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INDUCTION MOTORS 1

INDUCTION MOTORS 1. ROTOR CIRCUIT MODEL. The rotor current : I R = E R / [R R +jX R ] (1) I R =E R / [R R +js X R0 ] or I R =E R0 / [R R /s +j X R0 ] (2)

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INDUCTION MOTORS 1

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  1. INDUCTION MOTORS 1

  2. ROTOR CIRCUIT MODEL • The rotor current: IR = ER/ [RR+jXR] (1) • IR=ER/ [RR+js XR0] or IR=ER0 / [RR /s +j XR0] (2) • Note:from last equation, can treat rotor effects due to varying rotor speedas caused by a varying impedance supplied from a constant voltage source ER0 • Equivalent rotor impedance from this point of view: ZR, eq = RR / s + jXR0 (3) rotor equivalent circuit using this convention shown next 

  3. ROTOR CIRCUIT MODEL • Rotor voltage is ER0 constant& rotor impedance ZR,eqcontains effects of varying slip

  4. ROTOR CIRCUIT MODEL • Plot of current flow in rotorfrom equations: (1) & (2)shown below:

  5. FINAL EQUIVALENT CIRCUIT • Note:at very low slips resistive term RR/s>>XR0  rotor resistance predominates & rotor current varies linearly with slip • at high slips, XR0 much larger than RR/s& rotor current approaches steady-state valueas slip becomes very large • To develop a single, per-phase, equivalent circuit for induction motor  should refer to rotor part of model over stator side • Rotor circuit model, referred to stator side (shown in last equivalent circuit of rotor) in which effect of speed variation is concentrated in impedance term

  6. FINAL EQUIVALENT CIRCUIT • Reminding: that in an ordinary transformer, voltages, currents and impedances on secondary sideof device can be referred to primary sideby means of turns ratio of transformer • Vp=V’s=a Vs • Ip=I’s=Is / a • Z’s = a^2 Zs • Same sort of transformation employed for induction motor’s rotor circuit • If the effective turns ratio of induction motor = aeff The transformed voltages are: E1=E’R=aeff ER0 • and rotor current is :I2=IR / aeff • And rotor impedance become:Z2 =aeff^2 (RR/ s +j XR0)

  7. FINAL EQUIVALENT CIRCUIT • If the following definitions employed: R2 = aeff^2 RR X2 =aeff^2 XR0 • the equivalent circuit of induction motor is shown as below: (per-phase equivalent circuit)

  8. FINAL EQUIVALENT CIRCUIT • Rotor resistance RR & locked-rotor reactance XR0are difficult or impossible to determinedirectly on cage rotors& effective turns ratio aeff also difficult to determine for cage rotors • fortunately,can make measurements that directly providereferred resistanceand reactance R2 & X2 , (though RR, XR0 and aeff not known separately) • measurement of these parameters covered later

  9. POWER & TORQUE IN INDUCTION MOTORS-LOSSES • since induction motoris a singly excited machine,its power & torque relationships is different from sync. machines • Losses & power-flow Diagram • the input is electric power and the output mechanical power (while rotor windings are short circuited) • As shown in power flow Figure (next), Pin is in form of 3 phase electric voltages & currents • 1st losses is stator winding losses I^2 R=PSCL • 2nd Hysteresis & Eddy currents loss in stator Pcore • Power remained at this point transferred to rotor through air gap: is called air-gap power PAG

  10. POWER & TORQUE IN INDUCTION MOTORS-LOSSES • Part of power transferred to rotor lost as : I^2R=PRCL& rest converted from electrical to mechanical form Pconv, friction & windage losses PF&W & stray losses Pmisc subtracted Pout

  11. POWER & TORQUE IN INDUCTION MOTORS-LOSSES • Note:in practice core loss is partially related to stator and partially to rotor, however since induction motor operates at a speed near synchronous speed, relative motion of magnetic field over rotor surface is quite low (frequency of induced voltage = s fe)& rotor core losses are very tiny • These losses in induction motor equivalent circuit represented by a resistor RC (or GC) , • If core losses are given as a number (X Watts)often lumped with mechanical losses& subtracted at point on diagram where mechanical losses are located

  12. POWER & TORQUE IN INDUCTION MOTORS-LOSSES • The higher the speed of an induction motor,the higher its friction, windage, and stray losses,while the lower the core lossessometimes these 3 categories of losses are lumped together and calledrotational losses • Since component losses of rotational losseschange in opposite directions with a change in speed,total rotational losses of a motor often considered constant • EXAMPLE : A 480, 60 Hz, 50 hp, 3 phase induction motor is drawing 60 A at 0.85 PF lagging The stator copper losses are 2 kW, and rotor copper losses are 700 W, friction & windage losses are 600 W, core losses 1800 W, and stray losses negligible. Find: • (a) PAG (b) Pconv (c) Pout (d) efficieny of motor

  13. POWER & TORQUE IN INDUCTION MOTORS-LOSSES EXAMPLE • (a) PAG=Pin-PSCL – Pcore Pin=√3 VT IL=√3 (480) (60) (0.85)=42.4 kW PAG=42.4-2-1.8= 38.6 kW • (b) Pconv=PAG-PRCL=38.6-700/1000=37.9 kW • (c) Pout=Pconv-PF&W-Pmisc=37.9-600/1000-0 = 37.3 kW = 37.3/ 0.746 = 50 hp • (d) efficiency of motor η =Pout/Pin x 100 % =88 %

  14. POWER & TORQUE IN INDUCTION MOTORS • Employing the equivalent circuit, power & torque equations can be derived • Input current I1= Vφ/ Zeq = R1 + jX1 + 1/{[GC-jBM +1/[R2/s +jX2]}

  15. POWER & TORQUE IN INDUCTION MOTORS stator copper losses, core losses and rotor copper losses can be found • stator copper losses (3 phase)=PSCL= 3 I1^2 R1 • core losses Pcore = 3 E1^2 GC  PAG=Pin-PSCL-Pcore • only element in equ. cct. where air gap power can be consumed is resistor R2/s , & air gap power can also be given: PAG=3 I2^2 R2/s (1) • Actual resistive losses in rotor circuit: PRCL=3 I2^2R2 (2) • Pconv=PAG- PRCL=3I2^2R2/s-3I2^2R2 = 3I2^2 R2 (1/s-1) Pconv= 3I2^2 R2 (1-s)/s

  16. POWER & TORQUE IN INDUCTION MOTORS • Note: • from equations (1) & (2)  rotor copper losses = air gap power x slip • The lower the slip the lower the lower rotor losses • And if rotor is stationary s=1 & air gap power is entirely consumed in rotor, this is consistent with the fact that output power in this case would be zero since ωm=0,  Pout=Tload ωm=0 Pconv=PAG-PRCL=PAG-sPAG=(1-s)PAG(3) • If friction & windage losses and stray losses are known, output power Pout=Pconv-PF&W- Pmisc

  17. POWER & TORQUE IN INDUCTION MOTORS • Induced torque Tindas : torque generated by internal electric to mechanical powerconversion • It differs from available torqueby amount equal to friction & windage torques in machine • Tind=Pconv/ωm also calleddeveloped torque of machine • Substituting for Pconv from (3) & forωm, (1-s) ωsync  Tind= (1-s)PAG/ [(1-s)ωsync]= PAG/ωsync (4) So (4) express induced torque in terms of air-gap power & sync. Speed which is constant  PAG yields Tind

  18. POWER & TORQUE IN INDUCTION MOTORS • SEPARATION of PRCL & Pconv in induction motor Eq. cct. • Part of power coming across air gapconsumed in rotor copper losses,& the other part converted to mechanical power to drive motor shaft • it is possible to separate these two different uses of air-gap power& present them separately in the equivalent circuit • Equation (1) is an expression for total air-gap power, while (2) gives actual rotor losses, the difference between these two is Pconv & must be consumed in an equivalent resistor • Rconv=R2/s-R2 = R2(1/s-1) =R2 (1-s)/s

  19. POWER & TORQUE IN INDUCTION MOTORS • The per-phase equivalent circuit with rotor copper losses& power converted to mech. form separated into distinct elementsshown below:

  20. POWER & TORQUE IN INDUCTION MOTORS – TORQUE EXAMPLE • a 460 V, 60 Hz, 25 hp, 4 pole, Y connected induction motor has following impedances in Ω /phase referred to stator circuit: R1=0.641 Ω R2=0.332 Ω X1 = 1.106 Ω X2 = 0.464 Ω XM=26.3 Ω total rotational losses are 1100 W, & assumed to be constant core loss is lumped in with rotational losses. For rotor slip of 2.2 % at rated voltage & rated frequency,find: • Speed (b) stator current (c) P.F. (d) Pconv & Pout (e) Tind & Tload (f) Efficiency

  21. POWER & TORQUE IN INDUCTION MOTORS – TORQUE EXAMPLE • (a) nsync=120 fe/p=120 x60/4=1800 r/min ωsync=1800 x 2π x 1/60= 188.5 rad/s rotor’s mechanical shaft speed: nm=(1-s) nsync=(1-0.022) x 1800=1760 r/min ωm= (1-s) ωsync= (1-0.022) x 188.5= 184.4 rad/s • (b) to find stator current, consider eq. impedance of cct.Then combine referred rotor impedance in parallel with magnetization branch, and add stator impedance to the combination in series • The referred rotor impedance is : Z2= R2/s + j X2 =0.332 / 0.022 + j0.464 =15.09+j0.464=15.1/_ 1.76◦ Ω combined magnetization plus rotor impedance is: Zf = 1/[1/(jXM) + 1/Z2] = 1/ [-j0.038 + 0.0662/_ -1.76◦]= 1/[0.0773/_ -31.1◦]=12.94/_31.1 ◦

  22. POWER & TORQUE IN INDUCTION MOTORS – TORQUE EXAMPLE • Total impedance : Ztot= Zstat+Zf = = 0.641+j1.106+12.94/_31.1◦ Ω = 11.72 + j7.79 =14.07 /_33.6◦ Ω • Resulting stator current: I1=Vφ/Ztot = 206/_0◦ / 14.07 /_33.6 = 18.88/_-3.6 A

  23. POWER & TORQUE IN INDUCTION MOTORS – TORQUE EXAMPLE • (c) motor power factor: PF=cos 33.6 = 0.833 lagging • (d) Input power of motor: Pin=√3 VT VL cos θ=√3 x 460 x 18.88 x 0.833= 12530 W • PSCL=3 (18.88)^2 (0.641)=685 W • air-gap power :PAG=Pin-PSCL=12530-685=11845 Wand the power converted is: • Pconv=(1-s)PAG =(1-0.022)(11845) =11585 W • Pout=Pconv-Prot=11585-1100=10485 W=14.1 hp

  24. POWER & TORQUE IN INDUCTION MOTORS – TORQUE EXAMPLE • (e) induced torque is: Tind=PAG/ ωsync = 11845 / 188.5 =62.8 N.m • output torque: Tload=Pout/ωm= 10485/184.4=56.9 N.m • (f) motor’s efficiency: η= Pout/ Pin x 100% =10485/12530 x 100% = 83.7 %

  25. INDUCTION MOTOR TORQUE CHARACTERISTIC • How does its torque change as load changes? • How much torque can supply at starting conditions?how much does the speed of induction motor drop as its shaft load increases? • it is necessary to understand the relationship among motor’s torque, speed, and power • the torque-speed characteristic examinedfirst from physical viewpoint of motor’s magnetic field& then a general equation for torque as function of slip derived

  26. INDUCTION MOTOR TORQUE CHARACTERISTIC • Induced Torque from a Physical Viewpoint • Figure shows a cage rotor of an induction motor • Initially operating at no load &  nearly sync. speed

  27. INDUCTION MOTOR TORQUE CHARACTERISTIC • Net magnetic field Bnet produced by magnetization current IM flowing in motor’s equivalent circuit • Magnitude of IM and Bnet directly proportional to E1 • If E1 constant, then Bnet constant • In practice E1 varies as load changes, because stator impedance R1 and X1 cause varying voltage drops with varying load • However, these drops in stator winding is relatively small so E1 ( hence IM & Bnet) approximately constant with changes in load • In Fig (a) motor is at no load, its slip is very small & therefore relative motion between rotor and magnetic field is very small & rotor frequency also very small

  28. INDUCTION MOTOR TORQUE CHARACTERISTIC • Consequently ER induced in rotor is very small, and IR would be small • So frequency is very small, reactance of rotor is nearly zero, and maximum rotor current IR is almost in phase with rotor voltage ER • Rotor current produces a small BR at an angle just slightly greater than 90◦ behind Bnet • Note:stator current must be quite large even at no load, since it supply most of Bnet • That is why induction motors have large no load currents compared to other types of machines

  29. INDUCTION MOTOR TORQUE CHARACTERISTIC • The induced torque, which keeps rotor running is given by: Tind = k BR x Bnet or Tind=k BR Bnet sinδ • Since BR is very small, Tind also quite small, enough just to overcome motor’s rotational losses • suppose motor is loaded (in Fig (b)) as load increase, motor slip increase, and rotor speed falls. Since rotor speed decreased, more relative motion exist between rotor & stator magnetic fields in machine • Greater relative motionproduces a stronger rotor voltage ERwhich in turn produces a larger rotor current IR

  30. INDUCTION MOTOR TORQUE CHARACTERISTIC • Consequently BR also increases, however angle of rotor current & BR changes as well • Since rotor slip get larger, rotor frequency increases fr=sfe and rotor reactance increases (ω LR) • Rotor current now lags further behind rotor voltage (as shown) & BR shift with current • Fig b, shows motor operating at a fairly high load • Note:at this situation, rotor current increased and δ increased • Increase in BR tends to increase torque, while increase in δ tends to decrease the torque (δ>90) • However since the effect of first is higher than the second in overall induced torque increasedwith load

  31. INDUCTION MOTOR TORQUE CHARACTERISTIC • Using: Tind=k BR Bnet sinδ derive output torque-versus-speed characteristic of induction motor • Each term in above equation considered separately to derive overall machine behavior • Individual terms are: 1. BR directly proportional to current flowing in rotor, as long as rotor is unsaturated Current flow in rotor increases with increasing slip (decreasing speed) it is plotted 2- Bnet magnetic field in motor is proportional to E1 & therefore approximately constant (E1 actually decreases with increasing current flow) this effect small compared to the other two & ignored in drawing 3- sinδ : δ is just equal to P.F. angle of rotor plus 90◦ δ=θR+90 And sinδ=sin(θR+90)=cos θR which is P.F. of rotor

  32. INDUCTION MOTOR TORQUE CHARACTERISTIC • Rotor P.F. angle can be determined as follows: θR =atan XR/RR = atan sXR0 / RR PFR = cos θR PFR=cos(atan sXR0/RR) • plot of rotor P.F. versus speed shown in fig (c) • Since induced torqueis proportionalto product of these 3 terms,torque-speed characteristic can be constructed from graphical multiplication of 3 previous plots Figs (a,b,c) and shown in fig (d)

  33. INDUCTION MOTOR TORQUE CHARACTERISTIC • Development of induction motor torque –speed

  34. Development of induction motor torque –speed

  35. Development of induction motor torque –speed

  36. Development of induction motor torque –speed

  37. INDUCTION MOTOR TORQUE CHARACTERISTIC • This characteristic curve can be divided into three regions • 1st region:is low-slip regionin which motor slip increases approximately linearlywith increase load & rotor mechanical speed decreases approximately linearly with load • In this regionrotor reactance is negligible, so rotor PF is approximately unity, while rotor current increases linearly with slip • The entire normal steady-state operating range of an induction motor is included in this linear low-slip region

  38. INDUCTION MOTOR TORQUE CHARACTERISTIC • 2nd region on curve called moderate-slip region • Inmoderate-slip regionrotor frequencyis higher than before,& rotor reactance is on the same order of magnitude as rotor resistance - In this region rotor current, no longer increases as rapidly as beforeand the P.F. starts to drop - peak torque(pullout torque)of motor occurs at point where, for an incremental increase in load, increase in rotor current is exactly balanced by decrease in rotor P.F.

  39. INDUCTION MOTOR TORQUE CHARACTERISTIC • 3rd region on curve is called high-slip region • In high-slip region, induced torque actually decreases with increased load, since the increase in rotor current is completelyovershadowed by decrease in rotor P.F. • For a typical induction motor, pullout torque is 200 to 250 % of rated full-load torque • And starting torque (at zero speed) is about 150% of full-load torque • Unlike synchronous motor, induction motor can start with a full-load attached to its shaft

  40. INDUCTION MOTOR INDUCED-TORQUE EQUATION • Equiv. circuit of induction motor & its power flow diagram used to derive a relation for induced torque versus speed • Tind=Pconv/ωm or Tind=PAG/ωsync • Latter useful, since ωsync is constant (for fe & and a number of poles)so from PAG  Tind • The PAG is equal to power absorbed by resistor R2/s , how can this power be determined?

  41. INDUCTION MOTOR INDUCED-TORQUE EQUATION • In this figure the air-gap power supplied to one phase is: PAG,1φ=I2^2 R2/s • for 3 phase: PAG=3I2^2 R2/s

  42. INDUCTION MOTOR INDUCED-TORQUE EQUATION • If I2 can be determined, air-gap power & induced torque are known • easiest way to determine Thevenin equivalent of the portion of circuit to left of arrow E1in eq. cct. figure VTH= Vφ ZM/ [ZM+Z1] = Vφ j XM / [R1+jX1+jXM] • Magnitude of thevenin voltage: VTH= Vφ XM / √[R1^2+(X1+XM)^2] VTH≈ Vφ XM / [X1+XM], ZTH = Z1ZM /[Z1+ZM] ZTH=RTH+jXTH = jXM(R1+jX1)/[R1+j(X1+XM)]

  43. INDUCTION MOTOR INDUCED-TORQUE EQUATION • Thevenin equivalent voltage of induction motor

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