Transcript Slide 1
AC DRIVES AC motor Drives are used in many industrial and domestic application, such as in conveyer, lift, mixer, escalator etc. The AC motor have a number of advantages : • Lightweight (20% to 40% lighter than equivalent DC motor) • Inexpensive • Low maintenance The Disadvantages AC motor : * The power control relatively complex and more expensive There are two type of AC motor Drives : 1. Induction Motor Drives 2. Synchronous Motor Drives INDUCTION MOTOR DRIVES Three-phase induction motor are commonly used in adjustable-speed drives (ASD). Basic part of three-phase induction motor : Three-phase windings Rotor windings • Stator Threephase supply • Rotor • Air gap Stator Air gap s Rotor m T Three-phase windings Rotor windings Threephase supply Stator Air gap s Rotor m T The stator winding are supplied with balanced three-phase AC voltage, which produce induced voltage in the rotor windings. It is possible to arrange the distribution of stator winding so that there is an effect of multiple poles, producing several cycle of magnetomotive force (mmf) or field around the air gap. The speed of rotation of field is called the synchronous speed s , which is defined by : ωs is syncronous speed [rad/sec] 2 Ns is syncronous speed [rpm] s or p is numbers of poles p ω is the supply frequency [rad/sec] f is the supply frequency [Hz] 120 f Ns Nm is motor speed p The motor speed m s (1 S ) The rotor speed or motor speed is : Where S is slip, as defined as : Three-phase windings m S S S Rotor windings Threephase supply Stator Air gap s Rotor m T Or S NS Nm NS Equivalent Circuit Of Induction Motor Three-phase windings Where : Rotor windings Rs is resistance per-phase of stator winding Threephase supply Rr is resistance per-phase of rotor winding Air gap Stator Rotor m s Is Xs Xs is leakage reactance per-phase of the winding stator Xs is leakage reactance per-phase of the winding rotor T Xr’ Rs Im Xm is magnetizing reactance Ir’ Rr’/s Vs Xm Stator Rm Air gap motor Rm is Core losses as a reactance Performance Characteristic of Induction Motor Is Xs Xr’ Rs Im Ir’ Rr’/s Vs Xm Stator Rm Air gap motor Stator copper loss : Ps cu 3 I s Rs Rotor copper loss : Pr cu 3 ( I r ) 2 Rr 2 ' 2 Core losses : 2 V V Pc 3 m 3 s Rm Rm ' Performance Characteristic of Induction Motor - Power developed on air gap (Power fropm stator to ' rotor through air gap) : ' 2 Rr Pg 3 ( I r ) S ' 2 - Power developed by motor : Pd Pg Pr cu 3 ( I r ) ' Pd Pg (1 S ) or - Torque of motor : or Td Pd m Pd 60 Td 2 N m or Pg (1 S ) S (1 S ) Pg s Rr (1 S ) S Performance Characteristic of Induction Motor Input power of motor : Pi 3Vs I s cosm Pc Ps cu Pg Output power of motor : Po Pd Pno load Pd Pno load Po Efficiency : Pi Pc Ps cu Pg Performance Characteristic of Induction Motor If Pg (Pc Ps cu ) and Pd Pno load so, the efficiency can calculated as : Pd Pg (1 S ) 1 S Pg Pg Performance Characteristic of Induction Motor Generally, value of reactance magnetization Xm >> value Rm (core losses) and also X m 2 ( Rs 2 X s 2 ) So, the magnetizing voltage same with the input voltage : Vm Vs Therefore, the equivalent circuit is ; Is Xs Ii Im Xm Xr’ Rs Rr’/s Xm Vs Rm Po Pi Stator Air gap motor Is=Ir’ Ir’ Im Ir’ Rr’/s Vs Xs Xr’ Rs Stator Air gap rotor Performance Characteristic of Induction Motor Total Impedance of this circuit is : Ii Xs Xr’ Rs Is=Ir’ ' Ir’ Im Rr’/s Xm Vs Po Pi Stator Air gap rotor Ir ' The rotor current is : R X m ( X s X r ) jX m ( Rs r ) S Zi ' R ' Rs r j ( X m X s X r ) S ' Vs 2 Rr' ' Rs X s X r S 2 1 2 Ii Xs Xr’ Rs Is=Ir’ Ir’ Im Td Rr’/s Vs Po Pi Stator Air gap 3 Rr' Vs2 Rr' S s Rs S 2 X s X r' rotor Tmax Td Tst TL Tm=TL Operating point S=1 Nm =0 Torque – speed Characteristic Smax S=Sm S=0 m s Nm Ns 2 Three region operation : 1. Motoring : 0 S 1 2. Regenerating : S 0 3. Plugging : 1 S 2 Torque Forward regeneration Forward motoring Reverse plugging Tmax m s m s m s Tst S=Sm -Smax s S=-1 s Smax S=0 Ns -Tmax s =0 S=1 Nm =0 s S=2 Performance Characteristic of Induction Motor Starting speed of motor is m = 0 or S = 1, Starting torque of motor is : Tst 3 Rr' Vs2 ' 2 Rr ' 2 s Rs X s X r S Slip for the maximum torque Smax can be found by setting : So, the slip on maximum torque is : S max d Td 0 dS Rr' R X X 2 s s 1 2 ' 2 r Performance Characteristic of Induction Motor Torque maximum is : Tmax 3 Vs2 2s Rs Rs X s X r' 2 2 And the maximum regenerative torque can be found as : Tmax 3 Vs2 2s Rs Rs X s X r' Where the slip of motor s = - Sm 2 2 Speed-Torque Characteristic : Td Rr' S s Rs S X For the high Slip S. (starting) So, the torque of motor is : 3 Rr' Vs2 s Td And starting torque (slip S=1) is : X ' 2 r 2 X s X r' R Rs S ' r 3 Rr' Vs2 S s X s X Tst 2 2 ' 2 r 3 Rr' Vs2 s X s X ' 2 r For low slip S region, the motor speed near unity or synchronous speed, in this region the impedance motor is : R' ' 2 X So, the motor torque is : s Xr r S 3Vs2 S Td s R 'r And the slip at maximum torque is : S max Rr' R X X 1 2 ' 2 r 2 s The maximum motor torque is : Rs Td s 3 Rr' Vs2 Rr' S s Rs S 2 X s X r' 2 Stator Voltage Control AC Variable Voltage Sources Controlling Induction Motor Speed by Adjusting The Stator Voltage Td IM Td Vs 3 Rr' Vs2 Rr' S s Rs S 2 X s X r' 2 Td Vs > Vs1 > Vs2 Ii Xs Xr’ Rs Tmax Is=Ir’ Ir’ Im Rr’/s Vs Po Pi Stator air gap Tst Tst1 Tst2 TL rotor S=1 Nm =0 2 1 S=0 s Ns Frequency Voltage Control Controlling Induction Motor Speed by Adjusting The Frequency Stator Voltage AC Variable Voltage Sources IM Vs f Td Td 3 Rr' Vs2 Rr' S s Rs S 2 ' 2 X s X r Td fs2 < fs1 < fs Tmax Ii Xs Xr ’ Rs Ir’ Im Vs Tst2 Is=Ir’ Rr’/s f TL Po Pi Stator Tst1 Tst Air gap rotor S=1 m =0 2 S=0 fs2 1 S=0 fs1 s S=0 fs If the frequency is increased above its rated value, the flux and torque would decrease. If the synchronous speed corresponding to the rated frequency is call the base speed b, the synchronous speed at any other frequency becomes: s b And : S b m 1 m b b The motor torque : Td Td 3 Rr' Vs2 Rr' S s Rs S 2 X s X r' 2 3 Rr' Vs2 Rr' S b Rs S 2 ' 2 X s X r If Rs is negligible, the maximum torque at the base speed as : 3 Vs2 Tmb 2S b X s X r' And the maximum torque at any other frequency is : 2 Vs 3 Tm 2S b X s X r' 2 Sm At this maximum torque, slip S is : Rr ' X s X r' 2 Normalizing : Vs 3 Tm 2S b X s X r' 2 3 Vs2 Tmb 2S b X s X r' Tm 1 2 Tmb And Tm 2 Tmb Example : A three-phase , 11.2 kW, 1750 rpm, 460 V, 60 Hz, four pole, Y-connected induction motor has the following parameters : Rs = 0.1W,Rr’ = 0.38W,Xs = 1.14W,Xr’ = 1.71W,and Xm = 33.2W.If the breakdown torque requiretment is 35 Nm, Calculate : a) the frequency of supply voltage, b) speed of motor at the maximum torque Solution : Input voltage per-phase : Vs Base frequency : b 2 f 2 x 3.14 x 60 377rad / s Base Torque : Tmb Motor Torque : 460 265volt 3 60Po 60 x11200 61.11Nm 2 N m 2 x 3.14 x1750 Tm 35 Nm a) the frequency of supply voltage : Tm 1 2 Tmb Tmb 61.11 1.321 Tm 35 Synchronous speed at this frequency is : s b s 1.321x 377 498.01rad / s or 60 x 498.01 N s Nb 4755.65 rpm 2 x p NS 4 x 4755.65 So, the supply frequency is : f s 158.52 Hz 120 b 120 b) speed of motor at the maximum torque : Rr ' X s X r' Rr’ = 0.38W, Xs = 1.14W, Xr’ = 1.71Wand 1.31 At this maximum torque, slip Sm is : So, Sm 0.38 0.101 1.3211.14 1.71 Sm or, Nm NS (1 S ) 4755.65(1 0.101) 4275rpm CONTROLLING INDUCTION MOTOR SPEED USING ROTOR RESISTANCE (Rotor Voltage Control) Wound rotor induction motor applications cranes CONTROLLING INDUCTION MOTOR SPEED USING ROTOR RESISTANCE (Rotor Voltage Control) Equation of Speed-Torque : Td 3 Rr' Vs2 Rr' S s Rs S In a wound rotor induction motor, an external three-phase resistor may be connected to its slip rings, RX Stator RX Rotor Three-phase supply RX ' 2 X s X r 2 3Vs2 S Td s R 'r These resistors Rx are used to control motor starting and stopping anywhere from reduced voltage motors of low horsepower up to large motor applications such as materials handling, mine hoists, cranes etc. The most common applications are: AC Wound Rotor Induction Motors – where the resistor is wired into the motor secondary slip rings and provides a soft start as resistance is removed in steps. AC Squirrel Cage Motors – where the resistor is used as a ballast for soft starting also known as reduced voltage starting. DC Series Wound Motors – where the current limiting resistor is wired to the field to control motor current, since torque is directly proportional to current, for starting and stopping. The developed torque may be varying the resistance Rx The torque-speed characteristic for variations in rotor resistance This method increase the starting torque while limiting the starting current. The wound rotor induction motor are widely used in applications requiring frequent starting and braking with large motor torque (crane, hoists, etc) The three-phase resistor may be replaced by a three-phase diode rectifier and a DC chopper. The inductor Ld acts as a current source Id and the DC chopper varies the effective resistance: Re R(1 k ) Where k is duty cycle of DC chopper The speed can controlled by varying the duty cycle k, (slip power) Id Ld Stator D1 D3 Vd Rotor Three-phase supply D5 D4 D6 D2 GTO R Vdc The slip power in the rotor circuit may be returned to the supply by replacing the DC converter and resistance R with a three-phase full converter (inverter) Three-phase supply Transformer Id Ld Stator D1 D3 Vd Rotor Slip Power D5 D4 D6 Diode rectifier D2 T1 T3 T5 T2 T4 T6 Vdc Controlled rectifier/ inverter Na:Nb Example: A three-phase induction motor, 460, 60Hz, six-pole, Y connected, wound rotor that speed is controlled by slip power such as shown in Figure below. The motor parameters are Rs=0.041 W, Rr’=0.044 W, Xs=0.29 W, Xr’=0.44 W and Xm=6.1 W. The turn ratio of the rotor to stator winding is nm=Nr/Ns=0.9. The inductance Ld is very large and its current Id has negligible ripple. The value of Rs, Rr’, Xs and Xr’ for equivalent circuit can be considered negligible compared with the effective impedance of Ld. The no-load of motor is negligible. The losses of rectifier and Dc chopper are also negligible. The load torque, which is proportional to speed square is 750 Nm at 1175 rpm. (a) If the motor has to operate with a minimum speed of 800 rpm, determine the resistance R, if the desired speed is 1050 rpm, (b) Calculate the inductor current Id. (c) The duty cycle k of the DC chopper. (d) The voltage Vd. (e) The efficiency. (f) The power factor of input line of the motor. Vs 460 265.58 volt 3 p 6 2 x 60 377rad / s s 2 x 377/ 6 125.66 rad / s The equivalent circuit : The dc voltage at the rectifier output is : Vd I d Re I d R (1 k ) Nr Er S Vs S Vs nm Ns and For a three-phase rectifier, relates Er and Vd as : Vd 1.65 x 2 Er 2.3394Er Nr E S V S Vs nm Using : r s Ns Vd 2.3394S Vs nm If Pr is the slip power, air gap power is : Pg Pr S Pr 3Pr (1 S ) S) Developed power is : Pd 3( Pg Pr ) 3( S S Because the total slip power is 3Pr = Vd Id and So, Pd TL m (1 S )Vd I d Pd TLm TLm (1 S ) S Substituting Vd from Solving for Id gives : Vd 2.3394S Vs nm TLs Id 2.3394Vs nm In equation : Pd above, so Which indicates that the inductor current is independent of the speed. Vd I d Re I d R (1 k ) and equation : Vd 2.3394S Vs nm I d R(1 k ) 2.3394S Vs nm From equation : So, Which gives : S I d R(1 k ) 2.3394S Vs n m I d R(1 k ) S 2.3394S Vs n m I d R(1 k ) m s (1 S ) s 1 2 . 3394 V n s m The speed can be found from equation : as : TLs R(1 k ) m s 1 2 ( 2 . 3394 V n ) s m Which shows that for a fixed duty cycle, the speed decrease with load torque. By varying k from 0 to 1, the speed can be varied from minimum value to s m 180 / 30 83.77 rad / s Kvm 2 800 750x 347.67 Nm 1175 From torque equation : T L 2 From equation : TLs Id The corresponding inductor current is : 2.3394Vs nm 347.67 x 125.66 Id 78.13 A 2.3394x 265.58 x 0.9 The speed is minimum when the duty-cycle k is zero and equation : I d R(1 k ) m s (1 S ) s 1 2 . 3394 V n s m 78.13 R 83.77 125.66(1 ) 2.3394x 265.58 x 0.9 And : R 2.3856W