m  X [ R  X  X  m  X ( X m  X )]  bl R22  ( X m  X ) 2   X2  Xm   R2  ( Rbl  R1 )   X  m    Xm X  (X  X ) bl  X m  X  X bl  X nl  X 11  X  X m  X m  X nl  X  X  ( X bl  X ) sites.google.com/site/ncpdhbkhn X nl  X X nl  X bl 37 Polyphase Induction Machines Introduction to Polyphase Induction Machines Currents and Fluxes in Polyphase Induction Machines Induction – Motor Equivalent Circuit Analysis of the Equivalent Circuit Torque and Power by Use of Thevenin’s Theorem Parameter Determination from No – Load and Blocked – Rotor Tests Effects of Rotor Resistance sites.google.com/site/ncpdhbkhn 38 Effects of Rotor Resistance 100 90 80 70 Torque 60 50 40 30 R2 = 0.1 R2 = 0.2 20 R2 = 0.5 R2 = 1.0 10 R2 = 1.5 0 200 400 600 800 1000 1200 1400 1600 1800 rpm sites.google.com/site/ncpdhbkhn 39 ... Introduction to Polyphase Induction Machines (5) sites.google.com/site/ncpdhbkhn Polyphase Induction Machines Introduction to Polyphase Induction Machines Currents and Fluxes in Polyphase Induction Machines. .. of Thevenin’s Theorem (4) 600 I1 P T 500 400 300 200 100 -1 00 -2 -1 .8 -1 .6 -1 .4 -1 .2 -1 s -0 .8 sites.google.com/site/ncpdhbkhn -0 .6 -0 .4 -0 .2 27 Ex Torque and Power by Use of Thevenin’s Theorem... sites.google.com/site/ncpdhbkhn 22 Polyphase Induction Machines Introduction to Polyphase Induction Machines Currents and Fluxes in Polyphase Induction Machines Induction – Motor Equivalent Circuit

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